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1400	https://www.theforage.com/blog/skills/liabilities	Skip to content Table of contents Liabilities Definition Types of Liabilities Liabilities Examples Liabilities vs. Assets Showing You Understand Liabilities on Resumes What Are Liabilities? Definition and Examples Forage puts students first. Our blog articles are written independently by our editorial team. They have not been paid for or sponsored by our partners. See our full editorial guidelines. Table of Contents Liabilities Definition Types of Liabilities Liabilities Examples Liabilities vs. Assets Showing You Understand Liabilities on Resumes Liabilities are debts or obligations a person or company owes to someone else. For example, a liability can be as simple as an I.O.U. to a friend or as big as a multibillion dollar loan to purchase a tech company. In business, liabilities are building blocks of a company’s finances, often used to fund operations and expansions. Liabilities Definition A liability is simply a debt or obligation. Most people have liabilities in their day-to-day lives: car payments, rent, and credit card bills. In corporate finance, liabilities are similar, just on a much larger scale. Liabilities for a business may be long-term loans used to fund operations, money owed to vendors or suppliers, or leases for warehouse spaces. If a company has an obligation to pay someone or for something, it’s a liability. Who Deals With These Debts? Liabilities are a core part of business and finance — business owners and members of a company’s financial team are responsible for understanding what liabilities their company has and how they affect the company as a whole. Accountants also need a strong understanding of how liabilities function within an organization’s finances. Accounting processes often involve examining the relationships between liabilities, assets, and equity and how these things affect a business’s profitability and performance. Even in corporate finance, like investment banking and private equity, understanding the role of liabilities in a company’s financial structure is key to understanding a company’s financial position as a whole. >>MORE: See the accounting skills you need on your resume. Types of Liabilities Liabilities are typically categorized by expiration or due date: current liabilities are pressing debts and obligations, while non-current liabilities are important but don’t require immediate action. Current Current liabilities are short-term debts and obligations due within one year. Some common examples of current liabilities are: Accounts payable: Money the company owes to lenders, clients, customers, and suppliers Short-term loans: Loans with repayment periods of one year or less Payments on long-term debt: Payments due on larger loans with repayment periods of more than one year, and the remainder of the loan owed is listed as a non-current liability Interest: Unpaid interest on loans Accrued liabilities: Any unpaid short-term liabilities from previous accounting periods Income taxes: Federal, state, and local income taxes that have not been paid yet Deferred revenue or unearned revenue: Money the company has received for goods or services it has not yet provided to the customer Commercial paper: Unsecured promissory notes with fixed interest rates used by companies to fund very pressing liabilities like payroll Citi Finance Explore the ins and outs of business finances, like KPIs, financial risks, and sales numbers in this free job simulation from Citi. Enroll now Avg. Time: 5 to 6 hours Skills you’ll build: Fact finding, data analysis, communication, presentation, Basel III regulation, judgment, commercial awareness Enroll now Non-Current Non-current or long-term liabilities are debts and obligations due in the future but not in the next year. Some types of non-current liabilities are: Bonds: A type of marketable security that has a specified maturity date (when payment is due in full) and interest rate Deferred tax: Federal, state, and local taxes owed, though not due immediately Long-term debt: Loans and other debts that are not due within the year, including remaining principal amounts on loans paid in increments Mortgages: Agreements with lenders that give the lender the right to repossess the purchased property if the loan is not repaid Leases: Payments for the use of another person’s property or assets, such as machinery, office spaces, and vehicles Pensions: Retirement funds for employees PwC Tax Learn fundamental skills in accounting, like the differences between assets and liabilities, in this free job simulation from PwC. Enroll now Avg. Time: 3 to 4 hours Skills you’ll build: Reviewing a trial balance, chart of account mapping, book to tax difference (BTD), calculating M-adjustments, research, workpaper preparation, data analysis, tax research Enroll now Contingent Liabilities Contingent liabilities are a special type of debt or obligation that may or may not happen in the future. These liabilities are contingent (or dependent on) certain events. The most common example of a contingent liability is legal costs related to the outcome of a lawsuit. For example, if the company wins the case and doesn’t need to pay any money, the company doesn’t incur the contingent liability. However, if the company loses the lawsuit and needs to pay the other party, the contingent liability takes effect and the company must cover it. Another example of a contingent liability is a warranty. If a company’s product is faulty or needs to be repaired or replaced for the customer, the company needs to have the funds available to honor that warranty agreement. Not every possible event can be planned for, though. So, when it comes to reporting a company’s finances, only certain contingent liabilities need to be reported. According to the generally accepted accounting principles (GAAP), only probable liabilities need to be listed on a company’s balance sheet — these events are very likely to happen, and the cost can be reasonably estimated. The other two types of contingent liabilities — possible and remote — don’t need to be stated in the balance sheet because they’re less likely to occur and much harder to estimate. Possible contingent liabilities should at least be noted in the footnotes of the company’s financial statements, though. Liabilities Examples Using Apple’s balance sheet from 2023, we can see how current and non-current liabilities commonly appear on financial statements. Some companies may group certain liabilities under “other current/non-current liabilities” because the liabilities may not be common enough to warrant an entire line item. For instance, if a company rarely uses short-term loans, it may group those with other current liabilities under an “other” category. Liabilities vs. Assets The flip side of liabilities is assets — resources the company uses to generate income. Assets include inventory, machinery, savings account balances, and intellectual property. Typically, gaining an asset means incurring a liability. For example, buying new equipment may mean taking out a loan to finance the purchase. Assets and liabilities are two parts that make up a company’s finances, and the third part is equity or money put into the company by founders or private investors. These three accounts, or aspects of a company’s finances, cover nearly every type of transaction or business decision a company can make. Additionally, accountants use a formula called the accounting equation based on assets, liabilities, and equity, that ensures accurate reporting of a company’s finances. Find Your Career Fit Figure out which career is right for you with our free quiz! Get started Showing You Understand Liabilities on Resumes Liabilities are a core part of accounting roles and many other careers in finance. The easiest way to show employers you understand liabilities and how they affect a company’s finances is by referencing relevant core skills in accounting and finance on your resume. Some skills you can list in the skills section or description of work or internship experiences on your resume include: Ensuring compliance with U.S. generally accepted accounting principles (GAAP) Creation and utilization of financial statements such as balance sheets Calculating financial health metrics like the current ratio and net working capital Ready to learn new and in-demand skills? Check out Forage’s free accounting job simulations. Image credit: Canva Written by McKayla Girardin → Writer Read more from McKayla Girardin McKayla Girardin is a NYC-based writer with Forage. She is experienced at transforming complex concepts into easily digestible articles to help anyone better understand the world we live in. Related Posts 6 Negotiation Skills to Level Up Your Work Life How to Build Conflict Resolution Skills: Case Studies and Examples What Is GitHub? Uses and Getting Started Share on Twitter Share on Facebook Share on LinkedIn Related Posts 6 Negotiation Skills to Level Up Your Work Life How to Build Conflict Resolution Skills: Case Studies and Examples What Is GitHub? Uses and Getting Started Upskill With Forage Financial Analyst New York Jobs CEO Council Underwriting AIG Build the business and finance skills recruiters are looking for. Start now Privacy Preference Center We are committed to your privacy. Our Privacy Preference Center provides you with tools to allow you to control what personal information this website collects. We use cookies and similar technologies on this website, as almost all websites do. Cookies and pixels allow us, among other things, to store your preferences and settings, authenticate and log you in, provide you with a personalized experience, keep our websites secure, and analyze how our websites and online services are performing. 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1401	https://pubmed.ncbi.nlm.nih.gov/28778401/	Postnatal maturation of the fovea in Macaca mulatta using optical coherence tomography - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Postnatal maturation of the fovea in Macaca mulatta using optical coherence tomography Nimesh B Patel1,Li-Fang Hung2,Ronald S Harwerth2 Affiliations Expand Affiliations 1 University of Houston College of Optometry, United States. Electronic address: npatel@central.uh.edu. 2 University of Houston College of Optometry, United States. PMID: 28778401 PMCID: PMC5628152 DOI: 10.1016/j.exer.2017.07.018 Item in Clipboard Postnatal maturation of the fovea in Macaca mulatta using optical coherence tomography Nimesh B Patel et al. Exp Eye Res.2017 Nov. Show details Display options Display options Format Exp Eye Res Actions Search in PubMed Search in NLM Catalog Add to Search . 2017 Nov:164:8-21. doi: 10.1016/j.exer.2017.07.018. Epub 2017 Aug 2. Authors Nimesh B Patel1,Li-Fang Hung2,Ronald S Harwerth2 Affiliations 1 University of Houston College of Optometry, United States. Electronic address: npatel@central.uh.edu. 2 University of Houston College of Optometry, United States. PMID: 28778401 PMCID: PMC5628152 DOI: 10.1016/j.exer.2017.07.018 Item in Clipboard Full text links Cite Display options Display options Format Abstract Changes in the foveal anatomy during infancy are an important component in early development of spatial vision. The present longitudinal study in rhesus monkeys was undertaken to characterize the postnatal maturation of the fovea. Starting at four weeks after birth, the retinas of the left eyes of sixteen infant monkeys were imaged using spectral domain optical coherence tomography (SD OCT). Retinal scans were repeated every 30 days during the first year of life and every 60 days thereafter. Volume scans through the fovea were registered, scaled using a three surface schematic eye, and analyzed to measure foveal pit parameters. The individual layers of the retina were manually segmented and thicknesses were measured over a transverse distance of 1250 microns from the center of the foveal pit. Based on infrared scanning laser ophthalmoscope (IR SLO) images acquired with the SD OCT system, there were significant changes in the extent of the retina scanned as the eyes matured. Using a three-surface schematic eye, the length of each scan could be computed and was validated using image registration (R 2 = 0.88, slope = 1.003, p < 0.05). Over the first 18 months of life, the mean retinal thickness at the pit center had increased by 21.4% with a corresponding 20.3% decrease in pit depth. The major changes occurred within the first 120 days, but did not stabilize until a year after birth. In Macaca mulatta infants, the primary anatomical maturation of the fovea occurs within the first few months of life, as determined by longitudinal data from SD OCT measurements. The timelines for maturation of the fovea correspond well with the normal development of the lateral geniculate nucleus, cortical neurophysiology, and spatial resolution in monkeys. Keywords: Development; Fovea; Ocular magnification; Optical coherence tomography. Copyright © 2017 Elsevier Ltd. All rights reserved. PubMed Disclaimer Figures Figure 1 A. Thirty degree IR-SLO images… Figure 1 A. Thirty degree IR-SLO images centered on the fovea from 51 days (dashed… Figure 1 A. Thirty degree IR-SLO images centered on the fovea from 51 days (dashed outline) and 289 days of age in registration. B. IR-SLO images of the optic nerve in registration from the same animal and scan dates as in A. C. Comparison of image scaling using image registration and the three surface schematic eye. D. Validation of image registration technique, comparing IR-SLO image length from corresponding fovea and optic nerve registered scans. Figure 2 A. Total retinal thickness map… Figure 2 A. Total retinal thickness map for a region 15° × 20° centered on… Figure 2 A. Total retinal thickness map for a region 15° × 20° centered on the fovea (256 day old infant), B. Pit characteristics were determined using interpolated thickness data from the total thickness maps. For example, the top plot illustrates the thickness profile for a horizontal radial section through the thickness map as illustrated by the black line though A. Along with the thickness profiles from radial sections centered on the pit, slope data (bottom B) were used to determine pit characteristics and construct a skeleton to describe pit morphology. The dashed lines in B indicate the maximal slope location and top of pit location. C. Illustrated are the thickness map along with a skeletal reconstruction of the pit, including points used to fit the top and bottom of the pit. Figure 3 Top row: Registered SLO images… Figure 3 Top row: Registered SLO images and corresponding total retinal thickness maps for one… Figure 3 Top row: Registered SLO images and corresponding total retinal thickness maps for one animal (A – 33 days, B – 67days, C – 137days, D – 468 days, and E – 649 days of age). Each scan is in registration, and the montaged IR-SLO image is illustrated in F. Bottom row G–J: Change in total retinal thickness compared to baseline measures at 33 days of age. The series illustrates the increase in total retinal thickness in the central foveal region with age. Notice that as the animal ages, the portion of the retina imaged also increases. A scaled grid was used to determine average thickness measures within quadrants of concentric circular regions were used to determine overall retinal thickness changes with age. Figure 4 A. OCT section through fovea… Figure 4 A. OCT section through fovea of a 10 day old infant with corresponding… Figure 4 A. OCT section through fovea of a 10 day old infant with corresponding histology stained with Cresyl violet from the same animal. B. OCT section through the foveal region, of a 1.5 year old animal. C. Each layer was manually identified on the OCT B-scan with the assistance of reflectivity profiles, through the region of interest. The reflectivity profile illustrated is generated from the gray vertical line shown in B. D. For each eye, layer thicknesses at 89 points illustrated on the thickness plot were measured and monitored over time. Figure 5 The pit depth decreased rapidly,… Figure 5 The pit depth decreased rapidly, reaching adult like depths prior to a year… Figure 5 The pit depth decreased rapidly, reaching adult like depths prior to a year of age. Each subject is represented by a different symbol. Figure 6 A. The average retinal thickness… Figure 6 A. The average retinal thickness within 50μm of the pit increased by 20.9%… Figure 6 A. The average retinal thickness within 50μm of the pit increased by 20.9% with the majority of change occurring within the first 150 days after birth. The trend was similar for all animals and was best fit with an exponential to maximum function. Each subject is represented by a different symbol. B. The figure illustrates total retinal thickness change for each quadrant at the eccentricities illustrated in figure 3E. C. A surface plot illustrating changes in average thickness as a function of age and eccentricity up to 500 μm. Figure 7 Scaled SD OCT images for… Figure 7 Scaled SD OCT images for six subjects at baseline (left) and endpoint (right)… Figure 7 Scaled SD OCT images for six subjects at baseline (left) and endpoint (right) through the fovea center. A well defined foveal pit is seen in all animals at the first scan session. The overall shape of the foveal pit does not change, but there is an increase in outer retinal thickness on the endpoint scans. Figure 8 At each eccentricity, the RPE… Figure 8 At each eccentricity, the RPE thickness measured increased linearly, reaching peak thickness just… Figure 8 At each eccentricity, the RPE thickness measured increased linearly, reaching peak thickness just prior to a year of age. Thickness after this point decreased linearly. A. Changes in RPE thickness are illustrated for each animal at 700μm eccentricity. The best fit to the data illustrates an increase in thickness till around 289 days of age followed by a reduction in thickness. B. Differences in thickness with age and eccentricity illustrate this trend with age at each eccentricity. The horizontal line through the plot illustrates the location for data presented in plot A. Figure 9 Outer retinal layers including the… Figure 9 Outer retinal layers including the outer segments, inner segment/outer segment junction, inner segment,… Figure 9 Outer retinal layers including the outer segments, inner segment/outer segment junction, inner segment, and outer nuclear layer, have significant changes in thickness with age. Plots A, C, E, and G, illustrate thickness changes for each of the 16 animals at an eccentricity of 87 μm. Although each of these layers show significant change in thickness, they all mature at different rates, with the outer segment layer thickness stabilizing at an earlier time point. Plots B, D, F, and H illustrate the relationship between thickness, eccentricity and age for each of the layers. The horizontal line through the plots indicate the location of the sampled data in A, C, E, and G. All figures (9) See this image and copyright information in PMC Similar articles Evaluation of normal human foveal development using optical coherence tomography and histologic examination.Dubis AM, Costakos DM, Subramaniam CD, Godara P, Wirostko WJ, Carroll J, Provis JM.Dubis AM, et al.Arch Ophthalmol. 2012 Oct;130(10):1291-300. doi: 10.1001/archophthalmol.2012.2270.Arch Ophthalmol. 2012.PMID: 23044942 Free PMC article. Maturation of the human fovea: correlation of spectral-domain optical coherence tomography findings with histology.Vajzovic L, Hendrickson AE, O'Connell RV, Clark LA, Tran-Viet D, Possin D, Chiu SJ, Farsiu S, Toth CA.Vajzovic L, et al.Am J Ophthalmol. 2012 Nov;154(5):779-789.e2. doi: 10.1016/j.ajo.2012.05.004. Epub 2012 Aug 13.Am J Ophthalmol. 2012.PMID: 22898189 Free PMC article. Repeatability of Foveal Measurements Using Spectralis Optical Coherence Tomography Segmentation Software.Ctori I, Huntjens B.Ctori I, et al.PLoS One. 2015 Jun 15;10(6):e0129005. doi: 10.1371/journal.pone.0129005. eCollection 2015.PLoS One. 2015.PMID: 26076457 Free PMC article. The primate fovea: Structure, function and development.Bringmann A, Syrbe S, Görner K, Kacza J, Francke M, Wiedemann P, Reichenbach A.Bringmann A, et al.Prog Retin Eye Res. 2018 Sep;66:49-84. doi: 10.1016/j.preteyeres.2018.03.006. Epub 2018 Mar 30.Prog Retin Eye Res. 2018.PMID: 29609042 Review. Adaptation of the central retina for high acuity vision: cones, the fovea and the avascular zone.Provis JM, Dubis AM, Maddess T, Carroll J.Provis JM, et al.Prog Retin Eye Res. 2013 Jul;35:63-81. doi: 10.1016/j.preteyeres.2013.01.005. Epub 2013 Mar 15.Prog Retin Eye Res. 2013.PMID: 23500068 Free PMC article.Review. See all similar articles Cited by The development of and recovery from form-deprivation myopia in infant rhesus monkeys reared under reduced ambient lighting.She Z, Hung LF, Arumugam B, Beach KM, Smith EL 3rd.She Z, et al.Vision Res. 2021 Jun;183:106-117. doi: 10.1016/j.visres.2021.02.004. Epub 2021 Mar 30.Vision Res. 2021.PMID: 33799131 Free PMC article. Inter-individual differences in foveal shape in a scavenging raptor, the black kite Milvus migrans.Potier S, Mitkus M, Lisney TJ, Isard PF, Dulaurent T, Mentek M, Cornette R, Schikorski D, Kelber A.Potier S, et al.Sci Rep. 2020 Apr 9;10(1):6133. doi: 10.1038/s41598-020-63039-y.Sci Rep. 2020.PMID: 32273526 Free PMC article. In Vivo Imaging of the Retina, Choroid, and Optic Nerve Head in Guinea Pigs.Jnawali A, Beach KM, Ostrin LA.Jnawali A, et al.Curr Eye Res. 2018 Aug;43(8):1006-1018. doi: 10.1080/02713683.2018.1464195. Epub 2018 Apr 23.Curr Eye Res. 2018.PMID: 29641938 Free PMC article. Long-term histological changes in the macaque primary visual cortex and the lateral geniculate nucleus after monocular deprivation produced by early restricted retinal lesions and diffuser induced form deprivation.Takahata T, Patel NB, Balaram P, Chino YM, Kaas JH.Takahata T, et al.J Comp Neurol. 2018 Dec 15;526(18):2955-2972. doi: 10.1002/cne.24494. Epub 2018 Nov 14.J Comp Neurol. 2018.PMID: 30004587 Free PMC article. Maturation and refinement of the maculae and foveae in the Anolis sagrei lizard.Wahle MA, Kim HQ, Menke DB, Lauderdale JD, Rasys AM.Wahle MA, et al.Exp Eye Res. 2023 Sep;234:109611. doi: 10.1016/j.exer.2023.109611. Epub 2023 Aug 1.Exp Eye Res. 2023.PMID: 37536437 Free PMC article. See all "Cited by" articles References Abbott CJ, Grunert U, Pianta MJ, McBrien NA. Retinal thinning in tree shrews with induced high myopia: optical coherence tomography and histological assessment. Vision Res. 2011;51:376–385. - PubMed Abbott CJ, McBrien NA, Grunert U, Pianta MJ. Relationship of the optical coherence tomography signal to underlying retinal histology in the tree shrew (Tupaia belangeri) Investigative ophthalmology & visual science. 2009;50:414–423. - PubMed Anger EM, Unterhuber A, Hermann B, Sattmann H, Schubert C, Morgan JE, Cowey A, Ahnelt PK, Drexler W. Ultrahigh resolution optical coherence tomography of the monkey fovea. Identification of retinal sublayers by correlation with semithin histology sections. Exp Eye Res. 2004;78:1117–1125. - PubMed Banks MS, Bennett PJ. Optical and photoreceptor immaturities limit the spatial and chromatic vision of human neonates. Journal of the Optical Society of America. A, Optics and image science. 1988;5:2059–2079. - PubMed Bengtsson B, Krakau CE. Correction of optic disc measurements on fundus photographs. 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1402	https://www.clinicsinsurgery.com/open-access/glasgow-coma-scale-technique-and-intepretation-3009.pdf	Remedy Publications LLC., \| Clinics in Surgery 2017 \| Volume 2 \| Article 1575 1 Glasgow Coma Scale: Technique and Intepretation OPEN ACCESS Correspondence: Bhaskar S, Department of Neurosurgery, PGIMER & Dr RML Hospital, New Delhi-110001, India, Tel: 011-23365525 (ext-4714); E-mail: bhaskar.nsurg@yahoo.com Received Date : 24 May 2017 Accepted Date : 28 Jul 2017 Published Date : 03 Aug 2017 Citation: Bhaskar S. Glasgow Coma Scale: Technique and Intepretation. Clin Surg. 2017; 2: 1575. Copyright © 2017 Bhaskar S. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. Surgical Technique Published: 03 Aug, 2017 Abs tract The Glasgow Coma Scale (GCS) was designed to objectively, easily, methodically evaluate the neurological status of patients with impaired consciousness especially after traumatic brain injury. The score is used to grade and plan treatment of patients with head injury. It can also be used to monitor neurological status of critically ill patients. There are three parameters used- eye opening, verbal response and best motor response with the score ranging from 3-15. The technique, interpretation and certain practical issues concerning the GCS are discussed. Keywords: Glasgow coma scale; Head injury; Interpretation Bhaskar S Department of Neurosurgery, PGIMER & Dr RML Hospital, India Introduction The Glasgow Coma Scale was introduced in 1974 . The acronym stands for both Glasgow Coma Scale (individual components) and Score (total). The scale is used to decide the management issues of individual patients. The total is used to group patients into various categories so that management protocols can be designed and outcome measured. It has three components for assessment that are totalled and a combined score is given. The total score is used in classifying head injuries into mild (3-8), moderate (9-12) and severe (13-15) . The GCS has made a subjective assessment of head injury (coma, semi coma, drowsy, stupor, altered sensorium, locked in state) into an objective method [3,4]. It has significant advantages and also certain drawbacks that need to be understood when applying it. The basic techniques of how to perform the GCS and the issues involved in its interpretation are discussed. Technique Key Messages (Provide appropriate messages of about 35-50 words to be printed in centre box). There are three parameters that are assessed Eye opening (E) : used as a reflection of the intensity of impairment of activating functions. Verbal response (V ): Index of higher cortical function. Best motor response (M) : Knowing the integrity of the nervous system in patients who are not speaking. The original score had total score of 14, which was increased to 15 (Table 1). GCS assessment method [5,6] Check: To check for any factors that might interfere with the assessment like sedation, muscle relaxants, metabolic disturbances, fever, hemodynamic disturbances, eye swelling, airway injuries, limb injuries, intoxicants Observe: To see for spontaneous patient actions (eg: eye opening). Stimulate: Once it is decided that there is no spontaneous response then to stimulate and check for responses. Rate: After the stimulation for various parameters the score is recorded and totalled to arrive at the GCS score. Eye opening E4 (Spontaneous) : The patient will be opening his/her eyes without any external stimulus and this indicates that arousal mechanisms in the brainstem are intact. In the rare case of persistent vegetative state the patient will have a spontaneous opening without awareness. Bhaskar S Clinics in Surgery - General Surgery Remedy Publications LLC., \| 2017 \| Volume 2 \| Article 1575 2 E3 (Eye opening to command) : Next a verbal stimulus is given; this includes any loud sound that will make the patient open eyes and not necessarily a command to open eyes. The sound should be loud enough to evoke a response. E2 (Eye opening to pain) : The painful stimulus is given at either the nail bed with a pen/pencil or pinching the trapezius muscle or rubbing the sternum. The sternal region has just skin over the periosteum and hence is very sensitive. The stimulus has to be a firm rubbing movement and not excessive downward force that may fracture ribs or even the sternum in undiagnosed chest wall injuries. The stimulus is sustained for 10 seconds (if required) before concluding that there is no response to pain. E1 (None) : No Eye Opening to even painful stimulus: Patient might have closed the eyes as a result of the stress due to the trauma or at times even sleeping after the exhaustion following the event! So one should not give a painful stimulus before a verbal one. Pressing over the supraorbital notch is not recommended for eye opening as it may elicit an erroneous response by a grimace reflex. The body’s natural reflex to any painful stimulus in the facial region is to close eye and protect them from that stimulus! Another issue to be noted is to make sure that there are no facial injuries that might exacerbate by applying a stimulus in the facial region. If there is extensive periorbital swelling that precludes eye opening then it can be written as Not Assessed or Ep or E edema . Whatever is the format it should be uniformly followed by all caregivers in a given institution? Any confusion, then it should be explicitly communicated as to what state the patient is in. This would avoid any confusion in assessment and subsequent management. Verbal response V5 (Oriented) : Patient is asked who, where and what time (at least in terms of year, month and day of the week) if the response to all three questions are correct then it is V6. The thing to be remembered is some patients are not told that they have been admitted to that particular hospital, so if they tell the name of some other health centre then it should be corroborated with the relatives about the correctness. It might be the case that the patient was taken to that centre before being shifted to the current place of evaluation! If the patient only answers some of these questions and not all correctly then can be sub categorized into sub groups based on the number answered correctly. This will only be useful for research purposes and should not influence management decisions. V4 (Confused) : Patient does engage in conversation that includes appropriate sentences and words but is not answering appropriately the above three questions. V3 (Inappropriate Speech): Here it is a few exclamatory words (more often swearing) and does not result in sustained conversation. V2 (Incomprehensible sounds): The patient moans and groans, most often to painful stimulus. V1 (Absent): No verbal response. The patient who is intubated or tracheostomized cannot be assessed for an appropriate verbal response. Patients with focal damage to speech areas (eg: Broca area) can have an impaired speech response but may otherwise be alert. Many centres give a score of 1 in the verbal response to patients whose airway is secured by a definitive airway. All caregivers in that institution should uniformly apply this. Otherwise it will create an error in the evaluation. For example if a patient has been intubated for a faciomaxillary injury (eg mandibular fracture) that is compromising the airway, the GCS will be E4V1M6. This becomes moderate head injury whereas it might not be so. It is advisable to use the abbreviation of V ET or V T to denote an intubated or tracheostomized patient respectively. This way the confusion stated in the previous scenario would be avoided. This drawback with the GCS has become more in the recent times as the intubation and ventilation care has significantly increased. Best Motor response The motor response becomes very important in a person who is not conversing or at least is at a confused level. M6 (Obeying commands): This is tested by asking the patient to move fingers, wriggle toes, show tongue/open or close eyes (especially in suspected spinal injury). This response ensures that there is an intact arc of receiving stimulus (command), processing at the cortical level and executing via motor function (response). The stimulus should be loud and clear before inferring lack of response and proceeding to painful stimulus. One should remember that some motor response might be elicited by simple grasp reflex, startle response or even to postural changes. These should not be interpreted as M6. If in doubt then specificity of the response should be crosschecked eg: ask to release hand in case of suspected grasp reflex, holding limbs to verbal command. Once it is decided that the patient does not respond to verbal command then a painful stimulus is given in a standardized manner and maintained for appropriate time to see for motor response. The sites for a painful stimulus are- pinch the trapezius, firm rubbing movement (not pressing) over the sternum or press over the supra-orbital notch. These stimuli are sustained for a period of 10 seconds to make sure that the stimulus is adequate. The nail bed stimulus is not applied in this situation because the other upper limb has to reach over and localize plus it will be erroneous in case of focal brain injuries and spinal cord injuries. M5 (Localizing to painful stimulus): This response will be when the hand reaches the site of stimulus i.e. the sternum or trapezius. In the case of the supra-orbital area the patient’s hand even goes above the level of the clavicle not necessarily reaching the site of stimulus. A common misconception about the GCS is the equation of abnormal flexion and extensor response to decorticate and decerebrate rigidity respectively (these are based on the Sherrington experiments corresponding to the level of lesion, mid-brain or brainstem) . In head i njury the severity correlates with the GCS score irrespective of Original GCS-14 Expanded GCS-15 Eye Opening Spontaneous-4 To sound-3 To pain-2 Absent-1 Eye Opening Spontaneous-4 To sound-3 To pain-2 Absent-1 Verbal Response Orientated-5 Confused-4 Inappropriate words-3 Incomprehensible sounds-2 Absent-1 Verbal Response Orientated-5 Confused-4 Inappropriate words-3 Incomprehensible sounds-2 Absent-1 Best motor Response Obeys commands-5 Localizes to pain-4 Flexion to pain-3 Extension to pain-2 None-1 Best motor Response Obeys commands-6 Localizes to pain-5 Normal flexion (withdrawal) to pain-4 Abnormal flexion-3 Extension to pain-2 None-1 Table 1: The original score had total score of 14, which was increased to 15. Bhaskar S Clinics in Surgery - General Surgery Remedy Publications LLC., \| 2017 \| Volume 2 \| Article 1575 3 the site. A poor motor response can be as a result of severe cortical or hemispheric lesions. M3 & 4 (Normal/Abnormal flexion): What is important to document among these responses is if the patient demonstrates any degree of flexion. The patients who show any degree of flexion response on a persistent basis do well overall than those who have extensor posturing. One can also see a flexion response in the lower limbs. Experienced personnel can make the distinction between the two responses but inexperienced staffs particularly in the field have difficulty in differentiating between these responses. It is not of clinical importance to make a definite diagnosis between M3 and M4 as decisions like intubation, CT scan etc are unlikely to be influenced only on this parameter. Patient who is not localizing to painful stimulus is more likely to be in the severe than moderate head injury group. A range of movement is possible in a patient who is not localizing to painful stimulus. It can vary from a “Normal Flexion” which is rapid withdrawal, abduction of shoulder with external rotation. In “Abnormal Flexion” there is adduction, internal rotation of shoulder (classical decorticate posture). Between these two extremes of movements there may be varied patterns and also both types of movements may be seen at the same time in a patient. One simple and practical way to sort out between the two responses is done by observing the position of the forearm. If the forearm is in pronation then it is labelled as M3. The forearm is in supination without localizing to the painful stimulus then it is labelled as M4. In the situation of a difference in the motor score, it is best to manage as per the lower score. When there is a difference of 1, a moderate head injury may be diagnosed as severe (GCS- 8 or 9). It is only a mathematical probability and an unlikely clinical situation that the patient has E4V5M4. Such gross differences in the individual parameters can occur with focal brain injuries. This way the management will not be hampered by this difference in eliciting motor response. Depending on the patient’s response to the treatment initiated one can decide the future course of action. One needs to take into account that a patient’s response can be varied when tested by two different people or at different times. It is not always a stereotyped response that is uniformly seen in these patients. M2 (Extension response): The classic “decerebrate” posturing. Shoulder adducted, elbow extended, wrist hyerpronated, hip and knee extended with ankle plantar flexed. M1 (None): This is when there is no response to a painful stimulus. One should be very careful when deciding this response especially in spinal injuries with quadriparesis. The patient if conscious can be asked to show his/her tongue to document a M6 response. One should take into account of any muscle relaxants or sedative agents before concluding that there is no motor response. What is important during the assessment is that patients, who localize to painful stimulus and those who do not, should be identified without ambiguity at the earliest. It is possible that a patient is M5 but falls in the severe head injury. The GCS of E1V1M5, E2V1M5 or E1V2M5 are not commonly encountered in clinical scenario. More often than not, patients who are M5 fall in the moderate head injury group. When GCS scale like E1V1M6 is encountered, it is likely to be due to observer error rather than a real clinical situation. In situations like hypotension, hypoxia, intoxicants, sedation/relaxants and other confounding factors it is important to document these at the time of GCS assessment. Later when these are corrected one can correlate the GCS with that. The low GCS ma y be entirely due to this factor eg: hypotension which once corrected the GCS improves. It may also be that this is just one of the elements involved. One should also document the neurological status, pupils at the time of intubation so that a follow up can be done to see the progress. GCS should be calculated of the full score including all 3 parameters and not of two. For example GCS-9 out of 10 in an intubated patient will convey an inaccurate picture than E4V ET M5. How much change is significant? On clinical monitoring a change in GCS of 1 is a sign that after some time it is to be checked. Further decrease in the score merits an investigation and intervention if indicated. A drop in the GCS of 2 or more means that one needs to revaluate the patient, plan a repeat investigation and medical or surgical intervention as and when indicated. How frequently should it be assessed? The issue of how frequent should the assessment be done can be resolved by using common sense on a case-by-case basis. There cannot be fixed criteria for this. A simple measure is a patient who comes sooner after the injury and is a poorer grade should be monitored more closely than a patient who has reported after a significant time after injury and is in a good grade. The frequency of monitoring also depends on the number of personnel and number of patients being monitored. How to communicate GCS? The acronym stands for both Glasgow Coma Scale (individual components) and Score (total). The scale is used to decide the management of individual patients. The score will help in measuring outcome of group of patients with same score. It is always advisable to mention the complete score with its individual parameters than just the total score. If one parameter is not assessable due to any cause (periorbital swelling precluding eye opening response assessment) then this should be mentioned as such and the total should not be made excluding this parameter. A GCS of 9/10 when Verbal response cannot be assessed will not convey the right meaning and can cause errors in management. When in serious doubt, the actual response can be described verbatim so that the message is not lost in translation. Confounding factors that make testing of GCS parameters untenable [8,9] Drugs- anaesthetics, sedatives, neuromuscular blockades, anti psychotics Cranial nerve injuries Spinal cord injuries Intoxicants (alcohol or drugs) Hearing impairment Limb injuries with or without splints Dysphasia Pre-existing illnesses eg: dementia, psychiatric conditions Ocular injuries Language and Cultural barriers Conclusion The GCS is a very simple, fast, easy and objective method of assessment of neurological condition especially in head injury. One Bhaskar S Clinics in Surgery - General Surgery Remedy Publications LLC., \| 2017 \| Volume 2 \| Article 1575 4 needs to be aware of the fact that the GCS assessment does have certain lacunae in terms of overall picture of the patient. There is no mention of vitals, pupils or any other parameters. One needs to look at the GCS in context to the overall clinical picture of the patient. Adding too many other parameters to circumvent this, will take away the simple objective nature of GCS assessment. References Teasdale G, Jennett B. Assessment of coma and impaired consciousness. A practical scale. Lancet. 1974;2(7872):81-4. Teasdale G, Jennett B, Murray L, Murray G. Glasgow coma scale: to sum or not to sum. Lancet. 1983;2(8351):678. Marshall LF, Becker DP, Bowers SA, Cayard C, Eisenberg H, Gross CR, et al. The National Traumatic Coma Data Bank. Part 1: Design, purpose, goals, and results. J Neurosurg. 1983;59(2):276-84. Laureys S, Piret S, Ledoux D. Quantifying consciousness. Lancet Neurol. 2005;4(12):789-90. American College of Surgeons. Advanced Trauma Life Support. 9 th edition. 2012. Teasdale GM. The Glasgow Coma and Outcome Scales: Practical Questions and Answers in Practical Handbook of Neurosurgery. Sindou M, editor. Springer Wien New York. 2010;395-409. Marike Zwienenberg-Lee, J. Paul Muizelaar. Clinical Pathophysiology of Traumatic Brain Injury. In Clinical Youmans Neurological Surgery Fifth Edition. H. Richard Winn (ed). Philadelphia, WB Saunders, 5039-64. Graham Teasdale, Andrew Maas, Fiona Lecky, Geoffrey Manley, Nino Stocchetti, Gordon Murray. The Glasgow Coma Scale at 40 years: standing the test of time. Lancet Neurol. 2014;13(8):844-54. Steven Laureys, Olivier Bodart, Olivia Gosseries. The Glasgow Coma Scale: time for critical appraisal? Lancet Neurol. 2014;13:755-57.
1403	https://www.youtube.com/watch?v=_N7Sf7aDLH0	Permutations with restrictions Joel Speranza Math 26100 subscribers 108 likes Description 11588 views Posted: 16 Sep 2019 Simple, easy to understand math videos aimed at High School students. Want more videos? I've mapped hundreds of my videos to the Australian senior curriculum at my website 9 comments Transcript: all right so this time we're going to do permutations with restrictions or hoping this is going to be a quick video now something we maybe didn't talk about and really the key to the whole thing is this when considering permutations with restrictions we deal with the restrictions first so we're gonna run through four examples here and each time we're going to remember that consider the permutations restrictions deal with the restrictions first one how many arrangements of the word Darwin begin and end with Val alright so this is one of those cases we're drawing a little box method is going to help you out but it's not the end of the game all right so I'm gonna write arrangements is equal to all right so the first box has to have a vowel in it so that's two there now there's four more boxes and then this last one has to have the other vowel in it because there's an A in and I okay and the other box is well there's going to be four four three two one all right now that I've drawn up my box method I'm just going to sort of neaten this up a little bit and I'm gonna say that it's two times I guess four pick four which is just four factorial times one we're gonna answer there 48 all right let's look at our next question here using the digits 0 1 2 3 4 & 5 without repetition how many odd four-digit numbers can you form alright so they have to be odd so in order for them to be odd let's do our little little box method again alright so first of all their four digit numbers maybe I'll use like a line method instead of a box now how many odd for did you know so the last digit there needs to be a 1 a 3 or a 5 so there's three options for that one and now we need to consider the rest of the rest of these so there's our one two three four five six digits in total so I wouldn't at first glance I would think that you can write the number five in here as sort of this five options there but I don't think it can be 0 and one of these odd numbers is sitting there alright so now that I know that one of the odd numbers is sitting there and this digit can't be 0 there are only 1 2 3 4 options available to me at that specific point so I've used up two of my numbers here I can use any of the rest for that for the other two so there were six digits in total I've used two so I can have four here and then I've used three now there was six so I can have three different ones there so I say the answer is 4 times 4 times 3 times 3 which is equal to 144 all right so those are some restrictions let's look at ones where the items are grouped together where Bob and Jill will need to see you know rowing boat next to each other that seem to be a classic one let's see what questions we've got here that are similar all right how many arrangements of the word equals are there if the vales are kept together ok so this is a these sort of a bit tricky but the way to think about it is to think about the word equals the three that are going to get grouped together the either you and the a you're going to get grouped together so EU a becomes like a its own letter all to itself so now I have it's gonna be equal to so I'll just do e you a there so that means that there's one letter EU a there's another letter Q there's another letter L and there's another letter yes and if I'm going to arrange four letters that's gonna be four factorial so that's the arrangement of four letters but one of the letters can be arranged in multiple ways EU a could be EU a it could be EI you it could be a ue a e u it can be arranged in three factorial ways because there's three items that I'm going to arrange all of them okay so four factorial times three factorial it's gonna give me some in 144 is the answer and I've putting the little textbook explanation of how they've described how they do that sort of question as well so if I didn't make a lot of sense there you can you can read through that one as well his next question here really kicks it up a notch it says how many ways can two chemistry for physics and followed biology books be arranged on a shelf if the books of each subject are kept together so let's look at the books first of all two chemistry books there's four physics books and there's five biology books and I've put them in their little brackets here because I'm treating them as a single units right so the way that so I've got three objects there so the way to arrange three objects is just to you 3 factorial sorry which is 6 so that's right that's right in our Arrangements arrangements is equal to so the ways to do these three blocks three factorial but then we could put the chemistry books this way or that way right so the arrangements of two objects is two factorial which is just two the physics textbooks can be arranged in a lot of different ways for factorial ways finally the biology textbooks can be arranged in five factorial ways all right so I've just sort of made some notes here for you while I was doing that so the ways that these three blocks can be arranged as three factorial the ways that the first block of me arranged is two factorial ways the second block of the arrange is four factorial and the ways that the third block can be arranged is five factorial which is a large number that's a thirty four thousand five hundred and sixty all right so this has been permutations with some sort of restrictions the best thing that you can do is consider your restrictions first
1404	https://divisible.info/Divisors/Divisors-of-57.html	Divisors of 57 Divisible Divisors of 57 Here we will define what "the divisors of 57" means and show you how to find the divisors of 57. First, note that in a division problem like x divided by y equals z, x is the dividend, y is the divisor, and z is the quotient as illustrated here: Dividend / Divisor = Quotient Divisors of 57 are all the unique whole number divisors that make the quotient a whole number if you make the dividend 57: 57 / Divisor = Quotient To find all the divisors of 57, we first divide 57 by every whole number up to 57 like so: 57 / 1 = 57 57 / 2 = 28.5 57 / 3 = 19 57 / 4 = 14.25 etc... Then, we take the divisors from the list above if the quotient was a whole number. This new list is the Divisors of 57. The Divisors of 57 are as follows: 1, 3, 19, and 57. Divisors Calculator Enter another number to get all its divisors: Divisors of 58 Here is the next number that we calculated all the divisors for. Copyright\|Privacy Policy\|Disclaimer\|Contact
1405	https://medium.com/@georgije.stanisic00/why-do-we-use-integrating-factors-in-solving-differential-equations-f292561e4a24	Why do we use Integrating Factors in Solving Differential Equations? \| by Georgije Stanisic \| Medium Sitemap Open in app Sign up Sign in Write Search Sign up Sign in Why do we use Integrating Factors in Solving Differential Equations? Georgije Stanisic Follow 5 min read · Jul 25, 2020 22 Listen Share Integrating factors are a very important part of the study of linear first-order ordinary differential equations (ODE’s). The remainder of this article will be talking about linear first-order ODE’s, so allow me to omit the first-order part in every mention of them, but don’t forget about this condition. Although a lot of ODE’s in the real world are unable to be expressed in terms of elementary functions, there is something really satisfying about having a method that works to produce a ‘neat’ solution to this important branch of linear first order ODE’s . This is where the integrating factor (I.F), as well as other methods which I won’t touch on here, come in. I.F’s are important in many physical models, including electrical circuits, heat transfer, and diffusion. When students are learning about differential equations for the first time, the integrating factor is sometimes unfortunately just given as an equation to memorize and use to solve linear ODE’s. What’s missed in this first encounter of I.F’s is the beauty of them. In this short article, I am going to try and give some intuition behind why I.F’s are used. To start, let’s assume we have a linear first-order ODE in standard form. Equation (1) For the remainder of the article, I will not specify that p and q are functions of x for clarity, but this is very important to remember. Even if they are constant functions, they are still functions of x. What we want to do is derive a formula that lets us compute the solution to this ODE without too much effort. Let’s get into that. We want to multiply both sides of Equation (1) by a function, call it u(x). This will be our integrating factor. Upon doing this, we have the following equation. Equation (2) If we go back to Calculus I, where we learned about the Chain Rule, you might see that we could make a really handy choice of u(x) here so that we can simplify equation (2). I will touch on how we figure out what this ‘handy’ choice is in a minute, but for now just imagine that we know there is some u(x) that allows us to rewrite equation (2) in the follow form: Equation (3) We can see that the entire left side of equation (2) can be re-written as the derivative of the product of u and y. We then have the much simpler looking differential equation : Equation (4) where the product uq is the right hand side of equation (2). We can simplify this even further by integrating both sides with respect to x: Finally, dividing both sides by u yields the very nice looking solution to our ODE : Equation (5) This is all well and good, but I still haven’t even mentioned what u(x) is! This was just to give you some intuition on why the I.F is used; if there exists some function u(x) that we can use to rewrite equation (3) into the right hand side, we can have the nice equation (5). Get Georgije Stanisic’s stories in your inbox Join Medium for free to get updates from this writer. Subscribe Subscribe Now what is this u(x)?! Let’s go back to equation (3) for a second. Let us carry through the differentiation on the right side. Using the chain rule from Calculus, we obtain : Equation (6) Now comparing the left side of equation (3) and the right side of equation (6) we see that we get yet another (albeit relatively simple) differential equation : Equation (7) This is actually the most fundamental differential equation (exponential growth/decay), and I believe the people reading this recognize that the solution to this equation is : Equation (8) It is important to note here that there are an infinite amount of solutions here, depending on our choice of C, but the good news here is that all of them (besides the trivial solution with C=0) will work for our substitution into equation (5). Therefore, it is standard to choose C=1 and have our integrating factor be : Equation (9) Finally, substituting equation (9) into equation (5) from earlier, we yield the beautiful solution to our differential equation : Equation (10) Although this equation might not look as nice as some other famous equations we encounter in mathematics like : I still find it beautiful to be able to have such a strong, concise way of solving equation (1) which is found in tons of applications in many sciences and engineering. I hope this helps students know why they are learning about I.F’s , and that they aren’t just some arbitrary equations they need to memorize for one exam and then forget. Differential Equations Mathematics Engineering Science 22 22 Follow Written by Georgije Stanisic ---------------------------- 18 followers ·10 following Follow No responses yet Write a response What are your thoughts? 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1406	https://www.si.edu/newsdesk/releases/inka-road-remains-monumental-achievement-engineering-after-500-years-continuous-use	About About Inka Road Remains a Monumental Achievement in Engineering After 500 Years of Continuous Use Road Expanded the Power and Reach of the South American Empire News Release May 14, 2015 Versión en español The Smithsonian’s National Museum of the American Indian will present “The Great Inka Road: Engineering an Empire,” the first major bilingual exhibition on one of the greatest civilizations in South America, June 26 through June 1, 2018. It will explore why and to what end the Inka Road was built more than 500 years ago, and how its construction, without the use of metal or iron, the wheel or stock animals to pull heavy loads, stands as one of the greatest engineering feats. The museum uses the Inka spelling rather than the more common Inca because it is consistent with the usage of the original traditional South American language of Quechua and in line with the museum’s policy on Native-language preservation. The paved road is more than 24,000 miles in length, runs north to south crossing through Argentina, Bolivia, Chile, Colombia, Ecuador and Peru. The Inka Road engaged impressive engineering strategies in response to the challenges presented by the rugged Andes mountains. This landscape ranges from coastal desert to high plateau and uneven cordillera (a system of mountain ranges) separated by valleys characterized by biodiversity and many unique ecosystems. Through images, maps, models and 140 objects, including a ceramic Chavín stirrup spout bottle (the oldest item in the exhibition, ca. 800–100 B.C.), impressive gold ornaments, necklaces made from shells from the Lambayeque region, stone carvings, silver and gold figurines, and various textiles made from camelid hair and cotton. These items illustrate important concepts in Andean cosmology, the principles of duality, reciprocity and integration, and offer examples of the road’s infrastructure and spirituality. The Inka Empire, the final period of autonomy and pure indigenous tradition in South American history, began in the 14th century and flourished until the Spanish invasion in 1532. Throughout its 100 years of use, the extensive road served as a complex network and major axis for communication, transportation, expansion, administration and political control of vast and varied territories throughout the Inka Empire. After the Spanish invasion, the road lost its original symbolism and its political meaning, but it never lost its significance as a symbol and sacred space to indigenous people in the region. Contemporary descendants of the Inka continue using the road system and millions of people still speak the native languages of Quechua and Aymara. The Inka Empire ultimately transformed the world through the dissemination of important crops, minerals and medicines. The exhibition will unfold the history of the Inka Empire in 11 sections: Exhibition curators Ramiro Matos (Quechua) and José Barreiro (Taíno) have spent the past six years researching, traveling and documenting the Inka Road in preparation for this exhibition. “The Inka are one of the primary examples of the achievements of the indigenous people in the Western Hemisphere,” said Kevin Gover (Pawnee), director of the museum. “Their knowledge, their understanding of their environment, their agriculture and, of course, their engineering all remain infinitely interesting and instructive particularly in a world that is grappling with some quite near existential challenges of sustainability. There is knowledge to be gained from the examination of these indigenous cultures not simply for the sake of gathering knowledge, but for its application in our contemporary lives.” Symposium A two-day symposium, “The Great Inka Road: Engineering an Empire,” will be held Thursday, June 25, at 1:30 p.m. and Friday, June 26, at 9 a.m. in the Rasmuson Theater. It will look at the material, political, economic and religious structures that joined more than 100 Native nations and millions of people in the powerful Native confederation known as the Tawantinsuyu. Noted international scholars, writers and engineers will discuss how the Inka organized the Andean world of the 15th and early 16th centuries, using the Qhapaq Ñan (the Road of the Lord), the empire’s 24,000-mile sacred roadway, to connect vast territories that covered the whole or parts of six modern republics: Argentina, Bolivia, Chile, Colombia, Ecuador and Peru. Qhapaq Ñan is a monumental engineering achievement that UNESCO recently recognized as a World Heritage site. Many sections of the Inka Road are still used today and continue to be revered as sacred spaces and symbols of cultural continuity. Publication The exhibition is accompanied by a book, The Great Inka Road: Engineering an Empire, which is edited by Matos and Barreiro and features 23 essays on four main themes: the Andean landscape and the city of Cusco; Inka engineering; the four regional divisions of the empire and the ways the Inka Road system linked them; and the “living” road from the colonial era to the present. Beautifully designed and featuring more than 175 full-color illustrations, the book offers a fascinating view of an enduring symbol of the Inka Empire’s strength and adaptability. For more information about the book or to purchase a copy, visit www.nmaistore.si.edu or call 800-242-NMAI (6624). Museum members receive a 20 percent discount. Folklife Festival Marketplace In conjunction with the Smithsonian’s 2015 Folklife Festival program “Perú: Pachamama,” the Festival Marketplace will be open from Wednesday, June 24, through Sunday, July 12, in the museum’s Potomac Atrium. This is a once-a-year opportunity to share the inspired—and inspiring—sights and sounds of the Smithsonian Folklife Festival. Shoppers will find beautiful works created by the Festival participants demonstrating their craft on the National Mall, including jewelry, textiles, baskets, books, toys, pottery, clothing, sculpture and paper arts. The Inka Road project is organized by the National Museum of the American Indian and is made possible by federal support from the Latino Initiatives Pool, administered by the Smithsonian Latino Center, and internal Smithsonian Institution funds from the Consortium for Valuing World Cultures. Support for the exhibition is provided by the National Council of the National Museum of the American Indian and the ESA Foundation. For more information, visit www.AmericanIndian.si.edu. To join the conversation, follow the museum’s Twitter feed, @SmithsonianNMAI, and use the hashtags #InkaRoad or #Peru2SI. Caravan of Memory In 2010, twenty-five Quechua families took a caravan of llamas from their home village, Chawaytiri, in the highlands of Peru, to the city of Pisac. The participants invited the National Museum of the American Indian to document the journey and the cultural knowledge and ceremonies that surround it. The community's message? "Let us be careful not to lose the ways and customs of our ancestors, so that nothing useful disappears." In Quechua with English subtitlesto come. Produced by the National Museum of the American Indian, Smithsonian Institution, with the Museo Comunitario P´isaq and the Municipalidad Distrital de P´isaq. Music: Grupo Aymara; "Río de Chawaytiri," Irma Alvarado Yucra; "El Hijo del Sol" & Canta Pajarillo." Grupo Amaru Pumaq Kuntur, Música Tribal Andina, Cusco-Perú. To read more about the museum's collaborative research with the people of Chawaytiri, and the larger Inka Road project, please see # SI-182-2015 Photos Image Inka Road Image Inka Road Image Inka Road: Arybalo Image Inka Road: Coca bag Image Inka Road: Gold figure Image Inka Road: Khipu Image Inka Road: Manta Image Inka Road: Qero cups Image Inka Road: Spondylus princeps (mullu) belt
1407	https://en.wikipedia.org/wiki/Combinatorial_number_system	Jump to content Search Contents (Top) 1 Ordering combinations 2 Place of a combination in the ordering 3 Finding the k-combination for a given number 3.1 Example 4 National Lottery example 5 See also 6 References 7 Further reading Combinatorial number system Português Edit links Article Talk Read Edit View history Tools Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Print/export Download as PDF Printable version In other projects Wikidata item Appearance From Wikipedia, the free encyclopedia Concept in mathematics {0,3,4,6,9} and {0,1,3,7,9} In mathematics, and in particular in combinatorics, the combinatorial number system of degree k (for some positive integer k), also referred to as combinadics, or the Macaulay representation of an integer, is a correspondence between natural numbers (taken to include 0) N and k-combinations. The combinations are represented as strictly decreasing sequences ck > ... > c2 > c1 ≥ 0 where each ci corresponds to the index of a chosen element in a given k-combination. Distinct numbers correspond to distinct k-combinations, and produce them in lexicographic order. The numbers less than correspond to all k-combinations of {0, 1, ..., n − 1}. The correspondence does not depend on the size n of the set that the k-combinations are taken from, so it can be interpreted as a map from N to the k-combinations taken from N; in this view the correspondence is a bijection. The number N corresponding to (ck, ..., c2, c1) is given by : . The fact that a combination corresponds to a non-negative integer was observed by Lehmer (1964). Indeed, a greedy algorithm finds the k-combination corresponding to N: take ck maximal with , then take ck−1 maximal with , and so forth. Finding the number N, using the formula above, from the k-combination (ck, ..., c2, c1) is also known as "ranking", and the opposite operation (given by the greedy algorithm) as "unranking"; the operations are known by these names in most computer algebra systems, and in computational mathematics. The term "combinatorial representation of integers" was shortened to "combinatorial number system" by Knuth (2011). He also references Pascal (1887). The term "combinadic" is introduced by James McCaffrey. Unlike the factorial number system, the combinatorial number system of degree k is not a mixed radix system: the part of the number N represented by a "digit" ci is not obtained from it by simply multiplying by a place value. The main application of the combinatorial number system is that it allows rapid computation of the k-combination that is at a given position in the lexicographic ordering, without having to explicitly list the k-combinations preceding it; this allows for instance random generation of k-combinations of a given set. Enumeration of k-combinations has many applications, among which are software testing, sampling, quality control, and the analysis of lottery games. Ordering combinations [edit] A k-combination of a set S is a subset of S with k (distinct) elements. The main purpose of the combinatorial number system is to provide a representation, each by a single number, of all possible k-combinations of a set S of n elements. Choosing, for any n, {0, 1, ..., n − 1} as such a set, it can be arranged that the representation of a given k-combination C is independent of the value of n (although n must of course be sufficiently large); in other words considering C as a subset of a larger set by increasing n will not change the number that represents C. Thus for the combinatorial number system one just considers C as a k-combination of the set N of all natural numbers, without explicitly mentioning n. In order to ensure that the numbers representing the k-combinations of {0, 1, ..., n − 1} are less than those representing k-combinations not contained in {0, 1, ..., n − 1}, the k-combinations must be ordered in such a way that their largest elements are compared first. The most natural ordering that has this property is lexicographic ordering of the decreasing sequence of their elements. So comparing the 5-combinations C = {0,3,4,6,9} and C′ = {0,1,3,7,9}, one has that C comes before C′, since they have the same largest part 9, but the next largest part 6 of C is less than the next largest part 7 of C′; the sequences compared lexicographically are (9,6,4,3,0) and (9,7,3,1,0). Another way to describe this ordering is view combinations as describing the k raised bits in the binary representation of a number, so that C = {c1, ..., ck} describes the number (this associates distinct numbers to all finite sets of natural numbers); then comparison of k-combinations can be done by comparing the associated binary numbers. In the example C and C′ correspond to numbers 10010110012 = 60110 and 10100010112 = 65110, which again shows that C comes before C′. This number is not however the one one wants to represent the k-combination with, since many binary numbers have a number of raised bits different from k; one wants to find the relative position of C in the ordered list of (only) k-combinations. Place of a combination in the ordering [edit] The number associated in the combinatorial number system of degree k to a k-combination C is the number of k-combinations strictly less than C in the given ordering. This number can be computed from C = {ck, ..., c2, c1} with ck > ... > c2 > c1 as follows. From the definition of the ordering it follows that for each k-combination S strictly less than C, there is a unique index i such that ci is absent from S, while ck, ..., ci+1 are present in S, and no other value larger than ci is. One can therefore group those k-combinations S according to the possible values 1, 2, ..., k of i, and count each group separately. For a given value of i one must include ck, ..., ci+1 in S, and the remaining i elements of S must be chosen from the ci non-negative integers strictly less than ci; moreover any such choice will result in a k-combinations S strictly less than C. The number of possible choices is , which is therefore the number of combinations in group i; the total number of k-combinations strictly less than C then is and this is the index (starting from 0) of C in the ordered list of k-combinations. Obviously there is for every N ∈ N exactly one k-combination at index N in the list (supposing k ≥ 1, since the list is then infinite), so the above argument proves that every N can be written in exactly one way as a sum of k binomial coefficients of the given form. Finding the k-combination for a given number [edit] The given formula allows finding the place in the lexicographic ordering of a given k-combination immediately. The reverse process of finding the k-combination at a given place N requires somewhat more work, but is straightforward nonetheless. By the definition of the lexicographic ordering, two k-combinations that differ in their largest element ck will be ordered according to the comparison of those largest elements, from which it follows that all combinations with a fixed value of their largest element are contiguous in the list. Moreover the smallest combination with ck as the largest element is , and it has ci = i − 1 for all i < k (for this combination all terms in the expression except are zero). Therefore ck is the largest number such that . If k > 1 the remaining elements of the k-combination form the k − 1-combination corresponding to the number in the combinatorial number system of degree k − 1, and can therefore be found by continuing in the same way for and k − 1 instead of N and k. Example [edit] Suppose one wants to determine the 5-combination at position 72. The successive values of for n = 4, 5, 6, ... are 0, 1, 6, 21, 56, 126, 252, ..., of which the largest one not exceeding 72 is 56, for n = 8. Therefore c5 = 8, and the remaining elements form the 4-combination at position 72 − 56 = 16. The successive values of for n = 3, 4, 5, ... are 0, 1, 5, 15, 35, ..., of which the largest one not exceeding 16 is 15, for n = 6, so c4 = 6. Continuing similarly to search for a 3-combination at position 16 − 15 = 1 one finds c3 = 3, which uses up the final unit; this establishes , and the remaining values ci will be the maximal ones with , namely ci = i − 1. Thus we have found the 5-combination {8, 6, 3, 1, 0}. National Lottery example [edit] For each of the lottery combinations c1 < c2 < c3 < c4 < c5 < c6 , there is a list number N between 0 and which can be found by adding See also [edit] Factorial number system (also called factoradics) Primorial number system Asymmetric numeral systems - also e.g. of combination to natural number, widely used in data compression References [edit] ^ Applied Combinatorial Mathematics, Ed. E. F. Beckenbach (1964), pp.27−30. ^ Generating Elementary Combinatorial Objects, Lucia Moura, U. Ottawa, Fall 2009 ^ "Combinations — Sage 9.4 Reference Manual: Combinatorics". ^ Knuth, D. E. (2005), "Generating All Combinations and Partitions", The Art of Computer Programming, vol. 4, Fascicle 3, Addison-Wesley, pp. 5−6, ISBN 0-201-85394-9. ^ Pascal, Ernesto (1887), Giornale di Matematiche, vol. 25, pp. 45−49 ^ McCaffrey, James (2004), Generating the mth Lexicographical Element of a Mathematical Combination, Microsoft Developer Network Further reading [edit] Huneke, Craig; Swanson, Irena (2006), "Appendix 5", Integral closure of ideals, rings, and modules, London Mathematical Society Lecture Note Series, vol. 336, Cambridge, UK: Cambridge University Press, ISBN 978-0-521-68860-4, MR 2266432 Caviglia, Giulio (2005), "A theorem of Eakin and Sathaye and Green's hyperplane restriction theorem", Commutative Algebra: Geometric, Homological, Combinatorial and Computational Aspects, CRC Press, ISBN 978-1-420-02832-4 Green, Mark (1989), "Restrictions of linear series to hyperplanes, and some results of Macaulay and Gotzmann", Algebraic Curves and Projective Geometry, Lecture Notes in Mathematics, vol. 1389, Springer, pp. 76–86, doi:10.1007/BFb0085925, ISBN 978-3-540-48188-1 Retrieved from " Categories: Combinatorics Factorial and binomial topics Hidden categories: CS1: long volume value Articles with short description Short description matches Wikidata Combinatorial number system Add topic
1408	https://human.libretexts.org/Bookshelves/Languages/Chinese/CHN101_Elementary_Mandarin_I_Textbook/03%3A_Origins_and_Language/3.08%3A_Lesson_2_Grammar_-_Yes-no_questions_with_ma	Skip to main content 3.8: Lesson 2 Grammar - Yes-no questions with 吗 (ma) Last updated : Aug 12, 2020 Save as PDF 3.7: Lesson 2 Grammar - Standard negation with 不 (bù) 3.9: Lesson 2 Homework Exercises Page ID : 66563 Carl Polley (裴凯) Kapi'olani Community College ( \newcommand{\kernel}{\mathrm{null}\,}) The question particle 吗 (ma) is a simple way to form questions in Chinese. The tone of this character is a neutral tone, which can be written in pinyin as ma or ma0 or ma5. By placing 吗 (ma) on the end of a statement, you convert it into a yes/no question (questions that could be answered with "yes" or "no" in English). Structure Any statement can be converted into a yes/no question with 吗 (ma). You could think of 吗 (ma) as being like a question mark you say out loud. So the basic structure is: [Statement] + 吗？ Examples 你喜欢咖啡。 Nǐ xǐhuan kāfēi. You like coffee. The above statement "You like coffee" can easily be converted into the question "Do you like coffee?" by adding 吗 (ma): 你喜欢咖啡吗？ Nǐ xǐhuan kāfēi ma? Do you like coffee? More examples of yes/no questions that revert to statements when you remove the 吗 (ma): 你是大学生吗？ Nǐ shì dàxuéshēng ma? Are you a college student? 他是老板吗？ Tā shì lǎobǎn ma? Is he the boss? 你喜欢她吗？ Nǐ xǐhuan tā ma? Do you like her? 你想家吗？ Nǐ xiǎng jiā ma? Do you miss home? 你们明天见面吗？ Nǐmen míngtiān jiànmiàn ma? Are you going to meet tomorrow? 你们也去吗？ Nǐmen yě qù ma? Are you also going? 他在你们学校学中文吗？ Tā zài nǐmen xuéxiào xué Zhōngwén ma? Does he study Chinese in your school? 妈妈会做饭吗？ Māma huì zuòfàn ma? Does mom know how to cook? It's important to remember that you do not normally add 吗 (ma) to a sentence that's already a question. For example: 你是谁吗？<< BAD EXAMPLE, DO NOT SAY IT THIS WAY! Nǐ shì shéi ma? [Note: 谁 (shéi) is already a question word.] 这是不是书吗？<< BAD EXAMPLE, DO NOT SAY IT THIS WAY! Zhè shì bu shì shū ma? [Note: 是不是 (shì bu shì) is already a question pattern.] These would be something like "Are you who are you?" and "Is this is a book?" in English, both obviously ungrammatical. Still, if you're not careful, you may find yourself throwing a 吗 (ma) onto the end of a question that doesn't need it. Many learners make this mistake, so don't worry if it happens every once in a while, just catch it and remember it the next time. How to Answer Questions with 吗 (ma) You can answer a Chinese yes-no question in one of two ways: Answer with 对 (duì "correct") or the more casual 嗯 (ǹg, like English "mm-hmm" as an affirmative) to affirm what was asked. Answer a clearer "yes" by simply repeating the positive form of the verb, or "no" by using the negative form of the verb. Here's the slightly tricky part: if you answer with 对 (duì) and the question is in the positive, then you're saying "yes" (and affirming the positive verb in the question). If you answer with 对 (duì) and the question is in the negative, then you're saying "no" (and affirming the negative verb in the question). Let's take a look at some examples of this sort. A: 你是大学生吗？ Nǐ shì dàxuéshēng ma? Are you a college student? [Note: Positive verb in the question.] B: 对。 Duì. Yes, I am. / That's correct, I am. [Note: 對 (duì) affirms the positive verb.] A: 你没有工作吗？ Nǐ méiyǒu gōngzuò ma? Do you not have a job? [Note: Negative verb in the question.] B: 对。 Duì. No, I don't. / That's correct, I don't. [Note: 對 (duì) affirms the negative verb.] Now let's try some answers that reuse the verb for a super-clear "yes" or "no," which works the same way regardless of whether it's a positive or a negative verb in the question. A: 你明天不来吗？ Nǐ míngtiān bù lái ma? You're not coming tomorrow? [Note: Negative verb in the question.] B: 来。 Lái. Yes, I'll come. [Note: Repeat the verb when responding.] A: 你明天不来吗？ Nǐ míngtiān bù lái ma? You're not coming tomorrow? [Note: Negative verb in the question.] B: 不来。 Bù lái. No, I won't come. [Note: The negative verb means "no."] Finally, a mix of the two ways to answer, where one person is asking questions of two different people. A: 你喜欢中国菜吗？ Nǐ xǐhuan Zhōngguó cài ma? Do you like Chinese food? [Note: Positive verb in the question.] B: 喜欢。 Xǐhuan. Yes, I do. [Note: Repeat the verb in the response.] A: 你不喜欢中国菜吗？ Nǐ bù xǐhuan Zhōngguó cài ma? Don't you like Chinese food? [Note: The negative verb is in the question.] C: 对。 Duì. No, I don't. / That's correct, I don't. [Note: 对 (duì) affirms the negative verb.] You may sometimes imagine that "yes" in Chinese is 是 (shì) and "no" is 不是 (bù shì). This can be true, but only when the main verb in the question is also 是 (shì). If the verb in the question is something else, like 喜欢 (xǐhuan), then whatever is the main verb becomes the word for "yes." It's been said that Chinese has hundreds of ways to say "yes," and this is why: every verb can be used to mean "yes" as a response to a question. More Advanced Usage However, this doesn't mean that a sentence can't ever have a question word and 吗 (ma). If a sentence contains verbs of understanding such as 知道 (zhīdào),了解 (liǎojiě), 明白 (míngbai), 认识 (rènshi), etc., then 吗 (ma) can still be added at the end of the question. You will learn later on more about such advanced yes-no questions involving "ma". [adapted from AllSet Learning Chinese Grammar Wiki, Creative Commons License BY-NC-SA 3.0] Any Questions? If you have any questions about this grammar point, please ask in the class forums! 3.7: Lesson 2 Grammar - Standard negation with 不 (bù) 3.9: Lesson 2 Homework Exercises
1409	https://www.reddit.com/r/askmath/comments/1d8nmhf/seeking_further_explanation_for_a_comment_on_math/	Seeking further explanation for a comment on Math Stack Exchange. : r/askmath Skip to main contentSeeking further explanation for a comment on Math Stack Exchange. : r/askmath Open menu Open navigationGo to Reddit Home r/askmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to askmath r/askmath r/askmath This subreddit is for questions of a mathematical nature. Please read the subreddit rules below before posting. 209K Members Online •1 yr. ago Go_1707 Seeking further explanation for a comment on Math Stack Exchange. Analysis I ran into this question on Math Stack Exchange. Someone commented on the original post (The first comment): There is no inverse function to sin(x)on(0,π). That is, yes, "something" cannot be found and it does matter. The comment received quite a lot of upvotes, so I suppose it was on point, but I struggled to understand what he meant. Why is the inverse function of sin(x) considered here? The OP suggested a substitution u = sin(x). Why do we need to 'extract' x from u? I scrolled up and saw a potentially related comment by user Mark Viola, where he made the substitution from x to f(u) (he used the variable t) instead of from u to f(x). I could follow this particular answer on its own, but I couldn't seem to use its reasoning to understand the comment mentioned above. Could someone please explain the quoted comment in more detail? I didn't post my question in there as the current commenters seem to understand each others' arguments and argue over advanced stuff so I just don't want to throw my potentially basic question into this discussion 🥲 Thank you! Read more Archived post. New comments cannot be posted and votes cannot be cast. Share Related Answers Section Related Answers Understanding limits using intuitive examples How to approach word problems in algebra Exploring the Fibonacci sequence in nature Strategies for mastering calculus derivatives Using matrices to solve systems of equations New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community 0 0 Top Posts Reddit reReddit: Top posts of June 5, 2024 Reddit reReddit: Top posts of June 2024 Reddit reReddit: Top posts of 2024 Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation
1410	https://www.chegg.com/homework-help/questions-and-answers/1-find-market-equilibrium-following-demand-supply-qd-128-9p-0-qs-32-7p-0-2-find-equilibriu-q85433490	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: 1. Find the market equilibrium for the following demand and supply. Qd – 128 +9P = 0 Qs + 32 – 7P = 0. 2. Find the equilibrium price and quantity for the following related market for two goods. Qdi = 82 – 3P1 + P2 Qs1 = -5 + 15P1. and Qd2 = 92 +2P - 4P2 Qs2 = –6 + 32P2. Market prices and resource allocation are determined by the basic economic concepts of supply and de... Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
1411	https://math.stackexchange.com/questions/5087606/maximize-product-given-weighted-sum-constraint	optimization - Maximize product given weighted sum constraint - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Maximize product given weighted sum constraint Ask Question Asked 1 month ago Modified1 month ago Viewed 73 times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. Let a 1,…,a k∈N∪{0}a 1,…,a k∈N∪{0} satisfy the constraint: ∑i i a i=k∑i 𝑖 a i=𝑘. I’m trying to prove that: ∏i(1+a i)≤2 O(k√)∏i(1+a i)≤2 O(k). The rough intuition is that the product is maximized when the a i a i s are as “spread out” as possible. Specifically, when each a i∈{0,1}a i∈{0,1}, and the indices with a i=1 a i=1 are smallest. In that case, there are at most O(k−−√)𝑂(k) such terms, and the product is 2 O(k√)2 𝑂(𝑘). I’m looking for a formal proof that this is indeed the worst case. optimization Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications asked Aug 3 at 12:35 user994347user994347 31 2 2 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 3 Save this answer. Show activity on this post. Your intuition is not quite correct. For example, if k=10 k=10, then your intuitive solution a i=1,i∈{1,2,3,4}a i=1,i∈{1,2,3,4} with product 2 4=16 2 4=16 is inferior to the solution a i=3,a 2=2,a 3=1 a i=3,a 2=2,a 3=1 with product 4×3×2=24 4×3×2=24. Still the O(k−−√)O(k) bound has enough cushion to accommodate this. First we observe that a i≥a j,i0 a n>0, then a i>0 a i>0 for i≤n i≤n, and so k=Σ n i a i≥Σ n i 1=n(n+1)/2 k=Σ i n a i≥Σ i n 1=n(n+1)/2 implying that n(n+1)≤2 k n(n+1)≤2 k or n+1≤2 k−−√n+1≤2 k. Hence n=O(k−−√)n=O(k). Now fix n=2 k−−√n=2 k (or a higher value) and consider the maximization problem max Π n i=1(1+a i)max Π i=1 n(1+a i) s.t. Σ n i i a i=k Σ i n i a i=k with the relaxation that a i a i are no longer constrained to be integers. This relaxation can only increase the product, and so any upper bound that we establish for this relaxed problem is also an upper bound for the original problem. We form the Lagrangian max Π n i=1(1+a i)−λ(Σ n i i a i−k)max Π i=1 n(1+a i)−λ(Σ i n i a i−k) Differentiating with respect to a j a j, we get (letting M=Π n i=1(1+a i)M=Π i=1 n(1+a i)) M 1+a j=λ j O R j(1+a j)=M λ M 1+a j=λ j O R j(1+a j)=M λ This implies that j(1+a j)j(1+a j) has the same value for all j j. We can now see why your intuition that all a j a j are equal is not quite correct. Smaller j j is associated with higher a j a j. Now we know that a n≤1 a n≤1. This is because a j≥a n∀j≤n a j≥a n∀j≤n and Σ j a j=k Σ j a j=k. So if a n≥1 a n≥1, we would have Σ j a j≥Σ j=n(n+1)/2>2 k−−√2 k−−√/2=k Σ j a j≥Σ j=n(n+1)/2>2 k 2 k/2=k by our choice of n n. Hence we have j(1+a j)=n(1+a n)≤2 n j(1+a j)=n(1+a n)≤2 n and so 1+a j≤2 n/j 1+a j≤2 n/j. We get Π n i=1(1+a i)≤Π n i=1 2 n j=2 n n n n!Π i=1 n(1+a i)≤Π i=1 n 2 n j=2 n n n n! Using the Stirling formula for the factorial the last expression becomes 2 n n n O(n−−√)n n e−n 2 n n n O(n)n n e−n Since e<4 e<4, we can bound this by 2 n 2 2 n=2 3 n=2 O(k√)2 n 2 2 n=2 3 n=2 O(k) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Aug 12 at 6:11 answered Aug 4 at 13:19 Jayanth R VarmaJayanth R Varma 4,372 1 1 gold badge 2 2 silver badges 4 4 bronze badges 2 Thanks for the detailed answer! I’m not sure I follow the step where you conclude that a n≤1 a n≤1 because a j≥a n a j≥a n for all j≤n j≤n and ∑j a j=k∑j a j=k.user994347 –user994347 2025-08-11 11:04:21 +00:00 Commented Aug 11 at 11:04 Have edited my answer to clarify this a bit. Also fixed some typos.Jayanth R Varma –Jayanth R Varma 2025-08-12 06:12:05 +00:00 Commented Aug 12 at 6:12 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions optimization See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 1Maximizing a variable sum 4Convex quadratic program with linear equality constraint and bounds on variables 0Lagrange multiplier for sum of products with sum constraint 0Thresholding for relaxed linear constraint to satisfy the combinatorial constraint 2Intuition for the variables x i x i that minimise ∑i a i x 2 i∑i a i x i 2 with constraint ∑i x i=1∑i x i=1 2Maximize ∑a 3 i∑a i 3 for positive a i a i's with ∑a i=∑a 2 i=n−1∑a i=∑a i 2=n−1 1Maximize Frobenius norm of product under Frobenius norm constraint 3How to maximize the product of areas in a disk? Hot Network Questions On being a Maître de conférence (France): Importance of Postdoc An odd question Another way to draw RegionDifference of a cylinder and Cuboid What were "milk bars" in 1920s Japan? 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1412	https://dspace.mit.edu/bitstream/handle/1721.1/34888/18-03Spring2004/OcwWeb/Mathematics/18-03Spring2004/Readings/detail/supnotes2.htm	MIT OpenCourseWare \| Mathematics \| 18.03 Differential Equations, Spring 2004 \| Readings \| 18.03 Notes and Exercises skip to content Search » Advanced Search Course Home Syllabus Calendar Readings Lecture Notes Recitations Assignments Exams Tools Video Lectures Download this Course »MIT OpenCourseWare»Mathematics»Differential Equations, Spring 2004 18.03 Notes and Exercises These notes and exercises were written by Prof. Arthur Mattuck and are designed to supplement the textbook. \| SECTIONS \| TOPICS \| --- \| \| Notes \| \| D \| Definite Integral Solutions (PDF) \| \| G \| Graphical and Numerical Methods (PDF) \| \| C \| Complex Numbers (PDF) \| \| IR \| Input-Response Models (PDF) \| \| O \| Linear Differential Operators (PDF) \| \| S \| Stability (PDF) \| \| I \| Impulse Response and Convolution (PDF) \| \| LT \| Laplace Transform (PDF) \| \| LS \| Linear Systems of ODE's - LS1 of 6 (PDF), LS2 of 6 (PDF), LS3 of 6 (PDF), LS4 of 6 (PDF), LS5 of 6 (PDF), LS6 of 6 (PDF) \| \| GS \| Graphing Systems (PDF) \| \| LC \| Limit Cycles (PDF) \| \| Exercises \| \| 1 \| First-order ODE's (PDF) \| \| 2 \| Higher-order ODE's (PDF) \| \| 3 \| Laplace Transform (PDF) \| \| 4 \| Linear Systems (PDF) \| \| 5 \| Graphing Systems (PDF) \| \| 6 \| Power Series (PDF) \| \| 7 \| Fourier Series (PDF) \| \| Solutions to Exercises \| \| 1 \| First-order ODE's (PDF) \| \| 2 \| Higher-order ODE's (PDF) \| \| 3 \| Laplace Transform (PDF) \| \| 4 \| Linear Systems (PDF) \| \| 5 \| Graphing Systems (PDF) \| \| 6 \| Power Series (PDF) \| \| 7 \| Fourier Series (PDF) \| Massachusetts Institute of Technology © 2005 MIT Privacy Legal Notices Your use of the MIT OpenCourseWare site and course materials is subject to the conditions and terms of use in our Legal Notices section.
1413	https://math.stackexchange.com/questions/3948685/splitting-a-square-into-equal-areas	geometry - Splitting a square into equal areas - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Splitting a square into equal areas Ask Question Asked 4 years, 9 months ago Modified4 years, 9 months ago Viewed 1k times This question shows research effort; it is useful and clear 3 Save this question. Show activity on this post. I am looking to split a square into 4 equal areas, not necessarily needing to be the same shapes or sizes . I need to find a way to use the least length of lines ( which could be straight or curved lines) inside the square to split and divide it up. What would be the optimal solution. I need to code an algorithm to do it too but I wanted to understand the mathematical concept behind it first. Also how would I use this for a rectangle say side height 1 and length 2. geometry Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Dec 14, 2020 at 20:48 MathIsFunMathIsFun asked Dec 14, 2020 at 18:09 MathIsFunMathIsFun 41 4 4 bronze badges 3 I can't imagine it would be anything different than ⊞⊞... I don't have a proof in mind, but its obvious that it uses less length than ⊠⊠.JMoravitz –JMoravitz 2020-12-14 18:15:28 +00:00 Commented Dec 14, 2020 at 18:15 It could also be something similar to what happens with minimum surface area where the shapes get more involved and exotic... but using these sort of minimum surface tension with soap films wouldn't respect the requirement that the regions be of equal area.JMoravitz –JMoravitz 2020-12-14 18:28:46 +00:00 Commented Dec 14, 2020 at 18:28 Yeah but the most obvious way is clearly not the most optimal.MathIsFun –MathIsFun 2020-12-14 20:49:12 +00:00 Commented Dec 14, 2020 at 20:49 Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 4 Save this answer. Show activity on this post. My calculations show that the obvious configuration is not quite optimal. Consider the square [0,1]×[0,1][0,1]×[0,1]. The division by straight horizontal and vertical lines has total length 2 2. Let α α be the root of the polynomial 768 z 8+64 z 6−72 z 4+1 768 z 8+64 z 6−72 z 4+1 approximately 0.4694760731 0.4694760731, and β=1/(4 α)β=1/(4 α). Partition the square using the points [α,α][α,α], [1−α,1−α][1−α,1−α], [0,β][0,β], [β,0][β,0], [1,1−β][1,1−β], [1−β,1][1−β,1], looking like this: If I'm not mistaken, this has equal areas and total length 4 α 2+(α−β)2−−−−−−−−−−−√+2–√(1−2 α)≈1.981089021<2 4 α 2+(α−β)2+2(1−2 α)≈1.981089021<2. I'm not claiming that this is optimal, either. You should do better by replacing the four straight line segments touching the boundary with circular arcs. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications edited Dec 14, 2020 at 19:00 answered Dec 14, 2020 at 18:55 Robert IsraelRobert Israel 472k 28 28 gold badges 376 376 silver badges 714 714 bronze badges 6 I was thinking the same thing, actually! I needed to be quicker from lunch :-S +1.Oscar Lanzi –Oscar Lanzi 2020-12-14 19:13:26 +00:00 Commented Dec 14, 2020 at 19:13 How would I do it for circular arcs though please.MathIsFun –MathIsFun 2020-12-14 20:36:19 +00:00 Commented Dec 14, 2020 at 20:36 Also would it be the exact same for a shape like a rectangle. Say height 1 and length 2.MathIsFun –MathIsFun 2020-12-14 20:37:21 +00:00 Commented Dec 14, 2020 at 20:37 1 Also how did you get that polynomial ?MathIsFun –MathIsFun 2020-12-14 20:44:38 +00:00 Commented Dec 14, 2020 at 20:44 Thanks for all the help btw.MathIsFun –MathIsFun 2020-12-14 20:50:11 +00:00 Commented Dec 14, 2020 at 20:50 \|Show 1 more comment You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Related 31Trying to understand why circle area is not 2 π r 2 2 π r 2 32Algorithm to get the maximum size of n squares that fit into a rectangle with a given width and height 7Can you divide a square into 5 equal area regions 0How to divide a rectangle into fewest equally sized nearly-square sub-rectangles 2How to calculate optimal sizes of rectangles for this type of array visualization? 15Dividing a unit square into rectangles 3Triangle Quadrilateral and pentagon whose areas form a set of consecutive positive integers. 2How to find the largest semi-circle that fits inside a polygon? 2How to divide a square into equal parts using straight lines Hot Network Questions alignment in a table with custom separator Xubuntu 24.04 - Libreoffice How can the problem of a warlock with two spell slots be solved? 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1414	https://www.youtube.com/watch?v=2eXC6s9X6Wc	Add and Subtract with Significant Figures (1.6) Tyler DeWitt 10840 likes 1150620 views 1 Aug 2010 To see all my Chemistry videos, check out How to do addition and subtraction with significant figures, also known as significant digits. 709 comments
1415	https://ecampusontario.pressbooks.pub/introstats/chapter/3-4-the-complement-rule/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 3.4 The Complement Rule LEARNING OBJECTIVES Calculate probabilities using the complement rule. The complement of an event [latex]A[/latex] is the set of all outcomes in the sample space that are not in [latex]A[/latex]. The complement of [latex]A[/latex] is denoted by [latex]A^C[/latex] and is read “not [latex]A[/latex].” EXAMPLE Suppose a coin is flipped two times. Previously, we found the sample space for this experiment: [latex]S={HH, HT, TH, TT}[/latex] where [latex]H[/latex] is heads and [latex]T[/latex] is tails. What is the complement of the event “exactly one head”? What is the complement of the event “at least one tail.” Solution: The event “exactly one head” consists of the outcomes [latex]HT[/latex] and [latex]TH[/latex]. The complement of “exactly one head” consists of the outcomes [latex]HH[/latex] and [latex]TT[/latex]. These are the outcomes in the sample space [latex]S[/latex] that are NOT in the original event “exactly one head.” The event “at least one tail” consists of the outcomes [latex]HT[/latex], [latex]TH[/latex], and [latex]TT[/latex]. The complement of “at least one tail” consists of the outcomes [latex]HH[/latex]. These are the outcomes in the sample space [latex]S[/latex] that are NOT in the original event “at least one tail.” TRY IT Suppose you roll a fair six-sided die with the numbers [latex]1, 2, 3, 4, 5, 6[/latex] on the faces. Previously, we found the sample space for this experiment: [latex]S={1,2,3,4,5,6}[/latex] What is the complement of the event “rolling a 4”? What is the complement of the event “rolling a number greater than or equal to 5”? What is the complement of the event “rolling a even number”? What is the complement of the event “rolling a number less than 4”? Click to see Solution The complement is [latex]{1,2,3,5,6}[/latex]. The complement is [latex]{1,2,3,4}[/latex]. The complement is [latex]{1,3,5}[/latex]. The complement is [latex]{1,2,3,4}[/latex]. The Probability of the Complement In any experiment, an event [latex]A[/latex] or its complement [latex]A^C[/latex] must occur. This means that [latex]\displaystyle{P(A)+P(A^C)=1}[/latex]. Rearranging this equation gives us a formula for finding the probability of the complement from the original event: [latex]\begin{eqnarray}P(A^C)&=&1-P(A)\\\end{eqnarray}[/latex] EXAMPLE An online retailer knows that 30% of customers spend more than $100 per transaction. What is the probability that a customer spends at most $100 per transaction? Solution: Spending at most $100 ($100 or less) per transaction is the complement of spending more than $100 per transaction. [latex]\begin{eqnarray} P(\mbox{at most \$100}) & = & 1-P(\mbox{more than \$100}) \ & = & 1-0.3 \ & = & 0.7 \end{eqnarray}[/latex] TRY IT At a local college, a statistics professor has a class of 80 students. After polling the students in the class, the professor finds out that 15 of the students play on one of the school’s sports team and 60 of the students have part-time jobs. What is the probability that a student in the class does not play on one of the school’s sports teams? What is the probability that a student in the class does not have a part-time job? Click to see Solution [latex]\displaystyle{P(\mbox{no sports team})=1-P(\mbox{sports team})=1-\frac{15}{80}=0.8125}[/latex] [latex]\displaystyle{P(\mbox{no part-time job})=1-P(\mbox{part-time job})=1-\frac{60}{80}=0.25}[/latex] Concept Review The complement, [latex]A^C[/latex], of an event [latex]A[/latex] consists of all of the outcomes in the sample space that are NOT in event [latex]A[/latex]. The probability of the complement can be found from the original event using the formula: [latex]\displaystyle{P(A^C)=1-P(A)}[/latex]. Attribution “3.1 Terminology” in Introductory Statistics by OpenStax is licensed under a Creative Commons Attribution 4.0 International License. License Introduction to Statistics Copyright © 2022 by Valerie Watts is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted. Share This Book
1416	https://books.google.com/books?id=gsv7HALW2jYC&lr=	Computational Geometry in C - Joseph O'Rourke - Google Books Sign in Hidden fields Try the new Google Books Books View sample Add to my library Try the new Google Books Check out the new look and enjoy easier access to your favorite features Try it now No thanks Try the new Google Books My library Help Advanced Book Search Get print book No eBook available Cambridge University Press Amazon.com Barnes&Noble.com Books-A-Million IndieBound Find in a library All sellers» My library My History Computational Geometry in C =========================== Joseph O'Rourke Cambridge University Press, Oct 13, 1998 - Computers - 376 pages This is the revised and expanded 1998 edition of a popular introduction to the design and implementation of geometry algorithms arising in areas such as computer graphics, robotics, and engineering design. The basic techniques used in computational geometry are all covered: polygon triangulations, convex hulls, Voronoi diagrams, arrangements, geometric searching, and motion planning. The self-contained treatment presumes only an elementary knowledge of mathematics, but reaches topics on the frontier of current research, making it a useful reference for practitioners at all levels. The second edition contains material on several new topics, such as randomized algorithms for polygon triangulation, planar point location, 3D convex hull construction, intersection algorithms for ray-segment and ray-triangle, and point-in-polyhedron. The code in this edition is significantly improved from the first edition (more efficient and more robust), and four new routines are included. Java versions for this new edition are also available. All code is accessible from the book's Web site ( or by anonymous ftp. More » Preview this book » Selected pages Title Page Table of Contents Index References Contents II 1 III 11 IV 16 V 24 VI 27 VII 32 VIII 44 X 47 XXXVII 169 XXXVIII 179 XXXIX 182 XL 191 XLI 193 XLII 194 XLIII 199 XLIV 201 More XI 51 XII 56 XIII 58 XIV 63 XV 64 XVI 66 XVII 68 XVIII 69 XIX 72 XX 87 XXI 88 XXII 91 XXIII 96 XXIV 101 XXVI 109 XXVII 117 XXVIII 146 XXIX 149 XXX 150 XXXI 153 XXXII 155 XXXIV 157 XXXV 161 XXXVI 165 XLVI 205 XLVII 209 XLVIII 218 XLIX 220 L 226 LI 239 LII 245 LIII 252 LIV 264 LV 266 LVI 269 LVII 272 LVIII 285 LIX 294 LX 295 LXI 300 LXII 302 LXIII 313 LXIV 322 LXV 347 LXVI351 LXVII361 Copyright Less Other editions - View all Computational Geometry in C Joseph O'Rourke Limited preview - 1998 Computational Geometry in C Joseph O'Rourke No preview available - 1994 Computational Geometry in C Joseph O'Rourke No preview available - 1994 Common terms and phrases a₁adjacentai+1anglearbitraryArea2arrangementaxisboolboundarycellChaptercirclecollinearcomplexitycomputational geometryconeconstructedconvex hullconvex polygoncoordinatescounterclockwisecrossingscubedata structuredefineDelaunay triangulationdeleteddetermineddiagonaldualEdelsbrunnerendpointsEquationexampleExercisefaceshalfplanehorizontalimplementationincremental algorithmindependent setinflagintegerinteriorLemmalengthlinearlooplower boundmedial axisMinkowski summonotone mountainsmonotone polygonmovingnearest neighbornodesnonconvexnumber of verticesO(logO(n logO(n²O(n³O(nlognobstaclesorthogonal polygonoutputp₁partitionPi+1planepointerpolyhedrapolyhedronpolytopeproblemprogrammingproofqueryreachabilityregionregular polytopesrightmostrobotSectionsegmentset of pointsshortest pathshown in Figuresortingstackstar polygonstangenttetrahedronTheoremthree dimensionstPointitrapezoidtreetwo-dimensionaltypedefvectorvertexvisibility graphVoronoi diagram Popular passages Page 351 - A. Aggarwal, LJ Guibas, J. Saxe, and PW Shor. A linear-time algorithm for computing the Voronoi diagram of a convex polygon. Discrete Comput.‎ Appears in 42 books from 1957-2008 Page 360 - An O (n log log n)- time algorithm for triangulating a simple polygon, SIAM J. Comput.‎ Appears in 6 books from 1975-1999 More Page 360 - New results in computational geometry relevant to pattern recognition in practice," in Pattern Recognition in Practice II, ES Gelsema & LN Kanal, Eds., North-Holland, 1986, pp.‎ Appears in 5 books from 1987-1999 Less References to this book Quantitative Geography: Perspectives on Spatial Data Analysis A Stewart Fotheringham,Chris Brunsdon,Martin Charlton Limited preview - 2000 Open Source GIS: A GRASS GIS Approach Markus Neteler,Helena Mitasova Limited preview - 2002 All Book Search results » Bibliographic information Title Computational Geometry in C Cambridge Tracts in Theoretical Computer Science Computational Geometry in C, Joseph O'Rourke AuthorJoseph O'Rourke Edition illustrated, reprint, revised Publisher Cambridge University Press, 1998 ISBN 0521649765, 9780521649766 Length 376 pages SubjectsComputers › Languages › General Computers / General Computers / Languages / C Computers / Languages / General Computers / Programming / Algorithms Mathematics / Geometry / General Export CitationBiBTeXEndNoteRefMan About Google Books - Privacy Policy - Terms of Service - Information for Publishers - Report an issue - Help - Google Home
1417	https://www.studyforfe.com/blog/logic-minimization/	Skip to content 🌟 Summer Special – 10% OFF on All Plans! 10 20 31 17 FE Electrical Course PE Power Course Home Blog FE Electrical Tips Logic Minimization in FE Electrical Exam Logic Minimization in FE Electrical Exam Wasim Asghar Published: Jul 5, 2024 Updated: Sep 30, 2024 Explore this crucial exam topic in depth to know all about logic minimization in the FE Electrical exam, covering SOP, POS, and K-Maps in detail. Logic minimization is a crucial concept related to digital circuit design in the FE Electrical exam. It aims to reduce the complexity and optimize the performance of logic circuits. Understanding how to reduce complex equations with logic minimization using algebraic methods and K-Maps is a crucial topic per the NCEES® exam guidelines for the exam. If you are gearing up for FE exam preparation, this is the topic you cannot afford to miss. Let’s explore this crucial exam topic in depth to learn about logic minimization techniques in the FE Electrical exam. Understanding Logic Gates Logic gates are fundamental building blocks of digital circuits that perform Boolean operations on one or more binary inputs to produce a single binary output based on a specific truth table. The basic logic gates include: AND Gate: Outputs true (1) only if all inputs are true. OR Gate: Outputs true (1) if at least one input is true. NOT Gate: Inverts the input; outputs true (1) if the input is false (0), and vice versa. NAND Gate: Combination of AND gate followed by NOT gate. NOR Gate: Combination of OR gate followed by NOT gate. XOR Gate: Outputs true (1) if the number of true inputs is odd. Need for Minimization of Logical Circuits Digital circuits can become complex quickly, especially in larger designs. Each additional logic gate introduces propagation delay, increases power consumption, and adds to the circuit’s physical size. Therefore, minimizing the number of logic gates required for a given logic function is essential to: Reduce Cost: Fewer gates mean lower production costs. Improve Performance: Minimized circuits operate faster due to reduced propagation delay. Enhance Reliability: Simplified circuits are less prone to errors and easier to debug. Logic Minimization Techniques Logic minimization involves simplifying Boolean expressions or truth tables to achieve an equivalent but simpler logic circuit. Two primary methods are used for logic minimization: Boolean Algebra The approach Utilizes algebraic rules to simplify Boolean expressions. The key rules include: Commutative: A+B = B+A Associative: (A+B) + C = A + (B+C) Distributive: A ⋅ (B+C) = (A⋅B) + (A⋅C) Addition: A + A = 1 A + 1 = 1 A + 0 = A Multiplication: A . 0 = 0 A . 1 = A A . A = 0 Karnaugh Maps (K-Maps) Graphical method to visualize and minimize Boolean functions for up to four variables. K-Maps exploit the adjacency of 1s in truth tables to identify prime implicants and minimize the number of terms in the simplified expression. SOP and POS in Boolean Algebra SOP (Sum of Products) SOP, or Sum of Products, represents a Boolean expression combining multiple terms (products). Each product term consists of ANDed variables or their complements. SOP is useful for expressing a logic function where we list all combinations of inputs that result in an output of 1. Example of SOP For a Boolean function F (A, B, C) = A’B’C + AB’C + ABC’, the terms A’B’C, AB’C, and ABC’ are combined. POS (Product of Sums) POS, or Product of Sums, represents a Boolean expression where multiple sums (ORed terms) are multiplied. Each sum term consists of ORed variables or their complements. POS is useful for expressing a logic function where we list all combinations of inputs that result in an output of 0. Example of POS For a Boolean function F(A, B, C) = (A + B + C’)(A’ + B’ + C)(A’ + B + C), the terms (A + B + C’), (A’ + B’ + C), and (A’ + B + C) are multiplied together. Simplification of SOP and POS with Boolean Algebra Step by Step SOP (Sum of Products) Minimization Consider the Boolean function F(A, B, C) given by the truth table: \| \| \| \| \| --- --- \| \| A \| B \| C \| F(A,B,C) \| \| 0 \| 0 \| 0 \| 1 \| \| 0 \| 0 \| 1 \| 0 \| \| 0 \| 1 \| 0 \| 1 \| \| 0 \| 1 \| 1 \| 1 \| \| 1 \| 0 \| 0 \| 0 \| \| 1 \| 0 \| 1 \| 1 \| \| 1 \| 1 \| 0 \| 0 \| \| 1 \| 1 \| 1 \| 1 \| Step 01: Identify minterms Minterms are the terms where the function F(A, B, C) equals 1. From the truth table: Minterms are 0, 2, 3, 5, 7. Step 02: Write the SOP expression Express F(A, B, C) as a sum of products (SOP) of minterms. F(A,B,C) = Σ(0,2,3,5,7) Convert minterms to their Boolean expressions: 𝑀(0) = 𝐴′𝐵′𝐶 𝑀(2) = 𝐴′𝐵𝐶 𝑀(3) = 𝐴𝐵′𝐶′ 𝑀5 = 𝐴𝐵𝐶′ 𝑀7 = 𝐴𝐵𝐶 F(A,B,C) = 𝐴′𝐵’𝐶 + 𝐴’𝐵𝐶 + 𝐴𝐵′𝐶′ + 𝐴𝐵𝐶’ + ABC Step 03: Simplify using Boolean algebra Apply Boolean algebraic rules to simplify the expression. F(A,B,C) = 𝐴′𝐵’𝐶 + 𝐴’𝐵𝐶 + 𝐴𝐵′𝐶′ + 𝐴𝐵𝐶’ + ABC Using Boolean algebra simplification techniques (associative, distributive, etc.), we can combine terms: Combine 𝐴BC and A′BC to factor BC Combine AB’C’ and ABC’ to factor AC’ Isolate A’B’C F(A,B,C) = [ABC+𝐴’𝐵𝐶] + [𝐴𝐵′𝐶′+𝐴𝐵𝐶’] + 𝐴′𝐵’𝐶 F(A,B,C) = [BC(A+𝐴’)] + [𝐴C'(𝐵′+B)] +𝐴′𝐵’𝐶 F(A,B,C) = [BC(A+𝐴’)] + [𝐴C'(𝐵′+B)] + 𝐴′𝐵’𝐶 By A+A’=1 and B+B’=1 F(A,B,C) = BC+𝐴C’ + 𝐴′𝐵’𝐶 Which is a simplified expression. Step-by-Step POS (Product of Sums) Minimization Consider the following truth table and extract its POS equation: \| \| \| \| \| --- --- \| \| A \| B \| C \| F(A,B,C) \| \| 0 \| 0 \| 0 \| 1 \| \| 0 \| 0 \| 1 \| 0 \| \| 0 \| 1 \| 0 \| 1 \| \| 0 \| 1 \| 1 \| 0 \| \| 1 \| 0 \| 0 \| 1 \| \| 1 \| 0 \| 1 \| 1 \| \| 1 \| 1 \| 0 \| 1 \| \| 1 \| 1 \| 1 \| 1 \| Step 01: Identify max terms Maxterms are the terms where the function F(A, B, C) equals 0. From the truth table: Maxterms are 1, 3. Step 02: Write the POS expression Express F(A, B, C) as a product of sums (POS) of max terms. F(A,B,C) = Π(1,3) Convert max terms to their Boolean expressions: 𝑀1 = (𝐴+𝐵+𝐶′) 𝑀4 = (A’+B+C) Step 03: Simplify using Boolean algebra Apply Boolean algebraic rules to simplify the expression. 𝐹(A,B,C) = (𝐴+𝐵+𝐶′)(A’+B+C) Using distributive law and combining terms: F(A,B,C) = A(A′+B+C)+B(A′ +B+C)+C′(A′+B+C) Apply distributive property again: F(A,B,C)=AA′ + AB + AC + BA′ + BB + BC + C′A′ + C′B + C′C Simplify using the complement property 𝐴𝐴′ = 0 𝐵𝐵 = 𝐵 and 𝐶𝐶′ = 0 F(A,B,C) = AB + AC + BA′ + B+BC + C′A′ + C′B Rearrange the terms F(A,B,C) = AB + BA′ + AC + B + BC + C′B + C′A′ F(A,B,C) = B(A+A′) + AC+B+B(C+C′) + C′A′ Using A+A’ = 1 and C+C’ = 1 F(A,B,C) = B + AC + B + B + C′A′ F(A,B,C) = B + B + B + AC + C′A′ Using B + B + B = B F(A,B,C) = B + AC + C′A′ Which is a simplified expression. Verification for Simplified POS or SOP Expressions To verify the minimization results for SOP and POS: Substitute A, B, and C values from the truth table into the simplified SOP and POS expressions. Ensure that F(A, B, C) evaluates to 1 for the minterms (0, 2, 3, 5, 7) and evaluates to 0 for the maxterms (1, 4). K-MAP – Everything You Need to Know Karnaugh Maps (K Maps) are graphical tools that simplify Boolean expressions. They are particularly useful for reducing Boolean functions to their simplest form, especially when dealing with 3 or 4 variables. Here’s a general step-by-step explanation for using K Maps for Sum of Products (SOP) with 3 and 4 variables: Step-by-Step Working for Karnaugh Maps (SOP) with 3 Variables Step 1: Construct the Karnaugh Map Dimensions: Create a K Map with 2 rows and 4 columns for 3 variables (A, B, C). Cell Arrangement: Arrange cells so that adjacent cells differ by only one variable (Gray code ordering). Step 2: Mark Minterms Identify Minterms: Mark cells corresponding to minterms where the Boolean function equals 1. Each cell corresponds to a unique combination of A, B, and C. Step 3: Group Adjacent 1s Grouping: Group adjacent 1s in powers of 2 (1, 2, 4… cells) to cover as many 1s as possible with the fewest groups. Rules for Grouping: Groups must be rectangular in shape and should ideally cover 1, 2, or 4 cells. Step 4: Derive Simplified Expression Read Grouped Terms: Each group represents a product term in the simplified Boolean expression. Combine Terms: Combine terms within each group using the variables that do not change across the group. Step 5: Verify and Complete Verify: Substitute the simplified expression into the K Map to ensure it covers all marked 1s. Check: Ensure the simplified expression matches the original Boolean function across all input combinations. For 3 Variables \| \| \| \| \| \| --- --- \| A/AB \| 00 \| 01 \| 11 \| 10 \| \| 0 \| 0 \| 1 \| 1 \| 0 \| \| 1 \| 0 \| 0 \| 1 \| 1 \| The terms involve: Group 01: A’B’C+A’BC = A’C(B+B’)=A’C Group 02: ABC+ABC’=AB(C+C’)=AB Final Expression: A’C+AB Step-by-Step Working for Karnaugh Maps (SOP) with 4 Variables Step 1: Construct the Karnaugh Map Dimensions: Create a K Map with 4 rows and 8 columns for 4 variables (A, B, C, D). Cell Arrangement: Arrange cells so that adjacent cells differ by only one variable (Gray code ordering). Step 2: Mark Minterms Identify Minterms: Mark cells corresponding to minterms where the Boolean function equals 1. Each cell corresponds to a unique combination of A, B, C, and D. Step 3: Group Adjacent 1s Grouping: Group adjacent 1s in powers of 2 (1, 2, 4, 8… cells) to cover as many 1s as possible with the fewest groups. Rules for Grouping: Groups must be rectangular in shape and should ideally cover 1, 2, 4, or 8 cells. Step 4: Derive Simplified Expression Read Grouped Terms: Each group represents a product term in the simplified Boolean expression. Combine Terms: Combine terms within each group using the variables that do not change across the group. Step 5: Verify and Complete Verify: Substitute the simplified expression into the K Map to ensure it covers all marked 1s. Check: Ensure the simplified expression matches the original Boolean function across all input combinations. For 4 Variables \| \| \| \| \| \| --- --- \| AB/CD \| 00 \| 01 \| 11 \| 10 \| \| 00 \| 0 \| 1 \| 1 \| 0 \| \| 01 \| 0 \| 1 \| 1 \| 0 \| \| 11 \| 0 \| 0 \| 1 \| 1 \| \| 10 \| 0 \| 0 \| 0 \| 1 \| Group 01: A’B’C’D + A’B’CD + A’BC’D + A’BCD D(A’B’C’ + A’B’C + A’BC’ + A’BC) D(A’B’(C’ + C) + A’B(C’ + C)) D(A’B’ + A’B) D(A’(B’ + B)) D(A’) A’D (Which are repeating variables in all 4 cells) Group 02: ABCD + ABCD’ ABC(D + D’) ABC(1) ABC (Which are repeating variables in all 2 cells) Group 03: AB’CD’ (We cannot deduct any variable as all exist for the single cell). (There is no need to solve with algebraic simplification; you just have to take the terms of their conjugate that repeats in all cells. For instance, you can directly count A’ and D and combine them in a single term as A’ and D repeats in all 4 cells of tetra selection where 1s occur). Guidelines for Working with Karnaugh Maps: Cell Ordering: Use Gray code ordering to arrange cells for variables to ensure adjacent cells differ by only one variable. Grouping: Groups should be rectangular and cover 1, 2, 4, or 8 cells (make the biggest selection possible for efficient simplification). Simplification: Combine terms within groups to form the minimized Boolean expression using common variables. Verification: Always verify the simplified expression against the original Boolean function to ensure correctness. By following these general steps and guidelines, you can effectively use Karnaugh Maps to simplify Boolean expressions with 3 or 4 variables using the Sum of Products (SOP) method. Conclusion Logic minimization in the FE Electrical exam is crucial from an exam standpoint. It also greatly helps in digital circuit design optimize performance, reduce complexity, and enhance reliability. Electrical and computer engineers can efficiently minimize logic circuits while maintaining functionality by employing Boolean algebra and Karnaugh maps. This approach ensures digital systems operate effectively with minimal resources. For more FE-related guides and exam preparation resources, explore Study for FE, your go-to place for exam preparation. Wasim Asghar - P.E, P.ENG, M.ENG Licensed Professional Engineer in Texas (PE), Florida (PE) and Ontario (P. Eng) with consulting experience in design, commissioning and plant engineering for clients in Energy, Mining and Infrastructure. Recent Posts Top Study Tips to Pass the FE Electrical Exam June 23, 2025 How Hard Is the PE Exam – What you should know! April 23, 2025 6 Benefits of Becoming a Licensed PE (Professional Engineer) April 15, 2025 How To Learn Electrical Engineering In 2025 February 10, 2025 PE Exam Passing Score 2025 January 16, 2025 How to Become an Electronics Engineer? January 10, 2025
1418	https://artofproblemsolving.com/wiki/index.php/Residue_class?srsltid=AfmBOopUbdmR5LwVfmgquJbb4uIv0fig9-Jdt_7S9_l1VXUb5F3mKnYH	Art of Problem Solving Residue class - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Residue class Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Residue class In modular arithmetic, a residue of an integer in modulo is the unique value of such that . In the context of division, a residue is simply a remainder. A residue class is a complete set of integers that are congruent modulo for some positive integer . In modulo , there are exactly different residue classes, corresponding to the possible residues Each residue class contains all integers in the form where is the corresponding residue. Retrieved from " Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1419	https://mathteachercoach.com/wp-content/uploads/2020/06/5-2-Slide-Show-Multiplying-Fractions-and-Whole-Number.pdf	Multiplication of Fractions and Whole Number Unit 5 Lesson 2 Math 5 Multiplication of Fractions and Whole Numbers Students will be able to: • Use fraction of a set to interpret fractions as division. • Use tape diagrams to multiply a whole number by a fraction. • Relate a fraction of a set to interpret fraction multiplication through repeated addition. • Solve problems involving multiplication of factions and whole number using visual models or equations. Multiplication of Fractions and Whole Numbers Key Vocabulary: Fractions Whole Numbers Multiplication Tape Diagram Arrays Numerator Denominator Multiplication of Fractions and Whole Numbers Finding Fractions of a Set A fraction is a division of the numerator by the denominator. A fraction can be expressed as division and vice versa. FRACTION DIVISION 𝑎 𝑏 𝑎÷ 𝑏 Multiplication of Fractions and Whole Numbers Finding Fractions of a Set In this lesson, you will multiply fractions and whole numbers using arrays of figures or shapes to find fractions of a set. If there are 6 cookies and you are to divide it into 3 parts equally, how many cookies will each part have? Multiplication of Fractions and Whole Numbers Finding Fractions of a Set Dividing the cookies, you’ll get: Multiplication of Fractions and Whole Numbers Finding Fractions of a Set The division sentence for the given problem is: 6 ÷ 3 2 The division sentence expressed as a fraction is: 6 3 2 If you want to know 1 third (1 3) of the cookies, how many cookies will there be in each part? Multiplication of Fractions and Whole Numbers Finding Fractions of a Set Each part/group of cookies divided equally is one-third of the entire cookies. 1 3 1 3 1 3 Multiplication of Fractions and Whole Numbers Finding Fractions of a Set This shows that when you divide 6 by 3, there are 2 in each group, therefore: 𝟏 𝟑 of 6 cookies is equal to 2 cookies. 𝟐 𝟑 of 6 cookies is equal to 4 cookies. 𝟑 𝟑 of 6 cookies is equal to 6 cookies. Multiplication of Fractions and Whole Numbers Example 1: What is 1 3 of 12? Make an array of 12 shapes (of the same kind) and divide it equally into three parts. Multiplication of Fractions and Whole Numbers The division sentence expressed as a fraction is: 12 3 4 Each part/group of the stars divided equally is 1 third of the 24 stars. 1 3 1 3 1 3 Multiplication of Fractions and Whole Numbers Therefore, 1 3 of 12 = 4 2 3 of 12 = 8 3 3 of 12 = 12 12 divided into 3 equal parts, each part or group is equal to 3. 1 3 of 12 = 4 Multiplication of Fractions and Whole Numbers Example 2: What is 3 4 of 16? Draw 16 any shape of the same kind, and divide it equally into 6 groups. 1 4 1 4 1 4 1 4 Multiplication of Fractions and Whole Numbers Here is three-fourths of 16: 1 4 1 4 1 4 1 4 Therefore, 3 4 of 16 = 12. Multiplication of Fractions and Whole Numbers The problem can also be solved this way. What is 3 4 of 16? ? 16 Multiplication of Fractions and Whole Numbers Multiplication of Fractions and Whole Numbers Example 3: Find the value of the following. Multiplication of Fractions and Whole Numbers Multiplication of Fractions and Whole Numbers Sample Problem 1: A.Draw an array of circles to find the value of the following. 1. 1 5 of 35 = ? 2. 4 6 of 36 = ? Multiplication of Fractions and Whole Numbers Sample Problem 1: B. Find the value of the following. Multiplication of Fractions and Whole Numbers Solution: A.Draw an array of circles to find the value of the following. Multiplication of Fractions and Whole Numbers Solution: A.Draw an array of circles to find the value of the following. Multiplication of Fractions and Whole Numbers Solution: B. Find the value of the following. Multiplication of Fractions and Whole Numbers Multiplying Fractions by Whole Numbers Using Tape Diagrams Previously, you created arrays of shapes to show how to find a part of a whole, which is exactly the same as multiplying a fraction to a whole number, example: 1 4 of 8 is the same as 1 4 8 Multiplication of Fractions and Whole Numbers Multiplying Fractions by Whole Numbers Using Tape Diagrams This time you will multiply fractions and whole numbers using tape diagrams. How is it done? Examples: a. Using a tape diagram, find 3 7 of 42. What is 3 7 of 42? Multiplication of Fractions and Whole Numbers Multiplying Fractions by Whole Numbers Using Tape Diagrams Step 1: Here, you need to find 3 sevenths of 42. First, determine the whole in the given problem. 3 7 of 42 = ? 42 is the whole! Multiplication of Fractions and Whole Numbers Multiplying Fractions by Whole Numbers Using Tape Diagrams Step 2: Draw a bar that will represent the whole. 42 Multiplication of Fractions and Whole Numbers Multiplying Fractions by Whole Numbers Using Tape Diagrams Step 3: Since we want 3 7 of 42, the denominator in 3 7 tells us how many units we need to cut the tape diagram. Thus, we need to cut the bar into 7 equal parts. 42 Multiplication of Fractions and Whole Numbers Multiplying Fractions by Whole Numbers Using Tape Diagrams Step 4: Determine how much 1 unit is. 42 ? 7 units = 42 1 unit = 42 ÷ 7 1 unit = 6 1 7 of 42 = 6 Multiplication of Fractions and Whole Numbers Multiplying Fractions by Whole Numbers Using Tape Diagrams Multiplication of Fractions and Whole Numbers Multiplying Fractions by Whole Numbers Using Tape Diagrams Multiplication of Fractions and Whole Numbers Multiplying Fractions by Whole Numbers Using Tape Diagrams Multiplication of Fractions and Whole Numbers Sample Problem 2: Solve the following using a tape diagram. Multiplication of Fractions and Whole Numbers Sample Problem 2: Solve the following using a tape diagram. Multiplication of Fractions and Whole Numbers Sample Problem 2: Solve the following using a tape diagram. Multiplication of Fractions and Whole Numbers Fraction Multiplication as Repeated Addition It is a known fact that multiplication is repeated addition. For instance: 7 6 42 is the same as 7 + 7 + 7 + 7 + 7 + 7 42 Multiplication of Fractions and Whole Numbers Fraction Multiplication as Repeated Addition How does repeated addition work for multiplication of fractions and whole number? For instance: 2 3 9 can be expressed as: 9 duplicates of 2 thirds or 2 thirds added altogether 9 times Multiplication of Fractions and Whole Numbers Using a Tape Diagram Repeated Addition 2 3 × 9 =? 3 units = 9 1 unit = 9 ÷ 3 1 unit = 𝟗 𝟑 1 unit = 3 2 units = 2 x 𝟗 𝟑 2 units = 𝟗 𝟑+𝟗 𝟑 2 units = 𝟏𝟖 𝟑 2 units = 6 2 3 × 9 =? = 2 3 + 2 3 + 2 3 + 2 3 + 2 3 + 2 3 + 2 3 + 2 3 + 2 3 = 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 3 2 3 × 9 = 18 3 2 3 × 9 = 6 9 2 3 ? ? Multiplication of Fractions and Whole Numbers Multiplication of Fractions and Whole Numbers In general, multiplying fractions and whole numbers is just a piece of cake! The visual models such as arrays and tape diagrams, plus the repeated addition discussed in this lesson gives you a better understanding of how and what happens when fractions and whole numbers are multiplied. So how is it done? Multiplication of Fractions and Whole Numbers Example: What is 3 4 of 44? Step 1: Remember that whole numbers when expressed as fractions has a denominator of 1. 44 = 44 1 Therefore, the multiplication sentence is: 3 4 × 44 1 Multiplication of Fractions and Whole Numbers Step 2: Multiply the numerators of the fractions. Multiply the denominators of the fractions. 3 4 44 1 3 44 4 1 At this point, you have 2 options; you can either multiply the numerators and the denominators of the fraction before simplifying the answer, or you can also reduce the fraction first before multiplying. Either way will give the same answer. Multiplication of Fractions and Whole Numbers Multiplying the numerators and denominators of the fraction before simplifying: Step 3: 3 4 44 1 3 44 4 1 132 4 3 4 44 1 33 Multiplication of Fractions and Whole Numbers Reducing the fractions by removing the common factors, before simplifying. Step 3: 3 4 44 1 3 44 4 1 3 4 11 4 1 3 4 44 1 3 11 1 33 1 33 Therefore, 3 4 of 44 = 33. Multiplication of Fractions and Whole Numbers Sample Problem 3: Multiply the following using two ways. The first one is done for you. Multiplication of Fractions and Whole Numbers Multiplication of Fractions and Whole Numbers Solution: Multiplication of Fractions and Whole Numbers Solution: Multiplication of Fractions and Whole Numbers Multiplication of Fractions and Whole Numbers - Word Problems This time you will be dealing with problems involving multiplication of fractions and whole numbers in real life situations. Arrays and tape diagrams can also be used to solve these problems. Example: Nica is 32 years old. She spent 3 8 of her life in Chicago. For how many years did Nica live in Chicago? Multiplication of Fractions and Whole Numbers Example: Nica is 32 years old. She spent 3 8 of her life in Chicago. For how many years did Nica live in Chicago? a. Use a tape diagram to show your solution: Multiplication of Fractions and Whole Numbers Example: Nica is 32 years old. She spent 3 8 of her life in Chicago. For how many years did Nica live in Chicago? b. Multiply directly to show the solution. Multiplication of Fractions and Whole Numbers Sample Problem 4: Patty baked 4 dozens of cookies. 5 6 are choco-chip cookies, and the rest are butter cookies. How many choco-chip and butter cookies are there? Show your solution using a tape diagram and using multiplication. Multiplication of Fractions and Whole Numbers Solution: a. 1 dozen = 12 4 dozens = 48 5 6 of 48 = ? There are 40 choco-chip cookies and 8 butter cookies. Multiplication of Fractions and Whole Numbers Solution: b. Multiply directly to show the solution. 5 6 × 48 1 = (5)(48) (6)(1) = 240 6 = 40 Name: _______ Period: __ Date: ______ Multiplication of Fractions and Whole Number G There are 40 choco-chip cookies and 8 butter cookies.
1420	https://www.quora.com/Why-is-hybridization-in-BF3-SP2-and-in-BF4-SP3	Why is hybridization in BF3 SP2 and in BF4 SP3? - Quora Something went wrong. Wait a moment and try again. Try again Skip to content Skip to search Sign In Chemistry Shape of BF3 SP2 Hybridization Boron Trifluoride BF4 Ion Electronic Geometry Hybridization Theory Molecular Structure Hybridization of NH3 5 Why is hybridization in BF3 SP2 and in BF4 SP3? All related (33) Sort Recommended Assistant Bot · 1y To understand the hybridization of BF₃ (boron trifluoride) and BF₄⁻ (tetrafluoroborate), we need to look at the geometry and bonding of each molecule. BF₃ (Boron Trifluoride) Molecular Structure: BF₃ has a trigonal planar geometry. Electron Domains: Boron (B) in BF₃ is surrounded by three fluorine (F) atoms and has no lone pairs. This results in three regions of electron density around the boron atom. Hybridization: To accommodate these three regions of electron density, one 2s orbital and two 2p orbitals from boron hybridize to form three equivalent sp² hybrid orbitals. These orbitals are arrange Continue Reading To understand the hybridization of BF₃ (boron trifluoride) and BF₄⁻ (tetrafluoroborate), we need to look at the geometry and bonding of each molecule. BF₃ (Boron Trifluoride) Molecular Structure: BF₃ has a trigonal planar geometry. Electron Domains: Boron (B) in BF₃ is surrounded by three fluorine (F) atoms and has no lone pairs. This results in three regions of electron density around the boron atom. Hybridization: To accommodate these three regions of electron density, one 2s orbital and two 2p orbitals from boron hybridize to form three equivalent sp² hybrid orbitals. These orbitals are arranged in a plane at 120° angles, which corresponds to the trigonal planar shape. BF₄⁻ (Tetrafluoroborate) Molecular Structure: BF₄⁻ has a tetrahedral geometry. Electron Domains: In BF₄⁻, boron is surrounded by four fluorine atoms and no lone pairs. This creates four regions of electron density around the boron atom. Hybridization: To accommodate these four electron pairs, one 2s orbital and three 2p orbitals from boron hybridize to form four equivalent sp³ hybrid orbitals. These orbitals are arranged in a tetrahedral shape with bond angles of approximately 109.5°. Summary BF₃: Hybridization is sp² due to three electron domains (trigonal planar). BF₄⁻: Hybridization is sp³ due to four electron domains (tetrahedral). This difference in hybridization is primarily due to the number of bonding pairs and lone pairs of electrons surrounding the boron atom in each molecule. Upvote · Related questions More answers below What is the hybridization of B in BF3 and BF4? How can we know if BF3 undergoes sp2 hybridisation? What is BF3 hybridization? What is the the type of hybridization of BF4-? What is the difference between SP, SP2, and SP3 hybridization? Anand Shankar Sahay Former Assoc. Professor of Chemistry(Retd) at TNB College (1978–2016) · Author has 232 answers and 1M answer views ·6y The valence shell electron pair repulsion principle(VSEPR) tells that the electron pairs in the valence shell of an atom ( which is part of a molecule) are oriented in space so as to minimise the repulsion between them. BF3 has 3 bond pairs of electrons and BF4– has 4 bond pairs. To minimise the electron pair repulsion, the bond pairs in BF3 and BF4– are oriented towards the apices of an equilateral triangle and a tetrahedron respectively. The angles between the bond pairs are, therefore, 120 and 109.5 degrees respectively in BF3 and BF4-. The atomic orbitals of boron, that overlap with those Continue Reading The valence shell electron pair repulsion principle(VSEPR) tells that the electron pairs in the valence shell of an atom ( which is part of a molecule) are oriented in space so as to minimise the repulsion between them. BF3 has 3 bond pairs of electrons and BF4– has 4 bond pairs. To minimise the electron pair repulsion, the bond pairs in BF3 and BF4– are oriented towards the apices of an equilateral triangle and a tetrahedron respectively. The angles between the bond pairs are, therefore, 120 and 109.5 degrees respectively in BF3 and BF4-. The atomic orbitals of boron, that overlap with those of fluorine atoms to form the bonds, must also be at the same angles in BF3 and BF4- to satisfy the principle of maximum overlap. Obviously, pure s and p orbitals cannot form the bonds because the p orbitals are mutually perpendicular and not at 120 or 109.5 degrees to each other. The hybrid orbitals sp2 , however, are at 120 degrees to each other so in BF3, the boron atom forms bonds with sp2 hybrid orbitals. Similarly , in BF4- boron forms bonds with sp3 hybrid orbitals which are at 109.5 degree with each other. In fact, for molecules/ ions AXn, the hybridisation state of the central atom A depends on the number of its valence shell electron pairs ( bond pairs + lone pairs). ( A is not a transition element.) If bp+lp = 2, sp hybridisation bp+lp = 3, sp2 hybridisation bp+lp = 4, sp3 hybridisation bp+lp = 5, sp3d hybridisation bp+lp =6, sp3d2 hybridisation Upvote · 9 1 9 1 Sponsored by Google Ads Google AI finds ready-to-buy shoppers this peak season. Turn browsing into sales this peak season with an AI-powered campaign from Google Ads. Sign Up 9 7 Igor Rudnyk M.Sc. in Organic Chemistry&Heterocyclic Chemistry, Taras Shevchenko National University of Kyiv (Graduated 1981) · Author has 351 answers and 155.6K answer views ·6y Boron has electronic configuration 1s22s22p1 in the normal state and configuration 1s22s12p2 in the activated state. In BF3 they form three sigma-bonds with sp2 hybridization. As well as there are no another electrones on the outer level molecule has plane structure (with minimum energy level). In BF4-ion there are four sigma bonds with sp3 hybridization. It has structure of pyramide (with minimum energy level). Upvote · Ali Hassan BS- chemistry from Govt. Murray College Sialkot (Graduated 2021) · Author has 64 answers and 282.5K answer views ·7y Related What is the hybridisation of BF3? Boron has three electrons in their valance shell . In BF3 central atom is Boron. B = 5 E.C = 1s2, 2s2, 2p1 All the bonds in BF3 are sigma bonds. It's shape is triangular. Continue Reading Boron has three electrons in their valance shell . In BF3 central atom is Boron. B = 5 E.C = 1s2, 2s2, 2p1 All the bonds in BF3 are sigma bonds. It's shape is triangular. Upvote · 99 10 Related questions More answers below What is the maximum lone pair of SP3 and SP4 hybridization? What is the difference between SP2 and SP3? What is the most common type of hybridization, sp3 or sp2, and why? What is the difference between SP, SP2, and SP3 hybridization? In what cases should we use one over another? When BF3 forms BF4, its hybridization changes from sp2 to sp3, but how? Can hybridization take place even after some hybrid orbitals have already formed covalent bonds? Grt Masti B.tech from Indian Institute of Technology, Delhi (Graduated 2016) ·8y Related What is the hybridisation of BF3? Upvote · 99 14 Sponsored by Grammarly 92% of professionals who use Grammarly say it has saved them time Work faster with AI, while ensuring your writing always makes the right impression. Download 999 207 Kalyan Jyoti Kalita Studied Chemistry at Tezpur University · Author has 143 answers and 637K answer views ·9y Related What is BF3 hybridization? sp2 hybridisation Electronic configuration of boron is (1s)2 (2s)2 (2px)1 (2py)0 (2pz)0. Valence orbitals for boron are 2s, 2px, 2py and 2pz. As we bring the three Fluorine atoms near Boron, the atomic orbitals on Boron change (or hybridize). Continue Reading sp2 hybridisation Electronic configuration of boron is (1s)2 (2s)2 (2px)1 (2py)0 (2pz)0. Valence orbitals for boron are 2s, 2px, 2py and 2pz. As we bring the three Fluorine atoms near Boron, the atomic orbitals on Boron change (or hybridize). Upvote · 99 24 9 1 Ernest Leung M.Phil. in Chemistry, The Chinese University of Hong Kong · Author has 11.9K answers and 5.8M answer views ·1y Related How do you explain SO3 hybridization? Why is it not sp3d2? How do you explain SO₃ hybridization? Why is it not sp³d²? The answer is as follows. A more detailed answer is as follows. Continue Reading How do you explain SO₃ hybridization? Why is it not sp³d²? The answer is as follows. A more detailed answer is as follows. Upvote · 9 2 Sponsored by All Out Kill Dengue, Malaria and Chikungunya with New 30% Faster All Out. Chance Mat Lo, Naya All Out Lo - Recommended by Indian Medical Association. Shop Now 999 621 Ashish Pyakurel MBBS from Nepal Medical College · Author has 97 answers and 196.9K answer views ·7y Related What is the hybridization of B in BF3 and BF4? You can see the photo and as you see its easy to calculate hybridisation but be aware to choose what central atom is Hope it helps Continue Reading You can see the photo and as you see its easy to calculate hybridisation but be aware to choose what central atom is Hope it helps Upvote · 99 11 Daniel James Berger PhD in organic/organosilicon chemistry · Upvoted by Antonello Frau , PhD Chemistry & Materials Science and Engineering, University of Houston (2009) · Author has 4.6K answers and 11.3M answer views ·Updated 7y Related By formula method, the hybridization of SO2 is sp3 but it comes under sp2 hybridization. Why is it so? Your citation confines itself to carbon. I’m not sure what you mean by a “formula method” for determining hybridization. We start by noting that, in S O 2 S O 2, sulfur has six valence electrons. S O 2 S O 2 can be depicted in either of two ways: an octet-based structure in which sulfur has a +2 formal charge, and an extended-octet structure in which sulfur has two double bonds. In both structures, sulfur has two oxygen ligands and a lone pair: In the left-hand structure, sulfur uses four electrons to make the two double bonds, leaving two for a lone pair. In the right-hand structure, sulfur u Continue Reading Your citation confines itself to carbon. I’m not sure what you mean by a “formula method” for determining hybridization. We start by noting that, in S O 2 S O 2, sulfur has six valence electrons. S O 2 S O 2 can be depicted in either of two ways: an octet-based structure in which sulfur has a +2 formal charge, and an extended-octet structure in which sulfur has two double bonds. In both structures, sulfur has two oxygen ligands and a lone pair: In the left-hand structure, sulfur uses four electrons to make the two double bonds, leaving two for a lone pair. In the right-hand structure, sulfur uses two electrons to make the two single bonds, and loses two for the formal charge, again leaving two for a lone pair. Qualitatively, hybridization can be estimated by a compound’s VSEPR electron geometry. Sulfur in S O 2 S O 2 has three electron groups, so its electron geometry is trigonal planar. The atom geometry is “bent,” of course. Trigonal planar electron geometries are s p 2 s p 2 hybridized. End of story. For practice, see Molecule Shapes from PhET. Upvote · 9 4 9 1 9 3 Sponsored by Dell Technologies Built for what's next. Do more, faster with Dell AI PCs powered by #IntelCoreUltra processors & ready for Windows 11. Learn More 9 8 Bhavika Vyas Chemistry Faculty at Hindustan Times Study Mate Centres · Author has 56 answers and 576.5K answer views ·7y Related What is the difference between d2sp3 and sp3d2 hybridisation? sp3d2 and d2sp3 are two types of hybridization for complexes having coordination no. 06. d2sp3 is inner shell complex which is also known as low spin complex, having participation of inner d orbitals in hybridization along with s and p orbitals. For Example: 3 d , 4s and 4p will combine and form d2sp3 . when inner shell’s two orbitals of 3d and one from 4s and rest 3 from 4p participate to form hybridization, such a type is called d2sp3. On the other hand, sp3d2 is outer shell complex or high spin complex which involves participation of s , p and outer d orbitals . If all the inner d orbitals are Continue Reading sp3d2 and d2sp3 are two types of hybridization for complexes having coordination no. 06. d2sp3 is inner shell complex which is also known as low spin complex, having participation of inner d orbitals in hybridization along with s and p orbitals. For Example: 3 d , 4s and 4p will combine and form d2sp3 . when inner shell’s two orbitals of 3d and one from 4s and rest 3 from 4p participate to form hybridization, such a type is called d2sp3. On the other hand, sp3d2 is outer shell complex or high spin complex which involves participation of s , p and outer d orbitals . If all the inner d orbitals are already occupied by electrons the formed complex will be high spin involving formation of complex by 4s, 4p and 3 orbitals of 4d . Although both the hybridization result into same octahedral geometry. Thank you. Upvote · 99 18 9 1 9 2 Bart Deloof Executive Director ·8y Related Why is ClO2 sp3 hybridized? I think you might be confused between Chlorite ion which is sp3 hybridized but it has an extra charge (ClO2-). and ClO2 or Chlorine dioxide is sp2 hybridized, although it is a bit tricky. Remember that hybridization is not the only model to represent bonding and such… and it doesn’t work equally well in every situation. Technically, for the most correct answer of the electron structure you should solve a shroedinger equation for this molecule and get all the info you need, but I’m probably a bit too nerdy there :). So what happens (or how you can look at it) is that Cl will keep one lone pair, m Continue Reading I think you might be confused between Chlorite ion which is sp3 hybridized but it has an extra charge (ClO2-). and ClO2 or Chlorine dioxide is sp2 hybridized, although it is a bit tricky. Remember that hybridization is not the only model to represent bonding and such… and it doesn’t work equally well in every situation. Technically, for the most correct answer of the electron structure you should solve a shroedinger equation for this molecule and get all the info you need, but I’m probably a bit too nerdy there :). So what happens (or how you can look at it) is that Cl will keep one lone pair, make a bond with each of the O atoms (that’s 4 electrons out of 7) and then it will make an extra 3-electron bond with an O atom, which it will resonate (the bond will alternate between the two oxygen atoms). From the quantum physics this will likely (I didn’t see the actual structure) look more like a molecular orbital that spans the whole molecule, which is why that is hard to represent in a lewis formula (There are several situations where the Lewis notation doesn’t work very well). Anyway, one lone pair and two bonds, means sp2… you’re then left with some weird 3 electron orbital for the rest of the electrons. But again, that’s one way of looking at it, so I hope you see that the difficulty stems from trying to map a special molecular situation on a model that isn’t very suited for it. The molecule is also not very stable, which I guess is related to it’s specialness. Upvote · 99 21 9 1 Alfredo Eladio Moreno B.A. in Latin American Studies&Chemistry, Pomona College (Graduated 2022) ·5y Related Why is the hybridization of oxygen in furan sp2, and not in sp3? The key here is that we think the lone pairs should occupy two sp3 hybrid orbitals because there are four regions of electron density around oxygen, however the reality is that one pair occupies one unhybridized p orbital and the other occupies a hybrid sp2 orbital (which is the definition of sp2 hybridization). WHY is that, though? Because the lone pair of electrons occupying the unhybridized p or Continue Reading The key here is that we think the lone pairs should occupy two sp3 hybrid orbitals because there are four regions of electron density around oxygen, however the reality is that one pair occupies one unhybridized p orbital and the other occupies a hybrid sp2 orbital (which is the definition of sp2 hybridization). WHY is that, though? Because the lone pair of electrons occupying the unhybridized p orbital would now be available to participate in delocalization across the ring, as a result of the overlap of adjacent p orbitals across the ENTIRE ring. This is more ... Upvote · 99 39 9 5 Daniel James Berger PhD in organic/organosilicon chemistry · Author has 4.6K answers and 11.3M answer views ·1y Related What determines if a carbon atom will become sp3, sp2 or sp hybridized? Hybridization is not a physical property; it’s a technique we use when modeling bonds in an existing molecular structure. It’s a mathematical technique for converting s and p valence orbital functions — also a model! — into orbitals that point in the correct direction for the bonds one is trying to model. Hybridization correlates to physical properties such as geometry, bond lengths and strengths, and Bronsted acid or base strength. That’s because it provides a model that allows us to construct descriptions of a molecule’s electron distribution that are good enough to allow us to predict chemic Continue Reading Hybridization is not a physical property; it’s a technique we use when modeling bonds in an existing molecular structure. It’s a mathematical technique for converting s and p valence orbital functions — also a model! — into orbitals that point in the correct direction for the bonds one is trying to model. Hybridization correlates to physical properties such as geometry, bond lengths and strengths, and Bronsted acid or base strength. That’s because it provides a model that allows us to construct descriptions of a molecule’s electron distribution that are good enough to allow us to predict chemical behavior. Carbon atoms do not “become … hybridized” in advance. They form bonds, and the resulting structures can then be described using hybridization. Common hydrocarbons can be described using hybridization pictures that correspond to an individual carbon atom’s VSEPR geometry: Linear carbons are s p s p hybridized. Trigonal planar carbons are s p 2 s p 2 hybridized. Tetrahedral carbons are s p 3 s p 3 hybridized. The resulting hybridization model allows us to predict the chemical properties of those carbon atoms pretty well. But it’s a model, not reality; and it’s not determined in advance. Upvote · 9 3 Related questions What is the hybridization of B in BF3 and BF4? How can we know if BF3 undergoes sp2 hybridisation? What is BF3 hybridization? What is the the type of hybridization of BF4-? What is the difference between SP, SP2, and SP3 hybridization? What is the maximum lone pair of SP3 and SP4 hybridization? What is the difference between SP2 and SP3? What is the most common type of hybridization, sp3 or sp2, and why? What is the difference between SP, SP2, and SP3 hybridization? In what cases should we use one over another? When BF3 forms BF4, its hybridization changes from sp2 to sp3, but how? Can hybridization take place even after some hybrid orbitals have already formed covalent bonds? What is the shape of SP3 hybridization without a lone pair? What is the basic difference between SP2 and SP3? Which, among the following hybridized orbitals, has the highest electronegativity: sp, sp2, or sp3? What is BF3 hybridization ki? What is hybridisation? Why there is sp, sp2, and sp3 hybridisation? Related questions What is the hybridization of B in BF3 and BF4? How can we know if BF3 undergoes sp2 hybridisation? What is BF3 hybridization? What is the the type of hybridization of BF4-? What is the difference between SP, SP2, and SP3 hybridization? What is the maximum lone pair of SP3 and SP4 hybridization? Advertisement About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1421	https://www.chegg.com/homework-help/questions-and-answers/differential-equation-damped-harmonic-oscillator-x-2-zeta-omegan-x-omegan-2-x-0-natural-fr-q15227405	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: The differential equation for a damped harmonic oscillator is x + 2 zeta omega_n x + omega_n^2 x = 0. Where the natural frequency for mass m and stiffness k can be calculated as: omega_n = Squareroot k/m and the damping ratio for damping constant c can be calculated as: zeta = c/2m omega_n M ATLAB provide tools for simulating such differential equations. This AI-generated tip is based on Chegg's full solution. Sign up to see more! To solve the first step of the question, begin by calculating the natural frequency using the formula . Auxillary function to dissociate second order ode to 2 first order ode: function index=a(t,x,Ji,w) index=zeros(2,1); index(1)=x(2); %index(1) is dx1/dt and index(2) is dx2/dt index(2)=… Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
1422	https://www.conceptsnrec.com/blog/flow-coefficient-and-work-coefficient	Flow Coefficient and Work Coefficient Search Sign In Find a Rep 24/7 Feedback Subscribe Engineering Figure out what you need Our engineers can design turbomachinery for you. ##### Integrated Tools We can supply software so you can design & manufacture your own turbomachinery. Software Release Notes ##### Make it for you Our shop can manufacture turbomachinery for you. ##### Teach you how to do it We teach you how to design turbomachinery with proven methods and the latest technology. Design & Development Engineering Review Services Laboratory Testing Fundamental Research Diffuser Consortium Rotordynamics & Bearing Services Software Figure out what you need Our engineers can design turbomachinery for you. ##### Integrated Tools We can supply software so you can design & manufacture your own turbomachinery. Software Release Notes ##### Make it for you Our shop can manufacture turbomachinery for you. ##### Teach you how to do it We teach you how to design turbomachinery with proven methods and the latest technology. Design Software Solutions Thermodynamic Cycle Analysis Axial Compressor Design Axial Turbine Design Centrifugal Compressor Design Radial Turbine Design Turbo Pump Design Industrial Pump Design Fans & Blowers Design Turbocharger Design Turboexpander Design Advanced Rotating Machinery Dynamics Manufacturing Software Solutions Turbomachinery CAM Solutions Manufacturing Figure out what you need Our engineers can design turbomachinery for you. ##### Integrated Tools We can supply software so you can design & manufacture your own turbomachinery. Software Release Notes ##### Make it for you Our shop can manufacture turbomachinery for you. ##### Teach you how to do it We teach you how to design turbomachinery with proven methods and the latest technology. Products VAROC Air Dynamometers Specialized Products Turbine Generators Rocket Turbopumps Services Custom Manufacturing Resources Precision Machining Prototypes Reverse Engineering Testing & Inspection Training Figure out what you need Our engineers can design turbomachinery for you. ##### Integrated Tools We can supply software so you can design & manufacture your own turbomachinery. Software Release Notes ##### Make it for you Our shop can manufacture turbomachinery for you. ##### Teach you how to do it We teach you how to design turbomachinery with proven methods and the latest technology. Engineering Courses & Seminars Software Workshops Textbooks Blog News Events Careers Resources Industries SpinOffs A blog on what's new, notable, and next in turbomachinery Flow Coefficient and Work Coefficient By Mark R. Anderson Mar 22, 2018 Share Two often used quantities to characterize turbomachinery are flow coefficient and work coefficient. The two are generally represented as Φ for flow coefficient and φ for work coefficient. The mathematical definition for the two quantities are as follows: C m is the meridional velocity (meridional velocity is the component of velocity in the radial and axial plane). U is the local rotational speed (radiusrotational velocity) and Δh0 is the change in total enthalpy or energy of the fluid. Although not immediately obvious, the two quantities are directly related to the shape of the velocity triangle. This becomes more clear when we consider the Euler turbomachinery equation: Where C θ is the tangential velocity of the fluid. Let's take a simple axial compressor example to help visualize the equations. The figure on the left is a simple compressor stage with the rotor (in blue) moving to the right and the stator (in red) which is not rotating.The absolute velocity (C) is in red. The relative velocity (W) is blue and represents what the rotor actually “sees” of the fluid. The absolute velocity is the vector sum of the relative velocity and the local wheel speed (U) shown in green. Since this is a purely axial example, there is no difference between the outflow and inflow radius and therefore the wheel speed U is also uniform. The Euler turbomachinery equation reduces to: Plugging this into the equation for work coefficient we get: If we normalize the fluid velocities by U and plot them together, we get a handy graphical representation of the coefficients: We can see immediately that the shape of the triangle and the resulting angles are determined by the coefficients. There is, actually, a third variable at play, which is the reaction of the stage (see blog: Reaction Verses Impulse). The reaction of the stage is the overall pressure split between the rotor and stator. Sometimes the reaction is defined in terms of energy, but the principle is the same. In this example, the reaction of the stage is 50% and the velocities that the rotor sees (relative) and the velocities that the stator sees (absolute) are identical. This gives two symmetrical triangles balanced about the center of the plot. The situation becomes a bit more complex with a radial machine. Since the wheel speeds are no longer equal, the coefficients can be defined differently. The typical convention is to take the largest radius (exit radius for a compressor or pump and inlet radius for a radial turbine) as the basis for the U value. The equation can be recast as: Where m is the mass flow, ρ is the density, and A is the area. A radial turbine with zero exit swirl (a reasonable design target) makes no contribution to the Euler work, thus the work coefficient is a function of the inlet condition alone. Plotting this in a similar basis as the axial compressor gives us the triangle distribution below. Flow and work coefficients are two of the most common means of classifying turbomachinery stages. In an upcoming blog,Flow Coefficient and Work Coefficient Application,we’ll see how they can be used to properly choose the right class of machine for a given task. Be sure to read my blog on Specific Speed Demystified! Tags:Turbines, Compressors, Engineering Related Articles Which Type of Turbomachinery is Aerodynamically Easier to Design, a Compressor or Turbine? By Daniel V. Hinch, Corporate VP Sales and Marketing, Concepts NREC Oct 29, 2021 In the world of aerodynamics, there are several branches and sub-branches of different types of aerodynamics. In the big picture the field of aerodynamics can be broken down into external and... Continue Preliminary Design Considerations of Zero Reaction Stages in Turbines By Oleg Dubitsky, Distinguished Corporate Fellow, Director of Corporate Technology Aug 27, 2021 Low reaction stages are often used as control stages of steam turbines, ORC turbines, drive and rocket turbopump turbines. Some of the benefits of low reaction stages vs higher reaction stages are: Continue How Many Pieces of Turbomachinery(Fans, Blowers, Compressors, Turbines, Pumps) Do You Have in Your House? Part 2 By Daniel V. Hinch, Corporate VP Sales and Marketing, Concepts NREC May 14, 2021 In my last blog I wrote about visiting a local middle school to give a talk on ‘What is Turbomachinery, and How Does It Work?’ The quiz at the end of the talk was for the students to list all the... 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1423	https://stackoverflow.com/questions/47368492/strategy-for-a-nim-variation-game-stonegamestrategist-srm-309	Skip to main content Stack Overflow About For Teams Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers Advertising Reach devs & technologists worldwide about your product, service or employer brand Knowledge Solutions Data licensing offering for businesses to build and improve AI tools and models Labs The future of collective knowledge sharing About the company Visit the blog Strategy for a Nim variation game - StoneGameStrategist - SRM 309 Ask Question Asked Modified 7 years, 9 months ago Viewed 734 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. I'm fairly new to game theory and have understood only normal nim game where you remove stones from piles with no condition and the last player to remove wins. But then I came across a nice problem while reading Game theory tutorial on Topcoder. The gist is as below: You and a friend are playing a game in which you take turns removing stones from piles. Initially, every pile has at least as many stones as the pile to its left. This property must be maintained throughout the game. On each turn, you remove one or more stones from a single pile. You and your friend alternate turns until it is no longer possible to make a valid move. The last player to have made a move wins the game. Note that if you remove all the stones from a pile, it is still considered a pile. You are said to have made a "winning move" if after making that move, you can eventually win no matter what your friend does. You are given a int[] piles representing the number of stones in each pile from left to right. It is your turn to move. Find a winning move and return it as a String formatted as "TAKE s STONES FROM PILE k" (quotes for clarity only), where s and k (a 0-based index) are each integers with no leading zeros. If there are multiple winning moves, choose the one that minimizes s. If there is still a tie, choose the one that minimizes k. If no winning move is possible, return the String "YOU LOSE" (quotes for clarity only). Removal of stones here has a condition such that you need to maintain the overall non-decreasing order, which is becoming a roadblock for me in coming up with a logic. I tried reading the editorial for that, but unfortunately couldn't grasp the idea behind it. Can anyone please explain the solution in a more simple terms? c++ game-theory olpc Share CC BY-SA 3.0 Improve this question Follow this question to receive notifications asked Nov 18, 2017 at 16:43 Surya KumarSurya Kumar 1122 bronze badges Add a comment \| 1 Answer 1 Reset to default This answer is useful 0 Save this answer. Show activity on this post. The editorial does not explain how to solve the original game of Nim, but only provides a link to the Wikipedia page (where the solution can be found). The editorial just explains how to map the Topcoder problem to that of a regular game of Nim: First, the game can be transformed into one where the piles have the difference between the original piles (so the 3 6 6 example becomes 3 3 0). Then the order of the piles is reversed (so the example becomes 0 3 3). Then a move in this new game becomes a two-step process: remove stones from one pile and add it to the previous one (in the example, the winning move takes 3 from the last and add them to the middle pile, becoming 0 6 0). Then if you just look at the odd-numbered piles (#1, #3, #5, etc), you get the regular game of Nim, and can apply a documented algorithm on that (so 0 3 3 is the same as a Nim position of 0 3). The given explanation for that is thus: any move on an odd-numbered pile becomes just like a move in the normal game of Nim; any move on an even-numbered pile can be negated by just moving the same number of stones from the receiving odd-numbered pile to the next even-numbered pile (so the same losing position can be imposed again on the player). Share CC BY-SA 3.0 Improve this answer Follow this answer to receive notifications answered Nov 18, 2017 at 20:04 ProteanProtean 48344 silver badges99 bronze badges 1 Did that make sense? Or do you need more clarification? – Protean Commented Nov 21, 2017 at 13:44 Add a comment \| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions c++ game-theory olpc See similar questions with these tags. 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1424	https://artofproblemsolving.com/wiki/index.php/Principle_of_Inclusion-Exclusion?srsltid=AfmBOor-JZr7fSTw_mkRoIGQNgOH8C-Y7woDhLp7ebgGkeGzUIXL-6Kx	Art of Problem Solving Principle of Inclusion-Exclusion - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Principle of Inclusion-Exclusion Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Principle of Inclusion-Exclusion The Principle of Inclusion-Exclusion (abbreviated PIE) provides an organized method/formula to find the number of elements in the union of a given group of sets, the size of each set, and the size of all possible intersections among the sets. Contents 1 Important Note(!) 2 Application 2.1 Two Set Example 2.2 Three Set Example 2.3 Four Set Example 2.3.1 Problem 2.3.2 Solution 2.4 Five Set Example 2.4.1 Problem 2.4.2 Solution 3 Statement 4 Proof 5 Remarks 6 Examples 7 See also Important Note(!) When using PIE, one should understand how to strategically overcount and undercount, in the end making sure every element is counted once and only once. In particular, memorizing a formula for PIE is a bad idea for problem solving. Application Here, we will illustrate how PIE is applied with various numbers of sets. Two Set Example Assume we are given the sizes of two sets, and , and the size of their intersection, . We wish to find the size of their union, . To find the union, we can add and . In doing so, we know we have counted everything in at least once. However, some things were counted twice. The elements that were counted twice are precisely those in . Thus, we have that: . Three Set Example Assume we are given the sizes of three sets, and , the size of their pairwise intersections, , and , and the size their overall intersection, . We wish to find the size of their union, . Just like in the Two Set Example, we start with the sum of the sizes of the individual sets . We have counted the elements which are in exactly one of the original three sets once, but we've obviously counted other things twice, and even other things thrice! To account for the elements that are in two of the three sets, we first subtract out . Now we have correctly accounted for them since we counted them twice originally, and just subtracted them out once. However, the elements that are in all three sets were originally counted three times and then subtracted out three times. We have to add back in . Putting this all together gives: . Four Set Example Problem Six people of different heights are getting in line to buy donuts. Compute the number of ways they can arrange themselves in line such that no three consecutive people are in increasing order of height, from front to back. (2015 ARML I10) Solution Let be the event that the first, second, and third people are in ordered height, be the event that the second, third, and fourth people are in ordered height, be the event that the third, fourth, and fifth people are in ordered height, and be the event that the fourth, fifth and sixth people are in ordered height. By a combination of complementary counting and PIE, we have that our answer will be . Now for the daunting task of evaluating all of this. For , we just choose people and there is only one way to put them in order, then ways to order the other three guys for . Same goes for , , and . Now, for , that's just putting four guys in order. By the same logic as above, this is . Again, would be putting five guys in order, so . is just choosing guys out of , then guys out of for . Now, is just the same as , so , is so , and is so . Moving on to the next set: is the same as which is , is ordering everybody so , is again ordering everybody which is , and is the same as so . Finally, is ordering everybody so . Now, lets substitute everything back in. We get a massive expression of . Five Set Example Problem There are five courses at my school. Students take the classes as follows: 243 take algebra. 323 take language arts. 143 take social studies. 241 take biology. 300 take history. 213 take algebra and language arts. 264 take algebra and social studies. 144 take algebra and biology. 121 take algebra and history. 111 take language arts and social studies. 90 take language arts and biology. 80 take language arts and history. 60 take social studies and biology. 70 take social studies and history. 60 take biology and history. 50 take algebra, language arts, and social studies. 50 take algebra, language arts, and biology. 50 take algebra, language arts, and history. 50 take algebra, social studies, and biology. 50 take algebra, social studies, and history. 50 take algebra, biology, and history. 50 take language arts, social studies, and biology. 50 take language arts, social studies, and history. 50 take language arts, biology, and history. 50 take social studies, biology, and history. 20 take algebra, language arts, social studies, and biology. 15 take algebra, language arts, social studies, and history. 15 take algebra, language arts, biology, and history. 10 take algebra, social studies, biology, and history. 10 take language arts, social studies, biology, and history. 5 take all five. None take none. How many people are in my school? Solution Let A be the subset of students who take Algebra, L-languages, S-Social Studies, B-biology, H-history, M-the set of all students. We have: Thus, there are people in my school. Statement If are finite sets, then: . Proof We prove that each element is counted once. Say that some element is in sets. Without loss of generality, these sets are We proceed by induction. This is obvious for If this is true for we prove this is true for For every set of sets not containing with size there is a set of sets containing with size In PIE, the sum of how many times these sets are counted is There is also one additional set of sets so is counted exactly once. Remarks Sometimes it is also useful to know that, if you take into account only the first sums on the right, then you will get an overestimate if is odd and an underestimate if is even. So, , , , and so on. Examples 2011 AMC 8 Problems/Problem 6 2017 AMC 10B Problems/Problem 13 2005 AMC 12A Problems/Problem 18 2001 AIME II Problems/Problem 9 2002 AIME I Problems/Problem 1 2020 AIME II Problems/Problem 9 2001 AIME II Problems/Problem 2 2017 AIME II Problems/Problem 1 See also Combinatorics Overcounting Retrieved from " Category: Combinatorics Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1425	https://www.chegg.com/homework-help/questions-and-answers/microwaves-travel-speed-light-c-3-x-108-m-s-frequency-10-ghz-waves-cause-water-molecules-b-q12723561	Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. Question: Microwaves travel with the speed of light, c= 3 x 108 m/s at a frequency of 10 GHz. These waves cause the water molecules in your burrito to vibrate. What is their wavelength =?. 5Pts. 1 GHz = 109 cycles/sec Microwaves travel with the speed of light, c= 3 x 108 m/s at a frequency of 10 GHz. These waves cause the water molecules in your burrito to vibrate. What is their wavelength =?. 5Pts. 1 GHz = 109 cycles/sec Given The speed of the microwaves is v=c=3×108 m/s. The frequency of the microwaves is f=10 GHz=10×109 Hz=1010 Hz. Not the question you’re looking for? Post any question and get expert help quickly. Chegg Products & Services CompanyCompany Company Chegg NetworkChegg Network Chegg Network Customer ServiceCustomer Service Customer Service EducatorsEducators Educators
1426	https://www.townandcountrymag.com/society/a20736482/british-royal-family-tree/	We use technologies that provide information about your interactions with this site to others for functionality, analytics, targeted advertising, and other uses. Learn more in our Privacy Notice. Skip to Content Designed by Michael Stillwell The British Royal Family Tree A comprehensive who's who of the royal family, from the first Windsors to Princess Lilibet, and every cousin in between. Designed by Michael Stillwell and Text by Emma Dibdin The House of Windsor as we know it today began in 1917 when the family changed its name from the German “Saxe-Coburg-Gotha” in the wake of the first World War. King Charles's great-grandfather, King George V, was the first Windsor monarch, and today's working royals are the descendants of King George and his wife, Queen Mary. Below follow the line of succession and explore the many branches of the family over which Charles now presides. King George V, 1865-1936 The grandson of Queen Victoria—and great-grandfather to King Charles III—George V was born third in the line of succession and did not expect to become king. That changed after his elder brother Prince Albert Victor died in 1892. George ascended the throne after the death of his father in 1910, serving as King of the United Kingdom and Emperor of India until his death in 1936. Print Collector//Getty Images From left to right: Prince Albert, Duke of York, Queen Mary, Edward Prince Wales, King George V, Prince Henry, Duke of Gloucester, and Mary, Princess Royal. Queen Mary, 1867-1953 King Charles's great-grandmother Queen Mary was royal by birth (her great-grandfather was King George III). Despite technically being a princess of the German Duchy of Teck, she was born and raised in England. She was first engaged to marry Prince Albert Victor, the eldest son of King Edward VII and her second cousin once removed, but after Albert’s sudden death in 1892, Mary agreed to marry his brother, the future King George V. The couple married in 1893, and had six children, two of whom would become reigning monarchs. She died in 1953, one year after her son, King George VI. King Edward VIII, 1894-1972 Bettmann//Getty Images The eldest son of George V and Queen Mary, Edward became king after his father’s death in 1936, but threw the country into crisis months later when he proposed to Wallis Simpson, an American divorcée. As monarch, Edward was head of the Church of England, which at the time did not allow divorced people with a living former spouse to remarry in the church, and thus the government opposed the marriage. Unable to marry Simpson and remain on the throne, Edward abdicated in December of 1936, and was succeeded by his younger brother Albert, Queen Elizabeth’s father, who would go on to become King George VI. Edward’s reign lasted just 326 days, one of the shortest in British history. After his abdication, he was named Duke of Windsor and married Simpson in 1937. They lived abroad until his death in 1972. Princess Mary, 1897-1965 The only daughter of George V and Queen Mary. During World War I, Mary devoted herself to charity work, visiting hospitals and launching fundraising campaigns to support British soldiers and sailors. She later trained as a nurse, and worked two days a week at the Great Ormond Street children’s hospital in London. In 1922, Mary married Viscount Lascelles, who later became Earl of Harewood; theirs was the first royal wedding to receive coverage in fashion magazines like Vogue. Those fans of the Downton Abbey movie will recognize Mary from her part in the plot. Prince John, 1905-1919 The youngest child of George V and Queen Mary, John was diagnosed with epilepsy at the age of four, and was sent to live at Sandringham House where he was cared for by his governess. He died in 1919 at the age of 13, following a severe seizure. His condition was not disclosed to the public until after his death. Prince Henry, Duke of Gloucester, 1900-1974 King George V and Queen Mary’s third son, Henry was the first child of a British monarch to be educated at school, rather than be tutored at home, and ultimately attended Eton College. He served in the British military and had ambitions to command a regiment, but his career was interrupted by royal responsibilities following the 1936 abdication of his brother Edward VIII. He married Lady Alice Montagu Douglas Scott in 1935, and the couple had two sons, Prince William and Prince Richard. Henry died in 1974 as the eldest surviving child of George V and Mary. Princess Alice, Duchess of Gloucester, 1901-2004 The wife of Prince Henry, Duke of Gloucester, Lady Alice was a direct descendant of Charles II through his illegitimate son, the nobleman James Scott, 1st Duke of Monmouth. She married Prince Henry in 1935, days after the death of her father, the 7th Duke of Buccleuch. The couple had two sons, Prince William and Prince Richard. Alice died at the age of 102 in 2004. Prince George, Duke of Kent, 1902-1942 The fourth son of George V and Queen Mary. Like his elder brother Henry, George was educated at school, and spent time in the Navy before becoming the first member of the royal family to work as a civil servant. In 1934, he married Princess Marina of Greece and Denmark, and the couple had three children: Prince Edward, Princess Alexandra, and Prince Michael. At the start of World War II, he returned to active military service in the Royal Navy and later the Royal Air Force. His death in 1942 in a military air crash marked the first time in more than 450 years that a member of the royal family died during active service. Princess Marina, Duchess of Kent, 1906-1968 The wife of Prince George, and a princess of the Greek royal house, Princess Marina was the daughter of Prince Nicholas of Greece and Denmark, and Grand Duchess Elena Vladimirovna of Russia. (Prince Philip is her first cousin.) In 1932, she met Prince George during a visit to London, and the couple married two years later; theirs was the first royal wedding to be broadcast by wireless radio. The couple had three children: Prince Edward, Princess Alexandra, and Prince Michael. Following her husband’s death in 1942, Marina remained an active member of the royal family and carried out many royal duties across the world, even representing the Queen at some events. She died in 1968 at the age of 61. King George VI, 1895 - 1952 Known publicly as Prince Albert until his accession, King George VI did not expect to inherit the throne because his elder brother Edward VIII was first in the line of succession. Mondadori Portfolio//Getty Images The British Royal Family appears on the Buckingham Palace balcony after the coronation of George VI in May of 1937. As the second son of George V and Queen Mary, he was made Duke of York in 1920, after serving in the Royal Navy and Royal Air Force during World War I. In 1923, he married Lady Elizabeth Bowes-Lyon, and the couple had two daughters: the future Queen Elizabeth and Princess Margaret. Following Edward’s abdication in 1936, Albert took the throne and assumed the name King George VI. The dissolution of the British Empire and formation of the British Commonwealth were finalized during George’s reign, so he was both the last Emperor of India and the first Head of the Commonwealth. George died in 1952 at the age of 56, and was succeeded by his daughter. Queen Elizabeth, The Queen Mother, 1900 - 2002 Lady Elizabeth Bowes-Lyon was born into British nobility, the 9th of 10 siblings. In 1923, she married Prince Albert, the Duke of York, having turned down several previous proposals because she had misgivings about royal life. When her brother-in-law abdicated in 1936, Albert became King George VI and Elizabeth became the Queen Consort of the United Kingdom. Upon her husband’s death in 1952, her elder daughter Elizabeth ascended to the throne, and she became known as the Queen Mother. She remained active in public life, even after her 100th birthday in 2000, and died at 101, seven weeks after the death of her younger daughter, Princess Margaret. Prince William of Gloucester, 1941-1972 As the eldest son of Prince Henry and Lady Alice, Prince William was highly educated, studying at Eton College, Cambridge University, and Stanford University. While he later held jobs in banking and in the British civil service, Queen Elizabeth’s first-cousin was also a licensed pilot, and regularly competed in air show races. It was that passion eventually lead to his untimely death. In 1972, at the age of 30, Prince William died in an airplane crash. Prince Richard, Duke of Gloucester, 1944- The younger son of Prince Henry and Lady Alice, Prince Richard initially had a career as an architect, but following the death of his older brother Prince William in 1972, he took on additional royal duties. Max Mumby/Indigo//Getty Images Prince Richard, Duke of Gloucester, Birgitte, Duchess of Gloucester, Prince Edward, Duke of Kent, and Princess Alexandra watch a flypast from the balcony of Buckingham Palace in 2018. That same year, he married Birgitte van Deurs (1946-) whom he met at Cambridge University, and just two years after that, Richard inherited the title of Duke of Gloucester from his father Prince Henry. Now in his 70s, Richard remains active in public life and carries out regular royal duties for his first cousin, the Queen. He and his wife have three children together–Alexander Windsor (1974-), Lady Davina Lewis (1977-) and Lady Rose Gilman (1980-)–and six grandchildren (Xan Windsor, Lady Cosima Windsor, Senna Lewis, Tāne Lewis, Lyla Gilman and Rufus Gilman). The couple resides in Kensington Palace. Prince Edward, Duke of Kent, 1935- The eldest child of Prince George, the Duke of Kent and Princess Marina, Prince Edward is directly related to both Prince Philip and the Queen. As a grandchild of George V and Queen Mary, he is the Queen’s first cousin, and since his mother was a first cousin to Prince Philip, Edward is also Philip’s first cousin once removed. Edward inherited the dukedom of Kent following his father’s death in a 1942 military air crash. Nearly two decades later, he married Katharine Worsley, and the couple have three children together–George Windsor, Earl of St Andrews (1962-), Lady Helen Taylor (1964-), Lord Nicholas Windsor (1970-)–and ten grandchildren (Lord Edward Windsor, Lady Marina Charlotte Windsor, Lady Amelia Windsor, Columbus Taylor, Cassius Taylor, Eloise Taylor, Estella Taylor, Albert Windsor, Leopold Windsor and Louis Windsor). Now in his 80s, Prince Edward regularly carries out royal duties on behalf of the Queen. He and his wife live on the grounds of Kensington Palace in the royal residence Wren House. Princess Alexandra, The Honourable Lady Ogilvy, 1936- Like her two brothers, Princess Alexandra is directly related to both Prince Philip and the Queen. As the eldest daughter of Prince George, the Duke of Kent and Princess Marina, she is both Queen Elizabeth’s first cousin and Prince Philip’s first cousin once removed. Princess Alexandra married the businessman Sir Angus Ogilvy in 1963, and the couple have two children–James Ogilvy (1964-) and Marina Ogilvy (1966-)–and four grandchildren (Alexander Charles Ogilvy, Flora Alexandra Ogilvy, Zenouska Mowatt and Christian Mowatt). Alexandra is reportedly quite close with the royal couple, and while Sir Angus Ogilvy passed away in 2004, she continues to be an active working royal and resides in St James’s Palace in London. Prince Michael of Kent, 1942- Like his brother Prince Edward and his sister Princess Alexandra, Prince Michael of Kent is directly related to both Prince Philip and the Queen. Max Mumby/Indigo//Getty Images Prince and Princess Michael of Kent As the youngest child of Prince George, the Duke of Kent and Princess Marina, he is both Queen Elizabeth’s first cousin and Prince Philip’s first cousin once removed. In 1978, he married Baroness Marie Christine von Reibnitz in a civil ceremony in Austria, and the couple have two children together: Lord Frederick Windsor (1979-) and Lady Gabriella Windsor (1981-). Michael takes on fewer royal responsibilities than his siblings, but he does sometimes represent the Queen at events in Commonwealth countries outside of the United Kingdom. In recognition of this work, the Queen provided Prince Michael and his wife with an apartment at Kensington Palace for a number of years, but after that proved controversial, they now pay rent. Queen Elizabeth II, 1926-2022 Born third in the line of succession, Elizabeth became the presumptive heir to the throne in 1936, following the abdication of her uncle Edward VIII and the ascension of her father, George VI. In 1947, she became engaged to Prince Philip of Greece and Denmark, whom she had first met at the age of 13. The couple were married the same year at Westminster Abbey, and had four children together. After her father died in 1952, Elizabeth ascended to the throne, going on to become both the longest-reigning and the longest-living British monarch in history, having reigned for 70 years. Her great-grandmother Queen Victoria, the second longest-reigning monarch, reigned for 63 years. Elizabeth died on September 8, 2022 at the age of 96 and was succeeded to the throne by her son, Charles. Keystone//Getty Images Queen Elizabeth with her husband Prince Philip on her coronation day. Princess Margaret, 1930-2002 Queen Elizabeth’s younger sister Margaret was 22 when her sister took the throne, and shortly afterwards became engaged to air force officer Peter Townsend. Because Townsend was divorced, the Church of England would not approve the marriage, and Margaret was famously forced to choose between ending the relationship and losing her royal privileges. She broke off her engagement with Townsend, and in 1960 married society photographer Antony Armstrong-Jones, who was given the title Earl of Snowdon. The couple had two children together, and ultimately divorced in 1978 after a tempestuous 20-year marriage. Margaret died in 2002, at the age of 71. Antony Armstrong-Jones, Lord Snowdon, 1930-2017 Antony Armstrong-Jones, a.k.a. Lord Snowdon, was the husband of Princess Margaret, and brother-in-law to Queen Elizabeth. Armstrong-Jones was a fashion and society photographer when he met Margaret in 1958, and they married two years later in 1960. The couple had two children together – David Armstrong-Jones (1961-) and Lady Sarah Chatto (1964-) – and four grandchildren (Charles Armstrong-Jones, Viscount Linley, Lady Margarita Armstrong-Jones, Samuel Chatto and Arthur Chatto), but divorced in 1978. Armstrong-Jones married his second wife Lucy Mary Lindsay-Hogg that same year, and they remained married until 2000. Armstrong-Jones died in 2017 at the age of 86. Prince Philip, Duke of Edinburgh, 1921-2021 Prince Philip was best known as Queen Elizabeth’s husband and consort, but he is also royal in his own right. He was born Prince Philip of Greece and Denmark, but Philip and his family were exiled from Greece during his childhood, and so he studied in France, Germany, and the United Kingdom before eventually serving in the British Royal Navy. He married then-Princess Elizabeth in 1947, during the reign of her father George VI, and the couple have four children together. When he passed away in 2021 at the age of 99, Prince Philip was not only the the longest-serving consort of a reigning British monarch, but also the longest-living male British royal in history. King Charles III, 1948- Charles is the current King of the United Kingdom and Head of the Commonwealth. The eldest child of Queen Elizabeth and Prince Philip, Charles was born in 1948 in Buckingham Palace. He went on to be educated at a number of institutions including Cheam and Gordonstoun Schools (which his father attended before him) and Cambridge University, before serving in the Royal Air Force and Royal Navy, making him the first heir to the British throne (and the first British monarch) ever to have a university degree. Tim Graham//Getty Images Charles and Diana with William and Harry in 1986. In 1981, Charles married Diana Spencer, and the couple had two sons, Prince William and Prince Harry, before divorcing in 1996. Charles later married his second wife Camilla Parker Bowles in 2005. Having held the title since 1958, Charles was the longest-serving Prince of Wales in history before he ascended the throne following his mother’s death in 2022. Charles was officially crowned on May 6, 2023. Diana, Princess of Wales, 1961-1997 Diana Spencer was born on July 1, 1961 into British nobility, as the third John Spencer, Viscount Althorp and Frances Roche's four children. She met Prince Charles when she was 16, and married him in July of 1981, becoming the Princess of Wales. Charles and Diana had two children together, Prince William and Prince Harry before divorcing in 1996. One year later, she tragically died in a car accident in Paris on August 31, 1997. Camilla, Queen Consort, 1947- The second wife of King Charles III, Camilla Rosemary Shand is the eldest daughter of military officer and businessman Major Bruce Shand and his wife Rosalind Shand. She is also the granddaughter of nobleman Roland Cubitt, 3rd Baron Ashcombe. In 1973, Camilla married her first husband Andrew Parker Bowles, and the couple had two children, Tom and Lisa, before divorcing in 1995. In 2005, Camilla married then-Prince Charles in a civil ceremony, and she became the Duchess of Cornwall. In 2022, upon the death of Queen Elizabeth II, she took on the title Queen Consort. Princess Anne, Princess Royal, 1950- The second child and only daughter of Queen Elizabeth and Prince Philip, Princess Anne is one of the hardest working members of the royal family. She is also an accomplished equestrian, and was even the first British royal to compete in the Olympic Games. In 1973, Anne married Captain Mark Phillips, and the couple had two children together before divorcing in 1992. Later that year, Anne married Vice Admiral Sir Timothy Laurence, her mother’s former equerry. She currently resides in St James’s Palace. Captain Mark Phillips, 1948- Princess Anne met her first husband, Captain Mark Phillips, at the 1972 Olympics in Munich, where he was part of the British equestrian team and also competed individually. The couple married in 1973, and had two children together before divorcing in 1992. Sir Timothy Laurence, 1955- The second husband of Anne, Princess Royal. A retired Royal Navy officer, Timothy met Anne in 1986 while he was serving as equerry to Queen Elizabeth. After her divorce from Captain Mark Phillips in 1992, Anne and Timothy married, and although he received no title upon the marriage, in 2008 he was appointed as a personal aide-de-camp to the Queen. Tim Graham//Getty Images Members of the royal family attend a dinner in celebration of the Queen and Prince Philip’s diamond wedding anniversary in 2007. Peter Phillips, 1977- Peter Phillips is the only son of Princess Anne and her first husband Captain Mark Phillips, and the eldest grandchild of Queen Elizabeth. Peter’s parents reportedly turned down the Queen’s offer of a royal title for their son, hoping instead to enable him to lead a more normal life. In 2008 he married Autumn Kelly, and the couple have two children together: Savannah Phillips (2010-) and Isla Phillips (2012-). Peter and Autumn announced their separation in February of 2020, and the following summer their divorce was finalized. Autumn Kelly, 1978- Originally born in Canada, Autumn Kelly was married to the Queen’s grandson Peter Phillips from 2008 until 2021. Peter and Autumn announced their separation in 2020, and finalized their divorce in June 2021. The couple have two children together, Savannah and Isla Phillips. Zara Tindall, 1981- Zara Tindall is the younger child of Princess Anne and Captain Mark Phillips and the eldest granddaughter of Queen Elizabeth. That said, she does not hold a royal title. Her parents reportedly turned down the Queen’s offer for one in hopes that Zara might lead a more normal life. Like her mother, Zara is an accomplished equestrian and Olympian, winning a silver medal at the 2012 Olympics in London, and she has been appointed Member of the Order of the British Empire for her services to equestrianism. Zara married Mike Tindall, a former rugby player, in 2011, and the couple have three children together: Mia Tindall (2014-), Lena Tindall (2018-), and Lucas Tindall (2021-). Mike Tindall, 1978- A former rugby player for the England championship team, Mike Tindall married Zara Phillips in July of 2011. The couple have since welcomed three children: Mia, Lena, and Lucas Tindall. Prince Andrew, Duke of York, 1960- The third child and second son of Queen Elizabeth and Prince Philip, Prince Andrew served in the Royal Navy for many years, including during the Falklands War in 1982, and holds the ranks of commander and vice admiral. He married Sarah Ferguson in 1986, and the couple had two daughters, Princesses Beatrice and Eugenie, before divorcing in 1996. In 2019, he stepped back from his working royal duties following enormous public criticism over his association with convicted sex offender Jeffrey Epstein. UK Press//Getty Images The York family in 2000. Sarah, Duchess of York, 1959- The former wife of Prince Andrew, Sarah Ferguson is widely known by the nickname “Fergie.” Sarah had known Andrew since childhood, and became engaged to him in 1986. The couple married at Westminster Abbey later that year, and went on to have two daughters. Sarah and Andrew announced their separation in 1992, and were divorced four years later in 1996, though by all accounts they still have an amicable relationship. Princess Beatrice of York, 1988- Princess Beatrice is the oldest daughter of Prince Andrew and Sarah Ferguson, and holds a place in the British line of succession even though she is not a working royal. The princess has a career outside the Palace, and currently works for a New York-based artificial intelligence company, but she also often attends major family events like Trooping the Colour and the annual Christmas church services. In July of 2020, she married her boyfriend Edoardo Mapelli Mozzi in a private wedding ceremony in Windsor, and became stepmother to his son, Wolfie. In October of 2021, Princess Beatrice and her husband welcomed a daughter, Sienna Elizabeth Mapelli Mozzi, and in January 2025, they welcomed their second child, daughter Athena Elizabeth Rose Mapelli Mozzi. Edoardo Mapelli Mozzi, 1983- In July of 2020, Edoardo Mapelli Mozzi married Princess Beatrice in a small private wedding ceremony in Windsor. He has a young son, Wolfie, from a previous relationship—making Beatrice an instant stepmother. Princess Eugenie, 1990- The younger daughter of Prince Andrew and Sarah, Duchess of York, and Queen Elizabeth’s granddaughter. Eugenie attended St George’s School with her elder sister Beatrice, and later graduated from Newcastle University. In October of 2018, Eugenie married her partner of seven years, Jack Brooksbank, in a ceremony at Windsor Castle. In February of 2021, Princess Eugenie gave birth to a baby son, August Philip Hawke Brooksbank. In 2023, she welcomed another son named Ernest George Ronnie Brooksbank. WPA Pool//Getty Images Prince Andrew walks his daughter Eugenie down the aisle at her 2018 wedding. Jack Brooksbank, 1986- Jack Brooksbank first met Princess Eugenie in Verbier, Switzerland, while on a ski vacation. The pair dated for approximately seven years before marrying in October of 2018 in front of friends and family in St George’s Chapel at Windsor Castle. Princess Eugenie and Jack welcomed their first baby in February of 2021, naming their son August Philip Hawke Brooksbank, and their second child on June 5, 2023, Ernest George Ronnie Brooksbank. Prince Edward, Duke of Edinburgh, 1964- The youngest child and third son of Queen Elizabeth and Prince Philip, Edward attended Cambridge University and later joined the Royal Marines, but dropped out after four months. In 1999 he married Sophie Rhys-Jones, and the couple have two children. Prince Edward is a full-time working royal, and after his father Prince Philip retired from public life in 2017, he took on many of his responsibilities. In 2023, he also inherited his father's title (he was previously known as the Earl of Wessex), which was conferred onto him by his brother, King Charles III, to mark Edward's 59th birthday Sophie, Duchess of Edinburgh, 1965- Sophie Helen Rhys-Jones met Prince Edward while she was working in radio, and the couple dated for six years before marrying in 1999. They have two children together, Lady Louise Windsor and James, Viscount Severn. While she previously had a career in public relations, Sophie is now a full-time working royal like her husband. Lady Louise Windsor, 2003- The elder child and only daughter of Prince Edward, Duke of Edinburgh, and Sophie, Duchess of Edinburgh, Lady Louise is the youngest granddaughter of Queen Elizabeth. She and her brother James embarked on their first royal engagement in 2015, accompanying their parents to South Africa. You might also recognize her as one of the bridesmaids from Will and Kate’s royal wedding in 2011. James, Earl of Wessex, 2007- The younger child and only son of Prince Edward, Duke of Edinburgh, and Sophie, Duchess of Edinburgh, James is the youngest grandchild of Queen Elizabeth. Both he and his older sister Louise embarked on their first royal engagement in 2015, accompanying their parents to South Africa. Though he was originally styled Viscount Severn, in 2023, he inherited the title of Earl of Wessex from his father, after Edward was named the new Duke of Edinburgh (a title previously held by Edward's late father Prince Philip.) Prince William, Prince of Wales, 1982- The elder son of King Charles III and Princess Diana, William is currently the heir apparent and first in line to the British throne. After attending Eton College and St Andrew’s University, he trained at the Royal Military Academy Sandhurst and served in the Royal Air Force, eventually becoming a search-and-rescue pilot. He has since left the military and is now a full-time working royal. In 2011, he married his longtime girlfriend, Catherine Middleton, whom he met at St Andrew’s, and the couple now have three children, Prince George, Princess Charlotte, and Prince Louis. WPA Pool//Getty Images William and Kate attend Christmas services with George and Charlotte in 2016. Catherine, Princess of Wales, 1982- After growing up in Chapel Row near Newbury as the oldest daughter of Carole and Michael Middleton, Kate met Prince William at the University of St Andrews in Scotland. After a long courtship, the couple married at Westminster Abbey in 2011 in a ceremony which was attended by celebrities, dignitaries, and royals from across Europe. She and William have three children together, Prince George, Princess Charlotte, and Prince Louis, and Kate now works as a full-time royal focusing on organizations which support young people and mothers, and that help to fight the stigma of mental health issues. DOMINIC LIPINSKI//Getty Images The Cambridge family at Prince Louis’s christening ceremony in 2018. Prince George, 2013- The first child and elder son of Prince William and Catherine, Princess of Wales, George was born on July 22, 2013 and is currently second in the line of succession. Princess Charlotte, 2015- The second child, and only daughter, of Prince William and Catherine, Princess of Wales, Charlotte was born on May 2, 2015. She is currently third in the line of succession. Prince Louis, 2018- The third child, and second son, of Prince William and Catherine, Princess of Wales, Prince Louis was born on April 23, 2018. He is currently fourth in the line of succession. Prince Harry, Duke of Sussex, 1984- The younger son of King Charles III and Princess Diana, Harry is currently fifth in the line of succession. After attending Eton College like his elder brother William, Harry trained at the Royal Military Academy Sandhurst, and served in the British Army where he was twice deployed to Afghanistan, making Harry the first royal to serve in a war zone since his uncle Prince Andrew. In May of 2018, Harry married American actress Meghan Markle in a widely-watched royal wedding. A year and a half later, he and Meghan announced their decision to step back from their roles as working roles, and have since carved out space for themselves in the private sector, inking a deal with Netflix and signing with a speaking agency. In May of 2018, Harry married American actress Meghan Markle in a widely-watched royal wedding. A year and a half later, he and Meghan announced their decision to step back from their roles as working roles, and have since carved out space for themselves in the private sector, inking a deal with Netflix and signing with a speaking agency. In May of 2019, they welcomed their first child, Prince Archie. Two years later, in June of 2021, the couple welcomed their baby daughter, Princess Lilibet 'Lili' Diana. The family of four currently lives in California. Chris Jackson//Getty Images Harry and Meghan shortly before they announced they are expecting their first child together. Meghan, Duchess of Sussex, 1981- The Duchess of Sussex broke the mold of the expected royal bride, as a biracial, California-born actress. Markle, who divorced her first husband in 2013, was reportedly set up on a blind date with Harry in 2016, and the rest is history. They married in May 2018 at Windsor Castle, and Meghan spent a year and a half as a working royal before she and Harry decided to step back from their roles. She now lives in Santa Barbara, California with Harry and their son, Prince Archie, who was born in May 2019, and their daughter, Princess Lilibet, born in June 2021. Prince Archie Harrison, 2019- The first child of the Duke and Duchess of Sussex, Archie Harrison Mountbatten-Windsor was born on May 6, 2019. At the time, he had no official royal title, but later earned the styling Prince Archie when his grandfather, King Charles, became reigning monarch. He is currently sixth in the line of succession. Archie became a big brother to newborn Lilibet in June of 2021, just days after his second birthday. Princess Lilibet 'Lili' Diana, 2021- The Duke and Duchess of Sussex’s second child, Lilibet 'Lili' Diana Mountbatten-Windsor was born on June 4, 2021 in Santa Barbara, California. She’s named after both her great-grandmother, Queen Elizabeth, and her grandmother Princess Diana—the Queen’s family nickname is Lilibet. Like her brother, she didn't inherit the title Princess until 2022 when her grandfather became king, but even then it was not clear she would use the title until the announcement of her christening in 2023. She is currently seventh in the line of succession to the throne. Michael Stillwell Michael Stillwell is the Digital Art Director for Town & Country, where he creates visuals for stories covering arts, travel, fashion, beauty, education, lifestyle, luxury, royals and celebrities. Before joining T&C exclusivly he covered several other Hearst brands for over a decade including Delish, Popular Mechanics, Esquire, House Beautiful, and Elle Décor. Follow him on Instagram @stillmike Watch Next Advertisement - Continue Reading Below Readers Also Read The British Line of SuccessionSee Photos From the Duchess of Kent’s FuneralThe Duchess of Gloucester’s Honeysuckle TiaraWho Was Katharine, Duchess of Kent? Advertisement - Continue Reading Below See All the Tiaras at the U.S. State BanquetWho Is the Duke of Kent?Queen Camilla Wore a Tiara With a Personal HistoryRoyal Family at the U.S. State Banquet Prince Harry on Bringing His Kids to the UKQueen Camilla's Top Tiara Moments, in PhotosDuchess of Edinburgh Meets 105-Year-Old WWII VetMeet Princess Märtha Louise’s Daughters Advertisement - Continue Reading Below
1427	https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/14%3A_Conjugated_Compounds_and_Ultraviolet_Spectroscopy/14.03%3A_Kinetic_vs._Thermodynamic_Control_of_Reactions	Skip to main content 14.3: Kinetic vs. Thermodynamic Control of Reactions Last updated : Sep 30, 2024 Save as PDF 14.2: Electrophilic Additions to Conjugated Dienes- Allylic Carbocations 14.4: The Diels-Alder Cycloaddition Reaction Page ID : 448690 OpenStax OpenStax ( \newcommand{\kernel}{\mathrm{null}\,}) Electrophilic addition to a conjugated diene at or below room temperature normally leads to a mixture of products in which the 1,2 adduct predominates over the 1,4 adduct. When the same reaction is carried out at higher temperatures, however, the product ratio often changes and the 1,4 adduct predominates. For example, addition of HBr to 1,3-butadiene at 0 °C yields a 71 : 29 mixture of 1,2 and 1,4 adducts, but the same reaction carried out at 40 °C yields a 15 : 85 mixture. Furthermore, when the product mixture formed at 0 °C is heated to 40 °C in the presence of HBr, the ratio of adducts slowly changes from 71 : 29 to 15 : 85. Why? To understand the effect of temperature on product distribution, let’s briefly review what we said in Section 6.7 about rates and equilibria. Imagine a reaction that can give either or both of two products, B and C. Let’s assume that B forms faster than C (in other words, ΔG‡B<ΔG‡C) but that C is more stable than B (in other words, ΔGoC>ΔGoB). An energy diagram for the two processes might look like that shown in Figure 14.3.1. Let’s first carry out the reaction at a lower temperature so that both processes are irreversible and no equilibrium is reached. Because B forms faster than C, B is the major product. It doesn’t matter that C is more stable than B, because the two are not in equilibrium. The product of an irreversible reaction depends only on relative rates, not on stability. Such reactions are said to be under kinetic control. Now let’s carry out the same reaction at some higher temperature so that both processes are readily reversible and an equilibrium is reached. Because C is more stable than B, C is the major product obtained. It doesn’t matter that C forms more slowly than B, because the two are in equilibrium. The product of a readily reversible reaction depends only on stability, not on relative rates. Such reactions are said to be under equilibrium control, or thermodynamic control. We can now explain the effect of temperature on the electrophilic addition reactions of conjugated dienes. At low temperature (0 °C), HBr adds to 1,3-butadiene under kinetic control to give a 71 : 29 mixture of products, with the more rapidly formed 1,2 adduct predominating. Because these low-temperature conditions don’t allow the reaction to reach equilibrium, the product that forms faster predominates. At higher temperature (40 °C), however, the reaction occurs under thermodynamic control to give a 15 : 85 mixture of products, with the more stable 1,4 adduct predominating. The higher temperature allows the addition process to become reversible, so an equilibrium mixture of products results. Figure 14.3.2 shows this situation in an energy diagram. The electrophilic addition of HBr to 1,3-butadiene is a good example of how a change in experimental conditions can change the product of a reaction. The concept of thermodynamic control versus kinetic control is a useful one that we can sometimes take advantage of in the laboratory. Exercise 14.3.1 The 1,2 adduct and the 1,4 adduct formed by reaction of HBr with 1,3-butadiene are in equilibrium at 40 °C. Propose a mechanism by which the interconversion of products takes place. Answer : Interconversion occurs by SN1 dissociation to a common intermediate cation. Exercise 14.3.2 Why do you suppose 1,4 adducts of 1,3-butadiene are generally more stable than 1,2 adducts? Answer : The double bond is more highly substituted. 14.2: Electrophilic Additions to Conjugated Dienes- Allylic Carbocations 14.4: The Diels-Alder Cycloaddition Reaction
1428	https://cks.nice.org.uk/topics/heart-failure-chronic/prescribing-information/ace-inhibitors/	Cookies on the NICE website and services Cookies are files saved on your phone, tablet or computer when you visit a website. We use cookies to store information about how you use the NICE website and services, such as the pages you visit. For more information, view our cookie statement. (Opens in a new window) Essential cookies These cookies enable basic functions like page navigation and access to secure areas of the website. Our website cannot function properly without these cookies and they can only be deactivated by changing your browser preferences. Preference cookies These cookies remember information that changes the way our website behaves or looks, like your preferred layout or previously viewed pages. Website usage cookies We use tools such as Google Analytics, Hotjar, Google Optimize and Loop11 to help us anonymously measure how you use our websites. This allows us to make improvements based on our users' needs. These tools set cookies that store anonymised information about how you got to the site, and how you interact with the site. Marketing and advertising cookies We use Google Ads, LinkedIn, Facebook, Twitter, Stackadapt, Amazon DSP, Xandr and Microsoft Ads to show adverts on external sites to promote NICE services, events and content. These services may use cookies to help make advertising more effective. These cookies are used for things like showing relevant adverts based on website visits, preventing the same ad from continuously re-appearing, or by measuring how many times people click on these adverts. Skip to content Accessibility help Sign in CKS is only available in the UK The NICE Clinical Knowledge Summaries (CKS) site is only available to users in the UK, Crown Dependencies and British Overseas Territories. CKS content is produced by Clarity Informatics Ltd (trading as Agilio Software \| Primary Care). It is available to users outside the UK via subscription from the Agilio \| Prodigy website. If you believe you are seeing this page in error please contact us .
1429	https://www.statpearls.com/point-of-care/20112	Systemic Sclerosis (Scleroderma) \| Treatment & Management \| Point of Care PhysicianPhysician Board ReviewsPhysician Associate Board ReviewsCMELifetime CMEFree CMEMATE and DEA ComplianceCPD StudentUSMLE® Step 1USMLE® Step 2USMLE® Step 3COMLEX-USA® Level 1COMLEX-USA® Level 2COMLEX-USA® Level 396 Medical School ExamsStudent Resource Center NCLEX - RNNCLEX - LPN/LVN/PN24 Nursing Exams Nurse PractitionerAPRN/NP Board ReviewsCNS Certification ReviewsCE - Nurse PractitionerFREE CE NurseRN Certification ReviewsCE - NurseFREE CE PharmacistPharmacy Board Exam PrepCE - Pharmacist AlliedAllied Health Exam PrepDentist ExamsCE - Social WorkerCE - Dentist Point of Care Institutional Sales Student Resources Search CME 0.0 Sign-upLogin PhysicianPhysician Board ReviewsPhysician Associate Board ReviewsCMELifetime CMEFree CMEMATE and DEA ComplianceCPD StudentUSMLE® Step 1USMLE® Step 2USMLE® Step 3COMLEX-USA® Level 1COMLEX-USA® Level 2COMLEX-USA® Level 396 Medical School ExamsStudent Resource Center NCLEX - RNNCLEX - LPN/LVN/PN24 Nursing Exams Nurse PractitionerAPRN/NP Board ReviewsCNS Certification ReviewsCE - Nurse PractitionerFREE CE NurseRN Certification ReviewsCE - NurseFREE CE PharmacistPharmacy Board Exam PrepCE - Pharmacist AlliedAllied Health Exam PrepDentist ExamsCE - Social WorkerCE - Dentist Point of Care Free CME/CE Tools Back To Search Results Topic Outline IntroductionEtiologyEpidemiologyPathophysiologyHistopathologyHistory and PhysicalEvaluationTreatment / ManagementDifferential DiagnosisPrognosisComplicationsDeterrence and Patient EducationEnhancing Healthcare Team Outcomes Looking for an easier read? Click here for a simplified versionMediaBilateral Signs of Raynaud Phenomenon in a Patient's Hands. A male patient is exhibiting triphasic color changes in the digits of his hands, indicative of Raynaud Phenomenon.Pursed Mouth in a Female Patient With Scleroderma. This image shows a female patient with a pursed mouth, indicative of the cutaneous manifestations of scleroderma. Topic Outline Back To Search Results Systemic Sclerosis (Scleroderma) PrintEmailFacebookLinkedInX PrintEmailFacebookLinkedInX Author: Rotimi AdigunAuthor: Amandeep GoyalEditor: Anis HarizUpdated: 4/5/2024 3:34:03 AM Introduction Systemic sclerosis, also known as scleroderma, is a rare connective tissue disorder with an unknown and complex pathogenesis. Scleroderma can be divided into 2 primary forms—localized scleroderma (including morphea, linear scleroderma, and scleroderma en coup de sabre) and systemic sclerosis. Systemic sclerosis can be further classified as limited systemic sclerosis (formerly known as CREST syndrome, characterized by calcinosis, Raynaud phenomenon, esophageal dysmotility, sclerodactyly, and telangiectasia) or diffuse systemic sclerosis based on clinical and serological criteria.Although significant progress has been made in understanding the pathophysiology of scleroderma over the past centuries, the disease continues to pose significant morbidity and mortality in patients. Localized scleroderma primarily affects the skin and subcutaneous tissue, leading to patches of thickened skin that, on biopsy, reveal dermal fibrosis similar to the histopathological changes seen in the thickened skin in systemic sclerosis. However, it is not associated with the Raynaud phenomenon, digital ischemic events, or internal organ involvement. Antinuclear antibodies may be present in up to 50% of cases of localized scleroderma; however, more specific autoantibodies such as anti-centromere, anti-Scl-70, and anti-RNA polymerase III are absent in this condition.Notably, localized scleroderma is not associated with increased mortality.On the other hand, systemic sclerosis is associated with several systemic manifestations and internal organ involvement, leading to increased mortality, and its classification is based on skin involvement. Limited cutaneous systemic sclerosis, previously known as CREST syndrome, is characterized by skin thickening distal to the elbows and knees and/or on the face without trunk involvement. On the other hand, diffuse cutaneous systemic sclerosis involves skin thickening that may affect areas proximal to the elbows, knees, face, and/or trunk. Both limited cutaneous systemic sclerosis and diffuse cutaneous systemic sclerosis are associated with several systemic manifestations and positive autoantibodies. Antinuclear antibodies may be present in more than 90% of cases of systemic sclerosis, and up to 70% of cases exhibit at least one of the more specific autoantibodies (anti-centromere, anti-Scl-70, and anti-RNA polymerase III). Scleroderma most commonly affects the skin, gastrointestinal tract, lungs, kidneys, skeletal muscle, and pericardium among affected organs.The manifestations of scleroderma may overlap extensively with those of other rheumatological or immunological diseases. The severity of the presentation may also vary depending on the timing of the systemic sclerosis diagnosis. Etiology Register For Free And Read The Full Article Get the answers you need instantly with the StatPearls Clinical Decision Support tool. 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Log in Etiology The exact etiology of systemic sclerosis is not completely understood, and it is believed that both genetic and environmental factors are thought to contribute to its development. Genetic Factors Systemic sclerosis shows familial clustering and is associated with clusters of multiple other autoimmune diseases within families. Genomewide association studies have confirmed the involvement of the major histocompatibility complex genetic region in systemic sclerosis, akin to other autoimmune disorders like systemic lupus erythematosus and rheumatoid arthritis. The association of specific human leukocyte antigens (HLA), including HLA DRB11104, DQA10501, and DQB10301, with systemic sclerosis has been long known.Further, non-HLA loci such as PTPN22, NLRP1, STAT4, and IRF5 have also been implicated in the etiology of systemic sclerosis. Environmental Factors Several environmental triggers are associated with the subsequent development of systemic sclerosis, including infectious agents such as Cytomegalovirus (CMV),Epstein-Barr virus, and parvovirus B19. Exposure to silica dust is also associated with systemic sclerosis, along with occasional exposure to other agents such as organic solvents, toluene, xylene, trichloroethylene, and polyvinyl chloride. Notably, cigarette smoking is not a proven risk factor for systemic sclerosis. Several scleroderma-like disorders, which can be distinguished from systemic sclerosis by clinical, histopathological, and laboratory features, are associated with environmental exposures such as contaminated rapeseed cooking oil causing toxic oil syndrome and L-tryptophan linked to eosinophilia-myalgia syndrome.In addition, certain drugs, such as bleomycin and cocaine, have been associated with developing systemic sclerosis-like illnesses. Epidemiology The rarity of systemic sclerosis contributes to sparse epidemiological data, with reported incidence and prevalence rates influenced by factors such as geography, case definition, and ascertainment methods. Overall, prevalence rates range between 38 and 341 cases per million persons, while incidence rates vary from 8 to 56 new cases per million persons annually globally. A study in the United States reported an estimated incidence of 19.3 new cases per million adults per year, with a prevalence of 242 cases per million adults in the Detroit area from 1989 to 1991.Another study from Quebec in 2003 found a prevalence of 443 cases per million adults.Interestingly, prevalence rates are higher in the United States and Australia compared to Europe and Asia (specifically, Japan and Taiwan).Notably, the Choctaw Native Americans from Oklahoma exhibited the highest prevalence of systemic sclerosis at 660 cases per million adults, highlighting geographical variations in the prevalence of this condition. Systemic sclerosis exhibits a female predominance with a female-to-male ratio of about 5:1. Females typically develop the condition at an earlier age than males. The peak age of onset is between 45 and 54 in African-American females and between 55 and 64 in European-American females. Notably, systemic sclerosis is rare among children and teenagers aged 15 or younger.Disease onset in individuals aged between 15 and 24 is also rare, with an incidence of 21.2 per million in African-American females and 11.16 per million in European-American females. Patients with African ancestry face a higher risk of developing systemic sclerosis, experiencing an earlier onset of the disease, and often developing more severe manifestations. Pathophysiology The pathophysiology of systemic sclerosis is intricate and not entirely understood. The disease is characterized by 3 main hallmarks—vascular insult, autoimmunity, and tissue fibrosis. Different clinical phenotypes of systemic sclerosis are attributed to variations in the contribution of these factors in each patient's disease pathogenesis. Various triggers, such as viruses, environmental exposures, autoantibodies, proteolytic enzymes, and inflammatory cytokines, can initiate the initial vascular insult in systemic sclerosis. This initial vascular insult leads to endothelial cell activation, followed by overexpression of adhesion molecules, leading to the activation of platelets and thrombotic and fibrinolytic cascades. Activated endothelial cells also release ET-1, a potent vasoconstrictor, and promote leukocyte adhesion, vascular smooth muscle cell proliferation, and fibroblast activation.These activated endothelial cells transdifferentiate into mesenchymal cells with prominent functional abnormalities, such as impaired responsiveness to vasodilators, including nitric oxide and prostacyclins. In addition, activated platelets release thromboxane-A2, platelet-derived growth factor, and transforming growth factor-β (TGF-β), as well as activate thrombin, causing coagulation, thrombosis, vasoconstriction, and fibroblast activation. The initial vascular insult ultimately results in tissue hypoxia. Patients with systemic sclerosis exhibit impaired compensatory vascular repair mechanisms, widespread microangiopathy, microvasculature loss, and diminished angiogenesis, contributing to chronic tissue hypoxia and oxidative stress. Immune dysregulation and inflammation play a crucial role in the pathogenesis of systemic sclerosis, with dysregulation of both innate and humoral immune systems. Activated T cells are predominant in the tissue and peripheral blood in systemic sclerosis patients. Patients with systemic sclerosis have an imbalance of types 1 and 2 helper T cells (Th1/Th2) cytokines with a predominant Th2 profile. This Th2 profile skewing contributes to increased fibrosis through enhanced collagen synthesis and myofibroblast transdifferentiation, driven by pro-fibrotic cytokines such as TGF-β, interleukins (ILs) 4, 5, and 13, and less anti-fibrotic cytokines such as interferon-γ. Activated macrophages, monocytes, and dendritic cells further promote vascular injury and fibrosis by activating T- and B cells and producing pro-fibrotic and pro-inflammatory cytokines. Nearly all patients with systemic sclerosis exhibit humoral autoimmunity and the presence of autoantibodies. Activated B cells are responsible for producing these autoantibodies, which are considered directly pathological and serve as important diagnostic markers for the disease. Furthermore, specific autoantibody profiles are associated with distinct disease phenotypes in systemic sclerosis. Vascular injury and inflammation resulting from autoimmunity contribute to tissue fibrosis through mesenchymal cell activation and differentiation. This process leads to irreversible extracellular matrix accumulation, uncontrolled fibroblast activation, the persistence of myofibroblasts, an increase in the microvascular pericyte compartment, and pathologic epithelial-mesenchymal transition. Histopathology The hallmark of systemic sclerosis is noninflammatory proliferative or obliterative vasculopathy followed by interstitial or vascular fibrosis. Perivascular inflammatory infiltrates of CD4+ T lymphocytes may be observed early in the disease but are absent in long-standing systemic sclerosis cases. The vasculopathy in systemic sclerosis is characterized by bland intimal proliferation, thickening of the basement membrane, capillary rarefaction, loss of vascular endothelial cadherin, platelet aggregation, and microthrombi formation. Vasculitis or immune complex deposition is rare. However, in the later stages of systemic sclerosis, extensive perivascular fibrosis, progressive luminal occlusion, and tissue fibrosis become evident. In early systemic sclerosis, dermal edema in the skin is prominent, and perivascular inflammatory infiltrates mainly consist of T lymphocytes and monocytes. Dermal fibrosis becomes the predominant feature as the disease progresses to later stages. Histopathological examination typically shows a loss of dermal capillaries and skin appendages such as hair follicles, sweat glands, and sebaceous glands.A marked dermal expansion is characterized by dense collagen and hyaluronic acid accumulation. Additionally, a loss of dermal lymphatics and the subcutaneous adipose layer is commonly observed. Pulmonary involvement is widespread in systemic sclerosis, mirroring the skin's pathology. Early in the disease, pulmonary involvement manifests with inflammatory changes, leading to fibrosis and vascular damage. Early inflammatory changes affect the alveolar walls, which are characterized by infiltration with lymphocytes, macrophages, and plasma cells. The most common pulmonary involvement in systemic sclerosis is nonspecific interstitial pneumonitis, which exhibits a fairly uniform distribution of fibrosis along with interstitial inflammation and type II pneumocyte hyperplasia. Another pattern, usual interstitial pneumonia, is characterized by a patchy distribution of fibrosis with scattered fibroblastic foci and carries a poorer prognosis. Pulmonary arterial hypertension (PAH) develops due to vasculopathy and the loss of pulmonary microvasculature secondary to progressive pulmonary fibrosis. Renal involvement in systemic sclerosis is relatively less common. Chronic ischemic changes are frequently observed, although glomerulonephritis is uncommon unless there is an overlapping syndrome. Scleroderma renal crisis, a rare yet life-threatening manifestation of systemic sclerosis, is associated with changes resembling those seen in thrombotic microangiopathy or malignant hypertension. Histologically, onion skinning (intimal proliferation and reduplication of the elastic lamina with luminal narrowing) of the small interlobular and arcuate renal arteries is a characteristic finding in such cases. Fibrosis is the predominant pathological feature in systemic sclerosis, affecting various organs beyond the skin. Fibrosis can occur in organs such as the gastrointestinal tract (from the mouth to the rectum), thyroid gland, salivary glands, and penile blood vessels. Inflammatory myositis may also manifest in systemic sclerosis alone or overlap syndromes. Cardiac involvement can lead to constrictive pericarditis, pericardial fibrosis or effusions, and patchy myocardial fibrosis as significant findings. History and Physical Systemic sclerosis is a complex disorder affecting multiple systems, and its clinical presentation varies widely among individuals. Diffuse cutaneous systemic sclerosis typically presents more severe symptoms and higher mortality rates compared to limited cutaneous systemic sclerosis, often involving internal organs more extensively and severely. Additionally, systemic sclerosis tends to manifest more severely in males, African Americans, and individuals with a later age of onset. Raynaud Phenomenon Raynaud phenomenon is an early and common manifestation observed in over 95% of patients with systemic sclerosis.The condition involves vasospasms triggered by exposure to cold, resulting in triphasic color changes, typically affecting the digits, especially in the upper extremities. These color changes may also manifest in areas such as the ears, nose, or tongue. The sequence of color changes includes initial pallor with well-defined white discoloration, followed by ischemic changes with a dusky bluish appearance, and finally, reactive hyperemia leading to red discoloration (seeImage. Bilateral Signs of Raynaud Phenomenon in a Patient's Hands). Notably, not all patients report all 3 color phases. Raynaud phenomenon can manifest years before the onset of visceral involvement, particularly in cases of limited cutaneous systemic sclerosis.However, in diffuse cutaneous systemic sclerosis, Raynaud phenomenon may occur simultaneously or shortly after the onset of skin changes. Notably, up to 15% of individuals in the general population experience the Raynaud phenomenon without underlying systemic sclerosis or connective tissue diseases. This is termed primary Raynaud, which is associated with a female predominance (4:1 ratio), early onset (usually occurring in individuals aged 20 or younger), symmetrical mild clinical features, normal nailfold capillaries, negative antinuclear antibodies, and a benign course without ischemic complications. However, secondary Raynaud, as observed in systemic sclerosis, can lead to severe complications. In systemic sclerosis, the vasospastic insult caused by the Raynaud phenomenon is complicated by the underlying non-vasculitic vasculopathy characterized by intimal hyperplasia and fibrosis. Additionally, abnormal regulation of local vasomotor control contributes to the loss of vessel flexibility. Consequently, digital manifestations such as pitting, ulcers, and tissue loss frequently occur in systemic sclerosis, often leading to digital ischemia and dry gangrene, which can necessitate autoamputation. Moreover, secondary infections are common complications. In systemic sclerosis, Raynaud phenomenon is further characterized by an abnormal nailfold capillary examination, revealing features such as capillary dilation, hemorrhages, and dropout. Cutaneous Manifestations Skin involvement is the most overt feature of systemic sclerosis and is present in almost all patients with systemic sclerosis, varying in degree and severity. The classification of systemic sclerosis into limited and diffuse cutaneous subtypes is determined by the degree of cutaneous involvement. Limited cutaneous systemic sclerosis is characterized by skin involvement distal to the elbows and knees, with the trunk typically spared. In contrast, diffuse cutaneous systemic sclerosis involves skin proximal to the elbows, knees, and/or the trunk. Facial involvement may occur in both limited and diffuse cutaneous systemic sclerosis presentations. During the initial "puffy finger phase," inflammation and non-pitting edema of the hands are prevalent and may persist for several months. Symptoms such as pruritus, burning pain, and erythema are frequently reported. The edema can result in compression of underlying structures, leading to common compression neuropathies such as carpal tunnel syndrome are common. Loss of skin appendages contributes to dry and uncomfortable skin. Subsequently, skin thickening and fibrosis develop during this initial phase, followed by the fibrotic phase. During this second prolonged fibrotic phase, skin fibrosis and thickening commence distal to the metacarpophalangeal joints, a condition termed sclerodactyly, and progressively advance proximally. The thick, leather-like skin and fibrosis of deeper subcutaneous structures contribute to permanent contractures and reduced mobility of the peripheral joints. Further loss of skin appendages and subcutaneous adipose tissue (lipodystrophy) may occur. Facial involvement manifests as a small oral aperture, known as fish-mouth or masked facies (see Image. Pursed Mouth in a Female Patient With Scleroderma). Skin ulcers may develop at sites of trauma, such as the extensor surfaces of the metacarpophalangeal, interphalangeal, or elbow joints. Additionally, a salt-and-pepper-like appearance, characterized by areas of depigmentation among normally pigmented skin, may be observed. The final skin-softening phase may occasionally occur many years after the initial presentation. During this phase, the skin, particularly on the trunk and upper arms, may soften and return to a clinically normal appearance, although the underlying subcutaneous tissue remains fibrotic. Telangiectasias, attributed to capillary dilatation, are a common feature of systemic sclerosis and are often observed on the hands, face, mucosal surfaces, and occasionally on the trunk. These telangiectasias blanch upon pressure application and may progressively increase over time. Their presence is associated with a heightened risk of PAH. Subcutaneous calcinosis, although prevalent in both diffuse and limited cutaneous systemic sclerosis, is more frequently encountered in the latter and among patients testing positive for anti-centromere antibodies. Calcinosis arises from localized deposits of calcium hydroxyapatite in the subcutaneous tissue and typically manifests over areas prone to trauma, such as finger pads and the extensor surfaces of elbows. This calcinosis predisposes to skin ulceration and secondary infection. Musculoskeletal Manifestations Musculoskeletal manifestations are prevalent in nearly all systemic sclerosis patients. Among these, arthralgia and myalgia are frequently reported. Inflammatory arthritis with true synovitis may occur, often resembling the pattern observed in rheumatoid arthritis, affecting joints such as the metacarpophalangeal, proximal interphalangeal, wrist, and ankle joints. Although erosions are uncommon, they may occur, particularly in association with periarticular calcinosis. While large joint involvement is rare, it can manifest in diffuse cutaneous systemic sclerosis cases. Additionally, there is evidence of systemic sclerosis overlapping with rheumatoid arthritis in up to 5% of cases. Distal bone resorption and osteolysis are seen in the late stages of diffuse cutaneous systemic sclerosis. While hand joint contractures are common, large joint contractures are rare in this form of the disease. Tendon friction rubs, resulting from inflammation, edema, and fibrosis of the tendon sheath, may occur in up to 30% of cases, particularly in diffuse cutaneous systemic sclerosis, and are associated with an unfavorable prognosis. Muscle involvement or myopathy in systemic sclerosis is multifactorial. Approximately 10% of patients exhibit an inflammatory myopathy akin to polymyositis, characterized by rapidly progressive proximal muscle weakness, elevated muscle enzymes, and biopsy findings consistent with inflammatory myositis. Additionally, decreased muscle strength can arise from malnutrition or deconditioning due to joint disease and skin fibrosis. In systemic sclerosis, direct muscle involvement can result in a fibrosing myopathy characterized by muscle fibrosis and atrophy. Patients with this condition may exhibit mildly elevated muscle enzyme levels, and muscle biopsy typically reveals fibrosis and atrophy with minimal or no inflammation. Unlike inflammatory myopathies, these patients usually do not respond to anti-inflammatory agents. Overall, myopathy in systemic sclerosis is associated with a poor prognosis. Gastrointestinal Manifestations Gastrointestinal involvement is almost universal in systemic sclerosis, encompassing diffuse and limited cutaneous systemic sclerosis. Symptoms can range from mild to severe, and any segment of the gastrointestinal tract may be affected. Oropharyngeal involvement manifests as perioral skin tightening, reduced oral aperture, periodontitis, and gingivitis. Dry mouth often arises from salivary gland fibrosis secondary to systemic sclerosis or inflammatory infiltrates associated with secondary Sjögren syndrome. Esophageal dysmotility, primarily due to fibrosis affecting the distal two-thirds of the esophagus, frequently results in dysphagia and heartburn, affecting up to 90% of systemic sclerosis patients. The lower esophageal sphincter becomes hypotonic, exacerbating reflux symptoms. Complications of esophageal dysmotility may include esophagitis, esophageal strictures, Barrett's esophagus, and bleeding. Gastric involvement may manifest as delayed gastric emptying (gastroparesis), resulting in early satiety, bloating, nausea, vomiting, and anorexia, potentially leading to malnutrition and weight loss. Gastric antral vascular ectasia, also known as "watermelon stomach," results from gastric telangiectasias and can lead to occult or large amounts of gastrointestinal bleeding. Intestinal dysmotility can cause diarrhea, constipation, and intestinal bacterial overgrowth. Mucosal telangiectasias can result in occult gastrointestinal bleeding. Intestinal dysmotility can lead to serious pseudo-obstruction. Wide-mouthed diverticula due to intestinal mucosal muscular atrophy are unique to systemic sclerosis. Reduced anal sphincter tone can result in rectal prolapse and incontinence. Hepatic involvement is not typically associated with systemic sclerosis, although primary biliary cirrhosis has been reported in some cases. Pulmonary Manifestations Pulmonary disease is the leading cause of mortality in patients with systemic sclerosis. The characteristic pulmonary involvement includes interstitial lung disease and/or PAH. Rarer pulmonary manifestations encompass pleuritis, obstructive airway disease, aspiration pneumonia, pulmonary hemorrhage, and cryptogenic organizing pneumonia. The spectrum of pulmonary disease in systemic sclerosis spans from clinically asymptomatic conditions to progressive respiratory failure, entailing severe morbidity. Interstitial lung disease manifests as bibasilar pulmonary fibrosis and is notably more prevalent and severe in diffuse cutaneous systemic sclerosis, African Americans, males, and individuals with anti–topoisomerase I antibodies. Clinically significant interstitial lung disease is present in about 50% of systemic sclerosis cases, although the condition may remain asymptomatic. Postmortem examinations have revealed interstitial lung disease in approximately 80% of systemic sclerosis patients. Interstitial lung disease usually manifests within the initial 4 to 5 years after a systemic sclerosis diagnosis. While nonspecific interstitial pneumonitis is more prevalent than usual interstitial pneumonia, a mixed pattern is also observable. Symptoms include dyspnea, fatigue, and, later in the disease course, a non-productive cough. Pulmonary function test reveals a restrictive pattern with decreased lung volumes and FEV1/FVC ratio. Diffusing capacity for carbon monoxide (DLCO) may be reduced, partly due to the underlying pulmonary vascular disease. High-resolution computed tomography (HRCT) of the chest is highly sensitive, particularly in the early stages of the disease, revealing increased subpleural lung attenuations in the bilateral posterior basal areas. Ground-glass opacities may indicate alveolitis, while additional findings include honeycombing, traction bronchiectasis, and thickening of the interlobular septa. PAH represents another common pulmonary manifestation, with a spectrum ranging from asymptomatic disease to severe PAH accompanied by right heart failure. Typically, PAH manifests late in the course of the disease, often occurring more than 10 years after the initial diagnosis of systemic sclerosis. This condition can affect 30% to 50% of individuals with systemic sclerosis, with a higher incidence observed in cases of limited cutaneous systemic sclerosis. Risk factors for PAH in systemic sclerosis include a later age at diagnosis, multiple telangiectasias, and the presence of anti–U3-RNP antibodies. Symptoms may include dyspnea initially, followed by chest pain, lower extremity swelling, lightheadedness, and syncope as the disease progresses. An accentuated "P component" of the second heart sound (S2) may be detectable. A pulmonary function test reveals an isolated reduction in diffusion capacity. Additionally, levels of NT-proBNP in the blood may be elevated. On echocardiography, an elevated peak right ventricular systolic pressure may be observed. When PAH is suspected, a right heart catheterization should be performed to confirm the diagnosis and exclude other potential causes. A diagnosis of PAH is typically confirmed by an elevation in mean pulmonary arterial pressure to 25 mm Hg or higher, along with a normal pulmonary capillary wedge pressure. Cardiac Manifestations Cardiac manifestations include pericarditis, pericardial effusion,dilated cardiomyopathy, and arrhythmias. Left ventricular diastolic dysfunction can occur secondary to PAH. Cardiac involvement is associated with an increased risk of infection and a poor prognosis. Pericardial effusions are usually small and exudative, and tamponade is uncommon. Pericardial effusions are more common in diffuse cutaneous systemic sclerosis and are predictive of scleroderma renal crisis. Dilated cardiomyopathy arises from patchy myocardial fibrosis. Fibrosis within the conduction pathways may precipitate arrhythmias, with premature ventricular contractions being the most common, although heart block, intraventricular conduction delays, and supraventricular tachycardia have also been reported. Renal Manifestations Renal manifestations in systemic sclerosis are exemplified by scleroderma renal crisis, historically almost always fatal and the primary cause of mortality in systemic sclerosis before the discovery of angiotensin-converting enzyme (ACE) inhibitors.Scleroderma renal crisis occurs in approximately 10% of systemic sclerosis patients, predominantly in those with diffuse cutaneous systemic sclerosis, and typically manifests within three years of diagnosis. Risk factors for the development of scleroderma renal crisis include African-American race, pericardial effusion, tendon friction rubs, new-onset anemia, diffuse cutaneous systemic sclerosis,anti-RNA polymerase III antibody, and high-dose (or chronic low-dose) corticosteroids. Indicators of poor outcomes in scleroderma renal crisis encompass male gender, advanced age at onset, and serum creatinine levels exceeding 3 mg/dL. The pathology of scleroderma renal crisis primarily involves vasculopathy, akin to the vascular changes observed in other organs affected by systemic sclerosis rather than glomerulonephritis. Clinically, it manifests with new-onset hypertension or malignant hypertension, often accompanied by renal insufficiency. Microangiopathy associated with this condition frequently results in hemolytic anemia and thrombocytopenia. While concurrent proteinuria may occur, microscopic hematuria is often observed but tends to be mild. Other Manifestations Other manifestations of systemic sclerosis include hypothyroidism, which occurs in up to 15% of patients, particularly those with limited cutaneous systemic sclerosis, often due to thyroid gland fibrosis. Moreover, autoimmune thyroid diseases, such as Hashimoto thyroiditis and Graves disease, are more prevalent among systemic sclerosis patients. These individuals also have a higher risk of developing other autoimmune conditions, such as primary biliary cirrhosis and secondary Sjögren syndrome. Additionally, thrombocytopenia has been observed in some cases. Overlap syndromes are common in systemic sclerosis and can involve conditions such as rheumatoid arthritis and polymyositis. Additionally, patients with systemic sclerosis have an elevated risk of psychological disorders such as depression, affecting up to 50% of individuals. Another variant, systemic sclerosis sine scleroderma, is characterized by internal organ manifestations of systemic sclerosis, particularly in the pulmonary and gastrointestinal systems, without skin thickening. These patients typically exhibit the Raynaud phenomenon, along with abnormal nailfold capillary exam results and positive antinuclear antibodies. Classification Criteria In 2013, the American College of Rheumatology published the updated classification criteria for systemic sclerosis, offering improved sensitivity and specificity compared to the older 1980 criteria. Although primarily designed for clinical study inclusion, these criteria can be applied cautiously in clinical settings. Utilizing a scoring system, a score of 9 or higher indicates classification as systemic sclerosis. Bilateral skin thickening proximal to metacarpophalangeal joints: 9 Skin thickening of fingers (counting only higher score) Between distal and proximal interphalangeal joints: 4 Puffy fingers: 2 Fingertip lesions(counting only the higher score) Fingertip pitting scars: 3 Digital tip ulcers: 2 Telangiectasia: 2 Abnormal nailfold capillaries: 2 Raynaud phenomenon: 3 Lung disease (maximum score of 2) ILD: 2 PAH: 2 Positive systemic sclerosis-specific antibodies (anti-centromere,anti–Scl–70, and anti–RNA polymerase III): 3 Evaluation Systemic sclerosis is diagnosed through clinical assessment. Early detection of the disease, evaluation of the extent of involvement, and ongoing surveillance for internal organ manifestations are essential for its effective management. Clinical Evaluation Assessment of skin thickness is performed using the modified Rodnan skin score, which assigns scores ranging from 0 to 3, with 0 indicating uninvolved areas and 3 indicating severe skin thickening. Monitoring changes in the score over time is important as it holds prognostic significance. Nailfold capillary examination is warranted in all patients presenting with Raynaud phenomenon and suspected systemic sclerosis. A comprehensive physical examination of patients targeting multiple organ systems is conducted during each visit to detect underlying organ involvement. Regular blood pressure monitoring, both in the clinic and at home, is highly recommended, particularly in individuals recently diagnosed with diffuse cutaneous systemic sclerosis, those experiencing new-onset hypertension, or those with a significant worsening of existing hypertension, as it may signal the onset of scleroderma renal crisis. Autoantibody Tests Autoantibodies are an essential diagnostic tool, offering predictive insights into disease phenotype and prognosis. Antinuclear antibodies, detected through direct immunofluorescence, are positive in over 90% of systemic sclerosis cases. In instances where antinuclear antibodies are negative, it is imperative to rule out other potential diagnoses before confirming systemic sclerosis. Specific autoantibodies, with a positivity rate of 60% to 70%, offer further diagnostic specificity. These antibodies, as mentioned below, which are mutually exclusive, can manifest several months to years before the clinical onset of systemic sclerosis. Anti-centromere antibody: Anti-centromere antibodies target 4 centromere antigens (CENP -B, -A, -C, D) and are predominantly observed in limited cutaneous systemic sclerosis, although they may occasionally occur in diffuse cutaneous systemic sclerosis. They can also manifest in conditions such as Sjögren syndrome and systemic lupus erythematosus. Anti-centromere antibodies are associated with an increased risk of PAH. Furthermore, they are linked to limited cutaneous involvement, a reduced likelihood of interstitial lung disease, and generally better survival rates compared to other autoantibodies. Anti-topoisomerase I (Scl-70) antibody: Anti-topoisomerase I (Scl-70) antibodies target the catalytic region of DNA helicase topoisomerase I. They are predominantly observed in diffuse cutaneous systemic sclerosis and infrequently in limited cutaneous systemic sclerosis. The presence of anti–Scl-70 antibodies is linked to an elevated risk of diffuse cutaneous involvement, interstitial lung disease, and cardiac involvement. Anti-RNA polymerase III antibody: Anti-RNA polymerase III antibodies target the eukaryotic RNA polymerase III. They are particularly associated with diffuse cutaneous systemic sclerosis and are linked to rapidly progressing and aggressive diffuse skin involvement, poor cutaneous outcomes, and scleroderma renal crisis. Additionally, these antibodies are also associated with a lower risk of interstitial lung disease and PAH. Some studies have indicated an association between anti-RNA polymerase III antibodies and malignancies in systemic sclerosis patients. Anti–U3-RNP (fibrillarin) antibody:Anti-U3-RNP (fibrillarin) antibodies are prevalent in African Americans and are associated with an overall poor prognosis in systemic sclerosis. Their presence correlates with increased internal organ involvement, diffuse cutaneous manifestations, interstitial lung disease, PAH, scleroderma renal crisis, myositis/myopathy, and cardiac complications. Other autoantibodies:Other autoantibodies include anti-Th/To antibodies that are associated with limited skin disease.Anti–PM/Scl antibodies associated with limited skin disease and overlap syndrome predispose to inflammatory myositis and interstitial lung disease. Anti–U1-RNP antibodies are more prevalent in African Americans and are associated with overlap syndrome and mixed connective tissue disease. This antibody profile is linked to limited cutaneous involvement and increased risk of inflammatory arthritis, myositis, lupus skin rashes, and lupus nephritis. Anti-Ku antibodies are also associated with overlap syndrome in systemic sclerosis, which is associated with more inflammatory arthritis and myositis. Laboratory Evaluation Complete blood counts to assess for anemia may be multifactorial.Renal function and 24-hour urine protein or urine protein:creatinine ratio should be closely monitored. Although inflammatory markers typically do not provide significant diagnostic significance, notable elevations may indicate active inflammatory myopathy or inflammatory arthritis. Muscle enzymes, including creatine kinase and aldolase, may be elevated, especially in the presence of inflammatory myopathy. Ancillary and Radiographic Evaluation X-rays of the extremities can reveal calcinosis and loss of distal phalanges. Periarticular erosions are rare and uncommon, although periarticular osteopenia can be seen. Musculoskeletal ultrasound can be used to show features of tenosynovitis. When muscle involvement is suspected, electromyography/nerve conduction velocity testing is the initial diagnostic choice. If abnormal results are obtained, a muscle biopsy may be warranted. High-resolution CT (HRCT) scan is the preferred imaging modality for assessing interstitial lung disease, as it can detect subtle findings that may not be visible on a chest x-ray. Pulmonary function tests, including spirometry, lung volumes, and diffusion capacity, can detect a restrictive pattern in interstitial lung disease very early in the disease process. A lung biopsy is of very limited value and should be avoided unless other diagnostic concerns are concurrent. When PAH is suspected, transthoracic echocardiography is usually initially performed, although right heart catheterization should follow to confirm the diagnosis and rule out other etiologies. Electrocardiography and Holter monitoring can be useful in detecting arrhythmias. Pericardial effusions can be well visualized on transthoracic echocardiography. In patients suspected of myocardial involvement, a cardiac MRI may be necessary. When esophageal and upper gastrointestinal involvement is suspected, upper gastrointestinal endoscopy, esophageal manometry, barium swallow studies, and a 24-hour pH probe can be utilized. A characteristic finding indicative of esophageal dysmotility is a dilated esophagus with excessive air observed on a CT scan. Treatment / Management No definitive treatment or universally accepted disease-modifying agent can alter the natural course of the disease. However, managing the affected system or systems has proven effective. Early diagnosis is crucial for achieving improved outcomes. Clinical evaluation and identification of affected organs and disease progression are critical for treatment efficacy. Furthermore, treatment goals must be holistic and tailored to optimize the quality of life for affected patients while also preventing further organ damage. Patient education about the disease and encouragement to engage in regular exercise, maintain a healthy diet and lifestyle, and seek emotional support should be considered for every individual with systemic sclerosis. Several agents have been investigated for different manifestations of systemic sclerosis, yet a generalized lack of large randomized controlled trials undermines their efficacy. Cyclophosphamide (for lung disease and skin disease), mycophenolate mofetil (for lung disease and skin disease), methotrexate (MTX; for skin disease, inflammatory arthritis, and myositis), azathioprine (for skin disease, lung disease, and myositis), and hydroxychloroquine (HCQ; for skin disease) stand out as some of the most frequently utilized immunosuppressive agents. Cyclosporine (for skin disease), infliximab (for inflammatory arthritis), and rituximab (for skin disease and lung disease) have limited available data. Corticosteroids should generally be avoided in systemic sclerosis due to the risk of precipitating scleroderma renal crisis. High-dose corticosteroids and long-term use of low to moderate-dose corticosteroids have been associated with the precipitation of scleroderma renal crisis. Therefore, these medications should only be considered in cases of absolute necessity, such as refractory inflammatory myositis, inflammatory arthritis, or active inflammatory alveolitis, and then used at the lowest possible dose for the shortest duration possible. Specific Therapies in Scleroderma Raynaud phenomenon:Raynaud phenomenon treatment aims to prevent digital ischemia and ulcers, which are easier to prevent than treat. Conservative management for the Raynaud phenomenon is still the cornerstone of management, and patients are advised to keep their extremities and body warm, avoid smoking and stress, and eliminate sympathomimetic medications when possible. In cases where β-blockers are used, switching to an alternative medication should be attempted if possible, as β-blockers may exacerbate Raynaud symptoms. Vasodilator therapy has demonstrated efficacy in Raynaud syndrome. Dihydropyridine calcium channel blockers, such as nifedipine (30-120 mg/d) or amlodipine (5-20 mg/d), are the first-line agents. In addition, vasodilators, including pentoxifylline, nitroglycerin, and phosphodiesterase inhibitors such as sildenafil (20 mg once daily to 3 times daily) may be considered. Prostacyclin analogs (iloprost) and endothelin receptor antagonists (bosentan) can be utilized for refractory cases with digital ischemic ulcers. Typically, iloprost is administered for 3 to 5 consecutive days at a continuous dose of 0.5 to 2 ng/kg/min, titrated to the maximum tolerated dose.43 The infusion can be continuous in the inpatient setting or administered over 8 hours in the outpatient setting. Bosentan is typically initiated at 62.5 mg twice daily for 4 weeks and may be increased to 125 mg twice daily. In refractory cases, proximal or distal sympathectomy may be considered. While botulinum toxin injections have shown efficacy in some small studies, their overall effectiveness remains limited. Vasodilator pumps may be utilized for patients unable to tolerate vasodilator therapies. Effective wound care is essential in the management of the condition in patients with digital ulcers. Skin disease:Several immunosuppressive agents have been used in sclerodactyly, including MTX, HCQ, mycophenolate mofetil, and cyclophosphamide. MTX is usually administered orally or subcutaneously at a dose of 15 to 25 mg per week. When systemic glucocorticoids are prescribed, they are usually initiated concurrently with MTX. The recommended regimen consists of either intravenous (IV) methylprednisolone at a dosage of 30 mg/kg/d for 3 consecutive days per month or oral prednisone at a dosage of 1 mg/kg/d. Patients are often treated with MTX for a duration of 1 to 2 years. Following 6 to 12 months of documented inactivity, gradual tapering of the MTX dose (reduction by 2.5 mg every 2 to 4 weeks) is typically initiated. However, relapses may occur following cessation of therapy.44 IV methylprednisolone is typically discontinued after 3 to 4 months of symptom resolution. Oral prednisone is tapered slowly over 3 to 4 months. While lacking robust data, most rheumatologists prefer mycophenolate mofetil for moderate-to-severe skin thickening. The typical dosage is 1 g twice daily, with a maximum of 1.5 g twice daily. Treatment is continued for 6 to 12 months after symptom resolution, tapered over several months. Cyclophosphamide is recommended for patients refractory to MTX (15-25 mg weekly) or mycophenolate mofetil (1.5-3 g daily) or with progressive disease. Antihistamines and topical moisturizing agents can help with pruritus. Topical therapy with corticosteroids,vitamin D analogs, and tacrolimus can also be used for limited skin disease. Telangiectasias usually do not require any treatment, although laser therapy can be considered for cosmetic purposes. No medical treatment has been proven effective in calcinosis, and surgical debulking may provide some relief in severe cases.(B3) Musculoskeletal involvement:Mild arthralgias in most patients often do not require treatment or respond well to nonsteroidal anti-inflammatory drugs (NSAIDs). Low-dose glucocorticoids (<10 mg/d of prednisone or equivalent) may be used for less than 2 to 4 weeks initially to control arthritis. Inflammatory arthritis can be managed with disease-modifying antirheumatic drugs such as HCQ (200-400 mg daily) and MTX. MTX is added if patients do not respond to HCQ. Antitumor necrosis factor agents can be used in refractory severe inflammatory arthritis.48 Physical and occupational therapy are crucial in preventing contractures. Management of inflammatory myositis in systemic sclerosis is similar to polymyositis, but high-dose corticosteroids should be avoided. Immunosuppressive agents, such as MTX and azathioprine, are effective in managing inflammatory myositis. Treating noninflammatory myopathy in systemic sclerosis is challenging, and physical therapy and exercises are more beneficial than immunosuppressive agents. Pulmonary involvement:Pulmonary involvement in systemic sclerosis is now the leading cause of mortality, necessitating interprofessional involvement from rheumatologists, pulmonologists, and cardiologists. Early detection is crucial for managing interstitial lung disease in systemic sclerosis. While cyclophosphamide has shown benefit for up to 18 months, its efficacy diminishes by 24 months.Some small studies have demonstrated the efficacy of cyclophosphamide followed by azathioprine for maintenance therapy. Mycophenolate mofetil (with a target dose of 1.5-3 g daily in 2 divided doses) has shown benefits in managing systemic sclerosis-related interstitial lung disease. Lung transplantation may be necessary for carefully selected individuals. Nintedanib, a tyrosine kinase inhibitor, was approved by the US Food and Drug Administration (FDA) for managing interstitial lung disease in systemic sclerosis in 2019, as clinical trials showed a slowing of the decline of pulmonary function tests in patients with systemic sclerosis with interstitial lung disease. Other investigational modalities include stem cell transplants and antifibrotic agents. When no contraindication exists, supplemental oxygen, diuretics, and anticoagulation are recommended, alongside patient education, a healthy lifestyle, and exercise as tolerated. Calcium channel blockers are generally ineffective. Nifedipine (starting dose 30 mg/d), amlodipine (starting dose 2.5 mg/d), and extended-release diltiazem (starting dose 120 mg/d) can be titrated up to the maximum tolerated dosages over weeks. Vasodilator therapy is recommended and may include phosphodiesterase-5 inhibitors (such as tadalafil 40 mg daily and sildenafil 20 mg orally 3 times daily), endothelin receptor antagonists (such as bosentan at 62.5-125 mg twice daily or sitaxsentan, ambrisentan, or macitentan), and/or prostacyclin analogs (such as epoprostenol, treprostinil, beraprost, and iloprost at 2.5-5 μg inhaled 6-9 times daily). Prostacyclin therapy is considered the most effective, although all these agents improve hemodynamics and quality of life. Combination therapy may be considered in patients who fail to improve on monotherapy or have severe PAH. Cardiac involvement:Arrhythmias are managed with antiarrhythmic agents and, occasionally, pacemaker placement. Currently, evidence supporting the efficacy of immunosuppressive agents or vasodilator therapy does not exist in scleroderma-related cardiac involvement. Gastrointestinal involvement:Exercise may aid in improving the oral aperture while maintaining good dental hygiene, which is essential to prevent caries and cavities due to sicca symptoms. Sugar-free lozenges and secretagogues such as pilocarpine or cevimeline can also alleviate dry mouth. For patients experiencing heartburn and gastroesophageal reflux, lifestyle and dietary adjustments, such as elevating the head of the bed, avoiding late and large meals, steering clear of spicy foods, and opting for small, frequent meals, are recommended. NSAIDs should be avoided. Proton pump inhibitors are preferred over H2 blockers, with high doses permissible, particularly for patients with erosive esophagitis. A combination of proton pump inhibitors and H2 blockers may be considered in severe cases. Motility agents such as metoclopramide and proton pump inhibitors may benefit patients with gastroparesis. Laser coagulation is a viable option for managing bleeding due to gastric antral vascular ectasia. Patients diagnosed with small intestinal bacterial overgrowth (SIBO) syndrome may require rotational antibiotics for effective management. Scleroderma renal crisis:ACE inhibitors represent the sole effective treatment for scleroderma renal crisis. Initiation should occur upon the earliest signs, with a maximum tolerated dose utilized. While any ACE inhibitor is viable, captopril is preferred due to its dosing flexibility. Currently, no data support the efficacy of angiotensin receptor blockers or renin inhibitors in scleroderma renal crisis. Renal function may initially decline, but continuation of the ACE inhibitor is crucial. With appropriate management, renal function can markedly improve, albeit over several months to years, potentially allowing for discontinuation of dialysis. However, prophylactic use of ACE inhibitors is not recommended, as it does not prevent scleroderma renal crisis and is associated with increased morbidity and mortality. Differential Diagnosis Systemic sclerosis is primarily a clinical diagnosis. However, several other diseases that can mimic systemic sclerosis should be considered when establishing a differential diagnosis. Eosinophilic Fasciitis Eosinophilic fasciitis is characterized by eosinophilic inflammation of the deep fascia, resulting in thickening and a woody induration of the upper and lower extremities, excluding the hands and feet. Raynaud phenomenon is not associated with eosinophilic fasciitis, and nailfold capillary examination typically appears normal. Antinuclear antibodies and specific autoantibodies are generally negative in eosinophilic fasciitis cases. Patients may develop contractures resembling those seen in systemic sclerosis. Eosinophilic fasciitis may be associated with underlying malignancies. Histologically, a skin biopsy often reveals eosinophilic infiltrates in the deep fascia. Scleromyxedema Scleromyxedema is usually seen in patients with monoclonal gammopathy or multiple myeloma and is characterized by papular waxy lesions on the face, neck, extremities, and fingers. Patients may also exhibit associated symptoms such as seizures and dementia. Raynaud phenomenon is not a feature of scleromyxedema, and nailfold capillary examination typically appears normal. Antinuclear antibodies and specific autoantibodies are typically negative in scleromyxedema cases. Histologically, a skin biopsy of scleromyxedema often reveals dermal fibrosis accompanied by perivascular inflammation and the deposition of mucin and fibrocytes, which are not typical findings in systemic sclerosis. Scleredema Scleredema can be associated with conditions such as diabetes mellitus, monoclonal gammopathy, fatigue, infections, and malignancies. This condition is characterized by the doughy, indurated appearance of the skin on the neck, back, and face, with digits typically being spared. Raynaud phenomenon is not associated with scleredema, and the nailfold capillary examination appears usually normal. Antinuclear antibodies and specific autoantibodies are generally negative in scleredema cases. Histologically, a skin biopsy of scleredema often reveals dermal fibrosis without perivascular inflammation, along with mucin deposition. Nephrogenic Systemic Fibrosis Nephrogenic systemic fibrosis is a rare phenomenon observed in patients with end-stage renal disease following exposure to gadolinium contrast. This condition is characterized by cobblestone-like nodular plaques on the extremities, trunk, hands, and feet while sparing the face. Raynaud phenomenon is not associated with nephrogenic systemic fibrosis, and the nailfold capillary examination typically appears normal. Antinuclear antibodies and specific autoantibodies are generally negative in nephrogenic systemic fibrosis cases. Histologically, a skin biopsy of nephrogenic systemic fibrosis often reveals dermal and epidermal fibrosis without perivascular inflammation, accompanied by the deposition of mucin and fibrocytes. Eosinophilia Myalgia Syndrome This syndrome was an epidemic associated with the use of L-tryptophan, resulting in severe myalgias, visceral involvement, and elevated CK levels. Toxic Oil Syndrome This syndrome was also an epidemic in Spain associated with the consumption of adulterated rapeseed oil, causing livedo reticularis, pulmonary infiltrates, and elevated levels of creatine kinase. Prognosis Systemic sclerosis is associated with high mortality, having the highest case-specific mortality among all collagen vascular disorders. Previously, scleroderma renal crisis was the most common cause of mortality before the advent of ACE inhibitors. The use of ACE inhibitors and increased awareness have significantly reduced scleroderma renal crisis-related mortality. Currently, pulmonary disease is the most common cause of mortality in systemic sclerosis patients. Over the past 30 years, the prognosis of systemic sclerosis has improved, with 5-year survival rates reaching up to 80%. However, patients with advanced PAH have a less than 50% 2-year survival rate. Additionally, patients with systemic sclerosis-related PAH have a poorer prognosis compared to those with idiopathic PAH. Before the discovery of ACE inhibitors, the 1-year survival rate for patients experiencing scleroderma renal crisis was less than 15%. However, with increased awareness and the use of ACE inhibitors, the 1-year survival rate in scleroderma renal crisis has improved significantly to over 85%. Patients with systemic sclerosis also face a heightened risk of developing malignancies, particularly lung cancer. In addition, an increased risk of esophageal adenocarcinoma exists in systemic sclerosis cases associated with chronic gastroesophageal reflux, which can lead to Barrett esophagus. Prognostic Factors and Associations in Systemic Sclerosis African-American race: Early disease onset, severe disease, interstitial lung disease, and scleroderma renal crisis. Late age of diagnosis: PAH and poor outcomes after scleroderma renal crisis. Diffuse cutaneous systemic sclerosis: Diffuse skin disease, large joint inflammatory arthritis, tendon friction rubs, joint contractures, distal bone osteolysis, scleroderma renal crisis, interstitial lung disease, and pericardial effusion. Limited cutaneous systemic sclerosis: Limited skin disease, calcinosis, PAH, and hypothyroidism. Multiple telangiectasias:PAH. Pericardial effusion: Scleroderma renal crisis. Tendon friction rubs:Scleroderma renal crisis and a poor overall prognosis. New-onset anemia:Scleroderma renal crisis. Corticosteroid use:Scleroderma renal crisis. Anti-centromere antibody: Limited cutaneous systemic sclerosis and PAH. Lower risk of interstitial lung disease. Anti–Scl-70 antibody: Diffuse cutaneous systemic sclerosis,interstitial lung disease, and cardiac involvement. Anti-RNA polymerase III antibody: Rapidly progressing and aggressive diffuse skin involvement, poor cutaneous outcomes, scleroderma renal crisis, and malignancies. Lower risk of interstitial lung disease and PAH. Complications Systemic sclerosis is associated with various complications primarily due to end-organ damage from fibrosis. Digital ischemia may progress to gangrene, necessitating amputation. Gastrointestinal complications can result in malnutrition. Pulmonary involvement is the primary cause of morbidity, causing irreversible pulmonary fibrosis. Patients with scleroderma renal crisis can have permanent renal damage, although the use of ACE inhibitors leads to recovery of renal function in most cases. Deterrence and Patient Education Patient education plays a crucial role in disease management, including lifestyle adjustments. Individuals with Raynaud phenomenon should be educated on maintaining body and extremity warmth, avoiding extreme cold exposure, vasoconstrictive agents, and trauma to the digits. Patients with systemic sclerosis should be advised to quit smoking and avoid exposure to secondhand smoke. Regular blood pressure monitoring at home can aid in early detection of scleroderma renal crisis. Enhancing Healthcare Team Outcomes Systemic scleroderma results in significant morbidity, causing severe disability without a known cure and only offering symptomatic treatment. Due to its impact on multiple organ systems, a collaborative interprofessional healthcare team approach is necessary. This team should include the patient's primary care clinician, rheumatologists, gastroenterologists, cardiologists, pulmonologists, nephrologists, and dermatologists. Furthermore, nurses and pharmacists are critical in patient care for systemic sclerosis. Pharmacists ensure that patients are not taking medications causing vasoconstriction and educate them on adhering to blood pressure medications. Patients with scleroderma require close follow-up by healthcare providers. Nursing staff are essential for patient education, monitoring, and follow-up care. They can also coordinate activities among healthcare professionals involved in the patient's treatment. Patient education about managing systemic sclerosis manifestations is vital for preventing long-term morbidity. Media (Click Image to Enlarge) Bilateral Signs of Raynaud Phenomenon in a Patient's Hands. A male patient is exhibiting triphasic color changes in the digits of his hands, indicative of Raynaud Phenomenon. DermNet New Zealand (Click Image to Enlarge) Pursed Mouth in a Female Patient With Scleroderma. 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Joint bone spine. 2023 Jul:90(4):105560. doi: 10.1016/j.jbspin.2023.105560. Epub 2023 Mar 17 [PubMed PMID: 36933783] van den Hoogen F, Khanna D, Fransen J, Johnson SR, Baron M, Tyndall A, Matucci-Cerinic M, Naden RP, Medsger TA Jr, Carreira PE, Riemekasten G, Clements PJ, Denton CP, Distler O, Allanore Y, Furst DE, Gabrielli A, Mayes MD, van Laar JM, Seibold JR, Czirjak L, Steen VD, Inanc M, Kowal-Bielecka O, Müller-Ladner U, Valentini G, Veale DJ, Vonk MC, Walker UA, Chung L, Collier DH, Csuka ME, Fessler BJ, Guiducci S, Herrick A, Hsu VM, Jimenez S, Kahaleh B, Merkel PA, Sierakowski S, Silver RM, Simms RW, Varga J, Pope JE. 2013 classification criteria for systemic sclerosis: an American College of Rheumatology/European League against Rheumatism collaborative initiative. Arthritis and rheumatism. 2013 Nov:65(11):2737-47. doi: 10.1002/art.38098. Epub 2013 Oct 3 [PubMed PMID: 24122180] Almaabdi K, Ahmad Z, Johnson SR. Advanced Autoantibody Testing in Systemic Sclerosis. Diagnostics (Basel, Switzerland). 2023 Feb 23:13(5):. doi: 10.3390/diagnostics13050851. Epub 2023 Feb 23 [PubMed PMID: 36899995] Zanatta E, Huscher D, Ortolan A, Avouac J, Airò P, Balbir-Gurman A, Siegert E, Matucci Cerinic M, Cozzi F, Riemekasten G, Hoffmann-Vold AM, Distler O, Gabrielli A, Heitmann S, Hunzelmann N, Montecucco C, Morovic-Vergles J, Ribi C, Doria A, Allanore Y, EUSTAR collaborators. Phenotype of limited cutaneous systemic sclerosis patients with positive anti-topoisomerase I antibodies: data from the EUSTAR cohort. Rheumatology (Oxford, England). 2022 Nov 28:61(12):4786-4796. doi: 10.1093/rheumatology/keac188. Epub [PubMed PMID: 35348643] Steen V, Domsic RT, Lucas M, Fertig N, Medsger TA Jr. A clinical and serologic comparison of African American and Caucasian patients with systemic sclerosis. Arthritis and rheumatism. 2012 Sep:64(9):2986-94. doi: 10.1002/art.34482. Epub [PubMed PMID: 22576620] Li Z, Xu D, Jiang X, Li T, Su Y, Mu R. Anemia Is an Indicator for Worse Organ Damage Trajectories in Patients with Systemic Sclerosis: A Retrospective Study. Journal of clinical medicine. 2022 Aug 26:11(17):. doi: 10.3390/jcm11175013. Epub 2022 Aug 26 [PubMed PMID: 36078943] Level 2 (mid-level) evidence Adarsh MB, Sharma SK, Prasad KK, Dhir V, Singh S, Sinha SK. Esophageal manometry, esophagogastroduodenoscopy, and duodenal mucosal histopathology in systemic sclerosis. JGH open : an open access journal of gastroenterology and hepatology. 2019 Jun:3(3):206-209. doi: 10.1002/jgh3.12138. Epub 2019 Mar 20 [PubMed PMID: 31276037] Pope J, Fenlon D, Thompson A, Shea B, Furst D, Wells G, Silman A. Iloprost and cisaprost for Raynaud's phenomenon in progressive systemic sclerosis. The Cochrane database of systematic reviews. 2000:1998(2):CD000953 [PubMed PMID: 10796395] Level 1 (high-level) evidence Kroft EB, Creemers MC, van den Hoogen FH, Boezeman JB, de Jong EM. Effectiveness, side-effects and period of remission after treatment with methotrexate in localized scleroderma and related sclerotic skin diseases: an inception cohort study. The British journal of dermatology. 2009 May:160(5):1075-82. doi: 10.1111/j.1365-2133.2008.09017.x. Epub 2009 Feb 4 [PubMed PMID: 19210503] Level 2 (mid-level) evidence Tehlirian CV, Hummers LK, White B, Brodsky RA, Wigley FM. High-dose cyclophosphamide without stem cell rescue in scleroderma. Annals of the rheumatic diseases. 2008 Jun:67(6):775-81 [PubMed PMID: 17974598] Kroft EB, Groeneveld TJ, Seyger MM, de Jong EM. Efficacy of topical tacrolimus 0.1% in active plaque morphea: randomized, double-blind, emollient-controlled pilot study. American journal of clinical dermatology. 2009:10(3):181-7. doi: 10.2165/00128071-200910030-00004. Epub [PubMed PMID: 19354332] Level 3 (low-level) evidence Avouac J, Clements PJ, Khanna D, Furst DE, Allanore Y. Articular involvement in systemic sclerosis. Rheumatology (Oxford, England). 2012 Aug:51(8):1347-56. doi: 10.1093/rheumatology/kes041. Epub 2012 Mar 30 [PubMed PMID: 22467084] Lam GK, Hummers LK, Woods A, Wigley FM. Efficacy and safety of etanercept in the treatment of scleroderma-associated joint disease. The Journal of rheumatology. 2007 Jul:34(7):1636-7 [PubMed PMID: 17611970] Level 3 (low-level) evidence Cole A, Ong VH, Denton CP. Renal Disease and Systemic Sclerosis: an Update on Scleroderma Renal Crisis. Clinical reviews in allergy & immunology. 2023 Jun:64(3):378-391. doi: 10.1007/s12016-022-08945-x. Epub 2022 Jun 1 [PubMed PMID: 35648373] Niklas K, Niklas A, Puszczewicz M. [Eosinophilic fasciitis]. Postepy higieny i medycyny doswiadczalnej (Online). 2015 Jan 2:69():488-95 [PubMed PMID: 25897110] Kim S, Park TH, Lee SM, Kim YH, Cho MK, Whang KU, Kim HS. Scleromyxedema with multiple systemic involvement: Successful treatment with intravenous immunoglobulin. Dermatologic therapy. 2020 May:33(3):e13378. doi: 10.1111/dth.13378. Epub 2020 Apr 15 [PubMed PMID: 32250023] Gambichler T, Susok L, Doerler M, Dickel H, Chatzipantazi M. Löfgren syndrome associated with scleroedema adultorum of Buschke. Clinical and experimental dermatology. 2023 Jan 20:48(1):39-40. doi: 10.1093/ced/llac021. Epub [PubMed PMID: 36669186] Steen VD, Medsger TA. Changes in causes of death in systemic sclerosis, 1972-2002. Annals of the rheumatic diseases. 2007 Jul:66(7):940-4 [PubMed PMID: 17329309] × StatPearls Is Part Of The Inc. 5000 Fastest Growing Companies Become a Contributor Information Help & FAQs About us Contact us Privacy Policy Legal Refund policy Editorial Policy Education Physician CME Nurse Practitioner CE Nurse CE FREE CME/CE Contact Institutional Sales Feel free to get in touch with us and send a message support@statpearls.com Copyright Â© 2025 StatPearls Use the mouse wheel to zoom in and out, click and drag to pan the image × ✓ Thanks for sharing! 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1430	https://medium.com/@jshaik2452/choosing-the-best-model-a-friendly-guide-to-aic-and-bic-af220b33255f	Sitemap Open in app Sign in Sign in Writing is for everyone.Register for Medium Day Choosing the Best Model: A Friendly Guide to AIC and BIC Jani Data Diaries 5 min readNov 7, 2024 Introduction: The Dilemma of Choosing the Right Model Imagine you’re a detective working on a case. You have several theories (models) about what might have happened, each with its own set of details. Some theories are simple, while others are more complex. You want to pick the best theory — the one that explains the evidence without overcomplicating things. In the world of machine learning and statistics, we often face the same dilemma. We want a model that explains our data well but doesn’t overfit or add unnecessary complexity. This is where AIC and BIC come in. These criteria help us balance accuracy with simplicity, guiding us to the model that best fits the data without over-complicating things. Let’s dive into the world of AIC and BIC and see how they help us make smart choices in model selection. Understanding the Trade-Off: Complexity vs. Fit Imagine you’re trying to explain a mystery to a friend. You could tell a quick, simple version, or you could go into extreme detail with every tiny clue. If you go too simple, your friend might not understand the whole story, but if you add too much detail, they might get lost. Similarly, when building models, we want the model to capture the data patterns without making it overly complex. AIC and BIC both aim to help us strike this balance: AIC (Akaike Information Criterion) rewards models that fit the data well while penalizing unnecessary complexity. BIC (Bayesian Information Criterion) also penalizes complexity but is stricter, especially for large datasets. What is AIC? Akaike Information Criterion (AIC) is a measure used to evaluate models by balancing fit and complexity. It’s named after the Japanese statistician Hirotugu Akaike, who developed it to help select the best model without overfitting. In simpler terms: AIC considers the model’s ability to fit the data. It penalizes the model for each additional parameter to discourage overfitting. Here’s the formula for AIC: where: Likelihood measures how well the model explains the data. k is the number of parameters in the model. Imagine telling a story where you want to balance completeness with simplicity. Each detail (parameter) you add improves the story, but if you add too many details, it gets cluttered. AIC helps you keep it balanced, preferring models that tell the “simplest, most complete” version of the story. What is BIC? Bayesian Information Criterion (BIC) is similar to AIC but takes a stricter approach to model complexity. It’s based on Bayesian statistics, adding a stronger penalty for models with more parameters, especially when the dataset is large. Get Jani Data Diaries’s stories in your inbox Join Medium for free to get updates from this writer. The formula for BIC is: BIC=−2ln⁡(Likelihood)+kln⁡(n)BIC = -2 \ln(\text{Likelihood}) + k \ln(n)BIC=−2ln(Likelihood)+kln(n) where: n is the number of observations in the dataset. The penalty term increases with the size of the dataset. Think of BIC as a stricter editor who prefers the simplest possible explanation. As the dataset grows, BIC becomes more cautious about adding new parameters, preferring a model that remains as simple as possible while still capturing the essential patterns. Comparing AIC and BIC: When to Use Which? Both AIC and BIC help us find the right model, but they have slightly different preferences: AIC is more forgiving, often favoring slightly more complex models. It’s commonly used when we want to avoid underfitting, and is preferred for smaller datasets. BIC is stricter, especially as the dataset grows. It tends to favor simpler models, as the penalty for extra parameters increases with sample size. Imagine AIC as a friend who says, “Add a few details, as long as they help,” while BIC says, “Only add details if they’re absolutely necessary.” In practice, if you have a small dataset, AIC might be more suitable, as it doesn’t penalize complexity as harshly. For larger datasets, BIC is often a better choice, as it will help prevent overfitting by preferring simpler models. Python Code Example: Comparing AIC and BIC Let’s see how to calculate and compare AIC and BIC using Python on a simple dataset. ``` from sklearn.datasets import make_regressionfrom sklearn.linear_model import LinearRegressionfrom sklearn.metrics import mean_squared_errorimport numpy as np# Step 1: Generate a sample datasetX, y = make_regression(n_samples=100, n_features=2, noise=10, random_state=42)# Step 2: Fit a linear regression modelmodel = LinearRegression()model.fit(X, y)# Step 3: Calculate AIC and BICn = len(y) # Number of observationsk = X.shape # Number of parameters (features)# Calculate log likelihoodpredictions = model.predict(X)mse = mean_squared_error(y, predictions)log_likelihood = -n / 2 np.log(2 np.pi mse) - (1 / (2 mse)) np.sum((y - predictions) 2)# AIC and BIC calculationsaic = 2 k - 2 log_likelihoodbic = np.log(n) k - 2 log_likelihoodprint(f"AIC: {aic:.2f}")print(f"BIC: {bic:.2f}") from import from import from import import as Step 1: Generate a sample dataset 100 2 10 42 Step 2: Fit a linear regression model Step 3: Calculate AIC and BIC len Number of observations 1 Number of parameters (features) Calculate log likelihood 2 2 1 2 sum 2 AIC and BIC calculations 2 2 2 printf"AIC: {aic:.2f}"{aic:.2f}.2 printf"BIC: {bic:.2f}"{bic:.2f}.2 ``` In this code: AIC and BIC are calculated based on the model’s log-likelihood and the number of parameters. These values give us a sense of which model (or version of the model) best balances fit and complexity. Interpreting AIC and BIC Scores When comparing models, the lower the AIC or BIC, the better the model. For instance, if you try several models with different numbers of parameters: Choose the model with the lowest AIC if you’re focusing on fit with slight flexibility in complexity. Choose the model with the lowest BIC if you’re aiming for simplicity, especially with larger datasets. Remember, AIC and BIC values are only useful when comparing models on the same dataset. They don’t have an absolute interpretation but work as relative measures to help you pick the best model. Conclusion: AIC and BIC as Friendly Guides AIC and BIC are like helpful advisors guiding us toward the best model. They balance the model’s ability to capture data patterns without going overboard with unnecessary complexity. AIC is like a supportive friend, encouraging a bit more detail, while BIC is a careful editor, ensuring that each addition truly adds value. Next time you’re stuck choosing between models, let AIC and BIC help you decide. With their guidance, you’ll be able to pick the model that’s not too simple, not too complex — but just right! Data Science AI Deep Learning Model Selection Machine Learning ## Written by Jani Data Diaries 13 followers ·33 following Hi, I’m Jani, a data science enthusiast with a passion for simplifying complex concepts. Join me as I break down the world of data, one blog post at a time! Responses (1) Write a response What are your thoughts? Iteichert Aug 19 ``` Author is incorrect, should be ChatGPT ``` More from Jani Data Diaries Jani Data Diaries ## Multi-Class Logistic Regression: A Friendly Guide to Classifying the Many Introduction: One Model, Many Choices Nov 5, 2024 2 1 Jani Data Diaries ## Mastering Data Cleaning: Your Guide to Cleaning Numerical and Categorical Data Introduction: Why Data Cleaning Matters Nov 3, 2024 Jani Data Diaries ## Data Cleaning for Images and Videos: Preparing Visual Data for AI Introduction: Clearing the Lens for Clearer Insights Nov 8, 2024 10 Jani Data Diaries ## TinyTroupe: The Ensemble of Lightweight Libraries Introduction: A Tiny Troupe with a Big Impact Nov 26, 2024 See all from Jani Data Diaries Recommended from Medium FATIMA-ZOHRA BEZZA ## Segmenting Mixed Datasets with Clustering — Part I: Methods and Metrics Standard clustering algorithms such as K-Means and DBSCAN are widely used for identifying structure in datasets, but they are not… May 15 51 In Artificial Intelligence in Plain English by Adith Narasimhan Kumar ## Statistics for Data Science 101 Series — Bayes Theorem Did you know that Bayes’ Theorem helped crack the Enigma code during World War II? Alan Turing and his team at Bletchley Park used Bayesian… Apr 24 3 1 Rahul Nair ## From Black Box to Fair Lending: Explaining Credit Risk Models with SHAP Why your AI’s ‘why’ is critical for banks, regulators, and customers. Jul 7 4 1 Navnoor Bawa ## Goldman’s Copper Options Catastrophe: When Political Risk Models Fail How Wall Street’s most sophisticated desk misread Trump’s tariff implementation and learned a $100M+ lesson in political complexity Aug 2 20 1 In Data Science Collective by Manousos Klados, PhD Sep 9 11 1 Irene Markelic, PhD Mar 26 145 4 See more recommendations Text to speech
1431	https://metatutor.co.uk/wp-content/uploads/PDFs/Worksheets/Foundation/STANDARD-FORM.pdf	STANDARD FORM – PRACTICE QUESTIONS 1. Convert each number from standard form. (a) 9 × 105 (b) 4.5 × 102 (c) 8.1 × 104 (d) 7.77 × 103 (e) 1.18 × 104 (f) 3.9 × 105 (g) 1.26 × 106 (h) 9.2 × 104 (i) 5.533 × 107 (j) 8 × 105 2. Convert each number from standard form. (a) 5 × 10-2 (b) 7.81 × 10-3 (c) 1.95 × 10-4 (d) 5.9 × 10-5 (e) 6.34 × 10-2 (f) 7 × 10-6 (g) 9.234 × 10-3 (h) 8.01 × 10-4 (i) 6.5 × 10-6 (j) 3.31 × 10-2 3. Convert each number into standard form. (a) 2000 (b) 4500 (c) 781000 (d) 660000 (e) 1250000 (f) 9800 (g) 1457000 (h) 90800 (i) 40000000 (j) 72500 (k) 2020 4. Convert each number into standard form. (a) 0.03 (b) 0.0065 (c) 0.000015 (d) 0.00456 (e) 0.0235 (f) 0.00000001 (g) 0.00803 (h) 0.0022 (i) 0.0003928 (j) 0.09099 (k) 0.00216 5. Convert each number from standard form. (a) 8 × 104 (b) 1.12 × 105 (c) 8.9 × 10-2 (d) 7.24 × 106 (e) 4.467 × 10-3 (f) 3 × 10-7 (g) 9.04 × 10-4 (h) 7.355 × 105 (i) 2.21 × 10-5 (j) 7.102 × 103 (k) 9.9999 × 108 6. Convert each number into standard form. (a) 0.067 (b) 120000 (c) 9180 (d) 0.00004 (e) 0.02907 (f) 562000000 (g) 0.001111 (h) 0.00000035 (i) 9800000 (j) 0.000455 (k) 67550000 7. Put these numbers into order, smallest to largest. 6.2 × 105 6.3 × 103 6.4 × 10-2 6.5 × 104 8. Put these numbers into order, smallest to largest. 1.55 × 102 1.56 × 103 1.57 × 103 1.58 × 102 9. Put these numbers into order, smallest to largest. 6.14 × 104 6.145 × 105 6.144 × 104 6.4 × 103 10. Put these numbers into order, smallest to largest. 3 × 10-2 3.1 × 10-3 3.01 × 10-4 2.9 × 10-3 11. Put these numbers into order, smallest to largest. 1.11 × 106 1.01 × 105 1.001 × 106 1.1 × 104 12. Put these numbers into order, smallest to largest. 5.5 × 10-1 56 × 10-3 55.6 × 10-2 560 × 10-5 13. Put these numbers into order, smallest to largest. 0.7 × 106 7.1 × 105 705 × 103 71.2 × 103 14. Put these numbers into order, smallest to largest. 9.09 × 10-6 908 × 10-7 91 × 10-6 9.009 × 10-6 15. Put these numbers into order, smallest to largest. 7156 × 10-2 0.0071 × 105 7.16 × 102 0.761 × 103 16. Put these numbers into order, smallest to largest. 0.0004 × 108 4001 × 10-2 0.41 × 104 410000 × 10-5 17. Work out (6.5 × 105) + (2 × 104). Give your answer in standard form. 18. Work out (9.01 × 105) + (6 × 103). Give your answer in standard form. 19. Work out (4.3 × 107) – (6 × 105). Give your answer in standard form. 20. Work out (8.2 × 105) + (4.714 × 107). Give your answer in standard form. 21. Work out (5.75 × 109) – (4 × 106). Give your answer in standard form. 22. Work out (3 × 103) × (2 × 104). Give your answer in standard form. 23. Work out (8 × 107) ÷ (4 × 105). Give your answer in standard form. 24. Work out (9 × 103) × (2 × 103). Give your answer in standard form. 25. Work out (3 × 1010) ÷ (4 × 103). Give your answer in standard form. 26. Work out (4 × 1010) ÷ (5 × 103). Give your answer in standard form. 27. A floppy disk can store 1,600,000 bytes of data. Steve needs to store 4.8 × 107 bytes of data. How many floppy disks would Steve need to store the data? 28. The surface area of Earth is 4.95 × 108 km2. The surface area of Mars is 6.7 × 106 km2. What is the difference in surface area between Earth and Mars? Give your answer in standard form. 29. The thickness of a piece of paper is 1.25 × 10-4 metres. 5 × 105 pieces of paper are piled on top of each other. How tall is the pile of paper? Give your answer in centimetres. 30. A snail moves at a speed of 8 × 10-3 metres per second. How many seconds would it take the snail to travel 200 metres?
