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188500 | https://www.youtube.com/watch?v=ndnJ7CSQ6v8 | Factorising quadratics by finding factor pairs
mattam66
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Description
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Posted: 22 Sep 2014
To watch many more free videos visit website :
where I have sorted the videos in different playlists easy to navigate.
Transcript:
Let's look at some uh examples of factorizing quadratic expression. Now what is a quadratic? So let me write a quadratic. A quadratic would look somewhat like this. Say you got x^2 + 7 x + 6. So you want to factoriize you want to factoriize this. The question is to factoriize. So how do you factoriize? Factoriize means basically you have to put this in two brackets. So this is two brackets in the form you start with x and x. Now factorization is just the opposite of expansion. So in expanding two brackets uh you start with x and and you got a number here and then you got a number here. So you're thinking backwards. So here what you're doing is you're finding the two factors of six or pairs of factors of six. Factors of six or factor pairs. So let me write not factor factor pairs of six factor pairs of six. So you have to find the two factors of six which adds to give you seven. We'll see why later. So the factor pairs are you can have one and six can be one factor one pair. You can have two and three. You can have of course six and one. This is the same thing repeated or three and two. You can also have negative. Okay, I'm not going into negative. Yeah, it can also be -1 and -6. So basically what we are saying is 1 6 is 6. 2 3 is 6. 6 1 is 6. 3 1 is 6. So as there are no negatives, we are not concerned about negative. So which are the two f factors which adds to give you seven? Well, this seems to be right because 1 6 is 6. That's what we want. And 1 + 6 is 7. So the factors are or you put + 1 here and + 6 here. Does it make sense? It does make sense because if you expand this out, you should get this. So let's expand this out. So this is also a good practice for expansion. So this is what we saw in the last two lessons. So expanding this out to check this is just to check whether we have got the factors right. So this is x x x + 6 + 1 x + 6. Okay. Okay. So, if you expand this out further, so this is x x + 1 sorry x x + 1 x + 1 x. So, this is what factorization is. So, this just explain this in a reverse order. So, this is x 2 + 6 x + 1 x + 6. So can you see this is 1 6 and this is 1 + 6 and that's why we are looking for two numbers or two pairs or two factors of six or the pairs of factor pairs of six which adds to give you seven and this is your answer. That's all. Okay, let's do one more. Okay. So what would happen if you have x^2 + uh say 8 x + 12. Well, we are we are thinking of pairs factor pairs of 12. Factor pairs of 12. Okay. So well let's start from one. 1 6 sorry 1 12. So if you add 1 and 12 will it give you will it give 8? No. So let's cross that out. You have 2 6. So that 2 6 is 12 but and 2 + 6 is also 8. So this seems to be right. And so the factors are x + 2 x + 6. Now to check your answer, expand this out. And if you expand this out, you should get this back. Okay. What would happen if you have x^2 + 7 x + 12? Okay. So same. So let's find the factors of 12 or factor pairs of 12. So that's let's start with 1 12. 1 and 12 doesn't give you 7. So we'll cross that out. 2 6 is 12, but that doesn't give you 7. We're looking for this. These two numbers should add to give you seven. So that doesn't. Now you got 3 4. Well, that does seem to work. Why? because 3 + 4 is 7. So the answer is x + 3 x + 4. Okay. So let's do a few more. What would happen? Let me make it a slightly difficult. What would happen if you have x 2? This is a bit challenging. x^2 + 3x - 10. So now factor pairs of 10 factors. So let's start with 1 -10. We're looking for -10. So 1 + 1 + -10 is -9. So this is not right. What if you take 2 -5? 2 5 is 10, but 2 + -5 is -3. So, that's not working. What if I swap it around? 5 -2. 5 -2 is also -10 + 5 + -2 is + 3. So the factors so this is right 5 and -2. So the factors are x + 5 x - 2. Always expand this out and check whether you're getting the |
188501 | https://cldr.unicode.org/translation/date-time/date-time-patterns | Date/Time Patterns
Last updated: 2025-Apr-01
Patterns Introduction
Pre-requisite topics to read:
Date/Time Symbols
Date/Time Names
A date-time pattern is a string of characters in which fields representing date and time symbols are combined together with necessary “literal” strings that are used as is.
There are two types of substrings that are combined in a pattern:
Date/time fields, which are placeholders that represent particular calendar values such as month, weekday, year, etc..
“Literal” strings used as-is. These are necessary to make the pattern flow naturally in a given language.
For example:
| Skeleton format | English Pattern | English Example | Japanese Pattern | Japanese Example |
--- ---
| yMMMd | MMM e, y | Dec 5, 2010 | y年M月d 日 | 2010年12月5日 |
There are three aspects to patterns:
A letter in the set {a-z; A-Z} indicates the type of calendar field: See Date/Time Symbols.
For example:
M for month
d for week day
y for year
The number of letters indicate the form of the names to be used.
Example use for month would be:
Numeric representation: M or MM for 9 or 09, with leading 0. This is intended to be used in conjunction with a day number; thus, 9/12 or 09/12
Abbreviated form: MMM for Sep
Full form: MMMM for September
Narrow form: MMMMM (S)
When formatting a particular date, non-numeric values (e.g. month names from Gregorian calendar) are substituted into the appropriate pattern substrings using the Date Format Symbols.
For example:
M replaced by “3” for March
MMMM replaced by “March”.
When parsing a date string, the pattern substrings will be converted into the appropriate numeric calendar data.
See Date/Time Symbols for a brief summary of letters for different calendar fields, or Date Field Symbol Table for a more complete and detailed list.
The “literal” text in the date-time pattern are used as-is when formatting, and are expected when parsing a date string. To include a single straight quote as part of the literal text, use two of them together: ‘’ (either inside or outside a section of literal text enclosed in single straight quotes).
To demonstrate how this work with examples, see the spec Date Format Pattern Examples.
💡 Translation Tips
For “literal” text that should be included in patterns, enclose it in single straight quotes if it includes letters A-Za-z or a single straight quote. For example, to include “o’clock” in a long time pattern (hh ‘o’‘clock’ a, zzzz). If you need to include the single straight quote itself in literal text, you can do that by doubling it as in the previous example.
In order to provide the correct date and time patterns for your language, you may need to reorder the symbols. For example, you would need to reorder the U.S. date pattern d/M/y to MM/dd/y if your locale puts the month first in short date patterns.
For bidi scripts (e.g. Arabic and Hebrew), you may need to add directionality markers (U+200E ( LEFT-TO-RIGHT MARK, U+200F RIGHT-TO-LEFT MARK, U+061C ARABIC LETTER MARK). Typically these are most commonly needed in short (numeric) dates, usually before separators such as ‘/’ or ‘-‘.
For date formats in bidi scripts, the Survey Tool shows examples in both a right-to-left context and a neutral context. In the future it may show examples in a left-to-right contex as well.
The hightest priority is to make the formats look correct in a right-to-left context; the next most important is to make them look correct in a a neutral cotext. If possible it is also a good idea to make them work in a left-to-right context.
Always look at examples! Examples are in the Information Pane or hover over. When working with date and time formats, there are many substitution and dependencies to different calendar; therefore, the example are best representation to validate your intention on the end result.
Synchronizing Date/Time Names and Patterns
There is a tight coupling between the date/time patterns and the names that are used for date/time elements, described in Date/Time Names. Therefore, when supplying the data, it requires coordination between what you supply as the names and your intention on which name to use in patterns.
When to use Standalone vs. Formatting
Some types of names have two styles and these are differentiated in the survey tool section headings a Formatting and Standalone.
Using the month names as examples:
format style: Use formatting-style month names represented with “M” when month names is displayed with a day-of-the-month number (e.g. d MMMM y: 5 Monday January 1999).
stand-alone style: Use the stand-alone month names represented with “L” when the month name is without a day-of-the month. (e.g. LLLL y: Monday 1999)
Following are examples of differences and implications based on language:
| | wide format months (Use pattern: MMMM) | wide stand-alone months Use pattern: LLLL) | pattern for full date, e.g.in e-mail log: “d. MMMM y”; thus, intended to use the format style names as examples below. | pattern for month + year, e.g. in calendar: “LLLL y”; thus intended to use the stand-alone names as examples below: | pattern for month name by itself, e.g. in menu: “LLLL” ; thus, intended to use the stand-alone names as examples below: |
--- --- --- |
| 7 | heinäkuuta | heinäkuu | 5. heinäkuuta 2018 | heinäkuu 2018 | heinäkuu |
| 8 | elokuuta | elokuu | 10. elokuuta 2018 | elokuu 2018 | elokuu |
| 9 | syyskuuta | syyskuu | 15. syyskuuta 2018 | syyskuu 2018 | syyskuu |
| 10 | lokakuuta | lokakuu | 20. lokakuuta 2018 | lokakuu 2018 | lokakuu |
Some language (e.g. Catalan) use a preposition to combine the month and day number, e.g. “11 de setembre” (11 of September). If the month name begins with a vowel, the preposition is contracted, e.g. “12 d’octubre”.
For these languages:
The format month names should include the preposition in its correct form (contracted or not)
And the stand-alone month names should NOT include the preposition
And the patterns using the stand-alone should not include the preposition for month
In some languages, the weekday name format forms (e.g. EEEE) and stand-alone forms (e.g. cccc) forms may also differ, and may require similar coordination between names and patterns.
Below examples are with month names.
| | wide format months Use pattern (MMMM) | wide stand-alone months Use pattern (LLLL) | pattern for full date, e.g.in e-mail log: “d MMMM ‘de’ y”; thus, intended to use the format style names as examples below: | pattern for month + year, e.g. in calendar: “ LLLL ‘de’ y”; thus intended to use the stand-alone names as examples below: | pattern for month name by itself, e.g. in menu: “ LLLL ”; thus, intended to use the stand-alone names as examples below: |
--- --- --- |
| 7 | de juliol | juliol | 5 de juliol de 2018 | juliol de 2018 | juliol |
| 8 | d’agost | agost | 10 d’agost de 2018 | agost de 2018 | agost |
| 9 | de setembre | setembre | 15 de setembre de 2018 | setembre de 2018 | setembre |
| 10 | d’octubre | octubre | 20 d’octubre de 2018 | octubre de 2018 | octubre |
| Format String | Date | Result |
---
| LLLL | 2008-1-14 | gener |
| | 2008-4-14 | abril |
| d MMMM ‘de’ y | 2008-1-14 | 14 de gener de 2008 |
| | 2008-4-14 | 14 d’abril de 2008 |
💡 Translation Tips
When working with patterns, look at the examples in the right information pane to validate your intention between the name and the pattern.
In order to get the right formats for your language, you may need to change the ordering of the Date/Time Symbols, and change the text around them.
You must match the names that should be used in patterns with intended forms of the names (wide or full, abbreviated and/or short, narrow forms)
For example:
If you specify the time format “h:mm a”, by using “a” your intention is to use the Abbreviated formatting name for AM/PM; thus, the end result being 2:37 PM.
If you specify the time format “h:mm aaaaa”, by using “aaaaa”, your intention is to use the Narrow formatting name for AM/PM; thus, the end result being 2:37 p.
Understand the difference between formatting and standalone date and time names in your language and its intended usage in patterns. For example, if you are working in Catalan (a locale that uses prepositions in formatting month names), and you provide “setembre” for the formatting month name instead of “de setembre,” then the pattern d MMMM will display as “12 septembre” instead of the correct pattern “12 de setembre.
Some languages (e.g. Finnish and many Slavic languages) use a month name in nominative case when it is displayed without a day number, and use a different case (genitive, partitive, etc.) when the month is displayed with a day number. For these languages:
The stand-alone month names should be in nominative case,
And the format month names should be in genitive or a related case.
Even for language that do not require such different forms it is a good idea, for consistency, to use ‘L’ for months in patterns without ‘d’, and ‘M’ for months in patterns with ‘d’.
Basic Time Formats
The standard formats include four basic time formats. See Survey Tool Formats-Standard-Time formats.
full: Include hour (h/H), minute (mm), second (ss), and zone (zzzz).
long: Include hour, minute, second, and zone (z)
medium: Include hour, minute, second.
short: Include hour, minute.
Using 24 hour vs 12 hour symbols: The symbols you use in time format patterns depend on whether the main country/region of the language uses 12-hour time or 24 hour format.
If the default country/region of the language use the 12-hour time clock, use patterns like “h:mm a” or “hh:mm a”
h to mean a 12-hour clock cycle running 1 through 12 (midnight plus 1 minute is 12:01) with hh indicating leading zero to 2 digits
K to mean a 12-hour clock cycle running 0 through 11 (midnight plus 1 minute is 0:01).
a to get the equivalent of AM/PM
b to add special representation of noon/midnight
B to use day periods like “in the afternoon” instead of AM/PM.
If the default country/region of the language uses the 24-hour time clock, use a pattern like “H:mm” or “HH:mm”
H to mean a 24-hour clock cycle running 0 through 23 (midnight plus 1 minute is 0:01) with HH indicating zero-padding to 2 digits
k to mean a 24-hour clock cycle running 1 through 24 (midnight plus 1 minute is 24:01).
a, b, and B are same as usage in 12 hour formats, but are not normally used with 24-hour time.
If a non-default country/region use a different time format than the default country/region for the language, then this difference should be added for the sub-locale. Also see Regional variants.
For example, if es (= es_ES, Spanish as used in Spain) uses 24-hour time formats, but es_US (Spanish as used in United States) uses 12 hour time formats, then es would use “HH:mm/H:mm” and es_US would use the “h” formats as described above. .
💡 Translation Tips
Languages should always have translations for AM/PM/noon even if every country/region that the language is used in has 24 hour time. Many computing systems provide the flexibility for users to specify either 12 or 24 hour time setting on their system (also known as User preference overrides). This is why CLDR provide the flexible formats for both 12 and 24 hour systems.
Basic Date Formats
The standard Date formats include four basic formats. See Survey Tool Formats-Standard-Date formats. (Note that the Month symbol especially may vary (M vs MM, or MMM vs LLL). For more information, see Date/Time Symbols.)
| Form | Include | English pattern examples |
---
| Full | long-weekday (EEEE), year (y), long-month (MMMM), day In some non-Gregorian calendars, it will also contain G representing the era name, such as “AH” or “Heisei” (i.e. “平成”). For some languages, long months are not used; for example, in Japanese the numeric month is used in patterns, in front of the character 月. | EEEE, MMMM d, y (e.g. “Tuesday, September 14, 1999”) |
| Long | year (y), long-month (MMMM), day (d). | MMMM d, y (e.g. “September 14, 1999”.) |
| Medium | year (y), abbreviated-month (MMM), day (d). For languages that do not use abbreviated months , use the numeric month (MM/M). For example, “y/MM/dd” , corresponding to a date like “1999/09/14”. | MMM d, y (e.g. “Sep 14, 1999”.) |
| Short | year, numeric-month (MM/M), and day. | M/d/yy (e.g. “9/14/99”) |
Additional Date-Time Formats
The basic formats as described in above sections provide a small subset of the combinations. To expand to fully cover all variations of date and time formats, a skeleton format is provided that uses the Date/Time Symbols (without order or punctuation) to indicate what fields are desired.
An example usage of the flexible formats would be: a software program only needs the year and month information, the flexible pattern yMMM would be used to provide the desired year and month formatting as shown in the English and Japanese examples in this table.
| Skeleton | English Pattern | English Example | Japanese Pattern | Japanese Example |
--- ---
| yMMM | MMM y | Dec 2010 | y年M月 | 2010年12月 |
| Skeleton | Hypothetical English case | | For ja@calendar=chinese | | |
--- --- --- |
| | Pattern | Example | | | |
| supplied | MMMEd | E MMM d | Tue Apr 18 | MMMd日(E) | 四月18日(火) |
| inferred 1 | MMMMEd | E MMMM d | Tue April 18 | MMMMd日(E) | 四月18日(火) |
| inferred 2 | MMMEEEEd | EEEE MMM d | Tuesday Apr 18 [incorrect] | MMMd日(EEEE) | 四月18日(火 曜日) [incorrect] |
| inferred 3 | MMMMEEEEd | EEEE MMMM d | Tuesday April 18 [incorrect] | MMMMd日(EEEE) | 四月18日(火曜日) [incorrect] |
Standalone vs. Formatting in Flexible formats
In languages that distinguish between nominative and genitive (or related form such as partitive), the use of symbols in flexible formats also specify the use of nominative forms or genitive (or related) forms of month and day names.
month names when day is not included (e.g. skeleton: yMMM)
Specify the nominative forms in the standalone month names.
Use the symbol “LLL” or “LLLL” in the patter, therefore, “LLLLy”
month names when day is included (e.g. skeleton: MMMd)
Specify the genitive form (or a related form such as partitive) in format month names
Use the symbol “MMM” or “MMMM”; therefore resulting in “d MMMM”
day names may also vary for your language depending on the elements included in the pattern (e.g. skeleton E for day only, or skeleton MEd for month, day, and numeric day)
Use symbols “ccc” or “cccc” in patterns to use the standalone name
Use the symbol “E” or “EEEE” to use the format name
To understand which pattern characters are used for standalone forms (nominative) versus format forms (genitive or related, such as partitive), see Stand-Alone vs Format Styles, see also the discussion above in Synchronizing Date/Time Names and Patterns.
Supplied vs. Inferred patterns
Some patterns supplied for each skeleton may be extended from the abbreviated form to an inferred full form by developers who use CLDR.
Some Inferred patterns usage are incorrect, instead use the most closely associated skeletons that are available. Build on a chain of fallback options based on available patterns is a good practice.
Example expansions:
| | | | | | |
--- --- --- |
| supplied | MMMEEEEd | EEEE, MMM d | Tuesday, Apr 18 [correct] | MMMd日EEEE | 四月18日火曜日 [correct] |
| inferred 4 | MMMMEEEEd | EEEE, MMMM d | Tuesday, April 18 [correct] | MMMMd日EEEE | 四月18日火曜日 [correct] |
The correctness of inferred from abbreviated the full will differ by locale.
In the examples above table, Inferred 2 and Inferred 3 examples are incorrect, because:
English examples are incorrect, because you’d want a comma after the full day name EEEE (Tuesday, Apr 18).
Japanese examples are incorrect, because the full day name should not be in parenthesis.
Some languages do not use abbreviated months; therefore, a skeleton containing abbreviated month MMM is mapped to a pattern that uses numeric month M or MM. For example, skeleton yMMMd may be supplied with the numeric month d.M.y.
| Skeleton | Pattern | Example | |
--- --- |
| supplied | yMMMd | d.M.y | 18.4.2015 |
| supplied | yMMMMd | d. MMMM y | 18. huhtikuuta 2015 |
| without the second entry, would have: inferred | yMMMMd | d.M.y | 18.4.2015 |
💡 Translation Tips
In the Survey tool, the skeleton formats are under Formats-Flexible-Date formats and Formats-Flexible-12/24 Time formats.
The skeletons are listed in the Code column in Survey Tool.
For each skeleton formats, provide the ordering and punctuation that would be used in your locale. For example:
ordering could be, whether it is common to write month first (MMMy) or year first (yMMM)
punctuation could be, a comma (,) that may be needed between month and day
Follow the patterns in the survey tool and use the examples to understand the skeletons. Provide data for all flexible patterns.
: If your language has different grammatical forms for date symbols such as month and day names, the nominative forms of the names should be in the Survey Tool section marked “Standalone”, and the genitive or related forms should be in the Survey Tool section marked “Formatting”.
Day period patterns
There are three pattern characters that can you can use to indicate the day period marker in 12-hour-cycle time formats:
| Symbol | Meaning | English example |
---
| a (abbreviated) aaaa (full) aaaaa (narrow) | AM and PM | 12:00 AM, 10:00 AM, 12:00 PM, 7:00 PM |
| b (abbreviated) bbbb (full) bbbbb (narrow) | AM, PM, noon and midnight, | 10:00 AM, 12:00 noon, 7:00 PM |
| B (abbreviated) BBBB (full) BBBBB (narrow) | day periods with locale-specific ranges. If a locale does not have data for such day period ranges, this is equivalent to ‘b’. | 2:00 at night, 10:00 in the morning, 12:00 in the afternoon, 7:00 in the evening. |
With skeletons (listed in the Code column in Survey Tool) that specify ‘h’ but no day period, the corresponding patterns are specified with a day period, normally ‘a’ or ‘b’. The patterns can be used to infer how to handle a request for a different day period. For example:
| Code | English pattern (example) | |
---
| supplied | hm | h:mm a |
| inferred | ahm | h:mm a |
| inferred | aaaahm | h:mm aaaa |
| inferred | bhm | h:mm b |
| inferred | bbbbhm | h:mm bbbb |
| inferred | Bhm | h:mm B [see notes below] |
| inferred | BBBBhm | h:mm BBBB [see notes below] |
Beginning in CLDR 32, there are 5 additional time patterns that indicate how times should be formatted using day period ranges ‘B’. This allows locales to have formats for day period ranges that can display the range symbol in a different position than might be used for symbols for AM, PM, noon, or midnight:
| Code | English pattern (example) | |
---
| supplied | Bh | B h |
| supplied | Bhm | B h:mm |
| supplied | Bhms | B h:mm:ss |
| supplied | EBhm | B h:mm, E |
| supplied | EBhms | B h:mm:ss, E |
| inferred | BBBBhm | BBBB h:mm |
| inferred | … | … |
These new patterns are available in Gregorian and Generic calendars, In these new fields, provide locale data by:
💡 Translation Tips
Examples in the Survey Tool show some day periods for the locale as shown in this screenshot below. If the examples for day period ranges show them in the wrong position, then the time formats specific to using day period ranges may be updated.