1432	https://math.stackexchange.com/questions/3776765/in-any-triangle-the-side-opposite-to-the-larger-angle-is-longer	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams In any triangle, the side opposite to the larger angle is longer Ask Question Asked Modified 4 years, 3 months ago Viewed 1k times 2 $\begingroup$ I want to prove the following known theorem (2) In $\triangle ABC$, if $\angle B>\angle C$, then $AC>AB$ I saw some proofs and they used indirect proof (by contradiction) using the following theorem which seems easier (1) In $\triangle ABC$, if $AC\leq AB$, then $\angle B\leq\angle C$ Also I saw some proofs using triangle inequality: "sum of each two sides is larger than the other one" which again uses (1). However I like to prove (2) directly with more elementary methods without using (1) or Pythagorean theorem. Is it possible to do this? Any comments appreciated. geometry euclidean-geometry triangles triangle-inequality Share edited Jun 11, 2021 at 11:58 Clemens Bartholdy 14.3k44 gold badges2929 silver badges7777 bronze badges asked Aug 1, 2020 at 15:33 asadasad 97777 silver badges1414 bronze badges $\endgroup$ 1 $\begingroup$ It follows directly from the sine rule $\endgroup$ Mike Daas – Mike Daas 2020-08-01 15:38:48 +00:00 Commented Aug 1, 2020 at 15:38 Add a comment \| 2 Answers 2 Reset to default 2 $\begingroup$ You certainly do not need the Pythagorean theorem. A "direct proof" will depend on what you have previously proved. This is Euclid, Book I Proposition 19. The proof there is indirect, depending on the previous proposition. That's the proof you dislike. Share answered Aug 1, 2020 at 15:42 Ethan BolkerEthan Bolker 105k77 gold badges127127 silver badges223223 bronze badges $\endgroup$ Add a comment \| 2 $\begingroup$ Here, $AC=AP+PC=A'P+PC>A'C=AB$. Share answered Aug 1, 2020 at 15:47 Hagen von EitzenHagen von Eitzen 383k3333 gold badges378378 silver badges685685 bronze badges $\endgroup$ 1 $\begingroup$ Interesting, where did you use the hypothesis on angles ? $\endgroup$ Maman – Maman 2025-02-27 23:50:30 +00:00 Commented Feb 27 at 23:50 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions geometry euclidean-geometry triangles triangle-inequality See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Linked 2 Using law of sines to prove that larger angle is opposite to largest side in a triangle Related 1 Proof using properties of an isosceles or right-angle triangle 1 Finding the side and angle of a triangle. 1 Geometric inequality 1 Prove that the sides of the orthic triangle meet the sides of the given triangle in three collinear points. Comparing lengths in a triangle where the angle bisector makes an obtuse angle with the opposite side 1 Given $\triangle ABC$, with $AB Orthocenter: The "Bad Boy" of Distinguished Points in a Triangle Prove that a triangle with bisectral triangle being a right triangle always has a 120-degree angle 3 Proving directly that, for scalene $\triangle ABC$, the bisector of $\angle A$ and perpendicular bisector of $BC$ meet at a point on the circumcircle Hot Network Questions Lingering odor presumably from bad chicken Identifying a movie where a man relives the same day How do trees drop their leaves? How to locate a leak in an irrigation system? how do I remove a item from the applications menu Why do universities push for high impact journal publications? Why is the definite article used in “Mi deporte favorito es el fútbol”? My dissertation is wrong, but I already defended. How to remedy? Can I use the TEA1733AT for a 150-watt load despite datasheet saying 75 W? 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1433	https://fiskalloy.com/wire-facts/tensile-yield-and-elongation/	Wire Facts: Tensile, Yield and Elongation - Fisk Alloy We value your privacy We use cookies to enhance your browsing experience, serve personalised ads or content, and analyse our traffic. By clicking "Accept All", you consent to our use of cookies. Customise Reject All Accept All Customise Consent Preferences We use cookies to help you navigate efficiently and perform certain functions. You will find detailed information about all cookies under each consent category below. The cookies that are categorised as "Necessary" are stored on your browser as they are essential for enabling the basic functionalities of the site. ...Show more Necessary Always Active Necessary cookies are required to enable the basic features of this site, such as providing secure log-in or adjusting your consent preferences. These cookies do not store any personally identifiable data. 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1434	https://eepower.com/technical-articles/resonance-explained/	Log In Join Join the Community Register Log In Or sign in with Facebook Google GitHub Linkedin SIGN UP FOR OUR NEWSLETTER Technical Article Resonance Explained Join our Engineering Community! Sign-in with: Home Technical Articles Resonance Explained Resonance Explained This article introduces resonance and how it is related to impedance and admittance, the parallel and series RLC circuit resonant design, calculations of its parameters and electrical quantities, and the application of the resonance circuits. Technical Article Sep 29, 2022 by Simon Mugo In electric circuits, there are energy-storing elements like capacitors and inductors. The energy-storing elements are what cause resonance. Resonance is the driving concept behind the working principle of TV and radio receivers, allowing viewers to select desired frequencies for their TV channels. Image used courtesy of Adobe Stock Electrical Resonance For electrical resonance to occur, there must be a cancelation of admittances and impedances of the elements of the circuit at a given resonant frequency. In some special circuits, this can only occur when the impedance between the output and input of the system is about zero and its transfer function is closer to one. There exist two types of resonance that are series and parallel. These configurations exist based on how elements of the circuits are connected–in either series or parallel. This article presents a wider scope of series and parallel resonance. Series Resonance In this type of resonance, all the circuit passive elements are connected in series in what we refer to as a series RLC circuit. Look at the series RLC circuit in Figure 1 below with values represented in the phasor domain. Figure 1. Series Resonance Circuit. Image used courtesy of Simon Mugo The inductor, resistor, and capacitor which form part of the passive elements are connected in series with the sinusoidal voltage source which serves as the input; Apply Kirchoff’s Voltage Law around the given loop in the circuit: [V-V_{R}-V_{L}-V_{C}=0] Replacing the values of the inductor, resistor, and capacitor in the above equation gives: [V-IR-I(jX_{L})-I(-jX_{C})=0] Rearranging the equation: [V=IR+I(jX_{L})+I(-jX_{C})] Further rearrangement: (V=I[R+j(X_{L}-X_{C})]..........)(1) Note that equation 1 above is in the form of V = IZ and therefore, the impedance Z is shown by the equation below: [Z=R+j(X_{L}-X_{C})]] Electrical Quantities and Parameters at Resonance The series RLC resonance circuit will have several parameters and electrical quantities when at resonance and it shall be very significant if we derive them in this article. Resonant Frequency This is the frequency at which resonance occurs and is denoted as fr. When you have a series RLC circuit, resonance when the impedance Z term is equal to zero that is the difference between the value of XL – XC gives us zero. Equating the above to zero gives us: [X_{L}=X_{C}] Substituting (X_L = 2\pi fL) and (X_{C}=\frac{1}{2\pi fC}) in the equation above gives, [2\pi fL=\frac{1}{2\pi fC}] Making f the subject of the formula gives: [f=\frac{1}{(2\pi)\sqrt{LC}}] This translates to the resonance frequency of the series RLC circuit which can be written as: [f_{r}=\frac{1}{(2\pi)\sqrt{LC}}] Where C is the capacitor's capacitance and L is the inductor's inductance. While the series RLC resonant frequency is independent of the resistor R, it always depends on the capacitance C and inductance L. Impedance From earlier derivations, we found that impedance Z of the series RLC circuit is given by: [Z=R+j(X_{L}-X_{C})]] But [X_{L}=X_{C}] And substituting this in the impedance Z equation, we get [Z=R+j(0)] Hence [Z=R] This explains that the impedance Z of the series RLC circuit is equal to the resistance R in the circuit. Current I Flowing Through Series RLC Circuit Earlier we derived equation (1) as [V=I[R+j(X_{L}-X_{C})]] Let’s substitute XL=XC into the equation. Therefore, [V=I[R+j(0)]] [V=IR] Making I the subject of the form gives [I=\frac{V}{R}] And this is the current flowing through the series RLC circuit at resonance. In the series RLC circuit, the impedance arrives at the minimum value at resonance. Therefore, the maximum current of the series RLC circuit is attained at resonance. Voltage Across Resistor The voltage that goes across the resistor is given by the formula [V_{R}=IR] Let us substitute the value of I in the equation [V_{R}=(\frac{V}{R})R] Canceling R out we find that [V_{R}=V] And that gives as the voltage across the resistor Voltage Across Capacitor From the circuit above, the voltage across the capacitor is given by: [V_{C}=I(-jX_{C})] Substituting the current I value in the equation [V_{C}=\frac{V}{R}\Big(-jX_{C}\Big)] Opening the bracket [V_{C}=-j\Bigg(\frac{X_{C}}{R}\Bigg)V] But we know that (\frac{X_{C}}{R} = Q-Factor),Hence, the equation becomes [V_{C}=-jQV] And this is the voltage across the capacitor at resonance. The magnitude of the voltage across this capacitor at resonance can be found to be [\|V_{C}\|=QV] Voltage Across Inductor This is given by [V_{L}=I(jX_{L})] By substituting the value of I in the equation; [V_{L}=\frac{V}{R}(jX_{L})] Rearranging [V_{L}=j\Bigg(\frac{X_{L}}{R}\Bigg)V] Hence [V_{L}=jQV] And this gives the voltage across the inductor at resonance. The magnitude across the inductor at resonance will be [\|V_{L}\|=QV] It is significant to take note that series resonance RLC circuits are named voltage magnification circuits because both the magnitude of the voltage across the capacitor and inductor is equal that voltage is equal to Q times the input voltage V. Parallel Resonance This is the resonance that occurs on the parallel RLC circuit. Let’s look at the parallel RLC circuit below which has a phasor domain representation. Figure 2. Series Parallel Resonance Circuit. Image used courtesy of Simon Mugo In this circuit, it is evident all the passive elements’ connection is in parallel with the input sinusoidal current source. We have labeled node P in the same circuit and can write the nodal equation at this point. [-I+I_{R}+I_{L}+I_{C}=0] By introducing the value of currents across the elements into the equation. [-I+\frac{V}{R}+\frac{V}{jX_{L}}+\frac{V}{-jX_{C}}=0] Rearranging the above equation, we get: (I=V\Bigg[\frac{1}{R}+j\Bigg(\frac{1}{X_{C}}-\frac{1}{X_{L}}\Bigg)\Bigg]..........)Equation 1 The equation above is of form I = VY Where Y is the admittance of the parallel RLC circuit. Therefore, admittance will be given by: [Y=\frac{1}{R}+j\Bigg(\frac{1}{X_{C}}-\frac{1}{X_{L}}\Bigg)] Resonance Electrical Quantities and Parameters of the RLC Circuits Here, we will focus on the derivation of electrical quantities and parameters of the parallel RLC circuits at resonance. Resonant Frequency Here, resonance occurs when the term of the imaginary admittance is zero. [\frac{1}{X_{C}}-\frac{1}{X_{L}}=0] Rearranging you get [X_{L}=X_{C}] Note that the above resonance condition is similar to the RLC circuit, implying that the resonance frequency fx remains the same. [f_{r}=\frac{1}{(2\pi)\sqrt{LC}}] Admittance From the derivation of equation 1, note that admittance is given by: [Y=\frac{1}{R}+j\Bigg(\frac{1}{X_{C}}-\frac{1}{X_{L}}\Bigg)] But at resonance [X_{L}=X_{C}] Substituting it in the admittance equation we get [Y=\frac{1}{R}] Therefore, at resonance, we have the admittance Y being equal to the resistance reciprocal. Voltage Available Across Each Element We know that (\frac{1}{X_{C}}-\frac{1}{X_{L}}=0) and by replacing this into equation 1, find that [I=\frac{V}{R}] Rearranging [V=IR] This gives the voltage across the element in the parallel RLC resonance circuit. Current Through the Resistor This is given by the formula [I_{R}=\frac{V}{R}] But we already know the value of voltage as V=IR and can substitute it Hence, [I_{R}=I] And this gives us the current flowing through the resistor at resonance. Current Through Inductor at Resonance This is given by [I_{L}=\frac{V}{jX_{L}}] Let’s substitute the value of V in the equation above [I_{L}=\frac{VR}{jX_{L}}] On rearranging, we get [I_{L}=-j\Bigg(\frac{R}{X_{L}}\Bigg)I] But (\frac{R}{X_{L}}) is the Quality factor, Q of the system hence current flowing through the inductor becomes [I_{L}=-jQI] Therefore, the magnitude of this current flowing through the inductor will be given by: [\|I_{L}\|=QI] Current Through the Capacitor This is given by: [I_{C}=\frac{V}{-jX_{L}}] Substituting the value of V in the equation gives [I_{C}=j\Bigg(\frac{R}{X_{C}}\Bigg)I] Hence, [I_{C}=jQI] Which is the current flowing through the capacitor at resonance. The magnitude flowing through the capacitor is given by: [\|I_{C}\|=QI] The parallel RLC electrical circuits are also known as current magnification circuits because the magnitude of the electrical current that flows through the capacitor and inductor is equal to the quality factor times the current at the input. Applications of Resonance Television broadcasts and radio transmissions Voltage magnification especially the series RLC circuit Communication systems and signal processing Oscillator circuits Induction heating Key Takeaways of Resonance In circuits, there exist energy-storing elements such as capacitors and inductors. There exist two types of resonance that are series and parallel RLC circuit resonance. Resonance is about the cancellation of admittance and impedances in an electrical circuit. The article has introduced us to how to calculate the electrical quantities and parameters of both series and parallel RLC circuits. 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1435	https://fiveable.me/ap-stats/unit-5/sampling-distributions-for-sample-proportions/study-guide/Ezxev8MPpv3mFKjV4Gq3	printables 📊AP Statistics Unit 5 Review 5.5 Sampling Distributions for Sample Proportions 📊AP Statistics Unit 5 Review 5.5 Sampling Distributions for Sample Proportions Written by the Fiveable Content Team • Last updated September 2025 Verified for the 2026 exam Verified for the 2026 exam•Written by the Fiveable Content Team • Last updated September 2025 📊AP Statistics Unit & Topic Study Guides Unit 5 Overview: Sampling Distributions 5.1 Introducing Statistics: Why Is My Sample Not Like Yours? 5.2 The Normal Distribution, Revisited 5.3 The Central Limit Theorem 5.4 Biased and Unbiased Point Estimates 5.5 Sampling Distributions for Sample Proportions 5.6 Sampling Distributions for Differences in Sample Proportions 5.7 Sampling Distributions for Sample Means 5.8 Sampling Distributions for Differences in Sample Means Formulas You can usually tell if you will solve a problem using sample proportions if the problem gives you a probability or percentage. For a sample proportion with probability p, the mean of our sampling distribution is equal to the probability. All formulas in this section can be found on page 2 of the given formula sheet. more resources to help you study practice questionscheatsheetscore calculator Source: (NEW) AP Statistics Formula Sheet Large Counts Condition Before you can use a sampling distribution for sample proportions to make inferences about a population proportion, you need to check that the sample meets certain conditions. One of these conditions is the large counts condition, which states that the sample size should be large enough for the distribution of the sample proportion to be approximately normal. The large counts condition can be expressed as np ≥ 10 and n(1-p) ≥ 10, where n is the sample size and p is the sample proportion. This means that both the number of successes (np) and the number of failures (n(1-p)) in the sample should be at least 10. If these conditions are met, then you can assume that the sampling distribution for the sample proportion is approximately normal, and you can use statistical techniques that rely on normality, such as confidence intervals or hypothesis tests (we'll cover this in future sections!). ✔️ For the shape (normal) of distributions of means, you can check the Central Limit Theorem, but for proportions you must always check the Large Counts Condition. Practice Problem Suppose that you are conducting a survey to estimate the proportion of people in your town who support a new public transportation system. You decide to use a simple random sample of 1000 people, and you ask them whether or not they support the new system. After collecting the data, you find that 600 people out of the 1000 respondents support the system. a) Calculate the sample proportion of respondents who support the new system. b) Explain what the sampling distribution for the sample proportion represents and why it is useful in this situation. c) Suppose that the true population proportion of people who support the new system is actually 0.6. Describe the shape, center, and spread of the sampling distribution for the sample proportion in this case. d) Explain why the Central Limit Theorem applies to the sampling distribution for the sample proportion in this situation. BOLD_PLACEHOLDER_9 Calculate a 95% confidence interval for the population proportion of people who support the new system based on the sample data. (Optional for now, but feel free to answer if you already checked out the section on confidence intervals!) f) Discuss one potential source of bias that could affect the results of this study, and explain how it could influence the estimate of the population proportion." Answers a) The sample proportion of respondents who support the new system is 600/1000 = 0.6. b) The sampling distribution for the sample proportion represents the distribution of possible values for the sample proportion if the study were repeated many times. It is useful in this situation because it allows us to make inferences about the population proportion based on the sample data. c) If the true population proportion of people who support the new system is 0.6, the sampling distribution for the sample proportion would be approximately normal with a center at 0.6 and a spread that depends on the sample size and the variability of the population. d) The Central Limit Theorem applies to the sampling distribution for the sample proportion in this situation because the sample size (n = 1000) is large enough for the distribution to be approximately normal, even if the population is not normally distributed. BOLD_PLACEHOLDER_16 A 95% confidence interval for the population proportion of people who support the new system can be calculated as 0.6 +/- (1.96 sqrt((0.6(1-0.6))/1000)). This gives a confidence interval of (0.570, 0.630). f) One potential source of bias in this study could be nonresponse bias, which occurs when certain groups of individuals are more or less likely to respond to the survey. For example, if people who support the new system are more likely to respond to the survey, the sample could be biased toward higher levels of support and produce an overestimate of the population proportion. On the other hand, if people who do not support the new system are more likely to respond, the sample could be biased toward lower levels of support and produce an underestimate of the population proportion. Frequently Asked Questions What's the formula for the standard deviation of sample proportions? The standard deviation (standard error) of the sample proportion p̂ is σp̂ = sqrt[ p(1 − p) / n ]. This is for independent trials (sampling with replacement). If you sample without replacement from a finite population and your sample is more than about 10% of the population, multiply by the finite population correction √[(N − n)/(N − 1)] to get a smaller σp̂. For using the normal approximation on the AP exam, check the success–failure condition: np ≥ 10 and n(1 − p) ≥ 10. Mean of the sampling distribution is μp̂ = p. For a quick topic review see the Fiveable study guide ( and try practice problems ( to solidify this. How do I know when the sampling distribution of p-hat is approximately normal? You can treat the sampling distribution of p̂ as approximately normal when three conditions hold: 1. Independence—observations are independent (random sample or randomized experiment). If you sample without replacement, the sample size must be less than 10% of the population. 2. Success–failure (large enough n)—np ≥ 10 and n(1 − p) ≥ 10. This ensures enough expected successes and failures so the binomial is well approximated by a normal. 3. Use the sampling parameters: μp̂ = p and σp̂ = sqrt[p(1 − p)/n] (or the finite-population correction if sampling without replacement and n is not <10% of N). On the AP exam, you’ll verify those conditions before using z-approximations for proportions (CED UNC-3.L.1, UNC-3.K). For a quick refresher, check the Topic 5.5 study guide ( For extra practice, try problems at ( When do I use np ≥ 10 and n(1-p) ≥ 10 and what happens if these conditions aren't met? Use np ≥ 10 and n(1−p) ≥ 10 to check the success–failure condition for p̂. If both hold, the sampling distribution of the sample proportion is approximately normal with mean μp̂ = p and standard error √[p(1−p)/n] (CED UNC-3.L.1 and UNC-3.K.1). Also check independence: sample with replacement or n ≤ 10% of the population. If either np or n(1−p) < 10, the sampling distribution is noticeably skewed and the normal (z) approximation is unreliable. In that case don’t use z-based CI or tests—use the exact binomial model, a simulation/randomization method, or increase n so the condition holds. On the AP exam, you must verify these conditions before applying normal-approximation inference for proportions. More review and worked examples are in the Topic 5.5 study guide ( For extra practice, try problems at the Unit 5 page ( or the practice bank ( I'm confused about the difference between p and p-hat - can someone explain this? p is the population proportion—the true (but usually unknown) fraction of the whole population with a certain trait (a parameter). p̂ (read “p-hat”) is the sample proportion—the fraction in your sample that have the trait (a statistic, and your best point estimate of p). Key facts from the CED (Topic 5.5): - Sampling distribution: μp̂ = p (p̂ is an unbiased estimator of p). - Standard deviation (SE) of p̂ (for independent trials): σp̂ = √[p(1−p)/n]. If sampling without replacement and n > 10% of population, use the finite-population correction. - Normal approximation: p̂ is approx. Normal when np ≥ 10 and n(1−p) ≥ 10 (use these for z-tests/CIs). Interpretation tip: p is fixed (but unknown); p̂ varies from sample to sample. Use p̂ to make probabilistic statements about p via the sampling distribution. For examples and practice, see the Topic 5.5 study guide ( and more practice problems ( What's the mean of the sampling distribution for sample proportions and why is it equal to p? The mean of the sampling distribution of the sample proportion p̂ is μp̂ = p—the true population proportion. Why? Each sample proportion is just the average of n independent 0/1 “success” indicators (1 = success, 0 = failure). The expected value of each indicator is p, so the expected value of their average is also p. In other words, p̂ is an unbiased estimator of p (CED UNC-3.K.1). This holds when trials are independent (sampling with replacement or when n < 10% of the population). For large n you also get an approximately normal sampling distribution when np ≥ 10 and n(1 − p) ≥ 10 (UNC-3.L.1), which is why many inference procedures on proportions use the normal approximation. For a focused review see the Topic 5.5 study guide ( and try practice problems ( How do I calculate the standard deviation when sampling without replacement vs with replacement? If you sample with replacement (or your draws are independent), the sampling distribution of p̂ has mean μp̂ = p and standard deviation σp̂ = sqrt[p(1−p)/n]. If you sample without replacement from a finite population of size N, use the finite population correction (FPC): σp̂ = sqrt[p(1−p)/n] × sqrt((N−n)/(N−1)). That second square root shrinks the standard deviation because observations aren’t independent. If n < 0.10N, the FPC ≈ 1 and you can ignore it (that’s the 10% condition in the CED). Remember to check normality for using z-based methods: require np ≥ 10 and n(1−p) ≥ 10 (success–failure). For AP review see the Topic 5.5 study guide ( the Unit 5 overview ( and practice problems ( Can someone walk me through finding the probability that a sample proportion is within a certain range step by step? Step-by-step (use the normal approximation to the sampling distribution of p̂): 1. Identify p (population proportion) and n (sample size). 2. Check conditions: independent trials (10% condition if no replacement) and success–failure: np ≥ 10 and n(1−p) ≥ 10. If they fail, don’t use normal approx. 3. Mean of sampling distribution: μp̂ = p. Standard error: SE = sqrt[p(1−p)/n]. 4. Convert the proportion bounds to z-scores. For lower bound a and upper bound b compute: z_a = (a − p)/SE and z_b = (b − p)/SE. 5. Probability that p̂ is between a and b = P(z_a < Z < z_b). Use Z-table or calculator to find Φ(z_b) − Φ(z_a). 6. Interpret in context (e.g., “there’s about X% chance a random sample of n will have a sample proportion between a and b”). Quick numeric example: p=0.4, n=100, want P(\|p̂−0.4\|<0.05) ⇒ a=0.35, b=0.45. SE = sqrt(0.4·0.6/100)=0.049. z_a=(0.35−0.4)/0.049≈−1.02, z_b≈+1.02. P≈Φ(1.02)−Φ(−1.02)≈0.846−0.154≈0.692 (69.2%). For more AP-aligned notes and examples, see the Topic 5.5 study guide ( and tons of practice problems ( When is the 10% rule important and how does it affect my calculations? The 10% rule matters when you sample WITHOUT replacement from a finite population. AP’s formulas for the sampling distribution of p̂ (UNC-3.K.1) assume independent trials; if your sample is less than 10% of the population (n < 0.10N), those trials are “close enough” to independent and you can use σ_p̂ = sqrt[p(1−p)/n]. If n is ≥ 10% of N, independence fails and the true standard deviation is smaller by the finite population correction (FPC): σ_p̂ = sqrt{p(1−p)/n}·sqrt[(N−n)/(N−1)]. Example: from N = 10,000, samples under 1,000 meet the 10% rule. So the rule doesn’t change the mean (μ_p̂ = p) but it tells you whether to apply the FPC—otherwise you’ll slightly overestimate uncertainty. For AP review, see the Topic 5.5 study guide ( and more unit practice ( I don't understand why we need np ≥ 10 and n(1-p) ≥ 10 for normal approximation - what's the logic behind this? The np ≥ 10 and n(1−p) ≥ 10 rule is a practical check that the sampling distribution of p̂ will be roughly normal. p̂ comes from a binomial process (success/failure). The Central Limit Theorem says averages/proportions get approximately normal as sample size grows, but the binomial is skewed when p is near 0 or 1. Requiring at least about 10 expected successes (np) and 10 expected failures (n(1−p)) keeps both tails big enough so skew is small and the normal approximation works. Also remember independence (random sampling or 10% condition if sampling without replacement). On the AP, you need to verify these conditions before using z-based methods for proportions (CED UNC-3.L.1). For a quick review, see the Topic 5.5 study guide ( the Unit 5 overview ( and practice problems ( What's the formula σp̂ = √(p(1-p)/n) and when do I use it? σp̂ = √(p(1 − p)/n) is the standard deviation (standard error) of the sampling distribution of a sample proportion p̂. Use it when you’re modeling repeated independent trials of a categorical variable (binomial setting) and you want how much p̂ typically varies around the true population proportion p. Key points from the CED: μp̂ = p and σp̂ = √(p(1−p)/n) when sampling with replacement or when trials are independent. If sampling without replacement, the SD is a bit smaller—you can ignore that difference when n < 0.10·N. To use a normal approximation (so you can compute z-scores or CIs), check the success–failure condition: np ≥ 10 and n(1−p) ≥ 10. The AP exam provides relevant formulas/tables; make sure to report context and units when interpreting probabilities. For a focused walkthrough, see the Topic 5.5 study guide ( For more practice, try Fiveable’s AP Stats problems ( How do I solve problems where they ask for the probability that p-hat is greater than some value? You do it the same way you’d find any tail probability for a sampling distribution of p̂—check conditions, get the sampling mean and standard error, convert to a z-score, then use the normal table (or your calculator). Steps: 1. Check success–failure: np ≥ 10 and n(1 − p) ≥ 10 (if sampling without replacement, also check n < 0.10·N). 2. Mean: μp̂ = p. Standard error: SE = sqrt[p(1 − p)/n]. 3. For P(p̂ > a), compute z = (a − p)/SE. Then P(p̂ > a) = P(Z > z) = 1 − Φ(z). Example: p = 0.4, n = 100, a = 0.45 → SE = sqrt(.4·.6/100)=.049; z = (0.45−0.40)/.049 ≈ 1.02; P ≈ 1 − Φ(1.02) ≈ 0.153. Use your calculator’s normalcdf or a Z-table. AP tip: always state conditions and interpret probability in context. For a quick topic review see the Fiveable study guide ( and try practice problems ( What does it mean that the standard deviation is smaller when sampling without replacement? It means your sample proportions vary less when you draw without replacement because draws are no longer independent—once you pick a “success,” there are fewer successes left in the pool, so later draws are slightly less variable. The center of the sampling distribution stays the same (μ_p̂ = p), but the standard deviation (standard error) is reduced by the finite population correction (FPC): multiply √(p(1–p)/n) by √((N–n)/(N–1)). Practically, APs expect you to know that if n < 10% of N, the FPC ≈ 1 and you can ignore it (the 10% condition for independence). If n is a substantial fraction of N, use the FPC so your SE is smaller and confidence intervals/tests are narrower—i.e., more precise. For a quick review, see the Topic 5.5 study guide ( and more unit review at ( Practice problems: ( I missed class - what's the difference between population proportion and sample proportion in context problems? Population proportion p is the true (but usually unknown) fraction of the whole population with some trait—e.g., 0.35 of all teens prefer brand A. Sample proportion p̂ is the fraction you actually observe in your sample—e.g., 14/40 = 0.35. For Topic 5.5 the key ideas: the sampling distribution of p̂ has mean μp̂ = p and standard error σp̂ = sqrt[p(1−p)/n] (use finite-pop correction only when sample >10% of population). The CLT for proportions says p̂ is approximately Normal when np ≥ 10 and n(1−p) ≥ 10. So p is the parameter (what you’re trying to learn about) and p̂ is the statistic (your estimate, with variability you can quantify). On the AP exam you’ll be asked to identify these, compute the SE, check success–failure and 10% conditions, and interpret probabilities in context. For topic review check Fiveable’s study guide ( unit overview ( and practice problems ( How do I interpret the parameters of a sampling distribution in word problems about proportions? Think of the sampling distribution of p̂ as the long-run behavior of the sample proportion when you repeat the same random sampling process many times. Two parameters you’ll need to interpret in context: - Mean μp̂ = p: this equals the true population proportion. In words: “On average, the sample proportion will be p (e.g., 0.34 of adults).” Always name the population and the unit (proportion or percent). - Standard deviation (standard error) σp̂ = √[p(1−p)/n]: this measures how much p̂ typically varies from sample to sample. In words: “Typical difference between p̂ and p is about √[p(1−p)/n].” If sampling without replacement and n > 10% of the population, apply the finite-population correction. Also check normal approx conditions: np ≥ 10 and n(1−p) ≥ 10 (and independence). When answering AP questions (UNC-3.M) always state numbers, units, population, and whether probabilities refer to repeated samples. For a quick review see the Topic 5.5 study guide ( and more unit review ( For practice, try problems at ( Why does the sampling distribution of sample proportions become normal as sample size increases? Because p̂ is the average of many independent 0/1 trial outcomes (success = 1, failure = 0), the Central Limit Theorem (CLT) applies: as the number of independent trials n gets large, the distribution of that average becomes approximately normal. For proportions this means p̂ has mean μp̂ = p and standard error σp̂ = sqrt[p(1−p)/n]. The AP CED turns that into a simple rule you can use on the exam: the sampling distribution of p̂ is approximately normal when np ≥ 10 and n(1−p) ≥ 10 (the success–failure condition). Intuitively, larger n shrinks the SE and averages out the discrete binomial bumps so the histogram of many p̂’s looks bell-shaped. For a quick refresher and practice, check the Topic 5.5 study guide ( and hit more practice problems at (
1436	https://www.youtube.com/watch?v=BSKtQcLQLWU	Lagrange multiplier example, part 1 Khan Academy 9090000 subscribers 1382 likes Description 266417 views Posted: 15 Nov 2016 Courses on Khan Academy are always 100% free. Start practicing—and saving your progress—now: A Lagrange multipliers example of maximizing revenues subject to a budgetary constraint. 42 comments Transcript: so let's say you're running some kind of company and you guys produce widgets you produce some little trinket that people enjoy buying and the main costs that you have are labor you know the the workers that you have creating these and steel and let's just say that your labor costs are $20 per hour $20 each hour and then your steel costs are uh $2,000 keep the numbers kind of related to each other $2,000 for every ton of Steel and then you've had your analysts work a little bit on trying to model the revenues you can make with your widgets as a function of hours of Labor and tons of steel and let's say the revenue model that they've come up with the revenue as a function of you know hours of Labor and then s for steel let's say um is equal to about you know 100 times the hours of Labor to the^ 2/3 multiplied by the tons of steel to the power 1/3 if you put in a given amount of Labor and a given amount of Steel this is about how much money you're going to expect to earn and of course you want to earn as much as you can but let's say you actually have a budget for how much you're able to spend on all these things and your budget the budget is $220,000 you're willing to spend $20,000 and you want to make as much money as you can according to this model based on that now this is exactly the kind of problem that the lrange multiplier technique is made for we're trying to maximize some kind of function and we have a constraint now right now the constraint isn't written as a formula but we can pretty easily write it as a formula because what makes up our budget well it's going to be the number of hours of Labor multiplied by 20 so that's going to be $20 per hour multiplied by the number of hours you put in plus 2,000 doll per tons of steel times the tons of steel that you put in so the constraint is basically that you have to have these values equal $20,000 I mean you could say less than right you could say you're not willing to go any more than that but intuitively and in reality it's going to be the case that in order to maximize your revenues you should squeeze every dollar that you have available and actually hit this constraint so this right here is the constraint of our problem and let's go ahead and give this guy a name the function that we're dealing with a name and I'm going to call it G of HS which is going to be that guy and now if you'll remember in the last few videos The Way We visualize something like this is to think about the set of all possible inputs so in this case you know you might be thinking about the HS plane you know the number of hours of labor on one axis the number of tons of steel on another and this constraint well in this case it's a it's a linear function so this constraint is going to give us some kind of line that tells us which pairs of s and H are going to achieve that constraint and then the revenue function that we're dealing with will have certain Contours you know maybe revenues of $10,000 have a certain Contour that looks like this and revenues of $100,000 have a certain Contour that looks like this but what we want is to find which value is barely touching the constraint curve just tangent to it at a given point because that's going to be the contour line where if you up the value by just a little bit it would no longer intersect with that curve there would no longer be values of H s that satisfy this constraint and the way to think about finding that tangency is to consider the vector perpendicular to the tangent line to the curve at that point which fortunately is represented by let's see let me make some room for myself here uh represented by the gradient the gradient of our our R function the the function whose Contours this is the revenue and what it means for this to be tangent to the constraint line is that there's going to be another Vector the gradient of G of our constraint function that points in the same direction that's proportional to that and typically the way you write this is to say that the gradient of this function is proportional to proportional to the gradient of G and this proportionality constant is called our lrange multiplier is called the lrange multiplier so let's go ahead and start working it out let's first compute the gradient of R so the gradient of R is going to be the partial derivative of r with respect to its first variable which is H so partial derivative with respect to H and the second component is its partial derivative with respect to that second variable s with respect to S and in this case that first part derivative if we treat H as a variable and S as a constant then that 2/3 gets brought down so that'll be 100 2/3 H H to the power of well we've got to subtract 1 from 2/3 when we bring it down so that'll be - 1/3 multiplied s to the 1/3 and then the second component here the partial derivative with respect to S is going to be 100 well now by treating s is the variable we take down that 1/3 so that's 1/3 h^ the 2/3 just looks like a constant as far as s is concerned and then we take s to the 1/3 - 1 which is 2/3 - 2/3 great so that's the gradient of R and now we need the gradient of G and that one's a lot easier actually because G is just a linear function so when we take the gradient of G which is its partial derivative with respect to H partial H and its partial derivative with respect to S partial s well the partial with respect to H is just 20 the function looks like 20 h plus something that's a constant so that ends up being 20 and then the partial with respect to S likewise it's just 2,000 because it's just some constant multiplied by S Plus a bunch of other stuff that looks like constants so that's great and this means when we set the gradient of r equal to the gradient of G the pair of equations that we get and let me just write it all out again is we have this top one which I'll call 2003 times and let's go ahead and do a little simplifying while I'm rewriting things here so H to the 1/3 is really 1 over H to the 1/3 sorry is 1 over H 1/3 and that's s 1/3 so all of this that first component is being set equal to the first component of the gradient of G which is 20 Lambda time this lrange multiplier because we're not setting the gradients equal to each other we're just setting them proportional to each other so that's the first equation and then the second one I'll go ahead and do some simplifying while I rewrite that one also that's going to be 100/3 and then H to 2/3 so times H to the 2/3 / s 2/3 CU s the -2/3 is the same as 1/ s 2/3 all of that is equal to 2,000 2,000 Lambda and the important thing is it's that same Lambda because the entire Vector has to be proportional and I think right here is probably a pretty good point to stop and in the next video I'll go ahead and work through the details and we'll land on a solution
1437	https://ioer.ilsharedlearning.org/resource/138297/NISt_Chemical_Kinetics_Database	Illinois Open Educational Resource Detail Skip to Content ResourcesLibrariesLearning Lists / Learning Sets Contribute Resources Login/Register NISt Chemical Kinetics Database Description The NIST Chemical Kinetics Database includes essentially all reported kinetics results for thermal gas-phase chemical reactions. The database is designed to be searched for kinetics data based on the specific reactants involved, for reactions resulting in specified products, for all the reactions of a particular species, or for various combinations of these. Creator:kinetics.nist.gov Publisher:kinetics.nist.gov Submitter:National Science Digital Library (NSDL) Last Updated:7/8/2014 Rights:Unknown Usage Rights 11 Tags Language English Education Levels Grades 9-10 Grades 11-12 Adult Education Resource Type Manipulative Reference Material Media Type Webpage End User Adult Learner Professional Student Teacher/Education Specialist 0 Learning Standards This Resource has not been aligned to any learning standards. 5 Keywords EducationComputing and InformationComputational ScienceNSDLChemistry #Education #ComputingandInformation #ComputationalScience #NSDL #Chemistry 10 More Like This NIST Chemistry WebBook For information on thermodynamic data for over 5,000 chemical compounds and on ion-energetics data for over 10,000 chemical compounds, NIST's Chemistry WebBook is available for searching. Compounds can be searched by name, chemical formula, or CAS (Chemical Abstracts Service) registry number. Information on enthalpy of formation, heat capacity, entropy, ionization potential, and appearance potential is available. Note that this site works best on browsers that support tables. NIST also supplies reference databases in the area of physics. NIST Chemistry WebBook The NIST Chemistry WebBook provides access to data compiled and distributed by NIST under the Standard Reference Data Program. This includes thermochemical parameters and spectroscopic (Infrared, Ultraviolet-Visible, Mass Spectral) data. NIST Chemistry WebBook Several tutorials on the use of the NIST Chemistry WebBook describing how to obtain thermochemical, thermophysical, and ion energetics data compiled by NIST under the Standard Reference Data Program. Reversible Chemical Kinetics Model The Reversible Chemical Kinetics Model illustrates the time evolution of a reversible chemical reaction, from a given set of initial reactants concentration until chemical equilibrium is reached. The concentrations are plotted against time and the respective numerical values are periodically displayed at the data Table. As chemical equilibrium is attained, the concentrations become constant, the reaction quotient (Q) equalizes the equilibrium constant and Gibbs's energy change (G/RT) approaches zero. The Reversible Chemical Kinetics Model was developed using the Easy Java Simulations (EJS) modeling tool. It is distributed as a ready-to-run (compiled) Java archive. Double clicking the jar file will run the program if Java is installed. You can modify this simulation if you have EJS installed by right-clicking within the map and selecting "Open Ejs Model" from the pop-up menu item. Deterministic versus Stochastic Chemical Kinetics Traditional phenomenological kinetics uses firstorder ordinary differential equations ODEs to describe the rate of change of the species in a chemical reaction The solution to these equations provides expressions for the concentration of each species as a function of time The use of ODEs to describe rate laws implies that concentrations are continuous quantities As a consequence this model is deterministic and provides exact solutions with no uncertainty involved Because Avogadros number is so enormous even a nanomole of material involves over 100 trillion molecules so in a typical labscale reaction it is reasonable to assume that concentrations are continuous variables However when concentrations or volumes become tiny as in the inside of a human cell this assumption can break down In truth molecules are discrete entities onetenth of a water molecule does not exist in the physical world As a consequence chemical reactions involving small numbers of molecules are stochastic processes better described by probability methods Stochastic simulation algorithms provide an alternative approach to chemical kinetics in which trajectories consistent with these probability functions can be simulated via Monte Carlo methods The model allows you to compare the results predicted by a traditional phenomenological method to a stochastic simulation algorithm known as the Gillespie or direct method 1 kinetics This applet allows the user to vary rate constants and concentrations in a consecutive, first step reversible set of chemical reactions. Computational Chemistry The investigation of chemical structure and behavior through the application of computational chemistry techniques. NetLogo Models Library: Enzyme Kinetics Model page from the NetLogo Models Library. The page provides a description and screenshots of a mode of Enzme Kinetics produced using the NetLogo software. The page provides a link to a javascript version of the model that can be run in the browser, as well as a download link for the model file that can be opened, run and edited in NetLogo. This model demonstrates the kinetics of single-substrate enzyme-catalysis. The interactions between enzymes and substrates are often difficult to understand and the model allows users to visualize the complex reaction. Chemistry Experiment Simulation, Tutorials and Computer Animations for College Chemistry This page from Iowa State University includes a number of simulations and animations for introductory level college chemistry courses. Topics include electrochemistry, redox titration, thermochemistry, general equilibria, acid-base equilibria, stoichiometry, kinetics, predicting the products of chemical reactions in organic chemistry and understanding fundamental concepts of periodic trends and chemical bonding. After reading instructions on what computer programs will be needed, users can simulate chemical processes related to electrochemistry, acid-base chemistry & stoichiometry. Chemistry Review: Chemical Change This review provides many details about chemical reactions and the components involved in order to make one happen, such as energy, heat, density, and substance. Provided are animations that show molecular combustion that occurs with a chemical reaction. The role of atoms is also described, including Ionic bonds that occur or break in a chemical reaction. Simple experiments use substances such as vinegar, water, and baking soda to show a real chemical reaction in progress. 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1438	https://ocw.mit.edu/courses/8-321-quantum-theory-i-fall-2017/0fdad099997ad5e46aeae92c7d27cea7_MIT8_321F17_Pset1.pdf	8.321 Quantum Theory-I Fall 2017 Prob Set 1 1. (a) Show that the set of N × N matrices form a vector space of di mension N 2 . (b) Show that Tr(A†B) deﬁnes an inner product on this vector space. (c) Show that any 2 × 2 matrix A may be written as A = a0 + i( σ · ( a (1) i.e the four 2 × 2 matrices 1, ( σ provide a basis for the N = 2 case. Here ( σ are the three Pauli matrices. How can these matrices be used to form an orthonormal basis? (d) Express a0, ai(i = 1, 2, 3) in terms of Tr(A), Tr(σiA). Hence ob tain a0, ai in terms of the matrix elements of A. 2. Prove the Schwarz’s inequality \|(ψ\|φ)\|2 ≤(ψ\|ψ)(φ\|φ) and the triangle inequality \|\| (ψ + φ) \|\| ≤ \|\|ψ\|\| + \|\|φ\|\| from the deﬁnition of the inner product. Here the norm of a vector is deﬁned by \|\|ψ\|\| = (ψ\|ψ) 1 2 . 3. Sakurai Prob 1.5 4. Sakurai Prob 1.10 5. Prove that the trace of an operator Tr(A) = (φi\|A\|φi) is independent i of the particular orthonormal basis φi that is chosen for its evaluation. 6. Propose a low cost method for measuring Planck’s constant h ¯ with precision better than 10 percent. The goal of this problem is to review key experimental facts that lead to h ¯; in answering the question please outline your method and provide a rough cost estimate. 1 MIT OpenCourseWare 8.321 Quantum Theory I Fall 2017 For information about citing these materials or our Terms of Use, visit:
1439	https://math.stackexchange.com/questions/963839/finding-the-tangent-line-to-a-curve-at-a-given-point-stumped-by-simple-problem	Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Finding the tangent line to a curve at a given point? Stumped by simple problem. Ask Question Asked Modified 7 years, 9 months ago Viewed 205k times 0 $\begingroup$ Obtaining an equation for the tangent of a curve is a problem I've done many times in the past and should be fairly straightforward for simple problems like these. However, I've been graphing my answers to check if they are correct and things are just not adding up. Given I've got the equation: y = 3x2-x3 and am asked to find the tangent line at (1,2) I'm going to differentiate it to obtain the tangent line's slope at any given point. y` = 6x - 3x2 Now I'm going to substitute in the given points to obtain the B in the slope line: y = mx+b 2 = 6(1)-3(1)2+b 2 = 3+b b = -1 Therefore, the equation of the tangent at point(1,2) is: y=6x-3x2-1 But this just doesn't add up when graphed. Wouldn't the tangent line be almost completely horizontal? Such as y = 2, I don't understand what I'm doing wrong. calculus ordinary-differential-equations Share asked Oct 8, 2014 at 14:46 MicrifiedMicrified 14911 gold badge33 silver badges1515 bronze badges $\endgroup$ 2 2 $\begingroup$ The slope of the tangent line at $(1,2)$ is the derivative's value at $x=1$ (i.e, $6\cdot 1-3\cdot1^2$), not the derivative function. $\endgroup$ David Mitra – David Mitra 2014-10-08 14:48:58 +00:00 Commented Oct 8, 2014 at 14:48 $\begingroup$ Ah! Thank you. I see. $\endgroup$ Micrified – Micrified 2014-10-08 14:57:10 +00:00 Commented Oct 8, 2014 at 14:57 Add a comment \| 1 Answer 1 Reset to default 0 $\begingroup$ If we set $f(x)=3x^2-x^3$, the slope is $f'(1)=6\cdot 1-3\cdot 1^2=3$. Hence, the tangent line is $$y-2=3(x-1)\iff y=3x-1.$$ Share answered Oct 8, 2014 at 14:48 mathlovemathlove 154k1010 gold badges126126 silver badges313313 bronze badges $\endgroup$ 4 $\begingroup$ I think I shouldn't have used y = mx+b, that is useful when differentiating a squared equation not a cubed one? $\endgroup$ Micrified – Micrified 2014-10-08 14:51:02 +00:00 Commented Oct 8, 2014 at 14:51 $\begingroup$ @Owatch: Well, I added a bit. Is there anything unclear? $\endgroup$ mathlove – mathlove 2014-10-08 14:54:11 +00:00 Commented Oct 8, 2014 at 14:54 $\begingroup$ Your answer and another comment made above made it clear I was wrong in attempting to use the derivative of the function as the slope of the tangent. That was my mistake. How silly. However, while I can see how I'd get the slope and use it, and that your equation: y-2 = 3(x-1) does become y = 3x-1 (The slope). I am not sure how you got: y-2 = 3(x-1). That said, I do understand how to solve the problem by simply using the given points to obtain the slope, then substituting that in as 'm'. $\endgroup$ Micrified – Micrified 2014-10-08 15:02:33 +00:00 Commented Oct 8, 2014 at 15:02 $\begingroup$ @Owatch: In general, the line which passes through a point $(b,c)$ and whose slope is $a$ is represented as $y-c=a(x-b)$. By the way, if you want to use $y=mx+b$, we agree with $m=3$, right? So, we have $y=3x+b$. Then, this line passes through the point $(1,2)$, so you'll have $2=3\times 1+b\Rightarrow b=-1$. So, you'll get $y=3x-1$. $\endgroup$ mathlove – mathlove 2014-10-08 15:05:49 +00:00 Commented Oct 8, 2014 at 15:05 Add a comment \| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus ordinary-differential-equations See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Related 3 Finding the equation of a tangent line at a given point 2 Help with finding tangent to curve at a point 0 Finding Tangent Line and Slope of a curve Help with Implicit Differentiation: Finding an equation for a tangent to a given point on a curve 0 Calculate the rate of change of the slope of a tangent line of a graph, given the equation, value of x, and rate of change of x. 0 Finding points of a curve parallel to another line, in point-slope form? Equation of a tangent that passes through a point not on the curve 2 How to check the case of singular point with polar equation tangent lines 3 How find tangent line of the given curve at this point? 1 Finding equation of tangent line to $\sin^{-1}(x) + \sin^{-1}(y) = \frac{\pi}{6}$ at the point $(0,\frac{1}{2})$ Hot Network Questions Determine which are P-cores/E-cores (Intel CPU) Can a GeoTIFF have 2 separate NoData values? Is this commentary on the Greek of Mark 1:19-20 accurate? How big of a hole can I drill in an exterior wall's bottom plate? Lingering odor presumably from bad chicken What is this chess h4 sac known as? Data lost/Corrupted on iCloud How to locate a leak in an irrigation system? Verify a Chinese ID Number Another way to draw RegionDifference of a cylinder and Cuboid Can a Box have a lifetime less than 'static? My dissertation is wrong, but I already defended. How to remedy? 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1440	https://www.youtube.com/watch?v=AhlLLsb4QyQ	Word problem on inverse variation Joshua Helston 6250 subscribers 2 likes Description 1478 views Posted: 14 Nov 2016 Word problem on inverse variation Transcript: okay I got a great technique for this one when you're solving word problems that deal with inverse variation inverse variation in general is the one that says that y equals a constant over X but this means that x Y Must Be a constant so we're just going to look at products and see what we can get to to match up in this in problems like this so we have acceleration and mass as invert or inversely varying so that means we're going to be looking for some kind of acceleration multiplied by mass so we have a force acting on a mass of 2 kg and it's accelerating it by 12 and then we have the same Force being applied to a different object and it accelerated by six so the acceleration is six what is the mass of this new object so notice that 12 2 is 24 6 and the unknown mass of M and so divide both sides by six and it means that the mass mass of that object must have been four and 12 2 is 24 6 what is 24 that must be the number four so the fact that the product is held constant makes it simple for us to be able to figure out the missing number so why don't you try this one on your own pause the video and see if you can figure out the solution to this problem it says the volume of a gas varies inversely with the pressure it is under so we have volume varying inversely with pressure so we're going to look at the product of volume and pressure volume of 3 l happens to have a pressure of 125 kilopascals what would the pressure be the pressure is the unknown for a volume of five and there's my pressure that's unknown so 3 125 is 375 5 equal 5 P then we will divide both sides by five to get the number 75 and so what was the pressure on that second situation turns out the pressure must have been 75 kilopascals because 75 is what you have to multiply against five to give you the same constant product of 375
1441	https://stats.stackexchange.com/questions/137425/how-seriously-should-i-consider-the-effects-of-multicollinearity-in-my-regressio	correlation - How seriously should I consider the effects of multicollinearity in my regression model? - Cross Validated Join Cross Validated By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Cross Validated helpchat Cross Validated Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How seriously should I consider the effects of multicollinearity in my regression model? Ask Question Asked 10 years, 7 months ago Modified10 years, 7 months ago Viewed 5k times This question shows research effort; it is useful and clear 4 Save this question. Show activity on this post. I have a model y ~ x + z and the correlation between x and z is 0.2. This is only weakly positive. So, how seriously should I consider the effects of multicollinearity? Is it worth expressing a simpler model like y ~ x? correlation multiple-regression multicollinearity Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Improve this question Follow Follow this question to receive notifications edited Feb 12, 2015 at 15:55 gung - Reinstate Monica 150k 90 90 gold badges 417 417 silver badges 741 741 bronze badges asked Feb 12, 2015 at 15:09 rk567rk567 711 1 1 gold badge 9 9 silver badges 20 20 bronze badges 1 7 Correlation of 0.2 is really nothing to worry about.Richard Hardy –Richard Hardy 2015-02-12 15:10:17 +00:00 Commented Feb 12, 2015 at 15:10 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 15 Save this answer. Show activity on this post. As mentioned, 0.2 is nothing to worry about in your specific case. But to answer the question, "how seriously should I consider the effect of multicollinearity in my regression model?" recognize that multicollinearity does not reduce the ability of the model to be predictive (you may still do a good job predicting your response variable in the presence of multicollinearity). It does, however, affect the quality of the interpretation of the individual predictors. In other words, in the presence of multicollinearity you may not have reliable results (indicated by large standard errors) about any individual β β coefficient. This is intuitive if we keep in mind what the interpretation of a β i β i coefficient is. It is the effect of a 1-unit increase in it's corresponding independent variable, X i X i, holding all other variables constant. When X i X i is highly correlated with some of the other independent variables then it tends to change when the other variables change. In other words, the data doesnt tend to show us what happens when X i X i changes and all other variables are constant. Consequently, it becomes more difficult to say the effect of X i X i holding all other variables constant. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Feb 13, 2015 at 2:28 answered Feb 12, 2015 at 15:23 TrynnaDoStatTrynnaDoStat 8,289 3 3 gold badges 30 30 silver badges 43 43 bronze badges 8 -1, I don't really agree w/ this. I don't know how to interpret the claim "multicollinearity does not reduce the ability of the model to be predictive", nor the claim "in the presence of multicollinearity you may not have valid results about any individual β coefficient". I do think "it does... affect interpretation of the individual predictors" is wrong; the interpretation is identical. The fact that Xi is highly correlated does not mean that "it changes when the other variables change", only that higher values of Xi tend to be associated w/ higher (lower) values of Xj.gung - Reinstate Monica –gung - Reinstate Monica 2015-02-12 15:59:39 +00:00 Commented Feb 12, 2015 at 15:59 What if I said, "it does affect the quality of the interpretation of the individual coefficients". The interpretation of the β β stays the same it's just that the standard errors will go up in the presence of multicollinearity reducing the quality of interpreting the point estimate.TrynnaDoStat –TrynnaDoStat 2015-02-12 16:07:25 +00:00 Commented Feb 12, 2015 at 16:07 The SEs do increase, but what "the quality of the interpretation" is is rather opaque. There are several issues here (although if you can correct them, I am happy to remove my downvote).gung - Reinstate Monica –gung - Reinstate Monica 2015-02-12 16:12:08 +00:00 Commented Feb 12, 2015 at 16:12 Please let me know if you are satisfied. I kept the sentence "In other words, we don't observe situations in the data when X i X i changes and all other variables are constant" because the opposite of constant is change and I wanted to make this clear. Of course, not all change is a linear relationship but the OP seems to have an understanding of the precise definition of correlation.TrynnaDoStat –TrynnaDoStat 2015-02-12 16:25:02 +00:00 Commented Feb 12, 2015 at 16:25 Unless there is perfect multicollinearity, there is no reason why you cannot have a change in Xi w/ all other variables held constant. There is nothing necessary or causal about the relationship amongst the variables when they are correlated. You may be confusing multicollinearity w/, eg, an interaction term.gung - Reinstate Monica –gung - Reinstate Monica 2015-02-12 16:47:40 +00:00 Commented Feb 12, 2015 at 16:47 \|Show 3 more comments This answer is useful 8 Save this answer. Show activity on this post. If you have perfect multicollinearity (i.e., r x z=1 r x z=1), then there is no unique set of numbers (i.e., β^x β^x and β^z β^z) that will optimize the fit of the data. Moreover, the typical method software uses to fit a model will fail and you may simply get an error message in this case. People sometimes define multicollinearity as only perfect multicollinearity, and this restrictive definition lies behind claims that multicollinearity renders the preferred model invalid. On the other hand, most people do not restrict multicollinearity to perfect linear dependence. If the multicollinearity isn't perfect (which in your case simply means the correlation isn't perfect, since you have only two variables), then there will be a unique solution and the model can be fit. However, if you were to draw a new sample and re-run your study identically, you would find that the beta estimates vary more widely than you might otherwise suspect. The sampling distributions of your betas remain unbiased (meaning that they are centered on the true value), but the width (i.e., the standard error) expands. In simple cases like yours, it is easy to calculate how much the variance expands. Specifically, you can calculate the variance inflation factor: V I F=1 1−r 2 x z V I F=1 1−r x z 2 In addition, if x x and z z are correlated, the bivariate sampling distribution of β^x β^x and β^z β^z will have non-zero covariance. Furthermore, the more strongly x x and z z are correlated, the larger the covariance. This point is less well-known, however, and typically is not what people are primarily concerned about. In your specific case, the variances of the sampling deviations of your betas (both of them) will be 1.04 1.04 times as wide (i.e., the standard errors will be 1.02 1.02 times as wide) as they would have been if r x z=0.0¯r x z=0.0¯ and everything else about your study (e.g., N N) were the same. It strikes me as very unlikely that you would need to worry about this (although strictly speaking, I don't know). So in answer to your explicit question, I would not worry about multicollinearity in your case, and would not try a simpler model. This is just about all there is to say about multicollinearity—the alpha and the psy, as it were; it is what should be said about multicollinearity in the abstract / without knowing a lot of details about a particular study. For the sake of clarity, let me explicitly state some additional things, lest they be left ambiguous by dint of omission. Because of the existence of multicollinearity, your model will / will not have sufficient predictive utility. This cannot be said without more information. Your coefficients will not be sufficiently reliable. This cannot be said without more information. Your variables will be non-significant. This cannot be said without more information. The interpretation of the coefficients that would normally hold in a multiple regression model does not apply, or is invalid or problematic, when there is multicollinearity. The interpretation of a coefficient in a multiple regression model is the change in Y Y associated with a one-unit change in X i X i holding all else equal (there is more information here, if needed). This can remain true even when there are substantial correlations amongst the X X variables. Consider that, in the following plot, x 1 x 1 and x 2 x 2 are correlated r x 1 x 2=.99 r x 1 x 2=.99 and yet there are many cases where x 1 x 1 differs in value but x 2 x 2 is constant: To understand this more clearly, it may help to contrast it with some well-known cases where the standard interpretation above is invalid or problematic. Consider the coefficient for X i X i when there is an interaction term including X i X i, say X i X j X i X j. In that case, there is no such thing as "a one-unit change in X i X i holding all else equal", as it cannot be guaranteed that the interaction term can be held equal. Hence, the coefficient for X i X i turns out to be the change in Y Y associated with a one-unit change in X i X i when X j=0 X j=0 (and all other variables are held equal). Similarly, if there is an X 2 i X i 2 term in the model, it cannot be held equal when X i X i changes. Although I have argued that it is best to interpret all polynomial terms involving the variable together, the coefficient on X i X iis related to the location in X i X i of the global extremum of the parabola. The point of these contrasting examples is that multicollinearity is not necessarily like this. As the figure above shows, there is nothing about the existence of multicollinearity that necessitates it being impossible for X i X i to vary while X j X j is constant, and so the standard interpretation of the coefficients remains generally valid / tenable. (Note, however, that interaction terms and squared terms typically induce multicollinearity, so there can be situations with multicollinearity where the normal interpretation does not apply.) To further clarify these points, I simulated the model Y=2+.03 X 1−.03 X 2+ε Y=2+.03 X 1−.03 X 2+ε, with ε∼N(0,1)ε∼N(0,1) using N=225 N=225 vs. 22500 22500, and using the above pattern of x 1 x 1 and x 2 x 2 vs. having them sampled on a uniform from [0,50][0,50]. Plots of the bivariate sampling distributions of the estimated coefficients are below: Despite a V I F V I F of 85.3 85.3, with high N N, the power of both β^1 β^1 and β^2 β^2 approach 1.0 1.0. Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Apr 13, 2017 at 12:44 CommunityBot 1 answered Feb 12, 2015 at 15:53 gung - Reinstate Monicagung - Reinstate Monica 150k 90 90 gold badges 417 417 silver badges 741 741 bronze badges 5 1 With perfect multicollinearity the model can be fit--but the fit is not unique.whuber –whuber♦ 2015-02-12 17:29:37 +00:00 Commented Feb 12, 2015 at 17:29 @whuber, I suppose I'm conflating two claims there. 1) the design matrix is not invertable, so software will return an error instead of a fit (but there are ways around this), & 2) the model is unidentifiable / there is no unique solution for the parameter estimates. I can clarify this, if you think it best. I left this simpler for this OP.gung - Reinstate Monica –gung - Reinstate Monica 2015-02-12 17:35:36 +00:00 Commented Feb 12, 2015 at 17:35 I figured that's what you intended, Gung. Note, though, that most software will not return an error: it will attempt to resolve the issue by eliminating variables, and it usually succeeds. Identifiability is worth discussing, especially if a distinction is made based on the purpose of the regression: it's not as much of an issue when the regression is performed solely for prediction. I think @Trynna did a good job covering these issues in another answer in this thread.whuber –whuber♦ 2015-02-12 17:40:00 +00:00 Commented Feb 12, 2015 at 17:40 That's interesting, @whuber, I have several problems w/ Trynna's post, as I tried to point out in comments & it currently being discussed in chat.gung - Reinstate Monica –gung - Reinstate Monica 2015-02-12 17:48:07 +00:00 Commented Feb 12, 2015 at 17:48 2 I took a look and--although I upvoted that post in the belief it offered clear, concise, helpful advice--I do agree with the points you are making in chat. The "tends" language and the apparently causal analysis in terms of "if you change this, that will change" could deceive naive readers. I am aware of two effective explanations that avoid these solecisms: the geometric one (of OLS as an orthogonal projection) and the one offered by Tukey & Mosteller, who use sequential "matching" to describe multiple regression and explain how near-collinearity inflates the SEs of the coefficient estimates.whuber –whuber♦ 2015-02-12 17:55:39 +00:00 Commented Feb 12, 2015 at 17:55 Add a comment\| Your Answer Thanks for contributing an answer to Cross Validated! Please be sure to answer the question. Provide details and share your research! But avoid … Asking for help, clarification, or responding to other answers. Making statements based on opinion; back them up with references or personal experience. Use MathJax to format equations. MathJax reference. To learn more, see our tips on writing great answers. 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1442	https://www.education.com/common-core/CCSS.MATH.CONTENT.6.RP.A.3.C/	Choose an Account to Log In Log in with different email For more assistance contact customer service SKIP TO CONTENT Upgrade Now Log In Sign Up Notifications Delete all Common Core Sixth Grade Math CCSS.MATH.CONTENT.6.RP.A.3.C 6.RP.A.3.C Worksheets, Workbooks, Lesson Plans, and Games CCSS.MATH.CONTENT.6.RP.A.3.C : "Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent." These worksheets can help students practice this Common Core State Standards skill. Worksheets Calculating Sales Tax Worksheet Calculating Sales Tax Get a thorough explanation of sales tax, and practice with some fun kids' products. Worksheet Let's Go Shopping! Interactive Worksheet Let's Go Shopping! Practice percentages by finding the discount prices on this quality merchandise. Interactive Worksheet Calculating Percentage Part 1 Interactive Worksheet Calculating Percentage Part 1 Give your students practice calculating percents with part one of this sixth-grade math worksheet! 6th Grade Math Interactive Worksheet Finding Percentages #1 Interactive Worksheet Finding Percentages #1 Use this sixth-grade math worksheet to give students practice finding percentages using two numbers: the part and whole. 6th Grade Math Interactive Worksheet Introduction to Percents Worksheet Introduction to Percents What is a percent? How do you write a percent? How are percents related to fractions and decimals? Answer these questions and more with this worksheet! 6th Grade Math Worksheet Calculating Percentage Part 2 Interactive Worksheet Calculating Percentage Part 2 Students gain more practice calculating percentages with the second installment of this sixth-grade math worksheet. 6th Grade Math Interactive Worksheet See more worksheets Lesson Plans No plans found for this common core code. Workbooks No workbooks found for this common core code. Games No games found for this common core code. Exercises No exercises found for this common core code. We enable strictly necessary cookies to give you the best possible experience on Education.com.Cookie Policy Add to collection Create new collection Create new collection New Collection 0 New Collection 0 items Sign up to start collecting! Bookmark this to easily find it later. Then send your curated collection to your children, or put together your own custom lesson plan. Sign up Log in
1443	https://www.youtube.com/watch?v=7vomY6g-f3Q	Square Roots, Cube Roots, and Other Roots Professor Dave Explains 3980000 subscribers 3347 likes Description 156311 views Posted: 11 Sep 2017 What are these funny little symbols? They're roots! Taking the square root of something give you the number that, when squared, gives the original number. Cube roots are the same, but they are the opposite of cubing. I'm not explaining it very well right now, but that's what this clip is for, so watch it! Watch the whole Mathematics playlist: Classical Physics Tutorials: Modern Physics Tutorials: General Chemistry Tutorials: Organic Chemistry Tutorials: Biochemistry Tutorials: Biology Tutorials: EMAIL► ProfessorDaveExplains@gmail.com PATREON► Check out "Is This Wi-Fi Organic?", my book on disarming pseudoscience! Amazon: Bookshop: Barnes and Noble: Book Depository: 232 comments Transcript: Intro It’s Professor Dave, let’s learn about square roots. We’ve learned how to solve simple algebraic equations, we just look at what is happening to the variable we want to solve for, and we do the inverse of that operation to both sides. If we have X plus two, we subtract two. If we have three X, we divide by three. What if we have X squared equals nine? What is the inverse of squaring something? That would be taking the square root. Taking Square Roots When we take the square root of X squared, we get X, because squaring X is what would give us X squared. On the other side, the square root of nine is equal to three, because three squared, or three times three, gives us nine. So taking the square root of something is like asking what number times itself will give us the number inside the square root? We must point out something very important here. Negative three is also a valid answer for the square root of nine, because negative three times negative three also equals nine. Remember that a negative number times a negative number is positive, so both of these values work for the square root of nine. That means that the square root of nine is equal to positive three or negative three, which we sometimes label this way, with both the plus and minus sign next to the three, and we refer to this as plus or minus three. Any square root will do this, so there are always two answers when we take a square root. Since negative three squared equals nine, we should also note that the square root of negative nine is not negative three. In fact, there is no real solution to this term, so we can’t take the square root of negative numbers the way we take the square root of positive numbers. If we were to try to do so, we would get an imaginary answer, which is something we will investigate later in the series. For now, let’s just say that only positive numbers can have their square root taken. It’s very easy to get the square root of perfect squares. Taking the Square Root of Perfect Squares The square root of four is plus or minus two. Continuing down the list, nine, sixteen, twenty-five, thirty-six, all give very tidy answers when we take the square root. We should make some effort to memorize the first ten or even twenty perfect squares, because they will come in handy later. But the numbers in between these perfect squares also give a square root, it just won’t be an integer. Some of these roots give a decimal value that goes on forever, meaning they are irrational. We will talk more about this concept later, but for now let’s acknowledge that this is an interesting result, because it means that there are real numbers that can’t be expressed as fractions, no matter how big we make the denominator. We should also note that radical expressions, as we often call them, can sometimes be simplified. Simplifying Radical Expressions Take something like root eight. Eight is not a perfect square, so we can’t get a neat and tidy answer when we take its square root. But, we can change eight into four times two. The reason we would want to do this is that root four times two is the same as root four times root two, and root four does work out nice and tidy, since it’s equal to plus or minus two. So root eight can be simplified to become plus or minus two root two. We could say the same for root twenty-seven. Twenty-seven is nine times three, and root nine is plus or minus three, so we get plus or minus three root three. Just remember to look for any factors that are perfect squares in order to get these expressions to their simplest form. Are there other roots we can take besides the square root? Other Types of Roots Certainly, we can take the cube root. This is the opposite of cubing something, or raising it to the third power. If X cubed is twenty-seven, what is X? We will now have to take the cube root of both sides. This is like saying, what number raised to the third power will give the number inside this root. In this case, the answer is three, because three times three times three equals twenty-seven. Also notice that in this case, this is the only answer. Negative three doesn’t work, because negative three times negative three times negative three equals negative twenty-seven. So negative three is the cube root of negative twenty-seven. In this way, cube roots differ from square roots, first because you can take the cube root of a negative number, and second because any number will only have one unique cube root, rather than two. If we take the fourth root of something, we see that the rules are similar to square roots in that we can’t take the fourth root of a negative number, because multiplying a negative value four times gives us a positive value. Taking the fifth root sees us going back to rules like for the cube root. We alternate back and forth depending on whether the root is odd or even. Odd roots have one real solution and even roots have two real solutions. For now, we just have to accept this, but later we will get a more satisfying explanation. Roots as Fractional Exponents We now come to one of the most important things to understand about roots, we have to realize that they represent fractional exponents. Taking the square root is the same as raising something to the one half power. This is because, if the square root of something times itself should equal the original value, like how root X times root X equals X, then it must be the case that we can express this as X to the one half times X to the one half equals X to the one, since one half plus one half equals one, according to the rules of exponents. This is a notational choice, and it works for any exponent. Taking the cube root is the same as raising something to the one third power. We can also combine two different operations into one fraction. Raising something to the two thirds power means squaring it and taking the cube root, so eight to the two thirds power equals four, which we can verify by either squaring first and then taking the cube root, or taking the cube root and then squaring. In this way, X to the A over B power simply means to multiply X by itself A times and then take the B root. When we combine this with what we know about operations with terms raised to exponents, we can do seemingly tricky things. Take this expression here, the square root of X times the cube root of X divided by the fourth root of X. First let’s change this to X to the one half times X to the one third divided by X to the one fourth. Since X to the A times X to the B equals X to the A plus B, we simply add these exponents together on the top. To do so, we need a common denominator for these fractions, so we change them to three sixths and two sixths, and add them together to get five sixths. Then if dividing by the X to the one fourth we subtract, which means we need the lowest common denominator, which will be twelve. So ten twelfths minus three twelfths equals seven twelfths. And there you have the simplified version of the original expression, X to the seven-twelfths power. Let’s check comprehension.
1444	https://www.sec.gov/Archives/edgar/data/1288776/000119312512025336/d260164d10k.htm	10-K 1 d260164d10k.htm FORM 10-K Form 10-K Table of Contents UNITED STATES SECURITIES AND EXCHANGE COMMISSION Washington, D.C. 20549 FORM 10-K (Mark One) x ANNUAL REPORT PURSUANT TO SECTION 13 OR 15(d) OF THE SECURITIES EXCHANGE ACT OF 1934 For the fiscal year ended December 31, 2011 OR ¨TRANSITION REPORT PURSUANT TO SECTION 13 OR 15(d) OF THE SECURITIES EXCHANGE ACT OF 1934 Commission file number: 000-50726 Google Inc. (Exact name of registrant as specified in its charter) Delaware 77-0493581 (State or other jurisdiction of incorporation or organization)(I.R.S. 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DOCUMENTS INCORPORATED BY REFERENCE Portions of the registrant’s Proxy Statement for the 2012 Annual Meeting of Stockholders are incorporated herein by reference in Part III of this Annual Report on Form 10-K to the extent stated herein. Such proxy statement will be filed with the Securities and Exchange Commission within 120 days of the registrant’s fiscal year ended December 31, 2011. Table of Contents Google Inc. 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These statements include, among other things, statements regarding: •the growth of our business and revenue; •seasonal fluctuations in internet usage and traditional retail seasonality, which are likely to cause fluctuations in our quarterly results; •our plans to continue to invest in systems, facilities, and infrastructure, increase our hiring, provide competitive compensation programs, and continue our current pace of acquisitions; •the potential for declines in our revenue growth rate; •our expectation that growth in advertising revenues from our websites will continue to exceed that from our Google Network Members’ websites, which will have a positive impact on our operating margins; •our expectation that we will continue to pay most of the Google AdSense fees we receive from advertisers to our Google Network Members; •our expectation that we will continue to take steps to improve the relevance of the ads we deliver and to reduce the number of accidental clicks; •fluctuations in aggregate paid clicks and average cost-per-click; •our belief that our foreign exchange risk management program will not fully offset the exposure to fluctuations in foreign currency exchange rates; •the increase of costs related to hedging activities under our foreign exchange risk management program; •our expectation that our cost of revenues, research and development expenses, sales and marketing expenses, and general and administrative expenses will increase in dollars and may increase as a percentage of revenues; •our potential exposure in connection with pending investigations and proceedings; •our expectations about the timing of the consummation of our proposed acquisition of Motorola Mobility Holdings, Inc. (Motorola); •our expectation that our traffic acquisition costs will fluctuate in the future; •continued investments in international markets; •our future compensation expenses; •fluctuations in our effective tax rate; •the sufficiency of our sources of funding; •our payment terms to certain advertisers, which may increase our working capital requirements; and •fluctuations in our capital expenditures; as well as other statements regarding our future operations, financial condition and prospects, and business strategies. Forward-looking statements may appear throughout this report, including without limitation, the following sections: Item 1 “Business,” Item 1A “Risk Factors,” and Item 7 “Management’s Discussion and Analysis of Financial Condition and Results of Operations.” Forward-looking statements generally can be identified by words such as “anticipates,” “believes,” “estimates,” “expects,” “intends,” “plans,” “predicts,” “projects,” “will be,” “will continue,” “will likely result,” and similar expressions. These forward-looking statements are based on current expectations and assumptions that are subject to risks and uncertainties, which could cause our actual results to differ materially from those reflected in the forward-looking statements. Factors that could cause or contribute to 1 Table of Contents such differences include, but are not limited to, those discussed in this Annual Report on Form 10-K, and in particular, the risks discussed under the caption “Risk Factors” in Item 1A and those discussed in other documents we file with the Securities and Exchange Commission (SEC). We undertake no obligation to revise or publicly release the results of any revision to these forward-looking statements, except as required by law. Given these risks and uncertainties, readers are cautioned not to place undue reliance on such forward-looking statements. As used herein, “Google,” “we,” “our,” and similar terms include Google Inc. and its subsidiaries, unless the context indicates otherwise. “Google” and other trademarks of ours appearing in this report are our property. This report contains additional trade names and trademarks of other companies. We do not intend our use or display of other companies’ trade names or trademarks to imply an endorsement or sponsorship of us by such companies, or any relationship with any of these companies. 2 Table of Contents ITEM 1.BUSINESS Overview Google is a global technology leader focused on improving the ways people connect with information. We aspire to build products that improve the lives of billions of people globally. Our mission is to organize the world’s information and make it universally accessible and useful. Our innovations in web search and advertising have made our website a top internet property and our brand one of the most recognized in the world. We generate revenue primarily by delivering relevant, cost-effective online advertising. Businesses use our AdWords program to promote their products and services with targeted advertising. In addition, the third parties that comprise the Google Network use our AdSense program to deliver relevant ads that generate revenue and enhance the user experience. We were incorporated in California in September 1998 and reincorporated in Delaware in August 2003. Our headquarters are located at 1600 Amphitheatre Parkway, Mountain View, California 94043, and our telephone number is (650)253-0000. We completed our initial public offering in August 2004 and our Class A common stock is listed on the Nasdaq Global Select Market under the symbol “GOOG.” Corporate Highlights On April 4, 2011, Larry Page, our Co-Founder, assumed day-to-day operations as our Chief Executive Officer. Since then, Larry’s primary focus has been on “increasing Google’s velocity and execution”—making things much simpler for our users and improving the overall Google experience while continually creating new products with the potential to improve the lives of billions of people. •Android—The growth of our Android operating system in 2011 has been impressive. As of January 2012, over 250 million Android devices have been activated globally. •Improved Ad Formats—We made a number of ad improvements in 2011. For instance, with visual ads, not only can users find theater times for a new movie, users can watch the trailer directly in the ad. Location-aware search ads can help users find what they are looking for more easily by putting thousands of local businesses on the map—literally. With the +1 button, people can find businesses recommended by their friends. After all, ads are just more answers to users’ queries. •Google+—In June 2011, we launched Google+, a new way to share online just like users do in the real world, sharing different things with different people. Google+ has added new users every week since its launch. As of January 2012, over 90 million people have joined Google+. •Product Prioritization—In an effort to prioritize our product efforts, we decided to shut down a number of products in 2011, including Google Buzz, Google Desktop, and Google Labs. We learned a lot from these discontinued products and are putting that learning to work every day in new products such as Google+. •New and Improved Google Experience—Constant revision and improvement is part of our overarching philosophy. In 2011, we began work on improving the Google experience guided by three key design principles: •Focus—With the new design changes, we have brought forward the things that matter to users and moved all the other clutter out of their way so that users can better focus on what they need at the moment. •Elasticity—The new design allows users to seamlessly transition from one device to another from the various mobile devices, tablets, high-resolution monitors, and TVs to access the web and have a consistent visual experience. •Effortlessness—Our design philosophy is to combine power with simplicity. We want to keep our look simple and clean, but behind the seemingly simple design, use technologies like HTML5, WebGL, and the latest, fastest browsers to make sure users have all the power of the web behind them. 3 Table of Contents On August 15, 2011, we entered into an Agreement and Plan of Merger (Merger Agreement) with Motorola, a provider of innovative technologies, products and services that enable a range of mobile and wireline digital communication, information and entertainment experiences, under which we will acquire Motorola for $40 per share in cash, or a total of approximately $12.5 billion. The completion of this transaction is subject to customary closing conditions, including the receipt of certain regulatory approvals. Our business is primarily focused around the following key areas: search, advertising, operating systems and platforms, and enterprise. Search We maintain a vast index of websites and other online content, and make it available through our search engine to anyone with an internet connection. Our search technologies sort through an ever-growing amount of information to deliver relevant and useful search results in response to user queries. We integrate innovative features into our search service and offer specialized search services to help users tailor their search. In addition, we are constantly improving and adding to our products and services, to provide users with more relevant results so that users find what they are looking for faster. In January 2012, we launched Search plus Your World. Now, when a user performs a signed-in search on Google, the user’s results page may include Google+ content from people that the user is close to (or might be interested in following). Relevant Google+ profiles and Google+ pages related to a specific topic or area of interest may also appear on a user’s results page. Advertising Google Search. The goal of AdWords, our primary auction-based advertising program, is to deliver ads that are so useful and relevant to search queries or web content that they are a form of information in their own right. With AdWords, advertisers create simple text-based ads that then appear beside related search results or web content on our websites and on thousands of partner websites in our Google Network, which is the network of third parties that use our advertising programs to deliver relevant ads with their search results and content. Most of our AdWords customers pay us on a cost-per-click basis, which means that an advertiser pays us only when a user clicks on one of its ads. We also offer AdWords on a cost-per-impression basis that enables advertisers to pay us based on the number of times their ads appear on our websites and our Google Network Members’ websites as specified by the advertiser. Our AdSense program enables websites that are part of the Google Network to deliver ads from our AdWords advertisers that are relevant to the search results or content on their websites. We share the majority of the revenues generated from these ads with the Google Network Members that display the ads. The AdSense program enables advertisers to extend the reach of their ad campaigns, improves our partners’ ability to generate revenue from their content, and delivers relevant ads for their users. Google Display. Display advertising comprises the videos, text, images, and other interactive ads that run across the web on computers and mobile devices, including smart phones and handheld computers such as netbooks and tablets. The Google Display Network provides advertisers services related to the delivery of display advertising across publishers participating in our AdSense program, publishers participating in the DoubleClick Ad Exchange, and Google-owned sites such as YouTube and Google Finance. Through our DoubleClick advertising technology, we provide to publishers, agencies, and advertisers the ad serving technology, which is the infrastructure that enables billions of ads to be served each day across the web.Our DoubleClick Ad Exchange creates a real-time auction marketplace for the trading of display ad space. We aim to simplify display advertising so it is easier for advertisers and publishers to manage campaigns across different formats, on different websites, and for different devices. 4 Table of Contents In addition, YouTube provides a range of video, interactive, and other ad formats for advertisers to reach their intended audience. YouTube’s video advertising solutions give advertisers a way to promote their content to the YouTube community, as well as to associate with content being watched by their target audience. YouTube also offers analytic tools to help advertisers understand their audience and derive general business intelligence. In the past year, YouTube has experienced strong growth in mobile viewers and has established key partnerships with content companies to help monetize mobile video. Google Mobile. Mobile advertising is still in relative infancy, though the mobile device is quickly becoming the world’s newest gateway to information. Google is focused on developing easy-to-use ad products to help advertisers extend their reach, help create revenue opportunities for our publisher partners, and deliver relevant and useful ads to users on the go. Google Mobile extends our products and services by providing mobile-specific features to mobile device users. Our mobile-specific search technologies include search by voice, search by sight, and search by location. Google Mobile also optimizes a large number of Google’s applications for mobile devices in both browser and downloadable form. In addition, we offer advertisers the ability to run search ad campaigns on mobile devices with popular mobile-specific ad formats, such as click-to-call ads in which advertisers can include a phone number within ad text. In 2010, we acquired AdMob, Inc. (AdMob), which offers effective ad units and solutions for application developers and advertisers. We continue to invest in improving users’ access to Google services through their mobile devices. Google Local. Google is committed to providing users with relevant local information. We’ve organized information around more than 50 million places globally from various sources across the web. Users can find addresses, phone numbers, hours of operation, directions and more for millions of local queries like shops, restaurants, parks and landmarks right on Google.com, on Google Maps and on Google Maps for mobile. They can also discover more places that are right for them by rating the places they’ve been, and getting customized recommendations based on their tastes and those of their friends directly within Google Maps. Our products and services also help local business owners manage their online presence and connect with potential customers. Millions of business owners have verified their free business listings via Google Places to ensure that users have up-to-date information about their establishments, and to contribute additional details such as photos and products/services offered. Google Offers brings people daily deals from local and national businesses, redeemable for discounted goods or services. From restaurants to spa treatments to outdoor adventures, Google has deals from the best businesses a city has to offer as well as popular national brands. Operating Systems and Platforms Android. Working closely with the Open Handset Alliance, a business alliance of more than 75 technology and mobile companies, we developed Android, a free, fully open source mobile software platform that any developer can use to create applications for mobile devices and any handset manufacturer can install on a device. We believe Android will drive greater innovation and choice in the mobile device ecosystem, and provide consumers with a more powerful mobile experience. Google Chrome OS and Google Chrome. Google Chrome OS is an open source operating system with the Google Chrome web browser as its foundation. Both the Google Chrome OS and the Google Chrome browser are built around the core tenets of speed, simplicity, and security. Designed for people who spend most of their time on the web, the Google Chrome OS is a new approach to operating systems. We are working with several original equipment manufacturers to bring computers running Google Chrome OS to users and businesses. The Chrome browser runs on Windows, Mac, and Linux computers. Google+. In June 2011, we launched Google+, a new way to share online just like users do in the real world, sharing different things with different people. Google+ has added new users every week since its launch. As of January 2012, over 90 million people have joined Google+. 5 Table of Contents Google TV. Google TV is a platform that gives consumers the power to experience television and the internet on a single screen, with the ability to search and find the content they want to watch. The Google TV platform is based on the Android operating system and runs the Google Chrome browser. Google Books. The Google Books platform (including reading applications, an electronic bookstore (eBookstore), book search, and personal library management) is designed to help people discover, search, and consume content from printed books online. Through the Google eBookstore, we make available for sale popular books in electronic book format to complement our large collection of free public domain books. Enterprise Google’s enterprise products provide familiar, easy-to-use Google technology for business settings. Through Google Apps, which includes Gmail, Google Docs, Google Calendar, and Google Sites, among other features, we provide hosted, web-based applications that people can use on any device with a browser and an internet connection. In addition, we provide our search technology for use within enterprises through the Google Search Appliance (real-time search of business applications, intranet applications, and public websites), on their public-facing sites with Google Site Search (custom search engine), and Google Commerce Search (for online retail enterprises). We also provide versions of our Google Maps Application Programming Interface (API) for businesses (including fully interactive Google Maps for public and internal websites), as well as Google Earth Enterprise (a behind-the-company-firewall software solution for imagery and data visualization). Our enterprise solutions have been adopted by a variety of businesses, governments, schools, and non-profit organizations. Google Apps is the first cloud computing suite of message and collaboration tools to receive U.S. government security certification. Research We continue to develop new products and services and to enhance our existing ones through research and product development and the licensing and acquisition of third-party businesses and technology. Our product development philosophy is to launch innovative products early and often, and then iterate rapidly to make those products even better. We often post early stage products at test locations online or directly on Google.com. We then use data and user feedback to decide if and how to invest further in those products. Our research and development expenses were $2.8 billion, $3.8 billion, and $5.2 billion in 2009, 2010, and 2011, respectively, which included stock-based compensation expense of $725 million, $861 million, and $1.1 billion, respectively. We expect to continue to invest in building the employee and systems infrastructures needed to support the development of new products and services and to improve existing ones. Intellectual Property We rely on a combination of intellectual property laws, as well as confidentiality procedures and contractual provisions, to protect our proprietary technology and our brand. We have registered, and applied for the registration of, U.S. and international trademarks, service marks, domain names, and copyrights. Additionally, we have filed U.S. and international patent applications covering certain of our proprietary technology. Over time, we have assembled a portfolio of patents, trademarks, service marks, copyrights, domain names, and trade secrets covering our products and services. Our proprietary technology is not dependent on any single patent or copyright or groups of related patents or copyrights. We believe the duration of our patents is adequate relative to the expected lives of our products. Although we rigorously protect our proprietary technology, any significant impairment of, or third-party claim against, our intellectual property rights could harm our business or our ability to compete. Sales and Support We continue to develop and grow our sales and support infrastructure. We have over 85 offices in over 40 countries, the large majority of which include sales people.Our global sales and support infrastructure has 6 Table of Contents specialized teams across vertical markets.We bring businesses into our advertising network through direct, remote, and online sales channels,using technology and automation wherever possible to improve our customers’ experience and to grow our business cost-effectively. Our direct advertising and sales teams focus on building relationships with the largest advertisers and leading internet companies. We have built a multi-product sales force, with teams selling campaigns that include search, display (including DoubleClick and YouTube), and mobile advertising. We provide customer service to our advertiser base through our global support organization. Our global support organization concentrates on helping our advertisers and Google Network Members get the most out of their relationship with us. No individual customer or groups of affiliated customers represented more than 10% of our revenues in 2009, 2010, or 2011. Government Contracts No material portion of our business is subject to renegotiation of profits or termination of contracts or subcontracts at the election of the U.S. government. Marketing Google’s global brand is well known. We believe that building a trusted, highly recognized brand begins with providing high-quality products and services that make a notable difference in people’s lives. Marketing is responsible for generating advertiser revenue through marketing campaigns to small businesses, as well as providing thought leadership to chief marketing officers through industry insight, research, and analysis. Our marketing, promotional, and public relations activities are designed to promote Google’s brand image and differentiate it from competitors. Competition Our business is characterized by rapid change and converging, as well as new and disruptive, technologies. We face formidable competition in every aspect of our business, particularly from companies that seek to connect people with information on the web and provide them with relevant advertising. We face competition from: •General purpose search engines, such as Yahoo and Microsoft’s Bing. •Vertical search engines and e-commerce websites, such as Kayak (travel queries), Monster.com (job queries), WebMD (for health queries), and Amazon.com and eBay (e-commerce). Some users will navigate directly to such websites rather than go through Google. •Social networks, such as Facebook and Twitter. Some users are relying more on social networks for product or service referrals, rather than seeking information through general purpose search engines. •Other forms of advertising, such as television, radio, newspapers, magazines, billboards, and yellow pages, for ad dollars. Our advertisers typically advertise in multiple media, both online and offline. •Mobile applications on iPhone and Android devices, which allow users to access information directly from a publisher without using search engines. •Providers of online products and services. A number of our online products and services, including Gmail, YouTube, and Google Docs, compete directly with new and established companies, which offer communication, information, and entertainment services integrated into their products or media properties. 7 Table of Contents We compete to attract and retain users, for whom other products and services are literally one click away, primarily on the basis of the relevance and usefulness of our search results and the features, availability, and ease of use of our products and services. We also compete to attract and retain content providers (Google Network Members, as well as other content providers for whom we distribute or license content), primarily based on the size and quality of our advertiser base, our ability to help these partners generate revenues from advertising, and the terms of our agreements with them. Government Regulation We are subject to numerous domestic and foreign laws and regulations covering a wide variety of subject matter. New laws and regulations (or new interpretations of existing laws and regulations) may also impact our business. The costs of compliance with these laws and regulations are high and are likely to increase in the future and any failure on our part to comply with these laws may subject us to significant liabilities and other penalties. Culture and Employees We take great pride in our culture. We embrace collaboration and creativity,and encourage the iteration of ideas to address complex technical challenges. Transparency and open dialog are central to us, and we like to ensure that company news reaches our employees first through internal channels. Despite our rapid growth, we still cherish our roots as a startup and give employees the freedom to act on their ideas regardless of role or function within the company. We strive to hire the best employees, with backgrounds and perspectives as diverse as our global users. We provide an environment where these talented people can have fulfilling careers working on some of the biggest challenges in technology, and have a huge, positive impact on the world. At December 31, 2011, we had 32,467 full-time employees, consisting of 11,665 in research and development, 11,933 in sales and marketing, 4,651 in general and administrative, and 4,218 in operations. All of Google’s full-time employees are also equityholders, with significant collective employee ownership. Although we have works councils and statutory employee representation obligations in certain countries, our employees are not represented by a labor union and we consider our employee relations to be good. Competition for qualified personnel in our industry is intense, particularly for software engineers, computer scientists, and other technical staff. Global Operations and Geographic Data We provide our products and services in more than 100 languages and in more than 50 countries, regions, and territories. On www.google.com or one of our other Google domains, users can find information in many different languages and in many different formats. The United States accounted for approximately 46% of our revenues in 2011. Information regarding financial data by geographic areas is set forth in Item 7 and Item 8 of this Annual Report on Form 10-K. See Note 16 of Notes to Consolidated Financial Statements under Item 8. Seasonality Our business is affected by both seasonal fluctuations in internet usage and traditional retail seasonality. Internet usage generally slows during the summer months, and commercial queries typically increase significantly in the fourth quarter of each year. These seasonal trends have caused and will likely continue to cause, fluctuations in our quarterly results, including fluctuations in sequential revenue growth rates. Available Information Our website is located at www.google.com, and our investor relations website is located at The following filings are available through our investor relations website after we file 8 Table of Contents them with the SEC: Annual Reports on Form 10-K, Quarterly Reports on Form 10-Q, and our Proxy Statements for our annual meetings of stockholders, for the last three years. These filings are also available for download free of charge on our investor relations website. We also provide a link to the section of the SEC’s website at www.sec.gov that has all of our public filings, including Annual Reports on Form 10-K, Quarterly Reports on Form 10-Q, Current Reports on Form 8-K, all amendments to those reports, our Proxy Statements, and other ownership related filings. Further, a copy of this Annual Report on Form 10-K is located at the SEC’s Public Reference Room at 100 F Street, NE, Washington, D.C. 20549. Information on the operation of the Public Reference Room can be obtained by calling the SEC at 1-800-SEC-0330. We webcast our earnings calls and certain events we participate in or host with members of the investment community on our investor relations website. Additionally, we provide notifications of news or announcements regarding our financial performance, including SEC filings, investor events, press and earnings releases, and blogs as part of our investor relations website. Investors and others can receive notifications of new information posted on our investor relations website in real time by signing up for email alerts and RSS feeds. Further corporate governance information, including our certificate of incorporation, bylaws, governance guidelines, board committee charters, and code of conduct, is also available on our investor relations website under the heading “Corporate Governance.” The contents of our websites are not incorporated by reference into this Annual Report on Form 10-K or in any other report or document we file with the SEC, and any references to our websites are intended to be inactive textual references only. ITEM 1A.RISK FACTORS Our operations and financial results are subject to various risks and uncertainties, including those described below, which could adversely affect our business, financial condition, results of operations, cash flows, and the trading price of our Class A common stock. Risks Related to Our Business and Industry We face intense competition. If we do not continue to innovate and provide products and services that are useful to users, we may not remain competitive, and our revenues and operating results could be adversely affected. Our business is rapidly evolving and intensely competitive, and is subject to changing technology, shifting user needs, and frequent introductions of new products and services. We have many competitors in different industries, including general purpose search engines, vertical search engines and e-commerce sites, social networking sites, traditional media companies, and providers of online products and services. Our current and potential competitors range from large and established companies to emerging start-ups. Established companies have longer operating histories and more established relationships with customers and users, and they can use their experience and resources in ways that could affect our competitive position, including by making acquisitions, investing aggressively in research and development, aggressively initiating intellectual property claims (whether or not meritorious) and competing aggressively for advertisers and websites. Emerging start-ups may be able to innovate and provide products and services faster than we can. Our success depends on providing products and services that make using the internet a more useful and enjoyable experience for our users. Our competitors are constantly developing innovations in web search, online advertising, and web-based products and services. As a result, we must continue to invest significant resources in research and development, including through acquisitions, in order to enhance our web search technology and our existing products and services, and introduce new products and services that people can easily and effectively use. If we are unable to provide quality products and services, then our users may become dissatisfied and move to a competitor’s products and services. In addition, these new products and services may present new and difficult technology challenges, and we may be subject to claims if users of these offerings experience service disruptions or failures or other quality issues. Our operating results would also suffer if our innovations are not responsive to the needs of our users, advertisers, and Google Network Members, are not appropriately timed with market 9 Table of Contents opportunities, or are not effectively brought to market. As search technology continues to develop, our competitors may be able to offer search results that are, or that are seen to be, substantially similar to or better than ours. This may force us to compete in different ways and expend significant resources in order to remain competitive. If our competitors are more successful than we are in developing compelling products or in attracting and retaining users, advertisers, and content providers, our revenues and operating results could be adversely affected. Our ongoing investment in new business strategies and new products, services, and technologies is inherently risky, and could disrupt our ongoing businesses. We have invested and expect to continue to invest in new business strategies, products, services, and technologies. Such endeavors may involve significant risks and uncertainties, including distraction of management from current operations, insufficient revenues to offset liabilities assumed and expenses associated with these new investments, inadequate return of capital on our investments, and unidentified issues not discovered in our due diligence of such strategies and offerings. Because these new ventures are inherently risky, no assurance can be given that such strategies and offerings will be successful and will not materially adversely affect our reputation, financial condition, and operating results. We generate our revenues almost entirely from advertising, and the reduction in spending by or loss of advertisers could seriously harm our business. We generated 96% of our revenues in 2011 from our advertisers. Our advertisers can generally terminate their contracts with us at any time. Advertisers will not continue to do business with us if their investment in advertising with us does not generate sales leads, and ultimately customers, or if we do not deliver their advertisements in an appropriate and effective manner. If we are unable to remain competitive and provide value to our advertisers, they may stop placing ads with us, which would negatively affect our revenues and business. In addition, expenditures by advertisers tend to be cyclical, reflecting overall economic conditions and budgeting and buying patterns. Adverse economic conditions can have a material negative impact on the demand for advertising and cause our advertisers to reduce the amounts they spend on advertising, which could negatively impact our revenues and business. Our revenue growth rate could decline over time, and we anticipate downward pressure on our operating margin in the future. Our revenue growth rate could decline over time as a result of a number of factors, including increasing competition, the challenges in maintaining our growth rate as our revenues increase to higher levels, and the increasing maturity of the online advertising market and the other markets in which we participate. We believe our operating margin will experience downward pressure as a result of increasing competition and increased expenditures for many aspects of our business. Our operating margin will also experience downward pressure if a greater percentage of our revenues comes from ads placed on our Google Network Members’ websites compared to revenues generated through ads placed on our own websites or if we spend a proportionately larger amount to promote the distribution of certain products, including Google Chrome. The margin on revenues we generate from our Google Network Members is significantly less than the margin on revenues we generate from advertising on our websites. Additionally, the margin we earn on revenues generated from our Google Network Members could decrease in the future if we pay an even larger percentage of advertising fees to our Google Network Members. We are subject to increased regulatory scrutiny that may negatively impact our business. The growth of our company and our expansion into a variety of new fields implicate a variety of new regulatory issues, and we have experienced increased regulatory scrutiny as we have grown. In particular, we are cooperating with the U.S. Federal Trade Commission (FTC), the European Commission (EC) and several state 10 Table of Contents attorneys general in investigations they are conducting with respect to our business and its impact on competition. Legislators and regulators, including those conducting investigations in the U.S. and Europe, may make legal and regulatory changes, or interpret and apply existing laws, in ways that make our products and services less useful to our users, require us to incur substantial costs, expose us to unanticipated civil or criminal liability, or cause us to change our business practices. These changes or increased costs could negatively impact our business and results of operations in material ways. We are involved in legal proceedings that may result in adverse outcomes. We are regularly subject to claims, suits, government investigations, and other proceedings involving competition and antitrust (such as the pending investigations by the FTC and the EC), intellectual property, privacy, tax, labor and employment, commercial disputes, content generated by our users, goods and services offered by advertisers or publishers using our platforms, and other matters. Such claims, suits, government investigations, and proceedings are inherently uncertain and their results cannot be predicted with certainty. Regardless of the outcome, such legal proceedings can have an adverse impact on us because of legal costs, diversion of management resources, and other factors. Determining reserves for our pending litigation is a complex, fact-intensive process that requires significant judgment. It is possible that a resolution of one or more such proceedings could result in substantial fines and penalties that could adversely affect our business, consolidated financial position, results of operations, or cash flows in a particular period. These proceedings could also result in criminal sanctions, consent decrees, or orders preventing us from offering certain features, functionalities, products, or services, requiring a change in our business practices, or requiring development of non-infringing products or technologies, which could also adversely affect our business and results of operations. Our business depends on a strong brand, and failing to maintain and enhance our brand would hurt our ability to expand our base of users, advertisers, Google Network Members, and other partners. The brand identity that we have developed has significantly contributed to the success of our business. Maintaining and enhancing the “Google” brand is critical to expanding our base of users, advertisers, Google Network Members, and other partners. We believe that the importance of brand recognition will increase due to the relatively low barriers to entry in the internet market. Our brand may be negatively impacted by a number of factors, including data privacy and security issues, service outages, and product malfunctions. If we fail to maintain and enhance the “Google” brand, or if we incur excessive expenses in this effort, our business, operating results, and financial condition will be materially and adversely affected. Maintaining and enhancing our brand will depend largely on our ability to be a technology leader and continue to provide high-quality products and services, which we may not do successfully. Acquisitions and investments could result in operating difficulties, dilution, and other harmful consequences that may adversely impact our business and results of operations. Acquisitions are an important element of our overall corporate strategy and use of capital, and we expect our current pace of acquisitions to continue. These transactions could be material to our financial condition and results of operations. We also expect to continue to evaluate and enter into discussions regarding a wide array of potential strategic transactions. The process of integrating an acquired company, business, or technology has created, and will continue to create, unforeseen operating difficulties and expenditures. The areas where we face risks include: •Diversion of management time and focus from operating our business to acquisition integration challenges. •Implementation or remediation of controls, procedures, and policies at the acquired company. •Integration of the acquired company’s accounting, human resource, and other administrative systems, and coordination of product, engineering, and sales and marketing functions. •Transition of operations, users, and customers onto our existing platforms. 11 Table of Contents •Failure to obtain required approvals from governmental authorities under competition and antitrust laws on a timely basis, if it all, which could, among other things, delay or prevent us from completing a transaction, or otherwise restrict our ability to realize the expected financial or strategic goals of an acquisition. •In the case of foreign acquisitions, the need to integrate operations across different cultures and languages and to address the particular economic, currency, political, and regulatory risks associated with specific countries. •Failure to successfully further develop the acquired technology. •Cultural challenges associated with integrating employees from the acquired company into our organization, and retention of employees from the businesses we acquire. •Liability for activities of the acquired company before the acquisition, including patent and trademark infringement claims, violations of laws, commercial disputes, tax liabilities, and other known and unknown liabilities. •Litigation or other claims in connection with the acquired company, including claims from terminated employees, customers, former stockholders, or other third parties. Our failure to address these risks or other problems encountered in connection with our past or future acquisitions and investments could cause us to fail to realize the anticipated benefits of such acquisitions or investments, incur unanticipated liabilities, and harm our business generally. Future acquisitions could also result in dilutive issuances of our equity securities, the incurrence of debt, contingent liabilities, or amortization expenses, or write-offs of goodwill, any of which could harm our financial condition. Also, the anticipated benefit of many of our acquisitions may not materialize. A variety of new and existing U.S. and foreign laws could subject us to claims or otherwise harm our business. We are subject to numerous U.S. and foreign laws and regulations covering a wide variety of subject matters. New laws and regulations (or new interpretations of existing laws and regulations) may also impact our business. The costs of compliance with these laws and regulations are high and are likely to increase in the future. Any failure on our part to comply with these laws and regulations can result in negative publicity and diversion of management time and effort and may subject us to significant liabilities and other penalties. Furthermore, many of these laws were adopted prior to the advent of the internet and related technologies and, as a result, do not contemplate or address the unique issues of the internet and related technologies. The laws that do reference the internet are being interpreted by the courts, but their applicability and scope remain uncertain. For example, the laws relating to the liability of providers of online services are currently unsettled both within the U.S. and abroad. Claims have also been, or may be, threatened and filed against us under both U.S. and foreign laws for defamation, invasion of privacy and other tort claims, unlawful activity, patent, copyright and trademark infringement, or other theories based on the nature and content of the materials searched and the ads posted by our users, our products and services, or content generated by our users. Moreover, recent amendments to U.S. patent laws will become effective in 2012 and may affect our ability to protect our innovations and defend against claims of patent infringement. In addition, the Digital Millennium Copyright Act has provisions that limit, but do not necessarily eliminate, our liability for caching or hosting, or for listing or linking to, third-party websites that include materials that infringe copyrights or other rights, so long as we comply with the statutory requirements of this act. Any future legislation impacting these safe harbors may adversely impact us. Various U.S. and international laws restrict the distribution of materials considered harmful to children and impose additional restrictions on the ability of online services to 12 Table of Contents collect information from minors. In the area of data protection, many states have passed laws requiring notification to users when there is a security breach for personal data, such as California’s Information Practices Act. We face similar risks and costs as our products and services are offered in international markets and may be subject to additional regulations. We are, and may in the future be, subject to intellectual property or other claims, which are costly to defend, could result in significant damage awards, and could limit our ability to use certain technologies in the future. Internet, technology, and media companies own large numbers of patents, copyrights, trademarks, and trade secrets and frequently enter into litigation based on allegations of infringement or other violations of intellectual property rights. In addition, patent holding companies may continue to seek to monetize patents they have purchased or otherwise obtained. As we have grown, the intellectual property rights claims against us have increased and may continue to increase as we develop new products, services, and technologies. We have had patent, copyright, and trademark infringement lawsuits filed against us claiming that certain of our products, services, and technologies, including Android, Google Search, Google AdWords, Google AdSense, Google Books, Google News, Google Image Search, Google Chrome, Google Talk, Google Voice, and YouTube, infringe the intellectual property rights of others. Adverse results in these lawsuits may include awards of substantial monetary damages, costly royalty or licensing agreements, or orders preventing us from offering certain features, functionalities, products, or services, and may also cause us to change our business practices, and require development of non-infringing products or technologies, which could result in a loss of revenues for us and otherwise harm our business. In addition, many of our agreements with our customers and partners require us to indemnify them for certain intellectual property infringement claims against them, which would increase our costs as a result of defending such claims, and may require that we pay significant damages if there were an adverse ruling in any such claims. Furthermore, such customers and partners may discontinue the use of our products, services, and technologies, as a result of injunctions or otherwise, which could result in loss of revenues and adversely impact our business. Regardless of the merits of the claims, intellectual property claims are often time consuming, expensive to litigate or settle, and cause significant diversion of management attention. To the extent such intellectual property infringement claims are successful, they may have an adverse effect on our business, consolidated financial position, results of operations, or cash flows. Our intellectual property rights are valuable, and any inability to protect them could reduce the value of our products, services, and brand. Our patents, trademarks, trade secrets, copyrights, and other intellectual property rights are important assets for us. Various events outside of our control pose a threat to our intellectual property rights, as well as to our products, services and technologies. For example, effective intellectual property protection may not be available in every country in which our products and services are distributed or made available through the internet. Also, the efforts we have taken to protect our proprietary rights may not be sufficient or effective. Although we seek to obtain patent protection for our innovations, it is possible we may not be able to protect some of these innovations. Moreover, because of our long-term interests in open source, we may not have adequate patent protection for certain innovations that later turn out to be important. Furthermore, there is always the possibility, despite our efforts, that the scope of the protection gained will be insufficient or that an issued patent may be deemed invalid or unenforceable. We also seek to maintain certain intellectual property as trade secrets. The secrecy could be compromised by outside parties, or by our employees, which would cause us to lose the competitive advantage resulting from these trade secrets. 13 Table of Contents We also face risks associated with our trademarks. For example, there is a risk that the word “Google” could become so commonly used that it becomes synonymous with the word “search.” If this happens, we could lose protection for this trademark, which could result in other people using the word “Google” to refer to their own products, thus diminishing our brand. Any significant impairment of our intellectual property rights could harm our business and our ability to compete. Also, protecting our intellectual property rights is costly and time consuming. Any increase in the unauthorized use of our intellectual property could make it more expensive to do business and harm our operating results. We may be subject to legal liability associated with providing online services or content. We host and provide a wide variety of services and products that enable users to exchange information, advertise products and services, conduct business, and engage in various online activities both domestically and internationally. The law relating to the liability of providers of these online services and products for activities of their users is still somewhat unsettled both within the U.S. and internationally. Claims have been threatened and have been brought against us for defamation, negligence, breaches of contract, copyright or trademark infringement, unfair competition, unlawful activity, tort, including personal injury, fraud, or other theories based on the nature and content of information which we publish or to which we provide links or that may be posted online or generated by us or by third parties, including our users. In addition, we have been and may again in the future be subject to domestic or international actions alleging that certain content we have generated or third-party content that we have made available within our services violates laws in domestic and international jurisdictions. We also arrange for the distribution of third-party advertisements to third-party publishers and advertising networks, and we offer third-party products, services, or content. We may be subject to claims concerning these products, services, or content by virtue of our involvement in marketing, branding, broadcasting, or providing access to them, even if we do not ourselves host, operate, provide, or provide access to these products, services, or content. Defense of any such actions could be costly and involve significant time and attention of our management and other resources, may result in monetary liabilities or penalties, and may require us to change our business in an adverse manner. More people are using devices other than personal computers to access the internet. If users do not widely adopt versions of our web search technology, products, or operating systems developed for these devices, our business could be adversely affected. The number of people who access the internet through devices other than personal computers, including mobile phones, smart phones, handheld computers such as netbooks and tablets, video game consoles, and television set-top devices, has increased dramatically in the past few years. The lower resolution, functionality, and memory associated with some alternative devices make the use of our products and services through such devices more difficult and the versions of our products and services developed for these devices may not be compelling to users, manufacturers, or distributors of alternative devices. Each manufacturer or distributor may establish unique technical standards for its devices, and our products and services may not work or be viewable on these devices as a result. We have limited experience to date in operating versions of our products and services developed or optimized for users of alternative devices, such as Google Mobile and Android, or in designing alternative devices. As new devices and new platforms are continually being released, it is difficult to predict the problems we may encounter in developing versions of our products and services for use on these alternative devices and we may need to devote significant resources to the creation, support, and maintenance of such devices. If we are unable to attract and retain a substantial number of alternative device manufacturers, distributors, and users to our products and services, or if we are slow to develop products and technologies that are more compatible with alternative devices, we will fail to capture a significant share of an increasingly important portion of the market for online services, which could adversely affect our business. 14 Table of Contents Privacy concerns relating to our technology could damage our reputation and deter current and potential users from using our products and services. From time to time, concerns have been expressed about whether our products and services compromise the privacy of users and others. Concerns about our practices with regard to the collection, use, disclosure, or security of personal information or other privacy related matters, even if unfounded, could damage our reputation and operating results. While we strive to comply with all applicable data protection laws and regulations, as well as our own posted privacy policies, any failure or perceived failure to comply may result, and has resulted, in proceedings or actions against us by government entities or others, or could cause us to lose users and customers, which could potentially have an adverse effect on our business. In addition, as nearly all of our products and services are web-based, the amount of data we store for our users on our servers (including personal information) has been increasing. Any systems failure or compromise of our security that results in the release of our users’ data could seriously limit the adoption of our products and services, as well as harm our reputation and brand and, therefore, our business. We expect to continue to expend significant resources to protect against security breaches. The risk that these types of events could seriously harm our business is likely to increase as we expand the number of web-based products and services we offer, and operate in more countries. Regulatory authorities around the world are considering a number of legislative and regulatory proposals concerning data protection. In addition, the interpretation and application of consumer and data protection laws in the U.S., Europe and elsewhere are often uncertain and in flux. It is possible that these laws may be interpreted and applied in a manner that is inconsistent with our data practices. If so, in addition to the possibility of fines, this could result in an order requiring that we change our data practices, which could have an adverse effect on our business and results of operations. Complying with these various laws could cause us to incur substantial costs or require us to change our business practices in a manner adverse to our business. If our security measures are breached, or if our services are subject to attacks that degrade or deny the ability of users to access our products and services, our products and services may be perceived as not being secure, users and customers may curtail or stop using our products and services, and we may incur significant legal and financial exposure. Our products and services involve the storage and transmission of users’ and customers’ proprietary information, and security breaches could expose us to a risk of loss of this information, litigation, and potential liability. Our security measures may be breached due to the actions of outside parties, employee error, malfeasance, or otherwise, and, as a result, an unauthorized party may obtain access to our data or our users’ or customers’ data. Additionally, outside parties may attempt to fraudulently induce employees, users, or customers to disclose sensitive information in order to gain access to our data or our users’ or customers’ data. Any such breach or unauthorized access could result in significant legal and financial exposure, damage to our reputation, and a loss of confidence in the security of our products and services that could potentially have an adverse effect on our business. Because the techniques used to obtain unauthorized access, disable or degrade service, or sabotage systems change frequently and often are not recognized until launched against a target, we may be unable to anticipate these techniques or to implement adequate preventative measures. If an actual or perceived breach of our security occurs, the market perception of the effectiveness of our security measures could be harmed and we could lose users and customers. Web spam and content farms could decrease our search quality, which could damage our reputation and deter our current and potential users from using our products and services. “Web spam” refers to websites that attempt to violate a search engine’s quality guidelines or that otherwise seek to rank higher in search results than a search engine’s assessment of their relevance and utility would rank them. Although English-language web spam in our search results has been significantly reduced, and web spam in 15 Table of Contents most other languages is limited, we expect web spammers will continue to seek ways to improve their rankings inappropriately. We continuously combat web spam, including through indexing technology that makes it harder for spam-like, less useful web content to rank highly. We have also improved our ability to detect hacked websites, a major source of web spam in 2010. We face new challenges from low-quality and irrelevant content websites,including “content farms,” which are websites that generate large quantities of low-quality content to help them improve their search rankings. In 2011, we launched several algorithmic changes focused on low-quality websites. If web spam and content farms continue to increase on Google, this could hurt our reputation for delivering relevant information or reduce user traffic to our websites. In addition, as we continue to take actions to improve our search quality and reduce low-quality content, this may in the short run reduce our AdSense revenues, since some of these websites are AdSense partners. Interruption or failure of our information technology and communications systems could hurt our ability to effectively provide our products and services, which could damage our reputation and harm our operating results. The availability of our products and services depends on the continuing operation of our information technology and communications systems. Our systems are vulnerable to damage or interruption from earthquakes, terrorist attacks, floods, fires, power loss, telecommunications failures, computer viruses, computer denial of service attacks, or other attempts to harm our systems. Some of our data centers are located in areas with a high risk of major earthquakes. Our data centers are also subject to break-ins, sabotage, and intentional acts of vandalism, and to potential disruptions if the operators of these facilities have financial difficulties. Some of our systems are not fully redundant, and our disaster recovery planning cannot account for all eventualities. The occurrence of a natural disaster, a decision to close a facility we are using without adequate notice for financial reasons, or other unanticipated problems at our data centers could result in lengthy interruptions in our service. In addition, our products and services are highly technical and complex and may contain errors or vulnerabilities. Any errors or vulnerabilities in our products and services, or damage to or failure of our systems, could result in interruptions in our services, which could reduce our revenues and profits, and damage our brand. Our business and operations are experiencing rapid growth. If we fail to effectively manage our growth, our business and operating results could be harmed. We have experienced rapid growth in our headcount and operations, which has placed, and will continue to place, significant demands on our management, operational, and financial infrastructure. If we do not effectively manage our growth, the quality of our products and services could suffer, which could negatively affect our brand and operating results. Our expansion and growth in international markets heighten these risks as a result of the particular challenges of supporting a rapidly growing business in an environment of multiple languages, cultures, customs, legal systems, alternative dispute resolution systems, regulatory systems, and commercial infrastructures. To effectively manage this growth, we will need to continue to improve our operational, financial and management controls, and our reporting systems and compliance procedures. These systems enhancements and improvements will require significant capital expenditures and management resources. Failure to implement these improvements could hurt our ability to manage our growth and our consolidated financial position. Our international operations expose us to additional risks which could harm our business, operating results, and financial condition. Our international operations are significant to our revenues and net income, and we plan to further expand internationally. International revenues accounted for approximately 54% of our total revenues in 2011 and more than half of our user traffic has been coming from outside the U.S. In certain international markets, we have limited operating experience and may not benefit from any first-to-market advantages or otherwise succeed. In addition to risks described elsewhere in this section, our international operations expose us to additional risks, including the following: •Changes in local political, economic, social, and labor conditions, which may adversely harm our business. 16 Table of Contents •Restrictions on foreign ownership and investments, and stringent foreign exchange controls that might prevent us from repatriating cash earned in countries outside the U.S. •Import and export requirements that may prevent us from offering products or providing services to a particular market and may increase our operating costs. •Currency exchange rate fluctuations and our ability to manage these fluctuations through our foreign exchange risk management program. •Longer payment cycles in some countries, increased credit risk, and higher levels of payment fraud. •Uncertainty regarding liability for services and content, including uncertainty as a result of local laws and lack of legal precedent. •Different employee/employer relationships, existence of workers’ councils and labor unions, and other challenges caused by distance, language, and cultural differences, making it harder to do business in certain jurisdictions. In addition, compliance with complex foreign and U.S. laws and regulations that apply to our international operations increases our cost of doing business. These numerous and sometimes conflicting laws and regulations include internal control and disclosure rules, data privacy and filtering requirements, anti-corruption laws, such as the Foreign Corrupt Practices Act, and other local laws prohibiting corrupt payments to governmental officials, and antitrust and competition regulations, among others. Violations of these laws and regulations could result in fines and penalties, criminal sanctions against us, our officers, or our employees, prohibitions on the conduct of our business and on our ability to offer our products and services in one or more countries, and could also materially affect our brand, our international expansion efforts, our ability to attract and retain employees, our business, and our operating results. Although we have implemented policies and procedures designed to ensure compliance with these laws and regulations, there can be no assurance that our employees, contractors, or agents will not violate our policies. Furthermore, since we conduct business in currencies other than U.S. dollars but report our financial results in U.S. dollars, we face exposure to fluctuations in currency exchange rates. Although we hedge a portion of our international currency exposure, significant fluctuations in exchange rates between the U.S. dollar and foreign currencies may adversely affect our net income. Additionally, hedging programs are inherently risky and could expose us to additional risks that could adversely affect our financial condition and results of operations. If we were to lose the services of Larry, Sergey, Eric, or other key personnel, we may not be able to execute our business strategy. Our future success depends in a large part upon the continued service of key members of our senior management team. In particular, Larry Page and Sergey Brin are critical to the overall management of Google and the development of our technology. Along with our Executive Chairman Eric E. Schmidt, they also play a key role in maintaining our culture and setting our strategic direction. All of our executive officers and key employees are at-will employees, and we do not maintain any key-person life insurance policies. The loss of key personnel could seriously harm our business. We rely on highly skilled personnel and, if we are unable to retain or motivate key personnel, hire qualified personnel, or maintain our corporate culture, we may not be able to grow effectively. Our performance largely depends on the talents and efforts of highly skilled individuals. Our future success depends on our continuing ability to identify, hire, develop, motivate, and retain highly skilled personnel for all areas of our organization. Competition in our industry for qualified employees is intense, and certain of our competitors have directly targeted our employees. In addition, our compensation arrangements, such as our equity award programs, may not always be successful in attracting new employees and retaining and motivating our existing employees. Our continued ability to compete effectively depends on our ability to attract new employees and to retain and motivate our existing employees. 17 Table of Contents In addition, we believe that our corporate culture fosters innovation, creativity, and teamwork. As our organization grows, and we are required to implement more complex organizational management structures, we may find it increasingly difficult to maintain the beneficial aspects of our corporate culture. This could negatively impact our future success. Our business depends on continued and unimpeded access to the internet by us and our users. Internet access providers may be able to block, degrade, or charge for access to certain of our products and services, which could lead to additional expenses and the loss of users and advertisers. Our products and services depend on the ability of our users to access the internet, and certain of our products require significant bandwidth to work effectively. Currently, this access is provided by companies that have significant market power in the broadband and internet access marketplace, including incumbent telephone companies, cable companies, mobile communications companies, and government-owned service providers. Some of these providers have taken, or have stated that they may take, measures that could degrade, disrupt, or increase the cost of user access to certain of our products by restricting or prohibiting the use of their infrastructure to support or facilitate our offerings, or by charging increased fees to us or our users to provide our offerings. Such interference could result in a loss of existing users and advertisers, and increased costs, and could impair our ability to attract new users and advertisers, thereby harming our revenues and growth. New technologies could block our ads, which would harm our business. Technologies have been developed (including by us) that can block the display of our ads and that provide tools to users to opt out of our advertising products. Most of our revenues are derived from fees paid to us by advertisers in connection with the display of ads on web pages for our users. As a result, such technologies and tools could adversely affect our operating results. We are exposed to fluctuations in the market values of our investment portfolio. Given the global nature of our business, we have investments both domestically and internationally. Credit ratings and pricing of these investments can be negatively impacted by liquidity, credit deterioration or losses, financial results, or other factors. As a result, the value or liquidity of our cash equivalents and marketable securities could decline and result in a material impairment, which could materially adversely affect our financial condition and operating results. We may have exposure to greater than anticipated tax liabilities. Our future income taxes could be adversely affected by earnings being lower than anticipated in jurisdictions that have lower statutory tax rates and higher than anticipated in jurisdictions that have higher statutory tax rates, by changes in the valuation of our deferred tax assets and liabilities, as a result of gains on our foreign exchange risk management program, or changes in tax laws, regulations, or accounting principles, as well as certain discrete items. We are subject to regular review and audit by both domestic and foreign tax authorities. Any adverse outcome of such a review or audit could have a negative effect on our operating results and financial condition. In addition, the determination of our worldwide provision for income taxes and other tax liabilities requires significant judgment, and there are many transactions and calculations where the ultimate tax determination is uncertain. Although we believe our estimates are reasonable, the ultimate tax outcome may differ from the amounts recorded in our financial statements and may materially affect our financial results in the period or periods for which such determination is made. Our operating results may fluctuate, which makes our results difficult to predict and could cause our results to fall short of expectations. Our operating results may fluctuate as a result of a number of factors, many outside of our control. As a result, comparing our operating results on a period-to-period basis may not be meaningful, and you should not rely 18 Table of Contents on our past results as an indication of our future performance. Our quarterly, year-to-date, and annual expenses as a percentage of our revenues may differ significantly from our historical or projected rates. Our operating results in future quarters may fall below expectations. Any of these events could cause our stock price to fall. Each of the risk factors listed in this section and the following factors may affect our operating results: •Our ability to continue to attract users to our websites and satisfy existing users on our websites. •Our ability to monetize (or generate revenues from) traffic on our websites and our Google Network Members’ websites. •Our ability to attract advertisers to our AdWords program, and our ability to attract websites to our AdSense program. •The mix in our revenues between those generated on our websites and those generated through our Google Network. •The amount of revenues and expenses generated and incurred in currencies other than U.S. dollars, and our ability to manage the resulting risk through our foreign exchange risk management program. •The amount and timing of operating costs and expenses and capital expenditures related to the maintenance and expansion of our businesses, operations, and infrastructure. •Our focus on long-term goals over short-term results. •The results of our investments in risky projects, including new business strategies and new products, services, and technologies. •Our ability to keep our websites operational at a reasonable cost and without service interruptions. •Our ability to generate significant revenues from services in which we have invested considerable time and resources, such as Google Checkout. Because our business is changing and evolving, our historical operating results may not be useful to you in predicting our future operating results. In addition, advertising spending has historically been cyclical in nature, reflecting overall economic conditions, as well as budgeting and buying patterns. Also, user traffic tends to be seasonal. Our rapid growth has tended to mask the cyclicality and seasonality of our business. As our growth rate has slowed, the cyclicality and seasonality in our business has become more pronounced and caused our operating results to fluctuate. Risks Related to Ownership of Our Common Stock The trading price for our Class A common stock may continue to be volatile. The trading price of our Class A common stock has at times experienced substantial price volatility and may continue to be volatile. For example, in 2011, the price of our Class A common stock ranged from $473.02 per share to $646.76 per share. The trading price of our Class A common stock may fluctuate widely in response to various factors, some of which are beyond our control. These factors include: •Quarterly variations in our results of operations or those of our competitors. •Announcements by us or our competitors of acquisitions, new products, significant contracts, commercial relationships, or capital commitments. •Recommendations by securities analysts or changes in earnings estimates. •Announcements about our earnings that are not in line with analyst expectations, the risk of which is enhanced because it is our policy not to give guidance on earnings. •Announcements by our competitors of their earnings that are not in line with analyst expectations. 19 Table of Contents •Commentary by industry and market professionals about our products, strategies, and other matters affecting our business and results, regardless of its accuracy. •The volume of shares of Class A common stock available for public sale. •Sales of stock by us or by our stockholders (including sales by our directors, executive officers, and other employees). •Short sales, hedging, and other derivative transactions on shares of our Class A common stock (including derivative transactions under our TSO program). In addition, the stock market in general, and the market for technology companies in particular, have experienced extreme price and volume fluctuations that have often been unrelated or disproportionate to the operating performance of those companies. These broad market and industry factors may harm the market price of our Class A common stock, regardless of our actual operating performance. The concentration of our stock ownership limits our stockholders’ ability to influence corporate matters. Our Class B common stock has 10 votes per share and our Class A common stock has one vote per share. As of December 31, 2011, Larry, Sergey, and Eric beneficially owned approximately 92% of our outstanding Class B common stock, representing approximately 66% of the voting power of our outstanding capital stock. Larry, Sergey, and Eric therefore have significant influence over management and affairs and over all matters requiring stockholder approval, including the election of directors and significant corporate transactions, such as a merger or other sale of our company or our assets, for the foreseeable future. This concentrated control limits our stockholders’ ability to influence corporate matters and, as a result, we may take actions that our stockholders do not view as beneficial. As a result, the market price of our Class A common stock could be adversely affected. Provisions in our charter documents and under Delaware law could discourage a takeover that stockholders may consider favorable. Provisions in our certificate of incorporation and bylaws may have the effect of delaying or preventing a change of control or changes in our management. These provisions include the following: •Our certificate of incorporation provides for a dual class common stock structure. As a result of this structure, Larry, Sergey, and Eric have significant influence over all matters requiring stockholder approval, including the election of directors and significant corporate transactions, such as a merger or other sale of our company or our assets. This concentrated control could discourage others from initiating any potential merger, takeover, or other change of control transaction that other stockholders may view as beneficial. •Our board of directors has the right to elect directors to fill a vacancy created by the expansion of the board of directors or the resignation, death, or removal of a director, which prevents stockholders from being able to fill vacancies on our board of directors. •Our stockholders may not act by written consent. As a result, a holder, or holders, controlling a majority of our capital stock would not be able to take certain actions without holding a stockholders’ meeting. •Our certificate of incorporation prohibits cumulative voting in the election of directors. This limits the ability of minority stockholders to elect director candidates. •Stockholders must provide advance notice to nominate individuals for election to the board of directors or to propose matters that can be acted upon at a stockholders’ meeting. These provisions may discourage or deter a potential acquirer from conducting a solicitation of proxies to elect the acquirer’s own slate of directors or otherwise attempting to obtain control of our company. 20 Table of Contents •Our board of directors may issue, without stockholder approval, shares of undesignated preferred stock. The ability to issue undesignated preferred stock makes it possible for our board of directors to issue preferred stock with voting or other rights or preferences that could impede the success of any attempt to acquire us. As a Delaware corporation, we are also subject to certain Delaware anti-takeover provisions. Under Delaware law, a corporation may not engage in a business combination with any holder of 15% or more of its outstanding voting stock unless the holder has held the stock for three years or, among other things, the board of directors has approved the transaction. Our board of directors could rely on Delaware law to prevent or delay an acquisition of us. ITEM 1B.UNRESOLVED STAFF COMMENTS None. ITEM 2.PROPERTIES Our headquarters are located in Mountain View, California, where we own approximately 3.4 million square feet of office and building space and approximately seven acres of developable land to accommodate anticipated future growth.We also own a 2.9 million square feet office building in New York, New York and 556,000 square feet of office and building space in Paris and Dublin. We also operate and own data centers in the U.S., Europe, and Asia pursuant to various lease agreements and co-location arrangements. In addition, we lease approximately 3.8 million square feet of office space and approximately 61 acres of undeveloped land in and near our headquarters in Mountain View, California. We also lease additional research and development, and sales and support offices throughout the United States and maintain leased facilities internationally in countries around the world. Larger leased sites include properties located in Dublin, Ireland; Zurich, Switzerland; London, UK; Hyderabad, India; San Francisco, CA; Paris, France; Hamburg, Germany; Sao Paulo, Brazil; Ann Arbor, MI; Bothell, WA; Cambridge, MA; Chicago, IL; Kirkland, WA; Venice, CA; Seattle, WA; Sydney, Australia; Beijing, China; Bangalore, India; Gurgaon, India; Tokyo, Japan; and Singapore. We believe our existing facilities, both owned and leased, are in good condition and suitable for the conduct of our business. ITEM 3.LEGAL PROCEEDINGS For a description of our material pending legal proceedings, please see Note 12 “Commitments and Contingencies—Legal Matters” of the Notes to Consolidated Financial Statements included in Item 8 of this Annual Report on Form 10-K, which is incorporated herein by reference. 21 Table of Contents PART II ITEM 5.MARKET FOR REGISTRANT’S COMMON EQUITY, RELATED STOCKHOLDER MATTERS AND ISSUER PURCHASES OF EQUITY SECURITIES Our Class A common stock has been listed on the Nasdaq Global Select Market under the symbol “GOOG” since August 19, 2004. Prior to that time, there was no public market for our stock. The following table sets forth for the indicated periods the high and low sales prices per share for our Class A common stock on the Nasdaq Global Select Market. Fiscal Year 2011 Quarters Ended:High Low March 31, 2011$642.96$551.28 June 30, 2011 595.19 473.02 September 30, 2011 627.50 490.86 December 31, 2011 646.76 480.60 Fiscal Year 2010 Quarters Ended:High Low March 31, 2010$629.51$520.00 June 30, 2010 597.84 444.72 September 30, 2010 536.85 433.63 December 31, 2010 630.85 518.85 Our Class B common stock is neither listed nor traded. Holders of Record As of December 31, 2011, there were approximately 2,874 stockholders of record of our Class A common stock, and the closing price of our Class A common stock was $645.90 per share as reported by the Nasdaq Global Select Market. Because many of our shares of Class A common stock are held by brokers and other institutions on behalf of stockholders, we are unable to estimate the total number of stockholders represented by these record holders. As of December 31, 2011, there were approximately 85 stockholders of record of our Class B common stock. Dividend Policy We have never declared or paid any cash dividend on our common stock. We intend to retain any future earnings and do not expect to pay any dividends in the foreseeable future. 22 Table of Contents Stock Performance Graph This performance graph shall not be deemed “filed” for purposes of Section 18 of the Securities Exchange Act of 1934, as amended (the Exchange Act), or incorporated by reference into any filing of Google under the Securities Act of 1933, as amended, or the Exchange Act, except as shall be expressly set forth by specific reference in such filing. The following graph shows a comparison from December 31, 2006 through December 31, 2011 of the cumulative total return for our Class A common stock,the S&P 500 Index,the Nasdaq Composite Index, and the RDG Internet Composite Index. Such returns are based on historical results and are not intended to suggest future performance. Data for the S&P 500 Index,the Nasdaq Composite Index, and the RDG Internet Composite Index assume reinvestment of dividends. We have never paid dividends on our Class A common stock and have no present plans to do so. COMPARISON OF 5 YEAR CUMULATIVE TOTAL RETURN Among Google Inc., the S&P 500 Index, the NASDAQ Composite Index, and the RDG Internet Composite Index $100 invested on 12/31/06 in stock or index, including reinvestment of dividends. Fiscal year ending December 31. Copyright© 2012 S&P, a division of The McGraw-Hill Companies Inc. All rights reserved. 23 Table of Contents Results of Google’s Transferable Stock Option (TSO) Program Under our TSO program, which we launched in April 2007, eligible employees are able to sell vested stock options to participating financial institutions in an online auction as an alternative to exercising options in the traditional method and then selling the underlying shares. The following table provides information with respect to sales by our employees of TSOs during the three months ended December 31, 2011: Aggregate Amounts Weighted-Average Per Share Amounts Period(1)Number of Shares Underlying TSOs Sold Sale Price of TSOs Sold TSO Premium(2)Exercise Price of TSOs Sold Sale Price of TSOs Sold TSO Premium(2) (in thousands) October 1– 31 152,594$38,884$2,385$351.45$254.82$15.63 November 1 – 30 180,140 49,500 1,958 338.42 274.79 10.87 December 1 – 31 0 0 0 0 0 0 Total (except weighted-average per share amounts)332,734$88,384$4,343$344.40$265.63$13.05 (1)The TSO program is generally active during regular trading hours for the Nasdaq Global Select Market when Google’s trading window is open. However, we have the right to suspend the TSO program at any time for any reason, including for maintenance and other technical reasons. (2)The TSO premium is calculated as the difference between (a)the sale price of the TSO and (b)the intrinsic value of the TSO, which we define as the excess, if any, of the price of our Class A common stock at the time of the sale over the exercise price of the TSO. In April 2009, we amended our TSO program to allow participation by executive officers (other than Larry Page, Sergey Brin, and Eric E. Schmidt) in our TSO program. The following table provides information with respect to sales by our executive officers of TSOs during the three months ended December 31, 2011: Aggregate Amounts Executive Officer Number of Shares Underlying TSOs Sold Sale Price of TSOs Sold TSO Premium (in thousands) Nikesh Arora 3,947$1,097$3 Total 3,947$1,097$3 24 Table of Contents ITEM 6.SELECTED FINANCIAL DATA You should read the following selected consolidated financial data in conjunction with Item 7 “Management’s Discussion and Analysis of Financial Condition and Results of Operations” and our consolidated financial statements and the related notes appearing in Item 8 “Financial Statements and Supplementary Data” of this Annual Report on Form 10-K. The consolidated statements of income data for the years ended December 31, 2009, 2010, and 2011 and the consolidated balance sheet data at December 31, 2010, and 2011 are derived from our audited consolidated financial statements appearing in Item 8 of this Annual Report on Form 10-K. The consolidated statements of income data for the years ended December 31, 2007 and 2008, and the consolidated balance sheet data at December 31, 2007, 2008, and 2009, are derived from our audited consolidated financial statements that are not included in this Annual Report on Form 10-K. The historical results are not necessarily indicative of the results to be expected in any future period. Year Ended December 31, 2007 2008 2009 2010 2011 (in millions, except per share amounts) Consolidated Statements of Income Data: Revenues$16,594$21,796$23,651$29,321$37,905 Income from operations 5,084 6,632 8,312 10,381 11,742 Net income 4,204 4,227 6,520 8,505 9,737 Net income per share of Class A and Class B common stock Basic$13.53$13.46$20.62$26.69$30.17 Diluted$13.29$13.31$20.41$26.31$29.76 As of December 31, 2007 2008 2009 2010 2011 (in millions) Consolidated Balance Sheet Data: Cash, cash equivalents, and marketable securities$14,219$15,846$24,485$34,975$44,626 Total assets 25,336 31,768 40,497 57,851 72,574 Total long-term liabilities 611 1,227 1,746 1,614 5,516 Total stockholders’ equity 22,690 28,239 36,004 46,241 58,145 25 Table of Contents ITEM 7.MANAGEMENT’S DISCUSSION AND ANALYSIS OF FINANCIAL CONDITION AND RESULTS OF OPERATIONS The following discussion and analysis of our financial condition and results of operations should be read together with our consolidated financial statements and related notes included under Item 8 of this Annual Report on Form 10-K. Overview Google is a global technology leader focused on improving the ways people connect with information. Our innovations in web search and advertising have made our website a top internet property and our brand one of the most recognized in the world. Our mission is to organize the world’s information and make it universally accessible and useful. We serve three primary constituencies: •Users. We provide users with products and services that enable people to more quickly and easily find, create, and organize information that is useful to them. •Advertisers. We provide advertisers with cost-effective ways to deliver online and offline ads to customers across Google-owned websites and through the Google Network, which is the network of third parties that use our advertising programs to deliver relevant ads with their search results and content. •Google Network Members and Other Content Providers. We provide members of our Google Network with our Google AdSense programs. These include programs through which we distribute our advertisers’ AdWords ads for display on the websites of our Google Network Members. We share most of the fees these ads generate with our Google Network Members, thereby creating an important revenue stream for them. In addition, we have entered into arrangements with other content providers under which we distribute or license their video and other content, and we may display ads next to or as part of this content on the pages of our websites. We share most of the fees these ads generate with these content providers, thereby creating an important revenue stream for these partners. Recent Development On August 15, 2011, we entered into the Merger Agreement with Motorola, a provider of innovative technologies, products and services that enable a range of mobile and wireline digital communication, information and entertainment experiences, under which we will acquire Motorola for $40 per share in cash, or a total of approximately $12.5 billion in cash. The completion of this transaction is subject to customary closing conditions, including the receipt of certain regulatory approvals. How We Generate Revenue Advertising revenues made up 97% of our revenues in 2009 and 96% of our revenues in 2010 and 2011. We derive most of our additional revenues from our enterprise products, as well as our display advertising management services to advertisers, ad agencies, and publishers. Google AdWords is our auction-based advertising program that enables advertisers to place text-based and display ads on our websites and our Google Network Members’ websites. Display advertising comprises the videos, text, images, and other interactive ads that run across the web on computers and mobile devices, including smart phones and handheld computers such as netbooks and tablets. Most of our AdWords customers pay us on a cost-per-click basis, which means that an advertiser pays us only when a user clicks on one of its ads. We also offer AdWords on a cost-per-impression basis that enables advertisers to pay us based on the number of times their ads appear on our websites and our Google Network Members’ websites as specified by the advertisers. For advertisers using our AdWords cost-per-click pricing, we recognize as revenue the fees charged to advertisers 26 Table of Contents each time a user clicks on one of the ads that appears next to the search results or content on our websites or our Google Network Members’ websites. For advertisers using our AdWords cost-per-impression pricing, we recognize as revenue the fees charged to advertisers each time their ads are displayed on our websites or our Google Network Members’ websites. Our AdWords agreements are generally terminable at any time by our advertisers. Google AdSense refers to the online programs through which we distribute our advertisers’ AdWords ads for display on our Google Network Members’ websites, as well as programs to deliver ads on television broadcasts. Our AdSense programs include AdSense for search and AdSense for content. AdSense for search is our online service for distributing relevant ads from our advertisers for display with search results on our Google Network Members’ websites. To use AdSense for search, most of our AdSense for search partners add Google search functionality to their web pages in the form of customizable Google search boxes. When visitors to these websites search either the website or the internet using these customizable search boxes, we display relevant ads on the search results pages, targeted to match user search queries. Ads shown through AdSense for search are text ads. AdSense for content is our online service for distributing ads from our advertisers that are relevant to content on our Google Network Members’ websites. Under this program, we use automated technology to analyze the meaning of the content on the web page and serve relevant ads based on the meaning of such content. For example, a web page on an automotive blog that contains an entry about vintage cars might display ads for vintage car parts or vintage car shows. These ads are displayed in spaces that our AdSense for content partners have set aside on their websites. AdSense for content allows a variety of ad types to be shown, including text ads, image ads, Google Video Ads, link units (which are sets of clickable links to topic pages related to page content), themed units (which are regular text ads with graphic treatments that change seasonally and by geography), and gadget ads (which are customized “mini-sites” that run as ads on AdSense publisher websites). For our online AdSense program, our advertisers pay us a fee each time a user clicks on one of our advertisers’ ads displayed on our Google Network Members’ websites or, for those advertisers who choose our cost-per-impression pricing, as their ads are displayed. To date, we have paid most of these advertiser fees to our Google Network Members, and we expect to continue doing so for the foreseeable future. We recognize these advertiser fees as revenue and the portion of the advertiser fee we pay to our Google Network Members as traffic acquisition costs under cost of revenues. Google Network Members do not pay any fees associated with the use of our AdSense program on their websites. Our agreements with Google Network Members consist largely of uniform online “click-wrap” agreements that members enter into by interacting with our registration websites. The standard agreements have no stated term and are terminable at will. Agreements with our larger members are individually negotiated. Both the standard agreements and the negotiated agreements require us to share with the Google Network Member most of the advertiser fees generated by users clicking on ads on the Google Network Member’s website or, for advertisers who choose our cost-per-impression pricing, as the ads are displayed on the Google Network Member’s website. For example, under our standard agreements, we pay 51% and 68% of the fees collected from advertisers to our Google Network Members in AdSense for search and AdSense for content, respectively. We also offer display advertising management services such as media planning, buying, implementation, and measurement tools for advertisers and agencies, and forecasting and reporting tools for publishers. We recognize the related fees as other revenues in the period advertising impressions are delivered. In January 2010, we launched, and in July 2010, we discontinued, our direct-to-consumer web store channel for distributing our Nexus One mobile phone. We had recognized fees derived from the sale of these phones as other revenues in the period in which they were delivered. We have entered into arrangements with certain content providers under which we distribute or license their video and other content. Our agreements with content providers are typically standard agreements with no stated 27 Table of Contents term and are terminable at will. Agreements with our larger members are individually negotiated. Both the standard agreements and the negotiated agreements require us to pay the content providers for the content we license. In a number of these arrangements, we display ads on the pages of our websites from which the content is viewed and share most of the fees these ads generate with the content providers. We recognize these advertiser fees as revenue and the fees we pay to our content providers as content acquisition costs under cost of revenues. We believe the factors that influence the success of our advertising programs include the following: •The relevance, objectivity, and quality of our search results and the relevance and quality of ads displayed with each search results page. •The number of searches initiated at our websites and our Google Network Members’ websites and the underlying purpose of these searches (for instance, whether they are for academic research, to find a news article, or to find a product or service). •The number and prominence of ads displayed on our websites and our Google Network Members’ websites. •The number of visits to, and the content of, our Google Network Members’ websites and certain of our websites and the relevance and quality of the ads we display next to this content. •The advertisers’ return on investment from advertising campaigns on our websites or our Google Network Members’ websites compared to other forms of advertising. •The total advertising spending budgets of each advertiser. •The number of advertisers and the breadth of items advertised. •The amount we ultimately pay our Google Network Members, distribution partners, and our content providers for traffic, access points, and content, compared to the amount of revenues we generate. •Our ability to increase traffic on our websites and our Google Network Members’ websites via new and improved ad formats including the ones on mobile devices. Trends in Our Business Advertising transactions continue to shift from offline to online as the digital economy evolves. This has contributed to the rapid growth of our business since inception, resulting in substantially increased revenues, and we expect that our business will continue to grow. However, our revenue growth rate has generally declined over time, and it could do so in the future as a result of a number of factors, including increasing competition, challenges in maintaining our growth rate as our revenues increase to higher levels, and increasing maturity of the online advertising market and other markets in which we participate. In addition, if there is a further general economic downturn, this may result in fewer commercial queries by our users and may cause advertisers to reduce the amount they spend on online advertising, including the amount they are willing to pay for each click or impression, which could negatively affect the growth rate of our revenues. We plan to continue to invest aggressively in our core areas of strategic focus. The main focus of our advertising programs is to provide relevant and useful advertising to our users, reflecting our commitment to constantly improve their overall web experience. As a result, we expect to continue to take steps to improve the relevance of the ads displayed on our websites and our Google Network Members’ websites. These steps include not displaying ads that generate low click-through rates or that send users to irrelevant or otherwise low quality websites and terminating our relationships with those Google Network Members whose websites do not meet our quality requirements. We may also continue to take steps to reduce the number of accidental clicks by our users. These steps could negatively affect the growth rate of our revenues. Both seasonal fluctuations in internet usage and traditional retail seasonality have affected, and are likely to continue to affect, our business. Internet usage generally slows during the summer months, and commercial 28 Table of Contents queries typically increase significantly in the fourth quarter of each year. These seasonal trends have caused, and will likely continue to cause, fluctuations in our quarterly results, including fluctuations in sequential revenues, as well as aggregate paid click and average cost-per-click growth rates. The operating margin we realize on revenues generated from ads placed on our Google Network Members’ websites through our AdSense program is significantly lower than the operating margin we realize from revenues generated from ads placed on our websites because most of the advertiser fees from ads served on Google Network Members’ websites are shared with our Google Network Members. For the past five years, growth in advertising revenues from our websites has generally exceeded that from our Google Network Members’ websites. This trend has had a positive impact on our operating margins, and we expect that this will continue for the foreseeable future, although the relative rate of growth in revenues from our websites compared to the rate of growth in revenues from our Google Network Members’ websites may vary over time. We also continue to invest aggressively in our systems, data centers, corporate facilities, information technology infrastructure, and employees. We increased our hiring in 2011, and we may continue to do so and to provide competitive compensation programs for our employees. For instance, effective January 1, 2011, we increased base salaries for all of our non-executive employees by 10% and shifted a portion of the bonus into base salary. Our full-time employee headcount was 24,400 at December 31, 2010 and 32,467 at December 31, 2011. Acquisitions will also remain an important component of our strategy and use of capital, and we expect our current pace of acquisitions to continue.We expect our cost of revenues will increase in dollars and may increase as a percentage of revenues in future periods, primarily as a result of forecasted increases in traffic acquisition costs, data center costs, content acquisition costs, credit card and other transaction fees, and other costs. In particular, traffic acquisition costs as a percentage of advertising revenues may increase in the future if we are unable to continue to improve the monetization or generation of revenues from traffic on our websites and our Google Network Members’ websites. As we expand our advertising programs and other products to international markets, we continue to increase our exposure to fluctuations in foreign currency to U.S. dollar exchange rates. We have a foreign exchange risk management program that is designed to reduce our exposure to fluctuations in foreign currency exchange rates. However, this program will not fully offset the effect of fluctuations on our revenues and earnings. Results of Operations The following table presents our historical operating results as a percentage of revenues for the periods indicated: Year Ended December 31, 2009 2010 2011 Consolidated Statements of Income Data: Revenues 100.0%100.0%100.0% Costs and expenses: Cost of revenues 37.4 35.5 34.8 Research and development 12.0 12.8 13.6 Sales and marketing 8.4 9.5 12.1 General and administrative 7.0 6.8 7.2 Charge related to the resolution of Department of Justice investigation 0 0 1.3 Total costs and expenses 64.8 64.6 69.0 Income from operations 35.2 35.4 31.0 Interest and other income, net 0.3 1.4 1.5 Income before income taxes 35.5 36.8 32.5 Provision for income taxes 7.9 7.8 6.8 Net income 27.6%29.0%25.7% 29 Table of Contents Revenues The following table presents our revenues, by revenue source, for the periods presented (in millions): Year Ended December 31, 2009 2010 2011 Advertising revenues: Google websites$15,723$19,444$26,145 Google Network Members’ websites 7,166 8,792 10,386 Total advertising revenues 22,889 28,236 36,531 Other revenues 762 1,085 1,374 Revenues$23,651$29,321$37,905 The following table presents our revenues, by revenue source, as a percentage of total revenues for the periods presented: Year Ended December 31, 2009 2010 2011 Advertising revenues: Google websites 67%66%69% Google Network Members’ websites 30 30 27 Total advertising revenues 97 96 96 Google websites as % of advertising revenues 69 69 72 Google Network Members’ websites as % of advertising revenues 31 31 28 Other revenues 3%4%4% The increase in our revenues from 2010 to 2011 resulted primarily from an increase in advertising revenues generated by Google websites and Google Network Members’ websites. The increase in advertising revenues for Google websites and Google Network Members’ websites resulted primarily from an increase in the number of paid clicks through our advertising programs and, to a lesser extent, an increase in the average cost-per-click paid by our advertisers. The increase in the number of paid clicks generated through our advertising programs was due to an increase in aggregate traffic, certain monetization improvements including new ad formats, and the continued global expansion of our products, and our advertiser and user base, as well as an increase in the number of Google Network Members. The increase in the average cost-per-click paid by our advertisers was primarily driven by the increased spending from advertisers and a general weakening of the U.S dollar compared to foreign currencies (primarily the Euro, Japanese yen, and British pound), partially offset by the changes in geographical mix due to traffic growth in emerging markets, where the average cost-per-click is typically lower, compared to more mature markets. In addition, the increase in advertising revenues for Google Network Members’ websites from 2010 to 2011 was partially offset by the loss of a search partnership and, to a lesser extent, by a search quality improvement made during the first quarter of 2011. The increase in our revenues from 2009 to 2010 resulted primarily from an increase in advertising revenues generated by Google websites and Google Network Members’ websites and, to a lesser extent, an increase in other revenues, largely as a result of the launch of our mobile phone business in the first quarter of 2010. The increase in advertising revenue for Google websites and Google Network Members’ websites from 2009 to 2010 resulted primarily from an increase in the number of paid clicks through our advertising programs and, to a lesser extent, an increase in the average cost-per-click paid by our advertisers. The increase in the number of paid clicks generated through our advertising programs was due to an increase in aggregate traffic, certain monetization improvements, including new ad formats, and the continued global expansion of our products, and our advertiser and user base, as well as an increase in the number of Google Network Members. The increase in the average cost-per-click paid by our advertisers was primarily driven by the increased spending from advertisers, partially offset by the changes in geographical mix due to traffic growth in emerging markets, where the average cost-per-click is typically lower, compared to more mature markets. 30 Table of Contents Improvements in our ability to ultimately monetize increased traffic primarily relate to enhancing the end user experience, including providing end users with ads that are more relevant to their search queries or to the content on the Google Network Members’ websites they visit. For instance, these improvements include increasing site links to be full size links with the URL (uniform resource locator), moving a portion of the first line of the ad to the heading to better promote the content of the ad, providing an option to preview the ad,and moving the ad’s URL to a separate line below the heading for greater page format consistency. Aggregate paid clicks on Google websites and Google Network Members’ websites increased approximately 25% from 2010 to 2011 and approximately 16% from 2009 to 2010. Average cost-per-click on Google websites and Google Network Members’ websites increased approximately 3% from 2010 to 2011 and 5% from 2009 to 2010. The rate of change in aggregate paid clicks and average cost-per-click, and their correlation with the rate of change in revenues, has fluctuated and may fluctuate in the future because of various factors, including the revenue growth rates on our websites compared to those of our Google Network Members, advertiser competition for keywords, changes in foreign currency exchange rates, seasonality, the fees advertisers are willing to pay based on how they manage their advertising costs, changes in advertising quality or formats, and general economic conditions. In addition, traffic growth in emerging markets compared to more mature markets and across various advertising verticals and channels also contributes to these fluctuations. Changes in aggregate paid clicks and average cost-per-click may not be indicative of our performance or advertiser experiences in any specific geographic market, vertical, or industry. We believe that the increase in the number of paid clicks on Google websites and Google Network Members’ websites is substantially the result of our commitment to improving the relevance and quality of both our search results and the advertisements displayed, which we believe results in a better user experience, which in turn results in more searches, advertisers, and Google Network Members and other partners. Revenues by Geography The following table presents our domestic and international revenues as a percentage of consolidated revenues, determined based on the billing addresses of our customers: Year Ended December 31, 2009 2010 2011 United States 47%48%46% United Kingdom 13%11%11% Rest of the world 40%41%43% The growth in international revenues (other than the United Kingdom) as a percentage of consolidated revenues from 2010 to 2011 resulted largely from increased acceptance of our advertising programs, our continued progress in developing localized versions of our products for these international markets, and, to a lesser extent, a general weakening of the U.S dollar compared to foreign currencies (primarily the Euro and Japanese yen). The growth in international revenues (other than the United Kingdom) as a percentage of consolidated revenues from 2009 to 2010 resulted largely from increased acceptance of our advertising programs and our continued progress in developing localized versions of our products for these international markets. The growth in revenues from the United States as a percentage of consolidated revenues from 2009 to 2010 resulted largely from the recovery of certain key verticals such as retail, travel, finance, and insurance. The decline in revenues from the United Kingdom as a percentage of consolidated revenues from 2009 to 2010 resulted primarily from a less robust economic recovery relative to the United States and the rest of the world, as well as a decrease in hedging gains recognized during 2010. The general weakening of the U.S. dollar relative to certain foreign currencies (primarily the Euro, Japanese yen, and British pound) from 2010 to 2011 had a favorable impact on our international revenues. Had foreign 31 Table of Contents exchange rates remained constant in these periods, our revenues from the United Kingdom would have been $129 million, or 3.2%, lower and our revenues from the rest of the world would have been approximately $834 million, or 5.1%, lower in 2011. This is before consideration of hedging gains of $9 million and $34 million recognized to revenues from the United Kingdom and the rest of the world in 2011. The general strengthening of the U.S. dollar relative to certain foreign currencies (primarily the Euro) from 2009 to 2010 had an unfavorable impact on our international revenues. Had foreign exchange rates remained constant in these periods, our revenues from the United Kingdom would have been approximately $10 million, or 0.3% higher, and our revenues from the rest of the world would have been approximately $25 million, or 0.2% higher in 2010. This is before consideration of hedging gains of $42 million and $161 million recognized to revenues from the United Kingdom and the rest of the world in 2010. Although we expect to continue to make investments in international markets, these investments may not result in an increase in our international revenues as a percentage of total revenues in 2012 or thereafter. See Note 16 of Notes to Consolidated Financial Statements included in Item 8 of this Annual Report on Form 10-K for additional information about geographic areas. Costs and Expenses Cost of Revenues Cost of revenues consists primarily of traffic acquisition costs. Traffic acquisition costs consist of amounts ultimately paid to our Google Network Members under AdSense arrangements and to certain other partners (our distribution partners) who distribute our toolbar and other products (collectively referred to as access points) or otherwise direct search queries to our website (collectively referred to as distribution arrangements). These amounts are primarily based on the revenue share and fixed fee arrangements with our Google Network Members and distribution partners. Certain distribution arrangements require us to pay our partners based on a fee per access point delivered and not exclusively—or at all—based on revenue share. These fees are non-refundable. Further, these arrangements are terminable at will, although under the terms of certain contracts we or our distribution partners may be subject to penalties in the event of early termination. We recognize fees under these arrangements over the estimated useful lives of the access points (approximately two years) to the extent we can reasonably estimate those lives and they are longer than one year, or based on any contractual revenue share, if greater. Otherwise, the fees are charged to expense as incurred. The estimated useful life of the access points is based on the historical average period of time they generate traffic and revenues. Cost of revenues also includes the expenses associated with the operation of our data centers, including depreciation, labor, energy, and bandwidth costs, credit card and other transaction fees related to processing customer transactions including Google Checkout transactions, amortization of acquired intangible assets, as well as content acquisition costs. We have entered into arrangements with certain content providers under which we distribute or license their video and other content. In a number of these arrangements, we display ads on the pages of our websites from which the content is viewed and share most of the fees these ads generate with the content providers. To the extent we are obligated to make guaranteed minimum revenue share payments to our content providers, we recognize as content acquisition costs the contractual revenue share amount or on a straight-line basis, whichever is greater, over the terms of the agreements. 32 Table of Contents The following tables present our cost of revenues and cost of revenues as a percentage of revenues, and our traffic acquisition costs, and traffic acquisition costs as a percentage of advertising revenues, for the periods presented (dollars in millions): Year Ended December 31, 2009 2010 2011 Cost of revenues$8,844$10,417$13,188 Cost of revenues as a percentage of revenues 37.4%35.5%34.8% Year Ended December 31, 2009 2010 2011 Traffic acquisition costs related to AdSense arrangements$5,265$6,162$7,294 Traffic acquisition costs related to distribution arrangements 904 1,155 1,517 Total traffic acquisition costs$6,169$7,317$8,811 Traffic acquisition costs as a percentage of advertising revenues 27.0%25.9%24.1% Cost of revenues increased $2,771 million from 2010 to 2011. The increase was primarily related to an increase in traffic acquisition costs of $1,132 million resulting from more advertiser fees generated through our AdSense program. The increase was also related to an increase in traffic acquisition costs of $362 million from our distribution arrangements as a result of more traffic directed to our websites, as well as more distribution fees paid. The decrease in traffic acquisition costs as a percentage of advertising revenues was primarily due to an increase in the proportion of advertising revenues from our websites compared to our Google Network Members’ websites, more revenues realized from Google Network Members to whom we pay less revenue share, and, to a lesser extent, expiration of an AdSense arrangement under which we paid guaranteed minimum revenue share. In addition, there was an increase in data center costs of $784 million, primarily resulting from the depreciation of additional information technology assets and data center buildings and an increase in labor, energy, and bandwidth costs, and an increase in content acquisition costs of $236 million, primarily related to content displayed on YouTube, partially offset by a decrease in mobile phone costs. Cost of revenues increased $1,573 million from 2009 to 2010. The increase was primarily related to an increase in traffic acquisition costs of $897 million resulting largely from more advertiser fees generated through our AdSense program. The increase was also related to an increase in traffic acquisition costs of $251 million from our distribution arrangements as a result of more traffic directed to our websites, as well as more distribution fees paid. The decrease in traffic acquisition costs as a percentage of advertising revenues was primarily due to more revenues realized from Google Network Members to whom we pay less revenue share, and expiration of an AdSense arrangement under which we paid guaranteed minimum revenue share. In addition, there was an increase in content acquisition costs of $169 million primarily related to content displayed on YouTube and an increase in mobile phone costs. We expect cost of revenues will increase in dollar amount and may increase as a percentage of revenues in 2012 and in future periods, primarily as a result of forecasted increases in traffic acquisition costs, data center costs, credit card and other transaction fees, content acquisition costs, and other costs. Traffic acquisition costs as a percentage of advertising revenues may fluctuate in the future based on a number of factors, including the following: •The relative growth rates of revenues from our websites and from our Google Network Members’ websites. •Whether we are able to enter into more AdSense arrangements that provide for lower revenue share obligations or whether increased competition for arrangements with existing and potential Google Network Members results in less favorable revenue share arrangements. •Whether we are able to continue to improve the monetization of traffic on our websites and our Google Network Members’ websites. 33 Table of Contents •The relative growth rates of expenses associated with distribution arrangements and the related revenues generated, including whether we share with certain existing and new distribution partners proportionately more of the aggregate advertising fees that we earn from paid clicks derived from search queries these partners direct to our websites. Research and Development The following table presents our research and development expenses, and research and development expenses as a percentage of our revenues for the periods presented (dollars in millions): Year Ended December 31, 2009 2010 2011 Research and development expenses$2,843$3,762$5,162 Research and development expenses as a percentage of revenues 12.0%12.8%13.6% Research and development expenses consist primarily of compensation and related costs for personnel responsible for the research and development of new and existing products and services. We expense research and development costs as they are incurred. Research and development expenses increased $1,400 million from 2010 to 2011. This increase was primarily due to an increase in labor and facilities-related costs of $875 million, largely as a result of a 23% increase in research and development headcount, including headcount from acquisitions, as well as an increase in employee base salaries of approximately 10%. In addition, there was an increase in stock-based compensation expense of $200 million. Research and development expenses increased $919 million from 2009 to 2010. This increase was primarily due to an increase in labor-related costs of $578 million, largely as a result of a 28% increase in research and development headcount, including headcount from acquisitions. In addition, there was an increase in stock-based compensation expense of $136 million. We expect that research and development expenses will increase in dollar amount and may increase as a percentage of revenues in 2012 and future periods because we expect to continue to invest in building the necessary employee and systems infrastructures required to support the development of new, and improve existing, products and services. Sales and Marketing The following table presents our sales and marketing expenses, and sales and marketing expenses as a percentage of revenues for the periods presented (dollars in millions): Year Ended December 31, 2009 2010 2011 Sales and marketing expenses$1,984$2,799$4,589 Sales and marketing expenses as a percentage of revenues 8.4%9.5%12.1% Sales and marketing expenses consist primarily of compensation and related costs for personnel engaged in customer service, sales, and sales support functions, as well as advertising and promotional expenditures. Sales and marketing expenses increased $1,790 million from 2010 to 2011. This increase was primarily due to an increase in labor and facilities-related costs of $787 million, largely as a result of a 36% increase in sales and marketing headcount, including headcount from acquisitions, as well as an increase in employee base salaries of approximately 10%. In addition, there was an increase in advertising and promotional expense of $700 million. 34 Table of Contents Sales and marketing expenses increased $815 million from 2009 to 2010. This increase was primarily due to an increase in advertising and promotional expense of $387 million. In addition, there was an increase in labor-related costs of $254 million, largely as a result of a 20% increase in sales and marketing headcount, and higher commission expense. We expect that sales and marketing expenses will increase in dollar amount and may increase as a percentage of revenues in 2012 and future periods, as we expand our business globally, increase advertising and promotional expenditures in connection with new and existing products, and increase the level of service we provide to our advertisers, Google Network Members, and other partners. General and Administrative The following table presents our general and administrative expenses, and general and administrative expenses as a percentage of revenues for the periods presented (dollars in millions): Year Ended December 31, 2009 2010 2011 General and administrative expenses$1,668$1,962$2,724 General and administrative expenses as a percentage of revenues 7.0%6.8%7.2% General and administrative expenses consist primarily of compensation and related costs for personnel and facilities, and include costs related to our facilities, finance, human resources, information technology, and legal organizations, and fees for professional services. Professional services are principally comprised of outside legal, audit, information technology consulting, and outsourcing services. General and administrative expenses increased $762 million from 2010 to 2011. This increase was primarily due to an increase in labor and facilities-related costs of $350 million, primarily as a result of a 37% increase in general and administrative headcount and an increase in employee base salaries of approximately 10%, as well as an increase in expense related to professional services of $260 million, the majority of which were related to consulting services and legal costs. In addition, there was an increase in stock-based compensation of $116 million. General and administrative expenses increased $294 million from 2009 to 2010. This increase was primarily due to increases in expense related to professional services of $137 million, the majority of which were related to legal and temporary services costs. In addition, there was an increase in labor-related costs of $64 million largely as a result of a 14% increase in general and administrative headcount, and an increase in charitable contributions of $64 million. As we expand our business and incur additional expenses, we expect general and administrative expenses will increase in dollar amount and may increase as a percentage of revenues in 2012 and future periods. Charge Related to the Resolution of Department of Justice Investigation In connection with a resolution of an investigation by the United States Department of Justice into the use of Google advertising by certain advertisers, we accrued $500 million during the three months ended March 31, 2011, which was paid in August 2011 upon final resolution of that matter. Stock-Based Compensation The following table presents our stock-based compensation, and stock-based compensation as a percentage of revenues for the periods presented (dollars in millions): Year Ended December 31, 2009 2010 2011 Stock-based compensation$1,164$1,376$1,974 Stock-based compensation as a percentage of revenues 4.9%4.7%5.2% 35 Table of Contents Stock-based compensation increased $598 million from 2010 to 2011. This increase was largely due to additional stock awards issued to existing and new employees. Stock-based compensation increased $212 million from 2009 to 2010. This increase was largely due to additional stock awards issued to existing and new employees. We estimate stock-based compensation to be approximately $2.0 billion in 2012 and $2.4 billion thereafter. This estimate does not include expenses to be recognized related to employee stock awards that are granted after December 31, 2011 or non-employee stock awards that have been or may be granted.In addition, to the extent forfeiture rates are different from what we have anticipated, stock-based compensation related to these awards will be different from our expectations. Interest and Other Income, Net Interest and other income, net increased $169 million from 2010 to 2011. This increase was primarily driven by an increase in interest income of $233 million due to an increase in our cash and investment balances and higher yields, as well as an increase in net realized gains on sales of available-for-sale investments of $69 million, partially offset by an increase in interest expense of $53 million primarily related to our long-term debt program. In addition, we recorded an impairment charge of $110 million related to certain equity investments during the year ended December 31, 2011. Interest and other income, net increased $346 million from 2009 to 2010. This increase was primarily driven by an increase in interest income of $349 million due to an increase in our cash and investment balances and higher yields, as well as an increase in net realized gains on sales of available-for-sale investments of $88 million. These increases were partially offset by an increase in net foreign exchange related costs of $95 million primarily related to our foreign exchange risk management program. The costs of our foreign exchange hedging activities that we recognized to interest and other income, net are primarily a function of the notional amount of the option and forward contracts and their related duration, the movement of the foreign exchange rates relative to the strike prices of the contracts, as well as the volatility of the foreign exchange rates. As we expand our international business, we believe costs related to hedging activities under our foreign exchange risk management program may increase in dollar amount in 2012 and future periods. Provision for Income Taxes The following table presents our provision for income taxes, and effective tax rate for the periods presented (dollars in millions): Year Ended December 31, 2009 2010 2011 Provision for income taxes$1,861$2,291$2,589 Effective tax rate 22.2%21.2%21.0% Our provision for income taxes increased from 2010 to 2011, primarily as a result of increases in federal income taxes, driven by higher taxable income year over year, partially offset by proportionately more earnings realized in countries that have lower statutory tax rates.Our effective tax rate decreased from 2010 to 2011, primarily as a result of proportionately more earnings realized in countries that have lower statutory tax rates, a decrease in state income taxes, and an increase in federal research and development credits recognized in 2011, partially offset by recognition of a charge related to the resolution of an investigation by the Department of Justice which is not deductible for tax purposes. 36 Table of Contents Our provision for income taxes increased from 2009 to 2010, primarily as a result of increases in federal and state income taxes, driven by higher taxable income year over year.Our effective tax rate decreased from 2009 to 2010, primarily because we released certain tax reserves as a result of the settlement of our 2005 and 2006 tax audits in 2010. This decrease was partially offset by proportionately higher earnings in countries where we have higher statutory tax rates. Our effective tax rate could fluctuate significantly on a quarterly basis and could be adversely affected to the extent earnings are lower than anticipated in countries that have lower statutory rates and higher than anticipated in countries that have higher statutory rates. Our effective tax rate could also fluctuate due to the net gains and losses recognized by legal entities on certain hedges and related hedged intercompany and other transactions under our foreign exchange risk management program, by changes in the valuation of our deferred tax assets or liabilities, or by changes in tax laws, regulations, or accounting principles, as well as certain discrete items. In addition, we are subject to the continuous examination of our income tax returns by the Internal Revenue Service (IRS) and other tax authorities. We regularly assess the likelihood of adverse outcomes resulting from these examinations to determine the adequacy of our provision for income taxes. See Critical Accounting Policies and Estimates below for additional information about our provision for income taxes. A reconciliation of the federal statutory income tax rate to our effective tax rate is set forth in Note 15 of Notes to Consolidated Financial Statements included in Item 8 of this Annual Report on Form 10-K. Quarterly Results of Operations You should read the following tables presenting our quarterly results of operations in conjunction with the consolidated financial statements and related notes included in Item 8 of this Annual Report on Form 10-K. We have prepared the unaudited information on the same basis as our audited consolidated financial statements. You should also keep in mind that our operating results for any quarter are not necessarily indicative of results for any future quarters or for a full year. 37 Table of Contents The following table presents our unaudited quarterly results of operations for the eight quarters ended December 31, 2011. This table includes all adjustments, consisting only of normal recurring adjustments, that we consider necessary for fair presentation of our consolidated financial position and operating results for the quarters presented. Both seasonal fluctuations in internet usage and traditional retail seasonality have affected, and are likely to continue to affect, our business. Internet usage generally slows during the summer months, and commercial queries typically increase significantly in the fourth quarter of each year. These seasonal trends have caused and will likely continue to cause, fluctuations in our quarterly results, including fluctuations in sequential revenue growth rates. Quarter Ended Mar 31, 2010 Jun 30, 2010 Sep 30, 2010 Dec 31, 2010 Mar 31, 2011 Jun 30, 2011 Sep 30, 2011 Dec 31, 2011 (In millions, except per share amounts) (unaudited) Consolidated Statements of Income Data: Revenues$6,775$6,820$7,286$8,440$8,575$9,026$9,720$10,584 Costs and expenses: Cost of revenues 2,452 2,467 2,552 2,946 2,936 3,172 3,378 3,702 Research and development 818 898 994 1,051 1,226 1,234 1,404 1,298 Sales and marketing 607 629 661 902 1,026 1,091 1,204 1,268 General and administrative 410 461 532 559 591 648 676 809 Charge related to the resolution of Department of Justice investigation 0 0 0 0 500 0 0 0 Total costs and expenses 4,287 4,455 4,739 5,458 6,279 6,145 6,662 7,077 Income from operations 2,488 2,365 2,547 2,982 2,296 2,881 3,058 3,507 Interest and other income (expense), net 18 69 167 160 96 204 302(18) Income before income taxes 2,506 2,434 2,714 3,142 2,392 3,085 3,360 3,489 Provision for income taxes 551 594 547 599 594 580 631 784 Net income$1,955$1,840$2,167$2,543$1,798$2,505$2,729$2,705 Net income per share: Basic$6.15$5.78$6.80$7.95$5.59$7.77$8.44$8.34 Diluted$6.06$5.71$6.72$7.81$5.51$7.68$8.33$8.22 38 Table of Contents The following table presents our unaudited quarterly results of operations as a percentage of revenues for the eight quarters ended December 31, 2011: Quarter Ended Mar 31, 2010 Jun 30, 2010 Sep 30, 2010 Dec 31, 2010 Mar 31, 2011 Jun 30, 2011 Sep 30, 2011 Dec 31, 2011 Revenues 100.0%100.0%100.0%100.0%100.0%100.0%100.0%100.0% Costs and expenses: Cost of revenues 36.2 36.2 35.0 34.9 34.2 35.1 34.8 35.0 Research and development 12.1 13.2 13.6 12.5 14.3 13.7 14.4 12.3 Sales and marketing 9.0 9.2 9.1 10.7 12.0 12.1 12.4 12.0 General and administrative 6.0 6.7 7.3 6.6 6.9 7.2 6.9 7.6 Charge related to the resolution of Department of Justice investigation 0 0 0 0 5.8 0 0 0 Total costs and expenses 63.3 65.3 65.0 64.7 73.2 68.1 68.5 66.9 Income from operations 36.7 34.7 35.0 35.3 26.8 31.9 31.5 33.1 Interest and other income (expense), net 0.3 1.0 2.2 1.9 1.1 2.3 3.1(0.1) Income before income taxes 37.0 35.7 37.2 37.2 27.9 34.2 34.6 33.0 Provision for income taxes 8.1 8.7 7.5 7.1 6.9 6.4 6.5 7.4 Net income 28.9%27.0%29.7%30.1%21.0%27.8%28.1%25.6% Liquidity and Capital Resources In summary, our cash flows are as follows (in millions): Year Ended December 31, 2009 2010 2011 Net cash provided by operating activities$9,316$11,081$14,565 Net cash used in investing activities(8,019)(10,680)(19,041) Net cash provided by financing activities 233 3,050 807 At December 31, 2011, we had $44.6 billion of cash, cash equivalents, and marketable securities. Cash equivalents and marketable securities are comprised of time deposits, money market and other funds, including cash collateral received related to our securities lending program, highly liquid debt instruments of the U.S. government and its agencies, debt instruments issued by foreign governments, and municipalities in the U.S., corporate securities, and mortgage-backed securities. As of December 31, 2011, $21.2 billion of the $44.6 billion of cash, cash equivalents, and marketable securities was held by our foreign subsidiaries. If these funds are needed for our operations in the U.S., we would be required to accrue and pay U.S. taxes to repatriate these funds. However, our intent is to permanently reinvest these funds outside of the U.S. and our current plans do not demonstrate a need to repatriate them to fund our U.S. operations. Our principal sources of liquidity are our cash, cash equivalents, and marketable securities, as well as the cash flow that we generate from our operations. At December 31, 2011, we had unused letters of credit for approximately $46 million. We believe that our sources of funding will be sufficient to satisfy our currently anticipated cash requirements through at least the next 12 months. Our liquidity could be negatively affected by a decrease in demand for our products and services. In addition, we may make acquisitions or license products and technologies complementary to our business and may need to raise additional capital through future debt or equity financing to provide for greater flexibility to fund any such acquisitions and licensing activities. Additional financing may not be available at all or on terms favorable to us. 39 Table of Contents We have a debt financing program of up to $3.0 billion through the issuance of commercial paper. Net proceeds from this program are used for general corporate purposes. As of December 31, 2011, we had $750 million of commercial paper outstanding recorded as short-term debt, with a weighted-average interest rate of 0.1%. Average commercial paper borrowings during the year were $1.4 billion and the maximum amount outstanding during the year was $3.0 billion.In conjunction with this program, we have a $3.0 billion revolving credit facility expiring in July 2016. The interest rate for the credit facility is determined based on a formula using certain market rates. As of December 31, 2011, we were in compliance with the financial covenant in the credit facility. No amounts were outstanding under the credit facility as of December 31, 2011. Additionally, as of December 31, 2011, we had a $468 million secured promissory note outstanding recorded as short-term debt, with an interest rate of 1.0% that matures in December 2012. In May 2011, we issued $3.0 billion of unsecured senior notes in three equal tranches, due in 2014,2016, and 2021, with stated interest rates of 1.25%, 2.125%, and 3.625%. The net proceeds from the sale of the notes were used to repay a portion of our outstanding commercial paper and for general corporate purposes. As of December 31, 2011, the total carrying value and estimated fair value of these notes were $3.0 billion and $3.2 billion. The estimated fair value was based on quoted prices for our publicly-traded debt as of December 31, 2011. We are not subject to any financial covenants under the notes. During 2011, we paid $35 million in interest payments related to these notes. In August 2011, we entered into a Merger Agreement with Motorola, a provider of innovative technologies, products and services that enable a range of mobile and wireline digital communication, information and entertainment experiences, under which we will acquire Motorola for $40 per share in cash, or a total of approximately $12.5 billion. The completion of this transaction is subject to customary closing conditions, including the receipt of certain regulatory approvals.In the event the Merger Agreement is terminated due to a failure to obtain certain regulatory approvals,we would be required to pay Motorola a fee of $2.5 billion.The transaction is currently expected to close in early 2012. Cash provided by operating activities consist of net income adjusted for certain non-cash items, including amortization, depreciation, deferred income taxes, excess tax benefits from stock-based award activities, stock-based compensation expense, and impairment of equity investments, as well as the effect of changes in working capital and other activities. Cash provided by operating activities in 2011 was $14,565 million and consisted of net income of $9,737 million, adjustments for non-cash items of $4,198 million, and cash provided by working capital and other activities of $630 million. Adjustments for non-cash items primarily consisted of $1,974 million of stock-based compensation expense, $1,396 million of depreciation and amortization expense of property and equipment, $455 million of amortization of intangible and other assets, $343 million of deferred income taxes, and $110 million related to impairment of equity investments. In addition, the increase in cash from changes in working capital activities primarily consisted of an increase in accrued expenses and other liabilities of $795 million, a net increase in income taxes payable and deferred income taxes of $731 million, an increase in accrued revenue share of $259 million, an increase of $162 million in deferred revenue, and an increase of $101 million in accounts payable. These increases were partially offset by an increase in accounts receivable of $1,156 million due to the growth in fees billed to our advertisers, and an increase in prepaid revenue share, expenses and other assets of $262 million. The increase in income taxes payable and deferred income taxes reflected primarily additional tax obligations accrued, partially offset by estimated income taxes paid during 2011. In addition, we paid $500 million related to the resolution of a Department of Justice investigation during the year. Cash provided by operating activities in 2010 was $11,081 million, and consisted of net income of $8,505 million, adjustments for non-cash items of $2,675 million, and cash used in working capital and other activities of $99 million. Adjustments for non-cash items primarily consisted of $1,376 million of stock-based compensation expense, $1,067 million of depreciation and amortization expense on property and equipment, and $329 million of amortization of intangible and other assets, partially offset by $94 million of excess tax benefits from stock-based 40 Table of Contents award activities. In addition, the decrease in cash from changes in working capital activities primarily consisted of an increase of $1,129 million in accounts receivable due to the growth in fees billed to our advertisers and an increase of $414 million in prepaid revenue share, expenses and other assets. These increases were partially offset by an increase in accrued expenses and other liabilities of $745 million, an increase in accounts payable of $272 million, an increase in accrued revenue share of $214 million, an increase in deferred revenue of $111 million, and a net increase in income tax payable and deferred income taxes of $102 million, which includes the same $94 million of excess tax benefits from stock-based award activities included under adjustments for non-cash items. The increase in accrued expense and other liabilities, accounts payable, accrued revenue share, and deferred revenues are primarily a result of the growth in our business and headcount. The increase in net income taxes payable and deferred income taxes was primarily a result of additional tax obligations accrued, partially offset by the release of certain tax reserves as a result of the settlement of our tax audits for our 2005 and 2006 tax years. Cash provided by operating activities in 2009 was $9,316 million, and consisted of net income of $6,520 million, adjustments for non-cash items of $2,310 million, and cash provided by working capital and other activities of $486 million. Adjustments for non-cash items primarily consisted of $1,240 million of depreciation and amortization expense on property and equipment, $1,164 million of stock-based compensation expense, and $284 million of amortization of intangible and other assets, partially offset by $268 million of deferred income taxes on earnings and $90 million of excess tax benefits from stock-based award activities. In addition, changes in working capital activities primarily consisted of a decrease of $262 million in prepaid revenue share, expenses, and other assets, an increase in accrued expenses and other liabilities of $243 million which is a direct result of the growth of our business, and a net increase in income taxes payable and deferred income taxes of $217 million, which includes the same $90 million of excess tax benefits from stock-based award activities included under adjustments for non-cash items, and an increase in accrued revenue share of $158 million. These increases were partially offset by an increase of $504 million in accounts receivable due to the growth in fees billed to our advertisers. The increase in net income taxes payable and deferred income taxes was primarily a result of additional tax obligations accrued, partially offset by an increase in the amount of estimated income taxes we paid during the year. The increase in accrued revenue share was due to the growth in our AdSense and distribution programs and the timing of payments made to our partners. As we expand our business internationally, we have offered payment terms to certain advertisers that are standard in their locales but longer than terms we would generally offer to our domestic advertisers. This may increase our working capital requirements and may have a negative effect on cash provided by our operating activities. Cash used in investing activities in 2011 of $19,041 million was primarily attributable to net purchases of marketable securities of $12,926 million, capital expenditures of $3,438 million related principally to our facilities, data centers, and related equipment, and cash consideration used in acquisitions and other investments of $2,328 million, including $676 million paid in connection with the acquisition of ITA. Also, in connection with our securities lending program, we returned $354 million of cash collateral. See Note 3 of Notes to Consolidated Financial Statements included in Item 8 of this Annual Report on Form 10-K for further information about our securities lending program. Cash used in investing activities in 2010 of $10,680 million was primarily attributable to net purchases of marketable securities of $6,886 million, capital expenditures of $4,018 million of which $1.8 billion was for the purchase of an office building in New York City in December 2010, and remaining amounts related principally to our data centers and related equipment, and cash consideration used in acquisitions and other investments of $1,067 million. Also, in connection with our securities lending program, we received $2,361 million of cash collateral which was invested in reverse repurchase agreements. Of the $2,361 million, $1,611 million was classified as cash and cash equivalents, and $750 million was classified as receivable under reverse repurchase agreements in the accompanying Consolidated Balance Sheet. Cash used in investing activities in 2009 of $8,019 million was primarily attributable to net purchases of marketable securities of $7,036 million and capital expenditures of $810 million. 41 Table of Contents In order to manage expected increases in internet traffic, advertising transactions, and new products and services, and to support our overall global business expansion, we expect to make significant investments in our systems, data centers, corporate facilities, information technology infrastructure, and employees in 2012 and thereafter. However, the amount of our capital expenditures has fluctuated and may continue to fluctuate on a quarterly basis. In addition, we expect to spend a significant amount of cash on acquisitions and other investments from time to time. These acquisitions generally enhance the breadth and depth of our expertise in engineering and other functional areas, our technologies, and our product offerings. Cash provided by financing activities in 2011 of $807 million was primarily driven by net proceeds of $726 million of debt issued and excess tax benefits from stock-based award activities of $86 million. Cash provided by financing activities in 2010 of $3,050 million was primarily driven by $3,463 million of net cash proceeds from the issuance of commercial paper and a promissory note. This was partially offset by $801 million in stock repurchases in connection with our acquisitions of AdMob and On2 Technologies, Inc., as well as net proceeds from stock-based award activities of $294 million, and excess tax benefits from stock-based award activities of $94 million. Cash provided by financing activities in 2009 of $233 million was primarily due to net proceeds related to stock-based award activities of $143 million. In addition, there were excess tax benefits of $90 million from stock-based award activities during the period which represented a portion of the $260 million reduction to income taxes payable that we recorded in 2009 related to the total direct tax benefit realized from the exercise, sale, or vesting of these awards. Contractual Obligations as of December 31, 2011 Payments due by period Total Less than 1 year 1-3 years 3-5 years More than 5 years (in millions) Operating lease obligations, net of sublease income amounts$2,879$372$704$544$1,259 Purchase obligations 1,910 910 860 37 103 Long-term debt obligations 3,471 70 1,134 1,104 1,163 Other long-term liabilities reflected on our balance sheet 328 106 187 7 28 Total contractual obligations$8,588$1,458$2,885$1,692$2,553 The above table does not include future rental income of $726 million related to the leases that we assumed in connection with our building purchases. Operating Leases We have entered into various non-cancelable operating lease agreements for certain of our offices, land, and data centers throughout the world with original lease periods expiring primarily between 2012 and 2063. We are committed to pay a portion of the related operating expenses under certain of these lease agreements. These operating expenses are not included in the above table. Certain of these leases have free or escalating rent payment provisions. We recognize rent expense under such leases on a straight-line basis over the term of the lease. Certain leases have adjustments for market provisions. 42 Table of Contents Purchase Obligations Purchase obligations represent non-cancelable contractual obligations at December 31, 2011. These contracts are primarily related to distribution arrangements, video and other content licensing revenue sharing arrangements, as well as data center operations and facility build-outs. In addition, we had $2.8 billion of open purchase orders for which we have not received the related services or goods at December 31, 2011. This amount is not included in the above table because we have the right to cancel the purchase orders prior to the date of delivery. Long-term Debt Obligations Long-term debt obligations represent principal and interest payments to be made over the life of our unsecured senior notes issued in May 2011. Please see Note 4 of the Notes to Consolidated Financial Statements included in Item 8 of this Annual Report on Form 10-K for further details. Other Long-Term Liabilities Other long-term liabilities consist of cash obligations, primarily the legal settlement with the Authors Guild and the Association of American Publishers (AAP), and milestone and royalty payments owed in connection with certain acquisitions and licensing agreements. In addition to the amounts above, we recorded additional long-term taxes payable of $438 million in 2011 related to tax positions for which the timing of the ultimate resolution is uncertain. At this time, we are unable to make a reasonably reliable estimate of the timing of payments in individual years beyond 12 months due to uncertainties in the timing of tax audit outcomes. As a result, this amount is not included in the above table. Off-Balance Sheet Entities At December 31, 2011, we did not have any off-balance sheet arrangements, as defined in Item 303(a)(4)(ii) of Regulation S-K promulgated by the SEC, that have or are reasonably likely to have a current or future effect on our financial condition, changes in our financial condition, revenues, or expenses, results of operations, liquidity, capital expenditures, or capital resources that is material to investors. Critical Accounting Policies and Estimates We prepare our consolidated financial statements in accordance with accounting principles generally accepted in the U.S. (U.S. GAAP). In doing so, we have to make estimates and assumptions that affect our reported amounts of assets, liabilities, revenues, and expenses, as well as related disclosure of contingent assets and liabilities. In some cases, we could reasonably have used different accounting policies and estimates. In some cases, changes in the accounting estimates are reasonably likely to occur from period to period. Accordingly, actual results could differ materially from our estimates. To the extent that there are material differences between these estimates and actual results, our financial condition or results of operations will be affected. We base our estimates on past experience and other assumptions that we believe are reasonable under the circumstances, and we evaluate these estimates on an ongoing basis. We refer to accounting estimates of this type as critical accounting policies and estimates, which we discuss further below. We have reviewed our critical accounting policies and estimates with the audit committee of our board of directors. Income Taxes We are subject to income taxes in the U.S. and numerous foreign jurisdictions. Significant judgment is required in evaluating our uncertain tax positions and determining our provision for income taxes. 43 Table of Contents Although we believe we have adequately reserved for our uncertain tax positions, no assurance can be given that the final tax outcome of these matters will not be different. We adjust these reserves in light of changing facts and circumstances, such as the closing of a tax audit or the refinement of an estimate. To the extent that the final tax outcome of these matters is different than the amounts recorded, such differences will impact the provision for income taxes in the period in which such determination is made. The provision for income taxes includes the impact of reserve provisions and changes to reserves that are considered appropriate, as well as the related net interest. Our effective tax rates have differed from the statutory rate primarily due to the tax impact of foreign operations, state taxes, certain benefits realized related to stock-based award activities, and research and experimentation tax credits. The effective tax rates were 22.2%, 21.2%, and 21.0% for 2009, 2010, and 2011. Our future effective tax rates could be adversely affected by earnings being lower than anticipated in countries that have lower statutory rates and higher than anticipated in countries that have higher statutory rates, the net gains and losses recognized by legal entities on certain hedges and related hedged intercompany and other transactions under our foreign exchange risk management program, changes in the valuation of our deferred tax assets or liabilities, or changes in tax laws, regulations, or accounting principles, as well as certain discrete items. In addition, we are subject to the continuous examination of our income tax returns by the IRS and other tax authorities. We regularly assess the likelihood of adverse outcomes resulting from these examinations to determine the adequacy of our provision for income taxes. Loss Contingencies We are regularly subject to claims, suits, government investigations, and other proceedings involving competition and antitrust, intellectual property, privacy, tax, labor and employment, commercial disputes, content generated by our users, goods and services offered by advertisers or publishers using our platforms, and other matters. Certain of these matters include speculative claims for substantial or indeterminate amounts of damages. We record a liability when we believe that it is both probable that a loss has been incurred, and the amount can be reasonably estimated. We evaluate, on a monthly basis, developments in our legal matters that could affect the amount of liability that has been previously accrued, and make adjustments as appropriate. Significant judgment is required to determine both likelihood of there being and the estimated amount of a loss related to such matters. Until the final resolution of such matters, there may be an exposure to loss in excess of the amount recorded, and such amounts could be material. Should any of our estimates and assumptions change or prove to have been incorrect, it could have a material impact on our business, consolidated financial position, results of operations, or cash flows. See Note 12 of Notes to Consolidated Financial Statements included in Item 8 of this Annual Report on Form 10-K for additional information regarding contingencies. Stock-Based Compensation Our stock-based compensation expense for stock options is estimated at the grant date based on the award’s fair value as calculated by the Black-Scholes-Merton (BSM) option pricing model and is recognized as expense over the requisite service period. The BSM model requires various highly judgmental assumptions including expected volatility and expected term. If any of the assumptions used in the BSM model changes significantly, stock-based compensation expense may differ materially in the future from that recorded in the current period. In addition, we are required to estimate the expected forfeiture rate and only recognize expense for those shares expected to vest. We estimate the forfeiture rate based on historical experience and our expectations regarding future pre-vesting termination behavior of employees. To the extent our actual forfeiture rate is different from our estimate, stock-based compensation expense is adjusted accordingly. Impairment of Marketable and Non-Marketable Securities We periodically review our marketable securities and our non-marketable equity securities for impairment. If we conclude that any of these investments are impaired, we determine whether such impairment is other-than-temporary. Factors we consider to make such determination include the duration and severity of the impairment, 44 Table of Contents the reason for the decline in value, the potential recovery period, and our intent to sell, or whether it is more likely than not that we will be required to sell, the investment before recovery. If any impairment is considered other-than-temporary, we will write down the asset to its fair value and take a corresponding charge to our Consolidated Statements of Income. Recent Accounting Pronouncements In June 2011, the Financial Accounting Standards Board (FASB) issued an amendment to an existing accounting standard which requires companies to present net income and other comprehensive income in one continuous statement or in two separate, but consecutive, statements.In addition, in December 2011, the FASB issued an amendment to an existing accounting standard which defers the requirement to present components of reclassifications of other comprehensive income on the face of the income statement.We adopted both standards in the fourth quarter of 2011. In September 2011, the FASB issued an amendment to an existing accounting standard, which provides entities an option to perform a qualitative assessment to determine whether further impairment testing on goodwill is necessary. Specifically, an entity has the option to first assess qualitative factors to determine whether it is necessary to perform the current two-step test.If an entity believes, as a result of its qualitative assessment, that it is more-likely-than-not that the fair value of a reporting unit is less than its carrying amount, the quantitative impairment test is required.Otherwise, no further testing is required. This standard is effective for annual and interim goodwill impairment tests performed for fiscal years beginning after December 15, 2011.We adopted this standard in the first quarter of 2012 and the adoption will not have a material impact on our financial statements. In May 2011, the FASB issued a new accounting standard update, which amends the fair value measurement guidance and includes some enhanced disclosure requirements. The most significant change in disclosures is an expansion of the information required for Level 3 measurements based on unobservable inputs.The standard is effective for fiscal years beginning after December 15, 2011.We adopted this standard in the first quarter of 2012 and the adoption will not have a material impact on our financial statements and disclosures. ITEM 7A.QUANTITATIVE AND QUALITATIVE DISCLOSURES ABOUT MARKET RISK We are exposed to financial market risks, including changes in currency exchange rates and interest rates. Foreign Currency Exchange Risk Economic Exposure We transact business in various foreign currencies and have significant international revenues, as well as costs denominated in foreign currencies. This exposes us to the risk of fluctuations in foreign currency exchange rates. We purchase foreign exchange option contracts to reduce the volatility of cash flows related to forecasted revenues denominated in certain foreign currencies. The objective of the foreign exchange contracts is to better ensure that the U.S. dollar-equivalent cash flows are not adversely affected by changes in the U.S. dollar/foreign currency exchange rates. These contracts are designated as cash flow hedges. The gain on the effective portion of a cash flow hedge is initially reported as a component of accumulated other comprehensive income (AOCI) and subsequently reclassified into revenues when the hedged revenues are recorded or as interest and other income, net, if the hedged transaction becomes probable of not occurring. Any gain after a hedge is de-designated or related to an ineffective portion of a hedge is recognized as interest and other income, net, immediately. At December 31, 2010, the notional principal and fair value of foreign exchange contracts to purchase U.S. dollars with Euros were€3.0 billion (or approximately $4.1 billion) and $227 million; the notional principal and fair value of foreign exchange contracts to purchase U.S. dollars with British pounds were £1.5 billion (or approximately $2.3 billion) and $97 million; and the notional principal and fair value of foreign exchange contracts to purchase U.S. dollars with Canadian dollars were C$407 million (or approximately $382 million) and $6 million. At 45 Table of Contents December 31, 2011, the notional principal and fair value of foreign exchange contracts to purchase U.S. dollars with Euros were €2.8 billion (or approximately $3.8 billion) and $232 million; the notional principal and fair value of foreign exchange contracts to purchase U.S. dollars with British pounds were £1.4 billion (or approximately $2.2 billion) and $80 million; and the notional principal and fair value of foreign exchange contracts to purchase U.S. dollars with Canadian dollars were C$504 million (or approximately $490 million) and $17 million. These foreign exchange contracts have maturities of 36 months or less. We may enter into similar contracts in other foreign currencies in the future. We considered the historical trends in currency exchange rates and determined that it was reasonably possible that changes in exchange rates of 20% for our foreign currencies instruments could be experienced in the near term. If the U.S. dollar weakened by 20%, the amount recorded in AOCI before tax effect would have been approximately $140 million and $132 million lower at December 31, 2010 and 2011, and the total amount of expense recorded as interest and other income, net, would have been approximately $134 million and $138 million higher in the years ended December 31, 2010 and 2011. If the U.S. dollar strengthened by 20%, the amount recorded in accumulated AOCI before tax effect would have been approximately $1.2 billion higher at both December 31, 2010 and 2011, and the total amount of expense recorded as interest and other income, net, would have been approximately $175 million and $202 million higher in the years ended December 31, 2010 and 2011. Transaction Exposure Our exposure to foreign currency transaction gains and losses is the result of certain net receivables due from our foreign subsidiaries and customers being denominated in currencies other than the functional currency of the subsidiary, primarily the Euro and the British pound. Our foreign subsidiaries conduct their businesses in local currency. We have entered into foreign exchange contracts to offset the foreign exchange risk on certain monetary assets and liabilities denominated in currencies other than the local currency of the subsidiary. The notional principal of foreign exchange contracts to purchase U.S. dollars with foreign currencies was $1.0 billion and $2.3 billion at December 31, 2010 and 2011. The notional principal of foreign exchange contracts to sell U.S. dollars for foreign currencies was $84 million and $472 million at December 31, 2010 and December 31, 2011. The notional principal of foreign exchange contracts to purchase Euros with other foreign currencies was €991 million (or approximately $1.3 billion) and €711 million (or approximately $929 million) at December 31, 2010 and 2011. The notional principal of foreign exchange contracts to sell Euros for other foreign currencies was €6 million (or approximately $8 million) at December 31, 2010 and no such contracts were outstanding at December 31, 2011. We considered the historical trends in currency exchange rates and determined that it was reasonably possible that adverse changes in exchange rates of 20% for all currencies could be experienced in the near term. These changes would have resulted in an adverse impact on income before income taxes of approximately $20 million and $27 million at December 31, 2010 and 2011. The adverse impact at December 31, 2010 and 2011 is after consideration of the offsetting effect of approximately $467 million and $503 million from foreign exchange contracts in place for the months of December 2010 and December 2011. These reasonably possible adverse changes in exchange rates of 20% were applied to total monetary assets and liabilities denominated in currencies other than the local currencies at the balance sheet dates to compute the adverse impact these changes would have had on our income before income taxes in the near term. Interest Rate Risk We invest our excess cash primarily in highly liquid debt instruments of the U.S. government and its agencies, municipalities in the U.S., debt instruments issued by foreign governments, time deposits, money market and other funds, mortgage-backed securities, and corporate debt securities. By policy, we limit the amount of credit exposure to any one issuer. 46 Table of Contents Investments in both fixed rate and floating rate interest earning securities carry a degree of interest rate risk. Fixed rate securities may have their fair market value adversely impacted due to a rise in interest rates, while floating rate securities may produce less income than predicted if interest rates fall. Due in part to these factors, our income from investments may decrease in the future. However, we use certain interest rate derivative contracts to hedge interest rate risk of our fixed income securities. We considered the historical volatility of short-term interest rates and determined that it was reasonably possible that an adverse change of 100 basis points could be experienced in the near term. A hypothetical 1.00% (100 basis points) increase in interest rates would have resulted in a decrease in the fair values of our marketable securities of approximately $895 million and $934 million at December 31, 2010 and 2011, after taking into consideration the offsetting effect from interest rate derivative contracts outstanding as of December 31, 2011. 47 Table of Contents ITEM 8.FINANCIAL STATEMENTS AND SUPPLEMENTARY DATA Google Inc. INDEX TO CONSOLIDATED FINANCIAL STATEMENTS Page Reports of Independent Registered Public Accounting Firm49 Financial Statements: Consolidated Balance Sheets51 Consolidated Statements of Income52 Consolidated Statements of Comprehensive Income53 Consolidated Statements of Stockholders’ Equity54 Consolidated Statements of Cash Flows55 Notes to Consolidated Financial Statements56 The supplementary financial information required by this Item 8 is included in Item 7 under the caption “Quarterly Results of Operations.” 48 Table of Contents REPORT OF INDEPENDENT REGISTERED PUBLIC ACCOUNTING FIRM The Board of Directors and Stockholders of Google Inc. We have audited the accompanying consolidated balance sheets of Google Inc. as of December 31, 2010 and 2011, and the related consolidated statements of income, comprehensive income, stockholders’ equity, and cash flows for each of the three years in the period ended December 31, 2011. Our audits also included the financial statement schedule listed in the Index at Item 15(a)2. These financial statements and schedule are the responsibility of the Company’s management. Our responsibility is to express an opinion on these financial statements and schedule based on our audits. We conducted our audits in accordance with the standards of the Public Company Accounting Oversight Board (United States). Those standards require that we plan and perform the audit to obtain reasonable assurance about whether the financial statements are free of material misstatement. An audit includes examining, on a test basis, evidence supporting the amounts and disclosures in the financial statements. An audit also includes assessing the accounting principles used and significant estimates made by management, as well as evaluating the overall financial statement presentation. We believe that our audits provide a reasonable basis for our opinion. In our opinion, the financial statements referred to above present fairly, in all material respects, the consolidated financial position of Google Inc. at December 31, 2010 and 2011, and the consolidated results of its operations and its cash flows for each of the three years in the period ended December 31, 2011, in conformity with U.S. generally accepted accounting principles. Also, in our opinion, the related financial statement schedule, when considered in relation to the basic financial statements taken as a whole, presents fairly in all material respects the information set forth therein. We also have audited, in accordance with the standards of the Public Company Accounting Oversight Board (United States), Google Inc.’s internal control over financial reporting as of December 31, 2011, based on criteria established in Internal Control—Integrated Framework issued by the Committee of Sponsoring Organizations of the Treadway Commission and our report dated January 26, 2012 expressed an unqualified opinion thereon. /s/E RNST& Y OUNG LLP San Jose, California January 26, 2012 49 Table of Contents REPORT OF INDEPENDENT REGISTERED PUBLIC ACCOUNTING FIRM The Board of Directors and Stockholders of Google Inc. We have audited Google Inc.’s internal control over financial reporting as of December 31, 2011, based on criteria established in Internal Control—Integrated Framework issued by the Committee of Sponsoring Organizations of the Treadway Commission (the COSO criteria). Google Inc.’s management is responsible for maintaining effective internal control over financial reporting, and for its assessment of the effectiveness of internal control over financial reporting included in the accompanying Management’s Report on Internal Control over Financial Reporting. Our responsibility is to express an opinion on the Company’s internal control over financial reporting based on our audit. We conducted our audit in accordance with the standards of the Public Company Accounting Oversight Board (United States). Those standards require that we plan and perform the audit to obtain reasonable assurance about whether effective internal control over financial reporting was maintained in all material respects. Our audit included obtaining an understanding of internal control over financial reporting, assessing the risk that a material weakness exists, testing and evaluating the design and operating effectiveness of internal control based on the assessed risk, and performing such other procedures as we considered necessary in the circumstances. We believe that our audit provides a reasonable basis for our opinion. A company’s internal control over financial reporting is a process designed to provide reasonable assurance regarding the reliability of financial reporting and the preparation of financial statements for external purposes in accordance with generally accepted accounting principles. A company’s internal control over financial reporting includes those policies and procedures that (1)pertain to the maintenance of records that, in reasonable detail, accurately and fairly reflect the transactions and dispositions of the assets of the company; (2)provide reasonable assurance that transactions are recorded as necessary to permit preparation of financial statements in accordance with generally accepted accounting principles, and that receipts and expenditures of the company are being made only in accordance with authorizations of management and directors of the company; and (3)provide reasonable assurance regarding prevention or timely detection of unauthorized acquisition, use, or disposition of the company’s assets that could have a material effect on the financial statements. Because of its inherent limitations, internal control over financial reporting may not prevent or detect misstatements. Also, projections of any evaluation of effectiveness to future periods are subject to the risk that controls may become inadequate because of changes in conditions, or that the degree of compliance with the policies or procedures may deteriorate. In our opinion, Google Inc. maintained, in all material respects, effective internal control over financial reporting as of December 31, 2011, based on the COSO criteria. We also have audited, in accordance with the standards of the Public Company Accounting Oversight Board (United States), the consolidated balance sheets of Google Inc. as of December 31, 2010 and 2011, and the related consolidated statements of income, comprehensive income, stockholders’ equity, and cash flows for each of the three years in the period ended December 31, 2011 of Google Inc. and our report dated January 26, 2012 expressed an unqualified opinion thereon. /s/E RNST& Y OUNG LLP San Jose, California January 26, 2012 50 Table of Contents Google Inc. CONSOLIDATED BALANCE SHEETS (In millions, except share and par value amounts which are reflected in thousands, and par value per share amounts) As of December 31, 2010 As of December 31, 2011 Assets Current assets: Cash and cash equivalents$13,630$9,983 Marketable securities 21,345 34,643 Total cash, cash equivalents, and marketable securities (including securities loaned of $4,031 and $2,778)34,975 44,626 Accounts receivable, net of allowance of $101 and $133 4,252 5,427 Receivable under reverse repurchase agreements 750 745 Deferred income taxes, net 259 215 Prepaid revenue share, expenses and other assets 1,326 1,745 Total current assets 41,562 52,758 Prepaid revenue share, expenses and other assets, non-current 442 499 Deferred income taxes, net, non-current 265 0 Non-marketable equity securities 523 790 Property and equipment, net 7,759 9,603 Intangible assets, net 1,044 1,578 Goodwill 6,256 7,346 Total assets$57,851$72,574 Liabilities and Stockholders’ Equity Current liabilities: Accounts payable$483$588 Short-term debt 3,465 1,218 Accrued compensation and benefits 1,410 1,818 Accrued expenses and other current liabilities 961 1,370 Accrued revenue share 885 1,168 Securities lending payable 2,361 2,007 Deferred revenue 394 547 Income taxes payable, net 37 197 Total current liabilities 9,996 8,913 Long-term debt 0 2,986 Deferred revenue, non-current 35 44 Income taxes payable, non-current 1,200 1,693 Deferred income taxes, net, non-current 0 287 Other long-term liabilities 379 506 Commitments and contingencies Stockholders’ equity: Convertible preferred stock, $0.001 par value per share, 100,000 shares authorized; no shares issued and outstanding 0 0 Class A and Class B common stock and additional paid-in capital, $0.001 par value per share: 9,000,000 shares authorized; 321,301 (Class A 250,413, Class B 70,888) and par value of $321 (Class A $250, Class B $71) and 324,895 (Class A 257,553, Class B 67,342) and par value of $325 (Class A $258, Class B $67) shares issued and outstanding 18,235 20,264 Accumulated other comprehensive income 138 276 Retained earnings 27,868 37,605 Total stockholders’ equity 46,241 58,145 Total liabilities and stockholders’ equity$57,851$72,574 See accompanying notes. 51 Table of Contents Google Inc. CONSOLIDATED STATEMENTS OF INCOME (In millions, except per share amounts) Year Ended December 31, 2009 2010 2011 Revenues$23,651$29,321$37,905 Costs and expenses: Cost of revenues (including stock-based compensation expense of $47, $67, $249)8,844 10,417 13,188 Research and development (including stock-based compensation expense of $725, $861, $1,061)2,843 3,762 5,162 Sales and marketing (including stock-based compensation expense of $231, $261, $361)1,984 2,799 4,589 General and administrative (including stock-based compensation expense of $161, $187, $303)1,668 1,962 2,724 Charge related to the resolution of Department of Justice investigation 0 0 500 Total costs and expenses 15,339 18,940 26,163 Income from operations 8,312 10,381 11,742 Interest and other income, net 69 415 584 Income before income taxes 8,381 10,796 12,326 Provision for income taxes 1,861 2,291 2,589 Net income$6,520$8,505$9,737 Net income per share of Class A and Class B common stock: Basic$20.62$26.69$30.17 Diluted$20.41$26.31$29.76 See accompanying notes. 52 Table of Contents Google Inc. CONSOLIDATED STATEMENTS OF COMPREHENSIVE INCOME (In millions) Year Ended December 31, 2009 2010 2011 Net income$6,520$8,505$9,737 Other comprehensive income (loss): Change in foreign currency translation adjustment 77(124)(107) Available-for-sale investments: Change in net unrealized gains 93 232 348 Less: reclassification adjustment for net gains included in net income(91)(151)(115) Net change (net of tax effect of $6, $52, $54)2 81 233 Cash flow hedges: Change in unrealized gains(9)196 39 Less: reclassification adjustment for gains included in net income(192)(120)(27) Net change (net of tax effect of $138, $52, $2)(201)76 12 Other comprehensive income (loss)(122)33 138 Comprehensive income$6,398$8,538$9,875 See accompanying notes. 53 Table of Contents Google Inc. CONSOLIDATED STATEMENTS OF STOCKHOLDERS’ EQUITY (In millions, except for share amounts which are reflected in thousands) Class A and Class B Common Stock and Additional Paid-In Capital Accumulated Other Comprehensive Income Retained Earnings Total Stockholders’ Equity Shares Amount Balance at January 1, 2009 315,114$14,450$227$13,562$28,239 Common stock issued 2,658 350 0 0 350 Stock-based compensation expense 1,164 0 0 1,164 Stock-based compensation tax benefits 60 0 0 60 Tax withholding related to vesting of restricted stock units(207)0 0(207) Net income 0 0 6,520 6,520 Other comprehensive loss 0(122)0(122) Balance at December 31, 2009 317,772 15,817 105 20,082 36,004 Common stock issued 5,126 1,412 0 0 1,412 Common stock repurchased(1,597)(82)0(719)(801) Stock-based compensation expense 1,376 0 0 1,376 Stock-based compensation tax benefits 72 0 0 72 Tax withholding related to vesting of restricted stock units(360)0 0(360) Net income 0 0 8,505 8,505 Other comprehensive income 0 33 0 33 Balance at December 31, 2010 321,301 18,235 138 27,868 46,241 Common stock issued 3,594 621 0 0 621 Stock-based compensation expense 1,974 0 0 1,974 Stock-based compensation tax benefits 60 0 0 60 Tax withholding related to vesting of restricted stock units(626)0 0(626) Net income 0 0 9,737 9,737 Other comprehensive income 0 138 0 138 Balance at December 31, 2011 324,895$20,264$276$37,605$58,145 See accompanying notes. 54 Table of Contents Google Inc. CONSOLIDATED STATEMENTS OF CASH FLOWS (In millions) Year Ended December 31, 2009 2010 2011 Operating activities Net income$6,520$8,505$9,737 Adjustments: Depreciation and amortization of property and equipment 1,240 1,067 1,396 Amortization of intangible and other assets 284 329 455 Stock-based compensation expense 1,164 1,376 1,974 Excess tax benefits from stock-based award activities(90)(94)(86) Deferred income taxes(268)9 343 Impairment of equity investments 0 0 110 Other(20)(12)6 Changes in assets and liabilities, net of effects of acquisitions: Accounts receivable(504)(1,129)(1,156) Income taxes, net 217 102 731 Prepaid revenue share, expenses and other assets 262(414)(262) Accounts payable 34 272 101 Accrued expenses and other liabilities 243 745 795 Accrued revenue share 158 214 259 Deferred revenue 76 111 162 Net cash provided by operating activities 9,316 11,081 14,565 Investing activities Purchases of property and equipment(810)(4,018)(3,438) Purchases of marketable securities(29,139)(43,985)(61,672) Maturities and sales of marketable securities 22,103 37,099 48,746 Investments in non-marketable equity securities(65)(320)(428) Cash collateral received (returned) from securities lending 0 2,361(354) Investments in reverse repurchase agreements 0(750)5 Acquisitions, net of cash acquired, and purchases of intangible and other assets(108)(1,067)(1,900) Net cash used in investing activities(8,019)(10,680)(19,041) Financing activities Net proceeds (payments) from stock-based award activities 143 294(5) Excess tax benefits from stock-based award activities 90 94 86 Repurchase of common stock in connection with acquisitions 0(801)0 Proceeds from issuance of debt, net of costs 0 5,246 10,905 Repayment of debt 0(1,783)(10,179) Net cash provided by financing activities 233 3,050 807 Effect of exchange rate changes on cash and cash equivalents 11(19)22 Net increase (decrease) in cash and cash equivalents 1,541 3,432(3,647) Cash and cash equivalents at beginning of year 8,657 10,198 13,630 Cash and cash equivalents at end of year$10,198$13,630$9,983 Supplemental disclosures of cash flow information Cash paid for interest$0$0$40 Cash paid for taxes$1,896$2,175$1,471 Non-cash financing activity: Fair value of common stock issued and vested options assumed in connection with acquisitions$0$750$0 See accompanying notes. 55 Table of Contents Google Inc. NOTES TO CONSOLIDATED FINANCIAL STATEMENTS Note 1.Google Inc. and Summary of Significant Accounting Policies Nature of Operations We were incorporated in California in September 1998. We were re-incorporated in the State of Delaware in August 2003. We generate revenues primarily by delivering relevant, cost-effective online advertising. Basis of Consolidation The consolidated financial statements include the accounts of Google Inc. and our wholly-owned subsidiaries. All intercompany balances and transactions have been eliminated. Use of Estimates The preparation of consolidated financial statements in conformity with U.S. Generally Accepted Accounting Principles (GAAP) requires us to make estimates and assumptions that affect the amounts reported and disclosed in the financial statements and the accompanying notes. Actual results could differ materially from these estimates. On an ongoing basis, we evaluate our estimates, including those related to the accounts receivable and sales allowances, fair values of financial instruments, intangible assets and goodwill, useful lives of intangible assets and property and equipment, fair values of stock-based awards, income taxes, and contingent liabilities, among others. We base our estimates on historical experience and on various other assumptions that are believed to be reasonable, the results of which form the basis for making judgments about the carrying values of assets and liabilities. Revenue Recognition The following table presents our revenues by revenue source (in millions): Year Ended December 31, 2009 2010 2011 Advertising revenues: Google websites$15,723$19,444$26,145 Google Network Members’ websites 7,166 8,792 10,386 Total advertising revenues 22,889 28,236 36,531 Other revenues 762 1,085 1,374 Revenues$23,651$29,321$37,905 Google AdWords is our auction-based advertising program that enables advertisers to place text-based and display ads on our websites and our Google Network Members’ websites. Display advertising comprises the videos, text, images, and other interactive ads that run across the web on computers and mobile devices, including smart phones and handheld computers such as netbooks and tablets. Most of our AdWords customers pay us on a cost-per-click basis, which means that an advertiser pays us only when a user clicks on one of its ads. We also offer AdWords on a cost-per-impression basis that enables advertisers to pay us based on the number of times their ads appear on our websites and our Google Network Members’ websites as specified by the advertisers. Google AdSense refers to the online programs through which we distribute our advertisers’ AdWords ads for display on our Google Network Members’ websites, as well as programs to deliver ads on television broadcasts. 56 Table of Contents We recognize as revenues the fees charged to advertisers each time a user clicks on one of the ads that appears next to the search results or content on our websites or our Google Network Members’ websites. For those advertisers using our AdWords cost-per-impression pricing, we recognize as revenues the fees charged to advertisers each time their ads are displayed on our websites or our Google Network Members’ websites. We report our Google AdSense revenues on a gross basis principally because we are the primary obligor to our advertisers. We recognize revenues when the services or products have been provided or delivered, the fees we charge are fixed or determinable, we and our advertisers or other customers understand the specific nature and terms of the agreed upon transactions, and collectability is reasonably assured. We record deferred revenue upon invoicing or when cash payments are received in advance of our performance in the underlying agreement on the accompanying Consolidated Balance Sheets. Cost of Revenues Cost of revenues consists primarily of traffic acquisition costs. Traffic acquisition costs consist of amounts ultimately paid to our Google Network members under AdSense arrangements and to certain other partners (our distribution partners) who distribute our toolbar and other products (collectively referred to as access points) or otherwise direct search queries to our website (collectively referred to as distribution arrangements). These amounts are primarily based on the revenue share and fixed fee arrangements with our Google Network Members and distribution partners. Certain distribution arrangements require us to pay our partners based on a fee per access point delivered and not exclusively—or at all—based on revenue share. These fees are non-refundable. Further, these arrangements are terminable at will, although under the terms of certain contracts we or our distribution partners may be subject to penalties in the event of early termination. We recognize fees under these arrangements over the estimated useful lives of the access points (approximately two years) to the extent we can reasonably estimate those lives and they are longer than one year, or based on any contractual revenue share, if greater. Otherwise, the fees are charged to expense as incurred. The estimated useful life of the access points is based on the historical average period of time they generate traffic and revenues. Further, we review the access points for impairment by distribution partner, type, and geography, and we have not made any impairment to date. Prepaid revenue share and distribution fees are included in prepaid revenue share, expenses, and other assets on the accompanying Consolidated Balance Sheets. Cost of revenues also includes the expenses associated with the operation of our data centers, including depreciation, labor, energy, and bandwidth costs, credit card and other transaction fees related to processing customer transactions including Google Checkout transactions, amortization of acquired intangible assets, as well as content acquisition costs. We have entered into arrangements with certain content providers under which we distribute or license their video and other content. In a number of these arrangements, we display ads on the pages of our web sites from which the content is viewed and share most of the fees these ads generate with the content providers. To the extent we are obligated to make guaranteed minimum revenue share payments to our content providers, we recognize as content acquisition costs the contractual revenue share amount or on a straight-line basis, whichever is greater, over the terms of the agreements. Stock-based Compensation We have elected to use the BSM option pricing model to determine the fair value of stock options on the dates of grant . Restricted stock units (RSUs) are measured based on the fair market values of the underlying stock on the dates of grant. Shares are issued on the vesting dates net of the minimum statutory tax withholding requirements to be paid by us on behalf of our employees. As a result, the actual number of shares issued will be fewer than the actual number of RSUs outstanding. Furthermore, we record the liability for withholding amounts to be paid by us as a reduction to additional paid-in capital when paid. Also, we recognize stock-based compensation using the straight-line method. 