Placing the pattern character (‘B’ in the example below) in the correct position.
If it is equally good in two different positions, favor the position used for the ‘a’ pattern character.
If the position needs to be different depending on the width of B, please file a ticket.
For more information on the day periods used in your language:
First see Day Periods (AM, PM, etc.) for a general discussion of how day periods work.
See the Day Periods chart for your language.
Date-Time Combined Formats
The date-time pattern shows how to combine separate patterns for date (represented by {1}) and time (represented by {0}) into a single pattern. It usually doesn’t need to be changed.
What you want to pay attention to are:
💡 Translation Tips
Remove a space if your language does not use spaces, which is common for many East Asian languages
Add a comma, or other punctuation that your language requires between the patterns
There are four format length styles: full, long, medium, and short. Each of these may come in up to three variants:
The standard variant (required), which is used for multiple purposes and typically does not include any literal text, for example “{1}, {0}”. Usage examples include:
Wall clock time: “Sunday, September 25, 1:30 PM”
Combining a date with a time range: “Sunday, September 25, 1:30 – 3:00 PM”
The “-atTime” variant, which specifies a fixed date at a particular time, typically for an event. In the longer formats (such as full and long),
this “-atTime” form may have a combining word linking the date and the time, for example “{1} ‘at’ {0}” to produce an English example like “Sunday, September 25 at 1:30 PM”).
If there is no “-atTime” variant, the standard variant will be used for this combination.
The “-relative” variant (new in CLDR 48), which specifies a relative date at a particular time, typically for an event. In the longer formats (such as full and long),
this “-relative” form may have a combining word linking the date and the time, for example “{1} ‘at’ {0}” to produce an English example like “tomorrow at 1:30 PM”.
If there is no “-relative” variant, the standard variant will be used for this combination.
Before CLDR 42, there was only one variant for these. In English that variant used the “-atTime” style, as did many other locales. For CLDR 42, that “-atTime” data was moved to the “-atTime” variants, and the standard data was initially extrapolated from the medium or short formats without literal text.
In CLDR 48, the “-relative” variant is split out of the “-atTime” variant to allow a different combining form to be used when the date part is relative (“tomorrow”) instead of a fixed date (“March 20”).
If you want the combination of a relative date and single time to be formatted as it was in CLDR 47, then you will need to copy the “-atTime” variant data to the new “-relative” variant.
The determination of which length style to use by developers using CLDR data is normally based on the date style, for example:
If the date has a full month and weekday name, use the full combining pattern.
If the date has numeric month, use the short version of the combining pattern.
Following are examples on how the data can be different by locale with different combinations of format length. (note: {1}=date format with {0}= time format)
Another way to look at the example with original patterns and combined result:
| Pattern | English | German | Japanese |
--- --- |
| Full {1} ‘at’ {0} {Full form of the date format} at {full form of the time format} | {1} ‘at’ {0} Sunday, September 5, 1999 at 1:25:59 PM Eastern Standard Time | {1} ‘um’ {0} Sonntag, 5. September 1999 um 13:25:59 Nordamerikanische Ostküsten-Normalzeit | {1} {0} 1999年9月5日日曜日 13時25分59秒 アメリカ東部標準時 |
| Medium {1}, {0} {medium form of the date format}, {medium form of the time format} Notice the comma and space between the date portion and the time portion. | {1}, {0} Sep 5, 1999, 1:25:59 PM | {1}, {0} 05.09.1999, 13:25:59 | {1} {0} 1999/09/05 13:25:59 |
| date pattern | time pattern | date-time combining pattern used | formatted example |
--- --- |
| MMMM d, y | h:mm a | [long] {1} ‘at’ {0} | September 14, 1999 at 1:25 PM |
| M/d/yy | h:mm a | [short] {1}, {0} | 9/14/99, 1:25 PM |
Week-Of Patterns
| Survey Tool field | English pattern | Pattern characters |
---
| yw-one | ‘week’ w ‘of’ Y (example: Week 37 of 2009) | w designates the number of the week within a year calculated for week-of year purposes and indicated using the pattern character Y (instead of the normal year designator y). The year indicated by Y typically begins on the locale’s first day of the week and ends on the last day of the week, so its transitions may differ by a few days from the standard year indicated by y. |
| yw-other | ‘week’ w ‘of’ Y | |
| MMMMW-one | ‘week’ W ‘of’ MMMM (example: Week 3 of April) | W designates the number of the week within the month |
| MMMMW-other | ‘week’ W ‘of’ MMMM | |
The week-of date patterns were introduced in CLDR 30 for enumerating week count in larger periods, e.g. “week 15 of 2016” or “week 4 of April”. The Survey Tool fields and corresponding English entries are shown below:
For developers who use CLDR data,
The pattern can be selected based on the plural form associated with the week number, in case the form depends on the number. (either -one or -other)
Currently these patterns only support cardinal numbers; in the future they may be extended to support ordinal numbers for usages such as “2nd week of April”. In this case the distinction by plural form associated with week number may become more relevant.
💡 Translation Tips
If your language has a grammatical difference in the month names(aside from just the simple addition of a prefix or suffix), localizers will need to use a work-around construction (file a ticket if this is the case for your language).
As described under Date/Time Names, CLDR also provides a relative week period pattern which is used for constructions such as “the week of April 11, 2016”. The English pattern that produces this is “the week of {0}”; the date format that replaces {0} is determined separately. Because the week-of patterns described here may appear in user interfaces that also show dates produced using the relative week period patterns, all of these patterns should be designed with consistent wording and structure.
Flexible - Timezone Append
timezone: {0} {1}
See in Survey Tool.
The flexible append format for Timezone is used to compose a time and a timezone, to get a result like “11:30 Pacific Time”.
{0} will be replaced by the localized time format.
{1} will be replaced by the localized timezone.
For almost all locales it is just “{0} {1}” , but some locales may change the order, add punctuation, or remove the space.
Date/Time Intervals
Interval patterns contain a start pattern and an end pattern (using the Date/Time Symbols). They are used for a range of dates or times, such as “Sept 10–12” (meaning the 10th of September through the 12th of September). The interval format is used where it is necessary to make the pattern as short as possible, and elide information that does not have to be repeated. For example, the pattern used to get “Sept 10–12” in English is “MMM d–d”.
Unlike simple Date/Time Patterns, these consist of two parts, typically separated by with some kind of punctuation mark (e.g. English uses en-dash “ – “). Also, some fields in the second part are omitted (e.g. “d – d MMM” omits repeat of MMM for the second part). The first field that comes from the second date is marked with red in the examples below.
Interval Formatting
| Format Pattern | Date 1 | Date 2 | Result |
--- --- |
| MMM d– d | 2008-09-13 | 2008-09-15 | Sept 13–15 |
| MMMM– MMMM , yyyy | 2008-09-01 | 2008-11-31 | September–November, 2008 |
Greatest Difference
Each combination of fields can be used with dates that differ by different amounts. For example, a format for the fields “yMMMd” (year, abbreviated month, and day) could be used with two dates that differ by year, month, or day.
As the examples below indicate, for a given skeleton, each type of difference might need a different pattern.
For example,
when the greatest difference is a year, no part of the second pattern is omitted;
when the greatest difference is a month, then the year is not repeated, since it would be the same for each pattern.
| Date 1 | Date 2 | Greatest Difference | Format Pattern | Shares |
--- ---
| 2008-09-13 | 2009 -09-15 | year (yMd) | MMM d, yyyy – MMM d, yyyy | nothing |
| 2008-09-01 | 2008- 11 -31 | month (Md) | MMM d – MMM d, yyyy | year |
| 2008-09-01 | 2008-09- 05 | day (d) | MMM d– d , yyyy | year and month |
Different Calendars
Date and Time patterns are also available for other calendars in addition to the Gregorian calendar. For example, the Buddhist, Islamic, Hebrew, or Japanese calendars.
Different calendars work with the data in Gregorian, and Generic in the following ways:
Basic time formats for all calendars are inherited from the Gregorian time formats.
In many locales, the main difference between date formats for Gregorian calendars and non-Gregorian calendars is that non-Gregorian calendars always include the calendar era with the year.
To avoid having to do this separately for each non-Gregorian calendar, CLDR has a special “Generic” calendar to specify standard date formats for non-Gregorian calendars.
If you specify the date formats for the “Generic” calendar, using year with a calendar era, then those data formats will be used for most non-Gregorian calendars unless those calendars explicitly specify their own date formats.
Because the Generic calendar does not have real names for months, weekdays and eras, the Survey Tool examples generated for this calendar may be confusing.
Calendars that do not inherit date formats from the Generic calendar are the East Asian lunar calendars: Chinese (lunar) and Dangi (Korean lunar). These have special formats involving cyclic names. The Dangu calendar inherits formats from the Chinese calendar data in the same locale, while the Chinese calendar inherits formats directly from the parent locale; that parent locale may be the root locale or inherit these formats directly from the root locale. For the lunar calendars, the root locale has formats that should be reasonable for use in most locales where the lunar calendars are not one of the primary calendars.
Year First Calendar
This is a variant of the Gregorian calendar whose formats always use year-month-day ordering and a 24-hour time cycle.
Note: the code is iso8601, but disregard that; it will be changed after submission.
| Page | Code | English |
---
| Date & Time / Gregorian YMD | yMMMd | y MMM d |
| Date & Time / Gregorian YMD | … | … |
| Locale Display Names / Keys | calendar-iso8601 | Gregorian Calendar (Year First) |
Please go though the Gregorian Calendar (Year First) fields.
You should change to what is most customary in your language:
punctuation (/, -, :, ., …) around numeric fields
other text separating non-numeric fields (era, day-of-week, day-periods, zones)
special forms (stand-alone vs. formatting) as needed: M → L, E → c
You should not change other features:
The ordering of fields in the pattern should be strictly the following order (for any that occur in a particular pattern):
era - year - month - day - day-of-week - hour - minute - second
Each numeric field should be consistently 2 digits: MM, dd, HH, hh, mm, ss
Each string field should match the Code in width: G, MMM, MMMM, E, EEE
But should use a different letter for stand-alone vs. formatting as needed.
It is helpful to have a second window up showing the Date & Time / Gregorian equivalents, so that you can be consistent (except for ordering) with the Gregorian calendar. |
188502 | https://tutorial.math.lamar.edu/classes/calciii/TICylindricalCoords.aspx | Paul's Online Notes
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Paul's Online Notes Home / Calculus III / Multiple Integrals / Triple Integrals in Cylindrical Coordinates
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Section 15.6 : Triple Integrals in Cylindrical Coordinates
In this section we want do take a look at triple integrals done completely in Cylindrical Coordinates. Recall that cylindrical coordinates are really nothing more than an extension of polar coordinates into three dimensions. The following are the conversion formulas for cylindrical coordinates.
[x = r\cos \theta \hspace{0.25in}y = r\sin \theta \hspace{0.25in}z = z]
In order to do the integral in cylindrical coordinates we will need to know what (dV) will become in terms of cylindrical coordinates. We will be able to show in the Change of Variables section of this chapter that,
[dV = r\;dz\,dr\,d\theta ]
The region, (E), over which we are integrating becomes,
[\begin{align}E & = \left{ {\left( {x,y,z} \right)|\left( {x,y} \right) \in D,\,\,\,{u_1}\left( {x,y} \right) \le z \le {u_2}\left( {x,y} \right)} \right}\ & = \left{ {\left( {r,\theta ,z} \right)|\alpha \le \theta \le \beta ,\,\,{h_1}\left( \theta \right) \le r \le {h_2}\left( \theta \right),\,\,\,{u_1}\left( {r\cos \theta ,r\sin \theta } \right) \le z \le {u_2}\left( {r\cos \theta ,r\sin \theta } \right)} \right}\end{align}]
Note that we’ve only given this for (E)’s in which (D) is in the (xy)-plane. We can modify this accordingly if (D) is in the (yz)-plane or the (xz)-plane as needed.
In terms of cylindrical coordinates a triple integral is,
[\iiint\limits_{E}{{f\left( {x,y,z} \right)\,dV}} = \int_{{\,\alpha }}^{{\,\beta }}{{\int_{{{h_1}\left( \theta \right)}}^{{{h_2}\left( \theta \right)}}{{\int_{{{u_1}\left( {r\cos \theta ,r\sin \theta } \right)}}^{{{u_2}\left( {r\cos \theta ,r\sin \theta } \right)}}{{r\,f\left( {r\cos \theta ,r\sin \theta ,z} \right)\,dz}}\,dr}}\,d\theta }}]
Don’t forget to add in the (r) and make sure that all the (x)’s and (y)’s also get converted over into cylindrical coordinates.
Let’s see an example.
Example 1 Evaluate ( \displaystyle \iiint\limits_{E}{{y\,dV}}) where (E) is the region that lies below the plane (z = x + 2) above the (xy)-plane and between the cylinders ({x^2} + {y^2} = 1) and ({x^2} + {y^2} = 4).
Show Solution
There really isn’t too much to do with this one other than do the conversions and then evaluate the integral.
We’ll start out by getting the range for (z) in terms of cylindrical coordinates.
[0 \le z \le x + 2\hspace{0.25in}\hspace{0.25in} \Rightarrow \hspace{0.25in}\hspace{0.25in}0 \le z \le r\cos \theta + 2]
Remember that we are above the (xy)-plane and so we are above the plane (z = 0)
Next, the region (D) is the region between the two circles ({x^2} + {y^2} = 1) and ({x^2} + {y^2} = 4) in the (xy)-plane and so the ranges for it are,
[0 \le \theta \le 2\pi \hspace{0.25in}\hspace{0.25in}1 \le r \le 2]
Here is the integral.
[\begin{align}\iiint\limits_{E}{{y\,dV}} & = \int_{0}^{{2\pi }}{{\int_{1}^{2}{{\int_{0}^{{r\cos \theta + 2}}{{\left( {r\sin \theta } \right)r\,dz}}\,dr}}\,d\theta }}\ & = \int_{0}^{{2\pi }}{{\int_{1}^{2}{{{r^2}\sin \theta \left( {r\cos \theta + 2} \right)\,dr}}\,d\theta }}\ & = \int_{0}^{{2\pi }}{{\int_{1}^{2}{{\frac{1}{2}{r^3}\sin \left( {2\theta } \right) + 2{r^2}\sin \theta \,dr}}\,d\theta }}\ & = \int_{0}^{{2\pi }}{{\left. {\left( {\frac{1}{8}{r^4}\sin \left( {2\theta } \right) + \frac{2}{3}{r^3}\sin \theta } \right)} \right|_1^2\,d\theta }}\ & = \int_{0}^{{2\pi }}{{\frac{{15}}{8}\sin \left( {2\theta } \right) + \frac{{14}}{3}\sin \theta \,d\theta }}\ & = \left. {\left( { - \frac{{15}}{{16}}\cos \left( {2\theta } \right) - \frac{{14}}{3}\cos \theta } \right)} \right|_0^{2\pi }\ & = 0\end{align}]
Just as we did with double integral involving polar coordinates we can start with an iterated integral in terms of (x), (y), and (z) and convert it to cylindrical coordinates.
Example 2 Convert (\displaystyle \int_{{\, - 1}}^{{\,1}}{{\int_{{\,0}}^{{\,\sqrt {1 - {y^2}} }}{{\int_{{\,{x^2} + {y^2}}}^{{\,\sqrt {{x^2} + {y^2}} }}{{xyz\,dz}}\,dx}}\,dy}}) into an integral in cylindrical coordinates.
Show Solution
Here are the ranges of the variables from this iterated integral.
[\begin{array}{c} - 1 \le y \le 1\ 0 \le x \le \sqrt {1 - {y^2}} \ {x^2} + {y^2} \le z \le \sqrt {{x^2} + {y^2}} \end{array}]
The first two inequalities define the region (D) and since the upper and lower bounds for the (x)’s are (x = \sqrt {1 - {y^2}} ) and (x = 0) we know that we’ve got at least part of the right half a circle of radius 1 centered at the origin. Since the range of (y)’s is ( - 1 \le y \le 1) we know that we have the complete right half of the disk of radius 1 centered at the origin. So, the ranges for (D) in cylindrical coordinates are,
[\begin{array}{c}\displaystyle - \frac{\pi }{2} \le \theta \le \frac{\pi }{2}\ 0 \le r \le 1\end{array}]
All that’s left to do now is to convert the limits of the (z) range, but that’s not too bad.
[{r^2} \le z \le r]
On a side note notice that the lower bound here is an elliptic paraboloid and the upper bound is a cone. Therefore, (E) is a portion of the region between these two surfaces.
The integral is,
[\begin{align}\int_{{\, - 1}}^{{\,1}}{{\int_{{\,0}}^{{\,\sqrt {1 - {y^2}} }}{{\int_{{\,{x^2} + {y^2}}}^{{\,\sqrt {{x^2} + {y^2}} }}{{xyz\,dz}}\,dx}}\,dy}} & = \int_{{\, - {\pi }/{2}\;}}^{{\,{\pi }/{2}\;}}{{\int_{{\,0}}^{{\,1}}{{\int_{{\,{r^2}}}^{{\,r}}{{r\left( {r\cos \theta } \right)\left( {r\sin \theta } \right)z\,dz}}\,dr}}\,d\theta }}\ & = \int_{{\, - {\pi }/{2}\;}}^{{\,{\pi }/{2}\;}}{{\int_{{\,0}}^{{\,1}}{{\int_{{\,{r^2}}}^{{\,r}}{{z{r^3}\cos \theta \sin \theta \,dz}}\,dr}}\,d\theta }}\end{align}]
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188503 | https://projecteuclid.org/journals/journal-of-the-mathematical-society-of-japan/volume-59/issue-4/Recursively-renewable-words-and-coding-of-irrational-rotations/10.2969/jmsj/05941199.pdf | doi: 10.2969/jmsj/05941199 Recursively renewable words and coding of irrational rotations By Shigeki AKIYAMA and Masayuki SHIRASAKA (Received Feb. 14, 2007) Abstract.
A word generated by coding of irrational rotation with respect to a general decomposition of the unit interval is shown to have an inverse limit structure directed by substitutions. We also characterize primitive substitutive rotation words, as those having quadratic parameters.
1.
Definitions and the results.
Let A ¼ f0; 1; . . . ; m 1g be a finite set of letters and A be the monoid over A generated by concatenation, having the identity element , the empty word. The set of right infinite words over A is denoted by A N.
A sturmian word z is an element of A N characterized by the property that pzðnÞ ¼ n þ 1, where pzðnÞ is the number of factors (i.e. subwords) of length n appears in z. The function pzðnÞ is called the complexity of z. Since pzð1Þ ¼ 2, we have A ¼ f0; 1g.
The sturmian word is known to have the lowest complexity among aperiodic words. The aperiodicity implies that exactly one of f00; 11g appears in z. Let us assume that 11 is forbidden in z. Then the sturmian word z ¼ z0z1 2 f0; 1g with z0 ¼ 0 allows a decomposition into a word over B ¼ f0; 01g. An important fact is that this new word over B is again a sturmian word. This property is effectively used to recode sturmian words by the continued fraction algorithm (see Chapter 6 in ). We wish to generalize this combinatorial property.
Let us come back to a general A ¼ f0; 1; . . . ; m 1g. An element z ¼ z0z1 2 A N is k-renewable if there is a finite set B A with #B k and B 6 A such that z is decomposed into an infinite word over B. For a given k, if the element z 2 A N allows infinitely many times this decomposition into k-blocks, then z is called recursively k-renewable. To be more precise, z ¼ z0z1 . . . is recursively k-renewable when there is a sequence of finite sets Bi ði ¼ 0; 1; . . .Þ with #Bi k, Biþ1 B i , Biþ1 6 Bi, B0 ¼ A and zðiÞ ¼ zðiÞ 0 zðiÞ 1 2 BN i is k-renewable by Biþ1 and decomposed into zðiþ1Þ ¼ zðiþ1Þ 0 zðiþ1Þ 1 2 BN iþ1 and the length of each zðiþ1Þ j as a word in A diverges1 as i ! 1.
Here we put zð0Þ i ¼ zi for i ¼ 0; 1; . . .. We also say that z 2 A N is recursively renewable if it is recursively k-renewable with some k.
2000 Mathematics Subject Classification.
Primary 37B10; Secondary 68R15, 37E05, 11J70, 11A63.
The first author is supported by the Japanese Ministry of Education, Culture, Sports, Science and Technology, Grant-in-Aid for fundamental research (No. 18540022, 2006–2008).
Key Words and Phrases.
continued fraction, irrational rotation, interval exchange, substitution.
1We must assume this divergence to remove trivial cases (a remark due to J. Cassaigne).
2007 The Mathematical Society of Japan J. Math. Soc. Japan Vol. 59, No. 4 (2007) pp. 1199–1234 EXAMPLE 1.
A purely periodic word vvv . . . with v 2 A is recursively 1-renew-able by taking Bi ¼ fv2ig. An eventually periodic word uvvv . . . with u; v 2 A is recursively 2-renewable by Bi ¼ fuv2i1; v2ig.
EXAMPLE 2.
The sturmian word is recursively 2-renewable.
Let C be a non empty finite set. A morphism is a monoid homomorphism from A to C that ðaÞ 6¼ for each a 2 A . Then naturally extends to a map from A N to C N.
A morphism is called letter to letter, if ðaÞ 2 C for each a 2 A . A substitution is a morphism from A to itself.
EXAMPLE 3.
Let be a morphism from A to C . The image by of a recursively k-renewable word in A N is recursively k-renewable in C N.