57 Table of Contents We include as part of cash flows from financing activities the benefits of tax deductions in excess of the tax-effected compensation of the related stock-based awards for options exercised and RSUs vested during the period. During the years ended December 31, 2009,December 31, 2010, and December 31, 2011, the amount of cash received from the exercise of stock options was $350 million, $656 million, and $621 million, and the total direct tax benefit realized, including the excess tax benefit, from stock-based award activities was $260 million, $355 million, and $451 million. We have elected to account for the indirect effects of stock-based awards—primarily the research and development tax credit—through the Consolidated Statements of Income. For the years ended December 31, 2009,December 31, 2010, and December 31, 2011, we recognized stock-based compensation and related tax benefits of $1,164 million and $264 million, $1,376 million and $314 million, and $1,974 million and $413 million. Certain Risks and Concentrations Our revenues are principally derived from online advertising, the market for which is highly competitive and rapidly changing. In addition, our revenues are generated from a multitude of vertical market segments in countries around the world. Significant changes in this industry or changes in customer buying or advertiser spending behavior could adversely affect our operating results. Financial instruments that potentially subject us to concentrations of credit risk consist principally of cash equivalents, marketable securities, foreign exchange contracts, and accounts receivable. Cash equivalents and marketable securities consist primarily of highly liquid debt instruments of the U.S. government and its agencies, municipalities in the U.S., debt instruments issued by foreign governments, mortgage-backed securities, corporate securities, time deposits, and money market and other funds, including cash collateral received related to our securities lending program. Foreign exchange contracts are transacted with various financial institutions with high credit standing. Accounts receivable are typically unsecured and are derived from revenues earned from customers located around the world. In 2009, 2010, and 2011, we generated approximately 47%, 48%, and 46% of our revenues from customers based in the U.S., with the majority of customers outside of the U.S. located in Europe and Japan. Many of our Google Network Members are in the internet industry. We perform ongoing evaluations to determine customer credit and we limit the amount of credit we extend, but generally we do not require collateral from our customers. We maintain reserves for estimated credit losses and these losses have generally been within our expectations. No individual customer or groups of affiliated customers represented more than 10% of our revenues in 2009, 2010, and 2011. Fair Value of Financial Instruments The carrying amounts of our financial instruments, including cash equivalents, accounts receivable, accounts payable, short-term debt, and accrued liabilities, approximate fair value because of their generally short maturities. Cash, Cash Equivalents, and Marketable Securities We invest our excess cash primarily in highly liquid debt instruments of the U.S. government and its agencies, municipalities in the U.S., debt instruments issued by foreign governments, mortgage-backed securities, corporate securities, time deposits, and money market and other funds, including cash collateral received related to our securities lending program. We classify all highly liquid investments with stated maturities of three months or less from date of purchase as cash equivalents and all highly liquid investments with stated maturities of greater than three months as marketable securities. We determine the appropriate classification of our investments in marketable securities at the time of purchase and reevaluate such designation at each balance sheet date. We have classified and accounted for our marketable securities as available-for-sale. We may or may not hold securities with stated maturities greater than 58 Table of Contents 12 months until maturity. After consideration of our risk versus reward objectives, as well as our liquidity requirements, we may sell these securities prior to their stated maturities. As we view these securities as available to support current operations, we classify securities with maturities beyond 12 months as current assets under the caption marketable securities in the accompanying Consolidated Balance Sheets. We carry these securities at fair value, and report the unrealized gains and losses, net of taxes, as a component of stockholders’ equity, except for unrealized losses determined to be other-than-temporary, which we record as interest and other income, net. We determine any realized gains or losses on the sale of marketable securities on a specific identification method, and we record such gains and losses as a component of interest and other income, net. Non-Marketable Equity Securities We have accounted for non-marketable equity securities primarily at cost because we do not have significant influence over the underlying investees. Impairment of Marketable and Non-Marketable Securities We periodically review our marketable securities, as well as our non-marketable equity securities, for impairment. If we conclude that any of these investments are impaired, we determine whether such impairment is other-than-temporary. Factors we consider to make such determination include the duration and severity of the impairment, the reason for the decline in value and the potential recovery period, and our intent to sell, or whether it is more likely than not that we will be required to sell, the investment before recovery. If any impairment is considered other-than-temporary, we will write down the asset to its fair value and take a corresponding charge to our Consolidated Statements of Income. Accounts Receivable We record accounts receivable at the invoiced amount and we do not charge interest. We maintain an allowance for doubtful accounts to reserve for potentially uncollectible receivables. We review the accounts receivable by amounts due by customers which are past due to identify specific customers with known disputes or collectability issues. In determining the amount of the reserve, we make judgments about the creditworthiness of significant customers based on ongoing credit evaluations. We also maintain a sales allowance to reserve for potential credits issued to customers. We determine the amount of the reserve based on historical credits issued. Property and Equipment We account for property and equipment at cost less accumulated depreciation and amortization. We compute depreciation using the straight-line method over the estimated useful lives of the assets, generally two to five years. We depreciate buildings over periods up to 25 years. We amortize leasehold improvements over the shorter of the remaining lease term or the estimated useful lives of the assets. Construction in progress is related to the construction or development of property (including land) and equipment that have not yet been placed in service for our intended use. Depreciation for equipment commences once it is placed in service and depreciation for buildings and leasehold improvements commences once they are ready for our intended use. Land is not depreciated. Software Development Costs We expense software development costs, including costs to develop software products or the software component of products to be marketed to external users, before technological feasibility of such products is reached. We have determined that technological feasibility was reached shortly before the release of those products and as a result, the development costs incurred after the establishment of technological feasibility and before the release of those products were not material, and accordingly, were expensed as incurred. 59 Table of Contents Long-Lived Assets Including Goodwill and Other Acquired Intangible Assets We review property and equipment and intangible assets, excluding goodwill, for impairment whenever events or changes in circumstances indicate the carrying amount of an asset may not be recoverable. We measure recoverability of these assets by comparing the carrying amounts to the future undiscounted cash flows the assets are expected to generate. If property and equipment and intangible assets are considered to be impaired, the impairment to be recognized equals the amount by which the carrying value of the asset exceeds its fair market value. We have made no material adjustments to our long-lived assets in any of the years presented. In addition, we test our goodwill for impairment at least annually or more frequently if events or changes in circumstances indicate that this asset may be impaired. Our tests are based on our single operating segment and reporting unit structure. We found no goodwill impairment in any of the years presented. Intangible assets with definite lives are amortized over their estimated useful lives. We amortize our acquired intangible assets on a straight-line basis with definite lives over periods ranging from one to 12 years. Income Taxes We recognize income taxes under the liability method. We recognize deferred income taxes for differences between the financial reporting and tax bases of assets and liabilities at enacted statutory tax rates in effect for the years in which differences are expected to reverse. We recognize the effect on deferred taxes of a change in tax rates in income in the period that includes the enactment date. Foreign Currency Generally, the functional currency of our international subsidiaries is the local currency. We translate the financial statements of these subsidiaries to U.S. dollars using month-end rates of exchange for assets and liabilities, and average rates of exchange for revenues, costs, and expenses. We record translation gains and losses in accumulated other comprehensive income as a component of stockholders’ equity. We recorded $77 million of net translation gains in 2009, $124 million of net translation losses in 2010, and $107 million of net translation losses in 2011. We record net gains and losses resulting from foreign exchange transactions as a component of interest and other income, net. These gains and losses are net of those realized on foreign exchange contracts. We recorded $8 million of net gains in 2009, $29 million of net losses in 2010, and $38 million of net losses in 2011. Advertising and Promotional Expenses We expense advertising and promotional costs in the period in which they are incurred. For the years ended December 31, 2009,December 31, 2010, and December 31, 2011, advertising and promotional expenses totaled approximately $353 million, $772 million, and $1,544 million. Recent Accounting Pronouncements In June 2011, the FASB issued an amendment to an existing accounting standard which requires companies to present net income and other comprehensive income in one continuous statement or in two separate, but consecutive, statements.In addition, in December 2011, the FASB issued an amendment to an existing accounting standard which defers the requirement to present components of reclassifications of other comprehensive income on the face of the income statement.We adopted both standards in the fourth quarter of 2011. In September 2011, the FASB issued an amendment to an existing accounting standard, which provides entities an option to perform a qualitative assessment to determine whether further impairment testing on goodwill is necessary. Specifically, an entity has the option to first assess qualitative factors to determine whether it is necessary to perform the current two-step test.If an entity believes, as a result of its qualitative assessment, that it is more-likely-than-not that the fair value of a reporting unit is less than its carrying amount, the quantitative 60 Table of Contents impairment test is required.Otherwise, no further testing is required. This standard is effective for annual and interim goodwill impairment tests performed for fiscal years beginning after December 15, 2011.We adopted this standard in the first quarter of 2012 and the adoption will not have a material impact on our financial statements. In May 2011, the FASB issued a new accounting standard update, which amends the fair value measurement guidance and includes some enhanced disclosure requirements. The most significant change in disclosures is an expansion of the information required for Level 3 measurements based on unobservable inputs.The standard is effective for fiscal years beginning after December 15, 2011.We adopted this standard in the first quarter of 2012 and the adoption will not have a material impact on our financial statements and disclosures. Note 2.Net Income Per Share of Class A and Class B Common Stock We compute net income per share of Class A and Class B common stock using the two-class method. Basic net income per share is computed using the weighted-average number of common shares outstanding during the period except that it does not include unvested common shares subject to repurchase or cancellation. Diluted net income per share is computed using the weighted-average number of common shares and the effect of potentially dilutive securities outstanding during the period. Potentially dilutive securities consist of stock options, warrants issued under the TSO program, restricted shares, restricted stock units, and unvested common shares subject to repurchase or cancellation. The dilutive effect of outstanding stock options, warrants, restricted shares, and restricted stock units is reflected in diluted earnings per share by application of the treasury stock method. The computation of the diluted net income per share of Class A common stock assumes the conversion of Class B common stock, while the diluted net income per share of Class B common stock does not assume the conversion of those shares. The rights, including the liquidation and dividend rights, of the holders of our Class A and Class B common stock are identical, except with respect to voting. Further, there are a number of safeguards built into our certificate of incorporation, as well as Delaware law, which preclude our board of directors from declaring or paying unequal per share dividends on our Class A and Class B common stock. Specifically, Delaware law provides that amendments to our certificate of incorporation which would have the effect of adversely altering the rights, powers, or preferences of a given class of stock (in this case the right of our Class A common stock to receive an equal dividend to any declared on our Class B common stock) must be approved by the class of stock adversely affected by the proposed amendment. In addition, our certificate of incorporation provides that before any such amendment may be put to a stockholder vote, it must be approved by the unanimous consent of our board of directors. As a result, the undistributed earnings for each year are allocated based on the contractual participation rights of the Class A and Class B common shares as if the earnings for the year had been distributed. As the liquidation and dividend rights are identical, the undistributed earnings are allocated on a proportionate basis. Further, as we assume the conversion of Class B common stock in the computation of the diluted net income per share of Class A common stock, the undistributed earnings are equal to net income for that computation. 61 Table of Contents The following table sets forth the computation of basic and diluted net income per share of Class A and Class B common stock (in millions, except share amounts which are reflected in thousands and per share amounts): Year Ended December 31, 2009 2010 2011 Class A Class B Class A Class B Class A Class B Basic net income per share: Numerator Allocation of undistributed earnings$4,981$1,539$6,569$1,936$7,658$2,079 Denominator Weighted-average common shares outstanding 241,575 74,651 246,168 72,534 253,862 68,916 Less: Weighted-average unvested common shares subject to repurchase or cancellation(5)0 0 0 0 0 Number of shares used in per share computation 241,570 74,651 246,168 72,534 253,862 68,916 Basic net income per share$20.62$20.62$26.69$26.69$30.17$30.17 Diluted net income per share: Numerator Allocation of undistributed earnings for basic computation$4,981$1,539$6,569$1,936$7,658$2,079 Reallocation of undistributed earnings as a result of conversion of Class B to Class A shares 1,539 0 1,936 0 2,079 0 Reallocation of undistributed earnings to Class B shares 0(13)0(26)0(27) Allocation of undistributed earnings$6,520$1,526$8,505$1,910$9,737$2,052 Denominator Number of shares used in basic computation 241,570 74,651 246,168 72,534 253,862 68,916 Weighted-average effect of dilutive securities Add: Conversion of Class B to Class A common shares outstanding 74,651 0 72,534 0 68,916 0 Unvested common shares subject to repurchase or cancellation 5 0 0 0 0 0 Employee stock options, including warrants issued under Transferable Stock Option program 2,569 114 3,410 71 2,958 46 Restricted shares and RSUs 621 0 1,139 0 1,478 0 Number of shares used in per share computation 319,416 74,765 323,251 72,605 327,214 68,962 Diluted net income per share$20.41$20.41$26.31$26.31$29.76$29.76 The net income per share amounts are the same for Class A and Class B common stock because the holders of each class are legally entitled to equal per share distributions whether through dividends or in liquidation. 62 Table of Contents Note 3.Cash and Investments Cash, cash equivalents, and marketable securities consisted of the following (in millions): As of December 31, 2010 2011 Cash and cash equivalents: Cash$4,652$4,712 Cash equivalents: Time deposits 973 534 Money market and other funds(1)7,547 4,462 U.S. government agencies 0 275 U.S. government notes 300 0 Foreign government bonds 150 0 Corporate debt securities 8 0 Total cash and cash equivalents 13,630 9,983 Marketable securities: Time deposits 307 495 U.S. government agencies 1,857 6,226 U.S. government notes 3,930 11,579 Foreign government bonds 1,172 1,629 Municipal securities 2,503 1,794 Corporate debt securities 5,742 6,112 Agency residential mortgage-backed securities 5,673 6,501 Marketable equity securities 161 307 Total marketable securities 21,345 34,643 Total cash, cash equivalents, and marketable securities$34,975$44,626 (1)The balances at December 31, 2010 and December 31, 2011 included $1.6 billion and $1.3 billion of cash collateral received in connection with our securities lending program, which was invested in reverse repurchase agreements maturing within three months. See below for further discussion on this program. The following tables summarize unrealized gains and losses related to our investments in marketable securities designated as available-for-sale (in millions): As of December 31,2010 Adjusted Cost Gross Unrealized Gains Gross Unrealized Losses Fair Value Time deposits$307$0$0$307 U.S. government agencies 1,864 1(8)1,857 U.S. government notes 3,950 30(50)3,930 Foreign government bonds 1,154 23(5)1,172 Municipal securities 2,492 16(5)2,503 Corporate debt securities 5,600 167(25)5,742 Agency residential mortgage-backed securities 5,649 56(32)5,673 Marketable equity security 150 11 0 161 Total$21,166$304$(125)$21,345 63 Table of Contents As of December 31,2011 Adjusted Cost Gross Unrealized Gains Gross Unrealized Losses Fair Value Time deposits$495$0$0$495 U.S. government agencies 6,211 15 0 6,226 U.S. government notes 11,475 104 0 11,579 Foreign government bonds 1,608 32(11)1,629 Municipal securities 1,775 19 0 1,794 Corporate debt securities 6,023 187(98)6,112 Agency residential mortgage-backed securities 6,359 147(5)6,501 Marketable equity securities 228 79 0 307 Total$34,174$583$(114)$34,643 Gross unrealized gains and losses on cash equivalents were not material at December 31, 2010 and December 31, 2011. We recognized gross realized gains of $212 million and $381 million for the years ended December 31, 2010 and 2011. We recognized gross realized losses of $27 million and $127 million for the years ended December 31, 2010 and 2011. We determine realized gains or losses on the sale of marketable securities on a specific identification method, and we reflect such gains and losses as a component of interest and other income, net, in our accompanying Consolidated Statements of Income. The following table summarizes the estimated fair value of our investments in marketable securities, excluding the marketable equity securities, designated as available-for-sale and classified by the contractual maturity date of the securities (in millions): As of December 31, 2011 Due in 1 year$13,242 Due in 1 year through 5 years 7,482 Due in 5 years through 10 years 5,525 Due after 10 years 8,087 Total$34,336 The following tables present gross unrealized losses and fair values for those investments that were in an unrealized loss position as of December 31, 2010 and 2011, aggregated by investment category and the length of time that individual securities have been in a continuous loss position (in millions): As of December 31,2010 Less than 12 Months Fair Value Unrealized Loss U.S. government agencies$831$(8) U.S. government notes 2,225(50) Foreign government bonds 129(5) Municipal securities 962(5) Corporate debt securities 1,061(25) Agency residential mortgage-backed securities 1,675(32) Total$6,883$(125) 64 Table of Contents As of December 31, 2010, we did not have any investments in marketable securities that were in an unrealized loss position for 12 months or greater. As of December 31,2011 Less than 12 Months 12 Months or Greater Total Fair Value Unrealized Loss Fair Value Unrealized Loss Fair Value Unrealized Loss Foreign government bonds$302$(11)$6$0$308$(11) Corporate debt securities 2,160(97)17(1)2,177(98) Agency residential mortgage-backed securities 716(3)19(2)735(5) Total$3,178$(111)$42$(3)$3,220$(114) Investment in a Marketable Equity Security During the fourth quarter of 2011, we recorded an other-than-temporary impairment charge of $88 million related to our investment in Clearwire Corporation. This amount was included in interest and other income, net in the accompanying Consolidated Statement of Income. Securities Lending Program From time to time, we enter into securities lending agreements with financial institutions to enhance investment income. We loan selected securities which are secured by collateral in the form of cash or securities. Cash collateral is invested in reverse repurchase agreements. We classify loaned securities as cash equivalents or marketable securities on the accompanying Consolidated Balance Sheets.We record the cash collateral as an asset with a corresponding liability. We classify reverse repurchase agreements maturing within three months as cash equivalents and those longer than three months as receivable under reverse repurchase agreements on the accompanying Consolidated Balance Sheets. For lending agreements collateralized by securities, we do not record an asset or liability as we are not permitted to sell or repledge the associated collateral. Note 4.Debt Short-Term Debt We have a debt financing program of up to $3.0 billion through the issuance of commercial paper. Net proceeds from this program are used for general corporate purposes. At December 31, 2010 and 2011, we had $3.0 billion and $750 million of commercial paper outstanding recorded as short-term debt with weighted-average interest rates of 0.3% and 0.1%. In conjunction with this program, we have a $3.0 billion revolving credit facility expiring in July 2016. The interest rate for the credit facility is determined based on a formula using certain market rates. At December 31, 2010 and 2011, we were in compliance with the financial covenant in the credit facility. No amounts were outstanding under the credit facility at December 31, 2010 and December 31, 2011. In December 2010, we issued a secured promissory note in the amount of $468 million with an interest rate of 1.0% and a one-year maturity date. Proceeds were used for the acquisition of an office building in New York City. In December 2011, we extended the maturity date of the note to December 2012. As of December 31, 2010 and 2011, the outstanding balance was $468 million. The estimated fair value of the short-term debt approximated its carrying value at December 31, 2010 and December 31, 2011. 65 Table of Contents Long-Term Debt In May 2011, we issued $3.0 billion of unsecured senior notes in three tranches as described in the table below (collectively, the Notes) (in millions): Outstanding Balance as of December 31, 2011 1.25% Notes due on May 19, 2014$1,000 2.125% Notes due on May 19, 2016 1,000 3.625% Notes due on May 19, 2021 1,000 Unamortized discount for the Notes above(14) Total$2,986 The effective interest yields of the 2014, 2016, and 2021 Notes were 1.258%, 2.241%, and 3.734%, respectively. Interest on the Notes is payable semi-annually in arrears on May 19 and November 19 of each year. We may redeem the Notes at any time in whole or from time to time in part at specified redemption prices. We are not subject to any financial covenants under the Notes. We used the net proceeds from the issuance of the Notes to repay a portion of our outstanding commercial paper and for general corporate purposes. The total estimated fair value of the Notes was approximately $3.2 billion, which is based on quoted prices for our publicly-traded debt as of December 31, 2011. At December 31, 2011, future principal payments for the Notes were as follows (in millions): Years ended 2012$0 2013 0 2014 1,000 2015 0 2016 1,000 Thereafter 1,000 Total$3,000 Note 5.Derivative Financial Instruments We enter into foreign currency contracts with financial institutions to reduce the risk that our cash flows and earnings will be adversely affected by foreign currency exchange rate fluctuations. We use certain interest rate derivative contracts to hedge interest rate exposures on our fixed income securities. Our program is not designated for trading or speculative purposes. We recognize derivative instruments as either assets or liabilities on the accompanying Consolidated Balance Sheets at fair value. We record changes in the fair value (i.e., gains or losses) of the derivatives in the accompanying Consolidated Statements of Income as interest and other income, net, as part of revenues, or to accumulated other comprehensive income (AOCI) in the accompanying Consolidated Balance Sheets. Cash Flow Hedges We use options designated as cash flow hedges to hedge certain forecasted revenue transactions denominated in currencies other than the U.S. dollar. We initially report any gain on the effective portion of a cash flow hedge as a component of AOCI and subsequently reclassify to revenues when the hedged revenues are 66 Table of Contents recorded or as interest and other income, net, if the hedged transaction becomes probable of not occurring. Further, we exclude the change in the time value of the options from our assessment of hedge effectiveness. We record the premium paid or time value of an option on the date of purchase as an asset. Thereafter, we recognize any change to this time value in interest and other income, net. At December 31, 2011, the effective portion of our cash flow hedges before tax effect was $154 million, of which $127 million is expected to be reclassified from AOCI to revenues within the next 12 months. The notional principal of foreign exchange contracts to purchase U.S. dollars with Euros was €3.0 billion (or approximately $4.1 billion) and €2.8 billion (or approximately $3.8 billion) at December 31, 2010 and December 31, 2011; the notional principal of foreign exchange contracts to purchase U.S. dollars with British pounds was £1.5 billion (or approximately $2.3 billion) and £1.4 billion (or approximately $2.2 billion) at December 31, 2010 and December 31, 2011; and the notional principal of foreign exchange contracts to purchase U.S. dollars with Canadian dollars was C$407 million (or approximately $382 million) and C$504 million (or approximately $490 million) at December 31, 2010 and December 31, 2011. These foreign exchange contracts have maturities of 36 months or less. Fair Value Hedges We use forward contracts designated as fair value hedges to hedge foreign currency risks for our investments denominated in currencies other than the U.S. dollar. Gains and losses on these contracts are recognized in interest and other income, net, along with the offsetting losses and gains of the related hedged items. We exclude changes in the time value for forward contracts from the assessment of hedge effectiveness and recognize them in interest and other income, net. The notional principal of foreign exchange contracts to purchase U.S. dollars with foreign currencies was $787 million and $1.0 billion at December 31, 2010 and December 31, 2011. Other Derivatives Other derivatives not designated as hedging instruments consist of forward and option contracts that we use to hedge intercompany transactions and other monetary assets or liabilities denominated in currencies other than the local currency of a subsidiary. We recognize gains and losses on these contracts as well as the related costs in interest and other income, net, along with the gains and losses of the related hedged items. The notional principal of foreign exchange contracts to purchase U.S. dollars with foreign currencies was $1.0 billion and $2.3 billion at December 31, 2010 and December 31, 2011. The notional principal of foreign exchange contracts to sell U.S. dollars for foreign currencies was $84 million and $472 million at December 31, 2010 and December 31, 2011. The notional principal of foreign exchange contracts to purchase Euros with other foreign currencies was €991 million (or approximately $1.3 billion) and €711 million (or approximately $929 million) at December 31, 2010 and December 31, 2011. The notional principal of foreign exchange contracts to sell Euros for other foreign currencies was €6 million (or approximately $8 million) at December 31, 2010 and no such contracts were outstanding at December 31, 2011. We also use exchange-traded interest rate futures contracts and “To Be Announced” (TBA) forward purchase commitments of mortgage-backed assets to hedge interest rate risks on certain fixed income securities. The TBA contracts meet the definition of derivative instruments in cases where physical delivery of the assets is not taken at the earliest available delivery date. Our interest rate futures and TBA contracts (together interest rate contracts) are not designated as hedging instruments. We recognize gains and losses on these contracts as well as the related costs in interest and other income, net. The gains and losses are generally economically offset by unrealized gains and losses in the underlying available-for-sale securities, which are recorded as a component of AOCI until the securities are sold or other-than-temporarily impaired, at which time the amounts are moved from AOCI into interest and other income, net. As of December 31, 2011, the total notional amounts of interest rate contracts outstanding were $100 million. 67 Table of Contents The fair values of our outstanding derivative instruments were as follows (in millions): Balance Sheet Location Fair Value of Derivative Instruments As of December 31, 2010 As of December 31, 2011 Derivative Assets Derivatives designated as hedging instruments: Foreign exchange contracts Prepaid revenue share, expenses and other assets, current and non-current$342$333 Derivatives not designated as hedging instruments: Foreign exchange contracts Prepaid revenue share, expenses and other assets, current 0 4 Total$342$337 Derivative Liabilities Derivatives designated as hedging instruments: Foreign exchange contracts Accrued expenses and other current liabilities$5$5 Derivatives not designated as hedging instruments: Foreign exchange contracts Accrued expenses and other current liabilities 3 1 Total$8$6 The effect of derivative instruments in cash flow hedging relationships on income and other comprehensive income is summarized below (in millions): Increase(Decrease)in Gains Recognized in AOCI on Derivatives Before Tax Effect(Effective Portion) Year Ended December 31, Derivatives in Cash Flow Hedging Relationship 2009 2010 2011 Foreign exchange contracts$(14)$331$54 Gains Reclassified from AOCI into Income(Effective Portion) Year Ended December 31, Derivatives in Cash Flow Hedging Relationship Location 2009 2010 2011 Foreign exchange contracts Revenues$325$203$43 Gains(Losses)Recognized in Income on Derivatives(Amount Excluded from Effectiveness Testing and Ineffective Portion)(1) Year Ended December 31, Derivatives in Cash Flow Hedging Relationship Location 2009 2010 2011 Foreign exchange contracts Interest and other income,net$(268)$(320)$(323) 1 Gains (losses) related to the ineffective portion of the hedges were not material in all periods presented. 68 Table of Contents The effect of derivative instruments in fair value hedging relationships on income is summarized below (in millions): Gains(Losses)Recognized in Income on Derivatives(2) Year Ended December 31, Derivatives in Fair Value Hedging Relationship Location 2009 2010 2011 Foreign exchange contracts Interest and other income,net$2$(35)$(2) Hedged item Interest and other income, net(2)29(12) Total$0$(6)$(14) 2 Losses related to the amount excluded from effectiveness testing of the hedges were $0, $6 million, and $14 million for the years ended December 31, 2009,December 31, 2010, and December 31, 2011. The effect of derivative instruments not designated as hedging instruments on income is summarized below (in millions): Gains(Losses)Recognized in Income on Derivatives Year Ended December 31, Derivatives Not Designated As Hedging Instruments Location 2009 2010 2011 Foreign exchange contracts Interest and other income,net$(78)$(40)$29 Interest rate contracts Interest and other income,net 0 0(19) $(78)$(40)$10 Note 6.Fair Value Measurements We measure our cash equivalents, marketable securities, auction rate securities (ARS), and foreign currency and interest rate derivative contracts at fair value. Fair value is an exit price, representing the amount that would be received to sell an asset or paid to transfer a liability in an orderly transaction between market participants. As such, fair value is a market-based measurement that should be determined based on assumptions that market participants would use in pricing an asset or a liability. A three-tier fair value hierarchy is established as a basis for considering such assumptions and for inputs used in the valuation methodologies in measuring fair value: Level 1—Observable inputs that reflect quoted prices (unadjusted) for identical assets or liabilities in active markets. Level 2—Include other inputs that are directly or indirectly observable in the marketplace. Level 3—Unobservable inputs that are supported by little or no market activities. The fair value hierarchy also requires an entity to maximize the use of observable inputs and minimize the use of unobservable inputs when measuring fair value. We classify our cash equivalents and marketable securities within Level 1 or Level 2. This is because we value our cash equivalents and marketable securities using quoted market prices or alternative pricing sources and models utilizing market observable inputs. We classify our investments in ARS within Level 3 because they are valued using valuation models with significant unobservable marketable inputs (see below). We classify our foreign currency and interest rate derivative contracts primarily within Level 2 as the valuation inputs are based on quoted prices and market observable data of similar instruments. 69 Table of Contents Assets and liabilities measured at fair value on a recurring basis are summarized below (in millions): Description As of December 31, 2010 Fair Value Measurement at Reporting Date Using Quoted Prices in Active Markets for Identical Assets (Level 1)Significant Other Observable Inputs (Level 2)Significant Unobservable Inputs (Level 3) Assets Cash equivalents: Time deposits$973$0$973$0 Money market and other funds 7,547 5,936 1,611(1)0 U.S. government notes 300 300 0 0 Foreign government bonds 150 0 150 0 Corporate debt securities 8 0 8 0 Marketable securities: Time deposits 307 0 307 0 U.S. government agencies 1,857 0 1,857 0 U.S. government notes 3,930 3,930 0 0 Foreign government bonds 1,172 0 1,172 0 Municipal securities 2,503 0 2,503 0 Corporate debt securities 5,742 0 5,742 0 Agency residential mortgage-backed securities 5,673 0 5,673 0 Marketable equity security 161 161 0 0 Derivative contracts 342 0 342 0 Auction rate securities 153 0 0 153 Total$30,818$10,327$20,338$153 Liabilities Derivative contracts$8$0$8$0 Total$8$0$8$0 Description Fair Value Measurement at Reporting Date Using As of December 31, 2011 Quoted Prices in Active Markets for Identical Assets (Level 1)Significant Other Observable Inputs (Level 2)Significant Unobservable Inputs (Level 3) Assets Cash equivalents: Time deposits$534$0$534$0 Money market and other funds 4,462 3,202 1,260(1)0 U.S. government agencies 275 0 275 0 Marketable securities: Time deposits 495 0 495 0 U.S. government agencies 6,226 0 6,226 0 U.S. government notes 11,579 11,579 0 0 Foreign government bonds 1,629 0 1,629 0 Municipal securities 1,794 0 1,794 0 Corporate debt securities 6,112 0 6,112 0 Agency residential mortgage-backed securities 6,501 0 6,501 0 Marketable equity securities 307 307 0 0 Derivative contracts 337 0 337 0 Auction rate securities 118 0 0 118 Total$40,369$15,088$25,163$118 Liabilities Derivative contracts$6$0$6$0 Total$6$0$6$0 70 Table of Contents (1)This balance represents cash collateral received in connection with our securities lending program, which was invested in reverse repurchase agreements maturing within three months. At December 31, 2011, we held $118 million of ARS. Historically, these securities have provided liquidity through a Dutch auction process. However, these auctions began to fail in the first quarter of 2008. To estimate their fair values at December 31, 2011, we used a discounted cash flow model based on estimated interest rates, timing and amount of cash flows, the credit quality of the underlying securities, and illiquidity considerations. At December 31, 2011, the estimated fair value of these ARS was $23 million below their costs. As we have no intent to sell these ARS and it is more likely than not that we will not be required to sell these ARS prior to recovery of our entire cost basis, we concluded the decline in the fair value was temporary and recorded the unrealized loss to AOCI on the accompanying Consolidated Balance Sheet at December 31, 2011. To the extent we determine that any impairment is other-than-temporary, we would record a charge to earnings. In addition, we have concluded that the auctions for these securities may continue to fail for at least the next 12 months and as a result, we classified them as non-current assets on the accompanying Consolidated Balance Sheet at December 31, 2011. The following table presents reconciliations for our assets measured and recorded at fair value on a recurring basis, using significant unobservable inputs (Level 3) (in millions): Level 3 Balance at January 1, 2009$197 Change in unrealized loss included in other comprehensive income 12 Net settlements(27) Balance at December 31, 2009 182 Change in unrealized loss included in other comprehensive income 4 Net settlements(33) Balance at December 31, 2010 153 Change in unrealized loss included in other comprehensive income(3) Net settlements(32) Balance at December 31, 2011$118 Note 7.Property and Equipment Property and equipment consisted of the following (in millions): As of December 31, 2010 As of December 31, 2011 Information technology assets$4,670$6,060 Land and buildings 3,969 5,228 Construction in progress 2,329 2,128 Leasehold improvements 738 919 Furniture and fixtures 65 65 Total 11,771 14,400 Less: accumulated depreciation and amortization 4,012 4,797 Property and equipment, net$7,759$9,603 71 Table of Contents Note 8.Acquisitions In April 2011, we completed the acquisition of ITA Software, Inc. (ITA), a privately-held flight information software company, for $676 million in cash, of which $394 million was attributed to acquired intangible assets, $323 million to goodwill, and $41 million to net liabilities assumed. During the year ended December 31, 2011, we completed 78 other acquisitions and purchases of intangible assets for a total cash consideration of approximately $1.3 billion, of which $795 million was attributed to goodwill, $593 million to acquired intangible assets, and $86 million to net liabilities assumed. These acquisitions generally enhance the breadth and depth of our expertise in engineering and other functional areas, our technologies, and our product offerings. Pro forma results of operations for these acquisitions have not been presented because they are not material to the consolidated results of operations, either individually or in the aggregate. Patents and developed technology have a weighted-average useful life of 6.3 years, customer relationships have a weighted-average useful life of 6.6 years and trade names and other have a weighted-average useful life of 4.2 years. The amount of goodwill expected to be deductible for tax purposes is $29 million. In August 2011, we entered into a Merger Agreement with Motorola, a provider of innovative technologies, products and services that enable a range of mobile and wireline digital communication, information and entertainment experiences, under which we will acquire Motorola for$40 per share in cash, or a total of approximately $12.5 billion.The completion of this transaction is subject to customary closing conditions,including the receipt of certain regulatory approvals.In the event the Merger Agreement is terminated due to a failure to obtain certain regulatory approvals,we would be required to pay Motorola a fee of $2.5 billion.The transaction is currently expected to close in early 2012. Note 9.Goodwill and Other Intangible Assets The changes in the carrying amount of goodwill for the year ended December 31, 2011 were as follows (in millions): Balance as of December 31, 2010$6,256 Goodwill acquired 1,118 Goodwill adjustment(28) Balance as of December 31, 2011$7,346 Information regarding our acquisition-related intangible assets is as follows (in millions): As of December 31, 2010 Gross Carrying Amount Accumulated Amortization Net Carrying Value Patents and developed technology$915$506$409 Customer relationships 950 400 550 Trade names and other 283 198 85 Total$2,148$1,104$1,044 As of December 31, 2011 Gross Carrying Amount Accumulated Amortization Net Carrying Value Patents and developed technology$1,451$698$753 Customer relationships 1,288 573 715 Trade names and other 359 249 110 Total$3,098$1,520$1,578 72 Table of Contents Patents and developed technology, customer relationships, and trade names and other have weighted-average useful lives from the date of purchase of 5.0 years, 6.1 years, and 4.3 years. Amortization expense of acquisition-related intangible assets for the years ended December 31, 2009, 2010, and 2011 was $266 million, $314 million, and $441 million. As of December 31, 2011, expected amortization expense for acquisition-related intangible assets for each of the next five years and thereafter was as follows (in millions): 2012$472 2013 364 2014 297 2015 147 2016 99 Thereafter 199 $1,578 Note 10.Interest and Other Income, Net The components of interest and other income, net were as follows (in millions): Year Ended December 31, 2009 2010 2011 Interest income$230$579$812 Interest expense 0(5)(58) Realized gains on available-for-sale investments, net 97 185 254 Impairment of equity investments 0 0(110) Foreign currency exchange losses, net(260)(355)(379) Other 2 11 65 Interest and other income, net$69$415$584 Note 11.Accumulated Other Comprehensive Income The components of accumulated other comprehensive income are as follows (in millions): As of December 31, 2010 2011 Foreign currency translation adjustment$(41)$(148) Net unrealized gains on available-for-sale investments, net of taxes 94 327 Unrealized gains on cash flow hedges, net of taxes 85 97 Accumulated other comprehensive income$138$276 Note 12.Commitments and Contingencies Operating Leases We have entered into various non-cancelable operating lease agreements for certain of our offices, land, and data centers throughout the world with original lease periods expiring primarily between 2012 and 2063. We are committed to pay a portion of the actual operating expenses under certain of these lease agreements. These operating expenses are not included in the table below. Certain of these arrangements have free or escalating rent payment provisions. We recognize rent expense under such arrangements on a straight-line basis. 73 Table of Contents At December 31, 2011, future minimum payments under non-cancelable operating leases, net of sublease income amounts, were as follows over each of the next five years and thereafter (in millions): Operating Leases Sub-lease Income Net Operating Leases 2012$389$17$372 2013 377 16 361 2014 357 14 343 2015 311 13 298 2016 256 10 246 Thereafter 1,264 5 1,259 Total minimum payments$2,954$75$2,879 Certain leases have adjustments for market provisions. Amounts in the above table represent our best estimates of future payments to be made under these leases. In addition, the above table does not include future rental income of $726 million related to the leases that we assumed in connection with our building purchases. Rent expense under operating leases, including co-location arrangements, was $323 million, $293 million, and $380 million in 2009, 2010, and 2011. Purchase Obligations We had $1.9 billion of other non-cancelable contractual obligations, primarily related to payments related to certain of our distribution arrangements and video and other content licensing revenue sharing arrangements, as well as data center operations and facility build-outs at December 31, 2011. In addition, we had $2.8 billion of open purchase orders for which we had not received the related services or goods at December 31, 2011. We have the right to cancel these open purchase orders prior to the date of delivery. Letters of Credit At December 31, 2011, we had unused letters of credit for $46 million. Indemnifications In the normal course of business to facilitate transactions of our services and products, we indemnify certain parties, including advertisers, Google Network Members, and lessors with respect to certain matters. We have agreed to hold certain parties harmless against losses arising from a breach of representations or covenants, or out of intellectual property infringement or other claims made against certain parties. Several of these agreements limit the time within which an indemnification claim can be made and the amount of the claim. In addition, we have entered into indemnification agreements with our officers and directors, and our bylaws contain similar indemnification obligations to our agents. It is not possible to make a reasonable estimate of the maximum potential amount under these indemnification agreements due to the unique facts and circumstances involved in each particular agreement. Additionally, we have a limited history of prior indemnification claims and the payments we have made under such agreements have not had a material adverse effect on our results of operations, cash flows, or financial position. However, to the extent that valid indemnification claims arise in the future, future payments by us could be significant and could have a material adverse effect on our results of operations or cash flows in a particular period. As of December 31, 2011, we did not have any indemnification claims that we considered to be probable or reasonably possible. 74 Table of Contents Legal Matters Antitrust Investigations On June 23, 2011, we received a Civil Investigative Demand (CID) from the U.S. Federal Trade Commission’s (FTC) Bureau of Competition and a subpoena from FTC’s Bureau of Consumer Protection relating to a review by the FTC of our business practices, including search and advertising. State attorneys general from the states of Texas, Ohio, and Mississippi have issued similar CIDs. We are cooperating with the FTC and the state attorneys general and responding to their information requests. The European Commission’s (EC) Directorate General for Competition has also opened an investigation into various antitrust-related complaints against us. On February 10, 2010, we received notification from the EC about three antitrust complaints filed by Ciao, Ejustice, and Foundem, respectively. On November 30, 2010, the EC formally opened proceedings against us. Since November 2010, 1plusV, parent company of Ejustice, and VfT, an association of business listings providers in Germany, have filed similar complaints against us. On March 31, 2011, Microsoft Corporation submitted a similar complaint to the EC against us. On the same day, the EC notified us of additional complaints filed by Elfvoetbal, Hotmaps, Interactive Labs, and nnpt.it, and on August 30, 2011 of a complaint by dealdujour.pro. On September 16, 2011, we responded to all of the allegations made against us. In addition, in December 2011, the Spanish Association of Daily Newspaper Publishers also submitted a complaint to the EC against us. We are cooperating with the EC and responding to its information requests. EPA Investigation In February 2009, we learned of a U.S. Environmental Protection Agency (EPA) investigation into an alleged release of refrigerant at one of our smaller data center facilities, which we acquired from DoubleClick, and the accuracy of related statements and records. We are cooperating with the EPA and have provided documents and other materials. Department of Justice Investigation (Advertising) In connection with the resolution of an investigation by the United States Department of Justice into the use of Google advertising by certain advertisers, we accrued $500 million during the three months ended March 31, 2011, which was paid in August 2011 upon final resolution of that matter. Patent and Intellectual Property Claims We have had patent, copyright, and trademark infringement lawsuits filed against us claiming that certain of our products, services, and technologies, including Android, Google Search, Google AdWords, Google AdSense, Google Books, Google News, Google Image Search, Google Chrome, Google Talk, Google Voice, and YouTube, infringe the intellectual property rights of others. Adverse results in these lawsuits may include awards of substantial monetary damages, costly royalty or licensing agreements, or orders preventing us from offering certain features, functionalities, products, or services, and may also cause us to change our business practices, and require development of non-infringing products or technologies, which could result in a loss of revenues for us and otherwise harm our business. In addition, many of our agreements with our customers and partners require us to indemnify them for certain intellectual property infringement claims against them, which would increase our costs as a result of defending such claims, and may require that we pay significant damages if there were an adverse ruling in any such claims. Furthermore, such customers and partners may discontinue the use of our products, services, and technologies, as a result of injunctions or otherwise, which could result in loss of revenues and adversely impact our business. Other We are also regularly subject to claims, suits, government investigations, and other proceedings involving competition and antitrust (such as the pending investigations by the FTC and the EC described above), intellectual 75 Table of Contents property, privacy, tax, labor and employment, commercial disputes, content generated by our users, goods and services offered by advertisers or publishers using our platforms, and other matters. Such claims, suits, government investigations, and other proceedings could result in fines, civil or criminal penalties, or other adverse consequences. Certain of our outstanding legal matters include speculative claims for substantial or indeterminate amounts of damages. We record a liability when we believe that it is both probable that a loss has been incurred, and the amount can be reasonably estimated. We evaluate, on a monthly basis, developments in our legal matters that could affect the amount of liability that has been previously accrued, and make adjustments as appropriate. Significant judgment is required to determine both likelihood of there being and the estimated amount of a loss related to such matters. With respect to our outstanding legal matters,based on our current knowledge, we believe that the amount or range of reasonably possible loss will not, either individually or in the aggregate, have a material adverse effect on our business, consolidated financial position, results of operations, or cash flows.However, the outcome of such legal matters is inherently unpredictable and subject to significant uncertainties. We expense legal fees in the period in which they are incurred. Income Taxes We are under audit by the Internal Revenue Service (IRS) and various other tax authorities. We have reserved for potential adjustments to our provision for income taxes that may result from examinations by, or any negotiated agreements with, these tax authorities, and we believe that the final outcome of these examinations or agreements will not have a material effect on our results of operations. If events occur which indicate payment of these amounts is unnecessary, the reversal of the liabilities would result in the recognition of tax benefits in the period we determine the liabilities are no longer necessary. If our estimates of the federal, state, and foreign income tax liabilities are less than the ultimate assessment, a further charge to expense would result. Note 13.Stockholders’ Equity Convertible Preferred Stock Our board of directors has authorized 100,000,000 shares of convertible preferred stock, $0.001 par value, issuable in series. At December 31, 2010 and December 31, 2011, there were no shares issued or outstanding. Class A and Class B Common Stock Our board of directors has authorized two classes of common stock, Class A and Class B. At December 31, 2011, there were 6,000,000,000 and 3,000,000,000 shares authorized and there were 257,552,401 and 67,342,362 shares outstanding of Class A and Class B common stock. The rights of the holders of Class A and Class B common stock are identical, except with respect to voting. Each share of Class A common stock is entitled to one vote per share. Each share of Class B common stock is entitled to 10 votes per share. Shares of Class B common stock may be converted at any time at the option of the stockholder and automatically convert upon sale or transfer to Class A common stock. We refer to Class A and Class B common stock as common stock throughout the notes to these financial statements, unless otherwise noted. Stock Plans We maintain the 1998 Stock Plan, the 2000 Stock Plan, the 2003 Stock Plan, the 2003 Stock Plan (No.2), the 2003 Stock Plan (No. 3), the 2004 Stock Plan, and plans assumed through acquisitions, all of which are collectively referred to as the “Stock Plans.” Under our Stock Plans, incentive and nonqualified stock options or rights to purchase common stock may be granted to eligible participants. Options are generally granted for a term 76 Table of Contents of 10 years. Options granted under the Stock Plans other than the 2004 Stock Plan may be exercised prior to vesting. Under the Stock Plans, we have also issued RSUs and restricted shares. An RSU award is an agreement to issue shares of our stock at the time of vest. Except for options granted pursuant to our stock option exchange program completed in March 2009 (the Exchange), options granted and RSUs issued to employees under the Stock Plans generally vest over four years contingent upon employment with us on the vesting date. At December 31, 2010 and December 31, 2011, there were 27,329,837 and 21,794,492 shares of common stock reserved for future issuance under our Stock Plans. We estimated the fair value of each option award on the date of grant using the BSM option pricing model. Our assumptions about stock-price volatility have been based exclusively on the implied volatilities of publicly traded options to buy our stock with contractual terms closest to the expected life of options granted to our employees. We estimate the expected term based upon the historical exercise behavior of our employees. The risk-free interest rate for periods within the contractual life of the award is based on the U.S. Treasury yield curve in effect at the time of grant. The following table presents the weighted-average assumptions used to estimate the fair values of the stock options granted (excluding options granted in connection with the Exchange discussed below) in the periods presented: Year Ended December 31, 2009 2010 2011 Risk-free interest rate 2.6%1.9%2.3% Expected volatility 37%35%33% Expected life (in years)5.8 5.4 5.9 Dividend yield 0 0 0 Weighted-average estimated fair value of options granted during the year$160.63$216.43$210.07 The following table summarizes the activities for our options for the year ended December 31, 2011: Options Outstanding Number of Shares Weighted- Average Exercise Price Weighted- Average Remaining Contractual Term (in years)Aggregate Intrinsic Value (in millions)(1) Balance at December 31, 2010 11,525,422$330.24 Granted 718,091$584.80 Exercised(2,181,355)$284.00 Forfeited/canceled(254,906)$377.73 Balance at December 31, 2011 9,807,252$357.92 5.7$2,825 Vested and exercisable as of December 31, 2011 6,244,783$314.38 5.4$2,070 Vested and exercisable as of December 31, 2011 and expected to vest thereafter(2)9,437,288$354.92 5.7$2,746 (1)The aggregate intrinsic value is calculated as the difference between the exercise price of the underlying awards and the closing stock price of $645.90 of our Class A common stock on December 31, 2011. (2)Options expected to vest reflect an estimated forfeiture rate. 77 Table of Contents The following table summarizes additional information regarding outstanding, exercisable, and exercisable and vested stock options at December 31, 2011: Options Outstanding Options Exercisable Options Exercisable and Vested Range of Exercise Prices Number of Shares Weighted- Average Remaining Life (in years)Weighted- Average Exercise Price Number of Shares Weighted- Average Exercise Price Number of Shares Weighted- Average Exercise Price $0.30–$94.80 297,360 2.6$28.06 296,392$27.85 278,071$24.73 $117.84–$198.41 412,843 2.9$179.12 412,843$179.12 412,843$179.12 $205.96–$298.86 505,760 3.2$275.21 502,038$275.12 502,038$275.12 $300.97–$399.00 5,478,754 5.0$309.68 4,012,895$309.71 4,012,895$309.71 $401.78–$499.07 935,944 6.6$439.16 514,417$437.28 514,417$437.28 $501.27–$595.35 1,892,340 8.4$532.44 509,780$524.69 509,780$524.69 $601.17–$699.35 284,171 9.1$610.18 14,659$612.09 14,659$612.09 $710.84 80 6.0$710.84 80$710.84 80$710.84 $0.30–$710.84 9,807,252 5.7$357.92 6,263,104$313.68 6,244,783$314.38 The above tables include approximately 1.2 million warrants held by selected financial institutions that were options purchased from employees under our TSO program, with a weighted-average exercise price of $336.64 and a weighted-average remaining life of 1.1 years. During 2011, the number of shares underlying TSOs sold to selected financial institutions under the TSO program was 635,047 at a total value of $167 million, or an average of $262.74 per share, including an average premium of $11.09 per share. The premium is calculated as the difference between (a)the sale price of the TSO and (b)the intrinsic value of the TSO, which we define as the excess, if any, of the price of our Class A common stock at the time of the sale over the exercise price of the TSO. The total grant date fair value of stock options vested during 2009, 2010, and 2011 was $690 million, $690 million, and $561 million. The aggregate intrinsic value of all options and warrants exercised during 2009, 2010, and 2011 was $566 million, $794 million, and $674 million. These amounts do not include the aggregate sales price of options sold under our TSO program. As of December 31, 2011, there was $583 million of unrecognized compensation cost related to outstanding employee stock options. This amount is expected to be recognized over a weighted-average period of 2.1 years. To the extent the actual forfeiture rate is different from what we have estimated, stock-based compensation related to these awards will be different from our expectations. The following table summarizes the activities for our unvested RSUs for the year ended December 31, 2011: Unvested Restricted Stock Units Number of Shares Weighted- Average Grant-Date Fair Value Unvested at December 31, 2010 6,671,971$509.04 Granted 5,260,421$531.20 Vested(2,691,549)$516.22 Forfeited/canceled(418,195)$538.01 Unvested at December 31, 2011 8,822,648$520.27 Expected to vest after December 31, 2011(1)7,903,328$520.27 (1)RSUs expected to vest reflect an estimated forfeiture rate. 78 Table of Contents As of December 31, 2011, there was $3,727 million of unrecognized compensation cost related to unvested employee RSUs. This amount is expected to be recognized over a weighted-average period of 2.9 years. To the extent the actual forfeiture rate is different from what we have estimated, stock-based compensation related to these awards will be different from our expectations. Note 14.401(k) Plan We have a 401(k) Savings Plan (401(k)Plan)that qualifies as a deferred salary arrangement under Section 401(k) of the Internal Revenue Code. Under the 401(k) Plan, participating employees may elect to contribute up to 60% of their eligible compensation, subject to certain limitations. Employee and our contributions are fully vested when contributed. We contributed approximately $83 million, $100 million, and $136 million during 2009, 2010, and 2011. Note 15.Income Taxes Income before income taxes included income from domestic operations of $3,579 million, $4,948 million, and $4,693 million for 2009, 2010, and 2011, and income from foreign operations of $4,802 million, $5,848 million, and $7,633 million for 2009, 2010, and 2011. Substantially all of the income from foreign operations was earned by an Irish subsidiary. The provision for income taxes consists of the following (in millions): Year Ended December 31, 2009 2010 2011 Current: Federal$1,531$1,657$1,724 State 450 458 274 Foreign 148 167 248 Total 2,129 2,282 2,246 Deferred: Federal(273)(25)452 State 13 47(109) Foreign(8)(13)(0) Total(268)9 343 Provision for income taxes$1,861$2,291$2,589 The reconciliation of federal statutory income tax rate to our effective income tax rate is as follows (in millions): Year ended December 31, 2009 2010 2011 Expected provision at federal statutory tax rate (35%)$2,933$3,779$4,314 State taxes, net of federal benefit 302 322 122 Stock-based compensation expense 63 79 105 Change in valuation allowance(41)(34)27 Foreign rate differential(1,339)(1,769)(2,001) Federal research credit(56)(84)(140) Tax exempt interest(15)(12)(10) Non-deductible legal settlement 0 0 175 Other permanent differences 14 10(3) Provision for income taxes$1,861$2,291$2,589 79 Table of Contents We have not provided U.S. income taxes and foreign withholding taxes on the undistributed earnings of foreign subsidiaries as of December 31, 2011 because we intend to permanently reinvest such earnings outside the U.S. If these foreign earnings were to be repatriated in the future, the related U.S. tax liability may be reduced by any foreign income taxes previously paid on these earnings. As of December 31, 2011, the cumulative amount of earnings upon which U.S. income taxes have not been provided is approximately $24.8 billion. Determination of the amount of unrecognized deferred tax liability related to these earnings is not practicable. Deferred Tax Assets Deferred income taxes reflect the net effects of temporary differences between the carrying amounts of assets and liabilities for financial reporting purposes and the amounts used for income tax purposes. Significant components of our deferred tax assets and liabilities are as follows (in millions): As of December 31, 2010 2011 Deferred tax assets: Stock-based compensation expense$299$288 State taxes 207 138 Capital loss from impairment of equity investments 292 285 Settlement with the Authors Guild and AAP 39 35 Depreciation and amortization 20 0 Vacation accruals 35 52 Deferred rent 34 43 Accruals and reserves not currently deductible 244 268 Acquired net operating losses 132 156 State tax credit 0 55 Other 0 11 Total deferred tax assets 1,302 1,331 Valuation allowance(292)(333) Total deferred tax assets net of valuation allowance 1,010 998 Deferred tax liabilities: Depreciation and amortization 0(479) Identified intangibles(308)(398) Unrealized gains on investments and other(56)(90) Other prepaids(95)(70) Other(27)(33) Total deferred tax liabilities(486)(1,070) Net deferred tax assets (liabilities)$524$(72) 80 Table of Contents As of December 31, 2011, our federal and state net operating loss carryforwards for income tax purposes were approximately $420 million and $310 million. If not utilized, the federal net operating loss carryforwards will begin to expire in 2018 and the state net operating loss carryforwards will begin to expire in 2014. The net operating loss carryforwards are subject to various annual limitations under Section 382 of the Internal Revenue Code. As of December 31, 2011, our California research and development credit carryforwards for income tax purposes were approximately $55 million that can be carried over indefinitely. We believe it is more likely than not that a portion of the state tax credit will not be realized. Therefore, we have recorded a valuation allowance on the state tax credit carryforward in the amount of $48 million. We will reassess the valuation allowance quarterly and if future evidence allows for a partial or full release of the valuation allowance, a tax benefit will be recorded accordingly. As of December 31, 2011, our federal and state capital loss carryforwards for income tax purposes were approximately $165 million and $422 million. We also have deferred tax assets for impairment losses that, if recognized, will be capital in nature. We believe that it is more likely than not that our deferred tax assets for capital losses and impairment losses will not be realized. Therefore, we have recorded a valuation allowance on both our federal and state deferred tax assets for these items in the amount of $285 million. We will reassess the valuation allowance quarterly and if future evidence allows for a partial or full release of the valuation allowance, a tax benefit will be recorded accordingly. Uncertain Tax Positions The following table summarizes the activity related to our gross unrecognized tax benefits from January 1, 2009 to December 31, 2011 (in millions): Balance as of January 1, 2009$721 Increases related to prior year tax positions 222 Decreases related to prior year tax positions(1) Increases related to current year tax positions 246 Balance as of December 31, 2009 1,188 Increases related to prior year tax positions 37 Decreases related to prior year tax positions(197) Decreases related to settlement with tax authorities(47) Decreases as a result of a lapse of applicable statute of limitation(97) Increases related to current year tax positions 256 Balance as of December 31, 2010 1,140 Increases related to prior year tax positions 77 Decreases related to prior year tax positions(9) Increases related to current year tax positions 361 Decreases related to settlement with tax authorities(5) Balance as of December 31, 2011$1,564 Our total unrecognized tax benefits that, if recognized, would affect our effective tax rate were $814 million, $951 million, and $1,350 million as of December 31, 2009, 2010, and 2011. As of December 31, 2010 and 2011, we had accrued $97 million and $129 million for payment of interest and penalties. Interest and penalties included in our provision for income taxes were not material in all the periods presented. We and our subsidiaries are routinely examined by various taxing authorities. Although we file U.S. federal, U.S. state, and foreign tax returns, our two major tax jurisdictions are the U.S. and Ireland. During the three months ended December 31, 2007, the IRS completed its examination of our 2003 and 2004 tax years. We have filed an 81 Table of Contents appeal with the IRS for certain issues related to this audit and no resolution of the issues has been achieved at this time, but we believe we have adequately provided for these items and any adverse results would have an immaterial impact on our unrecognized tax benefit balance within the next 12 months. The IRS is currently in examination of our 2007, 2008, and 2009 tax years. We do not expect the examination to be completed within the next 12 months. Therefore, we do not anticipate any significant impact to our unrecognized tax benefit balance in 2012, related to our 2007, 2008, and 2009 tax years. Our 2010 and 2011 tax years remain subject to examination by the IRS for U.S. federal tax purposes, and our 2003 through 2011 tax years remain subject to examination by the appropriate governmental agencies for Irish tax purposes. There are various other ongoing audits in various other jurisdictions that are not material to our financial statements. Note 16.Information about Geographic Areas Our chief operating decision-makers (i.e., chief executive officer and his direct reports) review financial information presented on a consolidated basis, accompanied by disaggregated information about revenues by geographic region for purposes of allocating resources and evaluating financial performance. There are no segment managers who are held accountable by our chief operating decision-makers, or anyone else, for operations, operating results, and planning for levels or components below the consolidated unit level. Accordingly, we consider ourselves to be in a single reporting segment and operating unit structure. Revenues by geography are based on the billing addresses of our customers. The following tables set forth revenues and long-lived assets by geographic area (in millions): Year Ended December 31, 2009 2010 2011 Revenues: United States$11,194$14,056$17,560 United Kingdom 2,986 3,329 4,057 Rest of the world 9,471 11,936 16,288 Total revenues$23,651$29,321$37,905 As of December 31, 2010 2011 Long-lived assets: United States$14,000$15,963 International 2,289 3,853 Total long-lived assets$16,289$19,816 82 Table of Contents ITEM 9.CHANGES IN AND DISAGREEMENTS WITH ACCOUNTANTS ON ACCOUNTING AND FINANCIAL DISCLOSURE None. ITEM 9A.CONTROLS AND PROCEDURES Evaluation of Disclosure Controls and Procedures Our management, with the participation of our chief executive officer and chief financial officer, evaluated the effectiveness of our disclosure controls and procedures pursuant to Rule 13a-15 under the Exchange Act, as of the end of the period covered by this Annual Report on Form 10-K. Based on this evaluation, our chief executive officer and chief financial officer concluded that, as of December 31, 2011, our disclosure controls and procedures are designed at a reasonable assurance level and are effective to provide reasonable assurance that information we are required to disclose in reports that we file or submit under the Exchange Act is recorded, processed, summarized, and reported within the time periods specified in the SEC’s rules and forms, and that such information is accumulated and communicated to our management, including our chief executive officer and chief financial officer, as appropriate, to allow timely decisions regarding required disclosure. Changes in Internal Control over Financial Reporting There were no changes in our internal control over financial reporting that occurred during the quarter ended December 31, 2011 that have materially affected, or are reasonably likely to materially affect, our internal control over financial reporting. Management’s Report on Internal Control over Financial Reporting Our management is responsible for establishing and maintaining adequate internal control over financial reporting, as defined in Rule 13a-15(f) of the Exchange Act. Our management conducted an evaluation of the effectiveness of our internal control over financial reporting based on the framework in Internal Control—Integrated Framework issued by the Committee of Sponsoring Organizations of the Treadway Commission. Based on this evaluation, management concluded that our internal control over financial reporting was effective as of December 31, 2011. Management reviewed the results of its assessment with our Audit Committee. The effectiveness of our internal control over financial reporting as of December 31, 2011 has been audited by Ernst& Young LLP, an independent registered public accounting firm, as stated in its report which is included in Item 8 of this Annual Report on Form 10-K. Limitations on Effectiveness of Controls and Procedures In designing and evaluating the disclosure controls and procedures, management recognizes that any controls and procedures, no matter how well designed and operated, can provide only reasonable assurance of achieving the desired control objectives. In addition, the design of disclosure controls and procedures must reflect the fact that there are resource constraints and that management is required to apply its judgment in evaluating the benefits of possible controls and procedures relative to their costs. ITEM 9B.OTHER INFORMATION None. 83 Table of Contents PART III ITEM 10.DIRECTORS, EXECUTIVE OFFICERS AND CORPORATE GOVERNANCE The information required by this item will be included under the caption “Directors, Executive Officers and Corporate Governance” in our Proxy Statement for the 2012 Annual Meeting of Stockholders to be filed with the SEC within 120 days of the fiscal year ended December 31, 2011 (2012 Proxy Statement) and is incorporated herein by reference. The information required by this item regarding delinquent filers pursuant to Item 405 of Regulation S-K will be included under the caption “Section 16(a) Beneficial Ownership Reporting Compliance” in the 2012 Proxy Statement and is incorporated herein by reference. ITEM 11.EXECUTIVE COMPENSATION The information required by this item will be included under the captions “Director Compensation,” “Executive Compensation” and “Directors, Executive Officers and Corporate Governance—Corporate Governance and Board Matters—Compensation Committee Interlocks and Insider Participation” in the 2012 Proxy Statement and is incorporated herein by reference. ITEM 12.SECURITY OWNERSHIP OF CERTAIN BENEFICIAL OWNERS AND MANAGEMENT AND RELATED STOCKHOLDER MATTERS The information required by this item will be included under the captions “Common Stock Ownership of Certain Beneficial Owners and Management” and “Equity Compensation Plan Information” in the 2012 Proxy Statement and is incorporated herein by reference. ITEM 13.CERTAIN RELATIONSHIPS AND RELATED TRANSACTIONS, AND DIRECTOR INDEPENDENCE The information required by this item will be included under the captions “Certain Relationships and Related Transactions” and “Directors, Executive Officers and Corporate Governance—Corporate Governance and Board Matters—Director Independence” in the 2012 Proxy Statement and is incorporated herein by reference. ITEM 14.PRINCIPAL ACCOUNTING FEES AND SERVICES The information required by this item will be included under the caption “Independent Registered Public Accounting Firm” in the 2012 Proxy Statement and is incorporated herein by reference. 84 Table of Contents PART IV ITEM 15.EXHIBITS, FINANCIAL STATEMENT SCHEDULES (a)We have filed the following documents as part of this Annual Report on Form 10-K: Consolidated Financial Statements Reports of Independent Registered Public Accounting Firm 49 Financial Statements: Consolidated Balance Sheets 51 Consolidated Statements of Income 52 Consolidated Statements of Comprehensive Income 53 Consolidated Statements of Stockholders’ Equity 54 Consolidated Statements of Cash Flows 55 Notes to Consolidated Financial Statements 56 Financial Statement Schedules Schedule II: Valuation and Qualifying Accounts Allowance for Doubtful Accounts and Sales Credits Balance at Beginning of Year Charged to Expenses/ Against Revenue Write-Offs, Net of Recoveries Balance at End of Year (In millions) Year ended December 31, 2009$80$149$(150)$79 Year ended December 31, 2010$79$200$(178)$101 Year ended December 31, 2011$101$214$(182)$133 Note:Additions to the allowance for doubtful accounts are charged to expense. Additions to the allowance for sales credits are charged against revenues. All other schedules have been omitted because they are not required, not applicable, or the required information is otherwise included. Exhibits See the Exhibit Index immediately following the signature page of this Annual Report on Form 10-K. 85 Table of Contents SIGNATURES Pursuant to the requirements of Section 13 or 15(d) of the Securities Exchange Act of 1934, the registrant has duly caused this Annual Report on Form 10-K to be signed on its behalf by the undersigned, thereunto duly authorized. Date: January 26, 2012 GOOGLE INC. By:/S/L ARRY P AGE Larry Page Chief Executive Officer POWER OF ATTORNEY KNOW ALL PERSONS BY THESE PRESENTS, that each person whose signature appears below constitutes and appoints Larry Page and Patrick Pichette, jointly and severally, his or her attorney-in-fact, with the power of substitution, for him or her in any and all capacities, to sign any amendments to this Annual Report on Form 10-K and to file the same, with exhibits thereto and other documents in connection therewith, with the Securities and Exchange Commission, hereby ratifying and confirming all that each of said attorneys-in-fact, or his or her substitute or substitutes, may do or cause to be done by virtue hereof. Pursuant to the requirements of the Securities Exchange Act of 1934, this Annual Report on Form 10-K has been signed below by the following persons on behalf of the registrant and in the capacities and on the dates indicated. Signature Title Date /S/L ARRY P AGE Larry Page Chief Executive Officer, Co-Founder and Director (Principal Executive Officer)January 26, 2012 /S/P ATRICK P ICHETTE Patrick Pichette Senior Vice President and Chief Financial Officer (Principal Financial and Accounting Officer)January 26, 2012 Eric Schmidt Executive Chairman /S/S ERGEY B RIN Sergey Brin Co-Founder and Director January 26, 2012 /S/L. J OHN D OERR L. John Doerr Director January 26, 2012 /S/D IANE B. G REENE Diane B. Greene Director January 26, 2012 /S/J OHN L. H ENNESSY John L. Hennessy Director January 26, 2012 /S/A NN M ATHER Ann Mather Director January 26, 2012 /S/P AUL S. O TELLINI Paul S. Otellini Director January 26, 2012 /S/K. R AM S HRIRAM K. Ram Shriram Director January 26, 2012 /S/S HIRLEY M. T ILGHMAN Director January 26, 2012 Shirley M. Tilghman Table of Contents EXHIBIT INDEX Exhibit Number Description Incorporated by reference herein Form Date 1.01 Form of Distribution Agreement, dated April 20, 2007, among Google Inc., Morgan Stanley & Co. Incorporated, Citigroup Global Markets Inc., Credit Suisse Securities (USA) LLC and UBS Securities LLC (Distribution Agreement)Current Report on Form 8-K (File No.000-50726)April 23,2007 1.01.1 Amendment No. 1 to the Distribution Agreement among Google Inc. and J.P. Morgan Securities Inc. entered into as of July 20, 2007 Quarterly Report on Form 10-Q (File No.000-50726)August 9,2007 1.01.2 Amendment Agreement, dated as of July 12, 2011, among Google Inc., Morgan Stanley& Co. LLC, Citigroup Global Markets Inc., Credit Suisse Management LLC, Credit Suisse Securities (USA) LLC, UBS AG, London Branch, and UBS Securities LLC Current Report on Form 8-K (File No.000-50726)July 12,2011 1.02 Form of Bidding Rules Agreement, dated April 20, 2007, among Google Inc., Morgan Stanley & Co. 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1445	https://pmc.ncbi.nlm.nih.gov/articles/PMC1100756/	(G–H)•–C and G–(C–H)• radicals derived from the guanine·cytosine base pair cause DNA subunit lesions - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Proc Natl Acad Sci U S A . 2005 Apr 6;102(19):6698–6703. doi: 10.1073/pnas.0408644102 Search in PMC Search in PubMed View in NLM Catalog Add to search (G–H)•–C and G–(C–H)• radicals derived from the guanine·cytosine base pair cause DNA subunit lesions Partha Pratim Bera Partha Pratim Bera 1 Center for Computational Chemistry, University of Georgia, Athens, GA 30602 Find articles by Partha Pratim Bera 1, Henry F Schaefer III Henry F Schaefer III 1 Center for Computational Chemistry, University of Georgia, Athens, GA 30602 Find articles by Henry F Schaefer III 1, Author information Article notes Copyright and License information 1 Center for Computational Chemistry, University of Georgia, Athens, GA 30602 To whom correspondence should be addressed. E-mail: hfs@uga.edu. Edited by Bruce J. Berne, Columbia University, New York, NY, and approved February 15, 2005 Series information Chemical Theory and Computation Special Feature Received 2004 Nov 15; Issue date 2005 May 10. Copyright © 2005, The National Academy of Sciences PMC Copyright notice PMCID: PMC1100756 PMID: 15814617 Abstract The radicals generated by the homolytic cleavage of an X–H bond from the guanine·cytosine (G·C) base pair were studied by using carefully calibrated theoretical methods. The gradient-corrected density functional B3LYP was applied in conjunction with double-ζ plus polarization and diffuse function basis sets. Optimized geometries, energies, and vibrational frequencies were obtained for all of the radicals considered. Structural perturbations along with energy relaxation due to radical formation were investigated. Dissociation energies of the G·C base pair and all of the radicals are predicted and compared with the dissociation energy of neutral G·C. The three lowest-energy base pair radicals all involve removal of an H atom from one of the N atoms in G·C. The lowest-energy base pair radical has the hydrogen atom removed from the guanine nitrogen atom used for the sugar phosphate linkage in DNA. This (G–H)•–C radical has a dissociation energy (to G–H• + C) of 30 kcal/mol, which may be compared with 27 kcal/mol for G·C. All of the radicals that are possible outcomes of direct ionizing radiation or oxidizing species were investigated for the presence of local minima with significant structural changes. Major structural deformations cause strain in the interstrand hydrogen bonding in the DNA double helix. Severe geometry changes were observed when the hydrogen was abstracted from interstrand hydrogen bonding sites, along with sizeable energy changes, indicating the potentially serious consequences to the G·C base pair. The purines and pyrimidines are foundational constituents of DNA and RNA, which are the translators of genetic information from generation to generation. The causes of damage to purine and pyrimidine bases have received significant attention from the scientific community. Both theoretical and experimental research on this important subject have been directed to understanding the underlying chemistry during and after damage to DNA (1). Yet little is known about the possible outcome of damage to the base pairs in terms of energetics and possible structural changes that can lead to strand breakage and therefore loss of genetic information. Sophisticated experimental techniques have been used to investigate the causes and effects of DNA damage and also to determine structural features, as well as delving into the electron-binding ability of the bases and base pairs (1–4). Some of the more advanced experimental and theoretical methods have also been applied to model systems in both the gas and liquid phases to estimate and predict the physical properties and to study hydrogen bonding in purine and pyrimidine base pairs (5–19). Over the years, experimental crystallographic data and theoretical work on the isolated bases and base pairs seemed to disagree on the hydrogen bond lengths and interaction energies of the base pairs. Guerra et al. (5) assign this apparent discrepancy to the effects of molecular environment. They report that solvating the base pairs by H 2 O and metal ions allows them to achieve agreement between experiment and theory with the BP86/TZ2P method (5). Nevertheless, highly complex double-stranded DNA helical patterns pose strong barriers to acquiring valuable experimental data from the DNA. Damage to DNA and RNA base pairs can occur in at least two elementary ways, one through direct radiation and the other through oxidative cleavage of bonds. There are numerous experimental reports available in this respect, documenting the mechanisms involved in the various channels (20, 21). Formation of the radicals from the neutral base pairs is possible due to direct exposure to radiation. Involvement of electronically excited triplet states is also indicated in the formation of radicals from closed-shell neutral molecules (22). The stacked nature of base pairs in DNA is also well suited for fast electron transfer by classes of mutagens causing damage to DNA (23–25). Radicals can bring about significant changes in the geometry of the base pair and overall shape of the DNA strand. Any significant structural change in the base pair can lead to strand breaking, possible mismatches in the pairing, and mutation. Investigations into the cause and effect of radical generation due to radiation and oxidative cleavage are necessary in terms of structural and energetic considerations. Experiments done by several groups prove that even very low-energy electrons can break DNA by various mechanisms (1). Electron attachment followed by single-strand break is one of the processes described (2). The purpose of the present research is to explore the possibility of formation of stable radicals with significant structural changes and also to understand the energy changes associated with such processes. This is an important subject not only because attachment of electrons to DNA base pairs or radical formation (due to high-energy electron impact and/or ionizing radiation) would lead to strand breaks and therefore deter the information flow. Conversely, to the advantage of living things, such processes may break up mutant and potentially dangerous tumor cells. Rearrangement of the radicals to a deformed structure will incorporate increased strain in the rigid DNA strand. The resultant H and base pair radicals may be quenched by thiols in biological systems (20). If not repaired quickly, the damaged DNA mutates, and it may ultimately cause cell death. The neutral single bases have been studied theoretically for their structures and energetic stability and predicted to be nonplanar in the gas phase. Complexation energies upon base pair formation have also been computed for both of the DNA base pairs (26–31), and electron affinities of the single bases (32) and base pairs have been evaluated (27, 33). Because of direct radiation damage and oxidation, hydrogen atoms present in the base pair can be removed, resulting in various neutral as well as cationic radicals. Hutter and Clark (26) predicted a facile proton shift in the G·C+ radical cation along its central hydrogen bond and also reported the ionization potential of the isolated bases. A recent study by Sponer et al. (34) investigated all possible combinations of base pairs in terms of their interaction energies and hydrogen bond lengths at the MP2 and CCSD(T) levels of theory by using a cc-pVTZ basis set. Radicals generated from guanine were also studied by Steenken et al. (35), Cullis et al. (36), and Melvin et al. (37) for their thermodynamic stability and role in strand breaking in DNA. The present research concentrates on the neutral radicals generated by removing one hydrogen atom from the G·C base pair, depicted in Fig. 1. Extracting a hydrogen-bonded H atom from the G·C base pair ruptures interstrand hydrogen bonds that hold the two strands together. Removing an H atom also leads to a radical at the 8 position (see Scheme 1) of the guanine. It is known that 8-oxoguanine is an important lesion that can change the conformational properties of the strand (3, 36, 38). Radicals are formed by reactions with reactive oxygen species in the liquid phase, generated by ionizing radiation, and by direct radiation damage in the gas phase (20). A radical at the 8 position can be a possible intermediate for the formation of the 8-oxo species. Radicals at position 8 both in the isolated guanine and Watson–Crick basepaired models are thus important species to investigate. Fig. 1. Open in a new tab Optimized geometry of the neutral G·C base pair at the B3LYP/DZP++ level of theory. All bond lengths are reported in angstroms. Scheme 1. Open in a new tab Numbering scheme for the G·C system. The present research is a contribution to the chemical physics of biomolecules in the gas phase. The decision to study the isolated base pair radicals is based on two considerations. The first is that any treatment of solvent effects for systems of the size of G·C radicals would degrade the reliability of the theoretical predictions. The second consideration is our conviction that the deepest understanding of biochemistry can result only when the full range of conditions is studied: namely (i) the isolated molecules; (ii) the same molecules in a microsolvated environment (e.g., a finite number of water molecules); and (iii) the fully solvated molecular species. Clearly, from this perspective, the first step is to have definite theoretical predictions for the isolated molecular systems of interest. Methods Optimized geometries, absolute energies, and vibrational frequencies were determined by using the generalized gradient approximation (GGA) exchange-correlation density functional on the Watson–Crick G·C base pair. The three-parameter hybrid Hartree-Fock/density functional theory (HF/DFT) exchange functional, B3 (39), was used in conjunction with the 1988 dynamical correlation functional of Lee, Yang, and Parr (LYP) (40). The dissociation energies of the G·C base pair systems were evaluated as follows. Dissociation Energy (D e) of G–C. Dissociation Energies of the Radicals. and A double-ζ quality basis set, DZP++, with polarization and diffuse functions was used throughout the optimization and vibrational frequency computations (41–44). This DZP++ set was constructed by augmenting the Huzinaga–Dunning set of contracted double-ζ Gaussian functions with one set of p polarization functions on the H atoms and also one set of five d polarization functions on each of the C, N, and O atoms. An even-tempered s diffuse function was added to each H atom and one s and one set of p type even-tempered diffuse function were added to the each of the heavy atoms (41). Therefore the final DZP++ basis set includes six functions per H atom (5s1p/3s/1p) and 19 functions per C, N, or O atom (10s6p1d/5s3p/1d), giving a total of 421 basis functions for the G·C base pair and 415 basis functions for the G·C radicals. Previous research has demonstrated that this level of theory is reliable for molecules for which higher level ab initio calculations are still too expensive, despite the improving speeds of computers (45). All our computations were done by using the q-chem 2.1 suite of density functional programs (46). Stationary points were obtained by analytic gradient methods. Results The crystallographic data (6, 8) available for the base pairs within DNA were compared with results predicted here (Table 1). However the structures of the possible radicals that can be generated from primary radiation damage have never been explored experimentally. Although the radical cations and other lesions have been investigated by using density functional levels of theory (26), neutral radicals have not previously been investigated. Table 1. Bond lengths for the neutral G·C base pair. \| \| Length, Å \| \| :---: \| Technique \| O6-N4 \| N1-N3 \| N2-O2 \| Source \| \| Experiment crystal structure \| 2.91 \| 2.95 \| 2.86 \| Ref. 6 \| \| B3LYP/DZP++ \| 2.76 \| 2.92 \| 2.91 \| This work \| Open in a new tab G·C. In the Watson–Crick model, the DNA double helices are formed from the coiling of two strands of purine and pyrimidine bases. Guanine pairs with cytosine, and adenine pairs with thymine in DNA. Interstrand hydrogen bonds are responsible for this pairing. The interatomic distances obtained in this work at C 1 symmetry by using the B3LYP/DZP++ level of theory do not compare well with the crystallographic conclusions (6, 8). We do find agreement with the theoretical structures reported by Guerra et al. (5) and Richardson et al. (27). All previous G·C computations have been in poor agreement with the experimental crystal structure. Guerra et al. (5) have discussed possible reasons for the discrepancies. Fig. 1 shows the optimized geometry of the G·C base pair with the B3LYP/DZP++ method. The apparent lack of agreement between theory and experiment for the hydrogen bond lengths has been analyzed by Guerra et al. (5). They conclude that the gas-phase geometry of G·C is inherently different from the observed crystal structure. The dissociation energies of G·C, (G–H)•–C, and G–(C–H)• are reported in Table 2. Our predicted G·C dissociation energy D e(G·C) = 27.2 kcal/mol is the same as that of Richardson et al. (27). The only experimental dissociation energy is the 1979 value D 0(G·C) = 21.0 kcal/mol of Yanson et al. (31). The reliability of this experiment is in doubt. The most satisfactory value of D e(G·C) is that provided by Sponer et al. (30). These authors estimate the Hartree–Fock limit to be 24.6 kcal/mol, the basis set limit for MP2 to be 25.8 kcal/mol, and the final estimated D e to be 26.3 kcal/mol. In the most recent paper on this subject, Sponer et al. (34) conclude that the true value of D e(G·C) may be slightly higher. The agreement with the present density functional theory prediction of 27.2 kcal/mol is sufficiently encouraging to have a degree of confidence in our radical dissociation energies, for which no previous results exist. Table 2. Energies of the isolated guanine and cytosine radicals and the G–C radicals; dissociation energies of the optimized G–C radicals. \| Radical \| Energy, hartrees \| Isolated guanine species \| Energy, hartrees \| Isolated cytosine species \| Energy, hartrees \| Dissociation energies (D e) of the radicals, kcal/mol \| :---: :---: :---: \| G·C \| -937.71968 \| G \| -542.66272 \| C \| -395.01372 \| 27.2 \| \| \| \| \| \| \| \| 27.2 (ref. 27), 23.0 (ref. 28), 23.8 (ref. 29), 26.3 (ref. 30), 21.0 (experiment, ref. 31) \| \| G(N9)•-C \| -937.06740 \| G(N9)• \| -542.00593 \| C \| -395.01372 \| 30.0 \| \| G(N2b)•-C \| -937.05049 \| G(N2b)• \| -542.00306 \| C \| -395.01372 \| 21.1 \| \| G(N1)•-C \| -937.04878 \| G(N1)• \| -542.00325 \| C \| -395.01372 \| 20.0 \| \| G-C(N1)• \| -937.04656 \| G \| -542.66272 \| C(N1)• \| -394.34180 \| 26.4 \| \| G(N2a)•-C \| -937.04331 \| G(N2a)• \| -542.01088 \| C \| -395.01372 \| 11.7 \| \| G-C(N4b)• \| -937.03130 \| G \| -542.66272 \| C(N4b)• \| -394.33784 \| 19.3 \| \| G-C(C6)• \| -937.03041 \| G \| -542.66272 \| C(C6)• \| -394.32708 \| 25.5 \| \| G(C8)•-C \| -937.02625 \| G(C8)• \| -541.96817 \| C \| -395.01372 \| 27.8 \| \| G-C(C5)• \| -937.02464 \| G \| -542.66272 \| C(C5)• \| -394.32042 \| 26.0 \| \| G-C(N4a)• \| -937.01560 \| G \| -542.66272 \| C(N4a)• \| -394.32926 \| 14.8 \| Open in a new tab One hartree = 4.4 × 10-18 J. Radicals.G(N1)–C. A generic label is used for each of the radicals investigated in this work. Radical G(N1)–C is generated by homolytically breaking the N1–H bond of guanine. G·C atoms are most conveniently labeled in Scheme 1. The energy of each base pair radical is first obtained without any change in the closed-shell neutral equilibrium geometry. However, upon optimization the energy relaxes by 20.7 kcal/mol and the hydrogen-bonding pattern drastically changes to give the structure shown in Fig. 2. The entire backbone shifts by ≈2.3 Å. Nevertheless, the hydrogen bond lengths do not change much because of optimization to the planar new structure with C s symmetry. This new radical structure is a true minimum in which two NH···N hydrogen bonds are favored over the two NH···O hydrogen bonds that occur for G·C. This major structural change is found despite the fact that O···H hydrogen bonds appear to be stronger than N···H hydrogen bonds. This surprising result may be due to extended conjugation of the rings. Both the N–C bonds adjacent to the radical center are shortened because of the radical formation. The resulting radical minimum is predicted to lie 11.7 kcal/mol above the lowest-energy radical G(N9)–C (see Fig. 5), produced by the hemolytic cleavage of the N9–H bond. An energy of 20.0 kcal/mol (Table 2) is necessary to dissociate the G(N1)–C radical into the G(N1) radical plus the cytosine molecule. The presence of two hydrogen bonds instead of three makes it easier to dissociate into the guanine radical and cytosine fragments, compared with the energy (27.2 kcal/mol) required to dissociate the neutral G·C base pair. Fig. 2. Open in a new tab Optimized geometry of the G(N1)–C radical at the B3LYP/DZP++ level of theory. Note that, relative to neutral G·C, the guanine H attached to the N1 has been removed; see Scheme 1 for atom numbering. Fig. 5. Open in a new tab Optimized geometry of the G(N9)–C radical at the B3LYP/DZP++ level of theory. All bond lengths are reported in angstroms. Note that, relative to G·C, the guanine H atom attached to N9 has been removed. G(N2a)–C. This radical, shown in Fig. 3, is generated from G·C by removing one of the H atoms bound to atom N2 of guanine. Removing the H from this N results in the loss of a hydrogen bond. The resulting radical relaxes by ≈7.9 kcal/mol when optimized at the B3LYP/DZP++ level of theory. The resulting minimum (Fig. 3) is predicted to lie 15.5 kcal/mol above the lowest-energy G(N9)–C (see Fig. 5) radical at the same level of theory (see Table 3). The geometrical perturbations seen in Fig. 3 are large compared with G·C because the three hydrogen bonds keep the backbone together in a plane. Once the H2a–O2 hydrogen bond is gone, the planarity is lost. The out-of-plane angle is ≈38°, as seen in Fig. 3 B. An out-of-plane angle so large will put significant strain on the closely stacked base pairs. The other two hydrogen bonds are slightly elongated (by 0.109 and 0.045 Å, respectively) compared with G·C. The dissociation energy of structure G(N2a)–C is reported in Table 2. The G(N2a)–C radical lies only 11.7 kcal/mol below the isolated G(N2a) radical plus cytosine molecule combined. A dissociation energy this small is indicative of the high strain, resulting from the 38° bond angle opening in this radical, in addition to the loss of the third hydrogen bond. Fig. 3. Open in a new tab Optimized structure of the G(N2a)–C radical at B3LYP/DZP++ level of theory. (A) All bond distances are reported in angstroms. Note that relative to neutral G·C base pair, the guanine atom H2a attached to N2 has been removed. (B) The G(N2a–C) radical from a different perspective, showing the opening angle between the guanine and cytosine rings. Table 3. Relative energies of the radicals at the B3LYP/DZP++ level of theory \| \| Relative energy, kcal/mol \| :---: \| \| Radical \| Energy relaxation upon structural optimization of the radicals \| Energy of optimized geometries of the radicals \| \| G—C \| \| \| \| G(N9)—C \| 12.95 \| 0.00 \| \| G(N2b)—C \| 5.96 \| 10.61 \| \| G(N1)—C \| 20.73 \| 11.68 \| \| G—C(N1) \| 5.23 \| 13.08 \| \| G(N2a)—C \| 7.94 \| 15.49 \| \| G—C(N4b) \| 3.05 \| 22.65 \| \| G—C(C6) \| 1.58 \| 23.21 \| \| G(C8)—C \| 1.32 \| 25.82 \| \| G—C(C5) \| 1.34 \| 26.83 \| \| G—C(N4a) \| 8.41 \| 32.15 \| Open in a new tab G(N2b)–C. Removing the guanine N2 hydrogen atom not used for making a hydrogen bond with cytosine gives the G(N2b)–C radical. Because this H atom is not involved in hydrogen bonding, removing it has little effect on the structure of the base pair. The hydrogen bond lengths are not expected to change in a major way. However, as depicted in Fig. 4, all three hydrogen bonds are lengthened (compared with closed-shell G·C) because of the radical generation. The energy change is only 6.0 kcal/mol upon optimization (relative to the closed-shell G·C structure), and the optimized radical is only 10.6 kcal/mol higher in energy than the lowest-energy radical, namely G(N9)–C. The binding energy (see Table 2) of the G(N2b)–C radical is –21.2 kcal/mol, which means that 21.2 kcal/mol energy is needed to dissociate the G(N2b)–C radical into the G(N2b) radical plus cytosine. Fig. 4. Open in a new tab Optimized structure of the G(N2b)–C radical at the B3LYP/DZP++ level of theory. Note that, relative to neutral G·C, guanine atom H2b attached to the N2 atom has been removed. G(N9)–C. The lowest-energy radical in this series is the radical generated by the abstraction of a hydrogen from atom N9 of guanine. The resulting G·C radical relaxes by 13.0 kcal/mol under geometry optimization and goes to a structure that is 10.6 kcal/mol lower in energy than the next-higher energy radical G(N2b)–C. Table 3 reports the relative energies of the radicals. Radical G(N9)–C is generated from a position that is used for the sugar phosphate linkage in the single-strand formation. Structural perturbations (relative to closed-shell G·C) are modest in this case, as shown in Fig. 5. Two of the three hydrogen bonds are shortened (by 0.033 and 0.066 Å, respectively, compared with G·C) in this case despite being away from the radical center. In both of these cases the N–H σ bond distances (not the hydrogen bonds) responsible for hydrogen bonding lengthen, although the elongation is not significant enough for an H shift toward cytosine. The dissociation energy (30.0 kcal/mol) of G(N9)–C is reported in Table 2. G(C8)–C. The radical generated by breaking the C–H bond at the C8 position (see Scheme 1) of guanine is particularly important. The 8-oxoguanine molecule has been found to resemble adenine and identified as a lesion generator (38). It is very important to investigate the properties of the radical G(C8)–C, because it may be an intermediate in the formation of 8-oxoguanine. The radical generated this way lies 25.8 kcal/mol above the lowest-energy radical G(N9)–C. The structure relaxes by only 1.3 kcal/mol compared with the frozen G·C geometry when the geometry of the radical is optimized. The dissociation energy for the G(C8)–C radical into G(C8) radical plus cytosine is 27.8 kcal/mol, comparable to that of G·C base pair (27.2 kcal/mol). G-C(N4a). Removing hydrogen atom H4a from the NH 2 group of cytosine results (Fig. 6) in the elimination of a hydrogen bond. This elimination brings about a change in the geometry of the base pair along with loss of planarity. This minimum hinders the other cytosine N4 hydrogen from collapsing to bridge the CO group of guanine and producing the G–C(N4b) structure (Fig. 7), which lies 9.5 kcal/mol lower in energy. It will be interesting to know the activation barrier for the interconversion of the G–C(N4a) and G–C(N4b) radicals. The G–C(N4a) radical is predicted to lie 32.2 kcal/mol above the lowest-energy G(N9)–C radical at the B3LYP/DZP++ level of theory. The G·C constrained hydrogen-abstracted radical releases 8.4 kcal/mol energy as it is optimized to the G–C(N4a) structure shown in Fig. 6. Planarity is lost because of the absence of the hydrogen bond, and the guanine and cytosine fragments move away from each other by ≈20°. The bond angle opens up (see Fig. 6 B), whereas the other two hydrogen bonds shorten in going from the frozen hydrogen-abstracted geometry to the optimized minimum. The geometry change and the 8.4-kcal/mol energy relaxation are significant in this case, because of the close stacking of base pairs in a double-stranded DNA helix. The loss of one hydrogen bond also helps to reduce the dissociation energy of this radical to 14.8 kcal/mol (Table 2). Puckering of the radical geometry (see Fig. 6 B) facilitates the dissociation of the GC(N4a) radical into guanine plus the C(N4a) radical. Fig. 6. Open in a new tab Optimized geometry of the G–C(N4a) radical at the B3LYP/DZP++ level of theory. (A) All bond distances are reported in angstroms. Note that, relative to G·C, the cytosine H atom attached to N4 has been removed. (B) A different perspective on the G–C(N4a) radical. Angle opening between the guanine and cytosine rings is ≈31°. Fig. 7. Open in a new tab Optimized geometry of the G–C(N4b) radical at the B3LYP/DZP++ level of theory. Note that, relative to G·C, the cytosine H atom attached to N4 has been removed. This structure is a conformer of G–C(N4a) with three rather than two hydrogen bonds. G–C(N4b). The second hydrogen attached to the cytosine NH 2 group is not hydrogen bonded, so removing this atom does not change the structural features of the base pair qualitatively. This G–C(N4b) radical when optimized releases only 3.1 kcal/mol energy and goes to a structure (Fig. 7) that is 22.7 kcal/mol above the lowest-energy G(N9)–C radical at the B3LYP/DZP++ level of theory. All three hydrogen bonds elongate compared with the G·C base pair, by 0.150, 0.064, and 0.049 Å, respectively. The dissociation energy of the G–C(N4b) radical is predicted to be 19.3 kcal/mol (Table 2). G–C(C5). Removal of the other (bonded to C5, C6, or N1) hydrogen atoms on the cytosine does not alter the geometry of the base pair greatly, except for small changes in the bond lengths due to resonance stabilization of the resulting radicals [e.g., G–C(N1)]. Structure G–C(C5) is predicted to lie 26.8 kcal/mol above the lowest-energy G(N9)–C radical. The energy relaxation with respect to the constrained G·C geometry is small (≈1.3 kcal/mol). The G–C(C5) radical requires 26.0 kcal/mol energy to dissociate into guanine plus the C(C5) radical. G–C(C6). This radical is generated by removing an H atom from atom C6 of cytosine. G–C(C6) lies 23.2 kcal/mol above the lowest-energy G(N9)–C radical at the B3LYP/DZP++ level of theory. There is no major change in the geometry relative to G·C, as expected from the position of the radical. Radicals generated far from the hydrogen bonds have little effect on the G·C dissociation energy. The dissociation energy predicted for this radical is 25.5 kcal/mol, rather close to the analogous G·C prediction of 27.2 kcal/mol. G–C(N1). The last radical structure to be considered is G–C(N1), generated by removing the H atom from N1 of cytosine. This radical is the lowest in energy among the radicals generated on cytosine in this series. The radical generated at atom N1 is stable because of resonance delocalization of the radical electron. In addition, the radical electron is qualitatively localized near an electronegative nitrogen atom. This radical is predicted to lie only 13.1 kcal/mol above the lowest-energy radical investigated, namely G(N9)–C. Fig. 8 shows the equilibrium geometry of the G–C(N1) radical. The NH···N and NH···O hydrogen bond lengths shorten somewhat (0.047 and 0.071 Å, respectively) as well as the N1–C6 bond (0.039Å) adjacent to the radical center. Again, the dissociation energy of G–C(N1) radical (–26.4 kcal/mol) is close that for the G·C base pair (27.2 kcal/mol), as reported in Table 2. Fig. 8. Open in a new tab Optimized structure of the G–C(N1) radical at the B3LYP/DZP++ level of theory. All bond lengths are reported in angstroms. Note that, relative to G·C, the cytosine H atom attached to N1 has been removed. Hydrogen bond lengths for all of the radicals considered in this work using the B3LYP gradient-corrected density functional in conjunction with the DZP++ basis set are reported in Table 4. Table 4. Hydrogen bond lengths of the G–C radicals. \| \| Geometry frozen \| Geometry optimized \| :---: \| Radical \| Bond \| Bond length, Å \| Bond \| Bond length, Å \| \| G·C neutral \| \| \| O···HN \| 1.717 \| \| \| \| \| NH···N \| 1.883 \| \| \| \| \| NH···O \| 1.886 \| \| G(N1)-C \| O···HN \| 1.717 \| N···HN \| 1.883 \| \| \| \| \| NH···N \| 1.865 \| \| \| NH···O \| 1.886 \| \| \| \| G(N2a)-C \| O···HN \| 1.717 \| O···HN \| 1.826 \| \| \| NH···N \| 1.883 \| NH···N \| 1.928 \| \| G(N2b)-C \| O···HN \| 1.717 \| O···HN \| 1.764 \| \| \| NH···N \| 1.883 \| NH···N \| 1.915 \| \| \| NH···O \| 1.886 \| NH···O \| 2.050 \| \| G(N9)-C \| O···HN \| 1.717 \| O···HN \| 1.780 \| \| \| NH···N \| 1.883 \| NH···N \| 1.850 \| \| \| NH···O \| 1.886 \| NH···O \| 1.819 \| \| G(C8)-C \| O···HN \| 1.717 \| O···HN \| 1.723 \| \| \| NH···N \| 1.883 \| NH···N \| 1.877 \| \| \| NH···O \| 1.886 \| NH···O \| 1.875 \| \| G-C(N4a) \| \| \| \| \| \| \| NH···N \| 1.883 \| NH···N \| 1.829 \| \| \| NH···O \| 1.886 \| NH···O \| 1.817 \| \| G-C(N4b) \| O···HN \| 1.717 \| O···HN \| 1.867 \| \| \| NH···N \| 1.883 \| NH···N \| 1.947 \| \| \| NH···O \| 1.886 \| NH···O \| 1.935 \| \| G-C(C5) \| O···HN \| 1.717 \| O···HN \| 1.708 \| \| \| NH···N \| 1.883 \| NH···N \| 1.909 \| \| \| NH···O \| 1.886 \| NH···O \| 1.889 \| \| G-C(C6) \| O···HN \| 1.717 \| O···HN \| 1.706 \| \| \| NH···N \| 1.883 \| NH···N \| 1.897 \| \| \| NH···O \| 1.886 \| NH···O \| 1.916 \| \| G-C(N1) \| O···HN \| 1.717 \| O···HN \| 1.724 \| \| \| NH···N \| 1.883 \| NH···N \| 1.836 \| \| \| NH···O \| 1.886 \| NH···O \| 1.817 \| Open in a new tab Distances are reported both for the frozen geometry of the neutral G·C base pair (Fig. 1) and for the optimized G-C radical geometries (Figs. 2, 3, 4, 5, 6, 7, 8). Biological Significance. Size complementarity in double stranded DNA ensures that the larger purines (adenine and guanine), pair with thymine and cytosine, the smaller pyrimidines, respectively (47, 48). Hydrogen bonded base pairs of correct lengths form the rungs of the ladder that is constrained by the uprights, made of the pentose sugar linked by phosphate groups. Size complementarity is probably more important than the hydrogen bonds itself (49–51). The magnitude of structural perturbations predicted here for some of the radicals described above indicates that they will inject significant amounts of strain into the ladder. Reorientation of the backbone on the G(N1)-C radical is particularly important given that the radical generated releases 21.2 kcal/mol energy while collapsing (relative to G·C) to a stable minimum. Other geometrical rearrangements are also significant due to the strain added by loss of planarity and puckering. Estimating the effects of this sort of geometry change in a long DNA ladder will be very interesting. Concluding Remarks. The theoretically obtained bond distances for the closed-shell neutral G·C agree quite well with previous theoretical predictions. As is well known, the gas-phase theoretical results at times do not coincide with the experimental crystallographic data. For a number of radicals studied in the present work the structural changes associated with H atom removal are major. These include base orientation angle changes, reconfiguration of the base pair, and elongation or shortening of the hydrogen bonds responsible for base pairing. The structural perturbations are most profound in the radicals generated at the hydrogen-bonding sites, because these break the structural integrity of base pairing. Similarly, in fact, even more profound structural changes can be expected in the adenine·thymine base pair, because it has only two hydrogen bonds. Breaking one of these bonds may cause severe structural disruption. Lesions generated in the G·C base pair will in turn cause strain in the closely stacked base pairing sequence. This strain is most significant (20.7 kcal/mol) in the G(N1)–C radical (Fig. 2), where the entire base pairing sequence changes because of a shift of the G·C backbone. In two other radicals (Figs. 3 and 6) the geometry changes in a major way because of openings of the angles between the guanine and cytosine backbones of between 20° and 31°. This deformation pushes one end of the radical [G(N2a)–C] almost 2.2 Å away from the planar geometry. In the other case [G–C(N4a)] the radical nitrogen atom protrudes out of the plane by ≈1.7 Å. It is to be emphasized that the range of radical relaxation energies is wide, from 1.3 to 20.7 kcal/mol. Hydrogen bond breakage due to radical formation also has an adverse effect on the radicals because of their reduced dissociation energies. Such a small dissociation energy indicates how easily a base pair radical can be separated into an isolated guanine or cytosine molecule plus the corresponding radical. This dissociation across the hydrogen bonds hints at the breakage of two complementary strands. In two of the radicals [G(N2a)–C and G–C(N4a)] where a radical is formed on a hydrogen-bonding site the dissociation energies are as low as 11.7 and 14.8 kcal/mol, less than half of the dissociation energy of neutral G·C base pair. This finding is further justified from the extent of deformation these two base pair radicals undergo. Overall, lesions in the G·C system may cause the base pair to undergo fatal structural perturbations leading to possible strand break and loss of genetic information. Acknowledgments P.P.B. thanks Brian Papas for help with q-chem. This work was supported by National Science Foundation Grant CHE-0451445 (to H.F.S.). 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1446	https://www.sciencedirect.com/science/article/pii/S0896627318304835	A Distance-Dependent Distribution of Presynaptic Boutons Tunes Frequency-Dependent Dendritic Integration - ScienceDirect Skip to main contentSkip to article Journals & Books ViewPDF Download full issue Search ScienceDirect Outline Highlights Summary Keywords Introduction Results Discussion STAR★Methods Acknowledgments Supplemental Information References Show full outline Cited by (30) Figures (3) Tables (1) Table Extras (3) Download all Document S1. Figures S1–S4 Data S1. Code for the Compartmental Model Used in Figures 3 and S4 Document S2. Article plus Supplemental Information Neuron ------ Volume 99, Issue 2, 25 July 2018, Pages 275-282.e3 Report A Distance-Dependent Distribution of Presynaptic Boutons Tunes Frequency-Dependent Dendritic Integration Author links open overlay panel Federico W.Grillo 1 2, Guilherme Neves 1 2, Alison Walker 1, Gema Vizcay-Barrena 3, Roland A.Fleck 3, Tiago Branco 4, Juan Burrone 1 2 5 Show more Outline Add to Mendeley Share Cite rights and content Under a Creative Commons license Open access Highlights •Presynaptic inputs decrease in size with distance along CA1 basal dendrites •Release probability decreases with distance along basal dendrites •Short-term facilitation increases with distance along basal dendrites •Increased synaptic facilitation offsets passive decay and boosts supralinear events Summary How presynaptic inputs and neurotransmitter release dynamics are distributed along a dendritic tree is not well established. Here, we show that presynaptic boutons that form onto basal dendrites of CA1 pyramidal neurons display a decrease in active zone (AZ) size with distance from the soma, resulting in a distance-dependent increase in short-term facilitation. Our findings suggest that the spatial distribution of short-term facilitation serves to compensate for the electrotonic attenuation of subthreshold distal inputs during repeated stimulation and fine-tunes the preferred input frequency of dendritic domains. Previous article in issue Next article in issue Keywords presynaptic terminal short-term plasticity dendritic integration synaptic transmission release probability active zone hippocampus Introduction Pyramidal neurons receive thousands of excitatory inputs on their extensive dendritic arbors. As a consequence, neurons need strategies to balance the strength of their synaptic inputs so that signals arriving at distal synapses, far from the soma, have a meaningful contribution to neuronal output. Active dendritic integration can greatly boost synaptic signals by amplifying local depolarizations onto dendritic spikes that travel to the soma. Much has been done to describe this phenomenon, and its regulation, by focusing on postsynaptic compartments. In particular, previous studies have characterized the distribution of the size and strength of postsynaptic spines within dendritic domains and have established their contributions to dendritic integration (Branco and Häusser, 2010, Magee, 2000, Spruston, 2008). Overall, these findings showed that whereas apical dendrites show an increase in the strength of synaptic inputs with distance from the soma (Magee and Cook, 2000), the thinner basal and apical oblique dendrites show a decrease in spine size with distance (Katz et al., 2009, Menon et al., 2013, Walker et al., 2017). What emerges is a complex picture of synapse distribution along dendritic arbors that appear to favor either local (dendritic) or global (cell-wide) integration, depending on dendrite identity. On the other hand, we know much less about the structural and functional distribution of presynaptic boutons (de Jong et al., 2012) and the role they play in shaping the integration of synaptic inputs on dendrites and their subdomains (Chabrol et al., 2015). Critically, whereas postsynaptic strength establishes the amount of local dendritic depolarization, changes in presynaptic structure and function determine the dynamics of neurotransmitter release, through short-term forms of plasticity. This feature, in turn, governs the type of information that is transmitted across a synapse (Branco and Staras, 2009, Fioravante and Regehr, 2011). To understand how dendrites integrate synaptic inputs, it is therefore important to first uncover how presynaptic boutons and the dynamics of neurotransmitter release are distributed along dendrites. A non-random distribution that echoes that of postsynaptic spines could lead to specialized dendritic domains that code for specific streams of information. Thus, short-term forms of plasticity that govern the release of neurotransmitter across different presynaptic boutons may have profound effects on how information is processed in postsynaptic dendrites (Abrahamsson et al., 2012). More globally, short-term plasticity (STP) has been shown to play many important roles in vivo (Regehr, 2012) and, in excitatory synapses of the hippocampus, STP is thought to contribute to the transmission of information about an animal’s place field (Kandaswamy et al., 2010, Klyachko and Stevens, 2006). Understanding the dynamics of neurotransmitter release in hippocampal presynaptic boutons will therefore also have implications for how information is encoded by CA1 neurons during active behaviors. Here, we show that presynaptic boutons decrease in size along the basal dendrites of CA1 hippocampal neurons, resulting in a decrease in release probability and an increase in short-term facilitation with distance from the soma. We reveal that this spatial distribution in STP tunes dendritic domains to specific information frequencies, introducing a further level of specialization to dendritic computations. Results We set out to map the distribution of the structure and function of presynaptic terminals in CA1 stratum oriens, a region of the hippocampus that receives ordered axonal inputs that are mostly perpendicular to the basal dendrites of pyramidal neurons (Andersen et al., 1980). We first performed Serial Block-Face Scanning Electron Microscopy (SBFSEM) of three regions of CA1 stratum oriens from a P22 brain, ranging from deep (close to the stratum pyramidale) to superficial (close to alveus) areas and reconstructed basal dendrites together with their synaptic inputs (Figures 1A–1C). Although dendritic segments showed a wide distribution in the size of excitatory inputs, we found, as expected, a strong correlation between morphological measures of presynaptic and postsynaptic compartments (Figure S1) (Holderith et al., 2012, Schikorski and Stevens, 1997). In agreement with the notion that basal dendrites taper toward tip ends (Menon et al., 2013), we also saw a decrease in dendrite diameter along the stratum oriens with increasing distance away from stratum pyramidale (Figure 1D; n= 35, mean= prox 0.75± 0.03 μm, med 0.65± 0.02 μm, dist 0.61± 0.01 μm, p< 0.001 Kruskal-Wallis test, prox-dist adjusted p< 0.001 Dunn’s multiple comparison test). Importantly, of the many structural features measured at the synapse (Figure S1), AZ area, a good predictor of release probability (Holderith et al., 2012), showed a strong decrease in size with distance along stratum oriens (Figure 1E; n= 604, mean= prox 0.074± 0.003 μm 2, med 0.058± 0.002 μm 2, dist 0.061± 0.003 μm 2, p< 0.0001 Kruskal-Wallis test; prox-med adjusted p< 0.001, prox-dist adjusted p< 0.001 Dunn’s multiple comparison test). Dendrite diameter, which can be taken as an indirect measure of distance from the soma along a tapering dendrite, also correlated well with AZ size (Figures 1F and 1G), providing further evidence that presynaptic terminals become smaller with distance along a basal dendrite. A reconstruction of proximal and distal dendrites performed on an adult brain (P100) showed a very similar distribution of presynaptic and postsynaptic properties with distance (Figures 1H–1K, Figure S1), indicating that the decrease in AZ size (and other synaptic parameters) was not unique to adolescent (P22) brains. Our dataset has some limitations, including the inability to follow the same dendrite from a single cell and to establish the identity of the inputs that arrive at different locations. We therefore turned to dissociated hippocampal neurons, where synaptic inputs can be labeled and followed along a single dendrite and where input identity is jumbled following the dissociation procedure. We find that measures of both presynaptic structure (vGlut labeling) and function (FM4-64 labeling of recycling vesicles) also decreased in a graded manner with distance along a dendrite (Figure S2). Together, these data suggest that the distance-dependent distribution of presynaptic inputs may be independent of the identity of the axon and is likely constrained by the properties of the postsynaptic dendrite. 1. Download: Download high-res image (1MB) 2. Download: Download full-size image Figure 1. Presynaptic Active Zone Size Scales with Dendrite Diameter and Distance from Cell Bodies (A) SBFSEM low-magnification image showing experimental design: 3 areas (red, proximal; purple, medial; blue, distal) at increasing distances from the pyramidal cell layer in the stratum oriens region of the CA1; S.O., stratum oriens; S.P., stratum pyramidale. Scale bar, 50 μm. (B) SBFSEM single section images (left panels), with corresponding 3D reconstructions (right panels), showing presynaptic (yellow) and postsynaptic (green) structures with PSDs (red) and AZs (light blue) highlighted. Top left panel shows a spine head (star) connected through the narrower spine neck to the main dendritic shaft (asterisk). Scale bars, 0.5 μm. (C) Two dendrites (thin dendrite left, thick dendrite right) reconstructed in 3D with spine heads in purple and boutons in green. Scale bar, 1 μm. (D–G) Data from a postnatal day 22 animal. (D) Cumulative fraction plot: dendrites reconstructed in the proximal area have larger diameters than distal and medial area dendrites; n= 35 dendrites, p< 0.001 ANOVA; proximal-distal adjusted p< 0.001; proximal-medial adjusted p< 0.01, Tukey’s multiple comparison test. (E) Cumulative fraction plot of AZ sizes, which are a larger in the proximal group; n= 604 AZs, p< 0.0001 Kruskal-Wallis test; proximal-medial adjusted p< 0.0001, proximal-distal adjusted p= 0.0001 Dunn’s multiple comparison test. (F) Smaller AZs tend to be found on thinner dendritic processes, n= 556 AZs, Spearman’s correlation; colors indicate area in which the dendrites were reconstructed. (G) Average AZ size (per dendrite) positively correlates with dendrite diameter, n= 35 dendrites, Spearman’s correlation. (H–K) Data from a postnatal day 100 animal. (H) Proximal dendrites are thicker than distal dendrites, cumulative fraction plot n= 26 dendrites, p<0.05 unpaired t test with Welch’s correction. (I) Cumulative fraction plot of AZ sizes, which are a larger in the proximal group; n= 505 AZs, p< 0.0001 Kolmogorov-Smirnov test. (J) Smaller AZs tend to be found on thinner dendritic processes, n= 505 AZs, Spearman’s correlation; colors indicate area in which the dendrites were reconstructed. (K) Average AZ size (per dendrite) positively correlates with dendrite diameter, n= 26 dendrites, Spearman’s correlation. See also Figure S1. Data are represented as mean± SEM. AZ size is thought to correlate well with release probability (P r) (Holderith et al., 2012, Murthy et al., 2001, Schikorski and Stevens, 1997). In addition, P r is tightly coupled to the dynamics of neurotransmitter release during a burst of action potentials, such that high P r synapses undergo short-term depression, whereas low P r synapses exhibit short-term facilitation (Dobrunz and Stevens, 1997). This correlation between P r and STP prompted us to measure neurotransmitter release dynamics along basal dendrites. We therefore patch clamped and filled CA1 pyramidal neurons with a fluorescent dye (Alexa 594) to visualize the basal dendritic arbor and measure synaptic inputs along its dendrites. Two stimulating pipettes were positioned along the basal dendrite, one at a site proximal to the soma and one distal to it (Figure 2A). In agreement with our structural findings (Figure 1), synaptic AMPA receptor currents measured by stimulating either proximal or distal axons independently showed that distal inputs facilitated more than proximal ones (Figures 2B and 2C). The paired-pulse ratio (PPR) of the first two stimuli delivered at 20 Hz was larger for distal inputs (mean= 1.59± 0.05) compared to proximal ones (mean= 1.33± 0.05) and remained higher throughout the 5 pulse stimulus train (Figures 2C and 2D). To establish whether the observed differences in STP were affected by postsynaptic properties, such as the inactivation or saturation state of AMPA receptors, we used cyclothiazide (CTZ, a drug that prevents AMPAR desensitization [Partin et al., 1993, Patneau et al., 1993]) or γDGG (a competitive antagonist of glutamate receptors [Liu et al., 1999]) to directly assess any contributions to STP by postsynaptic receptors (Figure 2E). Neither drug had any effect on the difference in STP between distal and proximal inputs (CTZ PPR= prox 1.18± 0.09, dist 1.51± 0.09, n= 9, p= 0.01; γDGG PPR= prox 1.36± 0.07, dist 1.64± 0.09, n= 13, p< 0.01; two-way ANOVA for control, CTZ, and γDGG, interaction p= 0.84), although there was a positive trend of γDGG on facilitation (Figure S3) that could reflect multivesicular release during the train (Christie and Jahr, 2006, Oertner et al., 2002, Wadiche and Jahr, 2001). To further corroborate that these effects were indeed presynaptic in origin, we performed similar measures of PPR in transgenic mice where synaptotagmin 7 (Syt 7) was knocked out. Syt 7−/− mice have been shown to lack any facilitation (Jackman et al., 2016) and should therefore abolish the differences in PPR measured here. We find that whereas WT mice show the same distance-dependent increase in PPR observed above, Syt7−/− littermates show neither facilitation nor any obvious difference in PPR between proximal and distal domains (Figure 2F). Together, our data point to a presynaptic origin in the difference in PPR along basal dendrites. 1. Download: Download high-res image (911KB) 2. Download: Download full-size image Figure 2. Increases in Short-Term Facilitation with Distance along a Dendrite Boost Distal Synaptic Integration (A) Whole-cell patch-clamp technique was used to record synaptic currents and fill CA1 pyramidal cells with a fluorescent dye to image its structure. Two stimulating pipettes were placed in the proximal (red) and distal (blue) extracellular domains of the basal dendritic tree of pyramidal cells to stimulate the local fibers. (B) Single cell example of average EPSC responses (average of 20 individual sweeps) to trains of 5 pulses at 20 Hz delivered to the proximal region (red trace) and the distal region (blue trace). The distal response shows greater facilitation compared to the proximal one. (C) Normalized average peak EPSC amplitudes for distal and proximal responses show greater sustained facilitation during a 5 pulse train (20 Hz) for distal synapses, n= 35 cells, multiple t tests with p values adjusted with the Holm-Sidak method, p< 0.05. (D) Paired-pulse ratios (PPRs) for each individual cell recorded at distal and proximal synapses. The majority (29/35) of cells display greater facilitation in the distal domain, n= 35 cells, p= 0.0003 two-tailed paired t test. (E) Distal increase in PPR is not ascribable to postsynaptic AMPA receptor desensitization (prevented by CTZ application) or to AMPA receptor saturation (avoided with γDGG application). Distal PPR is greater than proximal PPR with CTZ (n= 9 cells, p= 0.01), and γDGG (n= 13 cells, p< 0.01), two-tailed paired t test. Two-way ANOVA to test PPRs in control (D), CTZ, and γDGG conditions together shows no significant interaction, p= 0.84, indicating that the drugs have no effect on STP properties. (F) Full Synaptotagmin7 KO eliminates facilitation and proximo-distal STP differences. Triangles in lighter colors are from Syt7KO mice, n= 9 cells, circles in darker colors are littermate wild-type mice, n= 12 cells. For WT mice, proximal facilitation is lower than distal, multiple t tests, p< 0.05. WT facilitation is greater than Syt7KO facilitation, p< 0.01 multiple t tests. (G–I) Distal synaptic EPSCs take longer to reach the soma. (G) Left panel: normalized trace for a proximal and distal EPSC response showing the delayed kinetics of the distal compared to the proximal synaptic current. Right panel: the rise time constant of the EPSCs was significantly higher in distally triggered events, n= 49 cells, p< 0.001 Wilcoxon signed rank test. (H) Longer rise times correlate with the amount of facilitation, Spearman’s correlation. (I) PPR is higher when the stimulation electrode is placed further away from the soma, measured as distance along the dendrite, Spearman’s correlation. (J) Proximal synapses display greater P r than distal synapses. After MK-801 bath application, the normalized amplitude of EPSCs from proximally stimulated synapses decay faster, following successive stimulations, than distal ones; n= 8 cells. Data points were fit with a double exponential function (filled lines). Insets are example traces of 7 successive NMDA mediated EPSCs. (K) Frequency tuning curve showing PPRs for all frequencies tested. Distal PPRs (second stimulus only) increase significantly in the 20 Hz (n= 35) same as (D), and 50 Hz range (n= 20), p< 0.05, multiple t tests with Holm-Sidak adjusted p values. 5 Hz (n= 13 cells), 10 Hz (n= 16), 80 Hz (n= 9). (L–M) Short-term facilitation contributes to dendritic non-linear events in distal domains. (L) Current-clamp example traces (red proximal, blue distal stimulation) in response to a paired pulse, of increasing stimulus intensity (lighter color shades represent lower intensity). (M) The proportion of supra-linear events (at least 2 mV above the expected response) for distal synapses is greatly increased for the second pulse (P2) following facilitation at 20 Hz (0 events in P1, 8 events in P2, n= 10), while supra-linear events were detected to the first pulse (P1) for proximal stimulations (n= 10, 4 in P1, 5 in P2). At 5 Hz, where STP is absent, distal synapses had fewer supra-linear events (n= 5, 0 in P1, 2 in P2). See also Figures S2 and S3. Data are represented as mean± SEM. To confirm the location specificity of our stimulus pipettes, we measured the kinetics of synaptic transmission at the soma. In agreement with the electrotonic decay of signals along a dendrite (Mainen et al., 1996, Rall, 1962, Spruston et al., 1994), we confirmed that distal stimuli (mean= 1.66± 0.07 ms) elicited slower events than proximal stimuli (mean= 1.23± 0.09 ms; p< 0.001, Wilcoxon sign rank test; Figure 2G). Since both the rise time and time to peak of events are directly related to the distance from the recording pipette at the soma, we looked at their relationship with PPR across different cells. A positive correlation emerged between synaptic kinetics and PPR (Figure 2H and Figures S3F and S3G, n= 49), suggesting a gradual change in PPR along a dendrite. Indeed, a positive correlation was also observed when PPR was plotted against the absolute distance of the stimulating electrode from the soma (Figure 2I). Our structural observations (Figure 1) predict that distal synapses will have a lower release probability (P r) than proximal ones (Holderith et al., 2012), which may, in turn, help account for the increased levels of distal facilitation. To test for this, we used the irreversible open-channel blocker of NMDA receptors (NMDARs), MK-801, to measure P r at proximal and distal compartments (Figure 2J). We recorded NMDAR currents in the presence of MK-801 in response to successive stimuli to measure the gradual block of NMDAR channels. The decay was best fit by a double exponential function, with a fast and a slow phase, indicating at least two groups of synapses with high and low release probabilities, respectively (Hessler et al., 1993). The two time constants were very similar for proximal and distal dendrites (proximal, τ1: 1.44 and τ2: 14.64; distal, τ1: 1.37 and τ2: 16.14), but in distal dendrites a larger fraction of the decay was explained by the slower time constant (proximal: 49%; distal: 69%). The overall slower decay observed in distal dendrites (mean half-life, proximal: 7.28± 1.2, distal: 13.99± 3.03, n= 8, p= 0.027, paired t test) indicates that synaptic NMDARs experienced less neurotransmitter in response to successive stimuli and therefore belong to synapses with, on average, a lower P r than those found in proximal dendrites (Figure 2J). These findings mirror the structural correlation of presynaptic AZ size with dendrite diameter (Figures 1G and 1K) and strengthen our hypothesis that both the structure and function of presynaptic boutons display a distance-dependent distribution along basal dendrites. Furthermore, the lack of cross-inhibition by MK-801 between proximal and distal inputs is further support for the specificity of our stimulating electrodes in recruiting axons that form connections locally (Figure S3). Finally, we also established the frequency range over which inputs showed facilitation. We found that distal inputs showed significantly larger facilitation than proximal ones over a limited frequency range, between 10 and 50 Hz (Figures 2K and S3), with a peak at 20 Hz (PPR means= 5 Hz, prox 1.13± 0.04, dist 1.05± 0.04; 10 Hz, prox 1.18± 0.06, dist 1.35± 0.05; 20 Hz, see above; 50 Hz, prox 1.24± 0.07, p< 0.01 dist 1.53± 0.07, p< 0.01; 80 Hz, prox 1.17± 0.12, dist 1.22± 0.08). So far, our data show that distal inputs are tuned to band-pass frequencies in the near gamma frequency range, which curiously match the high-frequency discharges measured in vivo in hippocampal place cells when an animal passes through a place field (Huxter et al., 2003). To establish whether the biased distribution in short-term facilitation can influence dendritic integration and neuronal output, we performed current-clamp experiments and locally stimulated either proximal or distal afferents, as above. We found that stimulation of distal inputs at frequencies that showed increased facilitation (20 Hz), also showed a tendency to produce non-linear summations to multiple stimuli across the range of stimulation intensities (Figures 2L and 2M). This effect was less pronounced in proximal dendrites, where non-linear events were only observed for high stimulus intensities. The difference in integration properties between proximal and distal events has been shown to depend on the opening of NMDARs (Ariav et al., 2003, Branco and Häusser, 2011, Major et al., 2008, Schiller et al., 2000), which are more likely to be activated at the high impedance distal dendrites, where levels of depolarization to a given input are higher. We found that distal inputs are better suited to respond to multiple stimuli than to a single stimulus. This was apparent when comparing responses along a paired pulse. The second stimulus of a 20 Hz pair was consistently more likely to summate in a supralinear manner, when compared either to the first stimulus or to stimuli delivered at 5 Hz (Figures 2L and 2M). Together, our data show that the integration properties of distal domains are markedly different from proximal ones and depend on input frequency. Although there are many factors that likely play a role in driving these differences in dendritic integration, STP is an obvious candidate. To explore this further, we turned to a computational model where we could directly assess the role played by STP on the integration properties of different dendritic domains. We built a model consisting of a cluster of proximal and distal synapses (15 each; Figure 3A), where STP was adjusted to match our data (Figure 3B). This simple model aims to first determine the impact that a distance-dependent STP gradient can have on dendritic integration, when all other variables are the same. Activating either proximal or distal synapses separately with Poisson input trains delivered at different mean frequencies elicited non-linear responses (Figure 3C). Distal inputs, however, showed larger levels of membrane depolarization compared to proximal ones, over a large frequency range (continuous lines in Figure 3D). This distal amplification of synaptic inputs was partially lost when distal synapses were switched to proximal STP properties, suggesting that increased levels of facilitation enhance supralinear integration in distal dendrites. Interestingly, the difference in STP along a dendrite not only played a role in modulating dendritic non-linearities, it also contributed to distance-dependent input normalization; removal of active conductances from the model showed similar levels of membrane depolarization for distal and proximal inputs, which was lost when distal synapses were tuned to proximal STP properties (dashed lines in Figure 3D; note that the red and blue dashed lines overlap). This result suggests that the gradient in STP can boost distal inputs to counteract the effects of passive decay. Finally, increased levels of facilitation also resulted in an increased variance of membrane responses over a limited frequency domain, given by the short-term dynamics of the synapse (Figure 3E). This model shows that the STP gradient can have a significant impact on synaptic input integration, mainly exploiting differences in the recruitment of dendritic non-linearities. As such, it will act alongside other mechanisms that impact on non-linear integration, such as synaptic conductances, impedance, or release probability, all of which can also display gradients. However, models that include additional gradients of these other variables showed that the input-output curve remains biased toward larger depolarizations at higher frequencies in distal dendrites (Figure S4), indicating that the magnitude of the STP gradient that we observe is sufficient to have a significant impact on dendrite integration across a variety of conditions. Together, these simulations support the idea that distal inputs have an increased capacity for information transfer that is boosted by the short-term dynamics of neurotransmitter release from presynaptic inputs. 1. Download: Download high-res image (447KB) 2. Download: Download full-size image Figure 3. PPR Gradient Enhances Distal Supra-linear Integration (A) Compartmental model illustration with distal and proximal synapses along one dendritic branch (left) and respective single synapse AMPA conductance traces for 20 Hz stimulation (right). (B) Model data showing PPR dynamics for each pulse in a train of 20 Hz (left) and for PPR between 5 and 100 Hz for the first two pulses in the train. (C) Example simulation traces for Poisson input trains delivered independently at each synapse at two different rates. Top raster shows input times for all synapses, and traces below show responses to the same input for proximal (red) and distal (blue) synapses, and for distal synapses equipped with PPR identical to proximal synapses (purple). (D and E) (D) Mean depolarization during stimulation for different input frequencies and (E) respective standard deviation of the membrane potential, showing that increased PPF at distal synapses enhances supra-linear integration. Dashed lines in (D) and (E) are for passive models for each position (red proximal, blue distal, and purple distal with proximal PPR properties); note that in (D) the red and blue dashed lines overlap. See also Figure S4. Data are represented as mean± SEM. Discussion Our findings uncovered a gradient in the distribution of presynaptic terminals along dendrites that dictates the short-term dynamics of synaptic transmission. This spatial gradient in short-term facilitation serves to both normalize the distance-dependent decay in the amplitude of subthreshold inputs to repeated stimuli, as well as tune the preferred input frequency of different dendritic domains through supralinear integration. As a result, the type of information transmitted across a synapse will depend on its location along a dendrite, an arrangement that could be exploited by different input streams to achieve input-specific differential integration on the same dendrite. Previous studies have shown that pre- and post-synaptic compartments are well matched, both structurally and functionally (Kay et al., 2011, Murthy et al., 2001). Our findings of a spatial bias in the distribution of presynaptic boutons would predict that postsynaptic structures should follow suit. Indeed, work focusing on postsynaptic spines has shown that along the basal dendrites of CA1 pyramidal neurons, non-perforated spines, which represent the great majority of excitatory inputs, show a decrease in size (including PSD size) with dendritic distance (Menon et al., 2013, Walker et al., 2017). This reduction in size is thought to locally compensate for the increased impedance of thinner, distal dendrites, normalizing responses locally, rather than cell-wide (Katz et al., 2009). By including presynaptic boutons and neurotransmitter release dynamics, our findings uncover spatially segregated domains for the transfer of information within dendrites. However, our experiments provide no information on the identity of the axons that innervate different dendritic compartments along the stratum oriens. In general, it is thought that axons in this region of the hippocampus arrive mainly from pyramidal cells in areas CA3 and CA2, although other brain areas may be involved. There is, in fact, a biased topographic projection of CA3 pyramidal cells to the stratum oriens, such that CA3 cells whose soma lie closer to the dentate gyrus (DG) project preferentially to distal dendritic domains of CA1 neurons, whereas those CA3 cells further away from the DG project to more proximal dendritic domains (Ishizuka et al., 1990). It is therefore possible that the properties of distal and proximal boutons are dictated by the identity of the presynaptic neuron itself. However, our in vitro findings argue against this view. Using dissociated hippocampal neurons, we show that a similar distance-dependent decrease in presynaptic bouton properties also occurs in vitro (Figure S2) (de Jong et al., 2012) and suggests that the distribution of synapses observed in vivo may well be independent of the circuit or the identity of the synaptic input. It remains a likely possibility that the biased synaptic distribution along dendrites is therefore specified by the postsynaptic neuron in a cell-autonomous manner. Indeed, work in dissociated hippocampal neurons has shown that local dendritic depolarization is a major determinant of presynaptic release probability (P r), such that increases in depolarization induce a homeostatic decrease in P r (Branco et al., 2008). It is therefore tempting to speculate that the increase in impedance in distal dendrites, which would result in larger synaptic amplitudes (Spruston, 2008), would in turn act to reduce P r locally. In this way, dendritic impedance could act as a readout of dendritic distance. Possible molecular mechanisms may include retrograde messengers, some of which have been shown to modulate presynaptic function in response to postsynaptic membrane depolarization (Regehr et al., 2009). Our data show that synapses are distributed in a distance-dependent manner along dendrites and we provide multiple lines of evidence to suggest that this distribution is also graded with distance. First, we find a graded correlation between synapse morphology and dendrite diameter (Figures 1G and 1K), a measure that has been previously shown to be a good proxy for distance along a given dendrite (Walker et al., 2017). Second, measures of PPR are correlated to the time course of the EPSC (Figure 2H), reflecting the distance-dependent filtering suffered by an EPSC as it travels to the soma. However, these measures of distance are indirect and incur a certain amount of noise that likely arises from the fact that dendrites are heterogenous entities, especially across different cells, with variable degrees of tapering and where the passive electrotonic decay of synaptic events can be influenced by other factors. Importantly, direct measures of dendritic distance in either acute slices (Figure 2I) or in primary neuronal cells (Figure S2) both show a correlation with presynaptic properties. Together, our data support the notion of a graded distribution of synaptic structure and function along dendrites. Our study finds a strong correlation between presynaptic structure and function, where boutons with smaller AZs have a lower P r and an increased PPR. However, although PPR has been shown to correlate with the overall P r of a bouton (Dobrunz and Stevens, 1997), this correlation may well be driven by the release probability of individual vesicles (P vr) along an active zone. Differences in P vr can arise from a number of different scenarios, ranging from structural features (e.g., the relative distribution of vesicles and calcium channels) to molecular heterogeneity (e.g., the expression of specific proteins that influence exocytosis). Although we have not explored these possibilities in our study, they could also underlie some of the differences in PPR observed here. One other recent study has also shown a graded distribution of short-term plasticity along dendrites (Abrahamsson et al., 2012), although both the mechanism and direction of the gradient were opposed to that shown here. Stellate cells, a type of interneuron in the cerebellum, show a decrease in short-term facilitation along their thin dendrites (Abrahamsson et al., 2012). However, the effect is purely postsynaptic and arises from the large local depolarization in distal dendritic domains that reduces the driving force of postsynaptic receptors, resulting in the sub-linear integration of clustered inputs. As a result, stellate cells become ideal integrators of decorrelated inputs in both space and time, preferring the arrival of sparsely distributed, asynchronous events. This is clearly distinct form the mechanisms preferentially used by basal dendrites in the hippocampus and cortex, where clustered inputs drive local supralinear dendritic events (Branco and Häusser, 2011, Losonczy and Magee, 2006, Makara and Magee, 2013). Indeed, our data show that in CA1 basal dendrites, facilitation is distributed in the opposite direction to stellate cells, increasing with distance from the soma. Furthermore, according to our experimental data and computational model, increased distal facilitation is needed to trigger supralinear dendritic integration as this requires fast concomitant activation of multiple clustered synapses (Branco and Häusser, 2011), the likelihood of which is boosted by presynaptic facilitation. Collectively, these studies underscore the importance of short-term forms of plasticity on dendritic integration. Finally, the difference in STP in proximal versus distal dendrites will also have important consequences on the way each domain encodes information. Proximal synapses that facilitate less will be better suited to respond to more isolated or lower-frequency inputs, suggesting that they respond better to temporally decorrelated events. Distal dendrites, on the other hand, with their higher levels of facilitation will act as spatio-temporal filters that favor high-frequency, clustered inputs. These highly selective distal inputs may, therefore, carry salient information for working memory or place field location. Facilitation of synaptic transmission in excitatory hippocampal synapses has been proposed to act as an adaptive high-pass filter that transmits and amplifies the signals encoding place-field information (Kandaswamy et al., 2010, Klyachko and Stevens, 2006). Our data suggest that the distal domains of CA1 basal dendrites are better suited to perform this computation. Future work will need to establish the identity of the inputs that arrive along different dendritic locations and understand the type of information they encode. STAR★Methods Key Resources Table \| REAGENT or RESOURCE \| SOURCE \| IDENTIFIER \| --- \| Antibodies \| \| Rabbit polyclonal anti-VGlut1 \| Synaptic Systems \| Cat#153303; RRID: AB_887875 \| \| Chicken polyclonal anti-GFP \| Abcam \| Cat# ab13970; RRID: AB_300798 \| \| Chemicals, Peptides, and Recombinant Proteins \| \| AP-5 \| Cambridge Bioscience \| CAY14539; CAS: 79055-68-8 \| \| SR 95531 \| Cambridge Bioscience \| CAY14585; CAS: 104104-50-9 \| \| QX-314 \| Cambridge Bioscience \| CAY10011032; CAS: 24003-58-5 \| \| Alexa Fluor 594 \| Molecular Probes \| A10438 \| \| NBQX \| Santa Cruz Biotechnology \| sc-222048; CAS: 479347-86-9 \| \| MK-801 \| Cayman Chemical Company \| 10009019; CAS: 77086-22-7 \| \| CTZ \| Cambridge Bioscience \| CAY16335; CAS: 2259-96-3 \| \| γDGG \| Santa Cruz Biotechnology \| sc-203728; CAS: 6729-55-1 \| \| FM4-64 \| Molecular Probes \| T13320 \| \| CNQX \| Tocris \| 1045; CAS: 479347-85-8 \| \| TTX \| Alomone Labs \| T-550; CAS: 18660-81-6 \| \| Experimental Models: Organisms/Strains \| \| Mouse: Syt7 KO: B6.129S1-Syt7 tm1Nan/J \| The Jackson Laboratory \| JAX:004950 \| \| Mouse: wt: 129S2/SvPasCrl \| Charles River Laboratories \| 129 Mice \| \| Oligonucleotides \| \| Primers: Syt7mut Forward: CTT GGG TGG AGA GGC TAT TC, Reverse: AGG TGA GAT GAC AGG AGA TC Syt7wt Forward: CAT CCT CCA CTG GCC ATG AAT G Reverse: GCT TCA CCT TGG TCT CCA G \| N/A \| N/A \| \| Software and Algorithms \| \| TrakEM2 \| Cardona et al., 2012 \| \| \| Neuromorph \| Jorstad et al., 2015 \| \| \| NEURON \| Hines and Carnevale, 1997 \| \| Contact for Reagent and Resource Sharing As Lead Contact, Juan Burrone is responsible for all reagent and resource requests. Please contact Juan Burrone at juan.burrone@kcl.ac.uk with requests and inquiries. Method Details Animals All animal procedures were approved by the local ethics committee and licensed under the UK Animals (Scientific Procedures) Act of 1986. Male and Female SV-129 mice were housed grouped in standard cages and provided with ad libitum food and water. The Syt 7 knockout mice were obtained from The Jackson Laboratory (Chakrabarti et al., 2003). Sprague-Dawley rats were obtained from Charles River Laboratory. Dissociated hippocampal cultures Primary hippocampal cultures were prepared from embryonic day 18 Sprague-Dawley rats (Charles River Laboratory). Dissociated cells were plated onto 18 mm diameter coverslips (Menzel Gläser, Germany) pre-treated with poly-D-lysine (50 μg/mL) and laminin (20 μg/mL) at a density of 350 cells/mm 2 in Neurobasal media containing 1% fetal calf serum, 1% B-27 supplement, 0.5% glutamax and 0.5% penicillin/streptomycin. Neurons were kept for 17-21 days in vitro. Electron Microscopy Two mice (post-natal day 22 and 100) were transcardially perfused with 20 mL of ice-cold saline solution followed by 200 mL of ice-cold fixative (2% PFA and 0.2% glutaraldehyde mixture in 0.1 M phosphate buffer), followed by incubation overnight in fresh fixative at 4°C. Coronal vibratome sections (60 μm) were cut using a Leica VT1000S vibratome and further fixed in 1.5% potassium ferrocyanide: 2% osmium tetroxide in cacodylate buffer for 30 min at 4°C. Tissue was then thoroughly rinsed in distilled water and incubated in 1% aqueous thiocarbohydrazide for 4 min. After further rinsing, the samples were treated with 2% aqueous osmium tetroxide for 30 min, rinsed and en-bloc stained in 1% uranyl acetate for 2 hr. To further enhance contrasts in the samples, one last treatment with Walton’s Lead was carried out for 30 min at 60°C, before proceeding to dehydration in an ethanol series and infiltration with Durcupan ACM resin (Sigma). After embedding and curing, tissue blocks were mounted on Gatan 3View aluminum pins using conductive glue (CircuitWorks Conductive Epoxy) and trimmed accordingly. Before imaging, samples were gold coated to increase electron conductivity. The specimens were then placed inside a Jeol field emission scanning electron microscope (JSM-7100F) equipped with a 3View 2XP system (Gatan). Section thickness was set at 40 nm (Z resolution). Samples were imaged at 2.5kV under high vacuum using a 2048x2048 scan rate, which gave a final pixel size of 4.4 nm. Electron microscope images were registered and manually segmented using the ImageJ plugin TrakEM2 (Cardona et al., 2012). Extracted 3D structures were exported to the Blender software with the Neuromorph toolset (Jorstad et al., 2015), which was used to compute surface, volume, and length measurements and render 3D reconstructions shown in Figure 1. Electrophysiology Mice (21–33 days old) were sacrificed by decapitation following Isoflurane anesthesia, the brain was immediately extracted in ice cold high sucrose solution (in mM: 240 Sucrose, 5 KCl, 1.25 Na 2 PO 4, 2 MgSO 4, 1 CaCl 2, 26 NaHCO 3, 10 D-glucose, Saturated with 95% O 2 and 5% CO 2). In the same solution acute 300 μm thick coronal hippocampal slices were cut using a Leica vibratome (VT1000 S, Leica Microsystems). Slices were then transferred to a holding chamber containing room temperature ACSF (in mM: 125 NaCl, 5 KCl, 1.25 Na 2 PO 4, 1 MgSO 4, 2 CaCl 2, 26 NaHCO 3, 20 D-glucose) incubated for 1 hr and kept for up to 6 hr of experiments. Cells were visualized with a Scientifica two-photon microscope equipped with a water immersion 40X 0.8 numerical aperture Olympus lens. Dodt Gradient Contrast was used to approach and patch the neurons while a Chameleon femtosecond pulsed laser (Coherent) was used for two-photon imaging of the dendritic arbors. Whole-cell recordings were performed using a Multiclamp 700B amplifier (Molecular Devices), traces were filtered at 3 KHz and digitized at 50KHz. Series resistance was< 20 MΩ. Patch pipettes were pulled (Sutter Puller P-97; Sutter Instruments) from thick-walled borosilicate glass capillaries with an inner filament (1.5 mm outer diameter, 0.86 mm inner diameter; Sutter Instruments). Pipette resistance was 3–4 MΩ after fire polishing. Voltage clamp experiments were performed in the same extracellular ACSF with the addition of 25 μM AP-5 (Cambridge Bioscience) and 20 μM SR95531 (Cambridge Bioscience). The intracellular solution contained in mM: 135 CsMeSO 3, 10 HEPES, 10 Na 2-Phosphocreatine, 5 Glutathione, 4 MgCl 2, 4 Na 2 ATP, 0.4 NaGTP, 5 QX-314 (Cambridge Bioscience) and 20 μM Alexa Fluor 594 (Molecular Probes). Cells were held at −65 mV and visualized with the two-photon laser tuned at 840 nm, ≈10 min after membrane rupture to allow the dye to spread throughout. Two stimulating unipolar glass electrodes were placed in the distal and proximal dendritic region. Stimulus intensity ranged between 0.1 and 0.8 mA using an Iso-Flex stimulator (Intracel). Fibers were stimulated 20 times with 5 pulse trains every 30 s to calculate the average EPSC. For the NMDAR depletion experiments the extracellular solution contained the AMPAR channel blockerNBQX 10 μM (Santa Cruz Biotechnology) instead of AP-5. Baseline recordings of synaptic inputs, stimulated at proximal and distal sites, were obtained as above but with a single stimulation pulse repeated 20 times every 10 s. MK-801(40 μM, Cayman Chemical Company) was bath applied, and allowed to equilibrate for 5 min, after which the stimulation of either the proximal or distal site was repeated a further 40 times to obtain the curves in Figure 2J. For cross-depletion experiments in Figure S3, we continued the experiment and delivered another 40 stimuli to the site that had not been depleted. Care was taken to vary the site (proximal or distal) that was depleted first. In current clamp experiments the extracellular ACSF contained 20 μM SR95531 while the intracellular solution contained in mM: 115 K-MeSO 4, 20 KCl, 10 Na 2-Phosphocreatine, 10 HEPES, 2 MgATP, 2 Na 2 ATP, 0.4 mM Na 2 GTP, and 20 μM Alexa Fluor 594. The stimulus was delivered with a single bipolar glass theta electrode either distally or proximally with intensity ranging between 1 and 6.2 V. All recordings were performed at 30°C. Fibers were stimulated every 1 min with increasing voltage intensity consisting of 0.2 V steps. All electrophysiology experiments were analyzed in IGOR Pro software (Wavemetrics) with the NeuroMatic 2.7 package and MATLAB (MathWorks). To detect non-linear events, a linear fit was calculated for the gradual increase in EPSP amplitude in response to increasing current steps of the first pulse. The expected amplitude of the second pulse was calculated from this fit. If the recorded EPSP was 2 V above the expected amplitude it was then classified as a non-linearity. All chemicals were from Sigma unless otherwise stated. Modeling Simulations were performed with the NEURON simulation environment (Hines and Carnevale, 1997). The model consisted of a soma connected to one dendrite with length of one length constant distributed over 50 segments (diameter= 1 μm). Passive parameters were Cm= 1 μF/cm2, Rm= 10,000 Ω.cm2, Ra= 80 Ω·cm and a leak conductance with a reversal of −65 mV. Unless otherwise noted, active conductances in the dendrite were (in mS/cm2): voltage-activated sodium channels (4), voltage-activated potassium channels (0.8), M-type potassium channels (0.005), high-threshold voltage-activated calcium channels (0.05), low-threshold voltage-activated calcium channels (0.15× 10-3). Fifteen synapses containing AMPA and NMDA receptor conductances were distributed uniformly over the first or last 50% of the dendrite for the proximal and distal scenarios, respectively. Synaptic conductances were modeled as double exponential functions with gmax= 0.5 nS, τ1= 0.1 ms, τ2= 1 ms for AMPA and τ1= 1 ms, τ2= 20 ms for NMDA. For fitting short-term plasticity dynamics we initially used the Tsodyks and Markram model (Tsodyks and Markram, 1997), and fitted it by systematically varying the model parameters and comparing the model fits using the peak postsynaptic conductance for each presynaptic pulse on a train of 5 pulses according to ∑i=1 5\|m o d e l G i−d a t a G i\|, where modelG i and dataG i are the peak postsynaptic conductance for model and data respectively, for pulse i. While this model produced good fits for the first half of the pulse train, it failed to satisfactorily capture the profile of the remaining pulses. We then fitted the Varela et al. model (Varela et al., 1997) following the same procedure, which produced a good fit to all 5 pulses, mainly because of separately accounting for fast and slow depression kinetics. The Varela et al. model (Varela et al., 1997) was then used in all simulations. Short-term plasticity was implemented with time constants adjusted to match the experimental data and applied to both AMPA and NMDA conductances (facilitation: 2.5 distal, 1.6 proximal; fast depression: 0.4; slow depression: 0.94). For the passive scenario, active conductances were turned off and synapses only contained AMPA conductances. For the data shown in Figures 3A and 3B a single synapse was activated with a train of ten pulses at 5-100 Hz, and in Figures 3C–3E independent Poisson trains 500 ms long were delivered at each synapse with frequencies varying from 5-80 Hz. Mean depolarization and standard deviation in Figures 3D and 3E were measured during the stimulation period. All simulations were performed at 35°C. For the model variations shown in Figure S4, the following variables were changed for distal synapses: release probability was reduced by 30%, AMPA conductance was reduced by 19%, and dendrite diameter was reduced by 15% (all of the values were estimated from experimental data). The decrease in mean P r was taken from the double exponential fit in Figure 2J, where the time-constant is inversely proportional to P r (Hessler et al., 1993). We took the mean time-constant for a double exponential fit, taking into account the fraction of the curve each one accounted for (τ mean= τ fast∗fr fast+ τ slow∗fr slow, where τ fast and τ slow are the fast and slow time-constants and fr fast and fr slow are the fractions of the curve they account for), and calculated the change in τ mean (proximal: 8.09 ms; distal: 11.49 ms), which is inversely proportional to mean P r. We find that distal synapses show a ∼30% decrease in mean P r when compared to proximal compartments. The code for the mathematical model is provided as supplementary information (see supplementary text). Immuno-histochemistry and FM4-64 staining of primary neurons Dissociated hippocampal neurons expressing GCaMP3 were immunostained using the following primary antibodies: rabbit αVGlut1 (1:500 Synaptic Systems) and chicken αGFP (1:1000, Abcam). Neurons were fixed in 4% PFA for 20 min and permeabilized using 0.25% Triton X-100 (Sigma, UK) in PBS for 5 min. Cells were incubated with 10% goat serum (Sigma, UK) for 1 hr at room temperature, then incubated with primary antibodies in 2% goat serum overnight at 4°C, and finally with Alexa-conjugated secondary antibodies (1:1000, Molecular Probes) for 1 hr at room temperature. Coverslips were mounted onto glass microscope slides using mowiol. Imaging of immunostained neurons was performed using an Olympus FV1000 confocal microscope equipped with a 40X/0.8 NA water-immersion objective (Olympus). FM4-64 staining was performed by incubating neurons in high K+ HBS (78.5mM NaCl, 60mM KCl, 10mM HEPES, 10mM Glucose, 2mM CaCl, 1.3mM MgCl) supplemented with 10 μM FM4-64 (Molecular Probes), 1 μM TTX (Alamone Labs), 2.5nM AP-5 and 2nM CNQX (Tocris) for 90 s to load the entire releasable pool of vesicles. Cells were then washed twice for 4 min in HBS (139mM NaCl, 2.5mM KCl, 10mM HEPES, 10mM Glucose, 2mM CaCl, 1.3mM MgCl, 1 μM TTX) to remove FM4-64 from all external membranes. Neurons were then imaged using an Olympus IX71 inverted microscope with a CCD camera (Coolsnap HQ) controlled by Slidebook software (Intelligent Imaging Innovations), equipped with a 40X/1.0 NA oil-immersion objective (Olympus). The excitation light source was a xenon-arc lamp (Lambda LS; Sutter Instruments), in which light exposure was regulated by a rapid shutter (smartShutter; Sutter Instruments) controlled by a Sutter Instruments lambda 10-3 controller. Filtering was provided by a 470± 20 nm band pass excitation and 515± 20 nm band pass emission (Chroma Technology Corporation) filter set for GCaMP3 and a 565± 22 nm band pass excitation and 590-nm long pass dichroic plus 650± 36 nm band pass emission (Chroma Technology Corporation) filters for FM4-64. Statistics Statistical analysis was performed in MATLAB (MathWorks) and Prism (Graphpad), all the data analyzed with parametric tests was first tested for normality with the D’Agostino and Pearson normality test. Further information and requests for resources and reagents should be directed to and will be fulfilled by the Lead Contact, Juan Burrone (juan.burrone@kcl.ac.uk). Acknowledgments We would like to thank Alejandro Pan Vazquez, David Expósito Alonso and Erin Cocks for help with the EM image segmentation and reconstruction, Thomas Sainsbury for help with electrophysiology experiments. We would also like to thank Matthew Grubb, Oscar Marin, and members of the Burrone lab for critical feedback of the manuscript. This work was supported by an ERC Starter Grant (282047), a Wellcome Trust Investigator Award (095589/Z/11/Z), and an FP7 EC DESIRE grant (602531) to J.B.; a Wellcome Trust/Royal Society Henry Dale Fellowship (098400/Z/12/Z) and a Wellcome Trust and Gatsby Charitable Foundation SWC Fellowship to T.B., and a Wellcome Trust Multiuser Equipment grant (108461/Z/15/Z) to R.A.F. Author Contributions F.W.G. and J.B. designed the study and wrote the text. F.W.G. performed and analyzed the experiments with help from G.N. who also gave critical input. G.V.-B. prepared the EM samples and acquired EM images, R.A.F. provided EM support. A.W. performed and analyzed the experiments on dissociated neurons. T.B. produced the computational model. Declaration of Interests The authors declare no competing interests. Supplemental Information Download all supplementary files included with this articleWhat’s this? Download: Download Acrobat PDF file (5MB) Document S1. Figures S1–S4. Download: Download Acrobat PDF file (313KB) Data S1. Code for the Compartmental Model Used in Figures 3 and S4. Compartmental model of dendritic integration that implements synaptic short-term facilitation and depression as described previously (Varela et al., 1997). Simulations were performed with the NEURON simulation environment (Hines and Carnevale, 1997). Download: Download Acrobat PDF file (7MB) Document S2. Article plus Supplemental Information. Recommended articles References Abrahamsson et al., 2012T. Abrahamsson, L. Cathala, K. Matsui, R. Shigemoto, D.A. Digregorio Thin dendrites of cerebellar interneurons confer sublinear synaptic integration and a gradient of short-term plasticity Neuron, 73 (2012), pp. 1159-1172 View PDFView articleView in ScopusGoogle Scholar Andersen et al., 1980P. Andersen, H. 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Lowe, G. Vizcay-Barrena, R.A. Fleck, J. Burrone Distance-dependent gradient in NMDAR-driven spine calcium signals along tapering dendrites Proc. Natl. Acad. Sci. USA, 114 (2017), pp. E1986-E1995 View in ScopusGoogle Scholar Cited by (30) Astrocyte glun2c nmda receptors control basal synaptic strengths of hippocampal ca1 pyramidal neurons in the stratum radiatum 2021, Elife Show abstract Experience-dependent plasticity is a key feature of brain synapses for which neuronal N-Methyl-D-Aspartate receptors (NMDARs) play a major role, from developmental circuit refinement to learning and memory. Astrocytes also express NMDARs, although their exact function has remained controversial. Here, we identify in mouse hippocampus, a circuit function for GluN2C NMDAR, a subtype highly expressed in astrocytes, in layer-specific tuning of synaptic strengths in CA1 pyramidal neurons. Interfering with astrocyte NMDAR or GluN2C NMDAR activity reduces the range of presynaptic strength distribution specifically in the stratum radiatum inputs without an appreciable change in the mean presynaptic strength. Mathematical modeling shows that narrowing of the width of presynaptic release probability distribution compromises the expression of long-term synaptic plasticity. Our findings suggest a novel feedback signaling system that uses astrocyte GluN2C NMDARs to adjust basal synaptic weight distribution of Schaffer collateral inputs, which in turn impacts computations performed by the CA1 pyramidal neuron. ### Differential role of pre-and postsynaptic neurons in the activity-dependent control of synaptic strengths across dendrites 2019, Plos Biology Show abstract Neurons receive a large number of active synaptic inputs from their many presynaptic partners across their dendritic tree. However, little is known about how the strengths of individual synapses are controlled in balance with other synapses to effectively encode information while maintaining network homeostasis. This is in part due to the difficulty in assessing the activity of individual synapses with identified afferent and efferent connections for a synapse population in the brain. Here, to gain insights into the basic cellular rules that drive the activity-dependent spatial distribution of pre- and postsynaptic strengths across incoming axons and dendrites, we combine patch-clamp recordings with live-cell imaging of hippocampal pyramidal neurons in dissociated cultures and organotypic slices. Under basal conditions, both pre- and postsynaptic strengths cluster on single dendritic branches according to the identity of the presynaptic neurons, thus highlighting the ability of single dendritic branches to exhibit input specificity. Stimulating a single presynaptic neuron induces input-specific and dendritic branchwise spatial clustering of presynaptic strengths, which accompanies a widespread multiplicative scaling of postsynaptic strengths in dissociated cultures and heterosynaptic plasticity at distant synapses in organotypic slices. Our study provides evidence for a potential homeostatic mechanism by which the rapid changes in global or distant postsynaptic strengths compensate for input-specific presynaptic plasticity. ### The glutamatergic synapse: a complex machinery for information processing 2021, Cognitive Neurodynamics ### Distance-dependent regulation of NMDAR nanoscale organization along hippocampal neuron dendrites 2020, Proceedings of the National Academy of Sciences of the United States of America ### Homeostatic scaling of active zone scaffolds maintains global synaptic strength 2019, Journal of Cell Biology ### Circuit-specific control of the medial entorhinal inputs to the dentate gyrus by atypical presynaptic NMDARs activated by astrocytes 2019, Proceedings of the National Academy of Sciences of the United States of America View all citing articles on Scopus 5 Lead Contact © 2018 The Authors. Published by Elsevier Inc. 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1447	https://www.guidetopharmacology.org/GRAC/LigandDisplayForward?tab=clinical&ligandId=3963	hexamethonium \| Ligand page \| IUPHAR/BPS Guide to PHARMACOLOGY Home About About The Guide to PHARMACOLOGY About NC-IUPHAR Database Content Contributors Sponsors Citing Linking to us Disclaimer Privacy and Cookie Policy Targets GPCRs Ion channels LGICs VGICs Other channels Nuclear receptors Kinases Catalytic receptors Transporters Enzymes Other protein targets Target search Target search tools BLAST Ligands Ligand list Ligand families Ligand search Diseases Resources Help Tutorial FAQ Terms and symbols Nomenclature guidelines Publications Antibacterials Drug approvals Downloads Download data and reports Web services Slides and posters News Immunopharmacology Meeting 2018 Latest news Hot topics Latest pairings Concise Guide to PHARMACOLOGY Useful links Advanced search Target search Target search tools BLAST Ligand search Pharmacology search Immuno Portal Malaria Portal HomeAboutHelp GtoPdb is requesting financial support from commercial users. Please see our sustainability page for more information. Home Ligands hexamethonium hexamethonium GtoPdb Ligand ID: 3963 Compound class:Synthetic organic Comment: An antagonist at nicotinic acetylcholine receptors on sympathetic and parasympathetic ganglia. Ligand Activity Visualisation Charts These are box plot that provide a unique visualisation, summarising all the activity data for a ligand taken from ChEMBL and GtoPdb across multiple targets and species. Click on a plot to see the median, interquartile range, low and high data points. A value of zero indicates that no data are available. A separate chart is created for each target, and where possible the algorithm tries to merge ChEMBL and GtoPdb targets by matching them on name and UniProt accession, for each available species. However, please note that inconsistency in naming of targets may lead to data for the same target being reported across multiple charts. ✖ View interactive charts of activity data across species2D Structure ### 2D Structure An image of the ligand's 2D structure. For small molecules with SMILES these are drawn using the NCI/CADD Chemical Identifier Resolver. Click on the image to access the chemical structure search tool with the ligand pre-loaded in the structure editor. For other types of ligands, e.g. longer nucleotides and peptides, a manually drawn representation of the molecule may be provided. ✖ Physico-chemical Properties ### Physico-chemical Properties Calculated molecular properties are available for small molecules and natural products (not peptides). Properties were generated using the Chemistry Development Kit (CDK) (Willighagen EL et al. Journal of Cheminformatics vol. 9:33. 2017, doi:10.1186/s13321-017-0220-4; All properties were selected to enable the prediction of the Lipinski Rule-of-Five profile or ‘druglikeness’ for each ligand. For more info on each category see the help pages Hydrogen bond acceptors 0 Hydrogen bond donors 0 Rotatable bonds 7 Topological polar surface area 0 Molecular weight 202.24 XLogP 0.97 No. Lipinski's rules broken 0 Generated using the Chemistry Development Kit (CDK) (Willighagen EL et al. Journal of Cheminformatics vol. 9:33. 2017, doi:10.1186/s13321-017-0220-4; SMILES / InChI / InChIKey ### SMILES / InChI / InChIKey SMILES (Simplified Molecular Input Line Entry Specification) A specification for unambiguously describing the structure of chemical molecules using short ASCII strings. Canonical SMILES specify a unique representation of the 2D structure without chiral or isotopic specifications. Isomeric SMILES include chiral specification and isotopes. Standard InChI (IUPAC International Chemical Identifier) and InChIKey InChI is a non-proprietary, standard, textual identifier for chemical substances designed to facilitate linking of information and database searching. An InChIKey is a simplified version of a full InChI, designed for easier web searching. Generated using the Chemistry Development Kit (CDK) (Willighagen EL et al. Journal of Cheminformatics vol. 9:33. 2017, doi:10.1186/s13321-017-0220-4; ✖ Canonical SMILES CN+(C)C Isomeric SMILES CN+(C)C InChI InChI=1S/C12H30N2/c1-13(2,3)11-9-7-8-10-12-14(4,5)6/h7-12H2,1-6H3/q+2 InChI Key VZJFGSRCJCXDSG-UHFFFAOYSA-N Generated using the Chemistry Development Kit (CDK) (Willighagen EL et al. Journal of Cheminformatics vol. 9:33. 2017, doi:10.1186/s13321-017-0220-4; Summary Biological activity Clinical data Structure No information available. Summary of Clinical Use Hexamethonium was used to treat chronic hypertension, but use has been discontinued. No information regarding international approval status is available. Mechanism Of Action and Pharmacodynamic Effects Hexamethonium acts as negative allosteric modulator of nicotinic acetylcholine receptor activity in sympathetic and parasympathetic ganglia. Contact us Privacy and Cookie Policy Sponsors list This work is licensed under a Creative Commons Attribution-ShareAlike 4.0 International License
1448	https://nvlpubs.nist.gov/nistpubs/Legacy/circ/nbscircular11e2-1915.pdf	DEPARTMENT OF COMMERCE CALORIMETERS Issued July 1, 1915 oj WASHINGTON CIRCULARS 1. Verification of Standards and Measuring Instruments. 2. Measurements of Length and Area, Including Thermal Expansion. 3. Verification of Standards of Mass. 4. Verification of Standards of Capacity. 5. Testing of Clinical Thermometers. 6. Fees for Electric, Magnetic, and Photometric Testing. 7. Pyrometer Testing and Heat Measurements. 8. Testing of Thermometers. 9. Testing of Glass Volumetric Apparatus. 10. Legal Weights (in pounds) per Bushel of Various Commodities. 11. The Standardization of Bomb Calorimeters. 12. Verification of Polariscopic Apparatus. 13. Standard Specifications for the Purchase of Incandesent Lamps. 14. Samples of Analyzed Irons and Steels—Methods of Analysis. 15. A Proposed International Unit of Light. 16. The Testing of Hydrometers. 17. Magnetic Testing. 18. Standard Gauge for Sheet and Plate Iron and Steel. 19. Standard Density and Volumetric Tables. 20. Electrical Measuring Instruments. 21. Precision Measurements of Resistance and Electromotive Force. 22. Standard Specifications for Transformers, Oil-immersed, Self-cooled, 60-cycle, 2200 Volts. 23. Standardization of Electrical Practice in Mines. 24. Publications of the Bureau of Standards. 25. Standard Analyzed Samples—General Information. 26. Analyzed Iron and Manganese Ores—Methods of Analysis. 27. The Testing and Properties of Optical Instruments. 28. The Determination of the Optical Properties of Materials. 29. Announcement of a Change in the Value of the International Volt. 30. Lime: Its Properties and Uses. 31. Copper Wire Tables. 32. Standards for Gas Service. 33. United States Government Specification for Portland Cement. 34. The Relation of the Horsepower to the Kilowatt. 35. Melting Points of Chemical Elements. 36. The Testing and Properties of Electrical Condensers. 37. Electric Wire and Cable Terminology. 38. The Physical Testing of Mechanical Rubber Goods. 39. Specifications for and Measurement of Standard Sieves. 40. Sodium Oxalate as a Standard in Volumetric Analysis. 41. Testing and Properties of Textile Materials. 42. Metallographic Testing. 43. The Metric Carat. 44. The Testing of Materials. 46. Testing of Barometers. 47. Units of Weight and Measure; Definitions and Tables of Equivalents. 48. Standard Methods of Gas Testing. 49. Safety Rules to be Observed in the Operation of Electrical Equipment and Lines. 50. National Standard Hose Couplings and Fittings for Public Fire Service. 51. Measurement of Time and Tests of Timepieces. 52. Regulation of Electrotyping Solutions. 53. The Composition, Properties, and Testing of Printing Inks. 54. Proposed National Electrical Safety Code. 55. Measurements for the Household. [A complete list of Scientific Papers, Technologic Papers, and miscellaneous publications may be obtained free of charge on application to the Bureau of Standards, Washington, D. C.] DEPARTMENT OF COMMERCE Circular of THE Bureau of Standards S. W. STRATTON, Director No. 11 THE STANDARDIZATION OF BOMB CALORIMETERS [2d Edition] Issued July 1, 1915 WASHINGTON GOVERNMENT PRINTING OFFICE 1915 ADDITIONAL COPIES OF THIS PUBLICATION MAY BE PROCURED FROM THE SUPERINTENDENT OF DOCUMENTS GOVERNMENT PRINTING OFFICE WASHINGTON, D. C. AT 5 CENTS PER COPY THE STANDARDIZATION OF BOMB CALORIMETERS CONTENTS Page 1. Characteristics op bomb calorimeters 4 2. Methods op calibration 6 3. Standard materials 8 4. Precautions 9 () Oxygen 9 () Pressure 10 (c) Charge 10 (d) Firing . 10 (e) Firing current 10 (/) Water n (g) Stirring n (h) Temperature measurements 11 (i) Cooling correction 12 (j) Lag of the thermometer 12 (k) Nitric acid 12 5. Manipulation 13 (a) Preparation of the sample .... (b) Preparation of the bomb (c) Filling the bomb with oxygen (d) Calorimeter water (e) Temperature adjustments (/) Method of making observations 14 (g) Titration 14 6. Computation op results 14 7. Example op water equivalent determination 15 8. Basis for expressing results 16 9. Schedule of fees 17 Accurate determination of the heats of combustion 1 of solids, liquids, and gases are usually made by means of a calorimeter of 1 The heat of combustion at constant volume of a substance containing only the elements carbon, hydro-gen, and oxygen may be defined as the number of heat units liberated by the combination, in an inclosure of constant volume, of unit mass of the substance with oxygen to form carbon dioxide and water, the sub-stance and the oxygen being initially at the same temperature, the products of combustion being cooled to the initial temperature, and the water formed by combustion being condensed to the liquid state. To be exact, the temperature at which the reaction takes place should also be specified, but the change of the heat of combustion with change of initial temperature is so small that this is not necessary for a tempera-ture range between 15 0 and 30° C, temperatures at which experimental results are usually obtained. In practice also the initial and final temperatures are not usually the same, but differ by from 2 0 to 4 0 C. The effect of this difference is also small and is usually neglected, except that it is partly taken account of when the heat capacity of the combustible charge is included in that of the calorimeter. The term “ heat of combustion ” is also commonly used to denote the heat liberated by the oxidation of substances containing other oxidizable elements, such as sulphur, nitrogen, etc. When the term is so used, the definition must be extended so as to define the condition of the final oxidation products, as, for example, if sulphur is pres-ent, whether the resulting product is sulphur dioxide or sulphuric add. 3 4 Circular of the Bureau of Standards the Berthelot bomb type. In these calorimeters a known quan-tity or mass of the combustible is burned in oxygen under high pressure in a closed container or bomb. The heat given off during combustion is taken up by a known mass of water in which the bomb is immersed. Combustion in this case takes place at con-stant volume and the quantities to be measured are (a) the mass of combustible, (6) the mass of water, including the equivalent mass, or water equivalent of the bomb, container, and accesso-ries, (c) the rise of temperature, and (d) the so-called “radiation correction,” i. e., the correction for heat transfer or thermal leakage between the calorimeter and its surroundings. 1. CHARACTERISTICS OF BOMB CALORIMETERS Several makes of bomb calorimeters, including all the neces-sary equipment for measuring heats of combustion, are to be found on the market and a question often arises as to the relative accuracy of these equipments. The differences between the various makes are in general not of such a nature as to greatly affect the accuracy of the results obtained with them, but are rather differences in mechanical design which affect the conven-ience of manipulation, the cost of manufacture, etc. Certain suggestions may be useful in the design or selection of a calorimetric equipment. The three essential parts of a bomb calorimeter equipment are : () The bomb or container in which the combustible charge is burned. () The calorimeter proper or vessel containing water in which the bomb, thermometer, and stirring device are immersed. (c) The jacket used to protect the calorimeter from effects of variations in room temperature, air drafts, etc. The following conditions should be realized so far as practicable : (a) The bomb should be of sufficient strength to withstand a pressure of at least 500 or 600 atmospheres. The outer surface should be nickeled or otherwise finished so as to be protected from corrosion. The shape should be such as to allow free circulation of water between the bottom of the bomb and that of the calori-meter, as well as on all sides of the bomb. The inner surface of the bomb may be of platinum, gold, enamel, or other material not Standardization of Bomb Calorimeters 5 easily attacked by the products of combustion. Linings may for convenience be made removable, but such removable linings should not be used in bombs intended for combustions of gases unless they are so arranged that none of the combustible gases can find their way between the lining and the body of the bomb. Of the materials so far used platinum is perhaps the best and most durable, although it is also the most expensive. Gold has not been extensively used, but might prove as good as platinum. Enamel is satisfactory as long as it lasts, but it is liable to chip and crack from contact with hot particles of coal or fuse wire, which are sometimes thrown out of the crucible. These linings may last for many hundreds of combustions or may be injured badly in the first few combustions. They may be still usable for work not of the highest accuracy when badly chipped, as the errors introduced by the reaction between the small amount of exposed metal and the products of combustion may be still too small to be of importance. Nickel linings are attacked by acid formed during combustion and should only be used when the corrections for acid formed can be disregarded. Certain alloys, said to be very resistant to attack by products of combustion, have been used for the construction of bombs, which thus require no special linings. If the claims made for such bombs are justi-fied, these bombs should be entirely satisfactory. (b) The calorimeter should be as small as practicable while pro-viding for total immersion of the bomb and proper stirring of the water. Highest precision requires that the calorimeter be fitted with a floating cover or its equivalent, but the use of such a cover is inconvenient and may usually be omitted in fuel calorimetry. Two forms of stirrer are in common use, the reciprocating stirrer and the screw-propeller stirrer. A screw propeller, when mounted in a tube so as to give a positive circulation of the water, is more efficient than the other type and does not promote evaporation, as it has no reciprocating parts entering and leaving the water. Such a screw propeller should preferably have no bearing in the water of the calorimeter, and the metal parts, such as the shaft and the tube, which extend out of the calorimeter, should be sepa-rated from the metal parts above by some heat-insulating material to reduce the conduction of heat. It is important that the stirrer 6 Circular of the Bureau of Standards run at a constant speed, and this should be provided for in the design of the apparatus. (c) The jacket which surrounds the calorimeter should be a double-walled metallic vessel containing water between the walls. This water should preferably be stirred, so that its temperature can be measured and kept uniform. It is also advantageous to have this temperature controlled by means of a thermostat. There should be no material between the calorimeter and the inside of the jacket except the smallest practicable insulating supports for the calorimeter. The inner wall of the jacket should preferably be of polished metal and it, as well as the calorimeter surface, should be kept clean and bright. The jacket should have a cover, preferably of copper or aluminum, making good contact with the top of the jacket in order to prevent convection currents between the room and the space inclosing the calorimeter. A cover of such highly conducting metals, if of sufficient thick-ness and in good contact with the metal top of the calorimeter jacket, serves as a very fair approximation to a completely iso-thermal inclosure for the calorimeter. With some types of bombs, however, such a cover is not feasible on account of the long valve stem at the top of the bomb. When the highest accuracy is required, the cover should be double walled and so arranged that the jacket water circulates through it. 2. METHODS OF CALIBRATION The use of a bomb-combustion calorimeter for determining the heats of combustion of fuels, etc., requires a knowledge of the total heat capacity or “water equivalent” of the calorimeter as it is used. This water equivalent includes the heat capacity of the calorimeter itself (i. e., the vessel which contains the water), of the water which it contains, of the bomb and its contents, and of parts of the thermometer, stirring device, and supports for the calorimeter. If a floating cover is used on the calorimeter, this also is included in the water equivalent. The heat capacity of the water is its mass (observed weight corrected for buoyancy of air) times its specific heat. The specific heat depends somewhat upon the temperature and is here taken Standardization of Bomb Calorimeters 7 as unity at 20° C, so that the calorie is defined as the heat capacity of 1 gram of water per degree centigrade, at a temperature of 20° C. In the previous edition of this circular the calorie was defined as the heat capacity of 1 gram of water per degree centigrade at 1 5° C, instead of at 20° C as here used. There is a growing tend-ency to express results of calorimetric measurements in terms of the 200 calorie on account of the fact that this temperature more nearly represents average laboratory conditions and the fact that the heat capacity of water changes much less rapidly with tem-perature in this region. For these reasons the 20 0 calorie has been adopted for this and future editions of this circular. The numerical values of heats of combustion expressed in 26° calories are about 1 part in 1000 greater than when expressed in 1 0 calories. If the Btu were defined as the heat capacity of 1 pound mass of water at the same temperature (i. e., 20° C = 68° F), calories per gram would be convertible into Btu per pound by multiplying by the factor 1.8. In view of the facts that standard measurements in the English system of units are generally referred to 60 0 F (or 62° F in case of length standards), that weights of fuels are usually expressed as pounds weight (i. e., uncorrected for buoyancy of the air), and that common usage in the English system of measure-ments should perhaps have relatively more weight in definition than strict scientific accuracy, the Btu is here defined as the heat capacity of 1 pound weight of water per degree Fahrenheit at 6o° F. To convert calories per gram to Btu per pound, according to the foregoing definition, it is necessary to multiply by 1.798 or divide by 0.556. The water equivalent of a calorimetric outfit may be measured independently by several methods—as, for instance, by determining the heat capacity of its separate parts individually or by supplying heat, measured electrically, and determining the rise of tempera-ture of the calorimeter, bomb, water, etc., as used. But in practice it is much more convenient to determine the water equiv-alent by the combustion of known amounts of some material, the heat of combustion of which has previously been determined accu-rately in a calorimeter calibrated by one of the methods referred to above. 93136°—15 2 8 Circular of the Bureau of Standards The advantages of this procedure are that the calibration of one or two calorimeters can be performed with great care and to a high degree of accuracy in a laboratory possessing the equip-ment required for this kind of work, and after the heats of com-bustion of certain suitable materials have been thus determined these materials can be used in various laboratories under the same conditions which prevail in making fuel combustions. In this way the results obtained in different laboratories using the same standard samples are made more concordant and probably much more accurate than if each calorimeter were calibrated inde-pendently either at the laboratory in which it is to be used or even at a special testing laboratory—as, for instance, the Bureau of Standards. This is true because by the use of standard sam-ples in the laboratory where the calorimeter is to be used, and by the same observers who are to use it, the minor details of obser-vations are duplicated in both standardizations and fuel combus-tions, thus eliminating this source of small discrepancies; also because by the frequent use of standard samples it is possible to guard against errors such as might be introduced by changes in the apparatus or in the amount of water used and which might otherwise escape notice. 3. STANDARD MATERIALS Among the materials available for standard combustion samples, sucrose or cane sugar, benzoic acid, and naphthalene have been most generally adopted. While any of these substances as pro-cured from dealers in chemicals may be sufficiently pure for com-mercial calibrations, it is not safe to rely upon samples thus obtained, and the Bureau of Standards, after an extended study of the methods of preparation and the heats of combustions of these substances, is prepared to supply tested samples of any of these materials, as explained in the schedule of fees, page 17. Sucrose is neither volatile nor strongly hygroscopic, but is rather difficult to ignite and occasionally does not bum completely. It has a heat of combustion of about 3950 calories, or only about half that of the average coal. The more exact value for each sample will be given in the certificate. Standardization of Bomb Calorimeters 9 Benzoic acid is only slightly volatile, is not very hygroscopic, has a heat of combustion of about 6330 calories, and burns more readily than sugar. The results obtained by different observers with this substance are also more concordant than with either sucrose or naphthalene. Naphthalene is quite volatile, but not hygroscopic; it has a heat of combustion of about 9620 calories, a little higher than that of most coals, and it ignites and burns very readily. In its use some care is necessary to avoid errors due to sublimation. Of these materials, probably the most satisfactory for work of the highest accuracy is benzoic acid, but for calibration of commer-cial calorimeters to an accuracy of 0.1 per cent naphthalene has some advantages. The loss by sublimation from samples of naphthalene made into briquets will hardly exceed 0.1 or 0.2 per cent in an hour. The details of the methods of preparation of these standard combustion samples and of the determination of their heats of combustion have already been published. 2 The heats of com-bustion of other substances are being determined in the laborato-ries of the Bureau with a view to testing their suitability for use as standard combustion samples. 4. PRECAUTIONS (a) The oxygen used for combustion should be as pure as is obtainable commercially, and must be free from all combustible gases, such as hydrogen, carbon monoxide, and hydrocarbons. A small amount of nitrogen is usually present in the oxygen, and an additional 3 per cent to 4 per cent is added when oxygen at 30 to 20 atmospheres pressure is admitted to the bomb originally con-taining air. It has recently been shown 3 that the presence of at least 5 per cent of nitrogen is desirable, since it has the effect of promoting the oxidation of the sulphur which is contained to some extent in most coals. Where coals are high in sulphur this amount of nitrogen should be present. 2 Bulletin of the Bureau of Standards, 11 , p. 189; 1914 (Scientific Paper No. 230). 8 Regester, Jour, of Ind. and Eng. Chem., vol. 6, No. 10, p. 812; 1914. io Circular of the Bureau of Standards (b) The pressure to which the bomb should be filled is usually from 20 to 30 atmospheres, depending upon the volume of the bomb and the charge to be burned, but in no case should the total amount of oxygen be less than three times that which will combine with the charge. With a smaller amount of oxygen there is danger of incomplete combustion. The completeness of combus-tion can usually be determined sufficiently well by inspection, as the presence of a few tenths of a milligram of soot can be readily detected. Some bombs are so constructed that the gaseous products of combustion can be conveniently removed and analyzed. (c) The charge of a standard combustion sample should be pre-pared in the form of a briquet, both for ease of handling and to avoid danger of blowing the material out of the crucible when oxygen is admitted to the bomb. Failure to observe this pre-caution may lead to errors due to incomplete combustion. Sam-ples should be weighed in the crucible and placed in the bomb immediately afterwards, particularly if volatile. Naphthalene, in particular, requires briquetting or fusing into a solid mass to reduce the rate of sublimation. (d) Firing of the charge is usually accomplished by means of an electric fuse of either fine platinum or fine iron wire. Platinum wire will ignite coal, but it is not certain to ignite substances such as sugar, benzoic acid, or naphthalene, which melt readily. For these materials it is preferable to use iron wire, although a small correction must be made for its heat of combustion (1600 calories per gram). A piece of No. 34 Brown & Sharpe gauge iron wire from 1 to 3 cm long may be connected to somewhat heavier platinum leads. The longer wires (2 or 3 cm) should be wound into a small spiral and mounted just above the briquet to be fired. A straight wire 1 cm long will ignite naphthalene, but for sugar it is better to use 3 cm of wire. (e) The current for firing should be obtained from storage or dry cells and should not be of more than 10 or 15 volts potential. It is not advisable to use a high potential, such as that of a 1 10-volt lighting circuit, even though the current is cut down by lamps in series, since with a high voltage there is danger of arcing, accom-Standardization of Bomb Calorimeters n panied by evolution of heat within the bomb. In any case it is important to close the firing circuit with a switch for only about one second, or as short a time as will serve to ignite the charge. When possible, an ammeter should be used in the circuit to indicate how much current is used and when the fuse wire has burned out. Small ammeters which will serve for this purpose can be bought for from $i to $5. (/) The water should be weighed out in the calorimeter, and if the absolute heat capacity of this water is to be known, the weigh-ing must be reduced to vacuo. In practice, however, this weighing usually serves only to maintain the same mass in all the observations (both calibrations and fuel combustions) , and in this case it is not necessary to reduce to weight in vacuo. (g) The stirring of the calorimeter water during an observa-tion should be effected by some mechanical stirring device which can be made to run at a constant speed. Two types of stirrers in common use are described on page 5. The amount of energy supplied by stirring should not be exces-sive. This may be determined by adjusting the temperature of the calorimeter to equality with that of the jacket and allowing the stirrer to run continuously for 10 minutes. If the tempera-ture of the calorimeter rises more than about one hundredth of a degree in this length of time, the rate of stirring is excessive. (h Temperature measurements are usually more subject to error than any other parts of the calorimetric determination, and yet they often receive the least attention. Mercurial thermom-eters used for this purpose are usually either of the graduated-stem type, with a scale covering a io° to 15 0 C. range of temperature, graduated into o°.o5 or o°.o2 intervals; or of the Beckmann type, in which the scale covers a range of 5 0 or 6° C., graduated into o°.oi intervals, adjustable for use at different temperatures by remov-ing some of the mercury from the bulb to an upper reservoir. None of these thermometers can be relied upon to give true tem-perature differences unless they have been tested and corrections are applied for errors of the scale, for the emergent stem, and, in the case of Beckmann thermometers, for the amount of mercury removed from the bulb. 12 Circular of the Bureau of Standards The methods of making these corrections are explained in cer-tificates which accompany all thermometers tested at this Bureau ; also in Bureau Circular No. 8. If an accuracy greater than about one or two parts in a thousand is required, recourse must be had to some instrument more accurate than the ordinary mercurial thermometer. An electrical resistance thermometer equipment affording greater accuracy has been put on the market. Such an equipment is more expensive than a mercurial thermometer. Thermometers of this type, which have been constructed at the Bureau of Standards, have been described in a recent publication. 4 (i) A cooling correction or so-called radiation correction must be applied to calorimetric observations because of the transfer of heat between the calorimeter and its jacket. This heat transfer may take place by conduction, convection, radiation, or evapora-tion. Corrections are usually made on the assumption that Newton’s lav/ of cooling applies—i. e., that the rate of heat transfer is always proportional to the difference in temperature between the calorimeter and its jacket. This assumption is warranted provided the jacket is supplied with a cover (as described on p. 6) and kept at a constant temperature; that it contains only a neg-ligible amount of material, such as supports, between its inner surface and that of the calorimeter; and that the temperature of the calorimeter is not allowed to rise more than about a degree centigrade above that of the jacket, thereby avoiding any signifi-cant amount of evaporation from the calorimeter. A convenient method of computing the cooling correction is described under “Computation.” (/) The lag of the thermometer used in measuring temperatures has often been assumed to introduce an error requiring correction, but it may be shown that no such correction is required 5 under the usual conditions of combustion-bomb calorimetry. (k) The formation of nitric acid from the nitrogen contained in the oxygen used for filling the bomb supplies an amount of heat equivalent to about 230 calories per gram of acid formed, and this amount is to be subtracted from the apparent total amount of 4 Bulletin of the Bureau of Standards, 9, 483; 1913 (Scientific Paper No. 200). 6 W. P. White, Physical Review, 31 » p. 562, 1910; and Bulletin of the Bureau of Standards, 8, p. 659, 1912 (Scientific Paper No. 185). Standardization of Bomb Calorimeters 13 heat due to the combustion. The amount of nitric acid formed can be found by titration, provided no other acids are present, which is substantially true for observations made with the stand-ard combustion samples herein described. 5. MANIPULATION (a) Preparation of Sample.—The material to be used for stand-ardization should be removed from the container and compressed into a briquet of nearly the required weight, keeping the container closed except while removing the sample. The briquet should then be weighed (preferably in the crucible in which it is to be burned) If naphthalene is used, the briquet should be immediately placed in the bomb to avoid excessive sublimation. (b) Preparation of the Bomb.—The firing wire, if of iron, should be measured and connected between the platinum terminals, using, if necessary, a piece of platinum wire somewhat heavier than the iron wire, to make the connection. The platinum and the iron must both be clean to insure good electrical connections. About 0.5 cc of water should be placed in the bottom of the bomb to satu-rate with moisture the oxygen used for combustion. When the crucible is put in place in the bomb, the firing wire should touch the briquet but not the crucible. (c) Filling the Bomb with Oxygen.—Oxygen from the supply tank is to be slowly admitted until the required pressure of about 20 atmospheres for the larger bombs or about 30 atmospheres for the smaller bombs is reached. ( ) Calorimeter Water.—The amount of water to be used depends upon the type of calorimeter. Distilled water is prefer-able, and the amount should be determined by weighing. The amount must be kept the same as that used in fuel combustions, or a correction applied for the difference in weight. (e) Temperature Adjustments.—The initial temperature in the calorimeter should be so adjusted that the final temperature, after the combustion, will not be more than i° C., preferably about o?3 C, above that of the jacket, under which conditions the total correction for heat gained from or lost to the surroundings will be very small when the rise of temperature is 2 0 or 3 0 C. The effect of evaporation will also be small. 14 Circular of the Bureau of Standards (f) Method of Making an Observation.—The bomb, when ready for firing, is to be placed within the calorimeter, the firing wires connected, the cover put in place, and the stirrer and thermom-eter so placed as not to be in contact with the bomb or container. The stirrer is then started and after the rate of change of the thermometer reading has become uniform, not less than two minutes after the stirrer is started, temperatures are read at one-minute intervals for five minutes, and the charge is then fired, noting the exact time of firing. Observations of temper-ature are then made at intervals depending upon the method to be used for computing the cooling correction. When the rate of change of temperature has again become uniform, a series of thermometer readings is taken at one-minute intervals for five minutes to determine the cooling rate. (g) Titration.—After a combustion the bomb is to be opened, after allowing the gas to escape, and the inside examined for traces of unbumed material or sooty deposit. If these are found, the observation should be discarded. If the combustion appears complete, the bomb is to be rinsed out and the washings titrated to determine the amount of acid formed. A correction of 230 calories per gram of nitric acid should be subtracted from the total heat observed. 6. COMPUTATION OF RESULTS The following method of computation is recommended, to take the place of the Pfaundler or other similar formulas for computing the cooling correction (radiation correction). This method, de-scribed in full elsewhere 6 is there shown to give an accurate deter-mination of the cooling correction as based on the assumption of Newton’s law of cooling. The method is conveniently applied in the following manner : Observe (1) the rate of rise (rj of the calorimeter temperature in degrees per minute for five minutes before firing, (2) the time (tf) at which the last temperature reading is made and the charge fired, (3) the time (t2) when the rise of temperature has reached six-tenths of its total amount (this point can be determined by 6 Bulletin of the Bureau of Standards, 11 , p. 189; 1914. Standardization of Bomb Calorimeters 15 adding to the temperature observed at the time of firing 60 per cent of the expected temperature rise and noting the time when this point is reached, 7 (4) the time (t3) of a thermometer reading taken when the temperature change has become uniform some five minutes after firing, and (5) the final rate of cooling (r2) in degrees per minute for five minutes. The rate rx is to be multiplied by the time t2 — tt in minutes and tenths of a minute, and this product added (subtracted if the temperature were falling at the time tf) to the thermometer reading taken at the time tv The rate r2 is to be multiplied by the time t3 — 12 and this product added (subtracted if the temperature were rising at the time t3 and later) to the thermometer reading taken at time t3 . The difference of the two thermometer readings thus corrected, provided the corrections from the certificate have already been applied, gives the total rise of temperature due to the combustion. 7. EXAMPLE OF WATER EQUIVALENT DETERMINATION Weight of benzoic acid sample 8 (6329 cal20 per gram) grams. . 1. 2007 Weight of water in calorimeter 9 do 2700 Weight of water in bomb 9 do. . . . o. 5 Pressure of oxygen in bomb 9 atmospheres. . 30 Weight of iron fuse wire (2 cm at 0.065 S Per meter) grams . . o. 0013 Initial temperature observed i?46 Approximate rise of temperature 2?5o; 60 per cent of rise i?5o Temperature to observe at /2=i?46+i? 5 3?o Temperature of outer jacket (constant) 20^3 7 When the temperature rise is not approximately known beforehand, it is only necessary to take ther-mometer readings at 40, 50, 60 seconds (and possibly 70 seconds with some calorimeters) after firing, and from these observations to find when the temperature rise had reached 60 per cent of the total. Thus, if the temperature at firing was 2?i35, at 40 seconds 3^05, at 50 seconds 3^92, at 60 seconds 49 i 6, and the final temperature were 4?2oo, the total rise (temperature at firing minus final temperature) was 2?o7; 60 per cent of it was i924. The temperature to be observed was then 2? 14+ 1924=3938. Referring to the observations at 40 and 50 seconds, the temperatures were, respectively, 3905 and 3992. The time corresponding to the temperature of 3938 was therefore 4o+^o^_^o°^X 10=45 seconds. 8 The weight of the sample should be such as to give about the same temperature rise as is commonly uesd in fuel combustions. 9 These weights to be kept the same in all observations, both calibrations and fuel combustions. 1 6 Circular of the Bureau of Standards Temperature Observations by Beckmann Thermometer Time 1 Temperature 1 Thermometer corrections from certificate Corrected temperature h m s Degrees C Degrees C Degrees C 12 00 00 1.446 12 01 00 1.450 12 02 00 1.455 12 03 00 1.459 12 04 00 1.462 Charge fired (ti) 12 05 00 1.466 -0. 005 1.461 ('to') 12 05 48 3.0 o o CM T— 1 2 3.961 0. COO 3. 961 12 12 00 3.960 12 13 00 3.959 12 14 00 3.958 12 15 00 3.957 12 16 00 3.956 rx=o. 020-4-5 =0.0040 per minute; (^—^=0.8 minute), Initial temperature=i?46i-f0.004X0.8= i °464 r2=o. 005-^-5=0.001 per minute; (t3 — t2=5.2 minutes), Final temperature=3?9oi +0.001X5.2= 3? 966 Total rise of temperature (3^966— i?464) 2? 502 Stem and setting corrections to thermometer +o°. 002 Total corrected rise of temperature 2°. 504 Total heat supplied by benzoic acid sample, 6329X1.2007 7599 Heat supplied by formation of nitric acid 230X0.020 (0.020 found by titration) 4. 6 Heat supplied by combustion of iron firing wire, 1600X0.0013 2.1 Total heat supplied 7606 Water equivalent (7606-S-2.504) 3038 8 . BASIS FOR EXPRESSING RESULTS The heats of combustion at constant volume of the purest samples of sugar, benzoic acid, and naphthalene which have been produced at the Bureau were found to be as follows : Cane sugar 3949 ± 2 Benzoic acid 6329 ± 1 Naphthalene 9622 ± 2 200 calories per gram of the substance weighed in air against brass weights. Standardization of Bomb Calorimeters 17 These figures represent the total heat of combustion when all the water formed by combustion is condensed to liquid water at a temperature sensibly equal to that before combustion. Correc-tion has been made for the heat of formation of whatever nitric acid is produced by combustion, for the heat of combustion of the iron firing wire, as well as for the heat produced electrically in heating the wire up to its ignition point. 9 . SCHEDULE OF FEES Schedule 39 .—Calorimetry. (a) Standard combustion samples, of about 50 grams each, of sugar, naphtha-lene, benzoic acid, etc., accompanied by a certificate giving the heat of com-bustion of the sample, each $2. 00 Note.—On all orders for four or more combustion samples (alike or dif-ferent) a reduction of 10 per cent will be made. Orders for samples should be accompanied by a remittance, which may be by check, draft, or post-office order, and should be payable to the Bureau of Standards. (b) Test of a combustion calorimeter (type of Berthelot, Mahler, Atwater, Peters, Williams, Emerson, Parr, Junkers, Sargent, Simmance-Abady, Boys, Fery, etc.), including a determination of the water equivalent for calorime-ters of the bomb type and of the corrections to the accompanying thermome-ter 25. 00 (c) The fees for testing calorimetric and Beckmann thermometers will be found in Bureau of Standards Circular No. 8. S. W. Stratton, Director. Approved E. F. Sweet, Acting Secretary. ’
1449	https://www.scribd.com/document/649331846/Ebullioscopic	Ebullioscopic \| PDF \| Science & Mathematics Opens in a new window Opens an external website Opens an external website in a new window This website utilizes technologies such as cookies to enable essential site functionality, as well as for analytics, personalization, and targeted advertising. To learn more, view the following link: Privacy Policy Open navigation menu Close suggestions Search Search en Change Language Upload Sign in Sign in Download free for 30 days 0 ratings 0% found this document useful (0 votes) 124 views 3 pages Ebullioscopic The document lists ebullioscopic and cryoscopic constants for various compounds. The ebullioscopic constant relates a compound's boiling point elevation in solution. The cryoscopic constant … Full description Uploaded by Jesrel Austria Vidal Sarmiento AI-enhanced description Go to previous items Go to next items Download Save Save Ebullioscopic For Later Share 0%0% found this document useful, undefined 0%, undefined Print Embed Ask AI Report Download Save Ebullioscopic For Later You are on page 1/ 3 Search Fullscreen Ebullioscopic (Boiling Point Elevaon) Constants The ebullioscopic constant is a characterisc property of the solvent and not the solute in a soluon. C o m p o u n d E b u l l i o s c o p i c c o n s t a n t K b in units of [(°C·kg)/mol] or [°C/molal] A c ec a c i d 3.2 2 A c e t o n e 1.8 0 A c e t o n i t r i l e 1.4 4 A n i l i n e 3.8 2 A n i s o l e 4.2 0 B e n z a l d e h y d e 4.2 4 B e n z e n e 2.6 4 1-B u t a n o l 2.1 7 C a r b o n d i s u ld e 2.4 2 C h l o r o b e n z e n e 4.3 6 1-C h l o r o b u t a n e 3.1 3 C y c l o h e x a n e 2.9 2 C y c l o h e x a n o l 3.5 D e c a n e 6.1 0 D i c h l o r o m e t h a n e 2.4 2 D i e t h y l e t h e r 2.2 0 D i m e t h y l s u l f o x i d e 3.2 2 1,4-D i o x a n e 3.0 1 E t h a n o l 1.2 3 E t h y l a c e t a t e 2.8 2 E t h y l e n e g l y c o l 2.2 6 H e p t a n e 3.6 H e x a n e 2.9 0 I o d o m e t h a n e 4.3 1 M e t h a n o l 0.8 6 M e t h y l a c e t a t e 2.2 1 adDownload to read ad-free C o m p o u n d E b u l l i o s c o p i c c o n s t a n t K b in units of [(°C·kg)/mol] or [°C/molal] N -M e t h y l a n i l i n e 4.3 N -M e t h y l f o r m a m i d e 2.2 N i t r o b e n z e n e 5.2 N i t r o m e t h a n e 2.0 9 1-O c t a n o l 5.0 6 P h e n o l 3.5 4 1-P r o p a n o l 1.6 6 2-P r o p a n o l 1.5 8 P y r i d i n e 2.8 3 P y r r o l e 2.3 3 P y r r o l i d i n e 2.3 2 T e t r a c h l o r o e t h y l e n e 6.1 8 T e t r a c h l o r o m e t h a n e 5.2 6 T o l u e n e 3.4 0 T r i c h l o r o e t h y l e n e 4.5 2 T r i c h l o r o m e t h a n e 3.8 0 W a t e r 0.5 1 3 o-X y l e n e 4.2 5 Reference: hps://chem.libretexts.org/Ancillary_Materials/Reference/Reference_Tables/Bulk_Properes/B5%3A_Ebulli oscopic_(Boiling_Poi nt_Elevaon)_Constants?clid=IwAR00OIaQaS1T51L9dXjZEle_ihzEsq5u2V3AotHqSRgzXCZXVGQxXBDK3BU Cryoscopic (Melng Point Depression) Constants S o l v e n t N a m e K m CH 3 CO 2 H A c ec a c i d 3.9 0 CH 3 COCH 3 A c e t o n e 0.8 5 0 C 6 H 5 NH 2 A n i l i n e 5.8 7 C 6 H 6 B e n z e n e 4.9 0 adDownload to read ad-free S o l v e n t N a m e K m CS 2 C a r b o n d i s u ld e 3.8 3 CCl 4 C a r b o n t e t r a c h l o r i d e 3 0.0 CHCl 3 C h l o r o f o r m 4.7 0 C 6 H l2 C y c l o h e x a n e 2 0.0(C 2 H 5 ) 2 O D i e t h y l e t h e r 1.7 9 C 10 H 8 N a p h t a l e n e 6.8 0 C 6 H 5 NO 2 N i t r o b e n z e n e 7.0 0 C 6 H 5 O H P h e n o l 7.2 7 C 2 H 5 O H E t h a n o l 1.9 9 H 2 O W a t e r 1.8 6 Reference: hps://chem.libretexts.org/Ancillary_Materials/Reference/Reference_Tables/Bulk_Pro peres/B6%3A Cryosc opic(Me lng_P oint_Dep ression)_ Constant s?clid=I wAR2Fh uDxVCO-cSFEutDJU9eJjvkAG8pZy8CH4JYs75v4948Wt6pQLhxsA Share this document Share on Facebook, opens a new window Share on LinkedIn, opens a new window Share with Email, opens mail client Copy link Millions of documents at your fingertips, ad-free Subscribe with a free trial You might also like Parts of A Scientific Research Paper Using IMRaD No ratings yet Parts of A Scientific Research Paper Using IMRaD 22 pages Solution 3 No ratings yet Solution 3 26 pages Biomentors Classes Online, Mumbai No ratings yet Biomentors Classes Online, Mumbai 3 pages Chemistry No ratings yet Chemistry 18 pages Table of Cryoscopic and Ebullioscopic Constants No ratings yet Table of Cryoscopic and Ebullioscopic Constants 1 page Determination of Molar Mass by Boiling Point Elevetion BA Jan - 1 100% (1) Determination of Molar Mass by Boiling Point Elevetion BA Jan - 1 5 pages CHM 222 Module 4 No ratings yet CHM 222 Module 4 17 pages Rocks and Igneous Activities No ratings yet Rocks and Igneous Activities 25 pages Wa0008. 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1450	https://tenakill.closterschools.org/accnt_390190/site_398558/Documents/Grade-4-Math-Review.pdf	Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR1 Lesson 1 Reteach Add Dollars and Cents To add money amounts, line up the decimal points and then add as with whole numbers. Find the sum. $38.37 1 $41 .47 Step 1 Write the problem on grid paper. Align the digits by place value. Think of pennies as hundredths and dimes as tenths. Step 2 Add the hundredths. Regroup 14 hundredths as 1 tenth 4 hundredths. Write 1 in the tenths column. Then add the tenths. Step 3 Add the ones and then add the tens. Regroup if necessary. Write the decimal point and dollar sign. So, $38.37 1 $41 .47 5 $79.84. T O T H $ 3 8 . 3 7 + $ 4 1 . 4 7 T O T H 1 $ 3 8 . 3 7 + $ 4 1 . 4 7 8 4 T O T H 1 $ 3 8 . 3 7 + $ 4 1 . 4 7 $ 7 9 . 8 4 Find the sum. 1. $7.81 1 $5.09 __ 2. $35.06 1 $51.48 __ 3. $ 5.32 1 $85.44 __ 4. $40.36 1 $17.45 __ 5. $37.60 1 $ 9.04 __ 6. $80.26 1 $19.31 __ 7. $48.04 1 $64.65 __ 8. $52.66 1 $50.48 __ 9. $8.47 1 $7.33 __ 10. $69.19 1 $ 4.95 __ 11. $24.70 1 $62.33 __ 12. $10.00 1 $25.75 __ Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR2 Lesson 2 Reteach Subtract Dollars and Cents You can count up to find a difference. Find the difference. $48.32 2 $12.50 Step 1 Start with $12.50, the amount being subtracted. Count up until you reach $48.32. Record each amount that you use to count up. 1$0.50 1$7 1$28 1$0.32 $48.32 $12.50 $13 $20 $48 Step 2 Add the distances between counts to find the difference. $0.50 1 $7 .00 1 $28.00 1 $0.32 5 $35.82 So, $48.32 2 $12.50 5 $35.82. Find the difference. 1. $7.22 2 $4.02 __ 2. $36.06 2 $34.48 __ 3. $80.00 2 $35.75 __ 4. $98.36 2 $21.15 __ 5. $47.90 2 $ 8.34 __ 6. $60.24 2 $14.10 __ 7. $78.54 2 $ 9.62 __ 8. $52.00 2 $10.98 __ 9. $75.32 2 $24.32 __ 10. $85.09 2 $43.56 __ 11. $90.50 2 $76.80 __ 12. $12.13 2 $ 4.58 __ Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR3 Lesson 3 Reteach Algebra • Order of Operations The order of operations is a set of rules that gives the order in which calculations are done in an expression. Use the order of operations to find the value of the expression. Show each step. 8 1 (10 4 5) 2 4 Step 1 First divide. Think: 10 ÷ 5 = 2 8 1 (10 4 5) 2 4 8 1 2 2 4 Step 2 Then add and subtract from left to right. Think: 8 1 2 5 10 8 1 2 2 4 10 2 4 Step 3 Subtract. 10 2 4 5 6 So, 8 1 (10 4 5) 2 4 5 6. Write correct if the operations are listed in the correct order. If not correct, write the correct order of operations. 1. (9 4 3) 3 4 multiply, divide 2. 15 2 (8 4 2) subtract, divide 3. (36 1 10) 3 3 multiply, add 4. (16 2 4) 4 2 1 5 subtract, divide, add Follow the order of operations to find the value of the expression. Show each step. 5. (6 3 7) 1 3 6. (8 1 12) 4 4 7. (20 2 5) 3 3 1 4 8. 18 1 6 1 (16 4 4) Order of Operations 1. First, perform operations inside the parentheses. 2. Then, multiply and divide from left to right. 3. Last, add and subtract from left to right. Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR4 Lesson 4 Reteach Divide by Multiples of Ten You can use basic facts and patterns to divide by multiples of ten. Divide. Use a pattern to help. 6,000 4 30 Step 1 Look for a basic fact. 6,000 4 30 Think: 6 4 3 The basic fact is 6 4 3 5 2. Step 2 Use the basic fact to find a division sentence with the same divisor as the original problem. 6,000 4 30 divisor Think: 6 4 3 5 2, so 60 4 30 5 2. Step 3 Now look for a pattern. Think: If the number of zeros in the dividend increases, the number of zeros in the quotient increases by the same number. dividend 60 4 30 5 2 quotient 600 4 30 5 20 6,000 4 30 5 200 So, 6,000 4 30 5 200. Divide. Use a pattern to help. 1. 1,600 4 20 5 2. 2,400 4 80 5 3. 3,600 4 40 5 4. 1,200 4 30 5 5. 8,000 4 40 5 6. 2,000 4 50 5 7. 6,000 4 10 5 8. 4,900 4 70 5 9. 5,400 4 60 5 Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR5 Lesson 5 Reteach Model Division with 2-Digit Divisors You can use models to divide a whole number by a 2-digit divisor. Use base-ten blocks to find 143 4 13. Step 1 Use base-ten blocks to model the dividend, . Show 143 as _ hundred _ tens _ ones. Remember: Each large square represents 100, each line represents 10, and each small circle represents 1. Step 2 The divisor is . Divide the blocks equally between 13 groups. Since you cannot share the one-hundred square equally between the 13 groups, first break it into 10 tens. Then you will have _ tens, altogether. Share the tens equally among 13 groups. Step 3 After completing Step 2, you will have _ ten and _ ones left over. Since you cannot share the 10 equally between the 13 groups, break it into 10 ones. Then you will have _ ones, altogether. Share the 13 ones equally among the 13 groups. Each group contains _ ten and _ one, or 11. So 143 ÷ 13 = . Use base-ten blocks to divide. 1. 65 4 5 5 __ 2. 84 4 14 5 __ 3. 120 4 8 5 __ 4. 96 4 16 5 __ 5. 168 4 12 5 __ 6. 99 4 33 5 __ 143 1 4 3 13 14 1 13 1 1 11 3 Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR6 Lesson 6 Reteach Place Value Through Millions You can use a place-value chart to help you read and write numbers through millions. You can group the digits in a whole number into sections called periods. Each period has 3 digits. Each digit in a whole number has both a place and a value. In the place value chart below, the digit 3 is in the hundred thousands place. So its value is 3 3 100,000, or 300,000. Millions Thousands Ones Hundred Millions Ten Millions Millions Hundred Thousands Ten Thousands Thousands Hundreds Tens Ones 2 8 7 3 1 4 6 5 9 Use the place-value chart to read and write the number in standard form, word form, and expanded form. Standard Form: 287 ,314,659 Word Form: two hundred eighty-seven million, three hundred fourteen thousand, six hundred fifty-nine Expanded Form: 200,000,000 1 80,000,000 1 7 ,000,000 1 300,000 1 10,000 1 4,000 1 600 1 50 1 9 Read and write the number in two other forms. 1. sixty million, forty thousand, two hundred twenty-nine 2. 8,000,000 1 300,000 1 2,000 1 100 1 8 Periods Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR7 Lesson 7 Reteach Decimals and Place Value You can write decimals, like whole numbers, in standard form, word form, and expanded form. In a place-value chart, whole numbers are to the left of the decimal point. Decimal amounts are to the right of the decimal point. The value of each place is one-tenth, or 1 __ 10 , of the place to its left. When you write a decimal in word form, write the decimal point as “and.” Write the decimal 12.34 in word form and expanded form. Start by writing 12.34 in a place-value chart. First, align the decimal point with the decimal in the chart. Then place the digits. Hundreds Tens Ones . Tenths Hundredths 1 2 . 3 4 _ 3 10 _ 3 1 . _ 3 1 __ 10 _ 3 1 ___ 100 10 2 . 3 __ 10 4 ___ 100 Word form: 12.34 Two decimals indicate hundredths. __ and thirty-four ___ Expanded Form: Use the last row of the chart to help you write the decimal in expanded form. 12.34 5 10 1 _ 1 _ 1 0.04 Read and write the decimal in two other forms. 1. eight and seven tenths 2. 10 1 3 1 0.9 1 0.05 1 Twelve hundredths 2 3 4 2 0.3 Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR8 Lesson 8 Reteach Round Decimals You can use a number line to round whole numbers. You can also use a number line to round decimals. Round 1 .82 to the nearest whole number. Step 1 Find the whole numbers it is between. , 1.82 , Step 2 Plot the number on a number line. See which whole number it is closest to. 0 1 2 1.82 1.82 is closer to than . So, 1.82 rounded to the nearest whole number is . Note: If you are rounding a number with 5 in the tenths place, round the number to the greater whole number. 1.5 rounded to the nearest whole number is 2. However, 1.49 rounds to 1. Round to the nearest dollar or to the nearest whole number. 1. $1.23 2. 3.7 3. 7.12 4. $5.50 5. $2.89 6. 9.2 7. $4.49 8. 6.51 9. 8.5 10. $5.01 11. $1.89 12. $8.21 2 2 1 2 1 Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR9 Lesson 9 Reteach Place Value to Compare Decimals You can use a place-value chart to help you compare decimals. Use a place-value chart to compare the decimals. Write ,, ., or 5. 4.28 4.23 Step 1 Write both decimals in a place-value chart. Line up each place and the decimal. Step 2 Compare the numbers in each place, starting with the numbers in the ones place and working your way right. Ones . Tenths Hundredths 4 . 2 8 4 . 2 3 4 5 4 2 5 8 . Step 3 Since 8 is greater than 3, 4.28 is greater than 4.23. So, 4.28 . 4.23. 1. Use the place-value chart below to compare the decimals. Write ,, ., or 5. Ones . Tenths Hundredths 8 . 9 2 8 . 9 7 8 5 8 9 5 2 , So, 8.92 , 8.97. Compare the decimals. Write <, >, or =. 2. 6.87 6.80 3. 9.17 9.19 4. 5.73 5.78 5. 1.23 1.22 6. 2.56 2.5 7. 3.7 3.70 8. 7.22 7.2 9. 4.4 4.04 2 3 Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR10 Lesson 10 Reteach Decompose Multiples of 10, 100, 1,000 You can decompose a multiple of 10, 100, or 1,000 by finding its factors. • To decompose a multiple of 10: rewrite it as the product of 10 and another number. • To decompose a multiple of 100: rewrite it as the product of 100 and another number. • To decompose a multiple of 1,000: rewrite it as the product of 1,000 and another number. Decompose 3,200. One Way Use mental math and a pattern. 3,200 5 _ 3 1 3,200 5 _ 3 10 3,200 5 _ 3 100 So 3,200 5 32 3 100. Another Way Use place value. 3,200 5 32 hundreds 5 32 3 _ So 3,200 5 32 × 100. 1. Complete the exercise below to decompose 3,600. 3,600 5 ___ 3 1 3,600 5 ___ 3 10 3,600 5 ___ 3 100 2. Complete the exercise below to decompose 870. 870 5 ___ tens 5 ___ 3 ___ Decompose each number. 3. 90 5 ____ 4. 5,600 5 ____ 5. 3,000 5 ____ 3,200 320 32 100 Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR11 Lesson 11 Reteach Number Patterns You already know how to use a rule and the first term to write a sequence. Now you will use multiplication to describe a pattern. Stephen is saving his money to buy a car. The table shows how much money he has saved at the end of each month. If the pattern continues, how much money will Stephen have saved after months 5 and 6? Number of Months 1 2 3 4 Total Amount Saved ($) 15 30 60 120 Step 1 Describe the sequence. Think: How do I get from one term to the next? Try multiplying by 2, since 15 3 2 5 30. _ _ _____ _ _ _____ _______ 15, 30, 60, 120 Step 2 Write a rule that describes how much money Stephen has saved at the end of each month. Rule: ___ Step 3 Use the rule to find the next two terms in the sequence. 3 2 3 2 3 2 _ _ _____ _ _ _____ 15, 30, 60, 120 240, 480 So, at the end of month 5, Stephen will have saved . At the end of month 6, have will have saved . Describe the pattern. Then find the next two numbers in the pattern. 1. 2, 10, 50, , __ 2. 2, 6, 18, , __ $240 3 2 3 2 3 2 3 2 3 2 $480 Multiply by 2. Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR12 Lesson 12 Reteach Add Related Fractions You can add fractions with different denominators using a number line. First find an equivalent fraction so that both fractions have the same denominator. Add 1 _ 4 1 1 _ 2 . Use a number line to help. Step 1 Draw a number line from 0 to 1 . Divide it into 4 equal parts. Label the number line in fourths. Step 2 Draw another number line directly below the first number line. Line up the 0s and 1s. Divide the second number line into 2 equal parts. Label it. Step 3 Find how many fourths are equal to 1 _ 2 . 1 _ 2 5 2 _ 4 Step 4 Add. Use the equivalent fraction for 1 _ 2 . 1 __ 4 1 1 __ 2 5 1 __ 4 + 2 __ 4 5 3 _ 4 So, 1 _ 4 1 1 _ 2 5 3 _ 4 . 1 0 1 4 2 4 3 4 4 4 0 4 1 0 1 2 2 2 0 2 1 0 1 4 2 4 3 4 4 4 0 4 Add. Use a number line to help. 1. 3 __ 4 1 1 __ 8 5 _ 2. 1 __ 6 1 2 __ 3 5 _ 3. 2 __ 5 1 1 ___ 10 5 _ 4. 4 __ 8 1 1 __ 2 5 _ 5. 1 __ 2 1 1 __ 6 5 _ 6. 4 ___ 12 1 1 __ 4 5 _ Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR13 Lesson 13 Reteach Subtract Related Fractions You can subtract fractions with different denominators using a number line. First find an equivalent fraction so that both fractions have the same denominator. Subtract 3 _ 4 2 1 _ 8 . Use a number line to help. Step 1 Draw a number line from 0 to 1. Divide it into 4 equal parts. Label the number line in fourths. Step 2 Draw another number line directly below the first number line. Line up the 0s and 1s. Divide the second number line into 8 equal parts. Label it. Step 3 Find how many eighths are equal to 3 _ 4 . 3 _ 4 5 6 _ 8 Step 4 Add. Use the equivalent fraction for 3 _ 4 . 3 _ 4 2 1 _ 8 5 6 _ 8 2 1 _ 8 5 5 _ 8 So, 3 _ 4 2 1 _ 8 5 5 _ 8 . 1 0 1 4 2 4 3 4 4 4 0 4 1 0 8 8 0 8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 1 0 8 8 0 8 1 8 2 8 3 8 4 8 5 8 6 8 7 8 Subtract. Use a number line to help. 1. 5 __ 6 2 1 __ 3 5 _ 2. 1 __ 2 2 1 __ 4 5 _ 3. 5 __ 8 2 1 __ 2 5 _ 4. 6 ___ 10 2 2 __ 5 5 _ 5. 7 ___ 12 2 1 __ 3 5 _ 6. 6 __ 8 2 3 __ 4 5 _ Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR14 Lesson 14 Reteach Compare Fraction Products When a fraction less than one is multiplied by a whole number, is the product less than or greater than the fraction? Is the product of 3 _ 4 3 2 less than or greater than 3 _ 4 ? Step 1 Show two groups of 3 _ 4 . Step 2 Compare. The product 6 _ 4 is ___ 3 _ 4 . So, the product of 3 _ 4 3 2 is greater than 3 _ 4 . When a whole number is multiplied by a fraction less than one, is the product less than or greater than the whole number? Is the product of 3 3 3 _ 5 less than or greater than 3? Step 1 Show three groups of 3 _ 5 . Step 2 Compare. The product 9 _ 5 is ___ 3. So, the product of 3 3 3 _ 5 is less than 3. Complete each statement with greater than or less than. 1. 2 3 5 _ 6 will be _ ___ 5 _ 6 . 2. 3 _ 8 3 2 will be _ __ 2. 3. 3 3 2 _ 5 will be _ __ 3. 4. 2 _ 3 3 4 will be _ ___ 2 _ 3 . less than greater than The model shows 6 _ 4 shaded. The model shows 9 _ 5 shaded. 2 0 1 8 4 0 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 2 0 1 8 4 0 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 2 0 1 8 4 0 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 4 1 2 3 4 5 6 7 8 2 0 1 8 4 0 4 1 4 2 4 3 4 4 4 5 4 6 4 7 4 Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR15 Lesson 15 Reteach Use repeated subtraction to divide. 1. 3 4 1 __ 2 2. 2 4 1 __ 5 3. 1 4 1 __ 4 4. 4 4 1 __ 3 5. 2 4 1 __ 6 6. 2 4 1 __ 8 Repeated Subtraction with Fractions You can use repeated subtraction to divide whole numbers. You can also use repeated subtraction to divide a whole number by a fraction. Use repeated subtraction to find 2 4 1 _ 4 . Step 1 Draw a number line from 0 to 2. Divide it into fourths. Step 2 Start at 2. Count back by 1 _ 4 to subtract. Step 3 Keep subtracting 1 _ 4 until you reach 0 or get as close to it as possible. Step 4 Count the number of times you counted back by 1 _ 4 . You counted back 8 groups of 1 _ 4 . So, 2 4 1 _ 4 = 8. Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR16 Lesson 16 Reteach Fractions and Division You can use division to make equal shares or to make equal-sized groups. You can use a fraction to show division. Write the division problem as a fraction. 3 4 4 You can use fraction strips to model the relationship between division and fractions. Step 1 Begin with 3 wholes. Step 2 Think of each whole as 4 fourth-size pieces. Step 3 Arrange the fourth-size pieces into 4 equal groups. There are 3 fourth-size pieces in each equal group. So, 3 4 4 can be written as 3 _ 4 . Think of a division sign as a fraction bar. numerator 4 denominator numerator _ denominator Write the division problem as a fraction. Write each fraction greater than 1 as a whole number or mixed number. 1. 9 4 3 2. 1 4 6 3. 2 4 8 4. 5 4 4 5. 7 4 2 6. 12 4 8 Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR17 Lesson 17 Reteach Locate Points on a Grid A map has horizontal and vertical lines that make a grid. You can name a point on the grid using an ordered pair of numbers. The first number tells how many units to move right from zero. (1, 5) The second number tells how many units to move up from zero. Write the ordered pair for the location of the park. Step 1 Start at zero. Move right. Count the number of units until you are directly below the park. You move right 2 units. Step 2 Move up. Count the number of units until you reach the park. You move up 3 units. Step 3 You move right 2 units and up 3 units, so the ordered pair is (2, 3). So, the park is located at (2, 3) on the map. Use the grid. Write the ordered pair for each point. 1. A 2. B 3. C 4. D Use the grid. Write the point for each ordered pair. 5. (8, 0) 6. (9, 10) 7. (6, 7) 8. (2, 9) 5 6 5 6 4 4 3 3 2 Park 2 1 1 0 N W E S School Library Store Ty’s House 6 7 8 9 10 11 5 4 3 2 1 0 2 3 1 4 5 11 6 7 8 9 10 A C D E F G H B Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR18 Lesson 18 Reteach Area and Tiling In the model, whole tiles are shaded, and some half tiles are shaded. You can combine the areas of half tiles and whole tiles to find the total area. Find the area of the entryway. Write the area in square feet. Step 1 Count the number of whole tiles. There are 42 whole tiles. Step 2 Count the number of half tiles. There are 6 half tiles. Think: 2 half tiles 5 1 whole tile 6 half tiles 5 3 whole tiles Step 3 Use the total number of whole tiles to find the area. 42 1 3 5 45 whole tiles Think: 1 tile 5 4 square feet Multiply the number of whole tiles by 4 to find the area. 45 3 4 5 180 So, the area of Marta’s entryway is 180 square feet. Find the area of each shaded shape. Write the area in square units. 1. 1 square 5 4 square feet 2. 1 square 5 9 square meters 3. 1 square 5 16 square miles _ _ Marta’s Entryway 1 square = 4 square feet Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR19 4. (16 3 7) 3 3 5 5. 7 3 (18 3 6) 5 6. (12 3 8) 3 6 5 1. 3 3 (16 3 4) 5 2. (4 3 14) 3 6 5 3. 5 3 (13 3 5) 5 Lesson 19 Reteach Multiply Three Factors Step 1 Simplify the problem. Rewrite 2 3 (14 3 6) as a product of two factors. 2 3 (14 3 6) = 2 3 ( 3 14) Commutative Property = (2 3 ) 3 14 Associative Property = 3 14 So, 2 3 (14 3 6) = 12 3 14. Step 2 Multiply. 4 3 12 10 3 12 Add. So, 2 3 (14 3 6) 5 168. 6 6 12 Find each product. 48 1 120 _ 168 Remember Commutative Property of Multiplication You can multiply factors in any order and still get the same product. Example: 2 3 3 5 3 3 2 Associative Property of Multiplication You can group factors in any order and still get the same product. Example: 2 3 (3 3 4) 5 (2 3 3) 3 4 12 3 14 _ Name Reteach Grade 4 © Houghton Mifflin Harcourt Publishing Company GRR20 Lesson 20 Reteach Find Area of the Base A rectangular prism is a solid figure that has three-dimensions: length, width, and height. A rectangular prism has two bases. The bases are the same size and shape and are opposite each other. The base shape of a rectangular prism is a rectangle or a square. You can use the area formulas for a rectangle and a square to find the area of the base of a rectangular prism. Find the area of the base of the rectangular prism. Step 1 Identify the base shape. The length is _ feet. The width is _ feet. The base shape is a . Step 2 Find the area of the base shape. A = l 3 w Think: Use the area formula for a rectangle. = _ 3 _ = _ square feet So, the area of the base is __. 6 24 4 4 24 square feet Find the area of the base of the rectangular prism. 1. 3 in. 3 in. 4 in. __ 2. 8 ft 4 ft 3 ft _ 3. 6 m 5 m 5 m 4. 4 cm 9 cm 3 cm 5. 8 yd 8 yd 4 yd 6. 7 in. 7 in. 7 in. ___ width height length base base 6 ft 4 ft 5 ft 6 Remember Area of a rectangle: A 5 b 3 h or A 5 l 3 w Area of a square: A 5 s 3 s rectangle
1451	https://www.vedantu.com/question-answer/the-equation-of-the-tangent-to-the-circle-class-10-maths-cbse-5fd66d7ed6856f7ca9d4dd65	Talk to our experts 1800-120-456-456 The equation of the tangent to the circle $ {{x}^{2}}+{{y}^{2}}={{a}^{2}} $ , which makes a triangle of area $ {{a}^{2}} $ with the coordinate axes, is A. $ x\pm y=\pm a $ B. $ x\pm y=\pm a\sqrt{2} $ C. $ x\pm y=3a $ D. $ x\pm y=\pm 2a $ © 2025.Vedantu.com. All rights reserved
1452	https://en.wikipedia.org/wiki/Havel%E2%80%93Hakimi_algorithm	Havel–Hakimi algorithm - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 AlgorithmToggle Algorithm subsection 1.1 Proof 2 Examples 3 See also 4 Notes 5 References Havel–Hakimi algorithm [x] 6 languages Čeština Deutsch Français Magyar Русский 中文 Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia Algorithm in graph theory This article includes a list of references, related reading, or external links, but its sources remain unclear because it lacks inline citations. Please help improve this article by introducing more precise citations.(June 2020) (Learn how and when to remove this message) The Havel–Hakimi algorithm is an algorithm in graph theory solving the graph realization problem. That is, it answers the following question: Given a finite list of nonnegative integers in non-increasing order, is there a simple graph such that its degree sequence is exactly this list? A simple graph contains no double edges or loops. The degree sequence is a list of numbers in nonincreasing order indicating the number of edges incident to each vertex in the graph. If a simple graph exists for exactly the given degree sequence, the list of integers is called graphic. The Havel-Hakimi algorithm constructs a special solution if a simple graph for the given degree sequence exists, or proves that one cannot find a positive answer. This construction is based on a recursive algorithm. The algorithm was published by Havel (1955), and later by Hakimi (1962). Algorithm [edit] The Havel–Hakimi algorithm is based on the following theorem. Let A=(s,t 1,...,t s,d 1,...,d n){\displaystyle A=(s,t_{1},...,t_{s},d_{1},...,d_{n})} be a finite list of nonnegative integers that is nonincreasing. Let A′=(t 1−1,...,t s−1,d 1,...,d n){\displaystyle A'=(t_{1}-1,...,t_{s}-1,d_{1},...,d_{n})} be a second finite list of nonnegative integers that is rearranged to be nonincreasing. List A{\displaystyle A} is graphic if and only if list A′{\displaystyle A'} is graphic. If the given list A{\displaystyle A} is graphic, then the theorem will be applied at most n−1{\displaystyle n-1} times setting in each further step A:=A′{\displaystyle A:=A'}. Note that it can be necessary to sort this list again. This process ends when the whole list A′{\displaystyle A'} consists of zeros. Let G{\displaystyle G} be a simple graph with the degree sequence A{\displaystyle A}: Let the vertex S{\displaystyle S} have degree s{\displaystyle s}; let the vertices T 1,...,T s{\displaystyle T_{1},...,T_{s}} have respective degrees t 1,...,t s{\displaystyle t_{1},...,t_{s}}; let the vertices D 1,...,D n{\displaystyle D_{1},...,D_{n}} have respective degrees d 1,...,d n{\displaystyle d_{1},...,d_{n}}. In each step of the algorithm, one constructs the edges of a graph with vertices T 1,...,T s{\displaystyle T_{1},...,T_{s}}—i.e., if it is possible to reduce the list A{\displaystyle A} to A′{\displaystyle A'}, then we add edges {S,T 1},{S,T 2},⋯,{S,T s}{\displaystyle {S,T_{1}},{S,T_{2}},\cdots ,{S,T_{s}}}. When the list A{\displaystyle A} cannot be reduced to a list A′{\displaystyle A'} of nonnegative integers in any step of this approach, the theorem proves that the list A{\displaystyle A} from the beginning is not graphic. Proof [edit] The following is a summary based on the proof of the Havel-Hakimi algorithm in Invitation to Combinatorics (Shahriari 2022). To prove the Havel–Hakimi algorithm always works, assume that A′{\displaystyle A'} is graphic, and there exists a simple graph G′{\displaystyle G'} with the degree sequence A′=(t 1−1,...,t s−1,d 1,...,d n){\displaystyle A'=(t_{1}-1,...,t_{s}-1,d_{1},...,d_{n})}. Then we add a new vertex v{\displaystyle v} adjacent to the s{\displaystyle s} vertices with degrees t 1−1,...,t s−1{\displaystyle t_{1}-1,...,t_{s}-1} to obtain the degree sequence A{\displaystyle A}. To prove the other direction, assume that A{\displaystyle A} is graphic, and there exists a simple graph G{\displaystyle G} with the degree sequence A=(s,t 1,...,t s,d 1,...,d n){\displaystyle A=(s,t_{1},...,t_{s},d_{1},...,d_{n})} and vertices S,T 1,...,T s,D 1,...,D n{\displaystyle S,T_{1},...,T_{s},D_{1},...,D_{n}}. We do not know which s{\displaystyle s} vertices are adjacent to S{\displaystyle S}, so we have two possible cases. In the first case, S{\displaystyle S} is adjacent to the vertices T 1,...,T s{\displaystyle T_{1},...,T_{s}} in G{\displaystyle G}. In this case, we remove S{\displaystyle S} with all its incident edges to obtain the degree sequence A′{\displaystyle A'}. In the second case, S{\displaystyle S} is not adjacent to some vertex T i{\displaystyle T_{i}} for some 1≤i≤s{\displaystyle 1\leq i\leq s} in G{\displaystyle G}. Then we can change the graph G{\displaystyle G} so that S{\displaystyle S} is adjacent to T i{\displaystyle T_{i}} while maintaining the same degree sequence A{\displaystyle A}. Since S{\displaystyle S} has degree s{\displaystyle s}, the vertex S{\displaystyle S} must be adjacent to some vertex D j{\displaystyle D_{j}} in G{\displaystyle G} for 1≤j≤n{\displaystyle 1\leq j\leq n}: Let the degree of D j{\displaystyle D_{j}} be d j{\displaystyle d_{j}}. We know t i≥d j{\displaystyle t_{i}\geq d_{j}}, as the degree sequence A{\displaystyle A} is in non-increasing order. Since t i≥d j{\displaystyle t_{i}\geq d_{j}}, we have two possibilities: Either t i=d j{\displaystyle t_{i}=d_{j}}, or t i>d j{\displaystyle t_{i}>d_{j}}. If t i=d j{\displaystyle t_{i}=d_{j}}, then by switching the places of the vertices T i{\displaystyle T_{i}} and D j{\displaystyle D_{j}}, we can adjust G{\displaystyle G} so that S{\displaystyle S} is adjacent to T i{\displaystyle T_{i}} instead of D j.{\displaystyle D_{j}.} If t i>d j{\displaystyle t_{i}>d_{j}}, then since T i{\displaystyle T_{i}} is adjacent to more vertices than D j{\displaystyle D_{j}}, let another vertex W{\displaystyle W} be adjacent to T i{\displaystyle T_{i}} and not D j{\displaystyle D_{j}}. Then we can adjust G{\displaystyle G} by removing the edges {S,D j}{\displaystyle \left{S,D_{j}\right}} and {T i,W}{\displaystyle \left{T_{i},W\right}}, and adding the edges {S,T i}{\displaystyle \left{S,T_{i}\right}} and {W,D j}{\displaystyle \left{W,D_{j}\right}}. This modification preserves the degree sequence of G{\displaystyle G}, but the vertex S{\displaystyle S} is now adjacent to T i{\displaystyle T_{i}} instead of D j{\displaystyle D_{j}}. In this way, any vertex not connected to S{\displaystyle S} can be adjusted accordingly so that S{\displaystyle S} is adjacent to T i{\displaystyle T_{i}} while maintaining the original degree sequence A{\displaystyle A} of G{\displaystyle G}. Thus any vertex not connected to S{\displaystyle S} can be connected to S{\displaystyle S} using the above method, and then we have the first case once more, through which we can obtain the degree sequence A′{\displaystyle A'}. Hence, A{\displaystyle A} is graphic if and only if A′{\displaystyle A'} is also graphic. Examples [edit] Let 6,3,3,3,3,2,2,2,2,1,1{\displaystyle 6,3,3,3,3,2,2,2,2,1,1} be a nonincreasing, finite degree sequence of nonnegative integers. To test whether this degree sequence is graphic, we apply the Havel-Hakimi algorithm: First, we remove the vertex with the highest degree — in this case, 6{\displaystyle 6} — and all its incident edges to get 2,2,2,2,1,1,2,2,1,1{\displaystyle 2,2,2,2,1,1,2,2,1,1} (assuming the vertex with highest degree is adjacent to the 6{\displaystyle 6} vertices with next highest degree). We rearrange this sequence in nonincreasing order to get 2,2,2,2,2,2,1,1,1,1{\displaystyle 2,2,2,2,2,2,1,1,1,1}. We repeat the process, removing the vertex with the next highest degree to get 1,1,2,2,2,1,1,1,1{\displaystyle 1,1,2,2,2,1,1,1,1} and rearranging to get 2,2,2,1,1,1,1,1,1{\displaystyle 2,2,2,1,1,1,1,1,1}. We continue this removal to get 1,1,1,1,1,1,1,1{\displaystyle 1,1,1,1,1,1,1,1}, and then 0,0,0,0,0,0,0,0{\displaystyle 0,0,0,0,0,0,0,0}. This sequence is clearly graphic, as it is the simple graph of 8{\displaystyle 8} isolated vertices. To show an example of a non-graphic sequence, let 6,5,5,4,3,2,1{\displaystyle 6,5,5,4,3,2,1} be a nonincreasing, finite degree sequence of nonnegative integers. Applying the algorithm, we first remove the degree 6{\displaystyle 6} vertex and all its incident edges to get 4,4,3,2,1,0{\displaystyle 4,4,3,2,1,0}. Already, we know this degree sequence is not graphic, since it claims to have 6{\displaystyle 6} vertices with one vertex not adjacent to any of the other vertices; thus, the maximum degree of the other vertices is 4{\displaystyle 4}. This means that two of the vertices are connected to all the other vertices with the exception of the isolated one, so the minimum degree of each vertex should be 2{\displaystyle 2}; however, the sequence claims to have a vertex with degree 1{\displaystyle 1}. Thus, the sequence is not graphic. For the sake of the algorithm, if we were to reiterate the process, we would get 3,2,1,0,0{\displaystyle 3,2,1,0,0} which is yet more clearly not graphic. One vertex claims to have a degree of 3{\displaystyle 3}, and yet only two other vertices have neighbors. Thus the sequence cannot be graphic. See also [edit] Erdős–Gallai theorem Notes [edit] ^From Shahriari (2022, p. 48): "Definition 2.17 (Graphs & Subgraphs). A simple graph (or just a graph) G is a pair of sets (V, E) where V is a nonempty set called the set of vertices of G, and E is a (possibly empty) set of unordered pairs of distinct elements of V. The set E is called the set of edges of G. If the number of vertices of G is finite, then G is a finite graph (or a finite simple graph)." ^From Shahriari (2022, p. 355): "Definition 10.6 (The degree sequence of a graph; Graphic sequences). The degree sequence of a graph is the list of the degrees of its vertices in non-increasing order. A non-increasing sequence of non-negative integers is called graphic, if there exists a simple graph whose degree sequence is precisely that sequence.” References [edit] Havel, Václav (1955), "A remark on the existence of finite graphs", Časopis pro pěstování matematiky (in Czech), 80 (4): 477–480, doi:10.21136/CPM.1955.108220 Hakimi, S. L. (1962), "On realizability of a set of integers as degrees of the vertices of a linear graph. I", Journal of the Society for Industrial and Applied Mathematics, 10 (3): 496–506, doi:10.1137/0110037, MR0148049. Shahriari, Shahriar (2022), Invitation to Combinatorics, Cambridge U. Press. West, Douglas B. (2001). Introduction to graph theory. Second Edition. Prentice Hall, 2001. 45-46. Retrieved from " Category: Graph algorithms Hidden categories: Articles with short description Short description matches Wikidata Articles lacking in-text citations from June 2020 All articles lacking in-text citations CS1 Czech-language sources (cs) This page was last edited on 15 August 2025, at 22:11(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Havel–Hakimi algorithm 6 languagesAdd topic
1453	https://www.osti.gov/servlets/purl/15011584	UCRL-CONF-209023 A Detailed Chemical Kinetic Model for TNT W. J. Pitz, C. K. Westbrook January 14, 2005 2005 Joint Meeting of the U.S. Sections of The Combustion Institute Philadelphia, PA, United States March 20, 2005 through March 23, 2005 Disclaimer This document was prepared as an account of work sponsored by an agency of the United States Government. Neither the United States Government nor the University of California nor any of their employees, makes any warranty, express or implied, or assumes any legal liability or responsibility for the accuracy, completeness, or usefulness of any information, apparatus, product, or process disclosed, or represents that its use would not infringe privately owned rights. Reference herein to any specific commercial product, process, or service by trade name, trademark, manufacturer, or otherwise, does not necessarily constitute or imply its endorsement, recommendation, or favoring by the United States Government or the University of California. The views and opinions of authors expressed herein do not necessarily state or reflect those of the United States Government or the University of California, and shall not be used for advertising or product endorsement purposes. A Detailed Chemical Kinetic Model for TNT William J. Pitz and Charles K. Westbrook Lawrence Livermore National Laboratory, P.O. Box 808, Livermore, CA 94550 Abstract A detailed chemical kinetic mechanism for 2,4,6-tri-nitrotoluene (TNT ) has been developed to explore problems of explosive performance and soot formation during the destruction of munitions. The TNT mechanism treats only gas-phase reactions. Reactions for the decomposition of TNT and for the consumption of intermediate products formed from TNT are assembled based on information from the literature and on current understanding of aromatic chemistry. Thermodynamic properties of intermediate and radical species are estimated by group additivity. Reaction paths are developed based on similar paths for aromatic hydrocarbons. Reaction-rate constant expressions are estimated from the literature and from analogous reactions where the rate constants are available. The detailed reaction mechanism for TNT is added to existing reaction mechanisms for RDX and for hydrocarbons. Computed results show the effect of oxygen concentration on the amount of soot precursors that are formed in the combustion of RDX and TNT mixtures in N 2/O 2 mixtures. Introduction Destruction of outdated munitions is conventionally carried out by combustion, either via enclosed incineration, or by open burning or detonation. Soot emissions during this destructive combustion have become a serious problem for environmental reasons. There is also a continuing need from both civilian and military users of explosives to improve explosives performance and safety against accidental initiation. Just as in the case of combustion systems using conventional hydrocarbon fuels, the opportunity to use kinetic modeling can provide valuable tools to assist in solving these problems. Major obstacles exist for developing kinetic mechanisms for explosives. First, most such explosives include more types of atoms than are present in familiar hydrocarbons, with which kinetic modeling is most experienced. In particular, many explosives include N atoms in the form of nitro and amino groups, as well as N atoms substituted for C atoms in ring structures. Figure 1 shows some important explosives molecules and illustrates the considerable presence of N atoms. RDX HMX Figure 1 . Four common high explosives molecules in common use. Two additional H atoms (not shown) complete the aromatic ring for TNT. Each C in RDX and HMX has two additional H atoms not shown, and each N shown in RDX and HMX has an additional NO 2 group not shown here. These are large molecules, often based on aromatic hydrocarbon backbones, and the current frontier of kinetic combustion modeling is at this same level of simple aromatic hydrocarbons like benzene and toluene. Reaction mechanisms for these species are necessarily as large and complex as any currently being developed in conventional energy applications. The extreme energy content of these explosives molecules means that they react so rapidly and violently that careful laboratory experiments are extremely difficult or impossible, so the available experimental database is very small for comparisons with computed results. In the first generation of model results, sensible qualitative predictions are probably all that can be expected, but we have found that such qualitative and semi-quantitative results are very useful. Finally, all of the high explosives shown above are solids at room temperature and pressure, making their atmospheric combustion a multiphase technical problem. Such problems are also at the frontier of combustion simulations . Kinetic Model Development Previous high explosive kinetic reaction mechanisms have been developed for only a few gas phase and condensed phase explosives. Tieszen et al. [ 2] developed kinetic models for hexyl nitrate and nitroethane and used them to predict ignition and detonation cell sizes for use as a fuel/air explosive (FAE ). Melius developed the first mechanism for any of the compounds in Fig. 1 building on a relatively simple hydrocarbon mechanism by adding reactions of RDX itself and the subsequent decompositions of the radical products of RDX and used it to model the RDX flame of Ermolin et al. . Prasad, Yetter and Smooke further developed the kinetic model for RDX , modeling both the liquid phase and the gas phase kinetics. The resulting mechanism included 3 reactions in the liquid phase and 48 chemical species and 228 elementary reactions in the gas phase. They used their model to simulate the flame structure of a laser-supported deflagration of an RDX flame . The present modeling project was motivated by two factors, both of which led to the selection of TNT as a model high explosive fuel. Unlike RDX, TNT is based on an aromatic ring and resembles the purely hydrocarbon toluene, which has been the subject of our recent kinetic modeling attention [7 ] and thereby provides a core reaction mechanism on which to build a TNT model. In addition, while both RDX and TNT are important explosives in real applications, some particularly important practical explosives consist of mixtures of RDX and TNT; CompB3 is made of 40% TNT and 6 0% RDX, so the availability of a TNT reaction mechanism would make it also possible to simulate CompB3 kinetics as well as TNT and RDX individually. Therefore, the present TNT kinetic reaction mechanism has been built on our existing toluene mechanism [7 ] and the RDX mechanism of Prasad et al. . Addition of TNT eventually resulted in inclusion of 30 new chemical species and 47 new elementary reactions. Thermochemical parameters for the new species were estimated using principles of group additivity, using the THERM code of Ritter and Bozzelli [8,9 ] to calculate heats and enthalpy of formation, as well as temperature-dependent specific heats. Some groups were corrected to reflect heat of formation values in the literature. For example, the heat of formation for the CB/NO 2 group was corrected by 4.4 kcal/mole so that the group additivity estimate would give the heat of formation for nitrobenzene (16.38 kcal/mole) in the NIST database [1 0]. The NIST value agrees with the calculated BAC/MP4 value (14.18 kcal/mole) of Melius [1 1] after correcting a 2 kcal/mole BAC/MP4 systematic error for the benzene ring. However, for a considerable fraction of the 30 new species required, no previous thermochemical data were available. A representative sample of this data is summarized in Table I. SPECIES Hf SCp 300 400 500 600 800 1000 1500 TNT 1.55 112.64 52.12 62.40 70.74 77.52 87.81 94.75 104.70 TNBENZYL 38.45 110.08 52.87 63.00 70.87 77.10 86.40 92.57 101.33 TNTJ 62.45 114.12 51.71 61.22 68.81 74.83 84.06 90.27 99.46 DNT 4.97 101.15 43.10 52.66 60.62 67.16 77.10 83.92 93.87 DNTJ 65.87 102.63 42.69 51.48 58.69 64.47 73.35 79.44 88.63 NT 8.39 89.67 34.08 42.92 50.50 56.80 66.39 73.09 83.04 NT-2 8.39 89.67 34.08 42.92 50.50 56.80 66.39 73.09 83.04 DNTOH -37.13 108.52 48.06 58.02 66.18 72.69 82.48 89.02 98.65 DNT-OJ -2.73 109.82 46.48 55.78 63.51 69.76 79.27 85.61 94.84 TNPH 9.54 100.07 46.50 55.86 63.12 68.88 77.37 82.95 90.87 TNPHJ 70.44 103.73 46.09 54.68 61.19 66.19 73.62 78.47 85.63 MEDNCP 9.68 69.56 30.83 40.79 49.26 55.19 64.40 70.29 79.37 MEDNCPDJ 35.58 70.84 31.12 40.79 48.69 54.01 62.14 67.20 75.03 TNPHCO -19.16 117.70 54.30 64.16 71.86 78.00 87.25 93.05 .00 TNPHCJO 15.74 116.31 54.11 63.31 70.27 75.79 84.04 89.16 .00 TNPHOH -32.56 109.62 51.46 61.22 68.68 74.41 82.75 88.05 95.65 TNPHOJ 1.84 110.92 49.88 58.98 66.01 71.48 79.54 84.64 91.84 C#CCCNO2 35.45 77.15 22.60 26.97 30.25 32.76 36.47 39.20 43.15 C#CCCJN 94.55 78.54 22.41 26.22 28.89 30.84 33.65 35.71 38.62 Table I. Thermochemical data for some species included in TNT kinetic reaction mechanism. Hf=enthalpy of formation at 298K in kcal/mole, S=enthalpy at 298K in cal/mole-K, Cp = specific heat at constant pressure in cal/mole-K. (TNT = 2,4,6-tri-nitrotoluene, TNBENZYL = 2,4-6-tri-nitrobenzyl radical, TNTJ = 1-methyl-2,4,6-trinitrophenyl radical, DNT = 2 ,4-di-nitrotoluene, DNTJ = 1-methyl-4,6-di-nitrophenyl radical, NT = nitrotoluene, TN-2 = 2-nitrotoluene, DNTOH = 2-methyl-3,5-nitro-phenol , DNT-OJ = 2-methyl-3,5-nitro-phenoxy, TNPH = 1,3,5-tri-nitro-benzene, TNPHJ = 1,3,5-tri- nitro-phenyl, MEDNCP = 1,3-nitro-5-methyl-1,3-cyclopentadiene, MEDNCPDJ = 1,3-nitro-5-methyl-1,3-cyclopentadienyl radical, TNPHCJO = 2,4,6-tri-nitro-phenyl-formyl radical, TNPHOH = 2,4,6-tri-nitrophenol, TNPHOJ = 2,4,6-tri- nitrophenoxy, C#CCCNO2 = 1-nitro-2-ethynyl-ethene, C#CCCJN = 1-nitro-2-ethynyl-vin-1-yl) In a very similar fashion, the rates of elementary reactions were estimated when possible on the basis of known reactions for related species, drawing heavily on the work of Brill and James . For example, the principal initiation reaction for TNT involves breaking a C - N bond to produce NO 2. The rate constant used in the present model, k TNT , was based on the measured rate constants of the analogous reaction of 2-nitro-toluene (2-NT ) and 4-nitro-toluene (4-NT) measured by Tsang et al. in a shock tube [13 ]. Since TNT has two groups ortho to the methyl and one group para to the methyl , the rate constant was assume to be: kTNT = 2k 2-nt + k 4-nt . This reaction was found to be the primary reaction consuming TNT in the simulations performed in this study. Additionally, reactions were added to the reaction mechanism for isomerization of the - NO 2 group to -ONO and for breaking a C-CH 3 bond. These reactions were minor contributors to TNT consumption under the conditions of this study. A similar example of a rate constant estimate is one of the subsequent reactions of the methyl-d i-nitrophenyl radical (D NTJ) produced by the initiation step. An important reaction for the DNTJ radical is its reaction with NO 2 that leaves an O atom on the phenyl radical site and an NO product, as shown in Fig. 2. Figure 2 . Reaction of dinitrobenzyl radical with NO 2 (left), and analogous reaction (right) of phenyl radical studied by Preidel and Zellner [13 ]. The analogous reaction of phenyl and NO 2 was studied by Preidel and Zellner [14 ], and this rate was used for the new reaction in the TNT mechanism. The reaction of DNTJ and NO 2 was the primary Rates of H atom abstraction reactions in TNT were estimated as equal to H atom abstractions by the same radical in toluene, again corrected for the relative numbers of available H atoms. Other reaction rates, particularly for radical decomposition reactions and for complex addition/decomposition reactions, were estimated based on the rates of the reverse addition reactions and the relevant equilibrium constants. Examples of selected reactions with their modified Arrhenius rate coefficients are shown in Table II. AnEa (cal) tnt+o2=tnbenzyl+ho2 9.30E+08 1.3 40939.0 tnt=dntj+no2 8.54E+14 0.0 61470.0 tnt=tnphj+ch3 7.94E+16 0.0 104000.0 tnt=dnt-ono 1.00E+13 0.0 55980.0 dnt-oj+no=dnt-ono 5.44E+13 -0.7 0.0 tnt=tnbenzyl+h 3.10E+15 0.0 89210.0 tnt+h=dnt+no2 7.57E+18 -1.7 6410.0 dnt+h=nt+no2 5.05E+18 -1.7 6410.0 nt+h=c6h5ch3+no2 2.52E+18 -1.7 6410.0 tnt+h=tnph+ch3 7.57E+18 -1.7 6410.0 tnt+oh=tnbenzyl+h2o 5.19E+09 1.0 874.0 tnt+h=tnbenzyl+h2 4.00E+02 3.4 3120.0 tnt+ch3=tnbenzyl+ch4 2.21E+00 3.5 5675.0 tnt+o=tnbenzyl+oh 6.00E+10 0.7 7632.0 tnt+ho2=tnbenzyl+h2o2 1.02E+04 2.5 12340.0 tnt+no2=tnbenzyl+hono 1.20E+13 0.0 30000.0 tnt+h=dntj+hono 3.18E+15 0.0 15700.0 Table II . Modified Arrhenius coefficients for selected reactions in TNT mechanism. Units are cal-mole-sec. (d nt-ono = 2- methyl-3,5-nitro-phenylnitrite. The other species are identified in Table I). Figure 3 Model Applications The specific application chosen to test the TNT reaction mechanism originated from the problem of open destruction of old munitions. In a typical case, a number of munitions, with their high explosive content included, are placed in an open pit in the ground and detonated almost simultaneously. Two typical arrays are shown in Figure 3. This is a very complex combustion problem if it were to be treated in its entirety, since the explosives are present in discrete units, and the cases and other peripheral materials are also involved in the overall explosion. However, in our simulation, we idealize or assume almost everything in this problem, with the exception of the detailed reactions of the explosive fuel. We begin by assuming that all of the predictable combustion will occur in the gas phase. We also disregard all of the metal components including the munitions cases. We then treat the high explosives combustion as a homogeneous ignition problem. The explosive combustion is initiated by some sort of igniter that brings the explosive to a somewhat elevated temperature at atmospheric pressure. Once the explosive is ignited, it burns very rapidly, using the limited amount of oxygen already in the explosive molecule, together with oxidizer in any air which is assumed to be entrained into the reacting explosive material. The amount of air entrainment is another quantity that is assumed for the ignition calculations. The overall amount of fuel converted to final products is limited by the amount of oxygen in the entrained air, and for the purposes of these munitions destruction calculations, the amount of available air is insufficient to convert everything to H 2O, CO 2 and N 2. In principle, the problem of explosive rupture of a single isolated shell, the reactive flow and entrainment of air into the reacting gases, and the eventual reaction quenching as the reactants and products expand could all be computed, but this would be a very ambitious 3D or possibly 2D axisymmetric calculation in which detailed chemical kinetics could probably not included. Instead, making these rather sweeping simplifications, it is possible to examine the chemical kinetic factors involved in the combustion and develop a general understanding of the chemical features of the problem. The limitations of this approach suggest that we not put too much faith in the details of the calculations and use the results only for general guidance. The same overall assumptions have been used very productively in simulating diesel ignition and then in understanding the role of added oxygen on soot formation in diesel engine combustion [15 -18]. In those problems, just as in the present munitions simulations, the details of how the air is entrained into the fuel-rich reaction zone are not as important as the fact that air is entrained and not enough is entrained to consume all of the fuel. With these simple limitations, the evolution of the fuel-rich combustion leads to some very tangible and realistic chemical predictions. Soot Production Chemistry A considerable body of kinetic research has established that soot is produced by the growth of large polycyclic aromatic hydrocarbon species that grow by addition of small, usually unsaturated hydrocarbon fragments and molecules. Acetylene and hydrogen, together with vinyl and propargyl radicals are particularly effective growth species for soot growth. Many sooting fuels already have aromatic species present in the fuel itself, as is the case with conventional diesel and jet fuels, which greatly facilitates sooting with these fuels. If the fuel initially contains no aromatic hydrocarbons, the post-ignition fuel components have been shown to manufacture these aromatic soot seed molecules by a variety of kinetic processes in which acetylene molecules and propargyl and cyclopentadienyl radicals have been shown to be especially effective [19-21 ]. We have demonstrated that ignition under diesel conditions takes place under very fuel-rich conditions, and that the products of this fuel-rich ignition then react to produce soot precursors and participate in soot growth kinetics. The addition of oxygen in the form of oxygenated diesel fuels [16-18] or excess air [22 ] changes the kinetics of this ignition environment and reduces the amount of soot produced. Kinetic modeling has shown that the extra oxygen reduces the post-ignition concentrations of those same hydrocarbon fragments that produce initial aromatic species and then participate in soot growth reactions. Soot Production in TNT Combustion Our simulations of TNT ignition and burning followed exactly the same logical pathway as for diesel ignition. We know that reaction ignition starts combustion of the explosives, and that the reacting fuel then entrains some amount of air. We assumed that the overall equivalence ratio in the reacting explosive mixture was approximately the same as that in the diesel engine at the time of ignition, although there is no direct basis for assuming it being either smaller or larger that in the diesel environment. One factor involved in making our estimates was the fact that large quantities of soot were really observed in the actual destruction experiments. As a result, we assumed as initial conditions an amount of air mixed with the gaseous TNT at an initial temperature of between 900K and 1400K, making an overall TNT-rich equivalence ratio of about 3.0. Of course, even at 100% TNT, there is still some oxygen in the TNT molecule, so the definition of an overall equivalence ratio is still a bit ambiguous. The ignition of the TNT/air mixture was followed effectively to conclusion in each computation, and at the end of each computation, we calculated the sum of the concentrations of each species that we identified with soot production, including acetylene, benzene, ethylene, toluene, and propene. An example of the results of such a calculation is shown in Figure 4, showing the soot precursor level as a function of time with pure TNT and no air entrainment. Figure 4 . Time dependence of soot precursors during Figure 5 . Computed variation in residual soot precursors ignition of TNT, with no added oxygen. Ignition occurs with added oxygen in TNT ignition. Note that the value at 0.4 seconds, residual soot precursors remain at for no added oxygen is the residual, late time value extremely high level of about 1% following ignition. shown in Figure 4. As pure O 2 is added to the TNT and the ignition calculation is repeated, the level of post ignition soot precursors steadily declines, as shown in Figure 5. It is interesting to note that similar calculations under diesel conditions [16 ] as well as actual diesel engine experiments both showed that soot and soot precursor production completely disappeared by the time the fuel consisted of 25 - 30% O 2, rather astonishingly similar to the results summarized for TNT ignition in Figure 5. When the same calculations were repeated with RDX as the explosives fuel, the computed soot precursor concentrations were effectively zero, even without any O 2 addition to the RDX. This was found to be in good agreement with actual experimental observations that RDX is found to produce little or no soot when burned or detonated in the open atmosphere. An examination of the structure of RDX in Figure 1 shows that there are no C - C bonds in RDX, so the production of species such as acetylene, ethylene and any others with C - C single or double bonds would be expected to be very small. More detailed kinetic analysis of the results of the RDX calculations showed that a considerable fraction of the C atoms remain permanently bonded to N atoms, making them unable to participate in building aromatic or other soot precursor species. Comparative analysis of the TNT molecule in Figure 1 shows that the fuel is already based on an aromatic structure, and there are nearly limitless possibilities for production of small fragment species with single, double and triple C - C bonds as soot precursors and growth species. The computed results showing large amounts of soot production in TNT combustion and virtually none in RDX combustion is completely consistent with experimental observations and operational experience, and the kinetic analysis traces these tendencies directly to the molecular structures of the two explosives molecules. A final observation on the amount of oxygen required to eliminate soot production by TNT is that the 30% O 2 indicated in Figure 5 is a molar percentage, which can be converted to a 1:1 volumetric ratio if the oxygen is provided at 80 - 100 atm pressure, which is commonly available in pressurized laboratory gas containers. An array such as those in Fig. 3, with pressurized oxygen containers alternated with TNT munitions of similar overall size, might provide enough mixing between TNT and oxygen to significantly suppress soot production. This concept is awaiting experimental testing, although the required amounts of oxygen might be more conveniently or more economically provided in other ways than in the form of pressurized containers. Conclusions A detailed chemical kinetic reaction mechanism has been developed for TNT, based on existing models for toluene, nitrobenzene, and other related species. The mechanism was used to study soot production during open combustion of TNT munitions, and the model predicted soot precursor levels consistent with experimental observations. Comparable calculations for a different high explosive, RDX, showed that unlike TNT, RDX produced no soot or soot precursors, again in excellent agreement with experimental observations. This combination of computed results gives good credibility to the new TNT reaction mechanism, but many additional model tests are necessary to validate and improve the reaction mechanism. Acknowledgments The authors thank Dr. C. F. Melius for helpful discussions on the chemical kinetic mechanism. The work was supported by the Joint DoD/DOE Munitions Technology Program and performed under the auspices of the U.S. Department of Energy by the University of California, Lawrence Livermore National Laboratory under Contract No.W-7405-Eng-48. The authors thank Jim Wheeler, Director of t he U. S. Army’s Defense Ammunition Center, and Bruce Watkins, LLNL p rogram manager, for their support of this work. 0.0E+00 5.0E-03 1.0E-02 1.5E-02 2.0E-02 2.5E-02 3.0E-02 3.5E-02 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 Time (sec) Soot precursor fraction 0.0E+00 2.0E-03 4.0E-03 6.0E-03 8.0E-03 1.0E-02 1.2E-02 0510 15 20 25 30 %O2 in TNT/O2 mixture Soot precursor fraction References Westbrook, C.K., Mizobuchi, Y., Poinsot, T., Smith, P.A., and Warnatz, J., “Computational Combustion”, Proc. Combust. Inst. 30 , 125-157 (2004). Tieszen, S. R., Stamps, D. W., Westbrook, C. K., and Pitz, W. J., "Gaseous Hydrocarbon-Air Detonations", Combustion and Flame 84, 376-390 (1990). Melius, C. F., (1990). Thermochemical Modeling: II Application to Ignition and Combustion of Energetic Materials. In Bulusu, S. N. (edited) Chemistry and Physics of Energetic Materials. The Netherlands, Kluwer Academic Publishers, The Netherlands, pp. 51-78. N. E. Ermolin, O. P. Korobeinichev, L. V. Kuibida, and V. M. Fomin, Fiz. Goreniya Vzryva 22, 54 (1986) and data presented therein by O. P. Korobeinichev et al. in Mass Spectrometry and Chemical Kinetics, Moscow (1984). Prasad, K., Yetter, R. A. and Smooke, M. D., "An Eigenvalue Method for Computing the Burning Rates of RDX Propellants." Combust. Sci. Technol . 124, 35-82 (1997). Hanson-Parr. D. and Parr, T. “RDX Flame Structure”, Proceedings of the Combustion Institute 25 , 1635-1643 (1994). Pitz, W. J., Seiser, R., Bozzelli, J. W., Seshadri, K., C.-J. Chen, I. D. C., Fournet, R., Billaud, F., Battin-Leclerc, F. and Westbrook, C. K., “Chemical Kinetic Study of Toluene Oxidation under Premixed and Nonpremixed Conditions”, Lawrence Livermore National Laboratory, UCRL-CONF-201575, (2003). Ritter, E. R. and Bozzelli, J. W., "THERM - Thermodynamic Property Estimation for Gas-Phase Radicals and Molecules." International Journal of Chemical Kinetics 23, 767-778 (1991). Lay, T. H., Bozzelli, J. W., Dean, A. M. and Ritter, E. R., J. Phys. Chem. 99, 14514-14527 (1995). NIST Chemistry WebBook, NIST Standard Reference Database Number 69, Eds. P.J. Linstrom and W.G. Mallard, March 2003, National Institute of Standards and Technology, Gaithersburg MD, 20899 ( ). Reference cited in NIST database is Pedley, J.B., Naylor, R.D., Kirby, S.P., Thermochemical Data of Organic Compounds, Chapman and Hall, New York, 1986. 11 . Melius, C. F., in Chemistry and Physics of Energetic Materials (S. N. Bulusu, Ed.), Kluwer Academic Publishers, The Netherlands, 1990, pp.21-49. Brill, T.B., and James, K.J., “Kinetics and Mechanisms of Thermal Decompostion of Nitroaromatic Explosives”, Chem. Rev. 93 , 2667-2692 (1993). 13 . Tsang, W., Robaugh, D. and Mallard, W. G., “Single-Pulse Shock-Tube Studies on C-NO2 Bond Cleavage during the Decomposition of Some Nitro Aromatic Compounds”. J. Phys. Chem. 90, 5968-5973 (1986). 14 . Preidel, M. and R. Zellner, "A cw laser absorption study of the reactions of phenyl radicals with NO, NO 2, O 2 and selected organics between 298-404K." Ber. Bunsenges. Phys. Chem. 93, 1417 (1989). 15 . Flynn, P.F., Durrett, R.P., Hunter, G.L., zur Loye, A.O., Akinyemi, O.C., Dec, J.E., and Westbrook, C.K., “Diesel Combustion: An Integrated View Combining Laser Diagnostics, Chemical Kinetics, and Empirical Validation”, SAE Trans. 108, 587-600 (1999). 16 . Curran, H. J., Fisher, E. M., Glaude, P.-A., Marinov, N. M., Pitz, W. J., Westbrook, C. K., Layton, D. W., Flynn, P. F., Durrett, R. P., zur Loye, A. O., Akinyemi, O. C., and Dryer, F. L., "Detailed Chemical Kinetic Modeling of Diesel Combustion with Oxygenated Fuels," SAE Trans. 110 , 514-521 (2001). 17 . Mueller, C.J., Pitz, W.J., Pickett, L.M., Martin, G.C., Siebers, D.L., and Westbrook, C.K., “Effects of Oxygenates on Soot Processes in DI Diesel Engines: Experiments and Numerical Simulations”, Society of Automotive Engineers SAE 2003- 01-1791 (2003). 18 . Buchholz, B. A., Mueller, C. J., Upatnieks, A., Martin, G. C., Pitz, W. J. and Westbrook, C. K., "Using Carbon-14 Isotope Tracing to Investigate Molecular Structure Effects of the Oxygenate Dibutyl Maleate on Soot Emissions from a DI Diesel Engine," Society of Automotive Engineers SAE 2004-01-1849 (2004). 19 . Wang, H., and Frenklach, M., Combust. Flame 110, 173-221 (1997). Miller, J.A. and C.F. Melius, “Kinetic and thermodynamic issues in the formation of aromatic compounds in flames of aliphatic fuels” Combustion and Flame 91, 21-39 (1992). 21 . Marinov, N. M., Pitz, W. J., Westbrook, C. K., Vincitore, A. M., Castaldi, M. J., Senkan, S. M., and Melius, C. F., “Aromatic and Polycyclic Aromatic Hydrocarbon Formation in a Laminar Premixed n-Butane Flame,” Combustion and Flame 114 , 192-213 (1998). 22 . Siebers, D. and Higgins, B., “Flame Lift-Off on Direct-Injection Diesel Sprays under Quiescent Conditions,” Society of Automotive Engineers, SAE 2002-01-0530, (2001). 23 . Miyamoto, N., Ogawa, H., Nurun, N.M., Obata, K., and Arima, T., “Smokeless, Low NOx, High Thermal Efficiency, and Low Noise Diesel Combustion with Oxygenated Agents as Main Fuel”, Society of Automotive Engineers Paper SAE- 980506 (1998).