EXAMPLE 4.
Let be a substitution on A . If ð0Þ ¼ 0w with w 6¼ , then there is a unique fix point z of in A N which begins with the letter 0. The word z is successively approximated by nð0Þ ðn ¼ 1; 2; . . .Þ and we write z ¼ limn nð0Þ. The fix point z is recursively m-renewable by Bi ¼ fið0Þ; . . . ; iðm 1Þg.
In other words, z is recursively k-renewable if there is a sequence figi¼1;2;... of substitutions on f0; 1; . . . ; k 1g that z 1 z2 2 z3 3 z4 4 . . .
with zi 2 f0; 1; . . . ; k 1gN and the length of 12 . . . iðaÞ diverges for each letter a as i ! 1, i.e., z lies in lim if0; 1; . . . ; k 1gN, the inverse limit directed by fig.
We recall the definition of general rotation words. Take 2 ½0; 1Þ n Q and 2 ½0; 1Þ.
Start with a decomposition of the unit interval I ¼ ½0; 1Þ ¼ [ k1 i¼0 ½!i; !iþ1Þ ð1Þ with 0 ¼ !0 < !1 < < !k1 < !k ¼ 1 and put Ii ¼ ½!i; !iþ1Þ. We identify ½0; 1Þ with the torus T ¼ R=Z and define J : T ! f0; 1; . . . ; k 1g by x 2 IJðxÞ for x 2 T. Then the general rotation word of an angle and an initial value with respect to k-block decomposition (1) is defined by JðÞJð þ ÞJð þ 2Þ 2 f0; 1; . . . ; k 1gN: The classical rotation word comes from the decomposition ½0; 1Þ ¼ ½0; 1 Þ [ ½1 ; 1Þ.
It is well known (cf. , , , ) that the set of all classical rotation word coincides with that of all sturmian words.2 For classification of words of complexity 2n, Rote used the generalized rotation word with respect to 2-blocks and also gave 2Precisely, we need to consider another decomposition ð0; 1 ¼ ð0; 1 [ ð1 ; 1 in order to show the equivalence to sturmian words.
1200 S. AKIYAMA and M. SHIRASAKA a combinatorial characterization of the words with respect to ½0; 1=2Þ [ ½1=2; 1Þ. Further connection between general rotation words and sturmian words had been studied.
Didier characterized the general rotation words in terms of sturmian words and cellular automata. Berstel and Vuillon showed a way to recode generalized rotation words with respect to k-blocks into k-tuples of sturmian words.
In this paper, we first prove the following theorem.
THEOREM 1.
A general rotation word of an angle and an initial value with respect to k-block decomposition (1) is recursively ðk þ 1Þ-renewable.
There are recursively renewable words which can not be a general rotation word.
For examples, take a fix point z of the substitution ð0Þ ¼ 001; ð1Þ ¼ 111 introduced by Rote as a concrete word with pzðnÞ ¼ 2n. Example 4 says that z is recursively 2-renewable but z contains factors 1n for all n 1, which is impossible for a general rotation word. Another example is the fix point g of the Rauzy substitution: ð0Þ ¼ 01; ð1Þ ¼ 02; ð2Þ ¼ 0; having its complexity pgðnÞ ¼ 2n þ 1. Then g is recursively 3-renewable but not a general rotation word. Indeed, g has arbitrary large special factors w that w0; w1; w2 are also the factors of g, but general rotation words can not have this property. The maximal pattern complexity also tells apart that g can not be a general rotation word (cf. ). It is known that g is a coding of 6-interval exchange of T and moreover a natural coding of a rotation on T2 (cf. , ). It remains a problem to characterize generalized rotation words among recursively renewable words. In addition, authors got to know a relevant result with Theorem 1 after submission of this paper.
An element z 2 A N is primitive substitutive if it is an image of a morphism of a fixed point of a primitive substitution.3 Among recursively renewable words viewed as elements of the inverse limit, primitive substitutive words correspond to eventually periodic sequences figi¼1;2;... of substitutions. Durand and Holton-Zamboni independently4 gave a combinatorial characterization of primitive substitutive words using return words. With the help of their result and the idea of the proof of Theorem 1, we can characterize primitive substitutive rotation words; THEOREM 2.
A general rotation word of an angle and an initial value with respect to the decomposition (1) is primitive substitutive if and only if is quadratic irrational, 2 QðÞ and !i 2 QðÞ for all i.
Note that the last condition ‘ 2 QðÞ and !i 2 QðÞ for all i’ is equivalent to ‘!i 2 QðÞ for all i’.
3It is equivalent to being an image of a letter to letter morphism of a fixed point of a primitive substitution.
See Proposition 3.1 in .
4From p.13 and p.34 of .
Recursively renewable words and coding of irrational rotations 1201 Theorem 2 generalizes Theorem 7.8 of , the same result for ½0; !1Þ [ ½!1; 1Þ under a condition !1 = 2 Z þ Z, and Proposition 2.11 of which treated all sturmian case, i.e., ½0; 1 Þ [ ½1 ; 1Þ. Theorem 2 may also be expected from number of quadratic type results on characterization of sturmian words fixed by substitutions (cf. , , , and ). The proof of Theorem 2 seems more number theoretical than those in and , and Section 6–Section 10 are devoted to it. A basic idea is to show directly that return words with respect to a long prefix give a coding of a certain three interval exchange. Ostrowski’s numeration system in Section 9 is unconventionally used to control induced discontinuities and to deduce unique ergodicity of three interval exchanges in Section 10.
The authors wish to thank P. Arnoux, V. Berthe ´, J. Cassaigne, S. Ferenczi, T.
Komatsu, T. Kamae, H. Rao, M. Yoshida and L. Q. Zamboni for helpful discussion and supplying relevant references.
2.
Negative continued fraction and induced rotations.
Let bxc be the maximum integer not greater than x, and dxe ¼ bxc, i.e., the minimum integer not less than x. Define a map S : ð0; 1Þ ! ð0; 1Þ by SðxÞ ¼ d1=xe 1=x and set an ¼ d1=Sn1ðxÞe and xn ¼ SnðxÞ for x 2 ð0; 1Þ n Q. Then we have x ¼ 1 a1 1 a2 1 a3 ..
.
1 an xn ð2Þ with an 2 and also an infinite continued fraction: x ¼ 1 a1 1 a2 1 a3 ..
.
: ð3Þ Both of them are called the negative continued fraction of x. There are infinitely many n’s such that an > 2 in the infinite continued fraction. Let 0 ¼ 1 and 1 ¼ . We define nþ1 2 ½0; nÞ inductively by: nþ1 ¼ ann n1 with a positive integer an. The choice of an is unique since nþ1 2 ½0; nÞ and = 2 Q implies n 6¼ 0. We can easily show 1202 S. AKIYAMA and M. SHIRASAKA ¼ 1 a1 1 a2 1 a3 1 an nþ1 n ð4Þ and an 2. The expansion (2) of clearly coincides with this expansion with xn ¼ SnðÞ ¼ nþ1=n. Define integer sequences by Pnþ1 ¼ anPn Pn1 Qnþ1 ¼ anQn Qn1 ð5Þ for n 1 with initial values ðP0; P1; Q0; Q1Þ ¼ ð1; 0; 0; 1Þ. A useful matrix representa-tion: Pn Pnþ1 Qn Qnþ1 !
¼ 0 1 1 a1 !
0 1 1 a2 !
0 1 1 an !
allows us to show x ¼ Pnþ1 Pnxn Qnþ1 Qnxn ð6Þ with Pnþ1Qn Qnþ1Pn ¼ 1. It is easily shown by induction that Qn Pn; Qn Qn1 Pn Pn1 ð7Þ for n 1 and n Qn (mod ZÞ. Qn is uniquely determined by this congruence. By using (6) and (7), we have 0 < Pn Qn ¼ n=n1 QnðQn ðn=n1ÞQn1Þ < 1 QnðQn Qn1Þ ð8Þ which gives an equality n ¼ Qn Pn and also guarantees the convergence of (3).
An interval exchange transform x 7! x þ if x 2 ½0; 1 Þ x 7! x þ 1 if x 2 ½1 ; 1Þ ( ð9Þ gives the rotation x 7! x þ on the torus. The induced dynamics on ½0; 1Þ ’ R=1Z is given by the first return map Recursively renewable words and coding of irrational rotations 1203 x 7! x þ 2 if x 2 ½0; 1 2Þ x 7! x þ 2 1 if x 2 ½1 2; 1Þ ( : In the similar manner, we have successive induced systems acting on ½0; nÞ: nþ1 : x 7! x þ nþ1 if x 2 ½0; n nþ1Þ x 7! x þ nþ1 n if x 2 ½n nþ1; nÞ ( with n ¼ 0; 1; . . . which give rotations of the smaller tori ½0; nÞ ’ R=nZ of an angle nþ1 with an initial value 0. The first return is described as: nþ1ðxÞ ¼ an n ðxÞ if x 2 ½0; n nþ1Þ an1 n ðxÞ if x 2 ½n nþ1; nÞ ( which gives a dynamical interpretation of the negative continued fraction. One may confirm the orbit of 0 by: nþ1ð0Þ ¼ nðan1 n ð0ÞÞ ¼ nððan 1ÞnÞ ¼ ðan 1Þn þ n n1 ¼ nþ1 and the structure of successive induced systems reveals a dynamical meaning of the number Qnþ1, that is, the smallest positive integer M such that M (mod ZÞ falls into ½0; nÞ.
There is a simple way to convert the regular continued fraction x ¼ b1 þ 1 b2 þ 1 b3þ ..
.
: into (3) and vice versa (c.f. Proposition 1 in ). This is given by a rewriting rule of infinite words which transforms b1b2 . . . into a1a2 . . . from left to right: b1 ! b1 þ 1 b2j ! 2b2j1 b2jþ1 ! b2jþ1 þ 2 8 > < > : where j 1, 2k ¼ 2 . . . 2 zfflffl}|fflffl{ k and 20 indicates the empty word . The converse rule is clearly: a1 ! a1 1 2k ! k þ 1 an 3 ! an 2 8 > < > : 1204 S. AKIYAMA and M. SHIRASAKA for n 2. Especially, in view of the Lagrange theorem on regular continued fractions, the negative continued fraction expansion of x is eventually periodic if and only if x is real quadratic irrational.
The transformation S has an infinite invariant measure dx=ð1 xÞ on ½0; 1 and consequently Diophantine approximation by negative continued fraction is slow, which is also apparent from (8). However it matches better the induced system and the first return map than the regular continued fraction algorithm.
3.
The case ¼ 0.
We first prove Theorem 1 for ¼ 0. In this case the proof follows naturally by successive recoding the orbit m 2 T ðm ¼ 0; 1; . . .Þ into the induced rotation x 7!
x þ nþ1 acting on ½0; nÞ. The clue of the proof is the fact that the number of necessary decomposition of intervals ½0; nÞ does not increase for n 1.
Set B0 ¼ A , J0 ¼ J and we recall that the rotation 2 : x 7! x þ 2 on ½0; 1Þ is an induced system of the rotation 1 : x 7! x þ 1 on ½0; 1Þ by: 2ðxÞ ¼ a1 1 ðxÞ if x 2 ½0; 1 2Þ a11 1 ðxÞ if x 2 ½1 2; 1Þ ( : Therefore we define the recoding map J1 : ½0; 1Þ ! A by J1ðxÞ ¼ J0ðxÞJ0ðx þ 1Þ . . . J0ðx þ ða1 1Þ1Þ if x 2 ½0; 1 2Þ J0ðxÞJ0ðx þ 1Þ . . . J0ðx þ ða1 2Þ1Þ if x 2 ½1 2; 1Þ ( and put B1 ¼ J1ð½0; 1ÞÞ. We have B1 6 B0, since J1ð0Þ is of length a1 > 1 as a word over B0. The function J1 naturally extends to J1 : R=1Z ! B1 by periodicity. As the map J0 is discontinuous5 at the set fwi j i ¼ 1; . . . ; kg, the set of discontinuity of J1 is given as a set f!ð1Þ j j j ¼ 0; 1; . . . ; k1 þ 1g with 0 ¼ !ð1Þ 0 < !ð1Þ 1 < !ð1Þ 2 < < !ð1Þ k11 < !ð1Þ k1 < !ð1Þ k1þ1 ¼ 1 and each !ð1Þ j ði ¼ 1; . . . ; k1Þ has a form !u Nu1 ðu ¼ 1; 2; . . . ; kÞ where Nu is the non negative integer that !u Nu1 2 ½0; 1Þ. From the definition of the first return map, this Nu is a unique non negative integer less than a1 that ! Nu1 (mod 1Þ falls into ½0; 1Þ, that is, Nu ¼ b!u=1c < a1. Note that there exists j such that !ð1Þ j ¼ 1 2.
Indeed !k ða1 1Þ1 ¼ 1 ða11 1Þ ¼ 1 2. Thus we have k1 k and the map J1 has at most k1 þ 1 images, i.e., #B1 k1 þ 1.
According to the usual convention, interval ½!ð1Þ j ; !ð1Þ j Þ is called IJ1ðyÞ for y 2 ½!ð1Þ j ; !ð1Þ jþ1Þ since the name J1ðyÞ does not depend on the choice of y. Then recalling that 2 : x 7! x þ 2 on ½0; 1Þ is an induced system of the rotation 1 : x 7! x þ 1 on ½0; 1Þ.
The original rotation word 5This just means that its output word changes.
Recursively renewable words and coding of irrational rotations 1205 J0ð0ÞJ0ð1ÞJ0ð21Þ . . .
is decomposed into J0ð0ÞJ0ð1Þ . . . J0ðða1 1Þ1ÞJ0ð2ÞJ0ð2 þ 1Þ ¼ J1ð0ÞJ1ð2ÞJ1ð22Þ 2 BN 1 : We proceed in the similar manner. Assume n 2 and we already defined Jn1, Bn1 and the decomposition of ½0; n1Þ into 0 ¼ !ðn1Þ 0 < !ðn1Þ 1 < !ðn1Þ 2 < < !ðn1Þ kn11 < !ðn1Þ kn1 < !ðn1Þ kn1þ1 ¼ n1 with kn1 kn2. Further we may assume that there is a j such that !ðn1Þ j ¼ n1 n.
Since the rotation x 7! x þ nþ1 on ½0; nÞ is the induced system of the rotation x 7!
x þ n on ½0; n1Þ through: nþ1ðxÞ ¼ an n ðxÞ if x 2 ½0; n nþ1Þ an1 n ðxÞ if x 2 ½n nþ1; nÞ, ( we define the recoding map Jn : ½0; nÞ ! B n1 by JnðxÞ ¼ Jn1ðxÞJn1ðx þ nÞ . . . Jn1ðx þ ðan 1ÞnÞ if x 2 ½0; n nþ1Þ Jn1ðxÞJn1ðx þ nÞ . . . Jn1ðx þ ðan 2ÞnÞ if x 2 ½n nþ1; nÞ ( ð10Þ and put Bn ¼ Jnð½0; nÞÞ. It follows Bn 6 Bn1 from an 2. Extend Jn to a function from R=nZ to Bn by periodicity. Now Jn1 has kn1 þ 1 points of discontinuity f!ðn1Þ j j j ¼ 1; . . . ; kn1 þ 1g. The discontinuity of Jn arises from these kn1 þ 1 points together with n nþ1 and the end point n. The number of point of discontinuity of Jn might increase to kn1 þ 3 but it turns out that one can save 2 points. In fact such points are written down in a way: 0 ¼ !ðnÞ 0 < !ðnÞ 1 < !ðnÞ 2 < < !ðnÞ kn1 < !ðnÞ kn < !ðnÞ knþ1 ¼ n where each !ðnÞ j ði ¼ 1; . . . ; knÞ has a form !ðn1Þ u Nun ðu ¼ 1; 2; . . . ; kn1 þ 1Þ and Nu is the non negative integer that !ðn1Þ u Nun 2 ½0; nÞ, i.e., Nu ¼ b!ðn1Þ u =nc < an. Firstly there is an index s such that !ðnÞ s ¼ n nþ1 since n nþ1 ¼ n1 ðan 1Þn ¼ !ðn1Þ j ðan 2Þn. Secondly !ðn1Þ j ¼ n1 n and !ðn1Þ kn1þ1 ¼ n1 gives the same discontinuous point !ðnÞ s of Jn. These two coincidences show that kn kn1 and #Bn kn þ 1.
Set IJnðyÞ ¼ ½!ðnÞ j ; !ðnÞ jþ1Þ for y 2 ½!ðnÞ j ; !ðnÞ jþ1Þ since the name JnðyÞ does not depend on the choice of y. As nþ1 : x 7! x þ nþ1 on ½0; nÞ is an induced system of n : x 7! x þ n on ½0; n1Þ. The word 1206 S. AKIYAMA and M. SHIRASAKA Jn1ð0ÞJn1ðnÞJn1ð2nÞ . . .
is decomposed into Jn1ð0ÞJn1ðnÞ . . . Jn1ððan 1ÞnÞJn1ðnþ1ÞJn1ðnþ1 þ nÞ . . .
¼ Jnð0ÞJnðnþ1ÞJnð2nþ1Þ 2 BN n : The case ¼ 0 is completed.
For the later use, we study in detail the map Jn and the set Bn. Let us introduce an order <lex in A for two elements u ¼ u1 . . . us and v ¼ v1 . . . vt. If v is a proper prefix u then u <lex v. When v is not a proper prefix of u and u is not a proper prefix of v, then take the first i such that ui 6¼ vi whenever u 6¼ v. In this case if ui < vi (resp. ui > vi), then we say u <lex v (resp. v <lex u). The order <lex is just a lexicographical order for words of the same length. The symbol juj stands for the length of the word u and u lex v means u <lex v or u ¼ v. Then we can show a PROPOSITION 1.
The map Jn : ½0; nÞ ! Bn preserves the order, i.e., JnðxÞ lex JnðyÞ for 0 x y n and consequently Jnð!ðnÞ i Þ 1 since otherwise we have nothing to do. Put q ¼ bð!iþ1 !iÞ=1c. Then for intervals ½!i þ ðj 1Þ1; !i þ j1Þ for j ¼ 1; . . . ; q, the induced rotation is written as an interval exchange: 1208 S. AKIYAMA and M. SHIRASAKA x 7! x þ 2 if x 2 ½!i þ ðj 1Þ1; !i þ j1 2Þ x 7! x þ 2 1 if x 2 ½!i þ j1 2; !i þ j1Þ ( : For simplicity, take i ¼ 0 and consider the interval ½0; !1Þ with q ¼ b!1=c. Then the function J0 1 is defined in the similar manner as J1: J0 1ðxÞ ¼ J0ðxÞJ0ðx þ 1Þ . . . J0ðx þ ða1 1Þ1Þ if x 2 ½ðj 1Þ1; j1 2Þ J0ðxÞJ0ðx þ 1Þ . . . J0ðx þ ða1 2Þ1Þ if x 2 ½j1 2; j1Þ ( : The set of discontinuity of J0 1 is just a shifted set of those of J1: f!ð1Þ u þ ðj 1Þ1 j u ¼ 1; . . . ; k1 þ 1g and we have 00 . . . 0 zfflfflffl}|fflfflffl{ j1 J0 1ðxÞ ¼ J1ðx ðj 1Þ1Þ 00 . . . 0 zfflfflffl}|fflfflffl{ j1 : ð12Þ This means that if x 2 ½ðj 1Þ1; j1Þ then z is ðk1 þ 1Þ-renewable ðk1 kÞ over the words 0101 . . . 01 zfflfflfflfflfflfflfflfflfflffl}|fflfflfflfflfflfflfflfflfflffl{ j1 B1 00 . . . 0 zfflfflffl}|fflfflffl{ j1 where the symbol 01 indicates the removal of 0 from the prefix. Now we have shown that if lies in [ k1 i¼0 [ bð!iþ1!iÞ=c j¼1 !i þ ðj 1Þ1; !i þ j1 ½ Þ; then z is ðk1 þ 1Þ-renewable. The remainder set to be covered is [ k1 i¼0 !i þ !iþ1 !i 1 1; !iþ1 : This job is completed by considering the induced system on ½!iþ1 1; !iþ1Þ for i ¼ 0; . . . ; k 1. The construction is done in the similar manner. The induced rotation is given as an interval exchange: x 7! x þ 2 if x 2 ½!iþ1 1; !iþ1 2Þ x 7! x þ 2 1 if x 2 ½!iþ1 2; !iþ1Þ ( : Take i ¼ k for simplicity. We may assume !k1 < 1 1 since otherwise the covering is already over. One can similarly define Recursively renewable words and coding of irrational rotations 1209 J00 1 ðxÞ ¼ J0ðxÞJ0ðx þ 1Þ . . . J0ðx þ ða1 1Þ1Þ if x 2 ½1 1; 1 2Þ J0ðxÞJ0ðx þ 1Þ . . . J0ðx þ ða1 2Þ1Þ if x 2 ½1 2; 1Þ ( and the set of discontinuity of J00 1 is just a shifted set of those of J1: f!ð1Þ u 1 2 T j u ¼ 1; . . . ; k1 þ 1g: Similarly as (12), for each x 2 ½1 1; 1Þ, we have b 2 A such that J00 1 ðxÞb ¼ ðk 1ÞJ1ðx þ 1Þ and b is determined by x þ 1 2 ½!b; !bþ1Þ. Thus in the same way, when 2 ½!iþ1 1; !iþ1Þ, z is ðk1 þ 1Þ-renewable and we have shown that for any 2 T the general rotation word z is ðk þ 1Þ-renewable. Note that this proof shows that there are two choices of the set B1 A for 2 !iþ1 !i; !i þ !iþ1 !i 1 1 : Indeed, usually we have several choices of induced systems. The only requirement is that the interval should be decomposed into at most k þ 1 parts.