1454	https://math.stackexchange.com/questions/5077501/how-does-the-gcd-of-integer-vectors-transform-under-integer-matrices-with-determ	linear algebra - How does the gcd of integer vectors transform under integer matrices with determinant $k^{n-2}?$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community Mathematics helpchat Mathematics Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Teams Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Try Teams for freeExplore Teams 3. Teams 4. Ask questions, find answers and collaborate at work with Stack Overflow for Teams. Explore Teams Teams Q&A for work Connect and share knowledge within a single location that is structured and easy to search. Learn more about Teams Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more How does the gcd of integer vectors transform under integer matrices with determinant k n−2?k n−2? [closed] Ask Question Asked 3 months ago Modified3 months ago Viewed 54 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. Closed. This question needs to be more focused. It is not currently accepting answers. Want to improve this question? Guide the asker to update the question so it focuses on a single, specific problem. Narrowing the question will help others answer the question concisely. You may edit the question if you feel you can improve it yourself. If edited, the question will be reviewed and might be reopened. Closed 3 months ago. Improve this question Let L∈Z n×n L∈Z n×n be an integer matrix with determinant det(L)=k m det(L)=k m for some integer k>0 k>0. For a vector v∈Z n v∈Z n, define u=L v u=L v. I know that if d e t(L)=±1 d e t(L)=±1, then gcd(u)=gcd(v), where gcd(v)gcd(v) means the greatest common divisor of all entries of v v. My questions are: Is it true that in general for det(L)=k m det(L)=k m, we have gcd(u)=k⋅gcd(v) ? Could someone provide a proof or a reference for this fact? How does this relate to the Smith normal form or lattice theory? Any clarifications, examples, or references to textbooks would be greatly appreciated! Thank you! linear-algebra matrices number-theory modules smith-normal-form Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Jun 22 at 0:50 Mohyeedden SweidanMohyeedden Sweidan asked Jun 21 at 22:06 Mohyeedden SweidanMohyeedden Sweidan 11 2 2 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 0 Save this answer. Show activity on this post. We can say L∈M n(Z)L∈M n(Z) has Smith Normal Form L=S D T L=S D T where S,T S,T are invertible integer matrices and D=d i a g(d 1,⋯,d n)D=d i a g(d 1,⋯,d n) is diagonal. Note this means S−1,T−1 S−1,T−1 are also integer matrices and the elementary divisors satisfy d 1∣⋯∣d n d 1∣⋯∣d n. Let k=det L=d 1⋯d n k=det L=d 1⋯d n. For v=(v 1,⋯,v n)v=(v 1,⋯,v n) denote gcd(v):=gcd(v 1,⋯,v n)gcd(v):=gcd(v 1,⋯,v n). Then gcd(L v)=gcd(S D T v)=gcd(D T v)gcd(L v)=gcd(S D T v)=gcd(D T v) divides d n gcd(T v)=k gcd(v)d n gcd(T v)=k gcd(v) and is divisible by gcd(d 1 T v)=d 1 gcd(T v)=d 1 gcd(v)gcd(d 1 T v)=d 1 gcd(T v)=d 1 gcd(v). That is, d 1 gcd(v)∣gcd(L v)∣d n gcd(v).d 1 gcd(v)∣gcd(L v)∣d n gcd(v). This is the best we can say in general. Theorem. For any n>1 n>1 and g,d 1,⋯,d n,d∗g,d 1,⋯,d n,d∗ with d 1∣⋯∣d n d 1∣⋯∣d n and d 1∣d∗∣d n d 1∣d∗∣d n there is an L L with elementary divisors d 1,⋯,d n d 1,⋯,d n and a v v for which gcd(v)=g gcd(v)=g and gcd(L v)=d∗g gcd(L v)=d∗g. Proof. Suppose d i∣d∗∣d i+1 d i∣d∗∣d i+1 and d∗<d i+1 d∗<d i+1. Let D=d i a g(d 1,⋯,d n)D=d i a g(d 1,⋯,d n) and set v=g(d∗d i e i+e i+1)v=g(d∗d i e i+e i+1) where e 1,⋯,e n e 1,⋯,e n are the standard basis vectors of Z n Z n. In particular, there exist L,v≠0 L,v≠0 with det L=k det L=k and gcd(L v)=d∗gcd(v)gcd(L v)=d∗gcd(v) iff d∗∣k d∗∣k. This applies for all k k; whether or not it is an n n th or (n−2)(n−2)th power doesn't change anything. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Jun 22 at 0:10 coisocoiso 4,889 2 2 gold badges 5 5 silver badges 15 15 bronze badges Add a comment\| Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions linear-algebra matrices number-theory modules smith-normal-form See similar questions with these tags. 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1455	https://wp.uthscsa.edu/echo/wp-content/uploads/sites/59/2023/11/Anticoagulants-and-dental-extractions-copy.pdf	DENTAL EXTRACTIONS IN PATIENTS ON ANTICOAGULATION THERAPY Dr. Andrea Bordoy, DDS Assistant Professor Oral and Maxillofacial Surgery Department School of Dentistry OBJECTIVES o To understand the principles of anticoagulation therapy and its impact on dental procedures. o To identify the common anticoagulant medications and their management during dental extractions. o To learn about the precautions, techniques, and protocols for safe dental extractions in patients on anticoagulation therapy. o To discuss case studies and best practices in managing bleeding and complications in such patients during and after dental extraction HEMOSTASIS HEMOSTASIS COAGULATION CASCADE PTT PT Kaplovitch E, Dounaevskaia V. Treatment in the dental practice of the patient receiving anticoagulation therapy. J Am Dent Assoc. 2019 Jul;150(7):602-608. doi: 10.1016/j.adaj.2019.02.011. Epub 2019 May 17. PMID: 31109636. INDICATIONS ANTITHROMBOTIC DRUGS Antiplatelets ANTICOAGULANTS ANTITHROMBOTIC DRUGS Anticoagulants ANTICOAGULATION DRUGS-TARGET Richard M. Jay, Philip Lui. How anticoagulants work. Techniques in Regional Anesthesia and Pain Management, Volume 10, Issue 2,2006. o Binding to antithrombin III, which increases the ability of antithrombin III to inactivate factor Xa and factor II o They are indirect inhibitors of thrombin o Treatment and prevention of DVT, treatment of PE o PTT is used to monitor anticoagulation with heparin (UFH) UNFRACTIONATED HEPARIN (UFH) & LOW MOLECULAR WEIGHT HEPARIN (LMWH) WARFARIN (COUMADIN®) o Vitamin K antagonist (VKAs): block the function of the vitamin K epoxide reductase enzyme in the liver. o The half-life of warfarin is generally 20 to 60 hours (mean: 40 hours) o PT (INR) is used to monitor warfarin therapy because it measures three of the vitamin K– dependent coagulation proteins: factors VII and X and prothrombin. DIRECT ORAL ANTICOAGULANTS (DOACS) oFactor Xa inhibitors 1. No routine monitoring 2. Are considered safer and more predictable because of their direct effect on the coagulation cascade (targeting 1 specific factor) 3. The shortened half-life DIRECT ORAL ANTICOAGULANTS (DOACS) o Factor Xa inhibitors Rivaroxaban (Xarelto), Apixaban (Eliquis), and Edoxaban which have renal excretion rates of 25 to 35% and half-lives of 5 to 9, 8 to 15, and 10 to 14 hours, respectively. o Dabigatran, a direct thrombin inhibitor that has an 80 to 85% renal excretion rate and a half-life of 12 to 17 hours ANTIPLATELETS ANTIPLATELET DRUGS-TARGET Richard M. Jay, Philip Lui. How anticoagulants work. Techniques in Regional Anesthesia and Pain Management, Volume 10, Issue 2,2006. ANTIPLATELET DRUGS-TARGET THROMBOXANE A2 INHIBITORS o Acetylsalicylic acid (ASA) irreversibly acetylates and inactivates platelet cyclooxygenase (Cox-1), a crucial enzyme in the Prostaglandin synthesis of Thromboxane A2 and Prostacyclin. o This leads to permanent inhibition of Thromboxane A2 production for the lifespan of the platelet. o The plasma half-life of aspirin is only 20 minutes; however, because platelets cannot generate new COX, the effects of aspirin last for the duration of the life of the platelet (≈10 days) ADP RECEPTOR ANTAGONISTS o Ticlopidine and clopidogrel are thienopyridine derivatives that irreversibly block the binding of ADP to the platelet receptor P2Y12, thus inhibiting platelet aggregation responses by various platelet agonists: thrombin, collagen, ADP, and epinephrine. o Half life of is 6-8 hours but the effect of clopidogrel on the platelets may last for up to five days. GP IIB-IIIA RECEPTOR ANTAGONISTS o They block GP IIb-IIIa receptors on activated platelets, preventing fibrinogen and vWF binding, the final common pathway of platelet aggregation. o This blockage results in more than 80% inhibition of platelet aggregation, irrespective of the platelet agonist(s). Jay RM, Lui P. How anticoagulants work. Techniques in Regional Anesthesia and Pain Management. 2006 Apr 1;10(2):30-9. Protamine Vit K Jay RM, Lui P. How anticoagulants work. Techniques in Regional Anesthesia and Pain Management. 2006 Apr 1;10(2):30-9. No specific antidote Jay RM, Lui P. How anticoagulants work. Techniques in Regional Anesthesia and Pain Management. 2006 Apr 1;10(2):30-9. No specific antidote Jay RM, Lui P. How anticoagulants work. Techniques in Regional Anesthesia and Pain Management. 2006 Apr 1;10(2):30-9. PREOPERATIVE EVALUATION PREOPERATIVE EVALUATION o Comprehensive medical history o Patient’s individual bleeding risk: past and current usage and dosage of medications, including vitamins, herbal remedies, over-the-counter drugs, and prescription medications o Assess bleeding risk ( intra-op and post-op) based on medical conditions, medications, previous surgical history or complications, screening labs (as needed), and the type of surgery planned o Medical consult? Kaplovitch E, Dounaevskaia V. Treatment in the dental practice of the patient receiving anticoagulation therapy. J Am Dent Assoc. 2019 Jul;150(7):602-608. doi: 10.1016/j.adaj.2019.02.011. Epub 2019 May 17. PMID: 31109636. SCREENING LABS GUIDELINES AND RECOMMENDATIONS CHEST 2022; 162(5):e207-e243 VKA (WARFARIN) CHEST 2022; 162(5):e207-e243 VKA (WARFARIN) o Among four randomized trials comparing VKA continuation vs interruption, none showed a significant increase in bleeding with VKA continuation o One meta-analysis comparing continuing vs interrupting VKAs found no significant increased intra-procedural bleeding or post-procedural bleeding with VKA continuation Continuing VKAs is associated with a low (approximately 5%) risk for any bleeding VKA (WARFARIN) INR value: 1.09 to 2.84 with a group average of 1.97. No statistically significant differences for any of the comparisons concerning anticoagulation management (continued, bridged with LMWH or reduced) were detected. Oral Surg Oral Med Oral Pathol Oral Radiol 2013;116:709-716 Oral Surg Oral Med Oral Pathol Oral Radiol 2013;116:709-716 VKA Oral Surg Oral Med Oral Pathol Oral Radiol 2013;116:709-716 VKA (WARFARIN) VS. DOACS o No significant difference in number of post-extraction bleeding events (defined as hemorrhage lasting longer than 20 minutes) between the 2 groups. o 91.7% of the bleeding events were mild and could be controlled by applying pressure to the bleeding site with gauze. The remaining 8.3% required re-approximating the wound margins, applying a fibrin sealant and then re-suturing. Dental extractions and less-invasive procedures may be safely performed using local hemostatic measures, if necessary, without modifying or interrupting ongoing oral anticoagulant therapy (Oral Surg Oral Med Oral Pathol Oral Radiol 2016;122:e146-e155) DOACS (DIRECT ORAL ANTICOAGULANTS) CHEST 2022; 162(5):e207-e243 DOACS DOACS o The administration of NOACs does not need to be interrupted in most dental treatments, and the application of a hemostatic agent is normally sufficient to achieve hemostasis when postoperative bleeding events do occur. o The authors recommend at least 1 day of discontinuance in cases of the first stage of implant surgery , multiple tooth extraction and deep scaling with local inflammation, based on consideration of the medication half-life and the renal clearance rate. o Patients taking DOACs can be safely managed, even when affected by comorbidities. o The 2 causes possibly favoring excessive bleeding can in fact be controlled by the dentist. The first is the timing of the extraction, which must respect an interval of at least 4 hours after the last DOAC intake. The second is the extraction of 2 or 3 contiguous premolars and molars, which should be avoided by scheduling more than 1 session. o If 1 or both of these precautions cannot be satisfied, the collaboration of the patient’s cardiologist needs to be sought. CHEST 2022; 162(5):e207-e243 DOACS ANTIPLATELETS CHEST 2022; 162(5):e207-e243 ANTIPLATELET DRUGS Guideline Implementation Considerations: Patients who are receiving dual antiplatelet therapy with ASA and a P2Y12 inhibitor can continue ASA and interrupt the P2Y12 inhibitor Conditional Recommendation, Very Low Certainty of Evidence In patients receiving an antiplatelet drug (ASA or P2Y12 inhibitor) who are undergoing a minor dental procedure, we suggest continuing the antiplatelet drug (ASA or the P2Y12 inhibitor) over stopping the antiplatelet agent before the procedure ? o Dental extractions can be performed while patients continue the DAPT. o Although local factors such as periodontal disease or periapical lesions may lead post-extraction bleeding increasing, but local hemostasis measures are effective. o There is no need to interrupt antiplatelet therapy before extraction because the risk is significantly greater than its benefit. ANTIPLATELET DRUGS ANTIPLATELET DRUGS Oral Surg Oral Med Oral Pathol Oral Radiol 2013;116:709-716 PRE-OP, INTRA-OP AND POST-OP MEASURES PREOPERATIVE MEASURES OPERATIVE MEASURES OPERATIVE MEASURES Antifibrinolytic effects by blocking lysine binding sites on plasminogen molecules, inhibiting the interaction of plasminogen with formed plasmin and fibrin. Resorbable oxidized cellulose Absorbable Gelatin Sponge Porcine Gelatin POSTOPERATIVE MEASURES MANAGEMENT ALGORITHMS JADA 2019:150(7):602-608 Drug Class Drug Names Anticoagulant Warfarin (Coumadin®) Antiplatelet Agents Clopidogrel (Plavix®) Ticlodipine (Ticlid®) Prasugrel (Effient®) Ticagrelor (Brilinta®) Aspirin Direct-acting (or novel-acting) oral anticoagulants (Direct Thrombin Inhibitor) Dabigatran (Pradaxa®) Direct-acting (or novel-acting) oral anticoagulants (Factor Xa inhibitors) Rivaroxaban (Xarelto®) Apixaban (Eliquis®) Edoxana (Savays® in US, Lixiana® in Europe, Japan, elsewhere) Low Bleeding Risk Moderate bleeding Risk High bleeding Risk Local anesthetic administration Supragingival scaling Single tooth extraction Simple restorations Extractions of 2-4 teeth Gingival surgery of ≤5 teeth Dental implants ≤ 3 sites Abscess incision Apex Resection Extractions of ≥4 teeth Gingival surgery ≥5 teeth Dental implants ≥4 sites Determined by provider Jason P. Jones, D.D.S., M.D. Patient on Anticoagulant or Antiplatelet or DOAC/NOAC Anticoagulant (Warfarin) Therapeutic INR (<3) Low-Moderate Bleeding Risk Continue Anticoagulant (Local Hemostatic Measures Alone) High Bleeding Risk Consult with prescribing MD/DO Supratherapeutic INR (>3) Consult with prescribing MD/DO Antiplatelet (Clopidogrel, Ticlodipine, Prasugrel, Ticagrelor, ASA) Monotherapy or Dual Antiplatelet Therapy Low-Moderate Bleeding Risk Continue Anticoagulant (Local Hemostatic Measures Alone High Bleeding Risk Consult with prescribing MD/DO DOAC/NOAC (Dabigatran, Rivaroaban, Apixaban, Edoxaban) Low-Moderate Bleeding Risk Continue Anticoagulant (Local Hemostatic Measures Alone) High Bleeding Risk Consult with prescribing MD/DO Strong Evidence Limited Evidence Jason P. Jones, D.D.S., M.D. TAKE HOME POINTS o Acquiring a patient's past medical history, including a list of medications, is the initial step to assess whether a treatment is feasible or if clearance is required. o It is crucial to evaluate the risk of bleeding associated with planned procedures (#teeth extractions, contiguous teeth, alveoloplasty, etc.) to determine the necessity of clearance for a drug holiday. o If the potential risks of discontinuing antithrombotic medications outweigh the benefits, it is advisable to stage dental procedures taking appropriate intraoperative and postoperative precautions. TAKE HOME POINTS o For patients on Vitamin K antagonists, minor dental procedures with minimal bleeding risk can proceed without modifying medication if the INR is <3. o For patients on DOACs or dual antiplatelet therapy, minor dental procedures with minimal bleeding risk can generally be performed without discontinuing medication. THANK YOU Los Roques, Venezuela bordoysoto@uthscsa.edu
1456	https://langeek.co/en/grammar/course/102/linking-verbs	"Linking Verbs" in English Grammar \| LanGeek Premium\|Mobile App\|Dictionary LanGeek VocabularyGrammarExpressionsPronunciationReading Sign inToggle navigation ### Langeek EN Leaderboard Night Mode Learn Expressions Grammar Vocabulary Pronunciation Reading Mobile AppPremiumDictionary Login intermediate advanced Linking Verbs Share Dive into linking verbs like 'remain' and 'feel'. Detailed explanations and a quiz to test your knowledge. Home Grammar verbs and voices linking verbs complementslinking verbstransitivity What Are Linking Verbs? Linking verbs (also known as copula or copular verbs or complex intransitive verbs) simply link the subject of a sentence to a word or phrase that provides information about the subject (which is known as the subject complement). These verbs do not denote an action and do not take objects. Common English Linking Verbs The most common English linking verbs are: 1. be, become 2. seem, look, look like, appear 3. feel, sound, taste, smell 4. get, grow 5. remain, stay 6. turn, prove The Continuous Tense Normally, linking verbs are not used in the continuous tenses. However, a few of them can be used progressively to show a temporary or changing state, such as "be," "stay," "remain," "look," "feel," "become," "get," "grow," "turn," etc. Example She's becoming more confident. You're being mean. What Is a Subject Complement? A word or phrase that comes after a linking verb and modifies or describes the subject is called a subject complement. A subject complement can be: a predicative adjective Example Mike is handsome. In this example, 'handsome' is not an object to the verb 'is'. 'Handsome' is the subject complement to the subject 'he'. It means that 'he' and 'handsome' both refer to the same person. a noun phrase Example Mike is a dentist. 'A dentist' is the complement of the sentence and 'is' stands for a linking verb. a noun clause Example The truth is that Mike is a good dentist. The truth is whatever Mike says. a prepositional phrase Example Mike is behind the curtain. an adverb Example Mike is here. Warning Keep in mind that only adverbs of place and adverbs of time can be used as subject complements. Other types of adverb are not used as complements. Example He seems nice. (not '~~nicely~~') Here, 'nice' is an adjective and you cannot use 'nicely' instead. This pizza tastes delicious. (not '~~deliciously~~') Notice that you cannot say 'deliciously' because it comes after a linking verb. Linking Verbs vs. Stative Verbs As mentioned at the beginning of the lesson, linking verbs do not denote an action. This makes them seem similar to stative verbs. However, bear in mind that while all linking verbs are stative verbs, not all stative verbs are linking verbs. What distinguishes the two is that linking verbs only take subject complements, while stative verbs can take objects. Compare the examples: Example I feel sick. This is a linking verb. The adjective "sick" is describing the subject "I." I love chocolate. This is a stative verb. The noun "chocolate" does not refer to the subject, but shows the subject's preference. Review Linking verbs just link the subject of a sentence to the subject complement, which is a word or phrase that modifies or describes the subject. The subject complement can be: a noun (phrase) a noun clause an adjective (phrase) an adverb (phrase) a prepositional phrase Common English Linking Verbs be seem, look, look like, appear feel, sound, taste, smell :These verbs can be both state and action verbs. get, become: Normally a linking verb doesn't take a continuous tense (-ing form). But, the verbs 'get' and 'become' can take both the simple or continuous forms. Quiz: 1. Select all linking verbs from the list. (Select five.) become think love turn understand call get consider sound remain 2. Which sentence is incorrect? A You're being unreasonable. B She is becoming a doctor. C I am feeling a little down today. D She is seeming tired today. 3. Match each sentence with the correct type of subject complement. The problem is that we're out of time. The cake smells delicious. The keys are under the mat. Her dream is to travel the world. The children are upstairs. Noun Clause Prepositional Phrase Adverb Adjective To-Infinitive Phrase 4. Identify the sentence with a valid subject complement. A The class was boringly. B He appeared suddenly. C The meeting was at noon. D The milk tastes sourly. 5. Identify whether the verb of each sentence is Linking verb or Stative. Hint They own a restaurant. The soup tastes salty. The flowers smell sweet. He understands physics. I prefer tea. She grew impatient. Stative Verb Linking Verb Submit Comments (0) Loading Recaptcha... Send What Are Linking Verbs? Common English Linking Verbs What Is a Subject Complement? Linking Verbs vs. Stative Verbs Review Common English Linking Verbs Quiz ? Share on : Learn English Vocabulary Start learning categorized English vocabulary on Langeek. Click to start Recommended ### Light Verbs Have you ever repeated a word too much that made you think How boring it got! You can use light verbs instead of repeated verbs. ### Transitivity 'She smiled beautifully'. 'She started a rumor'. One of these sentences has an intransitive verb and one has a transitive one. Want to know the difference? ### Intransitive Verbs Some verbs do not need any complements to express a complete meaning. They show what the subject is doing without identifying a receiver for the action. ### Mono-transitive Verbs Mono-transitive verbs are a group of verbs that take a single verb complement, and show that the subject is performing the action of the verb on the complement. ### Ditransitive Verbs Ditransitive verbs are transitive verbs that take two objects. A direct object and an indirect object. Follow the article to read more about them. ### Complex Transitive Verbs There are five basic types of verbs: intransitive, linking, mono-transitive, di-transitive and complex-transitive verbs. In this lesson, we'll discuss the last. Download LanGeek app Download Langeek LanGeek is a language learning platform that makes your learning process faster and easier. info@langeek.co Quick access Quick access Home About Us Contact Us Help Center Vocabulary Vocabulary Level-based Topic-related Proficiency Tests Most Common See more ... 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1457	https://cdn.kutasoftware.com/Worksheets/Alg2/Systems%20of%20Two%20Equations.pdf	©3 R2z0F1v2f lKZuhtlaq oSLocfXtHwxaSrde0 YLpLBCN.H r vALlXlp 2riivgihNtDsN vr8eLsReOrCv5eqdE.B A 7M5aadHeB gwXirtuhn VIInxfkirnIijtSeV eAgllg8eHbErda8 g2D.q Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Period_ Date___ Systems of Two Equations Solve each system by graphing. 1) y = −3 x + 4 y = 3 x − 2 −5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 2) y = x + 2 x = −3 −5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 3) x − y = 3 7 x − y = −3 −5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 4) 4 x + y = 2 x − y = 3 −5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 Solve each system by substitution. 5) y = 4 x − 9 y = x − 3 6) 4 x + 2 y = 10 x − y = 13 7) y = −5 5 x + 4 y = −20 8) x + 7 y = 0 2 x − 8 y = 22 -1-©m u2P0p1J2H mKZuAtoav 2SuotfjtuwkaZrjeb tL6LWCB.k Y CAPlSlt mrdirgyhRtWsK 1rjecsrenrOvqeJdG.W v 7MXaxdjeC 9wli2tbhM SIXnTfYiYnZi0tme2 kAolVgleeb1r0a3 D2S.c Worksheet by Kuta Software LLC 9) 6 x + 8 y = −22 y = −5 10) −7 x + 2 y = 18 6 x + 6 y = 0 11) 7 x + 2 y = −19 − x + 2 y = 21 12) 3 x − 5 y = 17 y = −7 13) −7 x + 4 y = 24 4 x − 4 y = 0 14) 4 x − y = 20 −2 x − 2 y = 10 Solve each system by elimination. 15) 8 x − 6 y = −20 −16 x + 7 y = 30 16) 6 x − 12 y = 24 − x − 6 y = 4 17) −8 x − 10 y = 24 6 x + 5 y = 2 18) −24 − 8 x = 12 y 1 + 5 9 y = −7 18 x 19) −4 y − 11 x = 36 20 = −10 x − 10 y 20) −9 + 5 y = −4 x −11 x = −20 + 9 y 21) 0 = −2 y + 10 − 6 x 14 − 22 y = 18 x 22) −16 y = 22 + 6 x −11 y − 4 x = 15 23) −16 + 20 x − 8 y = 0 36 = −18 y − 22 x 24) −5 7 − 11 7 x = − y 2 y = 7 + 5 x Critical thinking questions: 25) Write a system of equations with the solution (4, −3). -2-©X I2e0s152Z XKZuOtGaI fSEoyfEtewLayrKev MLkL3CQ.L i lAWl2lV Xr4iogShBtjsh trceRsBeorVvseid5.K S TM4aPdWee KwmiptNhV UIGndf3ihnyiAt3ed HA1l9gTeNbPrSal s2S.B Worksheet by Kuta Software LLC Kuta Software - Infinite Algebra 2 Name_____ Period Date__ Systems of Two Equations Solve each system by graphing. 1) y = −3 x + 4 y = 3 x − 2 −5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 (1, 1) 2) y = x + 2 x = −3 −5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 (−3, −1) 3) x − y = 3 7 x − y = −3 −5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 (−1, −4) 4) 4 x + y = 2 x − y = 3 −5 −4 −3 −2 −1 0 1 2 3 4 5 −5 −4 −3 −2 −1 1 2 3 4 5 (1, −2) Solve each system by substitution. 5) y = 4 x − 9 y = x − 3 (2, −1) 6) 4 x + 2 y = 10 x − y = 13 (6, −7) 7) y = −5 5 x + 4 y = −20 (0, −5) 8) x + 7 y = 0 2 x − 8 y = 22 (7, −1) -1-©E h2p0Y132b CKZugtsau ZShoMfGtmwUaBrLe7 2LQLSCW.T t 5AalxlK 0rhiwgthft5sA 7rPe3ssezrQvFeZdb.9 N 7Mnazdiez nw2iRtehi JIBnwfJiwnjiktgeL 1Acl6gneYbTrtaJ v2D.O Worksheet by Kuta Software LLC 9) 6 x + 8 y = −22 y = −5 (3, −5) 10) −7 x + 2 y = 18 6 x + 6 y = 0 (−2, 2) 11) 7 x + 2 y = −19 − x + 2 y = 21 (−5, 8) 12) 3 x − 5 y = 17 y = −7 (−6, −7) 13) −7 x + 4 y = 24 4 x − 4 y = 0 (−8, −8) 14) 4 x − y = 20 −2 x − 2 y = 10 (3, −8) Solve each system by elimination. 15) 8 x − 6 y = −20 −16 x + 7 y = 30 (−1, 2) 16) 6 x − 12 y = 24 − x − 6 y = 4 (2, −1) 17) −8 x − 10 y = 24 6 x + 5 y = 2 (7, −8) 18) −24 − 8 x = 12 y 1 + 5 9 y = −7 18 x (6, −6) 19) −4 y − 11 x = 36 20 = −10 x − 10 y (−4, 2) 20) −9 + 5 y = −4 x −11 x = −20 + 9 y (1, 1) 21) 0 = −2 y + 10 − 6 x 14 − 22 y = 18 x (2, −1) 22) −16 y = 22 + 6 x −11 y − 4 x = 15 (−1, −1) 23) −16 + 20 x − 8 y = 0 36 = −18 y − 22 x (0, −2) 24) −5 7 − 11 7 x = − y 2 y = 7 + 5 x (−3, −4) Critical thinking questions: 25) Write a system of equations with the solution (4, −3). Many answers. Ex: x + y = 1, 2 x + y = 5 -2-Create your own worksheets like this one with Infinite Algebra 2. Free trial available at KutaSoftware.com
1458	https://www.ams.org/tran/1972-165-00/S0002-9947-1972-0328633-9/S0002-9947-1972-0328633-9.pdf	TRANSACTIONS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 165, March 1972 A GENERALIZED WEYL EQUIDISTRIBUTION THEOREM FOR OPERATORS, WITH APPLICATIONS BY J. R. BLUMO) AND V. J. MIZEL(2) Abstract. The present paper is motivated by the observation that Weyl's equi-distribution theorem for real sequences on a bounded interval can be formulated in a way which is also meaningful for sequences of selfadjoint operators on a Hubert space. We shall provide general results on weak convergence of operator measures which yield this version of Weyl's theorem as a corollary. Further, by combining the above results with the von Neumann ergodic theorem, we will obtain a Cesàro convergence property, equivalently, an "ergodic theorem", which is valid for all (projection-valued) spectral measures whose support is in a bounded interval, as well as for the more general class of positive operator-valued measures. Within the same circle of ideas we deduce a convergence property which completely characterizes those spectral measures associated with "strongly mixing" unitary transformations. The final sec-tions are devoted to applications of the preceding results in the study of complex-valued Borel measures as well as to an extension of our results to summability methods other than Cesàro convergence. In particular, we obtain a complete charac-terization, in purely measure theoretic terms, of those complex measures on a bounded interval whose Fourier-Stieltjes coefficients converge to zero. 1. Introduction. The present paper is motivated by the observation that Weyl's equidistribution theorem for real sequences on a bounded interval can be formu-lated in a way which is also meaningful for sequences of selfadjoint operators on a Hubert space. Namely, if {x,} is a sequence on [0, 27r] then the (equivalent) assertions I n-l (El) lim - 2 exP (imXj) = 0, m = ± 1, ±2,..., n-»oo n ,=0 and (E2) lim- 2 ! = (b-a)/27T for all [a, b]<=[0,2n] n-»oo " j<n;x,<=[a,b] can be regarded as assertions about the (selfadjoint) multiplication operators {x,} on the space of complex numbers. In fact, by writing (E2) in the equivalent form (E2') lim I "j Xta.MÍ,) - ib-a)l2n for all [a, b]c [0, 2n] Received by the editors January 20, 1971 and, in revised form, June 30, 1971. AMS 1969 subject classifications. Primary 4730, 2850, 6030; Secondary 2813. Key words and phrases. Weak convergence of operator measures, ergodic theorem for spectral measures, convergence methods, generalized Riemann-Lebesgue Lemma. O Research supported in part by NSF Grant No. GP-8290. (2) Research supported in part by NSF Grant No. GP-11679. Copyright © 1972, American Mathematical Society 291 292 J. R. BLUM AND V. J. MIZEL [March it is clear that the usual functional calculus makes equations (El) and (E2') mean-ingful even when {x¡} denotes a sequence of selfadjoint operators on a Hubert space. The question then arises as to whether, with an appropriate notion of convergence, the equations remain equivalent in this new context. We shall provide general results on weak convergence of operator measures which yield this version of Weyl's theorem as a corollary. Further, by combining the above results with the von Neumann ergodic theorem, we will obtain a Cesàro convergence property, equivalently, an "ergodic theorem", which is valid for all (projection-valued) spectral measures whose support is in a bounded interval, as well as for the more general class of positive operator-valued measures. Within the same circle of ideas we deduce a convergence property which completely charac-terizes those spectral measures associated with "strongly mixing" unitary trans-formations. The final sections are devoted to applications of the preceding results in the study of complex-valued Borel measures as well as to an extension of our results to summability methods other than Cesàro convergence. In particular, we obtain a complete characterization, in purely measure theoretic terms, of those complex measures on a bounded interval whose Fourier-Stieltjes coefficients converge to zero. 2. Weak convergence for operator measures. Hereafter 77 will denote a separable Hubert space and will denote the space of bounded linear transformations on 77. The symbol vn£+v will denote the statement (l/n) 2?=o vi^ v, convergence being taken in the relevant linear topological space containing the vn and v. For simplicity of presentation all our considerations will be taken relative to the interval [0, 2tt] whenever possible. The scaling needed to apply our results to other bounded real intervals will be obvious. For any real interval J, âS(J) denotes the class of Borel subsets of J. Finally, Q denotes the subset of [0, 27r) consisting of rational multiples of 7T. Let us recall that a sequence {vn} of real or complex-valued (hence totally finite) Borel measures on an interval Jc( — co, co) is said to converge weakly to a measure v0 provided (wO) fdvn-^-\ fdvQ V bounded continuous real/. Moreover when the {vn] are given as equibounded it suffices for the existence of a weak limit v0 that (wO') \| exp (i(2irml\J)d)vn(dd) -• a limit, m = 0, ±1, ±2,..., where \|J\| denotes the length of J. Our first result is, in effect, the assertion in the case of the weak operator topology that certain alternate formulations of weak convergence of measures which are equivalent to (wO) for positive real Borel measures remain equivalent for positive operator Borel measures. 1972] GENERALIZED WEYL EQUIDISTRIBUTION THEOREM 293 Theorem 1. Let H be a separable Hilbert space and let {Fk}k ¿, be a sequence of weakly countably additive Bor el measures on [0, 2n) whose values are positive operators on H. Suppose that for each x, ye H the sequence of complex measures defined by vx¡y(B) = (Fk(B)x, y), B e ®[0, 2tt), k ä 1, has equibounded variation () Var ,£,, â Kx,y, k ^ 1. Then the following conditions are equivalent, convergence being taken in weak operator topology: (wl) C eimeFk(d6) -> Cm, m = 0, ± 1, ± 2,..., (w2) there exists a unique weakly countably additive positive-operator-valued Borel measure F0 such that Fk(B) —¡- F0(B) whenever B e ¿$[0, 2n) is an Fa-con-tinuity set, i.e. F0(B — E) = 0. Moreover F0 and the sequence {Cm} are related as follows: Cm = J eimeF0(dd), w = 0, ±1, ±2,.... Remarks. 1. It will be seen in the course of the proof that, for each positive-operator-valued measure, the collection of continuity sets includes every interval whose boundary is disjoint from the countable collection of one-point atoms for this measure. The same arguments also yield this conclusion for other ("signed") operator-valued measures. However, the implication (wl) => (w2) is generally false for sequences of "signed" operator measures, even in one dimension. 2. The equiboundedness of variation prescribed in () is, by the uniform boun-dedness principle, equivalent to the condition \|\|Fk(Ä)\|\| fíKíor BeâS[0, 2n), k^l. 3. The arguments given below can be extended to the case in which [0, 27r) is replaced by a second countable locally compact abelian group A and {eime}mâl is replaced by Ä, the character group of A. Equation (wl) is then replaced by the assertion that J X~(X)Fk(dX) - C%, uniformly on compact subsets of Ä. (See , for example.) Proof. From () we obtain as in Remark 2 the appraisal (1) Yaxvlvè4K\xl\y\, fc fc 1. We show now that (wl) implies (w2). Given x,yeHwe have J eimevkXty(dO) -> (Cmx, v), m = 0, ± 1, ± 2. which, in view of (1) and (w0'), ensures the existence of a unique complex Borel measure vxy to which the sequence {vx_y}kil converges weakly. It follows that (2) j e™v°x,y(d6) = (Cmx, y), m = 0, ± 1, ± 2,..., 294 J. R. BLUM AND V. J. MIZEL [March and (3) Varvx,yi4K\x\\y\, while the positiveness of the measures {v%iX} ensures that each vxx is a positive measure. Obviously vx¡y is also bilinear as a function of x, y. Thus for each B 6 @[0, 2tt) we may write (4) vly(B) = (F0(B)x,y) where F0(B) e , by (3), and F0(B) is positive since v°¡x(B)^0 for all xeH. The weak countable additivity of this set function F0 is clear. Now let S<= 77 be a denumerable dense subset of H which we can take to be closed under addition and under scalar multiplication by i. Given x e S the weak convergence of the positive measures vXtX to vxx implies by Alexandroff's theorem that vkx¡x(B) -> v°x,xiB) for each ^-continuity set B: vxxiB-e) = 0. Now for each positive measure p on [0, 2n) there exists a denumerable subset 7>w such that all intervals whose boundary is in the complement of Du are ^-continuity sets. Thus it follows from the denumerability of S that the collection of Borel sets which are simultaneous ¡/^-continuity sets for all x e S includes every interval whose boundary is in the complement of a fixed countable set. We denote the collection of simultaneous v°x-continuity sets by a?! so that (5) »x.xiB) -> v°x x, yt -■ y one has, uniformly in k, (8) Var (»£.„-<„,)-0, £=1,2,..., and similarly Var O^.i,—£,,„,)-»- 0. This ensures the validity of (7) for arbitrary x, y e H and hence the validity of (w2). 1972] GENERALIZED WEYL EQUIDISTRIBUTION THEOREM 295 To prove the converse we note that the arguments of the preceding paragraph imply that for any operator measure F the /"-continuity sets consist of precisely those sets which are simultaneous continuity sets for some countable family of measures of the form {(Fx,x)}xeS, and that this collection contains all intervals whose endpoints are disjoint from a fixed countable subset of [0, 27r). From this it follows, in particular, that each function eime, being a bounded continuous function on [0, 27r], is a uniform limit of simple functions based on the class 3S-, of F0-continuity sets, thus, given x, ye H we have \| eim%Fk(dd)x, y) - f etm°(F0(dd)x, y) í j(eims-s(e)yx,y(d9) + (eim°-s(d)yx,y(de) + \s(eyXiy(dd)-\s(eyx in the uniform norm and by choosing k sufficiently large. This completes the proof. Our second result shows that a similar extension is available in the strong operator topology—provided that the {Fk} have values which are mutually commuting. Theorem 2. Let H and the {Fk} be as in Theorem 1 and suppose in addition that the {Fk} commute: (■) Fk(B)Fk,(B') = Fk.(B')Fk(B), k,k'Zl,B,B'e @[0, 2n). Then the conditions (wl) and (w2) are equivalent for convergence in strong operator topology. Proof. By a theorem of von Neumann on commuting families of operators on a separable space [11, p. 360], all operators in the family {Fk(B)}kE1¡Be&l0t2n) are functions of some fixed bounded hermitian operator A. Thus if E denotes the spectral measure for A and J denotes the smallest closed interval supporting E, then there exists for each pair k, B, k ^ 1, B e @[0, 2tt), an element/"^, • ) e £?2(E, J) such that (9) Fk(B) = J" p(B, o)E(do), k^l,Be@[0,2-n). By (l),fk(B, ■) is actually in ifx(E,J), and \fk(B, .)\|\|t- ú K, k^l,Be3S[0,2n). Moreover the positivity of the operator Fk(B) implies that fk(B, -)^0 E-a.e. We now observe that there exists a version {fk(B, •)} of the family {fk(B, •)} possessing the following properties : 296 J. R. BLUM AND V. J. MIZEL [March (i) 0^/(5,<7)gA:foralI/0 in L°° if and only if/„ ^/0 uniformly on J. If we replace each/fc(7i, •) by the equivalent function fkiB, ■) then (i) and (iii) are obvious, while the countable additivity of/"(-, a) required in (ii) follows from the fact that the positive operator measures Fk are actually countably additive and hence the classes fk(B) in LM(£, J) are countably additive in the norm topology. Since fk is, roughly speaking, a transition probability we can write (wl) in the following form : eimeFkidd) = f eimÍkid6, a)j E(d) - Cm, m = 0,±l,..., where Cm, as a limit of functions of A, is itself a function of A: Cm = ¡ cmi<j)Eid «•„(•) in L2(E,J), m = 0, ± 1, ±2,.... Equation (10) implies that every subsequence of {k} possesses a subsequence, {kt} say, such that (11) \| eimefk<(d6, a) -> cm(a) E-a.e., m = 0, + 1, ±2,.... Now the equiboundedness of the positive measures /"(•, (?) together with (11) ensures that for each a eJ, excluding an £-null set N, there exists a positive measure f°(-, o) to which thefk>(-, <?) converge weakly: (12) fk<-,o)^f°(-,o) weakly, for a <£ N. In particular, f°(-,a)£K for a $ N, and cm(a) = J" eimef°(d6, <?), <?<£A, while by Alexandroff's theorem we have for each a $ N (13) fiB,o)->f°(B,o) whenever B is an/°(-, <r)-continuity set: f°(B—È, <j)=0. We wish next to define a set function FQ on á?[0, 2tt) as follows: F0(B) = [f°(B, a)E(do), B e @[0, 2tr). 1972] GENERALIZED WEYL EQUIDISTRIBUTION THEOREM 297 Hence we must verify that f°(B, ■) is £-summable for all B e 38[0,2tt). It will suffice by (i) to show that/°([0, t), •) is £-summable for all t e (0, 2w]. Now by (12) it follows that /°([0, i), ) ^ lim inf/.([0, 0, ) ^ ''m sup/<([0, t], a) ef°([0,t],o), te(0,2n\a^N. Hence for a $ N (15) /°([0, t), o) = lim/Mfl), t], a) whenever (o, t) is not a point of one of the sets A = {(a, t) : lim inf/<([0, f), a) # lim sup/.([0, «1, ), a £ iV}, ¿' = \|(CT, t) : lim lim sup/^í 0, t--Y o\ < lim inf/fc<([0, t), o),o£N Now ^ and A' are á?[0,2w) x á?(/)-measurable since the positive functions fk([0, t), a) are measurable in o and monotone and left continuous in t, hence jointly measurable. Moreover (12) implies that for each a $ N the subset A u A' meets [0, 2tt) x a in a subset which is denumerable, hence of Lebesgue measure zero. It now follows by Fubini's theorem and (14) that/°([0, t), ■) is £-summable for Lebesgue almost all t. Using the monotonicity and left continuity in t we see that /°([0, /), ■) is in fact summable for all t e [0, 2ir), and after modification on the null set [0, 27r) x N is in fact jointly measurable. We will henceforth suppose the latter modification to have been made. By the definition of F0 and the monotone convergence theorem it follows that F0 is strongly countably additive and that the collection of F0-continuity sets is given by ®o = {B:f°iB-È,a) = 0 E-a.e.}<=âÏÏ[0, 2ir). Moreover by (12) and the dominated convergence theorem we deduce that every subsequence of {k} possesses a subsequence {fc¡} such Fki(B) = J f\B, c)E(do) -+jf°iB, o)E(da) = F0(B), for B e a0, convergence being in the strong operator topology. From this we conclude that in fact (16) Fk(B)-^F0(B) fortfe^o, which completes the proof that (wl) implies (w2). For the converse we observe that for each m the function gm(0) = eime is bounded and uniformly continuous on [0, 27r). Denote by vma=f°(-, a) o g-1 the positive Borel measure on the unit circumference in the complex plane associated to 298 J. R. BLUM AND V. J. MIZEL [March f(-,a) by gm. The image under g-1 of each vm,CT-continuity set is an f°(-,a)-continuity set: (17) vm,a(c-c) = o ^fXigñKC))- -(gmWy, a) = o. Moreover the fmtr-continuity sets include all arcs on the unit circumference whose endpoints lie in the complement of the countable subset Pm,a=gm(A„), Aa being the set of atoms of/°(-, a) on [0, 2tt). It will now suffice to show that the collection 36m = {B : B = g-\C), vm,a(C-C) = 0 E-a.e. in <?} has the property that gm is in the uniform closure of the class of simple functions based on 36m. For by (17), (13) and the dominated convergence theorem one has \s(6)Fkl(dd) = f \ \s(6)fkidd, a)]E(da) (18) J \ J J ->J [\s(6)f\d6, o^Eido) = js(8)F0(dd) whenever s is a simple function based on 38m, while by (1) i (gmie)-sie))Fkidd) \gms\co\|\|^([0, 2^)11 í K\gm-sU, k = 0,1,.... Combining these appraisals one obtains (19) ( eimeFkt(dd) -> j eimeF0(dd). Since every subsequence of {k} possesses a subsequence {fc¡} for which (19) holds, (w2) is then proved. Finally, we shall show that 3Sm has the requisite property by noting that the 36[0, 2n) x^(^-measurability of/°([0, t), a) implies that the set {(i, a) : t e Pm¡a} is á?[0, 2tt) x á?(./)-measurable and hence an / x E-null set, / being normalized Lebesgue measure on [0, 27r). It follows by Fubini's theorem that there exists a subset Tm of total /-measure on the unit circumference such that zeTm => z <£ Pm-a E-a.e. in a. This clearly implies that 3$m has the required property, even if one restricts attention to simple functions based on arcs, and thus completes the proof. We are now able to deduce the desired results on equidistribution of operators as corollaries. Corollary 1. Let {Sk} be a sequence of positive self adjoint operators on the separable Hubert space 77 with spectra contained in [0, 27r). Let the associated unitary operators and spectral measures be denoted by Uk = eiS and {Ek}, respectively. Then the following conditions are equivalent, convergence being taken in the weak operator topology : (wl) U¡?UCm, m = 0, ±1, ±2,..., (w2) there exists a positive-operator-valued measure F0 on [0, 2tt) such that 1972] GENERALIZED WEYL EQUIDISTRIBUTION THEOREM 299 Ek(B) £. F0(B)for each F0-continuity set B. If the Sk are mutually permutable then (wl) and (w2) are also equivalent for con-vergence in the strong operator topology. Proof. The corollary follows from Theorems 1 and 2 provided we take Fk = (Vk)2kiEk,k=l,2,.... Corollary 2. Let H and {Sk} be as in Corollary 1. Then the following conditions are equivalent, convergence being taken in the weak operator topology or strong operator topology according to whether the {Sk} commute: (wl) U¡?U0, m=±l, ±2,..., (w2) Ek(J) -^ /(/)/ for each interval /<= [0, 2n), where I denotes normalized Lebesgue measure on [0, 2n). This last result suggests that the desired extension of Weyl's notion of equi-distribution to operators {Sk} with spectrum in [0, 2n) consists in the requirement (20) \ Uf Ek([a, b)) = \ nf xia,b)(Sk) -> ^ /, whenever [a, ¿>)c[0, 27r). In this context Weyl's criterion becomes the formula (21) rfw^O, m=±l,±2,.... " k = 0 It should be noted that there is an analogous notion of equidistribution for positive operators {Sk} which are not uniformly bounded. Here we must replace Ek([a, b)) by ]>rœ= 0 Ek([a, b) + r2ir) = Ek([a, by), where [a, by is the periodic extension of [a, b) by 2tt. This corresponds to the application of the usual Weyl equidistribu-tion theorem to the fractional parts of a sequence of real numbers. 3. An ergodic theorem for operator measures. Our next result is a direct out-growth of the above ideas together with the von Neumann ergodic theorem. Let us denote by {rk}kil the following set of transformations of the Borel sets in [0, 2tt): rk(A) = lj ft^+x)' WhCre lA+lc~ = {' I ' = t'^ + JT' ' £ "4" It is then clear that for any Borel measure F on [0, 2ir) the set function F ° rk defined by (F= rk)(A) = F(rk(A)) is also a Borel measure on [0, 27r). Theorem 3. Let E be a spectral measure on [0, 2tt) having as its values projections in a not necessarily separable Hilbert space H. Then the following relation holds for every Borel set B whose boundary contains only points incommensurable with n, convergence being taken in the strong operator topology: (22) E o rk(B) ^ ¡(B)(I-E(Q)) + f ls(B)E(Bs), s=l 300 J. R. BLUM AND V. J. MIZEL [March where Bi = {0}, B,= U \|^)> s > 1, and where I denotes normalized Lebesgue measure, with ls,s^l, denoting the discrete measure of total mass 1 equally distributed among the points {2ttj/s}oú, dE-Î E(Am) = jXAm(0) dE, m = 0, ±1, ±2,..., where A0 = [0,2n), Am = \JlJnJ,ô1 {2nj/\m}, convergence again being in the strong operator topology. This follows from the formula (24) l:! etkme ^ xaM Be [0,2), m = 0, ±1, ±2,..., " k = 0 since the convergence in (24) is obviously bounded. We may write (24) as (25) [ e» xaJW, 6 e [0, 2n), m = 0, ± 1, ± 2,..., where /xnS denotes the discrete measure assigning mass l/n to each of the points {k9 (mod 2n)}oikSn_1. It follows from (25) by (w0') that for each 6 the positive measures {p.n,e} converge weakly to a positive measure p.e which satisfies (26) Jeim"pLe(dq>) = XAm(e)> e e P. 2»). m = 0, ±1, ±2,..., and (27) p.n,e(B) = - 2 XB((k9)) -> neiB) for all /x9-continuity sets B, n k = 0 where (/) e [0, 2tt) and (k6)=kd mod 2tt. By the use of (26) it is easily verified that the measures p,B, 6 e [0, 2n), are as follows : p.e = /, 0/ir irrational, (28) = /„ 6 = 2jtr/s,(j,s) = 1. In particular, all Borel sets whose boundary is in the complement of the set Q of rational multiples of 77 are simultaneous ^„-continuity sets for all 6. Moreover, by 1972] GENERALIZED WEYL EQUIDISTRIBUTION THEOREM 301 writing (27) in the form Xb((^)) -^> Pe(B)> we deduce that, in the strong operator topology, \xBiike))E(de)^\pe(B)E(d6) for all B which are simultaneous ^„-continuity sets. This last equation can be simplified by noting that Xii((k0))=xk(B)(6)- Thus we obtain by (27) (E o rk)(B) = f x^MEm -± i 1(B) dE+ I ls(B)E(Bs), J Jl0,2n)-Q s=l as claimed. Remark 4. If 77 is taken to be separable then Theorem 3 is essentially a direct corollary of Theorem 2. We simply observe that equation (23) above can be rewritten as (23') Ukm = f eime dEork-?-> E(Am) after which the existence of a weakly countably additive measure F0 such that E o rk(B) i^. 7^0(73) for Tvcontinuity sets B is assured. The actual calculation of F0 is as above. The above theorem can be extended to positive operator measures as well. As shown by Naimark each positive operator measure has a dilation to a spectral measure on a larger Hubert space. Thus if F is a positive operator measure on [0, 277) with F([0, 2tt)) = I, then there exists a spectral family £ on a larger Hubert space H' such that F=PEP where P is the orthogonal projection of 77' on 77. This implies that bounded pointwise convergence of a sequence of Borel functions fn to / ensures convergence in the strong operator topology of the corresponding operators Tn=J fn(ew) dF to T=¡f(ei?)dF. With this observation the proof of Theorem 3 also yields the following result. Theorem 3A. Let F be a positive operator measure on [0, 2-n) such that T^O, 27r)) = 7. Then the following relation holds for every Borel set B whose boundary contains only points outside Q, convergence being taken in the strong operator topology: Fo rkiB) -Î+ ¡(B)(I-F(Q))+ J ls(B)F(Bs). s=l As our final theorem of this section we obtain a form of Theorem 3 which characterizes "strongly mixing" spectral measures. Theorem 4. Let E be a spectral measure on [0, 27r) having as its values projections in a separable Hubert space H. In order that E be the spectral measure of a unitary operator U which is strongly mixing in the sense (29) Uk -> 0 weakly as k -> 00, 302 J. R. BLUM AND V. J. MIZEL [March it is necessary and sufficient that E satisfy the following condition, with convergence in the weak topology: (30) E o rkiJ) -> ¡iJ)I for all intervals J <= [0, 2n). Proof. We recall that the formula Ukm=j e,kmB dE^0,m^0, can be rewritten as Tjkm = fgime dE °rk^0, m¥=0. Then by Theorem 2 we may deduce that there exists an F0 such that E o TkiB) -■ F0iB) for 7r0-continuity sets B, and \| eime dF0 = 0, mïO, Fo([0, 2n)) = 7. This shows that T^o = II and hence that all subintervals of [0, 2n) are F0-continuity sets, so that (30) holds. The converse is along the lines of Theorem 2 and will be omitted. 4. Applications to complex measures. In this section we apply the theorems of the preceding sections to obtain several results concerning complex measures on a bounded interval and their Fourier-Stieltjes coefficients. As before we normalize to the consideration of complex measures on [0, 2tt). The various results are based on the simple observation that each such measure v can be realized as a spectral type for some spectral resolution E on [0, 2tt), i.e. dv=diEx,y) for a suitable pair of elements x and y. It should also be noted that in most instances the results ob-tained here via §3 may be verified in a direct analytic way. Throughout, if p, v, etc. are complex measures on [0, 27r) then we shall denote their Fourier-Stieltjes coefficients by ßim), vim), etc., w = 0, ±1, ±2.We recall the definition of the sets Bs : B1={0}, Bs={2irj/s : 0<j 1. Proposition 1. Given a complex measure p. let {p, ° rk}k S1 be defined as before. Then P o TkiB) -i pi[0, 2n)-Q)liB)+ I piBMB) s=l far every Borel set B<= [0, 2-n) whose boundary is in [0, 2tr) — Q. This result follows at once from Theorem 3 by applying our observation regard-ing spectral types. By restating matters in terms of Fourier-Stieltjes coefficients we obtain a different formulation. This utilizes the easily verified fact that P o Tkim) = fiikm), m - 0, +1, + 2,... ; k = 1,2. (An alternate proof would follow directly from the von Neumann ergodic theorem.) 1972] GENERALIZED WEYL EQUIDISTRIBUTION THEOREM 303 Proposition 1A. Given a complex measure p., fi(km)-^v(m), m = 0, ±1, ±2,..., where v is the limiting measure in Proposition 1. Now suppose U is a unitary operator which is strongly mixing. Then for any complex measure v which is a spectral type for U, i.e. dv=d(Ex, y), one has by (29) the property v(m) = (Umx, y)^0 as \m\ -> oo. That is, this property holds for every measure v absolutely continuous with respect to a maximal spectral type for U . We shall write v«/x to denote absolute continuity. The remark above suggests that whenever p, satisfies fi(m) -■ 0 and v«p,, then î>(m) -> 0. This result is actually valid having first been proved in a slightly weaker form, by Rajchmann . It can be regarded as a generalization of the Riemann-Lebesgue lemma. Proposition 2 (Generalized Riemann-Lebesgue Lemma). If the complex measure p. satisfies ß(m) -^ 0, \m\ -> oo, then v«p, implies v(m) -> 0, \m\ -> oo. Proof. By the Radon-Nikodym theorem Íeimev(d6) = f eimef(d)p,(d0) where feL1(\p,), \p.\ being the total variation measure for p.. Hence using the fact that the finite trigonometric polynomials are dense in Z-HIH) we may write íefm9K¿0)\| =í í eimePr(6)p,(d6) + í eime(f(0)-Pr(6))p,(dO) íeimePr(0)ix(dd) + \f~ ■PrlU1(laD» where Pr denotes a finite trigonometric polynomial. Since the second term on the right can be made arbitrarily small, and the first term on the right approaches zero with large m, the result follows. Using the above proposition we obtain at once, Proposition 3. A unitary operator U is strongly mixing if and only if ß(m) -■ 0, \m\ -> oo, where p. is any maximal spectral type for U. We observe next that the following result, which is a consequence of Proposition 3 and Theorem 4, fully characterizes measures for which ß(m) -»> 0. Proposition 4. A complex measure p. satisfies ß(m) -■ 0, \m\ -> oo, if and only if p. satisfies p. o Tn(J) -> p.([0, 2tt))1(J), for every interval /■= [0, 2n). lip. is a probability measure, then we may restate Proposition 4 in an illuminating probabilistic way. Namely, let Y be the random variable associated with p. and let 304 J. R. BLUM AND V. J. MIZEL [March in 7) be as before the fractional part of the random variable n Y modulo 27r. Then we have Proposition 4A. Let Y be a random variable on [0, 27r) and p be its probability measure. Then lim^,^ fi(m) = 0 if and only if the distribution ofinY) converges to normalized Lebesgue measure. Proposition 4A permits us to exhibit simple examples of singular measures on [0, 2-7r) whose Fourier-Stieltjes coefficients do not tend to zero. For let {x,:j= 1,2,...} be a stationary stochastic process with x, assuming values 0, 1,..., r— 1, where r ^ 2. Let Y= 2f= i xjr1 and let p be the probability measure associated with Y. Then it is known that p is either Lebesgue measure or a singular measure. But from the stationarity of the process it follows that the distribution of (n Y) is the same as the distribution of Y whenever « is a power of r. Thus we have Proposition 5. Let p be as above. Then p itself is either Lebesgue measure or else a singular measure. In the latter case fiim) -j> 0, \m\ -> oo. Another simple consequence of Proposition 4A is that if 7 is a random variable on [0, 27r), and we expand y=2™=i X;lr' to various bases r=2, 3,..., then either Y is uniformly distributed or else some such expansion (indeed some such expansion with r a prime) does not yield a stationary process {x¡ : j= 1, 2,...}. We exclude here the trivial case 7=0 with probability one. 5. General summability methods. In this section we examine the effect of em-ploying modes of convergence generalizing convergence £>. There will be no diffi-culty in formulating our results for two-sided sequences as easily as for one-sided sequences, and for definiteness we shall state matters only in the former context. We denote by Z the set of all integers, positive, negative and zero. Definition. A-convergence. Let A = (af)jeZ,n¿1 denote an infinite matrix of complex numbers whose rows are uniformly bounded in A(Z) : (31) Iklk«) ^ K. We shall say that A is a convergence method provided that the limit (32) lim 2 <felitt = 9>(«) n-oo jez exists for each a e [0, 2-n). If, in addition, (33) 9(a) = 0, a ^ 0, v will denote the statement 2 af^j^v, n->co, ieZ 1972] GENERALIZED WEYL EQUIDISTRIBUTION THEOREM 305 convergence being taken in the linear topological space containing v and the two-sided sequence {v,}. It is easily seen that, in the case of a generalized Cesàro method, all the results obtained earlier which employ convergence £> also remain valid when convergence £> is employed. In fact the same proofs used before apply, mutatis mutandis. The only new features are that Corollaries 1 and 2 now refer to two-sided sequences {Sk}, and that Theorems 3 and 3A utilize Tk also for k < 0. Here the definition of Tk is as follows : \k-l I 1 2ttA ^-¡¿AwA'+w k<0' where A'= — A + 2-n. In the case of a convergence method A which is not a generalized Cesàro method, Theorems 3 and 3A are modified. We have instead the following. Theorem 3. Let Abe a convergence method with <p as the associated limit func-tion, and let E be a spectral measure on [0, 27r) whose values are projections on a separable Hubert space H. Then there exists a weakly countably additive positive operator measure FQ such that (eime dF0 = Í F0(B), convergence being in the strong operator topology. Moreover, if <p is continuous then F0 is given by the formula F0 = ][fceZ ckE o Tfc where ck = ¡ e~ike<p(6) d6 are the Fourier coefficients of <p. Proof. By the definition of Cm= \| <p((m0)) dE, convergence being in the strong operator topology by the reasoning in Theorem 3. By Theorem 2, equation (35) ensures the existence of a strongly countably additive measure FQ satisfying the first conclusion of the theorem. To obtain the second conclusion, notice that (34) implies (in the presence of (31)) that for each 6 there exists a measure p.e on [0, 27r) such that both (36) ße(m) = p-e(B) for all ^-continuity sets B. 306 J. R. BLUM AND V. J. MIZEL [March In fact we can identify pe in the following manner. Define <p„ to be the function " = !> which is clearly continuous by virtue of (37). Then by definition <p satisfies (38) <p(0), 0e [0,2,7). Now for each fixed 6 it is easily verified that (39) <Pn((mB)) = Pe.n(m), m = 0, ± 1, ± 2,..., where pe¡n is the discrete measure on [0, 27r) given by (40) Pe.n = 2 <»(-/». Clearly the total variation of the {pg,n} is bounded uniformly in 6 and n by the constant K occurring in equation (31). It follows from (38) and (39) that the meas-ures {pe}, being weak limits of the {/¿9,„}, are also of uniformly bounded variation. Finally, we note that (38) ensures that the Fourier coefficients c¡ of the function <P form an absolutely convergent series. In fact by the dominated convergence theorem Cj = fe-ii$<p(6) dd = lim (e-"eco -M Therefore if <p is continuous then it coincides everywhere with the sum of its Fourier series: <p(d) = ~2JszCjeiie, 6e[0,2n). In this case we have the following formula for the measures {p$} : (41) pe = ^cA_m, 6 e [0,2,7). In order to complete our proof we proceed as follows. Í Í <p((m6)) dE = Í 2 <ym'8 dE = i eime dfe c¡E o t^ ■ Since (39) assures the strong convergence J \ <p((md)) dE, we deduce that T^o and the operator measure 2 CjE ° T¡ have the same Fourier-Stieltjes co-efficients, which completes the proof. Remark. The hypothesis that <p is continuous is essential, for when A is a generalized Cesàro method, then (0)= 1, and therefore all Fourier coefficients of <p are zero while T^o is not the zero measure. 1972] GENERALIZED WEYL EQUIDISTRIBUTION THEOREM 307 An analogous modified form for Theorem 3A is available, but we will omit the statement here. The arguments above imply in particular the following result. Corollary 3. Every function ((>"0))}mez is the Fourier-Stieltjes series of a measure, for each 0 e [0, 2tt). Acknowledgments. This work was partially supported by the National Science Foundation under Grants GP8290 and GP11697; an earlier version pre-pared while we were on Sabbatical leave at Technion-Israel Institute of Technology appeared as Technion Preprint Series No. MT-58. The unfailing consideration and stimulating mathematical atmosphere afforded us at the Technion Mathematics Department are herewith gratefully acknowledged. Bibliography 1. N. I. Ahiezer and I. M. Glazmann, The theory of linear operators in Hilbert space, GITTL, Moscow, 1950; German transi., Akademie-Verlag, Berlin, 1954; English transi., Ungar, New York, 1961. MR 13, 358; MR 16, 596. 2. A. D. Aleksandrov, Additive set functions in abstract spaces, Mat. Sb. 8 (50) (1940), 307-348; ibid. 9 (51) (1941), 563-628; ibid. 13 (55) (1943), 169-238. MR 2, 315; MR 3, 207; MR 6, 275. 3. J. Dieudonné, Sur le théorème de Lebesgue-Nikodym. IV, J. Indian Math. Soc. (N. S.) 15 (1951), 77-86. MR 13, 447. 4. N. Dunford and J. T. Schwartz, Linear operators. I : General theory, Pure and Appl. Math., vol. 7, Interscience, New York and London, 1958. MR 22 #8302. 5. A. Ionescu-Tulcea and C. Ionescu-Tulcea, Topics in the theory of lifting, Ergebnisse der Mathematik und ihrer Grenzegebiete, Band 48, Springer-Verlag, New York, 1969. 6. J. V. Neumann, Algebraische Repräsentanten der Funktionen "bis auf eine Menge vom Masse Null," J. Reine Angew. Math. 161 (1931), 109-115. 7. M. A. Nalmark, On a representation of additive operator set functions, C. R. (Dokl.) Acad. Sei. URSS 41 (1943), 359-361. MR 6, 71. 8. J. Neveau, Bases mathématiques du calcul des probabilités, Masson, Paris, 1964; English transi., Holden-Day, San Francisco, Calif., 1965. MR 33 #6659; MR 33 #6660. 9. K. R. Parthasarathy, Probability measures on metric spaces, Probability and Math. Statist., no. 3, Academic Press, New York, 1967. MR 37 #2271. 10. A. Rajchman, Une classe de séries trigonométriques qui convergent presque partout vers zéro, Math. Ann. 101 (1929), 686-700. 11. F. Riesz and B. Sz-Nagy, Leçons d'analyse fonctionnelle, 2nd ed., Akad. Kiadó, Budapest, 1953; English transi., Ungar, New York, 1955. MR 15, 132; MR 17, 175. 12. F. Tops0e, Topology and measure, Lecture Notes in Math., vol. 133, Springer-Verlag, New York, 1970. Department of Mathematics, University of New Mexico, Albuquerque, New Mexico 87106 Department of Mathematics, Carnegie-Mellon University, Pittsburgh, Pennsyl-vania 15213
1459	https://www.ams.org/books/chel/384/chel384-endmatter.pdf	INTRODUCTION TO NUMBER THEORY DANIEL E. FLATH AMS CHELSEA PUBLISHING Introduction to Number Theory Introduction to Number Theory Daniel E. Flath AMS CHELSEA PUBLISHING 10.1090/chel/384.H 2010 Mathematics Subject Classiﬁcation. Primary 11-01. For additional information and updates on this book, visit www.ams.org/bookpages/chel-384 Library of Congress Cataloging-in-Publication Data Names: Flath, Daniel E., author. Title: Introduction to number theory / Daniel E. Flath. Other titles: Number theory Description: [2018 edition]. \| Providence, Rhode Island : American Mathematical Society, 2018. \| Series: AMS Chelsea Publishing [series] ; 384 \| Originally published: New York : Wiley, 1989. \| Includes bibliographical references and indexes. Identiﬁers: LCCN 2018014214 \| ISBN 9781470446949 (alk. paper) Subjects: LCSH: Number theory. \| AMS: Number theory – Instructional exposition (textbooks, tutorial papers, etc.). msc Classiﬁcation: LCC QA241 .F59 2018 \| DDC 512.7–dc23 LC record available at Copying and reprinting. Individual readers of this publication, and nonproﬁt libraries acting for them, are permitted to make fair use of the material, such as to copy select pages for use in teaching or research. Permission is granted to quote brief passages from this publication in reviews, provided the customary acknowledgment of the source is given. Republication, systematic copying, or multiple reproduction of any material in this publication is permitted only under license from the American Mathematical Society. Requests for permission to reuse portions of AMS publication content are handled by the Copyright Clearance Center. For more information, please visit www.ams.org/publications/pubpermissions. Send requests for translation rights and licensed reprints to reprint-permission@ams.org. c ⃝1989 held by Daniel E. Flath. All rights reserved. Reprinted by the American Mathematical Society, 2018 Printed in the United States of America. ⃝ ∞The paper used in this book is acid-free and falls within the guidelines established to ensure permanence and durability. Visit the AMS home page at 10 9 8 7 6 5 4 3 2 1 23 22 21 20 19 18 To Laura Preface When I came to Singapore I was given a fourth year undergraduate Honours Number Theory course. I decided to teach Gauss’s immortal Disquisitiones Arithmeticae. This book is the result. On historical and mathematical grounds alike number theory has earned a place in the curriculum of every mathematics student. This is a textbook for an advanced undergraduate or beginning graduate core course in the subject. Such a course should stick pretty close to the naive questions, which in number theory concern prime numbers and Diophantine equations. The emphasis in this book is on Diophantine equations, especially quadratic equations in two variables. My own conscious interest in Diophantine equations goes back to a long winter’s night in a St. Louis basement in 1962 when my father and I tried to solve the notorious problem of the monkey and the coconuts as presented by Martin Gardner. No one told me then that Diophantine equations belong to a subject called “number theory,” and I found little help in the public library. I needed a teacher trained in number theory. It pleases me that several of my students of Gauss are now teaching in the schools. I might particularly mention Mr. Lee Ah Huat with whom I discovered Gauss’s first proof of the law of quadratic reciprocity. This book is closely based on lectures I gave to able groups of students during three consecutive years at the National University of Singapore. I thank the students for constantly demanding “ the notes,” which was how the text began. I tried during the writing always to keep my students in mind, always to remember that I was writing a textbook. I have sought to avoid the twin traps of doing algebra to the exclusion of number theory and of doing only trivial number theory. I take it for granted that the material I have chosen is interesting. My supreme stylistic goal is clarity. By the time this book is published I shall have gone on from Singapore. Singapore has been part of my life for three and a half years, and I shall miss it. I have many friends here. I wish them all well. Viii PREFACE Singapore August 1987 Da n Fl a t h Acknowledgment I am very grateful to Jean-Pierre Serre and the Singapore Mathematical Society for permission to reprint the article “ A = b2 - Aac” which was first published in the Mathematical Medley. Da n ie l E. Fl a t h ix Contents 1. Prime Numbers and Unique Factorization 1 1. Introduction, 1 2. Prime Numbers, 2 3. Unique Factorization, 6 4. 4- #2 ^ 2 ' +QrX r = n, 11 5. The Distribution of the Primes, 17 2. Sums of Two Squares 24 1. Introduction, 24 2. Integers mod m, 25 3. Applications of Lemma 1.4, 32 4. Gaussian Integers, 34 5. Farey Sequences, 39 6. Minkowski’s Theorem, 47 7. Method of Descent, 50 8. Reduction of Positive Definite Binary Quadratic Forms, 55 3. Quadratic Reciprocity 63 1. Introduction, 63 2. Composite Moduli, 65 3. The Legendre Symbol, 70 4. The First Proof, 77 5. The Gauss Lemma, 80 xi xii CONTENTS 6. Gauss Sums, 83 7. The Ring Z[e2”i/n], 89 8. The Jacobi Symbol, 92 9. The Kronecker Symbol, 95 10. Binary Quadratic Forms, 99 4. Indefinite Forms 104 1. Introduction, 104 2. The Square Root of 2, 105 3. The Pell Equation, 108 4. a X 2 + bX Y + c Y 2 = m, 118 5. Automorphisms, 124 6. Reduction of Indefinite Forms, 128 7. Continued Fractions, 135 8. Reduction (II), 139 9. Automorphisms (II), 142 5. The Oass Group and Genera 147 1. Introduction, 147 2. The Class Group, 148 3. The Genus Group, 156 4. What Gauss Did, 161 5. Counting Ambiguous Classes, 164 6. The Ternary Form Y 2 - XZ, 168 7. The Duplication Theorem, 172 8. Sums of Three Squares, 175 Appendix A. A = b1 — 4 ac by Jean-Pierre Serre 180 Appendix B. Tables 190 Bibliography 204 Subject Index 207 Notation Index 211 Bibliography Number theory has been blessed with many excellent books. This bibliography is a list of those that were available to me during the writing of this book. I have drawn most heavily from , , , , , and . The reader who would learn more about rational and integral quadratic forms must begin by studying the fields of />-adic numbers. There are fine introductions in and . His ultimate goal must be a close study of , which is magnificent. A proof of Dirichlet’s Theorem on Primes in Arithmetic Progressions can be found in . For a superb introduction to the distribution of prime numbers, see . Auslander, L. and Tolimieri, R., Ring structure and the Fourier transform, The Mathematical Intelligencer 7, 49-52 (1985). Bachmann, P., Niedere Zahlentheorie y Chelsea, New York, 1968. Baker, A., A Concise Introduction to the Theory of Numbers, Cambridge University Press, Cambridge, 1984. Borevich, Z. I. and Shafarevich, I. R., Number Theory, Academic, New York, 1966. Cassels, J. W. S., Rational Quadratic Forms, Academic, London, 1978. Cohn, H., A Second Course in Number Theory, Wiley, New York, 1962. Davenport, H., The Higher Arithmetic, 5th ed., Cambridge University Press, Cambridge, 1982. Dickson, L. E., Introduction to the Theory of Numbers, The University of Chicago Press, Chicago, 1929. Dirichlet, P. G. L., Vorlesungen uber Zahlentheorie, herausgegeben von R. Dedekind, 4th ed., Chelsea, New York, 1968. Edwards, H. M., Fermat's Last Theorem, Springer, New York, 1977. 204 BIBLIOGRAPHY 205 Frei, G., Leonhard Euler’s Convenient Numbers, The Mathematical Intelligencer, 7, 55-58, 64 (1985). Gauss, C. F., Disquisitiones Arithmeticae, Springer, New York, 1986. Hardy, G. H. and Wright, E. M., An Introduction to the Theory of Numbers, 5th ed., Oxford University Press, Oxford, 1979. Hua, L. K., Introduction to Number Theory, Springer, New York, 1982. Ireland, K. and Rosen, M., A Classical Introduction to Modern Number Theory, Springer, New York, 1982. Jones, B. W., The Arithmetic Theory of Quadratic Forms, The Mathematical Association of America, Providence, R.I., 1950. Knuth, D. E., The Art of Computer Programming, Vol. 2, 2nd ed., Addison-Wesley, Reading, Mass., 1981. Landau, E., Elementary Number Theory, Chelsea, New York, 1958. Landau, E., Handbuch der Lehre von der Verteilung der Primzahlen, Chelsea, New York, 1953. LeVeque, W. J., Fundamentals of Number Theory, Addison-Wesley, Reading, Mass., 1977. LeVeque, W. J., Topics in Number Theory, Vol. II, Addison-Wesley, Reading, Mass., 1956. Mathews, G. B., Theory of Numbers, Part I, Deighton, Bell and Co., Cambridge, 1892. Mordell, L. J., Diophantine Equations, Academic, London, 1969. Nagell, T., Introduction to Number Theory, 2nd ed., Chelsea, New York, 1981. Narkiewicz, W., Number Theory, World Scientific, Singapore, 1983. Niven, I. and Zuckerman, H. S., An Introduction to the Theory of Numbers, 4th ed., Wiley, New York, 1980. Serre, J.-P., A Course in Arithmetic, Springer, New York, 1973. Shapiro, H. N., Introduction to the Theory of Numbers, Wiley, New York, 1983. Sierpinski, W., Elementary Theory of Numbers, Panstwowe Wydawnictwo Naukowe, Warsaw, 1964. Stark, H. M., An Introduction to Number Theory, MIT Press, Cambridge, Mass., 1978. Venkov, B. A., Elementary Number Theory, Wolters-Noordhoff Publishing, Groningen, 1970. Vinogradov, I. M., Elements of Number Theory, Dover, London, 1954. Weil, A., Number Theory: An Approach through History, Birkhauser, Boston, 1983. Weil, A., Number Theory for Beginners, Springer, New York, 1979. Subject Index Algebraic number theory, 64, 89 Ambiguous class, see Class group Ambiguous form, see Quadratic forms, binary Automorphism, see Quadratic forms Baker, A., 187 Binomial coefficients, 19, 20, 30 Birch and Swinnerton-Dyer conjecture, 188 Box principle, 48. See also Pigeon-hole principle Character, group, 88 Chebyshev, P. L., 1, 3,17, 19 Chinese remainder theorem, 64, 65-66 Class group, 102, 147, 152-156, 160,183. See also Class number; Quadratic forms, binary ambiguous class, 159, 160, 164-168 composition: of classes, 151-153 of forms, 149, 154 concordant forms, 149, 150,155 principal class, 152, 154 Class number, 133, 183-189. See also Class group equals one, 155-156, 184-185 finiteness, 58, 61-62, 129, 152, 181 Composite integer, 2, 3, 94 witness, 94 Composition, see Class group; Product identities Concordant forms, see Class group Congruences, 25, 26. See also Integers mod m biquadratic x4 = a, 34, 95, 103 composite modulus 65-67 linear, 66, 69, 127, 155 quadratic, see also Legendre symbol x1 a -1 , 25, 27-28, 30, 178 x2 a ±2, 31, 53-54, 76 x2 a -3 , 31, 38 X2 S 5, 31 x2 a a, 28, 63-64, 67-69, 70, 95, 97-98, 99, 102 x2 + y 2 a -1 , 52 x2 + y 2 a a, 55 others, 95, 175-177 Conjugation in quadratic ring, 111, 117 Continued fractions, 10, 135, 138 nearly simple, 135-136, 140 Convenient numbers, see Genus group Convex set, 50 Cyclic groups, 29, 31, 131 Cyclotomic ring, 89 Decimal fractions, 31 Dedekind domain, 89 Descent, 25, 50-52, 105 infinite, 52 Deuring, M., 186 Diophantine approximation, see Rational approximation Diophantine equations, vii. See also Sum of squares; Pell equation linear, 1, 11-16, 119,124 positive solutions, 16, 17 quadratic: x2 = 2y2, 52, 105 x2 + y 2 —n, see Sum of squares, two x2 ± 2y2 - n, 33-34, 50, 53-54 x2 + 3y2 = n, 38, 62 x2 + xv + y 2 = 11, 38, 60, 62 207 208 SUBJECT INDEX Diophantine equations, vii. (Continued) x 2 + 5y 2 - /i, 102, 148, 154, 160 x 2 + 36>>2 - p , 103, 161 x 2 + 64y 2 = p , 95, 161 y 2 - 22 + 1, 105-108 x 2 - dy2 — ±1, see Pell equation x 2 — dy2 — m, 123, 124 ax2 + bxy + cy2 = » m, 104-105, 118-123, 128,134. See also Prime numbers, representation by quadratic forms x 2 + y 2 — z2, 54 others: y -X3 + k , 34, 39 X2 + 3y 2 - r 3, 39 X3+ y 3 - r 3, 55 X3+ p y 5 + />2z3 - 0, 53 X3+ dy 3 + d 2z 3 - id xy z = 1, 118 y 2 -X3 - x, 55 X4 + - z4, 55 X4 + / = z2, 54 X4 -1 - / = z4, 54 Dirichlet, P. G. L., primes in arithmetic progressions, 4, 5, 23, 99, 158-159, 162, 172, 204 3k + 2, 5 4k 4- ft, 4, 22-23, 24, 32 5k - 1, 76 6 + 5, 5 8 + ft, 34, 76 10 + 1, 32 10 + 9, 76 p"k + 1, 32 Discriminant, 95,180. See a/so Quadratic forms, binary fundamental, 183,184 Divisibility, 2, 25 divisor a product, 10 of factors in product, 8, 9 Division algorithm, 6, 28, 35 Divisors: common, 6 greatest common (GCD), 1, 6, 7, 9, 12, 13, 35-36 number, 10 sum, 10, 11 Duplication theorem, see Genus group Egyptian fraction, 47 Elementary divisors, 16 Elliptic curve, 187-188 Equivalence of forms, see Quadratic forms Eratosthenes, 3 Euclid, 2, 17, 34 Euclidean algorithm, 6, 36,119 Euclidean domain, 36 Euclid’s lemma, 8 Euler, L., 17, 22, 23, 52, 63, 73, 161,185 Euler product, 17-18, 22-23 Factorization, 119 of factorials, 20 uniqueness, 36, 89 Gaussian integers, 25, 34-36 integers, 1, 8-9, 18 other rings, 38, 39 Farey sequences, 39-45, 47 Father, vii Fermat, P., 5, 24, 50, 51 last theorem, 54, 55, 89 little theorem, 26-27, 30 Fibonacci numbers, 9, 118 Fourier transform, 84-89 Frobenius, G., 185 Fundamental theorem of arithmetic, see Factorization, uniqueness Fundamental unit, see Order, units in Gardner, Martin, vii Gauss, C. F., 1, 17, 34, 63-64, 72, 77, 80, 83, 102,129,147,149,157,162,163,178,180, 181, 183, 184 Gaussian form, see Quadratic forms, binary Gaussian integers, 34-38 generalizations, 89,111-112. See also Order Gauss lemma, 64, 80-83 Gauss sums, 64, 83, 89, 90-91 Gauss symbol, see Genus group Genus group, 102,147-148,157-158. See also Class group convenient numbers, 161 duplication theorem, 159,160,162,168,172, 175 Gauss symbol, 157 principal genus, 157, 160,175 Geometry of numbers, see Minkowski’s theorem Goldfeld, D. M., 187, 188 Greatest common divisor, see Divisors, greatest common Gronwall, 186 Gross, B. H., 188 Hadamard, J., 17 Heegner, K., 186, 187 Heilbronn, H., 186 Hensel’s lemma, 65, 69 SUBJECT INDEX 209 Identity, see Product identities Infinite descent, see Descent, infinite Integers, see also Factorization; Prime numbers nonsquare, 99, 109 relatively prime, 7 squarefree, 11 Integers mod m, 25, 26. See also Congruences invertible, 26, 30 primitive roots, 29-32, 54, 74, 75 roots of polynomials, 28, 30, 67, 69 Irrational numbers, 11,105,109, 111, 116 Jacobi symbol, 92-93 Kronecker symbol, 64, 95-99, 100, 157-158, 160 Lagrange, J. L., 52, 74 Landau, E., 23, 186 Lattices: integer, 40-43 in R2, 47 fundamental parallelogram, 47 Least common multiple, 11 Lee Ah Huat, vii Legendre, A. M., 17 Legendre symbol, 64, 70-73, 80, 92, 93. See also Quadratic reciprocity evaluation, see also Congruences, quadratic ( - 1 /P \ 71, 81 (2 /p ), 71-72, 81, 91, 93 (5//>), 74-75 L series, 23, 187 Mestre, J.-F., 188 Minkowski’s theorem, 25, 47-50 Modular forms, 187 Modular functions, 185, 186 Module of binary form, See Quadratic forms, binary Montgomery, H. L., 187,188-189 Negative Pell equation, see Pell equation, negative Neighbor of binary form, see Reduction, binary forms Newton’s method, see Hensel’s lemma Nonsquare integers, see Integers, nonsquare Norm in quadratic ring, 34,111-112,117,120 Order, 112,117, 122, 183 square discriminants, 113, 117 units in, 105, 112-114, 120. See also Pell equation fundamental unit, 114,117 Partial tractions, 16 Pell equation, 104-117,119,124-126, 145 and automorphisms of forms, 105,124-126, 142, 145 negative, 109, 116, 127,168 Pell form, 108 and units in orders, 105,113-114,117 Pell form, see Pell equation, Pell form Perfect numbers, 11 Period of quadratic form, see Reduction, binary forms Pi, 23, 45 Pigeon-hole principle, 110. See also Box principle Primality tests, 3, 65, 93-94 Prime numbers, vii, 1, 2, 93. See also Factorization in arithmetic progressions, see Dirichlet distribution, 1, 3, 17, 19, 21, 204 generated by polynomial, 6 infinitely many, 1, 2, 5,17,18 largest known, 5 representation by quadratic forms, 56, 59, 63, 64, 99-103, 148, 159-160 size of nth, 21 sum of reciprocals, 17-19, 22 Prime number theorem, 17 Primitive roots, see Integers mod m, primitive roots Principal class, see Class group Principal form, see Quadratic forms, binary Principal genus, see Genus group Product identities, 33, 52,102, 106, 112, 148, 154 Pythagoras, 105 Pythagorean triples, see Diophantine equations, x 2 4- y 2 » z2 Quadratic field, 111, 183,185, 186 Quadratic forms, 204. See also Diophantine equations; Reduction binary, 5, 55. See also Class group; Genus group ambiguous, 165, 168 special, 165, 168 automorphisms, 105, 124 improper, 125, 126-127, 134 proper, 105,125-126,142,144-145,166. See also Pell equation 210 SUBJECT INDEX Quadratic forms (Continued) discriminant, 55, 60. See also Discriminant equivalence, 57, 61, 100-102 proper, 57, 61-62, 102, 105, 128, 147, 151-153, 181 gaussian, 152 integral, 55 module, 119 Pell form, see Pell equation, Pell form positive definite, 55, 61 primitive, 101, 183 principal, 152 representation of integers by, 55,156. See also Prime numbers; Diophantine equations proper, 102, 124, 128, 134 roots, 139-140, 143 square discriminants, 61-62,118-119,123 ternary, 168-169 classically integral, 172 determinant, 169, 171 equivalence, 169 matrix, 169 y 2 - xz, 169, 171, 174-175 Quadratic nonresidue, 70, 74, 76-77 Quadratic reciprocity, 63-65, 72-73, 98, 101 proofs, 64, 76, 78-80, 82-83, 84, 91,163-164 supplement, 71, 81. See also Congruences, quadratic; Legendre symbol, evaluation Quadratic residue, 70, 76-77. See also Congruences, quadratic; Legendre symbol Rational approximation, 25,45-47, 50, 105-108, 110, 118 Reduced form, see Reduction, binary forms Reduction, see also Quadratic forms binary forms, 169 negative discriminant, 25, 58-61,181-183 reduced form, 59,168,183-184 positive nonsquare discriminant, 61, 105, 128-132, 141-142 and automorphisms, 144 neighbor, 129, 132, 134, 140 modified, 142 and Pell equation, 145 period, 130 reduced form, 129, 130-131, 139,142 positive square discriminant, 61-62 zero discriminant, 62 ternary forms, 169-171 Representations of integers, see Diophantine equations; Prime numbers; Quadratic forms Riemann, G. F. B., 17 Riemann hypothesis (GRH), 186, 189 Roots of quadratic forms, see Quadratic forms, binary Schur, I., 84, 85 Siegel, C. L., 186 Solovay, R., 93-94 Squarefree integers, see Integers, squarefree Stark, H. M., 187 Strassen, V., 93-94 Sum of squares: two, 5, 24, 25, 33, 36-37, 38, 46-47, 49-51, 59, 63, 76, 95,102-103,133-134,168,178 three, 147, 148, 159, 172, 177-178 four, 52-53, 179 Symmetric set, 50 Triangular numbers, 179 Unique factorization, see Factorization, uniqueness Unique factorization domain, 35, 36 de la Valine Poussin, C. J., 17 Vandermonde matrix, 87, 88 Visible point, 40 Weinberger, P. J., 187,188-189 Wilson’s theorem, 27, 28 generalized, 31 Witness, see Composite integer, witness Zagier, D., 188 Zeta functions, 186 Notation Index z The ring of integers. C (Z /«) 84 Q The field of rational numbers. Ff 85 R The field of real numbers. G 88 C The field of complex numbers. Z [ e 2n i/"] 89 Ax The group of units in a g D 90 ring A . °P ( a / m ) 92 A IX ] The ring of polynomials in one VA 95 variable with coefficients A A A 108 in a ring A . 109 G L „(R ) The group of invertible n X n 109 matrices with entries in R. r x 111 G L„(Z ) The group of n X n matrices Q (v/A) 111 with entries in Z and b\a 2 determinant ±1. GCD 7, 13 S L /f(Z) The group of n X n matrices 7 T ( X ) 17 with entries in Z and Z /m 25 determinant +1. vm 26 diag(-) The block diagonal matrix X 26 with specified blocks. s 26 'g The transpose of a matrix. Z[/] 34 det, tr The determinant and trace of N 34, 111, 117 a square matrix. A ( F ) 55 Im , Re The imaginary and real parts ( « / p ) 70 of a complex number. m 73 ker The kernel of a S (n ) 83 homomorphism. o, N 111, 117 im The image of a map. P a 112, 117 deg The degree of a nonzero 113 polynomial. £ i <S£ + 113 iff If and only if, used in a 114 definition. 119 sign The sign ( ± 1 ) of a nonzero 125 real number. 125 P I The number of elements in a a f 125 finite set S. 127 M The greatest integer less than [.a , b % c ] 128 or equal to a real f 129 number x. u 132 211 212 NOTATION INDEX ( a a, . . . < a „ ) 135 T V 135 »(/)■ «(/) 139 r ( « ) , s(fc) 143 a ’ t j Z . « 143 [a , ft, J 149 f g 149 f ' g 149 % 152 fo 152 <& {(& ) 153 156 «A 157 157 157 159 y ? ( A ) 159 160 M ( F ) 169 B ( F ) 169 M 169 ° g 174 A(A) 181 ft(A) 183 G R H 186. Errata for Introduction to Number Theory by Daniel E. Flath Page Line Correction 24 −3 qcs 1 should read qcs s 162 −2 ﬁrst Cl(∆) should be in the same font as the (correct) last Cl(∆) 211 20 tg should read tg 212 2 γy should read γy 212 6 [a, b,∗] should read [a, b, ∗] 212 7 f ∗g should read f ∗g 212 8 f ∼g should read f ∼g Updates There has been a great deal of progress in number theory, both theoretical and computational, since this book was published 1989, including a proof of Fermat’s Last Theorem. We mention just one result, the resolution by Mark Watkins of the class number 100 problem presented in Appendix A by Serre. Watkins found the complete list of all fundamental negative discriminants with class number 100 or less, considering all negative discriminants D with \|D\| ≤ 2, 383, 747. See Mark Watkins, Mathematics of Computation 73, 2004, pp. 907— 938. 1 Growing out of a course designed to teach Gauss’s Disquisitiones Arithmeticae to honors-level undergraduates, Flath’s Introduction to Number Theory focuses on Gauss’s theory of binary quadratic forms. It is suitable for use as a textbook in a course, or self-study by advanced undergraduates or graduate students who possess a basic familiarity with abstract algebra. The text treats a variety of topics from elementary number theory including the distribution of primes, sums of squares, continued factions, the Legendre, Jacobi and Kronecker symbols, the class group and genera. But the focus is on quadratic reciprocity (several proofs are given including one that highlights the p−q symmetry) and binary quadratic forms. The reader will come away with a good understanding of what Gauss intended in the Disquisitiones and Dirichlet in his Vorlesungen. The text also includes a lovely appendix by J. P. Serre titled Δ = b2 −4ac. The clarity of the author’s vision is matched by the clarity of his exposition. This is a book that reveals the discovery of the quadratic core of algebraic number theory. It should be on the desk of every instructor of introductory number theory as a source of inspiration, motivation, examples, and historical insight. CHEL/384.H
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1461	https://tsymbal.unl.edu/sites/unl.edu.cas.physics.tsymbal/files/media/file/Section%2007_Free_Electron_Model.pdf	Physics 927 E.Y.Tsymbal 1 Section 7: Free electron model A free electron model is the simplest way to represent the electronic structure of metals. Although the free electron model is a great oversimplification of the reality, surprisingly in many cases it works pretty well, so that it is able to describe many important properties of metals. According to this model, the valence electrons of the constituent atoms of the crystal become conduction electrons and travel freely throughout the crystal. Therefore, within this model we neglect the interaction of conduction electrons with ions of the lattice and the interaction between the conduction electrons. In this sense we are talking about a free electron gas. However, there is a principle difference between the free electron gas and ordinary gas of molecules. First, electrons are charged particles. Therefore, in order to maintain the charge neutrality of the whole crystal, we need to include positive ions. This is done within the jelly model, according to which the positive charge of ions is smeared out uniformly throughout the crystal. This positive background maintains the charge neutrality but does not exert any field on the electrons. Ions form a uniform jelly into which electrons move. Second important property of the free electron gas is that it should meet the Pauli exclusion principle, which leads to important consequences. One dimension We consider first a free electron gas in one dimension. We assume that an electron of mass m is confined to a length L by infinite potential barriers. The wavefunction ( ) n x ψ of the electron is a solution of the Schrödinger equation ( ) ( ) n n n H x E x ψ ψ = , where En is the energy of electron orbital. Since w can assume that the potential lies at zero, the Hamiltonian H includes only the kinetic energy so that 2 2 2 2 ( ) ( ) ( ) ( ) 2 2 n n n n n p d H x x x E x m m dx ψ ψ ψ ψ = = − = . (7.1) Note that this is a one-electron equation, which means that we neglect the electron-electron interactions. We use the term orbital to describe the solution of this equation. Since the ( ) n x ψ is a continuous function and is equal to zero beyond the length L, the boundary conditions for the wave function are (0) ( ) 0 n n L ψ ψ = = . The solution of Eq.(7.1) is therefore ( ) sin n n x A x L π ψ = , (7.2) where A is a constant and n is an integer. Substituting (7.2) into (7.1) we obtain for the eigenvalues 2 2 2 n n E m L π = . (7.3) These solutions correspond to standing waves with a different number of nodes within the potential well as is shown in Fig.1. Physics 927 E.Y.Tsymbal 2 Fig.1 First three energy levels and wave-functions of a free electron of mass m confined to a line of length L. The energy levels are labeled according to the quantum number n which gives the number of half-wavelengths in the wavefunction. The wavelengths are indicated on the wavefunctions. Now we need to accommodate N valence electrons in these quantum states. According to the Pauli exclusion principle no two electrons can have their quantum number identical. That is, each electronic quantum state can be occupied by at most one electron. The electronic state in a 1D solid is characterized by two quantum numbers that are n and ms, where n describes the orbital ( ) n x ψ , and ms describes the projection of the spin momentum on a quantization axis. Electron spin is equal to S=1/2, so that there (2S+1)=2 possible spin states with ms = ±½. Therefore, each orbital labeled by the quantum number n can accommodate two electrons, one with spin up and one with spin down orientation. Let nF denote the highest filled energy level, where we start filling the levels from the bottom (n = 1) and continue filling higher levels with electrons until all N electrons are accommodated. It is convenient to suppose that N is an even number. The condition 2nF = N determines nF, the value of n for the uppermost filled level. The energy of the highest occupied level is called the Fermi energy EF. For the one-dimensional system of N electrons we find, using Eq. (7.3), 2 2 2 2 F N E m L π = . (7.4) In metals the value of the Fermi energy is of the order of 5 eV. The ground state of the N electron system is illustrated in Fig.2a: All the electronic levels are filled upto the Fermi energy. All the levels above are empty. The Fermi distribution This is the ground state of the N electron system at absolute zero. What happens if the temperature is increased? The kinetic energy of the electron gas increases with temperature. Therefore, some energy levels become occupied which were vacant at zero temperature, and some levels become vacant which were occupied at absolute zero. The distribution of electrons among the levels is Physics 927 E.Y.Tsymbal 3 usually described by the distribution function, f(E), which is defined as the probability that the level E is occupied by an electron. Thus if the level is certainly empty, then, f(E) = 0, while if it is certainly full, then f(E) = 1. In general, f(E) has a value between zero and unity. Fig. 2 (a) Occupation of energy levels according to the Pauli exclusion principle, (b) The distribution function f(E), at T = 0°K and T> 0°K. It follows from the preceding discussion that the distribution function for electrons at T = 0°K has the form , 1 ( ) , 0 F F E E f E E E < = > . (7.5) That is, all levels below EF are completely filled, and all those above EF are completely empty. This function is plotted in Fig. 2(b), which shows the discontinuity at the Fermi energy. When the system is heated (T>0°K), thermal energy excites the electrons. However, all the electrons do not share this energy equally, as would be the case in the classical treatment, because the electrons lying well below the Fermi level EF cannot absorb energy. If they did so, they would move to a higher level, which would be already occupied, and hence the exclusion principle would be violated. Recall in this context that the energy which an electron may absorb thermally is of the order kBT ( = 0.025 eV at room temperature), which is much smaller than EF, this being of the order of 5 eV. Therefore only those electrons close to the Fermi level can be excited, because the levels above EF are empty, and hence when those electrons move to a higher level there is no violation of the exclusion principle. Thus only these electrons - which are a small fraction of the total number - are capable of being thermally excited. The distribution function at non-zero temperature is given by the Fermi distribution function. The derivation is presented in the appendix D of the textbook. The Fermi distribution function determines the probability that an orbital of energy E is occupied at thermal equilibrium ( )/ 1 ( ) e 1 B E k T f E µ − = + . (7.6) This function is also plotted in Fig. 2(b), which shows that it is substantially the same as the distribution at T = 0°K, except very close to the Fermi level, where some of the electrons are excited from below EF to above it. The quantity µ is called the chemical potential. The chemical potential can be determined in a way Physics 927 E.Y.Tsymbal 4 that the total number of electrons in the system is equal to N. At absolute zero F E µ = . Three dimensions The Schrödinger equation in the three dimensions takes the form 2 2 2 2 2 2 2 2 2 2 ( ) ( ) ( ) ( ) ( ) 2 2 2 p H E m m m x y z ψ ψ ψ ψ ψ ∂ ∂ ∂ = = − ∇ = − + + = ∂ ∂ ∂ r r r r r . (7.7) If the electrons are confined to a cube of edge L, the solution is the standing wave ( ) sin sin sin y x z n n n A x y z L L L π π π ψ = r , (7.8) where nx, ny, and nz are positive integers. In many cases, however, it convenient to introduce periodic boundary conditions, as we did for lattice vibrations. The advantage of this description is that we assume that our crystal is infinite and disregard the influence of the outer boundaries of the crystal on the solution. We require then that our wavefunction is periodic in x, y, and z directions with period L, so that ( , , ) ( , , ) x L y z x y z ψ ψ + = , (7.9) and similarly for the y and z coordinates. The solution of the Schrödinger equation (7.7) which satisfies these boundary conditions has the form of the traveling plane wave: ( ) exp( ) A i ψ = ⋅ r k r , (7.10) provided that the component of the wavevector k are determined from 2 2 2 ; ; y x z x y z n n n k k k L L L π π π = = = , (7.11) where nx, ny, and nz are positive or negative integers. If we now substitute this solution to Eq.(7.7) we obtain for the energies of the orbital with the wavevector k ( ) 2 2 2 2 2 2 2 2 x y z k E k k k m m = = + + k . (7.12) The wavefunctions (7.10) are the eigenfunctions of the momentum i = −∇ p , which can be easily seen by differentiating (7.10) : ( ) ( ) ( ) i ψ ψ ψ = −∇ = k k k p r r k r . (7.13) The eigenvalue of the momentum is k . The velocity of the electron is defined by / / m m = = v p k . In the ground state a system of N electrons occupies states with lowest possible energies. Therefore all the occupied states lie inside the sphere of radius kF. The energy at the surface of this sphere is Physics 927 E.Y.Tsymbal 5 the Fermi energy EF. The magnitude of the wavevector kF and the Fermi energy are related by the following equation: 2 2 2 F F k E m = . (7.14) The Fermi energy and the Fermi wavevector (momentum) are determined by the number of valence electrons in the system. In order to find the relationship between N and kF, we need to count the total number of orbitals in a sphere of radius kF which should be equal to N. There are two available spin states for a given set of kx, ky, and kz. The volume in the k space which is occupies by this state is equal to 3 (2 / ) L π . Thus in the sphere of 3 (4 /3) F k π the total number of states is 3 3 3 2 4 /3 2 (2 / ) 3 F F k V k N L π π π = = , (7.15) where the factor 2 comes from the spin degeneracy. Then 1/3 2 3 F N k V π = , (7.16) which depends only of the electron concentration. We obtain then for the Fermi energy: 2/3 2 2 3 2 F N E m V π = . (7.17) and the Fermi velocity 1/3 2 3 vF N m V π = . (7.18) A few estimates for Na: Na has bcc structure with cubic lattice parameter a=4.2Å and one valence electron per atom. Since there are 2 atoms in a unit cell, the electron concentration is N/V = 2/(4.2Å3) = 31022cm-3. Then, the Fermi momentum is 22 3 1/3 8 1 1 (3 10 3 10 ) 10 1Å F k cm cm − − − ≈ ⋅ ⋅⋅ ≈ = . The Fermi energy is given by 2 2 2 2 2 0 2 0 13.6 0.25 3.5 2 2 F F F k E k a eV eV m ma = = ≈ ⋅ ≈ . The Fermi temperature is defined / 40000 F F B T E k K = ≈ (1eV corresponds to 1.16104K). The Fermi velocity can be found from 2 2 3 0 0 0 2 0 v 1 0.5 4 10 137 F F F F me e k a k a k a c mca mc c − = = = ≈ ≈ ⋅ , which results in 8 v 10 F cm s ≈ . We see that the velocity of electrons is relatively large. An important quantity which characterizes electronic properties of a solid is the density of states, which is the number of electronic states per unit energy range. To find it we use Eq.(7.17) and write the total number of orbitals of energy < E : Physics 927 E.Y.Tsymbal 6 3/ 2 2 2 2 ( ) 3 V mE N E π = . (7.19) The density of states is then 3/ 2 1/ 2 2 2 2 ( ) 2 dN V m D E E dE π = = , (7.20) or equivalently 3 ( ) 2 N D E E = . (7.21) So within a factor of the order of unity, the number of states per unit energy interval at the Fermi energy, ( ) F D E , is the total number of conduction electrons divided by the Fermi energy, just we would expect. The density of states normalized in such a way that the integral 0 ( ) F E N D E dE = , (7.22) gives the total number of electrons in the system. At non-zero temperature we should take into account the Fermi distribution function so that 0 ( ) ( ) N D E f E dE ∞ = , (7.23) This expression also determines the chemical potential. Fig.3 Density of single-particle states as a function of energy, for a free electron gas in three dimensions. The dashed curve represents the density f(E,T)D(E) of filled orbitals at a finite temperature, but such that kT is small in comparison with EF. The shaded area represents the filled orbitals at absolute zero. The average energy is increased when the temperature is increased from 0 to T, for electrons are thermally excited from region 1 to region 2. Physics 927 E.Y.Tsymbal 7 Heat capacity The question that caused the greatest difficulty in the early development of the electron theory of metals concerns the heat capacity of the conduction electrons. Classical statistical mechanics predicts that a free particle should have a heat capacity of 3/2kB, where kB is the Boltzmann constant. If N atoms each give one valence electron to the electron gas, and the electrons are freely mobile, then the electronic contribution to the heat capacity should be 3/2NkB, just as for the atoms of a monatomic gas. But the observed electronic contribution at room temperature is usually less than 0.01 of this value. This discrepancy was resolved only upon the discovery of the Pauli exclusion principle and the Fermi distribution function. When we heat the specimen from absolute zero not every electron gains an energy ~kBT as expected classically, but only those electrons, which have the energy within an energy range kBT of the Fermi level, can be excited thermally. These electrons gain an energy, which is itself of the order of kBT, as in Fig. 3. This gives a qualitative solution to the problem of the heat capacity of the conduction electron gas. If N is the total number of electrons, only a fraction of the order of kBT/EF can be excited thermally at temperature T, because only these lie within an energy range of the order of kBT of the top of the energy distribution. Each of these NkBT/EF electrons has a thermal energy of the order of kBT. The total electronic thermal kinetic energy U is of the order of U≈(NkBT/EF)kBT. The electronic heat capacity is Cel=dU/dT≈NkB(kBT/EF) and is directly proportional to T, in agreement with the experimental results discussed in the following section. At room temperature C is smaller than the classical value ≈ NkB by a factor 0.01 or less. We now derive a quantitative expression for the electronic heat capacity valid at low temperatures kBT << EF. The total energy of a system of N electrons at temperature T is 0 ( ) ( , ) U ED E f E T dE ∞ = , (7.24) where f(E,T) is the Fermi distribution function and D(E) is the density of states. The heat capacity can be found by differentiating this equation with respect to temperature. Since only the distribution function depends on temperature we obtain: 0 ( , ) ( ) el dU df E T C ED E dE dT dT ∞ = = . (7.25) It is more convenient to represent this result in a different form: 0 ( , ) ( ) ( ) el F df E T C E E D E dE dT ∞ = − . (7.26) Eq. (7.27) is equivalent to Eq. (7.25) due to the fact which follows from Eq. (7.22): 0 ( , ) 0 ( ) F F dN df E T E E D E dE dT dT ∞ = = . (7.27) Physics 927 E.Y.Tsymbal 8 Since we are interested only temperatures for which kBT << EF the derivative df/dT is large only at the energies which lie very close to the Fermi energy. Therefore, we can ignore the variation of D(E) under the integral and take it outside the integrand at the Fermi energy, so that 0 ( , ) ( ) ( ) el F F df E T C D E E E dE dT ∞ = − . (7.28) We also ignore the variation of the chemical potential with temperature and assume that F E µ = , which is good approximation at room temperature and below. Then ( ) ( ) / 2 2 / ( , ) e e 1 F B F B E E k T F E E k T B E E df E T dT k T − − − = + . (7.29) Eq.(7.28) can, then, be rewritten as ( ) ( ) ( ) ( ) 3 2 / 2 2 2 2 2 / 0 / ( ) e e ( ) ( ) e 1 e 1 F B F B F B E E k T x B F el F F x E E k T B B E k T x k T E E C D E dE D E dx k T k T ∞ ∞ − − − − = = + + . (7.30) Taking into account that EF>>kBT, we can put the low integration limit to minus infinity and obtain ( ) 2 2 2 2 2 e ( ) ( ) 3 e 1 x el F B F B x x C D E k T dx D E k T π ∞ −∞ = = + . (7.31) For a free electron gas we should use Eq.(7.21) for the density of states to finally obtain 2 2 el B F T C Nk T π = , (7.32) where we defined the Fermi temperature / F F B T E k = . This is similar to what we expected to obtain according to the qualitative arguments given in the beginning of this section. Experimentally the heat capacity at temperatures much below both the Debye temperature and the Fermi temperature can be represented in the form: 3 el ph C C C T T α β = + = + . (7.33) The electronic term is dominant at sufficiently low temperatures. The constants α and β can be obtained by fitting the experimental data.
1462	https://avibase.bsc-eoc.org/species.jsp?avibaseid=9A684C997C1AF1BA	Limosa lapponica (Bar-tailed Godwit) - Avibase Avibase -The World Bird Database Bird checklists - taxonomy - distribution - maps - links AvibaseAvibase homeAbout AvibaseMastodonBirding webcamsCompare taxonomiesAvibase Flickr GroupBird of the day archivesPeter's Checklist DatabaseAvibase citationsBirdlinksTrip reports myAvibaseMyAvibase allows you to create and manage your own lifelists, and produce useful reports to help you plan your next birding excursion. myAvibase HomeManage lifelistsManage observationsmyAvibase reports ChecklistsThere are more than 20,000 regional checklists in Avibase, offered in 9 different taxonomies, including synonyms more than 175 languages. Each checklist can be viewed with photos shared by the birding community, and also printed as PDF checklists for field use. Avibase checklistsCompare Regions SearchAvibase searchBrowse by familiesBrowse by authorsBrowse by publicationsAvibase Taxonomic Concepts ContributeThere are a few ways by which you can help the development of this page, such as joining the Flickr group for photos or providing translations of the site in addition languages. Contribute to AvibaseAcknowledgementsFlickr groupMedia statsFlickr group membersMedia wantedContribute a better translation EnglishChange language Sign Up Logout Sign In Your login: Password: [x] Stay signed in Enter your login name or your email address and click on Send reminder to receive a reminder by email. Welcome Guest Bar-tailed Godwit Limosa lapponica (Linnaeus, C 1758) summary) taxon grid) synonyms) map) life history) eBird) Wikipedia) NatureServe) ITIS) Flickr) Audio) More links) The bar-tailed godwit is a large wader in the family Scolopacidae, which feeds on bristle-worms and shellfish on coastal mudflats and estuaries. It has distinctive red breeding plumage, long legs, and a long upturned bill. Bar-tailed godwits breed on Arctic coasts and tundra from Scandinavia to Alaska, and overwinter on coasts in temperate and tropical regions of the Old World, Australia, and New Zealand. The migration of the subspecies Limosa lapponica baueri across the Pacific Ocean from Alaska to New Zealand is the longest known non-stop flight of any bird, and also the longest journey without pausing to feed by any animal. The round-trip migration for this subspecies is over 29,000 km (18,020 mi). Source: Wikipedia) (0 votes) Photo powered by flickr.com. Classification Order: Charadriiformes Family: Scolopacidae Genus: Limosa Scientific: Limosa lapponica Original description Citation: (Linnaeus, C 1758) Reference: Systema Naturae ed. 10 1 p.147 Protonym: Scolopax lapponica Type locality: Lapland. Link: Avibase identifiers Avibase ID: 9A684C997C1AF1BA Avibase ID: avibase-9A684C99 Short link: Taxonomic Serial Number: TSN: 176687 NEAR THREATENED Geographic range: Limosa lapponica: breeds patchily in low Arctic tundra of northern Palearctic from far northern Scandinavia to Chukotskiy and Anadyr peninsulas (northeastern Russia) and northern and western Alaska; winters mainly temperate to tropical coasts from British Isles, southern Europe, western Mediterranean, Africa, and Malagasy region through Middle East, Indian subcontinent, southeastern Asia, Ryukyu Islands (southern Japan), southeastern China, Taiwan, Philippines, Indonesian Archipelago, Australasia from New Guinea region to Melanesia and New Zealand, and Fiji (southwestern Polynesia) Limosa lapponica lapponica: breeds from northern Scandinavia eastward to the Kanin Peninsula (northwestern Russia); winters from Lapland eastern through Kola and Kanin peninsulas to western Yamal (north-central Russia); winters from the North Sea, western Iberia, and northwestern Africa south to western South Africa Limosa lapponica taymyrensis: breeds north-central Siberia, roughly from the lower Yenisei River Valley eastward to the lower Anabar River Valley; winter range incompletely known but primarily coasts of West Africa Limosa lapponica yamalensis: breeds on the northern West-Siberian Plain, including the Yamal Peninsula and the lower Ob River Valley; winter range incompletely known, but extends from Oman eastward to western India, and probably the east coast of Africa, perhaps south to South Africa Limosa lapponica baueri: breeds western Alaska; winters coastal China to New Zealand Limosa lapponica anadyrensis: breeds eastern Siberia (Chukotka and Anadyr lowlands); winters in Australia, possibly also New Zealand Limosa lapponica menzbieri: northern Siberia; winters southeastern Asia to coastal Australia including Tasmania Show more... Source: Clements checklist English: Bar-tailed Godwit Catalan: Tètol cuabarrat Czech: břehouš rudý Danish: Lille Kobbersneppe German: Pfuhlschnepfe Spanish: Aguja Colipinta Spanish (Spain): Aguja Colipinta Finnish: punakuiri French: Barge rousse Croatian: Riđa muljača Hungarian: Kis goda Icelandic: Lappajaðrakan Italian: Pittima minore Japanese (romaji): oosorihashishigi Japanese: オオソリハシシギ Lithuanian: Laplandinis griciukas Dutch: Rosse Grutto Norwegian Nynorsk: Lappspove Norwegian: Lappspove Polish: szlamnik Portuguese: Fuselo Portuguese (Brazil): fuselo Portuguese (Portugal): Fuselo Russian: Малый веретенник Slovak: brehár hrdzavý Serbian: Laponska muljača Swedish: myrspov Ukrainian: Грицик малий Chinese: 斑尾塍鹬 Chinese (Traditional): 斑尾鷸 Authorities recognizing this taxonomic concept: African Bird Club Checklist 2004: Bar-tailed Godwit ( Limosa lapponica) [version 1]) African Bird Club Checklist August 2005: Bar-tailed Godwit ( Limosa lapponica) [version 1]) African Bird Club Checklist March 2007: Bar-tailed Godwit ( Limosa lapponica) [version 1]) African Bird Club Checklist January 2008: Bar-tailed Godwit ( Limosa lapponica) [version 1]) African Bird Club Checklist May 2009: Bar-tailed Godwit ( Limosa lapponica) [version 1]) African Bird Club Checklist March 2010: Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 1st edition (1886): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 1st edition (incl. 1st suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 1st edition (incl. 2nd suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 1st edition (incl. 3rd suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 1st edition (incl. 4th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 1st edition (incl. 5th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 1st edition (incl. 6th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 1st edition (incl. 7th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 2nd edition (1895): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 2nd edition (incl. 8th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 2nd edition (incl. 9th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 2nd edition (incl. 10th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 2nd edition (incl. 11th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 2nd edition (incl. 12th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 2nd edition (incl. 13th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 2nd edition (incl. 14th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 2nd edition (incl. 15th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 3rd edition (1910): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 3rd edition (incl. 16th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 3rd edition (incl. 17th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 3rd edition (incl. 18th suppl.): Pacific Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (1931): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 19th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 20th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 21st suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 22nd suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 23rd suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 24th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 25th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 26th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 27th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 28th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 29th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 30th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 4th edition (incl. 31st suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 5th edition (1957): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 5th edition (incl. 32nd suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 5th edition (incl. 33rd suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 5th edition (incl. 34th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 6th edition (1983): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 6th edition (incl. 35th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 6th edition (incl. 36th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 6th edition (incl. 37th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 6th edition (incl. 38th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 6th edition (incl. 39th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 6th edition (incl. 40th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 6th edition (incl. 41st suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (1998): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 42nd suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 43rd suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 44th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 45th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 46th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 47th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 48th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 49th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 50th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 51st suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 52nd suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 53rd suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 54th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 55th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 56th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Union 7th edition (incl. 57th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Society 7th edition (incl. 58th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Society 7th edition (incl. 59th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Society 7th edition (incl. 60th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Society 7th edition (incl. 61st suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Society 7th edition (incl. 61st suppl. and addendum): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Society 7th edition (incl. 62nd suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Society 7th edition (incl. 63rd suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Society 7th edition (incl. 64th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Society 7th edition (incl. 65th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) American Ornithologists' Society 7th edition (incl. 66th suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avibase taxonomic concepts (current): Bar-tailed Godwit ( Limosa lapponica) Avibase taxonomic concepts v. 2023 (May 2023): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avibase taxonomic concepts v. 2024 (Oct 2024): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avibase taxonomic concepts v. 01 (August 2013): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avibase taxonomic concepts v. 02 (May 2014): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avibase taxonomic concepts v. 03 (March 2015): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avibase taxonomic concepts v. 04 (Aug 2016): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avibase taxonomic concepts v. 05 (Jan 2017): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avibase taxonomic concepts v. 06 (Feb 2018): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avibase taxonomic concepts v. 07 (Feb 2020): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avibase taxonomic concepts v. 08 (Feb 2021): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Avilist 2025: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 00: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 01: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 02: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 03: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 04: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 05 (Jun 2012): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 05.1 (Oct 2012): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 06 (Nov 2013): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 06.1 (Feb 2014): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 07 (Jul 2014): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 08 (Oct 2015): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 09 (Dec 2016): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Birdlife checklist version 09.1 (Jun 2017): Bar-tailed Godwit ( Limosa lapponica) [version 1]) HBW and BirdLife Taxonomic Checklist v2 (Dec 2017): Bar-tailed Godwit ( Limosa lapponica) [version 1]) HBW and BirdLife Taxonomic Checklist v3 (Nov 2018): Bar-tailed Godwit ( Limosa lapponica) [version 1]) HBW and BirdLife Taxonomic Checklist v4 (Dec 2019): Bar-tailed Godwit ( Limosa lapponica) [version 1]) HBW and BirdLife Taxonomic Checklist v5 (Dec 2020): Bar-tailed Godwit ( Limosa lapponica) [version 1]) HBW and BirdLife Taxonomic Checklist v6 (Dec 2021): Bar-tailed Godwit ( Limosa lapponica) [version 1]) HBW and BirdLife Taxonomic Checklist v6b (Jul 2022): Bar-tailed Godwit ( Limosa lapponica) [version 1]) HBW and BirdLife Taxonomic Checklist v7 (Dec 2022): Bar-tailed Godwit ( Limosa lapponica) [version 1]) HBW and BirdLife Taxonomic Checklist v8 (Dec 2023): Bar-tailed Godwit ( Limosa lapponica) [version 1]) HBW and BirdLife Taxonomic Checklist v9 (Oct 2024): Bar-tailed Godwit ( Limosa lapponica) [version 1]) British Ornithologists' Union Checklist (7th edition): Bar-tailed Godwit ( Limosa lapponica) [version 1]) British Ornithologists' Union Checklist (7th edition, incl. Jan 2009 suppl.): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Christidis and Boles (2008): Systematics and Taxonomy of Australian Birds: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Commission internationale pour les noms français des oiseaux (1993, révision 2009): Barge rousse ( Limosa lapponica) [version 1]) Commission internationale pour les noms français des oiseaux (1993): Barge rousse ( Limosa lapponica) [version 1]) Clements 1st edition: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements, version 2021: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements, version 2022: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements, version 2023: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements, version 2024: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 3rd edition: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 4th edition: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 5th edition (as published): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 5th edition (incl. 2000 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 5th edition (incl. 2001 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 5th edition (incl. 2002 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 5th edition (incl. 2003 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 5th edition (incl. 2004 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 5th edition (incl. 2005 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 6th edition: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 6th edition (incl. 2007 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 6th edition (incl. 2008 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 6th edition (incl. 2009 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 6th edition (version 6.5 incl. 2010 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 6th edition (version 6.6 incl. 2011 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 6th edition (version 6.7 incl. 2012 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 6th edition (version 6.8 incl. 2013 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements 6th edition (version 6.9 incl. 2014 revisions): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements, version 2015: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements, version 2016: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements, version 2017: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements, version 2018: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Clements, version 2019: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 1.50: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 1.52: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 1.53: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 1.54: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 1.55: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 2015: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 2016: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 2017: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 2018: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 2019: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 2021: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 2022: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 2023: Bar-tailed Godwit ( Limosa lapponica) [version 1]) eBird version 2024: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 2nd edition: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 2nd edition (incl. ? suppl): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 3rd edition (as published): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 3rd edition (incl. corrigenda 1.2): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 3rd edition (incl. corrigenda 2.1): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 3rd edition (incl. corrigenda 3.1): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 3rd edition (incl. corrigenda 4): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 3rd edition (incl. corrigenda 5): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 3rd edition (incl. corrigenda 6): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 3rd edition (incl. corrigenda 7): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 3rd edition (incl. corrigenda 8): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 4th edition (vol. 1-2): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Howard and Moore 4th edition (incl. corrigenda vol.1-2): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Handbook of the Birds of the World (vol 1-16): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Handbook of the Birds of the World Alive (31/01/2015): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Handbook of the Birds of the World Alive (03/07/2017): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Handbook of the Birds of the World and Birdlife (Dec 2017): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Handbook of the Birds of the World and Birdlife (Dec 2018): Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 1.0: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 1.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 1.5: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 1.6: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 1.7: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.0: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.3: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.4: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.5: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.6: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.7: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.8: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.9: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.10: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 2.11: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 3.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 3.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 3.3: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 3.4: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 3.5: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 4.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 4.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 4.3: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 4.4: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 5.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 5.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 5.3: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 5.4: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 6.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 6.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 6.3: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 6.4: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 7.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 7.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 7.3: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 8.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 8.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 9.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 9.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 10.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 10.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 11.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 11.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 12.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 12.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 13.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 13.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 14.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 14.2: Bar-tailed Godwit ( Limosa lapponica) [version 1]) IOC World Bird Names, version 15.1: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Morony, Bock and Farrand: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Oriental Bird Club: Bar-tailed Godwit ( Limosa lapponica) [version 1]) Peters' Check-list of the Birds: Limosa lapponica[version 1]) Peters' Check-list of the Birds (original): Limosa lapponica[version 1]) Peters' Check-list of the Birds (2nd edition): Limosa lapponica[version 1]) Sibley and Monroe (1993): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Sibley and Monroe (1993, including corrections up to 1998): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Sibley and Monroe, Birds of the World Version 2.0: Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (24/10/2003): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (14/09/2005): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (16/07/2006): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (26/09/2007): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (01/05/2008): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (30/09/2009): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (19/08/2010): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (8/08/2011): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (8/02/2012): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (4/09/2012): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (17/03/2013): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (9/02/2014): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (16/02/2015): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (28/07/2016): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (22/04/2017): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (4/06/2018): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (6/06/2019): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (20/02/2020): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (8/06/2020): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (24/08/2021): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (24/07/2022): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (28/09/2023): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (27/07/2024): Bar-tailed Godwit ( Limosa lapponica) [version 1]) South American Classification Committee (31/07/2025): Bar-tailed Godwit ( Limosa lapponica) [version 1]) Zoonomen - Zoological Nomenclature Resource: Limosa lapponica[version 1]) Zoonomen - Zoological Nomenclature Resource: Limosa lapponica[version 1]) Taxonomic status: Species status: full species Your sightings No sightings Add to a lifelist: Date: You must be logged in to view your sighting details. To register to myAvibaseclick here. Related taxa Highlight taxa in a checklist (shown in red) Other related concepts Calidris canutus x Limosa lapponica (hybrid) Limosa [lapponica x haemastica] (hybrid) Limosa [lapponica or limosa] (taxon group) Other synonyms See also the taxonomic synonyms;) table. Afrikaans:Bandstertgriet, Hudsonbaaigriet Arabic:بقويقة سلطانية مخططة الذيل, بقويقة مخططة الذيل, بقويقة مخططة الذيل رهيز Asturian:Aguya Roxa, Punzón de raupintu Azerbaijani:kiçik ox cüllüt, Kiçik oxcüllüt Bashkir:бәләкәй ҡыҙғатау Belarusian:Грыцук малы, Малы грыцук Bulgarian:Пъстроопаш крайбрежен бекас, Пъстроопашат крайбрежен бекас Bengali:দাগিলেজ জৌরালি, দাচগমরজ হজৌযাচর Bengali (transliteration):dāgilēja jaurāli Breton:Ar marc'heg ruz, Marc'heg ruz Catalan:Cegall de mosson coabarrat, Tètol cuabarrat, Tètol cuanegre Catalan (Balears):Cegall de mosson coabarrat Valencian:Tètol cuanegre Montenegrin:laponska muljača, лапонска муљача Montenegrin (Cyrillic):лапонска муљача Czech:bøehouš rudý, Brehouš rudý, Břehouš rudý Chuvash:пĕчĕк вăрăмсăмса Welsh:Cyffylog y mor, Gïach pengafr, Rhostog coch, Rhostog goch, Rhostog gynffonfraith, Rhostog gynffonfrith Danish:Lille kobbersneppe German:Amerikanische Uferschnepfe, Pfuhlschnepfe Greek:Ακτοτούρλι, Θαλασσολιμόζα, Λαπωνική Βαλτομπεκάτσα Greek (Cypriot):Λαπωνική Βαλτομπεκάτσα English:Bartailed Godwit, Bar-tailed Godwit, Pacific Godwit, White-rumped Godwit English, HBW:Bar-tailed Godwit English (IOC):Bar-tailed Godwit English (Kenya):Bar-tailed Godwit English (United States):Bar-tailed Godwit Esperanto:lapona limozo, strivosta limozo Spanish:Aguja Colipinta, Aguja de Cola Pintada, Picopando Cola Barrada, picopando cola-barrada, Picopando Colibarrado Spanish (Chile):Zarapito de cola barrada Spanish (Spain):Aguja Colipinta Spanish (HBW):Aguja colipinta Spanish (Mexico):picopando cola barrada, picopando cola-barrada, Picopando Colibarrado Spanish (Puerto Rico):Barga Colipinta Spanish (Venezuela):Aguja Cola Rayada Estonian:Vöötsaba-vigle Basque:Kuliska gorri, Kuliska gorria, Tètol cuabarrat Persian:گیلانشاه حنایی Finnish:Punakuiri Faroese:Koparsnípa, Lónspógvi French:Barge de Laponie, Barge hudsonienne, Barge rousse, Borge rousse Friulian:pivuic ros Frisian:Heawylp Irish:Guilbneach Stríocearrach Gaelic:Cearra-Ghob Mhòr, Roid-Ghuilbneach Galician:Fuselo de rabo pinto, Mazarico de rabo pinto, Mazarico rubio, Tètol cuabarrat Manx:Feddag hraie, Gob-kiare mooar Hebrew:לימוזה חומת־בטן, לימוזה חומת-בטן Croatian:Riđa muljača, Smeđa muljača Hungarian:Kis goda Armenian:[Pokr Ilikakttsar ], Փոքր Իլիկակտցար Indonesian:Biru-laut ekor-blorok, Birulaut Ekor-blorok, Burung Biru-laut Ekor-blorok Icelandic:Lappajaðrakan Italian:Pittima minore Japanese (romaji):ō sori-hashi shigi, Oh-sorihashi-shigi, oosorihashishigi, Oosorihashi-shigi, O-sorihashi-shigi, Ou-sorihashi-shigi Japanese:オオソリハシシギ Japanese (Kanji):大反嘴鴫, 大反嘴鷸 Georgian:ზოლიანკუდა ლია Mongolian, Halh:бутын сараалж, морин цууцал, морин цууцаль Kazakh:Кіші шырғалақ, кішкентай шырғалақ, Үлкен дүпілдек Kazakh (Transliteration):kişi şırğalaq Korean:큰뒷부리도요 Korean (Transliteration):keun-dwit-buri-doyo North Karelian:pienikuiri Cornish:Godwyt Cough Kirghiz:кидик шиштумшук, кичи найза тумшук Scientific:Limosa foxii, Limosa lapponica, Limosa lapponica lapponica, Limosa lapponica novaezealandiae, Limosa rufa, Limosa uropygialis, Scolopax aegocephala, Scolopax lapponica, Tringa gregaria Luxembourgish:Pullschnepp Lithuanian:Laplandinis griciukas Latvian:Sarkanā puskuitala, Sarkanais kluitēns Maori:kuaka, kūaka Macedonian:Лапонска блатарка Malayalam:വരവാലൻ സ്നാപ്പ് Mongolian:бутын сараалж, морин цууцал, морин цууцаль, хар сүүлт морин зараг, Хурган цууцал, Хурган цууцаль Mongolian (Bichig, Inner Mongolia):ᠮᠣᠷᠢᠨ ᠴᠣᠣᠴᠠᠯᠢ Mongolian (Transliteration):khurg aŋ tsuutsal, khurg aŋ tsuutsal' Moldavian:Sitar de mal nordic Malay:Burung Kedidi Berjalur, Burung Kedidi Berjalur-jalur, Burung Kedidi Ekor Berjalur, Kedidi Ekor Berjalur Maltese:Girwiel Denbu bl-Istrixxi, Girwiel Nordiku Mongolian, Chakhar (Peripheral):морин цууцаль, хар сүүлт морин зараг Dutch:Rosse Grutto Norwegian Nynorsk:Lappspove Norwegian:Lappspove Polish:Rycyk, szlamik rdzawy, szlamnik, szlamnik (zwyczajny), Szlamnik rdzawy, Szlamnik zwyczajny, szlimik Pinyin:bān-wěi chéng-yù, bān-wěi yù Portuguese:Fuselo Portuguese (Brazil):Fuselo Portuguese (Portugal):Fuselo Rarotongan:Kōvē Romansh:Becassina da ruina Rhaeto-Romance:Becassina da ruina Romanian:Sitar de mal nordic Russian:Maly Veretennik, Веретенник малый, Малый веретенник Yakut:куралай, кыра кураҕаччы Scots:Cearra ghob mhor Northern Sami:Ruksesguškil Sinhalese (Transliteration):Waira-penda Gohuduwiththa Slovak:Brehár hrdzavý Slovenian:progastorepi kljunac, progastorepi kljunač Samoan:tuli Albanian:Gjelëza e vogël bishtvizuar, Gjelëza e vogël e bishtvijëzuar Serbian:crnorepa muljaca, crnorepa muljača, Laponska muljača, Mala (smedja) muljaca, Mala muljaca, Smedja muljaca, Мала (Смеђа) мулјача, Мала муљача, Смеда муљача Swedish:Myrspov Swahili:Msese Mkia-miraba Tamil:Pattaivaal Mookkaan Thai:นกปากแอ่นหางลาย Thai (Transliteration):nók pàak-àeen hăaŋ-laay Turkmen:kiçi ikburun Turkish:Kıyı Çamur Çulluğu, Kıyı Çamurçullğu, Kıyı Çamurçullşu, Kıyı çamurçulluğu, Kızıl çamurçulluğu, kyyy çamurçulluğu, Кіші шырғалақ Tatar:кечкенә веретенник Tuvinian:Бичии ээргииш Ukrainian:Грицик малий , Малий Веретенник, Малий грицик Uzbek:кичик веретенник Vietnamese:Chim Choắt mỏ thẳng đuôi vằn, Choắt mỏ thẳng đuôi vằn Chinese:[ban-wei cheng-yu], [ban-wei yu], 斑尾塍鷸, 斑尾塍鹬, 斑尾鷸 Chinese (Traditional):斑尾塍鷸, 斑尾鷸, 斑尾鷸〔斑尾塍鷸〕 Chinese (Taiwan, Traditional):斑尾鷸 Chinese (Taiwan):[ban-wei yu], 斑尾鷸, 斑尾鹬 Avibase has been visited 444,789,283 times since 24 June 2003. © Denis Lepage \| Privacy policy
1463	https://www.youtube.com/shorts/uqAv4RYXAfU	YouTube Back Skip navigation Search Search with your voice Sign in Home HomeShorts ShortsSubscriptions SubscriptionsYou YouHistory History @YATAQi Subscribe ### Does this Polish puzzle have a solution? What’s the maximum number of wins in this tournament? #maths#mathematics#puzzle#tournament#games 3K I like this Dislike I dislike this 43 Comments Share Share Remix Remix Sign in to confirm you’re not a bot This helps protect our community. Learn more Sign in Comments 43 Top commentsNewest first Description What’s the maximum number of wins in this tournament? #maths#mathematics#puzzle#tournament#games 3K Likes 50,428 Views Jun 21 2025 In a rock, paper, scissors tournament with 55 participants, how many wins can the tournament winner get under their belt under the specified condition?…...more ...more Show less How this content was made Auto-dubbed Audio tracks for some languages were automatically generated. Learn more What’s the maximum number of wins in this tournament? #maths #mathematics #puzzle #tournament #games @YATAQi Next video Search Info Shopping Tap to unmute 2x If playback doesn't begin shortly, try restarting your device. • You're signed out Videos you watch may be added to the TV's watch history and influence TV recommendations. To avoid this, cancel and sign in to YouTube on your computer. Cancel Confirm Share - [x] Include playlist An error occurred while retrieving sharing information. Please try again later. Watch later Share Copy link 0:00 / •Watch full video Live • • NaN / NaN [](
1464	https://math.libretexts.org/Bookshelves/PreAlgebra/Prealgebra_(Arnold)/01%3A_The_Whole_Numbers/1.02%3A_Adding_and_Subtracting_Whole_Numbers	Skip to main content 1.2: Adding and Subtracting Whole Numbers Last updated : Aug 30, 2024 Save as PDF 1.1: An Introduction to the Whole Numbers 1.3: Multiplication and Division of Whole Numbers Page ID : 22460 David Arnold College of the Redwoods ( \newcommand{\kernel}{\mathrm{null}\,}) In the expression 3 + 4, which shows the sum of two whole numbers, the whole numbers 3 and 4 are called addends or terms. We can use a visual approach to find the sum of 3 and 4. First, construct a number line as shown in Figure 1.2. To add 3 and 4, proceed as follows. Start at the number 0, then draw an arrow 3 units to the right, as shown in Figure 1.2. This arrow has magnitude (length) three and represents the whole number 3. Draw a second arrow of length four, starting at the end of the first arrow representing the number 3. This arrow has magnitude (length) four and represents the whole number 4. The sum of 3 and 4 could be represented by an arrow that starts at the number 0 and ends at the number 7. However, we prefer to mark this sum on the number line as a solid dot at the whole number 7. This number represents the sum of the whole numbers 3 and 4. The Commutative Property of Addition Let’s change the order in which we add the whole numbers 3 and 4. That is, let’s find the sum 4 + 3 instead. As you can see in Figure 1.3, we start at zero then draw an arrow of length four, followed by an arrow of length three. However, the result is the same; i.e., 4 + 3 = 7. Thus, the order in which we add three and four does not matter; that is, 3 + 4=4 + 3. This property of addition of whole numbers is known as the commutative property of addition. The Commutative Property of Addition Let a and b represent two whole numbers. Then, a + b = b + a. Grouping Symbols In mathematics, we use grouping symbols to affect the order in which an expression is evaluated. Whether we use parentheses, brackets, or curly braces, the expression inside any pair of grouping symbols must be evaluated first. For example, note how we first evaluate the sum in the parentheses in the following calculation. (3 + 4) + 5 = 7 + 5 = 12 The rule is simple: Whatever is inside the parentheses is evaluated first. Writing Mathematics When writing mathematical statements, follow the mantra: One equal sign per line. We can use brackets instead of parentheses. 5 + [7 + 9] = 5 + 16 = 21 Again, note how the expression inside the brackets is evaluated first. We can also use curly braces instead of parentheses or brackets. {2+3} + 4 = 5 + 4 = 9 Again, note how the expression inside the curly braces is evaluated first. If grouping symbols are nested, we evaluate the innermost parentheses first. For example, 2 + [3 + (4 + 5)] = 2 + [3 + 9] = 2 + 12 = 14. Grouping Symbols Use parentheses, brackets, or curly braces to delimit the part of an expression you want evaluated first. If grouping symbols are nested, evaluate the expression in the innermost pair of grouping symbols first. The Associative Property of Addition Consider the evaluation of the expression (2+3)+4. We evaluate the expression in parentheses first. (2 + 3) + 4 = 5 + 4 = 9 Now, suppose we change the order of addition to 2 + (3 + 4). Then, 2 + (3 + 4) = 2 + 7 = 9. Although the grouping has changed, the result is the same. That is, (2 + 3) + 4 = 2 + (3 + 4). This property of addition of whole numbers is called the associate property of addition. Associate Property of Addition Let a, b, and c represent whole numbers. Then, (a + b) + c = a + (b + c). Because of the associate property of addition, when presented with a sum of three numbers, whether you start by adding the first two numbers or the last two numbers, the resulting sum is the same. The Additive Identity Imagine a number line visualization of the sum of four and zero; i.e., 4 + 0. In Figure 1.4, we start at zero, then draw an arrow of magnitude (length) four pointing to the right. Now, at the end of this arrow, attach a second arrow of length zero. Of course, that means that we remain right where we are, at 4. Hence the shaded dot at 4 is the sum. That is, 4 + 0 = 4. The Additive Identity Property The whole number zero is called the additive identity. If a is any whole number, the a + 0 = a. The number zero is called the additive identity because if you add zero to any number, you get the identical number back. Adding Larger Whole Numbers For completeness, we include two examples of adding larger whole numbers. Hopefully, the algorithm is familiar from previous coursework. Example 1 Simplify: 1, 234 + 498. Solution Align the numbers vertically, then add, starting at the furthest column to the right. Add the digits in the ones column, 4 + 8 = 12. Write the 2, then carry a 1 to the tens column. Next, add the digits in the tens column, 3 + 9 = 12, add the carry to get 13, then write the 3 and carry a 1 to the hundreds column. Continue in this manner, working from right to left Therefore, 1, 234 + 498 = 1, 732 Exercise Simplify: 1,286 + 349. Answer : 1635 Add three or more numbers in the same manner. Example 2 Simplify: 256 + 322 + 418. Solution Align the numbers vertically, then add, starting at the furthest column to the right. Add the digits in the ones column, 6 + 2 + 8 = 16. Write the 6, then carry a 1 to the tens column. Continue in this manner, working from right to left. Therefore, 256 + 322 + 418 = 996. Exercise Simplify: 256 + 342 + 283 Answer : 881 Subtraction of Whole Numbers The key idea is this: Subtraction is the opposite of addition. For example, consider the difference 7 - 4 depicted on the number line in Figure 1.5. If we were adding 7 and 4, we first draw an arrow starting at zero pointing to the right with magnitude (length) seven. Then, to add 4, we would draw a second arrow of magnitude (length) 4, attached to the end of the first arrow and pointing to the right. However, because subtraction is the opposite of addition, in Figure 1.5 we attach an arrow of magnitude (length) four to the end of the first arrow, but pointing in the opposite direction (to the left). Note that this last arrow ends at the answer, which is a shaded dot on the number line at 3. That is, 7 − 4 = 3. Note that subtraction is not commutative; that is, it make no sense to say that 7 − 5 is the same as 5 − 7. Subtraction is not associative. It is not the case that (9 − 5) − 2 is the same as 9 − (5 − 2). On the one hand, (9 − 5) − 2 = 4 − 2 = 2, but 9 − (5 − 2) = 9 − 3 = 6. Subtracting Larger Whole Numbers Much as we did with adding larger whole numbers, to subtract two large whole numbers, align them vertically then subtract, working from right to left. You may have to “borrow” to complete the subtraction at any step. Example 3 Simplify: 1, 755 − 328. Solution Align the numbers vertically, then subtract, starting at the ones column, then working right to left. At the ones column, we cannot subtract 8 from 5, so we borrow from the previous column. Now, 8 from 15 is 7. Continue in this manner, working from right to left. Therefore, 1, 755 − 328 = 1, 427. Exercise Simplify: 5,635 - 288. Answer : 5,347 Order of Operations In the absence of grouping symbols, it is important to understand that addition holds no precedence over subtraction, and vice-versa. Perform all additions and subtractions in the order presented, moving left to right. Let’s look at an example. Example 4 Simplify the expression . Solution This example can be trickier than it seems. However, if we follow the rule (perform all additions and subtractions in the order presented, moving left to right), we should have no trouble. First comes fifteen minus eight, which is seven. Then seven plus four is eleven. Exercise Simplify: . Answer : 23 Caution! Incorrect answer ahead! Note that it is possible to arrive at a different (but incorrect) answer if we favor addition over subtraction in Example 4. If we first add eight and four, then 15 − 8 + 4 becomes 15 − 12, which is 3. However, note that this is incorrect, because it violates the rule “perform all additions and subtractions in the order presented, moving left to right.” Applications — Geometry There are any number of applications that require a sum or difference of whole numbers. Let’s examine a few from the world of geometry. Perimeter of a Polygon In geometry a polygon is a plane figure made up of a closed path of a finite sequence of segments. The segments are called the edges or sides of the polygon and the points where two edges meet are called the vertices of the polygon. The perimeter of any polygon is the sum of the lengths of its sides. Example 5 A quadrilateral is a polygon with four sides. Find the perimeter of the quadrilateral shown below, where the sides are measured below. Solution To find the perimeter of the quadrilateral, find the sum of the lengths of the sides. Perimeter = 3 + 3 + 4 + 5 = 15 Hence, the perimeter of the quadrilateral is 15 yards. Exercise A quadrilateral has sides that measure in 4 in., 3 in., 5 in., and 5 in. Find the perimeter. Answer : 17 inches Example 6 A quadrilateral (four sides) is a rectangle if all four of its angles are right angles. It can be shown that the opposite sides of a rectangle must be equal. Find the perimeter of the rectangle shown below, where the sides of the rectangle are measured in meters. Solution To find the perimeter of the rectangle, find the sum of the four sides. Because opposite sides have the same length, we have two sides of length 5 meters and two sides of length 3 meters. Hence, Perimeter = 5 + 3 + 5 + 3 = 16. Thus, the perimeter of the rectangle is 16 meters. Exercise A rectangle has length 12 meters and width 8 meters. Find its perimeter. Answer : 40 meters Example 7 A quadrilateral (four sides) is a square if all four of its sides measures 18 centimeters. Find its perimeter. are equal and all four of its angles are right angles. Pictured below is a square having a side of length 12 feet. Find the perimeter of the square. Solution Because the quadrilateral is a square, all four sides have the same length, namely 12 feet. To find the perimeter of the square, find the sum of the four sides. Perimeter = 12 + 12 + 12 + 12 = 48 Hence, the perimeter of the square is 48 feet. Exercise A square has a side that measures 18 centimeters. Find its perimeter. Answer : 72 centimeters Application — Alternative Fuels Automobiles that run on alternative fuels (other than gasoline) have increased in the United States over the years. Example 8 Table 1.2 show the number of cars (in thousands) running on compressed natural gas versus the year. Create a bar chart showing the number of cars running on compressed natural gas versus the year. \| \| \| \| \| \| \| \| \| \| \| --- --- --- --- --- \| \| Year \| 1992 \| 1993 \| 1994 \| 1995 \| 1996 \| 1997 \| 1998 \| 1999 \| 2000 \| \| Number \| 23 \| 32 \| 4` \| 50 \| 60 \| 73 \| 78 \| 89 \| 101 \| Table 1.2: Number of vehicles (in thousands) running on compressed natural gas. Solution Place the years on the horizontal axis. At each year, sketch a bar having height equal to the number of cars in that year that are running on compressed natural gas. Scale the vertical axis in thousands. Exercise The following table shows the number of hybrid cars (in thousands) by country. \| \| \| --- \| \| Country \| Number \| \| U.S. \| 279 \| \| Japan \| 77 \| \| Canada \| 17 \| \| U.K. \| 14 \| \| Netherlands \| 11 \| Create a bar chart showing the number of cars versus the country of use. Example 9 Using the data in Table 1.2, crease a table that shows the differences in consecutive years, then create a line plot of the result. In what consecutive years did the United States see the greatest increase in cars powered by compressed natural gas? Solution Table 1.3 shows the differences in consecutive years. Next, craft a line graph. Place consecutive years on the horizontal axis. At each consecutive year pair, plot a point at a height equal to the difference in alternative fuel vehicles. Connect the points with straight line segments. Note how the line graph makes it completely clear that the greatest increase in vehicles powered by compressed natural gas occurred in the consecutive years 1996-1997, an increase of 13,000 vehicles. Exercise The following table show Alphonso's percentage scores on his examinations in mathematics. \| \| \| --- \| \| Exam \| Percentage \| \| Exam #1 \| 52 \| \| Exam #2 \| 45 \| \| Exam #3 \| 72 \| \| Exam #4 \| 889 \| \| Exam #5 \| 76 \| Construct a line graph of Alphonso's exam scores versus exam number. Exercises Sketch a number line diagram depicting the sum 3 + 2, as shown in Figure 1.2 in the narrative of this section. Sketch a number line diagram depicting the sum 3 + 5, as shown in Figure 1.2 in the narrative of this section. Sketch a number line diagram depicting the sum 3 + 4, as shown in Figure 1.2 in the narrative of this section. Sketch a number line diagram depicting the sum 2 + 4, as shown in Figure 1.2 in the narrative of this section. Sketch a number line diagram depicting the sum 4 + 2, as shown in Figure 1.2 in the narrative of this section. Sketch a number line diagram depicting the sum 4 + 3, as shown in Figure 1.2 in the narrative of this section. Sketch a number line diagram depicting the sum 2 + 5, as shown in Figure 1.2 in the narrative of this section. Sketch a number line diagram depicting the sum 4 + 5, as shown in Figure 1.2 in the narrative of this section. Sketch a number line diagram depicting the sum 4 + 4, as shown in Figure 1.2 in the narrative of this section. Sketch a number line diagram depicting the sum 3 + 3, as shown in Figure 1.2 in the narrative of this section. In Exercises 11-28, determine which property of addition is depicted by the given identity 28 + 0 = 28 53 + 0 = 53 24 + 0 = 24 93 + 0 = 93 (51 + 66) + 88 = 51 + (66 + 88) (90 + 96) + 4 = 90 + (96 + 4) 64 + 39 = 39 + 64 68 + 73 = 73 + 68 (70 + 27) + 52 = 70 + (27 + 52) (8 + 53) + 81 = 8 + (53 + 81) 79 + 0 = 79 42 + 0 = 42 10 + 94 = 94 + 10 55 + 86 = 86 + 55 47 + 26 = 26 + 47 62 + 26 = 26 + 62 (61 + 53) + 29 = 61 + (53 + 29) (29 + 96) + 61 = 29 + (96 + 61) Sketch a number line diagram depicting the difference 8−2, as shown in Figure 1.5 in the narrative of this section. Sketch a number line diagram depicting the difference 8−4, as shown in Figure 1.5 in the narrative of this section. Sketch a number line diagram depicting the difference 7−2, as shown in Figure 1.5 in the narrative of this section. Sketch a number line diagram depicting the difference 9−5, as shown in Figure 1.5 in the narrative of this section. Sketch a number line diagram depicting the difference 7−4, as shown in Figure 1.5 in the narrative of this section. Sketch a number line diagram depicting the difference 6−4, as shown in Figure 1.5 in the narrative of this section. Sketch a number line diagram depicting the difference 9−4, as shown in Figure 1.5 in the narrative of this section. Sketch a number line diagram depicting the difference 6−5, as shown in Figure 1.5 in the narrative of this section. Sketch a number line diagram depicting the difference 8−5, as shown in Figure 1.5 in the narrative of this section. Sketch a number line diagram depicting the difference 9−3, as shown in Figure 1.5 in the narrative of this section. In Exercises 39-50, simplify the given expression. 16 − 8+2 17 − 3+5 20 − 5 + 14 14 − 5+6 15 − 2+5 13 − 4+2 12 − 5+4 19 − 4 + 13 12 − 6+4 13 − 4 + 18 15 − 5+8 13 − 3 + 11 In Exercises 51-58, the width W and length L of a rectangle are given. Find the perimeter P of the rectangle. W = 7 in, L = 9 in W = 4 in, L = 6 in W = 8 in, L = 9 in W = 5 in, L = 9 in W = 4 cm, L = 6 cm W = 5 in, L = 8 in W = 4 cm, L = 7 cm W = 4 in, L = 9 in In Exercises 59-66, the length s of a side of a square is given. Find the perimeter P of the square. s = 25 cm s = 21 in s = 16 cm s = 10 in s = 18 in s = 7 in s = 3 in s = 20 in In Exercises 67-86, find the sum. 3005 + 5217 1870 + 5021 575 + 354 + 759 140 + 962 + 817 472 + (520 + 575) 318 + (397 + 437) 274 + (764 + 690) 638 + (310 + 447) 8583 + 592 5357 + 9936 899 + 528 + 116 841 + 368 + 919 (466 + 744) + 517 (899 + 996) + 295 563 + 298 + 611 + 828 789 + 328 + 887 + 729 607 + 29 + 270 + 245 738 + 471 + 876 + 469 (86 + 557) + 80 (435 + 124) + 132 In Exercises 87-104, find the difference. 3493 − 2034 − 227 3950 − 1530 − 2363 8338 − 7366 2157 − 1224 2974 − 2374 881 − 606 3838 − (777 − 241) 8695 − (6290 − 4233) 5846 − 541 − 4577 5738 − 280 − 4280 3084 − (2882 − 614) 1841 − (217 − 28) 2103 − (1265 − 251) 1471 − (640 − 50) 9764 − 4837 − 150 9626 − 8363 − 1052 7095 − 226 4826 − 1199 Water Subsidies. Since the drought began in 2007, California farms have received 439 million. How much total water subsidies have farmers received? Associated Press Times-Standard 4/15/09 War Budget. The 2010 Federal budget allocates 130 billion for the nation’s two wars. How much will the Department of Defense receive altogether? Associated Press Times-Standard 5/8/09 Sun Frost. Arcata, CA is home to Sun Frost, a manufacturer of highly efficient refrigerators and freezers. The AC model RF12 refrigerator/freezer costs 3,017. How much more does the R16 model cost? Source: www.sunfrost.com/retail pricelist.html Shuttle Orbit. The space shuttle usually orbits at 250 miles above the surface of the earth. To service the Hubble Space Telescope, the shuttle had to go to 350 miles above the surface. How much higher did the shuttle have to orbit? Earth’s Orbit. Earth orbits the sun in an ellipse. When earth is at its closest to the sun, called perihelion, earth is about 147 million kilometers. When earth is at its furthest point from the sun, called aphelion, earth is about 152 million kilometers from the sun. What’s the difference in millions of kilometers between aphelion and perihelion? Pluto’s Orbit. Pluto’s orbit is highly eccentric. Find the difference between Pluto’s closest approach to the sun and Pluto’s furthest distance from the sun if Pluto’s perihelion (closest point on its orbit about the sun) is about 7 billion kilometers and its aphelion (furthest point on its orbit about the sun) is about 30 billion kilometers. Sunspot Temperature. The surface of the sun is about 10,000 degrees Fahrenheit. Sunspots are darker regions on the surface of the sun that have a relatively cooler temperature of 6,300 degrees Fahrenheit. How many degrees cooler are sunspots? Jobs. The Times-Standard reports that over the next year, the credit- and debitcard processing business Humboldt Merchant Services expects to cut 36 of its 80 jobs, but then turn around and hire another 21. How many people will be working for the company then? Times-Standard 5/6/09 Wild tigers. The chart shows the estimated wild tiger population, by region. According to this chart, what is the total wild tiger population worldwide? Associated Press-Times-Standard 01/24/10 Pressure mounts to save the tiger. \| \| \| --- \| \| Region \| Tiger population \| \| India, Nepal and Bhutan \| 1650 \| \| China and Russia \| 450 \| \| Bangladesh \| 250 \| \| Sumatra (Indonesia) \| 400 \| \| Malaysia \| 500 \| \| other SE Asia \| 350 \| Pirate Attacks. The following bar chart tracks pirate attacks off the coast of Somalia. Source: ICC International Maritime Bureau, AP Times-Standard, 4/15/2009 a) How many pirate attacks were there in 2003, 2004, and 2005 combined? b) How many pirate attacks were there in 2006, 2007, and 2008 combined? c) How many more pirate attacks were there in 2008 than in 2007? Emily shows improvement on each successive examination throughout the term. Her exam scores are recorded in the following table. \| \| \| --- \| \| Exam \| Score \| \| Exam #1 \| 48 \| \| Exam #2 \| 51 \| \| Exam #3 \| 54 \| \| Exam #4 \| 59 \| \| Exam #5 \| 67 \| \| Exam #6 \| 70 \| a) Create a bar plot for Emily’s examination scores. Place the examination numbers on the horizontal axis in the same order shown in the table above. b) Create a table that shows successive differences in examination scores. Make a line plot of these differences. Between which two exams did Emily show the greatest improvement? Jason shows improvement on each successive examination throughout the term. His exam scores are recorded in the following table. \| \| \| --- \| \| Exam \| Score \| \| Exam #1 \| 34 \| \| Exam #2 \| 42 \| \| Exam #3 \| 45 \| \| Exam #4 \| 50 \| \| Exam #5 \| 57 \| \| Exam #6 \| 62 \| a) Create a bar plot for Jason’s examination scores. Place the examination numbers on the horizontal axis in the same order shown in the table above. b) Create a table that shows successive differences in examination scores. Make a line plot of these differences. Between which two exams did Jason show the greatest improvement? Answers 3 + 2 = 5. 3 + 4 = 7. 4 + 2 = 6. 2 + 5 = 7. 4 + 4 = 8. Additive identity property of addition. Additive identity property of addition. Associative property of addition Commutative property of addition Associative property of addition Additive identity property of addition. Commutative property of addition Commutative property of addition Associative property of addition 8 − 2 = 6. 7 − 2 = 5. 7 − 4 = 3. 9 − 4 = 5. 8 − 5 = 3. 10 29 18 11 10 18 P = 32 in P = 34 in P = 20 cm P = 22 cm P = 100 cm P = 64 cm P = 72 in P = 12 in 8222 1688 1567 1728 9175 1543 1727 2300 1151 723 1232 972 600 3302 728 816 1089 4777 6869 $518 million $738 5 million kilometers 3,700 degrees Fahrenheit 3600 a) Bar chart. b) Line plot of consecutive differences. The line plot of consecutive examination score differences. The largest improvement was between Exam #4 and Exam #5, where Emily improved by 8 points. 1.1: An Introduction to the Whole Numbers 1.3: Multiplication and Division of Whole Numbers
1465	https://www.wineverygame.com/words/meander.html	Meander - Meaning, Examples - Meander in a sentence WinEveryGame Scrabble Finder Lexicon Foreword Menu Meander verb To take a long, twisting, curving path To move about without focus or purpose; to roam or change easily and aimlessly noun A twist or curve in a river or other path (usually plural) A long, roundabout excursion, often with little purpose Usage Ah, the Lazy River: the most wonderful waterpark ride for the indolent among us. After a long day of riding gigantic slides, standing on even more gigantic lines, and trying not to swallow questionable pool water, the Lazy River's meanders are more welcome than the softest bedsheets. There's no fear involved, no skill, no complications; you just give yourself over to the current and let your inner tube meander where it will. And the best part? While you drift along past grinning tikis and sublimely fake palm trees, your thoughts are free to meander too, as aimless and relaxed as the Lazy River itself. We'll try not to get sidetracked here: meander commonly functions as a verb to refer to the taking of a twisty, purposelessly circuitous track. A path or road that meanders stretches out in bends, curves, and spirals, taking strange or unnecessary detours and maybe even doubling back on itself. Winding, lazily flowing rivers (even those found at waterparks) are among the most common things said to meander in this way, but the word can also be used to talk about the courses of things like country roads and mountain passes. Unless you're navigating a river, though, you'd probably encounter meander in reference to aimless, nonlinear roaming. If you set out on a walk with the intent to meander, you've already kind of defeated yourself, as the word implies a movement without any sort of purpose, motivation, or logic. Meandering is also often thought of as slow and easy, perhaps a result of a relaxed or confused mind. This is the kind of ambling you might do when you've got absolutely no clear thoughts; you're just sort of heading wherever the mood strikes, stopping to smell any flower that attracts you and letting your feet do the thinking. Meander is also commonly used a little more figuratively to describe rambling in speech or thought. When giving an important speech or trying to write a poignant letter, it's probably not a good idea to meander from topic to topic, as doing so might make your message boring or nonsensical. However, allowing your mind to meander from time to time can be both relaxing and stimulate creative thinking, as everyone who's ever tried freewriting knows. So when you're actually composing that important speech or letter, or even just trying to come up with ideas for fun, don't be afraid to let your thoughts wander in circles! Meander can sometimes function as a noun, too, although this type of usage isn't especially widespread. You might hear someone describe the bends and detours of a river or trail as its meanders (technically, you could call just one such bend a meander, but here the word's mostly used in the plural to refer to curvature as a whole). A meander can also be a long, unfocused trip taken at a gentle pace, full of distractions, changes in course, and general aimlessness. If you were to take your dog to the park and just follow him as he sniffed around, you'd be engaged in a meander. Example: Unaware that the trail would meander so much, we took much longer to finish our hike than we expected. Example: Watching the bear at the zoo meander around its enclosure made Jane feel a little sorry for it. Example: The long hallway's frustrating meanders seem to defy the laws of space and time. Example: To his surprise, Frank's meander brought him to a section of town where he'd never been. Origin There's a pretty good reason why meander is so often related to rivers. The word's inspiration comes from the Greek Maiandros, which refers to a long, twisting river in present-day Turkey now known as the River Menderes. The Romans would then derive from Maindros the Latin word meander, which meant "a bending, circuitous path." By the 1570s, English-speakers were using meander as a noun to figuratively refer to complexities and illogical, confusing developments, a usage which has since fallen out of fashion (though its essence lives on in the word's verb form). Meander's first uses in English as a verb are attributed to the early 17 th century, when it was used in reference to rivers to mean "to run in a bending, intricate course." Interestingly, by the 1820s this application of meander had engendered another use of the word as a noun, this time meaning "a person who moves on (or down) a river." Although that meaning would grow obsolete, it did, by the 1830s, give rise to the modern verb meaning of "moving about in loops and curves or without purpose or thought." Derivative Words Meandrous: This adjective characterizes something as wandering and undirected either in path or in focus. Example: Following the elderly professor's meandrous lecture became difficult. Example: Having spent an hour and a half driving over the meandrous country roads, I pulled out a map to make sure I was going in the right direction. Meanders: This simple present form of the verb meander is used when a singular third-person subject wanders aimlessly or follows an intricately bending path. Example: The stream meanders through woods and plains before reaching a lake. Meandered: This preterit form of meander is used when a subject rambled or moved in a twisting, intricate fashion in the past. Example: The lexicographer's various examples meandered from subject to subject. Meandering: The progressive form of meander is used when a subject is currently moving in a circuitous fashion or roaming without clear purpose. Example: Every morning finds the rooster meandering contentedly about his pen. In Literature From Thomas Hardy's Tess of the d'Urbervilles - A Pure Woman: Her journey back was rather a meander than a march. It had no sprightliness, no purpose; only a tendency. Along the tedious length of Benvill Lane she began to grow tired, and she leant upon gates and paused by milestones. In this description, Hardy uses meander to describe Tess's long, disillusioned walk after having her boots taken by strangers (who thought they had been left by a tramp). Comments Though also the name of a river, the ancient Greeks sometimes used the term Maiandros to refer to intricate, interlocking patterns, such as those made with bundles of string. This convention still has a little influence on the modern English meander: although somewhat uncommon, the word is sometimes used as a noun that describes a decorative arrangement of lines or curving shapes. Meander is also sometimes given as an alternate name for the Greek river god Maiandros (or Maeander). Mnemonic Meander wanders without a plan Tags Rivers, Water, Mythology, Greek References River God Meander Meander as ornamental pattern Bring out the linguist in you! What is your own interpretation of meander. Did you use meander in a game? Provide an example sentence or a literary quote. Meander Pronunciation /mee-YAAN-dehr/ Plural Meanders Synonyms verb: Wander, Wind, Twist, Roam, Amble, Mosey, Dawdle, Digress, Ramble, Shift noun: Twist, Turn, Curve, Bend, Arc, Shift, Stroll, Jaunt Antonyms verb: Aim, Focus, Concentrate, Fixate noun: Straightaway, Beeline, Straight Line Meander in Games Scrabble: Valid; 10 points Words With Friends: Valid; 12 points Lexicon Mawkish Overly and disgustingly emotionalPrevious WordMeanderNext WordMellifluous Smooth and lovely (as in sound) Connect with us Quick Feedback Email: wineverygame @ gmail Word of the day: Qi - From traditional Chinese culture, the quintessential, intrinsic force which flows continuously through all things. Scrabble Finder Words Foreword Like us on FacebookFollow us on Twitter Copyright K Publishing
1466	https://askfilo.com/user-question-answers-algebra-2/let-a-and-b-real-numbers-with-use-the-floor-and-or-ceiling-35383430353135	Question asked by Filo student Let a and b real numbers with a<b . Use the floor and /or ceiling functions to express the number of integers n that satisfy the inequality a≤n≤b Views: 5,672 students Updated on: Oct 25, 2023 Text SolutionText solutionverified iconVerified Step 1. The ceil function returns the smallest nearest integer which is greater than or equal to the specified number, whereas the floor function returns the largest nearest integer which is less than or equal to a specified value. Step 2. Let a and b be real numbers with a < b. Using the floor and/or ceiling functions, we can express the number of integers n that satisfy the inequality a ≤ n ≤ b as: a b a < b n a ≤ n ≤ b ⌊b⌋−⌈a⌉+1 where floor(b) is the largest integer less than or equal to b, and ceil(a) is the smallest integer greater than or equal to a. This formula is valid only when a < b. floor(b) b ceil(a) a a < b Students who ask this question also asked Views: 5,387 Topic: Algebra 2 View solution Views: 5,277 Topic: Algebra 2 View solution Views: 5,295 Topic: Algebra 2 View solution Views: 5,377 Topic: Algebra 2 View solution Stuck on the question or explanation? Connect with our tutors online and get step by step solution of this question. \| \| \| --- \| \| Question Text \| Let a and b real numbers with a<b . Use the floor and /or ceiling functions to express the number of integers n that satisfy the inequality a≤n≤b \| \| Updated On \| Oct 25, 2023 \| \| Topic \| All Topics \| \| Subject \| Algebra 2 \| \| Class \| High School \| \| Answer Type \| Text solution:1 \| Are you ready to take control of your learning? Download Filo and start learning with your favorite tutors right away! Questions from top courses Explore Tutors by Cities Blog Knowledge © Copyright Filo EdTech INC. 2025
1467	https://www.floridamuseum.ufl.edu/discover-fish/species-profiles/largemouth-bass/	Skip to main content Outreach Learning Resources Life on Earth Cultural Heritage Fossils + Paleontology The Value of Museums Discover Fishes Largemouth Bass Micropterus salmoides Not all fish called bass are in the same family, and the largemouth bass is in the sunfish family, Centrarchidae. Originally the largemouth bass were localized to the Eastern United States, but because of their popularity as a sport fish and their ability to thrive in a variety of environments, they have been successfully exported and populated, making them one of the most widely distributed fish in the world. Bass fishing is a $60 billion dollar industry. Order – Perciformes Family – Centrarchidae Genus – Micropterus Species – salmoides Common Names Common names in the English language are largemouth bass, bass, American black bass, bayou bass, bigmouth bass, black bass, bucket mouth, chub, green bass, green trout, lake bass, large-mouth bass, largemouth, largemouth black bass, line side, marsh bass, northern largemouth bass, Oswego bass, slough bass, southern largemouth, trout, and welchman. Other common names include achiga (Portuguese), achigan (French), achigan à grande bouche (French), bas dehanbozorg (Persian), bass wielkgebowy (Polish), biban cu gura mare (Rumanian), black-bass à grande bouche (French), bol’sherotyi chernyi okun’ (Russian), buraku basu (Japanese), fekete sügér (Hungarian), forelbaars (Dutch), forellenbarsch (German), huro (Spanish), isobassi (Finnish), khorshid mahi baleh kuchak (Persian), lakseabbor (Norwegian), lobina negra (Spanish), okounek pstruhový (Czech), Okuchibasu (Japanese), ostracka (Czech), ostracka lososovitá (Slovak), perca americana (Spanish), perche d’amérique (French), perche noire (French), perche truite (French), perche truitée (French), persico trota (Italian), stormundet black bass (Danish), stormundet ørredaborre (Danish), tam suy lo ue (Cantonese), zwarte baars (Dutch), and öringabborre (Swedish). Importance to Humans Some surveys list bass fishing as the number one participation sport, while others place it in the upper five. According to U.S. Fish and Wildlife Service statistics, 43 percent of freshwater anglers fish for bass, 34 percent fish for panfish, 30 percent for catfish/bullhead, 30 percent for trout and 28 percent for crappie. Fishing is big business and a major American institution. America’s 30 million bass anglers are the basis of a $60 billion industry with one in every five fishing every year. Fishing The largemouth bass is among the most sought after species of freshwater game fish in the world. Due to its broad range and popularity as a sport fish, numerous fishing methods have been devised to catch this fish. Numerous artificial lures have been developed to entice these fish into biting including plastic worms, jigs, plugs, and spinners. Among the most exciting methods of fishing for the largemouth bass involves surface plugs which can incite vicious strikes on the surface where they are visible to the angler. Live bait fishing is also popular among bass fishers. Night crawlers, leeches, crawfish, and minnows are all commonly used live baits in different regions of the U.S. Evening and morning times are usually the best times when fishing for the largemouth bass. The largemouth’s liking for heavy cover makes it a challenge to land. A hooked largemouth usually heads for the surface, the opens it mouth wide, shaking its head or jumping in an attempt to throw, or get rid, of the hook. Once it has done this, it will dive under the cover and begin to wrap the line around logs and weeds located in the water. Largemouth bass flesh is moderately firm and has a mild flavor. If taken from lakes where the predominant cover is weeds, the flesh may have a grassy taste. Conservation The largemouth bass is widely distributed and as such, is not listed as endangered or vulnerable with the World Conservation Union (IUCN). The IUCN is a global union of states, governmental agencies, and non-governmental organizations in a partnership that assesses the conservation status of species. > Check the status of the largemouth bass at the IUCN website. Geographical Distribution This bass is one of the most widely distributed fishes in the world. This is due in large part to the popularity of bass as a sport fish which has lead to the introduction of largemouth bass populations into many areas where they are not native. Their current range includes the U.S., South Africa, Europe, Guam, Japan, Lebanon, New Zealand, and the Philippines. The original range of the largemouth bass is most of the eastern half of the United States, however it is now found generally in the majority of the United States, including swamps, ponds, lakes, reservoirs, creeks and large rivers. Habitat The largemouth bass lives in all types of water, including swamps, ponds, lakes, reservoirs, creeks and large rivers. The bass can even be found in estuaries. It prefers weedy oxbows and clears floodplain lakes. Since it is generally a warm water (81-86°F, (27.2°-30°C)) fish, it is seldom found at depths of more than 18.8 feet. During the winter, largemouth bass generally will move into deeper waters. In the spring, largemouth’s migrate into bays that have warmed up sooner than that of the main body of water. During the day, largemouth bass may cruise above aquatic plants at depths of 3.1-9.4 ft, or lie under lily pads or in the shade of overhanging trees, piers, or brush. In the evening hours, largemouth bass tend to move into shallow water to feed. After night falls, they return to deeper water, where they rest on the bottom under logs or trees. In cold temperate climates, largemouth bass generally move into deeper waters during the winter months followed by movement to warmer, shallow waters in the springtime. Biology Distinctive FeaturesThe largemouth bass has a large, slightly sloping mouth. Its body is slender to robust, slightly compressed laterally, and oval in cross section. The corner of the mouth extends past the eye, hence its common name. The two recognized subspecies of the largemouth bass are the northern largemouth (Micropterus salmoides salmoides) and the Florida largemouth (Micropterus salmoides floridanus). The Florida largemouth has slightly smaller scales than the largemouth bass in relation to the size of its body. It has 69 to 73 scales along the lateral line, compared to 59 to 65 on the northern largemouth. It grows faster and reaches larger sizes in warm waters than the northern largemouth, leading to its popularity in stocking efforts across the U.S. The species most often confused with the largemouth bass is the spotted bass (Micropterus punctulatus). Sometimes referred to as the Kentucky bass, its range overlaps with that of the largemouth bass and has similar coloration patterns. It can be distinguished from the largemouth bass in several ways. The spotted bass has its spiny and soft dorsal fins fused. Also, its upper jaw is noticeably smaller than that of the largemouth bass and does not extend beyond the eye. Largemouth bass. Photo © George Burgess Florida largemouth bass. Photo © George Burgess Spotted bass. Photo © Noel Burkhead Smallmouth bass. Photo © Noel Burkhead Smallmouth bass. Photo © Noel Burkhead The smallmouth bass, M. dolomieu, is another popular sport fish. It is restricted to a more northern range than the largemouth with its southernmost populations in northern Alabama and Oklahoma. The smallmouth bass may be distinguished from the largemouth by several characteristics including its bronze to brown coloration, its mottled pattern of dark bands, and its noticeably smaller mouth. Other similar species found in regions of the southeastern U.S. include shoal bass (M. cataractae), red-eyed bass (M. coosae), and Suwannee bass (M. notius), while in Texas there is a population of the Guadeloupe bass (M. treculi). The mouths of all of these species are smaller than the mouth of the largemouth bass. These four species also tend to be smaller in length with the largest species reaching a maximum length of 18.5 inches (47 cm). ColorationThe back and head are dark green to light green in color with lighter sides and a whitish belly and underside of the head. A prominent lateral stripe may be seen running from the snout through eye to the base of the tail. Towards the tail, there is a series of blotches of varying size. These blotches evolve into a solid, even stripe on the caudal peduncle. The eye is golden brown. Vertical fins lightly pigmented, paired fins generally clear; caudal fin alike in young and adult. Adults from mud-bottom lakes are dark olive brown to black, with markings hardly distinguishable. Males in breeding condition tend to be darker in overall color. DentitionThe largemouth bass has well-developed pharyngeal jaws consisting of six major pads of caniform teeth in the upper pharynx and two pads in the lower pharynx. There are no teeth present on the tongue. Size, Age, and GrowthThe average length is 18 inches but the largemouth bass may attain a length of 24 inches or more. The world record largemouth bass was caught in Montgomery Lake, Georgia in 1932. The fish weighed 22 pounds, 4 ounces (10.09 kg). Females may reach a maximum age of 9 years, while the males reach a maximum of 6 years. Food HabitsLargemouth bass may consume small fish, insects, mosquitoes, blackfly larvae, mayfly nymphs, worms, adult insects, mussels, crayfish, snails, tadpoles, frogs, small fish, salamanders, mice, turtles. In general largemouth bass feed at all hours, but most often in the early morning or late in the day. In some cases, the prey is not completely swallowed up initially; it is caught and held in the jaws and then it is sucked in. ReproductionThe largemouth bass generally does not spawn at temperatures much below 64°F (17.8°C). Optimal spawning conditions is when the water temperature is around 18.9°-20.0°C. which correlates to between February (for the extreme southern end of its range) and July (in northern latitudes). The selection of nest sites by the male generally begins when water temperatures reach 60°F (15.6°C). The male largemouth bass begins the spawning process by clearing out a small depression in the bottom substrate. The nest are cleared in fairly shallow water on bottoms composed of sand, gravel, or pebbles. The male then attracts a gravid (egg laden) female and together they pass over his bed, releasing sperm and eggs. She may release from 2,000-145,000 eggs with much of this variation attributed to her physical size. Shortly after spawning, the female departs and the male is left to guard the developing eggs. The male will fast while he defends the eggs from predators and debris until they hatch 3-7 days later. Upon hatching, young largemouth bass fry are transparent and 3 mm in length. After hatching, the young remain in the bottom of the nest until the yolk is absorbed from the eggs, which usually lasts a period of 6 to 7 days. They then rise from the nest, and begin feeding on zooplankton and schooling, where they are 5.9- 6.3 mm long and have changed color to become pale green. As young largemouth bass begin to grow, their diet changes to insects and other small fishes and then eventually switches to larger fishes, salamanders, crayfish, and a variety of other small organisms. The largemouth bass may form hybrid fish by spawning with the smallmouth bass, rock bass, bluegill, warmouth, and black crappie. PredatorsLargemouth bass larva and juveniles may fall prey to yellow perch (Perca flavescens), walleye (Stizostedion vitreum), northern pike (Esox lucieus), and musky as well as water birds such as great blue herons (Ardea herodias) and kingfishers (Ceryle alcyon). However, adult largemouth bass are considered top predators in many habitats where they reside and rarely become prey items due to their size, swimming speed, and protective dorsal spines. ParasitesAn enemy of the largemouth bass is the bass tapeworm, which attacks the reproductive organs, resulting in sterile bass. The ectoparasitic protozoan, Scyphidia tholiformis, is one of the most common parasites. The fungus Saprolegnia may attack dead eggs. Parasites such as protozoa, copepods, roundworms, tapeworms, flatworms and leeches are common in the largemouth bass. Taxonomy Lacépède was the first to describe this fish in 1802. He gave it the scientific name Labrus salmoides. In 1876, Nelson described this fish as Micropterus nigricans and in 1878 Jordan described it as Micropterus pallidus. Finally in 1884, Forbes described the fish as belonging to Micropterus salmoides, which is currently the valid scientific name. There are two recognized subspecies the northern largemouth bass (Micropterus salmoides salmoides) and the Florida largemouth bass (Micropterus salmoides floridanus). In 1965, Smith described the subspecies Micropterus salmoides salmoides, while in 1949 Bailey and Hubbs described a second subspecies as Micropterus salmoides floridanus. Prepared by: Andrew Bridges and Cathleen Bester 1 / 5
1468	https://headconf.org/wp-content/uploads/pdfs/5301.pdf	Using COCA to Foster Students’ Use of English Collocations in Academic Writing Mansour, Deena Mohammad The American University in Cairo, Egypt. Abstract Using Corpora in foreign language teaching is on of the very efficient ways in proposing solutions for pedagogical issues, either through adopting a corpus-based teaching approach or through using the corpus in developing material. Developing academic writing competence is a challenging pedagogical issue for English teachers as well as learners. Second language learners come across considerable challenges as they are developing their academic writing competence. One of these challenges is the use of English collocations since that learning about collocations is of a crucial importance. Hence, due to the wide variety of collocations and their different types, one of the challenges that face both L2 learners of English as well as teachers is how to use English collocations successfully in academic writing due to the fact that collocational awareness and proficiency is a necessity for competent written discourse especially in academic writing. Therefore, academic writing proficiency is one of the major problematic areas that university students encounter. This paper focuses on how to direct students to a more effective use of Corpus of Contemporary American English (COCA) in order to foster their learning process of different English collocations, and use English collocations accurately in their academic writings. This paper illustrates search mechanisms and search strings that could be implemented effectively on COCA to get more representative and accurate collocation results. Keywords: corpus, academic writing, collocations, COCA 3rd International Conference on Higher Education Advances, HEAd’17 Universitat Politecnica de Val encia, Valencia, 2017 DOI: This work is licensed under a Creative Commons License CC BY-NC-ND 4.0 Editorial Universitat Polit ecnica de Val` encia 600 600 Using COCA to Foster Students’ Use of English Collocations in Academic Writing 1. Introduction: Online Corpora & COCA The existence of online corpora in the field of applied linguistics significantly changed all the perceptions and approaches of most researchers in the modern age. Using online corpora in teaching and developing material is no longer a luxury as people have thought; it became a necessity and an efficient way to take second language pedagogy into the new digital era. Mark Davis’ 520-million-word Corpus of Contemporary American English (COCA) is the richest freely searchable online corpus for its wide range of texts that are from different genres. It can be efficiently used to propose solutions for many pedagogical challenges that face language teachers and leaners with respect to English collocations. As far as academic writing is concerned, English learners find difficulties when using collocations. There are two common mistakes university students tend to make in academic writing: incorrect or inaccurate use of collocations or using a collocation that is too informal to be used in an academic writing. In this paper, the researcher will introduce different search mechanisms and search strings that enhanced university students’ ability to analyze their collocation errors and find accurate collocates in the academic sub corpus of COCA. 2. Literature Review Teaching academic writing has been a major concern of academic research recently. Some studies suggested new frameworks and techniques for teaching academic writing, and some other studies directed the attention to the use of corpora in teaching writing in general, and academic writing in particular. Some studies conducted experimental research to test the corpus-based teaching approach on a number of students and observe the impact of using the corpora on their language proficiency. Other studies used the corpus to develop material for teaching English and created word lists to help students improve their language competence. Alex Gilmore (2009) found that free online corpora helped students have more natural writing. He investigated whether training a number of intermediate Japanese university students to use the BNC and COBUILD online corpora will effectively enhance writing second drafts of their academic essays or not. He also explored their reactions and preferences after using these online corpora. In this study, the corpora were used basically to help students correct grammatical problems occurred in their first draft. The results showed that students produced more natural texts in the second draft, and the problematic areas highlighted in the first draft were mostly corrected. Gilmore concluded that the use of the corpora can have an effective role in enhancing writing skills. He also found that using the online corpora does appeal more to visual students who are more analytic and logical. However, L2 learners should receive an appropriate and sufficient training in order to be able to use the corpora effectively. 601 Mansour, D. M. Applying the use of corpus-based teaching approach on an experimental group and investigating the impact on students’ performance is a significant turning point in the literature of corpus linguistics. Jafarpour et al (2013) compared the effect of a corpus-based teaching of collocations applied on an experimental group to that of traditional approach applied on a control group. The researchers focused on teaching collocations of near-synonymous pairs. They concluded that the corpus-based approach in teaching collocations outperform the traditional ways. The researchers used concordance materials retrieved from the BNC corpus, and collocations have been taught in context. Jafarpour et al found that their corpus-based teaching approach had a significant effect on the students’ knowledge and competence of collocations in the experimental group. According to Jafarpour et al (2013), student in both groups achieved a considerable progress; however, students in the experimental group showed a significant improvement in their writing proficiency after using the corpus producing more authentic language. Likewise, Yoon (2008) investigated the progress in students’ writing performance with corpus use, and how their writing competence has been influenced. Yoon investigated six case studies of L2 students in an English for Academic Purposes course. The results showed that the use of the corpus had a significant and instant impact on the students’ writing competence and helped them find immediate solutions for language problems they might have encountered. The participants’ writing process went through some minor changes after using the corpus, yet they are effective. In addition, this corpus-based approach developed the students’ perceptions of lexicogrammar and language awareness. Another significant finding in Yoon’s study was that the students became more autonomous learners taking more responsibility of their learning process and correcting their own mistakes due to their corpus experience. Focusing on English for academic purposes (EAP), Lee and Swales (2006) conducted their qualitative research testing the use of corpora in a customized EAP course for doctoral students using a special corpus for academic writing and speaking. Students in this study compared their own academic writing to those of more experienced academic writers, and they provided reports discussing their findings of these comparisons and how the use of the corpus can influence their academic writing in the future. As the authors stated, this approach can be specifically of benefit for academic students who are writing their theses and need to develop their styles and academic competence through looking up an academic specialized corpus of their field (Lee & Swales, 2006). The participants’ feedback was very positive that they found that using the corpus is ‘confidence-building and empowering’, and they also found the corpora much more useful and effective than reference and grammar books. The study proved to be very efficient that most of the participants bought their own versions of Wordsmith concordancing software to use it after the course. 602 602 Using COCA to Foster Students’ Use of English Collocations in Academic Writing In a more field-oriented study, Walker (2011) used the Bank of English corpus (BoE) and the British National Commercial Corpus (BNCc) for teaching business English. Walker stated that business English teachers encounter many questions that require answers and conducting some research. Walker suggested that such questions can be answered through investigating the ‘collocational behavior’ of specific key lexis; this can reveal all the different senses of a word and the other words that co-occur with it in a specific context. Applying a corpus-based teaching approach in two case studies, Walker reaches a conclusion that most collocations are not merely random combinations of words; it is the significantly frequent collocates that show the different meanings and uses of a word. In addition, using corpus here helps to reveal whether there are any negative connotations associated with the collocation (Walker, 2011). Hence, combining collocation lists would play an invaluable role in improving learners’ language competence. Very specifically, Daskalovska (2015) investigated the activities based on corpus for learning verb-adverb collocations in a comparison to the course book regular activities. The results of the study show that the students who used the online concordancer performed much better in the test they took. The findings of this study highlighted how motivating and engaging the corpus-based activities were to the students. The researcher concluded here that the online corpus is a very dynamic and influential tool that has its positive impact on the L2 learners. Literature illustrates how influential is using online corpora in teaching English as a second language. However, this paper takes a different and more practical approach regarding corpus-based teaching which is exploring different mechanisms and search strings that could effectively retrieve more accurate results when looking up a collocate in an academic context. 3. Collocates Search on COCA COCA search options give users multiple chances to check different types of collocations. The most common colocations that are usually looked up: verb-noun collocations, adjective-noun collocations, adverb-verb collocations, verb-preposition collocations, and adverb-adjective collocations. The current paper discusses a number of search mechanisms that are used on COCA and yielded positive results. The search mechanisms proposed in the current paper have been examined in two different academic language classes. They have been first examined for error analysis in translation classes, and second in an academic essay writing class. There are different sub corpora within the COCA; the one that I focus on in the present paper is the academic sub corpus. The suggested search strings in the present paper raised my translation students’ awareness of their errors and helped them correct these errors and find accurate collocations when they translate academic texts from 603 Mansour, D. M. Arabic into English. Likewise, using the proposed search strings fostered the students’ writing performance in my academic writing class. There are two ways of conducting collocate searches on COCA: either to use the default list option or to use the collocates option to limit the results. Table 1 illustrates possible search strings that can be conducted to check different types of collocations using the collocates option and the (POS) list which is available in the search engine on the corpus to specify the part of speech that will to be looked up. Academic sub corpus must be checked first in order to get specific results from the academic corpus. Table 1. Proposed search strings using the list display option Collocation Search String Examples verb-noun adjective-noun adverb-verb adverb-adjectve verb-preposition [verb].[nn] ex. decline _nn [adjective].[nn] ex. _j results [adverb].[v] ex. _r update [adverb].[j] ex. _r significant [verb].[prep] ex. rise _i (decline) with age/(decline) over time/(decline) the offer (conduct) research/ investigations/ interviews current/preliminary/ significant (results) important/ profound/ long-term (implications) regulary/ constantly/ peri-odically (update) critically/ carefully/ properly (evaluate) statistically/particularly/potentially (significant) perfectly/ barely/minimally (adequate) (rise) to/in/from (express) in/about/to 604 604 Using COCA to Foster Students’ Use of English Collocations in Academic Writing The search mechanisms and search strings proposed in the current paper have been practiced with my translation students as well as the academic essay writing students. Applying the suggested search stings on COCA enhanced translation students’ performance in their translations from Arabic into English. They started to use accurate collocates in their translations of academic texts. Similarly, the academic essay writing students started to use accurate collocates which are also more academic-related. The errors of using English collocates have been significantly declined in both groups after implementing the suggested search mechanisms on COCA. Another mechanism could be implemented to look up collocations using the collocates display option available on the corpus. Using this option, there is a tab for the target word which the student/learner needs to look up its collocates. The part of speech of the target word should be specified using the (POS) list. Likewise, the part of speech of the collocates that would be looked up should be specified using the (POS) list as it is displayed in Figure 1. The valuable option here is that the student/learner could limit the search to either the collocates occurring before the word or the words occurring after, or both. In Figure 1, I specified the search to the two words occurring after the target word. Figure 1. looking up collocations on COCA using the collocates display option There is another option that many students find valuable when looking up collocates. Many students find difficulty figuring out whether a word/phrase could be used academically or not. Using square brackets and the equal sign (=), learners could get synonyms for the word they need with a specific collocate and limit the search to the academic sub corpus. Figure 2 displays the search for synonyms for the word awesome, which should not be used in an academic writing, and the target synonyms should collocate with the word result. The 605 Mansour, D. M. academic sub corpus is selected to limit the results. Figure 3 displays the results of the search showing synonymous words for awesome which can collocate with results. Figure 2. Using the equal sign (=) to look up synonyms and their collocates Figure 3. The results of using the search string [=awesome] results 4. Conclusion The use of online corpora as a tool in teaching languages is unlimited; new applications and practices come up every day to serve language pedagogy. COCA is one the richest online resources that could be efficiently used to propose solutions for many pedagogical challenges that face English language teachers and leaners regarding the use of collocations. This papers focused on practical use of search mechanisms and search strings with translation students and academic essay writing students. Both groups have shown significant development of using collocations when they translate academic texts from Arabic into English, or when they write an academic essay. 606 606 Using COCA to Foster Students’ Use of English Collocations in Academic Writing References Daskalovska, N. (2015). Corpus-based versus Traditional Learning of Collocation. Computer Assisted Language Learning,28(2), 130-144. Gilmore, A. (2009;2008;). Using online corpora to develop students' writing skills. ELT Journal, 63(4), 363-372. doi:10.1093/elt/ccn056 Jafarpour, A. A., Hashemian, M., & Alipour, S. (2013). A Corpus-Based Approach toward Teaching Collocation of Synonyms. Theory and Practice in Language Studies, 3(1), 51-60. doi:10.4304/tpls.3.1.51-60 Lee, D., & Swales, J. (2006). A corpus-based EAP course for NNS doctoral students: Moving from available specialized corpora to self-compiled corpora. English for Specific Purposes, 25(1), 56-75. doi:10.1016/j.esp.2005.02.010 Walker, C. (2011). How a corpus-based study of the factors which influence collocation can help in the teaching of business English. English for Specific Purposes,30(2), 101-112. doi:10.1016/j.esp.2010.12.003 Yoon, H. (2008). More Than a Linguistic Reference: The Influence of Corpus Technology on L2 Academic Writing. Language Learning & Technology, 12(2), 31-48. 607
1469	https://en.wikipedia.org/wiki/Total_ring_of_fractions	Total ring of fractions - Wikipedia Jump to content [x] Main menu Main menu move to sidebar hide Navigation Main page Contents Current events Random article About Wikipedia Contact us Contribute Help Learn to edit Community portal Recent changes Upload file Special pages Search Search [x] Appearance Appearance move to sidebar hide Text Small Standard Large This page always uses small font size Width Standard Wide The content is as wide as possible for your browser window. Color (beta) Automatic Light Dark This page is always in light mode. Donate Create account Log in [x] Personal tools Donate Create account Log in Pages for logged out editors learn more Contributions Talk [x] Toggle the table of contents Contents move to sidebar hide (Top) 1 Definition 2 Examples 3 The total ring of fractions of a reduced ring 4 Generalization 5 Citations 6 References Total ring of fractions [x] 4 languages Deutsch Español 日本語 Українська Edit links Article Talk [x] English Read Edit View history [x] Tools Tools move to sidebar hide Actions Read Edit View history General What links here Related changes Upload file Permanent link Page information Cite this page Get shortened URL Download QR code Expand all Edit interlanguage links Print/export Download as PDF Printable version In other projects Wikidata item From Wikipedia, the free encyclopedia \| Algebraic structure → Ring theory Ring theory \| \| show Basic concepts Rings• Subrings• Ideal• Quotient ring• Fractional ideal• Total ring of fractions• Product of rings•Free product of associative algebras• Tensor product of algebras Ring homomorphisms • Kernel• Inner automorphism• Frobenius endomorphism Algebraic structures • Module• Associative algebra• Graded ring• Involutive ring• Category of rings• Initial ringZ{\displaystyle \mathbb {Z} }• Terminal ring0=Z/1 Z{\displaystyle 0=\mathbb {Z} /1\mathbb {Z} } Related structures • Field• Finite field• Non-associative ring• Lie ring• Jordan ring• Semiring• Semifield \| \| show Commutative algebra Commutative rings• Integral domain• Integrally closed domain• GCD domain• Unique factorization domain• Principal ideal domain• Euclidean domain• Field• Finite field• Polynomial ring• Formal power series ring Algebraic number theory • Algebraic number field• Integers modulo n• Ring of integers• p-adic integersZ p{\displaystyle \mathbb {Z} {p}}• p-adic numbersQ p{\displaystyle \mathbb {Q} {p}}• Prüfer p-ringZ(p∞){\displaystyle \mathbb {Z} (p^{\infty })} \| \| show Noncommutative algebra Noncommutative rings• Division ring• Semiprimitive ring• Simple ring• Commutator Noncommutative algebraic geometry Free algebra Clifford algebra • Geometric algebraOperator algebra \| \| v t e \| In abstract algebra, the total quotient ring or total ring of fractions is a construction that generalizes the notion of the field of fractions of an integral domain to commutative ringsR that may have zero divisors. The construction embedsR in a larger ring, giving every non-zero-divisor of R an inverse in the larger ring. If the homomorphism from R to the new ring is to be injective, no further elements can be given an inverse. Definition [edit] Let R{\displaystyle R} be a commutative ring and let S{\displaystyle S} be the set of elements that are not zero divisors in R{\displaystyle R}; then S{\displaystyle S} is a multiplicatively closed set. Hence we may localize the ring R{\displaystyle R} at the set S{\displaystyle S} to obtain the total quotient ring S−1 R=Q(R){\displaystyle S^{-1}R=Q(R)}. If R{\displaystyle R} is a domain, then S=R−{0}{\displaystyle S=R-{0}} and the total quotient ring is the same as the field of fractions. This justifies the notation Q(R){\displaystyle Q(R)}, which is sometimes used for the field of fractions as well, since there is no ambiguity in the case of a domain. Since S{\displaystyle S} in the construction contains no zero divisors, the natural map R→Q(R){\displaystyle R\to Q(R)} is injective, so the total quotient ring is an extension of R{\displaystyle R}. Examples [edit] For a product ringA × B, the total quotient ring Q(A × B) is the product of total quotient rings Q(A) × Q(B). In particular, if A and B are integral domains, it is the product of quotient fields. For the ring of holomorphic functions on an open setD of complex numbers, the total quotient ring is the ring of meromorphic functions on D, even if D is not connected. In an Artinian ring, all elements are units or zero divisors. Hence the set of non-zero-divisors is the group of units of the ring, R×{\displaystyle R^{\times }}, and so Q(R)=(R×)−1 R{\displaystyle Q(R)=(R^{\times })^{-1}R}. But since all these elements already have inverses, Q(R)=R{\displaystyle Q(R)=R}. In a commutative von Neumann regular ring R, the same thing happens. Suppose a in R is not a zero divisor. Then in a von Neumann regular ring a=axa for some x in R, giving the equation a(xa−1)=0. Since a is not a zero divisor, xa=1, showing a is a unit. Here again, Q(R)=R{\displaystyle Q(R)=R}. In algebraic geometry one considers a sheaf of total quotient rings on a scheme, and this may be used to give the definition of a Cartier divisor. The total ring of fractions of a reduced ring [edit] Proposition—Let A be a reduced ring that has only finitely many minimal prime ideals, p 1,…,p r{\displaystyle {\mathfrak {p}}{1},\dots ,{\mathfrak {p}}{r}} (e.g., a Noetherian reduced ring). Then Q(A)≃∏i=1 r Q(A/p i).{\displaystyle Q(A)\simeq \prod {i=1}^{r}Q(A/{\mathfrak {p}}{i}).} Geometrically, Spec⁡(Q(A)){\displaystyle \operatorname {Spec} (Q(A))} is the Artinian scheme consisting (as a finite set) of the generic points of the irreducible components of Spec⁡(A){\displaystyle \operatorname {Spec} (A)}. Proof: Every element of Q(A) is either a unit or a zero divisor. Thus, any properidealI of Q(A) is contained in the set of zero divisors of Q(A); that set equals the union of the minimal prime ideals p i Q(A){\displaystyle {\mathfrak {p}}{i}Q(A)} since _Q(A) is reduced. By prime avoidance, I must be contained in some p i Q(A){\displaystyle {\mathfrak {p}}{i}Q(A)}. Hence, the ideals p i Q(A){\displaystyle {\mathfrak {p}}{i}Q(A)} are maximal ideals of Q(A). Also, their intersection is zero. Thus, by the Chinese remainder theorem applied to Q(A), Q(A)≃∏i Q(A)/p i Q(A){\displaystyle Q(A)\simeq \prod {i}Q(A)/{\mathfrak {p}}{i}Q(A)}. Let S be the multiplicatively closed set of non-zero-divisors of A. By exactness of localization, Q(A)/p i Q(A)=A[S−1]/p i A[S−1]=(A/p i)[S−1]{\displaystyle Q(A)/{\mathfrak {p}}{i}Q(A)=A[S^{-1}]/{\mathfrak {p}}{i}A[S^{-1}]=(A/{\mathfrak {p}}{i})[S^{-1}]}, which is already a field and so must be Q(A/p i){\displaystyle Q(A/{\mathfrak {p}}{i})}. ◻{\displaystyle \square } Generalization [edit] If R{\displaystyle R} is a commutative ring and S{\displaystyle S} is any multiplicatively closed set in R{\displaystyle R}, the localizationS−1 R{\displaystyle S^{-1}R} can still be constructed, but the ring homomorphism from R{\displaystyle R} to S−1 R{\displaystyle S^{-1}R} might fail to be injective. For example, if 0∈S{\displaystyle 0\in S}, then S−1 R{\displaystyle S^{-1}R} is the trivial ring. Citations [edit] ^Matsumura 1980, p.12. ^Matsumura 1989, p.21. References [edit] Matsumura, Hideyuki (1980), Commutative algebra (2nd ed.), Benjamin/Cummings, ISBN978-0-8053-7026-3, OCLC988482880 Matsumura, Hideyuki (1989), Commutative ring theory, Cambridge University Press, ISBN978-0-521-36764-6, OCLC23133540 Retrieved from " Categories: Commutative algebra Ring theory This page was last edited on 29 January 2024, at 16:20(UTC). Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply. By using this site, you agree to the Terms of Use and Privacy Policy. Wikipedia® is a registered trademark of the Wikimedia Foundation, Inc., a non-profit organization. Privacy policy About Wikipedia Disclaimers Contact Wikipedia Code of Conduct Developers Statistics Cookie statement Mobile view Edit preview settings Search Search [x] Toggle the table of contents Total ring of fractions 4 languagesAdd topic
1470	https://www.aproged.pt/biblioteca/planeandsolidgeometry.pdf	PLANE AND SOLID GEOMETRY BY C. A. HART INSTRUCTOR OF MATHEMATICS, WADLEIGH HIGH SCHOOL, NEW YORK CITY AND DANIEL D. FELDMAN HEAD OF DEPARTMENT OF MATHEMATICS, ERASMUS HALL HIGH SCHOOL, BROOKLYN WITH THE EDITORIAL COOPERATION OF J. H. TANNER AND VIRGIL SNYDER PROFESSORS OF MATHEMATICS IN CORNELL UNIVERSITY NEW YORK ":" CINCINNATI ":" CHICAGO AMERICAN BOOK COMPANY H-3 Copyright, 1911, 1912, by AMERICAN BOOK COMPANY. Entered at Stationers' Hall, London. H.-F. PLANK AND SOLID GEOMETRY. W. P. I v" PREFACE This book is the outgrowth of an experienceof many years in the teaching of mathematics in secondary schools. The text has been used by many different teachers, in classes of all stages of development, and under varying conditions of sec-ondary school teaching. The proofs have had the benefit of the criticisms of hundreds of experienced teachers of mathe-matics throughout the country. The book in its present form is therefore the combined product of experience,classroom test, and severe criticism. The following are some of the leading features of the book : Tlie student is rapidly initiated into the subject. Definitions are given only as needed. The selection and arrangement of theorems is such as to meet the generaldemand of teachers, as expressed through the Mathe-matical Associations of the country. Most of the proofs have been given in full. In the Plane Geometry, proofs of some of the easier theorems and construc-tions are left as exercises for the student, or are given in an incomplete form. In the Solid Geometry, more proofs and parts of proofs are thus left to the student ; but in every case in which the proof is not complete, the incompleteness is specifically stated. The indirect method of proof is consistentlyapplied. The usual method of proving such propositions,for example, as Arts. 189 and 415, is confusing to the student. The method used here is convincing and clear. The exercises are carefully selected. In choosing exercises, each of the following groups has been given due importance : (a) Concrete exercises,including numerical problems and problems of construction. (6) So-called practicalproblems, such as indirect measure-ments of heights and distances by means of equal and similar triangles, drawing to scale as an application of similar figures, ..problems from physics, from design, etc, 304026 iv PREFACE (c)Traditional exercises of a more or less abstract nature. The arrangement of the exercises is pedagogical Compara-tively easy exercises are placedimmediatelyafter the theorems of which they are applications, instead of being grouped to-gether without regard to the principles involved in them. For the benefit of the brighter pupils,however, and for review classes, long lists of more or less difficultexercises are grouped at the end of each book. The definitions of closed figures are unique. The student's natural conception of a plane closed figure, for example, is not the boundary line only, nor the plane only, but the whole figure composed of the boundary line and the plane bounded. All definitions of closed figures involve this idea, which is entirely consistent with the higher mathematics. The numerical treatment of magyiitudesis explicit, the funda-mental principlesbeing definitely assumed (Art. 336, proof in Appendix, Art. 595). This novel procedure furnishes a logical,as well as a teachable, method of dealing with incommensurables. The area of a rectangleis introduced by actuallymeasuring it, thereby obtainingits measure-number. This method permits the same order of theorems and corollaries as is used in the parallelogram and the triangle. The correlation with arithmetic in this connection is valuable. A similar method is employed for introducingthe volume of a parallelopiped. Proofs of the superpositiontheorems and the concurrent line theorems will be found exceptionally accurate and complete. The many historical notes ivill add lifeand interest to the work. The carefullyarranged summaries throughout the book, the collection of formulas of Plane Geometry, and the collection of formulas of Solid Geometry, it is hoped, will be found helpful to teacher and student alike. Argument and reasons are arranged in parallel form. This arrangement gives a definite model for proving exercises, ren-ders the careless omission of the reasons in a demonstration impossible, leads to accurate thinking,and greatly lightens the labor of reading papers. PREFACE v Every construction figure contains all necessary construction lines. This method keeps constantlybefore the student a model for his construction work, and distinguishes between a figure for a construction and a figurefor a theorem. TJie mechanical arrangement is such as to give the student every possible aid in comprehending the subject matter. The followingare some of the specialfeatures of the Solid Geometry : The vital relation of the Solid Geometry to the Plane Geometry is emphasized at every point. (SeeArts. 703,786, 794, 813, 853, 924, 951, 955, 961, etc.) The student is given every possible aid in forming his early space concepts. In the early work in Solid Geometry, the average student experiences difficulty in fullycomprehending space relations,that is,in seeing geometric figuresin space. The student is aided in overcoming this difficulty by the intro-duction of many easy and practical questionsand exercises, as well as by being encouraged to make his figures. (See " 605.) As a further aid in this direction,reproductions of models made by students themselves are shown in a group (p.302) and at various pointsthroughout Book VI. Tlie student's knowledge of the thingsabout him is constantly draivn upon. Especially is this true of the work on the sphere, where the student's knowledge of mathematical geography has been appealed to in making clear the terms and the relations of figures connected with the sphere. The same logical rigor that characterizes the demonstrations in the Plane Geometry is used throughout the Solid. The treatment of the polyhedral angle (p. 336), of the prism (p. 345), and of the pyramid (p.350) is similar to that of the cylinder and of the cone. This is in accordance with the recommendations of the leading Mathematical Associations throughout the country. The gratefulacknowledgment of the authors is due to many friends for helpfulsuggestions; especiallyto Miss Grace A. Bruce, of the Wadleigh High School, New York; to Mr. Edward B. Parsons, of the Boys' High School,Brooklyn ; and to Professor McMahon, of Cornell University. CONTENTS PLANE GEOMETRY PAGE SYMBOLS AND ABBREVIATIONS viii INTRODUCTION 1 BOOK I. RECTILINEAR FIGURES 19 Necessity for Proof 21 Polygons. Triangles 22 Superposition 25 Measurement of Distances by Means of Triangles ... 32 Loci 42 Parallel Lines 65 Quadrilaterals. Parallelograms 84 Concurrent Line Theorems 100 Construction of Triangles 106 Directions for the Solution of Exercises 110 Miscellaneous Exercises Ill BOOK II. THE CIRCLE 113 Two Circles . 131 Measurement 133 Miscellaneous Exercises 156 BOOK III. PROPORTION AND SIMILAR FIGURES . . 161 Similar Polygons 176 Drawing to Scale 188 Miscellaneous Exercises 206 BOOK IV. AREAS OF POLYGONS 209 Transformation of Figures 223 Miscellaneous Exercises . . . . .... 239 BOOK V. REGULAR POLYGONS. MEASUREMENT OF THE CIRCLE 245 Measurement of the Circumference and of the Circle . . 258 Miscellaneous Exercises 274 vi CONTENTS vii PAGE FORMULAS OF PLANE GEOMETRY . . . . .282 APPENDIX TO PLANE GEOMETRY 284 Maxima and Minima 284 Variables and Limits. Theorems 291 Miscellaneous Theorems ........ 294-298 SOLID GEOMETRY BOOK VI. LINES, PLANES, AND ANGLES IN SPACE . 299 Lines and Planes . . . 301 Dihedral Angles . 322 Polyhedral Angles . .. .336 BOOK VII. POLYHEDRONS . . ... . .343 Prisms 345 Pyramids 350 Mensuration of the Prism and Pyramid 354 Areas 354 Volumes . . 358 Miscellaneous Exercises 381 BOOK VIII. CYLINDERS AND CONES 383 Cylinders 383 Cones " 388 Mensuration of the Cylinder and Cone 392 Areas 392 Volumes 406 Miscellaneous Exercises 414 BOOK IX. THE SPHERE 417 Lines and Planes Tangent to a Sphere 424 SphericalPolygons 429 Mensuration of the Sphere 444 Areas . . , 444 Volumes 457 Miscellaneous Exercises on Solid Geometry .... 467 FORMULAS OF SOLID GEOMETRY . . . . .471 APPENDIX TO SOLID GEOMETRY 474 SphericalSegments 474 The Prismatoid 475 Similar Polyhedrons 477 .INDEX 481 SYMBOLS AND ABBREVIATIONS = equals, equal to, is equal to. =fc does not equal. " greater than, is greater than. " less than, is less than. =0= equivalent, equivalent to, is equiva-lent to. ~ similar, similar to, is similar to. S^ is measured by. JL perpendicular, perpendicular to, is perpendicular to. Js perpendiculars. \|\| parallel, parallel to, is parallel to. \|\|s parallels. . . . and so on (sign of continuation). v since. .-, therefore. /-n arc ; AB, arc AB. "7, OD parallelogram, parallelograms. O, (D circle,circles. Z, A angle, angles. A, A triangle, triangles. q.e.d. Quod erat demonstrandum, which was to be proved. q.e.f. Quod erat faciendum, which was to be done. The signs +, " , X , + have the same meanings as in algebra. viii PLANE GEOMETRY INTRODUCTION 1. The Subject Matter of Geometry. In geometry, although we shall continue the use of arithmetic and algebra, our main work will be a study of what will later be defined (" 13) as geometric figures. The student is already familiar with the physical objects about him, such as a ball or a block of wood. By a careful study of the following exercise, he may be led to see the relation of such physical solids to the geometric figures with which he must become familiar. Exercise. Look at a block of wood (or a chalk box). Has it weight ? color ? taste ? shape ? size ? These are called properties of the solid. What do we call such a solid ? A physical solid. Can you think of the properties of this jj , solid apart from the block of wood ? Imagine the block removed. Can you imagine the space which it occupied ? What name would you give to this space ? A geometric solid. What properties has it that the block possessed ? Shape and size. What is it that separates this geometric solid from surrounding space ? How thick is this surface ? How many surfaces has the block ? Where do they intersect ? How many intersections are there ? How wide are the intersections ? how long ? What is their name ? They are lines. Do these lines intersect ? where ? How wide are these intersections ? how thick ? how long ? Can you say where this one is and so distinguish from where that one is ? What is its name ? It is a point. If you move the block through space, what will it generate as it moves ? What will the surfaces of the block generate ? all of them ? Can you move a surface so that it will not generate a solid ? Yes, by moving it along itself. What will the edges of the block generate ? Can you move an edge so that it will not generate a surface ? What will the corners generate ? Can you move a point so that it will not generate a line ? PLANE GEOMETRY FOUR FUNDAMENTAL GEOMETRIC CONCEPTS 2. The space in which we live,although boundless and unlimited in extent, may be thought of as divided into parts. A physicalsolid occupies a limited portion of space. The portion of space occupied by a physical solid is called a geometric solid. 3. A geometric solid has length,breadth, and thickness. It may also be divided into parts. The boundary of a solid is called a surface. 4. A surface is no part of a solid. It has length and breadth, but no thickness. It may also be divided into parts. The boundary of a surface is called a line. 5. A line is no part of a surface. It has length only. It may also be divided into parts. The boundary or extremity of a line is called a point. A point is no part of a line. It has neither length,nor breadth,nor thickness. It cannot be divided into parts. It is position only. THE FOUR CONCEPTS IN REVERSE ORDER 6. As we have considered geometric solid independently of surface,line,and point,so we may consider point indepen-dently, and from it build up to the solid. A small dot made with a sharp pencilon a sheet of paper represents approximately a geometric point. 7. If a point is allowed to move in space, the path in which it moves will be a line. A pieceof fine wire, or. a line drawn on paper with a sharp pencil, represents approximately a geometric line. This, how-ever fine it may be, has some thickness and is not therefore an ideal, or geometric, line. 8. If a line is allowed to move in space, its path in general will be a surface. INTRODUCTION 3 9. If a surface is allowed to move in space, its path in generalwill be a geometric solid. 10. A solid has threefold extent and so is said to have three dimensions ; a surface has twofold extent and is said to have two dimensions ; a line has onefold extent or one dimension ; a point has no extent and has therefore no dimensions. 11. The following may be used as working definitions of these four fundamental concepts : A geometric solid is a limited portionof space. A surface is that which bounds a solid or separates it from an adjoiningsolid or from the surrounding space. A line is that which has length only. A point is position only. DEFINITIONS AND ASSUMPTIONS 12. The primary object of elementarygeometry is to deter-mine, by a definite process of reasoning that will be introduced and developed later, the propertiesof geometric figures. In all logical arguments of this kind, just as in a debate,certain fundamental principles are agreed upon at the outset, and upon these as a foundation the argument is built. In ele-mentary geometry these fundamental principles are called definitions and assumptions. The assumptions here mentioned are divided into two classes, axioms and postulates. These, as well as the definitions, will be given throughout the book as occasion for them arises. 13. Def. A geometric figure is a point,line,surface, or solid, or a combination of any or all of these. 14. Def. Geometry is the science which treats of the properties of geometric figures. 15. Def. A postulate may be defined as the assumption of the possibility of performinga certain geometric operation. Before giving the next definition, it will be necessary to introduce a nostulate. 4 PLANE GEOMETRY 16. Transference postulate. Any geometric figure may be moved from one position to another without change of size or shape. 17. Def. Two geometric figures are said to coincide if, when either is placed upon the other,each point of one lies upon some point of the other. 18. Def. Two geometric figuresare equal if they can be made to coincide. 19. Def. The process of placingone figure upon another so that the two shall coincide is called superposition. This is an imaginary operation, no actual movement taking place. LINES 20. A line is usuallydesignated by two capitalletters, as line AB. It may be designated . " also by a small letter placed somewhere on the line, as line a. , Fig. 1. 21. Straight Lines. In " 7 we learned that a piece of fine wire or a line drawn on a sheet of paper representedapproximately a geometric line. So also a geometric straightline may be representedapproximatelyby a stringstretched taut between two points, or by the line made by placing a ruler (alsocalled a straightedge) on a flat surface and drawing a sharp pencil along its edge. 22. Questions. How does a gardener test the straightness of the edge of a flower bed ? How does he get his plants set out in straight rows ? How could you test the straightness of a wire ? Can you think of a wire not straight, but of such shape that you could cut out a piece of it and slipit along the wire so that it would always fit? If you re-versed this piece,so that its ends changed places,would it stillfit along the entire length of the wire ? If you turned it over, would it fit? Would the piece cut out fit under these various conditions if the wire were straight ? INTRODUCTION 23. Def. A straight line is a line such that,if any portion of it is placed with its ends in the line, the entire portion so placedwill lie in the line, howrever it may be applied. Thus, if AB is a straight line,and if any portionof AB, as CD, is placed on any other part of AB, with its ends in AB, A S 5 9 every pointof CD will lie in AB. FlG-2-A straightline is called also a right line. The word line, unqualified, is understood to mean straight line. 24. Straight line postulate. A straight line may be drawn from any one point to any other. 25. Draw a straightline AB. Can you draw a second straight line from A to B ? If so, where will every pointof the second line lie ("23)? It then follows that : Only one straight line can be drawn between two points; i.e. a straight line is determined by two points. 26. Draw two straight lines AB and CD intersecting in point P. Show that AB and CD cannot have a second point in com-mon ("23). It then follows that: Two intersecting straight lines can have only one point in com-mon; i.e.two intersecting straight lines determine a point. 27. Def. A limited portion of a straight line is called a line segment, or simply a line, or a segment. Thus, in Fig.2,AC, CD, and DB are line segments. 28. Def. Two line segments which lie in the same straight line are said to be collinear segments. 29. Def. A curved line (or curve) is a line no portionof which is straight, as GH. 30. Def. A broken line is a line made up of different succes-sive straight lines, as KL. 4 Fig. 3. 6 TLANE GEOMETRY 31. Use of Instruments. Only two instruments are permitted in the constructions of plane geometry: the ruler or straight-edge for drawing a straight line, already spoken of in " 21 ; and the compasses, for constructing circles or arcs of circles, and for transferring line segments from one position to another. Thus to add two lines, as AB and CD, draw, with a ruler, a straight line OX. Place one leg of the compasses at A and the other at B. Next place one leg at 0 and cut off segment OM equal to AB. In a similar manner lay off MN equal to CD. Then AB + CD " 0M+ MN = ON. Show how to subtract AB from ON. What is the remainder? Ex. 1. Can a straight line move so that its path will not be a surface ? If so, how ? Ex. 2. Can a curved line move in space so that its path will not be a surface ? If so, how ? Ex. 3. Can a broken line move in space so that its path will not be a surface ? Ex. 4. Draw three lines as AB, CD, and ^ EF. Construct the sum of AB and CD ; of AB and EF; of AB, CD, and EF. C Ex. 5. Construct : (") the difference be-E tween AB and CD ; (6) the difference between FlG 5 CD and EF; (c) the difference between AB and EF. Add the results obtained for (a) and (6) and see whether the sum is the result obtained for (c). Ex. 6. Draw a line twice as long as AB (the sum of AB and AB) ; Jhree times as long as AB. INTRODUCTION SURFACES 32. Plane Surface. It is well known that the carpenter's straightedgeis appliedto surfaces to test whether they are flat and even. If,no matter where the straightedge is placed on the surface, it always fits, the surface is called a plane. Now if we should use a powerful magnifier, we should doubt-less discover that in certain places the straightedgedid not exactly fit the surface on which it was placed. A sheet of fine plateglassmore nearlyapproaches the ideal. 33. Questions. Test the surface of the blackboard with a ruler to see whether it is a plane. How many times must you apply the ruler ? Can you think of a surface such that the ruler would fit in some positions (a great many) but not in all ? Can you think of a surface not plane but such that a piece of it could be cut out and slippedalong the rest so that it would fit? Would it fitif turned over (inside out) ? Fig. 6. 34. Def. A plane surface (or plane) is a surface of unlimited extent such that whatever two of its points are taken,a straightline joiningthem will lie wholly in the surface. 35. Def. A curved surface is a surface no portion of which is plane. 36. Def. A plane figure is a geometric figureall of whose pointslie in one plane. Plane Geometry treats of plane figures. 37. Def. A rectilinear figure is a plane figureall the lines of which are straight lines. Ex. 7. How can a plane move in space so that its path will not be a solid ? Ex. 8. Can a curved surface move in space so that its path will not be a solid ? If so, how ? PLANE GEOMETRY ANGLES 38. Def. An angle is the figure formed by two straight lines which divergefrom a point. The point is the vertex of the angle and the lines are its sides. 39. An angle may be designated by a number placed within it, as angle 1 and angle 2 in Fig. 7, and angle 3 in Fig. 8. Or L M H Fig. 9. three letters may be used, one on each side and one at the vertex, the last being read between the other two; thus in Fig.7, angle 1 may be read angle ABC, and angle 2, angle CBD. An angle is often designatedalso by the singleletter at its vertex, when no other angle has the same vertex, as angle F in Fig. 8. 40. Revolution postulate. A straight line may revolve in a plane,about a point as a pivot,and when it does revolve contin-, uouslyfrom one position to another,itpasses once and only once throughevery intermediate position. 41. A clear notion of the magnitude of an angle may be obtained by imagining that its two sides were at firstcollinear, and that one of them has revolved about a point common to the two. Thus in Fig.8. we may imagine FG first to have been in the position FE and then to have revolved about F as a pivot to the position FG. 42. Def. Two angles are adjacent if they have a common vertex and a common side which lies between them ; thus in Fig. 7, angle 1 and angle 2 are adjacent;also in Fig. 9, angle JIMK and angle KML are adjacent. 10 PLANE GEOMETRY 46. Def. If two lines meet, hut are not perpendicular to each other,they are said to be oblique to each other. 47. Def. An acute angle is an angle that is less than a rightangle ; as angle 1, Fig. 13. 48. Def. An obtuse angle is an angle that is greater than a rightangle and less than two right angles; as angle 2,Fig. 14. 49. Def. A reflex angle is an angle that is greater than two rightangles and less than four right angles; as angle 2, Fig. 15. 50. Note. Two lines diverging from the same point, as BA and BC, Fig. 15, always form two positiveangles, as the acute angle 1 and the reflex angle 2. Angle 1 may be thought of as formed by the revolution of a line counter-clockwise from the position BA to the position BC, and should be read angle ABC. Angle 2 may be thought of as formed by the revolution of a line counter-clockwise from the position BC to the position BA, and should be read angle CBA. 51. Def. Acute, obtuse, and reflex angles are sometimes called oblique angles. Ex. 9. (a) In Fig. 16, if angle 1 equals angle 2, what kind of angles are they ? (b) Make a statement with regard to the lines AB and CD. (c) If angle 3 does not equal angle 4, what kind of angles are they ? (eZ)Make a statement with regard to the lines AB and CE. D Fig. 16. 7? Ex. 10. A plumb line is suspended from the top of the blackboard. What kind of angles does it make with a horizontal line drawn on the blackboard ? with a line on the blackboard neither horizontal nor vertical ? INTRODUCTION 11 Ex. 11. Suppose the minute hand of a clock is at twelve. Where may the hour hand be so that the two hands make with each other : (a) an acute angle ? ("") a rightangle ? (c) an obtuse angle ? Ex. 12. Draw: (a) a pair of adjacent angles; (6) a pair of non-adjacent angles. Ex. 13. Draw two adjacent angles such that : (a) each is an acute angle ; (") each is a rightangle ; (c) each is an obtuse angle ; (e?)one is acute and the other right; (e) one is acute and the other obtuse. Ex. 14. In Fig. 17, angle 1 -f angle 2 = ? angle 3 + angle 4 = ? angle BAD + angle DAF = ? angle DAF " angle 3 = ? angle 2 + angle DAF " angle 4 = ? angle 4 + angle BAE " angle 1 = ? Ex. 15. Name six pairs of adjacent angles in Fig. 17. Ex. 16. Draw two non-adjacent angles that have : (a) a common vertex ; (") a common side ; (c) a common vertex and a common side. 52. Def. A line is said to be bisected if it is divided into two equal parts. FlG' 17' 53. Def. The bisector of an angle is the line which divides the angle into two equal angles. Ex. 17. Draw a line AB, neither horizontal nor vertical, (a) Draw freehand a line perpendicular to AB and not bisecting it : (6) a line bisecting AB and not perpendicular to it. ASSUMPTIONS 54. 1. Things equal to the same thing, or to equal things, are equal to each other. 2. If equalsare added to equals,the sums are equal. 3. If equals are subtracted from equals,the remainders are equal. 4. If equals are added to unequals,the sums are unequal in the same order. 5. If equals are subtracted from unequals,the remainders are unequal in the same order. ' The proofthat every angle has but one bisector will be found in the Appendix, " 599, 12 PLANE GEOMETRY 6. If unequals are subtracted from equals, the remainders are unequal in the reverse order. 7. (a) If equals are multipliedby equals,the products are equal; (6) if unequals are multiplied by equals, the products are unequal in the same order. 8. (a)If equalsare divided by equals, the quotients are equal; (b)if unequals are divided by equals, the quotients are unequal in the same order. 9. If unequals are added to unequals,the less to the less and the greaterto the greater,the sums are unequal in the same order. .10. If three magnitudes of the same kind are so related that the first is greater than the second and the second greater than the third, then the first is greater than the third. 11. The whole is equal to the sum of all itsparts. 12. The whole is greater than any of itsparts. 13. Like powers of equal numbers are equal,and like roots of equal numbers are equal. 14. Transference postulate. Any geometric figure may be moved from one position to another without change of size or shape. (See " 16.) 15. Straight line postulate I. A straight line may be drawn from any one point to any other. (See " 24.) 16. Straight line postulate H. A line segment may be pro-longed indefinitely at either end. 17. Revolution postulate. A straight line may revolve in a plane, about a point as a pivot,and when it does revolve continu-ously from one position to another,it passes once and only once throughevery intermediate position. (See" 40.) 55. A ssumptions 1-13 are usuallycalled axioms. That is, an axiom may be denned as a statement whose truth is as-sumed. Ex. 18. Illustrate the firstfive assumptions above by using arithmeti-cal numbers only. Ex. 19. Illustrate the next five by using generalnumbers (letters) only See Appendix, " 600. INTRODUCTION 13 DEMONSTRATIONS 56. It has been stated ("12) that the fundamental princi-ples agreed upon at the outset as forming the basis of the logi-cal arguments in geometry are called definitions, axioms, and postulates. Every new proposition advanced, whether it is a statement of a truth or a statement of something to be per-formed, must by a process of reasoning be shown to depend upon these fundamental principles. This process of reasoning is called a proof or demonstration. After the truth of a state-ment has thus been established, it in turn may be used to establish new truths. The propositions here mentioned are divided into two classes, theorems and problems. 57. Def. A theorem is a statement whose truth is required to be proved or demonstrated. For example, " If two anglesof a triangle are equal,the sides opposite are equal" is a theorem. There are two parts to every theorem : the hypothesis, or the conditional part ; and the conclusion, or the part to be proved. In the theorem just quoted, " If two a7igles of a tri-angle are equal " is the hypothesis ; and " the sides opposite are equal " is the conclusion. . Ex. 20. Write out carefullythe hypothesis and the conclusion of each of the following : (a) If you do your duty at all times,you will be rewarded. (6) If you try to memorize your proofs,you will never learn geometry. (c) You must suffer if you disobey a law of nature. (d) Things equal to the same thing are equal to each other. (e) All rightangles are equal. 58. Def. A corollary is a statement of a truth easily de-duced from another truth. Its correctness, like that of a theorem, must be proved. 59. Def. A problem, in general, is a questionto be solved. As applied to geometry, problems are of two kinds,namely, problems of construction and problems of computation. 14 PLANE GEOMETRY 60. From the revolution postulate(" 40) and from the definition of a perpendicular("45) we may deduce the follow-ing corollaries : 61. Cor. I. At every point in a straight line there exists a perpendicular to the line. Given line AB and point C in AB. To prove that there exists a JL to AB at a as en. c Fig. 18. B Let CE (Fig.18) meet AB at O, so that Z 1 " Z 2. Then if CE is revolved about C as a pivot toward position CA, Z 1 will continuouslyincrease and Z 2 will continuously decrease ("40)-.-. there must be one position of CE, as CD, in which the two A formed with AB are equal. In this position CD"AB ("45). 62. Cor. II. At every point in a straight line there exists only one perpendicular to the line. Given CD A. AB at C, i.e. Z BCD = /.DCA. To prove CD the only JL to AB at C. E J) i \ " \ \ C Fig. 19. B Let CE (Fig 19) be any line from C A other than CD. Let C# fall between CD and (M Then ZBCE"ZBCD; i.e." ZDCA. And ZECA "ZDCA (" 54, 12). .-. A BCE and EGA are not equal and CE is not " to J.i? (" 45). 63. ""61 and 62 may be combined in one statement as follows : At every point in a straightline there exists one and only one perpendicular to the line. A similar proof may be given if we suppose CE to fall between CB and CD. INTRODUCTION 15 64. Cor. m. All right angles are equal, B D Fig. 20. Given AAOB and Z CPD, any two rt. A. To prove AAOB " A CPD. Place Z CPD upon Z A OB so that point P shall fall upon point O, and so that PC shall be colli near with OA. Then PD and OB, both being _L to OA at 0, must be collinear (" 62). .-. the two A coincide and are equal ("18). 65. Cor. IV. If one straight line meets another straight line,the sum of the two adjacent angles is two rightangles. Given str. line CD meeting str. line AB at C, forming A BCD and DC A. To prove Z BCD + Z DC A " 2 rt. A. Let CE be " to AB at C (" 63). Then Z BCD + Z DCE = 1 rt. Z ; .-. Z BCD = 1 rt. Z -ZDCE. Again Z DC A = \rt.Z + ZDCE. /. Z BCD + Z 2XL4 = 2 rt. A. E C Fig. 21. B 66. Cor. V. T/^e sum of all the angles about a point on one side of a straight line passing through that point equals two right angles. Given A 1,2, 3, 4, 5, and 6, all the A about point 0 on one side of str. line CA. To prove Z1+Z2 + Z3 + Z4 + Z 5 + A6 = 2 rt. A. . The sum of the six A (Fig.22) is equal to ZAOB + Z BOC. 16 PLANE GEOMETRY 67. Cor. VI. The sum of all the angles about a point equals four right angles. Prolong one of the lines through the vertex and apply Cor. V. 68. Def. Two angleswhose sum is one right Fig. 23. angle are called complementary angles, as angles 1 and 2,Fig.25. Either of two such angles is said to be the complement of the other. 69. Def. Two angles whose sum is two right angles are called supplementary angles, as angles1 and 2,Fig.26. Either of two such anglesis said to be the supplement of the other. An angle that is equal to the sum of two right angles is sometimes called a straight angle, as angle ABC, Fig. 26. 70. Def. Two angles are said to be vertical if the sides of each are the pro-longations of the sides of the other; thus,in Fig.24, angles 1 and 2 are ver-tical angles, and angles3 and 4 are like-wise vertical angles. 71. Def. An angle of one degree is one nine-tieth of a rightangle. A rightangle, therefore, contains 90 angle degrees. Fig. 24. Ex. 21. In Fig. 25, angle ABC is a right angle. If angle 1 = 40", how many degrees are there in angle " "" 2 ? How many degrees,then, are there in the comple-ment of an angle of 40" ? Ex. 22. How many degrees are there in the complement of 20" ? of 35" ? of a0 ? of $ rightangles ? of k rightangles? Ex. 23. In Fig.26, angle 1 + angle 2, or angle i. ABC, equals 2 right angles. If angle 1 = 40", C B how many degrees are there in angle 2 ? How FlG-"" many degrees,then, are there in the supplement of an angle of 40" 18 PLANE GEOMETRY Ex. 36. (a) Can two angles be supplementary and not adjacent? Illustrate. (6) Can two angles be adjacent and not supplementary ? Illustrate. (c) Can two angles be both supplementary and adjacent ? Illustrate. (d) Can two angles be neither supplementary nor adjacent ? Illustrate. (e) Can two angles be both complementary and adjacent ? Illustrate. (/) Can two angles be both complementary and supplementary ? Illustrate. 73. From the definitions of complementary and supple-mentary angles may be deduced three additional corollaries as follows : 74. Cor. I. Complements of the same angle or of equal angles are equal. 75. Cor. II. Supplements of the same angle or of equal angles are equal. 76. Cor. III. If two adjacent angles are supplemen-tary, their exterior sides are eollinear. Given Z ABC + Z CBD = 2 rt. A. To prove BD the prolongation of AB. If BD is not the prolongation of AB, then some other line from B, as BE, must D B A be its prolongation. Fig. 29. Then ZABC+Z CBD -2 rt. A (By hyp.). Also ZABC + Z CBE = 2 rt. A (" 65). .-.ZCBD = ZCBE ("75). This is impossible (" 54, 12) ; i.e. the supposition that BE is the pro-longation of AB leads, by correct reasoning, to the impossible conclusion that Z CBD = Z CBE. Hence this supposition itself is false. .-. BD is the prolongation of AB. BOOK I RECTILINEAR FIGURES Proposition I. Theorem 77. If two straight lines intersect, the vertical angles are equal. Given two intersecting str. lines,forming the vertical A, 1 and 3, also 2 and 4. To prove Z 1 = Z3 and Z2 = Z4. . Argument Zl + Z2 = 2rt. A. Z3 + Z2 = 2rt. A. .\Z1 + Z2 = Z3 + Z2. .Z1 = Z3. 5. Likewise Z 2 = Z 4. Q.E.D. Reasons 1. If one str. line meets an-other str. line, the sum of the two adj. A is 2 rt. A. " 65. 2. Same reason as 1. 3. Things equal to the same thing are equal to each other. " 54, 1. 4. If equals are subtracted from equals, the remain-ders are equal. " 54, 3. 5. By steps similar to 1, 2, 3, and 4. 78. Cor. If two straight lines intersect so that any two adjacent angles thus formed are equal, all the angles are equal, and each angle is a right angle. 19 20 PLANE GEOMETRY Ex. 37. If three straight lines intersect at a common point,the sum of any three angles, no two of which are adjacent, as 1, 3, and 5, in the accompanying diagram, is equal to two rightangles. Ex. 38. In the same diagram, show that : (a) Angle 1 + angle 3 = angle 7. (6) Angle 7 " angle 4 = angle 6. (c) Angle 7 + angle 4 " angle 3 = twice angle 1. (d) Angle 2 + angle 4 + angle 6 = 2 rightangles. Ex. 39. The line which bisects one of two vertical angles bisects the other also. Ex. 40. The bisectors of two adjacent complementary angles form an angle of 45". Ex. 41. Show the relation of the bisectors of two adjacent supple-mentary angles to each other. Ex. 42. The bisectors of two pairsof vertical angles are perpendicu-lar to each other. 79. The student will observe from Prop. I that the complete solution of a theorem consists of four distinct steps: (1) The statement of the theorem,including a general state-ment of the hypothesis and conclusion. (2) The drawing of a figure (as general as possible), to satisfy the conditions set forth in the hypothesis. (3) The application, i.e. the restatement of the hypothesis and conclusion as appliedto the particular figurejustdrawn, under the headings of " Given " and " To prove." (4) The proof, or demonstration, the argument by which the truth or falsityof a statement is established, with the reason for each step in the argument. 80. The student should be careful to quote the reason for every step in the argument which he makes in proving a theorem. The only reasons admissible are of two kinds : 1. Something assumed, i.e. definitions, axioms, postulates, and the hypothesis. 2. Something previously proved, i.e. theorems, corollaries, and problems. BOOK I 21 81. The necessity for proof. Some theorems seem evident by merely looking at the figure, and the student will doubtless think a proof unnecessary. The eye, however, cannot always detect error, and reasoning enables us to be sure of our conclu-sions. The danger of trustingthe eye is illustrated in the followingexercises. Ex. 43. In the diagrams given below, tell which line of each pair is the longer,a or 6, and verifyyour answer by careful measurement. 3 c /^k7 E1^ b Ex. 44. In the figures below, are the lines everywhere the same dis-tance apart ? Verify your answer by using a ruler or a slipof paper. ////////////// wwww A7Wt"tL/////// Ex. 45. In the figuresbelow, tell which lines are prolongations of other lines. Verify your answers. A 7 / z 7 These diagrams are taken by permission from the Report of the Committee on Geom" etry, Proceedings of the Eighth Meeting of the Central Association of Science and Mathematics Teachers. 22 PLANE GEOMETRY POLYGONS. TRIANGLES 82. Def. A line on a plane is said to be closed if it sepa-rates a finite portionof the planefrom the remaining portion. 83. Def. A plane closed figure is a planefigure composed of a closed line and the finite portionof the plane bounded by it. 84. Def. A polygon is a plane closed figure whose boundary is composed of straight lines only. The pointsof intersection of the lines are the vertices of the polygon, and the segments of the boundary lines included between adjacent vertices are the sides of the polygon. 85. Def. The sum of the sides of a polygon is its perimeter. 86. Def. Any angle formed by two consecutive sides and found on the right in passing clockwise around the perimeter of a polygon is called an interior angle of the polygon,or, for brevity, an angle of the polygon. In Fig. 1, A ABC, BCD, CDE, BE A, and EAB are interior angles of the polygon. 87. Def. If any side of a polygon is prolonged through a vertex,the angle formed by the prolonga-tion and the adjacentside is called an exterior angle of the polygon. In Fig.1, A 1,2,3,4, and 5 are exterior angles. 88. Def. A line joiningany two non-adjacent vertices of a polygon is called a diagonal ; as AC, Fig. 1. 89. Def. A polygon which has all of its sides equal is an equilateral polygon. 90. Def. A polygon which has all of its anglesequalis an equiangular polygon. BOOK I 23 91. Def. A polygon which is both equilateral and equi-angular is a regular polygon. 92. Def. A polygon of three sides is called a triangle; one of four sides, a quadrilateral ; one of five sides, a pentagon ; one of six sides, a hexagon ; and so on. TRIANGLES CLASSIFIED WITH RESPECT TO SIDES 93. Def. A trianglehaving no two sides equal is a scalene triangle. 94. Def. A triangle having two sides equal is an isosceles triangle. The equal sides are spoken of as the sides of the triangle. The angle between the equal sides is the vertex angle, and the side oppositethe vertex angle is called the base. 95. Def. A trianglehaving its three sides equal is an equilateral triangle. Scalene Isosceles Equilateral TRIANGLES CLASSIFIED WITH RESPECT TO ANGLES 96. Def. A right triangle is a trianglewhich has a right angle. The side oppositethe rightangle is called the hypote-nuse. The other two sides are spoken of as the sides t of the triangle. 97. Def. An obtuse triangle is a triangle which has an obtuse angle. 98. Def. An acute triangle is a trianglein which all the angles are acute. The equal sides are sometimes called the arms of the isosceles triangle. This term will be used occasionally in the exercises. t Sometimes the sides of a right triangle including the right angle are called the arms of theftriangle. This term will be found in the exercises. 24 PLANE GEOMETRY Ex. 46. Draw a scalene trianglefreehand : (a) with all its angles acute and with its shortest side horizontal; (b) with one right angle. Ex. 47. Draw an isosceles triangle : (a) with one of its arms hori-zontal and one of its angles a right angle ; (") with one angle obtuse. 99. Def. The side upon which a polygon is supposed to stand is usually called its base ; however, since a polygon may be supposed to stand upon any one of its sides, any side may be considered as its base. The angle oppositethe base of a triangle is the vertex angle, and the vertex of the angle is called the vertex of the triangle. 100. Def. The altitude of a triangle is the perpendicular to its base from the opposite vertex. In general any side of a trianglemay be considered as its base. Thus in triangle EFG, g ^\ if FG is taken as base,EH is the ^ altitude;if GE is taken as base, f/' .Fif will be the altitude;if EF /^ is taken as base,the third alti-^--gr tude can be drawn. Thus every FlG# L triangle has three altitudes. It will be proved later that one perpendicular, and only one, can be drawn from a point to a line. 101. The sides of a triangleare often designated by the small letters corresponding to the capitalsat the opposite vertices ; as, sides e,/, and g, Eig. 1. Ex. 48. Draw an acute triangle ; draw its three altitudes freehand. Do they seem to meet in a point ? Where is this point located ? Ex. 49. Draw an obtuse triangle ; draw its three altitudes freehand. Do they meet in a point ? Where is this point located ? Ex. 50. Where do the three altitudes of a righttriangle meet ? 102. Def. The medians of a triangle are the lines from the vertices of the triangle to the mid-pointsof the oppositesides. 26 PLANE GEOMETRY Proposition II. Theorem 105. Two triangles are equal if a side and the two adjacent angles of one are equal respectively to a side and the two adjacent angles of the other. Given A ABC and DEF, AC=DF" Z A To prove A ABC= A DEF. Z D, and Z C = Zf. Argument L. Place A ABC upon A DEF so that A C shall fall upon its equal DF, A upon D, C upon F. 2. Then AB will become col-linear with DE, and B will fall somewhere on DE, or on its prolonga-tion. 3. Also CB will become col-linear with FE, and B will fall somewhere on FE, or on its prolonga-tion. 4. .. point B must fall on point E. 5. .'. A ABC = A DEF. Q.E.D. Reasons Any geometric figuremay be moved from one posi-tion to another without change of size or shape. "64, 14. Z A = Z D, by hyp. 3. Z'C=ZF, by hyp. 4. 5. Two intersecting str. lines determine a point. " 26. Two geometric figures are equal if they can be made to coincide. " 18. BOOK I 27 Ex. 53. In the figurefor Prop. II, assume AB = DE, angle A =p angle Z", and angle B = angle E ; repeat the proof by superposition, marking the lines with colored crayon as soon as their positionsare determined. Ex. " 54. Place polygon I upon polygon II so that some part of I shall fall upon its equal in II. Discuss the resulting positions of the remaining parts of the figure. Ex. 55. If two quadrilaterals have three sides and the included angles of one equal respectively to three sides and the included angles of the other, and arranged in the same order, are the quadrilaterals equal ? Prove. 106. Note. The method of superposition should be used for proving fundamental propositions only. In proving other propositions it is neces-sary to show merely that certain conditions are present and to quote theorems, previously proved, which state conclusions regarding such conditions. Ex. 56. If at any point in the bisector of an angle a perpendicularto the bisector is drawn meeting the sides of the angle,the two triangles thus formed will be equal. Ex. 57. If equal segments, measured from the point of intersection of two lines, are laid off on one of the lines, and if perpendicularsto this line are drawn at the ends of these segments, two " equal triangles will be formed. Ex. 58. If at the ends of a straightline per-pendiculars to it are drawn, these perpendiculars will cut off equal segments upon any line which bisects the given line and is not perpendicular to it. Fig. 1. Fig. Ex. 59. If two angles of a triangleare equal,the bisectors of these angles are equal (Fig.1). Ex. 60. If two triangles are equal,the bisector of any angle of one is eqjjal to the bisector of the corresponding angleof the other (Fig.2), 28 PLANE GEOMETRY Proposition III. Theorem 107. Two triangles are equal if two sides and the in-cluded angle of one are equal respectively to two sides %nd the included angle of the other. Given A ABC and MNO, AB = MN, AG = MO, and Z A = Z M. To prove A ABC = A MNO. Argument 1. Place A ABC upon A MNO so that AC shall fall upon its equal MO, A upon M, C upon 0. 2. Then AB will become col-linear with MN. 3. Point B will fall on point N. 4. . . B C will coincide with NO. 5. A ABC=A MNO. Q.E.D. Reasons 1. Any geometric figuremay be moved from one posi-tion to another without change of size or shape. " 54, 14. 2. ZA = ZM, by hyp. 3. AB = MN, by hyp. 4. Only one str. line can be drawn between two points. " 25. 5. Two geometric figures are equal if they can be made to coincide. " 18. 108. Cor. Two right triangles are equal if the two sides including the right angle of one are equal respectively to the two sides including the right angle of the other. Ex. 61. Prove Prop. HI by placingAB upon MN, BOOK I 29 109. Def. In equal figures, the points, lines, and angles in one which, when superposed,coincide respectively with points, lines,and angles in the other, are called homologous parts. Hence : 110. Homologous parts of equal figuresare equal. Ex. 62. If two straight lines bisect each other, the lines joiningtheir extremities are equal in pairs. Hint. To prove two lines or two anglesequal,try to find two triangles, each containingone of the lines or one of the angles. If the triangles can be proved equal,and the two lines or two angles are homologous parts of the triangles, then the lines or angles are equal. The parts given equal may be more easily remembered by marking them with the same symbol, or with colored crayon. Ex. 63. In case the lines in Ex. 62 are perpendicular to each other, what additional statement can you make ? Prove its correctness. Ex. 64. If equal segments measured from the vertex are laid off on the sides of an angle,and if their extremities are joined to any point in the bisector of the angle,two equal triangles will be formed. Ex. 65. If two medians of a triangleare perpendicular to the sides to which they are drawn, the triangle is equilateral. Ex. 66. If equal segments measured from the vertex are laid off on the arms of an isosceles triangle, the lines joining the ends of these segments to the oppositeends of the base will be equal (Fig.1). Fig. 1. Fie. 2. Ex. 67. Extend Ex. 66 to the case in which the equal segments are laid off on the arms prolonged through the vertex (Fig.2). 30 PLANE GEOMETRY Proposition IV. Theorem 111. The base angles of an isosceles triangle are equal. B Given isosceles A ABC, with AB and BC its equal sides. To prove Z A = Z C. 112. Cor. I. The bisector of the angle at the vertex of an isosceles triangle is perpendicular to the base and bisects it. 113. Cor. n. An equilateral triangle is also equi-angular. Ex. 68. The bisectors of the base angles of an isosceles triangle are eaual. BOOK I 31 114. Historical Note. Exercise 69 is known as the pons asi-norum, or bridge of asses, since it has proved difficult to many beginners in geometry. This proposition and the proof here suggested are due to Euclid, a great mathematician who wrote the first systematic text-book on geometry. In this work, known as Euclid's Elements, the exer-cise here given is the fifth prop-osition in Book I. Of the life of Euclid there is but littleknown except that he was gentle and modest and " was a Greek who lived and taught in Alexandria about 300 b.c." To him is attributed the saying, " There is no royal road to geometry. ' ' His appreciation of the culture value of geometry is shown in a story related by Stobaeus (which is probably authentic). "A lad who had just begun geometry asked, " What do I gain by learning all this stuff ? ' Euclid called his slave and said, ' Give this boy some coppers, since he must make a profit out of what he learns.' " Euclid Ex. 69. By using the accompanying diagram prove that the base angles of an isosceles triangle are equal. Hint. Prove A ABE = A BBC. Then prove A ACE = ABAC. Ex. 70. (a) If equal segments measured from the vertex are laid off on the arms of an isosceles triangle, the lines drawn from the ends of the segments to the foot of the bisector of the vertex angle will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the arms prolonged through the vertex. Ex. 71. (a) If equal segments measured from the ends of the base are laid off on the arms of an isosceles triangle, the lines drawn from the ends of the segments to the foot of the bisector of the vertex angle will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the arms prolonged through the ends of the base. 32 PLANE GEOMETRY Ex. 72. (a) If equal segments measured from the ends of the base are laid off on the base of an isosceles triangle, the lines joiningthe vertex of the triangle to the ends of the segments will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the base prolonged (Fig.1) . Fig. 1. Fig. 2. Ex. 73. (a) If equal segments measured from the ends of the base are laid off on the arms of an isoscles triangle, the lines drawn from the ends of the segments to the opposite ends of the base will be equal. (6) Extend (a) to the case in which the equal segments are laid off on the arms prolonged through the ends of the base (Fig.2). Ex. 74. Triangle ABC is equilateral, and AE = BF = CD. Prove triangle EFD equi-lateral. 115. Measurement of Distances by Means of Triangles. The theorems which prove trianglesequal are appliedpractically in measuring distances on the surface of the earth. Thus, if it is desired to find the distance between two places, A and B, which are separated by a pond or other obstruction, place a stake at some point accessible to both A and B, as F. Measure the distances FA and FB ; then, keeping in line with F and B, measure CF equal to FB, and, in line with F and A, measure FE equalto FA. Lastly measure CE, and the distance from A to B is thus obtained, since AB is equal to CE. Can this method be used when A and B are on opposite sides of a hill and each is invisible from the other ? 34 PLANE GEOMETRY Proposition V. Theorem 116. Two triangles are equal if the three sides of one are equal respectively to the three sides of the other. " Given A ABC and RST, AB=RS, BC = ST, and CA = TR. To prove A ABC = A RST. 2. Argument Place A RST so that the longestside RT shall fall upon its equal AC, R upon A, T upon C, and so that S shall fall oppo-site B. Draw BS. 1. 2. Reasons Any geometric figuremay-be moved from one posi-tion to another without change of size or shape. " 54, 14. A str. line may be drawn from any one point to any other. " 54,15. AB = RS, by hyp. The base A of an isosceles A are equal. " 111. BC"ST, by hyp. Same reason as 4. If equals are added to equals, the sums are equal. " 54?2. BOOK I 35 Argument 8. .. Z ABC=/-CSA. 9. .'. A ABC = A CSA; i.e. A ABC = A RST. Q.E.D. Reasons 8. The whole = the sum of all its parts. " 54, 11. 9. Two A are equal if two sides and the included Z of one are equal respec-tively to two sides and the included Z of the other. " 107. Ex. 81. (a) Prove Prop. V, using two obtuse triangles and applying the shortest side of one to the shortest side of the other. (b) Prove Prop. V, using two righttriangles and applying the shortest side of one to the shortest side of the other. 117. Question. Why is not Prop. V proved by superposition? SUMMARY OF CONDITIONS FOR EQUALITY OF TRIANGLES a side and the two adjacent angles two sides and the included angle three sides a side and the two adjacent angles two sides and the included angle three sides 118. Two triangles are equalif of one are equal respectively to of the other. . Ex. 82. The median to the base of an isosceles trianglebisects the angle at the vertex and is perpendicularto the base. Ex. 83. In a certain quadrilateral two adjacent sides are equal ; the other two sides are also equal. Find a pair of triangles which you can prove equal. 36 PLANE GEOMETRY Ex. 84. If the oppositesides of a quadrilateral are equal,the opposite angles also are equal. Ex. 85. If two isosceles triangles have the same base, the line joining their vertices bisects each vertex angle and is perpendicular to the -com-mon base. (Two cases.) Ex. 86. In what triangles are the three medians equal ? Ex. 87. In what triangles are two medians equal ? Ex. 88. If three rods of different lengths are put together to form a triangle, can a different trianglebe formed by arranging the rods in a different order ? Will the angles opposite the same rods always be the same ? Ex. 89. If two sides of one triangleare equal respectively to two sides of another, and the median drawn to one of these sides in the first is equal to the median drawn to the correspondingside in the second, the triangles are equal. 119. Def. A circle is a plane closed figure whose boundary is a curve such that all straight lines to it from a fixed point within are equal. The curve which forms the boundary of a circle is called the circumference. The fixed point within is called the center, and a line joining the center to any point on the circumference is a radius. 120. It follows from the definition of a circle that : All radii of the same circle are equal. 121. Def. Any portion of a circumference is called an arc. 122. Assumption 18. Circle postulate. A circle may be con-structed having any point as center, and having a radius equal to any finite line. 123. The solution of a problem of construction consists of three distinct steps : (1) Tlie construction, i.e. the process of drawing the required figure with ruler and compasses. (2) The proof a demonstration that the figureconstructed fulfills the given conditions. (3) The discussion, i.e. a statement of the conditions under which there may be no solution, one solution, or more than one. BOOK I 37 Proposition VI. Problem 124. To construct an equilateral triangle, with a given line as side. A Given line BC. To construct an equilateral triangle on BC. I. Construction 1. With B as center and BC as radius, construct circle CAB. 2. With C as center and BC as radius, construct circle BMA. 3. Connect pointA, at which the circumferences intersect, with B and C. 4. A ABC is the required triangle. Argument 1. AB = BC 3Hid CA " BC. 2. .'. AB = BC=CA. II. Proof 2. 3. .'.A ABC is equilateral. Q.E.D. Reasons All radii of the same circle are equal. " 120. Things equal to the same thing are equal to each other. " 54, 1. A A having its three sides equalis equilateral. " 95. III. Discussion This construction is always possible, and there is onlyone solution. (See" 116.) 38 PLANE GEOMETRY Proposition VII. Problem 125. With a given vertex and a given side, to construct an angle equal to a given angle. Given vertex A, side AB, and Z CDE. To construct an Z equal to Z CDE and having A as vertex and AB as side. I. Construction 1. With D as center,and with any convenient radius,de-scribe an arc intersecting the sides of Z D at F and G, respec-tively. 2. With A as center,and with the same radius, describe the indefinite arc IH, cuttingAB at /. 3. With I as center,and with a radius equal to str. line FG, describe an arc intersecting the arc IH at K. 4. Draw AK. 5. Z BAK = Z CDE, and is the Z required. II. Proof Argument 1. Draw FG and IK. 2. In A FDG and IAK, DF"AI. 3. DG " AK. 4. FG = IK. Reasons 1. A str. line may be drawn from any one point to any other. " 54, 15. 2. By cons. 3. By cons. 4. By cons. BOOK I 39 Argument 5. . . A FDG = A IAK. 6. ." . ZBAK = Z.CDE. Q.E.D. Eeasons 5. Two A are equal if the three sides of one are equal respectively to the three sides of the other. "116. 6. Homol. parts of equal figures are equal. " 110. III. Discussion This construction is always possible, and there is only one solution. Ex. 90. Construct a triangle, given two sides and the included angle. Ex. 91. Construct an isosceles triangle, given the vertex angle and Ex. 92. Construct a triangle, given a side and the two adjacent angles. Ex. 93. Construct an isosceles triangle, given an arm and one of the equal angles. Ex. 94. How many parts determine a triangle ? Do three angles determine it ? Explain. Ex. 95. Construct an isosceles triangle, given the base and an arm. Ex. 96. Construct-a scalene triangle^ given the three sides. 126. Def. The bisector of an angle of a triangle is the line from the vertex of the angle bisecting the angle and limited by the opposite side of the triangle. a Ex. 97. In what triangles are the three bisectors equal ? Ex. 98. In what triangles are two bisectors, and only two, equal ? Ex. 99. In what triangles are the medians, the bisectors, and the altitudes identical ? 40 PLANE GEOMETRY Proposition VIII. Problem 127. To construct the bisector of a given angle. Given Z.ABC. To construct the bisector of Z ABC. I. Construction 1. With B as center, and with any convenient radius; describe an arc intersecting BA at E and BC at D. 2. With D and E as centers, and with equal radii, describe arcs intersecting at F. 3. Draw BF. 4. BF is the bisector of Z ABC. II. Proof 42 PLANE GEOMETRY LOCI 129. In many geometric problems it is necessary to locate all pointswhich satisfycertain prescribed conditions, or to determine the path traced by a point which moves according to certain ijxed laws. Thus, the pointsin a plane two inches from a given point are in the circumference of a circle whose center is the given point and whose radius is two inches. Again, let it be required to find all points in a plane two inches from one fixed point and three inches from another. All points two inches from the fixed point P are in the cir-cumferen of the circle LMS, having P for center and having a radius equal to two inches. All points three inches from the fixed point Q are in the cir-cumference of the circle LET, having Q for center and having a radius equal to three inches. If the two circles are wholly outside of each other, there will be no pointssatisfying the two prescribed conditions ; if the two circumferences touch, but do not intersect, there will be one point; if the two circum-ferences intersect, there will be two points. It will be proved later ("324) that there cannot be more than two pointswhich satisfy both of the given conditions. 130. Def. A figure is the locus of all pointswhich satisfy one or more given conditions, if all points in the figuresatisfy the given conditions and if these conditions are satisfied by no other points. A locus, then, is an assemblage of points which obey one or more definite laws. It is often convenient to locate these pointsby thinking of them as the path traced by a moving pointthe motion oi which is controlled by certain fixed laws. BOOK I 43 131. In plane geometry a locus may be composed of one or more points or of one or more lines, or of any combination of pointsand lines. 132. Questions. " What is the locus of all points in space two inches from a given point? What is the locus of all points in space two inches from a given plane ? What is the locus of all points in space such that perpendiculars from them to a given plane shall be equal to a given line ? What is the locus of all points on the surface of the earth midway between the north and south poles? 23\|" from the equator ? 23"" from the north pole? 00" from the equator ? What is the locus of a gas jetfour feet from the ceiling of this room ? four feet from the ceiling and five feet from a side wall ? four feet from the ceiling, five feet from a side wall, and six feet from an end wall ? Ex. 109. Given an unlimited line AB and a point P. Find all points in AB which are also : (a) three inches from P; (b) at a given distance, a, from P. Ex. 110. Given a circle with center 0 and radius six inches. State, without proof,the locus : (a) of all points four inches from 0 ; (") of all pointsfive inches from the circumference of the circle, measured on the radius or radius prolonged. Ex. 111. Given the base and one adjacentangle of a triangle, what is the locus of the vertex of the angle oppositethe base ? (Statewithout proof.) Ex. 112. Given the base and one other side of a triangle, what is the locus of the vertex of the angle oppositethe base ? (Statewithout proof.) Ex. 113. Given the base and the other two sides of a triangle, what is the locus of the vertex of the angle oppositethe base ? Ex. 114. Given the base of a triangle and the median to the base, what is the locus of the end of the median which is remote from the base ? Ex. 115. Given the base of a triangle, one other side, and the median to the base, what is the locus of the vertex of the angle opposite the base ? 