Now we have shown that for any the general rotation word z is ðk þ 1Þ-renewable.
However this procedure can be repeated recursively to the subinterval (strictly speaking, the subsystem) to which belongs. As we have seen that the set of discontinuity of J0 1 and J00 1 are given by translations of those of J1, by the same reason as the case ¼ 0, the number of decomposition of subintervals does not exceed k þ 1.
Therefore z is recursively ðk þ 1Þ-renewable and the proof is finished.
The recursive covering of ½0; 1Þ constructed in this Section 4 will be reused in Section 10.
5.
An Example.
Let us show a sample computation of a general rotation word of an angle ¼ 22=3 and an initial value ¼ 0 with respect to a decomposition ½0; 1Þ ¼ ½0; 1=3Þ [ ½1=3; 1Þ 0101101101011011011110110110101101101011011011110110110 . . .
following the proof of Theorem 1.
½0; 1Þ ¼ ½0; 1=3Þ [ ½1=3; 1 22=3Þ [ ½1 22=3; 22=3Þ ¼ I01 [ I11 [ I1 ½0; 2Þ ¼ ½0; 4=3 21=3Þ [ ½4=3 21=3; 2 3 22=3Þ [ ½2 3 22=3; 1 þ 21=3Þ ¼ I01011 [ I01111 [ I011 1210 S. AKIYAMA and M. SHIRASAKA ½0; 3Þ ¼ ½0; 4=3 21=3Þ [ ½4=3 21=3; 2 3 22=3Þ [ ½2 3 22=3; 3 þ 5 22=3Þ ¼ I01011011 [ I01111011 [ I011 ½0; 4Þ ¼ ½0; 12 19 22=3Þ [ ½12 19 22=3; 19=3 5 21=3Þ [ ½19=3 5 21=3; 5 þ 4 21=3Þ ¼ I010110110101101101111011011 [ I0101101101011011011 [ I0101101101111011011 In fact, this method gives a rapid algorithm for computation of general rotation words. The recursive construction algorithm for 6¼ 0 in Section 4 may not be suitable in reality because we have to switch to other systems, keeping in memory all systems appear in the process. It is easier to shift the origin at the expense of one more subdivision, as explained in the beginning of Section 4.
6.
Quadratic rotations are primitive substitutive.
In this section, one direction of Theorem 2 is treated. We prove that a general rotation word of a quadratic irrational angle and an initial value with respect to the decomposition (1) is primitive substitutive provided !i 2 QðÞ for all i. By the same technique stated in the beginning of Section 4, it is sufficient to prove this fact for ¼ 0 and !i 2 QðÞ since it just amounts to increasing by 1 the number of subdivisions.
Moreover, we may assume that !i ¼ 1 for some i by the same reason. Under these assumptions, the proof of Theorem 1 reads that the irrational rotation x 7! x þ 1 on T gives rise to induced rotations x 7! x þ nþ1 on ½0; nÞ for n ¼ 1; 2; . . . and the set of discontinuity f!i j i ¼ 1; 2; . . . ; kg is transformed into f!ðnÞ i j i ¼ 1; 2; . . . ; kg7 with !ðnÞ i ¼ n nþ1 for some i. Each !ðnÞ i has a form !ðn1Þ u b!ðn1Þ u =ncn ðu ¼ 1; 2; . . . ; kÞ. Let us renormalize the sizes of the rotations and observe the orbit of !i. The n-th induced system ½0; nÞ 3 x 7! x þ nþ1 2 ½0; nÞ is renormalized into: ½0; 1Þ 3 x 7! x þ xnþ1 2 ½0; 1Þ with xnþ1 ¼ nþ1=n and the discontinuous points must be ðnÞ i ¼ !ðnÞ i =n. Therefore ðnÞ i should have a form ðn1Þ u =xn bðn1Þ u =xnc for some u. Moreover, we have xn ¼ SnðÞ ðn ¼ 0; 1; . . .Þ by the negative continued fraction map S as explained in Section 2.
For > 1, we define the -transform by T : ½0; 1Þ 3 x 7! x bxc 2 ½0; 1Þ: Then the set of discontinuity of n-th renormalized induced rotation is written as T1=xn1 . . . T1=x0ð!iÞ i ¼ 1; 2; . . . ; k n o : Let us denote by nðyÞ ¼ T1=xn1 . . . T1=x0ðyÞ for n ¼ 1; 2; . . . with xn ¼ SnðÞ. The clue of the proof is to show a 7The cardinality of f!ðnÞ i j i ¼ 1; 2; . . . ; kg could be less than k by coincidences !ðnÞ i ¼ !ðnÞ iþ1.
Recursively renewable words and coding of irrational rotations 1211 PROPOSITION 2.
If 2 ð0; 1Þ is quadratic irrational and y 2 QðÞ then the sequence ð nðyÞÞn¼1;2;... is eventually periodic.
In view of Proposition 4 given in Section 9, this is a generalization of the result in on Ostrowski expansions. Once Proposition 2 is established, it is easy to see that the quadratic rotation word is primitive substitutive. Indeed, taking the least common multiple L of periods of k sequences ð nð!iÞÞn¼0;1;..., there exists an integer m 0 such that the m-th renormalized induced system and ðm þ L 1Þ-th renormalized induced system is exactly the same. Hence the associated rotation word zðmÞ 2 BN m is successively decomposed ðL 1Þ-times into k-blocks and gives rise to zðmþL1Þ 2 BN mþL which has the same structure as zðmÞ. Therefore zðmÞ is a fixed point of the substitution on Bm. This substitution is primitive, because repeated application of the substitution to Bm yields sufficiently long words generated by the induced rotation x 7! x þ mþ1, each of which must contain all letters of Bm, by the minimality of the irrational rotation. This shows that zð1Þ is primitive substitutive.
We give two proofs of Proposition 2. A technical difficulty arises from the fact that each 1=xi is not necessary an algebraic integer. In the first proof, we choose periodically varying Z-bases to overcome it.
6.1.
The first proof.
According to the last remark of Section 2, ðxnÞn¼0;1;... is eventually periodic.
Without loss of generality, we may assume that the negative continued fraction of is purely periodic, i.e., there is a positive integer L that xn ¼ xnþL for n 0. By (6), there is ðPL; PLþ1; QL; QLþ1Þ 2 N 4 depending on xn such that xn ¼ ðPLþ1 PLxnÞ=ðQLþ1 QLxnÞ: ð13Þ Put hðyÞ ¼ QLy2 ðQLþ1 þ PLÞy þ PLþ1. By (7), we see hðxnÞ ¼ 0; hð0Þ ¼ PLþ1 > 0; hð1Þ ¼ ðPLþ1 PLÞ ðQLþ1 QLÞ 0 and the conjugate x0 n can not be less than 1.
x0 n = 2 Q implies x0 n > 1. Let ¼ minn x0 n > 1. The general term of the sequence is nðyÞ ¼ y Pn1 k¼0b kðyÞ=xkcx0x1 . . . xk x0x1 . . . xn1 : ð14Þ Note that by periodicity of ðxnÞn¼0;1;..., b kðyÞ=xkc b1=xkc are uniformly bounded by a positive constant L. Therefore jð nðyÞÞ0j jy0j n þ X n1 k¼0 L nk1 jy0j þ L 1 : This shows that ð nðyÞ; ð nðyÞÞ0Þ 2 R2 is bounded. On the other hand there exist a positive integer M and ðpn; qnÞ 2 Z2 such that nðyÞ ¼ ðpn þ qnxnÞ=M. Indeed, the case n ¼ 0 is trivial and 1212 S. AKIYAMA and M. SHIRASAKA nþ1ðyÞ ¼ nðyÞ xn cn ¼ 1 M pn xn þ qn cn ¼ 1 M ðpnðan xnþ1Þ þ qnÞ cn ¼ pnþ1 þ qnþ1xnþ1 M with cn ¼ b nðyÞ=xnc. Again by periodicity of ðxnÞn¼0;1;... and the boundedness of ð nðyÞ; ð nðyÞÞ0Þ, we see that ðpn; qnÞ 2 Z2 is bounded and this implies that the sequence ð nðyÞÞn¼1;2;... is eventually periodic.
The second proof is based on and where they treated beta expansions in a Pisot number base.
6.2.
The second proof.
We may assume that there is a positive integer L that xn ¼ xnþL for n 0. By (6), x ¼ PLþ1 xPL QLþ1 xQL with PLþ1QL QLþ1PL ¼ 1. Putting ¼ QLþ1 xQL, we have 2 ðQLþ1 PLÞ þ 1 ¼ 0: From (5) and an ¼ xn þ 1=xn1, we deduce Qnþ1 xnQn ¼ Qn xn1Qn1 xn1 ¼ ¼ Q1 x0Q0 x0x1 . . . xn1 ¼ 1 x0x1 . . . xn1 : Therefore ¼ QLþ1 xQL ¼ 1 x0x1 . . . xL1 ð15Þ is a quadratic unit with > 1 and j 0j < 1, i.e., a quadratic Pisot unit. Take a finite set D ¼ fPL1 k¼0 ckx0 . . . xk j 1=xk > ck 2 Zg and choose a positive integer U such that all dU are algebraic integers for d 2 D. Then in view of (14), we can write ML1ðyÞ ¼ My X M1 i¼0 di i with di 2 D. Then ðU ML1ðyÞÞM¼1;2... is a sequence of algebraic integers and one can easily show that ðU ML1ðyÞ; ðU ML1ðyÞÞ0Þ is bounded in R2 in a similar manner as in the first proof. Therefore the sequence ð ML1ðyÞÞM¼1;2;... is eventually periodic and this proves that ð nðyÞÞn¼1;2;... is eventually periodic.
Recursively renewable words and coding of irrational rotations 1213 In his undergraduate text book ([35, p.212]), Takagi emphasized that the denominator like of the modular equivalence equation (13) becomes a unit. The product relation (15) seems interesting in its own right.
7.
Recovering parameters of rotation words.
From a given general rotation word z on f0; 1; . . . ; k 1g with k 2, there is a way to recover all parameters, i.e, the angle , the initial value , and the decomposition of the unit interval ½0; 1Þ. For the later use, we briefly describe this method in this section.
A lot of methods had been discussed for sturmian words in relation to continued fraction expansions. One can find a nice survey in Chapter 6 of in which a dynamical and algorithmic way of this recoding is discussed in detail. Here we only mention a ‘transcendental’ way to recover such parameters, which might be a folklore.
First of all, from unique ergodicity of the irrational rotation, the decomposition of ½0; 1Þ is recovered immediately since it amounts to computing frequencies of letters of f0; 1; . . . ; k 1g. Thus we have given (1) with 0 ¼ !0 < !1 < < !k1 < !k ¼ 1 and we may assume that the angle is in ½1=2; 1=2Þ. Choose an arbitrary j 2 f1; 2; . . . ; k 1g and recode the word z ¼ z0z1 . . . into z0 ¼ z0 0z0 1 . . . where z0 i ¼ 0 for zi < j and z0 i ¼ 1 for zj j. Assume first that !j > 1=2. Then we observe in this rotation word z0 at most two different longest runs of 0’s, say, 0M and 0Mþ1. This implies that !j=ðM þ 1Þ !j=M with M 1. We get frequencies of 0M and 0Mþ1 which are denoted by M and Mþ1 (one of them could be zero). When 2 ½0; 1=2Þ, consider a decomposition ½0; Þ ¼ ½0; !j MÞ [ ½!j M; Þ ¼ IMþ1 [ IM: Then þ n (mod 1Þ 2 IMþ1 is the beginning of the M þ 1 runs of zeroes and þ n (mod 1Þ 2 IM is the beginning of the M runs of zeroes. Therefore, we have Mþ1 M ¼ !j M ðM þ 1Þ !j : From this equality, we got to know the angle: ¼ !jð M þ Mþ1Þ ðM þ 1Þ Mþ1 þ M M : The case 2 ½1=2; 0Þ is likewise and the final formula is the same. Secondly if !j < 1=2, then we observe in z0 at most two different runs of 1’s. The later discussion is the similar and we obtain: ¼ ð1 !jÞð M þ Mþ1Þ ðM þ 1Þ Mþ1 þ M M : Our final task is to get the initial value. For z ¼ z0z1 . . ., take an increasing integer sequence of occurrences of the letter 0, i.e., 0 ¼ n0 < n1 < . . . with znj ¼ 0. Then we have 1214 S. AKIYAMA and M. SHIRASAKA þ nj (mod 1Þ 2 ½0; !1Þ and hence 2 \ 1 j¼0 ½nj; !1 njÞ (mod 1Þ The intersection should be a single point, since it is non empty and n (mod 1Þ for n ¼ 1; 2; . . . are uniformly distributed in T. Therefore we finally recovered the parameter .
It might be a problem that without information of the natural order of letters, this recovered parameters are unique or not.
8.
Uniqueness of decomposition.
As we stated in the last part of Section 4, the choice of k þ 1 blocks of general rotation words is not unique. On the other hand, in this section we show that once we have chosen k þ 1 blocks as in the proof of Theorem 1, the way to decompose the general rotation word into these blocks Bn is unique when n is large. Such Bn is called a code.
This uniqueness will be used in the proof of the remaining direction of Theorem 2.
PROPOSITION 3.
Let Jnð!ðnÞ u Þ ¼ s1 . . . sQnþ1 2 Bn be a long word in the sense of Proposition 1. Assume that there are i; j such that si 6¼ sj. The we have ½!ðnÞ u ; !ðnÞ uþ1Þ ¼ \ Qnþ1 i¼1 ðJ1ðsiÞ ði 1ÞÞ ¼ \ Qnþ1 i¼1 ½!si ði 1Þ; !siþ1 ði 1ÞÞ in T: Clearly, if n is sufficiently large, the long word will contain two letters and satisfy the assumption of Proposition 3.
PROOF.
Note that y 2 T‘ i¼1ðJ1ðsiÞ ði 1ÞÞ is equivalent to Jðy þ ði 1ÞÞ ¼ si for i ¼ 1; . . . ; ‘. Hence by construction of !ðnÞ i , the inclusion ½!ðnÞ u ; !ðnÞ uþ1Þ \ Qnþ1 i¼1 ðJ1ðsiÞ ði 1ÞÞ is obvious. By the assumption, ðJ1ðsiÞ ði 1ÞÞ \ ðJ1ðsjÞ ðj 1ÞÞ is an interval, since the sum of length of two intervals does not exceed 1. Thus we can put \ Qnþ1 i¼1 ðJ1ðsiÞ ði 1ÞÞ ¼ ½t1; t2Þ; t1 !ðnÞ u !ðnÞ uþ1 t2: Recursively renewable words and coding of irrational rotations 1215 Let us show t2 ¼ !ðnÞ uþ1. Recall that !ðnÞ uþ1 is a discontinuous point of the n-th induced rotation on ½0; nÞ and Jnð!ðnÞ u Þ jJnð!ðnÞ uþ1Þj. Reviewing the proof of Theorem 1, this happens only when !ðnÞ uþ1 ¼ n nþ1 and pðJnð!ðnÞ u ÞÞ lex Jnð!ðnÞ uþ1Þ.
In this case, consider words JðxÞJðx þ Þ . . . Jðx þ ðQnþ1 1ÞÞ for x 2 ½!ðnÞ u ; !ðnÞ uþ1. If x 2 ½!ðnÞ u ; !ðnÞ uþ1Þ, the words are the same and Jð!ðnÞ uþ1 þ ðQnþ1 QnÞÞ ¼ Jðn nþ1 ðnþ1 nÞÞ ¼ Jð0Þ ¼ 0 and therefore k 1 ¼ Jð!ðnÞ uþ1 þ ðQnþ1 QnÞ "Þ 6¼ Jð!ðnÞ uþ1 þ ðQnþ1 QnÞÞ ¼ 0 for a sufficiently small positive ".8 This shows that Jð!ðnÞ u ÞJð!ðnÞ u þ Þ . . . Jð!ðnÞ u þ ðQnþ1 1ÞÞ 6¼ Jð!ðnÞ uþ1ÞJð!ðnÞ uþ1 þ Þ . . . Jð!ðnÞ uþ1 þ ðQnþ1 1ÞÞ and t2 !ðnÞ uþ1.
Finally we prove that t1 ¼ !ðnÞ u . Note that !ðnÞ 0 ¼ 0 for all n. Therefore if u ¼ 0, then by using J1ð0Þ ¼ ½0; !1Þ, we have t1 0 ¼ !ðnÞ 0 and the proof is completed for u ¼ 0.
When u > 0 by the proof of Proposition 1, the predecessor of a long word in Bn must be long and we have jJnð!ðnÞ u1Þj ¼ jJnð!ðnÞ u Þj. Thus we have Jnð!ðnÞ u1Þ 6¼ Jnð!ðnÞ u Þ and t1 !ðnÞ u in a similar manner.
Let us extend general rotation words to bi-infinite words. Fix 2 ½0; 1Þ n Q, 2 ½0; 1Þ and the decomposition (1). Then a bi-infinite general rotation word is naturally defined by z ¼ . . . Jð 2ÞJð ÞJðÞJð þ ÞJð þ 2Þ 2 f0; 1; . . . ; k 1gZ: Then z is ðk þ 1Þ-renewable by blocks Bn ¼ Jnð½0; nÞÞ in the similar manner.
8Considering the order <lex on these words, there should be some j < Qnþ1 Qn such that Jð!ðnÞ u þ jÞ < Jð!ðnÞ uþ1 þ jÞ indeed.
1216 S. AKIYAMA and M. SHIRASAKA THEOREM 3.
Assume that each long word of Bn has at least two letters of A , k Qnþ1 Qn and n !v 1 n þ nþ1 with some v 2 f1; . . . ; k 1g. Then there is only one way to decompose a generalized rotation word z into Bn.
From this Proposition, the decomposition of the original rotation word is unique, because for a given one-sided general rotation word, the extension to a bi-infinite rotation word is unique (cf. Section 7).
The length of short words of Bn is not less than Qn i¼1ðai 1Þ. Hence assumptions of Theorem 3 are fulfilled9 for a sufficiently large n. The condition k Qnþ1 Qn assures that the length of the short word is not less than the cardinality of short words. This condition is necessary. For instance, B1 ¼ f01; 0; 1g holds for the decomposition ½0; 1=2Þ [ ½1=2; 1Þ with > 1=2.
PROOF.
We say that two words ziziþ1 . . . ; ziþ‘ and zjzjþ1 . . . ; zjþ‘0 overlapped if there are i; j such that i < j < i þ ‘. Write z ¼ . . . z2z1z0z1z2 . . .
with zi ¼ Jð þ iÞ 2 A . First we give a standard algorithm to decompose z into Bn.
Put ‘ ¼ Qnþ1 1 for simplicity. Consider a set L ¼ fi 2 Z j ziziþ1 . . . ; ziþ‘ is a long word of Bng Z: Since ziziþ1 . . . ; ziþ‘ ¼ s is a long word, by Proposition 3 we have þ i (mod 1Þ 2 ½!ðnÞ u ; !ðnÞ uþ1Þ with s ¼ Jnð!ðnÞ u Þ. This implies that þ i (mod 1Þ 2 ½!ðnÞ u ; !ðnÞ uþ1Þ ½0; nÞ and the first return map to the interval ½0; nÞ is realized by x ! x þ ð‘ þ 1Þ. This means that þ k (mod 1Þ = 2 ½0; nÞ for k ¼ 1; 2; . . . ; ‘ and hence j 6¼ k for k ¼ 1; 2; . . . ; ‘.
Therefore there are no overlap of long words in z. In this manner we can first decide all locations of long words in z. Let zi . . . ziþ‘ and zj . . . zjþ‘ be above decided two long words with i þ ‘ < j and assume that there are no k such that i þ ‘ < k < j and k 2 L. If i þ ‘ þ 1 6¼ j, consider the word ziþ‘þ1 . . . zj1. þ ði þ ‘ þ 1Þ 2 ½0; nÞ implies that Jnð þ ði þ ‘ þ 1ÞÞ is a prefix of ziþ‘þ1 . . . and we must have i þ ‘ þ jJnð þ ði þ ‘ þ 1ÞÞj j 1 since otherwise this short word Jnð þ ði þ ‘ þ 1ÞÞ and the long word zj . . . zjþ‘ overlap and we would have þ ði þ ‘ þ jJnð þ ði þ ‘ þ 1ÞÞj þ 1Þ (mod 1Þ 2 ½0; nÞ, contradicting the property of the first return map again. Iterating this, ziþ‘þ1 . . . zj1 is decomposed into short words. Therefore we have a decomposition of z into Bn: z ¼ . . . x2x1x0x1x2 . . . with xk 2 Bn. Let us say that this is the standard decomposition.
We claim that any decomposition z ¼ . . . y2y1y0y1y2 . . . with yi 2 Bn and any K > 0, there exist i K and j K such that yi and yj are long words of Bn. In fact, for e.g., assume that yi is a short word for all i K and yK ¼ zj . . . zjþ‘. This would mean that any factor of length ‘ þ 1 appears in the left infinite word . . . zjþ‘1zjþ‘ must be a short word, since 9Recall that there are infinitely many i’s with ai > 2.
Recursively renewable words and coding of irrational rotations 1217 f þ ðj ð‘ þ 1ÞtÞ (mod 1Þ j t ¼ 0; 1; . . .g is dense in ½0; 1Þ. However, because the cardinality of short words is not greater than ‘ þ 1, the word . . . zjþ‘1zjþ‘ must be periodic (for e.g., see Proposition 1.1.1. in ).
Then would be rational, a contradiction. This proves the claim. Define Cð. . . x2x1x0x1x2 . . .Þ ¼ fi 2 Z j 9t xt ¼ zi . . . zj is a long wordg and Cð. . . y2y1y0y1y2 . . .Þ ¼ fi 2 Z j 9t yt ¼ zi . . . zj is a long wordg: By the above claim, Cð. . . y2y1y0y1y2 . . .Þ is neither bounded from below nor from above. From the definition of the standard decomposition, Cð. . . y2y1y0y1y2 . . .Þ Cð. . . x2x1x0x1x2 . . .Þ: It is sufficient to show that these two sets are equal. Indeed, if yi ¼ zi1 . . . zi2 and yj ¼ zj1 . . . zj2 are adjacent long words with i < j, the decomposition of word zi2þ1 . . . zj11 into short words is trivially unique.
Assume that there exists k 2 Cð. . . x2x1x0x1x2 . . .Þ n Cð. . . y2y1y0y1y2 . . .Þ: Choose i 2 Cð. . . y2y1y0y1y2 . . .Þ with i < k and find a minimum k1 2 Cð. . . x2x1x0x1x2 . . .Þ n Cð. . . y2y1y0y1y2 . . .Þ with i < k1. Then the decomposition of ziþ‘þ1 . . . zk11 2 A into Bn is unique, since there are no occurrence of long words any more. This shows that the long word zk1 . . . zk1þ‘ has a short word prefix zk1 . . . zk1þ‘0. In view of Proposition 1, this happens only when zk1 . . . zk1þ‘ is a long word JnðxÞ with x 2 ½0; n nþ1Þ and zk1 . . . zk1þ‘0 is the smallest short words with respect to the order <lex. Thus zk1þ‘0þ1 ¼ Jðx þ ‘0Þ is the prefix of a short word. Since ‘0 ¼ Qnþ1 Qn, we have Jðx þ ‘0Þ ¼ Jðx þ ðQnþ1 QnÞÞ ¼ Jðx þ nþ1 nÞ v because x þ nþ1 n 2 ½n þ nþ1; 0Þ. However by the assumption, each short word of Bn begins with a letter less than v, which gives a contradiction.
9.
Ostrowski’s numeration system and induced discontinuities.
Ostrowski’s numeration system has deep connection to the distribution of ðnÞn¼1;2;... in T and combinatorics on words: a survey is found in . It is extensively 1218 S. AKIYAMA and M. SHIRASAKA used in recoding sturmian words in Chapter 6 of . A generalization to the decomposition ½0; !1Þ [ ½!1; 1Þ is studied in Section 5 of to deduce an ergodic invariant of the rotation. In this section, we introduce Ostrowski’s type numeration system associated to the negative continued fraction of to analyze induced discontinuities.
One understands what happens when !1 2 Z þ Z in from the result of this section.
Recall that an is the digit of the negative continued fraction of and Qn is defined by Qnþ1 ¼ anQn Qn1 with Q0 ¼ 0 and Q1 ¼ 1.
LEMMA 1.
Each element m 2 N is uniquely expanded in a form: m ¼ X ‘ i¼1 miQi with mi 2 f0; 1; . . . ; ai 1g and ðm‘; m‘1; . . . ; m1Þ is a label10 of a finite walk of the next graph: ◦ 0,...,ai−2 ai−1 ◦ 0,...,ai−3 ai−2 ð16Þ A word ðm‘; m‘1; . . . ; m1Þ 2 Q‘ i¼1f0; 1; . . . ; aig is admissible if this is a label of a walk on this graph. Conversely, for an admissible ðm‘; m‘1; . . . ; m1Þ, m ¼ P‘ i¼1 miQi gives a greedy expansion of m.
PROOF.
The expansion of m 2 N is computed by a greedy algorithm; (1) x m (2) Find k 1 with Qk x < Qkþ1 (3) mk bx=Qkc (4) x x mkQk (5) If x ¼ 0 then stop. Otherwise go back to (2).
We find an expression m ¼ P‘ i¼1 miQi with mi 2 f0; . . . ; ai 1g. Let us prove that this expression is admissible. Reading the graph (16), a word ðm‘; m‘1; . . . ; m1Þ 2 Q‘ i¼1f0; 1; . . . ; aig is admissible if and only if there are no forbidden subwords of the form: ðap 1; ap1 2; . . . ; aqþ1 2; aq 1Þ: To be precise, ðm‘; m‘1; . . . ; m1Þ is admissible if and only if it has no suffix of the form: ðap 1; ap1 2; . . . ; aqþ1 2; aq 1; mq1; . . . ; m1Þ: One can easily prove an equality: 10Read from left to right. Each move in (16) decreases the index i by one.
Recursively renewable words and coding of irrational rotations 1219 ðap 1ÞQp þ ðaq 1ÞQq þ X p1 i¼qþ1 ðai 2ÞQi ¼ Qpþ1 þ Qq1 ð17Þ for 1 q < p. Therefore the words of forbidden form do not appear by the greedy algorithm. This proves that above obtained expansion is admissible. To prove the remaining part, it suffices to show if m ¼ P‘ i¼1 miQi is admissible, then m P‘ i¼kþ1 miQi ¼ Pk i¼1 miQi < Qkþ1, since this means that the admissible expansion must coincides with the greedy one. The digits vector ðmi; . . . ; mjþ1; mjÞ is an admissible block if mj < aj 2 (mj aj 2 for j ¼ 1 or i ¼ j) and for k 2 N with j < k < i we have mk ¼ ak 2 and mi ¼ ai 1. In other words, we cut the digit vector into blocks whenever we come back to the left vertex of (16). Note that a length one vector ðmjÞ with mj aj 2 is an admissible block by this definition. The admissibility of ðmk; mk1; . . . ; m1Þ allows us to decompose this digits vector into admissible blocks: ðmk11; . . . ; mk2Þ; ðmk21; . . . ; mk3Þ; . . . ; ðmk 1; . . . ; m1Þ with k1 ¼ k þ 1 and k þ1 ¼ 1.
By (17), mki1Qki1 þ þ mkiþ1Qkiþ1 Qki þ Qkiþ11 2Qkiþ1 for i > and the last 2Qkiþ1 is substituted with Qkiþ1 for i ¼ . This inequality is valid for the block of length one with ki ¼ kiþ1 þ 1. Summing up we have X k i¼1 miQi Qkþ1 Qk2 þ Qk21 Qk3 þ Qk31 Q1 þ Q0 < Qkþ1 which shows the result.
For a given x 2 ½0; 1Þ, we have another expansion: x ¼ X 1 i¼1 xii with xi 2 f0; 1; . . . ; ai 1g. In this case, since 0 < nþ1 < n, the algorithm works in the opposite direction11, that first we subtract x11 and then x22 and so on by the greedy choice of digits x1; x2; . . .. The fact that Qi and i satisfies the recurrence of the same shape with symmetric forbidden words, the infinite vector ðx1; x2; . . .Þ is a label of the infinite walk on the graph (16). An infinite vector ðx1; x2; . . .Þ 2 Q1 i¼1f0; 1; . . . ; ai 1g is admissible if it is a label of an infinite walk of the graph (16) which visit the left vertex infinitely many times.
LEMMA 2.
Each element x > 0 is uniquely expanded in a form: x ¼ X 1 i¼1 xii 11In this case, transition to the next vertex in (16) increases the index i by one.
1220 S. AKIYAMA and M. SHIRASAKA with xi 2 f0; 1; . . . ; ai 1g and ðx1; x2; . . .Þ is admissible. Conversely, for an admissible vector ðx1; x2; . . .Þ, P‘ i¼1 xii is a greedy expansion in the above sense.
It is interesting that the numeration system works by the same greedy algorithm but in two different directions. We call this expansion of x, the dual Ostrowski expansion.
PROOF.
The expansion of x is computed by a greedy algorithm; (1) i 1 (2) xi bx=ic (3) x x xii (4) i i þ 1 and return to (1).
Clearly xi 2 f0; . . . ; ai 1g. Let us show that for an expression x ¼ P1 i¼1 xii, ðx1; x2; . . .Þ is admissible. Reading the graph (16), a word ðx1; x2; . . .Þ is admissible if and only if there are no subwords of the form: ðxp; . . . ; xqÞ ¼ ðap 1; apþ1 2; . . . ; aq1 2; aq 1Þ with q > p and no suffix of the form ðxp; xpþ1; . . .Þ ¼ ðap 1; apþ1 2; apþ2 2; . . .Þ: We start with an equality analogous to (17): ðap 1Þp þ ðaq 1Þq þ X q1 i¼pþ1 ðai 2Þi ¼ p1 þ qþ1 ð18Þ for p < q which proves that the above expansion contains no words of the form ðap 1; apþ1 2; ; aq1 2; aq 1Þ. Taking q ! 1 in (18), we have ðap 1Þp þ X 1 i¼pþ1 ðai 2Þi ¼ p1 which shows the there are no suffix of the form ðap 1; ap1 2; ap2 2; . . .Þ. Thus we have shown that ðx1; x2; . . .Þ is admissible.
We will show that if ðx1; x2; . . .Þ is admissible, then x Pk i¼1 xii ¼ P1 i¼kþ1 xii < k, since this means that the admissible expansion must coincides with the greedy one. The digits vector ðxi; . . . ; xjÞ is a dual admissible block if xj < aj 2 (xj aj 2 for i ¼ j) and for k 2 N with i < k < j we have xk ¼ ak 2 and xi ¼ ai 1. In short, from the opposite direction of Lemma 1, we cut the digit vector into blocks whenever we come to the left vertex of (16). The admissibility of ðxkþ1; xkþ2; . . .Þ allows us to decompose this digits vector into dual admissible blocks: ðxk1; . . . ; xk21Þ; ðxk2; . . . ; xk31Þ; . . . ; ðxk ; . . . ; xk 1Þ; . . . with k1 ¼ k þ 1. An infinite sequence ðxkþ1; xkþ2; . . .Þ can not stay eventually in the right vertex of (16), the dual admissible block decomposition is always Recursively renewable words and coding of irrational rotations 1221 possible. By (17), xkiki þ þ xkiþ11kiþ11 ki1 þ kiþ1 2kiþ11 for each i. This inequality is valid for the block of length one with kiþ1 ¼ ki þ 1.
Therefore we have X 1 i¼kþ1 xii k11 k21 þ k2 k31 þ k3 < k: For the proof of the remaining direction of Theorem 2 in Section 10, it is important to understand how the set of discontinuities vary when the n-th rotation in ½0; nÞ is induced to the ðn þ 1Þ-th. Let !½0 be one of the discontinuities of the original rotation x 7! x þ in ½0; 1Þ. Then the recurrence !½k ¼ !½k 1 !½k 1 k k ð19Þ gives a point of discontinuity of k-th induced rotation x 7! x þ kþ1 in ½0; kÞ arose from !½0.
PROPOSITION 4.
Let !½0 ¼ P1 i¼1 xii be the dual Ostrowski expansion associated with . Then we have !½k ¼ X 1 i¼kþ1 mii PROOF.
This is obvious from the greedy algorithm of the dual Ostrowski expansion.
The point of discontinuity of the form m (mod 1Þ with m 2 Z plays an exceptional role; for positive m it eventually disappears, and for negative m it eventually corresponds to the point k kþ1: PROPOSITION 5.
Let m 2 Z and !½0 ¼ m (mod 1Þ. If m 0, then the Ostrowski expansion m ¼ P‘ i¼1 miQi gives rise to a finite dual Ostrowski expansion m (mod 1Þ ¼ X ‘ i¼1 mii: Thus !½k ¼ P‘ i¼kþ1 mii and especially !½‘ ¼ 0. If m < 0, then we have an infinite dual Ostrowski expansion 1222 S. AKIYAMA and M. SHIRASAKA m (mod 1Þ ¼ X 1 i¼1 xii and xi ¼ ai 2 for sufficiently large i. Especially, !½k ¼ P1 i¼kþ1ðai 2Þi ¼ k kþ1 for sufficiently large k. Conversely, if x > 0 has an expansion x ¼ P1 i¼1 xii with xi ¼ ai 2 for sufficiently large i, then x ¼ m (mod 1Þ with a negative integer m.
PROOF.
The case m 0 follows from Lemma 1 and 2 since finite forbidden words for two greedy expansions coincide. By this correspondence of Ostrowski expansions and their duals, the positive x’s having finite dual Ostrowski expansions are characterized.
We prove by induction that for any integer m 1 there exists k 0 such that its dual Ostrowski expansion has a form: m (mod 1Þ ¼ X 1 i¼1 xii with xi ¼ ai 2 for i > k. The equality (18) implies P1 i¼kþ1ðai 2Þi ¼ k kþ1 for any integer k 0. Taking k ¼ 0, we have (mod 1Þ ¼ X 1 i¼1 ðai 2Þi ð20Þ which shows the case m ¼ 1. Assume that m (mod 1Þ ¼ X 1 i¼1 xii with xi ¼ ai 2 for i > k and ðx1; x2; . . .Þ is admissible. Take the largest integer s 0 such that xi ¼ 0 for i s. If s ¼ 0, then x1 > 0 and ðx1 1; x2; . . .Þ is admissible. This implies that ðm þ 1Þ (mod 1Þ has the required form. If s > 0, then we use the relation (18) in a form ða1 1Þ1 þ ðas 1Þs þ X s1 i¼2 ðai 2Þi ¼ 1 þ sþ1: ð21Þ Thus we have ðm þ 1Þ ¼ þ X 1 i¼sþ1 xii ¼ X s1 i¼1 ðai 2Þi þ ðas 1Þs þ ðxsþ1 1Þsþ1 þ X 1 i¼sþ2 xii in T Recursively renewable words and coding of irrational rotations 1223 and ða1 2; . . . ; as2 2; as1 1; xsþ1 1; xsþ2; xsþ3; . . .Þ is easily shown to be admis-sible. Conversely assume that x > 0 has an expansion x ¼ P1 i¼1 xii with xi ¼ ai 2 for i > k and choose a minimum k. If k ¼ 0, then (20) shows that x ¼ . If k > 0, then xk ak 3 since ðx1; x2; . . .Þ is admissible.
X 1 i¼1 xii ¼ k kþ1 þ X k i¼1 xii ¼ Qk Qkþ1 þ X k i¼1 xiQi !
(mod 1Þ: By the admissibility of ðx1; . . . ; xk1; xk þ 1Þ, we have Qkþ1 þ ðxk þ 1ÞQk þ X k1 i¼1 xiQi < Qk Qkþ1 < 0 by Lemma 1.
Define !0½i ði ¼ 0; 1; . . .Þ by the same recurrence (19) as !½i with an initial value !0½0. Using the dual Ostrowski expansion, one can discuss when two points of discontinuity coincide in the k-th induced rotation: PROPOSITION 6.
Assume that !½0 6¼ m (mod 1Þ for all m 2 Z and there is an integer u such that !0½0 !½0 ¼ u in T. Then there is an integer k > 0 such that !0½k ¼ !½k. Conversely if there exist an integer k > 0 and !0½k ¼ !½k, then there exists an integer u such that !0½0 !½0 ¼ u in T.
The assumption !½0 6¼ m (mod 1Þ is necessary. Consider the case !½0 ¼ 0 and !0½0 ¼ 1 ¼ (mod 1Þ in Proposition 5.
PROOF.
In view of Proposition 4 and i ¼ Qi (mod 1Þ, the later assertion is obvious. Assume that !0½0 !½0 ¼ u in T for an integer u. Without loss of generality, we may assume that u > 0. By the assumption and Proposition 5, both !½0 and !0½0 have infinite dual Ostrowski expansions !½0 ¼ P1 i¼1 xii and !0½0 ¼ P1 i¼1 x0 ii and both of them do not have a suffix of the form ðap 2; apþ1 2; . . .Þ. We wish to show that there exist k > 0 such that xi ¼ x0 i for i > k. It suffices to show this in the case u ¼ 1 since we can iterate u-times the procedure. Therefore our goal is to show that 1 þ P1 i¼1 xii can be rewritten into P1 i¼1 x0 ii without infinite carries. Let k1 ¼ 1 and ðxk1; . . . ; xk21Þ; ðxk2; . . . ; xk32Þ; . . . is the dual admissible block decomposition as in the proof of Lemma 2. When ðx1 þ 1; x2; x3; . . .Þ is admissible, we have nothing to prove.
If not, then x1 þ 1 ¼ a1 or x1 þ 1 ¼ a1 1.
Case x1 ¼ a1 1. By the admissibility of ðx1; x2; . . .Þ the first dual admissible block is of length greater than one, i.e., k2 > 2. Since a11 ¼ 1 þ 2, we have 1224 S. AKIYAMA and M. SHIRASAKA a11 þ X 1 i¼2 xii ¼ ðx2 þ 1Þ2 þ X 1 i¼2 xii in T: If k2 4 then x2 þ 1 ¼ a2 1 and we have a new dual admissible block decomposition ð0Þ; ðx2 þ 1; x3; . . . ; xk21Þ; ðxk2; . . . ; xk32Þ; . . .
which shows that xi ¼ x0 i for i 3. If k2 ¼ 3 then x2 a2 3. In this case, the same computation gives a new decomposition ð0Þ; ðx2 þ 1Þ; ðxk2; . . . ; xk32Þ; . . .
which shows the same xi ¼ x0 i for i 3.
Case x1 ¼ a1 2. By the assumption on ðx1; x2; . . .Þ, there exists a maximum integer s 2 such that xi ¼ ai 2 for i < s. If xs ai 3 then ðx1 þ 1; x2; . . . ; xs1; xsÞ forms the first dual admissible block and ðx1 þ 1; x2; . . .Þ is admissible. This was excluded at the beginning. Therefore we must have xs ¼ ai 1, which is the beginning of an dual admissible block ðxs; . . . ; x‘Þ with ‘ > s. Using (21), we have ðx1 þ 1Þ þ X 1 i¼2 xii ¼ ðxsþ1 þ 1Þsþ1 þ X 1 i¼sþ2 xii in T: If ‘ s þ 2 then xsþ1 ¼ asþ1 2 and the last expression gives rise to a dual admissible block decomposition: ð0Þ; . . . ; ð0Þ; ððasþ1 1Þ; xsþ2; . . . ; x‘Þ; . . .
which shows that xi ¼ x0 i for i s þ 2. If ‘ ¼ s þ 1 then xsþ1 asþ1 3 and the decomposition is: ð0Þ; . . . ; ð0Þ; ðxsþ1 þ 1Þ; . . .
and we also have xi ¼ x0 i for i s þ 2.
10.
Primitive substitutive rotation words are quadratic.
We prove that if the general rotation words is primitive substitutive, then is quadratic and !i 2 QðÞ, and complete the proof of Theorem 2.
Let z ¼ z0z1 2 A N be uniformly recurrent, i.e., for each factor v of z there exists k > 0 that all factor of length k must contain v as a factor. Clearly, each general rotation word is uniformly recurrent. Take a prefix u ¼ z1 . . . zn. One can decompose z into blocks as z ¼ x0x1 . . . with xi 2 Sk i¼1 A k such that u is a prefix of all xixiþ1 . . . and all the factor u of z appear as a prefix of xixiþ1 . . . for some i. Such xi is called the return words of u.
Recursively renewable words and coding of irrational rotations 1225 Put u ¼ #fxi j i ¼ 0; 1; . . .g and define a map z;u from fxi j i ¼ 0; 1; . . .g to f0; 1; . . . ; u 1g by the rank of first occurrence of xi in z. The word z;uðx0Þz;uðx1Þz;uðx2Þ . . . is the derived word of z by the prefix u. Then THEOREM 4 (, ).
A uniformly recurrent word z is primitive substitutive if and only if its derived words by all prefixes form a finite set.
Note that once this theorem is established, the map z;u could be replaced by any bijection from fxi j i ¼ 0; 1; . . .g to f0; 1; . . . ; u 1g. This change would at most M! times enlarge the number of different derived words, where M is the maximum of the cardinality of return words with respect to a prefix of z.
Consider a general rotation word z ¼ z0z1 . . . with respect to a decomposition (1) with an angle 2 ½0; 1Þ n Q and assume that z is primitive substitutive. Take a sufficiently large positive integer n0, so that for n n0, the presumptions of Proposition 3, Proposition 5, Theorem 3 and Proposition 6 are fulfilled, that is, . Each long word of length Qn contains two different letters.
. Decomposition into Bn is unique.
. If !ðnÞ i !ðnÞ j ¼ u (mod 1Þ with i 6¼ j for some integer u then f!ðnÞ i ; !ðnÞ j g f0; n nþ1; ng.
10.1.
The case ¼ 0.