133. Question. In which of the exercises above was a triangle determined ? In order to develop the imagination of the student the authors deem it advisable in this article to introduce questions involving loci in space. It should be noted that no proofs of answers to these questions are demanded. v 44 PLANE GEOMETRY Proposition IX. Theorem 134. Every point in the perpendicular bisector of a line is equidistant from the ends of that line. A C B Given line AB, its " bisector CD, and P any pointin CD. To prove PA PB. Argument 1. In A APC and CPB, AC as CB. 2. PC=PC. 3. Z PC A = Z BCP. 4. ..A APC = A CPB. 5. .\PA = PB. Q. E. D. REASON8 1. By hyp. 2. By iden. 3. All rt. A are equal. " 64. 4. Two A are equal if two sides and the included Z of one are equal re-spectively to two sides and the included Z of the other. " 107. 5. Homol. parts of equal fig-ures are equal. " 110. Ex. 116. Four villages are so located that B is 25 miles east of A, C 20 miles north of A, and D 20 miles south of A. Prove that B is as far from C as it is from D. Ex. 117. In a given circumference,find the pointsequidistantfrom two given points. A and B. BOOK I 45 135. Def. One theorem is the converse of another when the conclusion of the first is the hypothesis of the second, and the hypothesisof the first is the conclusion of the second. The converse of a truth is not always true ; thus, " All men are bipeds " is true, but the converse, " All bipeds are men," is false. "All rightangles are equal" is true, but "All equal angles are rightangles" is false. 136. Def. One theorem is the opposite of another when the hypothesis of the first is the contradiction of the hypothesis of the second, and the conclusion of the first is the contradic-tion of the conclusion of the second. The opposite of a truth is not always true ; thus, " If a man lives in the cityof New York, he lives in New York State," is true, but the opposite, " If a man does not live in the cityof New York, he does not live in Neiv York State," is false. 137. Note. If the converse of a propositionis true, the oppositealso is true ; so, too, if the oppositeof a proposition is true, the converse also is true. This may be evident to the student after a consideration of the fol-lowing type forms : (1) Direct (2) Converse (8) Opposite If ^lis.B, If CisD, If A is not B, Then C is D. Then A is B. Then C is not D. If (2) is true, then (3) must be true. Again, if (3) is true, then (2) must be true. 138. A necessary and sufficient test of the completenessof a definition is that its converse shall also be true. Hence a definition may be quoted as the reason for a converse or for an oppositeas well as for a direct statement in an argument. Ex. 118. State the converse of the definition for equal figures ; straightline ; plane surface. Ex. 119. State the converse of : If one straightline meets another straight line, the sum of the two adjacent angles is two rightangles. Ex. 120. State the converse and oppositeof Prop. IX. Ex. 121. State the converse of Prop. I. Is it true? 46 PLANE GEOMETRY Proposition X. Theorem (Converseof Prop. IX) 139. Every point equidistant from the ends of a line lies in the perpendicular bisector of that line. Given line RS, and point Q such, that QR = QS, To prove that Q lies in the L bisector of RS. Argument 1. Let QT bisect Z RQS. 2. QR = QS. 3. .'.A RQS is isosceles. 4. .'. QT is the _L bisector of RS. 5. .'. Q lies in the _L bisector of RS. Q.E.D. 5. Reasons Every Z has but one bi-sector. " 53. By hyp. A A having two sides equal is an isosceles A. " 94. The bisector of the Z at the vertex of an isosceles A is " to the base and bisects it. " 112. By proof. 140. Cor. I. Every point not in the perpendicular bi-sector of a line is not equidistant from the ends of the line. Hint. Use " 137, or contradict the conclusion and tell why the con-tradiction is false. BOOK I 47 141. Cor. n. The locus of all pointsequidistant from the ends of a given line is the perpendicularbisector of that line. (For model proofsof locus theorems, see pp. 297 and 298.) 142. Cor. in. Two pointseach equidistant from the ends of a line determine the perpendicular bisector of the line. Hint. Use " 139 and " 25. 143. In order to prove that a locus problemis solved it is necessary and sufficient to show two things : (1)That every point in the proposedlocus satisfies the prescribed conditions. (2)That every pointoutside of the proposedlocus does not satisfy the prescribed conditions. Instead of proving (2), it may frequently be more convenient to prove : (2') That every pointwhich satisfiesthe prescribed con-ditions lies in the proposed locus. 144. Note. In exercises in which the student is asked to "Find a locus, " it must be understood that he has not found a locus until he has given a proof with regardto it as outlined above. The proofmust be based upon a direct proposition and its opposite ; or, upon a direct propo-sition and its converse. Ex. 122. Find the locus of all pointsequidistant from two given points A and B. Ex. 123. In a given unlimited line AB, find a pointequidistant from two given pointsG and D not on this line. Ex. 124. Given a circle with center 0, also a point P. Find all points which lie in the circumference of circle O, and which are also (a) two inches from P ; (6) a distance of d from P. Ex. 125. Find allpoints at a distance of d from a givenpointP, and at the same time at a distance of m from a givenpoint Q. Ex. 126. Given a circle 0 with radius r. Find the locus of the mid-points of the radii of the circle. Ex. 127. Given two circles having the same center. State, without proof, the locus of a pointequidistant from their circumferences. Ex. 128. The perpendicular bisector of the base of an isoscelestri-angle passes through the vertex, -48 PLANE GEOMETRY Proposition XI. Problem 145. To construct the perpendicular bisector of a given straight line. V I I \ 'H IF x Fig. 1. B A Fig. 2. Given line AB (Fig.1). To construct the perpendicular bisector of AB. I. Construction 1. With A as center, and with a convenient radius greater than half AB, describe the arc EF. 2. With B as center, and with the same radius, describe the arc HK. 3. Let c and D be the pointsof intersection of these two arcs. 4. Connect points C and D. 5. CD is the _L bisector of AB. II. The proof and discussion are left as an exercise for the student. Hint. Apply " 142. 146. Note. The construction given in Fig.2 may be used when the positionof the given line makes it more convenient. 147. Question. Is it necessary that CA shall equal CB? that CA ghall equalDA (Fig. 1) ? Give the equations that must hold. 50 PLANE GEOMETRY Proposition XIII. Problem 149. From a point outside a line to construct a per-pendicula to the line. "-B Xq Given line AB and pointP outside of AB. To construct a _L from P to AB. I. Construction 1. With. P as center, and with a radius of sufficient length, describe an arc cuttingAB at points C and D. 2. With c and D as centers, and with any convenient radius, describe arcs intersecting at Q. 3. Draw PQ. 4. PQ is a _L from P to AB. , II. The proof is left as an exercise for the student. The discussion will be given in " 154. 150. Question. Must P and Q be on opposite sides of AB? Is it necessary that PC=QG? Ex. 138. Construct the three altitudes of an acute triangle. Do they seem to meet ? where ? Ex. 139. Construct the three altitudes of a righttriangle. Where do they seem to meet ? Ex. 140. Construct the three altitudes of an obtuse triangle.Where do they seem to meet ? Ex. 141. Construct a triangleABC, given two sides and the median drawn to one of them. Abbreviate thus : givena, br and m^. BOOK I 51 I 151. Analysis of a problem of construction. In the more difficult problems of construction a course of reasoningis some-times necessary to enable the student to discover the process of drawing the requiredfigure. This course of reasoningis called the analysis of the problem. It is illustrated in " 152 and is more fullytreated in the exercises following" 274. 152. Note. In such problems as Ex. 141, it is well first to imagine the problem solved and to sketch a figureto represent the desired con-struction. Then mark (with colored crayon, if convenient) the parts supposed to be given. By studying carefullythe relation of the given parts to the whole figure, try to find some part of the figure that you can construct. This will generally be a triangle. After this part is constructed it is usuallyan easy matter to complete the required figure. Thus : Problem. Let it be required to con-struct an isosceles triangle, given an arm and the altitude upon it. By studying the figure with the given parts marked (heavy or with colored crayon), it will be seen that the solution of the problem depends in this case upon the construc-tion of a righttriangle, given the hypotenuse and one arm. The righttriangle ABB may now be constructed, and it will be readilyseen that to complete the construction it is only necessary to prolong BD to C, making BC = AB, and to connect A and C. Ex. 142. Construct a triangle ABC, given two sides and an altitude to one of the given sides. Abbreviate thus : given a, b, and hi,. Ex. 143. Construct a triangleABC, given a side,an adjacent angle, and the altitude to the side opposite the given angle. Abbreviate thus : given a, B, and ft. Ex. 144. Construct an isosceles triangle, given an arm and the me-dian to it. Ex. 145. Construct an isosceles triangle, given an arm and the angle which the median to it makes with it. Ex. 146. Construct a triangle, given two sides and the angle which a median to one side makes : (a) with that side ; (") with the other side. Ex. 147. Construct a triangle, given a side,an adjacent angle, and its bisector. 52 PLANE GEOMETRY t Proposition XIV. Theorem 153. If one side of a triangle is prolonged,the exterior angle formed is greater than either of the remote in-terior angles. B F Given A ABC with AC prolonged to D, making exterior Z DCB. To prove Z DCB " Z ABCov Z CAB. Argument 1. Let E be the mid-point of BCj draw AE, and pro-long it to F, making EF = AE. Draw CF. 2. In A ABE and EFCy BE = EC. 3. AE = EF. 4. Z BE A = Z CEF. 5. .'. A ABE = A EFC. .-.Zb = Zfce. Reasons 1. A str. line may be drawn from any one point to any other. " 54, 15. 2. By cons. E is the mid-point of BC. 3. By cons. 4. If two str. lines inter-sect, the vertical A are equal. " 77. 5. Two A are equal if two sides and the included Z of one are equal re-spectively to two sides and the included Z of the other. " 107. 6. Homol. parts of equal fig-ures are equal. " 110. BOOK I 53 7. Argument Z DCB " Z FCE. 8. .-. Z DC" " Z 5. 9. Likewise, if BC is pro-longed to G, Z.ACG" Z CAB. 10. But Z DCB = Z .4CG. 11. .. Z Z)C5 " Z CAB. 12. Reasons 7. The whole " any of its parts. " 54, 12. 8. SubstitutingZ.B for its equal Z i^C^. 9. By bisecting line AC, and by steps similar to 1-8. 10. Same reason as 4. 11. SubstitutingA DCB for its equalZ.ACG. 12. By proof. ZDCB" A ABC or Z CAB. Q.E.D 154. Cor. From a point outside a line there exists only one perpendicular to the line. Hint. If there exists a second _L to AB from A 7^ " \f B P, as PC, then A PC A and Z. PKA are both rt. A and are therefore equal. But this is impossibleby " 153-155. "" 149 and 154 may be combined in one statement as follows : From a point outside a line there exists one and onlyone per-pendicula to the line, Ex. 148. A triangle cannot contain two rightangles. Ex. 149. In the figure of Prop. XIV, is angleDCB necessarily greater than angle BCA ? than angle B ? j" Ex. 150. In Fig. 1, prove that : (1) Angle 1 is greater than angle CAE or angle AEC ; (2) Angle 5 is greater than angle CBA or angleBAE ; (3) Angle EDA is greater than angle 3 ; (4) Angle 4 is greater than angle DAE. Ex. 151. In Fig. 2, show that angle 6 is greater than angle 7 ; also that angle 9 is greater than angle A. Fig. 2. 54 PLANE GEOMETRY Proposition XV. Theorem 156. If two sides of a triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less side. Given A ABC with BC " BA. To prove Z CAB " Z C. Argument Reasons 1. On BC lay off BD = AB. 1. Circle post. "" 122, 157. " 2. Draw AD. 2. Str. line post. I. " 54, 15. 3. Then Z. =Z2. 3. The base A of an isosceles A are equal. " 111. 4. If one side of a A is pro-longed, the ext. Z formed " either of the remote int. A. " 153. 5. Substituting Z 1 for its equal Z 2. 6. The whole " any of its parts. " 54, 12. 7. If three magnitudes of the same kind are so related that the first " the sec-ond and the second " the third,then the first q.e.d. " the third. " 54, 10. 157. Note. Hereafter the student will not be required to state postulates and definitions in full unless requested to do so by the teacher. 4. NowZ2 " Z.C. 5. r.Zl " Zc. (5. 7. But Z CAB " Z 1. /. ZCAB " ZC. BOOK I 55 158. Note. When two magnitudes are given unequal, the laying off of the less upon the greater will often serve as the initial step in de-veloping a proof. Ex. 152. Given the isosceles triangle EST, with ST the base and BT prolonged any length,as to K. Prove angle KSR greater than angle K. Ex. 153. If two adjacent sides of a quadrilateral are greater respec-tively than the other two sides, the angle included between the two shorter sides is greater than the angle between the two greater sides. Ex. 154. If from a point within a triangle lines are drawn to the ends of one of its sides,the angle between these lines is greater than the angle between the other two sides of the triangle. (See Ex. 151.) 159. The indirect method, or proof by exclusion, consists in contradictingthe conclusion of a proposition, then showing the contradiction to be false. The conclusion of the proposition is thus established. This process requires an examination of every possiblecontradiction of the conclusion. For example, to prove indirectly that A equals B it would be necessary to consider the onlythree suppositions that are admissible in this case, viz. : (i) A " B} (2) A " B, \ (3) A = B. \ By proving(1)and (2) false, the truth of (3)is established, i.e. A = B. This method of reasoning is called reductio ad absurdum. It enables us to establish a conclusion by showing that every contradiction of it leads to an absurdity. Props. XVI and XVII will be proved by the indirect method. 160. Question. Would it be possible to base a proof upon a contra-diction of the hypothesis? 161. (a) In the use of the indirect method the student should give, as argument 1, all the suppositionsof which the case he is considering admits, includingthe conclusion. As reason 1 the number of such possi-ble suppositions should be cited. (") As a reason for the last step in the argument he should state which of these suppositionshave been proved false. 56 PLANE GEOMETRY Proposition XVI. Theorem (Converse of Prop. IV) 162. If two angles of a triangle are equal, the sides opposite are equal. Given A RST with Z.R " Z.T. To prove r = t . Argument 1. r " tj r"t, or r = t. 2. First suppose r"t\ then Zi2"Zr. 3. This is impossible. 4. Next suppose r"t; then Z.R" Zr. 5. This is impossible. 6. .-. r = t. Q.E.D. Reasons In this case only three suppositions are admis-sible. If two sides of a A are unequal, the Z opposite the greater side " the Z oppositethe less side. "156. By hyp.,Zs = ZT. Same reason as 2. Same reason as 3. The two suppositions, r " t and r " t, have been proved false. 163. Cor. An equiangular triangleis also equilateral. Ex. 155. The bisectors of the base angles of an isosceles triangle form an isosceles triangle. 58 PLANE GEOMETRY Ex. 156. The sum of the altitudes of any triangleis less than the perimeter of the triangle. Ex. 157. Given the quadrilateral ABCD with B and D rightangles and BC greater than CD. Prove AD greater than AB. Ex.158. Given triangle ABC, with AG " BC. Let bisectors of angles A and B meet at 0. Prove AO " BO. Ex. 159. A line drawn from the vertex of an isosceles triangle to any point in the base is less than one of the equal sides of the triangle. Ex. 160. If ABC and ABB are two triangles on the same base and on the same side of it such that AC " BD and AD = BC, and if AC and BD intersect at 0, prove triangle AOB isosceles. Ex. 161. Prove Prop. XVI by using the figure and method of Ex. 69. Ex. 162. Upon a given base is constructed a triangle one of the base angles of which is double the other. The bisector of the.larger base angle meets the oppositeside at the point P. Find the locus of P. Ex. 163. If the four sides of a quadrilateral are equal, its diagonals bisect each other. Ex. 164. The diagonals of an equilateral quadrilateral are perpen-dicular to each other, and they bisect the angles of the quadrilateral. Ex. 165. If two adjacent sides of a quadrilateral are equal and the other two sides are equal, one diagonal is the perpendicularbisector of the other. Tell which one is the bisector and prove the correctness of your answer. Ex. 166. If,from a point in a perpendicular to a line, oblique lines are drawn cuttingoff equal segments from the foot of the perpendicular, the oblique lines are equal. Ex. 167. State and prove the converse of Ex. 166. Ex. 168. If,from a point in a perpendicularto a line, obliquelines are drawn cutting off unequal segments from the foot of the perpendicular, the obliquelines are unequal. Prove by laying off the less segment upon the greater. Then use Ex. 166, " 153, and " 164. Ex. 169. If, from any point in a perpendicularto a line, two unequal oblique lines are drawn to the line, the oblique lines will cut off unequal segments from the foot of the perpendicular. Prove by the indirect method. Ex. 170. By means of Prop. XIV, prove that the sum of any two anglesof a triangle is less than two rightangles. Ex. 171. Construct a triangleABC, given two sides, a and 6, and the altitude to the third side,c. (See " 152.) BOOK I 59 Proposition XVIII. Theorem 167. The sum of any two sides of a triangle is greater than the third side Given A ABC. To prove a + c " b. 168. Cor. I. Any side of a triangle is less than the sum and greater than the differenceof the other two. 169. Cor. II. Any straight line is less than the sum of the.parts of a broken line having the same extremities. 60 PLANE GEOMETRY 170. Note to Teacher. Teachers who preferto assume that " a straight line is the shortest line between two points " may omit Prop. XVIII entirely.Then Prop. XVII may be proved by a method similar to that used in Prop. XV. (SeeEx. 172.) Ex. 172. Prove Prop. XVII by using the hint contained in " 158. Ex. 173. If two sides of a triangle are 14 and 9, between what limit-ing values must the third side be ? Ex. 174. If the opposite ends of any two non-intersecting line seg-ments are joined,the sum of the joining lines is greater than the sum of the other two lines. ft Ex. 175. Given two points,P and R, and a line AB not passing through either. To find a point O, on AB, such that PO + OB shall be as small as possible. This exercise illustrates the law by which light is reflected from a mirror. The lightfrom the object, P, is reflected and appears to come from L, as far behind the mirror as P is in front of it. Ex. 176. If from any point within a trianglelines are drawn to the extremities of any side of the triangle, the sum of these lines is less than the sum of the other two sides of the triangle. Hint. Let ABC be the given triangle, D the point within. Prolong AD until it intersects BG at E. Apply Prop. XVIII. 171. Note to Teacher. Up to this pointall proofs given have been complete, including argument and reasons. In written work it is frequently convenient, however, to have students give the argument only. These two forms will be distinguished by calling the former a complete demonstration and the latter, which is illustrated in Prop. XIX, argument only. It is often a sufficienttest of a student's understanding of a theorem to have him state merely the main points involved in a proof. This may be given in enumerated steps,as in Prop. XXXV, or in the form of a paragraph, as in Prop. XLIV. This form will be called outline of proof. BOOK I 61 Proposition XIX. Theorem 172. If two triangles have two sides of one equal respec-tively to two sides of the other, but the included angle of the first greater than the included angle of the second, then the third side of the firstis greater than the third side of the second. Given two A ABC and DEF with AB = DEy BO = EF, but /.ABC" /-F. To prove AC " DF. Argument Only 1. Place A DEF on A ABC so that BE shall fall upon its equal AB, D upon A, E upon B. 2. EF will then fall between AB and BC. Denote A DEF in its new position by ABF. 3. Draw BK bisecting Z FBC and meeting AC at K. 4. Draw KF. 5. In A FBK and KBC, BC = BF. 6. BK = BK. 7. Zfbk = Zkbc. 8. .'. A FBK = A KBC. 9. .'.KF=KC. 10. AK+ KF"AF. 11. .'. AK--KC"AF. 12. That is,AC" AF. 13. ..'. AC"DF. Q.E.D. 62 PLANE GEOMETRY Ex. 177. (a) Draw a figureand discuss the case for Prop. XIX when F falls on AC; when F falls within triangleABC. (") Discuss Prop. XIX, taking AC (the base) = DF, AB=DE, and angle CAB greater than angle D. B ~ Ex. 178. Prove Prop. XIX by using Fig.1. Hint. In A A CE, A CEA " A BE A. ;. Z. CEA " Z. EAB "ZEAC. FlG. i Ex. 179. Given triangleABC with AB greater than BC, and let point P be taken on AB and point Q on CB, so that AP " CQ. Prove ^4^ greater than CP. Ex. 180. Would the conclusion of Ex. 179 be true if P were taken on AB pro-B longed and Q on CB prolonged ? Prove. Ex. 181. In quadrilateral ABCD if AB = CD, and angle OZM is greater than angle DAB, prove AC greater than BD. Proposition XX. Theorem (Converseof Prop. XIX) 173. If two triangles have two sides of one equal respec-tively to two sides of the other, but the third side of the first greater than the third side of the second, then the angle opposite the third side of the first is greater than the angls opposite the third side of the second. C D Given A ABC and DEF with AB = DE, BC = EF, but AC " DF. To prove /-B " Z.E. BOOK I 63 Argument 1. Zb"Ze, Zb = Ze, or Zb"Ze. 2. First suppose Zb "ZE\ then AC "DF. 3. This is impossible. 4. Next suppose Zb = Ze ; then AABC = ADEF. 5. .-. AC"BF. 6. This is impossible. 7. .-. Zb"Ze. Q.E.D. REASON8 1. In this case onlythree sup-positions are admissible. 2. If two A have two sides of one equalrespectively to two sides of the other, but the included Z of the first " the included Z of the second, then the third side of the first " the third side of the second. " 172. 3. By hyp.,AC " DF. 4. Two A are equal if two sides and the included Z of one are equal respec-tively to two sides and the included Z of the other. " 107. 5. Homol. parts of equal fig-ures are equal. " 110. 6. Same reason as 3. 7. The two suppositions, Zb " ZE and Zb = Ze, have been proved false. Ex. 182. If two opposite sides of a quadrilateral are equal,but its diagonals are unequal, then one angle opposite thegreater diagonal is greater than one angle oppositethe less diagonal. Ex. 183. If two sides of a triangle are unequal, the median drawn to the third side makes unequal angles with the third side. Ex. 184. If from the vertex S of an isosceles triangle EST a line is drawn to point P in the base BT so that HP is greater than PZ7, then angle BSP is greater than angle PST. Ex. 185. If one angle of a triangle is equal to the sum of the other tw", the triangle can be divided into two isosceles triangles. 64 PLANE GEOMETRY SUMMARY OP THEOREMS FOR PROVING ANGLES UNEQUAL 174. (a) When the angles are in the same triangle : If two sides of a triangle are unequal,the angle opposite the greater side is greater than the angle oppositethe less side. (6) When the angles are in different triangles: If two triangles have two sides of one equal respectively to two sides of the other,but the third side of the first greater than the third side of the second,then the angle opposite the third side of the first is greater than the angle opposite the third side of the second. (c)An exterior angle of a triangle is greater than either re-mote interior angle. SUMMARY OF THEOREMS FOR PROVING LINES UNEQUAL 175. (a) When the lines are in the same triangle: The sum of any two sides of a triangle is greater than the third side. Any side of a triangle is less than the sum and greater than the difference of the other two^ If two anglesof a triangle are unequal, the side opposite the greater angle is greater than the side oppositethe less angle. (b) When the lines are in different triangles: If two triangles have two sides of one equal respectively to two sides of the other,but the included angle of the first greater than the included angle of the second, then the third side of the first is greater than the third side of the second. (c)Every point not in the perpendicular bisector of a line is not equidistant from the ends of the line. The perpendicular is the shortest straight line from a point to a line. Any straightline is less than the sum of the parts of a broken line having the same extremities. In some texts Exs. 168 and 169 are given as theorems. 66 PLANE GEOMETRY Proposition XXI. Theorem 180. If two straight lines are parallel to a third straight line,they are parallel to each other. D Given lines a and b, each II c. To prove a IIb. 181. Def. A transversal is a line that intersects two or more other lines. 182. Defs. If two straightlines are cut by a transversal, of the eightanglesformed, 3,4,5, 6 are interior angles; 1,2, 7, 8 are exterior angles; 4 and 5, 3 and 6, are alternate interior angles; 1 and 8, 2 and 7, are alternate exterior angles; 1 and 5, 3 and 7, 2 and 6,4 and 8, are corresponding angles(called also exterior interior angles) BOOK I 67 Proposition XXII. Theorem 183. If two straight lines are cut by a transversal making a pair of alternate interior angles equal, the lines are parallel. M N "P Q B" Given two str. lines MN and OQ cut by the transversal AB in points C and D, making Z MOD = ZQDC. To prove MN IIOQ. Argument 1. MN and OQ are either II or not II. 2. Supposethat if JV is not II OQ; then they will meet at some point as P, forming, with line DC, A PDC. 3. Then /.MOD " ZQDC. 4. This is impossible. 5. /. MN IIOQ. Q.E.D. 3. Reasons 1. In this case only two sup-positions are admissible. 2. By def. of II lines. " 177. If one side of a A is pro-longed, the ext. Z formed " either of the remote int. A. " 153. Z.MCD = ZQDC, by hyp. The suppositionthat MN and OQ are not II has been proved false. 184. Cor. I. If two straight lines are cut by a trans-versal making a pair of corresponding angles equal, the lines are parallel. Bint. Prove a pair of alt. int. A equal, and apply the theorem. 68 PLANE GEOMETRY 185. Cor. II. If two straight lines are cut by a trans-versal making a pair of alternate exterior angles e-qual, the lines are parallel. (Hint. Prove a pair of alt. int. A equal.) 186. Cor. HI. If two straight lines are cut by a trans-versal making the sum of the two interior angles on the same side of the transversal equal to two right angles,the lines are parallel. ^ Given lines m and n cut by the J m transversal t making Zl + Z2 = 2rt.A jt To prove m \|\| n. Argument 1. Zl + Z2 = 2rt.A 2. Zl + Z3 = 2rt.Zs. 3. .". Zl + Z2 = Zl-fZ3. 4. .-. Z2 = Z3. 5. .. m II n. Q.E.D. Reasons 1. By hyp. 2. If one str. line meets an-other str. line,the sum of the two adj. A is 2 rt. A. " 65. 3. Things equal to the same thing are equal to each other. " 54, 1. 4. If equals are subtracted from equals, the remain-ders are equal. " 54,3. 5. If two str. lines are cut by a transversal making a pairof alt. int. A equal, the lines are II. " 183. 187. Cor. IV. If two straight lines are perpendicular to a third straight line,they are parallel to each other. Ex. 190. If two straight lines are cut by a transversal making the sum of the two exterior angles on the same side of the transversal equal to two rightangles,the lines are parallel. BOOK I 69 Ex. 191. In the annexed diagram, if angle " BGE = angle CHF, are AB and CD parallel ? V Trove. A ^" R Ex. 192. In the same diagram, if angle HGB Q " "" D = angle GHC, prove that the bisectors of these .tt angles are parallel. Ex. 193. If two straight lines bisect each other,the lines joiningtheir extremities are parallel in pairs. Ex. 194. In the diagram for Ex. 191, if angle BGE and angle FHD are supplementary, prove AB parallel to CD. Ex. 195. If two adjacent angles of any quadrilateral are supplemen-tary, two sides of the quadrilateral will be parallel. Proposition XXIII. Problem 188. Through a given point to construct a line parallel to a given line. A A ""Y /R K D Given line AB and point P. To construct, through, point P, a line IIAB, I. Construction 1. Draw a line through P cuttingAB at some point, as R. 2. With P as vertex and PS as side, construct Z. YPS = ZBRP. "125. 3. XY will be IIAB. II. The proof and discussion are left as an exercise for the student. Ex 196. Through a given point construct a parallel to a given line by using: (a) " 183 ; (6) " 185 ; (c) " 187. 70 PLANE GEOMETRY Proposition XXIV. Theorem (Converse of Prop. XXII) 189. If two parallel lines are cut by a transversal, the alternate interior angles are equal. Given \|\| lines AB and CD cut by the transversal EF at points G and H. To prove Z A GH = Z DHG. Argument 1. Either Z A GH = Z DHG, or ZAGHJ= ZDHG. 2. Suppose ZAGH^ZDHG, but that line XY, through G, makes ZXGH= ZDHG. 3. Then XY \|\| CD. 4. But AB \|\| CD. 5. It is impossiblethat AB and XY both are \|\| CD. 6. .'.ZAGH= ZDHG. Q.E.D. Reasons 1. In this case only two sup-positions are admissible. 2. With a given vertex and a given side,an Z may be constructed equal to a given Z. " 125. 3. If two str. lines are cut by a transversal making a pair of alt. int. A equal, the lines are \|\| . "183. 4. By hyp. 5. Parallel line post. " 178. 6. The supposition that ZAGH^Z DHG has been proved false. BOOK I 71 ^ 190. Cor. I. (Converseof Cor. I of Prop.XXII). // two parallellines are cut by a transversal, the corresponding angles are equal. A Given two \|\| lines XY and MN out by the transversal AB, form-ing correspondingA 1 and 2. ^ To prove Z 1 = Z2. Hint. Z 3 = A 2 by " 189. \" 191. Cor. n. (Converse of Cor. II of Prop. XXII). If two parallellines are cut by a transversal, the alter-nate exterior angles are equal. 192. Cor. m. (Converse of Cor. Ill of Prop. XXII). If two parallel lines are cut by a transversal, the sum of the two interior angles on the same side of the trans-versal is two right angles. Given two IIlines XY and MN X_ cut by the transversal AB, form-ing int. A 1 and 2. jf To prove Z 1 + Z 2 = 2 rt. A. Q.E.D. 193. Cor. IV. A straight line perpendicular to one of two parallelsis perpendicular to the other also. 194. Cor. V. (Opposite of Cor. Ill of Prop. XXII). If two straight lines are cut by a transversal making the sum of the two interior angles on the same side of the transversal not equal to two right angles, the lines are not parallel. (Hint. Apply " 137, or use the indirect method.) 72 PLANE GEOMETRY 195. Cor. VI. Two lines perpendicular respectively to two intersecting lines also intersect. Given two intersecting lines AB A and CD, and BE L AB, CF " CD. To prove that BE and CF also intersect. Argument Only 1. Draw CB. 2. Zfcd is a rt. Z. 3. .-. Zl." art.Z. 4. Likewise Z2 " a rt. Z. 5. .-. Z1 + Z2 " 2rt. A. 6. .-. BE and CF also intersect. q.e.d. 196. Def. Two or more lines are said to be concurrent if they intersect at a common point. Ex. 197. If two parallels are cut by a transversal so that one of the angles formed is 45",how many degrees are there in each of the other seven angles? Ex. 198. If a quadrilateral has two of its sides parallel, two pairs of its angleswill be supplementary. " -, Ex. 199. In the annexed diagram AB is % parallel to CD, . and EF is parallel to OH. Prove (a) angle 1 " angle 2 ; \ \S (") angle 1 + angle 3 = 2 rightangles. F H Ex. 200. If a line is drawn through any point in the bisector of an angle,parallel to one of the sides of the angle,an isosceles triangle will be formed. Ex. 201. Draw a line parallel to the base of a triangle, cutting the sides so that the sum of the segments adjacent to the base shall equal the parallel line. Ex. 202. If two parallel lines are cut by a transversal, the bisectors of a pair of corresponding angles are parallel. 74 PLANE GEOMETRY Proposition XXV. Theorem 198. Two angles whose sides are parallel,each to each, are either equal or supplementary. Given Zl and the A at E, with AB IIEF and A C IIDE. To prove Zl Z5 = 2rt.Zs. Z2=Z3, and Zl + Z4 = 2 rt. 4 Zl + Argument 1. Prolong ".4 and Z"J" until they intersect at some point as H. 2. Z1 = Z6. 3. Z2 = Z6. 4. .-. Z1 = Z2. 5. Z3=Z2. 6 .-. Z1 = Z3. 7. Z2 + Z4 = 2rt.A 8. .-. Z 1 + Z 4 = 2 rt. Zs. Reasons 1. Str.linepost.il. "54,16. 2. Corresponding A of \|\| lines are equal. " 190. 3. Same reason as 2. 4. Things equal to the same thing are equal to each other. " 54, 1. 5. If two str. lines intersect, the vertical A are equal. "77. 6. Same reason as 4. 7.. If one str. line meets an-other str. line,the sum of the two adj.A is 2 rt. A. " 65. 8. SubstitutingZ 1 for its equal Z 2. BOOK 1 75 Argument 9. Z5=Z4. 10. .-. Zl + Z5 = 2rt.A. Q.E.D. Reasons 9. Same reason as 5. 10. Substituting Z5 for its equal Z4. 199. Note. Every angleviewed from its vertex has a rightand a left side ; thus, in the annexed diagram, the rightside of ZA is r and the left side,I ; the rightside of Z.B is r and the left, I. Fig. 2, Fig. 3. In the diagram of Prop. XXV, Zl and Z2, whose sides are \|\| rightto right(ABto EF) and left to left (AC to EK) are equal ; while Zl and Z4, whose sides are \|\| right to left (AB to EF) and left to right {AG to ED) are supplementary. Hence : 200. (a) If two angles have their sides parallel right to right and left to left, they are equal (b) If two angleshave their sides parallel right to left and left to right, they are supplementary. Ex. 213. In Fig.3, above, show which isthe rightside of each of the four anglesabout O. Ex. 214. If a quadrilateral has its oppositesides parallel, its opposite angles are equal. Ex. 215. Given two equal angles having a side of one parallel to a side of the other,are the other sides necessarily parallel ? Prove. Ex. 216. If two sides of a triangleare parallelrespectively to two homologous sides of an equal triangle, the third side of the first is parallel to the third side of the second. Ex. 217. Construct a triangle, given an angle and its bisector anil th" 'altitude drawn from the vertex of the given angle. 76 PLANE GEOMETRY Proposition XXVI. Theorem 201. Two angles whose sides are perpendicular, each to each, are either equal or supplementary. H K C D Given Zl and the A at P, with OA _L EC and OB J HD. To prove Z1=Z2=Z3, Zl+Z4=2rt. Z",Z1+Z5=2 rt. A. Hint. Draw OE IICK and OFW DH. Prove OE " OA and OF" OB. Prove Z 7 = Z 1, and prove Z 7 = Z 2. 202. Note. It will be seen that Z\ and Z2, whose sides are " right to right (OA to PIT) and left to left (OB to PIT), are equal ; while Zl and Z4, whose sides are " right to left (0.4 to PC) and left to right (OB to P^T), are supplementary. Hence : 203. (a) If two angles have their sides perpendicular right to right and left to left, they are equal. (b)If two angles have their sides perpendicular rightto left and left to right, they are supplementary. Ex. 218. If from a point outside of an angle perpendiculars are drawn to the sides of the angle, an angle is formed which is equal to the given angle. Ex. 219. In a right triangleif a perpendicular is drawn from the vertex of the right angle to the hypotenuse, the right angie is divided into two angles which are equal respectively to the acute angles of the triangle. Ex. 220. If from the end of the bisector of the vertex angle of an isosceles triangle a perpendicularis dropped upon one of the arms, the perpendicularforms with the base an angle equal to half the vertex angle. BOOK I 77 Proposition XXVII. Theorem 204. The sum of the angles of any triangle is two right angles. Given A ABC. To prove A A + /.ABC + AC Argument Through B draw BE Al + /ABC+ A 2 = 2 rt. A AC. 3. Zl=Zi. = 2 rt. A. Reasons 1. Parallel line post. "179. 2. The sum of all the ^ about a point on one side of a str. line passing through that point = 2 rt. A. "66. 3. Alt. int. A of II lines are equal. " 189. 4. Same reason as 3. 5. Substituting for A 1 and 2 their equals, z" ^1 and C, respectively. 205. Cor. I. In a right triangle the two acute angles are complementary. 206. Cor. II. In a triangle there can be but one right angle or one obtuse angle. 207. Cor. in. If two angles of one triangle are equal respectively to two angles of another, then the third angle of the first is equal to the third angle of the second. 208. Cor. IV. If two right triangles have an acute angle of one equal to an acute angle of the other, the othgr acute angles are equal. 78 PLANE GEOMETRY 209. Cor. V. Two right triangles are equal if the hy-potenu and an acute angle of one are equal respectively to the hypotenuse and an acute angle of the other. 210. Cor. VI. Two right triangles are equal if a side and an acute angle of one are equal respectively to a side and the homologous acute angle of the other. 211. Cor. VII. Two right triangles are equal if the hypotenuse and a side of one are equal respectively to the hypotenuse and a side of tJie other. Given rt. A ABC and BEFy with AB = BE and BC=EF. To prove A AB C = A BEF. Hint. Prove DFK a str. line ; then A DEK is isosceles. 212. Cor. VIII. The altitude upon the base of an isosceles triangle bisects the base and also the vertex angle. 213. Cor. IX. Each angle of an equilateral triangle is one third of two right angles, or 60". 214. Question. Why is the word homologous used in Cor. VI but not in Cor. V ? Ex. 221. If any angle of an isosceles triangle is 60",what is the value of each of the two remaining angles? Ex. 222. If the vertex angle of an isosceles triangle is 20",find the angle included by the bisectors of the base angles. Ex. 223. Find each angle of a triangleif the second angle equals twice the first and the third equals three times the second. Ex. 224. If one angle of a triangle is m" and another angle Z",write an expression for the third angle. Ex. 225. If the vertex angle of an isosceles triangleis a",write an expressionfor each base angle. BOOK I 79 Ex. 226. Construct an angle of 60" ; 120" ; 30" ; 15". Ex. 227. Construct a righttriangle having one of its acute angles 60". How largeis the other acute angle ? Ex. 228. Construct an angle of 160". Ex. 229. above. Ex. 230. Ex. 231. Ex. 232. Prove Prop. XXVII by using each of the diagrams given Given two angles of a triangle, construct the third angle. Find the sura of the angles of a quadrilateral. The angle between the bisectors of two adjacent angles of a quadrilateral is equal to half the .sum of the two remaining angles. Ex. 233. The angle between the bisectors of the base angles of an isos-celes triangle is equal to the exterior angle formed by prolonging the base. Ex. 234. If two straight lines are cut by a transversal and the bisectors of two interior angles on the same side of the transversal are perpendicular to each other,the lines are parallel. Ex. 235. If in an isosceles triangleeach of the base angles is one fourth the angle at the vertex, a line drawn perpendicular to the base at one of its ends and meeting the opposite side prolonged will form with the adjacent side and the exterior portion of the opposite side an equilateral triangle. Ex. 236. Two angles whose sides are perpendicular each to each are either equal or supple-mentary. (Prove by using the annexed diagram.) Ex. 237. If at the ends of the hypotenuse of a right triangleper-pendiculars to the hypotenuse are drawn meeting the other two sides of the triangleprolonged, then the figurecontains five triangleswhich are mutually equiangular. Ex. 238. If one angle of a triangleis 50" and another angle is 70", find the other interior angle of the triangle ; also the exterior angles of the triangle. What relation is there between an exterior angle and the two remote interior angles of the triangle ? 80 PLANE GEOMETRY Proposition XXVIII. Theorem 215. An exterior angle of a triangle is equal to the sum of the two remote interior angles. B C Given A ABC with Z.DCB an exterior Z. To prove A BOB = A A -f AB. The proof is left as an exercise for the student. Hint. Draw CE II AB. Ex. 239. The bisector of an exterior angle at the vertex of an isosceles triangle is parallel to the base. Ex. 240. If the sura of two exterior angles of a triangle is equal to three rightangles,the triangle is a righttriangle. Ex. 241. The sum of the three exterior angles of a triangle is four rightangles. Ex. 242. What is the sum of the exterior angles of a quadrilateral ? Ex. 243. If the two exterior angles at the base of any triangleare bisected, the angle between these bisectors is equal to half the sum of the interior base angles of the triangle. Ex. 244. If BE bisects angle B of triangle ABC, and AE bisects the exterior angle DA C, angle E is equal to one half angle C. Ex. 245. D is any point in the base BC of isosceles triangle ABC. The side AC is prolonged through C to E so that CE = CD, and DE is drawn meeting AB at F. Prove angle EFA equal to three times angle AEF. Ex. 246. The mid-point of the hypotenuse of a right triangleis equidistantfrom the three vertices. Prove by laying off on the right angle either acute angle. 82 PLANE GEOMETRY Proposition XXX. Theorem 218. If the sides of any polygon are prolonged in succes-sion one way, no two adjacent sides being prolonged through the same vertex, the sum, of the exterior angles thus formed is four right angles. 4/D Given polygon P with Zl, Z2, Z3, Z4, ..-its successive ex-terior angles. To prove Z 1 + Z 2 + Z3 + Z4 + """ = 4 rt. A. Argument 1. Zl+^6 = 2 rt. 4 Z2 + Z7 = 2 rt. Z, and so on ; i.e. the sum of the int. Z and the ext. Z at one vertex = 2 rt. A, 2. .-. the sum of the int. and ext. zs at the n vertices = 2 n rt. A. 3. Denote the sum of all the interior A by J and the sum of all the ext. A by E\ then E+I=2 n rt. Z. 4. But/=2nrt.Z-4rt.Z. 5. .-.." = 4 rt. Z; t\e. Zl + Z2 + Z3+Z4+ ... = 4rt.Z. q.e.d. 1. 2. Reasons If one str. line meets an-other str. line, the sum of the two adj. Z is 2 rt. Z. " 65. If equalsare multipliedby equals, the products are equal. " 54, 7 a. 3. Arg. 2. 4. The sum of all the Z of any polygon = 2 n rt. A -4 rt. Z. " 216. 5. If equals are subtracted from equals, the remain-ders are equal. " 54, 3. BOOK I 83 219. Note. The formula 2 n rt. A " 4 rt. A (" 216) is sometimes more useful iu the form (;i " 2) 2 rt. A. "fix. 247. Find the sum of the angles of a polygon of 7 sides ; of 8 sides ; of 10 sides. Ex. 248. Prove Prop. XXIX by drawing as many diagonals as possible from one vertex. Ex. 249. How many diagonalscan be drawn from one vertex in a polygon of 8 sides ? of 50 sides ? of n sides ? Show that the greatest num-ber of diagonals possiblein a polygon of n sides (using all vertices) is n(n -3) , 2 Ex. 250. How many degrees are there in each angle of an equi-angular quadrilateral ? in each angle of an equiangularpentagon ? Ex. 251. How many sides has a polygon the sum of whose angles is 14 rightangles ? 20 rightangles? 540" ? Ex. 252. How many sides has a polygon the sum of whose interior angles is double the sum of its exterior angles ? Ex. 253. Is it possible for an exterior angle of an equiangularpoly-gon to be 70"? 72"? 140"? 144"? Ex. 254. How many sides has a polygon each of whose exterior angles equals 12" ? Ex. 255. How many sides has a polygon each of whose exterior angles is one eleventh of its adjacent interior angle ? Ex. 256. How many sides has a polygon the sum of whose interior angles is six times the sum of its exterior angles? Ex. 257. How many sides has an equiangular polygon if the sum of three of its exterior angles is 180" ? Ex. 258. Tell what equiangular polygons can be put together to make a pavement. How many equiangular trianglesmust be placed with a common vertex to fillthe angular magnitude around a point ? 84 PLANE GEOMETRY QUADRILATERALS. PARALLELOGRAMS QUADRILATERALS CLASSIFIED WITH RESPECT TO PARALLELISM 220. Def. A parallelogram is a quadrilateral whose oppo-site sides are parallel. 221. Def. A trapezoid is a quadrilateral having two of its oppositesides parallel and the other two not parallel. 222. Def. A trapezium is a quadrilateral having no two of its sides parallel. PARALLELOGRAMS CLASSIFIED WITH RESPECT TO ANGLES 223. Def. A rectangle is a parallelogram having one right angle. \ It is shown later that all the anglesof a rectangle are right angles. 224. Def. A rhomboid is a parallelogram having an oblique angle. It is shown later that all the angles of a rhomboid are oblique. 225. Def. A rectanglehaving two adjacentsides equal is a square. It is shown later that all the sides of a square are equal. 226. Def. A rhomboid having two adjacent sides equalis a rhombus. It is shown later that all the sides of a rhombus are equal. 227. Def. A trapezoidhaving its two non-parallel sides equal is an isosceles trapezoid. 228. Def. Any side of a parallelogram may be regarded as its base, and the line drawn perpendicularto the base from any point in the opposite side is then the altitude. 229. Def. The bases of a trapezoid are its parallel sides, and its altitude is a line drawn from any point in one base per-pendicula to the other. BOOK I 85 Proposition XXXI. Theorem 230. Any two opposite angles of a parallelogram are equal, and any two consecutive angles are supplementary. Given O ABCD. To prove : (a) ZA = Zc, and ZB = Zd ; (6) any two consecutive A, as A and B, sup. Argument 1. ZA = Z C and Zb=Zd. 2. A A and B are sup. Q.E.D. Reasons If two A have their sides IIrightto right and left to left, they are equal. " 200, a. If two II-lines are cut by a transversal,the sum of the two int.A on the same side of the transversal is two rt. A. " 192. 231. Cor. All the angles of a rectangle are right angles, and all the angles of a rhomboid are obliqueangles. Ex. 259. If the oppositeangles of a quadrilateral are equal, the figure is a parallelogram. Ex. 260. If an angle of one parallelogram is equal to an angle of another, the remaining angles are equal each to each. Ex. 261. The bisectors of the angles of a parallelogram (not a rhom-bus op a square)inclose a rectangle. 86 PLANE GEOMETRY Proposition XXXII. Theorem 232. The opposite sides of a parallelogram are equal. Given O ABCD. To prove AB = CD and BC=AD. The proof is left as an exercise for the student. 233. Cor. I. All the sides of a square are equal, and all the sides of a rhombus are equal. 234. Cor. II. Parallel lines intercepted between the same parallel lines are equal. 235. Cor. III. The perpendiculars drawn to one of two parallel lines from any two points in the other are equal. 236. Cor. rv. A diagonal of a parallelogram divides it into two equal triangles. Ex. 262. The perpendicularsdrawn to a diagonal of a parallelogram from the oppositevertices are equal. Ex. 263. The diagonals of a rhombus are perpendicularto each other and so are the diagonalsof a square. Ex. 264. The diagonalsof a rectangleare equal. Ex. 265. The diagonalsof a rhomboid are unequal. Ex. 266. If the diagonalsof a parallelogramare equal,the figureis a rectangle. Ex. 267. If the diagonals of a parallelogramare not equal,the figure is a rhomboid. Ex. 268. Draw a line parallel to the base of a triangleso that the portion interceptedbetween the sides may be equal to a given line. Ex. 269. Explain the statement : Parallel lines are everywhere equi-distant. Has this been proved ? BOOK I 87 Ex. 270. Find the locus of a pointthat is equidistant from two given parallel lines. Ex. 271. Find the locus of a point: (a) one inch above a given horizontal line ; (") two inches below the given line. Ex. 272. Find the locus of a point : (a) one inch to the rightof a given vertical line ; (6) one inch to the left of the given line. Ex. 273. Given a horizontal line OX and a line OY perpendicular to OX. Find the locus of a point three inches above OX and two inches to the rightof OY. 237. Historical Note. Rene" Descartes (1596-1650) was the firstto observe the importance of the fact that the position of a point in a plane is determined if its distances, say x and y, from two fixed lines in the plane,perpendic-ular to each other,are known. He showed that geometric fig-ures can be represented by algebraicequations, and de-veloped the subjectof analytic geometry, which is known by his name as Cartesian geome-try. Descartes was born near Tours in France, and was sent at eight years of age to the famous Jesuit school at La Fl"che. He was of good fam-ily, and since, at that time, most men of position entered either the church or the army, he chose the latter, and joined the army of the Prince of Orange. One day, while walking in a street in a Holland town, he saw a placard which challenged every one who read it to solve a certain geometric problem. Descartes solved it with little difficulty and is said to have realized then that he had no taste for militarylife. He soon resignedhis commission and spent five years in travel and study. After this he lived a short time in Paris, but soon retired to Holland, where he lived for twenty years, devoting his time to mathematics, philosophy, astronomy, and physics. His work in philosophy was of such importance as to give him the name of the Father of Modern Philosophy. Descartes 88 PLANE GEOMETRY Proposition XXXIII. Theorem 238. The diagonals of a parallelogram bisect each other. Given O A BCD with its diagonals AC and BB intersecting at 0. To prove AO = OC and BO = OB. Hint. Prove A OBC = A OB A. .. AO = OC and BO = OD. Ex. 274. If through the vertices of a triangle lines are drawn parallel to the opposite sides of the triangle, the lines which jointhe vertices of the triangle thus formed to the oppositevertices of the given triangle are bisected by the sides of the given triangle. Ex. 275. A line terminated by the sides of a parallelogramand passing through the point of intersection of its diagonals is bisected at that point. Ex. 276. How many parallelograms can be constructed having a given base and altitude ? What is the locus of the point of intersection of the diagonalsof all these parallelograms ? Ex. 277. If the diagonals of a parallelogram are perpendicular to each other, the figureis a rhombus or a square. Ex. 278. If the diagonals of a parallelogram bisect the angles of the parallelogram,the figureis a rhombus or a square. Ex. 279. Find on one side of a triangle the point from which straight lines drawn parallel to the other two sides, and terminated by those sides, are equal. (See " 232.) Ex. 280. Find the locus of a point at a given distance^ from a given finite line AB. Ex. 281. Find the locus of a point at a given distance from a given line and also equidistant from the ends of another given line. Ex. 282. Construct a parallelogram,given a side, a diagonal, and the altitude upon the given side. 90 PLANE GEOMETRY Ex. 284. Construct a rectangle, given the base and the altitude. Ex. 285. Construct a square, given a side. Ex. 286. Through a given point construct a parallel to a given line by means of Prop. XXXIV. Ex. 287. Construct a median of a triangleby means of a parallelo-gram, (1) using "" 239 and 238 ; (2) using "" 220 and 238. Ex. 288. An angle of a triangleis right, acute, or obtuse according as the median drawn from its vertex is equal to,greater than, or less than half the side it bisects. Proposition XXXV. Theorem 240. If two opposite sides of a quadrilateral are equal and parallel, the figure is a parallelogram. A D Given quadrilateral ABCD, with BC both equal and IIto AD. To prove ABCD a O. Outline of Proof 1. Draw diagonal BD. 2. Prove A BCD " A ABD. 3. Then Zcdb == /.ABD and AB II CD. 4. .'.ABCD is a O. Ex. 289. If the mid-points of two opposite sides of a parallelogram are joinedto a pair of opposite vertices, a parallelogramwill be formed. Ex. 290. Construct a parallelogram, having given a base,an adjacent angle, and the altitude,making your construction depend upon " 240. Ex. 291. If the perpendiculars to a line from any two points in an-other line are equal,then the lines are parallel. Ex. 292. If two parallelogramshave two vertices and a diagonal in common, the lines joiningthe other four vertices form a parallelogram. BOOK I 91 Proposition XXXVI. Theorem (Converseof Prop. XXXIII) 241. If the diagonals of a quadrilateral bisect each other, the figure is a parallelogram. B A D Given quadrilateral ABCD with its diagonalsAC and BD in-tersect at 0 so that AO = CO and BO = 1)0. To prove ABCD a O. Argument Only 1. In A OBC and ODA, BO = DO. 2. C0 = AO. 3. Z3 = Z4. 4. .-.A OBC=A OBA. 5. .'. BC=AB. 6. AlsoZl = Z2. 7. .. BC\ AD. 8. .-. ABCD is a O. q.e.d. Ex. 293. In parallelogram ABCD, let diagonal AC be prolonged through A and C to X and Y, respectively, making AX = CY. Prove XB YD a parallelogram. Ex. 294. If each half of each diagonal of a parallelogram is bisected, the lines joiningthe points of bisection form a parallelogram. Ex. 295. Lines drawn from the vertices of two angles of a triangle and terminating in the oppositesides cannot bisect each other. Ex. 296. State four independent hypotheses which would lead to the conclusion,"the quadrilateral is a parallelogram." Ex. 297. Construct a parallelogram, given its diagonals and an angle between them. 92 PLANE GEOMETRY Proposition XXXVII. Theorem 242. Two parallelograms are equal if two sides and the included angle of one are equal respectively to two sides and the included angle of the other. II Given ZU I and II with AB = EF, AD = EH, and Z.A= /.E. To prove OI = OII. 7. 8. Argument Place O I uponO II so that AB shall fall upon its equal EF, A upon E, B upon F. Then AD will become col-linear with EH. Point D will fall on H. Now DC II AB, and HG II EF. .-. DC and HG are both II AB. .: DC will become collinear with HG, and C will fall somewhere on HG or its prolongation. Likewise BC will become collinear with FG, and C will fall somewhere on FG or its prolongation. .. point C must fall on pointG. 9. .-.01 = 011. Q.E.D. Reasons 1. Transference post. "54,14. 2. /.A = ZE, by hyp. 3. AD = EH, by hyp. 4. By def . of a O. " 220. 5. AB and ^i coincide, Arg. 1. 6. Parallel line post. "179. 7. By steps similar to 4, 5, and 6. 8. Two intersecting str. lines can have only one point in common. " 26. 9. By def.of equalfigures. " 18. BOOK I 93 243. Quadrilaterals QUADRILATERALS CLASSIFIED 1. Opposite sides II : Parallelogram. (a) Right-angled: Rectangle. Two adj. sides equal: Square. (b) Oblique-angled: Khomboid. Two adj.sides equal : Rhombus. 2. Two sides II, other two non-ll : Trap-ezoid. (a) Two non-ll sides equal : Isosceles trapezoid. 3. No two sides II : Trapezium. Ex. 298. If it is requiredto prove a given quadrilateral a rectangle, show by reference to " 243 that the logical steps are to prove first that it is a parallelogram; then that it has one rightangle. Ex. 299. If a given quadrilateral is to be proved a square, show that the only additional step after those in Ex. 298 is to prove two adjacent sides equal. Ex. 300. If a given quadrilateral is to be proved a rhombus, what are the three steps corresponding to those given in Ex. 298 and Ex. 299 ? Ex. 301. Since rectangles and rhomboids are parallelograms,they possess all the general properties of parallelograms. What property dif-ferentiate rectanglesfrom rhomboids (1) by definition ? (2) by proof ? (See Ex. 266 and Ex. 267.) Ex. 302. (a) What two properties that have been proved distinguish squares from other rectangles ? (See Ex. 277 and Ex. 278.) (b) What two properties that have been proved distinguish rhombuses from other rhomboids? (See Ex. 277 and Ex. 278.) (c) Show that the two properties which distinguish squares and rhom-buses from the other members of their class are due to the common prop-erty possessed by squares and rhombuses by definition. Ex. 303. The mid-point of the hypotenuse of a righttriangle is equi-distant from the three vertices. Ex. 304. If a line AB of given lengthis moved so that itsends always touch the sides of a given right angle,what is the locus of the mid-point of%B? 94 PLANE GEOMETRY Proposition XXXVIII. Theorem 244. If three or more parallel lines interceptequal seg-ments on one transversal, they intercept equal segments on any other transversal. Given IIlines AG, BH, CJ, DK, etc., which intercept the equal segments AB, BC, CD, etc., on transversal AF, and which intercept segments GH, HJ, JK, etc.,on transversal GL. To prove GH = HJ = JK, etc. Argument 1. Draw GM, HN, JR, etc. II AF. 2. Now AG MB, BHNC, CJRD, etc., are HJ. 3. .'.GM = AB, HN = BC, JR = CD, etc. 4. And AB = BC = CD, etc. 5. .-. GM = HN = JR, etc. 6. Again GM, HN, JR, etc., are IIto each other. 7. .-. Zl = Z2 = Z3, etc. 8. And Z4 = Z5 = Z6, etc. Reasons 1. Parallel line post. " 179. 2. By def. of a O. " 220. 7. The opposite sides of a O are equal. " 232. By hyp. Ax. 1. " 54, 1. If two str. lines are IIto a third str. line, they are IIto each other. " 180. Corresponding A of II lines are equal. " 190. If two A have their sides IIrightto rightand left to left, they are equal. " 200, a. BOOK I 95 9. .-. A GHM = A HJN = AJKE, etc. 10. .'. GH= HJ = JK, etc. Q.E.D. 9. Two A are equal if a side and the two adj. A of one are equal respectively to a side and the two adj. A of the other. " 105. 10. Homol. parts of equal figures are equal. " 110. 245. Question. Are the segments that the parallels intercept on one transversal equal to the segments that they intercepton another trans-versal ? Illustrate. 246. Cor. I. The line bisecting one of the non-paral-lel sides of a trapezoid and parallel to the bases bisects the other of the non-parallel sides also. 247. Cor. n. The line joining the mid-points of the non-parallel sides of a trapezoid is (a) parallel to the bases; and (b) equal to one half their sum. Hint, (a) Prove EF WBC and AD by the indirect method. (6) Draw GH IICD. Prove AH=GB; then prove EF = GC = % (GC + HD) = KBC+AD). 248. Cor. m. The line bisect-ing one side of a triangle and parallel to another side bisects the third side. H 249. Cor. IV. The line joining the mid-points of two sides of a triangle is parallel to the third side and equal to one half the third side. . Hint. Draw CF II BA. Prove CF=AE = EB. 96 PLANE GEOMETRY Proposition XXXIX. Problem 250. To divide a given straight line into any number of equal parts. A H B Given straight line AB. To divide AB into n equalparts. I. Construction 1. Draw the unlimited line AX. 2. Take any convenient segment, as AR, and, beginning at A, lay it off n times on AX.' 3. Connect the nth point of division, as K, with B, 4. Through the precedingpointof division, as P, draw a line PH IIKB. " 188. 5. Then HB is one nth of AB. 6. .-. HB, if laid off successively on AB, will divide AB into n equal parts. II. The proof and discussion are left as an exercise for the student. Ex. 305. Divide a straight line into 7 equal parts. Ex. 306. Construct an equilateral triangle, having given the perim-eter. Ex. 307. Construct a square, having given the perimeter. 251. Def. The line joining the mid-points of the non-parallel sides of a trapezoid is called the median of the trape-zoid. Ex. 308. Show, by generalizing, that Cor. Ill,Prop. XXXVIII, may be obtained from Cor. I -and Cor. IV from Cor. II. 98 PLANE GEOMETRY Proposition XL. Theorem 252. The two perpendiculars to the sides of an angle from any point in its bisector are equal. B D A Given Z ABC', P any point in BR, the bisector of ZABC; PD and PE, the is from P to BA and BC respectively. To prove PD = PE. Argument Only 1. In rt. A DBP and PBE, PB = PB. 2. Z DBP " Z PBE. 3. .-. ADBP = APBE. 4. .'. PD =PE. Q.E.D. 253. Prop. XL may be stated as follows : Every point in the bisector of an angleis equidistant from the sides of the angle. Ex. 322. Find a point in one side of a trianglewhich is equidistant from the other two sides of the triangle. Ex. 323. Find a point equidistant from two given intersecting lines and also at a given distance from a fixed third line. Ex. 324. Find a point equidistant from two given intersecting lines and also equidistant from two given parallel lines. Ex. 325. Find a point equidistant from the four sides of a rhombus. Ex. 326. The two altitudes of a'rhombus are equal. Prove. Ex. 327. Construct the locus of the center of a circle of given radius, which rolls so that it always touches the sides of a givenangle. Do not prove, BOOK I 99 Proposition XLI. Theorem (Opposite of Prop. XL) 254. The two perpendiculars to the sides of an angle from any point not in its bisector are unequal. C B G D A Given /.ABC; P any pointnot in BR, the bisector of /ABC; PD and PE, J" from P to BA and BC respectively. To prove PD =" PE. Outline of Proof Draw FG J_ BA ; draw PG. Then FE = FG. Now PF--FG" PG. .'.PE"PG. But PG " PD. .'. PE " PD. 255. Prop. XLI may be stated as follows : Every point not in the bisector of an angle is not equidistant from the sides of the angle. 256. Cor. I. (Converseof Prop. XL). Every point equi-distant from the sides of an angle lies in the bisector of the angle. Hint. Prove directly, using the figure for " 262, or apply " 137. 257. Cor. II. The bisector of an angle is the locus of all points equidistant from the sides of the angle. Ex. 328. What is the locus of all points that are equidistant from a pair of intersecting lines ? 100 PLANE GEOMETRY .CONCURRENT LINE THEOREMS Proposition XLII. Theorem 258. The bisectors of the angles of a triangle are con-current in a point which is equidistant from the three sides of the triangle. Given A AB O with AF, BE, CD the bisectors of A A, B, and C respectively. To prove: (a) AF, BE, CD concurrent in some point as 0; (b)the point0 equidistant from AB, BC, and CA. 1. Argument BE and CD will intersect at some pointas 0. 2. 3. Draw OL, OH, and OG, J" from 0 to AB, BC, and CA respectively. V 0 is in BE, OL = OH. 0 is in CD, OG=OH. OL = OG; Reasons 1. If two str. lines are cut by a transversal making the sum of the two int. A on the same side of the transversal not equal to 2 rt. A, the lines are not II. " 194. 2. From a pointoutside a line there exists one and only one J_ to the line. " 155. 3. The two Js to the sides of an Z from any point in its bisector are equal. "252. 4. Same reason as 3. 5. Things equal to the same thing are equal to each other. " 54;1. book i aoi Argument 6. .. AF, the bisector of Z.CAB, passes through O. 7. .-. AF, BE, and CD are con-current in 0. 8. Also 0 is equidistantfrom AB, BC, and CA. q.e.d. Reasons 6. Every point equidistant from the sides of an Z lies in the bisector of the Z. " 256. 7. By def . of concurrent lines. "196. 8. By proof,OL = OH= OG. 259. Cor. The point of intersection of the bisectors of the three angles of a triangle is the locus of all points equidistant from the three sides of the triangle. Ex. 329. Is it always possible to find a point equidistant from three given straight lines ? from four given straight lines ? Ex. 330. Find a point such that the perpendicularsfrom it to three sides of a quadrilateral shall be equal. (Give geometric construction.) Ex. 331. Prove that if the sides AB and AG of a triangle ABC are prolonged to E and F, respectively, the bisectors of the three anglesBAC, EBC, and BCF all pass through a point which is equallydistant from the three lines AE, AF, and BC. Is any other point in the bisector of the angle BAC equally distant from these three lines? Give reason for your answer. Ex. 332. Through a given point P draw a straightline such that perpendicularsto it from two fixed points Q and B shall cut off on it equal segments from P. (Hint. See " 246.) Ex. 333. Construct the locus of the center of a circle of givenradius, which rolls so that it always touches the sides of a given triangle. Do not prove. Ex. 334. Find the locus of a point in one side of a parallelogram and equidistant from two other sides. In what parallelogramsis this locus a vertex of the parallelogram? Ex. 335. Find the locus of a point in one side of a parallelogramand equidistant from two of the vertices of the parallelogram. In what class of parallelogramsis this locus a vertex of the parallelogram? Ex. 336. Construct the locus of the center of a circle of given radius which rolls so that it constantlytouches a given circumference. Do not prove. PLANE GEOMETRY Proposition XLIII. Theorem 260. The perpendicular bisectors of the sides of a tri-angle are concurrent in a point which is equidistantfrom the three vertices of the triangle. Given A ABC with FG, HK, ED, the L bisectors of AB, BC, CA. To prove : (a)FG, HK, ED concurrent in some pointas 0 ; (")the point 0 equidistant from A, B, and C. Argument 1. FG and ED will intersect at some pointas 0. 2. Draw OA, OB, and OC. 3. v 0 is in FG, the _L bisector of AB, OB = OA; and v 0 is in DE, the _L bi-sector of CA, 0C= OA. 4. .-. OB = OC. 5. .-. HK, the _L bisector of BC, passes through 0. 6. .. FG, HK, and ED are con-current in 0. J. Also 0 is equidistant from A, B, and C. q.e.d. Reasons 1. Two lines X respectively to two intersecting lines also intersect. " 195. 2. Str. line post.I. " 54, 15. 3. Every point in the _L bi-sector of a line is equi-distant from the ends of that line. " 134. 4 Ax. 1. " 54, 1. 5. Every point equidistant from the ends of a line lies in the _L bisector of that line. " 139. 6. By def . of concurrent lines "196. 7o By proof,OA=OB = OC. BOOK I 103 261. Cor. The point of intersection of the perpendic-ular bisectors of the three sides of a triangle is the locus of all points equidistant from the three vertices of the triangle. Ex. 337. Is it always possible to find a point equidistant from three given points? from four given points ? Ex. 338. Construct the perpendicular bisectors of two sides of an acute triangle, and then construct a circle whose circumference shall pass through the vertices of the triangle. Ex. 339. Construct a circle whose circumference shall pass through the vertices of a righttriangle. Ex. 340. Construct a circle whose circumference shall pass through the vertices of an obtuse triangle. Proposition XLIV. Theorem 262. The altitudes of a triangle are concurrent. Given A ABC with its altitudes AD, BE, and CF, To prove AD, BE, and CF concurrent. Outline of Proof Through the vertices A, B, and C, of triangleABC, draw lines \|\| BC, AC, and AB, respectively.Then prove, by means of /I7, that AD, BE, and CF are the _L bisectors, respectively, of the sides of the auxiliary A HKL. Then, by Prop. XLIII, AD, BE^sad CF are concurrent. q.e.d. 104 PLANE GEOMETRY Proposition XLV. Theorem 263. Any two medians of a triangle intersect each other in a trisection point of each. Given A ABC with AD and CE any two of its medians. To prove that AD and CE intersect in a point O such that OD = \AD and OE = i CE. Outline of Proof 1. AD and CE will intersect at some point as 0. " 194. 2. Let R and S be the mid-points of AO and CO respectively. 3. Quadrilateral REDS is a O. 4. .-. AR= RO= OD and CS = SO = OE. 5. That is, OD = " AD and 0^ = J-CB. q.e.d. 264. Cor. T/^e ""ree medians of a triangle are con-current. 265. Def. The point of intersection of the medians of a triangle is called the median center of the triangle. It is also called the centroid of the triangle. This point is the center of mass or center of gravity of the triangle. Ex. 341. Draw a trianglewhose altitudes will intersect on one of its sides,and repeat the proof for Prop. XLIV. Ex. 342. Draw a trianglewhose altitudes will intersect outside of the triangle, and repeat the proof for Prop. XLIV. Ex. 343. Prove Prop. XLV by prolonging OD its own length and drawing lines to B and C from the end of the prolongation. Ex. 344. Construct a triangle, given two of its medians and the angle between them. 106 PLANE GEOMETRY CONSTRUCTION OF TRIANGLES 269. A triangle is determined, in general, when three parts are given,provided that at least one of the given parts is ar line. The three sides and the three angles of a triangle are called its parts ; but there are also many indirect parts; as the three medians, the three altitudes, the three bisectors, and the parts into which both the sides and the angles are divided by these lines. A A 270. The notation given in the annexed figuresmay be used for brevity : A, B, C, the anglesof the triangle ; in a righttriangle, angle C is the rightangle. a, b, c, the sides of the triangle; in a righttriangle, c is the hypotenuse. mai mb, mc, the medians to a, b,and c respectively. ha,hbi hc, the altitudes to a, b,arid c respectively. ta, tb, tc, the bisectors of A, B, and C respectively. lu, da, the segments of a made by the altitude to a, sa, ra, the segments of a made by the bisector of angle A. 271. The student should review the chief cases of construc-tion of trianglesalready given : viz. a, B, c ; A, b,C ; a,b,c: right triangles : a, b ; b,c ; b, B ; b,A ; c, A ; review also " 152. Ex. 347. State in words the first eight cases given in " 271. BOOK I 107 Proposition XLVII. Problem 272. To construct a triangle having two of its sides equal respectively to two given lines,and the angle oppo-site one of these lines equal to a given angle. Y B Fig. 1. Given lines a and b, and Z.B. To construct A ABC. I. Construction 1. Draw any line, as BX. 2. At any point in BX, as B, construct Z XBY = to the given A B. " 125. 3. On .BFlay off BC = a. 4. With C as center and with b as radius, describe an arc cutting BX at A. 5. A ABC is the requiredA. II. The proof is left as an exercise for the student. III. Discussion (1) b may be greater than a; (2) b may equal a ; (3) b may be less than a. (1) If b " a, there will be one solution, i.e. one A and only one can be constructed which shall contain the given parts. This case is shown in Fig. 1. (2) If b = a, the A will be isosceles. The construction will be the same as for case (1). 108 PLANE GEOMETRY A B (3)If b " a. Make the construction as for case (1). If b " the _L from C to BX, there will be two solutions, since A ABC and BBC, Fig. 2, both contain the requiredparts. If b equals the _L from Cto BX there will be one solution. The A will be a rt. A. If b " the _L from C to BX, there will be no solution. In the cases thus far considered, the given Z was acute. The discussion of the cases in which the given Z is a rt. Z and in which it is an obtuse Z is left to the student. 273. Question. Why is (1) the only case possible when the given angle is either rightor obtuse ? 274. The followingexercises are given to j I illustrate analysis of problems and to show the use of auxiliary triangles in constructions. Ex. 348. Construct a triangle, given a, ha, ma. Analysis. Imagine the problem solved as in Fig. 1, and mark the given parts with heavy lines. The triangle AHM is determined and may be made the basis of the construction. Ex. 349. Construct a triangle, given 6, ma, mc Analysis. From Fig. 2 it will be seen that tri-angle AOC may be constructed. Its three sides are known, since AO " \ma and CO " f mc. Ex. 350. Construct a triangle, given a, ha,hc. Analysis. In Fig. 3, right triangleCHB is determined. The locus of vertex A is a line parallelto CB, so that the distance between it and CB is equal to hw BOOK I 109 Ex. 351. Construct atriangle, given b ,c, ma. Analysis. Triangle ABK, Fig. 4, is deter-mined by three sides, b, c, and 2raa. Since ABKC is a parallelogram, AK bisects CB. Ex. 352.. Construct a triangle, given a, "na, \ / ^"' and the angle between ma and a. " Analysis. Triangle AM G is determined, as shown in Fig. 5. A ~?B K Fig. 4. rr~v M d S " N Fig. 6. Ex. 353. Construct a trapezoid, given its four sides. Analysis. Triangle EST, Fig. 6, is determined. Ex. 354. Construct a parallelogram, given q jj the perimeter, one base angle,and the altitude. 2"37 K M Construction. The two parallels, CH and AJE, Fig.7, may be drawn so that the distance A B-E between them equals the altitude ; at any point FlG-"" B construct angle EBC equal to the given R M base angle ; draw CA, bisectingangle FOB ; measure AE equal to half the given perimeter ; complete parallelogram CHEB. f S N Ex. 355. Construct a trapezoid, given the Fig. 8. two non-parallel sides and the difference between the bases. (See Fig.8.) Construct a triangle, given : Ex. 356. A, ha, ta. Ex. 360. B, ha b. Ex. 357. ha} la,da. Ex. 361. a", ta,ha. Ex. 358. ha, ma, la. Ex. 362. a, mb, C. Ex. 359. a, U, C. Ex. 363. a, U, B. Construct an isosceles trapezoid,given : Ex. 364. The two bases and the altitude. Ex. 365. One base, the altitude, and a diagonal. Ex. 366. A base, a diagonal,and the angle between them. 110 PLANE GEOMETRY DIRECTIONS FOR THE SOLUTION OF EXERCISES 275. I. Make the figures clear, neat, accurate, and general in character. II. Fix firmly in mind hypothesis and conclusion with ref-erence to the given figure. III. Recall fundamental propositions related to the propo-sition in question. \ IV. If you can find no theorem which helps you, contradict the conclusion in every possible way {reductio ad absurdum) and try to show the absurdity of the contradiction. V. Make frequent use of the method of analysis, which con-sists in assuming the propositionproved, seeing what results follow until a known truth is reached, and then retracingthe steps taken. VI. If it is required to find a point which fulfills two conditions,it is often convenient to find the point by the Intersection of Loci. By finding the locus of a point which satisfies each condition separately, it is possibleto find the points in which the two loci intersect;i.e. the points which satisfy both conditions at the same time. VII. See " 152 and exercises following" 274 for method of attackingproblems of construction. The method just described under V is a shifting of an uncertain issue to a certain one. It is sometimes called the Method of Successive /Substitutions. It may be illustrated thus : 1. A is true if .B.istrue. 3. But C is true. 2. B is true if C is true. 4. .-. A is true. This is also called the Analytic Method of proof. The proofs of the theorems are put in what is called the Synthetic form. But these were firstthought through analytically, then rearranged in the form in which we find them. BOOK I 111 MISCELLANEOUS EXERCISES Ex. 367. The perpendiculars drawn from the extremities of one side of a triangle to the median upon that side are equal. Ex. 368. Construct an angle of 75" ; of 97^". Ex. 369. Upon a given line find a point such that perpendicufars from it to the sides of an angle shall be equal. Ex. 370. Construct a triangle, given its perimeter and two of its angles. Ex. 371. Construct a parallelogram, given the base, one base angle, and the bisector of the base angle. Ex. 372. Given two lines that would meet if sufficiently prolonged. Construct the bisector of their angle,without prolonging the lines. Ex. 373. Construct a triangle, having given one angle, one adjacent side,and the difference of the other two sides. Case 1 : The side oppo-site the given angle less than the other unknown side. Case 2 : The side oppositethe given angle greater than the other unknown side. Ex. 374. The difference between two adjacent angles of a parallelo-gram is 90" ; find all the angles. Ex. 375. A straightrailway passes 2 miles from a certain town. A placeis described as 4 miles from the town and 1 mile from the railway. Represent the town by a point and find by construction how many places answer the description. Ex. 376. Describe a circle through two given pointswhich lie out-side a given line, the center of the circle to be in that line. Show when no solution is possible. Ex. 377. Construct a right triangle, given the hypotenuse and the difference of the other two sides. Ex. 378. If two sides of a triangle are unequal, the median through their intersection makes the greater angle with the lesser side. Ex. 379. Two trapezoidsare equal if their sides taken in order are equal, each to each. Ex. 380. Construct a righttriangle, having given its perimeter and an acute angle. Ex. 381. Draw a line such that its segment intercepted between two given indefinite lines shall be equal and parallel to a given finite line. Ex. 382. One angle of a parallelogramis given in positionand the point of intersection of the diagonalsis given ; construct the parallelo-gram. 112 PLANE GEOMETRY Ex. 383. third side. Construct a triangle, given two sides and the median to the Ex. 384. If from any point within a triangle lines are drawn to the three vertices of the triangle, the sum of these lines is less than the sum of the,sides of the triangle, and greater than half their sum. Ex. 385. Repeat the proof of Prop. XIX for two cases at once, using Figs. 1 and 2. Ex. 386. If the angle at the vertex of an isosceles triangle is four times each base angle, the perpendicular to the base at one end of the base forms Fig. 1. Fig. 2. with one side of the triangle, and the prolongation of the other side through the vertex, an equilateral triangle. Ex. 387. The bisector of the angle G of a triangle ABC meets AB in D, and DE is drawn parallel to AG meeting BO in E and the bisector of the exterior angle at G in F. Prove DE = EF. Ex. 388. Define a locus. Find the locus of the mid-points of all the lines drawn from a given point to a given line not passing through the point. Construct an isosceles trapezoid, given the bases and one Construct a square, given the sum of a diagonal and one Ex. 389. angle. Ex. 390. side, Ex. 391. The difference of the distances from any point in the base prolonged of an isosceles triangle to the equal sides of the triangleis constant. Ex. 392. Find a point X equidistant from two intersecting lines and at a given distance from a given point. Ex. 393. When two lines are met by a transversal, the difference of two corresponding angles is equal to the angle between the two lines. % Construct a triangle, given ", Ex. 394. A, ha, la. Ex. 398. Ex. 395. A, ta,sa. Ex. 399. Ex. 396. a, ha, la. Ex. 400. Ex. 397. a, 6 + c, A Ex. 401. a, m. mc, hCJB. 6, c, A, B, b -TcT 114 PLANE GEOMETRY Proposition I. Theorem 284. Every diameter of a circle bisects the circumfer-ence and the circle. M Given circle AMBN with center O, and AB, any diameter. To prove: (a) that AB bisects circumference AMBN) (b)that AB bisects circle AMBN. Argument Turn figure AMB on AB as an axis until it falls upon the plane of ANB. Arc AMB will coincide with arc ANB. .-. arc AMB = arc ANB ; i.e. AB bisects circumference AMBN. 4. Also figure AMB will coincide with figure ANB. 5. .'.figure AMB = figure ANB', i.e. AB bisects circle AMBN. Q.E.D. Reasons 1. " 54, 14. 2. " 279, a. 3. " 18. 4. " 279, a. 5. " 18. Ex. 402. A semicircle is described upon each of the diagonals of a rectangleas diameters. Prove the semicircles equal. Ex. 403. Two diameters perpendicularto each other divide a circum-ference into four equal arcs. Prove by superposition. Ex. 404. Construct a circle which shall pass through two given points. Ex. 405. Construct a circle having a given radius r, and passing through two given points A and B BOOK II 115 285. Def. A secant of a cir-cle is a straight line which cuts the circumference in two points, but is not terminated by the circumference, as MN. 286. Def. A straight line is tangent to, or touches, a cir-cle if,however far prolonged, it meets the circumference in but one point. This point is called the point of tangency. HK is tangent to circle 0 at point T, and T is the point of tangency. 287. Def. A sector of a circle is a plane closed figure whose boundary is composed of two radii and their interceptedarc, as sector SOB. 288. Def. A segment of a circle is a plane closed figure whose boundary is composed of an arc and the chord joining its extremi-ties, as segment DCE. 289. Def. A segment which is one half of a circle is called a semi-circle, as segment AMB. 290. Def. Ad arc which is half of a circumference is called a semicircumference, as arc AMB. 291. Def. An arc greater than a semicircumference is called a major arc, as arc DME\ an arc less than a semicir-cumference is called a minor arc, as arc DCE. 292. Def. A central angle, or angle at the center, is an angle whose vertex is at the center of a circle and whose sides are radii. Ex. 406. The line joiningthe centers of two circles is 6, the radii are 8 and 10,respectively.What are the relative positions of the two circles ? ..Ex. 407. A circle can have only one center. " 116 PLANE GEOMETRY Proposition II. Theorem 293. In equal circles, or in the same circle, if two cen-tral angles are equal, they interceptequal arcs on the cir-cumferenc conversely, if two arcs are equal, the central angles that interceptthem are equal. I. Given equal circles ABM and CDN, and equal central A G and Q, intercepting arcs AB and CD, respectively. To prove "b = CD. Argument 1. Place circle ABM upon circle CDN so that center O shall fall upon center Q, and QA shall be collinear with QC. 2. A will fall upon C. 3. OB will become collinear with QD. 4. .-. B will fall upon D. 5. .-. AB will coincide with CD. 6. .'. AB = CD. Q.E.D. Reasons 1. "54,14. 2. " 279, b. 3. By hyp. 4. " 279, 6. 5. " 279, b. 6. "18. II. Conversely : Given equal circles ABM and CDN, and equal arcs AB and CD, intercepted by A 0 and Q, respectively. To prove Zo = ZQ. BOOK II 117 2. Argument Place circle ABM upon circle CDN so that center 0 shall fall upon center Q. Kotate circle ABM upon O as a pivot until AB falls upon its equal CD, A upon C, B upon D. OA will coincide with QCand OJ? with QD. .-. Z 0 = Z Q. Q.E.D. Reasons 1. "54,14. 2. "54,14. 3. "17. 4. "18. 294. Cor. In equal circles, or in the sams circle, if two central angles are unequal, the greater angle inter-cepts the greater arc; conversely,if two arcs are unequal, the central angle that intercepts the greater arc is the greater. (Hint. Lay off the smaller central angle upon the greater.) 295. Def. A fourth part of a circumference is called a quadrant. From Prop. II it is evident that a right angle at the center intercepts a quadrant on the cir-cumference. Thus, two " diam-eters AB and CD divide the circumference into four quadrants, AC, CB, BD, and DA. 296. Def. A degree of arc, or an arc degree, is the arc intercepted by a central angle of one degree. 297. A right angle contains ninety angle degrees ("71); therefore, since equal central anglesintercept equal arcs on the circumference, a quadrant contains ninetyarc degrees. Again, four rightanglescontain 360 angle degrees, and four rightangles at the center of a circle intercept a complete cir-cumferen ; therefore, a circumference contains 360 arc degrees. Hence, a semicircumference contains 180 arc degrees. Ex. 408. Divide a given circumference into eightequal arcs ; sixteen equaJ .arcs. 118 PLANE GEOMETRY Ex. 409. Divide a given circumference into six equal arcs ; three equal arcs ; twelve equal arcs. Ex. 410. A diameter and a secant perpendicular to it divide a cir-cumference into two pairsof equal arcs. Ex. 411. Construct a circle which shall pass through two given points A and B and shall have its center in a given line c. Ex. 412. If a diameter and another chord are drawn from a point in a circumference,the arc interceptedby the angle between them will be bisected by a diameter drawn parallel to the chord. Ex. 413. If a diameter and another chord are drawn from a point in a circumference, the diameter which bisects their interceptedarc will be parallel to the chord. Proposition III. Theorem 298. In equal circles,or in the same circle,if two chords are equal, they subtend equal arcs; conversely, if two arcs are equal, the chords that subtend them are equal. I. Given equal circles 0 and Q, with equalchords AB and CD To prove AB = CD. Argument 1. Draw radii OA, OB, QC, QD. 2. In A OAB and QCD, AB = CD. 3. OA = QC and OB = QD. 4. .-.A OAB = A QCD. 5. .-. Z0 = Z Q. 6. .'. AB= CD. Q.E.D. Reasons 1. " 54, 15. 2. By hyp. 3. "279, b. 4. "116. 5. "110. 6. " 293, I. BOOK II 119 II. Conversely : Given equal circles 0 and Q, and equal arcs AB and CD. To prove chord AB = chord CD. Argument 1. Draw radii OA, OB, QC, QD. 2. AB = CD. 3. .. Zboa = Zdqc. 4. OA = QC and OB = QD. 5. .: A OAB = A QG'D. 6. .-. chord ^4i? = chord CD. Q.E.D. Keasons " 54, 15. By hyp. " 293, II. " 279, b. "107. "110. Ex. 414. If a circumference is divided into any number of equal arcs, the chords joiningthe pointsof division will be equal. Ex. 415. A parallelograminscribed in a circle is a rectangle. Ex. 416, If two of the opposite sides of an inscribed quadrilateral are equal,its diagonals are equal. Ex. 417. State and prove the converse of Ex. 416. 299. Def. A polygon is inscribed in a circle if all its vertices are on the circumference. Thus, polygon ABODE is an inscribed polygon. 300. Def. If a polygon is in-scribed in a circle, the circle is said to be circumscribed about the poly-gon. _^ Ex. 418. Inscribe an equilateral hexagon in a circle ; an equilateral triangle. Ex. 419. The diagonalsof an inscribed equilateral pentagon are equal. Ex. 420. If the extremities of any two intersecting diameters are joined, an inscribed rectangle will be formed. Under what conditions will the rectanglebe a square ? Ex. 421. State the theorems which may be used in proving arcs equal. StafceVthetheorems which may be used in proving chords equal. 120 PLANE GEOMETRY Proposition IV. Theorem 301. In equal circles, or in the same circle, if two chords are unequal, the greater chord subtends the greater minor arc ; conversely, if two minor arcs are unequal, the chord that subtends the greater arc is tliegreater. Reasons 1. " 54, 15. " 2. " 279, a. 3. By hyp. 4. " 173. 5. " 294. I. Given circle 0, with chord AB " chord CD. To prove AB " CD. Argument 1. Draw radii OA, OB, OC, OD 2. In A OAB and OCD, OA = 3. Chord AB " chord CD. 4. .". Zl " Z2. 5. .. AB " CD. Q.E.D. II. Conversely : Given circle 0, with AB " CD. To prove chord AB " chord CD. Argument Reasons 1. Draw radii OA, OB, OC, OD. 1. " 54, 15. 2. In A OAB and OCD, OA = OC, OB " OD. 2. " 279, a. 3. AB "/SU^ 3. By hyp. 4. .\^C" Z2. 4. "294. 5. .-. chord AB " chord CD. q.e.d. 5. " 172. Ex. 422. Prove the converse of Prop. IV by the indirect method 122 PLANE GEOMETRY Ex. 424. Through a given point within a circle construct a chord which shall be bisected at the point. Ex. 425. Given a line fulfilling any two of the five following condi-tions, prove that it fulfillsthe remaining three: 1. A diameter. 2. A perpendicularto a chord. 3. A bisector of a chord. 4. A bisector of the major arc of a chord. 5. A bisector of the minor arc of a chord. Ex. 426. Any two chords of a circle are given in positionand magni-tude ; find the center of the circle. Ex. 427. The line passing through the middle points of two parallel chords passes through the center of the circle. Ex. 428. Given an arc of a circle, find the center of the circle. Proposition VI. Problem 306. To bisect a given arc. A Given AB, an arc of any circle. To bisect AB. The construction,proof, and discussion are left as an exercise for the student. Ex. 429. Construct -an arc of 45" ; of 30". Construct an arc of 30", using a radius twice as long as the one previouslyused. Are these two 30" arcs equal ? Ex. 430. Distinguishbetween findingthe " mid-point of an arc " and the " center of an arc." BOOK II 123 Proposition VII. Theorem 307. In equal circles,or in the same circle,iftwo chords are equal, they are equally distant from the center; con-versely, if two chords are equally distant from the center, they are equal. I. Given circle O with chord AB = chord CD, and let OE and OF be the distances of AB and CD from center O, respectively. To prove OE = OF. Argument 1. Draw radii OB and OC. 2. E and F are the mid-pointsof AB and CD, respectively. .. in rt. A OEB and OCF, EB = CF. OB = OC. .'. A OEB = A OCF. .'. OE" OF. Q.E.D. II. Conversely : Given circle 0 with OE, the distance of chord AB from center 0, equal to OF, the distance of chord CD from center 0. To prove chord AB = chord CZ). Hint. Prove A OEB = A OCJl Reasons 1. " 54, 15. 2. " 302. 3. " 54, 8 a. 4. " 279, a. 5. " 211. 6. " 110. Ex. 431. If perpendicularsfrom the center of a circle to the sides of an inscribed polygon are equal,the polygon is equilateral. Ex. 432. If through any point in a diameter two chords are drawn making equal angles with the diameter, the two chords are equal. 124 PLANE GEOMETRY Proposition VIII. Theorem 308. In equal circles, or in the same circle, if two chords are unequal, the greater chord is at the less dis-tance from the center. Given circle 0 with chord AB " chord CD, and let OF and OH be the distances of AB and CD from center 0, respectively. To prove OF " OH. Argument 1. From A draw a chord AE, equalto DC. 2. From 0 draw OG"AE. 3. Draw FG. 4. AB " CD. 5. .'. AB " AE. 6. F and G are the mid-points of AB and AE, respectively. 7. .-. ^F " AG. 8. .-. Zl "Z2. 9. Z ^0 = Z 0"4. 10. .-. Z3 "Z4. 11. .'. OF" 0G. 12. 06? = OH. 13. .'. 0F" OH. Q.E.D. 309. Note. The student should give the full statement of the sub-stitution made ; thus, reason 5 above should be ; " SubstitutingAE for its equal CD." BOOK II 125 Proposition IX. Theorem (Converse of Prop. VIII) 310. In equal circles, or in the same circle, if two chords are unequally distant from the center, the chord at the less distance is the greater. Given circle 0 with. OF, the distance of chord AB from center 0, less than OH, the distance of chord CD from center O. To prove chord AB " chord CD. The proof is left as an exercise for the student. Hint. Begin with A OGF. 311. Cor. I. A diameter is greater than any other chord. 312. Cor. n. The locus of the mid-points of all chords of a circle equal to a given chord is the circumference hav-ing the same center as the given circle, and having for ra-dius the perpendicular from the center to the given chord. Ex. 433. Prove Prop. IX by the indirect method. Ex. 434. Through a given point within a circle construct the mini-mum chord. Ex. 435. If two chords are drawn from one extremity of a diameter, making unequal angles with it,the chords are unequal. Ex. 436. The perpendicularfrom the center of a circle to a side of an inscribed equilateral triangleis less than the perpendicular from the center of the circle to a side of an inscribed square. (See " 308.) 126 PLANE GEOMETRY Proposition X. Theorem 313. A tangent to a circle is perpendicular to the radius drawn to the point of tangency. A M T B Given line AB, tangent to circle O at T, and OT, a radius drawn to the point of tangency. To prove AB " OT. Argument Reasons 1. Let M be any point on AB other than T; 1. " 286. then M is outside the circumference. 2. Draw OM, intersecting the circumference 2. " 54, 15. at s. 3. OS " om. 3. " 54, 12. 4. OS = OT. 4. " 279, a. 5. .-. OT" OM. 5. " 309. 6. .-. OT is the shortest line that can be 6. Arg. 5. drawn from 0 to AB. 7. /. OT"AB ; i.e. AB" OT. Q.E.D. 7. "165. 314. Cor. I. (Converse of Prop. X). A straight line perpendicular to a radius at its outer extremity is tangent to the circle. Hint. Prove by the indirect method. In the figure for Prop. X, suppose that AB is not tangent to circle O at point T; then draw CD through T, tangent to circle O. Apply " 63. 315. Cor. II. A perpendicular to a tangent at the point of tangency passes through the center of the circle. BOOK II 127 316. Cor. m. A line drawn from the center of a circle perpendicular to a tangent passes through the point of tangency. 317. Def. A polygon is circum-scribed about a circle if each side of the polygon is tangent to the circle. In the same figure the circle is said to be inscribed in the polygon. Ex. 437. The perpendicularsto the sides of a circumscribed polygon at their pointsof tangency pass through a com-mon point. Ex. 438. The line drawn from any vertex of a circumscribed polygon to the center of the circle bisects the angle at that vertex and also the angle between radii drawn to the adjacent points of tangency. Ex. 439. If two tangents are drawn from a point to a circle, the bisector of the angle between them passes through the center of the circle. Ex. 440. The bisectors of the angles of a circumscribed quadrilateral pass through a common point. Ex. 441. Tangents to a circle at the extremities of a diameter are parallel. Proposition XI. Problem 318. To construct a tangent to a circle at any given point in the circumference. The construction, proof,and discussion are left as an exer-cise for the student. (See " 314.) Ex. 442. Construct a quadrilateral which shall be circumscribed about a circle. What kinds of quadrilaterals are circumscriptible ? Ex. 443. Construct a parallelogram which shall be inscribed in a circle. What kinds of parallelogramsare inscriptible ? Ex. 444. Construct a line which shall be tangent to a given circle and parallel to a given line. Ex. 445. Construct a line which shall be tangent to a given circle and 'perpendicular to a given line. 128 PLANE GEOMETRY 319. Def. The length of a tangent is the lengthof the seg-ment included between the point of tangency and the point from which the tangent is drawn j as TP in the following figure. Proposition XII. Theorem 320. If two tangents are drawn from any given point to a cirele,these tangents are equal. Given PT and PS, two tangents from pointP to circle 0. To prove PT=PS. The proofis left as an exercise for the student. Ex. 446. The sum of two oppositesides of a circumscribed quadrilat-eral is equal to the sum of the other two sides. Ex. 447. The median of a circumscribed trapezoidis one fourth the perimeter of the trapezoid. Ex. 448. A parallelogram circumscribed about a circle is either a rhombus or a square. Ex. 449. The hypotenuse of a right trianglecircumscribed about a circle is equal to the sum of the other two sides minus a diameter of the circle. Ex. 450. If a circle is inscribed in any triangle, and if three triangles are cut from the given triangleby drawing tangents to the circle, then the sum of the perimeters of the three triangles will equal the perimeter of the given triangle. 130 PLANE GEOMETRY Proposition XIV. Problem 323. To circumscribe a circle about a given triangle. Given A ABC. To circumscribe a circle about A ABC. The construction, proof, and discussion are left as an exercise for the student. 324. Cor. Three points not in the same straight line determine a circle. Ex. 455. Discuss the positionof the center of a circle circumscribed about an acute triangle ; a righttriangle ; an obtuse triangle. Ex. 456. Circumscribe a circle about an isosceles trapezoid. Ex. 457. Given the base of an isosceles triangle and the radius of the circumscribed circle, to construct the triangle. Ex. 458. The inscribed and circumscribed circles of an equilateral triangleare concentric. Ex. 459. If upon the sides of any triangle equilateral triangles are drawn, and circles circumscribed about the three triangles, these circles will intersect at a common point. Ex. 460. The two segments of a secant which are between two con-centric circumferences are equal. Ex. 461. The perpendicular bisectors of the sides of an inscribed quadrilateral pass through a common point. Ex. 462. The bisector of an arc of a circle isdetermined by the center of the circle and another point equidistantfrom the extremities of the chord of the arc. Ex. 463. If two chords of a circle are equal,the lines which connect their mid-points with the center of the circle are equal. BOOK II 131 TWO CIKCLES 325. Def. The line determined by the centers of two circles is called their line of centers or center-line. 326. Def. Concentric circles are circles which have the same center. Proposition XV. Theorem 327. If two circumferences meet at a -point which is not on their line of centers, they also meet in one other point, M N Given circumferences M and N meeting at P, a point not on their line of centers OQ. To prove that the circumferences meet at one other point, as R. Argument 1. Draw OP and QP. Rotate AOPQ about OQ as an axis until it falls in the positionQRO. OR = OP = a radius of circle M. .-. R is on circumference M. Also QR = QP = a radius of circle N. .\ R is on circumference N. .\ R is on both circumference M and circumference N: i.e. circumferences M and N meet at R. Q.E.D. Reasons 1. "54,15. 2. " 54, 14. 3. By cons. 4. " 279, a. 5. By cons. 6. " 279, a. 7. Args. 4 and 6. 328. Cor. I. If two circumferences intersect, their line of venters bisects their common chord at right angles. 132 PLANE GEOMETRY 329. Cor. II. If two circumferences meet at one point only,that point is on their line of centers. Hint. If they meet at a point which is not on their line of eenters, they also meet in another point (" 327). This contradicts the hypothesis. 330. Def. Two circles are said to touch or be tangent to each other if they have one and only one point in common. They are tangent internally or externally according as one circle lies within or outside of the other. Tangent externally. Tangent internally. Concentric. 331. From " 330, Cor. II may be stated as follows : If two circles are tangent to each other, their common point lies on their line of centers. 332. Cor. III. If two circles are tangent to each other, they have a common tangent line at their point of contact. Hint. Apply " 314. 333. Def. A line touching two circles is called an external common tangent if both circles lie on the same side of it ; the line is called an internal common tangent if the two circles lie on oppositesides of it. Fig. 1. Fig. 2. Thus a belt connecting two wheels as in Fig. 1 is an illus-tration of external common tangents, while a belt arranged as in Fig. 2 illustrates internal common tangents. BOOK II 133 334. Questions. In case two circles are tangent internally how many common tangents can be drawn ? in case their circumferences intersect ? in case they are tangent externally? in case they are wholly outside of each other ? in case one is wholly within the other ? Ex. 464. If two circles intersect, their line of centers bisects the angles between the radii drawn to the points of intersection. Ex. 465. If the radii of two intersectingcircles are 5 inches and 8 inches, what may be the length of the line joining their centers ? Ex. 466. If two circles are tangent externally,tangents drawn to them from any point in their common internal tangent are equal. Ex. 467. Two circles are tangent to each other. Construct their com-mon tangent at their point of contact. Ex. 468. Construct a circle passing through a given point and tan-gent to a given circle at another given point. Ex. 469. Find the locus of the centers of all circles tangent to a given circle at a given point. MEASUREMENT 335. Def. To measure a quantity is to find how many times it contains another quantity of the same kind. The result of the measurement is a number and is called the numerical measure, or measure-number, of the quantity which is measured. The measure employed is called the unit of measure. Thus, the length or breadth of a room is measured by find-ing how many feet there are in it ; i.e.tiow many times it con-tains a foot as a measure. 336. It can be shown that to every geometric, magnitude there correspondsa definite number called its measure-number. The proof that to every straight line segment there belongsa measure-number is found in the Appendix, " 595. The method of proofthere used shows that operations with measure-numbers follow the ordinarylaws of algebra. 337. Def. Two quantities are commensurable if there exists a measure that is contained an integral number of times in each. Such a measure is called a common measure of.the two quantities. 134 PLANE GEOMETRY Thus, a yard and a foot are commensurable, each containing an inch a whole number of times ; so, too, 12" inches and 18f inches are commensurable, each containinga fourth of an inch a whole number of times. 338. Questions. If two quantitieshave a common measure, how many common measures have they ? Name some common measures of 12" inches and 18\| inches. What is their greatest common measure ? What is their least common measure ? 339. Def. Two quantities are incommensurable if there exists no measure that is contained an integral number of times in each. It will be shown later that a diagonal and a side of the same square cannot be measured by the same unit,without a re-mainder; and that the diagonal is equal to V2 times the numerical measure of the side. Now V2 can be expressed only approximately as a simple fraction or as a decimal. It lies between 1.4 and 1.5,for (1.4)2 = 1.96,and (1.5)2 = 2.25. Again, it lies between 1.41 and 1.42, between 1.414 and 1.415, between 1.4142 and 1.4143,and so on. By repeated trials values may be found approximating more and more closely to V2, but no decimal number can be obtained that, taken twice as a factor, will give exactly 2. 340. When we speak of the ratio of one quantity to another, we have in mind their relative sizes. By this is meant not the difference between the two, but how many times one con-tains the other or some aliquotpart of it. In algebrathe ratio of two numbers has been defined as the indicated quotient of the first divided by the second. Since to each geometricmag-nitude there corresponds a number called its measure-number ("336), therefore : 341. Def. The ratio of two geometric magnitudes may be defined as the quotient of their measure-numbers, when the same measure is applied to each. The student should multiply to get the successive approximations. BOOK II 135 Thus, if the lengthof a room is 36 feet and the width 27 feet, the ratio of the length to the width is said to be the ratio of 36 to 27 ; i.e.ff, which is equal to f. The ratio of the width to the length is \|-J, which is equal to j. The term ratio is never appliedto two magnitudes that are unlike. 342. Def. If the two magnitudes compared are commen-surable, the ratio is called a commensurable ratio and can always be expressed as a simplefraction. 343. Def. If the two magnitudes compared are incommen-surable, the ratio is called an incommensurable ratio and can be expressedonly approximately as a simple fraction. Closer and closer approximations to an incommensurable ratio may be obtained by repeatedly using smaller and smaller units as measures of the two magnitudes to be compared and by finding the quotient of the numbers thus obtained. Two magnitudes,e.g.two line segments, taken at random are usuallyincommensurable, commensurabilitybeing compara-tively rare. 344. Historical Note. The discovery of incommensurable magni-tudes is ascribed to Pythagoras, whose followers for a long time kept the discoverya secret. It is believed that Pythagoras was the first to prove that the side and diagonal of a square are incommensurable. A more complete account of the work of Pythagoras will be found in " 510. Ex. 470. What is the greatest common measure of 48 inches and 18 inches ? Will it divide 48 inches " 18 inches ? 48 inches " 2 x 18 inches ? Ex. 471. Draw any two line segments which have a common meas-ure. Find the sum of these lines and, by laying off the common measure, show that it is a measure of the sum of the lines. Ex. 472. Given two lines, 5 inches and 4 inches long,respectively. Show by a diagram that any common measure of 5 inches and 4 inches is also a measure of 15 inches plus or minus 8 inches. Ex. 473. Find the greatest common divisor of 728 and 844 by division and^ point out the similarity of the process to that used in Prop. XVI 136 PLANE GEOMETRY Proposition XVI. Problem 345. To determine whether two given lines are commen-surable or not; and if they are commensurable, to find their common measure and their ratio. A F B i i i-i-j C E G D i i " i " i " i_i Given lines AB and CD. To determine : (a)whether AB and CD are commensurable and if so, (6)what is their common measure; and (c)what is the ratio of AB to CD. I. Construction 1. Measure off AB on CD as many times as possible.Sup-pose it is contained once, with a remainder ED. 2. Measure off ED on AB as many times as possible.Sup-pose it is contained twice,with a remainder FB. 3. Measure off FB on ED as many times as possible.Sup-pose it is contained three times,with a remainder GD. 4. Measure off GD on FB as many times as possible, and so on. 5. It is evident that this process will terminate only when a remainder is obtained which is a measure of the remainder immediatelypreceding. 6. If this process terminates, then the two given lines are commensurable, and the last remainder is their greatest com-mon measure. 7. For example,if GD is a measure of FB, then AB and CD are commensurable, GD is their greatest common measure, and the ratio of AB to CD can be found. II. Proof Argument 1. Suppose FB = 2 GD. 2. ED = EG+ GD. "Reasons 1. See I,7. 2. "54,11. 138 PLANE GEOMETRY Ex. 475. Consider an isosceles trianglewith constant base angles, but variable base. Tell what other constants and what other variables there would be in this case. Ex. 476. If through any point in the base of an isosceles triangle lines are drawn parallel to the equal sides of the triangle, a parallelo-gram will be formed whose perimeter will be constant ; i.e. the perimeter will be independent of the positionof the point. 347. Def. A magnitude is constant if it does not change throughout a discussion. 348. Def. A magnitude is variable if it takes a series of different successive values during a discussion. 349. Def. If a variable approaches a constant in such a way that the difference between the variable and the constant may be made to become and remain smaller than any fixed number previously assigned, however small,the constant is called the limit of the variable. 350. The variable is said to approach its limit as it becomes more and more nearlyequal to it. Thus, suppose a point to move from A toward B, by successive steps,under the A " ^ F ? restriction that at each step it must go over one half the segment between it and B. At the first step it reaches C, whereupon there remains the segment CB to be traveled over ; at the next step it reaches D, and there remains an equal segment to be covered. Whatever the number of steps taken,there must always remain a segment equal to the segment last covered. But the segment between A and the moving point may be made to differ from AB by as littleas we ptease, i.e.by less than any previouslyassignedvalue. For assign some value,say, half an inch. Then the point,continuing to move under its governing law, may approach B until there remains a segment less'than half an inch. Whatever be the value assigned,the variable segment from A to the moving pointmay be made to differ from the constant segment AB by less than the assignedvalue. BOOK II 139 Again, the numbers in the series 4,2, 1,^,\,\,etc., in which each term is one half of the preceding term, approach 0 as a limit as the number of terms in the series is increased. For if we assignany value,as y-g-^-g-, it is evident that a term of the series may be found which is less than y-oxro-g-; it is also evident that no term of the series can become 0. 351. In elementary geometry the variables that approach limits are usuallysuch that they cannot attain their limits. There are, however, variables that do attain their limits. The limitingvalues of algebraic expressionsare frequently of this kind ; thus,the expression " " -approacheslasa; approaches X ~\~ -L 0, and has the limit 1 when x becomes zero. 352. Def. Two variables are said to be related when one depends upon the other so that,if the value of one variable is known, the value of the other can be obtained. For example, the diagonal and the area of a square are re-lated variables, for there is a value for the area for any value which may be given to the diagonal, and vice versa. 353. Questions. On the floor is a bushel of sand. If we keep adding to this pileforever, how large will it become ? Does it depend upon the law governing our additions ? If we add one quart each hour, how large will it become ? If we add one quart the firsthour, a half quart the second hour, a fourth quart the third hour, etc., each hour adding one half as much as the preceding hour, how large will the pilebecome ? 354. Historical Note. Achilles and the Tortoise. One of the early Greek schools of mathematics, founded during the fifth century b.c, at Elea, Italy, and known as the Eleatic School, was famous for its investigations of problems involving infinite series. Zeno, one of the most prominent members, proposed this question: He "argued that if Achilles ran ten times as fast as a tortoise, yet if the tortoise had (say) 1000 yards start,it could never be overtaken : for, when Achilles had gone the 1000 yards, the tortdise would stillbe 100 yards in front of him ; by the time he had covered these 100 yards, it would still be 10 yards in front of him ; and so on forever : thus Achilles would get nearer and nearer to the tortoise but never overtake it." Was Zeno right? If not, ca%you find the fallacy in his argument ? 140 PLANE GEOMETRY Proposition XVII. Theorem 355. If two variables are always equal,and if each ap-proaches a limit, then their limits are equal. Given two variables,V and V',which are always equal and which approach as limits L and L\ respectively. To prove L = L\ Argument 1. Either L = L', or L=" L\ 2. Suppose that one limit is greater than the other,say L " L' ; then V, in ap-proaching L, may assume a value be-tween L' and L j i.e. V may assume a value " L\ 3. But v' cannot assume a value " L'. 4. .-. V may become " V'. 5. But this is impossible, since V and v' are always equal. 6. .. L = L\ q.e.d. Ke a sons 1. " 161, a. 2. " 349. 3. " 349. 4. Args. 2 and 3. 5. By hyp. 6. " 161, 6. 356. Question. In the above proof are V and Vf increasing or de-creasing variables ? The student may adapt the argument above to the case in which V and V are decreasingvariables. Ex. 477. Apply Prop. XVII to the accompanying figure, where variable V is represented by the line AB, variable V by the line CD, limit L by line AE, and limit V by line CF. B H 357. Note. It will be seen that, in the application of Prop. XVII, there are three distinct thingsto be considered : (1) Two variables that are always equal; (2) The limits of these two variables ; (3) The equality of these limits themselves. BOOK II 141 Proposition XVIII. Theorem 358. An angle at the center of a circle is measured by its intercepted arc. Given central ZAOB and AB intercepted by it; let ZCOE be any unit Z (e. g. a degree), and let CE, intercepted by the unit Z, be the unit arc. To prove the measure-number of ZAOB, referred to ZCOE, equal to the measure-number of AB, referred to CE. I. If ZAOB and ZCOE are commensurable. (a)Suppose that ZCOE is contained in ZAOB an integral number of times. Argument 1. Apply Z COE to Zaob as a measure. Suppose that ZCOE is contained in Z AOB r times. 2. Then r is the measure-number of Z AOB referred to ZCOE as a unit. -").Now the r equal central A which com-pose ZAOB intercept r equal arcs on the circumference, each equal to GE. 4. .\ r is the measure-number of AB re-ferred to CE as a unit. 5. .. the measure-number of ZAOB, re-ferred to ZCOE as a unit,equals the measure-number of AB, referred to " 4 CE as a unit. q.e.d. Reasons 1. " 335. 2. " 335. 3. " 293, 1. 4. " 335. 5. "54,1. 142 PLANE GEOMETRY I. If Z AOB and Z COE are commensurable. (b) Suppose that Z COE is not contained in Z AOB an integral number of times. The proof is left as an exercise for the student. Hint. . Some aliquot part of ZCOE must be a measure of /.AOB. (Why ?) Try i ZCOE, \ZCOE, etc. II. If Z AOB and Z COE are incommensurable. Argument Let Z 1 be a measure of Z COE. Ap-ply Z 1 to Z AOB as many times as possible. There will then be a re-mainder, Z.FOB, less than Zl. Z AOF and Z COS are commensurable. .-. the measure-number of A AOF, re-ferred to Z CO.E as a unit,equals the measure-number of AF} referred to CE as a unit. Reasons 1. "339. 2. "337. 3. " 358, I. BOOK II 143 Reasons 4. "335. 5. " 294. 6. Args.4and5. Argument 4. Now take a smaller measure of /.COE. No matter how small a measure of Z COE is taken, when it is applied as a measure to /AOB, the remainder, Z FOB, will be smaller than the Z taken as a measure. 5. Also FB will be smaller than the arc in-tercepted by the Z taken as a measure. 6. .. the difference between /AOF and Z AOB may be made to become and remain less than any previously as-signed Z, however small; and like-wise the difference between AF and AB, less than the arc interceptedby the assignedZ. 7. .. Z AOF approaches Z AOB as a limit, and AF approaches Xb as a limit. 8. Hence the measure-number of Z .40^ approaches the measure-number of /AOB as a limit, and the measure-number of AF approaches the meas-ure-number of AB as a limit. ' 9. But the measure-number of Z AOF is always equal to the measure-number of AF. 10. .. the measure-number of /.AOB, re-ferred to Z COE as a unit,equals the measure-number of AB, referred to CE as a unit. q.e.d. 359. If a magnitude is variable and approaches a limit, then, as the magnitude varies,the successive measure-numbers of the variable approach as their limit the measure-number of the limit of the magnitude. (This theorem will be found in the Appendix, " 597.) 7. "349. 8. "359. 9. Arg. 3. 10. "355. 144 PLANE GEOMETRY 360. Cor. In equal circles, or in the same circle,two angles at the center have the same ratio as their inter-cepted arcs. Hint. The measure-numbers of the angles are equal respectively to the measure-numbers of their intercepted arcs. Therefore the ratio of the angles is equal to the ratio of the arcs. Ex. 478. Construct a secant which shall cut off two thirds of a given circumference. Ex. 479. Is the ratio of two chords in the same circle equal to the ratio of the arcs which they subtend ? Illustrate your answer, using a semicircumference and a quadrant. 361. The symbol oc will be used for is measured by. oc is the symbol of variation, and the macron (") means long or length' Hence oc suggests varies as the length of. 362. From " 336 it follows directly that : (a) In equal circles, or in the same circle, equal a7igles are measured by equal arcs ; conversely, equal arcs measure equal a,ngles. (sum \p) The measure of the ( stii/m. the [difference of two angles is equal to (b) The measure of the sum difference (c)The measure of any multiple of an angle is equal to that same multipleof the measure of the angle. \ of the measures of the angles. 363. Def. An angle is said to be inscribed in a circle if its vertex lies on the circumference and its sides are chords. 364. Def. An angle is said to be inscribed in a segment of a circle if its vertex lies on the arc of the segment and its sides pass through the extremities of that arc. It is,of course, inaccurate to speak of measuring one magnitude by another magnitude of a different kind.; but, in this case, it has become a convention so general that the student needs to become familiar with it. More accurately, in Prop. XVII I, the measure' number of an angle at the center, referred to any unit angle, is the same as the measure-number of its interceptedarc when the unit arc is the arc interceptedby the unit angle. 146 PLANE GEOMETRY D Fig. 1. Fig. 2. 366. Cor. I. All angles inscribed in the same segment are equal. (SeeFig.1.) 367. Cor. II. Any angle inscribed in a semicircle is a right angle. (SeeFig. 2.) 368. Cor. III. The locus of the vertex of a right tri-angle having a given hypotenuse as base is the circum-ference having tiie same hypotenuse as diameter. 369. Cor. IV. Any angle inscribed in a segment less than a semicircle is an obtuse angle. (See Fig.3.) 370. Cor. V. Any angle inscribed in a segment greater than a semicircle is an acute angle. (See Fig.3.) Ex. 480. If an inscribed angle contains 24 angle degrees,how many arc degrees are there in the intercepted arc ? how many in the rest of the circumference ? Ex. 481. If an inscribed angle intercepts an arc of 70",how many degrees are there in the angle ? Ex. 482. How many degrees are there in an angle inscribed in a seg-ment whose arc is 140" ? Ex. 483. Construct any segment of a circle so that an angle inscribed in it shall be an angle of : (a) 60" ; (6) 45" ; (c) 30". Ex. 484. Repeat Ex. 483, using a given line as chord of the segment. How many solutions are there to each case of Ex. 483 ? how many to each case of Ex. 484 ? Ex. 485. The opposite angles of an inscribed quadrilateral are sup-plementary BOOK II 147 Ex. 486. If the diameter of a circle is one of the equal sides of an isosceles triangle, the circumference will bisect the base of the triangle. Ex. 487. By means of a circle construct a righttriangle, given the hypotenuse and an arm. Ex. 488. By means of a circle construct a right triangle, given the hypotenuse and an adjacent angle. Ex. 489. Construct a righttriangle, having given the hypotenuse and the altitude upon the hypotenuse. .a), the founder of the 371. Historical Note. Thales (640-646 earliest Greek school of mathe-matics, is said to have discov-ered that all triangles having a diameter of a circle as base, with their vertices on the cir-cumference, have their vertex angles right angles. Thales was one of the Seven Wise Men. He had much business shrewdness and sagacity, and was renowned for his practi-cal and political ability. He went to Egypt in his youth, and while there studied geom-etry and astronomy. The story is told that one day while viewing the stars, he fell into a ditch ; whereupon an old woman said, " How canst thou know what is doing in the heavens, when thou seest not what is at thy feet ? " According to Plutarch, Thales computed the height of the Pyramids of Egypt from measurements of their shadows. Plutarch givesa dialoguein which Thales is addressed thus, "Placing your stick at the end of the shadow of the pyramid, you made by the same rays two triangles, and so proved that the height of the pyramid was to the length of the stick as the shadow of the pyramid to the shadow of the stick." This compu-tation was regarded by the Egyptians as quite remarkable, since they were not familiar with applications of abstract science. The geometry of the Greeks was in general ideal and speculative, the Greek mind being more attracted by beauty and by abstract relations than 1by the practical affairs of everyday life. Thales 148 PLANE GEOMETRY Proposition XX. Problem 372. To construct a perpendicular to a given straight line at a given point in the line. (Secondmethod. For another method, see " 148.) A Q" M ,'P ^' Given line AB and P, a point in AB. To construct a J to AB at P. I. Construction 1. With 0, any convenient point outside of AB, as center, and with OP as radius,construct a circumference cutting AB at P and Q. ' 2. Draw diameter QM. 3. Draw PM. 4. PM is 1_AB at P. II. The proofand discussion are left as an exercise for the student. The method of " 372 is useful when the point P is at or near the end of the line AB. Ex. 490. Construct a perpendicular to line AB at its extremity P, without prolonging AB. Ex. 491. Through one of the pointsof intersection of two circumfer-ences a diameter of each circle is drawn. Prove that the line joiningthe ends of the diameters passes through the other point of intersection. BOOK II 149 Proposition XXI. Problem 373. To construct a tangent to a circle from a point out-side. Given circle 0 and pointP ontside the circle. To construct a tangent from P to circle 0. I. Construction 1. Draw PO. 2. With Q, the mid-pointof PO, as center,and with QO as radius, construct a circumference intersecting the circumference of circle 0 in pointsT and V. 3. Draw P T and PV. 4. PT and PFare tangents from P to circle 0. II. The proof and discussion are left as an exercise for the student. Hint. Draw OT and OF and apply " 367. Ex. 492. Circumscribe an isosceles triangleabout a given circle, the base of the isosceles trianglebeing equal to a given line. What restric-tion is there on the length of the base ? Ex. 493. Circumscribe a righttriangle about a given circle, one arm of the trianglebeing equal to a given line. What is the least length possible for the given line, as compared with the diameter of the circle ? Ex. 494. If a circumference M passes through the center of a circle B, the tangents to B at the points of intersection of the circles intersect on circumference M. Ex. 495. Circumscribe an isosceles triangle about a circle, the altitude upon ^iebase of the triangle being given. 150 PLANE GEOMETRY Ex. 496. Construct a common external tangent to two given circles T , V Given circles MTB and NVS. To construct a common external tangent to circles MTB and N VS. I. Construction 1. Draw the line of centers OQ. 2. Suppose radius OM " radius QN. Then, with O as center and with OL " OM " $7Vas radius,construct circle LPK. 3. Construct tangent QP from point Q to circle LPK. " 373. 4. Draw OP and prolong it to meet circumference MTB at T. 5. Draw QVWOT. " 188. 6. Draw TV. 7. TV is tangent to circles MTB and NVS. II. The proof and discussion are left as an exercise for the student. Construct a second common external tangent to circles MTB by same Ex. 497. and NVS method. How is the method modified if the two circles are equal ? Ex. 498. Construct a common internal tangent to two circles. Hint. Follow steps of Ex. 496 except step 2. Make OL = OM + QN. Ex. 499. By moving Q toward 0 in the precedin figure, show when there are four common tangents ; when only three ; when only two ; when only one ; when none. BOOK II 151 Proposition XXII. Problem 374. With a given line as chord, to construct a segment of a circle capable of containing a given angle. 4 B xC VI / Given line AB and Z M. C \^ ^"^"^^ \ / To construct, with AB as ^ S^XH chord, a segment of a circle capable of containingZ Jf. ]" I. Construction 1. Construct an Z, as ZCDE, equal to the given Z M. " 125. 2. With H, any convenient point on DE, as center and with AB as radius, describe an arc cuttingDC at some point, as K. 3. Draw UK. 4. Circumscribe a circle about AKDH. " 323. 5. Segment KDH is the segment capable of containing Z-M. II. The proof and discussion are left to the student. 375. Questions. Without moving AB, can you construct a A with AB as base and a vertex Z = Z M? What is the sum of the base A ? 376. Cor. The locus of the vertices of all triangles hav-ing a given base and a given angle at their vertices is the arc which forms, with the given base, a segment capable of containing the given angle. Ex. 500. On a given line construct a segment that shall contain an angle of 105" ; of 135". Ex. 501. Find the locus of the vertices of all triangles having a com-mon base 2 inches long and having their vertex angles equal to 60". Ex. 502. Construct a triangle, having given ", hi, and B. 152 PLANE GEOMETRY Proposition XXIII. Theorem 377. An angle formed by two chords which intersect within a circle is measured by one half the sum of the arc intercepted between its sides and the arc intercepted be-tween tJie sides of its vertical angle. tL B Given two chords AB and CD, intersecting at E. To prove that Zl"i (BD + AC). Argument Draw AD. Z1=Z2 + Z3. Z_2\Bb. Z 3 \2b. .'. Z 1 oc i (BD + AC). Q.E.D. Reasons " 54, 15. "215. "365. "365. " 362, ". "309. Ex. 503. One angle formed by two intersecting chords intercepts an arc of 40". Its vertical angle intercepts an arc of 60". How largeis the angle ? Ex. 504. If an angle of two intersecting chords is 40" and its inter-cepted arc is 30",how largeis the oppositearc ? Ex. 505. If two chords intersect at rightangles within a circumfer-ence, the sum of two oppositeintercepted arcs is equal to a semicircum-ference. Ex. 506. If M is the center of a circle inscribed in triangle ABC and if AM is prolonged to meet the circumference of the circumscribed circle at D, prove that BD ss DM = DC. 154 PLANE GEOMETRY Proposition XXV. Theorem 379. An angle formed by two secants intersecting out-side of a circumference, an angle formed by a secant and a tangent, and an angle formed by two tangents are each measured by one half the difference of the intercepted arcs. Fig. 2. Fig. 3. I. An angle formed by two secants (Fig. 1). Given two secants BA and BC, forming Z 1. To prove that Z 1 cc A (AC " BE). Argument Draw CD. 2. Z1 + Z2 = Z3. 3. .-. Z 1 = Z 3 -Z 2. 4. Z 3 cc -i-ac, and Z 2 " \ BE. 5. Reasons " 54, 15. "215. " 54, 3. " 365. 5. "362,6. 6. " 309. .-. Z3-Z2 " {ac-be). 6. .-. Z 1 QC 1 (AC -BE). Q.E.D. II. An angle formed by a secant and a tangent (Fig. 2). III. An angle formed by two tangents (Fig. 3). The proofsof II and III are left to the student. 380. Note. In the preceding theorems the vertex of the angle may be : (1) within the circle ; (2) on the circumference ; (3) outside the circle. Ex. 511. Tell how to measure an angle having its vertex in each of the three possible positionswith regard to the circumference. BOOK II 155 Proposition XXVI. Theorem 381. Parallel lines intercept equal arcs on a circum-ference. n " H Fig. 1. Fig. 2. I. If the two II lines are secants or chords (Fig. 1). Given IIchords AB and CZ),intercepting arcs AC and BD. To prove AC = BD. Argument 1. Draw CB. 2. Z1 = Z2. 3. Z.\^\ac, andZ2ocij"b. 4. .-. \AC=\BB. 5. .'. AC = i"d. Q.E.D. Reasons 1. " 54, 15. 2. " 189. 3. " 365. 4. " 362, a. 5. " 54, 7 a. II. If one of the II lines is a secant and the other a tangent (Fig.2). III. If the two II lines are tangents (Fig. 3). The proofs of II and III are left as exercises for the student. Ex. 512. In Fig. 1, Prop. XXV, if angle 1 equals 42" and arc BE equals 40", how many degrees are there in angle 3 ? in arc AC? Ex. 513. In Fig. 2, if arc DC equals60" and angle 3 equals 100",find the number of degrees in angle 1. Ex. 514. In Fig.3, if angle 1 equals65",find angle 2 and angle 3. Ex. 515. An angle formed by two tangents is measured by 180" minus the intercepted arc. Ex. 516. If two tangents to a circle meet at an angle of 40", how nxaftjr degrees of arc do they intercept ? 156 PLANE GEOMETRY Ex. 517. Is the converse of Prop. XXVI true ? Prove your answer. Ex. 518. If two sides of an inscribed quadrilateral are parallel, the other two sides are equal. Ex. 519. Is the converse of Ex. 518 true ? Prove your answer. Ex. 520. If, through the point where the bisector of an inscribed angle cuts the circumference,a chord is drawn parallel to one side of the angle,this chord will equal the other side of the angle. Ex. 521. If through the points of intersection of two circumferences parallels are drawn terminating in the circumferences, these parallels will be equal. Ex. 522. Prove that a trapezoidinscribed in a circle is isosceles. Ex. 523. If two pairsof sides of an inscribed hexagon are parallel, the other two sides are equal. Ex. 524. If the sides AB and BC of an inscribed quadrilateral ABGD subtend arcs of 60" and 130",respectively, and if angle AED, formed by the diagonalsAC and BD intersecting at E, is 75",how many degrees are there in arcs AD and DC ? how many degrees in each angle of the quad-rilateral ? MISCELLANEOUS EXERCISES Ex. 525. Equal chords of a circle whose center is C intersect at E. Prove that CE bisects the angle formed by the chords. Ex. 526. A common tangent to two unequal circles intersects their line of centers at a point P ; from P a second tangent is drawn to one of the circles. Prove that it is also tangent to the other. Ex. 527. Through one of the points of intersection of two circum-ferences draw a chord of one that shall be bisected by the other circumference. Ex. 528. The angle ABC is any inscribed angle in a given segment of a circle ; AC is prolonged to P, making CP equal to CB. Find the locus of P. Ex. 529. Given two points P and Q, and a straightline through Q. Find the locus of the foot of the perpendicularfrom P to the given line, as the latter revolves around Q. Ex. 530. If two circles touch each other and a line is drawn through the point of contact and terminated by the circumferences, the tangents at its ends are parallel. Ex. 531. If two circles touch each other and two lines are drawn through the point of contact terminated by the circumferences, the chords joiningthe ends of these lines are parallel. BOOK II 157 Ex. 532. If one arm of a righttriangle is the diameter of a circle, the tangent at the point where the circumference cuts the hypotenuse bisects the other arm. Ex. 533. Two fixed circles touch each other externallyand a circle of variable radius touches both externally. Show that the difference of the distances from the center of the variable circle to the centers of the fixed circles is constant. Ex. 534. If two circles are tangent externally, their common internal tangent bisects their common external tangent. Ex. 535. If two circles are tangent externallyand if their common external tangent is drawn, lines drawn from the point of contact of the circles to the points of contact of the external tangent are perpendicular to each other. Ex. 536. The two common external tangents to two circles meet their line of centers at a common point. Also the two common internal tan-gents meet the line of centers at a common point. Ex. 537. Two circles whose radii are 17 and 10 inches,respectively, are tangent externally. How long is the line joiningtheir centers ? how long if the same circles are tangent internally ? Ex. 538. If a rightangle at the center of a circle is trisected, is the intercepted arc also trisected ? Is the chord which subtends the arc tri-sected ? Ex. 539. Draw a line intersecting two given circumferences in such a way that the chords interceptedby the two circumferences shall equal two given lines. What restriction is there on the lengths of the given lines ? Ex. 540. Construct a triangle, given its base, the vertex angle,and the median to the base. Under what conditions will there be no solution ? Ex. 541. In the same circle, or in equal circles, two inscribed triangles are equal, if two sides of one are equal respectively to two sides of the other. Ex. 542. If through the points of intersection of two circumferences two lines are drawn terminating in the circumferences,the chords which join their extremities are parallel. Ex. 543. The tangents drawn through the vertices of an inscribed rectangle, which is not a square, form a rhombus. Ex. 544. The line joiningthe center of the square described upon the hypotenuse of a righttriangle, to the vertex of the rightangle,bisects the righf^angle. 158 PLANE GEOMETRY Ex. 545. If two common external tangents or two common internal tangents are drawn to two circles, the segments of these tangents inter-cepted between the points of contact are equal. Ex. 546. Through two given pointsdraw two parallel lines at a given distance apart. Ex. 547. In a given circle inscribe a chord of given lengthwhich pro-longed shall be tangent to another given circle. Ex. 548. Find the locus of the middle point of a chord drawn from a given point in a given circumference. Ex. 549. The locus of the intersections of the altitudes of triangles having a given base and a given angle at the vertex is the arc which forms with the base a segment capable of containing an angle equal to the supplement of the given angle at the vertex. Hint. Let ABO be one of the A and O the intersection of the altitudes. In quadrilateral FOEC, /.EOF is the supplement of Z FOE. .-. Z EOF is the supplement of Z BOA. Ex. 550. In a circle, prove that any chord which bisects a radius at rightangles subtends an angle of 120" at the center. Ex. 551. Construct an equilateral triangle, having given the radius of the inscribed circle. Ex. 552. The two circles described upon two sides of a triangleas diameters intersect upon the third side. Ex. 553. All triangles circumscribed about the same circle and mutually equiangular are equal. Ex. 554. If two cir-cumferences meet on their line of centers, the circles are tangent to each other. Fig. 1. Fig. 2. M between 0 and Q OP+PQ"OQ. OP= OM. .'.PQ"MQ. .'. P is not on circumference ". Outline of Proof Fig. 1. II. M not between O and Q, Fig. 2. 0Q+ QP"OP. .. OQ+ QP"OM. .-. QP" QM. .-. P is not on circumference S. BOOK II 159 Ex. 555. If two circumferences intersect, neither point of intersection is on their line of centers. Ex. 556. In any right triangleABG, right-angledat G, the radius of the inscribed circle equals {a + b " c) and the radius of the escribed circle tangent to c equals\|(a + b + c). Ex. 557. In the accom-. T B P panying figurea, b, c, are the ^^^~ jT ~"\ sy ^N. sides of triangle AB G. Prove: ^v"( 1$T \ 1. a = TP; Sr^f \ / 2. AP=l(a + b + c); ^^-^/ 3. TB = l(a + c-b). ""\ Ex. 558. Trisect a quadrant ; a semicircumference ; a circumference. Ex. 559. Describe circles about the vertices of a given triangleas centers, so that each shall touch the other two. Ex. 560. Construct within a given circle three equal circles, so that each shall touch the other two and also the given circle. Construct a triangle, having given : Ex. 561. ha, hc, a Ex. 562. A, B, and B, the radius of the circumscribed circle. Ex. 563. a, B, B. Ex. 564. C and the segments of c made by tc. Ex. 565. C and the segments of c made by hc. Ex. 566. r, the radius of the inscribed circle, and the segments of c made by tc" Construct a righttriangle ABC, right-angled at C, having given : Ex. 567. c, hc. Ex. 568. c and one segment of c made by hc. Ex. 569. The segments of c made by hc. Ex. 570. The segments of c made by tc. Ex. 571. c and a line I,in which the vertex of G must lie. Ex. 572. c and the perpendicularfrom vertex G to a line I. Ex. 573. c and the distance from G to a point P. Construct a square, having given: Ex. 574. The perimeter. Ex. 575. A diagonal. Ex. 576. The sum of a diagonal and a side. 160 PLANE GEOMETRY Construct a rectangle, having given : Ex. 577. Two non-parallel sides. Ex. 578. A side and a diagonal. Ex. 579. The perimeter and a diagonal. Ex. 580. A diagonal and an angle formed by the diagonals. Ex. 581. A side and an angle formed by the diagonals. Ex. 582. The perimeter and an angle formed by the diagonals. Construct a rhombus, having given : Ex. 583. A side and a diagonal. Ex. 584. The perimeter and a diagonal. Ex. 585. One angle and a diagonal. Ex. 586. The two diagonals. Ex. 587. A side and the sum of the diagonals. Ex. 588. A side and the difference of the diagonals. Construct a parallelogram, having given : Ex. 589. Two non-parallel sides and an angle. Ex. 590. Two non-parallel sides and a diagonal. Ex. 591. One side and the two diagonals. Ex. 592. The diagonalsand an angle formed by them. Construct an isosceles trapezoid, having given : Ex. 593. The bases and a diagonal. Ex. 594. The longer base, the altitude,and one of the equal sides. Ex. 595. The shorter base, the altitude, and one of the equal sides. Ex. 596. Two sides and their included angle. Construct a trapezoid, having given : Ex. 597. The bases and the anglesadjacent to one base. Ex. 598. The bases, the altitude, and an angle. Ex. 599. One base, the two diagonals, and their included angle. Ex. 600. The bases,a diagonal, and the angle between the diagonals. Construct a circle which shall : Ex. 601. Touch a given circle at P and pass through a given point Q Ex. 602. Touch a given line I at P. Ex. 603. Touch three given lines two of which are parallel. Ex. 604. Touch a given line I at P and also touch another line m, Ex. 605. Have its center in line Z,cut I at P, and touch a circleK 162 PLANE GEOMETRY Proposition I. Theorem 388. If four numbers are in proportion, the product of the extremes is equal to the product of the means. Given a : b = c : d. To prove ad = be. Argument 1 -" " b~ ' d 2. bd = bd. 3. .'.ad = bc. q.e.d. Reasons 1. By hyp. 2. Byiden. 3. "54, la. 389. Note. The student should observe that the process used here' is merely the algebraic"clearing of fractions,"and that as fractional equations in algebra are usuallysimplified by this process, so, also,pro-portions may be simplified by placing the product of the means equal to the product of the extremes. 390. Cor. I. The mean proportional between two num-bers is equal to the square root of their product. 391. Cor. II. If two proportionshave any three terms of one equal respectively to the three corresponding terms of the other, then the remaining term of the firstis equal to the remaining term of the second. Ex. 606. Given the equation m : r = d : c ; solve (1) for d, (2) for r, (3) for m, (4) for c. Ex. 607. Find the fourth proportionalto 4, 6,and 10 j to 4, 10, and 6; to 10,6, and 4. Ex. 608. Find the mean proportionalbetween 9 and 144 ; between 144 and 9. Ex. 609. Find the third proportional to f and f ; to f and f. 392. Questions. What rearrangement of numbers can be made in Ex. 607 without affecting the required term ? in Ex. 608 ? in Ex. 609 ? Ex. 610. Find the third proportional to a2 " 62 and a " b. Ex. 611. Find the fourth proportionalto a2 " b2,a" b, and a + 6. See " 382 for the three wdys of writinga proportion. BOOK III 163 Ex. 612. If in any proportion the antecedents are equal,then the consequents are equal and conversely. Ex. 613. If a : b = c : d, prove that ma : kb = mc : kd. Ex. 614. If I : k = b : f", prove that Ir : kr = be : mc. Ex. 615. If x : y = b : c, prove that dx:y = bd:c. Ex. 616. If x : y = 6 : c, is (fcc : y = 6 : cd a true proportion? Proposition II. Theorem (Converseof Prop. I) " 393 If the product of two numbers is equal to the prod-uct of two other numbers, either pair may be made the means and the other pair the extremes of a proportion. Given ad = be. To prove a:b = c:d. Reasons 1. By hyp. 2. By iden. 3. " 54, 8 a. a c t i -= -; i.e. a:o = c:d. q.e.d. b d The proofthat a and d may be made the means and b and c the extremes is left as an exercise for the student. 394. Note. The pupil should observe that the divisor in Arg. 2 above must be chosen so as to give the desired quotientin the first mem-h k ber of the equation : thus, if hi = kf,and we wish to prove that -= -, we must divide by fl ; then M = K i.e. ? =-. fl fl f I Ex. 617. Given pt = cr. Prove p : r = c : t ; also, c :p = t: r. Ex. 618. From the equation rs = Im, derive the followingeight pro-portions: ,77 7 r:l = m:s, s:l = m:r, I :r = s:m, m:r = s:l r :m = l :s, s :m = l:r, I: s = r : m, m:s = r :l. Ex. 619. Form a proportion from 7 x 4 = 3 x a ; from ft = gb. How can the proportionsobtained be verified ? 164 PLANE GEOMETRY Ex. 620. Form a proportionfrom (a + c)(a " b) = de. Ex. 621. Form a proportionfrom m2 " 2 mn + n2 = ab. Ex. 622. Form a proportionfrom c2 + 2 cd + d2 = a + ". Ex. 623. Form a proportion from (a + b)(a" 6) =4x, making a; (1) an extreme ; (2) a mean. Ex. 624.' If 7 a + 3 y : 12 = 2a + y : 3, find the ratio a : y. Proposition III. Theorem 395. If four numbers are in proportion, they are in proportion by inversion; that is,the second term is to the firstas tlxe fourth is to the third. Given a:b=c:d. To prove b : a = d : c. 396. If four numbers are in proportion, they are in proportion by alternation; that is, the first term is to the third as the second is to the fourth. Given a:b = c: d. To prove a:c " b:d. Ex. 625. If x : -1/ = y : \|,what is the value of the ratio x : y ? Ex. 626. Transform a : x = 4 : 3 so that x shall occupy in turn every place in the proportion. book in 165 397. Many transformations may be easily brought about by the followingmethod: (1) Reduce the conclusion to an equation in its simplest form ("388), then from this derive the hypothesis. (2) Begin with the hypothesisand reverse the steps of (1). This method is illustrated in the analysis of Prop. V. Proposition V. Theorem 398. If four numbers are in proportion, they are in proportion by composition; that is, the sum of the first two terms is to the first {or second) term as the sum" of the last two terms is to the third {orfourth) term. Given a:b = c:d. To prove : {a) a -+-b : a , = c 4- d : c ; (6) a -f b : b = c -f d : d. I. Analysis (1)The conclusions required above, when reduced to equa-tions in their simplestforms, are as follows : (a) (b) 1. ac +ad = ac + bc. 1. be + bd = ad + bd. 2. Whence ad = be. 2. Whence be = ad. 3. .. a: b " c:d. 3. .-. a:b = c:d. . (2) Now begin with the hypothesis and reverse the steps. II. Proof {a) Argument 1. a:b = c:d. 2. .-. ad = be. 3. ac = ac. 4. .-.ac--ad=ac--bc; i.e. a{c+ d)= c{a--b). 6. .-.a +b:a = c + d: c. q.e.d Reasons 1. By hyp. 2. " 388. 3. By iden. 4. "54,2. 5. "393. (fyThe proof of (6) is left as an exercise for the student. 166 PLANE GEOMETRY 627. If = ?,find-+Z; ^"1. y 5 x y Proposition VI. Theorem 399. iy four numbers are in proportion, they are in proportion by division; that is, the difference of the first two terms is to tlw first(or second) term as the differ-ence of the last two terms is to the third (orfourth)term. Given a:b = c:d. To prove : (a) a " b:a = c " dic\ (b) a " bib = c " did. I. The analysis is left as an exercise for the student. Reasons 1. By hyp. 2. "388. 3. By iden. 4. "54,3. Q.E.D. 5. "393. (b)The proofof (b)is left as an exercise for the student. Ex. 628. If -= -, find ^^ ; X-^UL. y 3 x y Proposition VII. Theorem 400. If four numbers are in proportion, they are in proportion by composition and division; that is, the sum of the first two terms is to their differenceas the sum of the last two terms is to their difference. Given a ib = ci d. To prove a --bia" b = c--d:c " d. 167 i.e. a + b:a " b = c + die " d. q.e.d. Reasons 1. By hyp. 2. "398. 3. "399. 4. "54, 8 a. Proposition VIII. Theorem 401. In a series of equal ratios the sum of any number of antecedents is to the sum of the corresponding conse-quents as any antecedent is to its consequent. Given a:b = c:d = e:f = g :h. To prove a -f-c + e -f g : b + d + / -J-h = a: b. I. Analysis Simplifying the conclusion above,we have : ab -\|-be + be + bg = ab -f-ad -f af + ah. The terms required for the first member of this equation, and their equivalentsfor the second member, may be obtained from the hypothesis. 168 PLANE GEOMETRY Ex. 629. Ex. 630. With the hypothesis of Prop. VIII, prove a + c + e:b + d+f=g:h. If ffl = " = ft = .g proveffl-" + -J" = g. r t g q r-t+g"q q Ex. 631. If "ll -1 find 5. Hint. Use Prop. VI. 2y 4 y Ex. 632. If .a : b = c : d, show that " " a:" + a = cl " c:c? + c. Ex. 633. Given the proportion a : 6 = 11 : 6. Write the proportions that result from taking the terms (1) by inversion ; (2) by alternation ; (3) by composition ; (4) by division ; (5) by composition and division. Proposition IX. Theorem 402. The products of the corresponding terms of any number of proportions form a proportion. Given a : b = c : d, e:f=g:h, i:j " k:l. To prove aei : bfj= cgk : did. 2. Argument b~d' e g . i _k aei __ cgk m i.e. aei : bfj = cgk: did. Q.E.D. Reasons By hyp. 2. "54, 7 c 403. Cor. If four numbers are in proportion, equi-multiple of the firsttwo and equimultiples of the last two are also in proportion. Hint. Given a :b = x:y. To prove am :bm = nx :yiy. Ex. 634. If r : s = m i t, isfr : qs = qm -.ft ? Prove your answer. Ex. 635. If a : b = 3 : 4 and x : y " 8 ;9, find the value of ax : by. 170 PLANE GEOMETRY Ex. 642. In the continued proportion a:b = b:c = c\ d = d:e, prove that: c 62' d b' ' e~6 Ex.643. If x:yB = 8:27, find ". 2/ Ex. 644. If Vm : 1 = Vw : 16, find ^. 406. Def. The segments of a line are the parts into which it is divided. The line AB is divided internally at C if this point is between the extremities of the line. The segments into which it is divided are A C and CB. D AC B I , 1 1 AB is divided externally at D if this point is on the prolonga-tion of the line. The segments are AD and DB. It should be noted that in either case the point of division is one end of each segment. 407. Def. Two straightlines are divided proportionally if the ratio of one line to either of its segments is equal to the ratio of the other line to its correspondingsegment. 408. In Prop. XI, II, the following theorems (Appendix. "" 586 and 591) will be assumed : (a) The quotientof a variable by a constant is a variable. (b) TJie limit of the quotientof a variable by a constant is the limit of the variable divided by the constant. Thus, if a; is a variable and 7c a constant : (1) j is a variable. K (2)If the limit of x is y, then the limit of -is ". Ex. 645. In the figureof " 409, name the segments into which AB is divided by D; the segments into which AD is divided bv B. BOOK III 171 Proposition XI. Theorem 409. A straightline parallel to one side of a triangle divides the other two sides proportionally. Fig. 1. Given A ABC with line BE II BC, AB AG To prove AB AE I. If AB and AB are commensurable (Fig. 1). Argument 1. Let AF be a common measure of AB and AB, and suppose that AF is con-tained in AB r times and in AB s times. 2. Then =1. AB s 3. Through the several points of division on AB, as F, G, etc., draw lines \|\| BC. 4. These lines are \|\| BE and to each other. 5. .-. AC is divided into r equal parts and AE into s equal parts. AC _r AE ~~ s AB _AC AB~ AE' O. .. Q.E.D. Reasons 1. "335. 2. " 341. 3. " 179. 4. " 180. 5. " 244. 6. " 341. 7. "54,1. 172 PLANE GEOMETRY II. If AB and AD are incommensurable (Fig. 2). KC 1. 6. Argument Let m be a measure of AD. Apply m as a measure to AB as many times as possible. There will then be a remainder,HB, less than m. .4# and AD are commensurable. Draw HK \|\| "C. AH _AK AD~ AE' , Now take a smaller measure of AD. No matter how small a measure of AD is taken,when it is appliedas a measure to AB, the remainder, HB, will be smaller than the measure taken. .-. the difference between AH and AB may be made to become and remain less than any previously assigned line, however small. .-. AH approaches AB as a limit. .. " approaches " as a limit. AD AD Likewise the difference between AK and A C may be made to become and remain less than any previously as-signed line, however small. Reasons 1. " 339. 2. " 337. 3. "179. 4. " 409, I. 5. "335. 6. Arg. 5. 7. " 349. 8. " 408, b. 9. Arg. 6. BOOK III 173 10. Argument \ AK approaches AC as a limit. 11. .. " approaches " as a limit. AE AE AH AK 12. But " is always equal to " . AD J AE 13 AB = AC ' AD~ AE' Q.E.D. Reasons 10. " 349. 11. "408,6. 12. Arg. 4. 13. "355. 410. Cor. A straightline parallel to one side of a tri-angle divides the other two sides into segments which are -proportional.Thus, iu the figuresfor Prop. XI, AB : DB = AE:EC. Hint. Prove by using division. Ex. 646. Using Fig. 2 of Prop. XI, prove (1) AB:AC = AD:AE; (2) AB : AC = DB : EC. Ex. 647. In the diagram at the right,w, w, and r are parallel to each other. Prove that a : b = c :d; also that a : c = b : d. Ex. 648. If two sides of a triangle are 12 inches and 18 inches,and if a line is drawn paral-lel to the third side and cuts off 3 inches from the vertex on the 12-inch side,into what segments will it cut the 18-inch side ? Ex. 649. If two lines are cut by any number of parallels, the two lines are divided into segments which are proportional : (a) if the two lines are parallel ; (") if the two lines are oblique. Ex. 650. If through the point of intersection of the medians of a trianglea line is drawn parallelto any side of the triangle, this line divides the other two sides in the ratio of 2 to 1. Ex. 651. A line can be divided at but one point into segments which have a given ratio (measured from one end). Ex. 652. A line parallel to the bases of a trapezoiddivides the other two sides and also the two diagonalsproportionally. Ex. 653. Apply the proof of Prop. XI to the case in which the par-allel to the base cuts the sides prolonged : (a) through the ends of the b"sV,; (6) through the vertex. 174 PLANE GEOMETRY Proposition XIT. Problem 411. To construct the fourth proportional to three given lines. T R " \M b \L S Given lines a, b, and c. To construct the fourth proportional to a, b, and c. I. Construction 1. From any point, as R, draw two indefinite lines RS and RT. 2. On RS lay off RM = a and ML = b. 3. On RT lay off RF = c. 4. Draw MF. 5. Through L construct LG IIMF. " 188. 6. FG is the fourth proportional to a, b, and c. II. The proof and discussion are left as an exercise for the student. 412. Question. Could the segments a, b, and c be laid off in any other order ? 413. Cor. I. To construct the third proportional to two given lines. 414. Cor. n. To divide a given line into segments pro-portional to two or more given lines. Ex. 654. Divide a given line into segments in the ratio of 3 to 5. Ex. 655. Divide a given line into segments proportional to 2, 3,and 4. BOOK III 175 Ex. 656. Construct two lines, given their sum and their ratio. Ex. 657. Construct two lines,given their difference and their ratio. Ex. 658. If a, b, and c are three given lines, construct x so that: (a) x:a = b:c, (") x = " " b Proposition XIII. Theorem (Converseof Prop. XI) 415. If a straight line divides two sides of a triangle proportionally, it is parallel to the third side. A F E C Given A ABC, and BE so drawn that " To prove BE IIBC. AB AE Argument 1. BE and BC are either IIor not II. 2. Suppose that BE is not IIBC, but that some other line through B, as BF, is II BC. AB _ AC AB ~ AF AB _ AC AB ~ AE' 3. Then 4. But 5. .-. AF= AE. 6. This is impossible. 7. .'. BE IIBC. Q.E.D. Reasons 1. " 161, a. 2. "179. 3. "409. By hyp. "391. " 54, 12. " 161, b. 416. Cor. If a straight line divides two sides of a tri-angle into segments which are proportional, it is parallel to.the third side. Thus, if AB-.BB = AE: EC, BE is IIBC. 176 PLANE GEOMETRY Ex. 659. In the diagram at the right, if AB = 15, AC = 12, AD = 10, and AE = 8, prove D," parallel to BC. Ex. 660. If AB = 50, #" = 15, AE = 28, and EC = 12, is Z"# parallel to BC ? Prove. Ex. 661. If D^ \|\| BC, AB = 25, Di? = 5, and.4C = 20, find AE. Ex. 662. If BE \|\|.BC, AB= 30, ^li" = 25, and EC = 4, find .4C. SIMILAR POLYGONS 417. Def. If the angles of one polygon, taken in order, are equalrespectively to those of another, taken in order,the poly-gons are said to be mutually equiangular. The pairsof equal angles in the two polygons,taken in order, are called homolo-gous angles of the two polygons. 418. Def. If the sides of one polygon,taken in order as antecedents, form a series of equal ratios with the sides of another polygon, taken in order as consequents, the polygons are said to have their sides proportional. Thus, in the accom-panying figure, if a : I = b : m = c : n = d : o = e : p, the two polygonshave their sides proportional. The lines forming any ratio are called homologous lines of the two polygons,and the ratio of two such lines is called the ratio of similitude of the polygons. 419. Def. Two polygons are similar if they are mutually equiangular and if their sides are proportional. 178 PLANE GEOMETRY 422. Cor. II. Two right triangles are similar if an acute angle of one is equal to an acute angle of the other. 423. Cor. in. If a line is drawn parallel to any side of a triangle, this line, with the other two sides, forms a triangle which is similar to the given triangle. Ex. 663. Upon a given line as base construct a triangle similar to a given triangle. Ex. 664. Draw a triangleABC. Estimate the lengthsof its sides. Draw a second triangleDEF similar to ABC and having DE equal to two thirds of AB. Compute DF and EF. Ex. 665. Any two altitudes of a triangleare to each other inversely as the sides to which they are drawn. Ex. 666. At a cer-tain hour of the day a tree,BC, casts a shadow, CA. At the same time a vertical pole, ED, casts a shadow, DF. What measurements arc necessary to determine the height of the tree ? E ^ Ex. 667. If CA is found to be 64 feet ; DF, 16 feet what is the height of the tree ? Ex. 668. To find the distance across a river from A to B, a point C was located so that BC was perpendicularto AB at B. CD was then measured off 100 feet in length and perpendicularto BC at C. The line of sightfrom D to A intersected BC at E. By measurement CE was found to be 90 feet and EB 210 feet. What was the distance across the river ? BOOK III 179 Ex. 669. Two isosceles triangles are similar if the vertex angle of one equals the vertex angle of the other,or if a base angle of one equals a base angle of the other. 424. It follows from the definition of similar polygons, " 419, and from Prop. XIV that: (1) Homologous anglesof similar triangles are equal. (2) Homologous sides of similar triangles are proportional. (3) Homologous sides of similar Mangles are the sides opposite equal angles. 425. Note. In case, therefore, it is desired to prove four lines pro-portional, try to find a pair of triangleseach having two of the given lines as sides. If,then, these trianglescan be proved similar,their homologous sides will be proportional. By marking xoith colored crayon the lines required in the proportion,the triangles can readilybe found. If it is desired to prove the product of two lines equal to the product of two other lines, prove the four lines proportionalby the method just sug-gested, then put the product of the extremes equal to the product of the means 426. Def. The length of a secant from an external pointto a circle is the length of the segment included between the point and the second pointof intersection of the secant and the circumference. Ex. 670. If two chords intersect within a circle, es-tablish a proportionality among the segments of the chords. Place the product of the extremes equal to the product of the means, and state your result as a theorem. Ex. 671. If two secants are drawn from any given point to a circle, what are the segments of the secants? Does the theorem of Ex. 670 still hold with regard to them? Ex. 672. Rotate one of the secants of Ex. 671 about fjg^88 "n. the point of intersection of the two until the rotating fC \ secant becomes a tangent. What are the segments of the f\ \| secant which has become a tangent ? Does the theorem of \ \ J Ex. 670 stillhold ? Prove. ^_ S By the product of two lines is meant the product of their measure-numbers. This wiH \p discussed aerain in Book IV. 180 PLANE GEOMETRY Proposition XV. Theorem 427. Two triangles which have their sides proportional are similar. E Given A BEF and ABC such that " = " = AB BO AG To prove A DEF ~ A ABC. Argument 1. On BE lay off DH " AB, and on DF lay off BK=AC. 2. Draw "" ' o BE _BFf AB~ AC . BE BF 4. .". = " " BH BK 5. .\ HK II EF. 6. .-. A DEF ~ A BHK. It remains to prove A BHK = A ^"(7. " BE _EF BH~ HK But BE EF I.e. BE BH EF BC AB BC 9. .'.HK=BC. 10. Now BH= AB and DK=AC 11. .\ A BHK=A ABC. 12. But A BEF~ A DiMr. 13. .-. A BEF ~ A ^BC. Q.E.D. Reasons 1. "54,14. 2. "54,15. 3. By hyp. 4. "309. 5. "415. 6. "423. 7. "424,2. 8. By hyp. 9. "391. 10. Arg. 1. 11. "116. 12. Arg. 6. 13. "309. Ex. 673. If the sides of two trianglesare 9, 12, 15, and 6, 8, 10, respectively, are the triangles similar ? Explain. BOOK III 181 Ex. 674. Construct a triangle that shall have a givenperimeter and shall be similar to a given triangle. Ex. 675. Construct a trapezoid,given the two bases and the two diagonals. Hint. How do the diagonalsof a trapezoid divide each other ? Proposition XVI. Theorem 428. If two triangles have an angle of one equal to an angle of tlie other, and the including sides propor-tional, the trianglesare similar. Given A ABO and DEF with Z A = Z D and " = A To prove A AB C ~ A DEF. DE BF Ex. 676. Two triangles are similar if two sides and the median drawn to one of these sides in one triangle are proportionalto two sides and the corKe'\|ponding median in the other triangle. 182 PLANE GEOMETRY a B E Fig. 2. Ex. 677. In triangles AB C and DBC, Fig.\,AB = AC and BD = BC. Prove triangle ABC similar to triangle DBG. Ex.678. In Fig.2, AB . AG = AE : AD. Prove triangle AB G simi-lar to triangle ADE. Ex. 679. If in triangle ABC, Fig.3, GA = BC, and if 7) is a point such that GA : AB = AB : .4Z",prove ^1Z? = BD. Ex. 680. Construct a triangle similar to a given triangle and having the sum of two sides equal to a given line. Proposition XVII. Theorem 429. Two triangles that have their sides parallel each to each, or perpendicular each to each, are similar. Fig. 1. Given A ABC and A!B'c', with AS, BC, and CA II(Fig. 1) or _L (Fig.2) respectively to a'b',b'c', and c'a'. To prove A AB C ~ A A'b'c'. Argument 1. AB, BC, and CA are IIor _L respectively to A'B\ B'c', and C'A'. 2. .. A A, B, and C are equalrespectively or are sup. respectively to A A', B ', and c\ Reasons 1. By hyp. 2. ""198,201. BOOK III 183 Argument 3. Three suppositions may be made, there-fore, as follows : (1) Za + Za'=2tLA,Zb+Zb' = 2 rt. A, Zc + Zc' = 2 rt. A. (2) Za=Za',Zb + Zb' = 2i"A, Z C + Z c" = 2 rt. A. (3) Z A = Z A', Zb = Zb'; hence, also, Za= Z ". 4. According to (1)and (2)the sum of the A of the two A is more than four rt. A. 5. But this is impossible. 6. .-. (3) is the onlysupposition admissible; i.e. the two A are mutually equi-angular. 7. .-. AABC~AA'B'c'. q.e.d. Reasons 3. "161, a. 4. "54,2. 5. "204. 6. 161,b. 7. "420. 430. Question. Can one pair of angles in Prop. XVII he supple-mentary and the other two pairsequal ? SUMMARY OP CONDITIONS FOR SIMILARITY OF TRIANGLES 431. I. Two triangles are similar if they are mutuallyequi-angular (a) Two triangles are similar if two angles of one are equal respectively to two angles of the other. (6) Two right triangles are similar if an acute angle of one is equal to an acute angle of the other. (c) If a line is drawn parallel to any side of a triangle, this line, with the other two sides, forms a triangle which is similar to the given triangle. II. Two triangles are similar if their sides are proportional. III. Two triangles are similar if they have an angle of one equal to an angle of the other,and the including sides pro-portional IV. Two triangles are similar if their sides are parallel each to ea^h, or perpendiculareach to each. 184 PLANE GEOMETRY Ex. 681. Inscribe a trianglein a circle and circumscribe about the circle a triangle similar to the inscribed triangle. Ex. 682. Circumscribe a triangleabout a circle and inscribe in the circle a similar triangle. Ex. 683. The lines joining the mid-points of the sides of a triangle form a second triangle similar to the given triangle. Ex. 684. ABC is a triangle inscribed in a circle. A line is drawn from A to P, any point of BC, and a chord is drawn from B to a point Q in arc BC so that angle ABQ equals angle ABC. Prove AB x AC = AQ x AB. Proposition XVIII. Theorem 432. The bisector of an angle of a triangle divides the opposite side into segments which are proportional to the other two sides. Given A ABC with BP the bisector of Z ABC. To prove AP:PC= AB:BC. 186 PLANE GEOMETRY 434. Def. A line is divided harmonically if it is divided internally and externally into segments whose ratios are numeric callyequal ; thus,if line AC is divided internally at P and ex-ternally at Q so that the ratio of \| f \ AP to PC is numerically equal to ^ PC Q the ratio of AQ to QC, AC is said to be divided harmonically. Ex. 689. The bisectors of the interior and exterior angles at any vertex of a triangle divide the opposite side harmonically. Ex. 690. Divide a given straightline harmonically in the ratio of 3 to 6 ; in the ratio of a to 6, where a and b are given straight lines. Proposition XX. Theorem 435. In two similar triangles any two homologous alti-tudes have the same ratio as any two homologous sides. Given two similar A ABC and DEFf with two corresponding altitudes AH and DK. To prove AH DK 4. But^ In rt. A ABH and DEK, Z B = AE. .-. AABH~ A DEK. , AH, opposite Z B _ AB, opposite Z.BHA ' ' DK, opposite Z.E DE,o])ipositeZ.EKD BC _ CA DE EF FD ' AH =AB = BC _ CA DK~ DE EF FD Q.E.D. Reasons 1. "424,1. 2. "422. 3. " 424, 2. 4. "424,2. 5. "54,1. book in 187 Proposition XXI. Theorem 436. If three or more straight lines drawn through a common point intersect two parallels,the corresponding segments of the parallels are proportional. Given lines PA, PB, PC, PD drawn through a common pointP and intersecting the II lines AB and A'd' at pointsA, B, C, D and A\ B',C', d',respectively. Ex. 691. If three or more non-parallelstraightlines intercept pro-portional segments on two parallels, they pass through a common point. 188 PLANE GEOMETRY Ex. 692. A man is ridingin an automobile at the uniform rate of 30 miles an hour on one side of a road, while on a footpath on the other side a man is walking in the oppositedirection. If the distance between the footpath and the auto track is 44 feet,and a tree 4 feet from the footpath continuallyhides the chauffeur from the pedestrian, does the pedestrian walk at a uniform rate ? If so, at what rate does he walk ? Ex. 693. Two sides of a triangle are 8 and 11, and the altitude upon the third side is 6. A similar triangle has the side homologous to 8 equal to 12. Compute as many parts of the second triangle as you can. Ex. 694. In two similar triangles, any two homologous bisectors are in the same ratio as any two homologous sides. Ex. 695. In two similar triangles, any two homologous medians are in the same ratio as any two homologous sides. 437. Drawing to Scale. Measure the top of your desk. Make a drawing on paper in which each line is -^ as long as the corresponding line of your desk. Check your work by measuring the diagonal of your drawing, and the corresponding line of your desk. This is called drawing to scale. Map draw-ing is a common illustration of this principle.The scale of the drawing may be represented : (1)by saying, " Scale, -j^" or " Scale, 1 inch to 12 inches " ; (2)by actually drawing the scale as indicated. B 12 24 36 Ex. 696. Using the scale above, draw lines on paper to represent 24 inches ; 3 feet 3 inches. Ex. 697. On the black-board draw, to the scale above, a circle whose diame-ter is 28 feet. Ex. 698. The figure represents a farm drawn to the scale indicated. Find the cost of putting a fence around the farm, if the fenc-ing costs $2.50 per rod. B 40 160 BOOK III 189 Proposition XXII. Theorem 438. If two polygons are composed of the same number of triangles,similar each to each and similarly placed, the polygons are similar. C E J Given polygons ABODE and FGHIJ with A ABC ~ A FGHf AACD ~ A FHI, A ABE ~ A FIJ. To prove polygon ABODE ~ polygon FGHIJ. 190 PLANE GEOMETRY 439. Cor. Any two similar polygons may be divided into the same number of triangles similar each to each and similarlyplaced. Proposition XXIII. Problem 440. Upon a line homologous to a side of a given poly-gon, to construct a polygon similar to the given polygon. C R Given polygon AD and line MQ homol. to side AE. To construct, on MQ, a polygon ~ polygon AD. I. Construction 1. Draw all possiblediagonalsfrom A, as AC and AD. 2. At M, beginning with MQ as a side, construct A 7, 8, and 9 equal respectively to A 1,2, and 3. " 125. 3. At Q, with MQ as a side, construct Z 10 equalto A 4, and prolong side QP until it meets ML at P. " 125. 4. At P, with PM as a side, construct Zll equalto Z5, and prolong side PO until it meets MR at 0. " 125. 5. At 0, with OM as a side, construct Z12 equal to A6, and prolong side ON until it meets MF at N. " 125. 6. MP is the polygon required. II. Proof Argument 1. A ADE ~ A MPQ, A ACD ~ A MOP, and A ABC ~ A MNO. 2. .-. polygon MP ~ polygon AD. q.e.d. Reasons 1. "421. 2. " 438. book in 191 III. The discussion is left as an exercise for the student. Proposition XXIV. Theorem 441. The perimeters of two similar polygons are to each other as any two homologous sides. Given ~ polygonsP and Q, with sides a, b, c, d, e, and / homol. respectively to sides k, I,m, n, o, andjp. _ perimeter of P a To prove " " = -. perimeter of Q k Argument a^ !L" -" f " / ^ I m n o p k 2. IV U IIV IV \J JJ a + b+ c +d+e+f=a k + I +m + ?i + o-\|-/) " ThatiS,PerimeterofP = " perimeter of Q k Q.E.D. Reasons 1. "419. 2. " 401. 3. " 309. Ex. 699. The perimeters of two similar polygons are 152 and 138 ; a side of the first is 8. Find the homologous side of the second. Ex. 700. The perimetersof two similar polygons are to each other as any two homologous diagonals. Ex. 701. The perimetersof two similar triangles are to each other as any two homologous medians. Ex. 702. If perpendiculars are drawn to the hypotenuse of a right triangleat its extremities, and if the other two sides of the triangleare prolongedto meet these perpendiculars, the figurethus formed contains fivf, triangles each of which #is similar to any one of the others. 192 PLANE GEOMETRY Proposition XXV. Theorem 442. In a right triangle,if the altitude upon the hypotenuse is drawn: I. The triangles thus formed are similar to the given triangle and to each other. II. The altitude is a mean proportional between the segments of the hypotenuse. III. Either side is a mean proportional between the whole hypotenuse and the segment of the hypotenuse adjacent to that side. Given rt. A ABC and the altitude CD upon the hypotenuse. To prove : I. A BCD ~ A ABC, A ADC ~ A ABC, and A BCD ~ A ADC. II. BD: CD= CD : DA. III. AB : BC=BC:DB and AB: AC= AC: AD. I. Argument 1. In rt. A B CD and ABC, Z.B = /.B. 2. .'. A BCD ~ A ABC. 3. In rt. A ADC and ABC, Za = A A. 4. .-. A ADC ~ A ABC. 5. .:Z1 = Zb. 6. .'. A BCD ~ A ADC. Q.E.D. II. The proof of II is left as an exercise for the student Hint. Mark (with colored chalk,if convenient) the lines required in the proportion. Decide which triangles will furnish these lines, and use the fact that homologous sides of similar triangles are proportional. 194 PLANE GEOMETRY Proposition XXVI. Problem 445. To construct a mean proportional between two given lines. / t' N f I !0 3 rr^rr^ Given two lines, m and w. To construct a line Z so that m:l = l:n. I. Construction 1. On any indefinite str. line, as AB, lay off AG = m and CZ) = n. 2. With 0, the mid-point of AD, as center and with a radius equal to OB, describe a semicircumference. 3. At G construct GEWAD, meeting the semicircumference at E. " 148. 4. CE is the requiredline I. II. The proofand discussion are left as an exercise for the student. Ex. 706. By means of " 444, II,construct a mean proportional be-tween two given lines by a method different from that given in Prop. XXVI. Ex. 707. Use the method of Prop. XXVI to construct a line equal to V3 ab, a and b being given lines. Analysis. Let x = the requiredline. Then x = V3ab. .-. x = 3 ab. ,\ 3 a : x = x : b. Ex. 708. Construct a line equal to a V3, where a is a given line. Ex. 709. Using a line one inch long as a unit,construct a line equal to VS ; V5 ; V6. Choosing your own unit, construct a line equal to 3 a/2,2\% 5\/5. book in 195 Proposition XXVII. Theorem 443. In any right triangle the square of the hypotenuse is equal to the sum of the squares of the othe? " two sides. 447. Cor. I. The square of either side of a right tri-angle is equal to the square of the hypotenuse minus the square of the other side. 448. Cor. n. The diagonal of a square is equal to its side multiplied by the square root of two. Outline of Proof 1. d2=s2 + s2=2s2. 2. .\ d = s-y/2. q.e.d. 449. Historical Note. The property of the right trianglestated in Prop. XXVII was known at a very early date,the ancient Egyptians, 2000 b.c, having made a right triangle by stretching around three pegs a cor measured off into 3, 4, and 5 units. See Note, Book IV, " 510. 196 PLANE GEOMETRY Ex. 710. By means of " 448, construct a line equal to V2 inches. Ex. 711. If a side of a square is 6 inches,find its diagonal. Ex. 712. The hypotenuse of a righttriangle is 15 and one arm is 9. Find the other arm and the segments of the hypotenuse made by the perpendicular from the vertex of the rightangle. Ex. 713. Find the altitude of an equilateral triangle whose side is 6 inches. Ex. 714. Find a side of an equilateraltrianglewhose altitude is 8 inches. Ex. 715. Divide a line into segments which shall be in the ratio of 1 to V2. Ex. 716. The radius of a circle is 10 inches. Find the lengthof a chord 6 inches from the center ; 4 inches from the center. Ex. 717. The radius of a circle is 20 inches. How far from the cen-ter is a chord 'whose length is 32 inches ? whose length is 28 inches ? Ex. 718. Tn a circle a chord 24 inches long is 5 inches from the center. How far from the center is a chord whose length is 12 inches ? 450. Def. The projection of a point upon a line is the foot of the perpendicular from the point to the line. 451. Def. The projection of a line segment upon a line is the segment of the second line included between the projections of the extremities of the first line upon the second. Thus, C is the projectionof A upon MN, D is the projection of B upon MN} and CD is the projection of AB upon MN. B B 9 A. M~C Irk M C D N M V A Ex. 719. In the figures above, under what condition will the projec-tion of AB on M2f be a maximum ? a minimum ? Will the projection CD ever be equal to AB ? greater than AB ? Will the projection ever be a point ? Ex. 720. In a right isosceles triangle the hypotenuse of which is 10 inches,find the length of the projection of either arm upon the hypotenuse. Ex. 721. Find the projectionof one side of an equilateral triangle upon another if each side is 6 inches. BOOK III 197 Ex. 722. Draw the projections of the shortest side of a triangle upon each of the other sides : (1) in an acute triangle ; (2) in a righttriangle ; (3) in an obtuse triangle. Draw the projectionsof the longest side in each case. Ex. 723. Two sides of a triangleare 8 and 12 inches and their included angle is 60". Find the projection of the shorter upon the longer. Ex. 724. In Ex. 723, find the projection of the shorter side upon the longer if the included angle is 30" ; 45". Ex. 725. Parallel lines that have equal projections on the same line are equal. Proposition XXVIII. Problem 452. In any triangle to find the value of the square of the side opposite an acute angle in terms of the other two sides and of the projection of either of these sides upon the other. A X D Fig. 1. Given ABAX, with Z.X acute; and p, the projection of b upon a. To find the value of a2 in terms of a, b,and p. Argument 1. In rt. A BAD, a? = AD2 + DB2. 2. But AD2 = b2 -p2. 3. And DB = a"p (Fig. 1)or p " a (Fig. 2) . 4. .-. DB = a2 " 2 ap + p2. 5. .-. a,2 = b2 " p2 -f a2 " 2 ap -f p2; i.e. x? = a2 + b2 " 2 ap. q.e.f. 453. Question. Why is it not necessary to include here the figure aad^discussion for a right triangle? Reasons 1. "446. 2. "447. 3. " 54, 11. 4. " 54, 13. 5. "309. 198 PLANE GEOMETRY 454. Prop. XXVIII may be stated in the form of a theorem as follows : In any triangle, the square of the side opposite an acute^ angle is equal to the sum of the squares of the other two sides dimin-ished by twice the product of one of these sides and the projec-tion of the other side upon it. Ex. 726. If the sides of a triangleare 7, 8, and 10, is the angle opposite 10 obtuse,right, or acute ? Why ? Ex. 727. Apply the statement of Prop. XXVIII to the square of an arm of a righttriangle. Ex. 728. Find x (inthe figurefor Prop. XXVIII) in terms of a and b and the projection of a upon b. Ex. 729. If the sides of a triangle are 13, 14, and 15, find the pro-jection of the firstside upon the second. Ex. 730. If two sides of a triangle are 4 and 12 and the projection of the firstside upon the second is 2, find the third side of the triangle Proposition XXIX. Theorem 455. In any obtuse triangle,the square of the side oppo-site the obtuse angle is equal to the sum of the squares of the other two sides increased by twice the product of one of these sides and the projection of the other side upon it. Given A BAX with Z.X obtuse, and p, the projection of b upon a. To prove (c2 = a2 + b2 + 2 ap BOOK 111 199 Argument 1. In rt. A BAD, a?=Al? + Bb\ 2. But AB2 = b2-p\ 3. And DB = a+p. 4. .\ BB2 = a2 + 2 ap +p\ 5 .-. ^ = 62-p2+a2 + 2ap+P2; i. e. x1 = a2 + b2 -f 2 ap. q.e.d. Reasons 1. "446. 2. "447. 3. "54,11. 4. " 54, 13. 5. "309. 456. From Props. XXVIII and XXIX, we may derive the following formulas for computing the projection of one side of a triangleupon another ; thus if a, 6, and c represent the sides of a triangle : From Prop.XXVIII, p = ^ + b "~ ^ (1) 2a From Prop.XXIX, -p = a + "2 ~ g (2) It is seen that the second members of these two equations are identical and that the first members differ only in sign. Hence, formula (1) may always be used for computing the length of a projection. It need only be remembered that if p is positive in any calculation, it indicates that the angle oppo-site c is acute; while if p is negative,the angle opposite c is obtuse. It can likewise be shown (seeProp. XXVII) that if p = 0, the angle opposite c is a rightangle. Ex. 731. Write the formula for the projection of a upon b. Ex. 732. In triangle ABC, a = 15, 6 = 20, c = 25; find the projec-tion of b upon c. Is angle A acute, right, or obtuse ? Ex. 733. In the triangle of Ex. 732, find the projection of a upon b. Is angle C acute, right, or obtuse ? Ex. 734. The sides of a triangleare 8, 14, and 20. Is the angle opposite the side 20 acute, right, or obtuse ? Ex. 735. If two sides of a triangle are 10 and 12, and their included angle is 120",what is the value of the third side ? Ex. 736. If two sides of a triangle are 12 and 16, and their included angle is 45",find the third side. 200 PLANE GEOMETRY Ex. 737. If in triangle ABC, angle C = 120", prove that AS2 = BC2 + AC2 + AC-BC. Ex. 738. If a line is drawn from the vertex C of an isosceles tri-angle ABC, meeting base AB prolonged at D, prove that dl"2-CBi = AB. BD. Proposition XXX. Theorem 457. In any triangle,the sum of the squares of any two sides is equal to twice the square of half the third side increased by twice the square of the median upon that side. Given A ABC with ma, the median to side a. To prove b2 + c2 = 2 (-Y + 2 ma\ Argument 1. Suppose 6 " c ; then Z ADC is obtuse and Z BDA is acute. 2. Let p be the projection of ma upon a. 3. Then from A A D C, 4. And from A ABD, " + ma2 (2) " """ 5. .-. 62-hc2=2^J+2ma2. p. Q.E.D. Reasons 1. " 173. 2. "451. 3. "455. 4. "454. 5. "54,2. 202 PLANE GEOMETRY Proposition XXXI. Theorem 460. If through a point within a circle two chords are drawn, the product of the two segments of one of these chords is equal to the product of the two segments of the other. Given P, a point within circle 0, and AB and CD, any two chords drawn through P. To prove PA " PB = PC " PD. The proof is left as an exercise for the student. Hint. Prove A APC ~ A PDB. Ex. 751. In the figurefor Prop. XXXI, if PA = 5, PB = 12, and PD = 6, find PC. Ex. 752. In the same figure, if PC = 10, PD = 8, and AB = 21, find PA and PB. Ex. 753. In the same figure, if PC = 6, DC = 22, and AB = 20, find AP and PB. Ex. 754. In the same figure, if PA = m, PC = n, and PD " r, find PB. Ex. 755. If two chords intersecting within a circle are of lengths8 and 10, and the second bisects the first, what are the segments of the second ? Ex. 756. By means of Prop. XXXI construct a mean proportional between two given lines. Ex. 757. If two chords intersect within a circle and the segments of one chord are a and b inches, while the second chord measures d inches, construct the segments of the second chord. Hint. Find the locus of the mid-points of chords equal to d. BOOK III 203 Ex. 758. If two lines AB and CD intersect at E so that AE " EB = CE " ED, then a circumference can be passed through the four points A, B, C, D. Proposition XXXII. Theorem 461. // a tangent and a secant are drawn from any given point to a circle, the tangent is a mean propor-tional between the whole secant and its external segment. Given tangent PT and secant PB drawn from the point P to the circle 0. m PB To prove " -: PT PT PC Argument 1. Draw crand BT. . 2. In A PBT and CTP, Zp = Zp. 3. Z2 = Z2'. 4. .-. APBT~ A CTP. 5 . PB, oppositeZ 3 in A PBT ' PT, oppositeZ 3' in A CTP _ PT, oppositeZ 2 in A PBT ~ PC, oppositeZ 2' in A CTP Q.E.D. Reasons 1. "54,15. 2. Byiden. 3. " 362, a. 4. "421. 5. "424,2. 462. Cor. I. If a tangent and a secant are drawn from any given point to a circle, the square of the tangent is equal to the product of the whole secant and its external segment. 204 PLANE GEOMETRY 463. Cor. II If two or more secants are drawn from any given point to a circle, the product of any secant and its external segment is constant. T^r' Given secants PD, PE, PF, drawn from pointP to circle O, P and let their external segments be denoted by PA, PB, PC, respectively. To prove PD " PA = PE " PB =PF " PC, Argument 1. From P draw a tangent to circle 0, as PT. 2. PT2 = PD " PA; PT2 = PE " PB; PT2 = PF " PC. 3. .'. PD-PA = PE"PB = PF-PC. Q.E.D. Ex. 759. If a tangent and a secant drawn from the same point to a circle measure 6 and 18 inches,respectively, how long is the external segment of the secant ? Ex. 760. Two secants are drawn to a circle from an outside point. If their external segments are 12 and 9,and the internal segment of the first secant is 8, what is the length of the second secant ? Ex. 761. The tangents to two intersecting circles from any point in their common chord (prolonged) are equal. Ex. 762. If two circumferences intersect, their common chord (pro-longed) bisects their common tangents. 464. Def. A line is said to be divided in extreme and mean ratio if it is divided into two parts such that one part is a mean proportional between the whole line and the other part. Thus, AB is divided in extreme and mean ratio at P if AB : AP = " AP : PB. This division is known as the golden section. BOOK III 205 Proposition XXXIII. Pi-oblem 465. To divide a line internallyin extreme and mean ratio. x ""w i ' t XV "S "? I I I I I f I / A P B Given line AB. To find, in ABf a pointP such that AB:AP = AP: PB. I. Construction 1. From B draw 50 _L ^4 5 and " \AB. " 148. 2. With O as center and with BO as radius describe a cir-cumfere 3. From A draw a secant through 6, cutting the circumfer-ence at C and D. 4. With A as center and with AC as radius draw CP, cutting AB at P. 5. AB : ^IP = AP : PP. II. Proof Argument 1. AB is tangent to circle O. 2. .-. AD : AB = AB : AC. 3. .-. AD " ^1P: AB = AB " AC: AC. 4. .-. AD" CD : AB = AB" AP : AP. 5. .'. AP : AB = PB : ^P. 6. .'. ^1P: AP = JP: PB. Q.E.D. Reasons 1. "314. 2. "461. 3. " 399. 4! " 309. 5. " 309. 6. " 395. HJ. The discussion is left as an exercise for the student. 206 PLANE GEOMETRY Ex. 763. Divide a line AB externallyin extreme and mean ratio. Hint. In the figurefor Prop. XXXIII prolong BA to P', making FA = AD. Then prove AB : P'^l = PA : P'B. Ex. 764. If the line I is divided internally in extreme and mean ratio, and if s is the greater segment, find the value of s in terms of I. Hint. I : s = s : I " s. Ex. 765. A line 10 inches long is divided internally in extreme and mean ratio. Find the lengthsof the two segments. Ex. 766. A line 8 inches long is divided externallyin extreme and mean ratio. Find the length of the longer segment. MISCELLANEOUS EXERCISES Ex. 767. Explain how the accompanying figure can be used to find the distance from A to B on oppositesides of a hill. CE = %BC, CD = \AC. ED is found by measurement to be 125 feet. What is the distance AB ? Ex. 768. A littleboy wished to obtain the height of a tree in his yard. He set up a ver-tical pole 6 feet high and watched until the shadow of the pole measured exactly6 feet. He then measured quickly the length of the tree's shadow and called this the height of the tree. Was his answer correct ? Draw figures and explain. Use this method for measuring the heightof your school buildingand flag pole. Ex. 769. If light from a tree, as AB, is allowed to pass through a small aperture O, in a window shutter W, and strike a white screen or wall, an inverted image of the tree,as CD, is formed on the screen. If the dis-tance OE= 30 feet,OF =8 feet, and the lengthof the tree AB=%5 feet, find th3 length of the image CD. Under what condition will the length of the image equal the length of the tree ? This exercise illustrates the principle of the photographer'scamera. Ex. 770. By means of Prop. XXXII construct a mean proportional between two given lines. Ex. 771. In a certain circle a chord 5V5 inches from the center is 20 inches in length. Find the length of a chord 9 inches from the center. BOOK III 207 Ex. 772. Compute the length of : (1) the common external tangent, (2) the common internal tangent, to two circles whose radii are 8 and 6, respectively, and the distance between whose centers is 20. Ex. 773. If the hypotenuse of an isosceles righttriangle is 16 inches, what is the length of each arm ? Ex. 774. If from a point a tangent and a secant are drawn and the segments of the secant are 4 and 12, how long is the tangent ? Ex. 775. Given the equationVOl+JIl - ?_5"; solve for x. X + c c Ex. 776. Find a mean proportionalbetween a2 + 2 ab + b2 and a2 -2 ab + b2. Ex. 777. The mean proportional between two unequal lines is less than half their sum, Ex. 778. The diagonals of a trapezoid divide each other into segments which are proportional. Ex. 779. ABC is an isosceles triangle inscribed in a circle. Chord BD is drawn from the vertex B, cutting Ay the base in any point, as E. Prove BD : AB"AB : BE. Ex. 780. In a triangleABC the side AB is 305 feet. If a line parallel to BC divides AC in the ratio of 2 to 3,what are the lengthsof the segments into which it divides AB ? Ex. 781. Construct,in one figure,four lines whose lengthsshall be that of a given unit multipliedby V2, V3, 2, V5, respectively. Ex. 782. Two sides of a triangle are 12 and 18 inches, and the perpen-dicular upon the first from the opposite vertex is 9 inches. What is the length of the altitude upon the second side ? Ex. 783. If a : b = c : d, show that K")!K6)=K")=K" Also translate this fact into a verbal statement. Ex. 784. If a constant is added to or subtracted from each term of a proportion,will the resultingnumbers be in proportion? Give proof. Ex. 785. li r : s = t : q, is3r + -:s=7t:2q? Prove your answer. u Ex. 786. One segment of a chord drawn through a point 7 units from the center of a circle is 4 units long. If the diameter of the circle is 15 units, what is the length of the other segment ? Ex. 787. The non-parallel sides of a trapezoid and the line joining the mid-pointsof the parallel sides, if prolonged, are concurrent. 208 PLANE GEOMETRY Ex. 788. Construct a circle which shall pass through two given points and be tangent to a given straight line. Ex. 789. The sides of a triangle are 10, 12,15. Compute the lengths of the two segments into which the least side is divided by the bisector of the opposite angle. " Ex. 790. AB is a chord of a circle, and CE is any A^^%^. R chord drawn through the middle point C of arc AB, cut-ting chord AB at D. Prove CE : CA = CA : CD. Ex. 791. Construct a righttriangle, given its perime-ter and an acute angle. ^-" Ex. 792. The base of an isosceles triangle is a, and the perpendicular let fall from an extremity of the base to the oppositeside is b. Find the lengths of the equal sides. Ex. 793. AD and BE are two altitudes of triangle CAB. Prove that AD:BE= CA : BC. Ex. 794. If two circles touch each other, their common external tangent is a mean proportionalbetween their diameters. Ex. 795. If two circles are tangent internally, all chords of the greater circle drawn from the point of contact are divided proportionally by the circumference of the smaller circle. Ex. 796. If three circles intersect each other,their common chords pass through a common point. Ex. 797. The square of the bisector of an angle of a triangle is equal to the product of the sides of this angle diminished by the product of the segments of the third side made by the bisector. Given A ABC with t, the bisector of ZB, dividing side b into the two segments s and r. To prove t2 = ac " rs. Outline of Proof 1. Prove that a : t = t+ m : c. 2. Then ac = P + tm = t2 + rs. 3. .. t2 = ac " rs. Ex. 798. In any triangle the product of two sides is equal to the product of the altitude upon the third side and the diameter of the circumscribed circle. Hint. Prove A ABB ~AEBC. Then prove ac = Jut. 210 PLANE GEOMETRY G U E H Fig. 2. Rectangle EG = 8 U+. times without a remainder (seeFig.2), then by taking a square which is an aliquotpart of U, as one fourth of U, and applying E H Fig. 3. Rectangle EG = " U+= U\ U+. it as a measure to the rectangle(seeFig. 3) a number will be obtained which, divided by four,will give another (and G Fig. 4. Rectangle AO = ^U+= 111 U+. usuallycloser) approximate area of the given rectangle. By proceeding in this way (see Fig.4),closer and closer approxi-mations of the true area may be obtained. It takes four of the small squares to make the unit itself. BOOK IV 211 469. If the sides of the given rectangleand of the unit square are commensurable, a square may be found which is an aliquotpart of U, and which is contained in the rectanglealso an integral number of times. 470. If the sides of the given rectangle and of the unit square are incommensurable, then closer and closer approxima-tions to the area may be obtained, but no square which is an aliquotpart of U will be also an aliquotpart of the rectangle (bydefinition of incommensurable magnitudes). There is, how-ever, a definite limit which is approached more and more closely by the approximations obtained by using smaller and smaller subdivisions of the unit square, as these subdivisions approach zero as a limit. 471. Def. The measure-number, or area, of a rectangle which is incommensurable with the chosen unit square is the limit which successive approximate measure-numbers of the rec-tangle approach as the subdivisions of the unit square approach zero as a limit. For brevity the expression,the area of a figure,is used to mean the measure-number of the surface of the figure with respect to a chosen unit. 472. Def. The ratio of any two surfaces is the ratio of their measure-numbers (basedon the same unit). 473. Def. Equivalent figures are figureswhich have the same area. The student should note that : Equal figures have the same shape and size; such figures can be made to coincide. Similar figures have the same shape. Equivalent figures have the same size. Ex. 799. Draw two equivalent figures that are not equal. For none of these approximations can exceed a certain fixed number, for example (h + )(b + 1), where the measure applied is contained in the altitude h times with a re-mainder less than the measure, and in the base b times with a remainder less than the 212 PLANE GEOMETRY Ex. 800. Draw two equal figureson the blackboard or cut them out of paper, and show that equal figures may be added to them in such a way that the resulting figures are not equal. Are they the same size ? Ex. 801. Draw figures to show that axioms 2, 7, and 8, when applied to equal figures, do not give results which satisfy the test for equal figures. Fig. 1. Fig. 2. Ex. 802. Fig. 1 above represents a card containing64 small squares, cut into four pieces, I,II,III, and IV. Fig. 2 represents these four piecesplaced together in different positionsforming, as it would seem, a rectanglecontaining 65 of these small squares. By your knowledge of similar triangles, try to explain the fallacyin the construction. 474. Historical Note. Geometry is supposed to have had its origin in land surveying,and the earliest traditions state that it had its beginning in Egypt and Babylon. The records of Babylon were made on clay tablets,and give methods for finding the approximate areas of several rectilinear figures, and also of the circle. The Egyptian records were made on papyrus. Herodotus states that the fact that the inundations of the Nile caused changes in the amount of taxable land, rendered it neces-sary to devise accurate land measurements. This work was done by the Egyptian priests, and the earliest manu-script extant is that of Ahmes, who lived about 1700 b.c. This manu-script, known as the Rhind papyrus, is preserved in the British Museum. It is called "Directions for knowing all dark things,"and is thought to be a copy of an older manuscript, dating about 3400 b.c. In addition to problems in arithmetic it contains a discussion of areas. Problems on pyramids follow,which show some knowledge of the properties of similar figures and of trigonometry,and which give dimensions, agreeing closely with those of the great pyramids of Egypt. The geometry of the Egyptians was concrete and practical, unlike that of the Greeks, which was logical and deductive, even from its beginning. BOOK IV 213 Proposition I. Theorem 475. The area of a rectangle is equal to the product of its base and its altitude. (See " 476.) B C U A D u Given rectangle ABCD, with base AD and altitude AB, and let U be the chosen unit of surface, whose side is u. To prove the area of ABCD = AD " AB. I. If AD and AB are each commensurable with u. (a) Suppose that u is contained in AD and AB each an in-tegral number of times. Argument 1. Lay off u upon AD and AB, respectively. Suppose that u is contained in AD r times, and in AB s times. 2. At the pointsof division on AD and on A B erect Js to AD and AB, respectively. 3. Then rectangle ABCD is divided into unit squares. 4. There are r of these unit squares in a row on AD, and s rows of these squares in rectangle ABCD. 5. .-. the area of ABCD = r "s. 6. But r and s are the measure-numbers of AD and AB, respectively, referred to the linear unit u. 7. .-'."fche area of A BCD = AD " AB. q.e.d. 1. Reasons "335. 2. " 63. 7. " 309. 214 PLANE GEOMETRY (b) If u is not a measure of AD and AB, respectively, but if some aliquotpart of u is such a measure. The proof is left to the student. U A QD u II. If AD and AB are each incommensura ble with u. Argument 1. Let m be a measure of u. Apply m as a measure to AD and iB as many times as possible. There will be a remainder, as QD, on AD, and a re-mainder, as PB, on AB, each less than m. 2. Through Q draw W_L ^-D,and through P draw PiW J AB. 3. Now J. Q and ^LP are each commensu-rable with the measure m, and hence commensurable with u. 4. .. the area of rectangle APMQ=A Q-A P. 5. Now take a smaller measure of u. No matter how small a measure of u is taken, when it is appliedas a measure to AD and AB, the remainders,QD and PP, will be smaller than the measure taken. 6. .-. the difference between AQ and AD may be made to become and remain less than any previouslyassigned segment, however small. Reasons 1. " 339. 2. " 63. 3. " 337. 4. " 475, I. 5. " 335. 6. Arg. 5. BOOK IV 215 Argument 7. Likewise the difference between AF and AB may be made to become and remain less than any previously as-signed segment, however small. 8. .-. A Q approaches AD as a limit, and AP approaches AB as a limit. 9. .-. AQ-AP approaches AD-AB as a limit. 10. Again, the difference between APMQ and A BCD may be made to become . and remain less than any previously assignedarea, however small. 11. .-. APMQ approaches ABCD as a limit. 1 2. But the area of APMQ is always equal to AQ " AP. 13. .-. the area of ABCD = AD-AB. Q.E.D. Reasons 7. Arg. 5. 8. " 349. 9. " 477. 10. Arg. 5. 11. " 349. 12. Arg. 4. 13. " 355. III. If AD is commensurable with u but AB incommensu-rable with u, The proof is left as an exercise for the student. 476. Note. By the product of two lines is meant the product of the measure-numbers of the lines. The proof that to every straightline segment there belongs a measure-number is given in " 595. 477. If each of any finitenumber of variables approaches a finitelimit, not zero, then the limit of their product is equal to the product of their limits. (See " 593.) 478. Cor. I. The area of a square is equal to the square of its side. 479. Cor. n. Any two rectangles are to each other as the products of their bases and their altitudes. Outline of Proof. Denote the two rectanglesby R and R\ their bases by b and b', and their altitudes by h and h', re-spectively Then R = b " h and R' = b '" h'. R Bf b-h b' " h'' 216 PLANE GEOMETRY 480. Cor. III. (a) Two rectangleshaving equal bases are to each other as their altitudes, and (b) two rectangles having equal altitudes are to each other as their bases. Outline op Proof (a\ " " ^ ' ^ " ^L (1\ ;?_ " fr " j _ " Ex. 803. Draw a rectanglewhose base is 7 units and whose altitude is 4 units and show how many unit squares it contains. Ex. 804. Find the area of a rectangle whose base is 12 inches and whose altitude is 5 inches. Ex. 805. Find the area of a rectanglewhose diagonal is 10 inches, and one of whose sides is 6 inches. Ex. 806. If the area of a rectangle is 60 square feet, and the base, 5 inches,what is the altitude ? Ex. 807. If the base and altitude of a rectangleare 2\ inches and \ inches, "respectively, find the area of the rectangle. Ex. 808. Find the area of a square whose diagonal is 8 V2 inches. Ex. 809. Find the successive approximations to the area of a rec-tangle if its sides are V 10 and V 5, respectively, using 3 times 2 for the first approximation, taking the square roots to tenths for the next, to hundredths for the next, etc. Ex. 810. Compare two rectangles if a diagonal and a side of one are d and s, respectively, while a diagonal and side of the other are d' and s'. Ex. 811. Construct a rectanglewhose area shall be three times that of a given rectangle. Ex. 812. Construct a rectanglewhich shall be to a given rectanglein the ratio of two given lines, m and n. Ex. 813. Compare two rectangles whose altitudes are equal, but whose bases are 15 inches and 3 inches, respectively. Ex. 814. From a given rectangle cut off a rectanglewhose area is two thirds that of the given one. Ex. 815. If the base and altitude of a certain rectangleare 12 inches and 8 inches, respectively, and the base and altitude of a second rectangle are 6 inches and 4 inches,respectively, compare their areas. 218 PLANE GEOMETRY Proposition III. Theorem 485. The area of a triangle equals one half the product of its base and its altitude. X B 1. A b C Given A ABC, with base b and altitude h. To prove area of A ABC =. "b-h. Argument Through A draw a line II CB, through B draw a line II CA. these lines intersect at X. 2. Then AXBC is a O. 3. /. A ABC = \|D AXBC. 4. But area of AXBC = b"h. 5. .-. area of A ABC = J b h. " 486. Cor. I. Triangles having equal bases and equal altitudes are equivalent. 487. Cor. II. Any two triangles are to each other as the products of their bases and their altitudes. Outline op Proof Denote the two A by T and T', their bases by b and b', and their altitudes by h and h', respectively. 1 \V .h'~ Then T \ b - h and f ""' ///. b' " h' 488. Cor. III. (a) Two triangleshaving equal bases are to each other as their altitudes, and (b) two triangles having equal altitudes are to each as their bases. BOOK IV 219 Outline of Proof / \ T b-h h ,b"" T ^b-h b W T'~b-h' ft" K ; t b'-h br Cor. IV. A triangle is equivalent to one half of a parallelogram having the same base and altitude. Ex. 816. Draw four equivalentparallelograms on the same base. Ex. 817. Find the area of a parallelogramhaving two sides 8 inches and 12 inches,respectively, and the included angle 60". Find the area if the included angle is 45". Ex. 818. Find the ratio of two rhombuses whose perimeters are 24 inches and 16 inches, respectively, and whose smaller base angles are 30". Ex. 819. Find the area of an equilateral trianglehaving a side equal to 6 inches. Ex. 820. Find the area of an equilateral triangle whose altitude is 8 inches. Ex. 821. Construct three or more equivalent triangles on the same base. Ex. 822. Find the locus of the vertices of all triangles equivalentto a given triangle and standing on the same base. Ex. 823. Construct a triangle equivalentto a given triangle and hav-ing one of its sides equal to a given line. Ex. 824. Construct a triangle equivalentto a given triangle and hav-ing one of its angles equal to a given angle. Ex. 825. , Construct a triangle equivalentto a given triangle and hav-ing two of its sides equal,respectively, to two given lines. Ex. 826. Divide a triangle into three equivalenttriangles by drawing lines through one of its vertices. Ex. 827. ' Construct a triangleequivalent to " of a given triangle ; \| of a given triangle. Ex. 828. Construct a triangle equivalentto a givensquare. Ex. 829. The area of a rhombus is equal to one half the product of its diagonals. Ex. 830. If from any point in a diagonal of a parallelogramlines are drawn to the oppositevertices, two pairsof equivalenttriangles are formed. Ex. 831. Two lines joining the mid-point of the diagonal of a quadri-lateral to the opposite vertices divide the figure into two equivalentparts. Ex. 832. Find the area of a triangle if two of its sides are 6.inches and.9 inches, respectively, and the included angle is 60", 220 PLANE GEOMETRY Proposition IV. Problem 490. To derive a formula for the area of a triangle in terms of its sides. C a B D C Given A ABC, with sides a, b,and c. To derive a formula for the area of A ABC in terms of a, b} and c. Argument Reasons 1. Let ha denote the altitude upon a, p the pro-1. " 485. jection of b upon a, and T the area of A ABC. Then T=-aha = -ha. 2 ' 2 a 2. ha2=b2-p2 = (b+p)(b-p). 3. Butp = a2 + 52-c2 (Fig. 1), or -" + -"? 2 a K " " 2 a (Fig. 2). 4..-.K "4 a2 + b2 rX-^ 2a6 + a2-h"2-c2 2a"-a2-"2 + c2 2a 2a = (a+6+c)(q-f 6" c)(c+a" 6)(c"a+6) 4a2 k , _^ /(a+" + c)(q+ft-c)(c+a-6)(c-a+6) o. . . K=y" _ t. 6. Now let a + 6 + c = 2 s. Then a + "-c = 2s-2c = 2(s " c); a-" + c = 2s-2" = 2(s " 6); 6 + c-a = 2s-2a = 2(s-a). 2. " 447. 3. " 456. 4. " 309. 5. " 54, 13. 6. " 54, 3. BOOK IV 221 Argument '.Then K=yj 2 s 2(s-a) 2(8-6) 2(s-c) 4 ft2 = - Vs(s " d)(s " b)(s " c). ft 8. /. T=-" - Vs(s -ft)(s -b)(s -c) = V"(s " ft)(s " ")(s " c). Q.E.F. Reasons 7. " 309. 8. " 309. Ex. 833. Find the area of a triangle whose sides are 7, 10, and 13. Ex. 834. If the sides of a triangle are a, 6, and c, write the formula for the altitude upon b ; upon c. (See Prop. IV, Arg. 7.) Ex. 835. In triangleABC, a = 8, b = 12, c = 16 ; find the area of triangle ABC ; the altitude upon b ; the altitude upon c. Proposition V. Theorem 491. The area of a triangle is equal to one half the product of its perimeter and the radius of the inscribed circle. A b Given A ABC, with area T, sides a, b, and c, and radius of inscribed circle r. To prove T= 1 (ft + b -f-c) r. Outline of Proof 1. Area of A OBC = \ a " r\ area of A OCA = i 6 . r; area of A OAB"^c-r. 2. .'. r = i(ft-r-"-fc)r, q.e.d. 492. Cor. The area of any polygon circumscribed about a circle is -equal to one half its perimeter multi-pU'"lby the radius of the inscribed circle. 222 PLANE GEOMETRY Ex. 836. If the area of a triangle is 15 V3 square inches and its sides are 3, 5, and 7 inches,find the radius of the inscribed circle. Ex. 837. Derive a formula for the radius of a circle inscribed in a triangle in terms of the sides of the triangle. Outline of Solution 1. T="(a + b + c)r = $ (2s)r = sr. Ex. 838. Derive a formula for the radius of a circle circumscribed about a triangle, in terms of the sides of the triangle. Outline of Solution (See figurefor Ex. 798.) 1. dh = ac; i.e. d or 2 i2 = " . h 2. ,'.M= = Q.E.F. 2 h 4 V s(8-a) (s -b) 0 -c) Ex. 839. If the sides of a triangle are 9, 10, and 11, find the radius of the inscribed circle ; the radius of the circumscribed circle. Ex.840. The sides of a triangleare 3a, 4 a, and 5 a. Find the radius of the inscribed circle ; the radius of the circumscribed circle. What kind of a triangleis it ? Verify your answer by comparing the radius of the circumscribed circle with the longestside. Ex. 841. Derive formulas for the bisectors of the angles of a triangle in terms of the sides of the triangle. Outline of Solution (See figurefor Ex. 797.) 1. tb2=ac " rs. 2. But a :c=r:s. 3. .-. a --c : a =b : r. 4. .-. r = -5-. 5. Likewise 8 = -^-. 6. .-. tf = ac-ahH a + c a + c (a--c)2 ac(a + b + c)(a-b+c)=4acs{s-bK 7" ... % =^_Vacs(s _ 6). (a + cy (a + cy " a + c K J 8. Likewise ta = y/bcs(s " a) and tc = Vabs(s " c) " b+c a + 6 K J Q-E-F-Find r, B, T, ha, ma, and ta,having given : Ex.842, a = 11, b = 9, c= 16. What kind of an angle is C? Ex. 848. a = 13, b = 15, c = 20. What kind of an angle is C? Ex. 8,44. a = 24, b = 10, c = 26. What kind of an angle is C ? BOOK IV 223 TRANSFORMATION OF FIGURES 493. Def. To transform a figure means to find another figure which is equivalentto it. Proposition VI. Problem 494. To construct a triangle equivalent to a given polygon. Given polygon ABGDEF. To construct a A =o= polygon ABCDEF. (a) Construct a polygon=c= ABCDEF, but having one side less. I. Construction 1. Join any two alternate vertices, as C and A. 2. Construct BG IICA, meeting FA prolonged at G. " 188. 3. Draw CG. 4. Polygon GCDEF =o polygon ABCDEF and has one side less. 1. II. Proof Argument A AGC and ABC have the same base CA, and the same altitude, the J be-tween the lisCA and BG. .-. A AGC =c= A ABC. But polygon A CDEF " polygon ACDEF. .-. polygon GCDEF o= polygon AB CDEF. Q.E.P. Reasons 1. "235. 2. "486. 3. By iden. 4. "54,2. 224 PLANE GEOMETRY (6) In like manner, reduce the number of sides of the new polygon GCDEF until A DHK is obtained. The construction, proof, and discussion are left as an exercise for the student. Ex. 845. Transform a scalene triangle into an isosceles triangle. Ex. 846. Transform a trapezoidinto a righttriangle. Ex. 847. Transform a parallelograminto a trapezoid. Ex. 848. Transform a pentagon into an isosceles triangle. Ex. 849. Construct a triangle equivalent to " of a given trapezium. Ex. 850. Transform $ of a given pentagon into a triangle. Ex. 851. Construct a rhomboid and a rhombus which are equivalent, and which have a common diagonal. Proposition VII. Theorem 495. The area of a trapezoid equals the product of its altitude and one half the sum of its bases. V A b D Given trapezoid AFED, with altitude h and bases b and b1. To prove area of AFED = I- (h 4- b^h. 226 PLANE GEOMETRY Proposition VIII. Theorem 498. Two triangles which have an angle of one equal to an angle of the other are to each other as the products of tTie sides including the equal angles. A G Given A ABC and BEF, with Z A = Z.B. A ABC AC " AB To prove A BEF BF " BE Argument 1. Let h be the altitude of A ABC upon side AC, and h' the altitude of A BEF upon side BF. Then, A ABC AC-h ACmh_ BF hr 2. 3. 4. 5. A BEF BF " h' In rt. A ABG and BEK, ZA = ZB. .'. A ABG ~ A BEK. h AB^ h'~~BE' A ABC AC AB _AC " AB BF BE~ A BEF BF -BE Q.E.D. Reasons 1. " 487. 2. By hyp. 3. " 422. 4. " 424,2. 5. " 309. Ex. 859. Draw two triangles upon the blackboard, so that an angle of one shall equal an angle of the other. Give a rough estimate in inches of the sides includingthe equal angles in the two triangles, and compute the numerical ratio of the triangles. Ex. 860. If two triangles have an angle of one supplementary to an angle of the other, the triangles are to each other as the products of the sides including the supplementary angles. BOOK IV 227 Proposition IX. Problem 499. To construct a square equivalent to a given tri-angle Given A ABC, with base b and altitude h. To construct a square =0= A ABC. I. Analysis 1. Let # = the side of the required square; then "c2 = area of requiredsquare. \ b " h = area of the given A ABC. 2. 3. .\x2 = "b"h. \b : x = x : h. 4. 5. .. the side of the required square will be a mean pro-portional between i b and h. II. Construction 1. Construct a mean proportional between J b and h. Call it x. " 445. 2. On x, as base,construct a square, S. 3. "S is the requiredsquare. Q.E.D. Reasons 1. By cons. 2. " 388. 3. " 478. 4. " 485. 5. " 54, 1. 228 PLANE GEOMETRY IV. The discussion is left as an exercise for the student. 500. Question. Could x he constructed as a mean proportional be-tween b and \| h ? 501. Problem. To construct a square equivalent to a given parallelogram. Ex. 861. Construct a square equivalentto a given rectangle. Ex. 862. Construct a square equivalentto a given trapezoid. Ex. 863. Upon a given base construct a triangleequivalent to a given parallelogram. Ex. 864. Construct a rectangle having a given base and equivalent to a given square. 502. Props. VI, VIII, and IX form the basis of a large class of important constructions. (a)Prop. VI enables us to construct a triangleequivalent to any polygon. It is then an easy matter to construct a trapezoid, an isosceles trapezoid, a parallelogram, a rectangle, or a rhombus equivalentto the triangle and hence equivalent to the given polygon. (b)Prop. VIII givesus a method for constructing an equi-lateral triangle equivalent to any giventriangle.(SeeEx. 865.) Hence Prop. VIII, with Prop. VI, enables us to construct an equilateral triangle equivalent to any given polygon. (c)Likewise Prop. IX, with Prop. VI, enables us to con-struct a square equivalent to any given polygon or to any fractional part or to any multiple of any given polygon. Ex. 865. (a) Transform triangle ABC into triangle DBC, retaining base BC and making angle DBC " 60". (6) Transform triangle DBC into triangle EBF, retaining angle DBC = 60" and making sides EB and BF equal. (Each will be a mean proportionalbetween DB and BC.) (c) What kind of a triangle is EBF? Ex. 866. Transform a parallelograminto an equilateral triangle. Ex. 867. Construct an equilateral triangle equivalentto f of a given trapezium. BOOK IV 229 Ex. 868. Construct each of the followingfigures equivalentto f of a given irregular pentagon : (1) a triangle ; (2) an isosceles triangle ; (3) a righttriangle ; (4) an equilateral triangle ; (5) a trapezium ; (6) a trape-zoid ; (7) an isosceles trapezoid; (8) a parallelogram; (9) a rhombus ; (10) a rectangle j (11) a square. Ex. 869. Transform a trapezoid into a right trianglehaving the hypotenuse equal to a given line. What restrictions are there upon the given line ? Ex. 870. Construct a triangleequivalent to a given trapezoid,and having a given line as base and a given angle adjacent to the base. Proposition X. Theorem 503. Two similar triangles are to each other as the squares of any two homologous sides, B Given two similar A ABC and A'b'c', with, b and b' two homol. sides. Reasons 1. By hyp. 2. " 424, 1. 3. " 498. A A'B'c' b' b' b'2 Q.E.D. 4. " 424,2. 5. " 309. 230 PLANE GEOMETRY 504. Cor. Two similar triangles are to each other as the squares of any two homologous altitudes. (See " 435.) Ex. 871. Two similar trianglesare to each other as the squares of two homologous medians. Ex. 872. Construct a triangle similar to a given triangle and having an area four times as great. Ex. 873. Construct a trianglesimilar to a given triangle and having an area twice as great. Ex. 874. Divide a given triangle into two equivalentparts by a line parallel to the base. Ex. 875. Prove Prop. X by using " 487. Ex. 876. Draw a line parallel to the base of a triangle and cuttingoff a triangle that shall be equivalent to one third of the given triangle. Ex. 877. In two similar triangles a pair of homologous sides are 10 feet and 6 feet,respectively. Find the homologous side of a similar tri-angle equivalentto their difference. Ex. 878. Construct an equilateral triangle whose area shall be three fourths that of a given square. Proposition XI. Theorem 505. Two similar polygons are to each other as the squares of any two homologous sides. Given two similar polygons P and P' in which a and a\b and b', etc., are pairs of homol. sides. P a2 To prove -=-2. BOOK IV 231 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. Argument Draw all possible diagonalsfrom any two homol. vertices, as V and V1. Then the polygonswill be divided into the same number of A ~ each to each and similarly placed, as A I and A I', A II and A II', etc. Then AI A I' All Air A III But A III' a_ a'' a'2' " AT' 'd'2' 1 " " e'~dr t.=JL e'2 d'2' A II A III Air . A I + A 11+ A III ' A I' + A II' + A III' " L = ?L ' ' P' a'2' A III' _AI a2 A I' ,'2' Q.E.D. Reasons " 54, 15. " 439. 506. Cor. Two similar polygons are to each other as the squares of any two homologous diagonals. Ex. 879. If one square is double another, what is the ratio of their sides ? Ex. 880. Divide a given hexagon into two equivalentparts so that one part shall be a hexagon similar to the given hexagon. Ex. 881. The areas of two similar rhombuses are to each other as the squares of their homologous diagonals. Ex. 882. One side of a polygon is 8 and its area is 120. The homol ogous s^de of a similar polygon is 12 ; find its area. 232 PLANE GEOMETRY Proposition XII. Theorem 507. The square described on the hypotenuse of a right triangle is equivalent to the sum of the squares described on the other two sideSo Given rt. A ABC, right-angled at C, and the squares described on its three sides. To prove square AD =c= square BF + square CH. Argument 1. From C draw CM " AB, cuttingAB at L and KD at M. 2. Draw CK and BH. 3. A ACG, BCA, and FCB are all rt. A. 4. .'. ACF and GCB are str. lines. 5. In A CAK and HAB, CA = HA, AK = AB. 6. Z CAB = Z CAB. 7. Abak=Zhac. 8. ' . . Z CAK = Z HAB. 9. .. ACAK= A HAB. 10. A CAK and rectangle AM have the same base AK and the same altitude, the " _L between the lis AK and CM. 234 PLANE GEOMETRY 510. Historical Note. Prop. XII isusuallyknown as the Pythago-rean Proposition, because it was discovered by Pythagoras. The proof given here is that of Euclid (about300 B.C.). Pythagoras (569-500 b.c), one of the most famous mathe-maticians of antiquity, was born at Samos. He spent his earlyyears of manhood study-ing under Thales and traveled in Asia Minor and Egypt and probably also in Babylon and India. He returned to Samos where he established a school that was not a great success. Later he went to Crotona in Southern Italy and there gained many adherents. He formed, with his closest fol-lowers, a secret society,the members of which possessed Pythagoras all things in common. They used as their badge the five-pointed star or pentagram which they knew how to construct and which they considered symbolical of health. They ate simple food and practicedsevere discipline, having obedience,temper-ance, and purity as their ideals. The brotherhood regarded their leader with reverent esteem and attributed to him their most important dis-coveries, many of which were kept secret. Pythagoras knew something of incommensurable numbers and proved that the diagonal and the side of a square are incommensurable. The first man who propounded a theory of incommensurables is saidto have suffered shipwreck on account of the sacrilege, since such numbers were thought to be symbolical of the Deity. Pythagoras, having incurred the hatred of his politicalopponents, was murdered by them, but his school was reestablished after his death and it flourished for over a hundred years. Ex. 884. Use the adjoining figure to prove the Pythagorean theorem. Ex. 885. Construct a triangle equivalent to the sum of two given triangles. B BOOK IV 235 Ex. 886. The figure represents a farm drawn to the scale indicated. Make accurate measure-ments and calculate ap-proximately the number of acres in the farm., Ex. 887. A farm XYZW, in the form of a trapezium, has the follow-ing dimensions : XF=60 rods, YZ = 70 rods, ZW = 90 rods, WX = 100 rods, and XZ = 66 rods. Draw a plot of the farm to the scale 1 inch = 40 rods,and calculate the area of the farm in acres. I I 1 20 40 TC0 511. Def. By the rectangle of two lines is meant the rectangle having these two lines as adjacent sides. Ex. 888. The square described on the sum of two lines is equivalent to the sum of the squares described on the lines plus twice their rectangle. Ex. 889. The square described on the difference of two lines is equivalent to the sum of the squares described on the lines diminished by twice their rectangle. Hint. Let AB and CB be the given lines. li Ex. 890. The rectangle whose sides are the sum and difference respectivelyof two lines is equivalentto the difference of the squares described on the lines. Hint. Let AB and BC be the given lines. B C Ex. 891. Write the three algebraicformulas corresponding to the last three exercises. 236 PLANE GEOMETRY Proposition XIII. Problem 512. To construct a square equivalent to the sum of two given squares. p Q Given squares P and Q. To construct a square =c= the sum of P and Q. I. Construction 1. Construct the rt. A ABC, having for its sides p and g, the sides of the given squares. 2. On r, the hypotenuseof the A, construct the square R. 3. R is the requiredsquare. II. The proof and discussion are left to the student. Ex. 892. Construct a square equivalent to the sum of three or more given squares. Ex. 893 . Construct a square equivalent to the difference of two squares. Ex. 894. Construct a square equivalentto the sum of a given square and a given triangle. Ex. 895. Construct a polygon similar to two given similar polygons and equivalentto their sum. (See " 509.) Ex. 896. Construct a polygon similar to two given similar polygons and equivalentto their difference. Ex. 897. Construct an equilateral triangleequivalent to the sum of two given equilateral triangles. Ex. 898. Construct an equilateral triangleequivalent to the differ-ence of two given equilateral triangles. BOOK IV 237 Proposition XIV. Problem 513. To construct a polygon similar to one of two given polygons and equivalent to the other. Given polygons P and Q, with s a side of P. To construct a polygon ~ P and =c= Q. I. Analysis 1. Imagine the problem solved and let R be the required polygon with side x homol: to s, a side of P. 2. Then P : R = s2 : x2 ; i.e.P : Q = s2 : x2, since Q o R. (1) 3. Now to avoid comparing polygons which are not similar, we may reduce P and Q to o= squares. Let the sides of these squares be m and 71,respectively ; then m2 =o= P and n2 =o= Q. 4. .. m2 : n2 = s2 : x2, from (1). 5. .\ m : n = s : x. 6. That is, x is the fourth proportional to m, n, and s. II. The construction, proof, and discussion are left as an exercise for the student. 514. Historical Note. This problem was firstsolved by Pythagoras about 550 b.c. Ex. 899. Construct a triangle similar to a given triangleand equiva-lent to a given parallelogram. Ex. 900. Construct a square equivalentto a given pentagon. Ex. 901. Construct a triangle, given its angles and its area (equalto that of a given parallelogram). Hint. See Prop. XIV. Ex. 902. Divide a triangle into two equivalentparts by a line drawn perpendicular to the base. Hint. Draw a median to the base, then apply Prop.'"IV. 238 PLANE GEOMETRY Ex. 903. Fig. 1 represents maps of Utah and Colorado drawn to the scale indicated. By carefully measuring the maps: (1) Calculate the perimeter of each state. (2) Calculate the area of each state. (3) Check your results for (2) by comparing with the areas given for these states in your geography. U T COLORADO SCALE OF MILES 200 300 400 Ex. 904. Fig.2 repre-"? 50 sents a map of Pennsylva-FlG j nia. A straightline from the southwest corner to the northeast corner is 300 miles long. ^s\ (1) Determine the scale to which the map is drawn. (2) Calculate the distance from Pittsburg to Harrisburg ; from Harrisburg to Philadelphia ; from Philadelphia to Scran ton ; from Scranton to Harrisburg. (3) Calculate approximately the area of the state in square miles. Ex. 905. Let 0 represent a can-dle ; A a screen 1 foot square and 1 yard from C; B a screen 2 feet square and 2 yards from C ; D a screen 3 feet square and 3 yards from C. If screen A were removed, the quantityof light it received would fall on B. What would happen if B were removed ? On which screen, then, would the light be the least intense ? From the figure, determine the law of intensityof light. Fig. 3. Scranton " PENNSYLVANIA . Pittsburg Harrisburg Philadelphia i Fig. 2. Ex. 906. Divide a triangle into two equivalentparts by a line drawn from a given point in one of its sides. " Hint. Let M be the given point in AC of triangleABC; then i AB " AC = AM " AX, where MX is the requiredline. BOOK IV 239 Ex. 907. A represents a station. Cars approach the station on track BA and leave the station on track AC. Construct an arc of a circle DE, with given radius r, connecting the two intersecting car lines, and so that each car line is tangent to the arc. This same principle is involved in designing a buildingbetween two streets forming at their A"' point of intersection a small acute angle, as sv the Flatiron Building in New York City. Ex. 908. Find the area of a rhombus if its diagonalsare in the ratio of 5 to 7 and their sum is 16. FlG-4-MISCELLANEOUS EXERCISES Ex. 909. Show that if a and " are two sides of a triangle, the area is " ab when the included angle is 30" or 150"; \ abVll when the included angle is 45" or 135"; \ ab V3 when the included angle is 60" or 120". Ex. 910. The sum of the perpendiculars from any point within a convex polygon upon the sides is constant. Hint. Join the point with the vertices of the polygon, and consider the sum of the areas of the triangles. Ex. 911. The sum of the squares on the segments of two perpendicular chords in a circle is equivalentto the square on the diameter. Ex. 912. The hypotenuse of a righttriangle is 20, and the projection of one arm upon the hypotenuse is 4. What is its area? Ex. 913. A quadrilateral is equivalent to a triangle if its diagonals and the angle included between them are respectively equal to two sides and the included angle of the triangle. Ex. 914. Transform a given triangle into another triangle containing two given angles. Ex. 915. Prove geometrically the algebraic formula (a + 6) (c -f d) = ac + be + ad + bd. Ex. 916. If in any triangle an angle is equal to two thirds of a straight angle ("69),then the square on the side oppositeis equivalentto the sum of the squares on the other two sides and the rectangle contained by them. Ex. 917. The two medians BK and SH of the triangleB8T inter-sect at P. Prove that the triangle BPS is equivalentto the quadrilateral HPKT. Ex. 918. Find the area of a triangle if two of its sides are 6 inches andv7inches and the included angle is 30", 210 PLANE GEOMETRY Ex. 919. By two different methods find the area of an equilateral triangle whose side is 10 inches. Ex. 920. The area of an equilateral triangle is 36 VS ; find a side and an altitude. Ex. 921. By using the formula of Prop. IV, Arg. 8, derive the formula for the area of an equilateral triangle whose side is a. ' Ex. 922. What does the formula for T in Prop. IV become if angle C is a rightangle ? Ex. 923. Given an equilateral triangleABG, inscribed in a circle whose center is O. At the vertex G erect a perpendicularto BG cutting the circumference at D. Draw the radii OD and OG. Prove that the triangle ODC is equilateral. Ex. 924. Assuming that the areas of two triangles which have an angle of the one equal to an angle of the other are to each other as the products of the sides including the equal angles,prove that the bisector of an angle of a triangle divides the opposite side into segments propor-tional to the adjacent sides. Ex. 925. A rhombus and a square have equal perimeters,and the altitude of the rhombus is three fourths its side ; compare the areas of the two figures. Ex. 926. The length of a chord is 10 feet, and the greatest perpen-dicular from the subtending arc to the chord is 2 feet 7" inches. Find the radius of the circle. Ex. 927. In any righttriangle a line from the vertex of the rightangle perpendicular to the hypotenuse divides the given triangle into two tri-angles similar to each other and similar to the given triangle. Ex. 928. The bases of a trapezoid are 16 feet and 10 feet,respec-tively, and each of the non-parallel sides is 5 feet. Find the area of the trapezoid. Also find the area of a similar trapezoid, if each of its non-parallel sides is 8 feet. Ex. 929. A trianglehaving a base of 8 inches is cut by a line paral-lel to the base and 6 inches from it. If the base of the smaller triangle thus formed is 5 inches, find the area of the largertriangle. Ex. 930. If the ratio of similitude of two similar triangles is 7 to 1, how often is the less contained in the greater? Hint. See "" 418, 503. Ex. 931. Construct a square equivalentto one third of a given square. Ex. 932. If the side of one equilateral triangleis equal to the alti-tude of another, what is the ratio of their areas ? Ex. 933. Divide a righttriangle into two isosceles triangles. 242 PLANE GEOMETRY Ex. 942. The feet of the perpendicularsdropped upon the sides of a triangle from any point in the circumference of the circumscribed circle are collinear. Outline of Proof. The circle having AP as diameter will pass through M and Q. .-. Zl=ZV andZ2 = Z2'. Similarly Z3 = Z3', and Z PKM = Z PBM. .: ZAPB = ZQPK. ..ZAPQ = ZBPK. .-. Z AMQ = Z BMK. .-. "M and MK form one str. line. Ex. 943. Given the base, the angle at the vertex, and the sum of the other two sides of a triangle;construct the triangle. Analysis. Imagine the problem solved and draw A ABC Prolong AB, making BE = BC, since the line c -f a is given. Since ZE=\ZABQ, A AEG can be constructed. Ex. 944. The hypotenuse of a righttriangle is given in magnitude and position ; find the locus of the center of the inscribed circle. Ex. 945. Prove Prop. VIII, Book IV, by using the following figure, in which A'D'E' is placed in the position ABE. Ex. 946. Prove Prop. VIII, Book IV, using two triangles such that one will not fall wholly within the other. Ex. 947. If two triangles have an angle of one supplementary to an angle of the other, the triangles are to each other as the products of the sides includingthe supplementary angles. (Prove by method similar to that of Ex, 945.) BOOK IV 243 Ex. 948. Given base, difference of sides, and difference of base angles; construct the triangle. Analysis. In the accompanying figure suppose c and EA, (b " "), to be given. A consideration or the figure will show that Z2 = Zl -)-ZA. Add Z 1 to both members of the equation ; then Zl + Z2 = 2 Zl + ZA. ButZl + Z2 = ZB. .-. ZB = 2Z1 + ZA. .-. Zl = l(ZB -ZA). The A BE A may now be constructed. The rest of the construction is left for the student. Ex. 949. the triangle. Analysis. Given base, vertex angle,and difference of sides, construct . AE = AB-BC. ZAEC . A AEC can be constructed. 90" Ex. 950. If upon the sides of any triangle equilateral triangles are constructed, the lines joiningthe centers of the equilateral triangles form an equilateral triangle. Hint. Circumscribe circles about the three a equilateral A. Join 0, the common point of intersection of the three circles, to A, B, and C, the vertices of the given A. Prove each Z at 0 the supplement of the Z opposite in an equilateral A, and also the supplement of the Z opposite in the A to be proved equi-lateral. Ex. 951. To inscribe a square in a semicircle. Analysis. Imagine the problem solved and ABCD the required square. Prove OA meeting OB prolonged at F. .-. EF=2 OE. _AB 2 FE . Draw EF" EO :EO = BA: AO. \B E A 0 D Ex. 952. To inscribe a square in a given triangle. Outline of Solution. Imagine the prob-lem solved and ABCD the required square. Draw SFW BT and construct square KSFH. Draw BF, thus determining point C. The cons, will be evident from the figure. To prove ABCD a square, prove BC ^ 'gD. BO : SF = CD : FIT. But SF = FH. .: BC = CD. KD T H 244 PLANE GEOMETRY Ex. 953. In a given square construct a square, having a given side, so that its vertices shall lie in the sides of the given square. Hint. Construct a rt. A, given the hypotenuse and sum of arms. Ex. 954. Construct a triangle, given ma, mc, /i". Analysis. Imagine the problem solved and that ABC is the required A. If BF were moved II to itself till it contained K, the rt. A AKT would be formed and KT would equal \ BF. Then A AKT can be made the basis of the required construction. Ex. 955. Construct a quadrilateral, given two of its opposite sides, its two diagonals, and the angle between them. Outline op Construction. Imagine the problem solved and that ABCD is the requiredquadrilateral, s, s', d, d',and Z DOC being given. By II motion of d and d' the parallelogramBKFD may be obtained. The required construction may be begun by draw-ing O BKFD, since two sides and the in-cluded Z are known. With K as center and s as radius,describe an arc ; with D as center and sr as radius,describe an arc intersecting the first, as at C. Construct EJ ABKC, or ED DACF, to locate A. Ex. 956. Between two circles draw a line which shall be parallel to the line of centers and equal to a given line Z. / Ex.957. Find x =^, d2 Hint. Here x where a, b, c, and d represent given lines. ab Let and construct y. Then construct x. ab C d d d Ex. 958. Transform any given triangleinto an equilateral triangle by a method different from that used in Ex. 8G5. Analysis. Call the base of the given triangle b and its altitude a. Let x = the side of the required equilateral triangle. Then ^VS = ~b-h. .-. lb:x = x:hV3. 4 2 BOOK V REGULAR POLYGONS. MEASUREMENT OF THE CIRCLE 515. Def . A regular polygon is one which is both equilateral and equiangular. Ex. 959. Draw an equilateral triangle. Is it a regular polygon ? Ex. 960. Draw a quadrilateral that is equilateral but not equiangular ; equiangular but not equilateral ; neither equilateral nor equiangular ; both equilateral and equiangular. Which of these quadrilaterals is a regular polygon? Ex. 961. Find the number of degrees in an angle of a regular do-decagon. 516. Historical Note. The following theorem presupposes the pos-sibility of dividingthe circumference into a number of equal arcs. The actual division cannot be obtained by the methods of elementary geom-etry, except in certain special cases which will be discussed later. As early as Euclid's time it was known that the angular magnitude about a point (and hence a circumference) could be divided into 2n, 2" " 3, 2n " 5, 2n " 15 equal angles. In 1796 it was discovered by Gauss, the n nineteen years of age, that a regularpolygon of 17 sides can be con-structed by means of ruler and compasses, and that in general it is possi-ble to construct all polygons having (2"+1) sides,n being an integer and (2n + 1) a prime number. The first four numbers satisfying this condition are 3, 5, 17, 257. Gauss proved also that polygons having a number of sides equal to the product of two or more different numbers of this series can be constructed. Gauss proved,moreover, that only a limited class of regularpolygons are constructive by elementary geometry. For a note on the life of Gauss, see " 52.0. 245 246 PLANE GEOMETRY Proposition I. Theorem 517. If the circumference of a circle is divided into any number of equal arcs : (a) the chords joining the points of division form a regular polygon inscribed in the circle; (b) tangents drawn at the points of division form a regular polygon circumscribed about the circle. Given circumference ACE divided into equalarcs AB, BC, CD. etc., and let chords AB, BC, CD, etc.,jointhe several pointsof division, and let the tangents GH, HK, KL, etc., touch the circum-ference at the several pointsof division. To prove A BCD " " " and GHKL " " " regularpolygons. BOOK V 247 Argument Reasons 11. .\ GH=^HK=KL=-" -. 11. "54, la. 12. .-. GHKL """ is a regular polygon, q.e.d. 12. " 515. 518. Questions. If,in the figureof " 517, the circumference is di-vided into six equal parts, how many arcs, each equal to arc AB, will arc CEA contain ? arc DFB ? How many will each contain if the circum-ference is divided into n equal parts ? In step 10, why does AG = GB ? 519. Cor. If the vertices of a regular inscribed poly-gon are joined to the mid-points of the arcs subtended by tl%e sides of the polygon, the joining lines ivill form a regular inscribed polygon of double tlie number of sides. Ex. 962. An equilateral polygon inscribed in a circle is regular. Ex. 963. An equiangular polygon circumscribed about a circle is regular. , 520. Historical Note. Karl Friedrich Gauss (1777-1855) was born at Brunswick, Germany. Although he was the son of a bricklayer, he was enabled to receive a liberal ed-ucation, owing to the recogni-tion of his unusual talents by a nobleman. He was sent to the Caroline Collegebut,at the age of fifteen, it was admitted both by professorsand pupils that Gauss already knew all that they could teach him. He became a student in the Unir versity of Gottingen and while there did some important work on the theory of numbers. On his return to Brunswick, he lived humbly as a private tutor, until 1807, when he was appointed professor of astronomy and director of the observatory at Gottingen did important work in physics as well as in astronomy, the telegraph independently of S. F. B. Morse. His lectures were unusually clear, and he is said to have given in them the analytic steps by which he developed his proofs ; while in his writings there is'no hint of the processes by which he discovered his results. Gauss While there he He also invented 248 PLANE GEOMETRY Proposition II. Problem 521. To inscribe a square in a given circle. Given circle 0. To inscribe a square in circle 0. I. The construction is left as an exercise for the student. II. Proof Argument 1. AB _L CD. 2. .-. ZAOC = 90". 3. .-. AC = 90", i.e. one fourth of the cir-cumference. 4. .-. the circumference is divided into four equal parts. 5. .-. polygon ACBD, formed by joining the points of division, is a square, q.e.d. III. The discussion is left as an exercise for the student. 522. Cor. The side of a square inscribed in a circle is equal to the radius multiplied by V@ ; the side of a square circumscribed about a circle is equal to twice the radius. Reasons 1. By cons. 2. "71. 3. "358. 4. Arg. 3. 5. "517, a. Ex. 964. Inscribe a regularoctagon in a circle. Ex. 965. Inscribe in a circle a regularpolygon of sixteen sides. Ex. 966 Circumscribe a square about a circle. 250 PLANE GEOMETRY 524. Cor. I. A regular inscribed triangleis formed by joining the alternate vertices of a r regular inscribed hexagon. 525. Cor. II. A side of a regu-lar inscribed triangle is equal to the radius of the circle multi-plied by VJ. Hint. A ABD is a righttriangle whose hypotenuse is 2 B and one side B. Ex. 971. Inscribe a regular dodecagon in a circle. Ex. 972. Divide a given circle into two segments such that any angle inscribed in one segment is live times an angle inscribed in the other. Ex. 973. Circumscribe an equilateral triangle about a circle. Ex. 974. Circumscribe a regularhexagon about a circle. Ex. 975. On a given line as one side, construct a regularhexagon. Ex. 976. On a given line as one side, construct a regulardodecagon. Ex. 977. If a is the side of a regular dodecagon inscribed in a circle whose radius is B, then a = B\2 " VS. Proposition IV. Problem 526. To inscribe a regular decagon in a circle. 4? j i " R x Fig. 1. Given circle 0. To inscribe in circle 0 a regulardecagon. Fig. 2. BOOK V 251 I. Construction 1. Divide a radius R, of circle 0, in extreme and mean ratio. " 465. (See Fig. 2.) 2. In circle O draw a chord AB, equal to s, the greater seg-ment of R. 3. AS is a side of the requireddecagon. II. Proof Argument 1. Draw radius OA. 2. On OA lay off OM = s. 3. Draw OB and BM. 4. Then OA : OM = OM : MA. 5. .-. OA : AB = AB : MA. 6. Also in A OAB and if^tf,Za " Za. 7. .'.A 0.45 ~ A MAB. 8. \ " A OAB is isosceles, A MAB is isosceles, and BM = ^45. 9. But ^LB = s = OM. 10. .-. BM = 03f. 11. .-. A BOM is isosceles and ZMB0 = Zo. 12. But Z^Bif=Zo. 13. .: ZMBO + Z ABM, or Z ABO, =2 Zo. 14. .-. Z MAB = 2 Z 0. 15. In A ABO, ZABO -f ZMAB + Z0 = 180". 1G. .-. 2 Zo + 2 Z 0 + Z 0, or 5 Z 0, = 180". 17.,.-. Zo = 36". 18. .-. ^LB = 36", i.e. one tenth of the cir-cumference. 19. .. the circumference may be divided into ten equal parts. 20. .. polygon ABCD " " " , formed by join-ing the pointsof division, is a regular ^ 'jnscribed decagon. q.e.d. Reasons 1 "54,15. 2. "54,14. 3. "54,15. 4. By cons. 5. "309. 6. By iden. 7. "428. 8. "94. 9. By cons. 10. "54,1. 11. "111. 12. "424,1. 13. " 54, 2. 14. "111. 15. "204. 16. "309. 17. " 54, 8 a. 18. "358. 19. Arg. 18. 20. " 517, a. 252 PLANE GEOMETRY III. The discussion is left as an exercise for the student. 527. Cor. I. A regular pentagon is formed by joining the alternate vertices of a regular inscribed decagon.^ Ex. 978. Construct a regularinscribed polygon of 20 sides. Ex. 979. The diagonals of a regularinscribed pentagon are equal. Ex. 980. Construct an angle of 36" ; of 72". Ex. 981. Divide a rightangle into five equal parts. Ex. 982. The eightdiagonalsof a regular decagon drawn from any vertex divide the angle at that vertex into eightequal angles. Ex. 983. Circumscribe a regularpentagon about a circle. Ex. 984. Circumscribe a regulardecagon about a circle. Ex. 985. On a given line as one side,construct a regularpentagon. Ex. 986. On a given line as one side,construct a regulardecagon. Ex. 987. The side of a regular inscribed decagon is equal to \| B (V5 " 1), where B is the radius of the circle. Hint. By cons., B : s = s : B " s. Solve this proportionfor s. Proposition V. Problem 528. To inscribe a regular pentedecagon in a circle, it Given circle O. To inscribe in circle 0 a regular pentedecagon. I. Construction 1. Prom A, any point in the circumference, lay off chord AK equalto a side of a regularinscribed hexagon. " 523. BOOK V 253 2. Also layoff chord AF equalto a side of a regularinscribed decagon. " 526. 3. Draw chord FE. 4. FE is a side of the required pentedecagon. II. Proof Argument Arc AE = i of the circumference. y1^ of the circumference. Reasons By cons. By cons. " 54, 3. 4. Arg. 3. 5. " 517, a. Arc AF .-. arc FE = i " jL,i.e. ^ of the cir-cumference. .. the circumference may be divided into fifteen equal parts. .. the polygon formed by joining the pointsof division will be a regular inscribed pentedecagon. q.e.d. III. The discussion is left as an exercise for the student. 529. Note. It has now been shown that a circumference can be divided into the number of equal parts indicated below : 2, 4, 8, 16,.-. 2"^ 3, 6, 12, 24, ... 3x2" 5, 10, 20, 40, -. 5x2" 15, 30, 60, 120, ... 15 x 2" J [n being any positive integer]. Construct an angle of 24". Circumscribe a regularpentedecagon about a given circle. On a given line as one side, construct a regular pente-Ex. 988. Ex. 989. Ex. 990. decagon! Ex. 991. Assuming that it is possible to inscribe in a circle a regularpolygon of 17 sides,show how it is possibleto inscribe a regular polygon of 51 sides. Ex. 992. If a regular polygon is inscribed in a circle, the tangents drawn at the mid-points of the arcs subtended by the sides of the in-scribed polygon form a circumscribed regular polygon whose sides are parallel to the sides of the inscribed polygon, and whose vertices lie on the prolongations of the radii drawn to the vertices of the inscribed polygon. 254 PLANE GEOMETRY Proposition VI. Theorem 530. A circle may be circumscribed about any regular polygon ; and a circle may also be inscribed in it. B^-~ " ^C Given regularpolygon ABCD " " " . To prove : (a)that a circle may be circumscribed about it ; (b) that a circle may be inscribed in it. (a) Argument 1. Pass a circumference through points A, B, and C. 2. Connect O, the center of the circle, with all the vertices of the polygon. 3. Then OB = OC. 4. .\ Z1 = Z2. 5. But Z ABC = Z BCD. 6. .-. Z3 = Z4. 7. Also AB = CD. 8. .. AABO = AOCD. 9. .. OA = OD and circumference ABC passes through D. 10. In like manner it may be proved that circumference ABC passes through each of the vertices of the regular polygon; the circle will then be circumscribed about the polygon. Q.E.D. Keasons 1. " 324. 2. "54,15. BOOK V 255 (P) 1. 2. 3. Argument Again AB, BC, CD, etc.,the sides of the given polygon,are chords of the circumscribed circle. Hence Js from the center of the circle to these chords are equal. .. with 0 as center, and with a radius equal to one of these Js, as OH, a circle may be described to which all the sides of the polygon will be tangent. 4. .. this circle will be inscribed in the polygon. q.e.d. Reasons 1. " 281. 2. " 307. 3. " 314. 4. "317. 531. Def. The center of a regularpolygon is the common center of the circumscribed and inscribed circles ; as 0, Prop. VI. 532. Def. The radius of a regularpolygon is the radius of the circumscribed circle, as OA. 533. Def. The apothem of a regular polygon is the radius of the inscribed circle, as OH. 534. Def. In a regular polygon the angle at the center is the angle between radii of the polygondrawn to the extremities of any side,as Z AOF. 535. Cor. I. The angle at the center is equal to four right angles divided by the number of sides of the polygon. 536. Cor. II. An angle of a regular polygon is the supplement of the angle at the center. Ex. 993. Find the number of degrees in the angle at the center of a regularoctagon. Find the number of degrees in an angle of the octagon. Ex. 994. If the circle circumscribed about a triangle and the circle inscribed in it are concentric, the triangle is equilateral. Ex. 995. How many sides has a regularpolygon whose angle at the center is 30" ? 256 PLANE GEOMETRY Proposition VII. Theorem 537. Regular polygons of the same number of sides are similar. C Given two regular polygons,ABODE and A'b'c'd'e', of the same number of sides. To prove polygon ABODE ~ polygon A'b'c'd'e'. Argument 1. Let n represent the number of sides of each polygon; then each angle of " each polygon equals (w-2)2rt. A n 2. .. the polygons are mutually equi-angular. 3. AB = BC = CD = " " " . 4. A'B' = B'C' = C'D' = " " " . ~ . j^_JW__C]? A'B'~ B'C'~ C'D'~ 6. .'. polygonABODE ~ polygon A'b'c'd'e'. Q.E.D. Reasons 1. "217. 2. Arg. 1, 3. "515. 4. "515. 5. " 54, 8 a. 6. "419. Ex. 996. Two homologous sides of two regular pentagons are 3 inches and 5 inches,respectively ; what is the ratio of their perimeters? of their areas ? Ex. 997. The perimeters of two regularhexagons are 30 inches and 72 inches, respectively ; what is the ratio of their areas ? 258 PLANE GEOMETRY MEASUREMENT OF THE CIRCUMFERENCE AND OF THE CIRCLE 540. The measure of a straightline,i.e. its length, is obtained by laying off upon it a straightline taken as a stand-ard or unit ("335). Since a straightline cannot be made to coincide with a curve, it is obvious that some other system of measurement must be adopted for the circumference. The following theorems will develop the principles upon which such measure-ment is based. Proposition IX. Theorem 541. I. The perimeter and area of a regular polygon inscribed in a circle are less, respectively, than the perime-ter and area of the regular inscribed polygon of twice as many sides. II. The perimeter and area of a regular polygon cir-cumscribe about a circle are greater, respectively, than the perimeter and area of the regular circumscribed poly-gon of twice as many sides. The proof is left as an exercise for the student. Hint. The sum of two sides of a triangle is greater than the third side. Ex. 1000. A square and a regular octagon are inscribed in a circle whose radius is 10 inches ; find : (") The difference between their perimeters. (6) The difference between their areas. BOOK V 259 Archimedes 542. Historical Note. Archimedes (285?-212 b.c.) found the circumference and area of the circle by a method similar to that given in this text. He was born in Syracuse, Sicily, but studied in Egypt at the Universityof Alexandria. Although, like Plato,he re-garded practical applications of mathematics as of minor im-portance, yet, on his return to Sicilyhe is said to have won the admiration of King Hiero by applying his extraordinary lalS ^W"1"?V' ^"l mechanical genius to the con-struction of war-engines with which great havoc was wrought on the Roman army. By means of largelenses and mir-rors he is said to have focused the sun's rays and set. the Roman shipson fire. Although this story may be untrue, never-theless such a feat would be by no means impossible. Archimedes invented the Archimedes screw, which was used in Egypt to drain the fields after the inundations of the Nile. A ship which was so large that Hiero could not get it launched was moved by a system of cogwheels devised by Archimedes, who remarked in this connection that had he but a fixed fulcrum, he could move the world itself. The work most prizedby Archimedes himself,however, and that which giveshim rank among the greatest mathematicians of all time, is his in-vestigati of the mechanics of solids and fluids, his measurement of the circumference and area of the circle, and his work in solid geometry Archimedes was killed when Syracuse was captured by the Romans. The story is told that he was drawing diagrams in the sand, as was the custom in those days, when the Roman soldiers came upon him. He begged them not to destroyhis circles, but they, not knowing who he was, and thinking that he presumed to command them, killed him with their spears. The Romans, directed by Marcellus,who admired his genius and had given orders that he should be spared, erected a monument to his memory, on which were engraved a sphere inscribed in a cylinder. The story of the re-discovery of this tomb in 75 b.c. is delightfully told by Cicero,who found it covered with rubbish, when visiting Syracuse. Archimedes is regarded as the greatest mathematician the world has knowh, with the sole exception of Newton. 260 PLANE GEOMETRY Proposition X. Theorem 543. By repeatedly doubling the number of sides of a regular polygon inscribed in a circle, and making the polygons always regular: I. The apothem can be made to differfrom the ra-dius by less than any assigned value. II. The square of the apothem can be made to differ from the square of the radius by less than any assigned value. Given AB the side, OC the apothem, and OB the radius of a regular polygon inscribed in circle AMB. To prove that by repeatedly doubling the number of sides of the polygon : I. OB " OC can be made less than any assignedvalue. II. OB2 " OC2 can be made less than any assigned value. I. Argument 1. By repeatedly doubling the number of sides of the inscribed polygon and making the polygons always regular, AB, subtended by one side AB of the polygon,can be made less than any previously assigned arc, however small. 2. ,\ chord AB can be made less than any previously assignedline segment, how-ever small. Reasons "519. 2. " 301. BOOK V 261 3, "447. I, Arg. 3. Argument , CB, which is ^ AB, can be made less than any previously assigned value, however small. 4. But OB " OC " CB. 5. .-. OB " OC, being always less than CB, can be made less than any previously assigned value, however small. Q.E.D. II. 1. Again, OB2 " OC2 = CB2. 2. But CB can be made less than any pre-viously assigned value, however small. 3. .. CB can be made less than any previ-ously assigned value, however small. 4. .". 0~B" " OC , being always equal to CB , can be made less than any previously assigned value, however small. Q.E.D. 544. If a variable can be made less than any assigned value, the quotient of the variable by any constant, except zero, can be made less than any assigned value. 545. If a variable can be made less than any assigned value, the square of that variable can be made less than any assigned value, (For proofs of these theorems see Appendix, "" 586 and 589.) 3. " 545. 4. "309. Ex. 1001. Construct the following designs: (1) on the blackboard, making each line 12 times as long as in the figure ; (2) on paper, making each line 4 times as long : 262 PLANE GEOMETRY Proposition XI. Theorem 546. By repeatedly doubling the number of sides of regular circumscribed and inscribed polygons vf the same number of sides,and making the polygons always regular : I. TJveir perimeters approach a common limit. II. Their areas approach a common limit. Given P and p the perimeters, R and r the apothems,and K and k the areas respectively of regular circumscribed and inscribed polygonsof the same number of sides. To prove that by repeatedly doubling the number of sides of the polygons, and making the polygons always regular : I. P and p approach a common limit. II. K and k approach a common limit. I. Argument 1. Since the two regular polygonshave the P_R -p r "p _ R " r same number of sides, P"p = P K But by repeatedly doubling the number of sides of the polygons, and making them always regular,R " r can be L Reasons "538. 2. " 399. "54, 7 a. " 543,I. BOOK V 263 Argument made less than any previously as-signed value,however small. 5. .-. ~~r can be made less than any pre-R viously assignedvalue, however small. 6. .-. P ~r can be made less than any R previouslyassigned value, however small,P being a decreasingvariable. D n. 7 -\ P"p, being always equal to P , R can be made less than any previously assignedvalue, however small. 8. .\ P andp approach a common limit. Q.E.D. Reasons 5. " 544. G. " 547. 7. " 309. 8. " 548. II. The proofof II is left as an exercise for the student. Hint. Since the two regularpolygons have the same number of sides, E-El (" 539). The rest of the proof is similar to steps 2-8, " 546, I. k r2 547. If a variable can be made less than any assignedvalue, the product of that variable and a decreasing value may be made less than any assignedvalue. 548. If tioo related variables are such that one is always greater than the other,arid if the greater continuallydecreases while the less continually increases, so that the difference between the two may be made as small as ive please, then the tivo variables have a common limit ivhich lies between them. (For proofs of these theorems see Appendix, "" 587 and 594.) 549. Note. The above proof is limited to regular polygons, but it can be shown that the limit of the perimeter of any inscribed (or circum-scribed) polygon is the same by whatever method the number of its sides is successively increased, provided that each side approaches zero as a limit. 264 PLANE GEOMETRY 550. Def. The length of a circumference is the common limit which the successive perimetersof inscribed and circum-scribed regularpolygons(of3, 4,5,etc., sides) approach as the number of sides is successively increased and each side ap-proaches zero as a limit. The term "circumference" is frequentlyused for "the length of a circumference." (SeeProp. XII.) 551. The length of an arc of a circumference is such a par" of the length of the circumference as the central angle which intercepts the arc is of 360". (See " 360.) 552. The approximate lengthof a circumference is found in elementarygeometry by computing the perimeters of a series of regularinscribed and circumscribed polygonswhich are ob-tained by repeatedlydoubling the number of their sides. The perimeters of these inscribed and circumscribed polygons, since they approach a common limit, may be made to agree to as many decimal placesas we please, according to the number of times we double the number of sides of the polygons. Proposition XII. Theorem 553. The ratio of the circumference of a circle to its diameter is the same for all circles. Given any two circles with circumferences Cand (f, and with radii R and R',respectively. c_= "_ 2R 2r' To prove "" = -" , 266 PLANE GEOMETRY Proposition XIII. Theorem 557. The area of a regular polygon is equal to one half the product of its perimeter and its apothem.-C \Pr. Given regular polygon ABCD apothem. To prove area of ABCD " Argument 1. In polygon ABCD " " " , inscribe a circle. 2. Then r, the apothem of regularpolygon ABCD " " " , is the radius of circle 0. 3. .-. area of ABCD ...""jPr, q.e.d. , P its perimeter, and r its Reasons 1. " 530. 2. " 533. 3. "492. Ex. 1005. Find the area of a regularhexagon whose side is 6 inches. Ex. 1006. The area of an inscribed regularhex-agon is a mean proportionalbetween the areas of the inscribed and circumscribed equilateral triangles. Ex. 1007. The figurerepresents a flower bed drawn to the scale of 1 inch to 20 feet. Find the number of square feet in the flower bed. 558. Def. The area of a circle is the common limit which the successive areas of inscribed and circumscribed regular polygons approach as the number of sides is successively in-creased and each side approaches zero as a limit. BOOK V 267 Proposition XIY. Theorem 559. The area of a circle is equal to one half the prod-uct of its circumference and its radius. Given circle 0, with radius R, circumference C, and area K. To prove K=\CR. Argument Circumscribe about circle O a regular polygon. Call its perimeter P and its area S. Then S " ^pr. As the number of sides of the regular circumscribed polygon is repeatedly doubled,P approaches C as a limit. .. i PR approaches\ CR as a limit. Also s approaches Zasa limit. But S is always equalto J PR. .'. K"\ CR. Q.E.D. Reasons 1. "517,6. 2. " 557. 3. "550. 4. " 561. 5. " 558. 6. Arg. 2. 7. " 355. 560. The product of a variable and a constant is a variable. 561. TJie limit of the product of a variable OMd a constant, not zero, is the limit of the variable multiplied by the constant. (Proofs of these theorems will be found in the Appendix, "" 585 and 590.) 562. Cor. I. The area of a circle is equal to irR2. Hint. K = \ C R = I " 2 ttR . R = iri?2. 268 PLANE GEOMETRY 563. Cor. II. The areas of two circles are to each other as the squares of their radii, or as the squares of their diameters. 564. Cor. III. The area of a sector whose angle is a" is (See" 551.) 360^ Ex. 1008. Find the area of a circle whose radius is 3 inches. Ex. 1009. Find the area of a sector the angle of which is (a) 45", (") 120",(c) 17",and the radius,5 inches. Find the area of the segment corresponding to (6). Hint. Segment = sector " triangle. Ex. 1010. If the area of one circle is four times that of another,and the radius of the firstis 6 inches,what is the radius of the second ? Ex. 1011. Find the area of the ring included between the circumfer-ences of two concentric circles whose radii are 6 inches and 8 inches. Ex. 1012. Find the radius of a circle whose area is equal to the sum of the areas of two circles with radii 3 and 4, respectively. Ex. 1013. In the figurethe diameter AB = 2B, AC = \AB, AD = \ AB, and E is any point on AB. (1) Find arc AC + arc CB ; arc AD + arc DB ; arc AE + arc EB. (2) Com-pare each result with semicircumference AB. 565. Def. In different circles similar arcs, similar sectors, and similar segments are arcs, sectors, and segments that cor-respond to equalcentral angles. Ex. 1014. Ex. 1015. radii. Ex. 1016. ments are to the squares of their radii. Hint. Prove sector AOB _ A AOB sector CPD A CPD apply "" 396 and 399. Similar arcs are to each other as their radii. Similar sectors are to each other as the squares of their Similar seg-each other as Then BOOK V 269 Proposition XV. Problem 566. Given a circle of unit diameter and the side of a regular inscribed polygon of n sides, to find the side of a regular inscribed polygon of 2n sides. Given circle ABF of unit diameter,AB the side of a regular inscribed polygon of n sides, aud CB the side of a re'gular in-scribed polygon of 2 n sides ; denote AB by s and CB by x. To find x in terms of s. Argument 1. Draw diameter CF ; draw BO and BF. 2. Z CBF is a rt. Z. 3. Also CF is the L bisector of AB. L .: CB2 = CF-CK. 5. Now CF = 1,.50 1 CO 6. .% C5 =#2 1 1 2' ' 2 1 . CK= CK=C0 " KO = --K0. 7. .-. x2 = -" -JW " ir# -Mtjr Vl-s2 /rr Q.E.F. Reasons 1. "54,15. 2. "367. 3. "142. 4. " 443, II. 5. By cons. 6. "309. 7. "447. 8. "54,13. 270 PLANE GEOMETRY Proposition XVI. Problem 567. Given a circle of unit diameter and the side of a regular circumscribed polygon of n sides,to find the side of a regular circumscribed polygon of 2n sides. Given circle O of unit diameter,AB half the side of a reg-ular circumscribed polygon of n sides,and CB half the side of a regular circumscribed polygon of 2 n sides ; denote AB by \|and CB by \|. To find x in terms of s. 2. Argument Draw CO and AO. A BOC = Z BOA. 3. .-. in A OAB, AC : CB = AO : BO. 4. But AC = AB " CB. 5. And 6. .'. AB " CB : CB "-"/ AB + J50 AB -[-BO : BO. 7. Substituting\| for ^J5, -for C", and -for 'BO, "HS vr s " x : x = yV + I : 1. Reasons 1. "54,15. 2. "517, 6. 3. "432. 4. " 54, 11. 5. "446. 6. "309. 7. "309. 8. "403? BOOK V 271 9. 10. Argument . s " x = x Vs2 + 1. .\ X = 1 + V s2 + 1 Q.E.F. Reasons 9. "388. 10. Solvingfor x. Ex. 1017. Given a circle of unit diameter and an inscribed and a circumscribed square ; compute the side of the regular inscribed and the regularcircumscribed octagon Proposition XVII. Problem 568. To compute the approximate value of the circum-ference of a circle in terms of its diameter ; i.e. to compute the value of ir. Given circle A BCD, with unit diameter. To compute approximately the circumference of circle ABCD in terms of its' diameter ; i.e.to compute the value of ir. Argument The ratio of tne circumference of a circle to its diameter is the same for all circles. Since the diameter of the given circle is unity,the side of an inscribed square I will be \| V2. Reasons 1. "553. 2. "522. 272 PLANE GEOMETRY Argument 3. By using the formula x = \^"-4. VI-the sides of regularinscribed polygons of 8, 16, 32, etc.,sides may be com-puted ; and by multiplying the length of one side by the number of sides, the length of the perimeter of each polygon may be obtained. The re-sults are given in the table below. Likewise if the diameter of the given circle is unity,the side of a circum-scribed square will be 1. By usingthe foimula x = 1 + Vi^TT the sides of regular circumscribed polygonsof 8,16,32, etc.,sides may be computed ; and by multiplying the lengthof one side by the number of sides, the length of the perimeter of each polygon may be obtained. The results are givenin the following table. Reasons 3 "566. 4. "522. 5. "567. 274 PLANE GEOMETRY Ex. 1022. The figurerepresents a belt from drive wheel 0 to wheel Q. The diameter of wheel 0 is 2 feet and of wheel Q 16 inches. If the drive wheel makes 75 revo-lutions per minute, how many rev-olutions per minute will the smaller wheel make ? Ex. 1023. The radius of a circle is 6 inches. What is the area of a segment whose arc is 60" ? Ex. 1024. The radius of a circle is 8 inches. What is the area of the segment subtended by the side of an inscribed equilateral triangle ? Ex. 1025. The diagonalsof a rhombus are 16 and 30 ; find the area of the circle inscribed in the rhombus. MISCELLANEOUS EXERCISES Ex. 1026. An equiangular polygon inscribed in a circle is regular if the number of sides is odd. Ex. 1027. An equilateralpolygon circumscribed about a circle is regular if the number of sides is odd. Ex. 1028. Find the apothem and area in terms of the radius in an equilateral triangle ; in a square ; in a regularhexagon. Ex. 1029. The lines joining the mid-pointsof the apothems of a regu-lar pentagon form a regular pentagon. Find the ratio of its area to the area of the original pentagon. Ex. 1030. Within a circular grass plotof radius 6 feet, a flower bed in the form of an equilateral triangle is inscribed. How many square feet of turf remain ? Ex. 1031. The area of a regular hexagon inscribed in a circle is 24 V3. What is the area of the circle ? Ex. 1032. From a circle of radius 6 is cut a sector whose central angle is 105". Find the area and perimeter of the sector. (ir = 2r2.) Ex. 1033. Prove that the following method of _ inscribing a regular pentagon and a regular decagon .in a circle is correct. Draw diameter CD perpen-dicular to diameter AB ; bisect OA and joinits mid-point to D ; take EF = ED and draw FD. FD will be the side of the required pentagon, and OF the side of the requireddecagon. BOOK V 275 Ex. 1034. Divide a given circle into two segments such that any angle inscribed in one segment is twice an angle inscribed in the other ; so that an angle inscribed in one segment is three times an angle inscribed in the other ; seven times. Ex. 1035. Show how to cut off the corners of an equilateral triangle so as to leave a regularhexagon ; of a square to leave a regularoctagon. Ex. 1036. The diagonalsof a regularpentagon form a regularpenta-gon. Ex. 1037. The diagonalsjoiningalternate vertices of a regularhexa-gon form a regularhexagon one third as large as the original one. Ex. 1038. The area of a regular inscribed octagon is equal to the product of the side of an inscribed square and the diameter. Ex. 1039. If a is the side of a regularpentagon inscribed in a circle whose radius is 2?,then a = -5-v 10 " 2V5 29 Ex. 1040. The area of a regularinscribed dodecagon is equal to three times the square of the radius. Ex. 1041. Construct an angle of 9". Ex. 1042. Construct a regularpentagon, given one of the diagonals. Ex. 1043. Through a given point construct a line which shall divide a given circumference into two parts in the ratio of 3 to 7 ; in the ratio of 3 to 5. Can the given point lie within the circle? Ex. 1044. Transform a given regularoctagon into a square. Ex. 1045. Construct a circumference equal to the sum of two given circumferences. Ex. 1046. Divide a given circle by concentric circumferences into four equivalentparts. Ex. 1047. In a given sector whose angle is a right angle inscribe a square. Ex. 1048. In a given sector inscribe a circle. Ex. 1049. If two chords of a circle are perpendicular to each other, the sum of the four circles having the four segments as diameters is equivalentto the given circle. Ex. 1050. The area of a ring between two concentric circumferences whose radii are B and Rr respectively is tt(B2 " B'2). Ex. 1051. The area of the surface between two concentric circles is equal to twice the area of the smaller circle. Find the ratio between theij radii. 276 PLANE GEOMETRY MISCELLANEOUS EXERCISES ON PLANE GEOMETRY Ex. 1052. If equilateral triangles are constructed on the sides of any given triangle, the lines joiningthe vertices of the given triangleto the outer vertices of the oppositeequilateral triangles are equal. Ex. 1053. If,on the arms of a righttriangle as diameters, semicircles are drawn so as to lie outside of the triangle, and if, on the hypotenuse as a diameter, a semicircle is drawn passing through the vertex of the right angle, the sum of the areas of the two crescents included between the semicircles is equal to the area of the given triangle. Ex. 1054. The area of the regular inscribed triangleis half that of the regularinscribed hexagon. Ex. 1055. From a given point draw a secant to a circle such that its internal and external segments.shall be equal. Ex. 1056. Show that the diagonals of any quadrilateral inscribed in a circle divide the quadrilateral into four triangles which are similar, two and two. Ex. 1057. Through a point P, outside of a circle, construct a secant VAB so that JJ?2 = PAx PB. Ex. 1058. The radius of a circle is 6 feet. What are the radii of the circles concentric with it whose circumferences divide its area into three equivalentparts ? B a ^ . Ex. 1059. Given parallelogram ABCD, 2? the mid-point of BC; prove 0T=\ TC. Ex. 1060. Given PTa tangent to a circle at point T, and two other tangents parallel to each other cuttingPT at A and B respectively ; prove that the radius of the circle is a mean proportional between AT and TB. Ex. 1061. Show that a mean proportional between two unequal lines is less than half their sum. Ex. 1062. Given two similar triangles, construct a triangle equiva-lent to their sum. Ex. 1063. The square of the side of an inscribed equilateral triangle is equal to the sum of the squares of the sides of the inscribed square and of the inscribed regularhexagon. Ex. 1064. Prove that the area of a circular ring is equal to the area of a circle whose diameter equalsa chord of the outer circumference which is tangent to the inner 0. BOOK V 277 Ex. 1065. It two chords drawn from a common point P on the cir-cumferenc of a circle are cut by a line parallel to the tangent through P, the chords and the segments of the chords between the two parallel lines are inversely proportional. Ex. 1066. Construct a segment of a circle similar to two given simi-lar segments and equivalentto their sum. Ex. 1067. The distance between two parallels is a, and the distance between two pointsA and B in one parallelis 2 b. Find the radius of the circle which passes through A and P, and is tangent to the other parallel. Ex. 1068. Tangents are drawn through a point 6 inches from the circumference of a circle whose radius is 9 inches. Find the length of the tangents and also the lengthof the chord joiningthe points of contact. Ex. 1069. If the perimeterof each of the figures, equilateral triangle, square, and circle, is 396 feet, what is the area of each figure ? Ex. 1070. The lengthsof two sides of a triangle are 13 and 15 inches, and the altitude on the third side is 12 inches. Find the third side,and also the area of the triangle. (Give one solution only.) Ex. 1071. If the diameter of a circle is 3 inches,what is the length of an arc of 80" ? Ex. 1072. AD and BC are the parallel sides of a trapezoidABCD, whose diagonalsintersect at E. If F is the mid-point of BC, prove that FE prolonged bisects AD. Ex. 1073. Given a square ABCD. Let E be the mid-point of CD, and draw BE. A line is drawn parallel to BE and cutting the square. Let P be the mid-point of the segment of this line within the square. Find the locus of P when the line moves, always remaining parallel to BE. Describe the locus exactly, and prove the correctness of your answer. Ex. 1074. Let ABCD be any parallelogram, and from any point P in the diagonalAC draw the straight line PM cuttingAB in M, BC in jV, CD in P, and AD in K. Prove that PM " PN= PK " PL. Ex. 1075. Find the area of a segment of a circle whose 'height is 4 inches and chord 8V3 inches. Ex. 1076. A square, whose side is 5 inches long, has its corners cut off in such a way as to make it into a regularoctagon. Find the area and the perimeter of the octagon. Ex. 1077. Into what numbers of arcs less than 15 can the circumfer-enQe of a circle be divided with ruler and compasses only ? 