First let us show that is quadratic. Choose m n0. From Proposition 3, an occurrence of zi . . . ziþQmþ11 ¼ Jmð0Þ is equivalent to i (mod 1Þ 2 ½0; !ðmÞ 1 Þ. Find a unique n m such that nþ1 < !ðmÞ 1 < n, which is always possible under our assumption. We write n ¼ nðmÞ when it is necessary. Let z ¼ zðnÞ 0 zðnÞ 1 . . . with zðnÞ i 2 Bn be the n-th renewed word by Theorem 1. Define a map fn : fzðnÞ i j i ¼ 0; 1; . . .g !
f0; 1; . . . ; kng with #Bn ¼ kn þ 1 having the order preserving property x lex y ¼ ) fnðxÞ fnðyÞ: Then by construction, fnðzðnÞ 0 ÞfnðzðnÞ 1 Þ . . . is a coding of the rotation x 7! x þ nþ1 of the torus ½0; nÞ with respect to the decomposition 0 ¼ !ðnÞ 0 < !ðnÞ 1 < < !ðnÞ kn < !ðnÞ knþ1 ¼ n: There is an index 1 q kn such that !ðmÞ 1 ¼ !ðnÞ q . Hence in this setting, the return word of z with respect to the prefix Jmð0Þ is a coding of the induced system ½0; !ðmÞ 1 Þ of the n-th rotation x ! x þ nþ1 on ½0; nÞ. The first return map gm : ½0; !ðmÞ 1 Þ ! ½0; !ðmÞ 1 Þ becomes an exchange of three intervals, which is explicitly given as follows: gmðxÞ ¼ x þ nþ1; x 2 ½0; !ðmÞ 1 nþ1Þ x þ ðbm þ 1Þnþ1 n; x 2 ½!ðmÞ 1 nþ1; n bmnþ1Þ x þ bmnþ1 n; x 2 ½n bmnþ1; !ðmÞ 1 Þ 8 > > < > > : 1226 S. AKIYAMA and M. SHIRASAKA with bm ¼ dðn !ðmÞ 1 Þ=nþ1e. Define a natural coding Jm : ½0; !ðmÞ 1 Þ ! B n by: JmðxÞ ¼ JnðxÞ; x 2 ½0; !ðmÞ 1 nþ1Þ JnðxÞJnðx þ nþ1Þ . . . Jnðx þ bmnþ1Þ; x 2 ½!ðmÞ 1 nþ1; n bmnþ1Þ JnðxÞJnðx þ nþ1Þ . . . Jnðx þ ðbm 1Þnþ1Þ; x 2 ½n bmnþ1; !ðmÞ 1 Þ.
8 > > < > > : The discontinuities f!ðnÞ i j i ¼ 0; 1; . . . ; kn þ 1g are transformed into discontinuities f ðmÞ i j i ¼ 0; 1; . . . ; m0 þ 1g of Jm: 0 ¼ ðmÞ 0 < ðmÞ 1 < < ðmÞ m0þ1 ¼ !ðmÞ 1 : Obviously we have f ðmÞ i g \ ½0; !ðmÞ 1 nþ1Þ ¼ f!ðnÞ i g \ ½0; !ðmÞ 1 nþ1Þ and other elements are of the form !ðnÞ j pnþ1 2 ½!ðmÞ 1 nþ1; !ðmÞ 1 Þ with an integer p with 0 p bm. By the assumption, the correspondence between f ðmÞ i g and f!ðnÞ i g is one to one and we have m0 ¼ kn. Each JmðxÞ with x 2 ½0; !ðmÞ 1 Þ is a return word having the prefix Jmð0Þ and the decomposition of z into return words is written as z ¼ Jmð0ÞJmðgmð0ÞÞJmðg2 mð0ÞÞ . . .
which clearly gives a dynamical interpretation of the return word. The discontinuities of the interval exchange gm satisfies the i.d.o.c. (infinite distinct orbit condition) of . In fact, the denseness of gm-orbits of each ðmÞ i follows from the minimality of the irrational rotation x 7! x þ nþ1 on ½0; nÞ. The assumption m n0 guarantees that two negative orbits of !ðnÞ i and !ðnÞ j with 1 i < j kn do not intersect in the irrational rotation x 7! x þ nþ1. This implies that no two negative orbits by gm of discontinuities f ðmÞ i g intersect. Note that in the case !ðmÞ 1 ¼ n nþ1, three interval exchange gm is degenerated into two interval exchange; the irrational rotation. Anyway gm satisfies i.d.o.c. and therefore ð½0; !ðmÞ 1 Þ; gmÞ is minimal and uniquely ergodic (see12 , , Section 5.4.1 in ). As a result, 1-dimensional Lebesgue measure is the unique invariant measure of gm.
In what follows, we fix a bijection m from Jmð½0; !ðmÞ 1 ÞÞ to f0; 1; . . . ; kng by mðxÞ ¼ i for x 2 ½ ðmÞ i ; ðmÞ iþ1Þ. Then the derived word of z by the prefix Jmð0Þ is mðJmð0ÞÞmðJmðgmð0ÞÞÞmðJmðg2 mð0ÞÞÞ 2 f0; 1; . . . ; kngN: By the definition of m, we have 12This is an unpublished result in the thesis of M. Boshernitzan (due to S. Ferenczi). Unique ergodicity of a minimal three interval exchange follows from the fact that it is an integral automorphism of an irrational rotation with bounded return time (c.f. [20, Chapter 1, Section 5]).
Recursively renewable words and coding of irrational rotations 1227 0 x y < !ðmÞ 1 ¼ ) mðJmðxÞÞ lex mðJmðyÞÞ: Note that such a derived word is uniquely determined for each m n0. We claim that from a derived word, the value nðmÞþ1=!ðmÞ 1 is retrieved uniquely. From a derived word, Theorem 3 and unique ergodicity of gm allow us to recover the set: ðmÞ iþ1 ðmÞ i !m 1 i ¼ 0; . . . ; kn ( ) by computing frequencies of letters in f0; 1; . . . ; kng.13 We have to know which letters correspond to the interval ½0; !ðmÞ 1 nþ1Þ. By the definition of m, these letters should be consecutive integers including 0. Let us denote this set by f0; . . . ; rg. From dynamics of the three interval exchange, we observe that in the derived word, there is a successor letter of r larger than a successor letter of r þ 1. Moreover r is the minimum letter having this property. Therefore one can compute nðmÞþ1 !ðmÞ 1 ¼ 1 X r i¼0 ðmÞ iþ1 ðmÞ i !ðmÞ 1 as claimed. This gives a well defined map to ½0; 1Þ from the set of derived words of z with respect to prefixes Jmð0Þ with m n0. Theorem 4 says that the set fnðmÞþ1=!ðmÞ 1 j m n0g must be finite. The same technique allows us to retrieve uniquely the value ðnðmÞ bnðmÞnðmÞþ1Þ=!ðmÞ 1 from the derived word. In other words, we uniquely retrieve ratios of three intervals from the derived word and the set nðmÞþ1 !ðmÞ 1 ; nðmÞ bnðmÞnðmÞþ1 !ðmÞ 1 !
2 R2 m n0 ( ) is finite. We deduce that nðmÞ nðmÞþ1 bnðmÞ m n0 ( ) is also a finite set. One can take two distinct positive integers m1; m2 such that nðm1Þ nðm1Þþ1 bnðm1Þ ¼ nðm2Þ nðm2Þþ1 bnðm2Þ: ð22Þ Therefore 13Approximate by continuous functions the characteristic function of an interval and apply a variant of Theorem 6.19 in for measure theoretical dynamical systems.
1228 S. AKIYAMA and M. SHIRASAKA Qnðm1Þ Pnðm1Þ Qnðm1Þþ1 Pnðm1Þþ1 ¼ Qnðm2Þ Pnðm2Þ Qnðm2Þþ1 Pnðm1Þþ2 þ bnðm1Þ bnðm2Þ: ð23Þ If Qnðm1Þ Qnðm1Þþ1 6¼ Qnðm2Þ Qnðm2Þþ1 þ bnðm1Þ bnðm2Þ; then (23) is a quadratic equation of . If Qnðm1Þ Qnðm1Þþ1 ¼ Qnðm2Þ Qnðm2Þþ1 þ bnðm1Þ bnðm2Þ; then from 0 < Qn < Qnþ1 we have bnðm1Þ bnðm2Þ ¼ 0. From (22), this implies that nðm1Þþ1 nðm1Þ ¼ nðm2Þþ1 nðm2Þ which shows that has an eventually periodic continued fraction. In both cases, we have proven that is quadratic.
Now we wish to show that !i 2 QðÞ for each i. There are integers Nm; Mm such that !ðmÞ 1 ¼ !i Nm Mm with i 2 f1; 0; 1; . . . ; k 1g where we put !1 ¼ 1 for simplicity.14 From the finiteness of fnðmÞþ1=!ðmÞ 1 j m n0g, one can take an increasing sequence ðmiÞi¼1;2;... of integers such that !ð1Þ mi ¼ !v Nmi Mmi with a fixed v 2 f1; 0; . . . ; k 1g and a constant s > 0 that s ¼ Qnðm1Þ Pnðm1Þ !v Nm1 Mm1 ¼ Qnðm2Þ Pnðm2Þ !v Nm2 Mm2 ¼ . . . : This gives s ðNmj Nm1Þ ðMmj Mm1Þ ¼ ðQnðmjÞ Qnðm1ÞÞ ðPnðmjÞ Pnðm1ÞÞ for j ¼ 2; 3; . . .. Since Nmj > Nm1, we have shown that s 2 QðÞ. This proves that both !v and !ð1Þ mi are in QðÞ. Switching to a subsequence corresponding to the same derived word, we may assume that ðm1Þ iþ1 ðm1Þ i !ðm1Þ 1 ¼ ðm2Þ iþ1 ðm2Þ i !ðm2Þ 1 ¼ . . .
14It is possible that !1 ¼ !i holds for some i 0.
Recursively renewable words and coding of irrational rotations 1229 for i ¼ 0; 1; . . . ; kn, which is equivalent to ðm1Þ i !ðm1Þ 1 ¼ ðm2Þ i !ðm2Þ 1 ¼ . . .
for i ¼ 1; 2; . . . ; kn, since ðmÞ 0 ¼ 0. As ðmjÞ i ! ði;jÞ R ði;jÞ (mod 1Þ, we can choose for each j a vector ð! ð1;jÞ; ! ð2;jÞ; . . . ; ! ðkn;jÞÞ from a finite set f!1; !0; . . . ; !k1gkn. Thus taking a subsequence again, we may additionally assume that ðmjÞ i ! ðiÞ R ði;jÞ (mod 1Þ where ðiÞ does not depend on the choice of j. Thus there are integers S ði;jÞ such that ! ðiÞ R ði;1Þ S ði;1Þ !v Nm1 Mm1 ¼ ! ðiÞ R ði;2Þ S ði;2Þ !v Nm2 Mm2 which shows that ! ðiÞ 2 QðÞ. According to the discussion in Section 9, we know that each !j is congruent to some ! ðiÞ modulo Z which finishes the proof for ¼ 0.
10.2.
The case 6¼ 0.
We shall skip the details in parts where the analogy to the case ¼ 0 works. In Section 4, we gave a covering the torus ½0; 1Þ by induced systems of the shape ½ ; þ Þ.
The value was carefully chosen in a form !i (mod Þ for some i, not to destroy the combinatorial structure of the assertion. To treat the case 6¼ 0, we need a similar but more flexible covering so that every points on the torus correspond to long words.
We explain the construction only for ½0; 1Þ, i.e., the first step of the recursive construction as in Section 4.
The original rotation is the interval exchange on ½0; 1Þ given by (9) with discontinuities f!i j i ¼ 0; 1; . . . ; k 1g as general rotation words. Its induced system on ½ ; þ Þ with 2 ½0; 1Þ is x 7! x þ 2 if x 2 ½ ; þ 2Þ x 7! x þ 2 if x 2 ½ þ 2; Þ ( and discontinuities are written as ¼ !ð1; Þ 0 < !ð1; Þ 1 < < !ð1; Þ k < !ð1; Þ kþ1 ¼ þ : Each !ð1; Þ i has either a form þ 2 or !j Njð Þ where Njð Þ is the smallest non negative integer that !j Njð Þ < þ with j 0. The later discontinuities are 1230 S. AKIYAMA and M. SHIRASAKA step functions on . The interval ½0; 2Þ already corresponds to long words. Hence if 2 ½0; 2Þ then there is an interval ½!ð1Þ i ; !ð1Þ iþ1Þ 3 which exactly corresponds to a long word J1ðÞ ¼ J1ð!ð1Þ i Þ in the sense of Proposition 3. However this is not flexible enough for the later purpose. For a general 2 ½0; Þ, we choose such that 2 ½ ; þ 2Þ and consider an induced system on ½ ; þ Þ. Then we find consecutive discontinuities !ð1; Þ i and !ð1; Þ iþ1 in this system having the property ½!ð1; Þ i ; !ð1; Þ iþ1 Þ ¼ \ a1 i¼1 ðJ1ðsiÞ ði 1ÞÞ ð24Þ where s1 . . . sa1 ¼ J0ðÞJ0ð þ Þ . . . J0ð þ ða1 1ÞÞ is a long word for this shifted system. From the shape of the right side of (24), we see that the interval ½!ð1; Þ i ; !ð1; Þ iþ1 Þ does not depend on the choice of 2 ð þ 2; . This explains that if 2 ½0; Þ, then there is a canonical way to associate an interval ½!ð1; Þ i ; !ð1; Þ iþ1 Þ ½ ; þ Þ which exactly correspond to a long word.
Recall that we had constructed a covering of ½0; 1Þ by induced systems ½ ; þ Þ with !i (mod Þ in Section 4. The induced systems gives a decomposition of the rotation word z into blocks of the shape like b1B1b or ðk 1ÞB1b1 with some b 2 A .
For each 2 ½0; 1Þ, there is such an with 2 ½ ; þ Þ. An advantage of this choice of is that it only requires minor changes of renewed words, and consequently we may use three assumptions at the beginning of Section 10 as there are.
Shifting in an appropriate way, we can associate an interval ½!ð1; Þ i ; !ð1; Þ iþ1 Þ containing which exactly correspond to a long word which satisfies (24). One can iterate this procedure recursively to cover the ðn 1Þ-th induced system ½ n1; n1 þ n1Þ by the n-th system ½ n; n þ nÞ such that 2 ½ n; n þ n nþ1Þ for all n. Thus we find an interval ½!ðn; nÞ i ; !ðn; nÞ iþ1 Þ ending at consecutive discontinuities of the n-th induced system on ½ n; n þ nÞ which satisfies: 2 ½!ðn; nÞ i ; !ðn; nÞ iþ1 Þ ¼ \ Qnþ1 i¼1 J1ðsiÞ ði 1Þ with the long word JðÞJð þ Þ . . . Jð þ ðQnþ1 1ÞÞ ¼ s1s2 . . . sQnþ1 produced by this system. Note that the choice of n does not affect the three assumptions because shifting of n are performed within the covering of Section 4.
Let us prove that is quadratic. Choose m n0 and find a unique n ¼ nðmÞ and i ¼ iðmÞ such that nþ1 < !ðm; mÞ iðmÞþ1 !ðm; mÞ iðmÞ < n. Write !ðmÞ i ¼ !ðm; Þ iðmÞ for simplicity. As stated above, we have some freedom of choice of m and it is possible to take m ¼ !ðmÞ i for simplicity. Following the same argument as Section 10.1, the return word is a coding of a three interval exchange transform acts on ½!ðmÞ i ; !ðmÞ iþ1Þ: gmðxÞ ¼ x þ nþ1; x 2 ½!ðmÞ i ; !ðmÞ iþ1 nþ1Þ x þ ðbm þ 1Þnþ1 n; x 2 ½!ðmÞ iþ1 nþ1; !ðmÞ i þ n bmnþ1Þ x þ bmnþ1 n; x 2 ½!ðmÞ i þ n bmnþ1; !ðmÞ iþ1Þ 8 > > < > > : Recursively renewable words and coding of irrational rotations 1231 with bm ¼ dðn !ðmÞ iþ1 þ !ðmÞ i Þ=nþ1e. The set of discontinuities of gm is written as f ðmÞ i j i ¼ 0; . . . ; kn þ 1g and each ðmÞ i has a form !j t where j 2 f1; 0; . . . ; kg and t 2 N. In the similar manner as Section 10.1, one can define a natural coding function Jm and an appropriate ordering among return words and show that the set nðmÞþ1 !ðmÞ iþ1 !ðmÞ i ; nðmÞ bnðmÞnðmÞþ1 !ðmÞ iþ1 !ðmÞ i !
2 R2 m n0 ( ) is finite and therefore is quadratic.
Our final task is to show that !i 2 QðÞ for all i. By a proof analogous to Section 10.1, there exists a sequence ðmjÞj¼1;2;... that !ðmjÞ iþ1 !ðmjÞ i 2 QðÞ. Further we may assume that derived words by the prefix of length mj are identical and ðm1Þ iþ1 ðm1Þ i !ðm1Þ iþ1 !ðm1Þ i ¼ ðm2Þ iþ1 ðm2Þ i !ðm2Þ iþ1 !ðm2Þ i ¼ . . .
for i ¼ 0; 1; . . . ; kn. This implies that up to renormalization, three interval exchanges ð½!ðmjÞ i ; !ðmjÞ iþ1 Þ; gmjÞ with j ¼ 1; 2 . . . have exactly the same shape including the relative location of discontinuities of Jmj. Let us call three intervals appear in the above systems as Ii ði ¼ 1; 2; 3Þ and define GðxÞ ¼ i for x 2 Ii. Then we claim that the map f from x 2 ½!ðmjÞ i ; !ðmjÞ iþ1 Þ to f1; 2; 3gN defined by fðxÞ ¼ GðxÞGðgmjðxÞÞGðg2 mjðxÞÞ . . .
is injective. Indeed similarly as in the last part of Section 7, one can show that the intersection of inverse images becomes a single point, using the fact that the three interval exchange satisfies i.d.o.c., and the gm-orbit of each discontinuity ðmjÞ i is dense in ½!ðmjÞ i ; !ðmjÞ iþ1 Þ.
Since derived words by the prefix of length mj are identical and fðÞ is an image of a morphism of the derived word, fðÞ is independent of j. Thus we see that must be located in the same relative position in the interval ½!ðmjÞ i ; !ðmjÞ iþ1 Þ, i.e., !ðm1Þ i !ðm1Þ iþ1 !ðm1Þ i ¼ !ðm2Þ i !ðm2Þ iþ1 !ðm2Þ i ¼ . . .
holds. By using the same technique as Section 10.1, we can show that !ðmjÞ i 2 QðÞ and ðmjÞ i 2 QðÞ. This implies that !i 2 QðÞ as desired.
11.
Open questions.
Theorem 1 and 2 are devoted to coding of k interval exchange transforms which are degenerated into two intervals. A natural question is to generalize these result to genuine interval exchange transforms. Ferenczi, Holton and Zamboni showed in 1232 S. AKIYAMA and M. SHIRASAKA and that an i.d.o.c. three interval exchange gives a primitive substitutive system if and only if the parameters are in the same quadratic field. A related result for k intervals is found in .
Let us say a word z ¼ z0z1 2 A N is primitive substitutive in arithmetic progression (PSAP), if zazaþbzaþ2b . . . are primitive substitutive for all a 0 and b > 0.
Theorem 2 implies that a primitive substitutive rotation word is PSAP. Can we characterize PSAP words among primitive substitutive words? Durand had shown in Proposition I.6 of that zazaþbzaþ2b . . . is an image of the morphism of the fixed point of the substitution, not necessarily primitive. Their primitivity seem to be a subtle question.
References [ 1 ] B. Adamczewski, Codages de rotations et phe ´nome nes d’autosimilarite ´, J. The ´or. Nombres Bordeaux, 14 (2002), 351–386.
[ 2 ] P. Arnoux and G. Rauzy, Repre ´sentation ge ´ome ´trique de suites de complexite ´ 2n þ 1, Bull. Soc. Math.
France, 119 (1991), 199–215.
[ 3 ] J. Berstel and L. Vuillon, Coding rotations on intervals, Theoret. Comput. Sci., 281 (2002), 99–107.
[ 4 ] V. Berthe ´, Autour du systeme de nume ´ration d’Ostrowski, Bull. Belg. Math. Soc. Simon Stevin, 8 (2001), 209–239.
[ 5 ] V. Berthe ´, N. Chekhova, and S. Ferenczi, Covering numbers: arithmetics and dynamics for rotations and interval exchanges, J. Anal. Math., 79 (1999), 1–31.
[ 6 ] V. Berthe ´, S. Ferenczi, and L. Q. Zamboni, Interactions between dynamics, arithmetics and combinatorics: the good, the bad, and the ugly, Algebraic and topological dynamics, Contemp. Math., 385 (2005), 333–364.
[ 7 ] V. Berthe ´, C. Holton and L. Q. Zamboni, Initial powers of sturmian sequences, Acta Arith., 122 (2006), 315–347.
[ 8 ] A. Bertrand, De ´veloppements en base de Pisot et re ´partition modulo 1, C. R. Acad. Sci. Paris Se ´r. A-B, 285 (1977), A419–A421.
[ 9 ] M. D. Boshernitzan and C. R. Carroll, An extension of Lagrange’s theorem to interval exchange transformations over quadratic fields, J. Anal. Math., 72 (1997), 21–44.
E. M. Coven and G. A. Hedlund, Sequences with minimal block growth, Math. Systems Theory, 7 (1973), 138–153.
D. Crisp, W. Moran, A. Pollington, and P. Shiue, Substitution invariant cutting sequences, J. Theor.
Nombres Bordeaux, 5 (1993), 123–137.
G. Didier, Combinatoire des codages de rotations, Acta Arith., 85 (1998), 157–177.
F. Durand, Contribution a l’e ´tude des suites et systems dynamiques substitutifs, Ph.D. thesis, Faculte ´ de Sciences de Luminy, Universite ´ de la Me ´diterrane ´e, Aix-Marseille II, 1996.
F. Durand, A characterization of substitutive sequences using return words, Discrete Math., 179 (1998), 89–101.
S. Ferenczi, C. Holton, and L. Q. Zamboni, Structure of three interval exchange transformations I: An arithmetic study, Ann. Inst. Fourier (Grenoble), 51 (2001), 861–901.
S. Ferenczi, C. Holton, and L. Q. Zamboni, Structure of three-interval exchange transformations II: A combinatorial description of the trajectories, J. Anal. Math., 89 (2003), 239–276.
S. Ferenczi, C. Holton, and L. Q. Zamboni, Structure of three-interval exchange transformations III: Ergodic and spectral properties, J. Anal. Math., 93 (2004), 103–138.
N. Pytheas Fogg, Substitutions in dynamics, arithmetics and combinatorics, Lecture Notes in Math.
1794, Springer-Verlag, Berlin, 2002.
C. Holton and L. Q. Zamboni, Descendants of primitive substitutions, Theory Comput. Syst., 32 (1999), 133–157.
I. P. Cornfeld, S. V. Fomin, and Ya. G. Sinai, Ergodic theory, Grundlehren der mathematischen Wissenschaften, 245, Splinger-Verlag, Berlin-Heidelberg-New York, 1982.
Sh. Ito and H. Nakada, Approximations of real numbers by the sequence fn g and their metrical theory, Acta Math. Hungar., 52 (1988), 91–100.
Sh. Ito and S. Yasutomi, On continued fractions, substitutions and characteristic sequences ½nx þ y Recursively renewable words and coding of irrational rotations 1233 ½ðn 1Þx þ y, Japan. J. Math. (N.S.), 16 (1990), 287–306.
T. Kamae and L. Q. Zamboni, Maximal pattern complexity for discrete systems, Ergodic Theory Dynam. Systems, 22 (2002), 1201–1214.
M. Keane, Interval exchange transformations, Math. Z., 141 (1975), 25–31.
M. Keane, Non-ergodic interval exchange transformations, Israel J. Math., 26 (1977), 188–196.
T. Komatsu and A. J. van der Poorten, Substitution invariant Beatty sequences, Japan. J. Math.
(N.S.), 22 (1996), 349–354.
C. Kraaikamp and H. Nakada, On normal numbers for continued fractions, Ergodic Theory Dynam.
Systems, 20 (2000), 1405–1421.
M. Lothaire, Algebraic combinatorics on words, Encyclopedia of Mathematics and its Applications, 90, Cambridge University Press, Cambridge, 2002.
K. Masui, F. Sugisaki, and M. Yoshida, Denjoy system and dimension groups, to appear in Ergodic Theory Dynam. Systems.
M. Morse and G. A. Hedlund, Symbolic dynamics II. Sturmian trajectories, Amer. J. Math., 62 (1940), 1–42.
B. Parvaix, Proprietes d’invariance des mots sturmiens, J. Theor. Nombres Bordeaux, 9 (1997), 351–369.
G. Rauzy, Nombres alge ´briques et substitutions, Bull. Soc. Math. France, 110 (1982), 147–178.
G. Rote, Sequences with subword complexity 2n, J. Number Theory, 46 (1994), 196–213.
K. Schmidt, On periodic expansions of Pisot numbers and Salem numbers, Bull. London Math. Soc., 12 (1980), 269–278.
T. Takagi, Lectures in elementary number theory, Kyoritsu Publ., 1971.
P. Walters, An introduction to Ergodic Theory, Lecture Notes in Math. 79, (1982).
S. Yasutomi, On sturmian sequences which are invariant under some substitutions, Number theory and its applications (Kyoto, 1997), Kluwer Acad. Publ., Dordrecht, 1999, pp. 347–373.
Shigeki AKIYAMA Department of Mathematics Faculty of Science Niigata University Ikarashi 2-8050 Niigata 950-2181, Japan E-mail: akiyama@math.sc.niigata-u.ac.jp Masayuki SHIRASAKA Niizaki 112-2, Niigata, Japan E-mail: aeneus aeneus@hotmail.com 1234 S. AKIYAMA and M. SHIRASAKA |
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Contents
1 Explanation
2 Formula
3 Forces and equilibrium
3.1 Simplified model
4 Refinements
5 Principle of flotation
6 Eureka
7 See also
8 References
9 External links
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From Wikipedia, the free encyclopedia
Buoyancy principle in fluid dynamics
This article is about fluid dynamics. For the algebraic axiom, see Archimedean property.
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| Fluid mechanics | Fluids | | Statics · Dynamics Archimedes' principle · Bernoulli's principle Navier–Stokes equations Poiseuille equation · Pascal's law Viscosity + (Newtonian · non-Newtonian) Buoyancy · Mixing · Pressure | | Liquids | | Adhesion Capillary action Chromatography Cohesion (chemistry) Surface tension | | Gases | | Atmosphere Boyle's law Charles's law Combined gas law Fick's law Gay-Lussac's law Graham's law | | Plasma | |
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| v t e |
Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially, is equal to the weight of the fluid that the body displaces. Archimedes' principle is a law of physics fundamental to fluid mechanics. It was formulated by Archimedes of Syracuse.
Explanation
In On Floating Bodies, Archimedes suggested that (c. 246 BC):
Any object, totally or partially immersed in a fluid or liquid, is buoyed up by a force equal to the weight of the fluid displaced by the object.
Archimedes' principle allows the buoyancy of any floating object partially or fully immersed in a fluid to be calculated. The downward force on the object is simply its weight. The upward, or buoyant, force on the object is that stated by Archimedes' principle above. Thus, the net force on the object is the difference between the magnitudes of the buoyant force and its weight. If this net force is positive, the object rises; if negative, the object sinks; and if zero, the object is neutrally buoyant—that is, it remains in place without either rising or sinking. In simple words, Archimedes' principle states that, when a body is partially or completely immersed in a fluid, it experiences an apparent loss in weight that is equal to the weight of the fluid displaced by the immersed part of the body(s).
Formula
Consider a cuboid immersed in a fluid, its top and bottom faces orthogonal to the direction of gravity (assumed constant across the cube's stretch). The fluid will exert a normal force on each face, but only the normal forces on top and bottom will contribute to buoyancy. The pressure difference between the bottom and the top face is directly proportional to the height (difference in depth of submersion). Multiplying the pressure difference by the area of a face gives a net force on the cuboid—the buoyancy—equaling in magnitude the weight of the fluid displaced by the cuboid. By summing up sufficiently many arbitrarily small cuboids this reasoning may be extended to irregular shapes, and so, whatever the shape of the submerged body, the buoyant force is equal to the weight of the displaced fluid.
The weight of the displaced fluid is directly proportional to the volume of the displaced fluid (if the surrounding fluid is of uniform density). The apparent weight of the object in the fluid is reduced, because of the force acting on it, which is called upthrust. In simple terms, the principle states that the buoyant force (Fb) on an object is equal to the weight of the fluid displaced by the object, or the density (ρ) of the fluid multiplied by the submerged volume (V) times the gravity (g)
We can express this relation in the equation:
where denotes the buoyant force applied onto the submerged object, denotes the density of the fluid, represents the volume of the displaced fluid and is the acceleration due to gravity. Thus, among completely submerged objects with equal masses, objects with greater volume have greater buoyancy.
Suppose a rock's weight is measured as 10 newtons when suspended by a string in a vacuum with gravity acting on it. Suppose that, when the rock is lowered into the water, it displaces water of weight 3 newtons. The force it then exerts on the string from which it hangs would be 10 newtons minus the 3 newtons of buoyant force: 10 − 3 = 7 newtons. Buoyancy reduces the apparent weight of an object. It is generally easier to lift an object through the water than it is to pull it out of the water.
For a fully submerged object, Archimedes' principle can be reformulated as follows:
then inserted into the quotient of weights, which has been expanded by the mutual volume
yields the formula below. The density of the immersed object relative to the density of the fluid can easily be calculated without measuring any volume is
(This formula is used for example in describing the measuring principle of a dasymeter and of hydrostatic weighing.)
Example: If you drop wood into water, buoyancy will keep it afloat.
Example: A helium balloon in a moving car. When increasing speed or driving in a curve, the air moves in the opposite direction to the car's acceleration. However, due to buoyancy, the balloon is pushed "out of the way" by the air and will drift in the same direction as the car's acceleration.
When an object is immersed in a liquid, the liquid exerts an upward force, which is known as the buoyant force, that is proportional to the weight of the displaced liquid. Consequently, the net force acting on the object is equal to the difference between the weight of the object, or 'down' force, and the weight of the displaced fluid, or 'up' force. Equilibrium, or neutral buoyancy, is achieved when these two weights and thus forces are equal.
Forces and equilibrium
The equation to calculate the pressure inside a fluid in equilibrium is:
where f is the force density exerted by some outer field on the fluid, and σ is the Cauchy stress tensor. In this case the stress tensor is proportional to the identity tensor:
Here δij is the Kronecker delta. Using this the above equation becomes:
Assuming the outer force field is conservative, that is it can be written as the negative gradient of some scalar valued function:
Then:
Therefore, the shape of the open surface of a fluid equals the equipotential plane of the applied outer conservative force field. Let the z-axis point downward. In this case the field is gravity, so Φ = −ρfgz where g is the gravitational acceleration, ρf is the mass density of the fluid. Taking the pressure as zero at the surface, where z is zero, the constant will be zero, so the pressure inside the fluid, when it is subject to gravity, is
So pressure increases with depth below the surface of a liquid, as z denotes the distance from the surface of the liquid into it. Any object with a non-zero vertical depth will have different pressures on its top and bottom, with the pressure on the bottom being greater. This difference in pressure causes the upward buoyancy force.
The buoyancy force exerted on a body can now be calculated easily, since the internal pressure of the fluid is known. The force exerted on the body can be calculated by integrating the stress tensor over the surface of the body which is in contact with the fluid:
The surface integral can be transformed into a volume integral with the help of the Gauss theorem:
where V is the measure of the volume in contact with the fluid, that is the volume of the submerged part of the body, since the fluid doesn't exert force on the part of the body which is outside of it.
The magnitude of buoyancy force may be appreciated a bit more from the following argument. Consider any object of arbitrary shape and volume V surrounded by a liquid. The force the liquid exerts on an object within the liquid is equal to the weight of the liquid with a volume equal to that of the object. This force is applied in a direction opposite to gravitational force, that is of magnitude:
where ρf is the density of the fluid, Vdisp is the volume of the displaced body of liquid, and g is the gravitational acceleration at the location in question.
If this volume of liquid is replaced by a solid body of exactly the same shape, the force the liquid exerts on it must be exactly the same as above. In other words, the "buoyancy force" on a submerged body is directed in the opposite direction to gravity and is equal in magnitude to
The net force on the object must be zero if it is to be a situation of fluid statics such that Archimedes principle is applicable, and is thus the sum of the buoyancy force and the object's weight
If the buoyancy of an (unrestrained and unpowered) object exceeds its weight, it tends to rise. An object whose weight exceeds its buoyancy tends to sink. Calculation of the upwards force on a submerged object during its accelerating period cannot be done by the Archimedes principle alone; it is necessary to consider dynamics of an object involving buoyancy. Once it fully sinks to the floor of the fluid or rises to the surface and settles, Archimedes principle can be applied alone. For a floating object, only the submerged volume displaces water. For a sunken object, the entire volume displaces water, and there will be an additional force of reaction from the solid floor.
In order for Archimedes' principle to be used alone, the object in question must be in equilibrium (the sum of the forces on the object must be zero), therefore;
and therefore
showing that the depth to which a floating object will sink, and the volume of fluid it will displace, is independent of the gravitational field regardless of geographic location.
: (Note: If the fluid in question is seawater, it will not have the same density (ρ) at every location. For this reason, a ship may display a Plimsoll line.)
It can be the case that forces other than just buoyancy and gravity come into play. This is the case if the object is restrained or if the object sinks to the solid floor. An object which tends to float requires a tension restraint force T in order to remain fully submerged. An object which tends to sink will eventually have a normal force of constraint N exerted upon it by the solid floor. The constraint force can be tension in a spring scale measuring its weight in the fluid, and is how apparent weight is defined.
If the object would otherwise float, the tension to restrain it fully submerged is:
When a sinking object settles on the solid floor, it experiences a normal force of:
Another possible formula for calculating buoyancy of an object is by finding the apparent weight of that particular object in the air (calculated in Newtons), and apparent weight of that object in the water (in Newtons). To find the force of buoyancy acting on the object when in air, using this particular information, this formula applies:
: Buoyancy force = weight of object in empty space − weight of object immersed in fluid
The final result would be measured in Newtons.
Air's density is very small compared to most solids and liquids. For this reason, the weight of an object in air is approximately the same as its true weight in a vacuum. The buoyancy of air is neglected for most objects during a measurement in air because the error is usually insignificant (typically less than 0.1% except for objects of very low average density such as a balloon or light foam).
Simplified model
A simplified explanation for the integration of the pressure over the contact area may be stated as follows:
Consider a cube immersed in a fluid with the upper surface horizontal.
The sides are identical in area, and have the same depth distribution, therefore they also have the same pressure distribution, and consequently the same total force resulting from hydrostatic pressure, exerted perpendicular to the plane of the surface of each side.
There are two pairs of opposing sides, therefore the resultant horizontal forces balance in both orthogonal directions, and the resultant force is zero.
The upward force on the cube is the pressure on the bottom surface integrated over its area. The surface is at constant depth, so the pressure is constant. Therefore, the integral of the pressure over the area of the horizontal bottom surface of the cube is the hydrostatic pressure at that depth multiplied by the area of the bottom surface.
Similarly, the downward force on the cube is the pressure on the top surface integrated over its area. The surface is at constant depth, so the pressure is constant. Therefore, the integral of the pressure over the area of the horizontal top surface of the cube is the hydrostatic pressure at that depth multiplied by the area of the top surface.
As this is a cube, the top and bottom surfaces are identical in shape and area, and the pressure difference between the top and bottom of the cube is directly proportional to the depth difference, and the resultant force difference is exactly equal to the weight of the fluid that would occupy the volume of the cube in its absence.
This means that the resultant upward force on the cube is equal to the weight of the fluid that would fit into the volume of the cube, and the downward force on the cube is its weight, in the absence of external forces.
This analogy is valid for variations in the size of the cube.
If two cubes are placed alongside each other with a face of each in contact, the pressures and resultant forces on the sides or parts thereof in contact are balanced and may be disregarded, as the contact surfaces are equal in shape, size and pressure distribution, therefore the buoyancy of two cubes in contact is the sum of the buoyancies of each cube. This analogy can be extended to an arbitrary number of cubes.
An object of any shape can be approximated as a group of cubes in contact with each other, and as the size of the cubes is decreased, the precision of the approximation increases. The limiting case for infinitely small cubes is the exact equivalence.
Angled surfaces do not nullify the analogy as the resultant force can be split into orthogonal components and each dealt with in the same way.
Refinements
Archimedes' principle does not consider the surface tension (capillarity) acting on the body. Moreover, Archimedes' principle has been found to break down in complex fluids.
There is an exception to Archimedes' principle known as the bottom (or side) case. This occurs when a side of the object is touching the bottom (or side) of the vessel it is submerged in, and no liquid seeps in along that side. In this case, the net force has been found to be different from Archimedes' principle, as, since no fluid seeps in on that side, the symmetry of pressure is broken.
Principle of flotation
Archimedes' principle shows the buoyant force and displacement of fluid. However, the concept of Archimedes' principle can be applied when considering why objects float. Proposition 5 of Archimedes' treatise On Floating Bodies states that
Any floating object displaces its own weight of fluid.
— Archimedes of Syracuse
In other words, for an object floating on a liquid surface (like a boat) or floating submerged in a fluid (like a submarine in water or dirigible in air) the weight of the displaced fluid equals the weight of the object. Thus, only in the special case of floating does the buoyant force acting on an object equal the objects weight. Consider a 1-ton block of solid iron. As iron is nearly eight times as dense as water, it displaces only 1/8 ton of water when submerged, which is not enough to keep it afloat. Suppose the same iron block is reshaped into a bowl. It still weighs 1 ton, but when it is put in water, it displaces a greater volume of water than when it was a block. The deeper the iron bowl is immersed, the more water it displaces, and the greater the buoyant force acting on it. When the buoyant force equals 1 ton, it will sink no farther.
When any boat displaces a weight of water equal to its own weight, it floats. This is often called the "principle of flotation": A floating object displaces a weight of fluid equal to its own weight. Every ship, submarine, and dirigible must be designed to displace a weight of fluid at least equal to its own weight. A 10,000-ton ship's hull must be built wide enough, long enough and deep enough to displace 10,000 tons of water and still have some hull above the water to prevent it from sinking. It needs extra hull to fight waves that would otherwise fill it and, by increasing its mass, cause it to submerge. The same is true for vessels in air: a dirigible that weighs 100 tons needs to displace 100 tons of air. If it displaces more, it rises; if it displaces less, it falls. If the dirigible displaces exactly its weight, it hovers at a constant altitude.
While they are related to it, the principle of flotation and the concept that a submerged object displaces a volume of fluid equal to its own volume are not Archimedes' principle. Archimedes' principle, as stated above, equates the buoyant force to the weight of the fluid displaced.
One common point of confusion[by whom?] regarding Archimedes' principle is the meaning of displaced volume. Common demonstrations involve measuring the rise in water level when an object floats on the surface in order to calculate the displaced water. This measurement approach fails with a buoyant submerged object because the rise in the water level is directly related to the volume of the object and not the mass (except if the effective density of the object equals exactly the fluid density).
Eureka
Main article: Eureka (word)
Archimedes reportedly exclaimed "Eureka" after he realized how to detect whether a crown is made of impure gold. While he did not use Archimedes' principle in the widespread tale and used displaced water only for measuring the volume of the crown, there is an alternative approach using the principle: Balance the crown and pure gold on a scale in the air and then put the scale into water. According to Archimedes' principle, if the density of the crown differs from the density of pure gold, the scale will get out of balance under water.
See also
Phragmen's voting rules – a ballot load balancing method analogous to the idea of Archimedes' principle.
Ship stability – Archimedes's theory of displacement of fluid is a core principle of ship stability.
List of eponymous laws
^ a b "What is buoyant force?". Khan Academy.
^ Acott, Chris (1999). "The diving "Law-ers": A brief resume of their lives". South Pacific Underwater Medicine Society Journal. 29 (1). ISSN 0813-1988. OCLC 16986801. Archived from the original on 27 July 2011. Retrieved 13 June 2009.
^ "The buoyant force". bu.edu. Retrieved 3 September 2023.
^ Falkovich, G.; Weinberg, A.; Denissenko, P.; Lukaschuk, S. (23 June 2005). "Floater clustering in a standing wave: Capillarity effects drive hydrophilic or hydrophobic particles to congregate at specific points on a wave" (PDF). Nature. 435 (7045): 1045–1046. doi:10.1038/4351045a. PMID 15973398.
^ Wilson, R. Mark (2012). "Archimedes's principle gets updated". Physics Today. 65 (9): 15. Bibcode:2012PhT....65i..15W. doi:10.1063/PT.3.1701.
^ Lima, F M S. (2012). "Using surface integrals for checking the Archimedes' law of buoyancy". European Journal of Physics. 33 (1): 101–113. arXiv:1110.5264. Bibcode:2012EJPh...33..101L. doi:10.1088/0143-0807/33/1/009. S2CID 54556860.
^ "The works of Archimedes". Cambridge, University Press. 1897. p. 257. Retrieved 11 March 2010. Any solid lighter than a fluid will, if placed in the fluid, be so far immersed that the weight of the solid will be equal to the weight of the fluid displaced.
^ Mohindroo, K. K. (1997). Basic Principles of Physics. Pitambar Publishing. pp. 76–77. ISBN 978-81-209-0199-5.
^ Redish, Edward F.; Vicentini, Matilde; fisica, Società italiana di (2004). Research on Physics Education. IOS Press. p. 358. ISBN 978-1-58603-425-2.
^ Proof of Concept carpeastra.co.uk
^ "The Golden Crown". physics.weber.edu.
^ "'Eureka!' – The Story of Archimedes and the Golden Crown". Long Long Time Ago. 16 May 2014. Archived from the original on 2 June 2019. Retrieved 30 May 2018.
External links
Media related to Archimedes' principle at Wikimedia Commons
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aarohan567 3 years ago Posted 3 years ago. Direct link to aarohan567's post “Best of luck for today ev...” more Best of luck for today everyone ! Do your best !!
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Glori Hallelujah 2 years ago Posted 2 years ago. Direct link to Glori Hallelujah's post “I imagine that it's arbit...” more I imagine that it's arbitrary. Usually, you use x and y, but since they are variables and the numbers are given, they are using h and k as constants.
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Monica Bandora 6 years ago Posted 6 years ago. Direct link to Monica Bandora's post “A point on the circle is ...” more A point on the circle is literally any point lying on the circumference. To get the radius, you would have to calculate the distance between that point and the circle's center. A point on the circle does NOT mean radius.
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katieleonard032 7 years ago Posted 7 years ago. Direct link to katieleonard032's post “HERE IS THE PROVIDED QUES...” more HERE IS THE PROVIDED QUESTION:
"A circle in the xy-plane contains the points (−1,1), (1,1), and (−1,−1). Which of the following is an equation of the circle?"
HERE IS THE PROVIDED ANSWER:
The formula for a circle with a center (h,k) is:
(x−h)^2+(y−k)^2=r^2
We could check which formula is satisfied by the three given points.