278 PLANE GEOMETRY Ex. 1078. Through a point A on the circumference of a circle chords are drawn. On each one of these chords a point is taken one third of the distance from A to the other end of the chord. Find the locus of these points, and prove that your answer is correct. Ex. 1079. In what class of triangles do the altitudes meet within the triangle ? on the boundary ? outside the triangle ? Prove. Ex. 1080. Given a triangleABC and a fixed point D on side AC ; draw the line through D which divides the triangleinto two parts of equal area. Ex. 1081. The sides of a triangleare 5, 12, 13. Find the radius of the circle whose area is equal to that of the triangle. Ex. 1082. In a triangle ABC the angle C is a right angle,and the lengthsof AC and BC are 5 and 12 respectively ; the hypotenuse BA is prolonged through A to a point D so that the length of AD is 4 ; CA is prolonged through A to E so that the triangles AED and ABC have equal areas. What is the length of AE? Ex. 1083. Given three points A, B, and C, not in the same straight line ; through A draw a straight line such that the distances of B and C from the line shall be equal. Ex. 1084. Given two straightlines that cut each other ; draw four circles of given radius that shall be tangent to both of these lines. Ex. 1085. Construct two straight lines whose lengthsare in the ratio of the areas of two given polygons. Ex. 1086. The radius of a regular inscribed polygon is a mean pro-portional between its apothem and the radius of the similar circumscribed polygon. Ex. 1087. Draw a circumference which shall pass through two given pointsand bisect a given circumference. Ex. 1088. A parallelogramis constructed having its sides equal and parallel to the diagonalsof a given parallelogram. Show that its diagonals are parallel to the sides of the given parallelogram. Hint. Look for similar triangles. Ex. 1089. If two cnords are divided in the same ratio at their point of intersection, the chords are equal. Ex. 1090. The sides AB and iO of a triangle ABC are bisected in D and E respectively. Prove that the area of the triangle BBC is twice that of the triangle DEB. Ex. 1091. Two circles touch externally. How many common tan-gents have they ? Give a construction for the common tangents. BOOK V 279 Ex. 1092. Prove that the tangents at the extremities of a chord of a circle are equally inclined to the chord. Ex. 1093. Two unequal circles touch externally at P ; line AB touches the circles at A and B respectively.Prove angle APB a rightangle. Ex. 1094. Find a point within a trianglesuch that the lines joining this point to the vertices shall divide the triangleinto three equivalent parts. Ex. 1095. A triangle ABC is inscribed in a circle. The angle B is equal to 50" and the angle C is equal to 60". What angle does a tangent at A make with BC prolonged to meet it ? Ex. 1096. The bases of a trapezoidare 8 and 12, and the altitude is 6. Find the altitudes of the two triangles formed by prolonging the non-parallel sides until they intersect. Ex. 1097. The circumferences of two circles intersect in the pointsA and B. Through A a diameter of each circle is drawn, viz. AC and AD. Prove that the straightline joining C and D passes through B. Ex. 1098. How many lines can be drawn through a given point in a plane so as to form in each case an isosceles triangle with two given lines in the plane ? Ex. 1099. The lengthsof two chords drawn from the same point in the circumference of a circle to the extremities of a diameter are 5 feet and 12 feet respectively. Find the area of the circle. Ex. 1100. Through a point 21 inches from the center of a circle whose radius is 15 inches a secant is drawn. Find the product of the whole secant and its external segment. Ex. 1101. The diagonalsof a rhombus are 24 feet and 40 feet re-spectively. Compute its area. Ex. 1102. On the sides AB, BC, CA of an equilateral triangle ABC measure off segments AD, BE, CF, respectively, each equal to one third the length of a side ; draw triangle DEF ; prove that the sides- of triangle DEF are perpendicularrespectively to the sides of triangle ABC. Ex. 1103. Construct as if(a) -= -; (6)x = a V" x 3 Ex. 1104. Find the area included between a circumference of radius 7 and an inscribed square. Ex. 1105. What is the locus of the center of a circle of given radius whose circumference cuts at rightangles a given circumference ? Ex. 1106. Two chords of a certain circle bisect each other. One of them js10 inches long ; how far is it from the center of the circle ? 280 PLANE GEOMETRY Ex. 1107. Show how to find on a given straightline of indefinite length a point O which shall be equidistant from two given points A and B in the plane. If A and B lie on a straightline which cuts the given line at an angle of 45" at a point 7 inches distant from A and 17 inches from jB,show that OA will be 13 inches. Ex. 1108. A variable chord passes, when prolonged, through a fixed point outside of a given circle. What is the locus of the mid-point of the chord ? Ex. 1109. A certain parallelogram inscribed in a circle has two sides 20 feet in length and two sides 15 feet in length. What are the lengthsof the diagonals? Ex. 1110. Upon a given base is constructed a triangleone of the base angles of which is double the other. The bisector of the largerbase angle meets the oppositeside at the point P. Find the locus of P. Ex. 1111. What is the locus of the point of contact of tangents drawn from a fixed point to the different members of a system of con-centric circles ? Ex. 1112. Find the locus of all points, the perpendiculardistances of which from two intersecting lines are to each other as 3 to 2. Ex. 1113. The sides of a triangleare a, 6, c. Find the lengthsof the three medians. Ex. 1114. Given two triangles;construct a square equivalent to their sum. Ex. 1115. In a circle whose radius is 10 feet, two parallel chords are drawn, each equal to the radius. Find the area of the portion be-tween these chords. Ex. 1116. A has a circular garden and B one that is square. The distance around each is the same, namely, 120 rods. Which has the more land, A or B ? How much more has he ? Ex. 1117. Prove that the sum of the anglesof a pentagram (a five-pointed star)is equal to two rightangles. Ex. 1118. AB and A'B' are any two chords of the outer of two con-centric circles ; these chords intersect the circumference of the inner circle in pointsP, Q and P', Qf respectively : prove that^lP-PB=A'P' " P'B'. Ex. 1119. A running track consists of two parallel straight por-tions joined togetherat the ends by semicircles. The extreme length of the plot inclosed by the track is 176 yards. If the inside line of the track is a quarter of a mile in length,find the cost of seeding this plot at \| cent a square yard, (tt = -2fa.) 282 PLANE GEOMETRY FORMULAS OF PLANE GEOMETRY 570. In addition to the notation given in " 270,the follow-ing will be used : a = side of polygon in general. b = base of a plane figure. b' = bases of a trapezoid. C = circumference of a circle. D = diameter of a. circle. E = sum of exterior angles of a polygon. h = altitude of a plane figure. I = sum of interior angles of a polygon. K = area of a figure in general. I = line in general. P = perimeter of polygon in general. p = projection of b upon a. B = radius of circle, or radius of regular polygon. r = apothem of regular polygon, or radius of inscribed circle. s " the longer of two segments of a line ; or s = %(a + b + c). X = angle in general. xa = side of a triangleopposite an acute angle. x0 = side of a triangleopposite an obtuse angle. FORMULAS 283 Polygon. Circle inscribed in triangle. DLA REFERENCE + 2 ma\ " 457. s. " 465 and Ex. 763. "475. "478. "481. "485. ha = -vs(s " a) (i a b)(s-c). K = y/s(s-a)(s-b)(i K=l(a + b + c)r. K=\Pr. c). "490. "490. "491. "492. _ \/g(s-q)(s-6)(8-c) t Ex" 837" Circle circumscribed about triangle. Bisector of angle of tri-angle. Trapezoid. Regular polygons of same number of sides. Circles. Regular polygon. Circle. B abc Circles. Sector. Segment. 4Vs(s" a)(s" 6)(s" c) 2 b + c K=(b + b')h P _B _r P' B' r' ' K' JK'2 r' C = C 2B 2B'' C = 2tB. a b1' K=lPr. K=\C-B. K=ttB\ K' B' D'2' ~_ central Z _ y/bcsis" a). B. 360= K= sector ^ triangle. Ex. 838. Ex. 841. "495. "538. "539. " 553. " 556. "557. "559. "562. "563. "564. Ex. 1009. APPENDIX TO PLANE GEOMETRY MAXIMA AND MINIMA 571. Def. Of all geometric magnitudes that satisfy given conditions, the greatest is called the maximum, and the least is called the minimum. 572. Def. Isoperimetric figures are figureswhich have the same perimeter. Proposition I. Theorem 573. Of all triangles having two given sides, that in which these sides include a right angle is the maximum. Given A ABC and AEC, with AB and AC equal to AE and AC respectively. Let Z CAB be a rt. Z and Z CAE an oblique Z. To prove A ABC" A AEC. Draw the altitude EF. A ABC and AEC have the same base, AC. Altitude AB " altitude EF. .'. A ABC " A AEC. Q.E.D. 574. Cor. I. Conversely, if two sides are given, and if the triangle is a maximum, then the given sides include a right angle. Hint. Prove by reductio ad absurdum. In later mathematics a somewhat broader use will be made of these terms. 284 APPENDIX 285 575. Cor. II. Of all parallelograms having given sides, the one that is rectangular is a maximum, and con-versely. Ex. 1135. Construct the maximum parallelogramhaving two lines of given lengthsas diagonals. Ex. 1136. What is the minimum line from a given point to a given line? Ex. 1137. Of all triangles having the same base and altitude, that which is isosceles has the minimum perimeter. Proposition II. Theorem 576. Of all equivalent triangles having the same base, that which is isosceles has the least perimeter. nCD A C Given equivalent A ABC and AEC with the same base AC, and let AB= BC and AE^EC. To prove AB + BC + CA " AE + EC + CA. Draw CF J_ AC and let CF meet the prolongation of AB at G. Draw EG and BE and prolongBE to meet GC at F. BF II AC. Z.CBF=Z.FBG. BF bisects CG and is X CG. .-. BC = BG and EC = EG. AB + BG " AE--EG. .'. AB+BC " AE + EC. ^ 'j. AB + BC+CA " AE + EC 4- CA. Q.E.D. 286 PLANE GEOMETRY Ex. 1138. Of all equivalent triangleshaving the same base, that which has the least perimeter is isosceles. (Prove by reductio ad absur-dum.) Ex. 1139. Of all equivalenttriangles, the one that has the minimum perimeter is equilateral. Ex. 1140. State and prove the converse of Ex. 1139. Proposition III. Theorem 577. Of all isoperimetric triangles on the same base, the isosceles triangle is tJis maximum. AH C Given isosceles A ABC and any other A as AEC having the same base and the same perimeter as A ABC. To prove A ABC " A AEC. Draw BH" AC, EF from E \AC, and draw AF and FC. A AFC is isosceles. ,\ perimeter of A AFC " perimeterof A AEC. .. perimeter of A AFC " perimeter of A ABC. .'. AF " AB. .-. FH" BH. .'. A AFC " A ABC. .: A AEC "AABC; i.e. A ABC " A AEC. Q.e.d. Ex. 1141. Of all triangles having a given perimeter and a given base, the one that has the maximum area is isosceles. Ex. 1142. What is the maximum chord of a circle ? What is the maximum and what the minimum line that can be drawn from a given exterior point to a given circumference ? APPENDIX 287 Ex. 1143. Of all triangleshaving a given perimeter, the one that has the maximum area is equilateral. Proposition IV. Theorem 578. Of all polygons having all their sides but one equal, respectively, to given lines taken in order, the maximum can be inscribed in a semicircle having the undetermined side as diameter. C A H Given polygon ABCEFH, the maximum of all polygonssub-ject to the condition that AB, BC, CE, EF, FH, are equal respec-tively to given lines taken in order. To prove that the semicircumference described with AH as diameter passes through B, C, E, and F. Suppose that the semicircumference with AH as diameter does not pass through some vertex, as E. Draw AE and EH. Then Z AEH is not a rt. Z. Then if the figures ABCE"nd EFH"re revolved about E until AEH becomes a rt. Z, A AEH will be increased in area. .-. polygon ABCEFH can be increased in area without chang-ing any of the given sides. But this contradicts the hypothesisthat polygon ABCEFH is a maximum. .. the supposition that vertex E is not on the semicircumfer-ence is false. .-. the semicircumference passes through E. In the same way it may be proved that every vertex of the polygon lies on the semicircumference. q.e.d. Ex. 1144. Given the hase and the vertex angle of a triangle, con-"%ictthe triangle so that its area shall be a maximum. 288 PLANE GEOMETRY Ex. 1145. Find the point in a given straight line such that the tan-gents drawn from it to a given circle contain a maximum angle. Proposition V. Theorem 579. Of all polygons that have their sides equal, re-spectively, to given lines taken in order, the polygon that can be circumscribed by a circle is a maximum. Given polygon ABCD which is circumscribed by a O, and polygon A'b'c'd' which cannot be circumscribed by a O, with AB " A'B', BC=B'C', CD = "D',and DA = D'A'. To prove ABCD " A'b'"D'. From any vertex as A draw diameter AE; draw E C and ED. On C'D', which equals CD, construct Ad'c'e' equal to ADCE; draw A'e'. The circle whose diameter is A'E' does not pass through all the pointsB',C', D'. (Hyp.) .. either ABCE or EDA or both must be greater,and neither can be less, than the correspondingpart of polygon A'b'c'e'd' (" 578). .-. ABCED " A'B'C'E'D'. But A DCE = A D'C'e\ .'. ABCD " A'B'C'D' Q.E.D. Ex. 1146. In a given semicircle inscribe a trapezoidwhose area is a maximum. Ex. 1147. Of all equilateral polygons having a given side and a given number of sides, the one that is regularis a maximum. 290 PLANE GEOMETRY Proposition VIT. Theorem 582. Of two isoperimetric regular polygons,that which has the greater number of sides has the greater area. Given the isoperimetric polygonsP and Q, and let P have one more side than Q. To prove P " Q. In one side of Q take any pointas H. EFHG may be considered as an irregular polygonhaving the same number of sides as P. .. P " EFHG ; i.e.P " Q. Q.E.D. Proposition VIII. Theorem 583. Of two equivalent regular polygons, that which has the greater number of sides has the smaller pe-rimeter. H Given square S =o= regularhexagon H. To prove perimeter of S " perimeterof H. Construct square R isoperimetric with H. APPENDIX 291 Area of H " area of R ; i.e. area of S " area of R. .-. perimeter of S " perimeter of R. .. perimeter of S " perimeter of H. q.e.d. 584. Cor. Of all polygons having a given number of sides and a given area, that which has a minimum perimeter is regular. Ex. 1149. Among the triangles inscribed in a given circle, the one that has a maximum perimeter is equilateral. Ex. 1150. Of all polygons having a given number of sides and in-scribed in a given circle, the one that has a maximum perimeter is regular. VARIABLES AND LIMITS. THEOREMS Proposition I. Theorem 585. If a variable can be made less than any assigned value, the product of the variable and any constant can be made less than any assigned value. Given a variable V, which can be made less than any previ-ously assignedvalue,however small, and let K be any constant. To prove that v " K may be made as small as we please, i.e. less than any assignedvalue. Assign any value, as a, no matter how small. Now a value for V may be found as small as we please. Take v " ~. Then v-K" a; i.e. V-K may be made less than any assignedvalue. q.e.d. 586. Cor. I. If a variable can be made less than any assigned value, the quotient of the variable by any con-stant, except zero, can be made less than any assigned value. Hint. " ;= " " V, which is the product of the variable and a constant. A K 587. Cor. II. If a variable can be made less than any assigned value, the product of that variable and a de-t 292 PLANE GEOMETRY creasing value may be made less than any assigned value. Hint. Apply the precedingtheorem, using as K a value greater than any value of the decreasingmultiplier. 588. Cor. III. The product of a variable and a variable may be a constant or a variable. 589. Cor. IV. If a variable can be made less than any assigned value, the square of that variable can be made less than any assigned value. (ApplyCor. II.) Ex. 1151. Which of the corollaries under Prop. I is illustrated by the theorem: " The product of the segments of a chord drawn through a fixed pointwithin a circle is constant " ? Proposition II. Theorem 590. The limit of the product of a variable and a con-stant, not zero, is the limit of the variable multiplied by the constant. Given any variable V which, approachesthe finite limit i, and let K be any constant not zero. To prove the limit of K " V= K " L. Let R = L " V; then V"L " R. .\ K-V=K-L " K-R. But the limit of K . R = 0. .-. the limit of K . V= the limit of (K " L " K " R) = K " L. Q.E.D. 591. Cor. The limit of the quotientof a variable by a constant is the limit of the variable divided by the constant. Hint. " = " " F, which is the product of the variable and a constant. K jBl Proposition III. Theorem 592. If two variables approach finite limits, not zero, then the limit of their product is equal to the product of their limits. APPENDIX 293 Given variables V and V1 which approach the finite limits L and L\ respectively. To prove the limit of V " V' == L " L1. Let R = L"V and R' = L' " r'. Then V= L " R and F' = "' " R!. .-. V " V' = L " L' "(L' " R + L " R' " R - R^. But the limit of (jJ " R + " " i?' -R " i?') = 0. .-. the limit of F " V' = the limit of [L " i' " (l' . i? + i " R' " R.R')] = L-L\ .-. the limit of F " V' = " " i'. q.e.d. 593. Cor. 7/ eac/z/ of any finite number of variables approaches a finitelimit, not zero, then the limit of their product is equal to the product of their limits. Proposition IV. Theorem 594. If two related variables are such that one is always greater than the other, and if the greater con-tinually decreases while the less continually increases, so that the differencebetween the two may be made as small as we please, then the two variables have a com-mon limit which lies between them. A P Q R L R' Q' P' Given the two related variables AP and AP',AP' greater than AP, and let AP and AP' be such that as AP increases AP' shall decrease,so that the difference between AP and AP' shall ap-proach zero as a limit. To prove that AP and AP1 have a common limit, as AL, which lies between AP and AP'. Denote successive values of AP by AQ, AR, etc., and denote the correspondingvalues of AP' by AQ',AR',etc. Since every value which AP assumes" is less than any value which AP' assumes (Hyp.) .-. AP " AR'. ^ j$ut AP is continually increasing. 294 PLANE GEOMETRY A P Q R L R' Q' P' Hence AP has some limit. (By def. of a limit," 349.). Since any value which AP assumes is greater than every value which AP assumes (Hyp.) .. AP' " AR. But AP1 is continually decreasing. Hence AP1 has some limit. (By def. of a limit, " 349.) Suppose the limit of AP =" the limit of APf. Then let the limit of AP be AK, while that of AP' is AK. Then AK and AK1 have some finite values,as m and ra', and their difference is a finite value,as d. But the difference between some value of AP and the cor-responding value of AP' cannot be less than the difference of the two limits AK and AK'. This contradicts the hypothesis that the difference between AP and AP1 shall approach zero as a limit. .. the limit of AP = the limit of AP' and lies between AP and ap\ as AL. q.e.d. 595. Theorem. With every straight line segment there is associated a number which may be called its measure-number. For line segments commensurable with the unit this theo-rem was considered in "" 335 and 336; we shall now consider the case where the segment is incommensurable with the chosen unit. Given the straightline segment a and the unit segment u; to express a in terms of u. Apply u (as a measure) to a as many times as possible, sup-pose t times, then v ' t.u"a"(t + l)u. Now apply some fractional part of u, say -, to a, and sup-APPENDIX 295 pose it is contained txtimes,then p p Then applysmaller and smaller fractional parts of u to a, say " -, " , ",""", and suppose them to be contained U. ts" tu """ times p2 ps p respectively, then h ^ ^to + 1 U ^ ^t3 + l -2--. u" a"^" " u, J. . u " a " -?J---u, pi pi p6 pi Now the infinite series of increasingnumbers t, J, " , """, p p2 none of which exceeds the finite number t + 1, defines a num-ber n (thelimit of this series) which we shall call the measure-number of a with respect to u. Moreover, this number n is unique,i.e.independent of p (thenumber of parts into which . the unit was divided), for if m is any number such that m " n, then m ' u " a, and if m " n, then m " u " a ; we are therefore justified in associating the number n with a, and in saying that n " u = a. 596. Note. Manifestly, the above procedure may be applied to any geometric magnitude whatever, i.e. every geometric magnitude has a unique measure-number. 597. Cor. If a magnitude is variable and approaches a limit, then, as the magnitude varies, the successive measure-numbers of the variable approach as their limit the measure-number of the limit of the magnitude. 598. Discussion of the problem : To determine wlwther two given lines are commen-surable or not; and if they are commensurable, to find their common measure and their ratio ("345). Moreover, GD is the greatestcommon measure of AB and CD. For every measure of AB is a measure of its multiple GE. Hence, every common measure of AB and GD is a common measure of GE and CD and therefore a measure of their differ-296 PLANE GEOMETRY ence ED, and therefore of AF, which is a multipleof ED Hence, every common measure of AB and CD is a common measure of AB and AF and therefore a measure of their differ-ence FB. Again, every common measure of ED and FB is a common measure of ED and EG (amultiple of FB) and hence of their difference GD. Hence, no common measure of AB and CD can exceed GD. Therefore, GD is the greatestcommon measure of AB and CD. Now, if AB and CD are commensurable, the process must ter-minate ; for any common measure of AB and CD is a measure of each remainder,and every segment appliedas a measure is less than the preceding remainder. Now, if the process did not terminate, a remainder could be reached which would be less than any assignedvalue,however small, and therefore less than the greatestcommon measure, which is absurd. If AB and CD are incommensurable, the process will not ter-minate ; for, if it did,the last remainder obtained would be a common measure of AB and CD, as shown above. .C 599. Theorem. An angle can be bisected by only one line. Given Z ABC, bisected by BT. ^"^^ To prove that no other ^ j, bisector of Z ABC exists. ^ ' "~ " Suppose that another Z?-"^-^^^^^ bisector of Z ABC exists, " 'A e.g. BF. Then Z ABF = Z.ABT. This is impossible. .-. no other bisector of A ABC exists. q.e.d. 600. Note on Axioms. The thirteen axioms (" 54) refer to num-bers and may be used when referring to the measure-numbers of geometric magnitudes. Axioms 2-9 are not applicable always to equal figures. (See Exs. 800 and 801.) Axioms 7 and 8 hold for positivenumbers only, but do not hold for negative numbers, for zero, nor for infinity ; axioms 11 and 12 hold only when the number of parts is finite. 298 PLANE GEOMETRY (b) The locus of the mid-points of all chords of a circle parallel to a given line is the diameter perpendicular to the line. ' R f(-"~ ( " a\j"" ~rp Given circle 0, line AB, and diameter RS _L AB. To prove RS the locus of the mid-points of all chords of circle 0 that are IIAB. Argument Let P be any point in diameter RS. Through P draw FG IIAB. Now RS J. AB. .'. RS _L FG. .-. P is the mid-point of FG, a chord II AB. Let Q be any point not in diameter RS. Through Q draw II K IIAB, intersecting RS in T. Then AS J HK. .-. T is the mid-point of HK, i.e. Q is not the mid-point of HK, a chord IIAB. .'. RS is the locus of the mid-points of all chords of circle 0 that are IIAB. Q.E.D. 8. " 130 SOLID GEOMETRY BOOK VI LINES, PLANES, AND ANGLES IN SPACE 602. Def. Solid geometry or the geometry of space treats of figures whose parts are not all in the same plane. (For defini-tion of plane or plane surface, see " 34.) 603. From the definition of a plane it follows that : (a) If two points of a straight line lie in a plane, the whole line lies in that plane. (b) A straight line can intersect a plane in not more than one point. 604. Since a plane is unlimited in its two dimensions (length and breadth) onlya portion of it can be shown in a figure. This is usuallyrepre-sented by a quad-rilateral drawn as a parallelogram. Thus MN represents a plane. Sometimes, however, conditions make it necessary to represent a plane by a figure other than a parallelogram, as in " 617. Ex. 1152. Draw a rectanglefreehand which is supposed to lie : (a) in a vertical plane ; (b) in a horizontal plane. May the four angles of the rectangleof (a) be drawn equal ? those of the rectangleof (5)? 299 300 SOLID GEOMETRY 605. Note. In the figures in solid geometry dashed lines will be used to represent all auxiliarylines and lines that are not supposed to be visible but which, for purposes of proof,are represented in the figure. All other lines will be continuous. In the earlier work in solid geometry the stu-dent may experience difficulty in imagining the figures. If so, he may find it a great help, for a time at least, to make the figures. By using pasteboard to represent planes,thin sticks of wood or stiffwires to repre-sent lines perpendicular to a plane, and strings to represent oblique lines, any figure may be actuallymade with a comparatively small expenditure of time and with practically no expense. For reproductions of models actuallymade by high school students,see group on p. 302 ; also "" 622, 633, 678, 756, 762, 770, 797. 606. Assumption 20. Revolution postulate. A plane may revolve about a line in it as an axis,and as it does so revolve, it can contain any particularpoint in space in one and only one position. 607. From the revolution postulateit follows that : Through a given straight line any number of planes may be passed. For, as plane MN revolves about AB as an axis ("606) it may occupy an unlimited num-ber of positions each of which will represent a different plane through AB. 608. Def. A plane is said to be determined by given condi-tions if that plane and no other plane fulfills those conditions. 609. From "" 607 and 608 it is seen that: A straight line does not determine a plane. Ex. 1153. How many planes may be passed through any two points in space ? why ? Ex. 1154. At a point P in a given straightline AB in space, con-struct a line perpendicularto AB. How many such lines can be drawn ? BOOK VI /^ 301 LINES AND PLANES Proposition I. Theorem 610. A plane is determined by a straight line and a point not in the line. ^ 1 N A-"" -- /_ ^ / f /j' ' Given line AB and P, a pointnot in AB. To prove that AB and P determine a plane. Argument 1. Through AB pass any plane, as MN. 2. Revolve planeMN about AB as an axis until it contains point P. Call the plane in this position RS. 3. Then plane RS contains line AB and pointP. 4. Furthermore, in no other position can plane MN, in its rotation about AB, contain pointP. 5. .-. RS is the onlyplanethat can contain AB and P. 6. .'. AB and P determine a plane. q.e.d. 611. Cor. I. A plane is determined by three points not in the same straightline. Hint. Let A, J5,and C be the three givenpoint??.Join A and B by a straight line, and apply " 610. 612. Cor. II. A plane is determined by two intersec-ting^ straightlines. 302 SOLID GEOMETRY 613. Cor. III. A plane is determined by two parallel straight lines. Ex. 1155. Given line AB in space, and P a point not in AB\ Con-struct, through P, a line perpendicular to AB. Ex. 1156. Hold two pencilsso that a plane can he passed through them. In how many ways can this be done, assuming that the pencilsare lines ? why ? Ex. 1157. Can two pencilsbe held so that no plane can be passed through them ? If so, how ? Ex. 1158. In measuring wheat with a half bushel measure, the meas-ure is first heaped, then a straightedge is drawn across the top. Why is the measure then even full ? Ex. 1159. Why is a surveyor's transit or a photographer's camera always supported on three legsrather than on two or four ? Ex. 1160. How many planes are determined by four straightlines, no three of which lie in the same plane, if the four lines intersect : (1) at a common point ? (2) at four different points? 614. Def. The intersection of two surfaces is the locus of all points common to the two surfaces. 615. Assumption 21. Postulate. Two planeshaving one point in common also have another point in common. Reproduced from Models made by High School Students BOOK VI 303 Proposition IT. Theorem 616. If two -planes intersect, their intersection is a straight line-S N Given intersecting planes MN and RS. To prove the intersection of MN and RS a str. line. Argument 1. Let A and B be any two pointscommon to the two planes MN and RS. Draw str. line AB. Since both A and B lie in plane MN, str. line AB lies in plane MN. Likewise str. line AB lies in plane RS. Furthermore no point outside of AB can lie in both planes. .-. AB is the intersection of planes MN and RS. But AB is a str. line. .-. the intersection of MN and RS is a str. line. q.e.d. Reasons 1. " 615. 2. " 54, 15. 3. " 603, a. 4. " 603, a. 5. " 610. 6. " 614. 7. Arg. 2. 8. Args. 6 and 7. Ex. 1161. Is it possiblefor more than two planes to intersect in a straight line ? Explain. Ex. 1162. By referring to "" 26 and 608, give the meaning of the expression,"Two planes determine a straight line." Ex. 1163. Is the statement in Ex. 11G2 always true ? Give reasons for.youranswer. 304 SOLID GEOMETRY Proposition III. Theorem 617. If three planes, not passing through the same line, intersect each other, their three lines of intersection are concurrent, or else they are parallel, each to each. M M N M Fig. 1 Fig. 2. Given planes MQ, PS, and RN intersectingeach other in lines MN, PQ, and RS-, also : I. Given MN and RS intersecting at 0 (Fig.1). To prove MN, PQ, and RS concurrent. Argument ." 0 is in line MN, it lies in plane MQ. ." O is in line RS, it lies in plane PS. ". O, lying in planes MQ and PS, must lie in their intersection, PQ. \ PQ passes through 0; i.e. MN, PQ, and RS are concurrent in 0. q.e.d. II. Given MN IIRS (Fig.2). To prove PQ IIMN and RS. Argument 1. PQ and MN are either \|\|or not II. 2. Suppose that PQ intersects MN ; then MN also intersects RS. 3. But this is impossible, for MN IIRS. 4. .-. PQ Reasons 1. " 603, a. 2. " 003, a. 3. " 614. 4. Arg. 3. MN. 5. Likewise PQ IIRS. Q.E.D. Reasons " 161, a. " 617, I. 3. By hyp. 4. " 161, b. 5. By steps sim-ilar to 1-4. 306 SOLID GEOMETRY Proposition IV. Theorem 622. If a straight line is perpendicular to each of two intersecting straight lines at their point of inter-section, it is perpendicular to the plane of those lines. Given str. line FB L AB and to BC at B, and plane MN con-taining AB and BC. To prove FB J plane MN. Outline of Proof 1. In plane MN draw AC; through J? draw any line, as BHy meeting AC at H. 2. Prolong FB to E SO that BE=FB', draw AF, HF, CF, AE, HE, CE. 3. AB and BC are then L bisectors of FE ; i.e. FA = AE, FC = CE. ) 4. Prove A AFC = A EAC ; then Z HAF = Z EAH. 5. Prove A HAF = A EAH ; then HF = ##. 6. .-.bhA-FE; i.e. FB"BH, any line in plane 3fiV passing through B. 7. .'. FB"MN. 623. Cor. All the perpendiculars that can be drawn to a straight line at a given point in the line lie in a plane perpendicular to the line at the given point. BOOK VI 307 Proposition V. Problem 624. Through a given point to construct a plane per-pendicular to a given line. Fig. 1 Given point P and line AB. To construct, through P, a plane _L AB, I. Construction 1. Through line AB and pointP pass a plane, as APD (in Fig.1, any plane through AB). "" 607, 610. 2. In plane APD construct PD, through P, J AB. "" 148,149. 3. Through AB pass a second plane, as ABC. " 607. 4. In plane ylPC, through the foot of PD, construct a L to AB (PC in Fig. 1, DC in Fig. 2). " 148. 5. Plane MN, determined by C,D, and P, is the planerequired. II. The proof is left as an exercise for the student. Hint. Apply " 623. III. The discussion will be given in " 625. Ex. 1164. Tell how to test whether or not a flagpole is erect. Ex. 1165. Lines AB and CD are each perpendicular to line EF. Are AB and CD necessarily parallel ? Explain. Do they necessarily lie jh the same plane? why or why not ? 308 SOLID GEOMETRY Proposition VI. Theorem 625. Through a given point there exists only one plane perpendicular to a given line. ^8 S^ ::::y /r"^^p/ M" 1 M-1 ' \B \B Fig. 1. Fig. 2. Given plane MN, through P, _L AB. To prove MN the only plane through P " AB. Argument Only 1. Either MN is the only plane through P " AB or it is not. 2. In MN draw a line through P intersecting line AB, as PR. 3. Let plane determined by AB and PR be denoted by APR. 4. Suppose that there exists another plane through P A. AB; let this second plane intersect plane APR in line PS. 5. Then AB _L PR and also PS; i.e. PR and PS are _L AB. 6. This is impossible. 7. .. MN is the onlyplanethrough P_L AB. q.e.d. 626. Question. In Fig.2, explain why AB JL PS. 627. "" 624 and 625 may be combined in one statement: Through a givenpoint there exists one and only one plane per-pendicula to a given line. 628. Cor. I. The locus of all points in space equidis-tant from the extremities of a straight line segment is the plane perpendicular to the segment at its mid-point. 629. Def. A straight line is parallel to a plane if the straight line and the plane cannot meet. 630. Def. A straight line is oblique to a planeif it is neither perpendicular nor parallel to the plane. 631. Def. Two planes are parallel if they cannot meet. BOOK VI 309 Proposition VII. Theorem 632. Two planes perpendicular to the same straight line are parallel. a M V N 7 Given planes MN and RS, each " line AB. To prove MN IIRS. Hint. Use indirect proof. Compare with " 187. Proposition VIII. Theorem 633. If a plane intersects two parallel planes, the lines of intersection are parallel. Given IIplanes MN and RS, and any plane PQ intersecting MN and RS in AB and CD, respectively. To prove AB IICD. Hint. Show that AB and CD cannot meet. 634. Cor. I. Parallel lines intercepted between the same parallel planes are equal. (Hint. Compare with " 234.) .Ex. 1166. State the converse of Prop. VIII. Is it true? 310 SOLID GEOMETRY Proposition IX. Theorem 635. If two angles, not in the same plane, have their sides parallelrespectively, and lying on the same sidle of the line joining their vertices,they are equal- Given Z ABC in plane MN and Z DEF in plane RS with BA and BC II respectively to ED and EF, and lyingon the same side of line BE. To prove Z ABC= Z BEF. Ex. 1167. Prove Prop. IX if the angles lie on opposite sides of BE. It will also be seen (" 645) that the planesof these angles are parallel. BOOK VI 311 Proposition X. Theorem 636. If one of two -parallel lines is perpendicular to a plane, the other also is perpendicular to the plane. A B^ ^E Given AB IICD and AB _L plane MN. To prove CD _L plane MN. Argument Reasons 1. Through D draw any line in plane MN, 1. " 54, 15. as DF. 2. Through B draw BE in plane MN IIDF. 2. " 179. 3. Then Z ABE = Z CDF. 3. " 635. 4. But Z XB# is a rt. Z. 4. " 619. 5. .'. Z CD.F is a rt. Z; i.e. CD _L D.F, any 5. " 54, 1. line in plane MN through D. 6. .\ CD X plane MN. Q-E-D-6. "619. Ex. 1168. In the accompanying diagram AB and CD lie in the same plane. Angle CBA = S5", angle BCD = 35", angle ABE= 90", .RE lying in plane Jf2V. Is CD necessarilyperpendicular to plane MN? Prove your answer. Ex. 1169. Can a line be perpendicular to each of two intersecting planes ? Prove. Ex. 1170. If one of two planes is per-pendicular to a given line,but the other is not, the planes are not parallel. Ex. 1171. If a straight line and a plane are each perpendicular to the "sj,me straight line, they are parallel to each other. 812 SOLID GEOMETRY Proposition XL Problem 637. Through a given point to construct a line per-pendicula to a given plane. R M rN Fig. 1. Fig. 2. Given pointP and plane JI/iVT. To construct, through P, a line _L plane MN. I. Construction In plane MN draw any convenient line, as AB. Through ^construct plane PQ J AB. " 624. Let plane PQ intersect plane MN in CD. " 616. In plane PQ construct a line through P J_ CD, as PP. "" 148,149. 5. Pi? is the perpendicular required. II. Proof Argument 1. Through the foot of PR (P in Fig. 1, P in Pig. 2) in plane MN, draw EF II4J3. 2. ^P J_ planePQ. 3. .-. EF A. planePQ. 4. .-. EF L PP ; t\e.PR _L #\ 5. But PR J_ CD. 6. .-. PR 1. plane ifi\T. q.e.d. III. The discussion will be givenin " 639. 314 SOLID GEOMETRY Proposition XIII. Theorem 640. Two straight lines perpendicular to the same plane are parallel. M rN Given str. lines A B and CD L planeMN. To prove AB IICD. The proof is left as an exercise for the student. Hint. Suppose that AB is not \|\|CD, but that some other line through J5,as BE, is IICD. Use " 638. Proposition XIV. Theorem 641. If a straight line is parallel to a plane, the in-tersecti of the plane with any plane passing through the given line is parallel to the given line. Ax \B Given line AB IIplane MN, and plane AD, through AB, inter-secting plane MN in line CD. To prove AB IICD. The proof is left as an exercise for the student. Hint. Suppose that AB is not IICD. Show that AB will then meet plane MN. BOOK VI 315 642. Cor. I. // a plane intersects one of two -parallel lines, it must, if suffi-ciently extended, inter-sect the other also. Hint. Pass a plane through AB and CD and let it intersect plane MN in EF. Now if MN does not intersect CD, hut is II to it, then EF IICD, " 641. Apply " 178. 643. Cor. n. If two in-tersecting lines are each parallel to a given plane, the plane of these lines is parallel to the given plane. Hint. If plane MN, determined by AB and CD, is not II to plane BS, it will intersect it in some line, as EF. What is the relation of EF to AB and CD ? 644. Cor. III. Prob-lem. Through a given R point to construct : (a) A line parallel to a given plane. (b) A plane parallel to a given plane. Hint, (a) Let A be a point outside of plane MN. Through A con-struct any plane intersecting plane MN in line CD. Complete the con-struction. 645. Cor. IV. If two angles, not in the same plane, have their sides parallel respectively, their planes are parallel. Ex. 1173. Hold a pointerparallel to the blackboard. Is its shadow on the blackboard parallel Lo the pointer ? why ? Ex. 1174. Find the locus of all straight lines passing through a given point and parallel to a given plane. 316 SOLID GEOMETRY Proposition XV. Theorem 646. If two straight lines are parallel, a plane con-taining one of the lines, and only one, is paraliel to the other. 4 , # Given IIlines AB and CD, and plane MN containingCD. To prove plane MN IIAB. Argument 1. Either plane MN is II A B or it isnot. 2. Suppose MN is not II AB ; then plane MN will intersect AB. 3. Then plane MN must also intersect CD. 4. This is impossible, for MN contains CD. 5. .-. plane MN IIAB. q.e.d. Reasons 1. " 161, a. 2. " 629. 3. " 642. 4. By hyp. 5. " 161, 6. 647. Cor. I. Problem. Through a given line to construct a plane parallelto another given line-Hint. Through E, any point in CD, construct a line HK II AB. 648. Cor. II. Problem. Through a given point to con-struct a plane parallel to any two given straight lines in space. BOOK VI 317 Proposition XVI. Theorem 649. If two straight lines are intersected by three par-allel planes, the corresponding segments of these lines are proportional-y " ytf / Given IIplanes MN, PQ, and RS intersectingline AB in A, E, B and line CD in C, H, D, respectively. " AE CH To prove " = " " EB HD Argument . Reasons 1. Draw AD intersecting plane PQ in F. 1. " 54, 15. 2. Let the plane determined by ^LBand^D 2. "" 612, 616. intersect PQ in EF and RS in BD. 3. Let the plane determined by AD and 3. "" 612, 616. DC intersect PQ in FH and MN in AC. 4. .\ EF IIBD and FH IIAC. AE AF , CH AF EB FD HD FD 6. AE EB CH HD Q.E.D. 4. " 633. 5. " 410. 6. "54,1. 650. Cor. If two straight lines are intersected by three parallel planes, the lines are divided proportionally. Ex. 1175. If any number of lines passing through a common point are cut by two or more parallel planes,their corresponding segments are proportional. Ex. 1176. In the figurefor Prop. XVI, AE = 6, EB = 8, AD = 21, CD. = 28. Find AF and HD. 318 SOLID GEOMETRY Proposition XVII. Theorem 651. A straight line perpendicular to one of two par-allel planes is perpendicular to the other also. R r N 7 B Given plane MN IIplane RS and line AB J plane RS. To prove line AB JL plane MN. Argument Only 1. In plane MN, through C, draw any line CD, and let the plane determined by ^Cy,and CD intersect plane RS in EF. 2. Then CD II EF. 3. But AB J_ EF. 4. .. AB _L CD, any line in plane MN passingthrough C. 5. .-. line AB JL plane MN. q.e.d. 652. Cor. I. Through a given point there exists only one plane parallel to a. given plane. (Hint. Apply ""638, 6446, 651, 625.) 653. "" 6446 and 652 may be combined in one statement : Through a given point there exists one and only one plane parallel to a given plane. 654. Cor. II. If two planes are each parallelto a third plane, they are parallel to each other. (Hint. See " 180.) 655. Def. The projection of a point upon a plane is the foot of the perpendicularfrom the point to the plane. 656. Def. The projection of a line upon a plane is the locus of the projections of all pointsof the line upon the plane. BOOK VI Proposition XVIII. Theorem 319 A 657. The projection upon a plane of a straight line not perpendicular to the plane is a straight line. Given str. line AB not _L plane MN. To prove the projection of AB upon MN a str. line. Argument 1. Through C, any pointin AB, draw CD _L plane MN. 2. Let the plane determined by AB and CD intersect plane MN in the str. line EF. 3. From H, any point in AB, draw HK, in planeAF, IICD. 4. Then HK _L plane MN. 5. .-. K is the projection of R upon plane MN. 6. .-. EF is the projection of AB upon plane MN. 7. .'. the projection of AB upon plane Jf^T is a str. line. q.e.d. Reasons 1. " 639. 2. "" 612, 616. 3. " 179. 4. " 636. 5. " 655. 6. " 656. 7. Args. 2 and 6. Ex. 1177. Compare the length of the projectionof a line upon plane with the length of the line itself: (a) If the line is parallel to the plane. (6) If the line is neither parallel nor perpendicular to the plane, v ^c)If the line is perpendicularto the plane. 320 SOLID GEOMETRY Proposition XIX. Theorem 658. Of all oblique lines drawn from a point to a plane : I. Those having equal projections are equal. II. Those having unequal projectionsare unequal, and the one having the greater projection is the longer. Given line PO J_ plane MN and : I. Oblique lines PA and PB with projection OA = projec-tion OB. II. Oblique lines PA and PC with projection OC" projec-tion OA. To prove : I. PB = PA; II. PC " PA. The proofis left as an exercise for the student. 659. Cor. I. (Converse of Prop. XIX). Of all oblique lines drawn from a point to a plane : I. Equal oblique lines have equal projections. II. Unequal oblique lines have unequal projections, and the longer line has the greater projection. 660. Cor. II. The locus of a point in space equidistant from all points in the circumference of a circle is a straight line perpendicular to the plane of the circle and passing through its center. 661. Cor. III. The shortest line from a point to a given plane is the perpendicular from that point to the plane. 662. Def. The distance from a point to a plane is the length of the perpendicular from the point to the plane. 322 SOLID GEOMETRY DIHEDRAL ANGLES 666. Defs. A dihedral angle is the figure formed by two planes that diverge from a line. The planes forming a di-hedral angle are called its faces, and the intersection of these planes, its edge. 667. A dihedral angle may be designated by reading in order the two planes forming the angle; thus,an angle formed by planesAB and CD is angle AB-CD, and is usually written angleA-BC-D. If there is no other dihedral angle having the same edge,the line forming the edge is a sufficient designation, as dihedral angle BC. 668. Def. Points, lines,or planes lying in the same plane are said to be coplanar. 669. A clear notion of the magnitude of a dihedral angle may be obtained by imaginingthat its two faces,considered as finite portions of planes, were at first coplanar and that one of them has revolved about a line com-mon to the two. Thus in the figure we may imagine face CD first to have been in the position of face AB and then to have revolved about 5Cas an axis to the position of face CD. 670. Def. The plane angle of a dihedral angle is the angle formed by two straightlines, one in each face of the dihedral angle, perpendicularto its edge at the same point. Thus if EF, in face AB, is L BC at F, and FH, in face CD, is J BC at F, then Z EFH is the plane Z of the dihedral Z A-BC-D. Ex. 1188. All plane angles of a dihedral angle are equal. Ex. 1189. Is the plane of angle EFH (" 667) perpendicular to the edge BC? Prove. State your result in the form of a theorem. Ex. 1190. Is Ex. 309 true if the quadrilateral is a quadrilateral in space, i.e. if the vertices of the quadrilateral are not all in the same plane ? Prove. BOOK VI 323 671. Def. Two dihedral angles are adjacent if they have a common edge and a common face which lies between them ; thus Z A-BC-D and Z B-CB-E are adj. dihe-D dral A. 672. Def. If one plane meets another so as to make two adjacent dihe-dral angles equal,each of these anglesis a right dihedral angle, and , the planes are said to be perpendicular to each other. Thus if plane HP meets plane LM so that dihedral A H-KL-M and M-LK-N are equal, each Z is a rt. dihedral Z, and planesHP and LM are _L to each other. Ex. 1191. By comparison with the definitions of the corresponding terms in plane geometry, frame exact definitions of the followingterms : acute dihedral angle ; obtuse dihedral angle ; reflex dihedral angle ; oblique dihedral angle ; vertical dihedral angles; complementary di-hedral angles ; supplementary dihedral angles; bisector of a dihedral angle ; alternate interior dihedral angles; corresponding dihedral angles. Illustrate as many of these as you can with an open book. Ex. 1192. If one plane meets another plane, the sum of the two adjacentdihedral angles is two rightdihedral angles. Hint. See proof of " 65. Ex. 1193. If the sum of two adjacent dihedral angles is equal to two rightdihedral angles,their exterior faces are coplanar. Hint. See proof of " 76. Ex. 1194. If two planes intersect, the vertical dihedral angles are equal. " 'Hint. See proof of " 77. 324 SOLID GEOMETRY Proposition XX. Theorem 673. If two dihedral angles are equal, their -plane angles are equal. Given two equaldihedral A BO and b'o' whose plane A are A MNO and M'n'o', respectively. To prove Z MNO = Z M'N'O'. Argument Reasons 1. Superposedihedral Z BO upon its equal, 1. " 54, 14. dihedral Z B'o', so that point N of edge BO shall fall upon point N' of edge B'O'. 2. Then MN and m'n',two lines in plane 2. " 670. AB, are _L 7iC at pointN. 3. .'. MN and J/V are collinear. 3. " 62. 4. Likewise NO and JV'O' are collinear. 4. "" 670,62. 5. .'.A MNO = Z M'N'O'. q.e.o. 5. "18. 674. Cor. I. The plane angle of a right dihedral angle is a rightangle. 675. Cor. II. If two intersecting planes are each per-pendicular to a third plane, their intersections with the third plane intersect each other. BOOK VI 325 Given planes AB and CD _L plane MN and intersectingeach other in line DB ; also let AE and FC be the intersections of planes AB and CD with plane MN. To prove that AE and FC intersect each other. Argument 1. Either AE IIFC or AE and FC intersect each other. 2. Suppose AE II FC. Then through 77,any point in DB, pass a plane J/fi7" _L 7A7, intersecting FC in if and AE in 2,. 3. Then plane HKL is _L ;(# also. 4. .. Z 7/"X is the plane Z of dihedral Z .FC, and Z ifLTJ is the plane Z of di-hedral Z "4#. 5. But dihedral A FC and AE are rt. dihe-dral A. .'.A HKL and JTL7J are rt. A. .'.A HKL contains two rt. A. But this is impossible. AE and FC intersect each other, q.e.d. Reasons 1. 161, a. 2. " 627. " 636. " 670. 5. " 672. "674. Arg. 6. "206. " 161, b. Ex. 1195. Find the locus of all points equidistantfrom two given points in space. Ex. 1196. Find the locus of all points equidistantfrom three given points in space. Ex. 1197. Are the supplements of equal dihedral angles equal? complements ? Prove your answer. Ex. 1198. If two planes are each perpendicular to a third plane, can they be parallel to each other? Explain. If they are parallelto ea'qh other, prove their intersections with the third plane parallel. 326 SOLID GEOMETRY Proposition XXI. Theorem (Converseof Prop. XX) 676. If the -plane angles of two dihedral angles are equal, the dihedral angles are equal. Given two dihedral A BC and b'c' whose plane A MNO and M'N'O' are equal. To prove dihedral A BC = dihedral Z B'o'. Argument 1. Place dihedral A BC upon dihedral Ab'c' so that plane A MNO shall be superposed upon its equal, plane Am'n'O'. 2. Then BC and B'c' are both L MN and NO at N. 3. .-. BC and B'c' are both _L plane MNO at N. 4. .-. BC and ^'c' are collinear. 5. .-. planes AB and jl'JJ', determined by MN and J5C, are coplanar; also planes CD and C'D',determined by BC and NO, are coplanar. 6. .-. dihedral A BC = dihedral Z #'c'. Q.E.D. Reasons 1. " 54, 14. 2. " 670. 3. " 622. 4. " 638. 5. " 612. 6. "18. BOOK VI 327 677. Cor. If the plane angle of a dihedral angle is a right angle, the dihedral angle is a right dihedral angle. Ex. 1199. Prove Ex. 1194 by applying " 676. Ex. 1200. If two parallelplanes are cut by a transversal plane,the alternate interior dihedral angles are equal. Hint. Let Z ABC be the plane Z of jj dihedral Z V-WX-Y. Let the plane XL determined by AB and BO intersect plane f JW MN in CD. Then AB and CD lie in / j"-47 the same plane and are II(" 633). Prove / JmL /y that Z DCB is the plane Z of dihedral -XI / ly Z M-Z T-X. / J jS V Ex. 1201. State the converse of Ex. /D f/C / 1200, and prove it by the indirect method. zf y1 Ex. 1202. If two parallelplanes are 1/ cut by a transversal plane, the correspond-ing dihedral angles are equal. (Hint. See proof of " 190.) Ex. 1203. State the converse of Ex. 1202, and prove it by the indirect method. Ex. 1204. If two parallelplanes are cut by a transversal plane, the sum of the two interior dihedral angleson the same side of the transversal plane is two rightdihedral angles. (Hint. See proof of " 192.) Ex. 1205. Two dihedral angles whose faces are parallel, each to each, are either equal or supplementary dihedral angles. (Hint. See proof of " 198.) Ex. 1206. A dihedral angle has the same numerical measure as its plane angle. (Hint. Proof similar to that of " 358.) Ex. 1207. Two dihedral angles have the same ratio as their plane angles. Ex. 1208. Find a point in a plane equidistant from three given points not lying in the plane. Ex. 1209. If a straightline intersects one of two parallel planes,it must, if sufficiently prolonged,intersect the other also. (Hint. Use the indirect method and apply "" 663 and 664.) Ex. 1210. If a plane intersects one of two parallel planes,it must, if sufficiently extended, intersect the other also. (Hint. Use the indirect lfte'^iod and apply " 652. ) 328 SOLID GEOMETRY Proposition XXII. Theorem 678. If a straight line is perpendicular to a plane, every plane containing this line is perpendicular to the given plane. El M Given str. line AB _L plane MN and plane PQ containingline AB and intersecting plane MN in CD. To prove planePQ _L plane MN. Argument 1. AB _L CD. 2. Through B, in planeMN, draw BE _L CD. 3. Then Z ABE is the plane Z of dihedral Z Q-CD-M. 4. But Z ABE is a rt. Z. 5. .-. dihedral Z Q-CD-M is a rt. dihedral Z, and plane PQJL plane ifJV. q.e.d. Eeasons 1. "619. 2. "63. 3. "670. "619. "677. Ex. 1211. If from the foot of a per-pendicular to a plane a line is drawn at rightangles to any line in the plane,the line drawn from the point of intersec-tion so formed to any point in the per-pendicular is perpendicular%o the line of the plane. Hint. Make KE = EH. Prove AK = AH, and apply " Ex. 1212. In the figure of Ex. 1211, if AB = 20, BE JEK= 10, find AK. 142. 4V11, and 330 SOLID GEOMETRY Proposition XXIV. Theorem 681. If each of two intersecting planes is perpendicu-lar to a third plane ; I. Their line of intersection intersects the third plane. II. Their line of intersection is perpendicular to the third plane. Given planes PQ and RS _L plane MN and intersecting each other in line AB. To prove : I. That AB intersects plane MN. II. AB J plane MN. I. Argument 1. Let planes PQ and RS intersect plane MN in lines PD and RE, respectively. 2. Then PD and RE intersect in a point as C. 3. .. AB passes through C; i.e. AB intersects plane MN. q.e.d. II. Argument 1. Either AB _L plane MN or it is not. 2. Suppose AB is not _L plane MN, but that some other line through C, the point common to the three planes, is _L plane MN, as line OF. 3. Then CF lies in plane PQ, also in plane RS. 4. .-. CF is the intersection of planes PQ and RS. 5. .. planesPQ and RS intersect in two str. lines, which is impossible. 6. .-. AB 1. plane MN. q.e.d. 6. " 161, b. BOOK VI 331 Proposition XXV. Problem 682. Through any straight line, not perpendicular to a plane, to construct a plane perpendicular to the given plane. ^,5 Given line AB not L plane MN. To construct, through AB, a plane_L plane MN. The construction, proof, and discussion are left as an exer-cise for the student. Hint. Apply " 678. For discussion, see " 683. 683. Cor. Through a straightline, not perpendicular to a plane, there exists only one plane perpendicular to the given plane. Hint. Suppose there should exist another plane through AB " plane MN. What would you know about AB ? 684. "" 682 and 683 may be combined in one statement as follows : Through a straight line,not perpendicularto a plane, there exists one and only one planeperpendicular to the givenplane. Ex. 1216. Apply the truth of Prop. XXIV : (a) to the planes that intersect at the corner of a room ; (6) to the planes formed by an open book placed perpendicularto the top of the desk. Ex. 1217. If a plane is perpendicular to each of two intersecting plaaes,it is perpendicularto their intersection. 332 SOLID GEOMETRY T^ Proposition XXVI. Problem 685. To construct a common perpendicular to any two straight lines in space. K X Given AB and CD, any two str. lines in space. To construct a line _L both to AB and to CD. I. Construction 1. Through CD construct plane MN IIAB. " 647. 2. Through AB construct plane AF _L plane MN intersecting MN in EF, and CD in H. " 682. 3. Through H construct HK, in plane AF, _L EF. " 148. 4. HK is J to both ^5 and CD and is the line required. Reasons 1. By cons. "641. By cons. "193. "679. " 619. 7. Args. 4 and 6. III. The discussion will be given in " 686. 686. Cor. Between two straight lines in space {not in the same plane) there exists only one common per-pendicular. BOOK VI 333 Hint. Suppose JI, in figureof " 685, a second JL to AB and CD. Through Y draw ZW \|\| AB. What is the relation of XY to AB? to ZW? to CD? to plane MN? Through X draw XB " EF. What is the relation of XB to plane MN ? Complete the proof. 687. "" 685 and 686 may be combined in one statement as follows : Between two straight lines in space (notin the same plane) there exists one and only one common perpendicular. Ex. 1218. A room is 20 feet long, 15 feet wide, and 10 feet high. Find the length of the shortest line that can be drawn on floor and walls from a lower corner to the diagonallyoppositecorner. Find the length of the line that extends diagonallyacross the floor, then along the intersec-tion of two walls to the ceiling. Ex 1219. If two equal lines are drawn from a given point to a given plane,the inclinations of these lines to the given plane are equal. If two unequal lines are thus drawn, which has the greater inclination ? Prove. Ex. 1220. The two planes determined by two parallellines and a point not in their plane,intersect in a line which is parallel to each of the given parallels. Ex. 1221. If two lines are parallel, their projections on a plane are either the same line,or parallel lines. Ex. 1222. If each of three planes is perpendicularto the other two : (a) the intersection of any two of the planes is perpendicularto the third plane ; (6) each of the three lines of intersection is perpendicular to the other two. Find an illustration of this exercise in the classroom. Ex. 1223. If two planes are parallel, no line in the one can meet any line in the other. Ex. 1224. Find all points equidistant from two parallel planes and equidistant from three points : (a) if the points lie in neither plane ; (b) if the pointslie in one of the planes. Ex. 1225. Find all points equidistantfrom two given points,equi-distant from two parallelplanes,and at a given distance d from a third plane. Ex. 1226. If each of two intersecting planes is parallel to a given line, the intersection of the planes is parallel to the line. Ex. 1227. Construct,through a point in space, a straightline that shall be parallel to two intersecting planes. 334 SOLID GEOMETRY Proposition XXVII. Theorem 688. Every point in the plane that bisects a dihedral angle is equidistant from the faces of the angle' Given plane BE bisecting the dihedral Z formed by planes AC and CD ; also PH and PK Js from P, any pointin planeBE, to faces AC and CD, respectively. To prove PH= PK. Argument 1. Through PH and PK pass plane MN intersecting plane AC in CH, plane CD in Off,plane BE in PC, and edge BC in C. 2. Then plane MN L planesAC and CD; i.e. planes^ C and CD are J planeMN. /%. .\ BC L plane UK, 4. .-. BC" CH, CP, and C^. 5. .-. Apch and jBTCP are the plane z" of the dihedral A E-BC-A and D-CB-E. 6. But dihedral Z E-BC-A = dihedral Z D-CB-E. 7. .-. Zpch=Zkcp. 8. Also PC = PC. 9. .. rt. A PC#= rt. A KCP. 10. .'. PH= PK. Q.E.D. Reasons 1. " 612, 616. 2. " 678. 3. "681,11. 4. " 619. 5. " 670. 6. By hyp. 7. " 673. 8. Byiden. 9. " 209. 10. " 110. BOOK VI 335 689. Cor. I. Every point equidistant from the two faces of a dihedral angle lies in the plane bisecting the angle. 690. Cor. n. The plane bisecting a dihedral angle is the locus of all points in space equidistant from the faces of the angle. 691. Cor. m. Problem. To construct the bisector of a given dihedral angle. Ex. 1228. Prove that a dihedral angle can be bisected by only one plane. Hint. See proof of " 599. Ex. 1229. Find the locus of all points equidistantfrom two inter-secting planes. Of how many planes does this locus consist ? Ex. 1230. Find the locus of all pointsin space equidistant from two intersecting lines. Of how many planes does this locus consist ? Ex. 1231. Find the locus of all points in space equidistant from two parallel lines. Ex. 1232. Find the locus of all points in space equidistant from two intersecting planes and equidistant from all points in the circumference of a circle. Ex. 1233. Find the locus of all points in space equidistant from two intersecting planes and equidistantfrom two fixed points. Ex. 1234. Find the locus of all points in space equidistant from two intersecting planes,equidistant from two parallel planes,and equidistant from two fixed points. Ex. 1235. If from any point within a dihedral angle lines are drawn perpendicularto the faces of the angle,the angle formed by the perpen-diculars is supplementary to the plane angle of the dihedral angle. Ex. 1236. Given two points, P and #, one in each of two intersecting planes,M and N. Find a point Xin the intersection of planes M and N such that PX+XQ is a minimum. Ex. 1237. Given two points,P and Q, on one side of a given plane MN. Find a point X in plane MN such that PX + XQ shall be a minimum. "fyNT. See Ex. 175. 336 SOLID GEOMETRY POLYHEDRAL ANGLES 692. Def. A polyhedral angle is the figuregeneratedby a moving straightline segment that continually intersects the boundary of a fixed polygon and one extremity of which is a fixed point not in the plane of the given polygon. A poly-hedral angle is sometimes called a solid angle. 693. Defs. The moving line is called the v generatrix, as VA ; the fixed polygon is called a the directrix, as polygon ABODE; the fixed point is called the vertex of the polyhedral , angle, as V. I 694. Defs. The generatrix in any position4/ J is an element of the polyhedralangle; the /\/ \Af elements through the vertices of the poly-Bf" tt( gon are the edges, as VA, VB, etc. ; the ( \ portions of the planes determined by the edges of the polyhedral angle,and limited by them are the faces, as AVB, BVC, etc.; the angles formed by the edges are the face angles, as A AVB, BVC, etc.; the dihedral angles formed by the faces are called the dihedral angles of the poly-hedral angle,as dihedral A VA, VB, etc. 695. Def. The face angles and the dihedral angles taken togetherare sometimes called the parts of a polyhedral angle. 696. A polyhedralangle may be designated by a letter at the vertex and one on each edge, as V-ABCDE. If there is no other polyhedral angle having the same vertex, the letter at the vertex is a sufficient designation, as V. 697. Def. A convex polyhedral angle is a polyhedral angle whose directrix is a convex polygon, i.e. a polygon no side of which, if prolonged,will enter the polygon; as V-ABCDE. In this text only convex polyhedral angles will be considered. 698. Defs. A trihedral angle is a polyhedralangle whose directrix is a triangle (tri-gon) ; a tetrahedral angle, a polyhe-dral angle whose directrix is a quadrilateral (tetra-gon) ; etc. 338 SOLID GEOMETRY Proposition XXIX. Theorem 704. Two trihedral angles are equal if the three face angles of one are equal respectivelyto the three face angles of the other, and the equal parts are arranged in the same order. Given trihedral A V-ABC and V'-A'B'C', Z AVB = Z A'V'b' , Z B VC=Z B'v'c', Z CVA = Z c'v'a', and the equalface angles arranged in the same order. To prove trihedral Z V-ABC= trihedral Z v'-A'b'c'. 1. Outline of Proof Since,by hyp., any two face A of V-ABC, as A AVB and BVC, are equal,respectively, to the two correspondingface A of v'-A'b'c', it remains only to prove the included dihe-dral A VB and v'b' equal. " 702, II. (Seealso " 705.) 2. Let face A AVB and BVC be obliqueA\ then from any point E in VB, draw ED and EF, in planesAVB and BVC, respec-tively, and _L VB. 3. Since A AVB and BVC are obliqueA, ED and EF will meet VA and VC in D and F, respectively. Draw FD. 4. Similarly, lay off VE' = VE and draw A D'e'f'. Prove rt. A DVE=vt. A D'V'E' ; then VD" V'd', ED=E'd'. Prove rt. A EVF= rt. A E'v'f'-, then VF= v'f', EF=E,f'. Prove A FVD = A f'v'd' ; then FD = f'd'. . A DEF = A D'E'F' : then Z DEF = Z D'E'F' BOOK VI 339 9. But A DEF and d'e'f' are the planeA of dihedral A VB and r'B', respectively. 10. .-. dihedral Z VB = dihedral Z v'b'. 11. .-. trihedral Z F-XBC== trihedral Z F'-^'.B'C'. q.e.d. 705. Note. If all the face A are rt. A, show that all the dihedral A are rt. dihedral A and hence that all are equal. If two face A of a trihe-dral Z are rt. A, show that the third face Z is the plane Z of the included dihedral Z, and hence that two homologous dihedral A, as VB and V'B', are equal. It remains to prove that Prop. XXIX is true if only one face Z of the first trihedral Z and its homologous face Z of the other are rt. A, or if all face A are oblique. 706. Questions. State the proposition in Bk. I that corresponds to " 704. What was the main step in the proof of that proposition ? Did that correspond to proving dihedral Z VB of " 704 = dihedral Z V'B' ? 707. Def. Two polyhedral angles are said to be sym-metrical if their corresponding parts are equalbut arranged in reverse order. By making symmetrical polyhedralangles and comparing them, the student can easilysatisfy himself that in general they cannot be made to coincide. 708. Def. Two polyhedralangles are said to be vertical if the edges of each are the prolongations of the edges of the other. It will be seen that two vertical, like two symmetrical, poly-hedral angles have their corresponding parts equal but arranged in reverse order. Two Equal Polyhedral Two Vertical Poly-Two Symmetrical Poly-." " Angles hedral Angles hedral Angles 340 SOLID GEOMETRY Proposition XXX. Theorem 709. Two trihedral angles are symmetrical : I. If a face angle and the two adjacent dihedral an-gles of one are equal respectively to a face angle and the two adjacent dihedral angles of the other; II. If two face angles and the included dihedral angle of one are equal respectively to two face angles and the included dihedral angle of the other ; III. If the three face angles of one are equal respec-tively to the three face angles of the other : provided the equal parts are arranged in reverse order. The proofsare left as exercises for the student. Hint. Let V and V' be the two trihedral A with parts equal but arranged in reverse order. Construct trihedral Z V" symmetrical to V. Then what will be the relation of V" to V ? of V to V? Ex. 1240. Can two polyhedral angles be symmetrical and equal ? vertical and equal ? symmetrical and vertical ? If two polyhedral angles are vertical, are they necessarilysymmetrical ? if symmetrical, are they necessarily vertical ? Ex. 1241. Are two trirectangular trihedral anglesnecessarily equal? Are two birectangular trihedral angles equal ? Prove your answers. Ex. 1242. If two trihedral angles have three face anglesof one equal respectively to three face angles of the other, the dihedral angles of the firstare equal respectively to the dihedral angles of the second. BOOK VI 341 / Proposition XXXI. Theorem 710. The sum, of any two face angles of a trihedral angle is greater than the third face angle-V c Given trihedral Z V-ABC in which the greatest face Z is AVB. To prove Z BVC + Z CVA " Z AVB. Outline of Proof 1. Iii face AVB draw VD making Z DVB = A BVC, and through D draw any line intersectingva in E and VB in F. 2. On VC lay off VG = VD and draw FG and GE. 3. Prove A FVG = ADVF; then ^(? = FD. 4. But FG + GE " FD + DE; .: GE " D#. 5. In A GVE and "FD, prove Z GVE " Z J"FD. 6. But Z FVG = Z.DVF. 7. .'. Z J^FG +Z (?FE "Z#F.Z) + ZZ)^; i.e. ZBVC + Z CTJ. " Z ^FS. Q.E.D. 711. Question. State the theorem in Bk. I that corresponds to Prop. XXXI. Can that theorem be proved by a method similar to the one used here ? If so, give the proof. Ex. 1243. If,in trihedral angle V-ABC, angle BVC = 60",and angle CVA = 80",make a statement as to the number of degrees in angle AVB. Ex. 1244. Any face angle of a trihedral angle is greater than the difference of the other two. 342 SOLID GEOMETRY Proposition XXXII. Theorem 712. The sum of all the face angles of any convex, poly-hedral angle is less than four right angles-V Given polyhedral Z V with n faces. To prove the sum of the face A at V Z 4 rt. A. Hint. Let a plane intersect the edges of the polyhedral Z in A, B, C, etc. From O, any point in polygon ABC . . ., draw OA, OB, OC, etc. How many A have their vertices at V? at 0 ? What is the sum of all the A of all the A with vertices at V? at O ? Which is the greater, Z ABV + Z FBO or Z ABO + Z 07? O ? Then which is the greater,the sum of the base A of A with vertices at F, or the sum of the base A of A with vertices at O ? Then which is greater, the sum of the face A about F, or the sum of the A about 0 ? 713. Question. Is there a proposition in plane geometry corre-sponding to Prop. XXXII ? If so, state it. If not, state the one that most nearly corresponds to it. Ex. 1245. Can a polyhedral angle have for its faces three equi-lateral triangles ? four ? five ? six ? Ex. 1246. Can a polyhedral angle have for its faces three squares ? four ? Ex. 1247. Can a polyhedral angle have for its faces three regular pentagons ? four ? Ex. 1248. Show that the greatest number of polyhedral anglesthat can possiblybe formed with regularpolygons as faces is five. Ex. 1249. Can a trihedral angle have for its faces a regulardecagon and two equilateral triangles? a regular decagon, an equilateral tri-angle, and a square ? two regular octagons and a square ? BOOK VII POLYHEDRONS 714. Def. A surface is said to be closed if it separates a finite portion of space from the remaining space. 715. Def. A solid closed figure is a figurein space composed of a closed surface and the finite portion of space bounded by it. 716. Def. A polyhedron is a solid closed figure whose bounding surface is composed of planes only. 717. Defs. The intersections of the bounding planes are called the edges ; //\ "i the intersections of the edges, the /wl"ffl^ vertices; and the portions of the /$lM bounding planes bounded by the edges, jSfflll II lllllfllfc the faces, of the polyhedron. jM^WtSSm^ 718. Def. A diagonal of a polyhe-B ^^M^^Mmm^W dron is a straightline joiningany two ^\|["W vertices not in the same face, as AB. "\|pi"11 719. Defs. A polyhedron of four " y faces is called a tetrahedron; one of six faces, a hexahedron; one of eight faces, an octahedron; one of twelve faces, a do-decahedron ; one of twenty faces,an icosahedron ; etc. Ex. 1250. How many diagonals has a tetrahedron ? a hexahedron ? Ex. 1251. What is the least number of faces that a polyhedron can have ? edges ? vertices ? Ex. 1252. How many edges has a tetrahedron ? a hexahedron ? an octahedron ? Ex. 1253. How many vertices has a tetrahedron? a hexahedron? an octahedron ? Ex. 1254. If E represents the number of edges, F the number of faces,and Fthe number of vertices in each of the polyhedrons mentioned in Exs. 1252 and 1253, show that in each case E+2 = V+F. This result is known as Euler's theorem. 343 344 SOLID GEOMETRY Ex. 1255. Show that in a tetrahedron 8 = (V"2) 4 rightangles, where 8 is the sum of the face angles and Fis the number of vertices. ' Ex. 1256. Does the formula, 8= (V"2) 4 right angles,hold for a hexahedron ? an octahedron ? a dodecahedron ? 720. Def. A regular polyhedron is a polyhedron all of whose faces are equal regularpolygons, and all of whose polyhe-dral angles are equal. 721. Questions. How many equilateral triangles can meet to form a polyhedralangle (" 712) ? Then what is the greatest number of regular polyhedrons possiblehaving equilateral triangles as faces ? What is the greatest number of regularpolyhedrons possible having squares as faces ? having regular pentagons as faces ? Can a regular polyhedron have as faces regular polygons of more than five sides ? why ? What, then, is the maximum number of kinds of regularpolyhedrons possible? 722. From the questionsin " 721, the student has doubtless drawn the conclusion that not more than five kinds of regular polyhedrons exist. He should convince himself that these five are possible by actually making them from cardboard as indicated below : Tetrahedron Hexahedron Octahedron Dodecahedron Icosahedron 346 SOLID GEOMETRY 728. Def. A right section of a prism is a section formed by a plane which is perpendicularto a lateral edge of the prism and which cuts the lateral edges or the edges prolonged. rA \ \ Right Prism Regular Prism Oblique Prism 729. Def. A right prism is a prism whose lateral edges are perpendicularto the bases. 730. Def. A regular prism is a rightprism whose bases are regularpolygons. 731. Def. An oblique prism is a prism whose lateral edges are obliqueto the bases. 732. Defs. A prism is triangular, quadrangular, etc., accord-ing as its bases are triangles, quadrilaterals, etc. 733. Def. The altitude of a prism is the perpendicularfrom any pointin the plane of one base to the plane of the other base. 734. The followingare some of the properties of a prism ; the student should prove the correctness of each : (a)Any two lateral edges of a "prism are parallel. (b) The lateral edges of a prism are equal. (c)Any lateral edge of a right prism is equal to the altitude. (d) TJie lateral faces of a prism are parallelograms. (e) The lateral faces of a right prism are rectangles. (f) The bases of a prism are equalpolygons. (g) TJie sections of a prism made by two parallel planes cutting all the lateral edges are equalpolygons. (h)Every section of a 2^'ism made by a plane parallel to the base is equal to the base. BOOK VII 347 Proposition I. Theorem 735. Two prisms are equal if three faces including a trihedral angle of one are equal respectively, and simi-larly placed,to three faces including a trihedral angle of the other A B A B Given prisms AI and A'l', face AJ = face A'j', face AG = face A'G', face AD = face A'd'. To prove prism AI= prism Af. Argument 1. A BAF, FAE, and BAE are equal, respec-tively, to A b'a'f', f'a'e', and b'a'e'. 2. . \ trihedral Z A = trihedral Z A1. 3. Place prism AI upon prism A'l1 so that trihedral Z A shall be superposed upon its equal, trihedral Z A'. 4. Faces A J, AG, and AD are equal,re-spectively, to faces A'j', A'G1, and A'D'. 5. .. J, F, and G will fall upon J\ F1, and G',respectively. G. CH and c'h' are both \|\| BG. 7. .. CiJand C'h' are collinear. 8. .-. H will fall upon#'. 9. Likewise 2" will fall upon l\ 10.^.-. prism AI" prism A'l'. Q.E.D. 5. 10. Reasons " no. "704. " 54, 14. By hyp. "18. " 734, a. " 179. " 603, b. By stepssim-ilar to 6-8. " 18. 348 SOLID GEOMETKY 736. Def. A truncated prism is the portion of a prism in-cluded between the base and a section of the prism made by a plane oblique to the base, but which cuts all the edges of the prism. 737. Cor. I. Two truncated prisms are equal if three faces including a trihedral angle of one are equal re-spectively to three faces including a trihedral angle of the other, and the faces are similarly placed. 738. Cor. H. Two right prisms are equal if tliey have equal bases and equal altitudes. Ex. 1257. Two triangularprisms are equal if their lateral faces are equal,each to each. Ex. 1258. Classifythe polyhedrons whose faces are : (a) four tri-angles ; (6) two trianglesand three parallelograms; (c) two quadri-laterals and four parallelograms ; (d) two quadrilaterals and four rectangles ; (e) two squares and four rectangles. Ex. 1259. Find the sum of the plane angles of the dihedral angles whose edges are the lateral edges of a triangular prism ; a quadrangular prism. (Hint. Draw a rt. section of the prism.) Ex. 1260. Every section of a prism made by a plane parallel to a lateral edge is a parallelogram. Ex. 1261. Every section of a prism made by a plane parallel to a lateral face is a parallelogram. Ex. 1262. The section of a parallelopiped made by a plane passing through two diagonallyoppositeedges is a parallelogram. "3 iiiiiiiiiiiiiiF Oblique Parallelopiped Right Parallelopiped Rectangular Cube Parallelopiped 739. Def. A parallelopiped is a prism whose bases are parallelograms. BOOK VII 349 740. Def. A right parallelopiped is a parallelopiped whose lateral edges are perpendicular to the bases. 741. Def. A rectangular parallelopiped is a right parallelo-piped whose bases are rectangles. 742. Def. A cube (i.e. a regularhexahedron)is a rectangu-lar parallelopiped whose edges are all equal. 743. The followingare some of the properties of a parallel-opiped ; the student should prove the correctness of each : (a) All the faces of a parallelopiped are parallelograms. (b) All the faces of a rectangular parallelopiped are rectangles. (c) All the faces of a cube are squares. (d) Any two opposite faces of a parallelopipedare equal and parallel. (e) Any two oppositefaces of a parallelopiped may be taken as the bases. Ex. 1263. Classify the polyhedrons whose faces are : (a) six paral-lelograms ; (") six rectangles ; (c) six squares ; (d) two parallelograms and four rectangles ; (e) two rectangles and four parallelograms ; (/) two squares and four rectangles. Ex. 1264. Find the sum of all the face angles of a parallelopiped. Ex. 1265. Find the diagonal of a cube whose edge is 8 ; 12 ; e. Ex. 1266. Find the diagonal of a rectangular parallelopiped whose edges are (5, 8, and 12 ; whose edges are a, ", and c. Ex. 1267. The edge of a cube : the diagonal of a face : the diagonal of the cube =l:x:y ; find x and y. Ex. 1268. Find the edge of a cube whose diagonal is 20 V3 ; d. Ex. 1269. The diagonals of a rectangularparallelopiped are equal. Ex. 1270. The diagonalsof a parallelopiped bisect each other. Ex. 1271. The diagonalsof a parallelopiped meet in a point. This point is sometimes called the center of the parallelopiped. Ex. 1272. Any straight line through the center of a parallelopiped, with its extremities in the surface, is bisected at the center. Ex. 1273. The sum of the squares of the four diagonalsof a rectan-gular parallelopiped is equal to the sum of the squares of the twelve edges. Ex. 1274. Is the statement in Ex. 1273 true for any parallelopiped ? 350 SOLID GEOMETRY PYRAMIDS 744. Def. A pyramidal surface is a surface generated by a mov-ing straight line that continually intersects a fixed broken line and that passes through a fixed point not in the planeof the broken line. 745. Defs. By re-ferring to "" 693 and 694, give the defini-tions of generatrix, directrix, vertex, and element of a pyramidal surface. Point these out in the figure. 746. Def. A pyram-idal surface consists of two parts lying on opposite sides of the vertex, called the up-per and lower nappes. 747. Def. A pyramid is a polyhedronwhose boundary con-sists of the portion of a pyramidal surface extending from its vertex to a plane cuttingall its elements,and the section formed by this plane. 748. Defs. By referring to " 727, the student may give the definitions of base, lateral faces, and lateral edges of a pyramid. The vertex of the pyramidal surface is called the vertex of the pyramid, as V. Point these out in the figure. In this text only pyramids whose bases are convex polygons will be considered. BOOK VII 351 749. Defs. A pyramid is triangular, quadrangular, etc.,ac-cording as its base is a triangle, a quadrilateral, etc. 750. Questions. How many faces has a triangular pyramid ? a tet-rahedron ? Can these terms be used interchangeably ? How many dif-ferent bases may a triangularpyramid have ? 751. Def. The altitude of a pyramid is the perpendicular from the vertex to the plane of the base, as VO in the figure below, and in the figure on precedingpage. V Regular Pyramid Truncated Frustum of Frustum of Pyramid Triangular Pyramid Regular Pyramid 752. Def. A regular pyramid is a pyramid whose base is a regular polygon, and whose vertex lies in the perpendicular erected to the base at its center. 753. Def. A truncated pyramid is the portion of a pyramid included between the base and a section of the pyramid made by a plane cuttingall the edges. 754. Def. A frustum of a pyramid is the portion of a pyra-mid included between the base and a section of the pyramid made by a plane parallel to the.base. 755. The followingare some of the properties of a pyramid; the student should prove the correctness of each : (a) TJie lateral edges of a regular pyramid are equal. (b)TJie lateral edgesof a frustum of a regular pyramid are equal. (c) The lateral faces of a regular pyramid are equal isosceles triangles. (d) The lateral faces of a frustum of a regularpyramid are equal. .isosceles trapezoids. 352 SOLID GEOMETRY Proposition II. Theorem 756. If a -pyramid is cut by a plane parallel to the base : I. The edges and altitude are divided proportionally. II. The section is a polygon similar to the base. Given pyramid V" ABODE and plane MN \|\| base AD cutting the lateral edges in F, G, H, I, and J and the altitude in P. VD_VC _ VQt VF VG~ VH~ ~ VP II. FGHIJ~ ABODE. To prove : i. " I. Argument 1. Through V pass plane RS II plane KL 2. Then plane RS IIplane MN. VA _ VB VB _ VO VO VF~ VO' VG~ VH' VH VA_VB VC _ VF~~ VG~ VH~ 3. 4. II. The proof of II is left as an exercise for the student. 757. Cor I. Any section of a pyramid parallel to the base is to the base as the square of its distance from the vertex is to the square of the altitude of the pyramid. Hint. Prove AB1 VB2 VO1 354 SOLID GEOMETRY Ex. 1281. Prove that parallelsections of a pyramid are to each other as the squares of their distances from the vertex of the pyramid. Do the results obtained in Ex. 1280 fulfillthis condition ? Ex. 1282. Each side of the base of a regular hexagonal pyramid is 6 ; the altitude is 15. How far -from the vertex must a plane be passed parallel to the base to form a section whose area is 12 V3 ? Ex. 1283. The areas of the bases of a frustum of a pyramid are 288 square feet and 450 square feet ; the altitude of the frustum is 3 feet. Eind the altitude of the pyramid of which the given figureis a frustum. Ex. 1284. The bases of a frustum of a regular pyramid are equi-lateral triangleswhose sides are 10 inches and 18 inches,respectively ; the altitude of the frustum is 8 inches. Find the alti-tude of the pyramid of which the given figure is a frustum. Ex. 1285. The sum of the lateral faces of any pyramid is greater than the base. Hint. In the figure, let VE be the altitude of face VAD and VO the altitude of the pyramid. Which is the greater, VE or OE ? MENSURATION OF THE PRISM AND PYRAMID Areas 760. turn of 761. lowing a, 6, c = B = b = E = H = h = L = a pyramid,or a frus-of its lateral faces. Def. The lateral area of a prism, a pyramid is the sum of the areas In the mensuration of the prism and pyramid the fol-notation will be used : dimensions of a rectangu-lar parallelopiped. area of base in generalor of lower base of a frustum. area of upper base of a frustum. lateral edge, or element, or edge of a tetrahedron in general. altitude of a solid. altitude of a surface. slant height. S = T = V = VltV2 ... = vertex of a pyramid. perimeter of right sec-tion or of the lower base of a frustum. perimeter of upper base of a frustum. lateral area. total area. volume in general. volumes of smaller sol-ids into which a larger solid is divided. BOOK VII 355 Proposition III. Theorem 762. The lateral area of a prism is equal to the product of the perimeter of a right section and a lateral edge. Given prism AK with MQ a rt. section, E a lateral edge,S the lateral area, and P the perimeterof rt. section MQ. To prove S = P " E. Argument Reasons 1. Rt. section MQ _L Al, CJ, etc. 1. " 728. 2. .'.MN"AI; NQ".CJ\ etc. 2. "619. 3. .-. MN is the altitude of O AJ\ NQ is 3. " 228. the altitude of O CK; etc. 4. .-. area of O AJ = JW-AI= MN-E-, area of CJ CK = NQ " CJ= NQ " E-, etc. 5. CJAJ+CJCK-] =s(JCir + JTQ +")" 6. .. 8 =c P " #. 763. Cor. The lateral area of a right prism is equal to the product of the perimeter of its base and its altitude. Hint. Thus, if P = perimeter of base and H = altitude, S = P " H. 764. Def. The slant height of a regular pyramid is the alti-tude of any one of its triangular faces. 765. Def. The slant height of a frustum of a regular pyra-mid is the altitude of any one of its trapezoidal faces. 4. "481. 5. "54,2. 6. "309. 356 SOLID GEOMETRY Proposition IV. Theorem 766. The lateral area of a regular pyramid is equal to one half the product of the perimeter of its base and its slant height. C D Given regularpyramid O-ACD " "" with the perimeter of its base denoted by P, its slant heightby L, and its lateral area by S. To prove S = "-P . L. 767. Cor. The lateral area of a frustum of a regular pyramid is equal to one half the product of the sum of the perimeters of its bases and its slant height. Hint. Prove 8 = (P + p) L. Ex. 1286. Find the lateral area and the total area of a regular pyramid each side of whose square base is 24 inches,and whose altitude is 16 inches. Ex. 1287. The sides of the bases of a frustum of a regularoctagonal pyramid are 15 centimeters and 24 centimeters, respectively, and the slant height is 30 centimeters. Find the number of square decimeters in the lateral area of the frustum. BOOK VII 357 Ex. 1288. Find the lateral area of a prism whose rightsection is a quadrilateral with sides 5, 7, 9, and 13 inches,and whose lateral edge is 15 inches. Ex. 1289.-Find the lateral area of a rightprism whose altitude is 16 inches and whose base is a triangle with sides 8, 11, and 13 inches. Ex. 1290. The perimeter of a rightsection of a prism is 45 deci-meters ; its altitude is 10 V3 decimeters ; and a lateral edge makes with the base an angle of 60". Find the lateral area. Ex. 1291. Find the altitude of a regular prism, one side of whose triangular base is 5 inches and whose lateral area is 195 square inches. Ex. 1292. Find the total area of a regular hexagonal prism whose altitude is 20 inches and one side of whose base is 10 inches. Ex. 1293. Find the total area of a cube whose diagonal is 8 VS. Ex. 1294. Find the edge of a cube if its total area is 294 square centimeters ; if its total area is T. Ex. 1295. Find the total area of a regular tetrahedron whose edge is 6 inches. Ex. 1296. Find the lateral area and total area of a regular tetra-hedron whose slant height is 8 inches. Ex. 1297. Find the lateral area and total area of a regularhexagonal pyramid, a side of whose base is 6 inches and whose altitude is 10 inches. Ex. 1298. Find the total area of a rectangularparallelopiped whose edges are 6, 8, and 12 ; whose edges are a, b, and c. Ex. 1299. Find the total area of a rightparallelopiped, one side of whose square base is 4 inches, and whose altitude is 6 inches. Ex. 1300. The balcony of a theater is supported by four columns whose bases are regular hexagons. Find the cost, at 2 cents a square foot, of paintingthe columns if they are 20 feet high and the apothems of the bases are 10 inches. Ex. 1301. In a frustum of a regular triangularpyramid, the sides of the bases are 8 and 4 inches,respectively, and the altitude is 10 inches. Find the slant height and a lateral edge. Ex. 1302. In a frustum of a regular hexangular pyramid, the sides of the bases are 12 and 8, respectively, and the altitude is 16. Find the lateral area. Ex. 1303. In a regulartriangularpyramid, the side of the base is 8 inches, and the altitude is 12 inches ; a lateral face makes with the base an angle of 60". Find the lateral area. 358 SOLID GEOMETRY Volumes 768. Note. The student should compare carefully"" 769-776 with the corresponding discussion of the rectangle, "" 466-473. 769. A solid may be measured by finding how many times it contains a solid unit. The solid unit most frequently chosen is a cube whose edge is of unit length. If the unit length is an inch,the solid unit is a cube whose edge is an inch. Such a unit is called a cubic inch. If the unit length is a foot, the solid unit is a cube whose edge is a foot, and the unit is called a cubic foot. Fig. 1. Rectangular Parallelopiped AD = 60 77. 770. Def. The result of the measurement is a number, which is called the measure-number, or numerical measure, or volume of the solid. 771. Thus, if the unit cube U is contained in the rectangular D K Fio. 2. Rectangular Parallelopiped AD = 24 IT-BOOK VII 359 parallelopiped AD (Fig. 1)60 times,then the measure-number or volume of rectangular parallelopiped AD, in terms of U, is 60. If the given unit cube is not contained in the given rectangu-lar parallelopiped an integral number of times without a re-mainder (Fig. 2), then by taking a cube that is an aliquotpart of U, as one eighth of U, and applyingit as a measure to the rectangular parallelopiped (Fig. 3), a number will be obtained Fig. 3. Rectangular Parallelopiped AD = "" U+ = 39\| U+. which, divided by 8, will give another (and usuallycloser) approximate volume of the given rectangularparallelopiped. By proceeding in this way (Fig.4),closer and closer approxi-mations to the true volume may be obtained. Fig. 4. Rectangular Parallelopiped AD = %%" U+ = 41T93 TJ+. " I It takes eightof the small cubes to make the unit cube itself. 360 SOLID GEOMETRY 772. If the edges of the given rectangularparallelopiped and the edge' of the unit cube are commensurable, a cube may be found which is an aliquotpart of U, and which will be con-tained in the rectangularparallelopiped an integral number of times. 773. If the edges of the given rectangularparallelopiped and the edge of the unit cube are incommensurable, then closer and closer approximations to the volume may be obtained, but no cube which is an aliquotpart of U will be also an aliquot part of the rectangularparallelopiped (by definition of incom-mensurable magnitudes). There is, however, a definite limit which is approached more and more closely by the approximations obtained by using smaller and smaller subdivisions of the unit cube, as these subdivisions approach zero as a limit. 774. Def. The volume of a rectangular parallelopiped which is incommensurable with the chosen unit cube is the limit which . successive approximate volumes of the rectangular parallelopiped approach as the subdivisions of the unit cube approach zero as a limit. For brevitythe expression the volume of a solid, or simply the solid, is used to mean the volume of the solid with respect to a chosen unit. 775. Def. The ratio of any two solids is the ratio of their measure-numbers, or volumes (based on the same unit). 776. Def. Two solids are equivalent if their volumes are equal. 777. Historical Note. The determination of the volumes of polyhedrons is found in a document as ancient as the Rhind papyrus, which is thought to be a copy of a manuscript dating back possiblyas far as 3400 b.c. (See " 474.) In this manuscript Ahmes calculates the con-sents of an Egyptian barn by means of the formula, V" a " b " (c + \ c), where a, b, and c are supposed to be linear dimensions of the barn. But unfortunately the exact shape of these barns is unknown, so that the accuracy of the formula cannot be tested. 362 SOLID GEOMETRY (b) Suppose that u is not a measure of AC, AF, and AG, re-spectively but that some aliquotpart of u is such a measure. The proof is left as an exercise for the student. II. If AC, AF, and AG are each incommensurable with u. Argument 1. Let m be a measure of u. Apply m as a measure to AC, AF, and AG, respec-tively, as many times as possible. There will be remainders, as MC, NF, and QG, each less than m. 2. Through M draw plane MK " AC, through. N draw plane NK _L AF, and through Q draw plane QK _L AG. 3. Now AM, AN, and AQ are each com-mensurable with the measure m, and hence with u, the linear unit. 4. .-. the volume of rectangularparallel o-piped AK = AM " AN " AQ. 5. Now take a smaller measure of u. No matter how small a measure of u is taken,when it is applied as a measure to AC, AF, and AG, the remainders, MC, NF, and QG, will be smaller than the measure taken. Reasons 1. "339. 2. "627. 3. "337. 4. " 778, I. 5. "335. BOOK VII 363 Argument 6. .-. the difference between AM and AC, the difference between AN and AF, and the difference between AQ and AG, may each be made to become and remain less than any previously assigned segment, however small. 7. .-. AM approaches AC as a limit,AN approaches AF as a limit,and AQ approaches AG as a limit. 8. .-. AM " AN-AQ approaches AC-AF - AG as a limit. 9. Again, the difference between rectan-gular parallelopipedAK and rec-tangular parallelopiped AD may be made to become and remain less than any previously assigned volume, however small. 10. .. the volume of rectangularparallelo-piped AK approaches the volume of rectangular parallelopiped AD as a limit. 11. But the volume of AK is always equal to AM-AN " AQ. 12. .-. the volume of AD=AC-AF"AG. q.e.d. Reasons 6. Arg. 5. 7. "349. 8. "593. Arg. 5. 10. "349. 11. Arg. 4. 12. "355. III. If AC is commensurable with u but AF and AG are in-commens with u. IV. If AC and AF are commensurable with u but AG is in-commens with u. The proofsof III and IV are left as exercises for the student. 779. Cor I. The volume of a cube is equal to the cube of its edge. HN". Compare with " 478. 364 SOLID GEOMETRY 780. Cor.,II. Any two rectangular parallelopipeds are to each other as the products of their three di-mensions (Hint. Compare with " 479. ) 781. Note. By the product of a surface and a line is meant the product of the measure-numbers of the surface and the line. 782. Cor. III. The volume of a rectangular parallelo-piped is equal to the product of its base and its altitude. 783. Cor. IV. Any two rectangular parallelopipeds are to each other as the products of their bases and their altitudes. (Hint. Compare with " 479.) 784. Cor. V. (a) Two rectangular parallelopipeds having equivalent bases are to each other as their altitudes; (b) two rectangular parallelopipeds having equal altitudes are to each other as their bases. (Hint. Compare with " 480.) 785. Cor. VI. (a) Two rectangular parallelopipeds hav-ing two dimensions in common are to each other as their third dimensions, and (b) two rectangular parallelopi-peds having one dimension in common are to each other as the products of their other two dimensions. 786. Questions. What is it in Book IV that corresponds to volume in Book VII ? to rectangular parallelopiped ? State the theorem and corol-laries in Book IV that correspond to "" 778, 779,782,783, and 784. Will the proofs given there, with the corresponding changes in terms, apply-here ? Compare the entire discussion of "" 466-480 with "" 769-785. Ex. 1304. Find the volume of a cube whose diagonal is 5\/3 ; d. Ex. 1305. The volume of a rectangularparallelopiped is V; each side of the square base is one third the altitude of the parallelopiped. Find the side of the base. Find the side of the base if V = 192 cubic feet. Ex. 1306. The dimensions of two rectangularparallelopipeds are 6, 8, 10 and 5, 12, 16, respectively. Find the ratio of their volumes. Ex. 1307 . The total area of a cube is300 square inches ; find its volume. Ex. 1308. The volume of a certain cube is V; find the volume of a cube whose edge is twice that of the given cube. Ex. 1309. The edge of a cube is a ; find the edge of a cube twice as large; {.e. containingtwice the volume of the given cube. BOOK VII 365 787. Historical Note. Plato (429-348b.c.)was one of the first to discover a solution to that famous problem of antiquity,the duplication of a cube, i.e. the findingof the edge of a cube whose volume is double that of a given cube. There are two legends as to the originof the problem. The one is that an old tragic poet rep-resented King Minos as wishing to erect a tomb for his son Glau-cus. The king being dissatis-fied with the dimensions (100 feet each way) proposed by his architect, exclaimed : " The in-closure is too small for a royal tomb ; double it, but fail not in the cubical form." The other legend asserts that the Athenians, who were suf-Plato feringfrom a plague of typhoid fever, consulted the oracle at Delos as to how to stop the plague. Apollo replied that the Delians would have to double the size of his altar, which was in the form of a cube. A new altar was constructed having its edge twice as long as that of the old one. The pestilencebecame worse than before, whereupon the Delians ap-pealed to Plato. It is therefore known as the Delian problem. Plato was born in Athens, and for eightyears was a pupil of Socrates. Plato possessed considerable wealth, and after the death of Socrates in 399 b.c. he spent some years in travelingand in the study of mathe-matics. It was during this time that he became acquainted with the members of the Pythagorean School, especially with Archytas, who was then its head. No doubt it was his association with these people that gave him his passionfor mathematics. About 380 b.c he returned to his native city, where he established a school. Over the entrance to his school was this inscription: "Let none ignorant of geometry enter my door." Later an applicant who knew no geometry was actually turned away with the statement : u Depart, for thou hast not the grip of philosophy." Plato is noted as a teacher, rather than an originaldiscoverer, and his contributions to geometry are improvements in its method rather than additions to its matter. He valued geometry mainly as a "means of ed-ucation in right seeing and thinking and in the conception of imaginary processes." It is stated on good authority that "Plato was almost as imporfcrat as Pythagoras to the advance of Greek geometry." 366 SOLID GEOMETRY Proposition VI. Theorem 788. An oblique prism is equivalent to a right prism, whose base is a right section of the oblique prism, and whose altitude is equal to a lateral edge of the oblique prism. o' B C Given obliqueprism ADf ; also rt. prism EN' with base EN a rt. section of ad', and with EE', LL',etc.,lateral edges of EN', equal to AA',bb', etc., lateral edges of Ad\ To prove oblique prism AD 'o rt. prism EN'. Outline of Proof 1. In truncated prisms AN and A'N', prove the A of face BK equal,respectively, to the A of face b'k'. 2. Prove the sides of face BK equal,respectively, to the sides of face b'k'. 3. .-. face BK= face b'k'. 4. Similarly face "ilf=face B'm',and face BE" face b'e'. 5. .-. truncated prism AN = truncated prism a'n' (" 737). 6. But truncated prism A' If = truncated prism A'N. 7. .-. obliqueprism AD'=ort. prism EN'. q.e.d. 789. Question. Is there a theorem in Book IV that corresponds to Prop. VI ? If not, formulate one and see if you can prove it true. BOOK VII 367 Proposition VII. Theorem 790. The volume of any parallelopiped is equal to the product of its base and its altitude. \2M m S:Kj"n Given parallelopiped I with its volume denoted by V, its base by B, and its altitude by H. To prove V = B " H. Argument 1. Prolongedge AC and all edges of I II AC. 2. On the prolongationof AC take DF = AC, and through D and F pass planes _L AF, forming rt. parallelopiped 77. 3. Then I =^ //. 4. Prolong edge FK and all edgesof II II FK. 5. On the prolongation of FK take MN = FK, and through M and N pass planes _L FN , forming rectangular parallelo-piped III. 6. Then II o III. 7. .-.IoIII. 8. Again, BoB' = b". 9. Also H, the altitude of 1,= the altitude of ///. 10. But the volume of 111= b" . H. 1-1 .-. v= B " H. Q.E.D. Reasons " 54, 16. " 627. " 788. " 54, 16. " 627. 368 SOLID GEOMETRY 791. Cor I. Parallelopipeds having equivalent bases and equal altitudes are equivalent. 792. Cor. II. J_ny two parallelopipeds are to' each other as the products of their bases and their altitudes. 793. Cor. III. (a) Two parallelopipeds having equiva-lent bases are to each other as their altitudes, and (b) two parallelopipeds having equal altitudes are to each other as their bases. 794. Questions. What expression in Book IV corresponds to vol-ume of a parallelopiped? Quote the theorem and corollaries in Book IV that correspond to "" 790-793. Will the proofs given there, with the corresponding changes, apply here ? Ex. 1310. Prove Prop. VI by subtractingthe equal truncated prisms of Arg. 5 from the entire figure. Ex. 1311. The base of a parallelopiped is a parallelogramtwo adja-cent sides of which are 8 and 15, respectively, and they include an angle of 30". If the altitude of the parallelopiped is 10,find its volume. Ex. 1312. Four parallelopipeds have equivalentbases and equal lat-eral edges. In the first the lateral edge makes with the base an angle of 30" ; in the second an angle of 45" ; in the third an angle of 60" ; and in the fourth an angle of 90". Find the ratio of the volumes of the four parallelopipeds. Ex. 1313. Find the edge of a cube equivalent to a rectangular par-allelopipe whose edges are 6, 10, and 15 ; whose edges are a, b, and c. Ex. 1314. Find the diagonal of a cube whose volume is 512 cubic inches ; a cubic inches. Ex. 1315. The edge of a cube is a. Find the area of a section made by a plane through two diagonallyoppositeedges. Ex. 1316. How many cubic feet of cement will be needed to make a box, includinglid,if the inside dimensions of the box are 2 feet 6 inches, 3 feet,and 4 feet 6 inches, if the cement is 3 inches thick ? Hint. In a problem of this kind, always find the volume of the whole solid, and the volume of the inside solid, then subtract. Ex. 1317. The volume of a rectangularparallelopiped is 2430 cubic inches,and its edges are in the ratio of 3, 5, and 6. Find its edges. Ex. 1318. In a certain cube the area of the surface and the volume have the same numerical value. Find the volume of the cube. 370 SOLID GEOMETRY Proposition IX. Theorem 797. The volume of a triangular prism is equal to the product of its base and its altitude. Given triangularprism A CD-X with its volume denoted by V, its base by B, and its altitude by H. To prove V= B - H. The proof is left as an exercise for the student. 798. Questions. What proposition in Book IV corresponds to Prop. IX above ? Can you apply the proof there given ? What is the name of the figure CZ in " 797 ? What is its volume ? What part of CZ is A CD-X (" 795)? Ex. 1319. The volume of a triangularprism is equal to one half the product of any lateral face and the perpendicular from any point in the oppositeedge to that face. Hint. The triangularprism is one half of a certain parallelopiped (" 795). Ex. 1320. The base of a coal bin which is 8 feet deep is a triangle with sides 10 feet, 15 feet, and 20 feet, respectively. How many tons of coal will the bin hold considering35 cubic feet of coal to a ton ? Ex. 1321. One face of a triangular prism contains 45 square inches ; the perpendicularto this face from a point in the oppositeedge is 6 inches. Find the volume of the prism. Ex. 1322. During a rainfall of \| inch, how many barrels of water will fall upon a ten-acre field, counting 7" gallonsto a cubic foot and 31\| gallonsto a barrel ? Ex. 1323. The inside dimensions of an open tank before liningare 6 feet, 2 feet 6 inches, and 2 feet, respectively, the latter being the height. Find the number of pounds of zinc required to line the tank with a coat-ing \ inch thick,a cubic foot of zinc weighing 6860 ounces. book vn 371 Proposition X. Theorem 799. The volume of any prism is equal to the product of its base and its altitude. ^ AC Given prism AM with its volume denoted by V, its base by B, and its altitude by H. To prove V = B . H. Argument 1. From any vertex of the lower base,as A, draw diagonalsAD, AF, etc. 2. Through edge AI and these diagonals pass planesAK, AM, etc. 3. Prism AM is thus divided into triangu-lar prisms. 4. Denote the volume and base of trian-gular prism ACD-J by vl and bx; of ADF-K by v2 and b2; etc. Then Vj = bxH ; v2 = b2H; etc. 5. .-. ^ + ^2+ ... = (b1 + b2+-~)H. 6. .-. Fs^.a " , 1. Reasons " 54, 15. 2. " 612. 3. Q.E.D. equivalent " 732. "797. " 54, 2. " 309. bases and 800. Cor. I. Prisms having equal altitudes are equivalent. 801. Cor. n. Any two prisms are to each other as the products of their bases and their altitudes. 802. Cor. m. (a) Two prisms having equivalent bases are to each other as their altitudes: (b) two prisms having equal altitudes are to each other as their bases. 372 SOLID GEOMETRY Proposition XI. Theorem 803. Two triangular pyramids having equivalentbases and equal altitudes are equivalent. 0 " /A / wN Given triangularpyramids O-ACD and O'-a'c'd' with base ACD =c= base A'c'd', with altitudes each equal to QR, and with volumes denoted by V and V', respectively. To prove V = V' . Argument 1. V = V, V " Vf, or V " V'. 2. Suppose V " V',so that v' " v=k, a constant. For convenience, place the two pyramids so that their bases are in the same plane,MN. 3. Divide the common altitude QR into n equal parts, as QX, XY, etc., and through the several points of division pass planes IIplane MN. 4. Then section FGU o section f'g'u', section JEW o section j'k' w',etc. 5. On FGU, JEW, etc., as upper bases,con-struct prisms with edges IIDO and with altitudes = QX. Denote these prisms by II, III, etc. G. On A'c'D', F'g'u', etc.,as lower bases, construct prisms with edges II D'o' and with altitudes = QX. Denote these prismsby I', II'? etc. Reasons 1. 161, a. 2. " 54, 14. 3. "653. 4. "759. 5. "726. 6. "726. BOOK VII 373 Argument 7. Then prism II =c= prism II', prism III =c= prism III', etc. 8. Now denote the sum of the volumes of prisms II, III, etc.,by 5; the sum of the volumes of prismsI', II', III', etc., by S' ; and the volume of prism I' by v'. Then S' " s = v'. 9. But V1 "Sf and 8 " V. 10. .-. V + S" V+ S'. 11. .-. V' " V" S' " S ; i.e. V1 " V"vf. 12. By making the divisions of the altitude QR smaller and smaller,prism I', and hence v' may be made less than any previously assigned volume, however small. 13. .. V' " F, which is " v', may be made less than any previouslyassigned volume, however small. 14. .-.the supposition that v' " V=k, a constant, is false ; i.e. V is not " r'. 15. Similarly it may be proved that F is not " V. 16. .-. r= V'. -Q.E.D. Reasons 7. "800. 8. "54,3. 9. "54,12. 10. "54,9. 11. "54,5. 12. 802, a. 13, " 54, 10. 14. Arg. 13. 15. By stepssim-ilar to 2-14. 16. "161,6. Ex. 1324. The volume of an obliqueprism is equal to the product of its rightsection and a lateral edge. Hint. Apply " 788. Ex. 1325. The volume of a regularprism is equal to the product of its lateral area and one half the apothem of its base. Hint. See Ex. 1319. Ex. 1326. The base of a prism is a rhombus having one side 29 inches and one diagonal 42 inches. If the altitude of the prism is 25 inches, find its volume. Ex. 1327. In a certain cube the area of the surface and the com-bined lengths of its edges have the same numerical value. Find the volmn^ of the cube. 374 SOLID GEOMETRY Proposition XII. Theorem 804. The volume of a triangular pyramid is equal to one third the product of its base and its altitude? C c Given triangularpyramid O-ACD with its volume denoted by V, its base by B, and its altitude by H. To prove F= \B " H. Argument Only 1. Construct prism AG with base ACD and lateral edge CO. 2. The prism is then composed of triangularpyramid O-ACD and quadrangularpyramid O-ADGF. 3. Through OB and OF pass a plane,intersecting ADGF in DF and dividing quadrangularpyramid O-ADGF into two tri-angular pyramids O-ABF and O-DGF. 4. iDGFisaO; .-. A ADF = ADGF. 5. .\ O-ADF o= O-DGF. / 6. But in triangular pyramid O-DGF, OGF may be taken as base and D as vertex ; then O-DGF = D-OGF =c= O-ACD. 7. But O-ACD + O-ADF + O-DGF =c= prism AG. 8. .. 3 times the volume of 0-ACD=the volume of prism AG. 9. .. V= i the volume of prism AG. 10. But prism AG = B.H; .".V="B-R. q.e.d. Ex. 1328. Find the volume of a regulartetra-hedron whose edge is 6. Hint. O, the foot of the _L from V to the plane of base ABC, is the center of A ABC (" 752). Hence QA = f of the altitude of A ABC, and a _L from O to any edge of the base, as OD = \ of OA. Ex. 1329. Find the volume of a regulartetra-hedron with slant height2 V3 ; with altitude a. BOOK VII 375 Proposition XIII. Theorem 805. The volume of any pyramid is equal to one third the product of its base and its altitude, O C D Given pyramid O-ACDFG with its volume denoted by V, its base by B, and its altitude, OQ, by H. To prove V= \ B - H. The proof is left as an exercise for the student. Hint. See proof of Prop. X. 806. Cor. I. Pyramids having equivalent bases and equal altitudes are equivalent. 807. Cor. II. Any two pyramids are to each other as the products of their bases and their altitudes. 808. Cor. III. (a) Two pyramids having equivalent bases are to each other as their altitudes, and (b) two pyramids having equal altitudes are to each other as their bases. Ex. 1330. In the figureof " 805, if the base = 250 square inches, OO = 18 inches,and the inclination of OG to the base is 60",find the volume. Ex. 1331. A pyramid and a prism have equivalentbases and equal altitudes ; find the ratio of their volumes. 809. Historical Note. The proof of the proposition that "every pyramid is the third part of a prism on the same base and with the same altitude " is attributed to Eudoxus (408-355 B.C.), a great mathematician of the Athenian School. In a noted work written by Archimedes (287-212 b.c), called Sphere and Cylinder,there is also found an expression for the surface and volume of a pyramid. (For a further account of Archimedes, see "" 542, 896, and 973.) Later a solution of this problem was-g^ven by Brahmagupta, a noted Hindoo writer bom about 598 a.d. 376 SOLID GEOMETRY Proposition XIV. Theorem 810. Two triangular pyramids, having a trihedral angle of one equal to a trihedral angle of the other, are to each other as the products of the edges including the equal trihedral angles. D Given triangular pyramids O-ACD and Q-FGM with tri-hedral Z.0 = trihedral Z Q, and with volumes denoted by V and v'jrespectively. V OA-00-01) To prove V1 QF-QG-QM Argument 1. Place pyramid Q-FGM so that trihedral Z. Q shall coincide with trihedral Z 0. Eepresent pyramid Q-FGM in its new position by O-F'g'm'. 2. From D and M' draw DJ and MK L plane OAO. A OAO " DJ _ A OAO DJ ~~A OF'G' M'K A OAO OA " 00 a Then " " = . V' A OF'G' " i/'iT But A A OF'G' OF1 " OG' Again let the plane determined by DJ and if 'ifintersect plane OA C in line OKJ. Then rt. A BJO ~ rt. A M'KO. DJ _ OP M'K~ OM'' V OA-OC OD OA-OC-OD Q.E.D. 1. Reasons " 54, 14. 8. " 309. V' OF' " OG' OM' QF.QG-QM 811. Def. Two polyhedrons are similar if they have the same number of faces similar each to each and similarly placed, and have their corresponding polyhedral anglesequal. 378 SOLID GEOMETRY Proposition XVI. Theorem 815. The volume of a frustum of any pyramid is equal to one third the product of its altitude and the sum of its lower base, its upper base, and the mean pro-portional between its two bases. C D Given frustum AM, of pyramid O-AF, with its volume de-noted by V, its lower base by B, its upper base by b, and its altitude by H. To prove V=\H(B + b+ V B " b). 1. 2. 3. 4. 5. Argument AM = pyramid O-AF minus Frustum pyramid O-RM. Let H' denote the altitude of O-RM. Then V= J b(h+ h')" "5 . H1 = "HB + -lH\B-b). It now remains to find the value of H(. b = B . y" = Whence H = if H' H + tf' HVb VJ3-V" VB-Vb = \HB + \Hb + "H^B-b; i.e. V=\ H(B -h 6+ Vtf " 6).q.e.d. Reasons 1. " 54, 11. 2. " 805. 3. " 757. 4. " 54,13. 5. Solvingfor H '. 6. " 309. BOOK VII 379 Proposition XVII. Theorem 816. A truncated triangular -prism is equivalent to three triangular pyramids whose bases are the base of the frustum and whose vertices are the three vertices of the inclined section. -^-n(? Given truncated triangular prisin ACD-FGK. To prove ACD-FGK =c= F-ACD + G-ACD + K-ACD. Argument Through A, D, F and K, D, F pass planes dividing frustum ACD-FGK into three triangularpyramids F-ACD, F-ADK, and F-DGK Since F-ACD is one of the required pyramids,it re-mains to prove F-ADK o K-ACD and F-DGK o G-ACD. CF II plane AG. .-.the altitude of pyramid F-ADK = the altitude of pyramid C-ADK. .-. F-ADK =c= C-ADK. But in C-ADK, A CD may be taken as base and K as vertex. .-. F-ADK o K-ACD. Likewise F-DGK =c= C-DGK = K-CDG ; and K-CDG =0= A-CDG = G-ACD. .'. F-DGK =c= G-ACD. .'. ACD " FGK =0 F-ACD + G-^CZ) + -' '""K-ACD. Q.E.D. 1. Reasons "611. 380 SOLID GEOMETRY 817. Cor. I. The volume of a truncated right triangu-lar prism is equal to one third the product of its base and the sum of its lateral edges. ^/I^ 818. Cor. n. The volume of any -^ I truncated triangular prism is equal I II to one third the product of a right sec-/V^-,Y / tion and the sum of its lateral edges. I jk/^ Hint. Rt. section ACD divides truncated / / / triangular prism QR into two truncated right Qi^---J--J triangularprisms. ^^"^^ Ex. 1334. The base of a truncated right triangular prism has for its sides 13,14,and 15 inches ; its lateral edges are 8, 11,and 13 inches. Find its volume. Ex. 1335. In the formula of "815: (1) put 6 = 0 and compare result with formula of " 805 ; (2) put b = B and compare result with formula of " 799. Ex. 1336. A frustum of a square pyramid has an altitude of 13 inches ; the edges of the bases are 2\| inches and 4 inches,respectively. Find the volume. Ex. 1337. The edges of the bases of a frustum of a square pyramid are 3 inches and 5 inches, respectively, and the volume of the frustum is 204" cubic inches. Find the altitude of the frustum. Ex. 1338. The base of a pyramid contains 144 square inches, and its altitude is 10 inches. A section of the pyramid parallel to the base divides the altitude into two equal parts. Find : (a) the area of the section ; (6) the volume of the frustum formed. Ex. 1339. A section of a pyramid parallel to the base cuts off a pyra-mid similar to the given pyramid. Ex. 1340. The total areas of two similar tetrahedrons are to each other as the squares of any two homologous edges. Ex. 1341. The altitude of a pyramid is 6 inches. A plane parallel to the base cuts the pyramid into two equivalentparts. Find the altitude of the frustum thus formed. Ex. 1342. Two wheat bins are similar in shape ; the one holds 1000 bushels, and the other 800 bushels. If the first is 15 feet deep, how deep is the second ? Ex 1343. A plane is passed parallel to the base of a pyramid cut-ting the altitude into two equal parts. Find : (a) the ratio of the section to the base ; (") the ratio of the pyramid cut off to the whole pyramid. BOOK VII 381 MISCELLANEOUS EXERCISES Ex. 1344. Find the locus of all points equidistantfrom the three edges of a trihedral angle. Ex. 1345. Find the locus of all points equidistantfrom the three faces of a trihedral angle. Ex. 1346. (a) Find the ratio of the volumes and the ratio of the total areas of two similar tetrahedrons whose homologous edges are in the ratio of 2 to 5. (6) Find the ratio of their homologous edges and the ratio of their total areas if their volumes are in the ratio of 1 to 27. Ex. 1347. (a) Construct three or more equivalentpyramids on the same base. (6) Find the locus of the vertices of all pyramids equivalent to a given pyramid and standing on the same base. Hint. Compare with Exs. 821 and 822. Ex. 1348. The altitude of a pyramid is 12 inches. Its base is a regularhexagon whose side is 5 inches. Find the area of a section paral-lel to the base and 4 inches from the base ; 4 inches from the vertex. Ex. 1349. A farmer has a corncrib 20 feet long, a cross section of which is representedin the figure, the numbers denoting feet. If the crib is entirely filled with corn in the ear, how many bushels of corn will it contain,counting 2 bushels of corn in the ear for 1 bushel of shelled corn. (Use the approxi-mation, 1 bushel = 1\ cubic feet. For the exact volume of a bushel, see Ex. 1439.) Ex. 1350. A wheat elevator in the form of a frus-tum of a square pyramid is 30 feet high ; the edges of its bases are 12 feet and 6 feet, respectively. How many bushels of wheat will it hold ? (Use the approximation given in Ex. 1349.) Ex. 1351. A frustum of a regularsquare pyramid has an altitude of 12 inches,and the edges of its bases are 4 inches and 10 inches, respec-tively. Find the volume of the pyramid of which the frustum is a part. Ex. 1352. In a frustum of a regularquadrangular pyramid, the sides of the bases are 10 and 6, respectively, and the slant height is 14. Find the volume. Ex. 1353. Find the lateral area of a regular triangularpyramid whose altitude is 8 inches,and each side of whose base is 6 inches. Ex. 1354. The edge of a cube is a. Find the edge of a cube 3 times as large; n times as large. 382 SOLID GEOMETRY Ex. 1355. A berry box supposed to contain a quart of berries is in the form of a frustum of a pyramid 5 inches square at the top, 4\| inches square at the bottom, and 2$inches deep. The United States djryquart contains 67.2 cubic inches. Does the box contain more or less than a quart ? Ex. 1356. The space left in a basement for a coal bin is a rectangle 8 x 10 feet. How deep must the bin be made to hold 10 tons of coal ? Ex. 1357. The figurerepresents a barn, the numbers denoting the dimensions in feet. Find the number of cubic feet in the barn. Ex. 1358. Let AB, BC, and BD, the dimensions of the barn in Ex. 1357, be denoted by a, 6, and c, respectively. Substitute the values of a, 6, and c in Ahmes' formula given in " 777. Compare your result with the result obtained in Ex. 1357. Would Ahmes' formula have been correct if the Egyptian barns had been similar in shape to the barn in Ex. 1357? Ex. 1359. How much will it cost to paint the barn in Ex. 1357 at 1 cent per square foot for lateral surfaces and 2 cents per square foot for the roof ? Ex. 1360. The barn in Ex. 1357 has a stone foundation 18 inches wide and 3 feet deep. Find the number of cubic feet of masonry if the outer surfaces of the walls are in the same planes as the sides of the barn. Ex. 1361. The volume of a regular tetrahedron is \6-V2.Find its edge, slant height,and altitude. Ex. 1362. The edge of a regular octahedron is a. Prove that the volume equals" V2. Ex. 1363. The planes determined by the diagonals of a parallelo-piped divide the parallelopiped into six equivalentpyramids. Ex. 1364. A dam across a stream is 40 feet long, 12 feet high, 7 feet wide at the bottom, and 4 feet wide at the top. How many cubic feet of material are there in the dam ? how many loads,counting 1 cubic yard to a load ? Give the name of the geometricalsolid representedby the dam. Ex. 1365. Given S the lateral area, and H the altitude, of a regular square pyramid, find the volume. Ex. 1366. Find the volume V, of a regular square pyramid, if its total surface is T, and one edge of the base is a. BOOK VIII CYLINDERS AND CONES CYLINDERS 819. Def. A cylindrical surface is a surface generatedby a moving straight line that continually intersects a fixed curve and remains parallel to a fixed straight line not coplanarwith the given curve. CD K Fig. 1. CylindricalSurface Fig. 2. Cylinder 820. Defs. By referring to "" 693 and 694, the student may give the definitions of generatrix,directrix,and element of a cylindrical surface. Point these out in the figure. The student should note that by changing the directrix from a broken line to a curved line, a prismatic surface becomes a cylindrical surface. 821. Def. A cylinder is a solid closed figure whose boun-dary consists of a cylindrical surface and two parallel planes cutting the generatrix in each of its positions, as DC. 383 384 SOLID GEOMETRY 822. Defs. The two parallel plane sections are called the bases of the cylinder, as AC and DF (Fig. 4) ; the portion of the cylindrical surface between the bases is the lateral -surface of the cylinder ; and the portion of an element of the cylin-drical surface included between the bases is an element of the cylinder, as MN. 823. Def . A right cylinder is a cylinder whose elements are perpendicularto the bases. Fig. 3. Right Cylinder Fig. 4. Oblique Cylinder 824. Def. An oblique cylinder is a cylinder whose elements are not perpendicular to the bases. 825. Def. The altitude of a cylinder is the perpendicular from any point in the plane of one base to the plane of the other base,as HE in Figs.3 and 4. 826. The following are some of the properties of a cylinder ; the student should prove the correctness of each : (a)Any two elements of a cylinder are parallel and equal ("" 618 and 634). (b)Any element of a right cylinder is equal to its altitude. (c)A line drawn through any point in the lateral surfaceof a cylinder parallel to an element, and limited. by the bases, is itself an element ("" 822 and 179). 386 SOLID GEOMETRY Ex. 1367. Every section of a cylinder made by a plane parallel to its base is a circle, if the base is a circle. Ex. 1368. If a line joins the centers of the bases of a cylinder, this line passes through the center of every section of the cylinder parallelto the bases, if the bases are circles. ' 830. Def. A right section of a cylinder is a section formed by a planeperpendicular to an element,as section EF. Fig. 1. Cylinder with Circular Base AB WB Fig. 2. Circular Cylinder Fig. 3. Right Cir-cular Cylinder 831. Def. A circular cylinder is a cylinder in which a right section is a circle ; thus, in Fig. 2, if rt. section EF is a ", cylinderAD is a circular cylinder. 832. Def. A right circular cylinder is a right cylinder whose base is a circle (Fig. 3). 833. Questions. In Fig. 1, is rt. section EF a O ? In Fig. 2, is base AB a O ? In Fig.3, would a rt. section be a O ? 834. Note. The theorems and exercises on the cylinderthat follow will be limited to cases in which the bases of the cylindersare circles. When the term cylinder is used, therefore,it must be understood to mean a cylinder with circular bases. See also " 846. Ex. 1369. Find the locus of all points at a distance of 6 inches from a straight line 2 feet long. Ex. 1370. Find the locus of all points: (a)2 inches from the lateral surface of a right circular cylinderwhose altitude is 12 inches and the radius of whose base is 5 inches ; (") 2 inches from the entire surface. Ex. 1371. A log is 20 feet long and 30 inches in diameter at the smaller end. Find the dimensions of the largest piece of square timber, the same size at each end, that can be cut from the log. BOOK VIII Proposition II. Theorem 387 835. Every section of a cylindermade by a plane pass-ing through an element is a parallelogram. (See " 834.) E Given cylinder AB with base AK, and CDEF a section made by a plane through element CF and some point, as D, not in CF, but in the circumference of the base. To prove CDEF a O. Argument 1. Through D draw a line in plane DF IICF. 2. Then the line so drawn is an element; i.e. it lies in the cylindrical surface. 3. But this line lies also in plane DF. 4. Reasons " 179. " 826, c. Arg. 1. "614. it is the intersection of plane DF with the cylindrical surface, and coin-cides with DE. 5. .'. DE is a str. line and is IIand= CF. 5. " 826, a. 6. Also CD and EF are str. lines. 6. " 616. 7. .-.CDEF is a O. " q.e.d. 7. "240. 836. Cor. Every section of a right circular cyliivder made by a plane passing through an element is a rec-tangle. " Ex. 1372. In the figure of Prop. II,the radius of the base is 4 inches, element CF is 12 inches, CD is 1 inch from the center of the base, and CF makes with CD an angle of 60". Find the area of section CDEF. Ex. 1373. Every section of a cylinder, parallel to an element, is a parallelogram. How is the base of this cylinderrestricted ? (See " 834.) 388 SOLID GEOMETRY CONES Conical Surface 837. Def. A conical surface is a surface generated by a moving straightline that continually intersects a fixed curve and that passes through a fixed point not in the plane of the curve. 838. Defs. By referring to "" 693, 694, and 746, the stu-dent may give the definitions of generatrix, directrix,vertex, element, and upper and lower nappes of a conical surface. Point these out in the figure. The student should observe that by changing the directrix from a broken line to a curved line, a pyramidal surface becomes a conical surface. 839. Def. A cone is a solid closed figure whose boundary consists of the portion of a conical surface extending from its vertex to a plane cutting all its elements, and the section formed by this plane. 840. Defs. By referringto "" 748 and 822, the student may give the definitions of vertex, base, lateral surface, and element of a cone. Point these out in the figure. 841. Def. A circular cone is a cone containinga circular section such that a line joining the vertex of the cone to the center of the section is perpendicular to the section. Thus in Fig.4,if section AB of cone V-CD is a O with center 0, such that vo is J_ the section, cone V-CD is a circular cone. 842. Def. The altitude of a cone is the perpendicularfrom its vertex to the plane of its base, as VC in Fig. 3 and VO in Fig.5\ Fio. 2. Cone BOOK VIII 389 843. Defs. In a cone with a circular base,if the line join-ing its vertex to the center of its base is perpendicular to the Fig. 3. Cone with Circular Base Fig. 4. Circular Cone V Fig. 5. Right Circu-lar Cone plane of the base, the cone is a right cir-cular cone (Fig. 5). If such a line is not perpendicularto the plane of the base, the cone is called an oblique cone (Fig. 3). 844. Def. The axis of a right circular cone is the line joiningits vertex to the center of its base, as vo, Fig. 5. 845. The following are some of the propertiesof a cone; the student should prove the correct-ness of each : (a) The elements of a right circular cone are equal. (b) TJie axis of a right circular cone is equal to its altitude. (c)A straight line drawn from the vertex of a cone to any point in the perimeter of its base is an element. 846. Note. The theorems and exercises on the cone that follow will be limited to cases in which the bases of the cones are circles, though not necessarily to circular cones. When the term cone is used, therefore, it must be understood to mean a cone with circular base. See also " 834. Ex. 1374. What is the locus of all points 2 inches from the lateral surface, and 2 inches from the base, of a right circular cone whose alti-tudes 12 inches and the radius of whose base is 5 inches ? 390 SOLID GEOMETRY Proposition III. Theorem 847. Every section of a cone made by a plane passing through its vertex is a triangle. V Given cone V-AB with base AB and section VCD made by a plane through V. To prove VCD a A. Argument From V draw str. lines to C and D. Then the lines so drawn are elements ; i.e. they lie in the conical surface. But these lines lie also in plane VCD. .-. they are the intersections of plane VCD with the conical surface, and coincide with VC and VD, respectively. Also CD is a str. line. .. VC, VD, and CD are str. lines and VCD is a A. q.e.d. Reasons 1. " 54, 15. 2. " 845, c. ' 3. " 603, a. 4. "614. 5. " 616. 6. "92. Ex. 1375. What kind of trianglein general is the section of a cone through the vertex, if the cone is oblique? if the cone is a rightcircular cone ? Can any section of an oblique cone be perpendicular to the base of the cone? of a rightcircular cone ? Explain. Ex. 1376. Find the locus of all straight lines making a given angle with a given straightline,at a given point in the line. What will this locus be if the given angle is 90" ? Ex. 1377. Find the locus of all straightlines making a given angle with a given plane at a given point. What will the locus be if the given angle is 90" ? BOOK VIII 391 Proposition IV. Theorem 848. Every section of a cone made by a plane parallel to its base is a circle. (See " 846.) V Given CD a section of cone V-AB made by a planeII base AB. To prove section CD a O. Outline of Proof 1. Let R and S be any two points on the boundary of section CD-, pass planes through OF and points R and S. 2. Prove A VOM ~ A VPR and A VON ~ A VPS. 3. Then "". VO A ON " and " VO OM i.e. " = " ON PR VP PS VP ' PR PS 4. But OM " ON-, .-. PS = PR-, i.e. P is equidistant from any two points on the boundary of section CD. 5. .-. section CD is a O. q.e.d. 849. Cor. Any section of a cone parallel to its base is to the base as the square of its distance from the vertex is to the square of the altitude of the cone. V By " 563, Prove Then Outline of Proof section CD PR base AB VF2 "F$r applications of "" 848 and 849,see Exs. 1380-1384, 392 SOLID GEOMETRY MENSURATION OF THE CYLINDER AND CONE Areas 850. Def. A plane is tangent to a cylinder ifit contains an element,but no other point, of the cylinder. 851. Def. A prism is inscribed in a cylinder if its lateral edges are elements of the cylinder, and the bases of the two figures lie in the same plane. 852. Def. A prism is circumscribed about a cylinder if its lateral faces are all tangent to the cylinder, and the bases of the two figures lie in the same plane. Ex. 1378. How many planes can be tangent to a cylinder? If -two of these planesintersect, the line of intersection is parallel to an element. How are the bases of the cylindersin "" 850-852 restricted ? (See " 834.) 853. Before proceeding further it might be well for the student to review the more important steps in the develop-ment of the area of a circle. In that development it was shown that : (1) The area of a regular polygon circumscribed about a circle is greater,and the area of a regularpolygon inscribed in a circle less, than the area of the regularcircumscribed or in-scribed polygon of twice as many sides (" 541). (2) By repeatedlydoubling the number of sides of regular circumscribed and inscribed polygons of the same number of sides,and making the polygons always regular,their areas approach a common limit (" 546). (3)This common limit is defined as the area of the circle ("558). (4) Finallyfollows the theorem for the area of the circle (" 559). It will be observed that precisely the same method is used throughout the mensuration of the cylinder and the cone. Compare carefully the four articles just cited with "" 854, 855,857, and 858. 394 SOLID GEOMETRY Proposition VI. Theorem 855. By repeatedly doubling the number of sides of the bases of regular prisms circumscribed about, and in-scribed in, a rightcircular cylinder, and making the bases always regular polygons, their lateral areas approach a common limit. Given H the common attitude, R and r the apothems of the bases, P and p the perimetersof the bases,and S and s the lateral areas, respectively, of regular circumscribed and in-scribed prisms whose bases have the same number of sides. Let the given figurerepresent the base of the actual figure. To prove that by repeatedly doubling the number of sides of the bases of the prisms,and making them always regular polygons,S and s approach a common limit. Reasons " 763. 2. " 54, 8 a. 3. " 538. 4. " 54, 1. 5. " 399. BOOK VIII 395 Argument 6. 9. s" s = s But by repeatedly doubling the num-ber of sides of the bases of the prisms, and making them always regularpolygons, R " r can be made less than any previously assigned value,however small. .. can be made less than any previously assigned value, however small. R s R can be made less than any previously assigned value, however small,S being a decreasingvariable. 10. .-. S " s, being always equal to S ~ , R can be made less than any previously assigned value,however small. 11. .-. S and s approach a common limit. Q.E.D. Reasons 6. " 54, 7 a. 7. " 543, I. 8. " 586. 9. " 587. 10. "309. 11. " 594. 856. Note. The above proof is limited to regular prisms, but it can be shown that the limit of the lateral area of any inscribed (or circum-scribed) prism is the same by whatever method the number of the sides of its base is successively increased,provided that each side approaches zero as a limit. (See also " 549.) Compare the proof of " 855 with that of " 546, I. 857. Def. The lateral area of a right circular cylinder is the common limit which the successive lateral areas of circum-scribed and inscribed regularprisms (havingbases containing 3, 4, 5, etc.,sides)approach as the number of sides of the bases is successively increased and each side approaches zero as a limit. 396 SOLID GEOMETRY Proposition VII. Theorem 858. The lateral area of a right circular cylinder is equal to the product of the circumference of its base and its altitude. Given a rt. circular cylinderwith its lateral area denoted by S, the circumference of its base by C, and its altitude by //. To prove S = C " H. Argument Circumscribe about the rt. circular cylinder a regular prism. Denote its lateral area by S',the perimeter of its base by P, and its altitude by h. Then S' = P " H. As the number of sides of the base of the regular circumscribed prism is repeatedlydoubled, P approaches O as a limit. .-. P " H approaches C " H as a limit. Also S' approaches S as a limit. But s' is always equal to P " H. .'. S=C " H. Q.E.D. Reasons 1. " 852. 2. " 763. 3. " 550. 4. " 590. 5. " 857. 6. Arg. 2. 7. " 355. 859. Cor. If S denotes the lateral area, T the total area, II the altitude, and B the radius of the base, of a right circular cylinder, BOOK VIII 397 S=2ttRH', T= 2 ttRH + 2ttR2 = 2 ttR{H+ r). 860. Note. Since the lateral area of an oblique prism is equal to the product of the perimeter of a right section and a lateral edge (" 762), the student would naturally infer that the lateral area of an oblique cylinder with circular bases is equal to the product of the perimeter of a rightsec-tion and an element. This statement is true. But the right section of an obliquecylinderwith circular base is not a circle. And since the only curve dealt with in elementary geometry is the circle,this theorem and its applications have been omitted here. Ex. 1385. Find the lateral area and total area of a right circular cylinderwhose altitude is 20 centimeters and the diameter of whose base is 10 centimeters. Ex. 1386. How many square inches of tin will be required to make an open cylindrical pail 8 inches in diameter and 10 inches deep, making no allowance for waste ? Ex. 1387. In a right circular cylinder, find the ratio of the lateral area to the sum of the two bases. What is this ratio if the altitude and the radius of base are equal ? Ex. 1388. Find the altitude of a right circular cylinderif its lateral area is S and the radius of its base B. Ex. 1389. Find the radius of the base of a right circular cylinderif its total area is T and its altitude is H. 861. Def. Because it may be generated by a rectanglere-volving about one of its sides as an axis, a right circular cylinder is sometimes called a cylinder of rev-olution. 862. Questions. What part of the cylinder will side CD, opposite the axis AB, generate? What will AD and BC generate ? What will the plane AC gen-erate ? What might CD in any one of its positionsbe called ? f^ZZ^wT"^ 863. Def. Similar cylinders of revolution are cylinders generated by similar rectangles revolvingabout homologous sides as axes. 398 SOLID GEOMETRY Proposition VIII. Theorem 864. The lateral areas, and the total areas, of two similar cylinders of revolution are to each other as the squares of their altitudes, and as the squares of the radii of their bases. H Given two similar cylinders of revolution with their lateral areas denoted by S and S1, their total areas by T and T',their altitudes by Hand H1, and the radii of their bases by i? and R', respectively. , N 8 H2 R2 To prove : (a) " = " -= " -" S' IV T H12 R12 Reasons 1. " 859. 2. " 54,8 a. 3. " 863. 4. " 419. 5. " 309. 6. " 309. 7. " 859. BOOK VIII 399 Argument " 2L " R(H+R) . J. . H+R " T'~ R\H' + R') ~ R' ' H' + R' 9. But, from Arg. 4, H + R II H' + R' Hf R! 10. 11. Also "=R H_ t H^__H^_ B ' H'~ H'2' R! Rf R" Q.E.D. Reasons 8. " 54,8 a. 9. " 401. 10. " 309. 11. " 309. Ex. 1390. The altitudes of two similar cylinders of revolution are 5 inches and 7 inches,respectively, and the total area of the first is 676 square inches. Find the total area of the second. Ex. 1391. The lateral areas of two similar cylinders of revolution are 320 square inches and 500 square inches,and the radius of the base of the largeris 10 inches. Find the radius of the base of the smaller. Ex. 1392. Two adjacent sides of a rectangle are a and b; find the lateral area of the cylindergenerated by revolvingthe rectangle : (1) about a. as an axis ; (2) about b as an axis. Put the results in the form of a general statement. Have you proved this general statement ? 865. Def. The slant height of a right circular cone is a straight line joining its vertex to any pointin the circumference of its base. Thus any element of such a cone is its slant height. 866. Def. A plane is tangent to a cone if it contains an element, but no other point, of the cone. 867. Def. A pyramid is inscribed in a cone if its base is inscribed in the base of the cone and its vertex coincides with the vertex of the cone. 868. Def. A pyramid is circumscribed about a cone if its base is circumscribed about the base of the cone and its vertex coincides with the vertex of the cone. 869. The student may state and prove the theorems on the rightcircular cone correspondingto those mentioned in " 854. Ex. 1393. How many planes can be tangent to a cone? Through what point must each of these planes pass ? Prove. How are the bases of th cones in "" 866-868 restricted ? (See " 846.) 400 SOLID GEOMETRY Proposition IX. Theorem 870. By repeatedlydoubling the number of sides of the bases of regular pyramids circumscribed about, and in-scribed in, a right circular cone, and making the bases always regular polygons, tl%eir lateral areas approach a common limit. Given H the common altitude, L and I the slant heights, R and r the apothems of the bases, P and p the perimetersof the bases,and S and s the lateral areas, respectively, of regular circumscribed and inscribed pyramids whose bases have the same number of sides. Let the given figurerepresent the base of the actual figure. To prove that by repeatedlydoubling the number of sides of the bases of the pyramids,and making them always regular polygons, S and s approach a common limit. Reasons 1. " 766. 2. " 54, 8 a. 3. " 538. 4. " 309. 5. " 399. RL 402 SOLID GEOMETRY Proposition X. Theorem 873. The lateral area of a right circular cone is equal to one half the product of the circumference of its base and its slant height. Given a rt. circular cone with its lateral area -denoted by S, the circumference of its base by C, and its slant height by L. To prove S = \ C " L. The proof is left as an exercise for the student. 874. Question. What changes are necessary in the proof of Prop. VII to make it the proof of Prop. X ? 875. Cor. If s denotes the lateral area, T the total area, L the slant height, and B the radius of the base, of a right circular cone, s = ttRL ; T = ttRL + 7T i?2 = tR(L + #). Ex. 1394. The altitude of a right circular cone is 12 inches and the radius of the base 8 inches. Find the lateral area and the total area of the cone. Ex. 1395. How many yards of canvas 30 inches wide will be required to make a conical tent 16 feet high and 20 feet in diameter, if 10% of the goods is allowed for cuttingand fitting ? Ex. 1396. The lateral area of a rightcircular cone is ^4o vgg SqUare inches,and the radius of the base is 10 inches. Find the altitude. 876. Def. Because it may be generatedby a right triangle revolvingabout one of its sides as an axis, a rightcircular cone is sometimes called a cone of revolution. 877. Def. Similar cones of revolution are cones generatedby similar right triangles revolving about homologous sides as axes. BOOK VIII 403 Proposition XI. Theorem 878. The lateral areas, and the total areas, of two simi-lar cones of revolution are to each other as the squares of their altitudes, as the squares of their slant heights, and as the squares of the radii of their bases. Given two similar cones of revolution with their lateral areas denoted by S and S', their total areas by T and T1, their altitudes by H and H\ their slant heights by L and Lf,and the radii of their bases by R and R',respectively. , N S H2 L2 R2 To prove : (a) " = " -= " = " . V / of rW2 t/2 t"I2 S' ir \ ) Tl H" Lf% RI2' The proof is left as an exercise for the student. Hint. Apply the method of proof used in Prop. VIII. 879. Def. A frustum of a cone is the portion of the cone included between the base and a section of the cone made by a plane parallel to the base. 880. Questions. What are the upper and lower bases of a frustum of a cone ? the altitude ? What kind of a figureis the upper base of a frustum of a rightcircular cone (" 848) ? 881. Def. The slant height of a frustum of a rightcircular cone is the length of that portion of an element of the cone included between the bases of the frustum. Ex. 1397. Every section of a frustum of a cone, made hy a plane passing through an element, is a trapezoid. 404 SOLID GEOMETRY Proposition XII. Theorem 882. The lateral area of a frustum of a right circular cone is equal to one half the product of the sum of the circumferences of its bases and its slant height. 0 Given frustum AM, of right circular cone lateral area denoted by S, the circumferences C and c, the radii of its bases by R and r, and byz. To prove S = i(C+c)L. Argument 1.5 = lateral area of cone O-AF minus lateral area of cone O-DM. 2. Let V denote the slant heightof cone 0 -DM. Then S="c(L + L') "CL' = $CL + \|2/(C-c). It now remains to find the value of l! . c r But A OKD ~ A OQA. " R_OA_L + L' r~ OD~ L . c L--L' c L': L' CL C-c S = "CL + " cL C"c ((7-c)=l(C+c)X. Q.E.D. O-AF, with its of its bases by its slant height Reasons . " 54, 11. 2. " 873. 3. " 556. "422. " 424, 2. " 54, 1. Solving for L\ " 309. BOOK VIII 405 883. Cor. I. If s denotes the lateral area, T the total area, L the slant height, and R and r the radii of the bases, of a frustum, of a right circular cone, S = ttL(R + r) ; T = ttL(R + r)+ tt^R2 + r2). 884. Cor. n. The lateral area of a frustum of a right circular cone is equal to the product of its slant height and the circumference of a section midway between its bases. Ex. 1398. If 8 denotes the lateral area, L the slant height,and C the circumference of a section midway between the bases,of a frustum of a rightcircular cone, then S = CL. Ex. 1399. In the formulas of " 883 : (a) make r= 0 and compare results with formulas of " 875 ; (") make b = B and compare results with formulas of " 859. Ex. 1400. The altitude of a frustum of a right circular cone is 16 inches, and the diameters of its bases are 20 inches and 30 inches, respectively. Find its lateral area and also its total area. Ex. 1401. In the figure,AB and CD are arcs of circles ; OA = 2 inches, OB = 5 inches, and ZBOC= 120". Cut figureABCB out of paper and form it into a frustum of a cone. Find its lateral area and also its total area. Ex. 1402. A frustum of a rightcircular cone whose altitude is 4 inches and radii of bases 4 inches and 7 inches, respectively, is made as in-dicated in Ex. 1401. Find the radius of the circle from which it must be cut. Ex. 1403. The sum of the total areas of two similar cylindersof rev-olution is 216 square inches,and one altitude is f of the other. Find the total area of each cylinder. Ex. 1404. A regular triangularand a regular hexangular pyramid are inscribed in a right circular cone with altitude 20 inches and with radius of base 4 inches. Find the difference between their lateral areas. Ex. 1405. Cut out of paper a semicircle whose radius is 4 inches, and find its area. Form a cone with this semicircle and find its lateral area by " 875. Do the two results agree ? Ex. 1406. The slant height,and the diameter of the base,of a right circular cone are each equal to L. Find the total area. 406 SOLID GEOMETRY VOLUMES Proposition XIII. Theorem 885. I. The volume of a prism whose base is a regular polygon and which is circumscribed about a cylinder is greater than the volume of the circumscribed prism whose, base is a regular polygon with twice as many sides. II. The volume of a prism whose base is a regular poly-gon and which is inscribed in a cylinder is less than the volume of the inscribed prism whose base is a regular polygon with twice as many sides. The figures and proofs are left as exercises for the student. Proposition XIV. Theorem 886. By repeatedly doubling the number of sides of the bases of prisms circumscribed about, and inscribed in, a cylinder,and making the bases always regular poly-gons, their volumes approach a common limit. Given H the common altitude, B and b the areas of the bases,and V and v the volumes, respectively, of circumscribed and inscribed prisms whose bases are regular and have the same number of sides. Let the given figure represent the base of the actual figure. BOOK VIII 40T To prove that by repeatedly doubling the number of sides of the bases of the prisms,and making them always regularpoly-gons, V and v approach a common limit. Argument V= B " H and v = b - H. . v "B ' H " B " v~ b " H~ b' V " v B " b V V"V = V 5. Similar to Arg. 7, " 855. 6. Similar to Arg. 8, " 855. 7. Similar to Arg. 9, " 855. 8. Similar to Arg. 10, " 855. 9. .. V and v approach a common limit. Q.E.D. 887. Note. The proof of " 886 is limited in the same manner as the proof of " 855. Read " 856. 1. Reasons " 799. 2. " 54,8 a. 3. " 399. 4. " 54,7 a. 5. " 546, II. 6. " 586. 7. " 587. 8. " 309. 9. " 594. Ex. 1407. The total area of a right circular cone whose altitude is 10 inches is 280 square inches. Find the total area of the cone cut off hy a plane parallel to the base and 6 inches from the base. Ex. 1408. The altitude of a right circular cone is 12 inches. What part of the lateral surface is cut off by a plane parallel to the base and 6 inches from the vertex ? Ex. 1409 . The altitude of a rightcircular cone is II. How far from the vertex must a plane be passed parallel to the base so that the lateral area and the total area of the cone cut off shall be one half that of the original cone ? one third ? one nth ? 888. Def. The volume of a cylinder is the common limit which the successive volumes of circumscribed and inscribed prisms approach as the number of sides of the bases is suc-cessively increased, and each side approaches zero as a limit. 408 SOLID GEOMETRY Proposition XV. Theorem The volume of a cylinder is equal to the product of its base and its altitude. Given a cylinder, with its volume denoted by V, its base by B, and its altitude by H. To prove V=B " H. Argument 1. Circumscribe about the cylinder a prism whose base is a regularpoly-gon. Denote its volume by V and the area of its base by B'. 2. Then V' = B' . H. 3. As the number of sides of the base of the circumscribed prism is re-peatedly doubled,B' approaches B as a limit. 4. .-. B' " H approachesB " H as a limit. 5. Also v' approaches V as a limit. 6. But Vf is always equal to B' " H. 7. .. V"B " H. Q.E.D. Reasons 1. " 852. 2. " 799. 3. "558. 4. " 590. 5. " 888. 6. Arg. 2. 7. " 355. 890. Cor. // v denotes the volume, H the altitude, and R the radius of the base, of a cylinder, V=R2H. 410 SOLID GEOMETRY Ex. 1411. A cylinder of revolution has an altitude of 12 inches and a base with a radius of 5 inches. Find the total area of a similar cylinder whose volume is 8 times that of the given cylinder. Ex. 1412. The dimensions of a rectangle are 6 inches and 8 inches, respectively. Find the volume of the solid generated by revolving the rectangle : (a) about its longer side as an axis ; (b) about its shorter side. Compare the ratio of these volumes with the ratio of the sides of the rectangle. Ex. 1413. Cylinders having equal bases and equal altitudes are equivalent. Ex. 1414. Any two cylinders are to each other as the products of their bases and their altitudes. Ex. 1415. (a) Two cylinders having equal bases are to each other as their altitudes, and (6) having equal altitudes are to each other as their bases. Ex. 1416. The volume of a right circular cylinder is equal to the product of its lateral area and one half the radius of its base. Ex. 1417. Cut out a rectangular piece of paper 6x9 inches. Roll this into a rightcircular cylinder and find its volume (two answers). Ex. 1418. A cistern in the form of a right circular cylinder is to be 20 feet deep and 8 feet in diameter. How much will it cost to dig it at 5 cents a cubic foot ? Ex. 1419. Find the altitude of a right circular cylinder if its volume is V and the radius of its base R. Ex. 1420. In a certain right circular cylinder the lateral area and the volume have the same numerical value, (a) Find the radius of the base. (6) Find the volume if the altitude is equal to the diameter of the base. Ex. 1421. A cylinder is inscribed in a cube whose edge is 10 inches. Find : (a) the volume of each ; (ft)the ratio of the cylinderto the cube. Ex. 1422. A cylindrical tin tomato can is 4T% inches high, and the diameter of its base is 4 inches. Does it hold more or less than a liquid quart, i.e. f cubic inches ? 892. The student may : (a) State and prove the theorems on the cone corresponding to those given in "" 885 and 886. (6) State,by aid of " 888, the definition of the volume of a BOOK VIII 411 Proposition XVII. Theorem 893. The volume of a cone is equal to one third the product of its base and its altitude. Given a cone with its volume denoted by V, its base by B, and its altitude by H. To prove V=\B " H. The proofis left as an exercise for the student. 894. Question. What changes must be made in the proof of Prop. XV to make it the proof of Prop. XVII ? 895. Cor. If v denotes the volume, H the altitude, and R the radius of the base of a cone, V=\ttEH. Ex. 1423. Any two cones are to each other as the products of their bases and altitudes. Ex. 1424. The slant height of a right circular cone is 18 inches and makes with the base an angle of 60" ; the radius of the base is 8 inches. Find the volume of the cone. Ex. 1425. The base of a cone has a radius of 12 inches ; an element of the cone is 24 inches long and makes with the base an angle of 30". Find the volume of the cone. Ex. 1426. The hypotenuse of a right triangleis 17 inches and one side is 15 inches. Find the volume of the solid generated by revolvingthe triangle about its shortest side as an axis. Ex. 1427. A cone and a cylinder have equal bases and equal alti-tudes. Find the ratio of their volumes. 896. Historical Note. To Eudoxus is credited the proof of the proposition that "every cone is the third part of a cylinderon the same bage and with the same altitude." Proofs of this proposition were also given later by Archimedes and Brahmagupta. (Compare with " 809.) 412 SOLID GEOMETRY Proposition XVIII. Theorem 897. The volumes of two similar cones of revolution are to each other as the cubes of their altitudes, as the cubes of their slant heights,and as the cubes of the ra-dii of their bases. Given two similar cones of revolution with their volumes denoted by V and V1, their altitudes by H and H\ their slant heights by L and L\ and the radii of their bases by R and R\ respectively. " V H3 L3 R To prove -= _ = _ = _. The proof is left as an exercise for the student. Hint. Apply the method of proof used in Prop. XVI. Ex. 1428. If the altitude of a cone of revolution is three fourths that of a similar cone, what other fact follows by definition ? Compare the circumferences of the two bases ; their areas. Compare the total areas of the two cones ; their volumes. Ex. 1429. If the lateral area of a rightcircular cone is 1T\ times that of a similar cone, what is the ratio of their volumes ? of their altitudes ? Ex. 1430. Through a given cone X two planes are passed parallel to the base ; let Y denote the cone cut off by the upper plane,and Z the entire cone cut off by the lower plane. Prove that Y and Z are to each other as the cubes of the distances of the planes from the vertex of the given cone X. Hint. See Ex. 1381. Ex. 1431. Show that " 897 is a special case of Ex. 1430. Ex. 1432. The lateral area of a cone of revolution is 144 square inches and the total area 240 square inches. Find the volume. BOOK VIII 413 Proposition XIX. Theorem 898. The volume of a frustum of a cone is equal to one third the product of its altitude and the sum of its lower base, its, upper base, and the mean proportional between its two bases. O Given frustum AM, of cone O-AF, with its volume denoted by F, its lower base by B, its upper base by b,and its altitude by H. To prove V=\H(B + b + V^6). The proofis left as an exercise for the student. Hint. In the proof of " 815, change "pyramid" to "cone." 899. Cor. If v denotes the volume, H the altitude, and R and r the radii of the bases of a frustum of a cone, Ex. 1433. Make a frustum of a right circular cone as indicated in Ex. 1401, and of the same dimensions. Find its volume. Ex. 1434. A tin pail is in the form of a frustum of a cone ; the diameter of its upper base is 12 inches, of its lower base 10 inches. How high must the pailbe to hold 2\| gallonsof water? (See Ex. 1422.) Ex. 1435. A cone 6 feet high is cut by a plane parallel to the base and 4 feet from the vertex ; the volume of the frustum formed is 456 cubic inches. Find the volume of the entire cone. . Ex. 1436. Find the ratio of the base to the lateral area of a right circular cone whose altitude is equal to the diameter of its base. 414 SOLID GEOMETRY MISCELLANEOUS EXERCISES Ex. 1437. The base of a cylinder is inscribed in a face of -a cube whose edge is 10 inches. Find the altitude of the cylinder if its volume is equal to the volume of the cube. ' Ex. 1438. A block of marble in the form of a regularprism is 10 feet long and 2 feet 6 inches square at the base. Find the volume of the largest cylindrical pillar that can be cut from it. Ex. 1439. The Winchester bushel, formerly used in England, was the volume of a right circular cylinder 18\| inches in internal diameter and 8 inches in depth. Is this the same volume as the bushel used in the United States (2150.42 cubic inches) ? Ex. 1440. To determine the volume of an irregularbody, it was immersed in a vessel containing water. The vessel was in the form of a rightcircular cylinderthe radius of whose base was 8 inches. On placing the body in the cylinder, the surface of the water was raised 10\| inches. Find the volume of the irregular solid. Ex. 1441. In draining a certain pond a 4-inch tiling(i.e.a tiling whose inside diameter was 4 inches) was used. In draining another pond, supposed to contain half as much water, a 2-inch tiling was laid. It could not drain the pond. What was the error made ? Ex. 1442. A grain elevator in the form of a frustum of a rightcir-cular cone is 25 feet high ; the radii of its bases are 10 feet and 5 feet, respectively ; how many bushels of wheat will it hold, counting \ cubic feet to a bushel ? Ex. 1443. The altitude of a cone with circular base is 16 inches. At what distance from the vertex must a plane be passed parallel to the base to cut the cone into two equivalentparts ? Ex. 1444. Two sides of a triangleincluding an angle of 120" are 10 and 20, respectively. Find the volume of the solid generated by revolv-ing the triangle about side 10 as an axis. Ex. 1445. Find the volume of the solid generated by revolvingthe triangle of Ex. 1444 about side 20 as an axis. Ex. 1446. Find the volume of the solid generated by revolving the triangle of Ex. 1444 about its longestside as an axis. Ex. 1447. The slant height of a rightcircular cone is 20 inches,and the circumference of its base 4tt inches. A plane parallel to the base cuts off a cone whose slant height is 8 inches. Find the lateral area and the volume of the frustum remaining. BOOK VIII 415 Ex. 1448. A cone has an altitude of 12.5 feet and a base whose radius is 8.16 feet ; the base of a cylinderhaving the same volume as the cone has a radius of 6.25 feet. Find the altitude of the cylinder. Ex. 1449. A log 20 feet long is 3 feet in diameter at the top end and 4 feet in diameter at the butt end. (a) How many cubic feet of wood does the log contain ? (b) How many cubic feet are there in the largestpiece of square timber that can be cut from the log ? (c) How many cubic feet in the largest pieceof square timber the same size throughout its whole length ? (d) How many board feet does the piece of timber in (c) contain,a board foot being equivalent to a board 1 foot square and 1 inch thick ? Hint. In (") the larger end is square ABCD. What is the smaller end ? In (c) one end is square EFGH. What is the other end ? Ex. 1450. The base of a cone has a radius of 16 inches. A section of the cone through the vertex, through the center of the base, and per-pendicular 'to the base, is a triangle two of whose sides are 20 inches and 24 inches, respectively. Find the volume of the cone. Ex. 1451. The hypotenuse of a righttriangleis 10 inches and one side 8 inches ; find the area of the surface generated by revolvingthe tri-angle about its hypotenuse as an axis. Ex. 1452. A tin pail in the form of a frustum of a right circular cone is 8 inches deep ; the diameters of its bases are 8^ inches and 10\| inches, respectively. How many gallons of water will it hold ? (One liquidgallon contains 231 cubic inches.) Ex. 1453. The altitude of a cone is 12 inches. At what distances from the vertex must planes be passed parallel to the base to divide the cone into four equivalentparts ? Hint. See Ex. 1430. Ex. 1454. Find the volume of the solid generated by an equilateral triangle, whose side is a, revolvingabout one of its sides as an axis. Ex. 1455. Regular hexagonal prisms are inscribed in and circum-scribed about a rightcircular cylinder. Find (a) the ratio of the lateral areas of the three solids ; (6) the ratio of their total areas ; (c) the ratio of'tfreir volumes. 416 SOLID GEOMETRY Ex. 1456. How many miles of platinum wire fo of an inch in diam-eter can be made from 1 cubic foot of platinum ? . Ex. 1457. A tank in the form of a right circular cylinder is 5 feet long and the radius of its base is 8 inches. If placed so that its axis is horizontal and filled with gasoline to a depth of 12 inches, how many gallonsof gasolinewill it contain ? Hint. See Ex. 1024. Ex. 1458. Find the weight in pounds of an iron pipe 10 feet long,if the iron is \| inch thick and the outer diameter of the pipe is 4 inches. (1 cubic foot of bar iron weighs 7780 ounces.) Ex. 1459. In a certain right circular cone whose altitude and radius of base are equal,the total surface and the volume have the same numeri-cal value. Find the volume of the cone. Ex. 1460. Two cones of revolution lie on opposite sides of a common base. Their slant heights are 12 and 5, respectively, and the sum of their altitudes is 13. Find the radius of the common base. Ex. 1461. The radii of the lower and upper bases of a frustum of a rightcircular cone are R and R', respectively. Show that the area of a section midway between them is Zl " " L . Ex. 1462. A plane parallel to the base of a rightcircular cone leaves three fourths of the cone's volume. How far from the vertex is this plane? How far from the vertex is the plane if it cuts off half the volume ? Answer the same questionsfor a cylinder. Ex. 1463. Is every cone cut from a rightcircular cone by a plane par-allel to its base necessarily similar to the originalcone ? why ? How is it with a cylinder? why ? Ex. 1464. Water is carried from a spring to a house, a distance of \| mile, in a cylindricalpipe whose inside diameter is 2 inches. How many gallonsof water are contained in the pipe ? Ex. 1465. A square whose side is 6 inches is revolved about one of its diagonals as an axis. Find the surface and the volume of the solid generated. Can you rind the volume of the solid generated by revolving a cube about one of its diagonals as an axis ? Hint. Make a cube of convenient size from pasteboard,pass a hat-pin through two diagonallyopposite vertices, and revolve the cube rapidly. Ex. 1466. Given V the volume, and R the radius of the base, of a rightcircular cylinder. Find the lateral area and total area. Ex. 1467. Given the total area T, and the altitude H, of a right circular cylinder. Find the volume. 418 SOLID GEOMETRY Proposition I. Theorem 903. Every section of a sphere made by a plane is a circle. Given AMBN a section of sphere 0 made by a plane. To prove section AMBN a O. Argument 1. From 0 draw OQ JL section AMBN. 2. Join Q to C and B, any two points on the perimeterof section AMBN. Draw OC and OB. 3. In rt. A OQC and OQB, OQ = OQ. 4. OC " OB. 5. .'. A OQC = A OQB. 6. .'. QC = QB -, i.e. any two points on the perimeterof section AMBN are equi-distant from Q. 7. .'. section AMBN is a O. q.e.d. Reasons 1. " 639. 2. " 54, 15. 3. By iden. 4. " 902, a. 5. "211. 6. " 110. 276. 904. Def. A great circle of a sphere is a section made by a plane S^m "--^ which passes through the center of A the sphere,as O CRBS. c/-"-"" ffjl 905. Def. A small circle of a \^^J" sphere is a section made by a plane \ " which does not pass through the center of the sphere, as O AMBN. BOOK IX 419 906. Def. The axis of a circle of a sphere is the diameter of the sphere which is perpendicular to the plane of the circle. 907. Def. The poles of a circle of a sphere are the extremi-ties of the axis of the circle. Ex. 1472. Considering the earth as a sphere, what kind of circles are the parallels of latitude ? the equator ? the meridian circles ? What is the axis of the equator ? of the parallels of latitude ? What are the polesof the equator ? of the parallels of latitude ? Ex. 1473. The radius of a sphere is 17 inches. Find the area of a section made by a plane 8 inches from the center. Ex. 1474. The area of a section of. a sphere 45 centimeters from the center is 784 ir square centimeters. Find the radius of the sphere. Ex. 1475. The area of a section of a sphere 7 inches from the center is 576 ir. Find the area of a section 8 inches from the center. 908. The following are some of the properties of a sphere ; the student should prove the correctness of each : (a) In equal sjjheres, or in the same sphere,if two sections are equal,theyare equallydistant from the center, and conversely. Hint. Compare with " 307. (b) In equal spheres, or in the same sphere,if two sections are unequal, the greater section is at the less distance from the center, and conversely. (Hint. See ""308, 310.) (c)In equal spheres, or in the same sphere, all great circles are equal. (Hint. See " 279, c.) ' (d) The axis of a small circle of a sphere passes through the center of the circle, and conversely. (e)Any two great circles of a sphere bisect each other. (/) Every great circle of a sphere bisects the surface and the sphere. (g) Through two points on the surfaceof a sphere, not the ex-tremities of a diameter,there exists one and only one great circle. (h) Through three points on the surfaceof a sphere there exists one and only one circle 909. Def. The distance between two points on the surface of a sphere is the length of the minor arc of the great circle joining them. 420 SOLID GEOMETRY Proposition II. Theorem 910. All points on the circumference of a circle of a sphere are equidistantfrom either pole of the circle. =^Q Given C and D any two points on the circumference, and P and R the poles, of O AMBK. To prove arc PC= arc PD and arc RC = arc RD. The proof is left as an exercise for the student. Hint. Apply " 298. 911. Def. The polar distance of a circle of a sphere is the distance between any point on its circumference and the nearer poleof the circle. 912. Cor. I. The polar distance of a great circle is a quadrant. 913. Cor. II. Jn equal spheres,or in the same sphere, the polar distances of equal circles are equal. Ex. 1476. What is the locus of all points on the surface of the earth at a quadrant's distance from the north pole ? from the south pole ? from the equator ? from a point P on the equator ? at a distance of 23 \" from the south pole ? 23\|"from the equator ? 180" from the north pole ? Ex. 1477. Considering the earth as a sphere with a radius of 4000 miles, calculate in miles the polar distance of : (a) the Arctic Circle ; (6) the Tropic of Cancer ; (c)the equator. Ex. 1478. State a postulate for the construction of a circle on the surface of a sphere corresponding to " 122, the postulatefor the construc-tion of a circle in a plane. BOOK IX Proposition III. Theorem 421 914. A -point on the surface of a sphere at the distance of a quadrant from each of tivo other points {not the extrem-ities of the same diameter) on the surface, is the pole of the great circle passing through these two points. Given PC and PD quadrants of great " of sphere 0, and ACDB a great O passingthrough points C and D. To prove P the pole of great O ACDB. Argument Draw OC, OD, and OP. PC = 90" and PD = 90". .'.A POC and POD are rt. A\ i.e. OP J OC and OD .'. OP _L the plane of O ACDB. .-. OP is the axis of O ACDB. .'. P is the pole of great O ACDB. q.e.d. Reasons 1. " 54, 15. 2. By hyp 3. " 358. 4. " 622. 5. " 906. 6. " 907. Ex. 1479. Assuming the chord of a quadrant of a great circle of a sphere to be given, construct with compasses an arc of a great circle through two given points on the surface of the sphere. Ez. 1480. If the planesof two great circles are perpendicularto each other, each passes through the poles of the other. Ex. 1481. Find the locus of all pointsin space equidistant from two given points and at a given distance d from a third given point. Ex. 1482. Find the locus of all points in space at a distance d from a given point and at a distance m from a given plane. Ex. 1483. Find the locus of all points in space equidistant from two givenpoints and also equidistant from two given parallel lines. 422 SOLID GEOMETRY Proposition IV. Theorem 915. The intersection of two spherical surfaces is the circwrriference of a circle. Given two sphericalsurfaces generated by intersecting cir-cumferen 0 and Q revolvingabout line MKf as an axis. To prove the intersection of the two spherical surfaces the circumference of a 0. Outline op Proof 1. Show that MN _L AB at its mid-point 0 ("328). 2. Show that AC, revolvingabout axis MN, generates a plane. 3. Show that A generates the circumference of a O. 4. The locus of A is the intersection of what (" 614)? 5. .. the intersection of the two spherical surfaces is the cir-cumferen of a O. Q.E.D. Ex. 1484. Find the locus of all points in space 6 inches from a given point P and 10 inches from another given point Q. Ex. 1485. The radii of two intersecting spheres are 12 inches and 16 inches,respectively. The line joiningtheir centers is 24 inches. Find the circumference and area of their circle of intersection. 916. Def. An angle formed by two intersecting arcs of circles is the angle formed by tangents to the two arcs at their point of intersection;thus the Z formed A by arcs AB and AC is plane A DAE. \sT\~ " " "^-F 917. Def. A spherical angle is an angle formed by two intersecting arcs of great circles of a sphere. g A different meaning is sometimes attached to the expression "spherical angle" in advanced mathematics. BOOK IX Proposition V. Theorem 423 918. A spherical angle is measured by the arc of a great circle having the vertex of the angle as a pole and interceptedby the sides of the angle, prolonged if nec-essary. P Given spherical Z APB, with CD an arc of a great O whose pole is P and which is intercepted by sides PA and PB of Z APB. To prove that Z APB oc CD. Outline of Proof 1. Draw radii OP, OC, and OB. 2. From P draw PR tangent to PA and PT tangent to PB. 3. Prove OC and OD each _L OP. 4. Prove OC IIPR, OD IIPT, and hence Z COD = Z RPT. 5. But Z COD oc CD; .. Z iZPT, i.e. Z ARB QC CD. Q.E.D. 919. Cor. I. A spherical angle is equal to the plane angle of the dihedral angle formed by the planes of the sides of the angle. 920. Cor. II. The sum of all the spherical angles about a point on the surface of a sphere equals four right angles. Ex. 1486. By comparison with the definitions of the corresponding terms in plane geometry, frame exact definitions of the followingclasses of spherical angles : acute, right, obtuse, adjacent,complementary, supple-mentary, vertical. Ex. 1487. Any two vertical sphericalangles are equal. Ex. 1488. If one great circle passes through the pole of another great-circle, the circles are perpendicularto each other. 424 SOLID GEOMETRY LINES AND PLANES TANGENT TO A SPHERE 921. Def. A straight line or a plane is tangent to a sphere if, however far extended,it meets the spherein one and only one point. 922. Def. Two spheres are tangent to each other if they have one and only one point in common. They are tangent internally if one sphere lies within the other,and externally if neither sphere lies within the other. Proposition VI. Theorem 923. A plane tangent to a sphere is perpendicular to the radius drawn to the point of tangency. Given plane AB tangent to sphere 0 at T, and or a radius drawn to the point of tangency. To prove plane AB A. OT. The proofis left as an exercise for the student. 924. Question. What changes are necessary in the proof of " 313 to make it the proof of " 923 ? 925. Cor. I. (Converse of Prop. VI). A plane perpen-dicular to a radius of a sphere at its outer extremity is tangent to the sphere. Hint. See " 314. ' Ex. 1489. A straight line tangent to a sphere is perpendicular to the radius drawn to the point of tangency. Ex. 1490. State and prove the converse of Ex. 1489, 426 SOLID GEOMETRY Proposition VII. Problem 930. To inscribe a sphere in a given tetrahedron. V Given tetrahedron V-ABC. To inscribe a sphere in tetrahedron V-ABC. I. Construction 1. Construct planesBABS, SBCT, and TCAB bisecting dihedral A whose edges are AB, BC, and CA, respectively. " 691. 2. From 0, the point of intersection of the three planes, construct OF 1. plane ABC. " 637. 3. The sphere constructed with 0 as center and OF as radius will be inscribed in tetrahedron V-ABC. II. Proof Argument 1. Plane RABS, the bisector of dihedral Z AB, lies between planes ABV and ABC; i.e. it intersects edge VC in some point as D. 2. .-. plane RABS intersects plane BCV in line BD and plane ACV in line AD. 3. Plane SBCT lies between planes BCV and ABC; i.e.it intersects planeRASB in a line through B between BA and BD, as BS. 4. Similarly plane TCAR intersects plane RABS in a line through A between Reasons By cons. " 616. By cons. By steps ilar to sim-1-3. BOOK IX 427 Argument AB and AD, as AR ; and plane SBCT intersects plane TCAR in a line through G as CT. 5. But AR and "S pass through the in-terior of A ABB. 6. .-. AR and BS intersect in some point as 0, within A ABB. 7. .-. AR, BS, and CT are concurrent in point 0. 8. From 0 draw OH, OK, and OL L planes VAB, VBC, and FCM, respectively. 9. v 0 is in plane OAB, 0F = OH. 10. v 0 is in plane OB c, OF= OK. 11. v 0 is in plane OCA, OF = OL. 12. .-. OF = OH = OK = OL. 13. .-. each of the four faces of the tetra-hedron is tangent to sphere 0. 14. .-. sphere O is inscribed in tetrahedron V-ABC. Q.E.D; Reasons 5. Args. 3 " 4. 6. " 194. 7. " 617, I. 8. " 639. 9. " 688. 10. " 688. 11. " 688. 12. "54,1. 13. " 925. 14. "" 926, 927. III. Discussion The discussion is left as an exercise for the student. Ex. 1501. The six planes bisecting the dihedral angles of a tetrahe-dron meet in a point which is equidistantfrom the four faces of the tetrahedron. Hint. Compare with " 258. Ex. 1502 . Inscribe a sphere in a given cube. Ex. 1503. The volume of any tetrahedron is equal to the product of its surface and one third the radius of the inscribed sphere. Hint. See "" 491 and 492. Ex. 1504. Find a point within a triangularpyramid such that the planes determined by the lines joining this point to the vertices shall di-vide the pyramid into four equivalentparts. 'rfiNT.Compare with Ex. 1094. 428 SOLID GEOMETRY Proposition VIII. Problem 931. To circumscribe a sphere about a given tetrahe-dron. V Given tetrahedron V-ABC. To circumscribe a sphere about tetrahedron V-ABC. I. Construction 1. Through D, the mid-point of AB, construct plane DO A. AB ; through E, the mid-point of BC, construct plane EOA.BC; and through F, the mid-point of VB, construct plane FO J VB. 2. Join O, the point of intersection of the three planes, to A. 3. The sphere constructed with 0 as center and OA as radius will be circumscribed about tetrahedron V-ABC. II. Outline of Proof 1. Prove that the three planes OB, OE, and OF intersect each other in three lines. 2. Prove that these three lines of intersection meet in a point, as O. 3. Prove that OA = OB = OC = OV. 4. .. sphere 0 is circumscribed about tetrahedron V-ABC. Q.E.D. 932. Cor. Four points not in the same plane determine a sphere. 933. Questions. Are the methods used in "" 930 and 931 similar to those used in "" 321 and 323 ? In " 930 could the lines forming the edges of the dihedral angles bisected be three lines meeting in one vertex ? In " 931 could the three edges bisected be three lines lyingin the same face ? BOOK IX 429 Ex. 1505. The six planes perpendicularto the edges of a tetrahedron at their mid-points meet in a point which is equidistant from the four ver-tices of the tetrahedron. Ex. 1506. The four lines perpendicularto the faces of a tetrahedron, and erected at their centers, meet in a point which is equidistant from the four vertices of the tetrahedron. Ex. 1507. Circumscribe a sphere about a given cube. Ex. 1508. Circumscribe a sphere about a given rectangular paral-lelopiped. Can a sphere be inscribed in any rectangularparallelopiped ? Explain. Ex. 1509. Find a point equidistantfrom four pointsin space not all in the same plane. SPHERICAL POLYGONS 934. Def. A line on the surface of a sphere is said to be closed if it separates a portionof the surface from the remain-ing portion. 935. Def. A closed figure on the surface of a sphere is a figurecomposed of a portion of the surface of the sphereand its bounding line or lines. 936. Defs. A spherical polygon is a closed figure on the surface of a sphere whose boundary is composed of three or more arcs of great circles, as ABCD. The bounding arcs are called the sides of the polygon,the pointsof intersection of the arcs are the vertices of the poly-gon, and the sphericalangles formed by the sides are the ari'gtes of the polygon. 430 SOLID GEOMETRY' 937. Def. A diagonal of a spherical polygon is an arc of a great circle joining any two non-adjacent vertices. 938. Def. A spherical triangle is a spherical polygon"having three sides. Ex. 1510. By comparison with the definitions of the corresponding terms in plane geometry, frame exact definitions of the following classes of sphericaltriangles : scalene,isosceles, equilateral, acute, right, obtuse, and equiangular. Ex. 1511. With a given arc as one side, construct an equilateral spherical triangle. Hint. Compare the cons., step by step, with " 124. Ex. 1512. With three given arcs as sides, construct a scalene spheri-cal triangle. 939. Since each side of a sphericalpolygon is an arc of a great circle ("936), the planes of these arcs meet at the center of the sphere and form at that point a polyhedral angle, as polyhedralZ O-ABCD. This polyhedral angleand the spheri-cal polygon are very closely related. The following are some of the more important relations; the student should prove the correctness of each : 940. (a) The sides of a spherical polygon have the same measures as the corresponding face angles of the poly-hedral angle. (b) The anglesof a spherical polygon have the same measures as the correspondingdihedral anglesof the polyhedral angle. Thus, sides AB, BC, etc.,of spherical polygon ABCD have the same measures as face AAOB, BOC, etc.,of polyhedral Z O-ABCD ; and sphericalA ABC, BCD, etc.,have the same measures as the dihedral A whose edges are OB, OC, etc. These relations make it possible to establish certain prop-erties of sphericalpolygons from the corresponding known properties of the polyhedral angle, as in "" 941 and 942. BOOK IX Proposition IX. Theorem 431 941. The sum of any two sides of a spherical triangle is greater than the third side. Given spherical A ABC. To prove AB + BC " CA. Argument 1. Z AOB + Z BOC " Z COA. 2. Z ^05 oc ^Z?, Z BOCKBC, Z CO^ QC (3. 3. .-. ^i? + "C" CM. Q.E.D. Reasons 1. "710. 2. " 940, a. 3. " 362, b. Proposition X. Theorem 942. The sum of the sides of any sphericalpolygon is less than 360". Given spherical polygon ABC """ with n sides. To prove AB + BC + """ " 360". " 'Hint. See "" 712 and 940, a. 432 SOLID GEOMETRY Ex. 1513. In sphericaltriangleABC, arc AB " 40" and arc BC = 80". Between what limits must arc CA lie ? Ex. 1514. Any side of a spherical polygon is less than the sum of the remaining sides. Ex. 1515. If arc AB is the perpendicular bisector of arc CD, every point on the surface of the sphere and not in arc AB is unequally distant from C and D. Ex. 1516. In a sphericalquadrilateral, between what limits must the fourth side lie if three sides are 60", 70", and 80"? if three sides are 40", 50",and 70" ? Ex. 1517. Any side of a spherical polygon is less than 180". Hint. See Ex. 1514. 943. If,with A, B, and C the vertices of any spherical tri-angle as poles, three great circles are constructed, as B'c'ED, C'A'd, and EA'b',the surface of the spherewill be divided into eight spherical triangles, four of which are seen on the hemi-sphere representedin the figure. Of these eightspherical tri-angles, ^'.B'C'is the one and only one that is so situated that A and A1 lie on the same side of BC, B and B' on the same side of AC, and C and C' on the same side of AB. This particular tri-angle a'b'c' is called the polar triangle of triangle ABC. 944. Questions. In the figure above, A A'B'C, the polar of A ABC, is entirely outside of A ABC. Can the two A be so constructed that : (a) A'B'C is entirely within ABC ? (") A'B'C is partlyoutside of and partly within ABC? Ex. 1518. What is the polar triangleof a sphericaltriangleall of whose sides are quadrants ? 434 SOLID GEOMETRY Proposition XII. Theorem 947. In two polar triangles each angle of one and that side of the other of which its vertex is the pole are together equal, numerically, to 180"-E a Given polar A ABC and A'b'c', with sides denoted by a, b,c, and a',b', c', respectively. To prove: (a) Z ^4-a' = 180", Z B+b' = l$0",Z c+c'= 180"; (6) ZA'+a=180", ZB'+b=180", Z C'+c=180". (a) Argument Only 1. Let arcs AB and AC (prolonged if necessary)intersect arc B'C' at D and E, respectively ; then c'D = 90" and EB' = 90". 2. .-. c'D + EB' = 180". 3. . " . C'E + ED + EB + ZXB' = 180" ; i.e.ED + a' = 180". 4. But ED is the measure of Z A. 5. .\Z^ + a' = 180". 6. Likewise Z 5 + "' = 180", and Z C + c' = 180". q.e.d. (6)The proof of (6)is left as an exercise for the student. Hint. Let BC prolonged meet A'B' at .ffand A'C at if. 948. Question. In the history of mathematics, why are polar tri-angles frequentlyspoken of as supplemental triangles ? Ex. 1520. The angles of a sphericaltriangle are 75",85",and 145". Find the sides of its polartriangle. Ex. 1521. If a sphericaltriangleis equilateral, its polartriangle is equiangular; and conversely. BOOK IX 435 Proposition XIII. Theorem 949. The sum of tiie angles of a spherical triangle is greater than 180" and less than 540". Given spherical A ABC with sides denoted by a, b, and c. To prove ZA + Zb + Zc" 180" and " 540". Argument 1. Let A A'b'c', with sides denoted by a', b', and c', be the polar of A ABC. 2. Then Z A + a' = 180", Z B +V ' = 180", Zc+c'= 180". 3. .:ZA + ZB+Zc + (a'+b'+c') = 54:0o. 4. But a' + b' + c' "360". 5. .:ZA + ZB+Z "7"180". 6. Again, a1 + b'+ c' " 0". 7. .-. Z^ + Z^B + Z c" 540". q.e.d. Reasons 1. " 943. 2. " 947. 3. " 54, 2. 4. " 942. 5. " 54, 6. 6. " 938. 7. " 54, 6. 950. Cor. In a spherical triangle there can be one, two, or three right angles ; there cajv be one, two, or three obtuse angles. 951. Note. Throughout the Solid Geometry the student's attention has constantly been called to the relations between definitions and theorems of solid geometry and the corresponding definitions and theorems of plane geometry. In the remaining portion of the geometry of the sphere there will likewise be many of these comparisons, but here the student must be particularly on his guard for contrasts as well as comparisons. For ex-436 SOLID GEOMETRY ample, while the sum of the angles of a plane triangle is equal to exactly 180",he has learned (" 949) that the sum of the angles of a sphericaltri-angle may be any number from 180" to 540" ; while a plane trianglecan have but one rightor one obtuse angle,a spherical triangle may have one, two, or three rightangles or one, two, or three obtuse angles("950). If the student will recall that, considering the earth as a sphere, the north and south poles of the earth are the poles of the equator, and that all meridian circles are great circles perpendicularto the equator, it will make his thinking about spherical triangles more definite. 952. Def. A spherical triangle containing two rightangles is called a birectangular spherical triangle. 953. Def. A spherical triangle having all of its anglesright angles is called a trirectangular spherical triangle. Thus two meridians, as NA and NB, making at the north pole an acute or an obtuse Z, form with the equator a birectangular spherical A. If the Z between NA and NB is made a rt. Z, A ANB becomes a trirectangular spherical A. Ex. 1522. What kind of arcs are NA and NB ? Then what arc measures spherical angle ANB ? Are two sides of any birectangular sphericaltriangle quadrants? What is each side of a trirectangular spherical triangle ? Ex. 1523. If two sides of a sphericaltriangleare quadrants, the triangle is birectangular. (Hint. Apply " 947.) Ex. 1524. What is the polar triangleof a trirectangular spherical triangle ? Ex. 1525. An exterior angle of a sphericaltriangle is less than the sum of the two remote interior angles. Compare this exercise with " 215. Make this new fact clear by applying it to a birectangularspherical triangle whose third angle is : (a) acute; (6) right; (c) obtuse. BOOK IX 437 Proposition XIV. Theorem 954. In equal spheres, or in the same sphere,two spher-ical triangles are equal : I. If a side and the two adjacent angles of one are equal respectively to a side and the two adjacent angles of the other; II. If two sides and the included angle of one are equal respectively to two sides and tlie included angle of the other; III. If the three sides of one are equal respectively to the three sides of the other : provided the equal parts are arranged in the same order. The proofs are left as exercises for the student. Hint. In each of the above cases prove the corresponding trihedral A equal ("" 702, 704) ; and thus show that the spherical A are equal. 955. Questions. Compare Prop. XIV, I and II,with " 702, I and II,and with "" 105 and 107. Could the methods used there be employed in " 954 ? Is the method here suggestedpreferable ? why ? 956. Def. Two spherical polygons are symmetrical if the correspondingpolyhedral angles are symmetrical. ^" 957. The followingare some of the properties of symmet-rical spherical triangles; the student should prove the correct-ness of each : (a)Symmetricalspherical triangles have their parts respectively equal,but arranged in reverse order. (b) Two isosceles symmetrical spherical triangles are equal. Hint. Prove (6) by superposition or by showing that the correspond-ing trihedral A are equal. 438 SOLID GEOMETRY Proposition XV. Theorem 958. In equal spheres, or in the same sphere, two sym-metrical sphericaltriangles are equivalent. A A' Given symmetrical spherical A ABC and A'B'c' in equal spheres O and O'. To prove spherical A ABC=^ spherical A A'b'C. Argument 1. Let P and P' be poles of small " through A, B, C, and A',B',C', respectively. 2. Arcs AB, BC, CA are equal,respectively, to arcs A'B', B'C',C'A'. 3. .-. chords AB, BC, CA, are equal, respec-tively, to chords A'b',b'c',c'a'. 4. .-. plane A ABC = planeA a'b'c. 5. .-. O ABC = O A'B'C'. 6. Draw arcs of great (D PA, PB, PC, p'a', p'b', and P'c'. . 7. Then PA = PB = PC= P'A'= p'b'= p'c'. 8. .-. isosceles spherical A APB and a'p'b' are symmetrical. 9. .'. A APB == A A'P'B'. 10. Likewise A BPC = A b'p'c' and A CPA = A C'P'A'. 11. .-. spherical A ABCo spherical Aa'b'c'. Q.E.D. Reasons 1. " 908, h. 2. " 957, a. 3. " 298, II. 4. "116. 5. " 324. 6. " 908, g. 7. " 913. 8. "956. 9. "957, b. 10. By stepssimi-lar to 8-9. 11. " 54, 2. BOOK IX 439 Proposition XVI. Theorem 959. In equal spheres, or in the same sphere, two spheri-cal triangles are symmetrical, and thereforeequivalent : I. If a side and the two adjacent angles of one are equal respectively to a side and the two adjacent angles of the other ; II. // two sides and the included angle of one are equal respectively to two sides and the included angle of the other; III. If the three sides of one are equal respectively to the three sides of the other : provided the equal parts are arranged in reverse order. The proofsare left as exercises for the student. Hint. In each of the above cases prove the corresponding trihedral A symmetrical ("709) ; and thus show that the spherical " are symmetrical. "Ex. 1526. The bisector of the angle at the vertex of an isosceles sphericaltriangle is perpendicular to the base and bisects it. Ex. 1527. The arc drawn from the vertex of an isosceles spherical triangle to the mid-point of the base bisects the vertex angle and is per-pendicular to the base. Ex. 1528. State and prove the theorems on the sphere correspond-ing to the followingtheorems on the plane : (1) Every point in the perpendicular bisector of a line is equidistant from the ends of that line (" 134). (2) Every point equidistant from the ends of a line lies in the perpen-dicular bisector of that line (" 139). (3) Every point in the bisector of an angle is equidistantfrom the sides of the angle (" 253). Hint. In the figurefor (3),corresponding to the figureof " 252, draw PD " AB and lay off BE = BD. Ex. 1529. The diagonalsof an equilateral sphericalquadrilateral are perpendicular to each other. Prove. State the theorem in plane geome-try that corresponds to this exercise. Ex. 1530. In equal spheres, or in the same sphere, if two spherical triangles are mutually equilateral, their polar triangles are mutually equi-angular ; and conversely. 440 SOLID GEOMETRY Proposition XVII. Theorem 960. In equal spheres, or in the same sphere, if two spherical triangles are mutually equiangular, they are mutually equilateral, and are either equal or symmet-rical. Given spherical A T and T in equalspheres, or in the same sphere, and mutually equiangular.. To prove T and Tf mutually equilateral and either equal or symmetrical. Argument Let A P and P ' be the polarsof A T and T', respectively. T and T' are mutually equiangular. Then P and P' are mutually equi-lateral. .. P and P' are either equal or sym-metrical, and hence are mutually equiangular. .. T and T' are mutually equilateral. .-. T and T' are either equal or sym-metrical. Q.E.D. Reasons "943. By hyp. "947. 4. "" 954, III, and 959,III. "947. Same as 4. reason Ex. 1531. In plane geometry, if two trianglesare mutually equi-angular, what can be said of them ? Are they equal ? equivalent? Ex. 1532. Find the locus of all points of a sphere that are equidis-tant from two given points on the surface of the sphere ; from two given pointsin space, not on the surface of the sphere. 442 SOLID GEOMETRY Proposition XX. Theorem 963. If two angles of a spherical triangle are unequal, the side opposite the greater angle is greater than the side opposite the less angle. Given spherical A ABC witli Z A " Z C. To prove BC" AB. Argument 1. Draw an arc of a great O AD making Z1=Z a 2. Then Td = 'dc. 3. But bd + a1"" AB. 4. . \ BD + DC " AB ; i.e.BO " AB. Q.E.D. 1. Reasons " 908, g. 2. " 962. 3. " 941. 4. " 309. Ex. 1533. In a birectangularspherical triangle the side included by the two right angles is less than, equal to, or greater than, either of the other two sides, according as the angle oppositeis less than, equal to, or greater than 90". Ex. 1534. An equilateral spherical triangle is also equiangular. Ex. 1535. If two face angles of a trihedral angle are equal, the dihedral angles oppositeare equal. Ex. 1536. State and prove the converse of Ex. 1534. Ex. 1537. State and prove the converse of Ex. 1535. Ex. 1538. The arcs bisecting the base angles of an isosceles spheri-cal triangle form an isosceles spherical triangle. Ex. 1539. The bases of an isosceles trapezoid are 14 inches and 6 inches and the altitude 3 inches ; find the total area and volume of the solid generated by revolvingthe trapezoidabout its longerbase as an axis. Ex. 1540. Find the total area and volume of the solid generated by revolvingthe trapezoidof Ex. 1530 about its shorter base as an axis. BOOK IX 443 Proposition XXI. Theorem 964. If two sides of a spherical triangle are unequal, the angle opposite the greater side is greater than the angle opposite the less side. B Given spherical A ABC with BC " AB. To prove Z A " Z C. The proof is left as an exercise for the student. Hint. Prove by the indirect method. 965. Questions. Could Prop. XXI have been proved by the method used in " 156 ? Does reason 4 of that proof hold in spherical A ? See Ex. 1525. Ex. 1541. If two adjacent sides of a spherical quadrilateral are greater, respectively, than the other two sides, the spherical angle included between the two shorter sides is greater than the sphericalangle between the two greater sides. Hint. Compare with Ex. 153. Ex. 1542. Find the total area and volume of the solid generated by revolvingthe trapezoidof Ex. 1539 about the perpendicularbisector of its bases as an axis. Ex. 1543. Pind the radius of the sphere in-scribed in a regulartetrahedron whose edge is a. Hint. Let 0 be the center of the sphere,A the center of face VED, and B the center of face EDF. Then OA = radius of inscribed sphere. Show that rt. A VAO and VBC are similar. Then VO : VC = VA : VB. VC, VA, and VB can be found (Ex. 1328). Find VO, then OA. Ex. 1544. Pind the radius of the sphere circumscribed about a regu-lar tejtrahedron whose edge is a. Hin. In the figureof Ex. 1543,draw AD and OD. 444 SOLID GEOMETRY MENSURATION OF THE SPHERE Areas Proposition XXII. Theorem 966. If an isosceles triangle is revolved about a straight line lying in its plane and passing through its vertex but not intersectingits surface, the area of the surface gener-ated by the base of the triangle is equal to the product of its projection on the axis and the circumference of a circle whose radius is the altitude of the triangle-x X X c OH D Y Y Y Fig. 1. Fig. 2. Fig. 3. Given isosceles A AOB with base AB and altitude OE, a str. line XY lying in the plane of A AOB passingthrough O and. not intersecting the surface of A AOB, and CD the projection of AB on J7; let the area of the surface generatedby AB be denoted by area AB. To prove area AB = CD " 2 irOE. I. If AB is not II XY and does not meet XY (Fig. 1). Argument Only 1. From E draw EH A. XY. 2. Since the surface generated by AB is the surface of a frustum of a rt. circular cone, area AB = AB " 2 -n-EH. 3. From A draw AK" BD. 4. Then in rt. A BAK and OEH, Z BAR = Z OEH. BOOK IX 445 5. .-. ABAK ~ AOEH. 6. .. AB.AK" OE: EH) i.e. AB " EH = AK " OE. 7. But AK= CD; .'. AB -EH= CD " OE. 8. .-. area AB " CD -2 -nOE. q.e.d. II. If AB is not II XY and pointA lies in XY (Fig.2). III. If ^B II XT (Fig. 8). The proofs of II and III are left as exercises for the student. Hint. See if the proof given for I will apply to Figs.2 and 3. 967. Cor. I. If half of a regular polygon with an even number of sides is circumscribed about a semicircle, the area of the surface generated by its semiperimeter as it revolves about the diameter of the semicircle as an axis, is equal to the product of the diameter of the regular polygon and the circumference of a circle whose radius is R, the radius of the given semicircle. Outline of Proof 1. Area A'b' = A'f' -2ttR; area B'c' = f'o' " 2 ttR; etc. 2. .-. area a'b'c'--= (a'f' + f'o' + ) 2 ttR = A'E' " 2 ttR. 968. Cor. n. If half of a regular polygon with an even number of sides is inscribed in a semicircle, the area of the surface generated by its semi-perimeter as it revolves about the diameter of the semicircle as an axis, is equal to the product of the diameter of the semicircle and the circumference of a circle whose radius is the apothem of the regular polygon. ' if int. Prove area ABC " " " = AE " 2 ira. 446 SOLID GEOMETRY 969. Cor. m. If halves of regular polygons with the same even number of sides are circumscribed about, and inscribed in, a semicircle, then by repeatedlydoubling the number of sides of these polygons,and making the poly-gons always regular, the surfaces generated by the semi-perimeters of the polygons as they revolve about the diameter of the semicircle as an axis approach a com-mon limit. Outline of Proof 1. If S and s denote the areas of the surfaces generated by the semi-perimetersA'b'c'--- and ABC---as they revolve about a'e' as an axis, then 6' = A'E' " 2 -# ("967); and s = AE " 2 ttcl ("968). 2. .-. -= S A'E1 " 2 ttR 8 AE " 2 7ra But polygon a'b'c'--A'E' a'b' A'E' AE polygon ABC--" ("438). = -(""419,435). AE AB a V S s a R2 a2 S Rz 7. .-. by steps similar to Args. 6-11 (" 855),S and s ap-proach a common limit. q.e.d. 970. Def. The surface of a sphere is the common limit Which the successive surfaces generated by halves of regular polygons with the same even number of sides approach, if these semipolygons fulfillthe followingconditions : (1)They must be circumscribed about,and inscribed in,a great semicircle of the sphere; (2)The number of sides must be successively increased, each side approaching zero as a limit. BOOK IX 447 Proposition XXIII. Theorem 971. The area of the surface of a sphere is equal to four times the area of a great circle of the spliere. A 1. Reasons " 517, a. Given sphere 0 with its radius denoted by R, and the area of its surface denoted by S. To prove S = 4 -irR2. Argument 1. In the semicircle ACE inscribe ABCDE, half of a regular polygon with an even number of sides. Denote its apothem by a, and the area of the surface generatedby the semiperime-ter as it revolves about AE as an axis bys'. 2. Then S' = AE " 2 ira ; i.e. S' = 2 R " 2 ttcl = 4 irRa. 3. As the number of sides of the regular polygon, of which ABCDE is half,is repeatedlydoubled, S' approaches S as a limit. 4. Also a approachesR as a limit. 5. .-. 4 nR " a approaches 4 irR " R, i.e.4 irR\ as a limit. 6. But S' is always equal to 4 7ri? " a. 1."V- 5 = 4 irR2. Q.E.D. 2. " 968. "970. " 543, I. " 590. Arg. 2. " 355. 448 SOLID GEOMETRY 972. Cor. I. The areas of the surfaces of two spheres are to each other as the squares of their radii and as the squares of their diameters. Outline of Proof 1. S = 47ri?2andS' = 47ri2'2; .-.?-= i^-= ". ' 8' 4tt2'2 R12 2 But R2 = 4 R2 = (2R^ = D " " -= " ' U R,2~ "R"2~ (2rJ~ D'2' '"' S'~D'2' 973. Historical Note. Prop. XXIII is given as Prop. XXXV in the treatise entitled Sphere and Cylinder by Archimedes, already spoken of in S 809. Ex. 1545. Find the surface of a sphere whose diameter is 10 inches. Ex. 1546. What will it cost to gildthe surface of a globe whose radius is 1\| decimeters,at an average cost of " of a cent per square centimeter ? Ex. 1547. The area of a section of a sphere made by a plane 11 inches from the center is 3000 ir square inches. Find the surface of the sphere. Ex. 1548. Find the surface of a sphere circumscribed about a cube whose edge is 12 inches. Ex. 1549. The radius of a sphere is R. Find the radius of a sphere whose surface is twice the surface of the given sphere ; one half ; one nth. Ex. 1550. Find the surface of a sphere whose diameter is 2 R, and the total surface of a rightcircular cylinderwhose altitude and diameter are each equal to 2 R. Ex. 1551. From the results of Ex. 1550 state, in the form of a theorem, the relation of the surface of a sphere to the total surface of the circumscribed cylinder. Ex. 1552. Show that, in Ex. 1550, the surface of the sphere is exactlyequal to the lateral surface of the cylinder. 974. Historical Note. The discovery of the remarkable property that the surface of a sphere is two thirds of the surface of the circum-scribed cylinder (Exs. 1550 and 1551) is due again to Archimedes. The discovery of this proposition,and the discovery of the corresponding propositionfor volumes (" 1001), were the philosopher's chief pride,and he therefore asked that a figure of this proposition be inscribed on his tomb. His wishes were carried out by his friend Marcellus. (For a further account of Archimedes, read also " 542.) 450 SOLID GEOMETRY 980. Cor. III. In equal spheres, or in the same sphere, the areas of two zones are to each other as their altitudes. 981. Question. In general,surfaces are to each other as the prod-ucts of two lines. Is " 980 an exception to this rule ? Explain. Ex. 1553. The area of a zone of one base is equal to the area of a circle whose radius is the chord of the arc generating the zone. Hint. Use "" 979 and 444, II. Ex. 1554. Show that the formula of " 971 is a special case of " 979. Ex. 1555. Find the area of the surface of a zone if the distance between its bases is 8 inches and the radius of the sphere is 6 inches. Ex. 1556. The diameter of a sphere is 16 inches. Three parallel planes divide this diameter into four equal parts. Find the area of each of the four zones thus formed. Ex. 1557. Prove that one half of the earth's surface lies within 30" of the equator. Ex. 1558. Considering the earth as a sphere with radius B, find the 7? 2 7? area of the zone adjoiningthe north pole,whose altitude is " ; Is the one area twice the other ? Ex. 1559. Considering the earth as a sphere with radius B, find the area of the zone extending 30" from the north pole ; 60" from the north pole. Is the one area twice the other ? Ex. 1560. Considering the earth as a sphere with radius i?,find the area of the zone whose bases are parallels of latitude: (a) 30" and 45c from the north pole ; (6) 30" and 45" from the equator. Are the two areas equal ? Explain your answer. Ex. 1561. How far from the center of a sphere whose radius is B must the eye of an observer be so that one sixth of the surface of the sphere is visible ? Hint. Let E be the eye of the observer. Then AB must = " . Find OB, then use " 443,II. 3 " w , Ex. 1562. What portion of the surface of a sphere can be seen if the distance of the eye of the observer from the center of the sphere is 2B ? SB? nB? Ex. 1563. The radii of two concentric spheres are 6 inches and 10 inches. A plane is passed tangent to the inner sphere. Find : (a) the area of the section of the outer sphere made by the plane ; (") the area of the surface cut off of the outer sphere by the plane, BOOK IX 451 982. Def. A lune is a closed figure on the surface of a sphere whose boundary is composed of two semicircumferences of great circles, as NASB. 983. Defs. The two semicircumfer-ences are called the sides of the lune, as NAS and NBS; the pointsof inter-section of the sides are called the vertices of the lune,as N and S-, the spherical angles formed at the vertices by the sides of the lune are called the angles of the lune, as A ANB and BSA. 984. Prove, by superposition, the following property of lunes : In equal spheres, or in the same sphere,two lunes are equal if their anglesare equal. 985. So far,the surfaces considered in connection with the sphere have been measured in terms of square units, i.e. square inches,square feet, etc. For example, if the radius of a sphere is 6 inches,the surface of the sphere is 4"7r62, i.e. 144 ir square inches. But, as the sides and angles of a lune and a spherical polygon are given in degrees and not in linear units, it will be necessary to introduce some new unit for determining the areas of these figures.For this purpose the entire surface of a sphere is thought of as being divided into 720 equal parts,and each one of these parts is called a spherical degree. Hence : 986. Def. A spherical degree is yj-g-of the surface of a sphere. Now if the area of a lune or of a spherical triangle can be ob-tained in sphericaldegrees, the area can easilybe changed to square units. For example, if it is found that the area of a spherical triangle is 80 spherical degrees, its area is -f^, i.e.^ of the entire surface of the sphere. On the sphere whose radius is 6 inches, the area of the given triangle will be -J-of 144 7r square inches, i.e. 16 " square inches. The following theorems are for the purpose of determining the areas of figures on- tye surface of a sphere in terms of spherical degrees. 452 SOLID GEOMETRY Proposition XXIV. Theorem 987. The area of a lune is to the area of the surface of tl%e sphere as the number of degrees in the angle of the lune is to 360. N N Q E Given lune NASB with the number of degreesin its Z denoted by Nf its area denoted by L, and the area of the surface of the sphere denoted by S ; let O EQ be the great O whose poleis N. m L N To prove -= 8 360 I. If arc AB and circumference EQ are commensurable (Fig. 1). Argument Reasons 1. Let m be a common measure of arc AB 1. " 335. and circumference EQ, and suppose that m is contained in arc AB r times and in circumference EQ t times. arc AB r ~t' Then 4. circumference EQ Through the several pointsof division on circumference EQ pass semicir-cumferences of great circles from N to S. Then lune NASB is divided into r lunes and the surface of the sphere into t lunes,each equalto lune NCSB. " 341. " 908, h. 4. "984. 453 II. If arc AB and circumference EQ are incommensurable (Fig. 2). The proof is left as an exercise for the student. Hint. The proof is similar to that of " 409, II. 988. Cor. I. The area of a lune, expressed in spherical degrees, is equal to twice the number of degrees in Us angle. L = JL ("987). 8 360 V J Outline of Proof L N 720 360 L = 2n. Ex. 1564. Find the area of a lune in sphericaldegrees if its angle is 35". What part is the lune of the entire surface of the sphere ? Ex. 1565. Find the area of a lune in square inches if its angle is 42" and the radius of the sphere is 8 inches. (Use tr = -2/.) Ex. 1566. In equal spheres,or in the same sphere, two lunes are to each other as their angles. Ex. 1567. Two lunes in unequal spheres, but with equal angles, are to each other as the squares of the radii of their spheres. Ex. 1568. In a sphere whose radius is R, find the altitude of a zone equivalentto a lune whose angle is 45". Ex. 1569. Considering the earth as a sphere with radius B, rind the area of the zone visible from a point at a height h above the surface of the earth. 989. Def. The spherical excess of a spherical triangleis thre^excess of the sum of its angles over 180". 454 SOLID GEOMETRY Proposition XXV. Theorem 990. The area of a spherical triangle, expressed in sphericaldegrees, is equal to its spherical excess. B i A r / i J -" A/ / \ : / /A Given spherical A ABC with its spherical excess denoted by E. To prove area of A ABC = E spherical degrees. Argument 1. Complete the circumferences of which AB, BC, and CA are arcs. 2. Aab'c' and A'BC are symmetrical. 3. .-. AAB'C' oAA'BC 4. .-. A ABC + A AB'C'o A ABC -{-A A'BC. 5. .-.,expressed in spherical degrees, A ABC + A AB'C' f"= lune A = 2 A ; A ABC + A AB'C = lune B =2 B-, A ABC + A ABC' = lune C = 2 C. .-. 2A^LBC-f-(A^BC+AAB'(7' + A^LB,C -f A ABC')= 2(A+ B+ C). But A ABC+ A AB'C'+ A AB'C+AABC' = surface of a hemisphere = 360. .-. 2 A ABC+ 360 = 2 (A + 5 + c). .-. A ^BC + 180 = ^ + 5 + a . ". A ^z?c = (i+B + 6')- 180, i.e. ^ spherical degrees.' q.e.d. 6. 8. 9. 10. Reasons " 908, g. " 956. " 958. " 54, 2. " 988. BOOK IX 455 991. In " 949 it was proved that the sum of the angles of a spherical triangleis greater than 180" and less than 540". Hence the sphericalexcess of a sphericaltrianglemay vary from 0" to 360", from which it follows (" 990) that the area of a spherical trianglemay vary from yf^ to ff" of the entire surface ; i.e. the area of a spherical trianglemay vary from nothing to \ the surface of the sphere. Thus in a spherical trianglewhose angles are 70",80",and 100",respectively, the spherical excess is (70"+ 80" + 100") -180" = 70" ; i.e. the area of the given triangle is -f^ of the surface of the sphere. 992. Historical Note. Menelaus of Alexandria (circ.98 a.d.) wrote a treatise in which he describes the properties of spherical triangles, although there is no attempt at their solution. The expression for the area of a sphericaltriangle, as stated in " 990, was first given about 1626 a.d. by Girard. (See also " 946.) This theorem was also discovered in-dependent by Cavalieri, a prominent Italian mathematician. Ex. 1570. If three great circles are drawn, each perpendicularto the other two, into how many trirectangular spherical triangles is the surface divided ? Then what is the area of a trirectangular spherical triangle in spherical degrees ? Test your answer by applying Prop. XXV. Ex. 1571. Find the area in spherical degrees of a birectangular spherical triangle one of whose angles is 70" ; of an equilateral spherical triangle one of whose angles is 80". What part of the surface of the sphere is each triangle ? Ex. 1572. The angles of a spherical triangle in a sphere whose sur-face has an area of 216 square feet are 95", 105",and 130". Find the number of square feet in the area of the triangle. Ex. 1573. In a sphere whose diameter is 16 inches,find the area of a triangle whose angles are 70",86",and 120". Ex. 1574. The angles of a spherical triangle are 60", 120", and 160", and its area is lOOf square inches. Find the radius of the sphere. (Use Ex. 1575. The area of a sphericaltriangle is 90 sphericaldegrees, and the angles are in the ratio of 2, 3, and 5. Find the angles. Ex. 1576. Find the angle (1) of an equilateralsphericaltriangle, (2) of a lune, each equivalentto one third the surface of a sphere. Ex. 1577. Find the angle of a lune equivalentto an equilateral spherical triangle one of whose angles is 84". 456 SOLID GEOMETRY Proposition XXVI. Theorem 993. The area of a spherical polygon, expressed in spherical degrees,is equal to the sum of its angles dimin-ished by 180" taken as many times less two as the poly-gon has sides. c Given sphericalpolygon ABCD " " " with n sides ; denote the sum of its anglesby T. To prove area of polygon ABCD " "", expressedin spherical degrees,= T" (n " 2)180. 1. 2. Argument From any vertex such as A, draw all possible diagonals of the polygon, forming n " 2 spherical A, I, II, etc. Then, expressed in spherical degrees, AI = (Z 1 + Z2 + Z3)-180; A 11= (Z 4 + Z 5 + Z 6)-180; etc. .-. A I + AII + ... = r-(n-2)180. .. area of polygon ABCD """ = T-(n-2) 180. q.e.d. Reasons " 937. 2. " 990. " 54, 2. " 309. Ex. 1578. Prove Prop. XXVI by using a figuresimilar to that used in " 216. Ex. 1579. Find the area of a spherical polygon whose angles are 80", 92", 120",and 140",in a sphere whose radius is 8 inches. Ex. 1580. Find the angle of an equilateral sphericaltriangle equiva-lent to a spherical pentagon whose angles are 90",100",110",130",and 140". Ex. 1581. Find one angle of an equiangular spherical hexagon equivalentto six equilateral sphericaltriangles each with angles of 70". 458 SOLID GEOMETRY Proposition XXYII. Theorem 995. If an isosceles triangle is revolved about a straight line lying in its plane and passing through its vertex but not intersecting its surface, the volume of the solid gener-ated is equal to the product of the surface generated by the base of the triangle and one third of its altitude. Y Fig. 2. Fig. 3. Given isosceles A AOB with altitude OE, and a str. line XY lying in the plane of A AOB, passingthrough O and not inter-secting the surface of A AOB-, let the volume of the solid gen-erated by AAOB revolving about XF as an axis be denoted by volume AOB. To prove volume AOB = area AB " 1 OE. I. If AB is not II XY and does not meet XY (Fig.1). Argument Only 1. Draw AC and BD _L XY. 2. Prolong BA to meet XY at F. 3. Then volume AOB = volume FOB " volume FOA. 4. Volume FOB " volume FDB + volume BOB. 5. .'. volume FOB =\tt ~BD2 " FD + \ 7T BD2 " DO = 1 7T BDZ(FD + DO) = \ttBD-BD " FO. BD " FO = twice area of A FOB " BF " OE. 6. But 7. .-. volume FOB = ^ tt BD " BF 8. But ttBD " BF as area FB. 9. .-. volume FOB = area FB OE = ttBD BF " I OE. iOK BOOK IX 459 10. Likewise volume FOA = area FA " \ OE. 11. .-. volume AOB = area FB " i OE " area FA " = (area FB " area FA) i Q# = area AB " J OE. iOE Q.E.D. II. If AB is not II XY and point A lies in XY (Fig. 2). The proof is left as an exercise for the student. Hint. See Arg. 9 or Arg. 10 of " 995, I. III. If AB IIXY (Fig.3). The proof is left as an exercise for the student. Hint. Volume AOB = volume ACDB " twice volume CO A. 996. The student may : . (a) State and prove the corollaries on the volume of a sphere corresponding to "" 967 and 968. (b) State and prove the theorem on the volume of a sphere corresponding to " 969. 1. Outline of Proof V= (areaA'b'c' -)$$ = A'E' " 2 ttR " 1 R = %7rR2"A'E' ("" 996, a and 967). v = (areaAB C " " " )\ a = AE " 2 -n-a " -J-a = f 7ra2XE (""996, a and 968). F_\|ir^""^J,_JZa f A'E' = i?3 2 XE ~ a3 TTcr " ^1J" 4. Proceed as in " 855, observing that since the limit of a = R (" 543, I), the limit of a3= R3 (" 593) ; i.e. R^^-a3 may be made less than any previouslyassigned value, however small. (c)State,by aid of " 970, the definition of the volume of a 460 SOLID GEOMETRY Proposition XXVIII. Theorem 997. The volume of a sphere is equal to the product of the area of its surface and one third its radius. Given sphere 0 with its radius denoted by R, the area of its surface by S, and its volume by V. To prove V= S " " R. Argument 1. In the semicircle ACE inscribe ABODE, half of a regularpolygon with an even number of sides. Denote its apothem by a, the area of the surface generated by the semiperimeter as it revolves about AE as an axis by S', and the vol-ume of the solid generated by semi-polygon ABODE by V'. Then r = 8' "i. 3. As the number of sides of the regular polygon, of which ABODE is half,is repeatedly doubled, v' approaches V as a limit. 4. Also S' approaches Sasa limit. 5. And a approaches R as a limit. 6. .'. S' " a approaches S " R as a limit. 7. .'. " " \a approaches S " i R as a limit. 8. But Vf is always equal to s' . J a. 9. .'. V=S-"R. Q.E.D. Reasons 1. " 517, a. "2. " 996, a. 3. " 996, c. BOOK IX 461 998. Cor. I. If v denotes the volume, R the radius, and D the diameter of a sphere, V = \| Trie3 = I wDs. 999. Cor. n. The volumes of two spheres are to each other as the cubes of their radii and as tlie cubes of their diameters. (Hint. See " 972.) 1000. Historical Note. It is believed that the theorem of " 999 was proved as early as the middle of the fourth century b.c. by Eudoxus, a great Athenian mathematician already spoken of in "" 809 and 896. Ex. 1588. Find the volume of a sphere inscribed in a cube whose edge is 8 inches. Ex. 1589. The volume of a sphere is 1774"ir cubic centimeters. Find its surface. Ex. 1590. Find the radius of a sphere equivalentto a cone with alti-tude a and radius of base 6. Ex. 1591. Find the radius of a sphere equivalent to a cylinder with the same dimensions as those of the cone in Ex. 1590. Ex 1592. The metal cone and cylinderin the figurehave their alti-tude and diameter each equal to 2 B, the diameter of the sphere. Place the sphere in the cylinder, then fill the cone with water and empty it into the cylinder. How nearly is the cylinder filled ? Next fillthe cone with water and empty it into the cylinderthree times. Is the cylinder filled ? Ex. 1593. From the results of Ex. 1592 state, in the form of a theorem, the relation of the volume of a sphere : (a) to the volume of a circumscribed cylinder; (6) to the volume of the corresponding cone. Prove these statements. 1001. Historical Note. The problem " To find a sphere equivalent to a given cone or a given cylinder" (Exs. 1590 and 1591),as well as the properties that the volume of a sphere is two thirds of the volume of the circumscribed cylinderand twice the volume of the corresponding cone (Exs. 1592 and 1593),are due to Archimedes. The importance attached to thfe by the author himself is spoken of more fullyin "" 542 and 974. 462 SOLID CxEOMETRY Ex. 1594. A bowl whose inner surface is an exact hemisphere is made to hold \| gallon of water. Find the diameter of the howl. Ex. 1595. A sphere 12 inches in diameter weighs 93 pounds. Find the weight of a sphere of the same material 16 inches in diameter. Ex. 1596. In a certain sphere the area of the surface and the volume have the same numerical value. Find the volume of the sphere. Ex. 1597. Find the volume of a sphericalshell 5 inches thick if the radius of its inner surface is 10 inches. Ex. 1598. A pine sphere 24 inches in diameter weighs 175 pounds. Find the diameter of a sphere of the same material weighing 50 pounds. Ex. 1599. The radius of a sphere is B. Find the radius of a sphere whose volume is one half the volume of the given sphere ; twice the vol-ume ; n times the volume. 1002. Defs. A spherical sector is a solid closed figuregen-erated by a sector of a circle revolvingabout a diameter of the circle as an axis. C Fig. 2 Fig. 3. The zone generatedby the arc of the circular sector is called the base of the spherical sector. 1003. Def. If one radius of the circular sector generating a sphericalsector is a part of the axis,i.e. if the base of the spherical sector is a zone of one base,the spherical sector is sometimes called a spherical cone. Thus if circular sector AOB (Fig.1)revolves about diameter CD as an axis, arc AB will generate a zone which will be the base of the spherical sector generated by circular sector AOB (Fig. 2). If circular sector BOG revolves about diameter CDy the spherical sector generated, whose base is the zone generated by arc BC, will be a spherical cone (Fig. 3). BOOK IX 463 1004. Cor. HI. The volume of a spherical sector is equal to the product of Us base and one third the radius of the sphere. Outline of Proof Let V denote the volume generatedby polygon OA'b'c1, v the volume generatedby polygon OABC, S the area of the surface generated by broken line A'b'c', s the area of the surface generated by broken line ABC, and z the base of the spherical sector generated by cir-cular sector AOC. Then V = S " i R, and V = s V S-"R S R \a. v 8 ha s a S R2 But -= ^?(Args.2-5, " 969). s a1 R2 R^R 2 n /-"" . V v a Then by a a" steps similar to " 996, b and c, and " 997, the volume of the spherical sector generated by circular sector AOC= Z"\R. 1005. Cor. IV. If v denotes the volume of a spherical sector, z the area of the zone forming its base, H the alti-tude of the zone, and R the radius of the sphere, V=z-\R ("1004) = {H-2ttR)\R ("979) = %vR'H. pole and whose altitude is Ex. 1600. Considering the earth as a sphere with radius B, find the volume of the sphericalsector whose base is a zone adjoining the north 7? 2 Tt " ; " " Is the one volume twice the other? 3 3 Compare your results with those of Ex. 1558. Ex. 1601. Considering the earth as a sphere with radius R, find the volume of the sphericalsector whose base is a zone extending : (a) 30" from the north pole ; (ft) 60" from the north pole. Is the one volume twijie the other ? Compare your results with those of Ex. 1559. 464 SOLID GEOMETRY Ex. 1602. Considering the earth as a sphere with radius i?,find the volume of the sphericalsector whose base is a zone lying between the parallels of latitude : (a) 30" and 45" from the north pole ; (") 30" and 45" from the equator. Are the two volumes equal ? Compare your results with those of Ex. 1560. Ex. 1603. Considering the earth as a sphere with radius i?, find the area of the zone whose bases are the circumferences of small circles, one 30" north of the equator, the other 30" south of the equator. What part of the entire surface is this zone ? Ex. 1604. What part of the entire volume of the earth is that por-tion included between the planes of the bases of the zone in Ex. 1603 ? " Hint. This volume consists of two pyramids and a spherical sector. Ex. 1605. A spherical shell 2 inches in thickness contains the same amount of material as a sphere whose radius is 6 inches. Find the radius of the outer surface of the shell. Ex. 1606. A sphericalshell 3 inches thick has an outer diameter of 16 inches. Find the volume of the shell. " Ex. 1607. Find the volume of a sphere circumscribed about a rec-tangular parallelopiped whose edges are 3, 4, and 12. Ex. 1608. Find the volume of a sphere inscribed in a cube whose volume is 686 cubic centimeters. Ex. 1609. The surface of a sphere and the surface of a cube are each equal to ". Find the ratio of their volumes. Which is the greater ? Ex. 1610. In a certain sphere the volume and the circumference of a great circle have the same numerical value. Find the surface and the vol-ume of the sphere. Ex. 1611. How many bullets \ of an inch in diameter can be made from a sphere of lead 10 inches in diameter ? from a cube of lead whose edge is 10 inches ? 1006. Defs. A spherical wedge is a solid closed figure whose bounding surface consists of a lune and the planesof the sides of the lune. The lune is called the base of the spherical wedge, and the angle of the lune the angle of the spherical wedge. 1007. Prove, by superposition, the following property of wedges : In equal spheres, or in the same sphere,two sphericalwedges are equal iftheir angles are equal. 466 SOLID GEOMETRY 1010. Defs. A spherical pyramid is a solid closed figure whose bounding surface consists of a sphericalpolygon and the planesof the sides of the spherical polygon. The sphericalpolygon is the base, and the center of the sphere the vertex, of the spherical pyramid. 1011. By comparison with " 957,b and " 958, prove the following prop-erty of spherical pyramids : In equalspheres, or in the same sphere, two triangular spherical pyramids whose bases are symmetrical sphericaltriangles are equivalent. 1012. Cor. n. The volume of a spherical triangular pyramid is equal to the product of its base and one third the radius of the sphere. Outline of Proof 1. Pyramid O-AB'c' =c= pyramid (y-A'BC ("1011). 2. .-. pyramid O-ABC + pyramid O-AB'C1 =c= wedge A = 2 A . i R ("1009); pyramid O-ABC + pyra-mid O-AB'C = wedge B = 2 B " ^ R; pyramid O-ABC + pyramid O-ABC' " wedge C = 2 c " \ R. 3. .-. twice pyramid O-ABC + hemispnere = 2 (A + B -f C)-\| R. 4. .-. twice pyramid O-ABC + 360 " i R = 2 (A + B -f c)i R. 5. .-. pyramid O-ABC = (A+ B + C" 180)J R = AABC "\R = K - \R. q.e.d. 1013. Cor. in. The volume of any spherical pyramid is equal to the product of its base and one third the ra-dius of the sphere. (Hint. Compare with "805.) Ex. 1613. Show that the formula of " 997 is a special case of "" 1004, 1009 and 1013. Ex. 1614. In a sphere whose radius is 12 inches,find the volume of a spherical pyramid whose base is a triangle with angles 70",80", and 90". BOOK IX 467 MISCELLANEOUS EXERCISES ON SOLID GEOMETRY Ex. 1615. A sphericalpyramid whose base is an equiangular penta-gon is equivalent to a wedge whose angle is 30". Find an angle of the base of the pyramid. Ex. 1616. The volume of a sphericalpyramid whose base is an equi-angular sphericaltriangle with angles of 105" is 128 ir cubic inches. Find the radius of the sphere. Ex. 1617. In a sphere whose radius is 10 inches, find the angle of a sphericalwedge equivalent to a sphericalsector whose base has an alti-tude of 12 inches. Ex. 1618. Find the depth of a cubical tank that will hold 100 gallons of water. Ex. 1619. The altitude of a pyramid is H. At what distance from the vertex must a plane be passed parallel to the base so that the part cut off is one half of the whole pyramid ? one third ? one nth ? Ex. 1620. Allowing 550 pounds of copper to a cubic foot,find the weight of a copper wire \ of an inch in diameter and 2 miles long. Ex. 1621. Disregarding quality,and consideringoranges as spheres, i.e. as similar solids,determine which is the better bargain, oranges averaging 2f inches in diameter at 15 cents per dozen, or oranges averaging 3^ inches in diameter at 30 cents per dozen. Ex. 1622. In the figure,B, C, and D are the mid-points of the edges of the cube meet-ing at A. What part of the whole cube is the pyramid cut off by plane BCD ? Hint. Consider ABC as the base and D as the vertex of the pyramid. Ex. 1623. Is the result of Ex. 1622 the same if the figureis a rectangularparallelopiped ? Ex. 1624. It is proved in calculus that in order that a cylindrical tin can closed at the top and having a given capacity may requirethe small-est possibleamount of tin for its construction, the diameter of the base must equal the height of the can. Find the dimensions of such a can holding 1 quart ; 2 gallons. Ex. 1625. A cylindrical tin can holding 2 gallonshas its height equal to the diameter of its base. Another cylindrical tin can with the same capacity has its height equal to twice the diameter of its base. Find the ratio of the amount of tin required for making the two cans. Is your answer consistent with the fact contained in Ex. 1624 ? any parallelopiped ? 468 SOLID GEOMETRY Ex. 1626. A cannon ball 12 inches in diameter is melted, and the lead is cast in the form of a cube. Find the edge of the cube. Ex. 1627. The cube of Ex. 1626 is melted, and the lead is cast in the form of a cone, the diameter of whose base is 12 inches. Find the altitude of the cone. Ex. 1628. Find the weight of the cannon ball in Ex. 1626 if a cubic foot of iron weighs 450 pounds. Ex. 1629. The planes determined by the diagonalsof a cube divide the cube into six equal pyramids. Ex. 1630. Let D, E, F, and G be the mid-points of VA, AB, BC, and CV, respectively, of triangularpyramid V-ABC. Prove DEFG a parallelogram. Ex. 1631. In the figure, is plane DEFQ par-yt allel to edge AC? to edge VB? Prove that any r yr l\ section of a triangularpyramid made by a plane Si~~"\A parallel to two oppositeedges is a parallelogram. yS \ I \ Ex. 1632. The three lines joining the mid-^^^Ijr" ,"7 \tr pointsof the oppositeedges of a tetrahedron bisect F^JrE each other and hence meet in a point. " Hint. Draw DF and EG. Are these two of the required lines ? Ex. 1633. In a White Mountain two-quart ice cream freezer, the can is 4f inches in diameter and 6\| inches high ; the tub is 6\| inches in diameter at the bottom, 8 inches at the top, and 9f inches high, inside measurements, (a) Does the can actually hold 2 quarts ? (6) How many cubic inches of ice can be packed about the can ? Ex. 1634. Find the total area of a regular tetrahedron whose alti-tude is a centimeters. Ex. 1635. The lateral faces of a triangular pyramid are equilateral triangles, and the altitude of the pyramid is 6 inches. Find the total area. Ex. 1636. In the foundation work of the Woolworth Building, a 55-story building on Broadway, New York City, it was necessary, in order to penetrate the sand and quicksand to bed rock, to sink the caissons that contain the huge shafts of concrete to a depth, in some instances, of 131 feet. If the largest circular caisson, 19 feet in diameter, is 130 feet deep and was filled with concrete to within 30 feet of the surface, how many loads of concrete were required,considering1 cubic yard to a load ? Ex. 1637. From A draw a line meeting line XFin B ; let G be the mid-point of AB. Find the locus of C as B moves in line XY. Ex. 1638. In Ex. 1637, let XY be a plane. Find the locus of C as B moves arbitrarily in plane XY. BOOK IX 469 Ex. 1639. A granite shaft in the form of a frustum of a square pyramid contains 16 If cubic feet of granite; the edges of the bases are 4 feet and \ feet,respectively. Find the height of the shaft. Ex. 1640. The volume of a regularsquare pyramid is 42" cubic feet ; its altitude is twice one side of the base, (a) Find the total surface of the pyramid ; (6) find the area of a section made by a plane parallel to the base and one foot from the base. Ex. 1641. Allowing'l cubic yard to a load, find the number of loads of earth in a railway cut \| mile in length,the average dimensions of a cross section being as represented in the figure, the numbers denoting feet. Give the name of the geometrical solid represented by the cut. 12 "7" Why is it not a frustum of a pyramid ? 16 Ex. 1642. For protectionagainstfire, a tank in the form of a frustum of a rightcircular cone was placed in the tower room of a certain public building. The tank is 16 feet in diameter at the bottom, 12 feet in di-ameter at the top, and 16 feet deep. If the water in the tank is never allowed to get less than 14 feet deep, how many cubic feet of water would be available in case of an emergency ? how many barrels,counting i\ cubic feet to a barrel ? Ex. 1643. A sphere with radius B is inscribed in a cylinder,and the cylinderis inscribed in a cube. Find : (a) the ratio of the volume of the sphere to that of the cylinder; (b) the ratio of the cylinder to the cube ; (c) the ratio of the sphere to the cube. Ex. 1644. A cone has the same base and altitude as the cylinderin Ex. 1643. Find the ratio of the cone : (a) to the sphere ; (") to the cylinder; (c) to the cube. Ex. 1645. In a steam-heated house the heat for a room was supplied by a series of 10 radiators each 3 feet high. gl The average cross section of a radiator is s\ TTTN shown in the figure, the numbers denoting yj 2 "L/ inches. It consists of a rectangle with a " semicircle at each end. Find the total radiating surface in the room. Ex. 1646. A coffee pot is 5 inches deep, 4J inches in diameter at the top, and 5^ inches in diameter at the bottom. How many cups of coffee will ithold, allowing6 cups to a quart ? (Answer to nearest whole number.) Ex. 1647. Any plane passing through the center of a parallelopiped divides it into two equivalentsolids. Are these solids equal ? Ex. 1648. From two points,P and B, on the same side of plane AB, two lines are drawn to point 0 in plane AB, making equal angles with the^plane. Find the locus of point O. (Hint. See Ex. 1237.) 470 SOLID GEOMETRY Ex. 1649. A factorychimney is in the form of a frustum of a regu-lar square pyramid. The chimney is 120 feet high, and the edges of its bases are 12 feet and 8 feet,respectively. The flue is 6 feet square throughout. How many cubic feet of material does the chimney contain? Ex. 1650. Find the edge of the largest cube that can be cut from a regularsquare pyramid whose altitude is 10 inches and one side of whose base is 8 inches, if one face of the cube lies in the base of the pyramid. Ex. 1651. Fig. 1 represents a granite monument, the numbers denoting inches. The main part of the stone is 5 feet high, the total height of the stone being 5 feet 6 inches. Fig. 2 represents a view of ,8 20 8 Fig. 3. the main part of the stone looking directlyfrom above. Fig. 3 repre-sents a view of the top of the stone looking directly from above. Calcu-late the volume of the stone. Hint. From Fig. 2 it is seen that the main part of the stone con-sists of a rectangularparallelopiped A, four righttriangular prisms B, and a rectangular pyramid at each corner. Fig. 3 shows that the top con-sists of a righttriangular prism and two rectangularpyramids. Ex. 1652. The monument in Ex. 1651 was cut from a solid rock in the form of a rectangularparallelopiped. How many cubic feet of granite were wasted in the cutting? Ex. 1653. In the monument of Ex. 1651 the two ends of the main part, and the top, have a rock finish, the front and rear surfaces of the main part being polished. Find the number of square feet of rock finish and of polishedsurface. Ex. 1654. The base of a regularpyramid is a triangleinscribed in a circle whose radius is B, and the altitude of the pyramid is 2B. Find the lateral area of the pyramid. Ex. 1655. Find the weight in pounds of the water requiredto fill the tank in Ex. 1323, if a cubic foot of water weighs 1000 ounces. BOOK IX 471 Ex. 1656. By using the formula obtained in Ex. 1543, find the vol-ume of the sphere inscribed in a regular tetrahedron whose edge is 12. Ex. 1657. By using the formula obtained in Ex. 1544,find the volume of the sphere circumscribed about a regulartetrahedron whose edge is 12. Ex. 1658. A hole 6 inches in diameter was bored through a sphere 10 inches in diameter. Find the vol-ume of the part cut out. Hint. The part cut out consists of two spherical cones and the solid generated by revolvingisosceles A BOC about X7as an axis. Ex. 1659. Check your result for Ex. 1658 by findingthe volume of the part left. Ex. 1660. Find the area of the sphericalsurface left in Ex. 1658. Ex. 1661. Four spheres,each with a radius of 6 inches,are placed on a plane surface in a triangularpile, each one being tangent to each of the others. Find the total height of the triangularpile. Ex. 1662. Find the total height of a triangular pileof spheres,each with radius of 6 inches,if there are three layers;four layers; n layers. FORMULAS OF SOLID GEOMETRY 1014. In addition to the notation given in " 761,the follow-ing will be used : A, B, C, """ = number of degrees in the angles of a sphericalpolygon . a, b, c, """ = sides of a spherical polygon. B = base of sphericalsec-tor, wedge, and pyramid. C = circumference of base in general or of lower base of frus-tum of cone, c = circumference of up-per base of frustum of cone. D = diameter of a sphere. E = sphericalexcess of a sphericaltriangle. H = K = L = N = S: T W: Z : altitude of zone or spherical sector, area of a sphericaltriangleor sphericalpolygon, area of lune. number of degrees in the angle of a lune or wedge, radius of base in general, of lower base of frustum of cone, or of.sphere, radius of upper base of frustum of cone, area of surface of a sphere, sum of the angles of a spheri-cal polygon, volume of a wedge, area of a zone. 472 SOLID GEOMETRY FIGURE I Prism. Right prism. Regular pyramid. Frustum of regularpyramid. Rectangular parallelopiped. Cube. Rectangular parallelopipeds. Rectangular parallelopiped. Rectangular parallelopipeds. Any parallelopiped. Parallelopipeds. Triangular prism. Any prism. Prisms. Triangular pyramid. Any pyramid. Pyramids. Similar tetrahedrons. Frustum of any pyramid. Truncated righttriangular prism. Right circular cylinder. Similar cylindersof revolution. Right circular cone. Similar cones of revolution. FORMULA 8 = P E. 8=P-H. S=%P.L. V=a-b-c. V=E\ V a-b-c V B' . H i V_=E^ V E's' V=}H(B + b+y/Tn"). V= \B(E + E' + E"). 8 = C-H. 8 = 2wBH. T=2ttB(H+B). S_= H _B T =BT _B T' H'2 B'2' 8 = \C-L. 8 = w BL. T=ttB(L + B). " __ H _ Z2 . B_ 8' H'2~ L'2~ B'2' REFERENCE "762. ' " 763. "766. "767. "778. "779. "780. "782. "783. "790. "792. "797. "799. "801. "804. "805. "807. "812. "815. "817. "858. " 859. "859. "864. "864. "873. "875. "875. "878. 474 SOLID GEOMETRY APPENDIX TO SOLID GEOMETRY SPHERICAL SEGMENTS 1015. Defs. A spherical segment is a solid closed figure whose bounding surface consists of a zone and two parallel planes. SphericalSegment of Two Bases SphericalSegment of One Base The sections of the sphereformed by the two parallel planes are called the bases of the spherical segment. 1016. Defs. State, by aid of "" 976 and 977, definitions of: (a)Altitude of a spherical segment. (6) Segment of one base. Proposition I. Problem 1017. To derive a formula for the volume of a spher-ical segment in terms of the radii of its bases and its altitude. E Given spherical segment generated by ABCD revolvingabout EF as an axis, with its volume denoted by r, its altitude by h, and the radii of its bases by rx and r2, respectively. To derive a formula for V in terms of i\, r2, and h. APPENDIX 477 . \ pyramid P-ADC = J H " mx+ } II " mx + \H " mj = \ H " 4 j%. .. the volume of all lateral pyramids = \ H"4tM. .'. V=\H"B + \H-b+\H."M="rH(B+b + 4:M). Q.E.F. Rs. 1665. By substituting in the prismatoid formula, derive the formula for : (a) the volume of a prism (" 799) ; (6) the volume of a pyramid (" 805) ; (c) the volume of a frustum of a pyramid (" 815). Ex. 1666. Solve Ex. 1651 by applying the prismatoid formula to each part of the monument. SIMILAR POLYHEDRONS 1022. The student should prove the following : (a) Any two homologous edges of two similar polyhedrons have the same ratio as any other two homologous edges. (b)Any two homologous faces of two similar polyhedrons have the same ratio as the squares of any two homologous edges. (c) The total surfaces of two similar polyhedrons have the same ratio as the squares of any two homologous edges. 1023. Def. The ratio of similitude of two similar polyhe-drons is the ratio of any two homologous edges. 1024. Def. If two polyhedrons AB CD """ and a'b'c'd' ... are so situated that lines from a point 0 to A',b',C', d'}etc., are divided by points g' C' A, B, C, D, etc., in such ^^^7--^/\ a manner that ^"^^^/\ ! \ oa[ _ q" _ q" _ qif _ ^ oa ~ ob ~ oc ~ od ~ E' the two polyhedrons are said to be radially placed. Ex. 1667. Construct two polyhedrons radially placed and so that point 0 lies between the two polyhedrons ; within the two polyhedrons. See " 811. In this discussion only convex polyhedrons will be considered. 478 SOLID GEOMETRY Proposition III. Theorem 1025. Any two radially placed polyhedrons are similar. (SeeFig. 2 below.) Given polyhedronsEC and 'E'c' radially placed with respect to point 0. To prove polyhedronEC ~ polyhedron E'c'. AB, BC, CD, and DA are IIrespectively to a'b',B'c', c'd\ and D'A'. " 415. .-. ABCD IIA'B'C'D', and is similar to it. " 756, II. Likewise each face of polyhedron EC is ~ to the correspond-ing face of polyhedronE'c', and the faces are similarly placed. Again, face All IIface a'h', and face AF II face A'F'. .-. dihedral Z AE = dihedral Z A'e'. Likewise each dihedral Z of polyhedron EC is equal to its correspondingdihedral Z of polyhedron E'c'. .-. each polyhedral Z of polyhedron EC is equal to its corre-sponding polyhedral Z of polyhedronE'c'. " 18. .-. polyhedron EC ~ polyhedron E'c'. "811. q.e.d. Proposition IV. Theorem 1026. Any two similar polyhedrons may be radially placed. q' Fig. i. Fig. 2. Given two similar polyhedronsXM and E'". To prove that XM and E'c' may be radially placed. APPENDIX 479 Outline of Proof 1. Take any point 0 within polyhedron E'C? and construct polyhedron EC so that it is radially placedwith respect to E'C1 and so that OA' : 0A = OB' : OB = "" = A'B' : KL. 2. Then polyhedronEC ~ polyhedronE'c'. " 1025. 3. Prove that the dihedral A of polyhedron EC are equal, respectively, to the dihedral A of polyhedron XM, each being equal,respectively, to the dihedral A of polyhedronE'c'. 4. Prove that the faces of polyhedron EC are equal,respec-tively, t6 the faces of polyhedronXM. 5. Prove, by superposition, that polyhedronEC = XM. 6. .-. polyhedron XM may be placed in the position of EC. 7. But EC and E'c' are radially placed. 8. .-. XM and E'c' may be radially placed. q.e.d. Proposition V. Theorem 1027. If a pyramid is cut by a plane parallel to its base : I. The pyramid cut off is similar to the given pyra-mid. II. The two pyramids are to each other as the cubes of any two homologous edges. O The proofs are left as exercises for the student. Hint. For the proof of II, pass planes through OB' and diagonals B'D', B'E', etc., dividingeach of the pyramids into triangularpyramids. Then pyramid O-BCD ~ pyramid O-B'C'D' ; pyramid O-EBD ~ pyra-mid O-E'B'D', etc. Use " 812 and a method similar to that used in " 505. 480 SOLID GEOMETRY Proposition VI. Theorem. 1028. Two similar polyhedrons are to each other as the cubes of any two homologous edges. C Given two similar poly-hedrons XM and E'c',with their volumes denoted by V and V',respectively, and with KL and A'b' two homol. edges. :3 To prove KL A'B1 Place XM in positionEC, so that XM and E'C} are radially placedwith respectto point0 within both polyhedrons. " 1026. Denote the volumes of pyramids O-ABCD, O-AEFB, etc.,by vx, v2, etc., and the volumes of pyramids O-A'b'c'd', O-A'e'f'b', etc., by v-iyv2', etc. -^ ; ^ = =3; etc-"1027,11. j%,= -("1022, a); AB A^B' A'E'3 ;i _ "k' AB IB1 Vi + V 4-AB ArB' "401. polyhedron EC __ AB3 . V __ KL3 polyhedron E'C' ^B13 ' V' J^3 Q.E.D. 1029. Note. Since " 1028 was assumed earlyin the text (see"814), the teacher will find plenty of exercises throughout Books VII, VIII, and IX illustrating this principle. INDEX (The numbers refer to articles.) Art. Acute triangle 98 Adjacent angles 42 Adjacent dihedral angles . . .671 Alternation 396 Altitude, of cone 842 of cylinder 825 of parallelogram .... 228 of prism 733 of prismatoid .... 1020 of pyramid 751 of spherical segment . . 1016 of trapezoid 229 of triangle 100 of zone 976 Analysis, of a proportion . . . 397 of constructions .... 151, 152, 176 (Ex. 187) , 274, 499 of exercises 274 Angle 38 acute 47 at center of circle .... 292 at center of regular polygon 534 bisector of . . . 53, 127, 599 central 292 degree of 71 dihedral 666 inscribed in circle .... 363 inscribed in segment . . . 364 magnitude of ... 41, 669 obtuse 48 of lune 983 of one degree 71 of spherical wedge . . . 1006 of two intersecting arcs . 916 polyhedral .692 reflex 49 right 45, 71 right and left side of . . .199 sides of 38 solid 692 spherical 917 straight 69 tetrahedral 698 trihedral 698 vertex of 38 Angles 38 added 43 adjacent 42 alternate exterior .... 182 ' alternate interior . . . .182 Art. Angles, complementary ... 68 corresponding 182 designation of . 39, 50, 667, 696 difference of 44 equal 44 exterior 87, 182 exterior interior 182 homologous . . . 417, 424 interior 86, 182 oblique 51 of a polygon 86 subtracted ...... 44 sum of 43 supplementary 69 supplementary-adjacent " " 72 vertical 70 Antecedents 384 Apothem 533 Application 79 Arc 121, 280 degree of 296 length of 551 major 291 minor 291 Arcs, similar 565 Area, of a figure 471 of circle 558 of rectangle .... 468-471 of surface 467 Argument only 171 Arms, of isosceles triangle . . 94 of right triangle 96 Assumptions 12, 54 Axiom 55 Axioms 54 note on 600 Axis, of circle of sphere . . . 906 of right circular cone . . . 844 Base, of cone 840 of isosceles triangle .... 94 of polygon 99 of pyramid 748 of spherical pyramid . . 1010 of spherical sector . . . 1002 Bases, of cylinder 822 of prism 727 of spherical segment . . . 1015 of zone 975 Birectangularsphericaltriangle . 952 481 482 INDEX Art. Bisector, of angle . . .53, 127, 599 of line . , . . . . 52, 145 Bisectors of angles of triangle . 126 Center, of circle .... 119, 278 of regular polygon . . . .531 of sphere 901 Center-line 325 Central angle . . ... . . 292 Centroid 265 Chord 281 Circle 119, 276 arc of 121, 280 area of 558 center of 119, 278 chord of 281 circumference of . 119, 277, 550 circumscribed 300 determined 324 diameter of 282 escribed 322 inscribed 317 radius of ..... 119, 278 secant of 285 sector of 287, 564 segment of 288 tangent to 286 Circles, concentric 326 tangent, externally . . . 330 tangent, internally . . . 330 Circular cone 841 Circular cylinder 831 Circumference . . 119, 277, 550 length of 550, 552 measurement of .... 540 Circumscribed circle .... 300 Circumscribed polygon . . . 317 Circumscribed polyhedron . . 926 Circumscribed prism .... 852 Circumscribed pyramid . . . 868 Circumscribed sphere .... 929 Clockwise motion ..... 86 Closed figure . 83, 714, 715, 934, 935 Closed line 82 Coincidence 17 Collinear segments .... 28 Commensurable quantities 337, 342, 469 Common measure 337 Compasses, use of 31 Complementary angles ... 68 Complete demonstration . . .171 Composition 398 Composition and division . . 400 Concentric circles 326 Conclusion 57 Concurrent lines 196 Cone 839 altitude of 842 Art. Cone, base of 840 circular 841 element of 840 frustum of 879 lateral surface of .... 840 oblique 843 of revolution 876 plane tangent to 866 right circular 843 spherical 1003 vertex of 840 volume of 892,6 Cones, similar 877 Conical surface 837 directrix of 838 element of 838 generatrix of 838 lower nappe of . . . . . 838 upper nappe of 838 vertex of 838 Consequents 384 Constant 346, 347 Construction 123 of triangles 269 Continued proportion .... 405 Converse theorem 135 Convex polyhedral angle . . . 697 Coplanar points, lines,planes . 668 Corollary 58 Counter-clockwise motion . . 50 Cube 742 Cylinder 821 altitude of 825 bases of 822 circular 831 element of 822 lateral surface of .... 822 oblique 824 of revolution 861 plane tangent to 850 right _ . -823 right circular 832 right section of 830 volume of 888 Cylinders, similar 863 Cylindrical surface 819 directrix of .820 element of 820 generatrix of 820 Definition, test of 138 Degree, of angle 71 of arc 296 spherical 986 Demonstration 56, 79 complete 171 Determined circle 324 Determined line and point . 25, 26 Determined plane 608 484 INDEX Art. Historical notes Lambert .. . 569 Lindemann 569 Menelaus of Alexandria . . 992 Metius of Holland . . . 569 Morse, S. F. B 520 Newton 542 Origin of geometry .... 474 Plato .... 268, 542, 787 Plutarch 371 Pons asinorum 114 Ptolemy . 569 Pythagoras 344,510,514,723,787 Richter 569 Seven Wise Men . . . .371 Shanks 569 Snell 946 Socrates 787 Squaring the circle . . . 569 Stobseus 114 Thales 371, 510 Zeno 354 Homologous parts .... 109, 110, 417,418,424 Hypotenuse 96 Hypothesis . .. 57 Icosahedron 719 Inclination of line to plane . . 665 Incommensurable quantities . . 339, 343, 344, 470 Indirect proof 159, 161 Inscribed angle 363 Inscribed circle 317 Inscribed polygon 299 Inscribed polyhedron . . . .928 Inscribed prism 851 Inscribed pyramid 867 Inscribed sphere 927 Instruments, use of 31 Intersection of two surfaces . . 614 Inversion 395 Isoperimetric figures .... 572 Isosceles trapezoid 227 Isosceles triangle 94 arms of 94 base of 94 sides of 94 vertex angle of 94 Lateral area, of frustum . . . 760 of prism 760 of pyramid 760 of right circular cone . . . 872 of right circular cylinder . 857 Lateral edges, of prism . . . 727 of pyramid 748 Lateral faces, of prism .... 727 of pyramid ..,,,, 748 Art. Lateral surface, of cone . . . 840 of cylinder 822 of frustum of pyramid ,m .760 Length, of arc 551 of circumference . . . 550, 552 of perpendicular 166 of secant 426 of tangent 319 Limits . . 346, 349-351, 590-594 Line 4, 7, 11, 27 bisected 52, 145 broken 30 closed 82 curved 29 determined 25 divided externally .... 406 divided harmonically . . . 434 divided in extreme and mean ratio 464, 465 (Ex. 763), 526 divided internally .... 406 inclination of 665 oblique to plane 630 of centers 325 parallelto plane 629 perpendicular to plane . . . 619 projection of 656 right 23 segments of ... . 27, 406 straight 21, 23 tangent to sphere . . . .921 Line segment 27 Lines 20 concurrent 196 difference of 31 divided proportionally . . 407 homologous 109, 110, 418, 424 oblique 46 parallel 177 perpendicular 45 product of . 425, 475, 476, 511 rectangle of 425, 475, 476, 511 Locus .... 129, 130, 131, 601 as an assemblage . . . .130 as a path 130 finding of a 144 of a point 130 of all points 130 problem, solution of . 143, 60,1 Lune 982 angle of 983 sides of 983 vertices of 983 Magnitude of an angle . . 41, 669 Major arc 291 Maximum 571 Mean proportional . . . . .386 Means -385 INDEX 485 Art. Measure, common 337 numerical . . . 335, 467, 770 unit of 335, 466 Measurement 335 of angle 358 of arc 358 of circle . . . . . . 540, 558 of circumference . . . 540, 552 of distances by means of triangles 115 of line 335, 595 of rectangle .... 468-471 of surface 466 Measure-number 335, 467, 471, 595, 770 Median, of trapeze id . . . .251 of triangle 102 Median center of triangle . . . 265 Methods of attack 103, 104, 106, 110 (Ex. 62), 115, 151, 152, 158, 159, 176 (Ex. 187), 269-274, 275, 397, 398, 425, 499, 502, 513 Minimum 571 Minor arc 291 Nappes, upper and lower . 746, 838 Numerical measure . 335, 467, 770 Oblique angles 51 Oblique cone 843 Oblique cylinder 824 Oblique lines 46 Obtuse triangle 97 Octahedron 719 One to one correspondence 595, 596 Opposite theorem 136 Optical illusions 81 Outline of proof 171 Parallel lines 177 Parallel planes 631 Parallelogram 220 altitude of 228 base of 228 Parallelograms classified . 223, 243 Parallelopiped 739 rectangular 741 right 740 Pentagon 92 Perimeter of polygon .... 85 Perpendicular 45, 166, 619, 620, 672 foot of 621 Perpendicular planes .... 672 Pi , 554 evaluation of 568 Plane 32, 33, 34 determined 608 ' ' perpendicular to line . . " 620 Art. Plane, tangent to cone .... 866 tangent to cylinder .... 850 tangent to sphere . . . .921 Plane angle of dihedral angle . 670 Plane figure 36 Plane geometry 36 Planes, parallel 631 perpendicular 672 postulate of 615 Point 5, 6, 11 determined 26 of tangency 286 Polar distance of circle . . .911 Polar triangle 943 Poles of circle 907 Polygon 84 angles of 86 base of 99 circumscribed 317 diagonal of 88 equiangular 90 equilateral . -89 exterior angles of .... 87 inscribed 299 perimeter of 85 regular 91, 515 sides of 84 spherical_-^-.v__!__i, " " .936 ^.-verticesof . . . . "" . . . 84 "Polygons, mutually equiangular 417 sides proportional . . . .418 similar 419 Polyhedral angle 692 convex 697 dihedral angles of ... . 694 edges of 694 element of 694 face angles of ..... 694 faces of ....... 694 parts of : 695 vertex of . 693 Polyhedral angles, symmetrical 707 vertical 708 Polyhedron 716 circumscribed about sphere . 926 diagonal of 718 edges of 717 faces of 717 inscribed in sphere . . . 928 regular 720 vertices of 717 Polyhedrons, radially placed . 1024 similar 811, 1022 Pons asinorum 114 Portraits Archimedes 542 Descartes 237 Euclid 114 Gauss 520 486 INDEX Art. Portraits Plato 787 Pythagoras 510 Thales . 371 Postulate 15 circle 122 of planes 615 parallelline .... 178, 179 revolution ... 40, 54, 606 straightline 24, 54 transference 16, 54 Prism 726 altitude of 733 bases of 727 circumscribed about cylinder 852 inscribed in cylinder . . .851 lateral area of 760 lateral edges of 727 lateral faces of 727 oblique 731 quadrangular 732 regular 730 right 729 right section of 728 triangular 732 truncated . 736 Prismatic surface 724 Prismatoid 1019 altitude of 1020 Problem defined 59 Problems of computation ... 59 Problems of construction . . . 59 analysis of . . 151, 152, 176, (Ex. 187), 274, 499 discussion of 123 proof of 123 solution of 123 Product of lines 425, 475, 476, 511 Projection, of line . . . 451, 656 of point 450, 655 Proof 79, 123 analytic method of ... . 275 by exclusion .... 159, 161 by successive substitutions 275 indirect . . . . . 159, 161 necessity for 81 outline of 171 reductio ad absurdum . 159, 161 synthetic method of . . . 275 Proportion 382 analysis of 397 antecedents of 384 by alternation 396 by composition 398 by composition and division 400 by division 399 by inversion 395 consequents of 384 continued .,.,,, 405 Art. Proportion, extremes of . . . 385 means of 385 terms of ....... 383 ways of writing .... 382 Proportional, fourth .... 387 mean 386 third 386 Proportions simplified . . 389, 397 Proposition 56 Pyramid 747 altitude of 751 base of 748 circumscribed about cone . 868 frustum of 754 inscribed in cone .... 867 lateral area of 760 lateral edges of 748 lateral faces of 748 quadrangular 749 regular . 752 slant height of 764 spherical 1010 triangular 749, 750 truncated . 753 vertex of . 748 Pyramidal surface 744 Quadrangular prism .... 732 Quadrangular pyramid . . . 749 Quadrant 295 Quadrilateral 92 Quadrilaterals classified . 220, 243 Radially placed polyhedrons . 1024 Radius, of circle . . . 119, 278 of regular polygon .... 532 of sphere 901 Ratio .... 340-343, 382, 384 antecedent of 384 consequent of 384 extreme and mean .... 464, 465 (Ex. 763), 526 of any two surfaces . . .472 of similitude .... 418, 1023 of two commensurables . . 342 of two incommensurables . 343 of two magnitudes . . . .341 of two solids 775 Rectangle 223 area of 468-471 of two lines 425, 475, 476, 511 Rectangular parallel opiped . -741 Rectilinear figure 37 Reductio ad absurdum . 159, 161 Regular polygon . . . . 91, 515 angle at center of ... . 534 apothem of 533 center of 531 radius of 532 Regular polyhedron , . , , 720 INDEX 487 Art. Regular prism 730 Regular pyramid 752 Related variables . . . 352, 594 Rhomboid 224 Rhombus 226 Right circular cone 843 axis of 844 lateral area of 872 slant height of 865 Right circular cylinder . . . 832 lateral area of 857 Right cylinder 823 Right parallelopiped .... 740 Right prism 729 Right section, of cylinder . . 830 of prism . 728 Right triangle 96 arms of 96 hypotenuse of 96 sides of 96 Ruler, use of ...... 31 Scale, drawing to 437 Scalene triangle 93 Secant 285 length of 426 Sector 287, 564 spherical 1002 Segment, of circle 288 of line 27 of sphere 1015 Segments, added 31 collinear 28 difference of 31 of a line 406 similar 565 subtracted 31 sum of 31 Semicircle 289 Semicircumference .... 290 Series of equal ratios . . 401, 405 Sides,of angle 38 of polygon 84 Similar arcs . 565 Similar cones of revolution . . 877 Similar cylinders of revolution . 863 Similar polygons 419 Similar polyhedrons . .811, 1022 Similar sectors 565 Similar segments 565 Similarity of triangles . . . .431 Similitude, ratio of . . . 418, 1023 Slant height,of frustum of cone . 881 of frustum of pyramid . . 765 of pyramid 764 of right circular cone . . . 865 Small circle of sphere .... 905 "Solid, geometric . . . . 2, 9, 11 $olid angle 692 Art. Solid geometry 602 Solution, of exercises .... 275 of locus problems .... 143 of problems 123 of theorems 79 Sphere 900 center of 901 circumscribed about poly-hedron 929 diameter of 901 great circle of 904 inscribed in polyhedron . . 927 line tangent to 921 plane tangent to .... 921 radius of 901 small circle of 905 surface of 970 volume of 996,c Spheres, tangent externally . . 922 tangent internally .... 922 tangent to each other . . . 922 Spherical angle 917 Spherical cone 1003 Spherical degree 983 Spherical excess 989 Spherical polygon 936 angles of 936 diagonal of 937 sides of 936 vertices of 936 Spherical polygons, symmetrical 956 Spherical pyramid .... 1010 base of 1010 vertex of 1010 Spherical sector 1002 base of 1002 Spherical segment . . . . 1015 altitude of 1016 bases of 1015 of one base 1016 Spherical surface 970 Spherical triangle 938 birectangular 952 trirectangular 953 Spherical wedge 1006 angle of 1006 base of 1006 Square 225 Squaring the circle 569 Straight angle 69 Straight line 21, 23 determined 25 oblique to plane 630 parallel to plane .... 629 perpendicular to plane . . 619 Straightedge 21 Summary, of divisions of cir-cumference 529 of equal triangle theorems . 118 488 INDEX Akt. Summary, of formulas of plane geometry 570 of parallel line theorems . . 197 of quadrilaterals .... 243 of similar triangle theorems . 431 of trapezoids . . Exs. 320, 321 of unequal angle theorems . 174 of unequal line theorems . . 175 Superposition .... 19, 103, 106 Supplemental triangles . . . 948 Supplementary angles .... 69 Supplementary-adjacent angles . 72 Surface 3, 8, 11 closed 714 conica 837 curved 35 cylindrical 819 of sphere 970 plane 32, 33, 34 prismatic 724 pyramidal 744 Surfaces 32 Symmetrical polyhedral angles . 707 Symmetrical spherical polygons . 956 Tangent 286 external common .... 333 internal common .... 333 length of 319 Tetrahedral angle 698 Tetrahedron 719, 750 Theorem 57 converse of 135 opposite of 136 solution of 79 Third proportional . . . . . 386 Transformation . . ... . . 493 Transversal 181 Trapezium 222 Trapezoid . . ... . . . .221 altitude of 229 bases of 229 isosceles 227 median of 251 Triangle 92 acute 98 altitudes of 100 bisectors of angles of . . .126 centroid of 265 construction of 269 determined .... 133, 269 equilateral 95 isosceles 94 median center of .... 265 medians of 102 notation of 270 obtuse 97 parts of 269 Art. Triangle, polar 943 right 96 scalene 93 spherical -. 938 vertex angle of .... 94, 99 vertex of 99 Triangles, classified .... 93-98 similarity of 431 supplemental 948 Triangular prism 732 Triangular pyramid . . . 749, 750 Trihedral angle 698 birectangular 699 isosceles 700 rectangular 699 trirectangular 699 Trirectangular spherical triangle 953 Truncated prism 736 Truncated pyramid .... 753 Unit, of measure .... 335, 466 of volume 769 Variable . . . 346, 348, 585-594 approaches its limit . . . 350 independent 346 limit of 349 reaches its limit 351 related 352, 594 Vertex, of angle 38 of cone 840 of conical surface .... 838 of polyhedral angle . . . 693 of pyramid 748 of pyramidal surface . . . 745 of spherical pyramid . . 1010 of triangle 99 Vertex angle of triangle . . 94, 99 Vertical angles 70_ Vertical polyhedral angles . . 708 Vertices, of lune 983 of polygon 84 of polyhe ron 717 of spherical polygon . . . 936 Volume, of cone 892, b of cylinder . . . . . .888 of rectangular parallelopiped 774 of solid 769, 770 of sphere 996, c unit of . 769 Wedge, spherical 1006 Zone 975 altitude of 976 bast of 975 of one base 977 MILNE'S STANDARD. ALGEBRA By WILLIAM J. MILNE, Ph.D., LL.D., President of the New York State Normal College, Albany, N. Y. $I.OO THE Standard Algebra conforms to the most recent courses of study. The inductive method of presentation is followed, but declarative statements and observations are used, instead of questions. Added to this kind of unfold-ing and development of the subject are illustrative problems and explanations to bring out specific points, the whole being driven home by varied and abundant practice. ^f The problems are fresh in character, and besides the tradi-tional problems include a largenumber drawn from physics, geometry, and commercial life. They are classified accord-ing to the nature of the equationsinvolved, not according to subject matter. The statement of necessary definitions and of principles is clear and concise, but the proofs of principles, except some important ones, are left for the maturer years of the pupil. ^[ Accuracy and self-reliance are encouraged by the use of numerous checks and tests, and by the requirement that re-sults be verified. The subject of graphs is treated after simple equations, introduced by some of their simple uses in repre-senting statistics, and in picturing two related quantities in the process of change, and again after quadratics. Later they are utilized in discussing the values of quadratic expressions. Factoring receives particular attention. Not only are the usual cases given fully and completely with plenty of practice, but the factor theorem is taught. ^f The helpful and frequent reviews are made up of pointed oral questions, abstract exercises, problems, and recent college entrance examination questions. The book is unusuallyhandy in size and convenient for the pocket. The page size is small. ROBBINS'S PLANE TRIGONOMETRY By EDWARD R. ROBBINS, Senior Mathematical Mas-ter, William Perm Charter School, Philadelphia, Pa. $0.6o THIS book is intended for beginners. It aims to givea thorough familiarity with the essential truths,and a satisfactory skill in operating with those processes. It is illustrated in the usual maimer, but the diagramsare more than usually clear-cut and elucidating. ^[ The work is sound and teachable, and is written in clear and concise language, in a style that makes it easily under-stood. Immediatelyafter each principle has been proved, it is applied firstin illustrativeexamples, and then further im-pressed by numerous exercises. Accuracy and rigor of treat-ment are shown in every detail, and all irrelevant and ex-traneous matter is excluded, thus givinggreater prominence to universal rules and formulas. "fj The references to Plane Geometry preceding the first chapter are invaluable. A knowledge of the principles of geometry needed in trigonometry is, as a rule, too freely taken for granted. The author givesat the beginning of the book a statement of the applied principles, with reference to the sections of his Geometry, where such theorems are provedin full. Cross references in the text of the Trigonometry to those theorems make it easy for the pupilto review or to supplement imperfect knowledge. ^\| Due emphasis is given to the theoretical as well as to the practical applications of the science. The number of ex-amples, both concrete and abstract, is far in excess of those in other books on the market. This book contains four times as many exercises as most books, and twice as many as that having the next lowest number. AMERICAN BOOK COMPANY (313)
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You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Display square matrix as square Ask Question Asked 4 years, 10 months ago Modified4 years, 10 months ago Viewed 725 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. First off, before anyone jumps over with the 'duplicate' attacks: I've searched the forum a lot, but none of the solutions I found actually work anymore (they seem to be outdated or the reason is otherwise beyond my comprehension). My biggest hopes and most thorough attempts were related to the suggestions here. Anyways, I'm currently using pdfLaTex with Overleaf. The title pretty much describes what I'm trying to do, but here is a MWE for visualization: ```latex \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{mathdots} \usepackage{amsmath} \begin{document} Initial and good looking: [ \Delta_9 = \begin{vmatrix} \frac{8}{9} & 8 & \dots&\dots&\dots & 8 & -63\ \frac{8}{8} & \vdots & & & \iddots & -56 & 8\ \frac{8}{7} & \vdots & & \iddots & -49 & \iddots & \vdots\ \vdots&\vdots&\iddots&\iddots&\iddots& &\vdots\ \frac{8}{3} & 8 & -21 & \iddots & & & \vdots\ \frac{8}{2} & -14 & \iddots & & & & \vdots\ -7 & 8 & \dots & \dots & \dots & \dots & 8\ \end{vmatrix}] After some elementary operations and ugly looking: [ \begin{vmatrix} -7.1 & 8 & 8 & \dots & \dots & \dots & 8\ \frac{8}{2} +7 & -7.2 - 8 & 0 & \dots & \dots & \dots & 0\ \frac{8}{3} +7 & 0 & -7.3 - 8 & \ddots & & & \vdots\ \vdots&\vdots&\ddots&\ddots&\ddots& &\vdots\ \frac{8}{n-2} +7 & \vdots & & \ddots & -7(n-2) - 8 & \ddots & \vdots\ \frac{8}{n-1} +7 & \vdots & & & \ddots & -7(n-1) - 8 & 0\ \frac{8}{n} +7 & 0 & \dots&\dots&\dots & 0 & -7n - 8 \end{vmatrix} ] \end{document} ``` Which produces: Very important for me is in this case, that this square matrix is actually displayed as a square. That is, every entry inside to get formatted to a square (and not a rectangle as it is happening by default) with respect to the biggest of the two dimensions (either height or length) of the highest/widest entry. I really don't mind the whole thing becoming too gigantic as long as it's nicely visually represented and comprehensible. I'd love for the formatting to happen dynamically, since I've got a lot of matrices and I really wouldn't want to format each one separately with custom spacing values. (which is among the suggestions I've come across most often) I need this because of the dots, which I use to intuitively visualize the structure of the matrix. So, if you've got any better suggestions for approaching comprehensible 'dotting' of rows, columns and diagonals, that's also an alternative solution. And in general, maybe there's a 'best' way of tackling this need for dynamic comprehensive dotting of matrices, which I'm unaware of, so any clarification in this regard is also welcome! That seems to be it. Thank you a lot in advance! spacing matrices Share Share a link to this question Copy linkCC BY-SA 4.0 Improve this question Follow Follow this question to receive notifications edited Nov 19, 2020 at 16:11 Bernard 277k 14 14 gold badges 163 163 silver badges 384 384 bronze badges asked Nov 19, 2020 at 14:57 D. PetrovD. Petrov 147 7 7 bronze badges Add a comment\| 1 Answer 1 Sorted by: Reset to default This answer is useful 2 Save this answer. Show activity on this post. Try this code. ```latex \documentclass{article} \usepackage[utf8]{inputenc} \usepackage{mathdots} \usepackage{amsmath} \usepackage{array}% added \begin{document} Initial and good looking: [ \Delta_9 = \begin{vmatrix} \frac{8}{9} & 8 & \dots&\dots&\dots & 8 & -63\ \frac{8}{8} & \vdots & & & \iddots & -56 & 8\ \frac{8}{7} & \vdots & & \iddots & -49 & \iddots & \vdots\ \vdots&\vdots&\iddots&\iddots&\iddots& &\vdots\ \frac{8}{3} & 8 & -21 & \iddots & & & \vdots\ \frac{8}{2} & -14 & \iddots & & & & \vdots\ -7 & 8 & \dots & \dots & \dots & \dots & 8\ \end{vmatrix}] After some elementary operations and better looking: \renewcommand{\arraystretch}{3} % added [ \begin{vmatrix} -7.1 & 8 & 8 & \dots & \dots & \dots & 8\ \frac{8}{2} +7 & -7.2 - 8 & 0 & \dots & \dots & \dots & 0\ \frac{8}{3} +7 & 0 & -7.3 - 8 & \ddots & & & \vdots\ \vdots&\vdots&\ddots&\ddots&\ddots& &\vdots\ \frac{8}{n-2} +7 & \vdots & & \ddots & -7(n-2) - 8 & \ddots & \vdots\ \frac{8}{n-1} +7 & \vdots & & & \ddots & -7(n-1) - 8 & 0\ \frac{8}{n} +7 & 0 & \dots&\dots&\dots & 0 & -7n - 8 \end{vmatrix} ] \end{document} ``` UPDATE I wrote a macro to squarefy the matrix without manual adjustments (\arraystretch is calculated). Requires the mathtools package. ```latex \documentclass{article} \usepackage{mathtools} % added \usepackage{calc} \usepackage{mathdots} \newcommand{\getlengthratio}{% \number\numexpr \dimexpr#1\relax / \dimexpr#2\relax \relax} \newcommand{\GetMatrix}{} \newlength{\MWi} \newlength{\MWii} \newcounter{MWiii} \newcommand{\squarefy}{% \renewcommand{\GetMatrix}{} \renewcommand{\arraystretch}{1} \renewcommand{\GetMatrix}{#1} \setlength{\MWi}{\widthof{$\displaystyle\GetMatrix$}} \setlength{\MWii}{\heightof{$\displaystyle\GetMatrix$}} \setcounter{MWiii}{\getlengthratio{0.5\MWi}{\MWii}} \renewcommand{\arraystretch}{\theMWiii} %% added [#2#1]} \begin{document} \squarefy{% \begin{vmatrix} \frac{8}{9} & 8 & \dots&\dots&\dots & 8 & -63\ \frac{8}{8} & \vdots & & & \iddots & -56 & 8\ \frac{8}{7} & \vdots & & \iddots & -49 & \iddots & \vdots\ \vdots&\vdots&\iddots&\iddots&\iddots& &\vdots\ \frac{8}{3} & 8 & -21 & \iddots & & & \vdots\ \frac{8}{2} & -14 & \iddots & & & & \vdots\ -7 & 8 & \dots & \dots & \dots & \dots & 8\ \end{vmatrix} }{\Delta_9 = } \squarefy{% \begin{vmatrix} -7.1 & 8 & 8 & \dots & \dots & \dots & 8\ \frac{8}{2} +7 & -7.2 - 8 & 0 & \dots & \dots & \dots & 0\ \frac{8}{3} +7 & 0 & -7.3 - 8 & \ddots & & & \vdots\ \vdots&\vdots&\ddots&\ddots&\ddots& &\vdots\ \frac{8}{n-2} +7 & \vdots & & \ddots & -7(n-2) - 8 & \ddots & \vdots\ \frac{8}{n-1} +7 & \vdots & & & \ddots & -7(n-1) - 8 & 0\ \frac{8}{n} +7 & 0 & \dots&\dots&\dots & 0 & -7n - 8 \end{vmatrix} }{} \end{document} ``` Share Share a link to this answer Copy linkCC BY-SA 4.0 Improve this answer Follow Follow this answer to receive notifications edited Nov 19, 2020 at 22:42 answered Nov 19, 2020 at 15:47 Simon DispaSimon Dispa 42.5k 2 2 gold badges 19 19 silver badges 42 42 bronze badges 3 I tried playing around with this approach, but doesn't it actually require a certain constant stretching factor (in your MWE, that's 3) which depends on the context and I'll have to set it manually for every matrix? Or is there some other magic happening behind the scenes that I'm unaware of?D. Petrov –D. Petrov 2020-11-19 16:01:03 +00:00 Commented Nov 19, 2020 at 16:01 1 I don't know other magic. On the other hand setting arraystretch manually is something that can be done case by case during the final tuning of the document. Other solutions, as measuring the matrix width and calculate the stretching involve much more typing. But maybe another expert can provide a macro command ...Simon Dispa –Simon Dispa 2020-11-19 18:10:22 +00:00 Commented Nov 19, 2020 at 18:10 @D. Petrov Please see the updated answer,Simon Dispa –Simon Dispa 2020-11-19 22:43:12 +00:00 Commented Nov 19, 2020 at 22:43 Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions spacing matrices See similar questions with these tags. 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1472	https://community.the-hospitalist.org/content/guide-management-adnexal-masses-pregnancy	A guide to management: Adnexal masses in pregnancy \| MDedge Skip to main content A guide to management: Adnexal masses in pregnancy Breadcrumb Home User login Username Password Reset your password Article Type News Changed Tue, 08/28/2018 - 10:54 Display Headline A guide to management: Adnexal masses in pregnancy Author(s) Mitchel S. Hoffman, MD Robyn A. Sayer, MD Fast Track The serum CA-125 level is typically elevated during the first trimester, but may be useful for assessment later in pregnancy Some ovarian cancers may present acutely, such as a rapidly growing germ-cell tumor or a ruptured and hemorrhaging granulosa-cell tumor During the first and second trimesters, laparoscopy is as safe as laparotomy In general, if malignancy is suspected, a vertical incision is preferred CASE 1 An enlarging cystic tumor A 20-year-old gravida 3 para 1011 visits the emergency department with persistent right flank pain. Although ultrasonography (US) shows a 21-week gestation, the patient has had no prenatal care. Imaging also reveals a right-sided ovarian tumor, 14×11×8 cm, that is mainly cystic with some internal echogenicity. At 30 weeks’ gestation, a gynecologic oncologist is consulted. Repeat US reveals the mass to be about 20 cm in diameter and cystic, without internal papillation. The patient’s CA-125 level is 12 U/mL. Based on this information, the physicians decide the likely finding is a benign ovarian cystadenoma. How should they proceed? The discovery of an adnexal mass during pregnancy isn’t as rare as you might think—depending on when and how closely you look, it occurs in about 1 in 100 gestations. In most cases, we have found, the mass is clearly benign (TABLE 1), warranting only observation. TABLE 1 Adnexal masses removed during pregnancy: Histologic profile \| HISTOLOGIC TYPE \| NUMBER (%) \| --- \| \| Cystadenoma \| 549 (33) \| \| Dermoid \| 451 (27) \| \| Paraovarian/paratubal \| 204 (12) \| \| Functional \| 237 (14) \| \| Endometrioma \| 55 (3) \| \| Benign stromal \| 28 (2) \| \| Leiomyoma \| 23 (1.5) \| \| Luteoma \| 8 (0.5) \| \| Miscellaneous \| 55 (3) \| \| Malignant \| 68 (4) \| \| Total \| 1,678 \| \| Data supplied by the authors from surgical experience \| In the case described above, the physicians followed the patient and removed the mass at term because it was cystic with no other indications of malignancy. At 37 weeks’ gestation, a cesarean section was performed through a midline laparotomy incision, followed by removal of the ovarian tumor, which was benign. The pathologist measured the tumor at 16×12×4 cm and determined that it was a corpus luteum cyst. Presence of mass raises questions Despite the rarity of malignancy, the discovery of an ovarian mass during pregnancy prompts several important questions: How should the mass be assessed? How can the likelihood of malignancy be determined as quickly and efficiently as possible, without jeopardy to the pregnancy? When is surgical intervention warranted? And when can it be postponed? Specifically, is elective operative intervention for a tumor that is probably benign appropriate during pregnancy? When is the best time to operate? And what is the optimal surgical route? In this article, we address these questions with a focus on intervention. As we’ll explain, only a small percentage of gravidas who have an adnexal mass require surgery during pregnancy. When surgery is necessary, it is usually indicated for an emergent problem or suspicion of malignancy. Even when ovarian cancer is confirmed, we have found that it is usually in its early stages and therefore has a favorable prognosis (TABLE 2). TABLE 2 Malignant adnexal masses removed during pregnancy \| HISTOLOGIC TYPE \| NUMBER (%) \| --- \| \| Epithelial \| 101 (28) \| \| Borderline epithelial \| 147 (40) \| \| Germ-cell dysgerminoma \| 47 (13) \| \| Other \| 34 (9) \| \| Stromal \| 24 (7) \| \| Undifferentiated \| 5 (1.4) \| \| Sarcoma \| 2 (0.5) \| \| Metastatic \| 4 (1.1) \| \| Total \| 364 \| \| Data supplied by the authors from surgical experience \| How should a mass be assessed? Ultrasonography and other imaging often reveal the presence of a mass and help determine whether it is benign or malignant. In fact, most adnexal masses discovered during pregnancy are incidental findings at the time of routine prenatal US. (see the most commonly found tumors.) Operative intervention is required in 3 situations: malignancy is suspected an acute complication develops the sheer size of the tumor is likely to cause difficulty. Common adnexal tumors found during pregnancy Corpus luteum A persistent corpus luteum is a normal component of pregnancy. Although it usually appears as a small cystic structure on ultrasonographic imaging, the corpus luteum of pregnancy can reach 10 cm in size. Other types of “functional” ovarian cysts may also be found during pregnancy. Most functional cysts resolve by the early second trimester.4,6 In rare cases, a cyst may develop complications such as torsion or rupture, causing acute pain or hemorrhage. Otherwise, a cystic tumor identified in the first trimester should be characterized and followed using ultrasonography (US). Benign neoplasm An adnexal mass that persists beyond the first trimester is more likely to be a neoplasm.3-5,10,11,22 Such a mass is generally considered clinically significant if it exceeds 5 cm in diameter and has a complex sonographic appearance. Usually such a neoplasm will be a benign cystadenoma or cystic teratoma.5,10-13,19,23,24 Benign cystic teratoma This tumor can be identified with a fairly high degree of specificity using a variety of imaging techniques, with management based on the presumptive diagnosis. This tumor is unlikely to grow substantially during pregnancy. When it is smaller than 6 cm, such a tumor can simply be observed.14 A larger tumor can occasionally rupture or lead to torsion or obstruction of labor, but such occurrences are rare. Benign cystadenoma In an asymptomatic patient with imaging that suggests a benign cystadenoma (see sonogram), benign cystic teratoma, or other benign tumor, observation is reasonable in most cases.4,6,7,9-11,14,19 Operative intervention is required when there is less certainty regarding the benign nature of the tumor, an acute complication develops, or the tumor is expected to pose problems because of its large size alone. Uterine leiomyoma It is rare for an ovarian tumor detected during pregnancy to have a solid appearance on US. When it does, it may be a uterine leiomyoma mimicking an adnexal tumor (see intraoperative photograph). It should be reevaluated with more detailed US or magnetic resonance imaging.25 Malignancy About 10% of adnexal masses that persist during pregnancy are malignant, according to recent series.4,5,7-10,12,13,24,26 Most of the ovarian cancers diagnosed during pregnancy are epithelial, and a substantial portion of these are low-malignant-potential (LMP) tumors.5,10,11,13,19,23,24,26,27 This ratio is in keeping with the age of these women, which also explains the stage distribution (most are stage 1) and the large percentage of germ-cell tumors detected. The majority of ovarian cancers discovered in pregnant women have a favorable prognosis. Benign-appearing cystadenoma A morphologically benign-appearing, large, cystic adnexal mass can be seen in association with an 11-week gestation. Leiomyoma mimics an ovarian tumor This 17-week gestation was marked by a large pedunculated leiomyoma that at fist appeared to be a right adnexal tumor. Appearance of adnexal masses on US A functional cyst such as a follicular cyst, corpus luteum cyst, or theca lutein cyst usually has smooth borders and a fluid center. Other cysts may sometimes contain debris, such as clotted blood, that suggests endometriosis or a simple cyst with bleeding into it. A benign cystic teratoma often has multiple tissue lines, evidence of calcification, and layering of fat and fluid contents. A benign cystadenoma usually has the appearance of a simple cyst without large septates, whereas a cystadenocarcinoma often contains septates, abnormal blood flow, increased vascularity, or all of these. However, it is impossible to definitively distinguish a cystadenoma from a cystadenocarcinoma using US imaging alone. Functional cysts usually resolve by the second trimester. A cyst warrants closer scrutiny when it persists, is larger than 5 cm in diameter, or has a complex appearance on US. CA-125 may be useful after the first trimester The serum CA-125 level is typically elevated during the first trimester, but may be useful during later assessment or for follow-up of a malignancy.1 A markedly elevated serum level of alpha-fetoprotein (fractionated in some cases) has been reported in some gravidas with an endodermal sinus or mixed germ-cell ovarian tumor.2 Alpha-fetoprotein should be measured when there is suspicion for a germ-cell tumor based on clinical or US findings. When a mass is discovered during cesarean section Occasionally, an adnexal mass is detected at the time of cesarean section (FIGURE 1).3 This phenomenon is increasingly common, given the large number of cesarean deliveries in the United States. To eliminate the need for future surgery and avoid a delay in the diagnosis of an ovarian malignancy, inspect the adnexa routinely after closing the uterine incision in all women who deliver by cesarean section. FIGURE 1 Mass discovered at cesarean section This cystic tumor was discovered at cesarean section that was undertaken for obstetric indications. CASE 2 LMP tumor is suspected A 36-year-old gravida 3 para 1011 makes a prenatal visit during the first trimester. Her previous delivery was a cesarean section through a Pfannenstiel incision for a breech presentation. US imaging reveals a 6-week, 5-day fetus and a complex left adnexal mass, 4.5×3.9×4.1 cm. Imaging is repeated 1 month later at a tertiary-care center and shows an 11-week viable fetus, a right ovary with a corpus luteum cyst, and a left ovary with a 6.6×4 cm cystic mass with extensive vascular surface papillations that is suspicious for a low-malignant-potential (LMP) tumor. In several sonograms prior to the pregnancy, this mass appeared to be solid and was 3 cm in size. When is surgery warranted? Surgery is indicated when physical examination or imaging of a pregnant woman reveals an adnexal mass that is suspicious for malignancy, but the physician must weigh the benefit of prompt surgery against the risk to the pregnancy. This equation can be complicated in several ways. For example, surgical staging of clinically early ovarian cancer is more difficult due to the pregnant uterus, which is more extensively manipulated during these procedures. In addition, an optimal operation sometimes necessitates removal of the uterus. At 13 weeks’ gestation, the patient described in case 2 underwent laparoscopy with peritoneal washings and left salpingo-oophorectomy, but the tumor ruptured during removal. Final pathology showed it to be a serous LMP tumor involving the surface of the left ovary. Washings were in line with this diagnosis. The pregnancy continued uneventfully, and a repeat cesarean section was performed at 37 weeks through the Pfannenstiel scar, followed by limited surgical staging. Exploration and all biopsies were negative, and the final diagnosis was a stage 1C serous LMP tumor of the ovary. The patient articulated a desire to preserve her fertility and was monitored with US imaging of the remaining ovary every 6 months. Does ‘indolent’ behavior of malignancy justify watchful waiting? LMP tumors comprise a relatively large percentage of ovarian “cancers” encountered during pregnancy. Some authors report the accurate identification of these tumors prospectively, based on ultrasonographic characteristics.4,5 When an LMP tumor is the likely diagnosis, serial observation during pregnancy may be appropriate because of the indolent nature of the tumor. Further studies are needed to refine preoperative diagnosis and determine the overall safety of this approach. When the problem is acute In rare cases, a pregnant patient will have (or develop during observation) an acute problem due to torsion or rupture of an adnexal mass. Some ovarian cancers may present acutely, such as a rapidly growing malignant germ-cell tumor or a ruptured and hemorrhaging granulosecell tumor. Emergent surgery is necessary to manage the acute adnexal disease and reduce the likelihood of pregnancy loss. These events are infrequent, occurring in less than 10% of women with a known, persistent adnexal mass during pregnancy.4-14 Furthermore, recent studies have not found a substantial pregnancy complication rate associated with such emergency surgeries. CASE 3 Suspicious mass, ascites signal need for surgery A 19-year-old gravida 1 para 0 seeks prenatal care at 17 weeks’ gestation, complaining of rapidly enlarging abdominal girth. The physical examination estimates gestational size to be considerably greater than dates, but US is consistent with a 17-week intrauterine pregnancy. Imaging also reveals a 12-cm heterogenous left adnexal mass and a large amount of ascites. Surgery is clearly warranted, but how extensive should it be? When a malignancy is detected, a thorough staging procedure may be justified, depending on gestational age, exposure, desires of the patient, and operative findings. A midline incision is preferred. Pregnant and nonpregnant women with stage 1A or 1C epithelial ovarian cancer who undergo fertility-preserving surgery (with chemotherapy in selected patients) have a good prognosis and a high likelihood of achieving a subsequent normal pregnancy.15 The same is true for women with a malignant germ-cell tumor of the ovary, even when disease is advanced.16 However, careful surgical staging is necessary. The most important consideration when deciding whether to continue the pregnancy is the need for adjuvant chemotherapy. Depending on the gestational age and diagnosis, a short delay (4 to 6 weeks) may be appropriate to allow the pregnancy to progress beyond the first trimester or to maturity. In case 3, a laparotomy was performed at 19 weeks’ gestation via a midline incision, and approximately 5.3 L of ascites was evacuated. A large, nonadherent left ovarian tumor was removed. The right ovary appeared to be normal, as did the gravid uterus, which was minimally manipulated. The rest of the surgical exploration was normal, and the distal portion of the omentum was excised. The frozen-section diagnosis was a malignant stromal tumor. Final pathology showed an 18×13.5×8.8 cm, poorly differentiated, SertoliLeydig-cell tumor with heterologous elements in the form of mucinous epithelium. The omentum was negative for tumor. Chemotherapy was initiated in the third trimester, based on the limited data available, with intravenous etoposide and platinum administered every 21 days. The patient received 3 cycles of chemotherapy prior to delivery. At 37 weeks’ gestation, labor was successfully induced. After delivery, bleomycin was added to the chemotherapy regimen, and 3 additional courses with all 3 agents were administered. The patient was lost to follow-up shortly after completing chemotherapy. Clearly, an informed discussion of the options with the patient is imperative before any surgery, especially when chemotherapy may be delayed. Pregnancy does not appear to alter the prognosis for the patient with an ovarian malignancy, and ovarian cancer has not been reported to metastasize to the fetus. Integrating evidence and experience Pregnant women have a very low rate of ovarian cancer Leiserowitz GS, Xing G, Cress R, Brahmbhatt B, Dalrymple JL, Smith LH. Adnexal masses in pregnancy: how often are they malignant? Gynecol Oncol. 2006;101:315–321. Ovarian malignancies are rare during pregnancy. When they do occur, they are likely to be early stage and to have a favorable outcome, according to this recent population-based study. Using 3 large databases containing records on 4,846,505 California obstetric patients between 1991 and 1999, Leiserowitz and colleagues identified 9,375 women who had an ovarian mass associated with pregnancy. Of these, 87 had ovarian cancer and 115 had a low-malignant-potential (LMP) tumor, for a cancer occurrence rate of 0.93%, or 0.0179 per 1,000 deliveries. Thirty-four of the 87 cancers were germ-cell tumors. Of the 87 ovarian cancers, 65.5% were localized, 6.9% regional, 23% remote, and 4.6% of unknown stage. The respective rates for LMP tumors were 81.7%, 7.8%, 4.4%, and 6.1%. Women with malignant tumors were more likely than pregnant controls without cancer to undergo cesarean delivery, hysterectomy, transfusion, and prolonged hospitalization. These women did not, however, have a higher rate of adverse neonatal outcomes. When cancer is advanced Few data shed light on whether a pregnancy should continue when ovarian cancer is advanced.17 The definitive surgical approach must be highly individualized. It is not always possible to make an accurate diagnosis based on a frozen section. In such a case, the pregnancy should be preserved until the time of definitive diagnosis. As always, the patient’s wishes and gestational age must be considered. How factors besides malignancy can influence care Most persistent adnexal masses move well out of the pelvis as pregnancy progresses. Occasionally, however, an ovarian tumor may be located in the posterior cul-de-sac even at term, a fact easily confirmed by examination or US.4,7 A tumor in the posterior cul-de-sac can obstruct delivery or rupture. When it has a benign cystic appearance on US, it may be decompressed via transvaginal aspiration. Otherwise, the best approach is cesarean section and concomitant management of the mass. When size alone is the problem Some ovarian tumors are so large they seem incompatible with an advancing pregnancy. Tumors up to 20 cm in diameter have been removed intact at the time of cesarean section (FIGURE 2).18 The tumor may accommodate in shape and become less problematic as it is gradually pushed into the upper abdomen (FIGURE 3). The ability of the peritoneal cavity to accommodate a tumor varies greatly among women. As pregnancy advances, the likelihood that a large cystic mass will rupture tends to increase. Depending on the circumstances, percutaneous aspiration7,18 or removal of a benign-appearing cystic tumor may be appropriate. FIGURE 2 Even a very large tumor may coexist with advancing pregnancy This benign serous cystadenoma was exteriorized at the time of cesarean section at term. FIGURE 3 Large ovarian tumor has accommodated to the pregnancy Laparotomy—performed at term for cesarean section and to manage this large tumor—revealed that the tumor had accommodated in shape between the enlarging pregnant uterus and the abdominal wall. When is the best time to operate? Surgery is generally not recommended during the first trimester.5-11 Among the reasons are the high likelihood of a corpus luteum cyst, the low likelihood of an invasive malignancy, the low risk of adnexal complications associated with observation, and the potential for pregnancy loss or teratogenicity. However, as pregnancy progresses beyond the first trimester, surgery poses other problems: Operative exposure diminishes and the need to manipulate the pregnant uterus increases. Integrating evidence and experience Can surgery be delayed when a mass is detected? Schmeler KM, Mayo-Smith WW, Peipert JF, Weitzen S, Manuel MD, Gordinier ME. Adnexal masses in pregnancy: surgery compared with observation. Obstet Gynecol. 2005;105:1098–1103. Close observation is a reasonable alternative to operative intervention during pregnancy, unless a malignancy is suspected. Schmeler and colleagues reviewed the cases of 59 women who had an adnexal mass larger than 5 cm in diameter detected during pregnancy, out of a total of 127,177 deliveries at a single institution between 1990 and 2003. Antepartum surgery was performed in 17 women (29%). Of these, 13 cases had ultrasonographic findings suggesting malignancy, and 4 had ovarian torsion. The remaining women were observed, with surgery delayed until the time of cesarean section or later. Twenty-five of the 59 masses (42%) were dermoid cysts. Cancer was diagnosed in 4 patients (6.8%), and 1 patient (1.7%) had an LMP tumor. All 5 cases (100%) involving a malignancy had a suspicious US appearance and were identified during antepartum surgery, whereas only 12 patients with a benign tumor (22%) underwent surgery prior to delivery. Surgery poses risks to the pregnancy Elective surgery for an adnexal mass any time during pregnancy increases the risk of pregnancy loss and the likelihood of intrauterine growth restriction (IUGR) and preterm delivery.5,7,10,13,19 A 1989 study from Sweden20 defined a cohort of 5,405 women (from 720,000 births) who were known to have a nonobstetric operation while pregnant, with the following results: Congenital malformation and stillbirth were not increased in the women undergoing surgery The number of very-low- and low-birth-weight infants did rise, however—the result of both prematurity and IUGR Also elevated was the incidence of infants born alive but dying within 168 hours; these risks increased regardless of trimester No specific type of anesthesia or operation was associated with adverse reproductive outcomes, and the cause of those adverse outcomes was not determined. Some recent data suggest that adnexal surgery during the late second or early third trimester poses the greatest risk of preterm delivery or IUGR, or both.13 Window of opportunity: early to mid- second trimester During this time frame, elective surgery for an adnexal mass still affords some pelvic exposure without the need for significant uterine manipulation and has been associated with a lower risk of pregnancy complications. The other window for operation is at the time of cesarean section. An elective cesarean section is sometimes performed specifically to manage a persistent adnexal mass. Among the factors that warrant consideration when contemplating this approach are the elective uterine incision (with its attendant implications for future pregnancies), the higher risks associated with cesarean delivery in general, the type of skin incision (a vertical incision is appropriate in the event of ovarian malignancy), the potential for better exposure or laparoscopy at a later date, the increased difficulty of ovarian cystectomy at the time of cesarean section, and the patient’s wishes. Laparoscopy or laparotomy? The data on laparoscopy during the first and second trimesters of pregnancy indicate that it is as safe as laparotomy. A 1997 Swedish study21 identified cohorts of 2,181 women undergoing laparoscopy and 1,522 women undergoing laparotomy (from a total of 2,015,000 deliveries) between the fourth and 20th weeks of pregnancy. In both groups there was an increased risk for the infant to weigh less than 2,500 g, to be delivered before 37 weeks, and to have IUGR. There were no differences between the 2 groups for these and other adverse outcomes. Small series of laparoscopic procedures to manage an adnexal mass during pregnancy suggest that this approach is most applicable during the first (for highly selected emergent cases) or early second trimester to manage masses less than 10 cm in diameter, particularly when adnexectomy is planned. Laparoscopy may be considered “minimally invasive” because it reduces manipulation of the pregnant uterus during adnexal surgery. However, it is more difficult to assess and remove ovarian cysts laparoscopically, although an early ovarian malignancy could be staged via laparoscopy by an experienced surgeon. Considerations during laparotomy When performing a laparotomy or cesarean section for an adnexal mass, the surgeon must take into account a number of variables when selecting the type of incision (ie, vertical vs transverse). In general, if malignancy is suspected, or if uterine manipulation is to be minimized, a vertical incision is best. Other considerations include a prior scar, body habitus, obstetric issues, and the patient’s wishes. The author reports no financial relationships relevant to this article. References 1. Spitzer M, Kaushal N, Benjamin F. Maternal CA-125 levels in pregnancy and the puerperium. J Reprod Med. 1998;43:387-392. 2. Aoki Y, Higashino M, Ishii S, Tanaka K. Yolk sac tumor of the ovary during pregnancy: a case report. Gynecol Oncol. 2005;99:497-499. 3. Koonings PP, Platt LD, Wallace R. Incidental adnexal neoplasms at cesarean section. Obstet Gynecol. 1988;72:767-769. 4. Zanetta G, Mariani E, Lissoni A, et al. A prospective study of the role of ultrasound in the management of adnexal masses in pregnancy. BJOG. 2003;110:578-583. 5. Sherard GB, 3rd, Hodson CA, Williams HJ, et al. Adnexal masses and pregnancy: a 12 year experience. Am J Obstet Gynecol. 2003;189:358-363. 6. Bernhard LM, Klebba PK, Gray DL, Mutch DG. Predictors of persistence of adnexal masses in pregnancy. Obstet Gynecol. 1999;93:585-589. 7. Platek DN, Henderson CE, Goldberg GL. The management of a persistent adnexal mass in pregnancy. Am J Obstet Gynecol. 1995;173:1236-1240. 8. Bromley B, Benacerraf B. Adnexal masses during pregnancy: accuracy of sonographic diagnosis and outcome. J Ultrasound Med. 1997;16:447-452. 9. Hill LM, Connors-Beatty DJ, Nowak A, Tush B. The role of ultrasonography in the detection and management of adnexal mass during the second and third trimesters of pregnancy. Am J Obstet Gynecol. 1998;179:703-707. 10. Agarwal N, Parul, Kriplani A, Bhatla N, Gupta A. Management and outcome of pregnancies complicated with adnexal masses. Arch Gynecol Obstet. 2003;267:148-152. 11. Schmeler KM, Mayo-Smith WW, Peipert JF, Weitzen S, Manuel MD, Gordinier ME. Adnexal masses in pregnancy: surgery compared with observation. Obstet Gynecol. 2005;105:1098-1103. 12. Coenen VH, Dunton C, Cardonick E, Berghella V. Persistent adnexal masses during pregnancy. Obstet Gynecol. 1999;93:66S.- 13. Whitecar P, Turner S, Higby K. Adnexal masses in pregnancy: a review of 130 cases undergoing surgical management. Am J Obstet Gynecol. 1999;181:19-24. 14. Caspi B, Levi R, Appelman Z, Rabinerson D, Goldman G, Hagay Z. Conservative management of ovarian cystic teratoma during pregnancy and labor. Am J Obstet Gynecol. 2000;182:503-505. 15. Schilder JM, Thompson AM, DePriest PD, et al. Outcome of reproductive age women with stage IA or IC invasive epithelial ovarian cancer treated with fertility-sparing therapy. Gynecol Oncol. 2002;87:1-7. 16. Tangir J, Zelterman D, Ma W, Schwartz PE. Reproductive function after conservative surgery and chemotherapy for malignant germ cell tumors of the ovary. Obstet Gynecol. 2003;101:251-257. 17. Ferrandina G, Distefano M, Testa A, De Vincenzo R, Scambia G. Management of an advanced ovarian cancer at 15 weeks of gestation: case report and literature review. Gynecol Oncol. 2005;97:693-696. 18. Caspi B, Ben-Arie A, Appelman Z, Or Y, Hagay Z. Aspiration of simple pelvic cysts during pregnancy. Gynecol Obstet Invest. 2000;49:102-105. 19. Usui R, Minakami H, Kosuge S, et al. A retrospective survey of clinical, pathologic, and prognostic features of adnexal masses operated on during pregnancy. J Obstet Gynaecol Res. 2000;26(2):89-93. 20. Mazze RI, Källén B. Reproductive outcome after anesthesia and operation during pregnancy; a registry study of 5,405 cases. Am J Obstet Gynecol. 1989;161:1178-1185. 21. Reedy MB, Källén B, Kuehl TJ. Laparoscopy during pregnancy: a study of five fetal outcome parameters with use of the Swedish health registry. Am J Obstet Gynecol. 1997;177:673-679. 22. Hermans RHM, Fischer D-C, van der Putten HWHM, et al. Adnexal masses in pregnancy. Onkologie. 2003;26:167-172. 23. Hoffman MS. Primary ovarian carcinoma during pregnancy. Clin Consul Obstet Gynecol. 1995;7:237-241. 24. Ueda M, Ueki M. Ovarian tumors associated with pregnancy. Int J Obstet Gynaecol. 1996;55:59-65. 25. Curtis M, Hopkins MP, Zarlingo T, et al. Magnetic resonance imaging to avoid laparotomy in pregnancy. Obstet Gynecol. 1993;82:833-836. 26. Leiserowitz GS, Xing G, Cress R, et al. Adnexal masses in pregnancy; how often are they malignant? Gynecol Oncol. 2006;101:315-321. 27. Rahman MS, Al-Sibai MH, Rahman J, et al. Ovarian carcinoma associated with pregnancy. A review of 9 cases. Acta Obstet Gynecol Scand. 2002;81:260-264. Article PDF 1903OBGM_Article2.pdf Author and Disclosure Information Mitchel S. Hoffman, MD Professor and Director, Division of Gynecologic Oncology, Department of Obstetrics and Gynecology, University of South Florida, Tampa, Fla Robyn A. Sayer, MD Assistant Professor, Division of Gynecologic Oncology, Department of Obstetrics and Gynecology, University of South Florida, Tampa, Fla Issue OBG Management - 19(03) Publications MDedge ObGyn OBG Management Page Number 27-44 Legacy Keywords adnexal masses; laparoscopy; laparotomy; mass during cesarean section; serum CA-125; ovarian cancers; Mitchel S. Hoffman;MD; Hoffman MS; Robyn A. Sayer;MD; Sayer RA; Sections Clinical Review Author(s) Mitchel S. Hoffman, MD Robyn A. Sayer, MD Author(s) Mitchel S. Hoffman, MD Robyn A. Sayer, MD Author and Disclosure Information Mitchel S. Hoffman, MD Professor and Director, Division of Gynecologic Oncology, Department of Obstetrics and Gynecology, University of South Florida, Tampa, Fla Robyn A. Sayer, MD Assistant Professor, Division of Gynecologic Oncology, Department of Obstetrics and Gynecology, University of South Florida, Tampa, Fla Author and Disclosure Information Mitchel S. Hoffman, MD Professor and Director, Division of Gynecologic Oncology, Department of Obstetrics and Gynecology, University of South Florida, Tampa, Fla Robyn A. Sayer, MD Assistant Professor, Division of Gynecologic Oncology, Department of Obstetrics and Gynecology, University of South Florida, Tampa, Fla Article PDF 1903OBGM_Article2.pdf Article PDF 1903OBGM_Article2.pdf Fast Track The serum CA-125 level is typically elevated during the first trimester, but may be useful for assessment later in pregnancy Some ovarian cancers may present acutely, such as a rapidly growing germ-cell tumor or a ruptured and hemorrhaging granulosa-cell tumor During the first and second trimesters, laparoscopy is as safe as laparotomy In general, if malignancy is suspected, a vertical incision is preferred CASE 1 An enlarging cystic tumor A 20-year-old gravida 3 para 1011 visits the emergency department with persistent right flank pain. Although ultrasonography (US) shows a 21-week gestation, the patient has had no prenatal care. Imaging also reveals a right-sided ovarian tumor, 14×11×8 cm, that is mainly cystic with some internal echogenicity. At 30 weeks’ gestation, a gynecologic oncologist is consulted. Repeat US reveals the mass to be about 20 cm in diameter and cystic, without internal papillation. The patient’s CA-125 level is 12 U/mL. Based on this information, the physicians decide the likely finding is a benign ovarian cystadenoma. How should they proceed? The discovery of an adnexal mass during pregnancy isn’t as rare as you might think—depending on when and how closely you look, it occurs in about 1 in 100 gestations. In most cases, we have found, the mass is clearly benign (TABLE 1), warranting only observation. TABLE 1 Adnexal masses removed during pregnancy: Histologic profile \| HISTOLOGIC TYPE \| NUMBER (%) \| --- \| \| Cystadenoma \| 549 (33) \| \| Dermoid \| 451 (27) \| \| Paraovarian/paratubal \| 204 (12) \| \| Functional \| 237 (14) \| \| Endometrioma \| 55 (3) \| \| Benign stromal \| 28 (2) \| \| Leiomyoma \| 23 (1.5) \| \| Luteoma \| 8 (0.5) \| \| Miscellaneous \| 55 (3) \| \| Malignant \| 68 (4) \| \| Total \| 1,678 \| \| Data supplied by the authors from surgical experience \| In the case described above, the physicians followed the patient and removed the mass at term because it was cystic with no other indications of malignancy. At 37 weeks’ gestation, a cesarean section was performed through a midline laparotomy incision, followed by removal of the ovarian tumor, which was benign. The pathologist measured the tumor at 16×12×4 cm and determined that it was a corpus luteum cyst. Presence of mass raises questions Despite the rarity of malignancy, the discovery of an ovarian mass during pregnancy prompts several important questions: How should the mass be assessed? How can the likelihood of malignancy be determined as quickly and efficiently as possible, without jeopardy to the pregnancy? When is surgical intervention warranted? And when can it be postponed? Specifically, is elective operative intervention for a tumor that is probably benign appropriate during pregnancy? When is the best time to operate? And what is the optimal surgical route? In this article, we address these questions with a focus on intervention. As we’ll explain, only a small percentage of gravidas who have an adnexal mass require surgery during pregnancy. When surgery is necessary, it is usually indicated for an emergent problem or suspicion of malignancy. Even when ovarian cancer is confirmed, we have found that it is usually in its early stages and therefore has a favorable prognosis (TABLE 2). TABLE 2 Malignant adnexal masses removed during pregnancy \| HISTOLOGIC TYPE \| NUMBER (%) \| --- \| \| Epithelial \| 101 (28) \| \| Borderline epithelial \| 147 (40) \| \| Germ-cell dysgerminoma \| 47 (13) \| \| Other \| 34 (9) \| \| Stromal \| 24 (7) \| \| Undifferentiated \| 5 (1.4) \| \| Sarcoma \| 2 (0.5) \| \| Metastatic \| 4 (1.1) \| \| Total \| 364 \| \| Data supplied by the authors from surgical experience \| How should a mass be assessed? Ultrasonography and other imaging often reveal the presence of a mass and help determine whether it is benign or malignant. In fact, most adnexal masses discovered during pregnancy are incidental findings at the time of routine prenatal US. (see the most commonly found tumors.) Operative intervention is required in 3 situations: malignancy is suspected an acute complication develops the sheer size of the tumor is likely to cause difficulty. Common adnexal tumors found during pregnancy Corpus luteum A persistent corpus luteum is a normal component of pregnancy. Although it usually appears as a small cystic structure on ultrasonographic imaging, the corpus luteum of pregnancy can reach 10 cm in size. Other types of “functional” ovarian cysts may also be found during pregnancy. Most functional cysts resolve by the early second trimester.4,6 In rare cases, a cyst may develop complications such as torsion or rupture, causing acute pain or hemorrhage. Otherwise, a cystic tumor identified in the first trimester should be characterized and followed using ultrasonography (US). Benign neoplasm An adnexal mass that persists beyond the first trimester is more likely to be a neoplasm.3-5,10,11,22 Such a mass is generally considered clinically significant if it exceeds 5 cm in diameter and has a complex sonographic appearance. Usually such a neoplasm will be a benign cystadenoma or cystic teratoma.5,10-13,19,23,24 Benign cystic teratoma This tumor can be identified with a fairly high degree of specificity using a variety of imaging techniques, with management based on the presumptive diagnosis. This tumor is unlikely to grow substantially during pregnancy. When it is smaller than 6 cm, such a tumor can simply be observed.14 A larger tumor can occasionally rupture or lead to torsion or obstruction of labor, but such occurrences are rare. Benign cystadenoma In an asymptomatic patient with imaging that suggests a benign cystadenoma (see sonogram), benign cystic teratoma, or other benign tumor, observation is reasonable in most cases.4,6,7,9-11,14,19 Operative intervention is required when there is less certainty regarding the benign nature of the tumor, an acute complication develops, or the tumor is expected to pose problems because of its large size alone. Uterine leiomyoma It is rare for an ovarian tumor detected during pregnancy to have a solid appearance on US. When it does, it may be a uterine leiomyoma mimicking an adnexal tumor (see intraoperative photograph). It should be reevaluated with more detailed US or magnetic resonance imaging.25 Malignancy About 10% of adnexal masses that persist during pregnancy are malignant, according to recent series.4,5,7-10,12,13,24,26 Most of the ovarian cancers diagnosed during pregnancy are epithelial, and a substantial portion of these are low-malignant-potential (LMP) tumors.5,10,11,13,19,23,24,26,27 This ratio is in keeping with the age of these women, which also explains the stage distribution (most are stage 1) and the large percentage of germ-cell tumors detected. The majority of ovarian cancers discovered in pregnant women have a favorable prognosis. Benign-appearing cystadenoma A morphologically benign-appearing, large, cystic adnexal mass can be seen in association with an 11-week gestation. Leiomyoma mimics an ovarian tumor This 17-week gestation was marked by a large pedunculated leiomyoma that at fist appeared to be a right adnexal tumor. Appearance of adnexal masses on US A functional cyst such as a follicular cyst, corpus luteum cyst, or theca lutein cyst usually has smooth borders and a fluid center. Other cysts may sometimes contain debris, such as clotted blood, that suggests endometriosis or a simple cyst with bleeding into it. A benign cystic teratoma often has multiple tissue lines, evidence of calcification, and layering of fat and fluid contents. A benign cystadenoma usually has the appearance of a simple cyst without large septates, whereas a cystadenocarcinoma often contains septates, abnormal blood flow, increased vascularity, or all of these. However, it is impossible to definitively distinguish a cystadenoma from a cystadenocarcinoma using US imaging alone. Functional cysts usually resolve by the second trimester. A cyst warrants closer scrutiny when it persists, is larger than 5 cm in diameter, or has a complex appearance on US. CA-125 may be useful after the first trimester The serum CA-125 level is typically elevated during the first trimester, but may be useful during later assessment or for follow-up of a malignancy.1 A markedly elevated serum level of alpha-fetoprotein (fractionated in some cases) has been reported in some gravidas with an endodermal sinus or mixed germ-cell ovarian tumor.2 Alpha-fetoprotein should be measured when there is suspicion for a germ-cell tumor based on clinical or US findings. When a mass is discovered during cesarean section Occasionally, an adnexal mass is detected at the time of cesarean section (FIGURE 1).3 This phenomenon is increasingly common, given the large number of cesarean deliveries in the United States. To eliminate the need for future surgery and avoid a delay in the diagnosis of an ovarian malignancy, inspect the adnexa routinely after closing the uterine incision in all women who deliver by cesarean section. FIGURE 1 Mass discovered at cesarean section This cystic tumor was discovered at cesarean section that was undertaken for obstetric indications. CASE 2 LMP tumor is suspected A 36-year-old gravida 3 para 1011 makes a prenatal visit during the first trimester. Her previous delivery was a cesarean section through a Pfannenstiel incision for a breech presentation. US imaging reveals a 6-week, 5-day fetus and a complex left adnexal mass, 4.5×3.9×4.1 cm. Imaging is repeated 1 month later at a tertiary-care center and shows an 11-week viable fetus, a right ovary with a corpus luteum cyst, and a left ovary with a 6.6×4 cm cystic mass with extensive vascular surface papillations that is suspicious for a low-malignant-potential (LMP) tumor. In several sonograms prior to the pregnancy, this mass appeared to be solid and was 3 cm in size. When is surgery warranted? Surgery is indicated when physical examination or imaging of a pregnant woman reveals an adnexal mass that is suspicious for malignancy, but the physician must weigh the benefit of prompt surgery against the risk to the pregnancy. This equation can be complicated in several ways. For example, surgical staging of clinically early ovarian cancer is more difficult due to the pregnant uterus, which is more extensively manipulated during these procedures. In addition, an optimal operation sometimes necessitates removal of the uterus. At 13 weeks’ gestation, the patient described in case 2 underwent laparoscopy with peritoneal washings and left salpingo-oophorectomy, but the tumor ruptured during removal. Final pathology showed it to be a serous LMP tumor involving the surface of the left ovary. Washings were in line with this diagnosis. The pregnancy continued uneventfully, and a repeat cesarean section was performed at 37 weeks through the Pfannenstiel scar, followed by limited surgical staging. Exploration and all biopsies were negative, and the final diagnosis was a stage 1C serous LMP tumor of the ovary. The patient articulated a desire to preserve her fertility and was monitored with US imaging of the remaining ovary every 6 months. Does ‘indolent’ behavior of malignancy justify watchful waiting? LMP tumors comprise a relatively large percentage of ovarian “cancers” encountered during pregnancy. Some authors report the accurate identification of these tumors prospectively, based on ultrasonographic characteristics.4,5 When an LMP tumor is the likely diagnosis, serial observation during pregnancy may be appropriate because of the indolent nature of the tumor. Further studies are needed to refine preoperative diagnosis and determine the overall safety of this approach. When the problem is acute In rare cases, a pregnant patient will have (or develop during observation) an acute problem due to torsion or rupture of an adnexal mass. Some ovarian cancers may present acutely, such as a rapidly growing malignant germ-cell tumor or a ruptured and hemorrhaging granulosecell tumor. Emergent surgery is necessary to manage the acute adnexal disease and reduce the likelihood of pregnancy loss. These events are infrequent, occurring in less than 10% of women with a known, persistent adnexal mass during pregnancy.4-14 Furthermore, recent studies have not found a substantial pregnancy complication rate associated with such emergency surgeries. CASE 3 Suspicious mass, ascites signal need for surgery A 19-year-old gravida 1 para 0 seeks prenatal care at 17 weeks’ gestation, complaining of rapidly enlarging abdominal girth. The physical examination estimates gestational size to be considerably greater than dates, but US is consistent with a 17-week intrauterine pregnancy. Imaging also reveals a 12-cm heterogenous left adnexal mass and a large amount of ascites. Surgery is clearly warranted, but how extensive should it be? When a malignancy is detected, a thorough staging procedure may be justified, depending on gestational age, exposure, desires of the patient, and operative findings. A midline incision is preferred. Pregnant and nonpregnant women with stage 1A or 1C epithelial ovarian cancer who undergo fertility-preserving surgery (with chemotherapy in selected patients) have a good prognosis and a high likelihood of achieving a subsequent normal pregnancy.15 The same is true for women with a malignant germ-cell tumor of the ovary, even when disease is advanced.16 However, careful surgical staging is necessary. The most important consideration when deciding whether to continue the pregnancy is the need for adjuvant chemotherapy. Depending on the gestational age and diagnosis, a short delay (4 to 6 weeks) may be appropriate to allow the pregnancy to progress beyond the first trimester or to maturity. In case 3, a laparotomy was performed at 19 weeks’ gestation via a midline incision, and approximately 5.3 L of ascites was evacuated. A large, nonadherent left ovarian tumor was removed. The right ovary appeared to be normal, as did the gravid uterus, which was minimally manipulated. The rest of the surgical exploration was normal, and the distal portion of the omentum was excised. The frozen-section diagnosis was a malignant stromal tumor. Final pathology showed an 18×13.5×8.8 cm, poorly differentiated, SertoliLeydig-cell tumor with heterologous elements in the form of mucinous epithelium. The omentum was negative for tumor. Chemotherapy was initiated in the third trimester, based on the limited data available, with intravenous etoposide and platinum administered every 21 days. The patient received 3 cycles of chemotherapy prior to delivery. At 37 weeks’ gestation, labor was successfully induced. After delivery, bleomycin was added to the chemotherapy regimen, and 3 additional courses with all 3 agents were administered. The patient was lost to follow-up shortly after completing chemotherapy. Clearly, an informed discussion of the options with the patient is imperative before any surgery, especially when chemotherapy may be delayed. Pregnancy does not appear to alter the prognosis for the patient with an ovarian malignancy, and ovarian cancer has not been reported to metastasize to the fetus. Integrating evidence and experience Pregnant women have a very low rate of ovarian cancer Leiserowitz GS, Xing G, Cress R, Brahmbhatt B, Dalrymple JL, Smith LH. Adnexal masses in pregnancy: how often are they malignant? Gynecol Oncol. 2006;101:315–321. Ovarian malignancies are rare during pregnancy. When they do occur, they are likely to be early stage and to have a favorable outcome, according to this recent population-based study. Using 3 large databases containing records on 4,846,505 California obstetric patients between 1991 and 1999, Leiserowitz and colleagues identified 9,375 women who had an ovarian mass associated with pregnancy. Of these, 87 had ovarian cancer and 115 had a low-malignant-potential (LMP) tumor, for a cancer occurrence rate of 0.93%, or 0.0179 per 1,000 deliveries. Thirty-four of the 87 cancers were germ-cell tumors. Of the 87 ovarian cancers, 65.5% were localized, 6.9% regional, 23% remote, and 4.6% of unknown stage. The respective rates for LMP tumors were 81.7%, 7.8%, 4.4%, and 6.1%. Women with malignant tumors were more likely than pregnant controls without cancer to undergo cesarean delivery, hysterectomy, transfusion, and prolonged hospitalization. These women did not, however, have a higher rate of adverse neonatal outcomes. When cancer is advanced Few data shed light on whether a pregnancy should continue when ovarian cancer is advanced.17 The definitive surgical approach must be highly individualized. It is not always possible to make an accurate diagnosis based on a frozen section. In such a case, the pregnancy should be preserved until the time of definitive diagnosis. As always, the patient’s wishes and gestational age must be considered. How factors besides malignancy can influence care Most persistent adnexal masses move well out of the pelvis as pregnancy progresses. Occasionally, however, an ovarian tumor may be located in the posterior cul-de-sac even at term, a fact easily confirmed by examination or US.4,7 A tumor in the posterior cul-de-sac can obstruct delivery or rupture. When it has a benign cystic appearance on US, it may be decompressed via transvaginal aspiration. Otherwise, the best approach is cesarean section and concomitant management of the mass. When size alone is the problem Some ovarian tumors are so large they seem incompatible with an advancing pregnancy. Tumors up to 20 cm in diameter have been removed intact at the time of cesarean section (FIGURE 2).18 The tumor may accommodate in shape and become less problematic as it is gradually pushed into the upper abdomen (FIGURE 3). The ability of the peritoneal cavity to accommodate a tumor varies greatly among women. As pregnancy advances, the likelihood that a large cystic mass will rupture tends to increase. Depending on the circumstances, percutaneous aspiration7,18 or removal of a benign-appearing cystic tumor may be appropriate. FIGURE 2 Even a very large tumor may coexist with advancing pregnancy This benign serous cystadenoma was exteriorized at the time of cesarean section at term. FIGURE 3 Large ovarian tumor has accommodated to the pregnancy Laparotomy—performed at term for cesarean section and to manage this large tumor—revealed that the tumor had accommodated in shape between the enlarging pregnant uterus and the abdominal wall. When is the best time to operate? Surgery is generally not recommended during the first trimester.5-11 Among the reasons are the high likelihood of a corpus luteum cyst, the low likelihood of an invasive malignancy, the low risk of adnexal complications associated with observation, and the potential for pregnancy loss or teratogenicity. However, as pregnancy progresses beyond the first trimester, surgery poses other problems: Operative exposure diminishes and the need to manipulate the pregnant uterus increases. Integrating evidence and experience Can surgery be delayed when a mass is detected? Schmeler KM, Mayo-Smith WW, Peipert JF, Weitzen S, Manuel MD, Gordinier ME. Adnexal masses in pregnancy: surgery compared with observation. Obstet Gynecol. 2005;105:1098–1103. Close observation is a reasonable alternative to operative intervention during pregnancy, unless a malignancy is suspected. Schmeler and colleagues reviewed the cases of 59 women who had an adnexal mass larger than 5 cm in diameter detected during pregnancy, out of a total of 127,177 deliveries at a single institution between 1990 and 2003. Antepartum surgery was performed in 17 women (29%). Of these, 13 cases had ultrasonographic findings suggesting malignancy, and 4 had ovarian torsion. The remaining women were observed, with surgery delayed until the time of cesarean section or later. Twenty-five of the 59 masses (42%) were dermoid cysts. Cancer was diagnosed in 4 patients (6.8%), and 1 patient (1.7%) had an LMP tumor. All 5 cases (100%) involving a malignancy had a suspicious US appearance and were identified during antepartum surgery, whereas only 12 patients with a benign tumor (22%) underwent surgery prior to delivery. Surgery poses risks to the pregnancy Elective surgery for an adnexal mass any time during pregnancy increases the risk of pregnancy loss and the likelihood of intrauterine growth restriction (IUGR) and preterm delivery.5,7,10,13,19 A 1989 study from Sweden20 defined a cohort of 5,405 women (from 720,000 births) who were known to have a nonobstetric operation while pregnant, with the following results: Congenital malformation and stillbirth were not increased in the women undergoing surgery The number of very-low- and low-birth-weight infants did rise, however—the result of both prematurity and IUGR Also elevated was the incidence of infants born alive but dying within 168 hours; these risks increased regardless of trimester No specific type of anesthesia or operation was associated with adverse reproductive outcomes, and the cause of those adverse outcomes was not determined. Some recent data suggest that adnexal surgery during the late second or early third trimester poses the greatest risk of preterm delivery or IUGR, or both.13 Window of opportunity: early to mid- second trimester During this time frame, elective surgery for an adnexal mass still affords some pelvic exposure without the need for significant uterine manipulation and has been associated with a lower risk of pregnancy complications. The other window for operation is at the time of cesarean section. An elective cesarean section is sometimes performed specifically to manage a persistent adnexal mass. Among the factors that warrant consideration when contemplating this approach are the elective uterine incision (with its attendant implications for future pregnancies), the higher risks associated with cesarean delivery in general, the type of skin incision (a vertical incision is appropriate in the event of ovarian malignancy), the potential for better exposure or laparoscopy at a later date, the increased difficulty of ovarian cystectomy at the time of cesarean section, and the patient’s wishes. Laparoscopy or laparotomy? The data on laparoscopy during the first and second trimesters of pregnancy indicate that it is as safe as laparotomy. A 1997 Swedish study21 identified cohorts of 2,181 women undergoing laparoscopy and 1,522 women undergoing laparotomy (from a total of 2,015,000 deliveries) between the fourth and 20th weeks of pregnancy. In both groups there was an increased risk for the infant to weigh less than 2,500 g, to be delivered before 37 weeks, and to have IUGR. There were no differences between the 2 groups for these and other adverse outcomes. Small series of laparoscopic procedures to manage an adnexal mass during pregnancy suggest that this approach is most applicable during the first (for highly selected emergent cases) or early second trimester to manage masses less than 10 cm in diameter, particularly when adnexectomy is planned. Laparoscopy may be considered “minimally invasive” because it reduces manipulation of the pregnant uterus during adnexal surgery. However, it is more difficult to assess and remove ovarian cysts laparoscopically, although an early ovarian malignancy could be staged via laparoscopy by an experienced surgeon. Considerations during laparotomy When performing a laparotomy or cesarean section for an adnexal mass, the surgeon must take into account a number of variables when selecting the type of incision (ie, vertical vs transverse). In general, if malignancy is suspected, or if uterine manipulation is to be minimized, a vertical incision is best. Other considerations include a prior scar, body habitus, obstetric issues, and the patient’s wishes. The author reports no financial relationships relevant to this article. Fast Track The serum CA-125 level is typically elevated during the first trimester, but may be useful for assessment later in pregnancy Some ovarian cancers may present acutely, such as a rapidly growing germ-cell tumor or a ruptured and hemorrhaging granulosa-cell tumor During the first and second trimesters, laparoscopy is as safe as laparotomy In general, if malignancy is suspected, a vertical incision is preferred CASE 1 An enlarging cystic tumor A 20-year-old gravida 3 para 1011 visits the emergency department with persistent right flank pain. Although ultrasonography (US) shows a 21-week gestation, the patient has had no prenatal care. Imaging also reveals a right-sided ovarian tumor, 14×11×8 cm, that is mainly cystic with some internal echogenicity. At 30 weeks’ gestation, a gynecologic oncologist is consulted. Repeat US reveals the mass to be about 20 cm in diameter and cystic, without internal papillation. The patient’s CA-125 level is 12 U/mL. Based on this information, the physicians decide the likely finding is a benign ovarian cystadenoma. How should they proceed? The discovery of an adnexal mass during pregnancy isn’t as rare as you might think—depending on when and how closely you look, it occurs in about 1 in 100 gestations. In most cases, we have found, the mass is clearly benign (TABLE 1), warranting only observation. TABLE 1 Adnexal masses removed during pregnancy: Histologic profile \| HISTOLOGIC TYPE \| NUMBER (%) \| --- \| \| Cystadenoma \| 549 (33) \| \| Dermoid \| 451 (27) \| \| Paraovarian/paratubal \| 204 (12) \| \| Functional \| 237 (14) \| \| Endometrioma \| 55 (3) \| \| Benign stromal \| 28 (2) \| \| Leiomyoma \| 23 (1.5) \| \| Luteoma \| 8 (0.5) \| \| Miscellaneous \| 55 (3) \| \| Malignant \| 68 (4) \| \| Total \| 1,678 \| \| Data supplied by the authors from surgical experience \| In the case described above, the physicians followed the patient and removed the mass at term because it was cystic with no other indications of malignancy. At 37 weeks’ gestation, a cesarean section was performed through a midline laparotomy incision, followed by removal of the ovarian tumor, which was benign. The pathologist measured the tumor at 16×12×4 cm and determined that it was a corpus luteum cyst. Presence of mass raises questions Despite the rarity of malignancy, the discovery of an ovarian mass during pregnancy prompts several important questions: How should the mass be assessed? How can the likelihood of malignancy be determined as quickly and efficiently as possible, without jeopardy to the pregnancy? When is surgical intervention warranted? And when can it be postponed? Specifically, is elective operative intervention for a tumor that is probably benign appropriate during pregnancy? When is the best time to operate? And what is the optimal surgical route? In this article, we address these questions with a focus on intervention. As we’ll explain, only a small percentage of gravidas who have an adnexal mass require surgery during pregnancy. When surgery is necessary, it is usually indicated for an emergent problem or suspicion of malignancy. Even when ovarian cancer is confirmed, we have found that it is usually in its early stages and therefore has a favorable prognosis (TABLE 2). TABLE 2 Malignant adnexal masses removed during pregnancy \| HISTOLOGIC TYPE \| NUMBER (%) \| --- \| \| Epithelial \| 101 (28) \| \| Borderline epithelial \| 147 (40) \| \| Germ-cell dysgerminoma \| 47 (13) \| \| Other \| 34 (9) \| \| Stromal \| 24 (7) \| \| Undifferentiated \| 5 (1.4) \| \| Sarcoma \| 2 (0.5) \| \| Metastatic \| 4 (1.1) \| \| Total \| 364 \| \| Data supplied by the authors from surgical experience \| How should a mass be assessed? Ultrasonography and other imaging often reveal the presence of a mass and help determine whether it is benign or malignant. In fact, most adnexal masses discovered during pregnancy are incidental findings at the time of routine prenatal US. (see the most commonly found tumors.) Operative intervention is required in 3 situations: malignancy is suspected an acute complication develops the sheer size of the tumor is likely to cause difficulty. Common adnexal tumors found during pregnancy Corpus luteum A persistent corpus luteum is a normal component of pregnancy. Although it usually appears as a small cystic structure on ultrasonographic imaging, the corpus luteum of pregnancy can reach 10 cm in size. Other types of “functional” ovarian cysts may also be found during pregnancy. Most functional cysts resolve by the early second trimester.4,6 In rare cases, a cyst may develop complications such as torsion or rupture, causing acute pain or hemorrhage. Otherwise, a cystic tumor identified in the first trimester should be characterized and followed using ultrasonography (US). Benign neoplasm An adnexal mass that persists beyond the first trimester is more likely to be a neoplasm.3-5,10,11,22 Such a mass is generally considered clinically significant if it exceeds 5 cm in diameter and has a complex sonographic appearance. Usually such a neoplasm will be a benign cystadenoma or cystic teratoma.5,10-13,19,23,24 Benign cystic teratoma This tumor can be identified with a fairly high degree of specificity using a variety of imaging techniques, with management based on the presumptive diagnosis. This tumor is unlikely to grow substantially during pregnancy. When it is smaller than 6 cm, such a tumor can simply be observed.14 A larger tumor can occasionally rupture or lead to torsion or obstruction of labor, but such occurrences are rare. Benign cystadenoma In an asymptomatic patient with imaging that suggests a benign cystadenoma (see sonogram), benign cystic teratoma, or other benign tumor, observation is reasonable in most cases.4,6,7,9-11,14,19 Operative intervention is required when there is less certainty regarding the benign nature of the tumor, an acute complication develops, or the tumor is expected to pose problems because of its large size alone. Uterine leiomyoma It is rare for an ovarian tumor detected during pregnancy to have a solid appearance on US. When it does, it may be a uterine leiomyoma mimicking an adnexal tumor (see intraoperative photograph). It should be reevaluated with more detailed US or magnetic resonance imaging.25 Malignancy About 10% of adnexal masses that persist during pregnancy are malignant, according to recent series.4,5,7-10,12,13,24,26 Most of the ovarian cancers diagnosed during pregnancy are epithelial, and a substantial portion of these are low-malignant-potential (LMP) tumors.5,10,11,13,19,23,24,26,27 This ratio is in keeping with the age of these women, which also explains the stage distribution (most are stage 1) and the large percentage of germ-cell tumors detected. The majority of ovarian cancers discovered in pregnant women have a favorable prognosis. Benign-appearing cystadenoma A morphologically benign-appearing, large, cystic adnexal mass can be seen in association with an 11-week gestation. Leiomyoma mimics an ovarian tumor This 17-week gestation was marked by a large pedunculated leiomyoma that at fist appeared to be a right adnexal tumor. Appearance of adnexal masses on US A functional cyst such as a follicular cyst, corpus luteum cyst, or theca lutein cyst usually has smooth borders and a fluid center. Other cysts may sometimes contain debris, such as clotted blood, that suggests endometriosis or a simple cyst with bleeding into it. A benign cystic teratoma often has multiple tissue lines, evidence of calcification, and layering of fat and fluid contents. A benign cystadenoma usually has the appearance of a simple cyst without large septates, whereas a cystadenocarcinoma often contains septates, abnormal blood flow, increased vascularity, or all of these. However, it is impossible to definitively distinguish a cystadenoma from a cystadenocarcinoma using US imaging alone. Functional cysts usually resolve by the second trimester. A cyst warrants closer scrutiny when it persists, is larger than 5 cm in diameter, or has a complex appearance on US. CA-125 may be useful after the first trimester The serum CA-125 level is typically elevated during the first trimester, but may be useful during later assessment or for follow-up of a malignancy.1 A markedly elevated serum level of alpha-fetoprotein (fractionated in some cases) has been reported in some gravidas with an endodermal sinus or mixed germ-cell ovarian tumor.2 Alpha-fetoprotein should be measured when there is suspicion for a germ-cell tumor based on clinical or US findings. When a mass is discovered during cesarean section Occasionally, an adnexal mass is detected at the time of cesarean section (FIGURE 1).3 This phenomenon is increasingly common, given the large number of cesarean deliveries in the United States. To eliminate the need for future surgery and avoid a delay in the diagnosis of an ovarian malignancy, inspect the adnexa routinely after closing the uterine incision in all women who deliver by cesarean section. FIGURE 1 Mass discovered at cesarean section This cystic tumor was discovered at cesarean section that was undertaken for obstetric indications. CASE 2 LMP tumor is suspected A 36-year-old gravida 3 para 1011 makes a prenatal visit during the first trimester. Her previous delivery was a cesarean section through a Pfannenstiel incision for a breech presentation. US imaging reveals a 6-week, 5-day fetus and a complex left adnexal mass, 4.5×3.9×4.1 cm. Imaging is repeated 1 month later at a tertiary-care center and shows an 11-week viable fetus, a right ovary with a corpus luteum cyst, and a left ovary with a 6.6×4 cm cystic mass with extensive vascular surface papillations that is suspicious for a low-malignant-potential (LMP) tumor. In several sonograms prior to the pregnancy, this mass appeared to be solid and was 3 cm in size. When is surgery warranted? Surgery is indicated when physical examination or imaging of a pregnant woman reveals an adnexal mass that is suspicious for malignancy, but the physician must weigh the benefit of prompt surgery against the risk to the pregnancy. This equation can be complicated in several ways. For example, surgical staging of clinically early ovarian cancer is more difficult due to the pregnant uterus, which is more extensively manipulated during these procedures. In addition, an optimal operation sometimes necessitates removal of the uterus. At 13 weeks’ gestation, the patient described in case 2 underwent laparoscopy with peritoneal washings and left salpingo-oophorectomy, but the tumor ruptured during removal. Final pathology showed it to be a serous LMP tumor involving the surface of the left ovary. Washings were in line with this diagnosis. The pregnancy continued uneventfully, and a repeat cesarean section was performed at 37 weeks through the Pfannenstiel scar, followed by limited surgical staging. Exploration and all biopsies were negative, and the final diagnosis was a stage 1C serous LMP tumor of the ovary. The patient articulated a desire to preserve her fertility and was monitored with US imaging of the remaining ovary every 6 months. Does ‘indolent’ behavior of malignancy justify watchful waiting? LMP tumors comprise a relatively large percentage of ovarian “cancers” encountered during pregnancy. Some authors report the accurate identification of these tumors prospectively, based on ultrasonographic characteristics.4,5 When an LMP tumor is the likely diagnosis, serial observation during pregnancy may be appropriate because of the indolent nature of the tumor. Further studies are needed to refine preoperative diagnosis and determine the overall safety of this approach. When the problem is acute In rare cases, a pregnant patient will have (or develop during observation) an acute problem due to torsion or rupture of an adnexal mass. Some ovarian cancers may present acutely, such as a rapidly growing malignant germ-cell tumor or a ruptured and hemorrhaging granulosecell tumor. Emergent surgery is necessary to manage the acute adnexal disease and reduce the likelihood of pregnancy loss. These events are infrequent, occurring in less than 10% of women with a known, persistent adnexal mass during pregnancy.4-14 Furthermore, recent studies have not found a substantial pregnancy complication rate associated with such emergency surgeries. CASE 3 Suspicious mass, ascites signal need for surgery A 19-year-old gravida 1 para 0 seeks prenatal care at 17 weeks’ gestation, complaining of rapidly enlarging abdominal girth. The physical examination estimates gestational size to be considerably greater than dates, but US is consistent with a 17-week intrauterine pregnancy. Imaging also reveals a 12-cm heterogenous left adnexal mass and a large amount of ascites. Surgery is clearly warranted, but how extensive should it be? When a malignancy is detected, a thorough staging procedure may be justified, depending on gestational age, exposure, desires of the patient, and operative findings. A midline incision is preferred. Pregnant and nonpregnant women with stage 1A or 1C epithelial ovarian cancer who undergo fertility-preserving surgery (with chemotherapy in selected patients) have a good prognosis and a high likelihood of achieving a subsequent normal pregnancy.15 The same is true for women with a malignant germ-cell tumor of the ovary, even when disease is advanced.16 However, careful surgical staging is necessary. The most important consideration when deciding whether to continue the pregnancy is the need for adjuvant chemotherapy. Depending on the gestational age and diagnosis, a short delay (4 to 6 weeks) may be appropriate to allow the pregnancy to progress beyond the first trimester or to maturity. In case 3, a laparotomy was performed at 19 weeks’ gestation via a midline incision, and approximately 5.3 L of ascites was evacuated. A large, nonadherent left ovarian tumor was removed. The right ovary appeared to be normal, as did the gravid uterus, which was minimally manipulated. The rest of the surgical exploration was normal, and the distal portion of the omentum was excised. The frozen-section diagnosis was a malignant stromal tumor. Final pathology showed an 18×13.5×8.8 cm, poorly differentiated, SertoliLeydig-cell tumor with heterologous elements in the form of mucinous epithelium. The omentum was negative for tumor. Chemotherapy was initiated in the third trimester, based on the limited data available, with intravenous etoposide and platinum administered every 21 days. The patient received 3 cycles of chemotherapy prior to delivery. At 37 weeks’ gestation, labor was successfully induced. After delivery, bleomycin was added to the chemotherapy regimen, and 3 additional courses with all 3 agents were administered. The patient was lost to follow-up shortly after completing chemotherapy. Clearly, an informed discussion of the options with the patient is imperative before any surgery, especially when chemotherapy may be delayed. Pregnancy does not appear to alter the prognosis for the patient with an ovarian malignancy, and ovarian cancer has not been reported to metastasize to the fetus. Integrating evidence and experience Pregnant women have a very low rate of ovarian cancer Leiserowitz GS, Xing G, Cress R, Brahmbhatt B, Dalrymple JL, Smith LH. Adnexal masses in pregnancy: how often are they malignant? Gynecol Oncol. 2006;101:315–321. Ovarian malignancies are rare during pregnancy. When they do occur, they are likely to be early stage and to have a favorable outcome, according to this recent population-based study. Using 3 large databases containing records on 4,846,505 California obstetric patients between 1991 and 1999, Leiserowitz and colleagues identified 9,375 women who had an ovarian mass associated with pregnancy. Of these, 87 had ovarian cancer and 115 had a low-malignant-potential (LMP) tumor, for a cancer occurrence rate of 0.93%, or 0.0179 per 1,000 deliveries. Thirty-four of the 87 cancers were germ-cell tumors. Of the 87 ovarian cancers, 65.5% were localized, 6.9% regional, 23% remote, and 4.6% of unknown stage. The respective rates for LMP tumors were 81.7%, 7.8%, 4.4%, and 6.1%. Women with malignant tumors were more likely than pregnant controls without cancer to undergo cesarean delivery, hysterectomy, transfusion, and prolonged hospitalization. These women did not, however, have a higher rate of adverse neonatal outcomes. When cancer is advanced Few data shed light on whether a pregnancy should continue when ovarian cancer is advanced.17 The definitive surgical approach must be highly individualized. It is not always possible to make an accurate diagnosis based on a frozen section. In such a case, the pregnancy should be preserved until the time of definitive diagnosis. As always, the patient’s wishes and gestational age must be considered. How factors besides malignancy can influence care Most persistent adnexal masses move well out of the pelvis as pregnancy progresses. Occasionally, however, an ovarian tumor may be located in the posterior cul-de-sac even at term, a fact easily confirmed by examination or US.4,7 A tumor in the posterior cul-de-sac can obstruct delivery or rupture. When it has a benign cystic appearance on US, it may be decompressed via transvaginal aspiration. Otherwise, the best approach is cesarean section and concomitant management of the mass. When size alone is the problem Some ovarian tumors are so large they seem incompatible with an advancing pregnancy. Tumors up to 20 cm in diameter have been removed intact at the time of cesarean section (FIGURE 2).18 The tumor may accommodate in shape and become less problematic as it is gradually pushed into the upper abdomen (FIGURE 3). The ability of the peritoneal cavity to accommodate a tumor varies greatly among women. As pregnancy advances, the likelihood that a large cystic mass will rupture tends to increase. Depending on the circumstances, percutaneous aspiration7,18 or removal of a benign-appearing cystic tumor may be appropriate. FIGURE 2 Even a very large tumor may coexist with advancing pregnancy This benign serous cystadenoma was exteriorized at the time of cesarean section at term. FIGURE 3 Large ovarian tumor has accommodated to the pregnancy Laparotomy—performed at term for cesarean section and to manage this large tumor—revealed that the tumor had accommodated in shape between the enlarging pregnant uterus and the abdominal wall. When is the best time to operate? Surgery is generally not recommended during the first trimester.5-11 Among the reasons are the high likelihood of a corpus luteum cyst, the low likelihood of an invasive malignancy, the low risk of adnexal complications associated with observation, and the potential for pregnancy loss or teratogenicity. However, as pregnancy progresses beyond the first trimester, surgery poses other problems: Operative exposure diminishes and the need to manipulate the pregnant uterus increases. Integrating evidence and experience Can surgery be delayed when a mass is detected? Schmeler KM, Mayo-Smith WW, Peipert JF, Weitzen S, Manuel MD, Gordinier ME. Adnexal masses in pregnancy: surgery compared with observation. Obstet Gynecol. 2005;105:1098–1103. Close observation is a reasonable alternative to operative intervention during pregnancy, unless a malignancy is suspected. Schmeler and colleagues reviewed the cases of 59 women who had an adnexal mass larger than 5 cm in diameter detected during pregnancy, out of a total of 127,177 deliveries at a single institution between 1990 and 2003. Antepartum surgery was performed in 17 women (29%). Of these, 13 cases had ultrasonographic findings suggesting malignancy, and 4 had ovarian torsion. The remaining women were observed, with surgery delayed until the time of cesarean section or later. Twenty-five of the 59 masses (42%) were dermoid cysts. Cancer was diagnosed in 4 patients (6.8%), and 1 patient (1.7%) had an LMP tumor. All 5 cases (100%) involving a malignancy had a suspicious US appearance and were identified during antepartum surgery, whereas only 12 patients with a benign tumor (22%) underwent surgery prior to delivery. Surgery poses risks to the pregnancy Elective surgery for an adnexal mass any time during pregnancy increases the risk of pregnancy loss and the likelihood of intrauterine growth restriction (IUGR) and preterm delivery.5,7,10,13,19 A 1989 study from Sweden20 defined a cohort of 5,405 women (from 720,000 births) who were known to have a nonobstetric operation while pregnant, with the following results: Congenital malformation and stillbirth were not increased in the women undergoing surgery The number of very-low- and low-birth-weight infants did rise, however—the result of both prematurity and IUGR Also elevated was the incidence of infants born alive but dying within 168 hours; these risks increased regardless of trimester No specific type of anesthesia or operation was associated with adverse reproductive outcomes, and the cause of those adverse outcomes was not determined. Some recent data suggest that adnexal surgery during the late second or early third trimester poses the greatest risk of preterm delivery or IUGR, or both.13 Window of opportunity: early to mid- second trimester During this time frame, elective surgery for an adnexal mass still affords some pelvic exposure without the need for significant uterine manipulation and has been associated with a lower risk of pregnancy complications. The other window for operation is at the time of cesarean section. An elective cesarean section is sometimes performed specifically to manage a persistent adnexal mass. Among the factors that warrant consideration when contemplating this approach are the elective uterine incision (with its attendant implications for future pregnancies), the higher risks associated with cesarean delivery in general, the type of skin incision (a vertical incision is appropriate in the event of ovarian malignancy), the potential for better exposure or laparoscopy at a later date, the increased difficulty of ovarian cystectomy at the time of cesarean section, and the patient’s wishes. Laparoscopy or laparotomy? The data on laparoscopy during the first and second trimesters of pregnancy indicate that it is as safe as laparotomy. A 1997 Swedish study21 identified cohorts of 2,181 women undergoing laparoscopy and 1,522 women undergoing laparotomy (from a total of 2,015,000 deliveries) between the fourth and 20th weeks of pregnancy. In both groups there was an increased risk for the infant to weigh less than 2,500 g, to be delivered before 37 weeks, and to have IUGR. There were no differences between the 2 groups for these and other adverse outcomes. Small series of laparoscopic procedures to manage an adnexal mass during pregnancy suggest that this approach is most applicable during the first (for highly selected emergent cases) or early second trimester to manage masses less than 10 cm in diameter, particularly when adnexectomy is planned. Laparoscopy may be considered “minimally invasive” because it reduces manipulation of the pregnant uterus during adnexal surgery. However, it is more difficult to assess and remove ovarian cysts laparoscopically, although an early ovarian malignancy could be staged via laparoscopy by an experienced surgeon. Considerations during laparotomy When performing a laparotomy or cesarean section for an adnexal mass, the surgeon must take into account a number of variables when selecting the type of incision (ie, vertical vs transverse). In general, if malignancy is suspected, or if uterine manipulation is to be minimized, a vertical incision is best. Other considerations include a prior scar, body habitus, obstetric issues, and the patient’s wishes. The author reports no financial relationships relevant to this article. References 1. Spitzer M, Kaushal N, Benjamin F. Maternal CA-125 levels in pregnancy and the puerperium. J Reprod Med. 1998;43:387-392. 2. Aoki Y, Higashino M, Ishii S, Tanaka K. Yolk sac tumor of the ovary during pregnancy: a case report. Gynecol Oncol. 2005;99:497-499. 3. Koonings PP, Platt LD, Wallace R. Incidental adnexal neoplasms at cesarean section. Obstet Gynecol. 1988;72:767-769. 4. Zanetta G, Mariani E, Lissoni A, et al. A prospective study of the role of ultrasound in the management of adnexal masses in pregnancy. BJOG. 2003;110:578-583. 5. Sherard GB, 3rd, Hodson CA, Williams HJ, et al. Adnexal masses and pregnancy: a 12 year experience. Am J Obstet Gynecol. 2003;189:358-363. 6. Bernhard LM, Klebba PK, Gray DL, Mutch DG. Predictors of persistence of adnexal masses in pregnancy. Obstet Gynecol. 1999;93:585-589. 7. Platek DN, Henderson CE, Goldberg GL. The management of a persistent adnexal mass in pregnancy. Am J Obstet Gynecol. 1995;173:1236-1240. 8. Bromley B, Benacerraf B. Adnexal masses during pregnancy: accuracy of sonographic diagnosis and outcome. J Ultrasound Med. 1997;16:447-452. 9. Hill LM, Connors-Beatty DJ, Nowak A, Tush B. The role of ultrasonography in the detection and management of adnexal mass during the second and third trimesters of pregnancy. Am J Obstet Gynecol. 1998;179:703-707. 10. Agarwal N, Parul, Kriplani A, Bhatla N, Gupta A. Management and outcome of pregnancies complicated with adnexal masses. Arch Gynecol Obstet. 2003;267:148-152. 11. Schmeler KM, Mayo-Smith WW, Peipert JF, Weitzen S, Manuel MD, Gordinier ME. Adnexal masses in pregnancy: surgery compared with observation. Obstet Gynecol. 2005;105:1098-1103. 12. Coenen VH, Dunton C, Cardonick E, Berghella V. Persistent adnexal masses during pregnancy. Obstet Gynecol. 1999;93:66S.- 13. Whitecar P, Turner S, Higby K. Adnexal masses in pregnancy: a review of 130 cases undergoing surgical management. Am J Obstet Gynecol. 1999;181:19-24. 14. Caspi B, Levi R, Appelman Z, Rabinerson D, Goldman G, Hagay Z. Conservative management of ovarian cystic teratoma during pregnancy and labor. Am J Obstet Gynecol. 2000;182:503-505. 15. Schilder JM, Thompson AM, DePriest PD, et al. Outcome of reproductive age women with stage IA or IC invasive epithelial ovarian cancer treated with fertility-sparing therapy. Gynecol Oncol. 2002;87:1-7. 16. Tangir J, Zelterman D, Ma W, Schwartz PE. Reproductive function after conservative surgery and chemotherapy for malignant germ cell tumors of the ovary. Obstet Gynecol. 2003;101:251-257. 17. Ferrandina G, Distefano M, Testa A, De Vincenzo R, Scambia G. Management of an advanced ovarian cancer at 15 weeks of gestation: case report and literature review. Gynecol Oncol. 2005;97:693-696. 18. Caspi B, Ben-Arie A, Appelman Z, Or Y, Hagay Z. Aspiration of simple pelvic cysts during pregnancy. Gynecol Obstet Invest. 2000;49:102-105. 19. Usui R, Minakami H, Kosuge S, et al. A retrospective survey of clinical, pathologic, and prognostic features of adnexal masses operated on during pregnancy. J Obstet Gynaecol Res. 2000;26(2):89-93. 20. Mazze RI, Källén B. Reproductive outcome after anesthesia and operation during pregnancy; a registry study of 5,405 cases. Am J Obstet Gynecol. 1989;161:1178-1185. 21. Reedy MB, Källén B, Kuehl TJ. Laparoscopy during pregnancy: a study of five fetal outcome parameters with use of the Swedish health registry. Am J Obstet Gynecol. 1997;177:673-679. 22. Hermans RHM, Fischer D-C, van der Putten HWHM, et al. Adnexal masses in pregnancy. Onkologie. 2003;26:167-172. 23. Hoffman MS. Primary ovarian carcinoma during pregnancy. Clin Consul Obstet Gynecol. 1995;7:237-241. 24. Ueda M, Ueki M. Ovarian tumors associated with pregnancy. Int J Obstet Gynaecol. 1996;55:59-65. 25. Curtis M, Hopkins MP, Zarlingo T, et al. Magnetic resonance imaging to avoid laparotomy in pregnancy. Obstet Gynecol. 1993;82:833-836. 26. Leiserowitz GS, Xing G, Cress R, et al. Adnexal masses in pregnancy; how often are they malignant? Gynecol Oncol. 2006;101:315-321. 27. Rahman MS, Al-Sibai MH, Rahman J, et al. Ovarian carcinoma associated with pregnancy. A review of 9 cases. Acta Obstet Gynecol Scand. 2002;81:260-264. References 1. Spitzer M, Kaushal N, Benjamin F. Maternal CA-125 levels in pregnancy and the puerperium. J Reprod Med. 1998;43:387-392. 2. Aoki Y, Higashino M, Ishii S, Tanaka K. Yolk sac tumor of the ovary during pregnancy: a case report. Gynecol Oncol. 2005;99:497-499. 3. Koonings PP, Platt LD, Wallace R. Incidental adnexal neoplasms at cesarean section. Obstet Gynecol. 1988;72:767-769. 4. Zanetta G, Mariani E, Lissoni A, et al. A prospective study of the role of ultrasound in the management of adnexal masses in pregnancy. BJOG. 2003;110:578-583. 5. Sherard GB, 3rd, Hodson CA, Williams HJ, et al. Adnexal masses and pregnancy: a 12 year experience. Am J Obstet Gynecol. 2003;189:358-363. 6. Bernhard LM, Klebba PK, Gray DL, Mutch DG. Predictors of persistence of adnexal masses in pregnancy. Obstet Gynecol. 1999;93:585-589. 7. Platek DN, Henderson CE, Goldberg GL. The management of a persistent adnexal mass in pregnancy. Am J Obstet Gynecol. 1995;173:1236-1240. 8. Bromley B, Benacerraf B. Adnexal masses during pregnancy: accuracy of sonographic diagnosis and outcome. J Ultrasound Med. 1997;16:447-452. 9. Hill LM, Connors-Beatty DJ, Nowak A, Tush B. The role of ultrasonography in the detection and management of adnexal mass during the second and third trimesters of pregnancy. Am J Obstet Gynecol. 1998;179:703-707. 10. Agarwal N, Parul, Kriplani A, Bhatla N, Gupta A. Management and outcome of pregnancies complicated with adnexal masses. Arch Gynecol Obstet. 2003;267:148-152. 11. Schmeler KM, Mayo-Smith WW, Peipert JF, Weitzen S, Manuel MD, Gordinier ME. Adnexal masses in pregnancy: surgery compared with observation. Obstet Gynecol. 2005;105:1098-1103. 12. Coenen VH, Dunton C, Cardonick E, Berghella V. Persistent adnexal masses during pregnancy. Obstet Gynecol. 1999;93:66S.- 13. Whitecar P, Turner S, Higby K. Adnexal masses in pregnancy: a review of 130 cases undergoing surgical management. Am J Obstet Gynecol. 1999;181:19-24. 14. Caspi B, Levi R, Appelman Z, Rabinerson D, Goldman G, Hagay Z. Conservative management of ovarian cystic teratoma during pregnancy and labor. Am J Obstet Gynecol. 2000;182:503-505. 15. Schilder JM, Thompson AM, DePriest PD, et al. Outcome of reproductive age women with stage IA or IC invasive epithelial ovarian cancer treated with fertility-sparing therapy. Gynecol Oncol. 2002;87:1-7. 16. Tangir J, Zelterman D, Ma W, Schwartz PE. Reproductive function after conservative surgery and chemotherapy for malignant germ cell tumors of the ovary. Obstet Gynecol. 2003;101:251-257. 17. Ferrandina G, Distefano M, Testa A, De Vincenzo R, Scambia G. Management of an advanced ovarian cancer at 15 weeks of gestation: case report and literature review. Gynecol Oncol. 2005;97:693-696. 18. Caspi B, Ben-Arie A, Appelman Z, Or Y, Hagay Z. Aspiration of simple pelvic cysts during pregnancy. Gynecol Obstet Invest. 2000;49:102-105. 19. Usui R, Minakami H, Kosuge S, et al. A retrospective survey of clinical, pathologic, and prognostic features of adnexal masses operated on during pregnancy. J Obstet Gynaecol Res. 2000;26(2):89-93. 20. Mazze RI, Källén B. Reproductive outcome after anesthesia and operation during pregnancy; a registry study of 5,405 cases. Am J Obstet Gynecol. 1989;161:1178-1185. 21. Reedy MB, Källén B, Kuehl TJ. Laparoscopy during pregnancy: a study of five fetal outcome parameters with use of the Swedish health registry. Am J Obstet Gynecol. 1997;177:673-679. 22. Hermans RHM, Fischer D-C, van der Putten HWHM, et al. Adnexal masses in pregnancy. Onkologie. 2003;26:167-172. 23. Hoffman MS. Primary ovarian carcinoma during pregnancy. Clin Consul Obstet Gynecol. 1995;7:237-241. 24. Ueda M, Ueki M. Ovarian tumors associated with pregnancy. Int J Obstet Gynaecol. 1996;55:59-65. 25. Curtis M, Hopkins MP, Zarlingo T, et al. Magnetic resonance imaging to avoid laparotomy in pregnancy. Obstet Gynecol. 1993;82:833-836. 26. Leiserowitz GS, Xing G, Cress R, et al. Adnexal masses in pregnancy; how often are they malignant? Gynecol Oncol. 2006;101:315-321. 27. Rahman MS, Al-Sibai MH, Rahman J, et al. Ovarian carcinoma associated with pregnancy. A review of 9 cases. Acta Obstet Gynecol Scand. 2002;81:260-264. Issue OBG Management - 19(03) Issue OBG Management - 19(03) Page Number 27-44 Page Number 27-44 Publications MDedge ObGyn OBG Management Publications MDedge ObGyn OBG Management Article Type News Display Headline A guide to management: Adnexal masses in pregnancy Display Headline A guide to management: Adnexal masses in pregnancy Legacy Keywords adnexal masses; laparoscopy; laparotomy; mass during cesarean section; serum CA-125; ovarian cancers; Mitchel S. Hoffman;MD; Hoffman MS; Robyn A. Sayer;MD; Sayer RA; Legacy Keywords adnexal masses; laparoscopy; laparotomy; mass during cesarean section; serum CA-125; ovarian cancers; Mitchel S. Hoffman;MD; Hoffman MS; Robyn A. Sayer;MD; Sayer RA; Sections Clinical Review Article Source PURLs Copyright Inside the Article IN THIS ARTICLE The most common masses detected during pregnancy When you discover a mass during cesarean section Laparoscopy or laparotomy? Article PDF Media Document 1903OBGM_Article2.pdf
1473	https://www.wikihow.com/Calculate-Ratios	Home Random Browse Articles TrendingNew Quizzes & Games All Quizzes Love Quizzes Personality Quizzes Fun Games Dating Simulator Learn Something New Forums Courses Happiness Hub Explore More Support wikiHow About wikiHow Log in / Sign up Terms of Use Categories Education and Communications Studying Mathematics How to Calculate Ratios Download Article Download Article Understanding Ratios \| Using Ratios \| Catching Mistakes \| Practice Problems \| Video \| Q&A Ratios are mathematical expressions that compare two or more numbers. They can compare absolute quantities and amounts or can be used to compare portions of a larger whole. Ratios can be calculated and written in several different ways, but the principles guiding the use of ratios are universal to all. Steps Practice Problems Calculate Ratios Practice Problems Calculate Ratios Practice Problems ANSWER KEY Part 1 Part 1 of 3: Understanding Ratios Download Article 1 Be aware of how ratios are used. Ratios are used in both academic settings and in the real world to compare multiple amounts or quantities to each other. The simplest ratios compare only two values, but ratios comparing three or more values are also possible. In any situations in which two or more distinct numbers or quantities are being compared, ratios are applicable. By describing quantities in relation to each other, they explain how chemical formulas can be duplicated or recipes in the kitchen expanded. After you get to understand them, you will use ratios for the rest of your life. X Research source 2. 2 Get to know what a ratio means. As noted above, ratios demonstrate the quantity of at least two items in relation to each other. So, for example, if a cake contains two cups of flour and one cup of sugar, you would say that the ratio of flour to sugar was 2 to 1. X Research source Ratios can be used to show the relation between any quantities, even if one is not directly tied to the other (as they would be in a recipe). For example, if there are five girls and ten boys in a class, the ratio of girls to boys is 5 to 10. Neither quantity is dependent on or tied to the other, and would change if anyone left or new students came in. The ratio merely compares the quantities. Advertisement 3. 3 Notice the different ways in which ratios are expressed. Ratios can be written out using words or can be represented using mathematical symbols. X Research source You will commonly see ratios represented using words (as above). Because they are used so commonly and in such a variety of ways, if you find yourself working outside of mathematic or scientific fields, this may the most common form of ratio you will see. Ratios are frequently expressed using a colon. When comparing two numbers in a ratio, you'll use one colon (as in 7 : 13). When you're comparing more than two numbers, you'll put a colon between each set of numbers in succession (as in 10 : 2 : 23). In our classroom example, we might compare the number of boys to the number of girls with the ratio 5 girls : 10 boys. We can simply express the ratio as 5 : 10. Ratios are also sometimes expressed using fractional notation. In the case of the classroom, the 5 girls and 10 boys would be shown simply as 5/10. That said, it shouldn't be read out loud the same as a fraction, and you need to keep in mind that the numbers do not represent a portion of a whole. Advertisement Part 2 Part 2 of 3: Using Ratios Download Article 1 Reduce a ratio to its simplest form. Ratios can be reduced and simplified like fractions by removing any common factors of the terms in the ratio. To reduce a ratio, divide all the terms in the ratio by the common factors they share until no common factor exists. However, when doing this, it's important not to lose sight of the original quantities that led to the ratio in the first place. X Research source In the classroom example above, 5 girls to 10 boys (5 : 10), both sides of the ratio have a factor of 5. Divide both sides by 5 (the greatest common factor) to get 1 girl to 2 boys (or 1 : 2). However, we should keep the original quantities in mind, even when using this reduced ratio. There are not 3 total students in the class, but 15. The reduced ratio just compares the relationship between the number of boys and girls. There are 2 boys for every girl, not exactly 2 boys and 1 girl. Some ratios cannot be reduced. For example, 3 : 56 cannot be reduced because the two numbers share no common factors - 3 is a prime number, and 56 is not divisible by 3. 2. 2 Use multiplication or division to "scale" ratios. One common type of problem that employs ratios may involve using ratios to scale up or down the two numbers in proportion to each other. Multiplying or dividing all terms in a ratio by the same number creates a ratio with the same proportions as the original, so, to scale your ratio, multiply or divide through the ratio by the scaling factor. X Research source For example, a baker needs to triple the size of a cake recipe. If the normal ratio of flour to sugar is 2 to 1 (2 : 1), then both numbers must be increased by a factor of three. The appropriate quantities for the recipe are now 6 cups of flour to 3 cups of sugar (6 : 3). The same process can be reversed. If the baker needed only one-half of the normal recipe, both quantities could be multiplied by 1/2 (or divided by two). The result would be 1 cup of flour to 1/2 (0.5) cup of sugar. 3. 3 Find unknown variables when given two equivalent ratios. Another common type of problem that incorporates ratios asks you to find an unknown variable in one ratio, given the other number in that ratio and a second ratio that is equivalent to the first. The principle of cross multiplication makes solving these problems fairly simple. Write each ratio in its fractional form, then set the two ratios equal to each other and cross multiply to solve. X Research source For example, let's say we have a small group of students containing 2 boys and 5 girls. If we were to maintain this proportion of boys to girls, how many boys would be in a class that contained 20 girls? To solve, first, let's make two ratios, one with our unknown variables: 2 boys : 5 girls = x boys : 20 girls. If we convert these ratios to their fraction forms, we get 2/5 and x/20. If you cross multiply, you are left with 5x=40, and you can solve by dividing both figures by 5. The final solution is x=8. EXPERT TIP Grace Imson, MA Math Instructor, City College of San Francisco Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. Grace Imson, MA Math Instructor, City College of San Francisco Look at the order of terms to figure out the numerator and denominator in a word problem. The first term is usually the numerator, and the second is usually the denominator. For example, if a problem asks for the ratio of the length of an item to its width, the length will be the numerator, and width will be the denominator. Advertisement Part 3 Part 3 of 3: Catching Mistakes Download Article 1 Avoid addition or subtraction in ratio word problems. Many word problems look something like this: "A recipe calls for 4 potatoes and 5 carrots. If you want to use 8 potatoes instead, how many carrots will you need to keep the ratio the same?" Many students try to add the same amount of each quantity. You actually need to use multiplication, not addition, to keep the ratio the same. Here's an example of the wrong and right to solve this example: Wrong method: "8 - 4 = 4, so I added 4 potatoes to the recipe. That means I should take the 5 carrots and add 4 to that too... wait! That's not how ratios work. I'll try again." Right method: "8 ÷ 4 = 2, so I multiplied the number of potatoes by 2. That means I should multiply the 5 carrots by 2 as well. 5 x 2 = 10, so I want 10 carrots total in the new recipe." 2. 2 Convert to the same units. Some word problems get tricky by switching to a different unit partway through. Convert to the same unit before finding the ratio. Here's an example problem and solution: X Research source A dragon has 500 grams of gold and 10 kilograms of silver. What is the ratio of gold to silver in the dragon's hoard? Grams and kilograms are not the same unit, so we'll need to convert. 1 kilogram = 1,000 grams, so 10 kilograms = 10 kilograms x = 10 x 1,000 grams = 10,000 grams. The dragon has 500 grams of gold and 10,000 grams of silver. The ratio of gold to silver is . 3. 3 Write your units in the problem. In ratio word problems, it's much easier to catch mistakes if you write the units after each value. Remember, the same unit on the top and bottom of a fraction cancels out. After you cancel out as much as you can, you should end up with the right units for your answer. X Research source Example problem: If you have six boxes, and in every three boxes there are nine marbles, how many marbles do you have? Wrong method: Wait, nothing cancels out, so my answer would be "boxes x boxes / marbles." That doesn't make sense. Right method: 18 marbles. EXPERT TIP Grace Imson, MA Math Instructor, City College of San Francisco Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. Grace Imson, MA Math Instructor, City College of San Francisco One common problem is knowing which number to use as a numerator. In a word problem, the first term stated is usually the numerator and the second term stated is usually the denominator. If you want the ratio of the length of an item to the width, length becomes your numerator and width becomes your denominator. Advertisement Community Q&A Search Add New Question Question How do you calculate if something cost $175, and two people shared the cost at a ratio of 2:3? Community Answer You add the numbers of the ratio: 2 + 3 = 5 You divide the total cost ($175) by 5. 175 / 5 = 35 You multiply this number by each of the numbers of the ratio: 35 x 2 = 70, and 35 x 3 = 105. Solution: one paid $70, and the other one, $105. Both numbers added give you the total of 175 dollars. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 160 Helpful 472 Question A small theater sold 72 tickets to a play. The ratio of adults to children is 4:1. The ratio of adults to seniors is 4:3. How do I determine how much of each were sold? Donagan Top Answerer The ratio of adults to seniors to children is 4:3:1. Add those three numbers, and divide the sum into 72. Multiply the quotient by 4, 3, and 1 to find the number sold of each kind of ticket. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 154 Helpful 434 Question How do you convert the ratio 1:4 to a decimal or a percent? Community Answer You can treat a ratio as a fraction or a division problem: 1:4 = 1 / 4 = 1 ÷ 4. Solve this problem with long division (or a calculator) and you'll get the answer as a decimal: 0.25. To make this a percent, just move the decimal point two spaces to the right: 0.25 = 25%. Thanks! We're glad this was helpful. Thank you for your feedback. If wikiHow has helped you, please consider a small contribution to support us in helping more readers like you. We’re committed to providing the world with free how-to resources, and even $1 helps us in our mission. Support wikiHow Yes No Not Helpful 135 Helpful 316 See more answers Ask a Question 200 characters left Include your email address to get a message when this question is answered. Submit Advertisement Video Tips Submit a Tip All tip submissions are carefully reviewed before being published Name Please provide your name and last initial Submit Thanks for submitting a tip for review! You Might Also Like How to Calculate Compression RatioHow to Make a Ratio How to Simplify a RatioHow to Cross Multiply FractionsHow to Do FractionsHow to Understand FractionsHow to Work Out a Fraction of an AmountHow to Survive Sixth Grade MathHow to Do StoichiometryHow to Find Equivalent FractionsHow to Solve ProportionsHow to Solve Fraction Questions in MathHow to Convert Units EasilyHow to Divide Fractions by Fractions: 4 Easy Steps Advertisement References ↑ ↑ ↑ ↑ ↑ ↑ ↑ ↑ About This Article Co-authored by: Grace Imson, MA Math Instructor, City College of San Francisco This article was co-authored by Grace Imson, MA. Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been viewed 3,247,178 times. 322 votes - 62% Co-authors: 59 Updated: June 11, 2025 Views: 3,247,178 Categories: Mathematics Article SummaryX To calculate a ratio, start by determining which 2 quantities are being compared to each other. For example, if you wanted to know the ratio of girls to boys in a class where there are 5 girls and 10 boys, 5 and 10 would be the quantities you're comparing. Then, put a colon or the word "to" between the numbers to express them as a ratio. In this example, you'd write "5 to 10" or "5:10." Finally, simplify the ratio if possible by dividing both numbers by the greatest common factor. To learn how to solve equations and word problems with ratios, scroll down! Did this summary help you? In other languages Spanish Italian Portuguese French German Dutch Chinese Indonesian Hindi Vietnamese Korean Czech Russian Thai Turkish Arabic Japanese Print Send fan mail to authors Thanks to all authors for creating a page that has been read 3,247,178 times. Reader Success Stories Stew Jackson Feb 9, 2021 "This website has most of the calculation problems encountered in my years of engineering activities. Much easier to come here than try to ferret out solutions in books and notes. Good examples and steps to a problem solution. Thanks!"..." more More reader stories Hide reader stories Share your story Did this article help you? Advertisement Cookies make wikiHow better. By continuing to use our site, you agree to our cookie policy. Co-authored by: Grace Imson, MA Math Instructor, City College of San Francisco 322 votes - 62% Click a star to vote % of people told us that this article helped them. Co-authors: 59 Updated: June 11, 2025 Views: 3,247,178 Stew Jackson Feb 9, 2021 "This website has most of the calculation problems encountered in my years of engineering activities. Much easier to come here than try to ferret out solutions in books and notes. Good examples and steps to a problem solution. Thanks!"..." more Maria Bustos May 3, 2021 "I'm 73 years old and I'm reviewing to help my 13 year old with his math. This is outlined so well that I can learn and teach my kid in simple form. Thank you so much."..." more Lynn Robinson Apr 17, 2017 "What helped me the most was the teacher-friendly, sequential order in which the information was given. All questions that I had were answered. Thanks a bunch!"..." more Amaris U. Aug 18, 2024 "I had to repeat a grade, and this website and article have helped me a lot. I used to have a lot of problems with math, but this really helps me."..." more Rated this article: Em P. 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1474	http://corpora.lancs.ac.uk/BNCweb/	\| \| \| --- \| \| \| BNCweb (CQP-Edition)A web-based interface to the British National Corpus \| \| \| \| \| \| \| Table of Contents: What is BNCweb? Features of BNCweb Corpus Linguistics with BNCweb - a Practical Guide. (Hoffmann, Evert, Smith, Lee & Berglund Prytz - Peter Lang, 2008) - Short-listed for the BAAL Book Prize 2009 Access to the BNC via BNCweb at Lancaster University What's the story behind BNCweb? Installation Requirements Download Errors and inconsistencies in BNC-XML \| \| What is BNCweb? BNCweb is a web-based client program for searching and retrieving lexical, grammatical and textual data from the British National Corpus (BNC). It relies on the Corpus Query Processor (CQP) of the IMS Open Corpus Workbench to provide a convenient interface between the user and the rich variety of annotated text in the 100-million word BNC in its most recent incarnation, the XML-version. \|\| Main advantages of BNCweb: It is very user-friendly. It is a web-based application - end users do not need to install any extra software on their computers. Any web-browser (on any platform) will do. It is fast! Have you ever wanted to calculate collocations for the noun lemma TIME? A collocation analysis of its 180,243 instances in the BNC takes just under 30 seconds on a version of BNCweb which is installed on an entry-level Apple MacBook. It is powerful and flexible: In addition to basic queries (available via an intuitive query syntax), the interface allows more complex searches using full-fledged CQP-syntax. It is optimized for use by larger groups of users: a cache system minimizes CPU-load and disk space usage when different users perform the same queries. It is absolutely free (its components are released under the GNU Public License). \|\| Features of BNCweb BNCweb offers a whole range of features for corpus analysis (concordance display, sort, collocations, distribution analysis, etc.). The following list is a basic overview of the different functions. \| \| \| --- \| \| Name \| Description \| \| Query result \| Standard concordance display (KWIC-view and sentence view) with one-click access to the larger context of query results and the relevant speaker and file information (such as 'Age of speaker', 'Domicile of author', 'Date of creation' etc.). Restrictions for written and spoken texts can be applied to your lexical search. Concordance lines can be displayed in random order. Immediate access to the n-th instance of your query result. No upper restrictions apply - you can for example look at the 5,456,345th instance of the. Query results always include normalised frequency information (instances per million words). This also applies to queries which are based on metatextual restrictions or subcorpus searches. \| \| Thin \| Reduces the number of lines in the concordance window, e.g. by creating a random subset of 1000 solutions. \| \| Distribution \| Creates descriptive statistics for your query result, e.g. its distribution over the 9 text domains in the written component of the BNC. Includes normalised frequency counts and information about the total number of words in each category. Crosstabulation over two categories is also possible. BNCweb also lists the files in which the query result occurs most/least frequently. \| \| Sort \| Sorts the query result alphabetically on any of five positions to the left or right of your node. POS-tag restrictions can be applied to a sorted result and a frequency list of the chosen position can be compiled (with absolute frequencies and percentage information.) \| \| Collocations \| Compiles a ranked list of collocates (both word forms and lemma forms) for your query result according to several statistical methods: Mutual information, MI3, Z-score, T-score, Chi-squared with Yates' correction, Log-likelihood, and Modified Dice coefficient. Various parameters can be set to allow users optimal flexibility in their analysis (e.g. flexible window-size, including asymmetrical windows, etc.). \| \| Save queries \| Query results can be saved for future access. This is useful for queries which were manually post-processed by the user (e.g. using the 'delete hits' function). \| \| Categorize query result \| Individual hits of a query result can be manually categorized according to a user-defined set of values. Once categorization is complete, all post-processing features of BNCweb (such as distribution analysis, collocations, etc.) can be applied to categorized data separately. \| \| Download \| Query results can be downloaded to your hard-disk in a tab-delimited format (including metatextual information, if required). This data can be imported into the spreadsheet program of your choice for manual annotation. By default, the corpus positions (i.e. internal references to corpus tokens) of all matches will also be downloaded. This information is required to re-import a query result (or parts of a query result) back to BNCweb (see below). \| \| Upload external data file \| This feature can be used to upload a file with corpus positions to the BNCweb server to create a new saved query. This function can be used to re-import a manually postprocessed (e.g. cleaned) query result to make it available for use with any of the automated post-processing features offered by BNCweb (collocations, distribution, etc.). \| \| Scan keywords/titles \| Retrieves a list of BNC text files on the basis of the classification contained in the and elements of the file headers. This list can be used to define a subcorpus. \| \| Explore genre labels \| Retrieves a list of BNC text files on the basis of David Lee's genre classification scheme encoded in the file headers. This list can be used to define a subcorpus. \| \| Create/edit subcorpora \| Offers the user several options to create (and edit) user-definable subcorpora: by manually entering a list of filenames or via the keyword/title/genre search features. \| \| Browse a file \| Provides access to the larger context of any ~~-unit in the BNC via its file ID and ~~-unit number. The user has a choice between POS-tagged and untagged output. In addition, word-class colouring is available.~~~~ \| \| Word lookup \| Produces alphabetically ordered lists of lexical items (-units) and lemma forms - e.g. all words (or lemmata) starting with help (e.g. help, help-desk, helplessness, etc.). \| \| Frequency list \| Displays a frequency list of items (-units or lemmata) in the whole BNC or any user-defined subcorpus. Users can define restrictions for frequency lists in various ways (POS-tags, regular expression patterns, frequency range, etc.). \| \| Keywords \| Compares frequency lists for two (user-definable) subcorpora in the BNC and identifies items that are particularly frequent or infrequent in one of the two lists. \| \| Query history \| Lists all queries performed by the user - including the date and time of the query and the number of retrieved hits. Queries can be re-executed by clicking on a link. \| \| User settings \| Influences standard output formats as desired by the user (e.g. size of context display in number of ~~-units, KWIC or standard sentence display, etc.).~~ \| \| Recursive use of features \| All features of BNCweb are available recursively, i.e. you can for example first get a distribution of your query result over 'Age of author', choose only those hits written by authors aged 25-34, sort them, get a frequency list on the second word to the right of the node, click on the most frequent item in the list and calculate collocations of your query result on the basis of the remaining sentences. There is no upper limit for the number of steps that can be combined. \| \| Cached queries \| All queries are cached (up to a sysadmin-definable maximum of disk space) and can be re-executed instantaneously. Cached queries (and MySQL tables created for the post-processing features of BNCweb) are available to all users on the same server - this radically reduces disk space usage when whole groups of users perform the same query. \| \|\| Corpus Linguistics with BNCweb \| \| \| Hoffmann, Sebastian, Evert, Stefan, Smith, Nicholas, Lee, David and Ylva Berglund Prytz. 2008. Corpus Linguistics with BNCweb - a Practical Guide. Frankfurt am Main: Peter Lang. Abstract:This book presents a richly illustrated, hands-on discussion of one of the fastest growing fields in linguistics today. The authors address key methodological issues in corpus linguistics, such as collocations, keywords and the categorization of concordance lines. They show how these topics can be explored step-by-step with BNCweb, a user-friendly web-based tool that supports sophisticated analyses of the 100-million-word British National Corpus. Indeed, the BNC and BNCweb have been described by Geoffrey Leech as "an unparalleled combination of facilities for finding out about the English language of the present day" (Foreword). The book contains tasks and exercises, and is suitable for undergraduates, postgraduates and experienced corpus users alike. This book was short-listed for the BAAL Book Prize 2009 Sample Chapter: Table of Contents and Chapter 1 Buying the book: The book is now available through amazon.co.uk, amazon.de and amazon.com. It can also be ordered directly online through Peter Lang. You can also fill in this order form and send it to Peter Lang. Availability on amazon.co.jp is currently restricted to resellers who charge a ridiculously high price. We are in contact with Peter Lang about this and hope that this situation will improve soon. \| \| Access to the BNC via BNCweb at Lancaster UniversityThe BNC can be accessed via a service hosted at Lancaster University. The service is free of charge and available to anybody who registers with a valid e-mail address. Since the BNC is a licensed product, certain access restrictions are implemented: Standard users do not have access to the larger context of individual concordance lines. Query results with more than 5000 matches are automatically downsampled to 5000 concordance lines. The selection is random but reproducible. The "Browse a Text" feature is disabled. Please note that these restrictions may change at any time without prior notice. Current license holders of the BNC (XML-version) may in the near future be able to get full access to the corpus via the Lancaster server. Registered users of BNCweb will be notified of this option by e-mail once details are available. URL for signing up: \| \| What's the story behind BNCweb? Who created it? The original BNCweb was created by Hans Martin Lehmann, Sebastian Hoffmann and Peter Schneider at the University of Zurich. It was mainly written because the English Department in Zurich had no Windows computers to run the SARA-client provided as part of the BNC distribution. What started as a quick hack soon became a fully-fledged corpus tool. The functionality of BNCweb initially relied on the SARA server software to access the information contained in the BNC, but the integration of the powerful relational database system MySQL meant that the range of available features could be greatly extended. BNCweb was first publicly released in May 2002. An evaluation of the features offered by this earlier version of BNCweb can be found in a review that was published on the Linguist List. The CQP-version of BNCweb no longer relies on the SARA server but instead uses the powerful Corpus Query Processor (CQP) of the Corpus Workbench. The current version of the interface was created by Stefan Evert (University of Osnabrück) and Sebastian Hoffmann (Lancaster University). Further information on the CQP-version of BNCweb can also be found in Hoffmann & Evert (2006). \| \| Installation requirements Users may wish to install BNCwen on their own server. It has a client-server architecture: it is designed to give a (potentially large) number of concurrent users access to a server-side BNCweb installation via their standard web browsers. As a result, BNCweb requires no special installation procedure or client program on the part of the end user - any web-browser (under any kind of operating system) can be used to access the BNC via the Internet or a local area network. Please note that it is not possible to use BNCweb in conjunction with the BNC Online service hosted by the British Library. On the server side, BNCweb consists of a set of Perl scripts that require installation on a UNIX system such as Linux, Mac OS X, Sun Solaris, etc. While the installation procedure has been much improved in comparison with earlier versions of BNCweb, some basic knowledge of UNIX system administration is still required. An installation manual (in text-only format) is available here. The following tools and libraries need to be installed on your server - many of which may already be pre-installed on your system: xsltproc (from the Gnome LibXSLT package) MySQL 4.1 or higher - 5.0 recommended ( Perl 5.8 (version 5.6 may work, too, but might require installation of additional modules) Perl modules: + DBI + DBD::mysql + HTML::Entities + Parse::RecDescent A full installation of BNCweb requires about 3GB of disk space. However, during the indexing process, quite a bit more space is needed (approx. 8-10 GB free disk space should be more than sufficient.) \| \| Download BNCweb is now freely available under the terms of the GNU General Public License, version 3. Download: BNCweb-distribution.zip (version 4.0, 23/11/2007)An updated version of the scripts will be available for download soon. Update (20.11.2008): Feedback from beta-testers has meant that some changes had to be made. A new round of testing is currently being carried out. Sorry for the delay! Please note: BNCweb requires a version of the Corpus Workbench that is not yet available via sourcefourge. Please use the source code provided as part of the BNCweb distribution. Also, please check this page from time to time for updates to BNCweb. \| \| Errors and inconsistencies in BNC-XML No corpus of the size of the BNC can be without errors. Many of the problems of the first release of the corpus have in the meantime been corrected. However, a number of issues (e.g. duplicate stretches of text) remain even in the XML version of the corpus. More importantly, a range of new errors were introduced as a result of the automated conversion routines from SGML to XML. This is particularly the case for the spoken component of the corpus, where thousands of tags indicating speaker overlap, unclear passages etc. have been lost. Since this may have an impact on at least some of the linguistic findings that can be made with the corpus, users of BNCweb may wish to look at a detailed list of these errors. For questions about BNCweb, please write to bncweb@mac.com. \| \| Last updated: 15.10.2008
1475	https://ora.ox.ac.uk/objects/uuid:0dce50c0-0d46-4f3a-9b77-125b778e7603/files/md634028ca0a9173dd8851ac5df8e6e69	Page 1 of 47 Green -top Guideline No. 27 a1 Final Draft – Winter 201 72 3 Placenta Praevia and Placenta Accreta: Diagnosis and Management 4 5 This is the fourth edition of this guideline. The first, published in 2001, was entitled Placenta Praevia: 6 Diagnosis and Management ; the second, published in 2005, was entitled Placenta Praevia and 7 Placenta Praevia Accreta: Diagnosis and Management ; and the third, published in 2011, was entitled 8 Placenta Praevia , Placenta Praevia Accreta and Vasa Praevia: Diagnosis and Management .9 10 Executi ve summary 11 12 Antenatal diagnosis and management of placenta praevia or a low -lying placenta 13 14 What are the risk factors for placenta praevia or a low -lying placenta ?15 16 Caesarean delivery is associated with an increased risk of placenta praevia in subsequent 17 pregnancies. This risk rises as the number of prior caesarean sections increases. [B] 18 19 Assisted reproductive technology and maternal smoking increase the risk of placenta praevia. [B] 20 21 Should we screen for placenta praevia or a low -lying placenta , if so , at what gestation and with what 22 follow -up? 23 24 The midpregnancy routine fetal anomaly scan should include placental localisation thereby 25 identifying women at risk of persisting placenta praevia or a low -lying placenta . [GPP] 26 27 The term placenta praevia should b e used when the placenta lies directly over the internal os. For 28 pregnancies at more than 16 weeks of gestation the term low -lying placenta should be used when 29 the placental edge is 2 0 mm or less from the internal os on transabdominal or transvaginal 30 scann ing (TVS ). [D] 31 32 If the placenta is thought to be low lying (less than 20 mm from the internal os) or praevia 33 (covering the os) at the routine fetal anomaly scan, a follow -up ultrasound examination including 34 a TVS is recommended at 32 weeks of gestation to diagnose persistent low -lying placenta and/or 35 placenta praevia. [D] 36 37 What is the role and what are the risks of TVS? 38 39 Clinicians should be aware that TVS for the diagnosis of placenta praevia or a low -lying placenta is 40 superior to transabdominal and trans perineal approaches and is safe. [GPP] 41 42 In women with a persistent low -lying placenta or placenta praevia at 32 weeks of gestation who 43 remain asymptomatic, an additional TVS is recommended at around 36 weeks of gestation to 44 inform discussion about mode of delivery. [D] 45 46 Cervical length measurement may help facilitate management decisions in asymptomatic women 47 with placenta praevia. A short cervical length on TVS before 34 weeks of gestation increases the 48 risk of preterm emergency delivery and massive haemor rhage at caesarean section. [D] 49 50 Where should women with a low -lying placenta or placenta praevia be cared for in the third 51 trimester? 52 53 Page 2 of 47 Women with a low -lying placenta or placenta praevia with recurrent bleeding 54 55 Tailor antenatal care, including hospital isation, to individual patient need and social 56 circumstances, e.g. distance between home and hospital and availability of transportation, 57 previous bleeding episodes, haematology laboratory results and acceptance of receiving donor 58 blood or blood products . [GPP] 59 60 It should be made clear to any woman being treated at home in the third trimester that she should 61 attend the hospital immediately if she experiences any bleeding, including spotting, contractions 62 or pain (including vague suprapubic period -like aches ). [GPP] 63 64 Women with asymptomatic placenta praevia or a low -lying placenta 65 66 Women with asymptomatic placenta praevia or a low -lying placenta in the third trimester should 67 be counselled about the risks of preterm delivery and obstetric haemorrhage, and thei r care 68 should be tailored to their individual needs. [GPP] 69 70 Women with asymptomatic placenta praevia confirmed at the 32 -week follow -up scan and 71 managed at home should be encouraged to ensure they have safety precautions in place, including 72 having someone available to help them as necessary and ready access to the hospital. [GPP] 73 74 Is there a place for cervical cerclage in placenta praevia or a low -lying placenta ?75 76 The use of cervical cerclage to reduce bleeding and prolong pregnancy is not supported by 77 suff icient evidence to recommend its use outside of a clinical trial. [GPP] 78 79 In what circumstances, and at what gestation, should women be offered antenatal corticosteroids? 80 81 A single course of antenatal corticosteroid therapy is recommended between 34 +0 and 3 6+0 weeks 82 of gestation for pregnant women with a low -lying placenta or placenta praevia and is appropriate 83 prior to 34 +0 weeks of gestation in women at higher risk of preterm birth. [ GPP ]84 85 Is there a place for the use of tocolytics in women presenting with a low -lying placenta or placenta 86 praevia and preterm labour? 87 88 Tocolysis for women presenting with symptomatic placenta praevia or a low -lying placenta may 89 be considered for 48 hours to facilitate administration of antenatal corticosteroids. [ C]90 91 Should d elivery be indicated based on maternal or fetal concerns, tocolysis should not be used in 92 attempt to prolong gestation. [ C]93 94 At what gestation should planned delivery occur? 95 96 Late preterm ( 35+1 to 36 +6 weeks of gestation) delivery should be considered for women 97 presenting with complicated placenta praevia or a low -lying placenta. [C] 98 99 In what situations is vaginal delivery appropriate for women with a low -lying placenta? 100 101 In women with a third trimester asymptomatic low -lying placenta, the mode of delivery should be 102 based on the clinical background, the woman’s preferences supplemented by ultrasound findings, 103 including the distance between the placental edge and the fetal head position relative to the 104 leading edge of the placenta on TVS. [D] 105 106 Page 3 of 47 Optimising the delivery of placenta praevia 107 108 Prior to delivery, all women with placenta praevia and their partners should have a discussion 109 regarding delivery. Indications for blood transfusion and hysterectomy should be reviewed and 110 concerns or plans to decline blood o r blood products should be discussed openly and documented. 111 [GPP] 112 113 Placenta praevia and anterior low -lying placenta carr y a higher risk of massive obstetric 114 haemorrhage and hysterectomy. Delivery should be arranged in a maternity unit with on -site 115 blood t ransfusion services and access to critical care. [D] 116 117 Women with atypical antibodies form a particularly high -risk group and the care of these women 118 should involve discussions with the local haematologist and blood bank. [D] 119 120 Prevention and treatment of a naemia during the antenatal period is recommended for women 121 with placenta praevia or a low -lying placenta as for any pregnant woman . [D] 122 123 Delivery for women with placenta praevia or a low -lying placenta 124 125 What grade of obstetrician and anaesthetist should attend the caesarean delivery for a placenta 126 praevia? 127 128 As a minimum requirement for a planned caesarean section for placenta praevia, the surgical 129 procedure should be carried out by an appropriately experienced operator. [GPP] 130 131 In cases of planned caesarea n section for placenta praevia or a low -lying placenta , a senior 132 obstetrician (usually a consultant) and senior anaesthetist (usually a consultant) should be present 133 within the delivery or theatre suite where the surgery is occurring. [GPP] 134 135 When an emerg ency arises, the senior obstetrician and senior anaesthetist should be alerted 136 immediately and attend urgently. [ GPP ]137 138 What anaesthetic procedure is most appropriate for caesarean section in placenta praevia or a low -139 lying placenta ?140 141 Regional anaesthesia is considered safe and associated with lower risks of haemorrhage than 142 general anaesthesia for caesarean delivery in women with placenta praevia or a low -lying 143 placenta . Women with anterior placenta praevia or a low -lying placenta should be advised that it 144 may be necessary to convert to general anaesthesia if required and asked to consent. [D] 145 146 What blood products should be available? 147 148 Close liaison with the hospital transfusion laboratory is essential for women presenting with 149 placenta praevia or a low -lyi ng placenta . [GPP] 150 151 Rapid infusion and fluid warming devices should be immediately available. [GPP] 152 153 Cell salvage is recommended for patients where the anticipated blood loss is great enough to 154 induce anaemia, in particular, in women who would decline bloo d products. [D] 155 156 What surgical approach should be used for placenta praevia or a low -lying placenta ?157 158 Consider vertical skin and/or uterine incisions when the fetus is in a transverse lie to avoid the 159 Page 4 of 47 placenta, particularly below 28 weeks of gestation. [GP P] 160 161 Consider using preoperative and/or intraoperative ultrasonography to precisely determine 162 placental location and the optimal place for uterine incision. [D] 163 164 If the placenta is transected during the uterine incision, immediately clamp the umbilical cord 165 after fetal delivery to avoid excessive fetal blood loss. [D] 166 167 If pharmacological measures fail to control haemorrhage, initiate intrauterine tamponade and/or 168 surgical haemostatic techniques sooner rather than later. Interventional radiological techniques 169 should also be urgently employed where possible. [C] 170 171 Early recourse to hysterectomy is recommended if conservative medical and surgical interventions 172 prove ineffective. [D] 173 174 Antenatal diagnosis and outcome of placenta accreta spectrum 175 176 What are the ris k factors for placenta accreta spectrum? 177 178 The major risk factors for placenta accreta spectrum are history of accreta in a previous 179 pregnancy, previous caesarean delivery and other uterine surgery, including repeated endometrial 180 curettage. This risk rises as the number of prior caesarean sections increases. [B] 181 182 Women requesting elective caesarean delivery for non -medical indications should be informed of 183 the risk of placenta accreta spectrum and its consequences for subsequent pregnancies. [GPP] 184 185 How can a placenta accreta spectrum be suspected and diagnosed antenatally? 186 187 Antenatal diagnosis of placenta accreta spectrum is crucial in planning its management and has 188 been shown to reduce maternal morbidity and mortality. [D] 189 190 Previous caesarean delivery and t he presence of an anterior low -lying placenta or placenta praevia 191 should alert the antenatal care team of the higher risk of placenta accreta spectrum. [D] 192 193 Ultrasound screening and diagnosis of placenta accreta spectrum 194 195 Ultrasound imaging is highly accur ate when performed by a skilled operator with experience in 196 diagnosing placenta accreta spectrum . [C] 197 198 Refer women with any ultrasound features suggestive of placenta accreta spectrum to a specialist 199 unit with imaging expertise. [B] 200 201 Women with a history o f previous caesarean section seen to have an anterior low -lying placenta 202 or placenta praevia at the routine fetal anomaly scan should be specifically screened for placenta 203 accreta spectrum . [D] 204 205 Is there a role for magnetic resonance imaging (MRI) in the d iagnosis of placenta accreta spectrum? 206 207 Clinicians should be aware that the diagnostic value of MRI and ultrasound imaging in detecting 208 placenta accreta spectrum is similar when perform ed by experts. [C] 209 210 MRI may be used to complement ultrasound imaging to assess the depth of invasion and lateral 211 extension of myometrial invasion, especially with posterior placentation and/or in women with 212 Page 5 of 47 ultrasound signs suggesting parametrial invasion. [GPP] 213 214 Where should women with placenta accreta spectrum be cared for? 215 216 Women diagnosed with placenta accreta spectrum should be cared for by a multidisciplinary team 217 in a specialist centre with expertise in diagnosing and managing invasive placentation . [GPP] 218 219 Delivery for w omen diagnosed with placenta accreta spectrum shou ld take place in a specialist 220 centre with logistic support for immediate access to blood products, adult intensive care unit and 221 neonatal intensive care unit ( NICU ) by a multidisciplinary team with expertise in complex pelvic 222 surgery . [D] 223 224 When should deli very be planned for women with placenta accreta spectrum ?225 226 In the absence of risk factors for preterm delivery and evidence of invasive placentation , planned 227 delivery at 35 +0 to 36 +6 weeks of gestation provides the best balance between fetal maturity and 228 the risk of unscheduled delivery. [GPP] 229 230 Planning delivery of a suspected placenta accreta spectrum 231 232 Once the diagnosis of placenta accreta spectrum is made, a contingency plan for emergency 233 delivery should be developed, including the use of an institutiona l protocol for the management of 234 maternal haemorrhage . [GPP] 235 236 What should be included in the consent form for caesarean section in cases of suspected placenta 237 accreta spectrum? 238 239 Any woman giving consent for caesarean section should understand the risks ass ociated with 240 caesarean section in general, and the specific risks of placenta accreta spectrum in terms of 241 massive obstetric haemorrhage, increased risk of lower urinary tract damage, the need for blood 242 transfusion and the risk of hysterectomy. [GPP] 243 244 Addi tional possible interventions in the case of massive haemorrhage should also be discussed, 245 including cell salvage and interventional radiology where available. [D] 246 247 What healthcare professionals should be involved? 248 249 The elective delivery of women with plac enta accreta spectrum should be managed by a 250 multidisciplinary team, which should include senior anaesthetists, obstetricians and 251 gynaecologists with appropriate experience in managing the condition and other surgical 252 specialties if indicated. In an emerge ncy, the most senior clinicians available should be involved. 253 [GPP] 254 255 What anaesthetic is most appropriate for delivery? 256 257 The choice of anaesthetic technique for caesarean section for placenta accreta spectrum should be 258 made by the anaesthetist conducting t he procedure in consultation with the patient in advance. 259 [GPP] 260 261 The woman should be informed that the surgical procedure can be performed safely with regional 262 anaesthesia but should be advised that it may be necessary to convert to general anaesthesia if 263 required and asked to consent. [D] 264 265 Page 6 of 47 Optimising the delivery of placenta accreta spectrum 266 267 What surgical approach should be used for placenta accreta spectrum ?268 269 Caesarean section hysterectomy with the placenta left in situ is preferable to attempting to 270 sep arate it from the uterine wall. [C] 271 272 When the extent of the placenta accreta is limited in depth and surface area, and the entire 273 placental implantation area is accessible and visualised (i.e. completely anterior, fundal or 274 posterior without deep pelvic in vasion), uterus -preserving surgery may be appropriate, including 275 partial myometrial resection. [GPP] 276 277 Uterus -preserving surgical techniques should only be attempted by surgeons working in teams 278 with appropriate expertise to manage such cases and after appr opriate counselling regarding risks 279 and with informed consent. [D] 280 281 There are currently insufficient data to recommend the routine use of ureteric stents in placenta 282 creta and increta . [C] 283 284 What surgical approach should be used for placenta percreta? 285 286 Ther e is limited evidence to support uterus -preserving surgery in placenta percreta and women 287 should be informed of the high risk of peripartum and secondary complications, including the 288 need for secondary hysterectomy. [D] 289 290 Expectant management (leaving the p lacenta in situ) 291 292 Elective peripartum hysterectomy may be unacceptable to women desiring uterine preservation 293 or considered inappropriate by the surgical team. In such cases, leaving the placenta in situ should 294 be considered. [D] 295 296 When the placenta is left in situ, local arrangements need to be made to ensure regular review, 297 ultrasound examination and access to emergency care should the woman experience 298 complications, such as bleeding or infection. [D] 299 300 Methotrexate ( MTX ) adjuvant therapy should not be used for expectant management as it is of 301 unproven benefit and has significant adverse effects, including a reported maternal death. [C] 302 303 When is interventional radiology indicated? 304 305 Larger studies are necessary to determine the safety and efficacy of interven tional radiology 306 before this technique can be advised in the routine management of placenta accreta spectrum . [D] 307 308 Women diagnosed with placenta accreta spectrum who decline donor blood transfusion should be 309 managed in a unit with an interventional radiolo gy service. [D] 310 311 How is unsuspected placenta accreta spectrum at delivery best managed? 312 313 If at the time of an elective repeat caesarean section, where both mother and baby are stable, it is 314 immediately apparent that placenta percreta is present on opening the abdomen, the caesarean 315 section should be delayed until the appropriate staff and resources have been assembled and 316 adequate blood products are available. This may involve closure of the maternal abdomen and 317 urgent transfer to a specialist unit for deli very. [GPP] 318 Page 7 of 47 319 In case of unsuspected placenta accreta spectrum diagnosed after delivery of the baby, the 320 placenta should be left in situ and an emergency hysterectomy performed. [D] 321 322 Purpose and scope 323 324 The purpose of this guideline is to describe the di agnostic modalities and review the evidence -based 325 approach to the clinical management of pregnancies complicated by placenta praevia and placenta 326 accreta .327 328 Introduction and background epidemiology 329 330 Placenta praevia and placenta accreta are associated w ith high maternal and neonatal morbidity and 331 mortality. 1–5 The rates of placenta pr ae via and accreta have increased and will continue to do so as a 332 result of rising rates of c aesarean deliveries , increased maternal age and use of assisted reproductive 333 tech nolog y (ART) , plac ing greater demands on maternity -related resources. The highest rates of 334 complication for both mother and newborn are observed when these conditions are only diagnosed 335 at delivery. 336 337 2.1 Placenta praevia 338 339 Determining placental location is one of the first aims of routine midpregnancy (18 +6 –21 +6 weeks of 340 gestation) transabdominal obstetric ultrasound examination. 6,7 Placenta praevia was originally 341 defined using transabdominal scan as a placenta developing within the lower uterine segment an d342 graded according to the relationship and/or the distance between the lower placental edge and the 343 internal os of the uterine cervix. Grade I or minor praevia was defined as a lower edge inside the 344 lower uterine segment; grade II or marginal praevia as a lower edge reaching the internal os; grade 345 III or partial praevia when the placenta partially covers the cervix ; and grade IV or complete praevia 346 when the placenta completely covers the cervix. Grades I and II are also often defined as ’minor ’347 placenta pra evia whereas grades III and IV are referred to as ’major ’ placenta praevia. 348 349 The introduction of transvaginal scanning (TVS) in obstetric s in the 1980s has allowed for a more 350 precise evaluation of the distance between the placental edge and the internal o s. A recent 351 multidisciplinary workshop of the American Institute of Ultrasound in Medicine (AIUM) 8 ha s352 recommended discontinuing the us e of the terms ‘partial ’ and ‘marginal ’, suggest ing that the term 353 ‘placenta praevia ’ is used when the placenta lies direc tly over the internal os. For pregnancies 354 greater than 16 weeks of gestation , the placenta should be reported as ‘low lying ’ when the 355 placental edge is less than 20 mm from the internal os and as normal when the placental edge is 20356 mm or more from the int ernal os on transabdominal or TVS . This new classification could better 357 define the risks of perinatal complications , such as antepartum haemorrhage and major postpartum 358 haemorrhage (PPH) ,9,10 and has the potential of improving the obstetric management of p lacenta 359 praevia . Recent a rticles reviewed in th is guideline refer to the AIUM classification. 360 361 The estimated incidence of placenta praevia at term is 1 in 200 pregnancies. 5,9 However, this is 362 dependent on the definition used and is likely to change with th e introduction of the AIUM 363 classification described above and with the rising incidence of the main risk factors , i.e. prior 364 caesarean delivery and pregnancies resulting from ART . The relationship between a low -lying 365 placenta or placenta praevia and a vela mentous insertion of the umbilical cord is presented and 366 discussed in sister Green -top Guideline no. 27 b: Vasa Praevia: Diagnosis and Management .367 368 2.2 Placent a accreta 369 370 Placenta accreta is a histopathological term first defined by Irving and Hertig in 193 7, as the 371 Page 8 of 47 “abnormal adherence of the afterbirth in whole or in parts to the underlying uterine wall in the 372 partial or complete absence of decidua ”.11 Irving and Hertig did not include abnormally invasive 373 placentation in their series and thus in their descr iption was limited to abnormally adherent 374 placenta . Depending o n the depth of villous tissue invasiveness, p lacenta accreta was subsequently 375 subdivided by modern pathologists into ‘creta ’ or ‘adherenta ’ whe re the villi adheres superficially to 376 the myometri um without inter posing decidua ; ‘increta ’ wher e the villi penetrate deeply into the 377 uterine myometrium down to the serosa ; and ‘percreta ’ where the villous tissue perforates through 378 the entire uterine wall and may invade the surrounding pelvic organ s, such as the bladder .12 -14 Cases 379 of placenta accreta are also often subdivided into total, partial or focal according to the amount of 380 placental tissue involved and the different depths of accreta placentation have been found to co -381 exist in the same case .12,1 5 Thus p lacenta accreta is a spectrum disorder ranging from abnormally 382 adherent to deeply invasive placenta l tissue. 383 384 Detailed data on clinical findings and, where possible, on histopathological examination are essential 385 when describing different diagnostic or management techniques .16,1 7 The diagnostic conundrum is 386 obvious at the abnormally adherent end of the spectrum where the differe ntial diagnosis between a 387 difficult manual removal and an abnormally adherent or placenta accreta may be impossible in the 388 absence of histopathological confirmation. Th ese diagnostic difficulties probably explain the current 389 wide variation in reported prevalence of placenta accreta ranging between 1 in 300 and 1 in 2000 390 pregnancies 1–5 and highlight the need for a standardised approach to imaging , clinical and 391 histopathological description s. In the last decade, even the condition itself has begun to be known 392 by many different names, with ‘morbidly adherent placenta’ becoming particularly popular. This 393 terminology was originally used in the 19 th century to describe the clinical complications associated 394 with a retained placenta . This terminology is misleading as ‘morbidly adherent’ does not encompass 395 the abnormally invasive end of the accreta spectrum (increta and percreta) , which u sually have the 396 worst clinical outcomes. 16,17 In order to overcome these difficulties, the term s ‘placenta accreta 397 spectrum ’ or ’abnormally adherent and invasive placenta ’ should be used to include both the 398 abnormally adherent and invasive forms of accreta placentation .18 In this guideline , the term 399 placenta accreta spectrum will be used. 400 401 In the 1990s, t he m aternal mortality of placenta percreta was reported to be as high as 7% of 402 cases. 19 More recent large series have reported lower rates of maternal deat h and this is likely to be 403 further improve d by screening for placenta accreta spectrum in women at high risk and in planning 404 the delivery in speciali st centres .20 –22 405 406 Identification and assessment of evidence 407 408 This guideline was developed in accordance with standard methodology for producing Royal College 409 of Obstetricians and Gynaecologists (RCOG) Green -top Guidelines. The Cochrane Library (including 410 the Cochrane Database of Systematic Reviews and the Database of Abstracts of Reviews of Effects 411 [DARE]), EMBASE, Trip, MEDLINE and PubMed (electronic databases) were searched for relevant 412 randomised controlled trials (RCT) , systematic reviews and meta -analyses. The search was restricted 413 to articles published between May 2009 and December 201 7 (the search for the previous Guideline 414 was up to May 2009). The databases were searched using the relevant Medical Subject Headings 415 (MeSH) terms, including all subheadings, and this was combined with a keyword search. Search 416 words included ‘placenta pr aevia’ , ‘ low lying placenta ’, ‘ placenta accreta’ , ‘placenta increta’ 417 ‘placenta percreta’, ‘abnorma lly adherent placenta’ and ‘abnorma lly invasive placenta’ . The search 418 was restricted to humans and the English language . The National Library for Health and the National 419 Guideli ne Clearinghouse were also searched for relevant guidelines and reviews. 420 421 Where possible, recommendations are based on available evidence. In the absence of published 422 evidence, these have been annotated as ‘good practice points’. Further information about the 423 assessment of evidence and the grading of recommendations may be found in Appendix I. 424 Page 9 of 47 425 Antenatal diagnosis and management of placenta praevia or a low -lying placenta 426 427 1 What are the risk factors for placenta praevia or a low -lying placenta ?428 429 Caesa rean delivery is associated with an increased risk of placenta praevia in subsequent 430 pregnancies . This risk rises as the number of prior c aesarean sections increases . [B ]431 432 ART and maternal smoking increase the risk of placenta praevia. [ B]433 434 In 1997, a meta -analysis of the association of placenta praevia with history of caesarean delivery 435 found a dose -response pattern for the relative risk (RR) of placenta praevia of 4.5 (95% CI 3.6 –5.5) 436 for one, 7.4 (95% CI 7.1 –7.7) for two, 6.5 (95% CI 3.6 –11.6) for three, and 44.9 (95% CI 13.5 –149.5) 437 for four or more prior c aesarean deliveries compared with vaginal delivery. 23 [Evidence level 2++] 438 439 A systematic review and meta -analysis of 22 studies including over 2 million deliveries indicated that 440 the incidence of placen ta praevia increases from 10 in 1000 deliveries with one previous c aesarean 441 delivery to 28 in 1000 with three or more caesarean deliveries. 24 A 2014 meta -analysis confirmed 442 these findings and reported an overall o dds ratio (OR) of 1.47 (95% CI 1.44 –1.51) f or placenta 443 praevia after caesarean section .25 [Evidence level 1+] 444 445 Cohort studies have also reported that a second pregnancy within 1 year of a caesarean section is 446 associated with an increased risk of placenta praevia (RR 1.7, 95% CI 0.9 –3.1) .26 Compared with 447 vaginal birth , a previous prelabo ur caesarean section is associated with an increased risk of placenta 448 praevia in the second delivery (adjusted OR [aOR] 2.62 , 95% CI 1.24 –5.56). 27 [Evidence level 2+ +]449 450 There have been contradictory reports regarding the incidence of placenta praevia in multiple 451 pregnancies. A retrospective cohort study of 1 172 405 twin live births and stillbirths in the USA 452 between 1989 and 1998 found no increased risk in twins. 28 A retrospective cohort of 67 895 453 singleton and twin pregnancies found that dichorionic (aOR 1.54, 95% CI 1.15 –2.06) and 454 monochorionic (RR 3.29, 95% CI 1.32 –8.21) twin pregnancies had an increased risk of placenta 455 praevia compared with singletons. 29 [Evidence level 2+] 456 457 ART is associated with a higher inciden ce of placenta praevia independently of the high rate of 458 multiple pregnancies generated by the technique used. 30,31 A 2016 meta -analysis of ART singleton 459 pregnancies reported a RR of 3.71 (95% CI 2.67 –5.16) for placenta praevia 32 that was confirmed by a 460 20 17 meta -analysis (OR 2.67, 95% CI 2.01 –3.34). 33 Furthermore, a 2017 meta -analysis of the impact 461 of maternal smoking on placental position 34 (OR 1.42, 95% CI 1.30 –1.50) has found an increased risk 462 of placenta praevia . [Evidence level 1+] 463 464 Advanced maternal age has been also associated with a slight increase in the risk of placenta praevia 465 (OR 1.08, 95% CI 1.07 –1.09) but this effect may be due to parity 35 . [Evidence level 2 –]466 467 4.2 Should we screen for placenta praevia or a low -lying placenta , if so , at wha t gestation and with 468 what follow -up ?469 470 The mid pregnancy routine fetal anomaly scan should include placental locali sation thereby 471 identify ing women at risk of persisting placenta praevia or a low -lying placenta . [GPP] 472 473 The term placenta praevia should be us ed when the placenta lies directly over the internal os. For 474 pregnancies at more than 16 weeks of gestation the term low -lying placenta should be used when 475 the placental edge is 2 0 mm or less from the internal os on transabdominal or TVS. [D] 476 477 Page 10 of 47 If the place nta is thought to be low lying (less than 20 mm from the internal os) or praevia 478 (covering the os) at the routine fetal anomaly scan , a follow -up ultrasound examination including 479 a TVS is recommended at 32 weeks of gestation to diagnose persistent low -lyin g placenta and/or 480 placenta praevia. [D] 481 482 Placenta praevia is a well -established complication of pregnancy associated with high maternal and 483 perinatal complication rates. 4–9 The UK National Screening Committee (UK NSC) does not 484 recommend a national screen ing program for placenta praevia, but it has supported current local 485 practices of identifying it at the routine midpregnancy (18 +6 –21 +6 weeks of gestation ) antenatal 486 screening ultrasound examination in women whose placenta extends onto the internal cervica l os 487 (www.screening.nhs.uk/policies ). 36 An update published in 2014 that included a literature search 488 covering the period between January 2008 and November 2012 concluded that this practice was not 489 suppo rted by new evidence, but that the placental site is routinely reported at the time of the 490 routine fetal anomaly scan . In turn, this routine study has become the main screening test for 491 placenta praevia. 37 [Evidence level 4] 492 493 Apparent placental ‘migration’ following the development of the lower uterine segment during the 494 third trimester of pregnancy results in the resolution of the low -lying placenta in 90% of the cases 495 before term. 38 –46 This is less likely to occur in women with a previous caesarean delive ry .39 [Evidence 496 level 4] 497 498 In twin pregnancies, the likelihood of persistence of placenta praevia is also dependent on the 499 gestational age at sonographic detection . Among those with placenta praevia diagnosed in the 500 second trimester the majority of cases re solve by 32 weeks of gestation .29,47 After 32 weeks of 501 gestation around 50% of the remaining placenta praevia will resolve, with no further changes after 502 36 weeks of gestation .29 [Evidence level 3] 503 504 The timing of a confirmatory ultrasound examination in th e third trimester has varied between 32 505 and 36 weeks of gestation depending on the extent of the placenta praevia over the internal cervical 506 os. It is based on the perceived risk of antenatal haemorrhage , but there is no strong evidence that it 507 makes a dif ference in the care of asymptomatic women. 37 The timing of the follow -up ultrasound 508 examination should also be tailored according to a previous history of caesarean delivery to exclude 509 an associated placenta accreta spectrum . [Evidence level 4] 510 511 3 What is the role and what are the risks of TVS ?512 513 Clinicians should be aware that TVS for the diagnosis of placenta praevia or a low -lying placenta is 514 superior to transabdominal and transperineal approaches and is safe . [GPP] 515 516 In women with a persistent low -lying placenta or placenta praevia at 32 weeks of gestation who 517 remain asymptomatic, an additional TVS is recommended at around 36 weeks of gestation to 518 inform discussion about mode of delivery . [D]519 520 Cervical length measurement may help facilitate management de cisions in asymptomatic women 521 with placenta praevia. A short cervical length on TVS before 34 weeks of gestation increases the 522 risk of preterm emergency delivery and massive haemorrhage at caesarean section. [D] 523 524 TVS improves the accuracy of placental loc ali sation particularly when the placenta is posterior or if 525 the transabdominal ultrasound is unclear, for example , due to maternal obesity or the presence of 526 large uterine fibroids. 5 [Evidence level 4]527 528 There is only one small (n = 38) RCT comparing transa bdominal scan and TVS for placenta praevia, 529 which supports this safety profile and reports superior views, especially for posterior placentas. 48 530 Page 11 of 47 [Evidence level 1+]531 532 If the distance between the internal os and the placental edge is 20 mm or more on TVS , th e533 placental location should be recorded as normal and managed as per routine . Studies have not 534 demonstrated an increased risk for c aesarean section due to h aemorrhage in these cases .4,5 By 535 contrast , if the placenta extends beyond the internal os on TVS dur ing the second trimester, it is 536 likely to be confirmed as placenta praevia at 32 weeks of gestation .48 –50 However, ‘migration ’ is still 537 possible after 32 weeks of gestation .50,51 [Evidence level 2+] 538 539 TVS will reclassify 26 –60% of placenta s diagnosed as lo w lying at the routine fetal anomaly scan .52 –54 540 Overall, TVS has a high accuracy (positive predictive value of 93.3%, negative predictive value of 541 97.6% and false -negative rate of 2.33%) in predicting placenta praevia in women suspected of having 542 a low -lyi ng placenta on transabdominal scan in the second and early third trimester , with a 543 sensitivity of 87.5% and a specificity of 98.8% .55 [Evidence level 2+] 544 545 TVS ha s also been used to measure the cervical length to predict preterm birth 56 and cohort studies 546 with low risks of confounding bias have shown that cervical length is a predictor of antepartum 547 bleeding and emergency preterm caesarean section in placenta praevia. 57 –60 A prospective cohort 548 study of 59 women presenting with placenta praevia covering the i nternal os has shown that the 549 best cut -off point for the identification of women at risk of h aemorrhage requiring a caesarean 550 delivery before 34 weeks of gestation is a cervical length of 31 mm or less (sensitivity of 83.3% and 551 specificity of 76.6 %). Women with a cervical length of less than 31 mm have a 16 times (OR 16.4; 552 95% CI 3.4 –75.9) higher risk of emergency caesarean section due to massive haemorrhage .57 553 Similarly , a prospective cohort study of 54 women with placenta praevia covering the internal os has 554 shown that combining a cervical length of less than 30 mm and measurement of the lower placental 555 edge thickness of more than 10 mm has a sensitivity of 83.3% and a specificity of 78.4%. 58 More 556 prospective studies using a standardised ultrasound definit ion of placental edge thickness are 557 required before this sign can be used in clinical practice. [Evidence level 2+] 558 559 Compared with women with a long cervical length , women with a short cervical length (less than 25 560 mm) have a RR of 7.2 (95% CI 2.3 –22.3) f or massive haemorrhage during caesarean section for 561 placenta praevia .59 [Evidence level 2+] 562 563 Serial TVS cervical length measurements from 26 weeks of gestation have indicated that when the 564 length of the cervix decreases rapidly to 35 mm or less there is a n increased risk of preterm 565 caesarean section due to massive haemorrhage. 60 [Evidence level 2 –]566 567 4 Where should women with a low -lying placenta or placenta praevia be cared for in the third 568 trimester? 569 570 4.1 Wom en with a low -lying placenta or placenta p raevia with recurrent bleeding 571 572 Tailor antenatal care , including hospitali sation , to individual patient need and social 573 circumstances , e. g. distance between home and hospital and availability of transport ation ,574 previous bleeding episodes , haematology labor atory results and acceptance of receiving donor 575 blood or blood products . [GPP] 576 577 Where hospital admission has been decided, a documented assessment of risk factors for VTE in 578 pregnancy should be performed as outlined in RCOG GTG No. 37a. [D] 579 580 It should be m ade clear to any woman being treated at home in the third trimester that she should 581 attend the hospital immediately if she experiences any bleeding , including spotting, contractions 582 or pain (including vague suprapubic period -like aches). [GPP] 583 Page 12 of 47 584 The Cochran e systematic review by Nielson on t he impact of an intervention in women diagnosed as 585 having, or being likely to have a placenta praevia , which has not been updated since October 2002, 586 includes only one small RCT (n = 53 ) comparing hospital versus home car e for symptomatic placenta 587 praevia .61 This trial found little evidence of any clear advantage or disadvantage to a policy of home 588 versus hospital care , and the only significant difference was a reduction in length of hospital stay .62 589 [Evidence level 1 –]590 591 Two large retrospective studies of women presenting with placenta praevia at the routine fetal 592 anomaly scan have proposed s cores to predict the risk of emergency caesarean section . The first 593 study (n = 250) found that the risk is increased if the first ( sen tinel ) vaginal bleeding episode occurs 594 before 29 weeks of gestation (OR 2.64 , 95% CI 1.17 –5.98) , and with the occurrence of three or more 595 episodes of antepartum haemorrhage (OR 2.53 , 95% CI 1.1 –5.86) .63 The second (n = 214) found that 596 independent predictor s for emergency delivery are a history of caesarean section (OR 4.7 , 95 CI 1.2 –597 12); antepartum haemorrhage on one (OR 7.5 , 95% C I 2.5 –23), two (OR 14 , 95% CI 4.3 –47), and 598 three or more occasions (OR 27 , 95% CI 8.3 –90) ; and need for antenatal blood transfus ion (OR 6.4 ,599 95% CI 1.7 –23). 10 A retrospective study of 214 women with singleton pregnancies found that the risk 600 of preterm emergency caesarean delivery increase s with the number of antepartum bleeding 601 episodes with one (OR 7.5, 95% C, 2.5 -23), two (OR 14, 95% CI 4.3 -47), and three or more (OR 27, 602 95% CI 8.3 -90) , as well as need for blood transfusion (OR 6.4, 95% C, 1.7 -23) .64 The results of these 603 studies suggest that predictors for emergency delivery in women with placenta praevia can be used 604 for individua li sed antenatal care regarding need for hospital admission, corticosteroid s605 administration and timing of delivery. [Evidence level 2 –]606 607 4.2 Wom en with asymptomatic placenta praevia or a low -lying placenta 608 609 Women with asymptomatic placenta praevia or a low -lying placenta in the third trimester should 610 be counselled about the risks of preterm delivery and obstetric haemorrhage, and their care 611 should be tailored to their individual needs. [GPP] 612 613 Women with asymptomatic placenta praevia confirmed at the 32 -we ek follow -up scan and 614 managed at home should be encouraged to ensure they have safety precautions in place, including 615 having someone available to help them as necessary and ready access to the hospital. [GPP] 616 617 Most women with asymptomatic placenta pr aevia (no bleeding or contractions) can be cared for as 618 outpatients with similar outcomes compared with hospitalisation and at lower cost. 5 Numerous 619 factors influence the chances of the placenta pr aevia persisting until delivery , such as prior 620 caesarean section ,43 the distance between the placental edge and the internal os , and the thickness 621 of the placenta l edge .4 These parameters can be useful in tailoring individual patient needs. 622 [Evidence level 4] 623 624 5 Is there a place for cervical cerclage in placenta praev ia or a low -lying placenta ?625 626 The use of cervical cerclage to reduce bleeding and prolong pregnancy is not supported by 627 sufficient evidence to recommend its use outside of a clinical trial. [GPP] 628 629 The Cochrane systematic review by Nielson 61 on the impact of cerclage in women diagnosed as 630 having, or being likely to have, placenta praevia included two small RCTs (n = 25 and 36) comparing 631 cervical cerclage versus no cerclage. There may be a reduction in preterm births before 34 weeks of 632 gestation (RR 0.45 , 95% CI 0.23 –0.87) , but this evidence is not robust enough to recommend its use 633 outside of clinical trials. [Evidence level 1 –]634 635 There have been no new trials looking at this issue since the last update of th is g uideline. 636 Page 13 of 47 637 6 In what circumstances, and at what gestation, should women be offered antenatal 638 corticosteroids? 639 640 A single course of antenatal corticosteroid therapy is recommended between 34 +0 and 36 +0 weeks 641 of gestation for pregnant women with a low -lying placenta or placenta praevia and is appropriate 642 prior to 34 +0 weeks of gestation in women at higher risk of preterm birth. [ GPP ]643 644 A large case –control study found that neonatal morbidities in women with placenta praevia include 645 an increased risk of lower 5 -min ute Apgar scores, neonatal intensive care un it ( NICU ) admission, 646 anaemia, respiratory distress syndrome, mechanical ventilation and intraventricular haemorrhage. 65 647 Neonates born to mothers with placenta praevia ha ve lower birthweight s (2806 v ersu s 3285 g) and 648 lower gestational ages (36 +2 versu s 38 +1 weeks ). These differences were not significant after 649 adjusting for confounders such as prematurity .66 [Evidence level 2++ ]650 651 Compared with placebo or no treatment with antenatal corticosteroids (betamethasone ,652 dexamethasone or hydrocortisone) , antenatal co rticosteroids are associated with a reduction in the 653 most serious adverse outcomes related to prematurity, including perinatal death (RR 0.72 , 95% CI 654 0.58 –0.89) , respiratory distress syndrome (average RR 0.66 , 95% CI 0.56 –0.77) , intraventricular 655 haemorrhag e (average RR 0.55, 95% CI 0.40 –0.76) and necrotising enterocolitis (RR 0.50 , 95% CI 656 0.32 –0.78) .67 [Evidence level 1+ ]657 658 The 2016 RCT has found that the administration of betamethasone to women with a singleton 659 pregnancy at risk for late preterm delivery (3 4+0 to 36 +5 weeks of gestation ) significantly reduces the 660 rate of neonatal respiratory complications. 68 [Evidence level 1+] 661 662 A decision analytic model designed to compar e total maternal and neonatal quality -adjusted life 663 years for delivery of women with pl acenta praevia at 34 +0 to 3 6+6 weeks of gestation indicated that 664 corticosteroids administration at 35 +5 weeks of gestation followed by planned delivery at 36 weeks 665 of gestation optimi ses maternal and neonatal outcomes. 69 [Evidence level 4] 666 667 7 Is there a place for the use of tocolytics in women presenting with a low -lying placenta or 668 placenta praevia and preterm labour ?669 670 Tocolysis for women presenting with symptomatic placenta praevia or a low -lying placenta may 671 be considered for 48 hours to facilitate ad ministration of antenatal corticosteroids. [ C]672 673 Should delivery be indicated based on maternal or fetal concerns, tocolysis should not be used in 674 attempt to prolong gestation. [ C]675 A systematic review to determine if the prolonged (48 hour s or more ) use of tocolytics in women 676 with symptomatic preterm placenta pr aevia improves perinatal outcome identified two 677 retrospective studies (total , n = 217) and one RCT (n = 60). 70 The r esults of the RCT show ed that 678 pregnancy can be prolonged for more than 7 days with continued tocolytics (OR 3.10 , 95% CI 1.38 –679 6.96). W hen combined with the data of retrospective studies , the results did not reach significance 680 (OR 1.19, 95% CI 0.63 –2.28). The RCT was judged inadequately compliant with the Consolidated 681 Standards of Reporti ng Trials statement. [Evidence level 1 –]682 683 A recent randomi sed, double -blind, placebo -controlled multicentr e trial including 109 women at 24 +0 684 to 33 +6 weeks with at least one episode of placenta pr aevia bleeding and intact membranes has 685 shown that there is no difference in the prolongation of pregnancy between the nifedipine (n = 54) 686 and placebo (n = 55) group s.71 Adverse perinatal outcomes were comparable between groups. 687 [Evidence level 1+] 688 689 Page 14 of 47 8 At what gestation should planned delivery occur? 690 691 Delivery tim ing should be tailored according to antenatal sympto ms and for women presenting 692 with uncomplicated placenta praevia delivery should be considered between 36 and 37 weeks of 693 gestation . [C] 694 695 As the risk of major haemorrhage increases rapidly after 36 weeks of gestation, expert opinions have 696 highlighted that decisions regarding timing of delivery must be individualised and suggested t hat on 697 the basis of limited data available , women with uncomplicated placenta praevia should undergo 698 scheduled birth by caesarean section between 36 and 37 weeks of gestation .69,72,73 [Evidence level 4] 699 700 The r isk s of bleeding, labo ur, or bleeding and labo ur leading to the need for emergency delivery 701 increase with advancing gestational age , whereas the risks of morbidity associated with prematurity 702 decrease. 4,5 The risk of an emergent bleed associated with place nta praevia has been reported to be 703 4.7% by 35 weeks of gestation , 15% by 36 weeks of gestation , 30% by 37 weeks of gestation and 59% 704 by 38 weeks of gestation .74 [Evidence level 2 –]705 706 A US population -based cohort study using the Centre for Disease Control a nd Prevention’s Linked 707 Birth -Infant Death data files has evaluated the effects of delivering placenta praevia at 35, 36 and 37 708 weeks of gestation on the risk of several neonatal outcomes. 75 Compared with neonates born at 38 709 weeks of gestation , those delive red at 35, 36 and 37 weeks of gestation have no greater odds of 710 meconium passage, fetal distress, fetal anaemia, neonatal seizures, increased ventilator needs or 711 infant death at 1 year . However, aOR odds of 5 -min ute Apgar scores of less than 7 are greater at 35 712 and 36 weeks of gestation (aOR 3.33 , 95% CI 1.71 –6.47 ; and aOR 2.17 , 1.11 –4.22, respectively) as 713 are odds of NICU admission rates (aOR 2.25 , 95% CI 2.01 –2.50 ; and aOR 1.57 , 1.38 –1.76, 714 respectively). [Evidence level 2+] 715 716 9 In what situations is vagi nal delivery appropriate for women with a low -lying placenta ?717 718 In women with a third trimester asymptomatic low -lying placenta , the mode of delivery should be 719 based on the clinical background, the woman’s preferences supplemented by ultrasound findings, 720 in cluding the distance between the placental edge and the fetal head position relative to the 721 leading edge of the placenta on TVS . [D] 722 723 Wom en presenting with a placental edge less than 2 0 mm from the internal os in the third trimester 724 are more likely to need delivery by caesarean section when the placental edge is thicker (over 1 0725 mm) 76,77 and/or contains a sponge -like echo 78 or m arginal ‘sinus ’.79 These additional ultrasound 726 features are poorly defined , not routinely assessed in UK practice and the success r ates of vaginal 727 delivery when the placental edge is between 10 and 20 mm from the internal os vary widely (56 %728 and 93 %, respectively ).80–83 The corresponding studies are small , observational and re trospective ,729 making a recommendation for a specific mode of delivery based on ultrasound findings difficult .730 [Evidence level 2 –]731 732 Optimising the delivery of a placenta praevia 733 734 Prior to delivery, all women with placenta praevia and their partners should have a discussion 735 regarding delivery . Indications for blo od transfusion and hysterectomy should be reviewed and 736 concerns or plans to decline blood or blood products should be discussed openly and documented .737 [GPP] 738 739 Placenta praevia and anterior low -lying placenta carr y a higher risk of massive obstetric 740 haemorr hage and hysterectomy . Delivery should be arranged in a maternity unit with on -site 741 blood transfusion services and access to critical care. [D] 742 Deleted: Late preterm (35+1to 3 6+6weeks of gestation ) 743 Deleted: for women presenting with complicated placenta 744 praevia or alow -lying placenta 745 Deleted: ¶ 746 Deleted: late preterm (34 +0 to 36 +6 weeks of gestation) 747 Page 15 of 47 748 Women with atypical antibodies form a particularly high -risk group and the care of these women 749 should involve discussions with the local haematologist and blood bank. [D] 750 751 Prevention and treatment of anaemia during the antenatal period is recommended for women 752 with placenta praevia or a low -lying placenta as for any pregnant woman . [D] 753 754 General procedures for dis cussing and obtaining consent for caesarean section are described in 755 detail in RCOG Consent Advice No.7: Caesarean section .84 [Evidence level 4] 756 757 Women having a caesarean section for placenta praevia are at increased risk of blood loss of more 758 than 1000 m l compared with women having a caesarean section for other indications (RR 3.97 , 95% 759 CI 3.24 –4.85). 85 Women with anterior placenta regardless of type of placenta praevia are at 760 increased blood loss. 86 Placenta praevia covering the internal cervical os and anterior placentation 761 are independent risk factors (OR 4.1 and OR 3.5, respectively) for massive haemorrhage during 762 caesarean section .86 A US case –control study from the National Institute of Child Health (NICH) and 763 Human Development Maternal -Fetal Medicin e Units (MFMU) Network Caesarean Section Registry 764 has shown that maternal h aemorrhagic morbidity is more common in women with pr aevia (19 %765 versus 7%, a djusted RR 2.6, 95% CI 1.9 -3.5) and the main factors associated with maternal 766 haemorrhage include pre -del ivery an aemia, thrombocytopenia, diabetes and magnesium use .87 767 [Evidence level 2++] 768 769 The risk of massive haemorrhage together with the possibility of needing a blood transfusion has 770 been estimated to be approximately 12 times more likely in caesarean sect ion for placenta praevia 771 than in caesarean delivery for other indications. 88,89 Similarly to uncomplicated pregnancies, women 772 with placenta pr aevia should be screen ed for anaemia and investigated if their haemoglobin levels 773 are outside the normal UK range (11 0 g/l at first visit and 105 g/l at 28 weeks) .36 Iron 774 supplementation should be implemented if indicated. [Evidence level 4] 775 776 For wom en at high risk of emergency transfusion , such as those presenting with placenta praevia 777 and with no clinically signific ant alloantibodies , it has been recommended that group and screen 778 samples should be sent once a week to exclude or identify any new antibody formation and to keep 779 blood available if necessary for delivery . However, this should be at the discretion of the t eam 780 responsible and managed according to local facilitites .89 [Evidence level 4] 781 782 Delivery for women with placenta praevia or a low -lying placenta 783 784 6.1 What grade of obstetrician and anaesthetist should attend the caesarean delivery fo r a placenta 785 praev ia ?786 787 As a minimum requirement for a planned caesarean section for placenta praevia, the surgical 788 procedure should be carried out by a n appropriately experienced operator . [GPP] 789 790 In cases of planned caesarean section for placenta praevia or a low -lying plac enta , a senior 791 obstetrician (usually a consultant) and senior anaesthetist (usually a consultant) should be present 792 within the delivery or theatre suite where the surgery is occurring . [GPP] 793 794 When an emergency arises, the senior obstetrician and senior an aesthetist should be alerted 795 immediately and attend urgently . [GPP ]796 797 Maternal complications at caesarean section increase when the primary surgeon is a 798 trainee/resident rather than an experienced surgeon. 90 Placenta praevia is often associated with 799 Page 16 of 47 additi onal including fetal malpresentation (transverse or breech presentation) requiring complex 800 intraoperative manoeuvres to deliver the baby. 91 [Evidence level 4] 801 802 6.2 What anaesthetic procedure is most appropriate for caesarean section in placenta praevia? 803 804 Regional anaesthesia is considered safe and associated with lower risks of haemorrhage than 805 general anaesthesia for caesarean delivery in women with placenta praevia or a low -lying 806 placenta . Women with anterior placenta praevia or a low -lying placenta shoul d be advised that it 807 may be necessary to convert to general anaesthesia if required and asked to consent. [D] 808 809 There is insufficient evidence to support one technique over another and there have been no new 810 trials since the previous version of this g uideli ne. 811 812 An RCT of regional versus general anaesthesia for placenta praevia, including women with placenta 813 accreta , has indicated that blood transfusion requirements (although not estimated blood loss) are 814 greater in the general anaesthetic group. 92 [Evidence level 1 –]815 816 A 4-year observational study at 19 US academic centres of women undergoing caesarean delivery 817 found that the risk factors for haemorrhage -related morbidity are increased in those undergoing 818 general anaesthesia. 93 [Evidence level 2 –]819 820 The recen t case –control study from the NICH D/MFMU Network Cesarean Section Registry found 821 general anaesthesia to be one of the main factors associated with maternal haemorrhage in women 822 with placenta pr aevia. 87 [Evidence level 2++] 823 824 6.3 What blood products should b e available ?825 826 Close liaison with the hospital transfusion laboratory is essential for women present ing with 827 placenta praevia or a low -lying placenta . [GPP] 828 829 Rapid infusion and fluid warming devices should be immediately available. [GPP] 830 831 Cell salvage is re commended for patients where the anticipated blood loss is great enough to 832 induce anaemia, in particular , in women who would decline blood products . [D] 833 834 Red cells, fresh frozen plasma , and cryoprecipitate or fibrinogen concentrate are all kept by blood 835 ba nks supplying obstetric units. If the haemoglobin is less than 70 g/l in the postoperative period, 836 where there is no ongoing or threat of bleeding, the decision to transfuse should be made on an 837 informed individual basis .89 In an extreme situation and when the blood group is unknown, group O 838 rhesus D -negative red cells should be given .89 Further recommendations are provided in Green -top 839 Guideline No.52 : Prevention and Management of Postpartum Haemorrhage . 88 [Evidence level 4] 840 841 There is no evidence to supp ort the use of autologous blood transfusion for placenta praevia. 90 842 [Evidence level 4] 843 844 Cell salvage was not often used previously in obstetrics because of the perceived risk of amniotic 845 fluid embolism or induction of maternal alloimmuni sation. No definite cases of amniotic fluid 846 embolism have been reported so far and the risks of cell salvage in the obstetric population parallel 847 those in the non -pregnant population .94,95 [Evidence level 4] 848 849 6.4 What surgical approach should be used for placenta praevia or a low -lying placenta ?850 851 Consider v ertical skin and/or uterine incisions whe n the fetus is in a transverse lie to avoid the 852 Page 17 of 47 placenta, particularly below 28 weeks of gestation. [GPP] 853 854 Consider using preoperative and/or intraoperative ultrasonography to precis ely determine 855 placental location and the optimal place for uterine incision. [D] 856 857 If the placenta is transected during the uterine incision, immediately clamp the umbilical cord 858 after fetal delivery to avoid excessive fetal blood loss. [D] 859 860 If pharmacologi cal measures fail to control haemorrhage, initiate intrauterine tamponade and/or 861 surgical haemostatic techniques sooner rather than later. Interventional radiological techniques 862 should also be urgently employed where possible . [C] 863 864 Early recourse to hyster ectomy is recommended if conservative medical and surgical interventions 865 prove ineffective. [D] 866 867 In cases of anterior placenta praevia , cutting through the placenta is often associated with increased 868 maternal bleeding. A retrospective cohort study found t hat avoiding incision of the anterior placenta 869 praevia after 24 weeks of gestation reduces the need for maternal blood transfusion during or after 870 caesarean delivery .96 [Evidence level 2 –]871 872 A ‘J’-shaped uterine incision has been evaluated in women presenti ng with placenta praevia in a 873 small retrospective study and shown to decrease intraoperative blood loss and facilitate the delivery 874 of the fetus .97 [Evidence level 2–]875 876 Intrauterine ba lloon tamponade, different types of compression sutures and uterine arte ry occlusion 877 techniques have been increasingly used since the previous version of the guideline in women with 878 placenta praevia to control, reduce or stop intraoperative bleeding and PPH. Case series on the use 879 of intrauterine hydrostatic balloon catheters , including the Bakri balloon ,98 –102 the BT -Cath® 880 balloon 103 or the Sengstaken –Blakemore tube ,104 in women with placenta praevia have reported 881 success in controlling PPH ranging from 75% to 88%. [Evidence level 3] 882 883 Factors associated with the failure of Bak ri balloon tamponade for placenta praevia include prior 884 caesarean section , anterior placentation, thrombocytopenia and/or coagulopathy at the time of 885 insertion, and a PPH volume of more than 500 ml within the first hour of placement. 100 [Evidence 886 level 2++ ]887 888 Uterine compressive and endouterine sutures are well established techniques for the control of 889 haemorrhage following atonic PPH. The best known suture technique was described by B -Lynch in 890 105 A combined method of B -Lynch suture and the intrauterin e balloon has also been 891 successfully used in preventing PPH in placenta praevia. 106 [Evidence level 3] 892 893 Intr aoperative i nterventional radiolog ical techniques , including transarterial embolisation 107 and 894 temporary balloon occlusion 108 of the internal iliac arteries , have also been successfully used to 895 prevent and control haemorrhage in placenta praevia and should be considered when available .896 Follow -up studies of women who ha ve undergone arterial embolisation for control of PPH suggest 897 that the intervention does not impair subsequent menstruation and fertility. 109 –111 [Evidence level 3] 898 899 Antenatal diagnosis and outcome of placenta accreta spectrum 900 901 7.1 What are the risk factors for placenta accreta spectrum ?902 903 The major risk factors for placenta accreta spe ctrum are history of accreta in a previous 904 pregnancy, previous caesarean delivery and other uterine surgery , including repeated endometrial 905 Page 18 of 47 curettage. This risk rises as the number of prior caesarean sections increases. [B ]906 907 Women requesting elective caesa rean delivery for non -medical indications should be informed of 908 the risk of placenta accreta spectrum and its consequences for subsequent pregnancies. [GPP] 909 910 All epidemiological studies of the last 2 decades have shown a direct association between the 911 incr ease in caesarean deliveries and the incidence of placenta accreta spectrum (abnormally 912 adherent and invasive placenta) in subsequent pregnancies worldwide. 112 –122 The 2016 Nordic 913 Obstetric Surveillance Study found that the risk of invasive placentation in crease s seven -fold after 914 one prior caesarean section .118 [Evidence level 2+] 915 916 A meta -analysis of five cohorts and 11 case –control studies reported a summary OR of 1.96 (95% CI 917 1.41 –2.74) for placenta accreta spectrum after a caesarean section .24 [Evidence level 2++] 918 919 The risk of placenta accreta spectrum increases with the number of previous caesarean sections . A 920 systematic review reported an increase in t he incidence of accreta placenta tion from 3.3 –4.0 % in 921 women with placenta praevia and no previous c aesa rean delivery , to 50 –67% in women with three 922 or more caesarean deliveries. 25 When stratified for the number of previous caesarean sections , the 923 OR for placenta accreta spectrum in a subsequent pregnancy range s between 8.6 (95% CI 3.536 –924 21.078) 112 and 17.4 (95% CI 9.0 –31.4) for two previous caesarean sections , and 55.9 (95% CI 25.0 –925 110.3) for three or more caesarean sections .121 [Evidence level 2++] 926 927 Placenta praevia is another important risk factor for placenta accreta spectrum (see Appendix I I). A928 large mu lticentre US cohort study noted that for women presenting with placenta praevia and prior 929 caesarean section the risk of accreta placentation was 3%, 11%, 40%, 61% and 67% for one , two ,930 th ree , four , and fi ve or more c aesarean deliveries, respectively. 113 Th e national case –control study 931 using the UK Obstetric Surveillance System found that the incidence of placenta accreta spectrum 932 increases from 1.7 per 10 000 women overall to 577 per 10 000 in women with both a previous 933 caesarean section and placenta praevi a. 114 [Evidence level 2+] 934 935 Other additional risk factors include maternal age 111,114,118,121 and ART , in particular in vitro 936 fertilisation .114,121,123 –126 Advanced maternal age (35 years or more ) in women without a previous 937 caesarean section increase s the aOR by 1.30 (95%CI 1.13 -1.50) for every 1-year increase in age. 114 938 [Evidence level 2-]939 940 Placenta accreta spectrum is not exclusively a consequence of caesarean delivery . Other surgical 941 trauma to the integrity of the uterine endometrium and/or superficial m yometrium , such as those 942 following uterine curettage, manual removal of the placenta, postpartum endometritis or 943 myomectomy , has been associated with accreta placenta tion in subsequent pregnancies. 1,12,13 944 Overall, the aOR for placenta accreta spectrum afte r previous uterine surgery is 3.40 (95% CI 1.30 –945 8.91). 114 [Evidence level 2+] 946 947 The development of placenta accreta spectrum has also been reported in women with no surgical 948 history but presenting with a uterine pathology , such as bicornuate uterus, adenomy osis, 949 submucous fibroids and myotonic dystrophy. 1,12,13 [Evidence level 3] 950 951 More recently there ha s been an increase in reports describing implantation into deficient caesarean 952 section scars and mounting evidence that a caesarean scar pregnancy diagnosed i n early pregnancy 953 can evolve into a n abnormally adherent or invasive placenta in the second half of pregnancy .127 –131 A954 caesarean scar pregnancy can be diagnosed using TVS from the second month of pregnancy using 955 specific ultrasound criteria. 130 ,131 In the last decade , the number of reported cases of caesarean scar 956 pregnancy has increased due to improved awareness of the condition, widespread use of ultrasound 957 scanning in early pregnancy and an increase in the number of prior caesarean sections . The outcome 958 Page 19 of 47 of caesarean scar pregnancy depends on the amount of definitive placenta developing inside the 959 scar and depth of villous invasion. Further data are required to establish the relationship between a 960 first trimester scar pregnancy and the development of inva sive placentation . [Evidence level 3] 961 962 7.2 How can a placenta accreta spectrum be suspected and diagnosed antenatally ?963 964 Antenatal diagnosis of placenta accreta spectrum is crucial in planning its management and has 965 been shown to reduce maternal morbidity a nd mortality . [D] 966 967 Previous caesarean delivery and the presence of an anterior low -lying placenta or placenta praevia 968 should alert the antenatal care team of the higher risk of placenta accreta spectrum . [D] 969 970 Maternal complications in placenta accreta spec trum are primarily the result of massive 971 haemorrhage. 5 Median estimated blood loss in cohorts of placenta accreta spectrum ranges from 972 2000 to 7800 m l and the median number of units of blood transfused is 5 units .132 Antenatal 973 diagnosis of placenta accreta spectrum reduces maternal peripartum haemorrhage and 974 morbidity. 20,133 –136 [Evidence level 4] 975 976 Population studies have shown that placenta accreta spectrum remains undiagnosed before delivery 977 in one -half 137 to two -thirds of cases .121 In a series from speci alist centres , approximately one -third of 978 cases of placenta accreta were not diagnosed during pregnancy. 138 [Evidence level 2+] 979 980 Multidisciplinary management in a maternity unit with access to maternal and neonatal intensive 981 care is often required for wome n with placenta accreta spectrum .21,22,136 ,139 For such care to be 982 organi sed, the diagnosis must be made antenatally . [Evidence level 4] 983 984 7.2.1 Ultrasound screening and diagnosis of placenta accreta spectrum 985 986 Ultrasound imaging is highly accurate when perf orm ed by a skilled operator with experience in 987 diagnosing placenta accreta spectrum . [C] 988 989 Refer w omen with any ultrasound features suggestive of placenta accreta spectrum to a specialist 990 unit with imaging expertise. [B] 991 992 Women with a history of previous ca esarean section seen to have an anterior low -lying placenta 993 or placenta praevia at the routine fetal anomaly scan should be specifically screened for placenta 994 accreta spectrum . [D] 995 996 Numerous ultrasound imaging techniques have been reported over the years , including grey scale 997 imaging and colour Doppler imaging (CDI ), and/or three -dimensional power Doppler 998 sonography. 16,17,140 –142 In 2016, t he European Working Group on Abnormally Invasive Placenta 999 proposed a standardi sed description of ultrasound signs (see A ppendix II I) used for the prenatal 1000 diagnosis of placenta accreta 141 and the International Abnormally Invasive Placenta Expert Group 1001 produced a proforma protocol for the ultrasound assessment .142 [Evidence level 4] 1002 1003 A systematic review and meta -analysis of 23 ultrasound studies including 3707 pregnancies at risk of 1004 placenta accreta found that the overall performance of ultrasound when performed by skilled 1005 operators was very good with a sensitivity of 90.72% (95% CI 87.2 –93.6), specificity of 96.94% (95% 1006 CI 96.3 –97.5) and d iagnostic OR of 98.59 (95%CI 48.8 –199.0). Among the different ultrasound signs, 1007 abnormality of the uterus –bladder interface ha d the best specificity of 99.75% (95% CI 99.5 –99.9) 1008 for the prediction of placenta accreta . Abnormal vasculature on CDI ha d the best predictive accuracy 1009 with a sensitivity of 90.74% (95% CI 85.2 –94.7 ), specificity of 87.68% (95% CI 84.6 –90.4 ) and 1010 diagnostic OR of 69.02 (95% CI 22.8 –208.9) .143 [Evidence level 2++] 1011 Page 20 of 47 1012 A 2017 systematic review and meta -analysis using the s tandardised ultrasound signs ( see Appendix 1013 III) has shown that in women presenting with placenta pr aevia and history of prior caesarean 1014 section , the performance of ultrasound for the antenatal detection of placenta accreta spectrum is 1015 even higher with a se nsitivity of 97.0% (95% CI 93.0 –99.0) , specificity of 97.0% (95 % CI 97.0 –98.0) 1016 and diagnostic OR of 228.5 (95% CI 67.2 –776.9) in prospective studies .144 Placental lacunae giv e the 1017 placenta a ‘moth -eaten ’ appearance on grey scale imaging and the increased va scularity of the 1018 placental bed with large feeder vessels entering the lacunae are the most common ultrasound signs 1019 associated with placenta accreta spectrum .16,17,143,144 [Evidence level 2+ +]1020 1021 Determining the depth and lateral extension of placental invasi on is helpful for planning the 1022 individual care of women diagnosed with placenta accreta spectrum .16,17,145 No ultrasound sign or a 1023 combination of ultrasound signs have so far be en found to be specific to the depth of placenta 1024 accreta spectrum and thus to p rovide with an accurate differential diagnosis between adherent and 1025 invasive accreta placentation .16 This may be due to the wide heterogeneity in terminology used to 1026 describe the grades of placenta accreta spectrum , differences in the study design with mos t studies 1027 not reporting detailed data on clinical diagnosis at birth and/or on histopathology examination , and 1028 many studies having includ ed cases of placental retention in their cohort with no evidence of 1029 abnormal villous adherence or invasion . [Evidence l evel 2++] 1030 1031 As the vast majority of placenta accreta spectrum are now the consequence of low placentation into 1032 a previous caesarean section scar , TVS has an important role in the early diagnosis, follow -up ,1033 differential diagnosis between adherent and invasi ve accreta placentation and management of 1034 placenta accreta spectrum .144 [Evidence level 4] 1035 1036 7.2.2 Is there a role for magnetic resonance imaging ( MRI ) in the diagnosis of placenta accreta 1037 spectrum ?1038 1039 Clinicians should be aware that the diagnostic value of M RI and ultrasound imaging in detecting 1040 placenta accreta spectrum is similar when perform ed by experts . [C] 1041 1042 MRI may be used to complement ultrasound imaging to assess the depth of invasion and lateral 1043 extension of myometrial invasion, especially with poste rior placentation and/or in women with 1044 ultrasound signs suggesting parametrial invasion . [GPP ]1045 1046 MRI has been increasingly used for the prenatal diagnosis of placenta accreta. 146 –150 The main MRI 1047 features of placenta accreta include abnormal uterine bulging , dark intraplacental bands on T2 -1048 weighted imaging, heterogeneous signal intensity within the placenta , disorgani sed vasculature of 1049 placenta and disruption of the uteroplacental zone . A systematic review has found that most studies 1050 are of a small sample si ze and thus , sensitivity and specificity of MRI in diagnosing placenta accreta 1051 varies widely between 75 % and 100% , and 65 % and 100% , respectively. 149 [Evidence level 2+ +]1052 1053 Two systematic reviews and meta -analyses have found that the diagnostic value of ult rasound 1054 imaging and MRI in detecting placenta accreta spectrum is similar. The first review 148 included 13 1055 studies and reported a sensitivity of 83% (95% CI 77 –88), specificity of 95% (95% CI 93 –96) and 1056 detection OR of 63.41 (95% CI 29.04 –138.48) for ultra sound, compared with a sensitivity of 82% 1057 (95% CI 72 –90), specificity of 88% (95% CI 81 –94) and detection OR of 22.95 (95% CI 3.19 –165.11) 1058 for MRI. The second review (2014) 149 included 18 studies and found that the overall diagnostic 1059 accuracy of MRI has a sensitivity of 94.4% (95% CI 86.0 –97.9), specificity of 84.0% (95% CI 76.0 –89.8) 1060 and diagnostic OR of 89.0 (95% CI 22.8 –348.1). The latter review also found that MRI has high 1061 predictive accuracy in assessing both the depth and topography of placental invas ion. [Evidence level 1062 2++] 1063 1064 Page 21 of 47 The use of intravenous gadolinium injection may increase the sensitivity and specificity of MRI in the 1065 diagnosis of the invasive forms of placenta accreta spectrum but the evidence on long -term fetal 1066 safety is limited. 150 Further more, the experience of the radiologists remains an independent factor in 1067 the diagnostic accuracy of MRI and access to expert radiologists is highly variable. [Evidence level 4] 1068 1069 7.3 Where should women with placenta accreta spectrum be cared for? 1070 1071 Women di agnosed with placenta accreta spectrum should be cared for by a multidisciplinary team 1072 in a specialist centre with expertise in diagnosing and managing invasive placentation . [GPP] 1073 1074 Delivery for w omen diagnosed with placenta accreta spectrum should take pl ace in a specialist 1075 centre with logistic support for immediate access to blood products, adult intensive care unit and 1076 NICU by a multidisciplinary team with expertise in complex pelvic surgery . [D] 1077 1078 More data have become available since the last version of th is guideline on the specific management 1079 of placenta accreta spectrum . Overall , women with accreta placenta tion should be cared for 1080 according to the risks of severe maternal bleeding and premature delivery. Placenta percret a can be 1081 associated with major prenatal complications from early in pregnancy , such as uterine rupture 151 –153 1082 and bladder involvement with associated life -threatening haemorrhage .154 –156 [Evidence level 4]1083 1084 A 2015 expert review has suggested that caesarean delivery of women at high risk and /or diagnosed 1085 prenatally with placenta accreta spectrum, in particular its invasive forms, should occur in a 1086 specialist centre with multidisciplinary expertise and experience in managing complex pelvic surgery, 1087 and with access to an adult intensive car e unit and NICU .136 [Evidence level 4] 1088 1089 A retrospective cohort study of 77 women with suspected placenta accreta found that women who 1090 delivered prior to a planned delivery date were significantly more likel y to have had vaginal bleeding 1091 and uterine activit y when compared with women who had a scheduled delivery. 20 Each episode of 1092 antenatal vaginal bleeding i s associated with an increased risk of unscheduled delivery (aOR 3.8 , 95% 1093 CI 1.8 –7.8) and the risk increase s when associated with preterm prelabour ruptu re of membranes. 1094 [Evidence level 2 –]1095 1096 Considering the higher frequency of placenta praevia in the accreta group ,144,157 these results are 1097 likely to be influence d by the perinatal complications of placenta praevia. Surveys of healthcare 1098 providers in the US and Canada have highlighted widely varied approaches to virtually every aspect 1099 of care for placenta accreta spectrum .158 –161 Similarly , a recent online survey completed by members 1100 of the expert panel for the perinatal management of placenta accreta spectru m disorders for the 1101 International Federation of Gynecology and Obstetrics (FIGO) has found wide variation in global 1102 practices. 162 [Evidence level 4] 1103 1104 There is increasing evidence from r etrospective cohort studies from the US A that women with 1105 placenta accre ta spectrum diagnosed prenatally , cared for by a speciali st multidisciplinary team , are 1106 less likely to require large volume blood transfusion and reoperation within 7 days of delivery for 1107 bleeding complications compared with women cared for by non -multidis ciplinary standard obstetric 1108 care without a specific protocol. 21,22,136,139,163,164 Women admitted at 34 weeks of gestation and 1109 delivered between 34 and 35 weeks of gestation by a specialist multidisciplinary team ha ve a1110 significantly lower emergency surge ry rate than those not cared for by such a team (23% v ersu s 64%; 1111 P = 0.001) despite a similar median gestational age at delivery ( 34 [16 –39] weeks versu s 34 [19 -40 ]1112 weeks ; P = 0.50 , respectively). 21 In addition, maternal outcomes are improved over time wit h1113 increasing experience within a well -established multidisciplinary team performing 2 –3 cases per 1114 month. 22 Very few of these studies provide with data on the differential clinical diagnosis between 1115 abnormally adherent and abnormally invasive accreta and de tailed pathologic confirmation of the 1116 depth and lateral extension of villous myometrial invasion . [Evidence level 2 –]1117 Page 22 of 47 1118 7.4 When should delivery be planned for women with placenta accreta spectrum ?1119 1120 In the absence of risk factors for preterm delivery and ev idence of invasive placentation , p lanned 1121 delivery at 35+0 to 36 +6 weeks of gestation provides the best balance between fetal maturity and 1122 the risk of unscheduled delivery . [GPP] 1123 1124 Similarly to placenta praevia , clinical factors should be considered when det ermining the timing of 1125 administration of antenatal corticosteroids and the optimal gestational age for delivery in women 1126 with placental accreta. 165,166 There are currently no RCTs or well -controlled observational studies to 1127 guide best practice in delivery timing of placenta accreta spectrum . [Evidence level 4]1128 1129 In cases of suspected placenta accreta spectrum , where significant blood loss and caesarean 1130 hysterectomy is anticipated, delivery at between 34 and 35 weeks of gestation has been proposed in 1131 order t o avoid emergen cy delivery , which still occurs about 20% of the time even in scheduled 1132 cases. 165,167 A 2010 decision analysis supports this approach based on the increasing likelihood of 1133 emergency delivery as pregnancy goes beyond 34 weeks of gestation .168 [Evidence level 4] 1134 1135 Th e data of th ree recent single institution retrospective cohort stud ies of women with prior 1136 caesarean delivery diagnosed prenatally with placenta accreta have indicated that in the absence of 1137 risk factors for preterm delivery, it is s afe to plan the delivery at 36 weeks of gestation . The first 1138 study included 103 women delivered between 1982 and 2002 and found that the mean gestational 1139 age at delivery is 33 +5 weeks of gesta tion in cases of deep placental invasion (increta and percreta) 1140 compared with 35 +2 weeks of gestation in the superficial adherent group. 169 The second study of 216 1141 women found that urgent delivery for bleeding decreased significantly with advancing gestation. 170 1142 Most women were delivered at 36 weeks of gestation or gre ater, with nearly 90% in the absence of 1143 bleeding complications. The third study of 84 women who had reached 34 +0 weeks of gestation with 1144 a suspected pr aevia accreta found that those with no risk factors for preterm birth are at low risk for 1145 an unscheduled delivery prior to 36 weeks of gestation .171 [Evidence level 2+] 1146 1147 Planning delivery of a suspected placenta accreta spectrum 1148 1149 Once the diagnosis of placenta accreta spectrum is made , a contingency plan for emergency 1150 delivery should be developed, includin g the use of an institutional protocol for the management of 1151 maternal haemorrhage . [GPP] 1152 1153 Due to a lack of RCT s or well -controlled observational studies , the optimal management of placenta 1154 accreta spectrum remains undefined and is determined by the experti se available , the depth and 1155 lateral exten sion of the accreta portion of the placenta , the presence of an associated placenta 1156 praevia, radiological findings, the medical and surgical comorbidities, and finally , the accessibility of 1157 a regional team focused o n these patients .1158 1159 The main risk associated with the delivery of placenta accreta spectrum is massive h aemorrhage and 1160 its associated complication s, such as coagulopathy, multisystem organ failure and death. M any 1161 women with placenta accreta spectrum requir e massive blood transfusion (8 units or more ) and 1162 their median platelet count is lowest compared with other causes of massive PPH .172,173 [Evidence 1163 level 2+] 1164 1165 A review of 34 studies published between 1977 and 2012, including a total number of 508 617 1166 deliv eries and 865 cases of confirmed placenta accreta, found that the most significant maternal 1167 risks associated with delivery are the need for postpartum transfusion due to haemorrhage and 1168 peripartum hysterectomy. Maternal mortality remains rare, but signific antly higher than among 1169 matched postpartum controls. 123 [Evidence level 4] 1170 Page 23 of 47 1171 Transfusions in placenta accreta spectrum should be guided by a national and/or institutional 1172 protocol for management of PPH .88,89 [Ev idence level 4] 1173 1174 8.1 What should be included i n the consent form for caesarean section in cases of suspected placenta 1175 accreta spectrum ?1176 1177 Any woman giving consent for caesarean section should understand the risks associated with 1178 caesarean section in general, and the specific risks of placenta accreta spectrum in terms of 1179 massive obstetric haemorrhage, increased risk of lower urinary tract damage, the need for blood 1180 transfusion and the risk of hysterectomy. [GPP] 1181 1182 Additional possible interventions in the case of massive haemorrhage should also be discuss ed, 1183 including cell salvage and interventional radiology whe re available. [ D]1184 1185 Any woman with suspected placenta accreta spectrum should meet with a senior obstetrician in the 1186 antenatal period. The different risks and treatment options should have been disc ussed and a plan 1187 agreed, which should be reflected clearly in the consent form and medical record. This should 1188 include standard discussion for the caesarean section procedure 84 and whether conservative 1189 management of the placenta or proceeding straight to h ysterectomy is preferred in the situation 1190 where increta or percreta is confirmed at surgery. [Evidence level 4] 1191 1192 Where available, cell salvage should be considered . I f the woman refuses donor blood transfusion , it 1193 is recommended 89 that she be transferred t o a unit with a cell saver. [Evidence level 4] 1194 1195 8.2 What healthcare professionals should be involved ?1196 1197 The elective delivery of wom en with placenta accreta spectrum should be managed by a 1198 multidisciplinary team , which should includ e senior an aesthetists , obstetrician s and 1199 gynaecologists with appropriate experience in managing the condition and other surgical 1200 specialties if indicated. In an emergency, the most senior clinicians available should be involved .1201 [GPP] 1202 1203 Following the previous version of the guidel ine , the National Patient Safety Agency in collaboration 1204 with the RCOG and the Royal College of Midwives set up an expert working group to develop a care 1205 bundle for placenta accreta. 174 Six elements of good care were agreed upon . The care bundle was 1206 then t ested in six units over a 5 -month pilot study period and it was found to be both achievable and 1207 practical. Clinical outcomes were monitored, confirming the high morbidity associated with this 1208 condition. [Evidence level 4] 1209 1210 The six elements considered to b e reflective of good care are :1211 1212 • Consultant obstetrician plann ing and directly supervising delivery .1213 • Consultant anaesthetist plann ing and directly supervising anaesthe sia at delivery . 1214 • Blood and blood products available . 1215 • Multidisciplinary involvement in preop erative planning . 1216 • Discussion and consent , includ ing possible interventions (such as hysterectomy, leaving the 1217 placenta in situ , cell salvage and intervention al radiology) . 1218 • Local availability of a level 2 critical care bed. 1219 1220 The 2015 MBRRACE report from the Confidential Enquiry into Maternal Deaths in the UK has 1221 indicated that despite increasing numbers of women at risk from placenta accreta spectrum 1222 following previous caesarea n section , only one death occurred in a woman who had a placenta 1223 Page 24 of 47 praevia percreta and history of two previous caesarean sections .17 5 There were no deaths from 1224 unexpected placenta accreta found at caesarean section , suggesting that previous recommendations 1225 regarding imaging and preparations for women with placenta praevia and a previous caesarean 1226 section have been followed. 17 6 [Evidence level 2++ ]1227 1228 A 2015 single cent re retrospective cohort study of the effectiveness of a standardi sed operative 1229 approach in 98 cases of histologically confirmed placenta accreta supports the early presence of a 1230 gyn aecological surgeon and oncologist at delivery and demonstrates that a ‘call if needed ’ approach 1231 is not acceptable for these complex cases .17 7 [Evidence level 2 +]1232 1233 The American College of Obstetricians and Gynecologists (ACOG ) guidelines highlight that to 1234 enhance patient safety, it is important that the delivery be performed by an experienced obstetric 1235 team that includes an obstetric surgeon, with other surgical special ists, such as urologists, general 1236 surgeons , and gyn aecologic al surgeons and oncologists, available if necessary. 166 [Evidence level 4] 1237 1238 8.3 What anaesthetic is most appropriate for delivery? 1239 1240 The choice of anaesthetic technique for caesarean section for pl acenta accreta spectrum should be 1241 made by the anaesthetist conducting the procedure in consultation with the patient in advance. 1242 [GPP] 1243 1244 The woman should be informed that the surgical procedure can be performed safely with regional 1245 anaesthesia but should be advised that it may be necessary to convert to general anaesthesia if 1246 required and asked to consent . [D] 1247 1248 Both general and regional anaesthetic techniques have been shown to be safe for surgical 1249 procedures required for the delivery of placenta accreta spe ctrum ; the judgment of which type of 1250 technique to be used should be made on an individual basis. 167 [Evidence level 4] 1251 1252 There is insufficient evidence to support one technique over another and there have been no new 1253 trials since the previous version of th is guideline. 1254 1255 8.4 Optimising the delivery of placenta accreta spectrum 1256 1257 There are no RCT s comparing different surgical approaches for placenta accreta spectrum suspected 1258 antenatally. Both conservative and radical surgical approaches can be associated with a high 1259 maternal morbidity although the value of an experienced team in a specialist centre decreases the 1260 risk significantly .21,22,136,139,163,164 [Evidence level 4] 1261 1262 8.4 .1 What surgical approach should be used for placenta accreta spectrum ?1263 1264 Caesarean se ction hysterectomy with the placenta left in situ is preferable to attempt ing to 1265 separate it from the uterine wall. [C] 1266 1267 When the exten t of the placenta accreta is limited in depth and surface area , and the entire 1268 placental implantation area is accessible and visuali sed (i.e. completely anterior, fundal or 1269 posterior without deep pelvic invasion) , uterus -preserving surgery may be appropriate, including 1270 partial myometrial resection. [GPP] 1271 1272 Uterus -preserving surgical techniques should only be attempted by surg eons working in teams 1273 with appropriate expertise to manage such cases and after appropriate counselling regarding ris ks 1274 and with informed consent. [D ]1275 1276 Page 25 of 47 There are currently insufficient data to recommend the routine use of ureter ic stents in placenta 1277 creta and increta . [C]1278 1279 The choice of surgical technique will depend on the position of the placenta, the depth of invasion, 1280 and the parametrial extension of the placenta accreta spectrum as assessed by ultrasound and/or 1281 MRI before delivery, the visual assessmen t of the uterus at the time of surgery and the presenting 1282 clinical symptoms, i.e. bleeding or no bleeding. 5 [Evidence level 4] 1283 1284 The ACOG recommends planned, preterm caesarean section hysterectomy with the placenta left in 1285 situ as removal of a placenta accr eta spectrum is associated with significant haemorrhagic 1286 morbidity. 166 In cases of high suspicion for accreta during caesarean delivery , the majority of 1287 members of the US Society of Maternal -Fetal Medicine ( SMFM ) and FIGO expert panel proceed with 1288 hysterec tomy. 158 –162 [Evidence level 4] 1289 1290 Similarly, in a 2017 systematic review and meta -analysis on the diagnosis and outcome of placenta 1291 accreta, an elective or emergen cy caesarean hysterectomy was performed in 208 out of 232 (89.7%) 1292 cases. 144 [Evidence level 2 ++] 1293 1294 A retrospective study of 57 cases of suspected accreta demonstrated significantly reduced short -1295 term morbidity if the placenta is left in place and hysterectomy performed electively compared with 1296 attempting to remove the placenta first .17 8 Attempting placental separation risks hysterectomy in up 1297 to 100% of cases as also confirmed by other authors. 17 8,17 9 [Evidence level 2++] 1298 1299 A case –control study of 49 women requiring a peripartum hysterectomy for massive h aemorrhage ,1300 including 20 women presenting wi th placenta accret a, reported that the use of a vessel sealing 1301 device during surgery decreases the estimated blood loss, the need for massive blood transfusions ,1302 and does not increase operative time or complication rate s.180 [Evidence level 2+ ]1303 1304 A systemat ic review found that u terus -preserving surgery resulted in a secondary hysterectomy in 1305 24/77 women (31%), maternal mortality in 2/55 women (4%), subsequent menstruation in 28/34 1306 women (82%) and subsequent pregnancy in 19/26 women (73%). 181 A more recent sy stematic 1307 review showed that uterus -preserving surgery is associated with a success rate of 48/76 women 1308 (63.2 %), a secondary hysterectomy in 23/76 women (30 .0 %), maternal mortality in 2/54 women 1309 (3.7%), subsequent menstruation in 20/37 women (81.1%) and sub sequent pregnancy in 21/27 1310 women (77.8%). 182 [Evidence level 2++] 1311 1312 A small cohort study has shown that the introduction of the Triple -P procedure (perioperative 1313 placental localization, pelvic devascularization and placental non -separation) involving delive ry of 1314 the fetus via transverse uterine incision above the upper border of the placenta , myometrial excision 1315 and reconstruction of the uterine wall reduces the rate of hysterectomy, PPH and duration of 1316 hospital stay in women with placenta accreta. 183 The in cidence of post -operative complications of 1317 the Triple -P procedure depends on comorbidities and in particular , the placental position and the 1318 depth of villous invasion. 184 Small case series have also reported on the successful use of 1319 compression sutures and on using the cervix as a natural tamponade by inverting it into the uterine 1320 cavity , and suturing the anterior and/or the posterior cervical lips into the anterior and/or posterior 1321 walls of the lower uterine segment .185 –188 [Evidence level 3] 1322 1323 A systematic review of peripartum surgical techniques used in placenta accreta spectrum has found 1324 that methotrexate (MTX) and uterus -preserving surgical techniques are associated with a 16% 1325 unintentional urinary tract injury rate as opposed to 57% for standard hystere ctomy and that use of 1326 ureteric stents reduces the risk of urologic injury .189 [Evidence level 2++] 1327 1328 There are no RCT s on the use of ureter ic stents in placenta accreta spectrum . Ureteric stents or 1329 Commented [EJ1]: Too Commented [EJ2]: Page 26 of 47 catheters are more commonly used pre -operatively in the US A where around 26% of the members 1330 of both the SM FM 159 and ACOG fellows 161 are using them in the management of suspected of 1331 abnormally invasive placenta . [Evidence level 4] 1332 1333 8.4.2 What surgical approach should be used for placenta percreta? 1334 1335 There is limit ed evidence to support uterus -preserving surgery in placenta percreta and women 1336 should be informed of the high risk of peripartum and secondary complications , including the 1337 need for secondary hysterectomy. [D] 1338 1339 The following f our approaches have been descr ibed .137,159 –161,165,167,190 1340 1341 Primary hysterectomy following delivery of the fetus , without attempting placental separation .1342 Delivery of the fetus avoiding the placenta, with repair of the incision leaving the placenta in 1343 situ .1344 Delivery of the fetus withou t disturbing the placenta, followed by partial excision of the uterine 1345 wall (placental implantation site) and repair of the uterus .1346 Delivery of the fetus without disturbing the placenta, and leaving it in situ, followed by elective 1347 secondary hysterectomy 3 –7 days following the primary procedure. 1348 1349 There are no well -controlled observational studies , and therefore, no firm recommendations can be 1350 made. 1351 1352 Women with placenta percreta a re more likely to req uire additional blood products and intensive 1353 care admiss ion than women with pl acenta creta or increta. 190 The incidence of urologic al 1354 complications is also increased , including cystotomy and ureteric injury. 191 [Evidence level 4] 1355 1356 When the urinary bladder is invaded by placental tissue , preoperative cystoscopy and the placement 1357 of ureteric stents have been recommended. 161,192 Planned cystotomy can prevent extensive 1358 muscularis damage and bleeding from attempts at dissection. 192 [Evidence level 4] 1359 1360 Filling the bladder to identify the bladder separation site , openi ng the bladder to identify percreta 1361 villous tissue and remov al of the involved bladder area ha ve also been recommended by different 1362 authors .161,165,193 [Evidence level 4]1363 1364 Uterus -preserving surgery is possible in placenta percreta as demonstrated in a coho rt study of 71 1365 women . A multidisciplinary stepwise surgical approach , including bilateral ligations of the anterior 1366 division of the iliac arteries before removing the placenta , was shown to be successful in controlling 1367 the bleeding and preserving the patie nt's uterus in around 90 % of the cases , with 14% of urinary 1368 tract complications , most of which can be identified and repaired during caesarean section .194 1369 [Evidence level 3] 1370 1371 A review of 119 placenta percreta cases published in the international literature has shown that 1372 expectant management with the placenta left in situ is associated with severe long -term 1373 complications of h aemorrhage and infections, including a 58% risk of secondary hysterectomy up to 1374 9 months after the delivery. Local resection appears to be associated with fewer complications 1375 within 24 h ours postoperatively compared with hysterectomy or leaving the placenta in situ .1376 However, a selection bias in the direction of less severe cases for the local resection technique m ay 1377 in part explain the l ower complic ation rates with that approach. 195 [Evidence level 4] 1378 1379 8.5 Expectant management (leaving the placenta in situ )1380 1381 Elective peripartum hysterectomy may be unacceptable to women desiring uterine preservation 1382 Page 27 of 47 or considered inappropriate by the surgi cal team . I n such cases , leaving the placenta in situ should 1383 be considered . [D] 1384 1385 When the placenta is left in situ, local arrangements need to be made to ensure regular review ,1386 ultrasound examination and access to emergency care should the woman experience 1387 complications , such as bleeding or infection. [D] 1388 1389 MTX adjuvant therapy should not be used for expectant management as it is of unproven benefit 1390 and has significant adverse effects , including a reported maternal death . [C]1391 1392 Conservative management in plac enta accreta spectrum , including in cases of placenta increta and 1393 percreta, is an option in women who desire to preserve their fertility . However, it is not 1394 recommended in women presenting with major bleeding as it is unlikely to be successful and risks 1395 de laying definitive treatment and increasing morbidity. 5 [Evidence level 4] 1396 1397 A retrospective multicentr e study examined 167 women treated conservatively for placenta accreta 1398 in tertiary university hospital cent res in France between 1993 and 2007. Conservativ e expectant 1399 management with part of the placenta left in situ was successful in 131 out of 167 cases (78.4%; 95% 1400 CI 71.4 –84.4). 19 6 One woman died of myelosuppression and nephrotoxicity related to MTX 1401 administration through the umbilical cord . Spontaneous p lacental resorption occurred in 87 out of 1402 116 cases (75.0% ; 95% CI 66.1 –82.6), with a median delay from delivery of 13.5 weeks (range 4 –60 1403 weeks). 196 [Evidence level 2+] 1404 1405 The patient should be warned of the risks of chronic bleeding, sepsis, septic shock, peritonitis, 1406 uterine necrosis, fistula, injury to adjacent organs, acute pulmonary oedema, acute renal failure, 1407 deep venous thrombosis or pulmonary embolism. 196 Prophylactic antibiotics may be helpful in the 1408 immediate postpartum period to reduce the risk o f infective complications. 197 [Evidence level 4] 1409 1410 An observational case series , including 24 women with placenta accreta left in situ after deliver y and 1411 treated with MTX , reported placental delivery in 33.3% of the cases (spontaneously in 55% , and in 1412 45% f ollowing dilatation and surgical evacuation ). 199 There was no control group of patients who did 1413 not receive MTX and so it is unknown whether or not the MTX was clinically helpful. One patient did 1414 suffer liver damage and the risks of this therapy must be ba lanced against the unproven benefit. 1415 [Evidence level 3] 1416 1417 The pattern of follow -up for the conservative management of placenta accreta spectrum is not 1418 supported by RCTs and not stratified for the according to the depth and lateral extension of villous 1419 myom etrial invasion . Some authors have reported cases where retained villous tissues were 1420 removed after conservative management using hysteroscopic resection 200,201 or high -intensity 1421 focused ultrasound. 202 In rare cases, a disseminated intravascular coagulatio n may develop requiring 1422 a secondary hysterectomy. 203 [Evidence level 3] 1423 1424 8.6 When is interventional radiology indicated? 1425 1426 Larger studies are necessary to determine the safety and efficacy of interventional radiology 1427 before this technique can be advised in the routine management of placenta accreta spectrum . [D] 1428 1429 Wom en diagnosed with placenta accreta spectrum who decline donor blood transfusion should be 1430 managed in a unit with an interventional radiology service. [D] 1431 1432 Since the publication of the last versi on of th is guideline there ha ve been several cohort studies 1433 describing the use of interventional radiology in assisting surgical and conservative management of 1434 placenta accreta with variable success . The main aim of th is procedure is to reduce the risks of 1435 Page 28 of 47 intraoperative haemorrhage during the caesarean delivery of pregnancies diagnosed antenatally 1436 with praevia increta or percreta . Various combin ation s have been proposed , including intraoperative 1437 internal iliac artery and /or postoperative uteri ne artery emb olis ation 204,205 and internal iliac artery 20 6– 1438 209 or abdominal balloon occlusion 210 –215 . The latter technique has been increasingly used in China 1439 but the methodology of these studies is very heterogeneous with no data on the diagnosis of the 1440 different grad es of villous invasion and variable confounding factors such as placental position and 1441 number of previous caesarean deliveries . Small cohort studies have also been published on the use 1442 of a tourniquet 216,217 and of surgical artery ligation .218 [Evidence le vel 3] 1443 1444 A single institution observational cohort study of 45 cases of placenta accreta describes the use of 1445 prophylactic lower abdominal aorta balloon occlusion and found a reduced need for blood 1446 transfusion .212 One of the case s was complicated by lower e xtremity arterial thrombosis and another 1447 by ischaemic injury to the femoral nerve . A comparative study of abdominal aortic occlusion versus 1448 internal iliac artery occlusion found that aortic balloon occlusion resulted in better clinical outcomes 1449 with less b lood loss, blood transfusion, balloon insertion time, fluoroscopy time and fetal radiation 1450 dose. 215 [Evidence level 2 –]1451 1452 A systematic review reported success rates of 159/177 (89.8 %) for arterial embolisation, with 1453 secondary hysterectomy being necessary i n 20/177 (11.3%) and subsequent menstruation occurring 1454 in 74/85 (87.1%). In 3/10 women (30%) a subsequent pregnancy occurred. Arterial balloon occlusion 1455 catheters have been associated with a success rate of 33/42 (78.6%) and the need for a secondary 1456 hyster ectomy in 8/42 (19%). 182 [Evidence level 2++] 1457 1458 The value of prophylactic placement of balloon catheters in the iliac arteries in cases of placenta 1459 accreta has been more controversial . This is mainly because of the higher risks of complications than 1460 embolis ation , including iliac artery thrombus or rupture , and ischaemic nerve injury .219 –222 [Evidence 1461 level 3] 1462 1463 A small RCT of women presenting with a prenatal diagnosis of placenta accreta was published in 1464 2015 .223 The women were randomi sed to either preoperati ve prophylactic balloon catheters (n = 13) 1465 or to a control group (n = 14). No difference was observed for t he number of women with blood loss 1466 greater than 2500 m l, number of plasma products transfused, duration of surgery, peripartum 1467 complications and hosp itali sation length. Reversible adverse effects related to prophylactic balloon 1468 catheter insertion were noted in 2 /13 (15.4%) cases . [Evidence level 1+] 1469 1470 8.7 How is unsuspected placenta accreta spectrum at delivery best managed? 1471 1472 If at the time of an electi ve repeat caesarean section , where both mother and baby are stable , it is 1473 immediately apparent that placenta percreta is present on opening the abdomen, the caesarean 1474 section should be delayed until the appropriate staff and resources have been assembled a nd 1475 adequate blood products are available. This may involve closure of the maternal abdomen and 1476 urgent transfer to a specialist unit for delivery. [GPP] 1477 1478 In case of unsuspected placenta accreta spectrum diagnosed after delivery of the baby , the 1479 placenta sho uld be left in situ and an emergency hysterectomy performed . [D] 1480 1481 If the placenta fails to separate with the usual measures, leaving it in place and closing, or leaving it 1482 in place, closing the uterus and proceeding to a hysterectomy are both associated w ith less blood 1483 loss than trying to separate it. Attempts at removing placenta accreta at caesarean section can lead 1484 to massive haemorrhage, high maternal morbidity and possible maternal death. These risks are 1485 particularly high when the caesarean section ta kes place in an environment with no emergency 1486 access to blood bank products and expertise in managing placenta accreta. 20,21,123,136 [Evidence level 1487 4] 1488 Page 29 of 47 1489 Clinical governance 1490 1491 9.1 Debriefing 1492 1493 Postnatal follow -up should include debriefing with an explanati on of what happened, why it 1494 happened and any implications for future pregnancy or fertility. In particular, women where 1495 conservative treatment of placenta accreta spectrum has been successful should be info rmed of the 1496 risk of recurrence. 1497 1498 9.2 Training 1499 1500 Raising the awareness about the clinical risk factors of placenta accreta spectrum should be pursued 1501 locally, including organising policies or guidelines for flagging up women at risk and arranging for 1502 them to see a specialist consultant when suspected. 1503 1504 The re should be a ppropriate training for ultrasound staff in the antenatal diagnosis of placenta 1505 accreta spectrum .1506 1507 9.3 Clinical incident reporting 1508 1509 Any lack of compliance with the care bundle by the clinical team for a woman with either placenta 1510 praevia or accreta should be investigated .1511 1512 There should be written protocols for identification of and planning further care of women 1513 suspected to have placenta accreta spectrum. 1514 1515 Recommendations for future research 1516 1517 • A large prospective study comparing the imp act on the management of the use of the ’low -lying 1518 placenta or placenta praevia ’ classification with the traditional grades 1 –4 classification at 1519 different gestation s is needed. 1520 • Prospective studies are needed to assess the role of third trimester ultrasoun d in evaluating the 1521 risks of haemorrhage and emergency caesarean section in low -lying placenta and determining 1522 the mode of delivery .1523 • Large prospective population -based studies are needed in order to assess whether ultrasound is 1524 a cost -effective screening t ool for placenta accreta spectrum in women with a history of 1525 caesarean section (s) presenting with a low -lying placenta or placenta praevia in the second 1526 trimester of pregnancy. 1527 • Prospective c omparative ultrasound imaging including transvaginal ultrasound an d MRI studies 1528 are needed to evaluate the diagnostic accuracy for evaluation of the depth and topography of 1529 villous invasion in adjacent organs. 1530 • RCTs of optimal timing of delivery for both conditions (placenta praevia and placenta accret a)1531 are needed. 1532 • RCTs of surgical and nonsurgical management strategies for placenta accreta spectrum 1533 (including interventional radiology) and comparing conventional versus conservative 1534 management , stratified according to the depth and lateral extension of villous myometrial 1535 invasion , are needed. 1536 • Future studies on the diagnosis and management of placenta accreta spectrum should use 1537 standardised evidence -based approach including a systematic correlation between ultrasound 1538 signs and detailed clinical diagnosis at delivery and pat hologic confirmation of grades of villous 1539 invasiveness when possible .1540 1541 Page 30 of 47 Auditable topics 1542 1543 Placenta praevia 1544 1545 • Antenatal diagnosis of placenta praevia (100%). 1546 • Antenatal detection and treatment of anaemia (100%). 1547 • Antenatal imaging performed according to hos pital policy (100%). 1548 • Appropriate antenatal delivery plan made and documented, to include discussion with woman 1549 and her partner, documentation that the risks and indications for blood transfusion and 1550 hysterectomy have been discussed and that concerns, queri es or refusals of treatments have 1551 been addressed (100%). 1552 • Involvement of local blood bank and haematologist in the care of women with placenta praevia 1553 and atypical antibodies (100%). 1554 • Appropriate personnel present at delivery (100%). 1555 • Appropriate site for del ivery (100%). 1556 • Appropriate surgical approaches performed (100%). 1557 • Early -term e lective delivery between 37 +0 and 37 +6 weeks of gestation for asymptomatic women 1558 with placenta praevia and no other risk factors (100%). 1559 • Antenatal steroid administration between 34 +0 and 36 +0 weeks of gestation (100%) .1560 • Women requesting elective caesarean section for nonmedical reasons are informed of the risk of 1561 placenta praevia and accreta spectrum , and its consequences in future deliveries (100%). 1562 1563 Placenta accreta spectrum 1564 1565 • Anten atal imaging performed according to hospital policy with diagnosis confirmed at birth 1566 (100%). 1567 • Appropriate antenatal delivery plan made and documented, to include discussion with woman 1568 and her partner, documentation that the risks and indications for blood transfusion and 1569 hysterectomy have been discussed and that concerns, queries or refusals of treatments have 1570 been addressed (100%). 1571 • All elements of the care bundle satisfied before elective surgery in women with placenta accreta 1572 spectrum (100%): 1573 o consultant o bstetrician planned and directly supervising delivery 1574 o consultant anaesthetist planned and directly supervising anaesthetic at delivery 1575 o blood and blood products available 1576 o multidisciplinary involvement in preoperative planning 1577 o discussion and consent includes possible interventions (such as hysterectomy, leaving the 1578 placenta in place , cell salvage and interventional radiology) and local availability of a level 2 1579 critical care bed. 1580 1581 12 . 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Conservative management of 2102 placenta accreta: hysteroscopic resection of retained tissues. J Minim Invasive Gynecol 2103 2014;21:910 –3. 2104 201 . Mazzon I, Favilli A, Grasso M, Horvath S, Gerli S. Is the cold loop hysteroscopic technique a 2105 myometrial sparing treatment for placenta accreta residuals in a puerperal uterus? J Matern Fetal 2106 Neonatal Med 2016;29:1613 –6. 2107 Bai Y, Luo X, Li Q, Yin N, F u X, Zhang H, et al. High -intensity focused ultrasound treatment of 2108 placenta accreta after vaginal delivery: a preliminary study. Ultrasound Obstet Gynecol 2109 2016;47:492 –8. 2110 Judy AE, Lyell DJ, Druzin ML, Dorigo O. Disseminated intravascular coagulation c omplicating 2111 the conservative management of placenta percreta. Obstet Gynecol 2015;126:1016 –8. 2112 Page 41 of 47 Teixidor Viñas M, Chandraharan E, Moneta MV, Belli AM. The role of interventional 2113 radiology in reducing haemorrhage and hysterectomy following caesarean sect ion for morbidly 2114 adherent placenta. Clin Radiol 2014;69:e345 –51. 2115 Bouvier A, Sentilhes L, Thouveny F, Bouet PE, Gillard P, Willoteaux S, et al. Planned 2116 caesarean in the interventional radiology cath lab to enable immediate uterine artery 2117 embolization f or the conservative treatment of placenta accreta. Clin Radiol 2012;67:1089 –94. 2118 Dilauro MD, Dason S, Athreya S. Prophylactic balloon occlusion of internal iliac arteries in 2119 women with placenta accreta: literature review and analysis. Clin Radiol 2012; 67:515 –20. 2120 Clausen C, Stensballe J, Albrechtsen CK, Hansen MA, Lönn L, Langhoff -Roos J. Balloon occlusion 2121 of the internal iliac arteries in the multidisciplinary management of placenta percreta. Acta 2122 Obstet Gynecol Scand 2013;92:386 –91. 2123 D'Souza DL, Kingdom JC, Amsalem H, Beecroft JR, Windrim RC, Kachura JR. Conservative 2124 management of invasive placenta using combined prophylactic internal iliac artery balloon 2125 occlusion and immediate postoperative uterine artery embolization. Can Assoc Radiol J2126 201 5;66:179 –84. 2127 Chou MM, Kung HF, Hwang JI, Chen WC, Tseng JJ. Temporary prophylactic 2128 intravascular balloon occlusion of the common iliac arteries before cesarean hysterectomy for 2129 controlling operative blood loss in abnormal placentation. Taiwan J Obstet Gynecol 2015;54:493 –2130 2131 Duan XH, Wang YL, Han XW, Chen ZM, Chu QJ, Wang L, et al. Caesarean section combined 2132 with temporary aortic balloon occlusion followed by uterine artery embolisation for the 2133 management of placenta accreta. Clin Radiol 2015;70:9 32 –7. 2134 Duan XH, Wang YL, Han XW, Chen ZM, Chu QJ, Wang L, et al . Caesarean section combined 2135 with temporary aortic balloon occlusion followed by uterine artery embolisation for the 2136 management of placenta accreta. Clin Radiol 2015;70:932 –7. 2137 Wei X, Z hang J, Chu Q, Du Y, Xing N, Xu X, et al. Prophylactic abdominal aorta balloon 2138 occlusion during caesarean section: a retrospective case series. Int J Obstet Anesth 2016;27:3 –8. 2139 Wu Q, Liu Z, Zhao X, Liu C, Wang Y, Chu Q, et al . Outcome of Pregnancies 2140 After Balloon Occlusion of the Infrarenal Abdominal Aorta During Caesarean in 230 Patients 2141 With Placenta Praevia Accreta. Cardiovasc Intervent Radiol 2016;39:1573 –9. 2142 Xie L, Wang Y, Luo FY, Man YC, Zhao XL. Prophylactic use of an infrarenal abdominal 2143 ao rta balloon catheter in pregnancies complicated by placenta accreta. J Obstet Gynaecol 2144 2017;37:557 –61. 2145 Wang YL, Duan XH, Han XW, Wang L, Zhao XL, Chen ZM, et al . Comparison of temporary 2146 abdominal aortic occlusion with internal iliac artery occlusion f or patients with placenta accreta -2147 a non -randomised prospective study. Vasa 2017;46:53 –7. 2148 Ikeda T, Sameshima H, Kawaguchi H, Yamauchi N, Ikenoue T. Tourniquet technique prevents 2149 profuse blood loss in placenta accreta cesarean section. J Obstet Gynaec ol Res 2005;31:27 –31. 2150 Meng JL, Gong WY, Wang S, Ni XJ, Zuo CT, Gu YZ. Two -tourniquet sequential blocking as a 2151 simple intervention for hemorrhage during cesarean delivery for placenta previa accreta. Int J 2152 Gynaecol Obstet 2017;138:361 –2. 2153 Iwata A, Murayama Y, Itakura A, Baba K, Seki H, Takeda S. Limitations of internal iliac 2154 artery ligation for the reduction of intraoperative hemorrhage during cesarean hysterectomy in 2155 cases of placenta previa accreta. J Obstet Gynaecol Res 2010;36:254 –9. 2156 Bishop S, Butler K, Monaghan S, Chan K, Murphy G, Edozien L. Multiple complications 2157 following the use of prophylactic internal iliac artery balloon catheterisation in a patient 2158 with placenta percreta. Int J Obstet Anesth 2011;20:70 –3. 2159 Gagnon J, Boucher L, K aufman I, Brown R, Moore A. Iliac artery rupture related to balloon 2160 insertion for placenta accreta causing maternal hemorrhage and neonatal compromise. Can J 2161 Anaesth 2013;60:1212 –7. 2162 Teare J, Evans E, Belli A, Wendler R. Sciatic nerve ischaemia after iliac artery occlusion 2163 balloon catheter placement for placenta percreta. Int J Obstet Anesth 2014;23:178 –81. 2164 Page 42 of 47 Matsueda S, Hidaka N, Kondo Y, Fujiwara A, Fukushima K, Kato K. External iliac artery 2165 thrombosis after common iliac artery balloon occlusion d uring cesarean hysterectomy for 2166 placenta accreta in cervico -isthmic pregnancy. J Obstet Gynaecol Res 2015;41:1826 –30. 2167 Salim R, Chulski A, Romano S, Garmi G, Rudin M, Shalev E. Precesarean prophylactic balloon 2168 catheters for suspected placenta accreta: A randomized controlled trial. Obstet Gynecol 2169 2015;126:1022 –8. 2170 2171 2172 2173 Page 43 of 47 Appendix I : Explanation of guidelines and evidence levels 2174 2175 Clinical guidelines are: ‘systematically developed statements which assist clinicians and patients in 2176 making decisions about approp riate treatment for specific conditions’. Each guideline is 2177 systematically developed using a standardised methodology. Exact details of this process can be 2178 found in Clinical Governance Advice No.1 Development of RCOG Green -top Guidelines (available on 2179 the RCOG website at -top -development). These recommendations are 2180 not intended to dictate an exclusive course of management or treatment. They must be evaluated 2181 with reference to individual patient needs, resources and limitations un ique to the institution and 2182 variations in local populations. It is hoped that this process of local ownership will help to 2183 incorporate these guidelines into routine practice. Attention is drawn to areas of clinical uncertainty 2184 where further research may be indicated. 2185 2186 The evidence used in this guideline was graded using the scheme below and the recommendations 2187 formulated in a similar fashion with a standardised grading scheme. 2188 2189 Classification of evidence levels 2190 1++ High -quality meta -analyses, systematic r eviews of randomised controlled trials or randomised controlled trials with a very low risk of bias 1+ Well -conducted meta -analyses, systematic reviews of randomised controlled trials or randomised controlled trials with a low risk of bias 1– Meta -analys es, systematic reviews of randomised controlled trials or randomised controlled trials with a high risk of bias 2++ High -quality systematic reviews of case –control or cohort studies or high -quality case –control or cohort studies with a very low risk of co nfounding, bias or chance and a high probability that the relationship is causal 2+ Well -conducted case –control or cohort studies with a low risk of confounding, bias or chance and a moderate probability that the relationship is causal 2– Case –control or cohort studies with a high risk of confounding, bias or chance and a significant risk that the relationship is not causal 3 Non -analytical studies, e.g. case reports, case series 4 Expert opinion 2191 Grades of Recommendation 2192 At least one meta -analysis, s ystematic reviews or RCT rated as 1++, and directly applicable to 2193 the target population; or a systematic review of RCTs or a body of evidence consisting 2194 principally of studies rated as 1+, directly applicable to the target population and 2195 demonstrating over all consistency of results 2196 2197 A body of evidence including studies rated as 2++ directly applicable to the target 2198 population, and demonstrating overall consistency of results; or 2199 Extrapolated evidence from studies rated as 1++ or 1+ 2200 2201 A body of evidence i ncluding studies rated as 2+ directly applicable to the target population, 2202 and demonstrating overall consistency of results; or 2203 Extrapolated evidence from studies rated as 2++ 2204 2205 Evidence level 3 or 4; or 2206 Extrapolated evidence from studies rated as 2+ 2207 2208 Good Practice Points 2209 2210 Recommended best practice based on the clinical experience of the guideline development 2211 A B C ü DPage 44 of 47 group 2212 2213 Page 45 of 47 Appendix I I: Flow diagram for ultrasound diagnosis and follow -up of placenta praevia and placenta accreta spectrum 2214 2215 2216 Abbreviation s: BMI body mass index; PAS placenta accreta spectrum; TAS transabdominal scan; TVS transvaginal scan. 2217 Page 46 of 47 Appendix I II: Ultrasound imaging sign s commonly used to diagnose placenta accreta spectrum (modified from Collins SL) 141 2218 2219 Ultrasound imaging sign s Desc ription 2D greyscale si gns Loss of the ‘clear zone’ Loss or irregularity of the hypoechoic plane in the myometrium underneath the placental bed (the ‘clear zone’ ). Abnormal placental lacunae Presence of numerous lacunae , including some that are large a nd irregular (Finberg grade 3 ), often containing turbulent flow visible in greyscale imaging . Bladder wall interruption Loss or interruption of the bright bladder wall (the hyperechoic band or ‘line’ between the uterine serosa and the bladder lumen) . My ometrial thinning Thinning of the myometrium overlying the placenta to less than 1 mm or undetectable. Placental bulge Deviation of the uterine serosa away from the expected plane, caused by an abnormal bulge of placental tissue into a neighboring organ, typically the bladder. The uterine serosa appears intact but the outline shape is distorted. Focal exophytic mass Placental tissue seen breaking through the uterine serosa and extending beyond it. Most often seen inside a filled urinary bl adder. 2D colo ur Doppler signs Uterovesical hypervascularity Striking amount of colour Doppler signal seen between the myometrium and the posterior wall of the bladder. This sign probabl y indicates numerous, closely packed, tortuous vessels in that region (demonstrati ng multidirectional flow and aliasing artifact). Subplacental hypervascularity Striking amount of colour Doppler signal seen in the placental bed. This sign probably indicates numerous, closely packed, to rtuous vessels in that region (demonstrating multid irectional flow and aliasing artifact). Bridging vessels Vessels appearing to extend from the placenta, across the myometrium and beyond the serosa into the bladder or other organs. Often running perpendicular to the myometrium. Placental lacunae feeder vessels Vessels with high velocity blood flow leading from the myometrium into the placental lacunae, causing turbulence upon entry. 3D colour Doppler signs Intraplacental hypervascularity (power Doppler) Complex, irregular arrangement of numerous plac ental vessels, exhibiting tortuous courses and varying calibers. 2220 Page 47 of 47 2221 This guideline was produced on behalf of the Royal College of Obstetricians and Gynaecologists by: 2222 Professor ERM Jauniaux FRCOG, London (Lead Developer) ; Professor Z Alfirevic FRCOG, Liv erpool, UK; Mr AG 2223 Bhide FRCOG, London, UK; Professor MA Belfort, Baylor College of Medicine, Houston, Texas, USA; Professor GJ 2224 Burton, University of Cambridge, UK; Professor SL Collins MRCOG, Oxford, UK; Dr S Dornan, Royal Jubilee 2225 Maternity Hospital, Belfa st, UK; Mr D Jurkovic FRCOG, London, UK; Professor G Kayem, Armand -Trousseau and 2226 Louis -Mourier University Hospitals, Paris, France; Professor John Kingdom, Mont Sinai, Toronto University, 2227 Canada; Professor R Silver, University of Utah, Salt Lake City, Utah , USA; Professor L Sentilhes, University Hospital 2228 Angers, France 2229 2230 and peer reviewed by: Professor ML Brizot, University of São Paulo, São Paulo, Brazil; Dr G Calì MSIEOG, ARNAS 2231 Civico Hospital, Palermo, Italy; Professor J Dashe, University of Texas Southw estern Medical Center, Dallas, TX , USA; 2232 Professor O Erez, Soroka University Medical Center, Beer Sheva, Israel; Dr D Fraser FRCOG, Norwich; Dr F Forlani, 2233 University Hospital "Paolo Giaccone", Palermo, Italy; Dr J Hasegawa, St. Marianna University School of Medicine, 2234 Kawasaki, Kanagawa, Japan; Dr YY Hu, Sichuan University, Chengdu, Sichuan, China; Dr N Lucas, Obstetric 2235 Anaesthetists' Association, London; Professor P Martinelli, Università di Napoli Federico II, Naples, Italy; Princess 2236 Royal Maternity Invasiv e Placenta Team, London; RCOG Women’s Network; Professor SC Robson MRCOG, FRCP, 2237 Newcastle University; Royal College of Anaesthetists; Dr R Salim, Emek Medical Center, Afula, Israel; Professor RM 2238 Silver, The University of Utah, Salt Lake City, UT, USA; Dr J T Thomas FRANZCOG, CMFM, Mater Mothers' Hospital, 2239 Brisbane , Australia ; The UK Vasa Praevia Raising Awareness Trust and the International Vasa Previa Foundation; Mr 2240 N Thomson, Society and College of Radiographers, London; Dr M Tikkanen, Women´s Clinic, Hels inki University 2241 Hosp ital Finland, Helsinki, Finland 2242 2243 Committee lead reviewers were: Dr A McKelvey MRCOG, Norwich and Mr RJ Fernando FRCOG, London 2244 2245 The c hair of the Guidelines Committee was : Dr AJ Thomson MRCOG, Paisley. 2246 2247 All RCOG guidance developers are asked to declare any conflicts of interest. A statement summarising any conflicts of 2248 interest for this guideline is available from: XXX .2249 2250 The final version is the res ponsibility of the Guidelines Committee of the RCOG. 2251 2252 The guideline will be considered for update 3 years after publication, with an intermediate assessment of the need 2253 to update 2 years after publication. 2254 2255 2256 DISCLAIMER 2257 2258 The Royal College of Obstetricians and Gynaecologists produces guidelines as an educational aid to good clinical 2259 practice. They present recognised methods and techniques of clinical practice, based on published evidence, for 2260 consideration by obstetricians and gynaecologists and other relevant health professionals. The ultimate judgement 2261 regarding a particular clinical procedure or treatment plan must be made by the doctor or other attendant in the 2262 light of clinical data presented by the patient and the diag nostic and treatment options available. 2263 2264 This means that RCOG Guidelines are unlike protocols or guidelines issued by employers, as they are not intended to 2265 be prescriptive directions defining a single course of management. Departure from the local prescri ptive protocols or 2266 guidelines should be fully documented in the patient’s case notes at the time the relevant decision is taken. 2267
1476	https://math-salamanders.s3-us-west-1.amazonaws.com/Measurement/Math-Conversion-Charts/Metric-to-Standard/UK/area-conversion-chart-uk.pdf	Name Date MEASURE CONVERSION CHART – AREAS (UK) METRIC CONVERSIONS 1 sq. centimetre = 100 sq. millimetres 1 sq. cm = 100 sq. mm 1 sq. metre = 10,000 sq. centimetres 1 sq. m = 10,000 sq. cm 1 hectare = 10,000 sq. metres 1 ha = 10,000 sq. m 1 sq. km = 100 hectares 1 sq. km = 100 ha 1 sq. km = 1 million sq. metres 1 sq. km = 1,000,000 sq. m IMPERIAL CONVERSIONS 1 sq. foot = 144 sq. inches 1 sq. ft = 144 sq. in 1 sq. yard = 1296 sq. inches 1 sq. yd = 1296 sq. in 1 sq. yard = 9 sq. feet 1 sq. yd = 9 sq. ft 1 acre = 4840 sq. yards 1 acre = 4840 sq. yd 1 sq. mile = 640 acres 1 sq. mi = 640 acres METRIC -> IMPERIAL CONVERSIONS 1 sq. centimetre = 0.15500 sq. inches 1 sq. cm = 0.15500 sq. in 1 sq. metre = 10.763 91sq. feet 1 sq. m = 10.763 91 sq. ft 1 sq. metre = 1.195 99 sq. yards 1 sq. m = 1.195 99 sq. yd 1 hectare = 2.47105 acres 1 ha = 2.47105 acres 1 hectare = 0.00386 sq. miles 1 ha = 0.00386 sq. mi IMPERIAL -> METRIC CONVERSIONS 1 sq. inch = 6.4516 sq. centimetres 1 sq. in = 6.4516 sq. cm 1 sq. foot = 929.0304 sq. centimetres 1 sq. ft = 929.0304 sq. cm 1 sq. foot = 0.09290 sq. metres 1 sq. ft = 0.09290 sq. m 1 sq. yard = 0.83613 sq. metres 1 sq. yd = 0.83613 sq. m 1 acre = 0.40469 hectares 1 acre = 0.40469 ha 1 sq. mile = 258.99881 hectares 1 sq. mi = 258.99881 ha
1477	https://math.stackexchange.com/questions/739133/limit-rule-lim-fxgx	calculus - Limit Rule $\lim f(x)^{g(x)}$ - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Limit Rule lim f(x)g(x)lim f(x)g(x) Ask Question Asked 11 years, 6 months ago Modified8 years, 11 months ago Viewed 29k times This question shows research effort; it is useful and clear 10 Save this question. Show activity on this post. I want to know the following is true : If c,d∈R,lim x→0 f(x)=c>0,lim x→0 g(x)=d>0 c,d∈R,lim x→0 f(x)=c>0,lim x→0 g(x)=d>0 then lim x→0 f(x)g(x)=c d lim x→0 f(x)g(x)=c d In calculus book such formula cannot be found. Consider the problem : lim x→0(1+sin 4 x)cot x lim x→0(1+sin⁡4 x)cot⁡x To find a limit, we must use l o g l o g and L'Hospital. But some student suggests that lim x→0(1+sin 4 x)1 sin 4 x sin 4 x sin x cos x=e 4 lim x→0(1+sin⁡4 x)1 sin⁡4 x sin⁡4 x sin⁡x cos⁡x=e 4 This argument is clear. But I know that it is informal. Is there a minus point in such way ? calculus Share Share a link to this question Copy linkCC BY-SA 3.0 Cite Follow Follow this question to receive notifications asked Apr 4, 2014 at 4:11 HK LeeHK Lee 20.6k 7 7 gold badges 40 40 silver badges 109 109 bronze badges 3 1 The result you have mentioned is true (as shown in the answer) but it is much better to take logs and then evaluate the logarithm. The formula is valid for all c>0 c>0 and any finite value of d d. But the technique of taking logs works in other cases too one of c,d c,d is infinite or if c=0 c=0.Paramanand Singh –Paramanand Singh♦ 2014-04-05 04:48:34 +00:00 Commented Apr 5, 2014 at 4:48 @ParamanandSingh Here how can we apply ln ln without knowing inner limit exist?Meet Patel –Meet Patel 2023-05-11 03:18:58 +00:00 Commented May 11, 2023 at 3:18 @MeetPatel: this works because ln ln is a continuous bijection. The interplay of ln ln and its inverse exp exp is needed here. See more details at paramanands.blogspot.com/2013/11/…Paramanand Singh –Paramanand Singh♦ 2023-05-11 04:22:18 +00:00 Commented May 11, 2023 at 4:22 Add a comment\| 3 Answers 3 Sorted by: Reset to default This answer is useful 12 Save this answer. Show activity on this post. Assuming f(x)>0 f(x)>0 for some open set surrounding x 0 x 0, we know that because the natural logarithm is continuous on (0,∞)(0,∞) that ln(lim x→x 0 f(x)g(x))=lim x→x 0 ln(f(x)g(x))=lim x→x 0 g(x)ln f(x)=(lim x→x 0 g(x))(lim x→x 0 ln f(x))=(lim x→x 0 g(x))ln(lim x→x 0 f(x))=ln[(lim x→x 0 f(x))(lim x→x 0 g(x))]ln⁡(lim x→x 0 f(x)g(x))=lim x→x 0 ln⁡(f(x)g(x))=lim x→x 0 g(x)ln⁡f(x)=(lim x→x 0 g(x))(lim x→x 0 ln⁡f(x))=(lim x→x 0 g(x))ln⁡(lim x→x 0 f(x))=ln⁡[(lim x→x 0 f(x))(lim x→x 0 g(x))] Then, because the natural logarithm is injective, we find that lim x→x 0 f(x)g(x)=(lim x→x 0 f(x))(lim x→x 0 g(x))lim x→x 0 f(x)g(x)=(lim x→x 0 f(x))(lim x→x 0 g(x)) Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 4, 2014 at 4:19 HaydenHayden 17.1k 1 1 gold badge 41 41 silver badges 70 70 bronze badges 1 1 The limit might now exist at the left of x=x 0 x=x 0.Then how do I assume that the log goes inside?user586228 –user586228 2020-10-14 05:47:18 +00:00 Commented Oct 14, 2020 at 5:47 Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Yes. Note that the function f(x,y)=x y=e y ln(x)f(x,y)=x y=e y ln⁡(x) is continuous for x,y>0 x,y>0, so if (x n,y n)→(x,y)(x n,y n)→(x,y) with x,y>0 x,y>0 then f(x n,y n)→f(x,y)f(x n,y n)→f(x,y). Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Apr 4, 2014 at 4:17 Alex BeckerAlex Becker 62.1k 10 10 gold badges 140 140 silver badges 198 198 bronze badges Add a comment\| This answer is useful 3 Save this answer. Show activity on this post. Because y(x)=l n(x)y(x)=l n(x) is continuous at x=c>0 x=c>0 lim x→0 f(x)=c lim x→0 f(x)=c according to the composition law, we have lim x→0 l n f(x)=l n lim x→0 f(x)=l n c lim x→0 l n f(x)=l n lim x→0 f(x)=l n c Because lim x→0 g(x)=d lim x→0 g(x)=d, we have lim x→0 g(x)l n f(x)=lim x→0 g(x)⋅lim x→0 l n f(x)=d l n c lim x→0 g(x)l n f(x)=lim x→0 g(x)⋅lim x→0 l n f(x)=d l n c Apply composition law again, we get lim x→0 f(x)g(x)=lim x→0 e g(x)l n f(x)=e lim x→0 g(x)l n f(x)=e d l n c=c d lim x→0 f(x)g(x)=lim x→0 e g(x)l n f(x)=e lim x→0 g(x)l n f(x)=e d l n c=c d Share Share a link to this answer Copy linkCC BY-SA 3.0 Cite Follow Follow this answer to receive notifications answered Oct 1, 2016 at 11:04 SeareneSearene 1,021 1 1 gold badge 8 8 silver badges 8 8 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions calculus See similar questions with these tags. Featured on Meta Introducing a new proactive anti-spam measure Spevacus has joined us as a Community Manager stackoverflow.ai - rebuilt for attribution Community Asks Sprint Announcement - September 2025 Report this ad Linked 8Why is 0 0 the largest value of a a satisfying lim x→1(sin(x−1)+a−a x sin(x−1)+x−1)1−x 1−x√=1 4 lim x→1(sin⁡(x−1)+a−a x sin⁡(x−1)+x−1)1−x 1−x=1 4? 2Find lim x→+∞(2 1−(1 2)x)lim x→+∞(2 1−(1 2)x) without using logarithms. 2Prove Power Rule for Limits: lim x→a f(x)g(x)=(lim x→a f(x))lim x→a g(x)lim x→a f(x)g(x)=(lim x→a f(x))lim x→a g(x) 1Computing the limit of lim x→∞(x 2+x+1 3 x 2+2 x+1)(3 x 2+x x−2)lim x→∞(x 2+x+1 3 x 2+2 x+1)(3 x 2+x x−2) Related 4A Tricky Limit: lim x→0 9 x−5 x x lim x→0 9 x−5 x x without L'Hospital 26When L'Hôpital's Rule Fails 6Looking for an elementary solution of this limit 3Finding Limit by Using L'Hospital 5How to find lim n→0 cos(π n sin n cos n)lim n→0 cos⁡(π n sin⁡n cos⁡n) and lim n→0 n sin(2 n)−cos(n 2)+1 lim n→0 n sin⁡(2 n)−cos⁡(n 2)+1 72When does L' Hopital's rule fail? 2L'Hospital Rule: Requirement that the limit exists 1Calculate limits without L'Hospital's Rule 1Simple limit with asymptotic approach. 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1478	https://aopsacademy.org/campus/frisco	Art of Problem Solving AoPS Online Beast Academy AoPS Academy A Problem Solving Community in Frisco, TX AoPS Academy is an enrichment program for grades 1–12, offering after-school and weekend classes for highly-motivated students. Students develop their creativity, critical thinking, and complex problem solving skills with outstanding peers and mentors. GET STARTED AS SEEN IN Campus Information AoPS Academy Frisco Campus Address AoPS Academy Frisco 5285 Independence Pkwy, Suite 500 Frisco, TX 75035 Campus Calendar Hours Mon - Fri: 12:00 pm - 8:30 pm Sat: 9:00 am - 6:00 pm Sun: 9:00 am - 8:00 pm Phone (469) 200-1010 Email frisco@aopsacademy.org Get Started Our Team Campus Leadership Our Programs Explore our academic year courses and summer camps in advanced math, language arts, science, and computer science. With a variety of programs for different interests and age groups, our curriculum combines learning and fun. ### Academic Year Courses Our academic year program runs classes after school and on weekends. Students learn how to ask meaningful questions, actively seek answers, and think critically — building a problem solving foundation they will use beyond the classroom.### Summer Camps Our two-week summer camps are designed to keep students engaged and sharpen their skills for the upcoming school year. ##### Mathematics Build a deep understanding of mathematics and develop the confidence to approach new, interesting challenges. Learn More##### Language Arts Enhance your communication skills by exploring literature, improving writing techniques, and understanding the power of language in expressing ideas. Learn More##### Science Ask questions about the physical world and develop observational, analytical, and scientific thinking skills. Learn More Lifelong Skills Critical Thinking Students focus on advanced problem solving and conceptual understanding — not rote memorization. Active Learning Students lead the discovery process, guided by motivating questions from expert instructors. Constructive Collaboration Students build strong relationships with their cohort as they solve challenging problems together. Effective Communication Students learn how to precisely explain their ideas, questions, and solutions across any subject and topic they tackle. TESTIMONIALS “I would say that AoPS is an advanced, challenging curriculum for students who are looking for new opportunities in problem solving, and the best possible preparation for advanced math contests later in life.” Nantana Rowan AOPS PARENT “Our daughter started at AoPS at the beginning of the year and has gone from ‘I hate math’ to ‘I love math.’ ” Senthil Hariramasamy AoPS Parent “I would highly recommend AoPS! It's a great place to learn and grow!” Isabelle Wang AOPS PARENT “I find all the resources very helpful and the teachers are great. AoPS helped me learn a lot of new things I had never seen before and I would like to continue learning here.” Aarifah F. AOPS Student “I admire how the materials are organized and helpful for students’ gradual learning of knowledge. The school frequently organizes some national competitions to encourage students to test their limits. The school staff are very attentive to communication with parents and respect their feedback.” Claire AOPS PARENT “This class helped me learn of things that I haven’t encountered in the world of mathematics. I’ll be sure to sign up for another class so that I can keep learning.” Waynstan A. AOPS Student “We love the philosophy around problem solving and that solving challenging problems can be a fun and rewarding experience for kids.” Arun Bodapati AOPS PARENT “AoPS develops grit and confidence like nothing else! ” Harshitha Chandra AOPS PARENT Request Information Contact us today to learn more or enroll in our program. GET STARTED
1479	https://www.youtube.com/watch?v=zXIzER7t9GA	The Vieta Jumping Trick That Solves an Impossible Olympiad Problem Logic Unlocked 861 subscribers 54 likes Description 1683 views Posted: 25 Aug 2025 Some math problems look simple but are deceptively hard. Others look impossible... until you discover the one trick that cracks them wide open. This is one of those problems, straight from the world of math competitions. At its heart, this video is a masterclass in a powerful problem-solving technique known as "Vieta Jumping." We'll take a seemingly unsolvable Diophantine equation and systematically dismantle it, revealing a hidden structure of infinite solutions. Join us to learn: ► The critical first step: How to reframe a complex fraction into a simple quadratic equation. ► The "Jump": A deep dive into Vieta's formulas and how they allow us to generate new solutions from existing ones. ► The "Aha!" Moment: How to use the concept of a "minimal solution" to trap our variables with inequalities. ► The elegant finale where we prove that all solutions fall into three, perfectly structured families. This isn't just about finding the answer; it's about learning a new way to think. If you've ever been stumped by a number theory problem, the Vieta Jumping technique is the tool you need in your arsenal. VietaJumping #MathOlympiad #NumberTheory #DiophantineEquation #ProblemSolving 4 comments Transcript: This looks like just another fraction, but hidden inside are infinite, wellorganized sets of integer solutions. We're on a quest to find all pairs of positive integers that make this expression a positive integer. First, let's set the expression equal to some variable where this variable represents any positive integer. Since our variable and the numerator are both positive, the denominator must also be positive. For our result to be a positive integer, two crucial conditions must be satisfied. The denominator must be a positive integer and it must divide the numerator evenly. Let's clear the fraction by multiplying both sides by the denominator. This gives us a much cleaner form to work with. Now we'll rearrange everything to one side. And here comes our first critical insight that will unlock everything. If we treat the other variables as fixed constants, this becomes a quadratic in terms of our first variable. This structure is the key to unlocking the entire problem. Before tackling the general problem, let's explore the simplest possible case when our second variable equals 1. We substitute this special value into our original expression. The denominator simplifies nicely. The constants cancel out leaving us with this clean fraction. And this simplifies to just our first variable / 2. Interestingly, when our second variable equals 1, the constant term in our quadratic becomes zero, giving us this simple factored form. For our result to be a positive integer, it's now crystal clear that our first variable must be a positive even integer. We can express any positive even integer as 2 another positive integer. This gives us our first infinite family of solutions expressed as ordered pairs. Let's do a quick sanity check to make sure this abstract result holds true in practice. Let's pick a specific value for our parameter. Our solution pair becomes 6, 1. Plugging this into the original expression gives us a clean calculation that results in three. It's a positive integer, confirming our solution works exactly as expected. Now for the more complex cases where our second variable is greater than one, we'll use a powerful technique involving Vieta's relationships to find partner solutions. Let's return to our quadratic. Assume we found one integer solution which we'll call our first root because this is a quadratic. If one integer root exists, there must be another root which we'll call our second root. From Vetta's relationships, we know the sum of the roots is the negative of the coefficient of the linear term. Since our constants and our first solution are all integers, the other root must also be an integer. The product of the roots is the constant term. From our quadratic, we can derive this key identity by rearranging terms. This immediately shows that our first variable divides the right side. So the partner root equals this quotient and is guaranteed to be an integer. Since our second variable is greater than one, we have the cubic minus one is positive. Our variable is positive. So this partner is also a positive integer. For every solution we find, we can generate another integer solution. Before we dive into complex inequalities, let me show you an elegant shortcut that will dramatically simplify our analysis. Let's call our denominator D and factor it in a clever way. This reveals a hidden modular structure. This means our denominator always leaves remainder 1 when divided by the square of our second variable. This is a powerful constraint. We can write our denominator in this explicit form where the parameter is a non-gative integer. If the parameter is at least one, then our denominator is at least 1 + the square of our second variable. This will be crucial for our contradiction argument. Now we use our lema strategically. fix any value greater than one for our second variable and let our first variable be the smallest positive integer that makes our expression an integer. This minimality assumption is crucial. It allows us to control the partner solution and derive tight bounds. Since our first variable is minimal, its partner must be greater than or equal to it. Multiplying both sides by our first variable gives us this inequality. Substituting our expression for the constant in terms of the denominator. Since the denominator is positive, we can divide and then multiply to get this key inequality. Now we substitute our factored form of the denominator. This gives us our key inequality that we need to analyze carefully. After some algebra, this rearranges to show that our first variable is strictly less than our second variable. But wait, let me prove this by contradiction to make it crystal clear. If we assume our first variable is at least as large as our second variable, then our derived inequality forces it to be strictly smaller. This is a contradiction. So we must have our first variable strictly less than our second variable. Now our lema delivers the knockout punch. We've established that our first variable is less than our second variable. So their squares satisfy this same relationship. This seems simple, but it's about to create a powerful contradiction when combined with our lema. Remember our denominator must divide the square of our first variable. And it has that special modular form we derived. If the parameter is at least 1, then our denominator is at least 1 + the second variable squared, which exceeds the first variable squared since we prove the first is less than the second. But the denominator must divide the first variable squared. Contradiction. The only way to avoid contradiction is if the parameter equals 0, which means the denominator equals exactly 1. When the denominator equals 1, we can solve directly for the relationship between our variables. Our first variable must equal exactly half of our second variable. Since our first variable must be a positive integer, this immediately tells us that our second variable must be even. Now we can solve the even case completely. Let our second variable be twice some positive integer. We'll represent any positive even integer as 2 another positive integer. Our first variable becomes this parameter and since the denominator is 1, our constant becomes the parameter squared. Let's verify this works in our quadratic by substituting these values. The substitution gives us this clean equation that confirms our constant must equal the parameter squared. A monic quadratic has at most two roots and we found both integer roots for this specific constant value. This guarantees completeness. We found all solutions for even second variables. So if a solution exists for an even second variable, it must be of this specific form. Let's test it. We substitute our proposed values into the original expression. First we resolve the exponents carefully. Then we multiply the terms in the denominator. The cubic terms cancel out completely leaving us with the parameter squared. Since our parameter is a positive integer, this squared value is also a positive integer. That's exactly what we needed. This confirms our second infinite family of solutions expressed as ordered pairs. But wait, remember Vieta jumping? For every solution, there is another. Let's find the jumped solution. Our first route is the parameter. Our second variable is twice the parameter. And we just found that our constant is the parameter squared. The other root is given by this veta relationship. Let's substitute our values carefully. Simplifying the term in the parenthesis gives us our final expression for the second root. And this reveals our third and final infinite family of solutions. What about odd values greater than one for our second variable? Our elegant analysis has already ruled them out completely. Our proof established that for any second variable greater than one, the first variable must equal exactly half of it. But if our second variable is odd, then half of it is not an integer. Since our first variable must be a positive integer, this is impossible. Therefore, there are no solutions when our second variable is an odd number greater than one. And so our journey is complete. We have rigorously proven that from one complicated fraction exactly three distinct families of integer solutions emerge. And these are the only ones that exist. Here is our complete theorem. The given expression equals a positive integer if and only if the solution pair belongs to one of exactly three families. Using our parameters as any positive integer, these three families capture every possible integer solution. No other solutions exist. Let's visualize our complete solution. The green points show our first family forming a horizontal line. The blue points show our second family, a perfect diagonal, and the red points show our third family, which grows explosively fast due to the cubic term. Let me verify one of our third family solutions. When we substitute these values, we get exactly one, confirming our approach works as expected. What began as a seemingly chaotic algebraic problem revealed itself through patient analysis to have a beautiful, completely determined structure. This problem showcases the power of elegant mathematical insight over brute force computation. A single modular observation cut straight to the heart of the matter. The quadratic structure, VA relationships, and our key lema worked in perfect harmony, revealing every solution and proving none others exist. Sometimes the most beautiful mathematics comes not from complex calculations, but from seeing the hidden patterns that make everything click into place. Thank you for joining me on this journey. If you enjoyed this video, please like and subscribe for more mathematical explorations.
1480	https://m.antpedia.com/news/2325154.html	微生物培养基的原理、制作和现象：尿素琼脂首页分析测试百科网推荐焦点政策标准洞见新品企业财经商机采购 Lab 科技成果综述会展厂商微信热榜更多喜欢作者打赏￥1 打赏￥2 打赏￥5 打赏￥66 打赏￥88 打赏￥520 其他金额￥微信支付 × 我要投稿成为作者点赞打赏发文微生物培养基的原理、制作和现象：尿素琼脂 289 2019.8.31 #### 王辉致力于为分析测试行业奉献终身成分蛋白胨 1g 氯化钠 5g 葡萄糖 1g 磷酸二氢钾 2g 0.4%酚红溶液 3mL 琼脂 20g 蒸馏水 1000mL 20%尿素溶液 100mL pH7.2±0.1 制法将除尿素和琼脂以外的成分配好，并校正pH，加入琼脂，加热溶化并分装烧瓶。121℃高压灭菌15min。冷至50～55℃，加入经除菌过滤的尿素溶液。尿素的最终浓度为2%，最终pH应为7.2±0.1。分装于灭菌试管内，放成斜面备用。试验方法挑取琼脂培养物接种，在36±1℃培养24h，观察结果。尿素酶阳性者由于产碱而使培养基变为红色。尿素琼脂互联网文章推荐磁性器件损耗的分析设计优化（三）磁性器件损耗的分析设计优化（二）高GC含量PCR样本的扩增困难与防止污染与降解的操作方法 Copyright ©2007-2025 ANTPEDIA, All Rights Reserved 京ICP备07018254号京公网安备1101085018 电信与信息服务业务经营许可证：京ICP证110310号
1481	https://mathoverflow.net/questions/30891/explicit-formula-for-the-j-invariant-of-binary-quartic-form	polynomials - Explicit formula for the j-invariant of binary quartic form - MathOverflow Join MathOverflow By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Visit Stack Exchange Loading… Tour Start here for a quick overview of the site Help Center Detailed answers to any questions you might have Meta Discuss the workings and policies of this site About Us Learn more about Stack Overflow the company, and our products current community MathOverflow helpchat MathOverflow Meta your communities Sign up or log in to customize your list. more stack exchange communities company blog Log in Sign up Home Questions Unanswered AI Assist Labs Tags Chat Users Hang on, you can't upvote just yet. You'll need to complete a few actions and gain 15 reputation points before being able to upvote. Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Explicit formula for the j-invariant of binary quartic form Ask Question Asked 15 years, 2 months ago Modified11 months ago Viewed 1k times This question shows research effort; it is useful and clear 6 Save this question. Show activity on this post. A binary quartic form a X 4+b X 3 Y+c X 2 Y 2+d X Y 3+Y 4 a X 4+b X 3 Y+c X 2 Y 2+d X Y 3+Y 4 decomposes as a product of linear factors Y−t j X Y−t j X, j=1,…,4 j=1,…,4. I would like to have an explicit formula for symmetrization of the crossratio of t j t j. polynomials rt.representation-theory Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Improve this question Follow Follow this question to receive notifications edited May 3, 2024 at 13:43 LSpice 13.9k 4 4 gold badges 47 47 silver badges 74 74 bronze badges asked Jul 7, 2010 at 13:51 David MarínDavid Marín 105 5 5 bronze badges 3 1 Do you mean the j-invariant of the elliptic curve y 2=a x 4+b x 3+c x 2+d x+1 y 2=a x 4+b x 3+c x 2+d x+1?Robin Chapman –Robin Chapman 2010-07-07 13:54:19 +00:00 Commented Jul 7, 2010 at 13:54 Yes, that is exactly what I am looking for.David Marín –David Marín 2010-07-07 14:00:22 +00:00 Commented Jul 7, 2010 at 14:00 4 If you have access to a computer algebra system, you can do the following. Let ξ ξ denote a solution to f(1,ξ)=0 f(1,ξ)=0 where f f is your quartic. Then f(X,Y+ξ X)=b′X 3 Y+⋯+Y 4 f(X,Y+ξ X)=b′X 3 Y+⋯+Y 4. The elliptic curve is now isomorphic to y 2=b′x 3+c′x 2+d′x+1 y 2=b′x 3+c′x 2+d′x+1. Transform it to the usual Weierstrass form and take the j j-invariant. Note that b′b′ etc. will have ξ ξ s in them, but they should all cancel out via the equation f(1,ξ)=0 f(1,ξ)=0 in the final result.Robin Chapman –Robin Chapman 2010-07-07 14:23:39 +00:00 Commented Jul 7, 2010 at 14:23 Add a comment\| 2 Answers 2 Sorted by: Reset to default This answer is useful 13 Save this answer. Show activity on this post. The j j invariant is j=S 3 S 3−27 T 2 j=S 3 S 3−27 T 2 where S=a−b d 4+c 2 12 S=a−b d 4+c 2 12 and T=a c 6+b c d 48−c 3 216−a d 2 16−b 2 16.T=a c 6+b c d 48−c 3 216−a d 2 16−b 2 16. For more details see my article "A computational solution to a question by Beauville on the invariants of the binary quintic", J. Algebra 303 (2006) 771-788. The preprint version is here. Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited May 3, 2024 at 13:44 LSpice 13.9k 4 4 gold badges 47 47 silver badges 74 74 bronze badges answered Jul 7, 2010 at 22:32 Abdelmalek AbdesselamAbdelmalek Abdesselam 23.6k 1 1 gold badge 73 73 silver badges 136 136 bronze badges 3 Please beware you are calculating J which is j/1728 Oliver –Oliver 2020-05-24 22:15:11 +00:00 Commented May 24, 2020 at 22:15 There are different conventions for what the proper normalization of j j is. Number theorists have their preferred choice. Folks working on moonshine matters have their own. Classical invariant theorists also. The is no universally canonical choice.Abdelmalek Abdesselam –Abdelmalek Abdesselam 2020-05-24 23:01:06 +00:00 Commented May 24, 2020 at 23:01 1 @AbdelmalekAbdesselam It's certainly true that there is no universally used choice. However, up to translation by an integer, there is a canonical choice if one wants a formula that has the correct properties in all characteristics, including characteristic 2 and 3. So that would seem to make the "number theorists'" version canonical, up to (as I said) translation by an integer. OTOH, if you're doing complex analysis, one could take the "j j-invariant" of X 3+A X+B X 3+A X+B to be (π e+i tan(π//2–√))⋅4 A 3 4 A 3+27 B 2,(π e+i tan⁡(π//2))⋅4 A 3 4 A 3+27 B 2, for example. :)Joe Silverman –Joe Silverman 2020-05-25 01:18:09 +00:00 Commented May 25, 2020 at 1:18 Add a comment\| This answer is useful 2 Save this answer. Show activity on this post. The j-invariant of the binary quartic form a X 4+b X 3 Y+c X 2 Y 2+d X Y 3+e Y 4 a X 4+b X 3 Y+c X 2 Y 2+d X Y 3+e Y 4 can be expressed by the two basic invariants I and J of the binary quartic form. j I J=I 3 Δ=I 3 I 3−27 J 2=∣∣∣a c 6 c 6 e∣∣∣−4∣∣∣∣b 4 c 6 c 6 d 4∣∣∣∣=a e−b d 4+c 2 12=∣∣∣∣∣a b 4 c 6 b 4 c 6 d 4 c 6 d 4 e∣∣∣∣∣=a c e 6−a d 2 16−b 2 e 16+b c d 48−c 3 216 j=I 3 Δ=I 3 I 3−27 J 2 I=\|a c 6 c 6 e\|−4\|b 4 c 6 c 6 d 4\|=a e−b d 4+c 2 12 J=\|a b 4 c 6 b 4 c 6 d 4 c 6 d 4 e\|=a c e 6−a d 2 16−b 2 e 16+b c d 48−c 3 216 An Introduction to Invariants and Moduli (Cambridge Studies in Advanced Mathematics, Series Number 81)(Page 27 Proposition 1.25., Page 427 Theorem 11.44.) Invariants of binary and ternary forms (Page 18) Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Improve this answer Follow Follow this answer to receive notifications edited Oct 30, 2024 at 5:46 answered May 3, 2024 at 7:55 user165013user165013 181 5 5 bronze badges Add a comment\| You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions polynomials rt.representation-theory See similar questions with these tags. 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1482	https://pubmed.ncbi.nlm.nih.gov/25676128/	Cost-Benefit Analysis for Optimization of Risk Protection Under Budget Constraints - PubMed Clipboard, Search History, and several other advanced features are temporarily unavailable. Skip to main page content An official website of the United States government Here's how you know The .gov means it’s official. Federal government websites often end in .gov or .mil. Before sharing sensitive information, make sure you’re on a federal government site. The site is secure. The https:// ensures that you are connecting to the official website and that any information you provide is encrypted and transmitted securely. 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Cost-Benefit Analysis for Optimization of Risk Protection Under Budget Constraints Olga Špačková1,Daniel Straub1 Affiliations Expand Affiliation 1 Engineering Risk Analysis Group, Faculty of Civil, Geo and Environmental Engineering, Technische Universität München, Munich, Germany. PMID: 25676128 DOI: 10.1111/risa.12310 Item in Clipboard Cost-Benefit Analysis for Optimization of Risk Protection Under Budget Constraints Olga Špačková et al. Risk Anal.2015 May. Show details Display options Display options Format Risk Anal Actions Search in PubMed Search in NLM Catalog Add to Search . 2015 May;35(5):941-59. doi: 10.1111/risa.12310. Epub 2015 Feb 10. Authors Olga Špačková1,Daniel Straub1 Affiliation 1 Engineering Risk Analysis Group, Faculty of Civil, Geo and Environmental Engineering, Technische Universität München, Munich, Germany. PMID: 25676128 DOI: 10.1111/risa.12310 Item in Clipboard Cite Display options Display options Format Abstract Cost-benefit analysis (CBA) is commonly applied as a tool for deciding on risk protection. With CBA, one can identify risk mitigation strategies that lead to an optimal tradeoff between the costs of the mitigation measures and the achieved risk reduction. In practical applications of CBA, the strategies are typically evaluated through efficiency indicators such as the benefit-cost ratio (BCR) and the marginal cost (MC) criterion. In many of these applications, the BCR is not consistently defined, which, as we demonstrate in this article, can lead to the identification of suboptimal solutions. This is of particular relevance when the overall budget for risk reduction measures is limited and an optimal allocation of resources among different subsystems is necessary. We show that this problem can be formulated as a hierarchical decision problem, where the general rules and decisions on the available budget are made at a central level (e.g., central government agency, top management), whereas the decisions on the specific measures are made at the subsystem level (e.g., local communities, company division). It is shown that the MC criterion provides optimal solutions in such hierarchical optimization. Since most practical applications only include a discrete set of possible risk protection measures, the MC criterion is extended to this situation. The findings are illustrated through a hypothetical numerical example. This study was prepared as part of our work on the optimal management of natural hazard risks, but its conclusions also apply to other fields of risk management. Keywords: Benefit-cost ratio; cost-benefit analysis; marginal cost criterion; natural hazards; risk protection optimization; risk-based decision making. © 2015 Society for Risk Analysis. PubMed Disclaimer Similar articles Risk management frameworks for human health and environmental risks.Jardine C, Hrudey S, Shortreed J, Craig L, Krewski D, Furgal C, McColl S.Jardine C, et al.J Toxicol Environ Health B Crit Rev. 2003 Nov-Dec;6(6):569-720. doi: 10.1080/10937400390208608.J Toxicol Environ Health B Crit Rev. 2003.PMID: 14698953 Review. Implementation of equity in resource allocation for regional earthquake risk mitigation using two-stage stochastic programming.Zolfaghari MR, Peyghaleh E.Zolfaghari MR, et al.Risk Anal. 2015 Mar;35(3):434-58. doi: 10.1111/risa.12321. Epub 2015 Feb 10.Risk Anal. 2015.PMID: 25676412 Budgeting in health care systems.Maynard A.Maynard A.Eff Health Care. 1984;2(2):41-9.Eff Health Care. 1984.PMID: 10269645 Cost-effectiveness analysis in relation to budgetary constraints and reallocative restrictions.Adang E, Voordijk L, Jan van der Wilt G, Ament A.Adang E, et al.Health Policy. 2005 Oct;74(2):146-56. doi: 10.1016/j.healthpol.2004.12.015. Epub 2005 Jan 26.Health Policy. 2005.PMID: 16153475 Health economics in developing countries.Abel-Smith B.Abel-Smith B.J Trop Med Hyg. 1989 Aug;92(4):229-41.J Trop Med Hyg. 1989.PMID: 2503621 Review. See all similar articles [x] Cite Copy Download .nbib.nbib Format: Send To Clipboard Email Save My Bibliography Collections Citation Manager [x] NCBI Literature Resources MeSHPMCBookshelfDisclaimer The PubMed wordmark and PubMed logo are registered trademarks of the U.S. Department of Health and Human Services (HHS). Unauthorized use of these marks is strictly prohibited. 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1483	https://zh.wikipedia.org/zh-hans/%E8%B4%9F%E4%BA%8C%E9%A1%B9%E5%88%86%E5%B8%83	负二项分布 - 维基百科，自由的百科全书跳转到内容 [x] 主菜单主菜单移至侧栏隐藏导航首页分类索引特色内容新闻动态最近更改随机条目帮助帮助维基社群方针与指引互助客栈知识问答字词转换 IRC即时聊天联络我们关于维基百科特殊页面搜索搜索 [x] 外观资助维基百科创建账号登录 [x] 个人工具资助维基百科创建账号登录目录移至侧栏隐藏序言 1 定义 2 概率质量函数开关概率质量函数子章节 2.1 帕斯卡分布 2.1.1 二项系数与负二项名称来源 2.1.2 概率质量函数对所有可能k值求和为1 2.2 几何分布 2.3 例子 3 相关分布 4 参见 5 参考文献 [x] 开关目录负二项分布 [x] 27种语言 Беларуская Català Deutsch Ελληνικά English Español Euskara فارسی Suomi Français Galego עברית Magyar Italiano 日本語 Nederlands Novial Polski Português Русский Slovenščina Sunda Svenska తెలుగు ไทย Türkçe Українська 编辑链接条目讨论 [x] 简体不转换简体繁體大陆简体香港繁體澳門繁體大马简体新加坡简体臺灣正體阅读编辑查看历史 [x] 工具工具移至侧栏隐藏操作阅读编辑查看历史常规链入页面相关更改上传文件固定链接页面信息引用此页获取短链接下载二维码打印/导出下载为PDF 打印版本在其他项目中维基共享资源维基数据项目外观移至侧栏隐藏维基百科，自由的百科全书负二项分布不同来源对负二项分布的定义略有差异：随机变量的最小可能取值可能是 k=0{\displaystyle k=0}（仅计失败的次数，或反之），亦可能是 k=r{\displaystyle k=r}（总次数，不论成败）；参数 p{\displaystyle p}可能表示每次试验成功的概率，也可能表示失败的概率；试验的终止条件可能是成功 r{\displaystyle r}次或失败 r{\displaystyle r}次。概率质量函数红线是平均值绿线是标准差参数 r>0{\displaystyle r>0!} (实) 0<p<1{\displaystyle 0<p<1!}（实）值域k∈{0,1,2,…}{\displaystyle k\in {0,1,2,\ldots }!} 概率质量函数Γ(r+k)k!Γ(r)p r(1−p)k{\displaystyle {\frac {\Gamma (r+k)}{k!\,\Gamma (r)}}\,p^{r}\,(1-p)^{k}!} 累积分布函数I p(r,k+1){\displaystyle I_{p}(r,k+1)} 期望值r 1−p p{\displaystyle r\,{\frac {1-p}{p}}!} 众数⌊(r−1)(1−p)/p⌋if r>1{\displaystyle \lfloor (r-1)\,(1-p)/p\rfloor {\text{ if }}r>1} 0 if r≤1{\displaystyle 0{\text{ if }}r\leq 1} 方差r 1−p p 2{\displaystyle r\,{\frac {1-p}{p^{2}}}!} 偏度2−p r(1−p){\displaystyle {\frac {2-p}{\sqrt {r\,(1-p)}}}!} 峰度6 r+p 2 r(1−p){\displaystyle {\frac {6}{r}}+{\frac {p^{2}}{r\,(1-p)}}!} 矩生成函数(p 1−(1−p)e t)r{\displaystyle \left({\frac {p}{1-(1-p)e^{t}}}\right)^{r}!} 特征函数(p 1−(1−p)e i t)r{\displaystyle \left({\frac {p}{1-(1-p)e^{i\,t}}}\right)^{r}!} 负二项分布（英语：Negative binomial distribution）是统计学上一种描述在一系列独立同分布的伯努利试验中，成功次数达到指定次数（记为 r{\displaystyle r}）时失败次数的离散概率分布。比如，如果我们定义掷骰子随机变量 x{\displaystyle x}值为 x=1{\displaystyle x=1}时成功，所有 x≠1{\displaystyle x\neq 1}为失败，这时我们反复掷骰子直到1出现3次（成功次数 r=3{\displaystyle r=3}），此时非1数字出现次数的概率分布即为负二项分布。帕斯卡分布（Pascal distribution，来自布莱兹·帕斯卡 (Blaise Pascal)）和波利亚分布（Polya distribution，又称罐子模型，来自乔治·波利亚 (George Pólya)）均是负二项分布的特例。在工程、气候等领域中经常用“负二项分布”或“帕斯卡分布”来描述变量 r{\displaystyle r}为整数的情况，而使用“波利亚分布”来描述 r{\displaystyle r}取到实数值 R{\displaystyle R}的情况。对于“相关的离散事件”（"associated discrete events"）的发生，例如龙卷风爆发，相比于泊松分布，波利亚分布由于允许其平均值和方差不同，而能够给出更精确的模型。在流行病学中，它已被用于模拟传染病的疾病传播，其中可能的继发感染数量可能因个体和环境而异。更一般地说，由于正协方差项，事件具有正相关的事件导致比独立事件更大的方差可能是合适的。 “负二项分布”与“二项分布”的区别在于：“二项分布”是固定试验总次数 N{\displaystyle N}的独立试验中，成功次数k的分布；而“负二项分布”是所有到r次成功时即终止的独立试验中，失败次数k的分布。术语“负二项式”可能是因为出现在分布的概率质量函数公式中的某个二项式系数可以用负数更简单地写出。定义 [编辑] 若每次伯努利试验有两种可能的结果，分别为成功或者失败。在每次试验中，成功的概率为 p{\displaystyle p}，失败的概率为 1−p{\displaystyle 1-p}。反复进行该伯努利试验，直到观察到第 r{\displaystyle r}次成功发生。此时试验失败次数 X{\displaystyle X} 的分布即为负二项分布（或称帕斯卡分布），那么：若随机变量 X{\displaystyle {\mathit {X}}}服从参数为 r{\displaystyle {\mathit {r}}}和 p{\displaystyle {\mathit {p}}}的负二项分布，则记为 X∼N B(r,p){\displaystyle X\sim NB(r,p)}. 在实际生活中，我们可以使用负二项分布描述某种机器在坏掉前，能够工作的天数的分布。此时，“成功”的事件可以指机器正常工作一天，“失败”的事件可以指机器故障的一天。如果我们使用负二项分布来描述运动员在获取r个奖牌前尝试的次数的分布，此时，“失败”的事件指运动员的一次尝试，“成功”的事件指运动员获取一枚奖牌。如果使用负二项分布来描述掷一枚硬币出现r次正面前，出现硬币反面的次数的分布，“成功”的事件指出现硬币的正面，“失败”的事件指出现硬币的反面。概率质量函数 [编辑] 帕斯卡分布 [编辑] 当 r{\displaystyle r} 是整数时的负二项分布又称帕斯卡分布，其概率质量函数为： f(k;r,p)≡Pr(X=k)=(k+r−1 r−1)p r(1−p)k for k=0,1,2,…{\displaystyle f(k;r,p)\equiv \Pr(X=k)={\binom {k+r-1}{r-1}}p^{r}(1-p)^{k}\quad {\text{for }}k=0,1,2,\dotsc } 其中 k{\displaystyle k} 是失败的次数， r{\displaystyle r} 是成功的次数， p{\displaystyle p} 是事件成功的概率。在负二项分布的概率质量函数中，由于 k+r{\displaystyle k+r} 次伯努利试验为独立同分布，每个成功 r{\displaystyle r} 次、失败 k{\displaystyle k} 次的事件的概率为 p r(1−p)k{\displaystyle p^{r}(1-p)^{k}}。由于第 r{\displaystyle r} 次成功一定是最后一次试验，所以应该在 k+r−1{\displaystyle k+r-1}次试验中选择 r−1{\displaystyle r-1}次成功，使用排列组合二项系数获取所有可能的选择数。二项系数与负二项名称来源 [编辑] 括号中为二项式系数表达式： (k+r−1 r−1)=(k+r−1)!k!(r−1)!=(k+r−1)(k+r−2)⋯(r)k!{\displaystyle {\binom {k+r-1}{r-1}}={\frac {(k+r-1)!}{k!\,(r-1)!}}={\frac {(k+r-1)(k+r-2)\dotsm (r)}{k!}}} 该表达式可以写成带负值参数的二项系数的形式，如下式所示，解释了“负二项”名称的来源： (k+r−1)⋯(r)k!=(−1)k(−r)(−r−1)(−r−2)⋯(−r−k+1)k!=(−1)k(−r k).{\displaystyle {\begin{aligned}&{\frac {(k+r-1)\dotsm (r)}{k!}}\[6pt]={}&(-1)^{k}{\frac {(-r)(-r-1)(-r-2)\dotsm (-r-k+1)}{k!}}=(-1)^{k}{\binom {-r}{k}}.\end{aligned}}} 概率质量函数对所有可能k值求和为1 [编辑] 帕斯卡分布概率质量函数 f(k;r,p){\displaystyle f(k;r,p)}对所有可能 k{\displaystyle k} 值求和，一定等于1： ∑k=0∞(k+r−1 k)p r q k=1{\displaystyle \sum _{k=0}^{\infty }{\binom {k+r-1}{k}}p^{r}q^{k}=1} 证明如下： 1=p r p−r=p r(1−q)−r=p r∑k=0∞(−r k)(−q)k=p r∑k=0∞(−1)k(−r k)q k=∑k=0∞(k+r−1 k)p r q k{\displaystyle 1=p^{r}p^{-r}=p^{r}(1-q)^{-r}=p^{r}\sum {k=0}^{\infty }{\binom {-r}{k}}(-q)^{k}=p^{r}\sum {k=0}^{\infty }(-1)^{k}{\binom {-r}{k}}q^{k}=\sum _{k=0}^{\infty }{\binom {k+r-1}{k}}p^{r}q^{k}} 其中第三步用到了二项序列展开。几何分布 [编辑] 取 r=1{\displaystyle r=1}，负二项分布等于几何分布。其概率质量函数为 f(k;1,p)=p⋅(1−p)k{\displaystyle f(k;1,p)=p\cdot (1-p)^{k}!}。例子 [编辑] 举例说，若我们掷骰子，掷到一即视为成功。则每次掷骰的成功率是 1 6{\displaystyle {\frac {1}{6}}}。要掷出三次一，所需的掷骰次数属于集合{ 3, 4, 5, 6, ... }。掷到三次一的掷骰次数是负二项分布的随机变数。要在第三次掷骰时，掷到第三次一，则之前两次都要掷到一，其机率为(1 6)3{\displaystyle ({\frac {1}{6}})^{3}}。注意掷骰是伯努利试验，之前的结果不影响随后的结果。若要在第四次掷骰时，掷到第三次一，则之前三次之中要有刚好两次掷到一，在三次掷骰中掷到2次1的机率为(3 3−1)(5 6)(1 6)2{\displaystyle {3 \choose 3-1}\left({5 \over 6}\right)\left({1 \over 6}\right)^{2}}。第四次掷骰要掷到一，所以要将前面的机率再乘 1 6{\displaystyle {\frac {1}{6}}}：((1+3)−1 3−1)(1 6)3(5 6){\displaystyle {(1+3)-1 \choose 3-1}\left({1 \over 6}\right)^{3}\left({5 \over 6}\right)}。相关分布 [编辑] 几何分布（在 { 0, 1, 2, 3, ... } 上）是负二项分布的一个特例，其中 Geom⁡(p)=NB⁡(1,1−p).{\displaystyle \operatorname {Geom} (p)=\operatorname {NB} (1,\,1-p).\,} 负二项分布是离散相型分布（英语：Discrete phase-type distribution）的一个特例。负二项分布是离散复合泊松分布的一个特例。参见 [编辑] 二项式分布几何分布参考文献 [编辑] ^DeGroot, Morris H. Probability and Statistics Second. Addison-Wesley. 1986: 258–259. ISBN 0-201-11366-X. LCCN 84006269. OCLC 10605205. ^e.g. J.O. Lloyd-Smith, S.J. Schreiber, P.E. Kopp, and W.M. Getz (2005), Superspreading and the effect of individual variation on disease emergence, Nature, 438, 355–359. doi:10.1038/nature04153 The overdispersion parameter is usually denoted by the letter k{\displaystyle k} in epidemiology, rather than r{\displaystyle r} as here. 3. ^Casella, George; Berger, Roger L. Statistical inference 2nd. Thomson Learning. 2002: 95. ISBN 0-534-24312-6. \| 查论编概率分布（列表（英语：List of probability distributions）） \| \| 离散单变量 \| \| 有限支集 \| 本福特定律伯努利分布 Β-二项式分布二项式分布 categorical（英语：categorical distribution）超几何分布 negative（英语：Negative hypergeometric distribution） Poisson binomial（英语：Poisson binomial distribution） Rademacher（英语：Rademacher distribution）孤子分布离散型均匀分布齐夫定律 Zipf–Mandelbrot（英语：Zipf–Mandelbrot law） \| \| 无限支集 \| beta negative binomial（英语：beta negative binomial distribution） Borel（英语：Borel distribution） Conway–Maxwell–Poisson（英语：Conway–Maxwell–Poisson distribution） discrete phase-type（英语：Discrete phase-type distribution） Delaporte（英语：Delaporte distribution） extended negative binomial（英语：extended negative binomial distribution） Flory–Schulz（英语：Flory–Schulz distribution） Gauss–Kuzmin（英语：Gauss–Kuzmin distribution）几何分布对数分布 mixed Poisson（英语：mixed Poisson distribution）负二项分布 Panjer（英语：(a,b,0) class of distributions） parabolic fractal（英语：parabolic fractal distribution）卜瓦松分布 Skellam（英语：Skellam distribution） Yule–Simon（英语：Yule–Simon distribution） zeta（英语：zeta distribution） \| \| \| \| 连续单变量 \| \| 紧支集 \| arcsine（英语：Arcsine distribution） ARGUS（英语：ARGUS distribution） Balding–Nichols（英语：Balding–Nichols model） Bates（英语：Bates distribution） Β分布 beta rectangular（英语：Beta rectangular distribution） continuous Bernoulli（英语：Continuous Bernoulli distribution）欧文–贺尔分布 Kumaraswamy（英语：Kumaraswamy distribution） logit-normal（英语：Logit-normal distribution） noncentral beta（英语：Noncentral beta distribution） PERT（英语：PERT distribution） raised cosine（英语：Raised cosine distribution） reciprocal（英语：Reciprocal distribution）三角形分布 U-quadratic（英语：U-quadratic distribution）连续型均匀分布维格纳半圆分布 \| \| 半无限区间支集 \| Benini（英语：Benini distribution） Benktander 1st kind（英语：Benktander type I distribution） Benktander 2nd kind（英语：Benktander type II distribution） beta prime（英语：Beta prime distribution） Burr（英语：Burr distribution） chi（英语：chi distribution）卡方分布 noncentral（英语：Noncentral chi-squared distribution） inverse（英语：Inverse-chi-squared distribution） scaled（英语：Scaled inverse chi-squared distribution） Dagum（英语：Dagum distribution） Davis（英语：Davis distribution）爱尔朗分布 hyper（英语：Hyper-Erlang distribution）指数分布亚指数分布 logarithmic（英语：Exponential-logarithmic distribution） F-分布 noncentral（英语：Noncentral F-distribution） folded normal（英语：folded normal distribution） Fréchet（英语：Fréchet distribution）伽玛分布 generalized（英语：Generalized gamma distribution） inverse（英语：Inverse-gamma distribution） gamma/Gompertz（英语：Gamma/Gompertz distribution） Gompertz（英语：Gompertz distribution） shifted（英语：shifted Gompertz distribution） half-logistic（英语：Half-logistic distribution） half-normal（英语：half-normal distribution） Hotelling's T-squared（英语：Hotelling's T-squared distribution）逆高斯分布广义逆高斯分布柯尔莫哥洛夫-斯米尔诺夫检验 Lévy（英语：Lévy distribution） log-Cauchy（英语：log-Cauchy distribution） log-Laplace（英语：log-Laplace distribution） log-logistic（英语：log-logistic distribution）对数正态分布 log-t（英语：log-t distribution） Lomax（英语：Lomax distribution） matrix-exponential（英语：matrix-exponential distribution）麦克斯韦-玻尔兹曼分布 Maxwell–Jüttner（英语：Maxwell–Jüttner distribution） Mittag-Leffler（英语：Mittag-Leffler distribution） Nakagami（英语：Nakagami distribution）帕累托分布 phase-type（英语：Phase-type distribution） Poly-Weibull（英语：Poly-Weibull distribution）瑞利分布 relativistic Breit–Wigner（英语：Relativistic Breit–Wigner distribution）莱斯分布 truncated normal（英语：truncated normal distribution） type-2 Gumbel（英语：type-2 Gumbel distribution）韦伯分布 discrete（英语：discrete Weibull distribution） Wilks's lambda（英语：Wilks's lambda distribution） \| \| 无限区间支集 \| 柯西分布 exponential power（英语：Generalized normal distribution）费雪z分布 Kaniadakis κ-Gaussian（英语：Kaniadakis Gaussian distribution） Gaussian q（英语：Gaussian q-distribution） generalized normal（英语：Generalized normal distribution） generalized hyperbolic（英语：Generalised hyperbolic distribution） geometric stable（英语：geometric stable distribution）耿贝尔分布 Holtsmark（英语：Holtsmark distribution） hyperbolic secant（英语：hyperbolic secant distribution） Johnson's S U（英语：Johnson's SU-distribution）朗道分布拉普拉斯分布 asymmetric（英语：Asymmetric Laplace distribution） logistic（英语：logistic distribution） noncentral t（英语：Noncentral t-distribution）正态分布 normal-inverse Gaussian（英语：normal-inverse Gaussian distribution） skew normal（英语：Skew normal distribution） slash（英语：Slash distribution）稳定分布司徒顿t分布 Tracy–Widom（英语：Tracy–Widom distribution） variance-gamma（英语：Variance-gamma distribution）福格特函数 \| \| 可变类型支集 \| generalized chi-squared（英语：Generalized chi-squared distribution） generalized extreme value（英语：generalized extreme value distribution） generalized Pareto（英语：Generalized Pareto distribution） Marchenko–Pastur（英语：Marchenko–Pastur distribution） Kaniadakis κ-exponential（英语：Kaniadakis Exponential distribution） Kaniadakis κ-Gamma（英语：Kaniadakis Gamma distribution） Kaniadakis κ-Weibull（英语：Kaniadakis Weibull distribution） Kaniadakis κ-Logistic（英语：Kaniadakis Logistic distribution） Kaniadakis κ-Erlangl（英语：Kaniadakis Erlang distribution） q-exponential（英语：q-exponential distribution） q-Gaussian（英语：q-Gaussian distribution） q-Weibull（英语：q-Weibull distribution） shifted log-logistic（英语：Shifted log-logistic distribution） Tukey lambda（英语：Tukey lambda distribution） \| \| \| 混合单变量 \| \| 连续离散 \| 修正高斯分布（英语：Rectified Gaussian distribution） \| \| \| 联合分布 \| Discrete: Ewens（英语：Ewens's sampling formula） multinomial（英语：multinomial distribution） Dirichlet（英语：Dirichlet-multinomial distribution） negative（英语：negative multinomial distribution） Continuous: 狄利克雷分布 generalized（英语：Generalized Dirichlet distribution） multivariate Laplace（英语：Multivariate Laplace distribution）多元正态分布 multivariate stable（英语：Multivariate stable distribution） multivariate t（英语：Multivariate t-distribution） normal-gamma（英语：normal-gamma distribution） inverse（英语：Normal-inverse-gamma distribution）随机矩阵 LKJ（英语：Lewandowski-Kurowicka-Joe distribution）矩阵正态分布 matrix t（英语：matrix t-distribution） matrix gamma（英语：matrix gamma distribution） inverse matrix gamma（英语：inverse matrix gamma distribution）威沙特分布 normal（英语：normal-Wishart distribution）逆威沙特分布 normal-inverse（英语：normal-inverse-Wishart distribution） \| \| 定向统计（英语：Directional statistics） \| 循环单变量定向统计（英语：Directional statistics）圆均匀分布univariate von Mises（英语：von Mises distribution）wrapped normal（英语：Wrapped normal distribution）wrapped Cauchy（英语：Wrapped Cauchy distribution）wrapped exponential（英语：Wrapped exponential distribution）wrapped asymmetric Laplace（英语：Wrapped asymmetric Laplace distribution）wrapped Lévy（英语：Wrapped Lévy distribution）球形双变量Kent（英语：Kent distribution）环形双变量bivariate von Mises（英语：bivariate von Mises distribution）多变量von Mises–Fisher（英语：Von Mises–Fisher distribution）Bingham（英语：Bingham distribution） \| \| 退化分布和奇异分布（英语：Singular distribution） \| 退化狄拉克δ函数奇异康托尔分布 \| \| 其它 \| Circular（英语：Circular distribution）复合泊松分布 elliptical（英语：Elliptical distribution） exponential（英语：Exponential family） natural exponential（英语：Natural exponential family） location–scale（英语：location–scale family） Maximum entropy（英语：Maximum entropy probability distribution） Mixture（英语：Mixture distribution） 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1484	http://euclideangeometry.weebly.com/uploads/5/3/8/4/53847747/nine_point_circle_article.pdf	NINE POINT CIRCLE tkhalid August 16, 2015 Abstract I am proud to present one of my ﬁrst articles concerning Olympiad Geometry. In particular this article is about the Nine Point Circle, some proofs of its exis-tence, properties pertaining to it, and beautiful problems to accompany it all. I hope you will enjoy the article and I wish you a happy reading! Existence definition: Consider a triangle ABC with orthocenter H. Let D, E, and F be the feet of the altitudes, let MA, MB, and MC be the midpoints of the sides, and let MAH, MBH, and MCH be the midpoints of AH, BH, and CH respectively, as shown. Then these 9 points are concyclic in the Nine Point Circle, as shown in Figure 1. 1 Figure 1: proof 1: We shall begin with perhaps the most elementary of proofs. The following proof requires no knowledge of advanced geometry, but rather near-trivial ideas taught in regular high school courses. Consider the following ﬁgure: 2 Figure 2: We know that because MC and MBH are midpoints of BA and BH, that MCMBH ∥AH and MCMBH = 1 2AH. Similarly we have MBMCH ∥AH and MBMCH = 1 2AH. Thus quadrilateral MCMBHMBMCH is a parallelo-gram. However we also know that MBHMCH ∥BC, so because MCMBH ∥AH and AH ⊥BC, we must have MCMBH ⊥MBHMCH. Thus quadrilateral MCMBHMBMCH is a rectangle. Similarly we ﬁnd MAMCMAHMCH and MAMBHMAHMB are rectangles. Thus they are all concylic in a circle with center N where N is the midpoint of MXMXH for X ∈{A, B, C}. Now we must prove that D, E, and F also lie on this circle. We know that MAMAH is a diame-ter of the circle, so any right angle that intercepts that arc must lie on the circle. Hence we have D lies on the circle, and using similar logic we see E and F also lie on the circle. Lastly, we want to prove that N is the midpoint of OH, where O is the circumcenter of △ABC. Note that OMA = 1 2AH = AMAH = MAHH. Thus OMAHMAH is a parallegram, and because N is the midpoint of MAMAH, it must also be the midpoint of HO. □ proof 2: To prove the existence of the Nine Point Circle, we shall consider the circumcircle of △ABC, and a few special points on it. 3 Figure 3: 4 We stick to the conﬁguration of Figure 1, but now a few things have been added in (as shown in Figure 3). Namely, these are the circumcircle of ABC (which I will denote with ω), and the points P, Q, R and X, Y, Z. Let P, Q, and R be the intersection of HMA, HMB, and HMC with ω respectively. Let X, Y, and Z be the intersection of HD, HE, and HF with ω respectively. As shown in the diagram above, we will prove that each red segment is the same length as it’s corresponding orange segment. For starters, we already deﬁned MAH, MBH, and MCH as the midpoints of AH, BH, and CH respectively, so those 3 segments are taken care of. Now we will use the method of Phan-tom Pointing. Consider the point T such that T is the reﬂection of H about BC. Then ∠BTC = ∠BHC = 180 −A. So ∠BTC + ∠BAC = 180, which implies that T lies on ω, and this further implies that T ≡X. So we know by deﬁnition of reﬂection that HD = DX. Similarly we have HE = EY and HF = FZ. Now let K be the reﬂection of H about MA. Then because HMA = MAK and BMA = MAC, we see that BHCK is a parallelogram, so ∠BKC = ∠BHC = 180 −A = ⇒∠BKC + ∠BAC = 180 = ⇒K ≡P. Thus indeed HMA = MAP. Similarly we ﬁnd that HMB = MBQ and HMC = MCR. Now we consider the homothethy with center H and factor 1 2. This pulls each of the nine points on ω to their respecitive points on the Nine Point Circle. In addition, we see that the Nine Point Center is simply the midpoint of HO. proof 3: There also exists a proof with complex numbers, which although generally I wouldn’t present, because it is fairly nice I will give an outline. Consider a triangle abc inscribed in the unit circle. Let N9 = a + b + c 2 . Then (using the same notation as in proof 2) we have H = a + b + c 2 . Also Ma = b + c 2 . Thus the distance from N9 to MA is simply a + b + c 2 −b + c 2 = c 2 = \|c\| 2 = 1 2 . Similarly we ﬁnd N9MB = N9MC = 1 2. Now we know MAH = a + b + c 2 , so N9MAH = a + b + c 2 −a −b + c 2 = a 2 = 1 2 . Similarly the distance from N9 to MBH and MCH is 1 2. Now we use the formula for the foot of an altitude. If K is the foot of the altitude in △AZB, where A and B are on the unit circle, then we have k = 1 2(a + b + z −ab¯ z) 5 . Using this we ﬁnd D = a + b + c 2 −ab 2c. So the distance from N9 to D is a + b + c 2 −a + b + c 2 + ab 2c = ab 2c = 1 2 . Similarly we ﬁnd that the distance from N9 to E and F is also 1 2. Thus, all 9 of the aforementioned points are exactly 1 2 away from N9, which implies they are concylic in a circle with center N9. And again, in addition we have that N9 is the midpoint of HO, because N9 = a + b + c 2 . proof 4: Finally we arrive at our last proof. It is well known that H and O are isogonal conjugates, so we introduce the following lemma: lemma Figure 4: proof We want to show (by power of a point) that AQC · APC = APB · AQB. From there we can apply similar logic to another set of four points to prove the concyclicity of all 6. We note that because P and Q are isogonal conjugates and ∠PPCA = ∠QQCA = ∠PPBA = ∠QQBA = 90◦, that AQC·APC = AP·sin ∠BAP·AQ·sin ∠BAQ = AP·sin ∠CAP·AQ·sin ∠CAQ = APB·AQB . So the points PC, QC, PB, and QB are concyclic. Repeating this with the other points we ﬁnd that the six points PA, PB, PC, QA, QB, and QC are concyclic. 6 Now the center of this circle is the perpendicular bisector of segment PCQC and segment PBQB, which is exactly O. Getting back to the problem: Now because H and O are isogonal conjugates we set P = H and Q = O. So we see that (using the same notation as proof 1) D, E, F, MA, MB, and MC are concyclic in a circle with center the midpoint of HO. Thus we need only prove that MAH, MBH, and MCH also lie on this circle. To do this we simply use power of a point. We want to show AMAH · AD = AE · AMB. (Note the other points follow similarly, so I will only show that MAH lies on the circle) Or equivalently AH · AD = AE · AC. But this is obviously true because quadrilateral DHEC is cyclic. □ Properties For this section, I will omit the proofs and leave them up to the reader! So go ahead and try to prove these properties as you read through this section, and in case you get stuck make sure to read the hints located towards the end of this article. Good luck! • The Nine Point Center lies on the Euler Line. • Given any point on the circumcircle of △ABC the segment formed by connecting that point to the orthocenter of △ABC is bisected by the Nine Point Circle. • Let N, H, and G be the Nine Point center, Orthocenter, and Centroid of △ABC respectively. Then HN = 3NG. • The Nine Point Circle of △ABC is internally tangent to incircle and ex-ternally tangent to the 3 exicrcles at the Feuerbach points of △ABC. • Consider a △ABC and let MA, MB, and MC be the midpoints of sides BC, CA, and AB. The tangents to the Nine Point Circle of △ABC at MA, MB, and MC bound a triangle that is similar to orthic triangle of △ABC. • If ABCD is a cyclic quadrilateral then the Nine Point Circles of triangles ABC, BCD, CDA, and DAB are concurrent at the anticenter of ABCD. Furthermore the Nine Point Centers of those triangles form a cyclic quadri-lateral that is homothetic to the reference quadrilateral ABCD with ratio −1 2. • Let l be a line passing through the circumcenter O of △ABC. Then the orthopole of l lies on the Nine Point Circle of △ABC. 7 Problems In this section you will encounter around 12 problems I have gathered. Most if not all of these problems are from Art of Problem Solving, written by some wonderful people. I do not take claim for any of these questions. There will be a section at the end with the links to the questions where you will be able to ﬁnd answers and solutions. And of course, there will be hints provided in the hints section. 1. Let ABC be a triangle and let ω1 be the circle with diameter BC. Let ω1 intersect sides CA and AB at points E and F respectively. Let ω2 be the circumcircle of △AFE. A random line through E intersects ω2 at X and ω1 at Y . Prove that the midpoint of XY lies on the Nine Point Circle of △ABC. 2. Consider a triangle ABC with circumcenter O. Let AD, BE, and CF be altitudes of △ABC and let MA, MB, and MC be the midpoints of sides BC, CA, and AB respectively. Let the tangent to the Nine Point Circle of △ABC at D intersect MAMB at D1. Deﬁne E1 and F1 similary. Prove that D1, E1, and F1 are collinear on a line perpendicular to the Euler line of △ABC. 3. Let ABC be a triangle with orthocenter H. The lines AH, BH, and CH intersect the circumcircle of ABC again at the points D, E, and F respectively. Let A1 be the reﬂection of A in the line EF. Deﬁne B1 and C1 similarly. The line B1C1 intersects the side BC again at X; the points Y and Z are similarly deﬁned. Show that X, Y, and Z are collinear on a line which is tangent to the Nine Point Circle of △DEF. 4. Let ABC be a triangle with orthocenter H and Nine Point Center N. Let l be the line passing through N perpendicular to AN. Let the feet of the perpendiculars from B, C, and H to l be X, Y, and Z. Prove that AN + HZ = BX + CY . 5. Let ABC be a triangle with orthocenter H. Let the centers of (ABH), (BCH), and (CAH) be C1, A1, and B1. Prove that △A1B1C1 ∼ = △ABC and that their Nine Point Circles coincide. 6. Show that a triangle ABC is right if and only if its Nine Point Circle and circumcircle are tangent. 7. Let ABC be a triangle with circumcenter O. D, E, and F are the centers of ⊙(OBC), ⊙(OCA), and ⊙(OAB), respectively. X, Y, and Z are the reﬂections of D, E, and F in BC, CA, and AB, respectively. Prove that the Nine Point center of △XY Z lies on the Euler line of △ABC. 8. Let ABCD be a cyclic quadrilateral with circumcircle (O). Let AD in-tersect BC at E, AB intersect CD at F, and AC intersect BD at G. 8 Prove that the Nine Point Circle of △GEF passes through the centroid of ABCD. Furthermore prove that it is the midcircle of (O) and (GEF). Note: Let (K) and (L) be two circles. Then the midcircle of (K) and (L) is deﬁned to be the circle centered at the midpoint of KL passing through the intersections of (K) and (L). 9. Given two points P and Q on the circumcircle of △ABC, prove that the Simson Lines of P and Q intersect on the Nine Point Circle of △ABC. 10. Let MA, MB, and MC be the midpoints of sides BC, CA, and AB respec-tively in △ABC. Let H be the orthocenter and let D, E, and F be the midpoints of AH, BH, and CH respectively. Let ω be the Nine Point Circle of △ABC. Let the tangents to ω at MA, MB, and MC bound a triangle PQR, and let the tangents to ω at D, E, and F bound a triangle XY Z. Show that the two bounded triangles are congruent. 11. Let I be the incenter of △ABC and let D and E be the intersection of BI and CI with AC and AB respectively. Prove that the Nine Point Center of △IDE lies on AI. bonus: 12. Let D, E, and F be the points of contact of the incircle with sides BC, CA, and AB of △ABC respectively. Let D′, E′, and F ′ be the reﬂections of D, E, and F about EF, FD, and DE respectively. Prove that the Nine Point Center of △ABC lies on the Euler line of △D′E′F ′. Hints Properties 1. What is the Nine Point Center the midpoint of? 2. Homothety 3. Recall that HG = 2GO. 4. This ones pretty hard, so see if this diagram helps: 9 10 5. Angle Chasing (what do you know about the medial triangle and tan-gents?) 6. This one’s also a little tricky. First thing to realize is that the Nine Point Circles have the same radius. Also, you know that the triangles all have the same circumcenter, so because the Nine Point Center is the midpoint of the segment connecting the orthocenter to the circumcenter, look instead at the orthocenters of the triangles in the problem and try proving those are concyclic. Problems 1. Try proving that the reﬂection H′ of H about the midpoint of XY lies on the circumcircle. You might want to use Spiral similarity. 2. Radical Axis is a very powerful tool 3. What exactly is the point of tangency? Could it possibly be the Feuerbach Point? 4. Consider the midpoints of AH and BC. 5. What do you know about the radii of (ABH), (BCH), and (CAH)? Try looking for parallelograms, and remember the full deﬁnition of Nine Point Circle. 6. Homothety 7. Try ﬁnding similar triangles, this should lead you to the conclusion that H is the incenter of △XY Z. 8. This question is very hard, I don’t think I can give a hint without it either being to trivial or it giving away too much of the problem. Try heading over to the link provided in the next section. 9. What do we know about two Simson Lines of diametrically opposite points? Recall that the Simson Line of a point P bisects the segment PH where H is the orthocenter of △ABC. 10. Angle chasing. 11. Introduce some points such as the orthocenter of △IDE into your dia-gram. 12. The question is bonus for a reason! I myself do not know the proof of this one. This was a question created by the AoPS user TelvCohl. Because it was such a beautiful question I decided to put it on this article. All credits to him of course. 11 Links to Problems 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. This question was actually proposed by me, although I don’t take credit for it or its originality because it most likely has been though of or posted somewhere before. Anyways, here’s the solution: Note that the medial triangle is congruent to △DEF. Thus the bounded triangles have the same incircle and congruent contact (or intouch) tri-angles, which means they themselves are congruent. Alternatively we can note that both bounded triangles are similar to the orthic triangle of △ABC, and they have the same incircle, so they must be congruent. 11. 12. The link is here: but as you can see no one has posted a solution yet. 12
1485	https://www.quora.com/What-is-the-most-tangible-way-to-introduce-the-Chinese-Remainder-Theorem-What-are-the-practical-and-really-interesting-examples-of-this-theorem-I-am-looking-for-examples-which-have-a-real-impact-on-students	Something went wrong. Wait a moment and try again. Real Life Examples Chinese Remainder Theorem Practical Applications Math in the Real World Application of Mathematic... Education in Mathematics Mathematical Examples Mathematics Number Theory 5 What is the most tangible way to introduce the Chinese Remainder Theorem? What are the practical and really interesting examples of this theorem. I am looking for examples which have a real impact on students. Daniel Martín Math enthusiast · Upvoted by Micah Pearce , M.S. in Mathematics Mathematics, Texas Tech University (2013) and Alon Amit , Lover of math. Also, Ph.D. · Author has 167 answers and 976.1K answer views · 11y I want to contribute with the introduction that my professor used when he explained the Chinese Remainder Theorem to me: "Imagine that you are in charge of the Chinese army and you are given the task to count your soldiers. Of course, you can count them one by one, but this will be very time consuming because there are many soldiers. One way to count them easily can be by grouping them in rectangles of a particular number of rows and see how many soldiers are left. Here's where this theorem applies: You notice that if you let them parade in rows of 3 soldiers, two soldiers will be left. If you l I want to contribute with the introduction that my professor used when he explained the Chinese Remainder Theorem to me: "Imagine that you are in charge of the Chinese army and you are given the task to count your soldiers. Of course, you can count them one by one, but this will be very time consuming because there are many soldiers. One way to count them easily can be by grouping them in rectangles of a particular number of rows and see how many soldiers are left. Here's where this theorem applies: You notice that if you let them parade in rows of 3 soldiers, two soldiers will be left. If you let them parade in rows of 5, 3 will be left, and in rows of 7, 2 will be left." In math, what you are trying to solve is this linear congruence system: x == 2 (mod 3) x == 3 (mod 5) x == 2 (mod 7) The theorem says that, because 3, 5 and 7 are coprime, the system has a unique solution modulo n = 3 5 7 = 105. Then you can proceed to introduce the theorem more formally and solve this particular example with it. Related questions What is the Chinese Remainder Theorem? When is it ever used in real life? How do you find the multiplicative inverse in the Chinese remainder theorem? How would you explain the Chinese Remainder Theorem to a beginner? Why is it called the Chinese Remainder Theorem? What do the Chinese call the "Chinese Remainder Theorem"? Laurel Farooqui I think the following is an interesting property: The Chinese Remainder Theorem (for ideals) is essentially equivalent to the statement that a degree n polynomial is uniquely determined by its value at n+1 points. This is a fairly easy statement to prove by explicit construction. I also find it interesting and useful since it means that one can prove that a complicated expression known to be a degree n polynomial (say ∑k=nk=0kn−1) is equivalent to a given polynomial by showing that they give the same value at n+1 points. It also tells us how to find such a polynomial given the data I think the following is an interesting property: The Chinese Remainder Theorem (for ideals) is essentially equivalent to the statement that a degree n polynomial is uniquely determined by its value at n+1 points. This is a fairly easy statement to prove by explicit construction. I also find it interesting and useful since it means that one can prove that a complicated expression known to be a degree n polynomial (say ∑k=nk=0kn−1) is equivalent to a given polynomial by showing that they give the same value at n+1 points. It also tells us how to find such a polynomial given the data. CRT for polynomials is a theorem that's used a fair amount in mathematics and outside. Andrew Lim data scientist at Custora · Author has 95 answers and 600.5K answer views · 11y In addition to Daniel Martín's answer, I suggest you check out the chapter "The Drive to Generality and Abstraction. The Chinese Remainder Theorem: A Case Study" from Davis and Hersh's The Mathematical Experience. It explores the Chinese Remainder Theorem as represented by mathematicians throughout history, starting with Sun Tzu and ending with some 20th century texts. (I wish I'd taken a class on math history in college!) Sun Tzu and Fibonacci state the theorem in terms of a concrete example with 3 pairwise coprime numbers (as Daniel does above) rather than general rules. Euler uses algebraic In addition to Daniel Martín's answer, I suggest you check out the chapter "The Drive to Generality and Abstraction. The Chinese Remainder Theorem: A Case Study" from Davis and Hersh's The Mathematical Experience. It explores the Chinese Remainder Theorem as represented by mathematicians throughout history, starting with Sun Tzu and ending with some 20th century texts. (I wish I'd taken a class on math history in college!) Sun Tzu and Fibonacci state the theorem in terms of a concrete example with 3 pairwise coprime numbers (as Daniel does above) rather than general rules. Euler uses algebraic notation, though still with an arbitrarily fixed number of coprimes (5). A 1967 text (Shockley) states the theorem in familiar modern algebraic terms. A 1976 computer science text (Prather) restates it in terms of the product of cyclic groups (!). And a 1963 text (Weiss) generalizes out of integers into fields. (This last one starts reaching beyond my pay grade; I've only learned basic algebra.) I thought this chapter was the best in a very good book. It could be transformed by a skillful teacher into some wonderful math lectures. Jeff Suzuki Mathematician and teacher of mathematics (20+ years). · Author has 4.4K answers and 9.9M answer views · 6y Related How can I understand the Chinese remainder theorem? What are some examples? The basic intuition behind the Chinese Remainder Theorem is this: If you’ve found one solution to x≡amodp, then you can add p and still have a solution. Consider the system of congruences x≡5mod7, x≡4mod11, and x≡4mod8. The first congruence has solution x = 5. But (since it’s a congruence mod 7), it will also have solutions x = 12, 19, 26, 33, and so on. Look among these for a solution to the second congruence: we see x = 26 solves the first and second congruences. To keep this as a solution, we need to add multiples of 7×11=77: 7, because addi The basic intuition behind the Chinese Remainder Theorem is this: If you’ve found one solution to x≡amodp, then you can add p and still have a solution. Consider the system of congruences x≡5mod7, x≡4mod11, and x≡4mod8. The first congruence has solution x = 5. But (since it’s a congruence mod 7), it will also have solutions x = 12, 19, 26, 33, and so on. Look among these for a solution to the second congruence: we see x = 26 solves the first and second congruences. To keep this as a solution, we need to add multiples of 7×11=77: 7, because adding a multiple of 7 won’t undo our solution to the first congruence; and 11, for the same reason. So solutions to the first two congruences are x = 26, 103, 180, and so on. We keep going until we find a solution to the third congruence: x = 180 solves the third congruence and, because of the way we’ve constructed it, it will also solve the other two. Promoted by Network Solutions® Lauren Robinson Senior Copywriter at Network Solutions (2010–present) · Jun 25 Is website is enough for the Business or not? Short answer: It’s a great start... but it’s kind of like showing up to a potluck with just napkins. Let me explain. Your Website = Your Digital Home Base 🏡 Your website is the place people go when they want to learn more about you, stalk your services, or double-check if you’re legit. It’s your digital handshake. And the good news? You don’t need to sell your soul to a developer or learn to code in the dead of night. Thanks to DIY website tools (like the one I help out with at Network Solutions 👋), building a sleek, functional site is totally doable. 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How is the Chinese remainder theorem applicable in maths? How can I understand this theorem? What are the exact steps to use the Chinese Remainder Theorem? What is the real life example of pythogarous theorem? How does the Chinese remainder theorem influence your daily living by providing examples? Joyneel Bepari Aspiring to compete in the official IMO... · Author has 739 answers and 211.9K answer views · Jan 2 Related How can I understand the Chinese remainder theorem? What are some examples? The CRT––––––––––– The Chinese Remainder Theorem states that the set of solutions to the system of linear modular congruence equations: x≡⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩r1(modn1)r2(modn2)⋮rk(modnk) is given by: x≡k∑i=1riNiMi(modN) where the following conditions are to be met: gcd(ni,nj)=1 N=n1n2…nk Ni=Nni \color{darkcyan}{M_iN_i \equiv 1\pmod{n_i}\text{, i.e.:}}\tag The CRT––––––––––– The Chinese Remainder Theorem states that the set of solutions to the system of linear modular congruence equations: x≡⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩r1(modn1)r2(modn2)⋮rk(modnk) is given by: x≡k∑i=1riNiMi(modN) where the following conditions are to be met: gcd(ni,nj)=1 N=n1n2…nk Ni=Nni MiNi≡1(modni), i.e.: Mi is the modular multiplicative inverse of Nimodni Example 1–––––––––––– The best way to understand this is using examples. Consider the system: x≡4(mod7) and x≡2(mod9) Define N=7×9=63 and each Ni: N1=637=9 and N2=639=7 We observe that M1=4 and M2=4. Hence, the CRT states that: x≡4×9×4+2×7×4≡11(mod63) i.e., x=63q+11 where q∈Z. We can, of course, test whether this works. Inputting x=11 gives: 11≡4(mod7) and 11≡2(mod9) Perhaps, x=63+11=74: 74≡4(mod7) and 74≡2(mod9) Also, x=63×2+11=137: 137≡4(mod7) and 137≡2(mod9) Example 2–––––––––––– Let’s attempt one last example. x≡⎧⎨⎩4(mod7)1(mod3)5(mod11) We have N=7×3×11=231 and: N1=2317=33 ; N2=2313=77 ; N3=23111=21 The modular multiplicative inverses are M1=3, M2=2, & M3=10 because these values of Mi result in: NiMi≡1(modni) Therefore, the solution to x is: x≡396+154+1050≡214(mod231) Now that seems really cool, doesn’t it? The CRT is a very organized way of solving a system of linear modular congruence equation rather than taking two equations at a time and working in pairwise equations, which would be quite a rigmarole for large k. If you’re further interested in why this works, then I’ve got you; we shall discuss it’s proof! Before that, it is important that I acquaint you with some lemmas. The Chinese Remainder Lemma––––––––––––––––––––––––––––––––––– A lot of documents discuss the CRL as if it is the CRT, but in my opinion, the CRL is more of a lemma for which the CRT is formed. If x≡r(modn1,n2,…,nk), then: x≡r(modN) where N=n1n2…nk and gcd(ni,nj)=1. I’m sure you recognize this! Many scriptures replace x,r with a,b. This lemma is the underlying reason for the set of solutions to x mentioned in the CRT. Proof–––––– By definition of modular congruency: x−r=n1q1=n2q2=⋯=nkqk where qi∈Z. We observe that: n1,n2,…,nk \| x−r Since each ni are pairwise coprime: N=n1n2…nk \| x−r ∴x≡r(modN) Modular Multiplicative Inverse Lemma––––––––––––––––––––––––––––––––––––––––––– If a,n are coprime positive integers, then there exists a unique integer b with 0<b<n for which: ab≡1(modn) This means that a will have a unique modular multiplicative inverse modulo n if gcd(a,n)=1. Proof of Existence–––––––––––––––––––– By Bezout’s Lemma, we know that there exists integers p,q such that: ap+nq=gcd(a,n)=1⟹ap≡1(modn) One of many values of p can be found via Euclid’s Division Algorithm. Supposing that b is the smallest positive integer value of p, then solving the Diophantine gives: p=b+ns⟹p≡b(modn) and obviously, b≠0, so 0<b<n. Proof of Uniqueness–––––––––––––––––––––– We also stated that b will be unique to a, i.e., for all a1≢a2(modn), b1≠b2. For the sake of contradiction, suppose that b1=b2. This means that: a1b1≡a2b1≡1(modn) Since the proof of existence (the Bezout’s Lemma part) tacitly suggests that b is coprime to n: a1≡a2(modn) is a contradiction! So, b1≠b2. Proof of the Chinese Remainder Theorem––––––––––––––––––––––––––––––––––––––––––––– With those two lemmas covered, we shall proceed with the proof for the CRT. With each pairwise ni being coprime, we are given the system of equations: x≡⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩r1(modn1)r2(modn2)⋮rk(modnk) Our aim is to convert this system into a format where the CRL can be applied. Since any ni is coprime to another nj, we can systematically define: Ni=Nni and observe that Ni is coprime to ni because Ni is the product of the other nj. By the modular multiplicative inverse lemma, Ni will have a (unique) modular inverse, say Mi, modulo ni. Thus, we can rewrite each equation in the system as: x≡ri≡riNiMi(modni) Lastly, we heed to the fact that any other Nj is divisible by ni. Thus: x≡k∑j=1rjNjMj(modni) And therefore, by the Chinese Remainder Lemma: x≡k∑j=1rjNjMj(modN) This completes the proof! ■ I must admit that Sun Zi thought of this really smartly! What this also says is that the smallest non-negative integer solution to x is always less than N. George Mathew Mathematics teacher at a government college in India. · Author has 1.3K answers and 1.8M answer views · Updated 9y Related How can I understand the Chinese remainder theorem? What are some examples? We say that two numbers (positive integers) a and b are relatively prime (prime to each other) , if they have no common prime factors. We say that the numbers m 1 , m 2 , . . . . m r , pair wise relatively prime if any two distinct numbers in that collection, are relatively prime. Chinese remainder theorem says that given any r pair wise relatively prime numbers m 1 , m 2 , . . . . m r , and any numbers b 1 , b 2 , b 3 , . . b r , we can always find a number M which leaves the remainders b 1 , b 2 , b 3 , . . b r when it is divided by m 1 , m We say that two numbers (positive integers) a and b are relatively prime (prime to each other) , if they have no common prime factors. We say that the numbers m1,m2,....mr, pair wise relatively prime if any two distinct numbers in that collection, are relatively prime. Chinese remainder theorem says that given any r pair wise relatively prime numbers m1,m2,....mr, and any numbers b1,b2,b3,..br, we can always find a number M which leaves the remainders b1,b2,b3,..br when it is divided by m1,m2,...mrrespectively. Any integer M+k.m1.m2.m3....mr ....................... (i) has the same property for any integer value of k. Thus Chinese remainder theorem discusses the solvability and method of solving a given system of linear congruences (modulo pair wise relatively prime numbers.) It is a standard result in number theory that If a and m are numbers prime to each other, then there exists a number a′ such that a.a′ leaves the remainder 1 when divided by m. ie a.a′≡1modm Note that a′ is unique modulo m. a′ is called the reciprocal of a modulo m. Chinese remainder theorem gives a method to solve the problem (i) ie Solution of the simultaneous linear congruences: x≡bjmodmj,j=1,2,...,r, where mj are pair wise relatively prime. .................() The number M=b1.M1.M′1+b2.M2.M′2+....+br.Mr.M′r ............................................... (ii) where Mj=m1.m2....mrmj, for each j=1,2,...,rand M′j is the reciprocal of Mj modulo mj. ie solution of the system () is x≡MmodN where N=m1.m2....mr Each integer M+k.m1.m2...mr is a solution for any integer value of k. Let us illustrate this by two examples. Problem 1: Find the solution of the system of congruences: x≡2mod4, x≡3mod5, x≡1mod3 (ie Find all integers which leave remainders 2,3,1 when divided by 4,5,3respectively) We apply Chinese remainder theorem. Here b1=2,b2=3,b3=1. m1=4,m2=5,m3=3. Hence we get M1=5×3=15 M2=4×3=12 M3=4×5=20 Note that M1=15≡3mod4 M2=12≡2mod5 M3=20≡2mod3 Hence M′1≡3mod4 M′2≡3mod5 M′3≡2mod3 Therefore M=2×15×3+3×12×3+1×20×2=90+108+40=238 .........................(iii) Here m1.m2.m3=4×5×3=60 Therefore from (iii) we get ,the solution is x≡58mod60. ie Each integer 58+k.60 for any integer value of k gives a solution. Problem 2: Find the integers which leave the remainders 4, 6, 7 when divided by 5, 9 and 11 respectively. We apply Chinese remainder theorem to solve this problem. Here b1=4,b2=6,b3=7. m1=5,m2=9,m3=11. Hence we get M1=9×11=99 M2=5×11=55 M3=5×9=45 Note that M1=99≡4mod5 M2=55≡1mod9 M3=45≡1mod11 Hence M′1≡4mod5 M′2≡1mod9 M′3≡1mod11 Therefore M=4×99×4+6×55×1+7×45×1=1584+330+315=2229 .........................(iv) Here m1.m2.m3=5×9×11=495 Therefore from (iv) we get ,the solution is x≡249mod495. ie Each integer 249+k.495 for any integer value of k gives a solution. Therefore , if we are interested in finding the smallest positive integer which leaves the remainders 4, 6, 7 when divided by 5, 9 and 11 respectively, the answer is 249. Note that the first problem can also be considered as the problem of finding the number which leaves the same remainder - 2 when divided by 4, 5 and 3. Therefore the answer can easily be found as k. L + - 2, where L is the least common multiple (LCM) of 4,5 and 3, k is any integer. But this special method will not work in the second problem. Chinese remainder theorem has got interesting applications in various fields. Sponsored by OrderlyMeds Is Your GLP-1 Personalized? Find GLP-1 plans tailored to your unique body needs. David Joyce Professor Emeritus of Mathematics at Clark Univerisity · Author has 9.9K answers and 68.1M answer views · Updated 7y Related What are the exact steps to use the Chinese Remainder Theorem? I’ll borrow from my notes Qin Jiushaos algorithm for finding one. This is the Chinese algorithm. A different algorithm was used in India. Qin Jiushao (1202-1261) was a Chinese mathematician who wrote Shushu jiuzhang (Mathematical Treatise in Nine Sections). In it he has a general method for solving simultaneous linear congruences (the Chinese Remainder Theorem). The theorem was already a thousand years old in China. The general process has several steps, the hardest one being the step called “finding one”. The situation is this. I’ll use modern notation. You have k linear congruences that you wa I’ll borrow from my notes Qin Jiushaos algorithm for finding one. This is the Chinese algorithm. A different algorithm was used in India. Qin Jiushao (1202-1261) was a Chinese mathematician who wrote Shushu jiuzhang (Mathematical Treatise in Nine Sections). In it he has a general method for solving simultaneous linear congruences (the Chinese Remainder Theorem). The theorem was already a thousand years old in China. The general process has several steps, the hardest one being the step called “finding one”. The situation is this. I’ll use modern notation. You have k linear congruences that you want a single solution. N≡r1(mod m1)N≡r2(mod m2)⋯⋯N≡rk(mod mk) Read the first congruence as “N is congruent to r1 modulo m1.” It means that when you divide the unknown integer N by m1 you get a remainder of r1. The numbers m1,m2,…,mk are called “moduli” and the remainders r1,r2…,rk are sometimes called “residues”. In order for this to always work, it is necessary that the moduli are pairwise relatively prime. If not, it’s not guaranteed to work. Compute the product M=m1m2⋯mk of all the moduli. For each modulus, compute the product of the remaining moduli. For the first modulus m1, the product is m2⋯mk which is also equal to M/m1. Reduce that product modulo the given modulus. That means to replace the product by the remainder when you divide the product by the modulus. Call these reduced numbers s1,s2,…,sk. “Finding one”. Details about this step are below. For each of those numbers found in step 3, find the reciprocal modulo the given modulus. For m1, we need to solve the congruence x so that s1x≡1(mod m1). If the modulus is small, you can just search for it, but if it’s large, that search is not efficient. Call these numbers t1,t2,…,tn. Compute the sum r1s1t1+r2s2t2+⋯rksktk. Reduce that sum modulo M from step 1. That gives you the answer N. Qin Jiushao’s method of finding one In step 4 above, we need to solve a congruence of the form sx≡1(mod m) where m and s are known, and relatively prime, but x is unknown. This is essentially the extended Euclidean algorithm. Draw a 2×2 box with 1 and 0 in the left column and s and m in the second column. Let’s do it with specific numbers so you can see what’s happening. Let s=168 and m=295. Look in the right column. Subtract the smaller number 168 from the larger 295, and in the left column add the number to the left of the smaller number, namely 1, to the number to the left of the larger number, namely 0. You’ll get Next subtract 127 from 168 while adding the 1 in the lower left to the 1 in the upper left. Keep repeating this. Besides subtraction, you can speed things up if you use division. For the next stage, divide 41 into 127. You get a quotient of 3 and a remainder of 4. Replace the 127 by the remainder 4, and multiply the quotient 3 by the upper left entry 2 to get 6, and add that 6 to the lower left entry 1 to get 7. Next, 41 divided by 4 gives a quotient of 10 and a remainder of 1. Replace the 41 by 1, and add the product of 7 and 10 to the 2 in the upper left corner to put 72 there. Continue on until you get a 1 in the upper right corner which we have now. This 1 be eventually found since s and m are relatively prime. The number to its left will be the solution x=72, that is, the reciprocal of s=168 modulo m=295. You can check this. Multiply 168 times 295, then divide that by 72 to get a remainder of 1. Brian Sittinger PhD in Mathematics, University of California, Santa Barbara (Graduated 2006) · Author has 8.4K answers and 20.7M answer views · 2y Related How do you use Chinese Remainder Theorem to find 3 302 mod 5005 ? Note that 5005=5⋅7⋅11⋅13. Hence, we compute 3302 modulo each of these prime factors one at a time and then use the Chinese Remainder Theorem to find the value of this expression modulo 5005. (i) Working modulo 5: We do not actually need Fermat’s Little Theorem for this calculation, because 32≡9≡−1mod5. Therefore, we see that 3302≡(32)151≡(−1)151≡−1≡4mod5. (ii) Working modulo 7: Similar to (i), note that 33≡27≡−1mod7. Therefore, we deduce that 3302≡(33)100⋅32≡(−1) Note that 5005=5⋅7⋅11⋅13. Hence, we compute 3302 modulo each of these prime factors one at a time and then use the Chinese Remainder Theorem to find the value of this expression modulo 5005. (i) Working modulo 5: We do not actually need Fermat’s Little Theorem for this calculation, because 32≡9≡−1mod5. Therefore, we see that 3302≡(32)151≡(−1)151≡−1≡4mod5. (ii) Working modulo 7: Similar to (i), note that 33≡27≡−1mod7. Therefore, we deduce that 3302≡(33)100⋅32≡(−1)100⋅9≡2mod7. (iii) Working modulo 11: This time, we apply Fermat’s Little Theorem (since 11 is prime) to find that 311−1≡310≡1mod11. Hence, we obtain 3302≡(310)30⋅32≡130⋅9≡9mod11. (iv) Working modulo 13: This time, Fermat’s Little Theorem yields 313−1≡312≡1mod13. Hence, we obtain 3302≡(312)25⋅32≡125⋅9≡9mod13. From the results from (i)-(iv), we see that 3302≡4≡9mod5, 3302≡2≡9mod7, 3302≡9mod11, and 3302≡9mod13. Since these four prime moduli are clearly pairwise relatively prime, we conclude (without using the Chinese Remainder Theorem algorithm) that 3302≡9(mod5⋅7⋅11⋅13=5005). Promoted by Grammarly Grammarly Great Writing, Simplified · Mon Which are the best AI tools for students? There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? Which tools are built for students and school—not just for clicks or content generation? And more importantly, which ones help you sharpen what you already know instead of just doing the work for you? That’s where Grammarly comes in. It’s an all-in-one writing surface designed specifically for students, with tools that help you brainstorm, write, revise, and grow your skills—without cutting corners. Here are five AI tools inside Grammarly’s document editor that are worth checking out: Do There are a lot of AI tools out there right now—so how do you know which ones are actually worth your time? 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Chinese remainder theorem was published by a Chinese mathematician Sun Tzu in 3rd to 5th centuries. It was later generalized with a complete soulution in Qin Jiushao's 1247 Mathematical Treatise in Nine Sections. The solution was then called as Dayanshu. The title of the article on the Chinese Wikipedia (zh.wikipedia.org/wiki/…) is 中国剩余定理, which is basically a translation of "Chinese remainder theorem." However, the article also gives eight older names: 「韓信點兵」、「孫子定理」、「鬼谷算」、「隔墻算」、「剪管術」、「秦王暗點兵」、「物不知數」, which I (roughly and hesitantly) translate as "Han Xin's inspection of troops", "Sun Tzu's The Chinese remainder theorem was published by a Chinese mathematician Sun Tzu in 3rd to 5th centuries. It was later generalized with a complete soulution in Qin Jiushao's 1247 Mathematical Treatise in Nine Sections. The solution was then called as Dayanshu. The title of the article on the Chinese Wikipedia (zh.wikipedia.org/wiki/…) is 中国剩余定理, which is basically a translation of "Chinese remainder theorem." However, the article also gives eight older names: 「韓信點兵」、「孫子定理」、「鬼谷算」、「隔墻算」、「剪管術」、「秦王暗點兵」、「物不知數」, which I (roughly and hesitantly) translate as "Han Xin's inspection of troops", "Sun Tzu's Theorem", "Guigu (Wang Xu)'s Counting", "Counting by partitioning", "Cut tube technique", "Qin Shi Huang inspecting troops in the dark", and "Objects with unknown number" May be some Chinese nationals (our fellow Quorans) should answer this question so as to be definitive about what they call "Chinese Remainder Theorem". Source: Wikipedia. Alf Salte Works at Schlumberger (company) · Author has 3.8K answers and 2.5M answer views · 4y Related What is the Chinese Remainder Theorem? When is it ever used in real life? Mathematically the chinese remainder theorem states that if you know the remainder of a number n when divided by several factors all of which are co-primes to each other (i.e. gcd of each pair of them is 1) then you know the remainder if you divide by the product of them. For example if I tell you there is a number which is such that when you divide it by 15 you get a remainder of 13 and if you divide it by 7 you get a remainder of 6, what is the remainder if you divide the number by 105? Note that 7 and 15 are co-primes, they have no shared factors. So n is clearly of the form n = 15k + 13 where Mathematically the chinese remainder theorem states that if you know the remainder of a number n when divided by several factors all of which are co-primes to each other (i.e. gcd of each pair of them is 1) then you know the remainder if you divide by the product of them. For example if I tell you there is a number which is such that when you divide it by 15 you get a remainder of 13 and if you divide it by 7 you get a remainder of 6, what is the remainder if you divide the number by 105? Note that 7 and 15 are co-primes, they have no shared factors. So n is clearly of the form n = 15k + 13 where k is some integer. It also must be n = 7j + 6 so the question is, if we divide this number by 105 what is the remainder? If we recognize that k is any integer so it can be written as 7m + r where m is the quotient and r is the remainder if you divide k by 7. Then n = 15(7m + r) + 13 = 105m + 15r + 13 If we divide this by 7 we get n = 7(15m + (7(2) + 1)r + 7 + 6 = 7(15m + 2r + 1) + r + 6 All the stuff in the parenthesis is part of the quotient and does not make up the remainder. So the remainder is r + 6 so clearly any r that is a multiple of 7 will give the proper remainder of 6 when divided by 7. So since r is 0..6 this means r must equal 0 so k is an integer that is a multiple of 7 so n = 105m + 13 I.e. the remainder when you divide the number by 105 must be 13. Now, in this example the remainder is the same as when we divided by 15 but in general that is not the case. However the theorem states that you can always find this remainder given the remainders of those factors. The only requirement is that the factors must be co-primes, i.e. they have no shared factors. For example if the number you divide by is 12 it won’t help to know the remainder when divided by 4 and the remainder when divided by 6 as 4 and 6 are not co-primes, they are both even numbers and have a factor of 2 in common. It can be used in real life in a number of ways. For example as a quick way to solve some set of linear equations. You can just do modulus some prime number just to get a solution that is easy to compute in your head. By doing it modulus multiple primes you can then use this theorem to find the solution to the original equation. The point is that modulus computation is generally easier to do than regular computation. In other situations you actually get the scenario of the Chinese remainder theorem straight up and you need to solve a problem using it. However, many people will not recognize it, partly because they are not used to modulus computations and partly because they don’t know about the Chinese remainder theorem. Also, the theorem states that there is a solution, it doesn’t really tell you how to find that solution, that has to be learned in addition to knowing the theorem itself. However, some times it is enough to know there is a solution there to know you can move forward even if you do not know what that solution is. In those cases the theorem is all you need. However, in practice it is generally not that hard to find a solution as I outlined above even with my oversimplified method where I didn’t use modulus calculations directly at all. If you know how to compute in modulus it is even easier. Anonymous Upvoted by David Joyce , Ph.D. Mathematics, University of Pennsylvania (1979) and Pramit Das , B.Stat and M.Stat Statistics & Mathematics, Indian Statistical Institute (2020) · 9y Related What is the Chinese remainder theorem? Can anyone explain me what exactly it is? For elementary purposes, the Chinese Remainder Theorem provides an algorithm to solve a system of modular congruence relations, i.e., to solve a system of the following form: ⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩x≡a1(modn1)x≡a2(modn2)⋮x≡ak(modnk) where the ni's are coprime to each other, i.e., gcd(ni,nj)=1 ∀ i≠j The solution is of the form x≡a(modn1n2⋯nk) To use this algorithm, you first need to know about multiplicative modular inverses. Basically, modular inverse of an element k coprime to n is a number x such that kx\equi For elementary purposes, the Chinese Remainder Theorem provides an algorithm to solve a system of modular congruence relations, i.e., to solve a system of the following form: ⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩x≡a1(modn1)x≡a2(modn2)⋮x≡ak(modnk) where the ni's are coprime to each other, i.e., gcd(ni,nj)=1 ∀ i≠j The solution is of the form x≡a(modn1n2⋯nk) To use this algorithm, you first need to know about multiplicative modular inverses. Basically, modular inverse of an element k coprime to n is a number x such that kx≡1(modn). Computing modular inverse can be done either by elementary modular arithmetic manipulations or by a systematic method known as Extended Euclidean Algorithm. Now, back to CRT, this theorem gives us the following algorithm: Let N=n1n2⋯nk. Define yi:=N/ni and zi to be the modular inverse of yi modulo ni, i.e., a number zi such that we have ziyi≡1(modni) Then, CRT says that the unique solution to our system of congruences is given by x≡k∑i=1aiyizi(modN) Ram Kushwah Up and coming Most viewed writer · Author has 6.6K answers and 15.7M answer views · Updated 1y Related What is the Chinese remainder theorem? Very good question. It was the third century. Sunji Suwanjing lived in China. He asked a question: Suppose you have some pieces. If you divide these into three pieces, then you have 2 pieces left. If you divide them into pieces of five, then you have 3 pieces left. If you divide them into seven pieces then you have 2 pieces left. How many pieces are there in total? We have shown this question on our computer: We request all the readers to answer this without thinking ahead and write honestly how many pieces are there in the comments. If we write this question according to mathematics, then we will write Very good question. It was the third century. Sunji Suwanjing lived in China. He asked a question: Suppose you have some pieces. If you divide these into three pieces, then you have 2 pieces left. If you divide them into pieces of five, then you have 3 pieces left. If you divide them into seven pieces then you have 2 pieces left. How many pieces are there in total? We have shown this question on our computer: We request all the readers to answer this without thinking ahead and write honestly how many pieces are there in the comments. If we write this question according to mathematics, then we will write like this. Suppose the total number of pieces is x. x = 2 (mod 3) x = 3 (mod 5) x = 2 (mod 7) Sunji did this to solve this: The first number is the multiplication of the mods of the second and third. The second number is the multiplication of the mod of the first and third. The third number is the multiplication of the mod of the first and the second. They got three numbers: 5 7 = 35 3 7 = 21 3 5 = 15 Multiply these numbers so long as there is only one left to do with modding. 35 mod 3 = 2. But if we double it, 70 mod 3 = 1 21 mod 5 = 1 15 mod 7 = 1 Multiply these numbers by the remainder. 2 70 = 140 3 21 = 63 2 15 = 30 Now add these: 140 + 63 + 30 = 233 Multiply the three mods by themselves = 3 5 7 = 105 Now take their mod = 233 mod 105 = 23 This is your answer. You have 23 + n 105 pieces. Meaning the answer is 23, 128, 233,… all. You can also check this: If its pieces are divided into three, then 2 gontia will survive. If its pieces are divided into five pieces, there will be 3 pieces left. If its pieces are reduced to seven, then 2 pieces will be left. This is called the Chinese remainder theorem. If mod is a prime number, we can answer any question in the same way. Today we use it in encryption. Sunji solved it but he did not understand much. Aryabhata, the great mathematics scholar of India, answered this and explained it with the proof of complete mathematics. He was born in Patna in the 5th century. He was born 1400 years before Ramanujan ji and made many rules of mathematics. We salute him! Translated from quora hindi Image: Google image and answer in hindi Pavan Yeddanapudi Studied at Mountain House High School (Graduated 2024) · Updated 4y Related What is the Chinese remainder theorem? I’ve heard it helps in problem solving but just have no clue what it is. It's basically a way to solve linear congruences. Here, I'll give you an example: x≡1(mod3) x≡6(mod7) In the first equation, we get that x=3y+1 If we substitute this into the second congruence and solve it, 3y+1≡6(mod7) 3y≡5(mod7) 15y≡25(mod7) y≡4(mod7) y=7z+4 We can now substitute this for x. x=3(7z+4)+1 x=21z+13 We can create a congruence for this x≡13(mod21) What’s special about this is that only solutions that satisfy this will satisfy the previous system of linear c It's basically a way to solve linear congruences. Here, I'll give you an example: x≡1(mod3) x≡6(mod7) In the first equation, we get that x=3y+1 If we substitute this into the second congruence and solve it, 3y+1≡6(mod7) 3y≡5(mod7) 15y≡25(mod7) y≡4(mod7) y=7z+4 We can now substitute this for x. x=3(7z+4)+1 x=21z+13 We can create a congruence for this x≡13(mod21) What’s special about this is that only solutions that satisfy this will satisfy the previous system of linear congruence. If you want to go more in-depth get AoPS: Intro to Number Theory. It will go very in-depth in this subject as well as many other number theory related topics. And we're done. Related questions What is the Chinese Remainder Theorem? When is it ever used in real life? How do you find the multiplicative inverse in the Chinese remainder theorem? How would you explain the Chinese Remainder Theorem to a beginner? Why is it called the Chinese Remainder Theorem? What do the Chinese call the "Chinese Remainder Theorem"? What are the Real life examples for fermat's last theorem? How is the Chinese remainder theorem applicable in maths? How can I understand this theorem? What are the exact steps to use the Chinese Remainder Theorem? What is the real life example of pythogarous theorem? How does the Chinese remainder theorem influence your daily living by providing examples? How does the Chinese Remainder Theorem work? What is the smallest x (x==3 mod5, x==5 mod7, x==1 mod9)? I've tried to follow similar problems online, but I never can get the right answer for this one. How do you use the principle of Chinese Remainder Theorem to find the value of x? What are some examples of how to use the remainder theorem in real life situations? What is the real life example for Green's theorem? What tangible, real-world benefits are there behind proving Fermat's last theorem? About · Careers · Privacy · Terms · Contact · Languages · Your Ad Choices · Press · © Quora, Inc. 2025
1486	https://artofproblemsolving.com/wiki/index.php/1982_USAMO_Problems/Problem_1?srsltid=AfmBOorpm7mqYtWSINyttwMex-opcRNO5dlYh8A-VqTq5Pql_d6iWdWM	Art of Problem Solving 1982 USAMO Problems/Problem 1 - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki 1982 USAMO Problems/Problem 1 Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search 1982 USAMO Problems/Problem 1 Problem In a party with people, among any group of four there is at least one person who knows each of the other three. What is the minimum number of people in the party who know everyone else? Solution We induct on to prove that in a party with people, there must be at least people who know everyone else. (Clearly this is achievable by having everyone know everyone else except three people , who do not know each other.) Base case: is obvious. Inductive step: Suppose in a party with people (with ), at least people know everyone else. Consider a party with people. Take of the people (leaving another person, , out) and apply the inductive step to conclude that at least people know everyone else in the -person group, . Now suppose that everyone in the group knows each other. Then take of these people and to deduce that knows a person , which means knows everyone else. Then apply the inductive step on the remaining people (excluding ) to find people out of them that know everyone else (including , of course). Then these people and , which enumerate people, know everyone else. Suppose that there exist two people who do not know each other. Because , there exist at least one person in , person , who knows everyone else in . Now, take and observe that because do not know each other, either or knows everyone else of (by the problem condition), so in particular and know each other. Then apply the inductive step on the remaining people (excluding ) to find people out of them that know everyone else (including , of course). Then these people and , which enumerate people, know everyone else. This completes the inductive step and thus the proof of this stronger result, which easily implies that at least people know everyone else. See Also 1982 USAMO (Problems • Resources) Preceded by First QuestionFollowed by Problem 2 1•2•3•4•5 All USAMO Problems and Solutions These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Retrieved from " Category: Olympiad Combinatorics Problems Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1487	https://thecorestandards.org/Math/Content/2/MD/C/8/	Grade 2 » Measurement & Data » Work with time and money. » 8 \| Common Core State Standards Initiative Video Map Search Filter by: ▾ Common Core State Standards Initiative Common Core State Standards Initiative Home About the Standards Development Process Frequently Asked Questions Myths vs. Facts Branding Guidelines Contact What Parents Should Know Standards in Your State Read the Standards English Language Arts Standards Mathematics Standards Other Resources Key Shifts in English Language Arts Key Shifts in Mathematics Statements of Support Grade 2 » Measurement & Data » Work with time and money. » 8 Print this page Solve word problems involving dollar bills, quarters, dimes, nickels, and pennies, using $ and ¢ symbols appropriately. Example: If you have 2 dimes and 3 pennies, how many cents do you have? Kindergarten-Grade 12 Standards for Mathematical Practice Introduction How to read the grade level standards Kindergarten Introduction Counting & Cardinality Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry Grade 1 Introduction Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry Grade 2 Introduction Operations & Algebraic Thinking Number & Operations in Base Ten Measurement & Data Geometry Grade 3 Introduction Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions¹ Measurement & Data Geometry Grade 4 Introduction Operations & Algebraic Thinking Number & Operations in Base Ten¹ Number & Operations—Fractions¹ Measurement & Data Geometry Grade 5 Introduction Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions Measurement & Data Geometry Grade 6 Introduction Ratios & Proportional Relationships The Number System Expressions & Equations Geometry Statistics & Probability Grade 7 Introduction Ratios & Proportional Relationships The Number System Expressions & Equations Geometry Statistics & Probability Grade 8 Introduction The Number System Expressions & Equations Functions Geometry Statistics & Probability High School: Number and Quantity Introduction The Real Number System Quantities The Complex Number System Vector & Matrix Quantities High School: Algebra Introduction Seeing Structure in Expressions Arithmetic with Polynomials & Rational Expressions Creating Equations Reasoning with Equations & Inequalities High School: Functions Introduction Interpreting Functions Building Functions Linear, Quadratic, & Exponential Models Trigonometric Functions High School: Modeling High School: Geometry Introduction Congruence Similarity, Right Triangles, & Trigonometry Circles Expressing Geometric Properties with Equations Geometric Measurement & Dimension Modeling with Geometry High School: Statistics & Probability Introduction Interpreting Categorical & Quantitative Data Making Inferences & Justifying Conclusions Conditional Probability & the Rules of Probability Using Probability to Make Decisions Note on courses & transitions Courses & Transitions Mathematics Glossary Mathematics Glossary Table 1 Table 2 Table 3 Table 4 Table 5 Standards by Domain Counting & Cardinality Operations & Algebraic Thinking Number & Operations in Base Ten Number & Operations—Fractions Measurement & Data Geometry Ratios & Proportional Relationships The Number System Expressions & Equations Functions Statistics & Probability Mathematics Appendix Mathematics Appendix A ADA Compliant Version Please click here for the ADA Compliant version of the Math Standards. This website is an unofficial resource and is not affiliated with the National Governors Association Center for Best Practices or the Council of Chief State School Officers. The text of the Common Core State Standards are © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers. The Council of Chief State School Officers provides content from their website under the Creative Commons license. The Common Core State Standards are provided on this website under the terms of the Public License for purposes that support the Common Core State Standards Initiative. The phrase "Common Core State Standards" is a trademark of the National Governors Association Center for Best Practices and the Council of Chief State School Officers, and is used in accordance with their Branding Guidelines for purposes that support the Common Core State Standards Initiative. The National Governors Association Center for Best Practices and the Council of Chief State School Officers retain all rights to any of their respective trademarks, service marks, and logos that appear on this site. Contact Terms of Use Public License Developers & Publishers © Copyright 2010. National Governors Association Center for Best Practices and Council of Chief State School Officers Close Close
1488	https://www.expii.com/t/angle-bisector-theorem-5503	Expii Angle Bisector Theorem - Expii The angle bisector theorem states that given any triangle ABC with the foot of the angle bisector of angle A at point D (so AD bisects ∠BAC), two specific ratios are the same: AC : CD = AB : BD. Written another way, DB : DC = AB : AC, meaning D splits side BC in the side ratio AB : AC. Geometry Bisector Properties Angle Bisector Theorem The angle bisector theorem states that given any triangle ABC with the foot of the angle bisector of angle A at point D (so AD bisects ∠BAC), two specific ratios are the same: AC : CD = AB : BD. Written another way, DB : DC = AB : AC, meaning D splits side BC in the side ratio AB : AC. Angle Bisector Theorem Go to Topic Explanations (2) Vivian Loh Text 3 The angle bisector theorem is a formula having to do with a triangle split in half by a line bisecting (halving) one of its interior angles and cutting the opposite side. This formula is the relationship between the line segments created by the angle bisector line and the other two sides. Report Share 3 Like Related Lessons Points on a Perpendicular Bisector Perpendicular Bisectors of a Triangle Meet Angle Bisectors of a Triangle Meet Intro to Scalene, Isosceles, and Equilateral Triangles View All Related Lessons Expii Guest Text 2 Report Share 2 Like You've reached the end TOP How can we improve? General Bug Feature Send Feedback
1489	https://www.chegg.com/homework-help/questions-and-answers/parable-bacteria-bottle-upon-time-precisely-11-00-pm-single-bacterium-placed-nutrient-fill-q73245628	Solved The Parable of the Bacteria in a Bottle': Once upon a \| Chegg.com Skip to main content Books Rent/Buy Read Return Sell Study Tasks Homework help Understand a topic Writing & citations Tools Expert Q&A Math Solver Citations Plagiarism checker Grammar checker Expert proofreading Career For educators Help Sign in Paste Copy Cut Options Upload Image Math Mode ÷ ≤ ≥ o π ∞ ∩ ∪           √  ∫              Math Math Geometry Physics Greek Alphabet Math Algebra Algebra questions and answers The Parable of the Bacteria in a Bottle': Once upon a time, at precisely 11:00 pm, a single bacterium was placed into a nutrient-filled bottle in a laboratory. The bacterium immediately began gobbling up nutrients, and after just one minute - making the time 11:01 - it had grown so much that it divided into two bacteria. These two ate until, one minute Your solution’s ready to go! Our expert help has broken down your problem into an easy-to-learn solution you can count on. See Answer See Answer See Answer done loading Question: The Parable of the Bacteria in a Bottle': Once upon a time, at precisely 11:00 pm, a single bacterium was placed into a nutrient-filled bottle in a laboratory. The bacterium immediately began gobbling up nutrients, and after just one minute - making the time 11:01 - it had grown so much that it divided into two bacteria. These two ate until, one minute Show transcribed image text Here’s the best way to solve it.Solution 100%(1 rating) Share Share Share done loading Copy link Here’s how to approach this question This AI-generated tip is based on Chegg's full solution. Sign up to see more! To model the number of bacteria in the bottle n minutes after 11:00 p.m., observe how the bacteria count doubles every minute and express this relationship mathematically. View the full answer Previous question Transcribed image text: The Parable of the Bacteria in a Bottle': Once upon a time, at precisely 11:00 pm, a single bacterium was placed into a nutrient-filled bottle in a laboratory. The bacterium immediately began gobbling up nutrients, and after just one minute - making the time 11:01 - it had grown so much that it divided into two bacteria. These two ate until, one minute later, they each divided into two bacteria, so that there were a total of four bacteria in the bottle at 11:02. The four bacteria grew and divided into a total of eight bacteria at 11:03, sixteen bacteria at 11:04, and so on. All seemed fine, and the bacteria kept on eating happily and doubling their number every minute, until the "midnight catastrophe." The catastrophe was this: At the stroke of 12:00 midnight, the bottle became completely full of bacteria, with no nutrients remaining - which meant that every single one of the bacteria was suddenly doomed to death. Adapted from Jeffrey Bennett's excellent book, Math for Life. a. Find an equation to model the number of bacteria in the bottlen minutes after 11:00p.m. b. In this situation, the quantity of bacteria is growing by what relative amount (percentage) each minute? c. This type of equation is called an equation and it models growth d. How many bacteria were in the bottle at 12:00 a.m.? e. Graph the two equations on the same screen in your graphing calculators using the given windows. On the screen above each window dimension, draw a representation of the graphs as shown on your calculator. Label the equations. Number of Bacteria Number of Bacteria Window [0,4,5) x[-1,8, 1) Window (0, 10, 5] x [-1, 100, 10] f. The graphs reveal that the equation grew at a much faster rate. We now turn to our questions, as we seek to draw lessons from the tragic dernise of the bacterial colony. Not the question you’re looking for? Post any question and get expert help quickly. Start learning Chegg Products & Services Chegg Study Help Citation Generator Grammar Checker Math Solver Mobile Apps Plagiarism Checker Chegg Perks Company Company About Chegg Chegg For Good Advertise with us Investor Relations Jobs Join Our Affiliate Program Media Center Chegg Network Chegg Network Busuu Citation Machine EasyBib Mathway Customer Service Customer Service Give Us Feedback Customer Service Manage Subscription Educators Educators Academic Integrity Honor Shield Institute of Digital Learning © 2003-2025 Chegg Inc. All rights reserved. Cookie NoticeYour Privacy ChoicesDo Not Sell My Personal InformationGeneral PoliciesPrivacy PolicyHonor CodeIP Rights Do Not Sell My Personal Information When you visit our website, we store cookies on your browser to collect information. 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1490	https://artofproblemsolving.com/wiki/index.php/Factoring?srsltid=AfmBOoqxb1wmdz09IWdnIIh-POkZiQwNW-4fPALIWPV5b8ThJ7rVUNu1	Art of Problem Solving Factoring - AoPS Wiki Art of Problem Solving AoPS Online Math texts, online classes, and more for students in grades 5-12. Visit AoPS Online ‚ Books for Grades 5-12Online Courses Beast Academy Engaging math books and online learning for students ages 6-13. Visit Beast Academy ‚ Books for Ages 6-13Beast Academy Online AoPS Academy Small live classes for advanced math and language arts learners in grades 2-12. Visit AoPS Academy ‚ Find a Physical CampusVisit the Virtual Campus Sign In Register online school Class ScheduleRecommendationsOlympiad CoursesFree Sessions books tore AoPS CurriculumBeast AcademyOnline BooksRecommendationsOther Books & GearAll ProductsGift Certificates community ForumsContestsSearchHelp resources math training & toolsAlcumusVideosFor the Win!MATHCOUNTS TrainerAoPS Practice ContestsAoPS WikiLaTeX TeXeRMIT PRIMES/CrowdMathKeep LearningAll Ten contests on aopsPractice Math ContestsUSABO newsAoPS BlogWebinars view all 0 Sign In Register AoPS Wiki ResourcesAops Wiki Factoring Page ArticleDiscussionView sourceHistory Toolbox Recent changesRandom pageHelpWhat links hereSpecial pages Search Factoring Factoring is an essential part of any mathematical toolbox. To factor, or to break an expression into factors, is to write the expression (often an integer or polynomial) as a product of different terms. This often allows one to find information about an expression that was not otherwise obvious. Contents [hide] 1 Differences and Sums of Powers 2 Vieta's/Newton Factorizations 3 Other Useful Factorizations 4 Practice Problems 5 Other Resources Differences and Sums of Powers Using the formula for the sum of a geometric sequence, it's easy to derive the general formula for difference of powers: If , this creates the difference of squares factorization, This leads to the difference of cubes factorization, In addition, if is odd: This also leads to the formula for the sum of cubes, Another way to discover these factorizations is the following: the expression is equal to zero if . If one factorizes a product which is equal to zero, one of the factors must be equal to zero, so must have a factor of . Similarly, we note that the expression when is odd is equal to zero if , so it must have a factor of . Note that when is even, , rather than 0, so this gives us no useful information. Vieta's/Newton Factorizations These factorizations are useful for problems that could otherwise be solved by Newton's sums or problems that give a polynomial and ask a question about the roots. Combined with Vieta's formulas, these are excellent factorizations that show up everywhere. Other Useful Factorizations Binomial theorem Simon's Favorite Factoring Trick Sophie Germain Identity Factor Theorem Practice Problems Prove that is never divisible by 121 for any positive integer . Prove that is divisible by 7. - USSR Problem Book Factor . Factor into two polynomials with real coefficients. Given that , prove that . Other Resources More Common Factorizations. Retrieved from " Categories: Algebra Definition Art of Problem Solving is an ACS WASC Accredited School aops programs AoPS Online Beast Academy AoPS Academy About About AoPS Our Team Our History Jobs AoPS Blog Site Info Terms Privacy Contact Us follow us Subscribe for news and updates © 2025 AoPS Incorporated © 2025 Art of Problem Solving About Us•Contact Us•Terms•Privacy Copyright © 2025 Art of Problem Solving Something appears to not have loaded correctly. Click to refresh.
1491	https://my.clevelandclinic.org/health/diseases/9129-pelvic-inflammatory-disease-pid	Abu Dhabi\|Canada\|Florida\|London\|Nevada\|Ohio\| Home/ Health Library/ Diseases & Conditions/ Pelvic Inflammatory Disease AdvertisementAdvertisement Pelvic Inflammatory Disease Pelvic inflammatory disease (PID) is an infection that occurs in your uterus, fallopian tubes or ovaries. Sexually transmitted infections typically cause it. Symptoms include stomach, lower abdominal pain and vaginal discharge. Prompt PID treatment, usually antibiotics, helps avoid complications such as infertility. Your partner should get tested and treated, too. ContentsOverviewSymptoms and CausesDiagnosis and TestsManagement and TreatmentOutlook / PrognosisPreventionLiving WithAdditional Common Questions Overview What is pelvic inflammatory disease (PID)? Pelvic inflammatory disease (PID) is a serious infection in your uterus, fallopian tubes and/or ovaries. It affects females. PID develops when certain types of bacteria spread from your vagina to your reproductive organs. Bacteria from untreated sexually transmitted infections (STIs) are the most common cause of PID. However, bacteria normally found in your vagina can also cause PID. Advertisement Cleveland Clinic is a non-profit academic medical center. Advertising on our site helps support our mission. We do not endorse non-Cleveland Clinic products or services. Policy When you have PID, you may feel pain in your lower abdomen (belly) or pelvis. You may also have unusual discharge (leaking) from your vagina. Severe PID can cause permanent damage your reproductive organs and prevent you from getting pregnant. Getting tested and treated for STIs is the best way to prevent PID. How do you get PID? Most people get PID through unprotected sex. Sex lets bacteria enter your reproductive system, where they can infect your organs. How common is pelvic inflammatory disease? Each year, more than 1 million females in the U.S. get PID. And more than 100,000 people become infertile because of it. PID occurs most frequently in females between 15 and 25 years old. Symptoms and Causes What are the most common signs of PID? You may not realize you have PID. Symptoms might be mild or unnoticeable. But symptoms of PID can also start suddenly. They can include: Pain or tenderness in your stomach or lower abdomen (belly), the most common symptom. Abnormal vaginal discharge, usually yellow or green with an unusual odor. Chills or fever. Nausea and vomiting. Pain during sex. Burning when you pee. Irregular periods or having spotting or cramping throughout the month. Where do you feel PID pain? Pelvic inflammatory disease pain is mainly felt in your lower abdomen or pelvic region. It may feel tender and sore or like a dull ache. You may also feel pain deep in your pelvis during sex. Advertisement What causes pelvic inflammatory disease (PID)? Bacteria entering your reproductive tract causes pelvic inflammatory disease. These bacteria are passed from your vagina, through your cervix and up into your uterus, fallopian tubes and ovaries. Normally, when bacteria enter your vagina, your cervix keeps them from spreading deeper to other reproductive organs. However, any type of infection can disrupt your cervix, preventing it from doing its job. Many types of bacteria can cause PID, but the two most common infections that cause PID are gonorrhea and chlamydia. You get both of these infections through unprotected sex. These two STIs cause about 90% of all PID cases. Less commonly, PID happens when normal bacteria gets into your reproductive organs. This can happen after: Childbirth. Pelvic surgery. Miscarriage. Getting an intrauterine device (IUD). The risk is highest in the few weeks after your provider inserts an IUD. How long does it take to have symptoms of PID? It can take anywhere from a few days to a few weeks to develop PID if untreated gonorrhea or chlamydia is the cause. If you get PID from something else, it may take several months to develop it. Does douching cause pelvic inflammatory disease (PID)? Most studies report only an association between douching and PID. What can be said is that douching can lead to bacterial vaginosis infections, but there’s only a potential association between douching and PID. Most healthcare providers advise against douching. Is PID contagious? Yes, PID spreads most often during direct sexual contact. Who’s at risk for PID? You’re at higher risk for pelvic inflammatory disease if you: Have a sexually transmitted infection (STI), especially gonorrhea or chlamydia. Have many sexual partners or have a partner who has had multiple partners. Have had PID in the past. Are sexually active and younger than 25. Have had tubal ligation or other pelvic surgery. Are there complications of PID? The longer you have PID, the more serious its effects can become. The infection can cause scar tissue to form inside your fallopian tubes. The scarring can lead to several problems, including: Chronic pelvic pain: Long-lasting pelvic pain is the most common complication. One study estimates about 20% of people develop chronic pelvic pain. Ectopic pregnancy: Scarring can prevent a fertilized egg from moving into your uterus. This can cause it to implant inside your fallopian tubes instead. The rate of ectopic pregnancy in people with PID is much higher than in people without PID. Infertility: Up to 10% of people with PID lose the ability to get pregnant because scar tissue blocks their fallopian tubes and prevents them from releasing an egg. Tubo-ovarian abscess (TOA): TOA is a pocket of infection in your pelvis that can make you extremely sick. Advertisement Diagnosis and Tests How is pelvic inflammatory disease diagnosed? If you feel symptoms of PID, see your healthcare provider right away. The sooner you get care, the greater your chances of successful treatment. There isn’t one specific test for PID. Usually, your healthcare provider can diagnose PID through: Asking about your medical history, including your general health, sexual activity and symptoms. A pelvic exam to examine your reproductive organs and feel for tenderness or abscesses (collections of pus). A vaginal culture to test your vaginal discharge for certain bacteria. What other tests might I need to diagnose PID? Your provider may also order: Blood tests. Urine test to rule out a urinary tract infection, which causes similar pelvic pain. Ultrasound to get clearer images of your reproductive system. Testing for STIs. In some cases, your provider may recommend: Endometrial biopsy: Your provider removes a small tissue sample from your uterine lining and tests it for diseases. Laparoscopy: Your provider makes small incisions in your pelvis, then inserts a lighted instrument to look more closely at your reproductive organs. Management and Treatment How is pelvic inflammatory disease (PID) treated? Your provider will prescribe antibiotics that you take by mouth, typically for 14 days. Make sure to take all your medicine, even if you start feeling better. Often, your symptoms improve before the infection goes away. Your provider may recommend you return a few days after starting the medicine. They can check that treatment is working. Advertisement Some people take antibiotics and still have symptoms. If that happens, you may need to go to the hospital to receive antibiotics through an IV. You may also need IV medication if you: Are pregnant. Have a severe infection and feel very sick. Have an abscess (collection of pus) in your fallopian tube or ovary. You shouldn’t have sex until you finish treatment. When you do have sex again, use condoms every time to prevent infections. Will I need surgery for pelvic inflammatory disease? Surgery is rare for PID but can help in some cases. If you still have symptoms or an abscess after taking antibiotics, talk to your healthcare provider about surgery. Does my partner need treatment for pelvic inflammatory disease? If you have pelvic inflammatory disease, tell your sexual partner(s). They should receive treatment. Otherwise, you may get PID again when you resume sex. Care at Cleveland Clinic Find Your Ob/Gyn Make an Appointment Outlook / Prognosis Is there a cure for pelvic inflammatory disease? If you get prompt diagnosis and treatment for an infection, antibiotics can cure PID. But treatment can’t reverse any damage that already happened to your reproductive organs. Don’t wait to get treated. See your provider right away so you can get the help you need. If I had pelvic inflammatory disease, will I have trouble getting pregnant? PID can affect fertility. Of the people who had PID, studies found that 1 in 8 had difficulty getting pregnant. Up to 1 in 10 people ultimately received a diagnosis of infertility. People who had repeat infections had a harder time getting pregnant. Advertisement How does PID affect fertility? Bacteria from PID can cause scarring on your fallopian tubes. This scar tissue makes it harder for an egg to get from your ovary to your fallopian tube, then down to your uterus. If an egg can’t get through your fallopian tube, sperm can’t fertilize it. Can I get pelvic inflammatory disease again? Yes, you can get PID again. Getting PID once doesn’t protect you from getting it again. If I had PID, when can I resume having sex? You and your partner should wait a week after finishing your antibiotics before resuming sex. Doing so will help prevent re-infection. Can women who have sex with women get PID? Yes, women who have sex with women (WSW) can get PID. This is because bacteria that cause PID can live on a vagina, penis or any object you place in your vagina. Prevention Can I prevent pelvic inflammatory disease? Sometimes, PID isn’t due to a sexually transmitted infection. It can come from normal vaginal bacteria traveling to your reproductive organs. Avoiding douching may lower the risk. Most of the time, though, PID happens because of unprotected sex. Take steps to practice safe sex. Ways to protect yourself from sexually transmitted infections (STIs) that can cause PID include: Limiting sexual partners: Your risk increases if you have multiple partners. Choosing barrier methods of birth control: These types of birth control include condoms and diaphragms. Combine a barrier method with spermicide, even if you take birth control pills. Seeking treatment if you notice symptoms: If you notice signs of PID or other STIs, get treatment right away. Symptoms include unusual vaginal discharge, pelvic pain or bleeding between periods. Getting regular checkups: Have regular gynecological exams and screenings. Often, providers can identify and treat cervical infections before they spread to reproductive organs. How can I lower my risk for PID? If you’re sexually active, talk to your healthcare provider about yearly testing for sexually transmitted infections. Providers often recommend testing for chlamydia and gonorrhea to help keep you safe. Also, before having sex with a new partner, it’s a good idea for both of you to get tested for STIs. Living With How can I take care of myself if I have pelvic inflammatory disease? If you feel symptoms of PID, see your healthcare provider right away. If you have PID, the most important thing you can do is get treatment. Other tips for taking care of yourself include: Avoid douching to prevent pushing bacteria upward from your vagina into your uterus and fallopian tubes. Return to your healthcare provider a few days after starting medication to make sure it’s working. Take all your medicine as directed. Use condoms or dental dams every time you have sex to protect yourself from infections. Wait one week after you (and your partner) have finished medication to resume your sex life. When should I see my healthcare provider? See your provider if you experience any symptoms of PID. Seek immediate medical care if you have: Intense lower belly pain. Smelly or discolored vaginal discharge. Severe vomiting. High fever. What questions should I ask my provider? If you have PID, ask your provider: What treatment will I need? Do I need to get rechecked? Will PID affect my ability to get pregnant? What are the possible complications of PID? When can I resume having sex? What can I do to prevent PID? Additional Common Questions Can you get PID without having an STD? Yes, it’s possible to get PID when the bacteria that normally exist in your vagina travel up into your reproductive organs. This is usually not the cause, though. Is pelvic inflammatory disease serious? Yes, PID can be serious. Early diagnosis and treatment are essential to reducing your risk of long-term complications like infertility. A note from Cleveland Clinic Pelvic inflammatory disease (PID) is an infection of your reproductive organs. A sexually transmitted infection, such as gonorrhea or chlamydia, typically causes it. If you notice symptoms of PID, such as pain in your lower abdomen, talk to your healthcare provider. The provider can diagnose PID and give you antibiotics to treat it. Early treatment is key to avoiding complications of PID such as infertility. Your partner(s) should get treated, as well. You can prevent PID by using a condom every time you have sex. Care at Cleveland Clinic Your Ob/Gyn is there for you throughout many different parts of your life. At Cleveland Clinic, our experts can help you throughout your entire journey. Find Your Ob/Gyn Make an Appointment Medically Reviewed Last reviewed on 02/08/2023. 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1492	https://www.sciencedirect.com/science/article/abs/pii/S0079661122000192	Spatiotemporal distribution of paralytic shellfish poisoning (PSP) toxins in Beagle Channel (South America) during 2005–2017 - ScienceDirect Skip to main contentSkip to article Journals & Books Access throughyour organization Purchase PDF Search ScienceDirect Article preview Abstract Introduction Section snippets References (88) Cited by (8) Progress in Oceanography Volume 204, June 2022, 102757 Spatiotemporal distribution of paralytic shellfish poisoning (PSP) toxins in Beagle Channel (South America) during 2005–2017 Author links open overlay panel A.M.Cadaillon a, G.O.Almandoz b c, M.P.Hernando d e, L.Saravia f a, S.Maldonado g, I.R.Schloss a h i Show more Add to Mendeley Share Cite rights and content Highlights •PSP outbreaks showed a high spatial variability in a small geographical scale. •Mussels from eastern Beagle Channel had relatively low PSP detoxification rates. •Magellan mussels presented higher PSP concentrations than blue mussels. •Higher PSP toxicity in the mussel culture areas were associated with aquaculture. •PSP outbreaks were related to hydrological-meteorological conditions. Abstract Toxigenic algal blooms have been reported since 1886 in the Beagle Channel, at the southern limit of South America, mainly associated with the occurrence of paralytic shellfish poisoning (PSP) toxins produced by dinoflagellates of the genus Alexandrium. Toxic outbreaks have serious impacts on marine wildlife and ecosystem services, particularly on fisheries and mussel aquaculture. Within the framework of the Shellfish Toxicity Monitoring Program carried out close to Almanza in Beagle Channel eastern coastal waters, here we analyzed the spatiotemporal patterns of PSP toxins in the Magellan mussel Aulacomya ater and the blue mussel Mytilus edulis, measured by the mouse bioassay for 12 years, between 2005 and 2017. PSP outbreaks occurred mostly during summer and their duration was significantly longer than for outbreaks detected during spring and winter. A marked interannual variability in their magnitude was also evident. The largest outbreaks were detected in summers 2010–2011, in extreme warm and low-wind conditions, reaching up to 5600 µg saxitoxin equivalent per 100 g of tissue, causing a closure of shellfish fisheries of up to 200 days. Small scale spatial variability in PSP was also found, the highest values being observed in the main mussel culture areas, and the lowest in the natural mussel beds. PSP toxicities for the Magellan mussel were significantly higher than for the blue mussel. Natural detoxification in mussels from Beagle Channel followed an exponential decay, with mean detoxification rates of 3.5% toxin day−1, lower than those from elsewhere in the world for similar PSP toxin concentrations, which means a longer time to attain concentration limits safe for human consumption. Understanding PSP dynamics is important to avoid large economic losses related to halting mussel production. Introduction Harmful algal blooms (HABs) represent a serious problem in the world oceans, with adverse effects not only for human health, shellfish aquaculture and coastal tourism activities, but also for natural populations of marine organisms (Shumway et al., 2003, Anderson et al., 2012, Broadwater et al., 2018). Phycotoxins produced by microalgae include a wide spectrum of substances of different chemical nature, action mechanism and biological activity. In particular, paralytic shellfish poisoning (PSP) toxins, i.e., saxitoxin (STX) and analogues are highly potent neurotoxins, produced by dinoflagellates of the genera Alexandrium, Gymnodinium, and Pyrodinium (Basti et al., 2018). In the Beagle Channel, a Subantarctic semi-estuarine environment located at the southern limit of South America, toxigenic algal blooms have been reported since 1886, when the native population near Ushuaia (Argentina) was severely intoxicated, and some people died after the ingestion of bivalves (Segers, 1908). The Beagle Channel was thereafter considered free of phycotoxins until 1992, favouring the commercial harvesting of molluscs with harvest volumes exceeding coastal fish catches (Goya and Maldonado, 2014). Then, in summer 1992 a major PSP outbreak occurred, in which PSP toxicity reached the highest values ever recorded worldwide, 127 000 µg saxitoxin equivalent per 100 g of tissue (µg STX eq 100 g−1 tissue), 160x the allowable regulatory level (RL) of 80 µg STX eq 100 g−1 tissue (Benavides et al., 1995). This exceptional PSP event, associated with a bloom of the dinoflagellate Alexandrium catenella, had serious health consequences, with several cases of human intoxication (some of them fatal), and detrimental impacts on marine wildlife and ecosystem services (Carreto and Benavides, 1993, Vinuesa, 1993). From that moment on and up to 2011, detectable PSP toxin levels have been recorded repeatedly in molluscs, often exceeding the safety limit for human consumption for extended periods mostly during summer seasons (Goya and Maldonado, 2014). A. catenella is considered the principal PSP toxin-producing dinoflagellate in the Beagle Channel (Carreto et al., 1985, Carreto et al., 1986, Carreto et al., 1990; Almandoz et al., 2011, Almandoz et al., 2019, Benavides et al., 2019), but little is known about the intensity, location, and duration of its blooms. The first mussel (Mytilus edulis) aquaculture initiatives in the Beagle Channel date from the early 1990s, with an increasing production up to the 2010s (Goya and Maldonado, 2014), due to the presence of natural high-quality beds, which can be easily accessed from intertidal and subtidal zones (Goya and Maldonado, 2014). In addition, artisanal fishing of the mussel Aulacomya ater is also important in the area. Because of the toxic blooms, bivalve production is sometimes interrupted for several months, hindering production and harvest for long periods (Álvarez, 2012, Bertolotti et al., 2014, Benavides et al., 2019). Due to an unusual persistence of high toxicity in mussels during summers 2010 and 2011, many areas remained closed even until winter, leading to significant economic losses for producers (Goya and Maldonado, 2014). Although aquaculture activities in the Beagle Channel are presently limited, a new and important investment in aquaculture is at its early stages during these days, so that it could once again become an intense zone of mussel production in Argentina in a near future. This warns about the need to continue with the monitoring efforts of HABs in this region (Benavides et al., 2019). The great interannual variability in the intensity and location of PSP outbreaks in the Beagle Channel highlights the importance to study the spatiotemporal patterns of these toxins as well as to describe their dynamics over time. Here we analyse the dynamics of 12 years PSP outbreaks in the Beagle Channel, based on data of the presence, spatial distribution, and temporal patterns, as well as peak evolution of PSP toxins in two mussel species (A. ater and M. edulis) during the period 2005–2017. Access through your organization Check access to the full text by signing in through your organization. Access through your organization Section snippets Methods The data for the present study were obtained in the frame of the Shellfish Toxicity Monitoring Program carried out in Beagle Channel coastal waters since 1985 by the Dirección de Laboratorio Ambiental (Environmental Laboratory of the Fisheries and Aquaculture Secretary of the Tierra del Fuego Province). At that time, samples were analyzed by the Regional Laboratory of SENASA, in Mar del Plata. From 1995 to 1998, the official laboratories at Ushuaia started to perform PSP analysis in parallel Meteorological characteristics Yearly average air temperature for the period 2006–2017 was 6.41±3.50°C. Temperatures show a typical seasonality. Both the highest and the lowest individual records (18.3°C and −5.6°C, respectively) were registered during 2015, in summer 2015 (2015–01-22) and in winter (2015–07-07), respectively. During the entire period of the study, average winds (59.76±21.41 km h−1) were predominantly from the SW, exhibiting a seasonal pattern with maximum values in spring - summer and minimum in Discussion The present study shows that the concentrations of PSP toxins in shellfish differs among seasons and in a short spatial scale, as shown for the four studied sites in the eastern Beagle Channel coast over 12 years, also varying between natural and aquaculture areas and species considered. Conclusion The present study shows for the first time the high temporal variability in PSP toxins in two shellfish species, for 12 years and among seasons, which is detectable even in a small geographic scale in the Subantarctic coastal waters of the Beagle Channel. The most intense PSP outbreaks were characterized by long detoxification periods. The low mean detoxification rate (3.5% toxin day−1) as compared to other places in the world could be probably explained by a lower metabolism in shellfish due Declaration of Competing Interest The authors declare that they have no known competing financial interests or personal relationships that could have appeared to influence the work reported in this paper. Acknowledgements We are grateful to the Dirección General de Laboratorio de Toxinas y Microbiología (Secretaría de Pesca y Acuicultura de la Provincia de Tierra del Fuego, Antártida e Islas del Atlántico Sur) for the data provided for the present study, especially to Juan Fosati for mussels sampling. We also thank the Secretary of Fisheries and Aquaculture (Secretaría de Pesca y Acuicultura), Mr. Carlos Cantú, Sub-Secretary of Fisheries and Aquiculture Mr. Diego Marzioni, the Director of Artisanal Fishery, Mr. Funding This work has received funding from the European Union’s Horizon 2020 research and innovation programme under grant agreement No 863034 ASTRAL – All Atlantic Ocean Sustainable, Profitable and Resilient Aquaculture – Recommended articles References (88) E.M. Acha et al. Marine fronts at the continental shelves of austral South America: physical and ecological processes J. Mar. Syst. (2004) G.O. Almandoz et al. Seasonal phytoplankton dynamics in extreme southern South America J. Sea Res. (2011) G.O. Almandoz et al. Toxic strains of the Alexandrium ostenfeldii complex in southern South America (Beagle Channel, Argentina) Harmful Algae (2014) G.O. Almandoz et al. 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In particular, paralytic shellfish toxins (PSTs) produced by marine dinoflagellates of the genera Alexandrium, Gymnodinium and Pyrodinium, include >50 related tetrahydropurine alkaloids (Cruz et al., 2018). Show abstract Gonyautoxins (GTXs), a group of potent neurotoxins belonging to paralytic shellfish toxins (PSTs), are often associated with harmful algal blooms of toxic dinoflagellates in the sea and represent serious health and ecological concerns worldwide. In the study, a highly selective and sensitive fluorescence nanoprobe was constructed based on photoinduced electron transfer recognition mechanism to rapidly detect GTXs in seawater, using specific entrapment of molecularly imprinted polymers (MIPs) combined with fluorescence analyses. The green emissive fluorescein isothiocyanate was grafted in a silicate matrix as a signal transducer and fluorescence intensity of the nanoprobe with a core-shell structure exhibited a strong enhancement due to efficient analyte blockage in a short response time. Under optimal conditions, the developed MIPs nanoprobe presented an excellent analytical performance for spiked seawater samples including a recovery from 94.44% to 98.23%, a linear range between 0.018 nmol L−1 and 0.36 nmol L−1, as well as good accuracy. Furthermore, the method had extremely high sensitivity, with limit of detection obtained as 0.005 nmol L−1 for GTXs and GTX2/3. Finally, the nanoprobe was applied for the determination of GTXs in seven natural seawater samples with GTXs mixture (0.035–0.058 nmol L−1) or single GTX2/3 (0.033–0.050 nmol L−1), and the results agreed well with those of a UPLC-MS/MS method. The findings of our study suggest that the constructed MIPs-based fluorescence enhancement nanoprobe was suitable for rapid, selective and ultrasensitive detection of GTXs, particular GTX2/3, in natural seawater samples. ### Multispecies mass mortality in the Beagle Channel associated with paralytic shellfish toxins 2024, Harmful Algae Citation Excerpt : These findings clearly demonstrate bivalves' role as filter feeders and concentrators of toxins, which is why they are considered sentinel organisms in monitoring marine environments (Bricelj and Shumway, 1998). They further highlight the magnitude of this toxic event, which is four times higher than the highest recorded in the last 12 years (Cadaillon et al., 2022). The differences observed between the results obtained through HPLC and MBA methods may be attributed to inherent errors in the bioassay itself, i.e., its low sensitivity and specificity (Fernández et al., 2002). Show abstract The Beagle Channel is a Subantarctic semi-estuarine environment at the southern tip of South America, where intoxication events associated with harmful algal blooms have been reported since 1886, including a world record in toxicity due to Alexandrium catenella in 1992. Toxic algae affect public health and ecosystem services, particularly mussel aquaculture and fisheries management. During the austral summer of 2022, an intense bloom of A. catenella (5×10 4 cells L−1) occurred in the Beagle Channel, leading to the second most toxic event in the area, with mussel toxicity reaching 197,266 µg STXeq kg−1. This event was synchronous with the mortality of marine organisms from different trophic levels and terrestrial fauna, i.e., two Fuegian red foxes and a southern caracara. Stomach content and liver samples from dead kelp gulls (Larus dominicanus), Magellanic penguins (Spheniscus magellanicus), papua penguins (Pygoscelis papua), and imperial cormorants (Leucocarbo atriceps), presented variable paralytic shellfish toxins (PST) levels (up to 3427 µg STXeq kg−1) as measured by high performance liquid chromatography (HPLC), suggesting that deaths were associated with high PST toxicity level. The different toxin profiles found in phytoplankton, zooplankton, squat lobsters (Grimothea gregaria), Fuegian sprat (Sprattus fuegensis), and seabirds evidenced possible toxin transformation along the food web and the possible transfer vectors. The unexpected detection of PST in terrestrial fauna (up to 2707 µg STXeq kg−1) suggested intoxication by scavenging on squat lobsters, which had high toxicity (26,663 µg STXeq kg−1). PST trace levels were also detected in a liver sample of a dead false killer whale (Pseudorca crassidens), an oceanic odontocete stranded on the coast during the bloom. Overall, our results denote the exceptional nature of the toxic, multispecies mortality event and that toxins may propagate to several levels of the food web in this Subantarctic environment. ### Harmful algal blooms in Patagonian fjords and channels systems: Recent advances, gaps, and priorities in a changing ocean 2023, Progress in Oceanography ### Alexandrium catenella dynamics and paralytic shellfish toxins distribution along the Beagle Channel (southern Patagonia) 2023, Journal of Marine Systems Citation Excerpt : Such a production could later in the season fuel mixotrophic and heterotrophic dinoflagellates growth. Moreover, on the northern coast of the central part of the channel, some impact of the salmon farms industry might have favored the development of toxic outbreaks via additional highly labile DOM, further explaining the recurrent toxic events in the area (Cadaillon et al., 2022). All the considerations addressed in the previous paragraphs are synthesized in Fig. 9. Show abstract The Beagle Channel at the southernmost tip of South America is an interoceanic passage connecting the Pacific and the Atlantic Oceans. It is characterized by intricate coastlines and changing bathymetry, a dominant West to East circulation pattern, and a strong longitudinal gradient of glacial fresh-water discharge. Harmful algal blooms (HAB) and toxic outbreaks have been detected along the channel for the last two decades and monitored by both Chilean and Argentinean agencies. This unique scenario was used to try to answer whether HABs propagate from West to East along the channel following the main water flow, so that a sequential, spatial pattern can be identified, or if local dynamics due to particular hydrographic characteristics might favor HAB formation in different areas along the channel. For this analysis, we selected data from three austral spring-summer seasons, 2009–2010, 2010–2011, and 2012–2013, when by means of the mouse bioassay important concentrations of saxitoxins (STX) derivatives were detected in shellfish from different stations along the channel. Relevant information on hydrographical (temperature, salinity, bathymetry, main currents), chemical (macronutrients), and biological (cell abundance of the main species responsible for the toxic outbreaks in the channel, i.e. Alexandrium catenella) characteristics are analyzed. Results show that during the years when toxicity was highest, there was no evidence of West to East (longitudinal) transport of toxins along the channel. Contrastingly, smaller-scale patterns could explain the observed dynamics associated with three sub-basins previously described for the Beagle Channel, identified as western, central, and eastern regions. We built a conceptual model based on own and published data on bathymetry, PAR, salinity, water residence time, silicate availability, and DOM to understand the differences in HAB presence among the sub-basins, which allows explaining the higher toxicity values registered in the central part of the channel as compared to the western or easternmost regions. ### New insights into the dynamics of causative dinoflagellates and the related contamination of molluscs by paralytic toxins in the southwestern Mediterranean coastal waters of Morocco 2022, Marine Pollution Bulletin Citation Excerpt : They affect mammalian cells by disabling the voltage-gated sodium channels, thereby generating multiple symptoms such as nausea, diarrhoea, vomiting, tingling sensation of the mouth, lips and tongue, numbness of the extremities, headache and, in extreme cases, death by asphyxia (Farabegoli et al., 2018). Paralytic shellfish toxins are produced by numerous dinoflagellates belonging to the genera Gymnodinium, Alexandrium and Pyrodinium (Costa et al., 2009; Cadaillon et al., 2022). The main species confirmed to be involved in PSP syndrome are Gymnodinium catenatum, Alexandrium pacificum, Alexandrium catenella, Alexandrium minutum and Pyrodinium bahamense (Beppu et al., 2008; Valbi et al., 2019); they are distributed globally (Lassus et al., 2016). Show abstract The distribution of the two potentially toxic dinoflagellates Gymnodinium catenatum and Alexandrium spp. was investigated in the Mediterranean Moroccan Sea from March 2018 to March 2019. The cockle Acanthocardia tuberculata and the smooth clam Callista chione were collected at four stations, and their toxin levels were assessed using the mouse bioassay. The toxin profile was analysed by LC-MS/MS in G. catenatum and in the bivalves harvested in M'diq and Djawn. The species G. catenatum was present throughout the year, whereas Alexandrium spp. was less abundant. The paralytic shellfish toxin (PST) level in cockles was, on average, six times above the sanitary threshold; GTX5 was the major contributor to the total PST level, followed by dc-STX and STX. The toxin level of the smooth clam was considerably lower than that of the cockle; GTX5 and C-toxins were the dominating analogues. Our results suggest the responsibility of G. catenatum for the recurrent PST contamination in the Moroccan Mediterranean Sea, with a west-east gradient. ### Increase in records of toxic phytoplankton and associated toxins in water samples in the Patagonian Shelf (Argentina) over 40 years of field surveys 2022, Harmful Algae Citation Excerpt : This excludes ichthyotoxic phytoplankton species that do not produce shellfish toxins (e.g. Chattonella marina), benthic species, cyanobacteria, and high biomass phytoplankton blooms that lead to hypoxia. In addition, phycotoxins measured in animal tissues like shellfish or marine mammals (e.g. D'Agostino et al., 2015; Goya et al., 2020; Cadaillón et al., 2022) were not considered in this assessment, because sampling and data interpretation are not comparative with algae/toxins concentration in water samples . A total of 112 downloaded articles were filtered for metadata collection (Fig. 2), keeping only those that were specifically focused on potentially toxic phytoplankton in water samples –collected with bottle, hose and/or with phytoplankton net (e.g. Fabro et al., 2017; Guinder et al., 2018; Tillmann et al., 2019)–, and those based on general phytoplankton community composition but with specific quantification of HAB species (e.g. Esteves et al., 1997; Antacli et al., 2018; Santinelli et al., 2018). Show abstract Historical records (1980–2018) of potentially toxic phytoplankton and phycotoxins on the Argentine Continental Shelf (35°S-56.5°S) and adjacent ocean waters were systematically reviewed from scientific literature to assess their abundance and diversity over spatial and temporal scales. Records increased from 124 in the period 1980–1992 to 638 in 2006–2018, and the scanned area expanded from coastal to offshore waters including the shelf-break front. Alexandrium was the most reported genus (54%) during 1980–1992 and Pseudo-nitzschia (52%) during 1993–2005. By 2006–2018, a higher diversity was documented: Alexandrium (20%), Dinophysis (32%), Pseudo-nitzschia (31%), and the most recently described potentially toxic dinoflagellates of the family Amphidomataceae (8%). Likewise, a wider spectrum of phycotoxins was documented in the last decade, with lipophilic (LSTs) and paralytic shellfish toxins (PSTs) as the most recorded. Increased records are related to intensified monitoring, more detailed taxonomic analyses and more sensitive chemical techniques for marine biotoxin detection. This quantitative assessment brings light to the widespread occurrence of HABs along contrasting areas of the Patagonian Shelf and sets the basis for ecosystem risk evaluation. Moreover, comparison of toxic phytoplankton reported in the SW Atlantic with those in similar temperate seas in the North Atlantic and the Pacific Ocean, disclose ocean basin differences in strain toxicity of A. ostenfeldii, D. tripos and Azadinium species. View all citing articles on Scopus View full text © 2022 Elsevier Ltd. All rights reserved. Recommended articles Prevalence of phycotoxin contamination in shellfish from the Northern Bering Sea and the Chukchi Sea Toxicon, Volume 167, 2019, pp. 76-81 Chunlei Gao, …, Chenlin Liu ### Determination of multiple toxins in whelk and clam samples collected from the Chukchi and Bering seas Toxicon, Volume 109, 2016, pp. 84-93 Aifeng Li, …, Haifeng Gu ### Toxicity equivalence factors for regulated and non-regulated marine toxins Current Opinion in Food Science, Volume 18, 2017, pp. 64-70 Maria Carmen Louzao, …, Natalia Vilariño ### Biogeography of resistance to paralytic shellfish toxins in softshell clam, Mya arenaria (L.), populations along the Atlantic coast of North America Aquatic Toxicology, Volume 202, 2018, pp. 196-206 Jennifer M.Phillips, …, Laurie B.Connell ### Paralytic shellfish toxin uptake, tissue distribution, and depuration in the Southern Rock Lobster Jasus edwardsii Hutton Harmful Algae, Volume 95, 2020, Article 101818 Alison Turnbull, …, Gustaaf Hallegraeff ### Accumulation of paralytic shellfish toxins by Southern Rock lobster Jasus edwardsii causes minimal impact on lobster health Aquatic Toxicology, Volume 230, 2021, Article 105704 Alison Turnbull, …, Quinn Fitzgibbon Show 3 more articles About ScienceDirect Remote access Contact and support Terms and conditions Privacy policy Cookies are used by this site.Cookie settings All content on this site: Copyright © 2025 Elsevier B.V., its licensors, and contributors. 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1493	https://personales.unican.es/junqueraj/javierjunquera_files/fisica-1/movimientos_circulares.pdf	Física Dinámica de la Partícula Página 1 MOV. CIRCULARES: Un aparato de un parque de atracciones consiste en un gran cilindro vertical que gira alrededor de su eje lo suficientemente rápido para que cualquier persona que se encuentre dentro de él se mantenga pegada contra la pared cuando se quita el piso. El coeficiente de rozamiento es µ = 0.4 y el radio del cilindro es R = 4 m. a) Encontrar el periodo máximo de revolución para evitar que la persona caiga. b) ¿Cuántas revoluciones por minuto realiza el cilindro? Solución: I.T.I. 93, 96, I.T.T. 00 Texto solución La masa de un péndulo de 2 m de longitud describe un arco de circunferencia en un plano vertical. Si en la posición de la figura la tensión de la cuerda es 2.5 veces el peso de la masa, hallar la velocidad y aceleración de la masa en dicho instante. Solución: I.T.I. 93, I.T.T. 00 Texto solución 30º Física Dinámica de la Partícula Página 2 Un bloque está sostenido por una mesa giratoria que, partiendo del reposo, gira de modo que la aceleración tangencial es de 1 m/s2. Si el coeficiente de rozamiento estático entre el bloque y la mesa es de 0.5 determinar el tiempo que tarda el bloque en deslizar y su velocidad en dicho instante. ¿Cuál será el menor tiempo para alcanzar una velocidad de 2 m/s sin deslizar? Distancia del bloque al centro d = 2 m. Solución: I.T.I. 00, 03, I.T.T. 04 Como parte del reposo y la aceleración tangencial es constante: at = dv dt ⇒ dv = atdt ⇒ dv 0 v ∫ = atdt 0 t ∫ ⇒ v = att (1) Con lo que la aceleración normal y el módulo de la aceleración del bloque valdrán: an = v 2 d = at 2 d t 2 ⇒ a = at 2 + an 2 = at 1 + at d 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ t 4 Esta aceleración es producida por la fuerza de rozamiento del bloque con la mesa giratoria (es la única fuerza horizontal): F roz.est. = ma = mat 1+ at d 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ t 4 En el instante tdesl. en que comienza a deslizar, la fuerza de rozamiento estática alcanza su valor máximo: F roz.est.máx. = ma = mat 1 + at d 2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ tdesl. 4 Froz.est.máx. = µ est.N = µest.mg ⎫ ⎬ ⎪ ⎪ ⎭ ⎪ ⎪ ⇒ Y la velocidad en dicho instante será: v tdesl. ( ) = at tdesl. = Si queremos alcanzar una velocidad v partiendo del reposo con una aceleración tangencial at el tiempo t que tardaremos lo podemos calcular a partir de la ecuación (1): t v ( ) = v at y será tanto más pequeño cuanto mayor sea la aceleración tangencial at. En todo caso este tiempo debe ser siempre inferior al tiempo que tarde el bloque en deslizar, ya que queremos conseguir dicha velocidad sin que el bloque deslice: tdesl. = d at ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 µ est.g at ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 −1 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 4 = 3.13 s at d ( ) 1 2 µ est.g at ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 −1 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 4 = 3.13 m/s Física Dinámica de la Partícula Página 3 v at ≤ d at ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 µ est.g at ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2 −1 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 4 ⇒ d 2at 2 −µest.gd 2at + v 4 ≤0 La inecuación anterior se cumple siempre que: 1 2 µest .g 1− 1− 4v 4 µest. 2 g2d2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ ≤at ≤1 2 µ est.g 1+ 1− 4v 4 µ est. 2 g2d2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ (basta con encontrar las raíces de la ecuación d 2at 2 −µest.gd 2at + v 4 = 0 y verificar que d 2at 2 −µest.gd 2at + v 4 ≤0 para at entre dichos valores). Como cuanto mayor es la aceleración tangencial menor es el tiempo necesario para alcanzar una determinada velocidad tenemos que el tiempo mínimo necesario para alcanzar una velocidad v será: tmín. v ( ) = v at ,máx = 2v µest.g 1 + 1− 4v 4 µest . 2 g2d2 ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ −1 Para la velocidad v1 = 2 m/s que nos dan en el enunciado: Una pequeña arandela de m = 100 g se desliza a lo largo de un alambre de radio R = 10 cm como el de la figura, que gira a razón de ω = 2 rev/s. Calcular el valor de θ para que la arandela quede en equilibrio. Solución: I.T.I. 04, I.T.T. 00 Dibujando el diagrama de fuerzas y planteando la segunda ley de Newton: θ ω R tmín. v1 ( ) = 0.5 s  N m g  a N cosθ −mg = 0 ⇒ N = mg cosθ N senθ = maN = mω 2Rgiro = mω 2Rsenθ ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ θ = Arcos g ω 2R ⎛ ⎝ ⎜ ⎞ ⎠ ⎟= 51.6º Física Dinámica de la Partícula Página 4 El objeto muy pequeño de la figura gira con velocidad angular ω constante y no resbala por la parte interior de un cono de semiángulo ϕ encontrándose a una distancia R del eje de giro. Si el rozamiento es despreciable determinar el valor que debe tener la frecuencia del movimiento circular para que esto ocurra. Solución: I.T.I. 05 Planteando la segunda ley de Newton y teniendo en cuenta que la aceleración es normal:  N + M g = M a a = ω 2R = 4π 2ν 2R ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ N cosϕ = 4π 2Mν 2R N senϕ −Mg = 0 ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ tgϕ = g 4π 2ν 2R ⇒ Una masa de 4 kg esta sujeta a una barra horizontal, como indica la figura. Las cuerdas están bajo tensión cuando la barra gira alrededor de su eje. Si la velocidad de la masa es constante e igual a 4 m/s. Calcular la tensión de las cuerdas cuando la masa está a) en su punto mas bajo, b) en la posición horizontal y c) en su punto mas alto. Si colocamos la barra vertical y la velocidad de la bola es ahora de 6 m/s, determinar: d) las tensiones de las cuerdas superior e inferior. Datos: l1 = 3 m, l2 = 2 m. Solución: I.T.I. 93, I.T.T. 00 a) Si dibujamos el diagrama de fuerzas y planteamos la segunda ley de Newton teniendo en cuenta que por simetría las dos tensiones van a ser iguales en magnitud:  T1 +  T2 + m g = m a , T1 = T2 = T ⇒ 2T senθ −mg = maN = m v2 R ⇒ T = m v2 R + g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2senθ = m v2 R + g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 2R / l2 l1 l2  T1 m g  a  T2 θ € m g ϕ €  a €  N ν = 1 2π g R tgϕ Física Dinámica de la Partícula Página 5 teniendo en cuenta que R = l2 2 −l1 / 2 ( ) 2 sustituyendo: ⇒ T == 1 2 ml2 v2 l2 2 −l1 / 2 ( ) 2 ( ) −1 + g l2 2 −l1 / 2 ( ) 2 ( ) −1/2 ⎡ ⎣ ⎢ ⎤ ⎦ ⎥= No está acabado b) asfasf c) asfas d) asfasf 66.2 N Física Dinámica de la Partícula Página 6 El cuerpo P tiene una masa de 5 kg y se encuentra sobre una superficie cónica lisa girando con una velocidad angular de 20 r.p.m. alrededor del eje € E ʹ′ E . Calcular la velocidad lineal del cuerpo y la reacción de la superficie sobre el cuerpo. Calcular la tensión del hilo y la velocidad angular para el caso en que la reacción del plano sea nula. Datos: θ = 45º, L = 0 5 m. Solución: I.T.I. 96, 99, 02 El bloque va a realizar un movimiento circular uniforme (ω constante) en el cual su aceleración es centrípeta, a = ω 2R = v 2 R , con R el radio de la trayectoria circular, que según la figura del enunciado: R = L senθ . Teniendo en cuenta el diagrama de fuerzas para el bloque y aplicando la segunda ley de Newton:  T + m g +  N = m a ⇒ T senθ −N cosθ = ma = mω 2R T cosθ + N senθ −mg = 0 ⎧ ⎨ ⎪ ⎩ ⎪ ⇒ N = m gsenθ −ω 2Rcosθ ( ) T = m gcosθ + ω 2Rsenθ ( ) ⎧ ⎨ ⎪ ⎩ ⎪ En el primer caso con los datos que nos dan: v = ωR = , N = En el segundo caso si imponemos que la normal N se anule: m gsenθ −ω 2Rcosθ ( ) = 0 ⇒ ω = g R tgθ = y la tensión del hilo en este caso valdría: T = 0.74 m/s E € ʹ′ E θ €  T € m g € θ €  a €  N 29.2 N 5.26 rad/s = 50.3 r.p.m. 40.1 N Física Dinámica de la Partícula Página 7 ¿Cuántas revoluciones por segundo ha de girar el aparato de la figura para que la cuerda forme un ángulo de 45º con la vertical? ¿Cuál será entonces la tensión en la cuerda? Datos: r = 10 cm, l = 20 cm, m = 200 g. Solución: I.T.I. 01, I.T.T. 01 La esfera va a realizar un movimiento circular uniforme (ω constante) en el cual su aceleración es centrípeta, a = ω 2R, con R el radio de la trayectoria circular, que según la figura del enunciado: R = r + l senθ. Teniendo en cuenta el diagrama de fuerzas para la esfera y aplicando la segunda ley de Newton:  T + m g = m a ⇒ T senθ = ma = mω 2R ⇒ ω = T senθ mR ⎛ ⎝ ⎞ ⎠ 1 2 = gtgθ r+ l senθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 = T cosθ −mg = 0 ⇒ T = mg cosθ = ⎧ ⎨ ⎪ ⎪ ⎪ ⎩ ⎪ ⎪ ⎪ m l r €  T € m g € θ €  a 2.77N € 6.37 rad s = 1.01rps Física Dinámica de la Partícula Página 8 Dos alambres están unidos a una esfera como se indica en la figura. Se hace girar la esfera de modo que describa una circunferencia horizontal a velocidad constante v. Determinar el intervalo de valores de v para los cuales ambos alambres permanecen tensos. ¿Para que velocidad ambos alambres soportan la misma tensión? ¿Cuál será el valor de dicha tensión? Datos: m = 5 kg, θ1 = 30º, θ2 = 60º, l = 1.2 m. Solución: I.T.I. 95, 97, 99, 01, 03, I.T.T. 97, 01 Si llamamos  T al vector suma de las dos tensiones  T 1 y  T 2, el ángulo θ que forma con la horizontal estará comprendido entre θ1 y θ2 . La aceleración de la esfera será una aceleración centrípeta dirigida hacia el centro de la trayectoria circular de radio R. Dibujando el diagrama de todas la fuerzas que actúan sobre la esfera y aplicando la segunda ley de Newton:  T + m g = m a  T =  T 1 +  T 2 ( ) ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ T senθ = mg T cosθ = ma = m v 2 R ⎧ ⎨ ⎪ ⎩ ⎪ Dividiendo estas dos ecuaciones podemos sacar información acerca de la variación de la velocidad de la esfera en función del ángulo θ : tgθ = g a = gR v 2 ⇒ v θ ( ) = gR tgθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 Esta ecuación nos indica que cuanto mayor sea el ángulo menor será la velocidad de la esfera, y al revés, cuanto menor sea el ángulo mayor será la velocidad. La velocidad mínima se alcanzará cuando el ángulo θ valga θ 2 (en esta situación la tensión T1 será nula) y la máxima cuando valga θ 1 (en este caso se anulará T2 ). El radio R de la trayectoria lo podemos sacar a partir de la figura y de los datos que nos dan en el enunciado del problema: l = R tgθ2 −tgθ1 ( ) ⇒ R = l tgθ2 −tgθ1 ( ) −1 La expresión para la velocidad en función del ángulo quedará: v θ ( ) = gl tgθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 tgθ2 −tgθ1 ( ) −1 2 La velocidad mínima será: vmín. = v θ2 ( ) = gl tgθ2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 tgθ2 −tgθ1 ( ) −1 2 = θ 1 θ2  T 1  T 2  Τ m g  a 2.42m /s € θ 1 m l € θ2 Física Dinámica de la Partícula Página 9 La velocidad máxima será: vmáx. = v θ 1 ( ) = gl tgθ1 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 tgθ2 −tgθ1 ( ) −1 2 = Si las tensiones T1 y T2 tuvieran un mismo valor (llamémosle € T ), la suma de ambas, €  T , formaría un ángulo con la horizontal intermedio entre θ1 y θ2 (llamémosle € θ ): θ = θ 1 +θ2 2 = 45º En esta situación la velocidad de la esfera sería: v θ ( ) = gl tgθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ 1 2 tgθ2 −tgθ1 ( ) −1 2 = 3.19m /s Aplicando la segunda ley de Newton para las componentes horizontales: T cosθ1 + T cosθ2 = m v θ ( ) 2 R = mg tgθ ⇒ T = mg tgθ cosθ1 + cosθ2 ( ) −1 = Una carretera tiene 13.8 m de ancho. Calcule la diferencia de nivel entre los bordes de la misma para que un coche sea capaz de tomar la curva a 90 km/h sin derrapar suponiendo que el coeficiente de rozamiento entre las ruedas y el suelo es µ = 0.1 y el radio de la curva es de 300 m. Solución: I.T.I. 95, 99, 04 El coche va a realizar un movimiento circular uniforme de radio R. Su aceleración será centrípeta, a = v 2 R. Si toma la curva con excesiva velocidad va a derrapar hacia el exterior. La fuerza de rozamiento con el suelo se opone a dicho movimiento. Dibujando el diagrama de fuerzas ejercidas sobre el automóvil y planteando la segunda ley de Newton:  N +  F roz. + M g = M a ⇒ N cosθ −Froz. senθ −Mg = 0 N senθ + Froz.cosθ = Ma = M v2 R ⎧ ⎨ ⎪ ⎩ ⎪ La fuerza de rozamiento es estática, no hay derrape. El movimiento del coche se produce a lo largo de una dirección perpendicular al plano de la figura, no en la dirección en la que actúa la fuerza de rozamiento. Esta fuerza tendrá un valor límite F roz.máx. = µN , con lo que sustituyendo en las ecuaciones anteriores: € 4.20m/s € 35.87N Física Dinámica de la Partícula Página 10 N cosθ −Mg = F roz.senθ ≤µNsenθ M v2 R = N senθ + F roz. cosθ ≤N senθ + µN cosθ ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ N cosθ −µsenθ ( ) ≤Mg 1 ( ) N µ cosθ + senθ ( ) ≥M v2 R 2 ( ) ⎧ ⎨ ⎪ ⎩ ⎪ Dividiendo (1) entre (2): cosθ −µsenθ µ cosθ + senθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟≤gR v2 ⇒ 1−µ tgθ µ + tgθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟≤gR v2 ⇒ tgθ ≥ 1−µ gR v2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ µ + gR v2 ⇒ θ ≥arctg v2 gR ⎛ ⎝ ⎜ ⎞ ⎠ ⎟−arctg µ ( ) = 6.29º Si llamamos l a la anchura de la carretera, el desnivel entre los bordes de ésta será: d = lsenθ ≥ 1.51 m Física Dinámica de la Partícula Página 11 Un automóvil de 1500 kg toma una curva de 35 m de radio. Si el coeficiente de rozamiento estático µest. entre los neumáticos y el suelo es de 0.5, calcular la máxima velocidad que puede llevar el automóvil sin derrapar. Resolver el mismo problema si la curva tiene un peralte de 10º. Solución: I.T.I. 02, 05, I.T.T. 03 Vamos a resolver el problema suponiendo un peralte y luego concretaremos para los casos que nos piden. El coche va a realizar un movimiento circular uniforme de radio R. Su aceleración será centrípeta, a = v 2 R. Si toma la curva con excesiva velocidad va a derrapar hacia el exterior. La fuerza de rozamiento con el suelo se opone a dicho movimiento. Dibujando el diagrama de fuerzas ejercidas sobre el automóvil y planteando la segunda ley de Newton:  N +  F roz. + M g = M a ⇒ N cosθ −Froz. senθ −Mg = 0 N senθ + Froz.cosθ = Ma = M v2 R ⎧ ⎨ ⎪ ⎩ ⎪ La fuerza de rozamiento es estática, no hay derrape. El movimiento del coche se produce a lo largo de una dirección perpendicular al plano de la figura, no en la dirección en la que actúa la fuerza de rozamiento. Esta fuerza tendrá un valor límite F roz.máx. = µN , que lo alcanzará cuando la velocidad sea máxima. En este caso las ecuaciones anteriores se transformarán en: N cosθ −µ senθ ( ) −Mg = 0 N senθ + µcosθ ( ) = Ma = M vmáx. 2 R ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ N = Mg cosθ −µsenθ vmáx. = senθ + µcosθ cosθ −µ senθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ gR ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 Vemos en la solución que la velocidad máxima con la que puede tomar el coche la curva no depende de su masa. Para los dos casos que nos piden: Peralte nulo, θ = 0: vmáx. = µ gR = Peralte θ = 10º: vmáx. = € M g €  N €  F roz. θ  a 13.1 m /s = 47.1 km /h 16.0 m /s = 57.4 km/h Física Dinámica de la Partícula Página 12 Una carretera está peraltada de modo que un coche desplazándose a 40 km/h puede tomar una curva de 30 m de radio incluso si existe una capa de hielo (equivalente a un coeficiente de fricción prácticamente nulo). Determinar el intervalo de velocidades a que un coche puede tomar esta curva sin derrapar si el coeficiente de fricción estática entre la carretera y las ruedas es de 0.3 Solución: I.T.I. 92, 98, 00, I.T.T. 96, 99, 00, 02, 05 El coche va a realizar un movimiento circular uniforme de radio R. Su aceleración será centrípeta, a = v 2 R. A la velocidad que nos indican toma la curva sin necesidad de rozamiento. Dibujando el diagrama de fuerzas ejercidas sobre el automóvil y planteando la segunda ley de Newton:  N + M g = M a ⇒ N cosθ −Mg = 0 N senθ = Ma = M v2 R ⎧ ⎨ ⎪ ⎩ ⎪ ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ θ = arctg v 2 Rg ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 22º 4 ʹ′ 7 N = Mg cosθ ⎧ ⎨ ⎪ ⎪ ⎩ ⎪ ⎪ Esto nos ha permitido calcular la inclinación del peralte. Si el automóvil toma la curva con excesiva velocidad (v > 40 km/h) va a derrapar hacia el exterior. La fuerza de rozamiento con el suelo se opone a dicho movimiento. Dibujando ahora el nuevo diagrama de fuerzas ejercidas sobre el automóvil y planteando la segunda ley de Newton:  N +  F roz. + M g = M a ⇒ N cosθ −Froz. senθ −Mg = 0 N senθ + Froz.cosθ = Ma = M v2 R ⎧ ⎨ ⎪ ⎩ ⎪ La fuerza de rozamiento es estática, no hay derrape. El movimiento del coche se produce a lo largo de una dirección perpendicular al plano de la figura, no en la dirección en la que actúa la fuerza de rozamiento. Esta fuerza tendrá un valor límite F roz.máx. = µN , que lo alcanzará cuando la velocidad sea máxima. En este caso las ecuaciones anteriores se transformarán en: € M g €  N θ  a M g €  N  F roz. θ  a Física Dinámica de la Partícula Página 13 N cosθ −µ senθ ( ) −Mg = 0 N senθ + µcosθ ( ) = Ma = M vmáx. 2 R ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ N = Mg cosθ −µ senθ vmáx. = senθ + µcosθ cosθ −µ senθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ gR ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 = 56.0 km /h Si toma la curva con poca velocidad (v < 40 km/h) va a derrapar hacia el interior. La fuerza de rozamiento con el suelo se opone a dicho movimiento. Dibujando el diagrama de fuerzas ejercidas sobre el automóvil y planteando la segunda ley de Newton:  N +  F roz. + M g = M a ⇒ N cosθ + Froz. senθ −Mg = 0 N senθ −F roz. cosθ = Ma = M v 2 R ⎧ ⎨ ⎪ ⎩ ⎪ Igual que en el caso anterior, la fuerza de rozamiento tendrá un valor límite F roz.máx. = µN , que lo alcanzará cuando la velocidad sea mínima. En este caso las ecuaciones anteriores se transformarán en: N cosθ + µsenθ ( ) −Mg = 0 N senθ −µ cosθ ( ) = Ma = M vmín. 2 R ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ N = Mg cosθ + µsenθ vmín. = senθ −µcosθ cosθ + µsenθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ gR ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 = 20.1 km/h Como se puede comprobar las soluciones se corresponden con las encontradas anteriormente salvo por el cambio de signo de los términos que contienen a µ. El rango de velocidades para el cual el coche puede tomar la curva sin derrapar será: € M g €  N €  F roz. θ  a 20.1 km /h ≤v ≤56.0 km /h Física Dinámica de la Partícula Página 14 Un automóvil da vueltas sobre una curva peraltada. El radio de curvatura de la carretera es R. El ángulo de peralte es θ, y el coeficiente de rozamiento es µ. a) Determinar la gama de velocidades que puede tener el vehículo sin derrapar. b) Determinar el valor mínimo de µ para que la rapidez mínima sea nula. c) Resolver el primer apartado si R = 100 m, θ = 15º y µ = 0.1 Solución: I.T.I. 01, 03, I.T.T. 01, 04 a) El coche va a realizar un movimiento circular uniforme de radio R. Su aceleración será centrípeta, a = v 2 R. Si toma la curva con excesiva velocidad va a derrapar hacia el exterior. La fuerza de rozamiento con el suelo se opone a dicho movimiento. Dibujando el diagrama de fuerzas ejercidas sobre el automóvil y planteando la segunda ley de Newton:  N +  F roz. + M g = M a ⇒ N cosθ −Froz. senθ −Mg = 0 N senθ + Froz.cosθ = Ma = M v2 R ⎧ ⎨ ⎪ ⎩ ⎪ La fuerza de rozamiento es estática, no hay derrape. El movimiento del coche se produce a lo largo de una dirección perpendicular al plano de la figura, no en la dirección en la que actúa la fuerza de rozamiento. Esta fuerza tendrá un valor límite F roz.máx. = µN , que lo alcanzará cuando la velocidad sea máxima. En este caso las ecuaciones anteriores se transformarán en: N cosθ −µ senθ ( ) −Mg = 0 N senθ + µcosθ ( ) = Ma = M vmáx. 2 R ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ N = Mg cosθ −µsenθ vmáx. = senθ + µcosθ cosθ −µ senθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ gR ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 Si toma la curva con poca velocidad va a derrapar hacia el interior. La fuerza de rozamiento con el suelo se opone a dicho movimiento. Dibujando el diagrama de fuerzas ejercidas sobre el automóvil y planteando la segunda ley de Newton:  N +  F roz. + M g = M a ⇒ N cosθ + Froz. senθ −Mg = 0 N senθ −F roz. cosθ = Ma = M v 2 R ⎧ ⎨ ⎪ ⎩ ⎪ M g  N  F roz. θ  a € M g €  N €  F roz. θ  a Física Dinámica de la Partícula Página 15 Igual que en el caso anterior, la fuerza de rozamiento tendrá un valor límite F roz.máx. = µN , que lo alcanzará cuando la velocidad sea mínima. En este caso las ecuaciones anteriores se transformarán en: N cosθ + µsenθ ( ) −Mg = 0 N senθ −µ cosθ ( ) = Ma = M vmín. 2 R ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ N = Mg cosθ + µsenθ vmín. = senθ −µcosθ cosθ + µsenθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ gR ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 Como se puede comprobar las soluciones se corresponden con las encontradas anteriormente salvo por el cambio de signo de los términos que contienen a µ. El rango de velocidades para el cual el coche puede tomar la curva sin derrapar será: b) Si la velocidad mínima es nula tendremos que: senθ −µ cosθ cosθ + µ senθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ gR ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ = 0 ⇒ senθ −µ cosθ = 0 ⇒ c) Con los datos que nos dan el rango de valores para la velocidad será: senθ −µ cosθ cosθ + µ senθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ gR ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 ≤v ≤ senθ + µ cosθ cosθ −µsenθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ gR ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 € µ = tgθ 12.66 m/s ≤v ≤18.82 m /s Física Dinámica de la Partícula Página 16 En el interior de una esfera hueca de radio R, que gira con una velocidad angular constante ω se halla un objeto pequeño como se indica en la figura. Si el coeficiente de rozamiento estático es µ y el ángulo θ es también conocido, determinar los valores de ω para que el objeto no se deslice. Solución: I.T.I. 98, 99, 00, 04, I.T.T. 97, 99, 02, 05 Dependiendo de si ω es demasiado grande, de forma que el objeto tendería a subir por la superficie esférica disminuyendo θ, o demasiado pequeña, el objeto tendería a bajar por la superficie esférica aumentando θ, la fuerza de rozamiento del objeto con la superficie se orientará de forma de evitar dicho movimiento intentando que el objeto mantenga su posición angular θ. Para una velocidad ω pequeña:  N +  F roz. + M g = M a ⇒ N senθ + Froz.cosθ −Mg = 0 N cosθ −Froz.senθ = Ma = Mω 2Rcosθ ⎧ ⎨ ⎪ ⎩ ⎪ La fuerza de rozamiento es estática, no hay derrape. El movimiento del objeto se produce a lo largo de una dirección perpendicular al plano de la figura, no en la dirección en la que actúa la fuerza de rozamiento. Esta fuerza tendrá un valor límite F roz.máx. = µN , que lo alcanzará cuando la velocidad sea mínima. En este caso las ecuaciones anteriores se transformarán en: N senθ + µcosθ ( ) −Mg = 0 N cosθ −µ senθ ( ) = Ma = Mωmín. 2 Rcosθ ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ cosθ −µ senθ senθ + µcosθ = ωmín. 2 Rcosθ g ⇒ Para una velocidad ω grande: €  ω θ € M g €  N  F roz. θ €  a ωmín. = 1−µtgθ senθ + µ cosθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ g R ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 Física Dinámica de la Partícula Página 17  N +  F roz. + M g = M a ⇒ N senθ −Froz.cosθ −Mg = 0 N cosθ + F roz. senθ = Ma = Mω 2Rcosθ ⎧ ⎨ ⎪ ⎩ ⎪ Igual que en el caso anterior, la fuerza de rozamiento tendrá un valor límite F roz.máx. = µN , que lo alcanzará cuando la velocidad sea máxima. En este caso las ecuaciones anteriores se transformarán en: N senθ −µcosθ ( ) −Mg = 0 N cosθ + µsenθ ( ) = Ma = Mωmín. 2 Rcosθ ⎫ ⎬ ⎪ ⎭ ⎪ ⇒ cosθ + µ senθ senθ −µcosθ = ωmín. 2 Rcosθ g ⇒ Para cualquier valor de ω entre estos dos valores el cuerpo no deslizará. € M g €  N €  F roz. θ €  a ωmín. = 1 + µtgθ senθ −µcosθ ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ g R ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ 1 2 Física Dinámica de la Partícula Página 18 Un piloto de masa m que vuela en un avión a reacción ejecuta una maniobra de rizar el rizo. Si el avión se mueve a una velocidad cte. de 225 m/s determinar el peso aparente del piloto en la parte superior e inferior del rizo si el radio de éste es de 1 km. Solución: I.T.I. 02, 05, I.T.T. 03 El movimiento del caza es un movimiento circular uniforme con lo que en todo momento su aceleración es normal a la trayectoria, a = v 2 R. Cuando se encuentre en la parte más alta del rizo el diagrama de las fuerzas que actúan sobre el piloto será: Nsup. + mg = ma = m v2 R ⇒ Nsup. = m v2 R −g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = Cuando se encuentre en la parte más baja del rizo el diagrama de las fuerzas que actúan sobre el piloto será: Ninf. −mg = ma = m v2 R ⇒ Ninf . = m v 2 R + g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = m g  a  N sup. € 4.2 mg m g  a  N inf. 6.2 mg Física Dinámica de la Partícula Página 19 El piloto de un avión se lanza en picado a la velocidad de 400 km/h y termina su descenso describiendo, a aquella velocidad, un arco de circunferencia situado en el plano vertical. ¿Cuál será el mínimo radio de esa circunferencia para que la aceleración en el punto más bajo no exceda de 7g?¿Cuál será entonces el peso aparente del aviador en el punto más bajo de la trayectoria? Analizar el problema tanto desde el punto de vista de un observador inercial como de uno no inercial. Solución: I.T.I. 94, I.T.T. 95, I.I. 94 Texto solución El protagonista de “Top Gun” está haciendo unas maniobras aéreas con su caza a reacción con una rapidez constante de 1000 km/h. Determinar: a) si realiza un rizo con R = 400 m ¿cual sería el peso del piloto en la parte superior e inferior del rizo si tuviese en ese mismo momento una balanza para pesarse? b) si estos pilotos de combate están entrenados para resistir aceleraciones de hasta 7g (para aceleraciones mayores perderían la consciencia) ¿cuál es el radio de giro mínimo del rizo que puede hacer con su avión? Solución: I.T.I. 92, 98, 99, 04, I.T.T. 96, 99, 02, 05 El movimiento del caza es un movimiento circular uniforme con lo que en todo momento su aceleración es normal a la trayectoria, a = v 2 R. a) Cuando se encuentre en la parte más alta del rizo el diagrama de las fuerzas que actúan sobre el piloto será: Nsup. + mg = ma = m v2 R ⇒ Nsup. = m v2 R −g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = Cuando se encuentre en la parte más baja del rizo el diagrama de las fuerzas que actúan sobre el piloto será: m g  a  N sup. 18.7 mg m g  a  N inf. Física Dinámica de la Partícula Página 20 Ninf. −mg = ma = m v2 R ⇒ Ninf . = m v 2 R + g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = b) Si nuestro piloto puede soportar “sólo” hasta siete veces su propio peso (en el ejemplo del apartado anterior el piloto hubiese muerto aplastado contra el asiento) y visto que se siente más pesado en la parte inferior del rizo el radio mínimo para dicha maniobra será: m v 2 R + g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ ≤7mg ⇒ El radio de una noria de feria mide 5 m y da una vuelta en 10 s. a) Hállese la diferencia entre los pesos aparentes de un pasajero en los puntos más bajo y más alto, expresada como fracción de peso. b) ¿Cuál debería ser el tiempo correspondiente a una vuelta para que el peso aparente en el punto más alto fuese nulo? c) ¿Cuál sería entonces el peso aparente en el punto inferior? Solución: I.T.I. 01, I.T.T. 01, 04 a) El pasajero está realizando un movimiento circular uniforme. La aceleración que sufre es una aceleración centrípeta: a = ω 2R = 2π T ⎛ ⎝ ⎞ ⎠ 2 R Cuando el pasajero se encuentre en la parte inferior, el diagrama de fuerzas es el que se muestra en la figura. Aplicando la segunda ley de Newton podemos calcular el valor de la fuerza normal, que mide el peso aparente de la persona:  N inf. + m g = m a ⇒ Ninf . −mg = ma = m 2π T ⎛ ⎝ ⎞ ⎠ 2 R ⇒ Ninf. = mg + m 2π T ⎛ ⎝ ⎞ ⎠ 2 R = 1 + 2π T ⎛ ⎝ ⎞ ⎠ 2 R g ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ mg =1.20 mg El pasajero se siente un 20% más pesado. En el caso de que el pasajero se encuentre en la parte superior: 20.7 mg R ≥Rmín. = v 2 6g =1312 m m g  a  N inf. m g  a  N sup. Física Dinámica de la Partícula Página 21  N sup. + m g = m a ⇒ mg −Nsup . = ma = m 2π T ⎛ ⎝ ⎞ ⎠ 2 R ⇒ Nsup. = mg−m 2π T ⎛ ⎝ ⎞ ⎠ 2 R = 1−2π T ⎛ ⎝ ⎞ ⎠ 2 R g ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ mg = 0.80 mg El pasajero se siente un 20% más ligero. La diferencia entre los dos pesos aparentes será: Ninf. −Nsup . = b) Tomando la expresión para Nsup.y anulándola para un periodo de rotación € ʹ′ T : 1−2π ʹ′ T ⎛ ⎝ ⎞ ⎠ 2 R g ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ mg = 0 ⇒ ʹ′ T = 2π R g = c) En este caso el peso aparente en la posición inferior será: ʹ′ N sup. = 1 + 2π ʹ′ T ⎛ ⎝ ⎞ ⎠ 2 R g ⎡ ⎣ ⎢ ⎢ ⎤ ⎦ ⎥ ⎥ mg = Una masa de 0.4 kg está atada al extremo de una cuerda, sujetado el otro extremo por una persona que hace girar la piedra en un círculo horizontal de radio 0.8 m a una velocidad angular de 80 r.p.m. ¿Cuál es la tensión de la cuerda? Si la cuerda se rompe cuando la tensión es de 500 N ¿cuál es la máxima velocidad angular posible? ¿Podemos ignorar el peso de la piedra? Solución: I.T.I. 95 Texto solución Un electrón de un átomo de hidrógeno gira alrededor del protón en una trayectoria casi circular de radio 0.5·10–10 m con una velocidad de 2.2·106 m/s. Calcular la fuerza entre ambas partículas. Solución: I.T.I. 95 Texto solución € 0.40 mg 4.49s 2.00 mg Física Dinámica de la Partícula Página 22 Un bloque de masa 1 kg situado en el extremo de una cuerda se hace dar vueltas en un circulo vertical de radio 1 m. Hallar la velocidad critica por debajo de la cual la cuerda se afloja al estar en el punto más alto. Para este caso, ¿cuál es la velocidad en la parte inferior del círculo? Solución: I.T.I. 02, 05, I.T.T. 04 El bloque va a realizar un movimiento circular de radio R. Dibujando el diagrama de fuerzas cuando se encuentra en la parte más alta del círculo vemos que su aceleración va a ser vertical y por lo tanto sólo tendrá componente normal an = v 2 R. Aplicando la segunda ley de Newton:  T + m g = m a ⇒ T + mg = ma = m v 2 R ⇒ T = m v 2 R −g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ Para la velocidad crítica la cuerda pierde su tensión: m vcrítica 2 R −g ⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = 0 ⇒ Si dibujamos el diagrama de fuerzas en un momento en el que la cuerda forme un ángulo θ con la vertical la componente tangencial de la aceleración será: mgsenθ = mat = mdv dt ⇒ dv dt = gsenθ Teniendo en cuenta la relación entre la velocidad lineal y la angular v = ωR: R dω dt = gsenθ ⇒ R dθ dt ⎛ ⎝ ⎞ ⎠ dω dθ = gsenθ ⇒ Rω dω dθ = gsenθ Separando variables e integrando teniendo en cuenta las condiciones iniciales del movimiento ωcrítica = vcrítica R para θ = 0 ⎛ ⎝ ⎞ ⎠ obtendremos la velocidad angular en función del ángulo: Rω dω ωcrítica ω ∫ = gsenθ dθ 0 θ ∫ ⇒ 1 2 Rω 2 ⎤ ⎦ ⎥ ωcrítica ω = −gcosθ]0 θ ⇒ ω θ ( ) = ωcrítica 2 + 2g R 1−cosθ ( ) ⎡ ⎣ ⎢ ⎤ ⎦ ⎥ 1/2 La velocidad lineal del bloque en función del ángulo será entonces: vcrítica = gR €  a € m g €  T € m g €  T  a θ Física Dinámica de la Partícula Página 23 v θ ( ) = ω θ ( )R = vcrítica 2 + 2gR 1−cosθ ( ) ⎡ ⎣ ⎤ ⎦ 1/2 Y cuando se encuentre en la parte inferior de su trayectoria, θ = 180º: vinferior = vcrítica 2 + 4gR [ ] 1/ 2 = 5gR
1494	https://math.stackexchange.com/questions/4085381/combination-permutation-on-seating-arrangements	combinatorics - Combination/permutation on seating arrangements - Mathematics Stack Exchange Join Mathematics By clicking “Sign up”, you agree to our terms of service and acknowledge you have read our privacy policy. Sign up with Google OR Email Password Sign up Already have an account? Log in Skip to main content Stack Exchange Network Stack Exchange network consists of 183 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. 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Upvoting indicates when questions and answers are useful. What's reputation and how do I get it? Instead, you can save this post to reference later. Save this post for later Not now Thanks for your vote! You now have 5 free votes weekly. Free votes count toward the total vote score does not give reputation to the author Continue to help good content that is interesting, well-researched, and useful, rise to the top! To gain full voting privileges, earn reputation. Got it!Go to help center to learn more Combination/permutation on seating arrangements Ask Question Asked 4 years, 6 months ago Modified4 years, 6 months ago Viewed 429 times This question shows research effort; it is useful and clear 1 Save this question. Show activity on this post. 32 32 people are to be seated in a square table where each side A,B,C,A,B,C, and D D has 10 10 chairs. If five of the the people want to sit at side A A and another six want to sit at side B B, how many ways can all of them be seated? I'm not sure on how to approach this problem. Initially I did (10 5)(10 6)(29 21)(10 5)(10 6)(29 21) but this was said to be incorrect. Can someone help me out? combinatorics permutations Share Share a link to this question Copy linkCC BY-SA 4.0 Cite Follow Follow this question to receive notifications edited Apr 1, 2021 at 11:03 N. F. Taussig 79.3k 14 14 gold badges 62 62 silver badges 77 77 bronze badges asked Apr 1, 2021 at 10:59 KumarKumar 21 5 5 bronze badges 12 Just fix some of the (10 5)(10 5) possibilities for side A A, and also fix some of the (10 6)(10 6) possibilities for side B B. How many possibilities are now to distribute the remained people on (one - say C C - or equivalently) both remained sides?dan_fulea –dan_fulea 2021-04-01 11:10:26 +00:00 Commented Apr 1, 2021 at 11:10 1 I'd read a claim like, "5 5 people want to sit on side A A" as saying that "at least 5 5 people must be on side A A". So you first need the number of 4−4−tuples (a,b,c,d)(a,b,c,d) of non-negative inegers with a+b+c+d=32 a+b+c+d=32 and 5≤a≤10,6≤b≤10,c≤10,d≤10 5≤a≤10,6≤b≤10,c≤10,d≤10. Also, it's not clear to me whether or not the order of the people on a given side matters. That needs to be specified.lulu –lulu 2021-04-01 11:11:39 +00:00 Commented Apr 1, 2021 at 11:11 You forgot the number of ways to choose the people from 32: (32 5)(27 6)(32 5)(27 6)Alex –Alex 2021-04-01 11:15:09 +00:00 Commented Apr 1, 2021 at 11:15 @lulu Oh, now it's clearer. What's the best way to accomplish this though without using generating functions?Kumar –Kumar 2021-04-01 11:16:16 +00:00 Commented Apr 1, 2021 at 11:16 The numbers are small, I'df just do case by case work. Letting A=a−5,B=b−6 A=a−5,B=b−6 we jsut want A+B+c+d=21 A+B+c+d=21. Stars and Bars basically works, but you have to deal with the caps. A little messy, but not really hard.lulu –lulu 2021-04-01 11:18:35 +00:00 Commented Apr 1, 2021 at 11:18 \|Show 7 more comments 1 Answer 1 Sorted by: Reset to default This answer is useful 1 Save this answer. Show activity on this post. Seat the "special" people for side A A in P(10,5)P(10,5) ways, the "special" people on side B B in P(10,6)P(10,6) ways, and permute the balance in P(29,21)P(29,21) ways This assumes that people don't care whether they are clustered together or not, and that the "special" people for sides A A and B B are known Share Share a link to this answer Copy linkCC BY-SA 4.0 Cite Follow Follow this answer to receive notifications answered Apr 1, 2021 at 13:01 true blue aniltrue blue anil 49.2k 4 4 gold badges 31 31 silver badges 64 64 bronze badges 7 This count fails to take into account the fact that none of the sides can take more than 10 10 people.lulu –lulu 2021-04-01 14:19:10 +00:00 Commented Apr 1, 2021 at 14:19 @lulu: I don't see how that comes into the picture. 5 5 seats are vacant on A, 4 4 on B and 10 10 each on C and D, totaling 29 29, with 21 21 people to be seated. Surely a vacant seat is a vacant seat, whichever side it is on ? So (29 21)21!(29 21)21! arrangements ?true blue anil –true blue anil 2021-04-01 14:45:42 +00:00 Commented Apr 1, 2021 at 14:45 I don;'t understand. Suppose the people are {a 1,⋯,a 32}{a 1,⋯,a 32} with a 1,⋯,a 5 a 1,⋯,a 5 sitting on A A and a 6,⋯,a 11 a 6,⋯,a 11 sitting on B B. My permutation is...just leave them all in numerical order. How should those people seat? What side does, say, a 12 a 12 sit on?lulu –lulu 2021-04-01 15:03:31 +00:00 Commented Apr 1, 2021 at 15:03 @lulu:a 12 a 12 is free to sit in any of the 29 29 vacant seats. The sides have only been specified for 5+6 5+6 people, the rest have no restrictions.true blue anil –true blue anil 2021-04-01 15:10:43 +00:00 Commented Apr 1, 2021 at 15:10 But then what difference does the permutation make? a 12 a 12 could be seated between a 1 a 1 and a 2 a 2 in Row A A, or a 12 a 12 could be seated next to a 11 a 11 in row B B, or many other possibilities. You haven't specified the seating arrangement in any way, nor have you taken the 10 10 seat cap into account.lulu –lulu 2021-04-01 15:12:49 +00:00 Commented Apr 1, 2021 at 15:12 \|Show 2 more comments You must log in to answer this question. Start asking to get answers Find the answer to your question by asking. Ask question Explore related questions combinatorics permutations See similar questions with these tags. 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1495	https://ua.pressbooks.pub/collegealgebraformanagerialscience/chapter/6-7-transformation-of-functions/	Skip to content Want to create or adapt books like this? Learn more about how Pressbooks supports open publishing practices. 6.7 Transformation of Functions Often when given a problem, we try to model the scenario using mathematics in the form of words, tables, graphs and equations in order to explain or solve it. When building models, it is often helpful to build off of existing formulas or models. Knowing the basic graphs of your tool-kit functions can help you solve problems by being able to model new behavior by adapting something you already know. Unfortunately, the models and existing formulas we know are not always exactly the same as the ones presented in the problems we face. Fortunately, there are systematic ways to shift, stretch, compress, flip and combine functions to help them become better models for the problems we are trying to solve. We can transform what we already know into what we need, hence the name, “Transformation of functions.” When we have a story problem, formula, graph, or table, we can then transform that function in a variety of ways to form new functions. Parent Functions Before we begin transforming them, let’s review the parent functions that everyone should know. \| \| \| \| \| --- --- \| \| Function \| Graph \| Domain \| Range \| \| Linear \| \| \| \| \| Absolute Value: \| \| \| \| \| Quadratic: \| \| \| \| \| Cubic: \| \| \| \| \| Square Root: \| \| \| \| \| Cube Root: \| \| \| \| \| Reciprocal: \| \| \| \| \| Reciprocal Squared: \| \| \| \| Knowing these functions, we can consider how to transform them, which creates a much easier way to graph, or sketch many different functions. Shifts If we want to look at the function , we know from graphing quadratic functions that this means the vertex is (0,4) instead of (0,0) which is the vertex of the parent function, . Notice that this moves up the vertex 4 units. Similarly, we could look at the other points, such as and , which are 1 and 5 respectively, we see that is just the graph moved up 4 units. This is an example of a vertical shift. The input values stay the same and only the output values change. The range also changes. The range of is and the range of is moved up 4 with the graph to Vertical Shift Given a function , if we define a new function as , where is a constant then is a vertical shift of the function , where all the output values have been increased by . If is positive, then the graph will shift up If is negative, then the graph will shift down Example of a Vertical Shift with a Function Defined as a Table A function is given as a table below. Create a table for the function \| \| \| \| \| \| --- --- \| x \| 2 \| 4 \| 6 \| 8 \| \| f(x) \| 1 \| 3 \| 7 \| 11 \| The formula tells us that we can find the output values of the g function by subtracting 3 from the output values of the f function. For example, Subtracting 3 from each value, we can complete a table of values for \| \| \| \| \| \| --- --- \| x \| 2 \| 4 \| 6 \| 8 \| \| g(x) \| -2 \| 0 \| 4 \| 8 \| As with the earlier vertical shift, notice the input values stay the same and only the output values change. Try it Now 1 The function gives the height h of a ball (in meters) thrown upwards from the ground after t seconds. Suppose the ball was instead thrown from the top of a 10 meter building. Relate this new height function b(t) to h(t), then find a formula for b(t). The vertical shift is a change to the output, or outside, of the function. We will now look at how changes to input, on the inside of the function, change its graph and meaning. Consider the function . In a similar, parent function, , there is a point in the center at (0,0). But, in order for to have an output of 0, would have to be . So we have a point at (4,0). Now consider, what we would need to have an output of 1. Take the cube root of both sides: So we have a point at (5,1) for and (1,1) for Notice that now the outputs stay the same, but the inputs change. is moved to the right 4 units as seen below: Horizontal changes or “inside changes” affect the domain of a function (the input) instead of the range and often seem counter-intuitive. The new function uses the same outputs as , but matches those outputs to inputs 4 units less than those of . Said another way, we must add 4 to the input of to find the corresponding output for : . Horizontal Shift Given a function , if we define a new function as , where is a constant then is a horizontal shift of the function If h is positive, then the graph will shift right If h is negative, then the graph will shift left Example of a Horizontal Shift with a Table A function is given as a table below. Create a table for the function \| \| \| \| \| \| --- --- \| x \| 2 \| 4 \| 6 \| 8 \| \| f(x) \| 1 \| 3 \| 7 \| 11 \| The formula tells us that the output values of g are the same as the output value of f with an input value three smaller. For example, we know that . To get the same output from the g function, we will need an input value that is 3 larger: We input a value that is three larger for g(x) because the function takes three away before evaluating the function f. \| \| \| \| \| \| --- --- \| x \| 5 \| 7 \| 9 \| 11 \| \| g(x) \| 1 \| 3 \| 7 \| 11 \| The result is that the function g(x) has been shifted to the right by 3. Notice the output values for g(x) remain the same as the output values for f(x) in the chart, but the corresponding input values, x, have shifted to the right by 3: 2 shifted to 5, 4 shifted to 7, 6 shifted to 9 and 8 shifted to 11. Example of a Horizontal Shift with a Graph The graph shown is a transformation of a parent function . Relate this new function g(x) to f(x), and then find a formula for g(x). Notice that the graph looks almost identical in shape to the function, but the x values are shifted to the right two units. The vertex used to be at (0, 0) but now the vertex is at (2, 0) . The graph is the basic quadratic function shifted two to the right, so Notice how we must input the value x = 2, to get the output value y = 0; the x values must be two units larger, because of the shift to the right by 2 units. We can then use the definition of the f(x) function to write a formula for g(x) by evaluating : Since and Notice how the domain and range are the same for this function. The domain is subject to change with horizontal shifts, but in this case, both are all real numbers or If you find yourself having trouble determining whether the shift is +2 or -2, it might help to consider a single point on the graph. For a quadratic, looking at the vertex is convenient. In the original function, . In our shifted function, . To obtain the output value of 0 from the f function, we need to decide whether a +2 or -2 will work to satisfy . For this to work, we will need to subtract 2 from our input values. When thinking about horizontal and vertical shifts, it is good to keep in mind that vertical shifts are affecting the output values of the function, while horizontal shifts are affecting the input values of the function. Example of an Application with a Shift The function G(m) gives the number of gallons of gas required to drive m miles. Interpret and G(m+10) . is adding 10 to the output gallons. This is 10 gallons of gas more than is required to drive m miles. So, this is the gas required to drive m miles, plus another 10 gallons of gas, or maybe the amount of gas required to drive m miles and have 10 gallons left in the tank. is adding 10 to the input, miles. This is the number of gallons of gas required to drive 10 miles more than m miles. So this might be what you want if you want to give yourself a 10 mile buffer before you run out of gas. Try it Now 2 Given the function graph the original function and the transformation . a. Is this a horizontal or a vertical change? b. Which way is the graph shifted and by how many units? c. Graph and on the same axes. d. What is the domain and range of ? Now that we have two transformations, we can combine them together. Vertical Shifts are outside changes that affect the output (vertical) axis values shifting the transformed function up or down. Up if the constant is added, down if subtracted. Horizontal Shifts are inside changes that affect the input (horizontal) axis values shifting the transformed function left or right. Right if the constant is subtracted and left if it is added. Examples of a Functions Shifted both Horizontally and Vertically Given , sketch a graph of and give the new domain and range. The function f is our parent absolute value function. We know that this graph has a V shape, with the point at the origin. The graph of h has transformed f in two ways: is a change on the inside of the function, giving a horizontal shift left by 1, then the subtraction by 3 in . is a change to the outside of the function, giving a vertical shift down by 3. Transforming the graph gives: The domain of the transformed function is still , but the range has changed to . b.Write a formula for the graph shown, a transformation of the parent square root function. The graph of the parent function starts at the origin, so this graph has been shifted 1 to the right, and up 2. In function notation, we could write that as . Using the formula for the square root function we can write Note that this transformation has changed the domain and range of the function. This new graph has domain and range . Reflections Another transformation that can be applied to a function is a reflection over the horizontal or vertical axis. Examples Reflect the graph of both vertically and horizontally. Reflecting the graph vertically, each output value will be reflected over the horizontal t axis: \| \| \| \| --- \| \| \| \| Since each output value is the opposite of the original output value, we can write: Notice this is an outside change or vertical change that affects the output values so the negative sign belongs outside of the function. Reflecting horizontally, each input value will be reflected over the vertical axis. Since each input value is the opposite of the original input value, we can write: Notice this is an inside change or horizontal change that affects the input values so the negative sign is on the inside of the function. Note that these transformations can affect the domain and range of the functions. While the original square root function has domain and range , the vertical reflection gives the function the range , and the horizontal reflection gives the function the domain . (Because in order to take the square root of -t, t must be negative, for example, -2, as . Reflections Given a function , if we define a new function as , then is a vertical reflection of the function , sometimes called a reflection about the x-axis or horizontal axis. If we define a new function as , then is a horizontal reflection of the function , sometimes called a reflection about the y-axis or vertical axis. Example of Reflections with Functions defined as Tables A function is given as a table below. Create a table for the function and \| \| \| \| \| \| --- --- \| x \| 2 \| 4 \| 6 \| 8 \| \| f(x) \| 1 \| 3 \| 7 \| 11 \| For , this is a vertical reflection, so the x values stay the same and each output value will be the opposite of the original output value: \| \| \| \| \| \| --- --- \| x \| 2 \| 4 \| 6 \| 8 \| \| g(x) \| -1 \| -3 \| -7 \| -11 \| For , this is a horizontal reflection, and each input value will be the opposite of the original input value and the values stay the same as the values: \| \| \| \| \| \| --- --- \| x \| -2 \| -4 \| -6 \| -8 \| \| h(x) \| 1 \| 3 \| 7 \| 11 \| Try it Now 3 Given the parent function , graph and . Do you notice anything surprising? Stretches and Compressions With shifts, we saw the effect of adding or subtracting to the inputs or outputs of a function. We now explore the effects of multiplying the inputs or outputs. Remember, we can transform the inside (input values) of a function or we can transform the outside (output values) of a function. Each change has a specific effect that can be seen graphically. Consider the function . In the parent function, , the point in the center at (0,0) is still there in . However, what about an input of 1, the parent function has the point (1,1), but multiplies it by 2, so the point is at (1,2). If we consider an input at 8, the parent function has a point at (8,2), but is (8,4). So the output is doubled each time. This is a vertical stretch as seen below: Vertical Stretch and Compression Given a function , if we define a new function as , where is a constant then is a vertical stretch or compression of the function . If , then the graph will be stretched If , then the graph will be compressed If , then there will be combination of a vertical stretch or compression with a vertical reflection Example of a Vertical Compression with a Table A function is given as a table below. Create a table for the function \| \| \| \| \| \| --- --- \| x \| 2 \| 4 \| 6 \| 8 \| \| f(x) \| 1 \| 3 \| 7 \| 11 \| The formula tells us that the output values of are half the output values of with the same inputs. For example, we know that , so \| \| \| \| \| \| --- --- \| x \| 2 \| 4 \| 6 \| 8 \| \| g(x) \| 1/2 \| 3/2 \| 7/2 \| 11/2 \| The result is that the function has been compressed vertically by ½. Each output value has been cut in half, so the graph would now be half the original height. Example of a Vertical Stretch/Compression of a Graph The graph shown is a transformation of the parent function . Relate this new function to , then find a formula for . When trying to determine a vertical stretch or shift, it is helpful to look for a point on the graph that is relatively clear. In this graph, it appears that With the basic cubic function at the same input, Based on that, it appears that the outputs of g are ¼ the outputs of the function f, since From this we can fairly safely conclude that: We can write a formula for by using the definition of the function Try it Now 4 Use the parent function to graph and give the domain and range of . Horizontal stretches and compressions are also possible, but we can usually re-write the function so that this is a vertical stretch or compression. For example, a horizontal compression of the function by ½ would result in a new function , but this can also be written as , a vertical stretch of by 4. When writing a formula for a transformed parent function, we only need to find one transformation that would produce the graph. Combining Transformations When combining transformations, it is very important to consider the order of the transformations. For example, vertically shifting by 3 and then vertically stretching by 2 does not create the same graph as vertically stretching by 2 then vertically shifting by 3. When we see an expression like , which transformation should we start with? The answer here follows nicely from order of operations, for outside transformations. Given the output value of , we first multiply by 2, causing the vertical stretch, then add 3, causing the vertical shift. (Multiplication before Addition) Combining Vertical Transformations When combining vertical transformations written in the form , first vertically stretch by a, then vertically shift by k. Horizontal transformations are a little trickier to think about. When we write for example, we have to think about how the inputs to the g function relate to the inputs to the f function. Suppose we know . What input to g would produce that output? In other words, what value of x will allow ? We would need 3-x=12. To solve for x, we would first subtract 3, resulting in horizontal shift, then multiply by -1, causing a horizontal reflection. Independence of Horizontal and Vertical Transformations Horizontal and vertical transformations are independent. It does not matter whether horizontal or vertical transformations are done first. Example of Several Transformations with a Table Given the table of values for the function below, create a table of values for the function \| \| \| \| \| \| --- --- \| x \| 6 \| 12 \| 18 \| 24 \| \| f(x) \| 10 \| 14 \| 15 \| 17 \| There are 3 steps to this transformation and we will work from the inside out. Starting with the horizontal transformations, is a horizontal shift to the left by 2, which means we subtract each x value by 2. \| \| \| \| \| \| --- --- \| x \| 4 \| 10 \| 16 \| 22 \| \| f(x+2) \| 10 \| 14 \| 15 \| 17 \| Looking now to the vertical transformations, we start with the vertical stretch, which will multiply the output values by 3. We apply this to the previous transformation. \| \| \| \| \| \| --- --- \| x \| 4 \| 10 \| 16 \| 22 \| \| 3f(x+2) \| 30 \| 42 \| 45 \| 51 \| Finally, we can apply the vertical shift, which will add 1 to all the output values. \| \| \| \| \| \| --- --- \| x \| 4 \| 10 \| 16 \| 22 \| \| 3f(x+2)+1 \| 31 \| 43 \| 46 \| 52 \| Example of Multiple Transformation of a Graph Use the parent function , sketch a graph of First, we see that 2 in the denominator and we did not cover horiztonal stretches, so we will change this slightly to . This makes a little easier to see that the -1/2 is just a vertical compression and reflection. See the animation below: We begin with our parent function: . Let’s start with our horizontal (inside) transformation and graph which is just a shift of 3 to the left. Then, we will do our vertical transformations, starting with a vertical compression by 1/2 and then a reflection across the x-axis. Finally we have the vertical shift of 1 unit down. Example Writing a Transformed Function from the Graph Write an equation for the transformed graph function shown and give the domain and range. This looks like a parabola so let’s try our parent function Observing the graph, we notice several transformations: The original parent function has been flipped over the x axis, a vertical stretch has occurred, and we can see a shift to the right 3 units and a shift up 1 unit. In total there are four operations: Vertical reflection, requiring a negative sign outside the function Vertical Stretch of 2 (remember that concave down parabolas start with over 1 down 1, this one is over 1, down 2) Horizontal Shift Right 3 units, which tells us to put on the inside of the function Vertical Shift up 1 unit, telling us to add 1 on the outside of the function So this takes our parent function, : First do the horizontal shift Now the Stretch: Now the Reflection: And now the vertical shift: This looks familiar, it is the vertex form of a parabola! If you had trouble figuring out the transformations, that’s okay. We can still find the formula for a parabola using the techniques we discussed in chapter 5. All we need is the vertex and another point. The domain is still but the range is now Try it Now 5 a. Try graphing and give the domain and range. b. Try to figure out the what the parent function (it’s one of the ones at the beginning of the section) and transformations are, figure out the domain and range and write an equation for this graph: Try it Now Answers a. Horizontal shift b. The function is shifted to the LEFT by 2 units c. d. Domain: Range: 2. Notice looks the same as 3. Domain: Range: 4. a. Domain: Range: b. The parent function is the square reciprocal: . It is shifted right 2 units, reflected on the x-axis and up 3: Domain: Range: Media Attributions linear absvalue square cubic sqroot cuberoot recip squaredrecip graphofverticalshift graphofhorizshift 66example6 Reminder is licensed under a Public Domain license 67example1 67example2 67example3 67example4 67example5 graphofvertstretch 67example7 67example6 TIN67 tinasnwer672 tinanswer673 tinanswer674 tinanswer675
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1497	https://www.khanacademy.org/math/ncert-class-12/x7ce8ca9c5869750b:determinants-ncert-new/x7ce8ca9c5869750b:untitled-852/e/determinants-of-scaled-matrices	Determinants of scaled matrices (practice) \| Khan Academy Skip to main content If you're seeing this message, it means we're having trouble loading external resources on our website. If you're behind a web filter, please make sure that the domains .kastatic.org and .kasandbox.org are unblocked. Explore Browse By Standards Explore Khanmigo Math: Pre-K - 8th grade Math: High school & college Math: Multiple grades Math: Illustrative Math-aligned Math: Eureka Math-aligned Math: Get ready courses Test prep Science Economics Reading & language arts Computing Life skills Social studies Partner courses Khan for educators Select a category to view its courses Search AI for Teachers FreeDonateLog inSign up Search for courses, skills, and videos Help us do more We'll get right to the point: we're asking you to help support Khan Academy. 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Skip to lesson content NCERT Math Class 12 Course: NCERT Math Class 12>Unit 4 Lesson 1: Determinants Determinant of a 2x2 matrix Determinant of a 2x2 matrix Determinant of a 3x3 matrix: standard method (1 of 2) Determinant of a 3x3 matrix: shortcut method (2 of 2) Determinant of a 3x3 matrix Evaluating determinants of scaled matrices Determinants of scaled matrices Evaluating determinants involving variables Evaluating determinants involving variables Evaluating determinants involving trigonometric functions Determinants involving trigonometric functions Finding area of a triangle using determinants Area of triangle using determinants Finding equation of a line using determinants Equation of a line using determinants Math> NCERT Math Class 12> Determinants> Determinants © 2025 Khan Academy Terms of usePrivacy PolicyCookie NoticeAccessibility Statement Determinants of scaled matrices Google Classroom Microsoft Teams Problem Let A‍ be a square matrix of order 3‍. If \|A\|=5‍ and \|k A\|=40‍, what is the value of k‍? Choose 1 answer: Choose 1 answer: (Choice A) 8/3‍ A 8/3‍ (Choice B) 2‍ B 2‍ (Choice C) 8‍ C 8‍ (Choice D) 2 2‍ D 2 2‍ Report a problem Do 4 problems Skip Check Use of cookies Cookies are small files placed on your device that collect information when you use Khan Academy. Strictly necessary cookies are used to make our site work and are required. Other types of cookies are used to improve your experience, to analyze how Khan Academy is used, and to market our service. You can allow or disallow these other cookies by checking or unchecking the boxes below. You can learn more in our cookie policy Accept All Cookies Strictly Necessary Only Cookies Settings Privacy Preference Center When you visit any website, it may store or retrieve information on your browser, mostly in the form of cookies. This information might be about you, your preferences or your device and is mostly used to make the site work as you expect it to. 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1498	https://pmc.ncbi.nlm.nih.gov/articles/PMC2980660/	Cisatracurium in different doses versus atracurium during general anesthesia for abdominal surgery - PMC Skip to main content An official website of the United States government Here's how you know Here's how you know Official websites use .gov A .gov website belongs to an official government organization in the United States. Secure .gov websites use HTTPS A lock ( ) or https:// means you've safely connected to the .gov website. Share sensitive information only on official, secure websites. Search Log in Dashboard Publications Account settings Log out Search… Search NCBI Primary site navigation Search Logged in as: Dashboard Publications Account settings Log in Search PMC Full-Text Archive Search in PMC Journal List User Guide PERMALINK Copy As a library, NLM provides access to scientific literature. Inclusion in an NLM database does not imply endorsement of, or agreement with, the contents by NLM or the National Institutes of Health. Learn more: PMC Disclaimer \| PMC Copyright Notice Saudi J Anaesth . 2010 Sep-Dec;4(3):152–157. doi: 10.4103/1658-354X.71571 Search in PMC Search in PubMed View in NLM Catalog Add to search Cisatracurium in different doses versus atracurium during general anesthesia for abdominal surgery A M El-Kasaby A M El-Kasaby 1 Department of Anaesthesia, Faculty of Medicine, Suez Canal University, Ismailia, Egypt Find articles by A M El-Kasaby 1,✉, H M Atef H M Atef 1 Department of Anaesthesia, Faculty of Medicine, Suez Canal University, Ismailia, Egypt Find articles by H M Atef 1, A M Helmy A M Helmy 1 Department of Anaesthesia, Faculty of Medicine, Suez Canal University, Ismailia, Egypt Find articles by A M Helmy 1, M Abo El-Nasr M Abo El-Nasr 1 Department of Anaesthesia, Faculty of Medicine, Suez Canal University, Ismailia, Egypt Find articles by M Abo El-Nasr 1 Author information Copyright and License information 1 Department of Anaesthesia, Faculty of Medicine, Suez Canal University, Ismailia, Egypt ✉ Address for correspondence: Dr. Hosam M. Atef, Department of Anaesthesiology, Faculty of Medicine, Suez Canal University, Ismailia, Egypt E-mail: hosamatef375@yahoo.com © Saudi Journal of Anaesthesia This is an open-access article distributed under the terms of the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. PMC Copyright notice PMCID: PMC2980660 PMID: 21189851 Abstract Background: Cisatracurium in clinical practice is devoid of histamine-induced cardiovascular effects. On the other hand, 2 ED 95 doses of cisatracurium (100 µ g/kg) do not create satisfactory intubating conditions such as those seen with equipotent doses of atracurium. The recommended intubating dose of cisatracurium is 3 ED 95. To understand this discrepancy better, we evaluated the potency and onset of atracurium and cisatracurium. Materials and Methods: The study designed as randomized controlled clinical trial to compare between atracurium (2×ED 95) and different doses of cisatracurium (2×ED 95, 4×ED 95, 6×ED 95) regarding onset time, duration of action, condition of intubation, hemodynamic effects, and sings of histamine release clinically. Sixty four patients were randomly assigned to one of four groups, the first group (group 1) received 2×ED 95 dose of atracurium, group 2 received 2×ED 95 dose of cisatracurium, group 3 received 4×ED 95 dose of cisatracurium, while group 4 received 6×ED 95 dose of cisatracurium. The Datex relaxograph (Type NMT-100-23-01, S/N: 37541) for neuromuscular monitoring was used. Results: HR, MABP was statistically significant increased post-intubation with administration of 2×ED 95 dose of atracurium in group 1 and the same dose of cisatracurium in group 2 but 5-20 min later was not statistically significant with administration of 4×ED 95 and 6×ED 95 doses of cisatracurium in groups 3 and 4, respectively. Onset time was found to be significantly lower with 2×ED 95 dose of atracurium than with the same dose of cisatracurium. At the same time, higher doses of cisatracurium (4×ED 95 and 6×ED 95) showed onset time and longer duration of action that was significantly lower than with atracurium and with lower dose of cisatracurium (2×ED 95). Only 6×ED 95 dose of cisatracurium showed statistically significant difference versus the atracurium dose with higher percentages of patients with excellent condition of intubation. 4×ED 95 and 6×ED 95 doses of cisatracurium were significantly better than 2×ED 95 dose of cisatracurium. 2×ED 95 dose of atracurium and 2×ED 95 dose of cisatracurium were similar, while 4×ED 95 and 6×ED 95 doses of cisatracurium were significantly better than atracurium and 2×ED 95 dose of cisatracurium. Conclusion: The same dose (2×ED 95 dose) atracurium is more effective neuromuscular blocking agent than cisatracurium, while higher doses of cisatracurium 4×ED 95 and 6×ED 95 provide more effective, more rapid neuromuscular blocking with longer duration of action, stable hemodynamic status, and no associated signs of histamine release clinically. Keywords:Cisatracurium, atracurium, different ED95 dose and neuromuscular monitoring INTRODUCTION Muscle relaxants rapidly became a routine part of the anesthesiologist’s drug arsenal. The neuromuscular blocking potency of cisatracurium (NIMBEX) is approximately three-fold that of atracurium besylate, the time to maximum block is up to 2 min longer for equipotent doses of NIMBEX compared to atracurium besylate. The clinically effective duration of action and rate of spontaneous recovery from equipotent doses of NIMBEX and atracurium besylate are similar. Although cisatracurium is more potent than the parent mixture (95% effective dose (ED 95) 0.05 mg/kg vs. 0.2 mg/kg), its pharmacodynamic profile is similar to that of atracurium, except for a reportedly slower onset. Cisatracurium unlike atracurium is devoid of histamine-induced cardiovascular effects. On the other hand, 2 ED 95 doses of cisatracurium (100 μg/kg) do not yield satisfactory intubating conditions such as those seen with equipotent doses of atracurium. The recommended intubating dose of cisatracurium is 3 ED 95.[4,5] Aim of the work This study was designed to compare between atracurium (2×ED 95) and different doses of cisatracurium (2×ED 95, 4×ED 95, 6×ED 95) regarding onset time, condition of intubation, duration of action, hemodynamic effects, and signs of histamine release. MATERIALS AND METHODS The study was carried out as a comparative clinical trial on patients of both sexes underwent elective abdominal surgery in Suez Canal University Hospital in the routine surgical theaters. Inclusion criteria ASA I and ASA II patients aged from 20 to 65 years old, both sexes scheduled for abdominal surgeries of an anticipated duration of at least 1 h and half included in the study. Exclusion criteria were any disorder of the cardiovascular, hepatic, renal, or neuromuscular systems known from history or clinical examination. Patients in whom difficult intubation was expected: pregnant or lactating women and patients on medication known to interact with neuromuscular blocking drugs e.g. Antibiotics (aminoglycosides and tetracycline), antidepressants, anticonvulsants antiarrhythmics (calcium channel blockers and quinidine) and magnesium sulfate. The detectable difference between the means of the group using the onset of action (time from end of injection to 90% neuromuscular block) and it equals 0.7 minutes. The calculated sample is 16 per group with total sample size 64 for the four groups of the study. The 64 patients were equally and randomly divided into four groups: Group 1: atracurium for 16 patients with initial dose of 0.5 mg/kg (2×ED 95). Group 2: cisatracurium for 16 patients with initial dose of 0.1 mg/kg (2×ED 95). Group 3: cisatracurium for 16 patients with initial dose of 0.2 mg/kg (4×ED 95). Group 4: cisatracurium for 16 patients with initial dose of 0.3 mg/kg (6×ED 95). Patients were randomly allocated using an online research randomizer ( into four equal groups. The techniques were explained to patients including benefits and complications of each and written consent was taken. The patient’s age, sex, ASA status, duration, and type of surgery were recorded. Patients were premedicated using 2 mg midazolam through IV route 20 min preoperatively. Monitoring equipments (Datex-Ohmeda™) were attached to the patient including three leads ECG, non-invasive blood pressure, pulse oximetry, capnography, and temperature probe. The Datex relaxograph (Type-NMT-100-23-01, S/N: 37541) for neuromuscular monitoring. The first response (T 1) of the train of four (TOF) stimulation was the parameter, which used for the pharmacodynamic measurements. The hand, wrist and half of the forearm were wrapped with crepe bandage to avoid hypothermia. Patients were preoxygenated with 100% oxygen for 3 min. General anesthesia was induced with fentanyl (1-1.5 μg/kg), followed 20 s later by propofol (2 mg/kg) intravenously. Anesthesia was maintained with a mixture of 50% N 2 O in O 2 and isoflurane (0.5%-1.5% vol%) and assisted ventilation. Neuromuscular monitoring was carried out after obtaining the control values by supramaximal stimulus (70 mA) from relaxograph (2 Hz/0.5 s; pulse width 0.2 ms) every 20s to stimulate the ulnar nerve via surface electrodes. After a stable base line period of at least 5 min, the muscle relaxant was given for patients according to the previously mentioned initial doses for each group and injected intravenously within 5-10 s. After 2 min, endotracheal intubation was attempted using proper size tube (male: 8-9, female: 7-7.5) and the condition of intubation was assessed and recorded according to the following: Excellent: Easy passage of the tube without coughing. Vocal cords relaxed and abducted. Good: Passage of the tube with slight coughing and/or bucking. Vocal cords relaxed and abducted. Poor: Passage of tubes with moderate coughing and/or bucking vocal cords moderately adducted. Not possible: Vocal cords not relaxed, tightly adducted. The onset time was determined as the interval from the end of muscle relaxant injection until the maximal suppression of T 1%. Anesthesia was maintained with a mixture of 50% N 2 O in O 2, isoflourane (1-1.5 MAC), boluses of the muscle relaxant (10% of the initial dose) with 25% recovery of response to T 1% and ventilation was controlled by the Datex-Ohmeda™ ventilator which will adjust end tidal CO 2 at (30-35 mmHg). Neuromuscular blockade after induction was monitored and recorded every 5 min by supramaximal train-of-four (TOF) stimuli. The duration of the muscle relaxant (time from the end of injection of the drug until 25% recovery of T 1%) was recorded. Patients were monitored for any signs of histamine release clinically through skin changes graded as flush (if redness lasted> 120 s), erythema, or wheals and presence of any hemodynamic changes or bronchospasm. Intra-operative hemodynamic changes were continuously displayed on the monitor including: heart rate (HR), mean arterial blood pressure (MABP) every 5 min, oxygen saturation (SO 2), and end tidal CO 2. Body temperature was maintained between 35 and 37°C by means of warmed IV fluids and warming blankets (body core temperature through nasopharyngeal probe and skin temperature probe). At the end of operation with 25% recovery of T 1%, reversal (induced recovery) was achieved by administration of neostigmine and atropine mixture (2.5 mg neostigmine: 1 mg atropine) through slow IV injection. TOF-ratio>0.9 was sufficient for safe extubation of the trachea. Statistical methods Data were processed using SPSS version 15 (SPSS Inc., Chicago, IL, USA). Quantitative data were expressed as means±SD while qualitative data were expressed as numbers and percentages (%). Student t test and ANOVA test were used to test significance of difference for quantitative variables (HR, BP) that follow normal distribution and chi square was used to test the significance of difference for qualitative variables. A probability value (P-value)<0.05 was considered statistically significant. RESULTS The studied patients were matched regarding age and sex with no statistically significant difference being recorded [Table 1]. Table 1. Demographic characteristics of the studied patients \| \| \| \| Atracurium group (n=16) 2×ED 95 \| Cisatracurium group (n=16) 2×ED 95 \| Cisatracurium group (n=16) 4×ED 95 \| Cisatracurium group (n=16) 6×ED 95 \| P-value \| :---: :---: :---: :---: \| \| Age \| Mean ± SD \| \| 43.3 ± 6.5 \| 45.7 ± 3.1 \| 42.9 ± 7.3 \| 39.9 ± 5.4 \| 0.05 \| \| \| Range \| \| (28 – 54) \| (31 – 56) \| (27 – 50) \| (29 – 53) \| (NS) \| \| Sex \| Male \| N (%) \| 9 (56.25%) \| 8 (50%) \| 6 (37.5%) \| 7 (43.75%) \| 0.7 \| \| \| Female \| N (%) \| 7 (43.75%) \| 8 (50%) \| 10 (62.5%) \| 9 (56.25%) \| (NS) \| Open in a new tab NS - No statistically significant difference There was a statistically significant increase in HR, MABP post intubation when compared to baseline and postinjection of 2×ED 95 dose of atracurium in group 1 and the same dose of cisatracurium in group 2. HR, MABP changes 5-20 minutes later were not statistically significant with administration of 4×ED 95 and 6×ED 95 doses of cisatracurium in groups 3 and 4, respectively [Tables 2 and 3]. Table 2. Heart rate changes before and after administration of atracurium or cisatracurium \| \| Heart rate (beat/min) \| :---: \| \| \| Baseline reading \| After injection of muscle relaxant \| After attempt of intubation \| 5 min \| 10 min \| 15 min \| 20 min \| \| Atracurium group (n=16) 2×ED 95 \| 69.8 ± 6.65 \| 73.8 ± 7.9 \| 83.1 ± 7.96 \| 73.9 ± 8.96 \| 73.3 ± 6.81 \| 74.1 ± 6.75 \| 73.6 ± 6.57 \| \| Cisatracurium group (n=16) 2×ED 95 \| 73.4 ± 5.61 \| 73.9 ± 6.3 \| 84.6 ± 547 \| 76.3 ± 6.51 \| 75.4 ± 5.87 \| 73.6 ± 5.78 \| 74.1 ± 5.65 \| \| Cisatracurium group (n=16) 4×ED 95 \| 70.8 ± 4.83 \| 71.2 ± 5.2 \| 75.2 ± 3.45 \| 74.9 ± 6.53 \| 76.7 ± 5.56 \| 73.1 ± 6.75 \| 72.6 ± 5.84 \| \| Cisatracurium group (n=16) 6×ED 95 \| 71.1 ± 6.54 \| 71.8 ± 7.3 \| 74.9 ± 5.54 \| 73.8 ± 6.45 \| 75.4 ± 6.85 \| 72.4 ± 4.75 \| 73.7 ± 5.38 \| Open in a new tab Statistically significant difference versus Baseline reading (P-value < 0.05) Table 3. Mean arterial blood pressure changes before and after administration of atracurium or cisatracurium \| \| Mean arterial blood pressure (mmHg) \| :---: \| \| \| Baseline reading \| After injection of muscle relaxant \| After attempt of intubation \| 5 min \| 10 min \| 15 min \| 20 min \| \| Atracurium group (n=16) 2×ED 95 \| 81.5 ± 8.37 \| 78.9 ± 9.4 \| 91.6 ± 6.4 \| 86.4 ± 7.28 \| 82.6 ± 6.89 \| 81.5 ± 8.75 \| 80.7 ± 7.54 \| \| Cisatracurium group (n=16) 2×ED 95 \| 80.9 ± 10.7 \| 79.8 ± 10.5 \| 92.6 ± 7.59 \| 88.4 ± 8.64 \| 84.6 ± 9.65 \| 85.7 ± 7.43 \| 83.5 ± 8.93 \| \| Cisatracurium group (n=16) 4×ED 95 \| 82.6 ± 9.5 \| 82.1 ± 8.5 \| 84.5 ± 6.73 \| 83.5 ± 7.52 \| 85.6 ± 8.42 \| 84.9 ± 8.64 \| 86.4 ± 9.24 \| \| Cisatracurium group (n=16) 6×ED 95 \| 81.7 ± 10.1 \| 80.1 ± 9.6 \| 85.3 ± 8.43 \| 86.4 ± 9.34 \| 84.9 ± 6.95 \| 85.9 ± 6.58 \| 82.6 ± 8.68 \| Open in a new tab Statistically significant difference versus Baseline reading (P-value < 0.05) Time onset was found to be significantly lower with 2×ED 95 dose of atracurium than with the same dose of cisatracurium. At the same time, higher doses of cisatracurium (4×ED 95 and 6×ED 95) showed onset time that was significantly lower than with atracurium and with lower dose of cisatracurium (2×ED 95). Regarding the duration of action, higher doses of cisatracurium (4×ED 95 and 6×ED 95) showed statistically significant longer duration of action than lower doses of cisatracurium and the atracurium (2×ED 95) [Table 4]. Table 4. Neuromuscular blockade after administration of atracurium and cisatracurium \| \| Onset time (Time to maximum suppression of T1%) (min) \| Duration of action (25% recovery T1%) (min) \| :---: \| Atracurium group (n=16) 2×ED 95 \| 3.24 ± 0.55 \| 44.4 ± 4.13 \| \| Cisatracurium group (n=16) 2×ED 95 \| 4.37 ± 0.46 \| 43.6 ± 4.15 \| \| Cisatracurium group (n=16) 4×ED 95 \| 2.9 ± 1.4 # \| 65.5 ± 10.5 # \| \| Cisatracurium group (n=16) 6×ED 95 \| 2 ± 1.2 # \| 78.4 ± 8.6 # \| Open in a new tab Statistically significant difference versus 2×ED95 dose of atracurium (P-value < 0.05); Statistically significant difference versus 2×ED95 dose of cisatracurium (P-value < 0.05) Only 6×ED95 dose of cisatracurium was statistically significant versus the atracurium dose with higher percentages of patients with excellent condition of intubation. 4×ED 95 and 6×ED 95 doses of cisatracurium were significantly better than 2×ED 95 dose of cisatracurium. No one of the studied patients in the four groups been reported as not possible intubation. Assessment of vocal cords, 2×ED 95 dose of atracurium and 2×ED 95 and 4×ED 95 doses of cisatracurium were similar while 6×ED 95 dose of cisatracurium was significantly better than atracurium and 2×ED 95 dose of cisatracurium [Table 5]. Table 5. Condition of intubation and vocal cords assessment as recoded after 2 minutes of administration of atracurium and cisatracurium \| \| \| \| Atracurium group (n=16) 2×ED 95 \| Cisatracurium group (n=16) 2×ED 95 \| Cisatracurium group (n=16) 4×ED 95 \| Cisatracurium group (n=16) 6×ED 95 \| :--- :---: :---: \| Vocal cords \| Open \| N (%) \| 9 (56.25) \| 6 (37.5) \| 11 (68.75) \| 16 # (100) \| \| \| Abducted \| N (%) \| 5 (31.25) \| 8 (50) \| 5 (31.25) \| 0 (0) \| \| \| Adducted \| N (%) \| 2 (12.5) \| 2 (12.5) \| 0 (0) \| 0 (0) \| \| Condition of intubation \| Excellent \| N (%) \| 6 (37.5) \| 2 (12.5) \| 10 # (62.5) \| 13 # (81.25) \| \| \| Good \| N (%) \| 8 (50) \| 9 (56.25) \| 5 (31.25) \| 3 (18.75) \| \| \| Poor \| N (%) \| 2 (12.5) \| 5 (31.25) \| 1 (6.25) \| 0 (0) \| \| \| Not possible \| N (%) \| 0 (0) \| 0 (0) \| 0 (0) \| 0 (0) \| Open in a new tab Statistically significant difference versus 2×ED 95 dose of atracurium (P-value < 0.05); Statistically significant difference versus 2×ED 95 dose of cisatracurium (P-value < 0.05); Figures in parenthesis are in percentage No signs of histamine release were noted with any doses of cisatracurium, while it was noted with atracurium (2 cases; 1 case showed flush and the other case showed erythema). DISCUSSION All patients were assessed for hemodynamic state (heart rate, blood pressure), onset time, duration of action, and signs of histamine release clinically, condition of intubations, and vocal cords assessment. The four groups of the study were matched regarding patients’ age and sex. Hemodynamic stability for both heart rate and mean arterial blood pressure were more evident among higher doses of cisatracurium (4×ED 95, 6×ED 95). There was a statistically significant increase in HR, MABP postintubation 120 s postinjection of the muscle relaxant when compared to baseline and postinjection of 2×ED 95 dose of atracurium in group 1 and the same dose of cisatracurium in group 2 because of stress intubation and the patients were not fully relaxed. However, changes in HR and MABP 5-20 minutes later were not statistically significant with administration of 4×ED 95 and 6×ED 95 doses of cisatracurium in groups 3 and 4, respectively. Lien _et al_., and Basta et al., concluded that the maximal MABP and HR changes of patients receiving cisatracurium were small and similar to those observed in patients receiving two times the ED 95 of atracurium. In his study no patient developed a decrease in blood pressure >20% or an increase in heart rate >20% that was attributable to muscle relaxant administration. Signs of histamine release were shown in one patient in this study in the form of transient facial flushing after the administration of atracurium; however, this patient did not experience hypotension or tachycardia. The onset time was determined as the interval from the end of muscle relaxant injection until the onset of the maximal suppression of T 1 and the duration of action of the muscle relaxant was defined as time from disappearance of TOF stimulation till 25% recovery of T 1. 2×ED 95 dose of atracurium had more rapid onset of action (with statistical significance) than the equivalent dose of cisatracurium (2×ED 95). But higher doses of cisatracurium (4×ED 95 and 6×ED 95) were found to be statistically significant more rapid onset of action and longer duration of action than 2×ED 95 dose of both atracurium and cisatracurium. Bluestein and colleagues, studied 80 ASA physical status I or II, 18 – 70 years of age whom were randomly assigned to four groups (A-D). Group A received cisatracurium 0.1 mg/kg (2×ED 95), group B received atracurium 0.5 mg/kg (2×ED 95). Patients in group C and group D were treated with cisatracurium 0.2 mg/kg (4×ED 95) and 0.15 mg/kg (3×ED 95), respectively. They assessed the mean time of onset, mean time of clinically effective duration, and condition of intubation. As regarding the mean time of onset and mean time of clinically effective duration there results were in accordance with ours. They reported that increasing the initial dose of cisatracurium (from 0.1 to 0.15 and 0.2 mg/kg), decreased the mean time of onset (from 4.6 to 3.4 and 2.8 min, respectively) and increased the mean time of clinically effective duration (45 to 55 and 61 min, respectively). Mellinghoff et al., studied 80 patients randomized to receive either cisatracurium (n=40) or atracurium (n=20) and compared the time course of neuromuscular block. Results obtained by Mellinghoff et al., were similar to our results. They estimated that onset times were 3.1±1.0 min with cisatracurium and 2.3±1.1 min with atracurium (P=0.008). After the infusion, the spontaneous recovery intervals were 25-75% of 18±11 min and 18±8 min for cisatracurium and atracurium (P=0.896) [Table 6]. Table 6. Histamine release among the studied patients \| \| \| Atracurium group (n=16) 2×ED 95 \| Cisatracurium group (n=16) 2×ED 95 \| Cisatracurium group (n=16) 4×ED 95 \| Cisatracurium group (n=16) 6×ED 95 \| P-value \| :---: :---: :---: \| Flush \| N (%) \| 1 (6.5) \| 0 (0) \| 0 (0) \| 0 (0) \| 0.5 \| \| Erythema \| N (%) \| 1 (6.5) \| 0 (0) \| 0 (0) \| 0 (0) \| (NS) \| \| Wheals \| N (%) \| 0 (0) \| 0 (0) \| 0 (0) \| 0 (0) \| \| Open in a new tab NS - No statistically significant difference; Figures in parenthesis are in percentage As regards the condition of intubation in our study, it was estimated that only 6×ED 95 dose of cisatracurium showed a statistically significant difference versus the atracurium dose with excellent condition of intubation. 4×ED 95 and 6×ED 95 doses of cisatracurium were significantly better than 2×ED 95 dose of cisatracurium. There was not any case of not possible intubation among the four studied groups. The assessment of vocal cords, 2×ED 95 dose of atracurium, and 2×ED 95 dose of cisatracurium were similar, while 4×ED 95 and 6×ED 95 doses of cisatracurium were significantly better than atracurium and 2×ED 95 dose of cisatracurium. Results found by Bluestein et al., were consistent with our results. They reported that intubation conditions were good or excellent in over 90% of patients in all treatment groups (2 min after approximately 2×ED 95 doses of cisatracurium or atracurium and 1.5 min after 3× and 4×ED 95 doses of cisatracurium). Mandal conducted a study in 60 adult patients of either sex, belonging to physical status ASA grade I or 11 to find out the minimum possible dose of cisatracurium for achieving excellent to good intubating conditions within 90 s of its administration under general anesthesia. Patients were divided into three groups according to their dosage schedule. After induction of anesthesia with the standardized method, group I (n=20) received 0.15 mg/kg, group II (n=20) received 0.20 mg/kg, and group III (n=20) received 0.25 mg/kg of cisatracurium. For each group laryngoscopy and intubation was tried at either 75 or 90s, thereby patients were further divided into six subgroups. Subgroup ‘a’ denotes the procedure at 75s and, ‘b’ denotes at 90s. For grading the intubating conditions, the ease of laryngoscopy, the position or movement of the vocal cords and the degree of coughing were evaluated. Excellent to good intubating conditions could be achieved only in group IIb (0.20 mg/kg at 90 s) and both subgroups (0.25 mg/kg at 7 5s and 90 s) in group III patients. Hence the minimum dose required to achieve excellent to good intubating conditions with cisatracurium is 0.20 mg/kg at 90 s after its administration. The adequacy of conditions for tracheal intubation is a function of several factors, such as the depth of anesthesia at the time of the intubation attempt and the level of neuromuscular block at the time of attempt. One of two intubating doses of cisatracurium may be chosen based on the desired time of intubation and the anticipated length of surgery. Doses of 0.15 mg/ kg (3×ED 95) and 0.2 mg/kg (4×ED 95) of cisatracurium, as components of a propofol /nitrous oxide/oxygen induction intubation technique, may produce generally good or excellent conditions of intubation in 2.0 and 1.5 min, respectively. The cisatracurium dose of 0.15 mg/kg (3×ED 95) is higher than the dose of atracurium 0.5 mg/kg (2×ED 95) required to produce clinically acceptable intubation conditions after 120 s. However, this dose of cisatracurium still provides neuromuscular block of intermediate duration and following rapid administration not associated with histamine-mediated cardiovascular effects. No signs of histamine release were noted in any doses of cisatracurium while it was noted with atracurium (2 cases; 1 case showed flush and the other case showed erythema). Signs of histamine release were shown in two patients with administration of 2×ED 95 dose of atracurium in group 1, one patient in the form of flushing at the site of injection in the ventral aspect of the forearm but the patient did not experience hypotension or tachycardia and the other patient had shown transient erythema after the administration of atracurium and tachycardia only (an increase in heart rate<20% of base line) without any decrease in blood pressure. As a benzylisoquinoline, atracurium has the potential for release of histamine. The syndrome becomes clinically evident when doses of 0.5 mg/kg (two times ED 95) or more are injected rapidly. When plasma histamine levels increase to over 1000 pg/ml, a transient decrease in blood pressure, together with facial erythema, may be noted. 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1499	https://www.reddit.com/r/askmath/comments/1e99idm/taylor_expansion_for_sqrt1x/	Taylor expansion for Sqrt(1+x) : r/askmath Skip to main contentTaylor expansion for Sqrt(1+x) : r/askmath Open menu Open navigationGo to Reddit Home r/askmath A chip A close button Log InLog in to Reddit Expand user menu Open settings menu Go to askmath r/askmath r/askmath This subreddit is for questions of a mathematical nature. Please read the subreddit rules below before posting. 207K Members Online •1 yr. ago PsychologicalHold121 Taylor expansion for Sqrt(1+x) Analysis Im trying to understand whats wrong with my estimate for the error of a Taylor expansion for f(x)=Sqrt(1+x) expanded at 0. Computing the remainder as in the picture should be 210^(-10) at x=1, but when I evaluate \|Sqrt(2)-T_10(1)\|, the error is 10^(-4). Any help is appreciated. Sqrt expansion Read more Archived post. New comments cannot be posted and votes cannot be cast. Share Related Answers Section Related Answers Understanding Taylor expansion for Sqrt(1+x) Taylor series expansion for Sqrt(1+x) Taylor series for square root functions Taylor expansion error estimation tips Taylor series for 1/(1-x) New to Reddit? Create your account and connect with a world of communities. Continue with Email Continue With Phone Number By continuing, you agree to ourUser Agreementand acknowledge that you understand thePrivacy Policy. Public Anyone can view, post, and comment to this community Top Posts Reddit reReddit: Top posts of July 22, 2024 Reddit reReddit: Top posts of July 2024 Reddit reReddit: Top posts of 2024 Trending topics today Alligator Alcatraz shut down Famine declared in Gaza Bungie CEO steps down Pulse memorial crosswalk erased 4chan rejects UK safety fines Mariah Carey gets Vanguard Apple TV+ price jumps to $13 Dafne Keen joins Percy Jackson Lil Nas X hospitalized in LA Trump fraud fine voided Brent Hinds dies at 51 Bulls to retire Rose's #1 Apollonia sues Prince estate The Pitt S2 teaser drops early Coppola, Dunst reunite Texas House passes new maps Gabbard to halve intel staff Judge Caprio dies at 88 Hurricane Erin triggers tropical storm warning in North Carolina KPop Demon Hunters breaks out Bears extend Bagent Reddit RulesPrivacy PolicyUser AgreementAccessibilityReddit, Inc. © 2025. All rights reserved. Expand Navigation Collapse Navigation

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