Alternatively, let's substitute the three points for (x,y)
First, substituting (1,1), we have:
(1−h)^2+(1−k)^2=r^2
Substituting (−1,1), we have:
(−1−h)^2+(1−k)^2=r^2
Finally, substituting (−1,−1), we have:
(−1−h)^2+(−1−k)^2=r^2
Next, let's subtract the second equation from the first. We get:
(1−h)^2−(−1−h)^2=0
OR
h^2−2h+1−(h^2+2h+1)=0
−4h=0
h=0
Let's subtract the third equation from the second. We get:
(1−k)^2−(−1−k)^2=0
This is the same equation as that above with
k instead of h. So k=0, and the center of the circle is (0,0). Now, let's find r.
Since (1,1) is a point on the circle, and the center is (0,0), the radius is squarerootof2.
The equation of the circle is
x^2+y^2=2x OR x^2+y^2−2=0
HERE ARE MY QUESTIONS:
I don't understand how the answer was found, and I also have a few general questions:
Is it assumed that none three given points are the center of the circle?
Because I do not know how to write a circle equation without knowing the center. If the center is known to not be one of the three listed points, then it is obviously (0,0), but if the center IS one of the listed points, then it is (-1,1). However, I do not know if the center is included in the listed points or not, so I couldn't figure much else out without that information....
The provided answer talks about subtracting equations from each other like it's a normal thing. I have never encountered a situation where an equation is subtracted form another when dealing with circles... elaborate?
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widanbadshah1956 10 months ago Posted 10 months ago. Direct link to widanbadshah1956's post “Another approach, We know...” more Another approach,
We know that the circle has it's center at (7,14), find an equation which gives you the point (7,14). Now there are two equations with those points, Option b and d, right. Now put the given point (7, 34) into those 2 equations to find the right one.
When you put 7 in the 2nd equation for x, that become zero and 34 for y and subtract 14, you'll be left with 20 and square it which would be 200 and it's not equal to 20. Now put the same value of point in d option and you'll get same 200, and the equation would be satisfied.
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[Instructor] A circle in the xy-plane has its center at seven comma 14. If the point seven comma 34 lies on the circle, which of the following is an equation of the circle? Let's just remind ourselves the general form of the equation of a circle. If any of this looks like, looks unusual to you, you haven't seen it before, I encourage you to watch the equation of a circle videos on Kahn Academy. The general form for the equation of a circle, if we have a circle that's centered at the point h comma k, the x-coordinate of the center is h. The y-coordinate of the center is k. If it has a radius of r, so the circle is gonna look something like that. I'm trying to draw it as circular as I can. Man, that's not so circular. I think you get the point. The radius is r. This equation, the equation of this circle, of all the points that are exactly r away from the point h comma k, is going to be x minus the x-coordinate of the center squared plus y minus the y-coordinate of the center squared is equal to the radius squared. Now in this case, we know what our h and k is. H is seven, and k is 14, and we just need to figure out the radius. We figure that out, we will be able to figure out the equation of the entire circle. Now, they tell us that the center's at seven comma 14, so that point right over there is seven comma 14, and then the point seven comma 34 also lies on the circle, so it has the same x-coordinate. Its y-coordinate is just higher, so this might be, that right over there might be the point seven comma 34. It would be right above it. We just increased our y without changing our x. So, seven comma 14 is the center. This point right over here is seven comma 34, lies on the circle, so the circle is gonna look something like this, and so the radius of this circle, what we have to figure out, is just the distance between these two points. These two points, you don't even have to really apply the distance formula or anything like this. The distance here is just our change in y. Our x doesn't change. It's gonna be 34 minus 14. So, the radius here is equal to 20. So, now we know. We know h is equal to seven. K is equal to 14, and r is equal to 20. So, it's gonna be x minus h, x minus seven squared plus y minus k, y minus 14 squared is equal to r squared. 20 squared is 400. So let's see, that is this choice right over here. X minus seven squared plus y minus 14 squared is equal to 400, and we're done.
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188506 | https://math.stackexchange.com/questions/1999533/why-are-the-domains-for-ln-x2-and-2-ln-x-different | Stack Exchange Network
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Why are the domains for $\ln x^2$ and $2\ln x$ different?
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If I have a function like this $f(x)=2 \ln(x)$ and I want to find the domain, I put $x>0$. But if I use the properties of logarithmic functions, I can write that function like $f(x)=\ln(x^2)$ and so the domain is all $\mathbb{R}$ and the graphic of function is different. Where is the mistake?
algebra-precalculus
logarithms
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edited Nov 5, 2016 at 2:01
Daniel W. Farlow
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asked Nov 4, 2016 at 18:17
L.G.A.G.L.G.A.G.
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$\begingroup$ You can't have a function without having a domain. You can have an expression however and then ask for what values is the expression meaningful. That seems to be what you're doing. $\endgroup$
zhw.
– zhw.
2016-11-04 21:31:23 +00:00
Commented Nov 4, 2016 at 21:31
$\begingroup$ My question is: why for example $ln(x)+ln(x-3)$ that is the same of $ln(x(x-3))$ have two different domains and graphs. $\endgroup$
L.G.A.G.
– L.G.A.G.
2016-11-04 22:13:43 +00:00
Commented Nov 4, 2016 at 22:13
1
$\begingroup$ @L.G.A.G. They are the same where they are both defined, but they need not be defined in the same places. This is a tricky thing about logarithms that I find precalculus references don't cover very well. $\endgroup$
Ian
– Ian
2016-11-05 02:23:52 +00:00
Commented Nov 5, 2016 at 2:23
$\begingroup$ just because they tell you in algebra that you can't have log(n) n<=0 doesn't mean that you can't... youtube.com/watch?v=IX_23EWpF5U also youtube.com/watch?v=soFDU-1knNE $\endgroup$
cmarangu
– cmarangu
2019-07-06 00:09:22 +00:00
Commented Jul 6, 2019 at 0:09
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First, you could use $\ln x$ to define functions with different domains as long as $\ln x$ is defined in that domain.
Second, the rule $\ln x^n=n\cdot \ln x$ is a bit sloppy. It should always be pointed out that $x>0$. Likewise, $\ln ab=\ln a+\ln b$, only if $a,b>0$.
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edited Nov 5, 2016 at 3:05
answered Nov 5, 2016 at 2:33
LehsLehs
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Note that: $$ \ln (x^2)=2\ln |x| \ne 2 \ln x $$
so the two functions are different and have different domains.
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answered Nov 4, 2016 at 18:20
Emilio NovatiEmilio Novati
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$\begingroup$ Thanks. But if I take $ln(x)+ln(x-3)$, this function is $ln(x(x-3))$ from the properties of logarithmic functions, but the domains are different. Why if they should be the same thing? $\endgroup$
L.G.A.G.
– L.G.A.G.
2016-11-04 18:26:58 +00:00
Commented Nov 4, 2016 at 18:26
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The functions $f(x) =2 \ln x$ and $g(x) = \ln x^2$ have different domains. The domain of $f$ is $(0,\infty)$, and the domain of $g$ is $\mathbb{R} - {0}$. But as you said, when $x$ is in the domain of $f$ and the domain of $g$, we have $f(x) = g(x)$.
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edited Nov 5, 2016 at 5:08
answered Nov 4, 2016 at 18:21
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The domain of a function is part of its definition. Restricting ourselves to functions from subsets of the real numbers to the real numbers, the logarithm function $x \mapsto \ln x$ is defined to have the domain $(0,\infty) \subset \mathbb R$. (I mention the restriction to $\mathbb R$ because there is also a function named $\ln$ whose domain is the non-zero complex numbers.)
There is nothing to stop you from defining a function with $f_1$ with domain $[17,23]$ such that $f_1(x) = \ln x$ whenever $17\leq x \leq23$. The new function $f_1$ does not have all of the nice properties of $\ln$, for example it is never true that $f_1(x) + f_1(y) = f_1(xy)$, because $x$, $y$, and $xy$ cannot all simultaneously be in the domain of $f_1$. Nevertheless, $f_1$ is a perfectly well-defined function, even if it is far less useful than $\ln$, just as the function $\ln$ with domain $(0,\infty)$ is a perfectly-well defined function despite being less useful (for some purposes) than the complex principal logarithm function.
Because the domain of $f_1$ is different from the domain of $\ln$, $f_1$ is not the same function as $\ln$.
Now you want to define a function with the formula $f(x) = 2 \ln x$, but the definition needs a domain. I would argue that there is no such thing as "the" domain for a function defined by that formula, since it is possible to use the mapping $x \mapsto 2 \ln x$ to define functions over many different domains such as $(0,10]$, $[17,71]$, or $(3,4]\cup[10,11)\cup{37}$, but there is a maximal domain for functions of that kind, namely, the union of the domains of all possible functions that can be defined by that formula. That domain is again $(0,\infty)$. So if someone asks me for "the" domain of $f(x) = 2 \ln x$ I would guess that they meant the domain $(0,\infty)$; it is the "best" choice for most purposes.
An alternative definition of the function $f(x) = 2 \ln x$ on the domain $(0,\infty)$ is to say that $f(x) = \ln(x^2)$ when $x \in (0,\infty)$. This is the same function because it has the same domain and takes the same value at each point in that domain.
It is also possible to define a function $g(x) = \ln(x^2)$ for all $x \in \mathbb R - {0}$. That is a perfectly well-defined function, but it is a different function than $f$ since it has a different domain.
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answered Nov 5, 2016 at 1:47
David KDavid K
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Natural log of x^2 can take negative values because they are squared before they are fed to the logarithm function.
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answered Nov 5, 2016 at 2:20
Jacob WakemJacob Wakem
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188507 | https://www.youtube.com/watch?v=QA_gm7a_ZDw | Three Mistakes Students make when finding the inverse of a function
Brian McLogan
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In this video I will show you the three most common mistakes students make when they are finding the inverse of a function
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Transcript:
Intro welcome in this video i am going to show you how to find the inverse of three different functions as well as cover the top three mistakes that i see my students make and i chose these three examples because this is what we recently did in my class and i wanted to make sure i could share it with you so you can see the top mistakes and avoid them when you're working on finding the First Mistake inverse so in the first example we have a linear equation and remember when we're finding the inverse we're going to replace our function notation with our y that's just going to make things a little bit easier to work with you can go ahead and swap the variables that's something that we made sense of by looking at the graphs of a function and its inverse and then from here you actually have to swap the variables and then from there now you're just going to go ahead and solve for y using our inverse operation and that's where our first mistake comes from so we're going to go ahead and subtract a 6 on both sides right we get x minus 6 equals a negative 1 3 y a lot of students still do not like fractions right so they think you know a a common mistake but not the major mistake is they you know need to get rid of this negative 1 3. i prefer to multiply by the reciprocal on both sides but be careful if you write it like me you just made that in the first mistake and what i mean by that is it's not negative 3 times x when you multiply negative 3 on both sides you're multiplying not x by negative 3 but the quantity x my 6. so that first major mistake students make is they do not use parentheses to represent the multiplication of the quantity now you could distribute it if you wanted to but i'm going to leave it outside just so i just could i can leave my example and see everything where it's at any time you have a number multiplied by its reciprocal those are going to multiply to one where you can equal y we can write them rewrite or rearrange them and then we can back introduce the f inverse of x or the f inverse notation so swap back out the y and there you go so make sure you're using those parentheses when you are solving Second Mistake in the next example we have a couple couple of issues with this one but again we're going to start with what we know which is first to replace the f of x with y then we swap the variables so we have x is equal to a y plus 1 all over a 2 y minus 1. so the mistake that students make in this one is either there's actually kind of two of them i'm just gonna embed them in kind of one but two well actually i'll get to the major mistake here in a second but two they try to solve for y in the denominator you can't solve for y in the denominator nor can you solve for two different y's in the same equation right so a couple things we have to do we have to get rid of the variable in the denominator and we have to combine these variables so to get my 2y minus 1 off the denominator i'm going to multiply by 2y minus 1. again i'm using parentheses because i don't want to make that first mistake right we don't want to make the mistake over here now i am going to distribute this and then these divide out because of the division property and that's going to leave me with a 2yx minus 1 or 2yx i'm sorry minus x equals y plus 1. all right so now we have a variable on both sides and if you recall we can only solve a variable when they're on the same side so the other mistake that students will make here is they will try to you know combine like terms here and or they get stuck they either make something up so we have 2y x minus y is equal to a i'm just going to rewrite this as an x plus one so there's a lot of mistakes that students will make here one they'll try to combine these they're not like terms so you can't combine them another one is they try to divide by y on both sides and that's great that gets rid of the y on the left hand side but guess but guess what that does on the right hand side that gives us a y on the right hand side so you cannot just divide by y to get rid of it and you cannot combine like terms so it's the second mistake is not knowing how to solve or how to isolate the y and there's many many different variations or mistakes that students will do and what you need to do is to solve this actually is just to factor so you're going to take the common term which is a y and factor it out now you're going to be left with a 2x minus 1 equals an x plus 1. and now you can see my variable y to isolate it i'm just going to divide by 2x minus 1. and therefore i'll reintroduce my inverse notation f inverse of x equals x plus 1 divided by 2x minus 1. and there you go Third Mistake all right the last mistake is it's going to be a big one and again it really kind of comes into part of the lesson as far as fully understanding um the domains of a function and the range of the function and its inverse but um but what we're going to do is we're just going to follow the operations first and then we'll kind of go back into the understanding of where this mistake comes from so first of all let's just replace with this y let's swap the variables okay and then let's keep on swapping your variables i'm making that mistake all the time and then we're going to go ahead and use our inverse operation so that's a negative x equals square root of 2y we're going to undo squaring by or under the square root by squaring both sides and therefore i get a negative squared equals 2y divide by 2 divide by 2 and i'm just going to rewrite this in this format 1 y hat negative x squared now again you could write that as a x positive x squared i get it but i'm just going to leave it from here now here's the point that i want you to make and again you have to understand the graphing portion and if you want to confirm my graphs go on to desmos you know an online graphing calculator and confirm but here's where the understanding of the graph comes from this is the graph if you look at the if you remember a function in its graph this graph has a reflection about the x-axis okay so originally actually there is the square root graph okay so if i'm going to apply my transformations to graph it i'm going to reflect about the x-axis and i'm going to have a horizontal compression of 2. so this graph is going to look something like this all right so that is going to be my f x now if this is my inverse which it is let me let me go and write this f inverse of x that is the inverse what does this graph look like well again let's go through the transformations that's why i left that there is because i wanted to see the transformations i did the same thing over here because i like to see the transformations in case i need to graph it but either way i have a horde a vertical compression of one half and a reflection about the y-axis so this graph you know it's going to look something like this Conclusion now remember they a function and its inverse the whole reason why we swap the x and the y is because they are symmetrical about the y equals x line so does this graph when you flip it across this dashed line does it produce a parabola no that doesn't even make any sense how does this produce now a double equation it doesn't it can't right so what that means is it only produces a portion of the function so our inverse is algebraically correct but what we need to do is we need to apply a restriction we need to either take should we take the positive version of the graph or sorry the negative version or the positive version and hopefully visually you can kind of see if we erase the positive version that now we have a function that a inverse that is a reflection right so that we can say this is the f inverse of x so the answer or the equation is correct but now we need to add our restriction which is x has to be less than or equal to zero so the third and final mistake that students will make is when we have a problem where there needs to be a restriction they will forget to add the restriction and again it's very helpful to know the graphs so you know when to expect that you need a restriction because remember for a function to be invertible or for a function being invertible it has to be one to one it has it has to pass the horizontal line test but there you go guys those are the three major mistakes that i covered um in the classroom this week and i hopefully that can be helpful for you see you guys in the next video cheers |
188508 | https://projecteuclid.org/journals/annals-of-statistics/volume-36/issue-2/Effect-of-mean-on-variance-function-estimation-in-nonparametric-regression/10.1214/009053607000000901.pdf | The Annals of Statistics 2008, Vol. 36, No. 2, 646–664 DOI: 10.1214/009053607000000901 © Institute of Mathematical Statistics, 2008 EFFECT OF MEAN ON VARIANCE FUNCTION ESTIMATION IN NONPARAMETRIC REGRESSION BY LIE WANG, LAWRENCE D. BROWN, T. TONY CAI1 AND MICHAEL LEVINE University of Pennsylvania, University of Pennsylvania, University of Pennsylvania and Purdue University Variance function estimation in nonparametric regression is considered and the minimax rate of convergence is derived. We are particularly interested in the effect of the unknown mean on the estimation of the variance function.
Our results indicate that, contrary to the common practice, it is not desirable to base the estimator of the variance function on the residuals from an optimal estimator of the mean when the mean function is not smooth. Instead it is more desirable to use estimators of the mean with minimal bias. On the other hand, when the mean function is very smooth, our numerical results show that the residual-based method performs better, but not substantial better than the first-order-difference-based estimator. In addition our asymptotic results also correct the optimal rate claimed in Hall and Carroll [J. Roy. Statist. Soc.
Ser. B 51 (1989) 3–14].
1. Introduction.
Consider the heteroscedastic nonparametric regression model yi = f (xi) + V 1/2(xi)zi, i = 1,...,n, (1) where xi = i/n and zi are independent with zero mean, unit variance and uni-formly bounded fourth moments. Both the mean function f and variance function V are defined on [0,1] and are unknown. The main object of interest is the vari-ance function V . The estimation accuracy is measured both globally by the mean integrated squared error R( ˆ V ,V ) = E 1 0 ˆ V (x) −V (x) 2 dx (2) and locally by the mean squared error at a point R( ˆ V (x∗),V (x∗)) = E ˆ V (x∗) −V (x∗) 2.
(3) We wish to study the effect of the unknown mean f on the estimation of the vari-ance function V . In particular, we are interested in the case where the difficulty in estimation of V is driven by the degree of smoothness of the mean f .
Received March 2006; revised April 2007.
1Supported in part by NSF Grant DMS-03-06576.
AMS 2000 subject classifications. 62G08, 62G20.
Key words and phrases. Minimax estimation, nonparametric regression, variance estimation.
646 EFFECT OF MEAN ON VARIANCE FUNCTION ESTIMATION 647 The effect of not knowing the mean f on the estimation of V has been studied before in Hall and Carroll (1989). The main conclusion of their paper is that it is possible to characterize explicitly how the smoothness of the unknown mean function influences the rate of convergence of the variance estimator. In associa-tion with this they claim an explicit minimax rate of convergence for the variance estimator under pointwise risk. For example, they state that the “classical” rates of convergence (n−4/5) for the twice differentiable variance function estimator is achievable if and only if f is in the Lipschitz class of order at least 1/3. More precisely, Hall and Carroll (1989) stated that, under the pointwise mean squared error loss, the minimax rate of convergence for estimating V is max n−4α/(2α+1),n−2β/(2β+1) (4) if f has α derivatives and V has β derivatives. We shall show here that this result is in fact incorrect.
In the present paper we revisit the problem in the same setting as in Hall and Carroll (1989). We show that the minimax rate of convergence under both the pointwise squared error and global integrated mean squared error is max n−4α,n−2β/(2β+1) (5) if f has α derivatives and V has β derivatives. The derivation of the minimax lower bound is involved and is based on a moment matching technique and a two-point testing argument. A key step is to study a hypothesis testing problem where the alternative hypothesis is a Gaussian location mixture with a special moment matching property. The minimax upper bound is obtained using kernel smoothing of the squared first order differences.
Our results have two interesting implications. First, if V is known to belong to a regular parametric model, such as the set of positive polynomials of a given order, the cutoff for the smoothness of f on the estimation of V is 1/4, not 1/2 as stated in Hall and Carroll (1989). That is, if f has at least 1/4 derivative then the minimax rate of convergence for estimating V is solely determined by the smoothness of V as if f were known. On the other hand, if f has less than 1/4 derivative then the minimax rate depends on the relative smoothness of both f and V and will be completely driven by the roughness of f .
Second, contrary to the common practice, our results indicate that it is often not desirable to base the estimator ˆ V of the variance function V on the residuals from an optimal estimator ˆ f of the mean function f when f is not smooth. Instead it is more desirable to use estimators of the mean f with minimal bias. The main reason is that the bias and variance of ˆ f have quite different effects on the estimation of V .
The bias of ˆ f cannot be removed or even reduced in the second stage smoothing of the squared residuals, while the variance of ˆ f can be incorporated easily. On the other hand, when the mean function is very smooth, our numerical results show that the residual-based method performs better, but not substantial better than the first-order-difference-based estimator.
648 WANG, BROWN, CAI AND LEVINE The paper is organized as follows. Section 2 presents an upper bound for the minimax risk while Section 3 derives a rate-sharp lower bound for the minimax risk under both the global and local losses. The lower and upper bounds together yield the minimax rate of convergence. Section 4 discusses the obtained results and their implications for practical variance estimation in the nonparametric re-gression. Section 5 considers finite sample performance of the difference-based method for estimating the variance function. The proofs are given in Section 6.
2. Upper bound.
In this section we shall construct a kernel estimator based on the square of the first order differences. Such and more general difference based kernel estimators of the variance function have been considered, for exam-ple, in Müller and Stadtmüller (1987, 1993). For estimating a constant variance, difference based estimators have a long history. See von Neumann (1941, 1942), Rice (1984), Hall, Kay and Titterington (1990) and Munk, Bissantz, Wagner and Freitag (2005).
Define the Lipschitz class α(M) in the usual way, α(M) = g :for all 0 ≤x,y ≤1, k = 0,...,⌊α⌋−1, g(k)(x) ≤M and g(⌊α⌋)(x) −g(⌊α⌋)(y) ≤M|x −y|α′, where ⌊α⌋is the largest integer less than α and α′ = α −⌊α⌋. We shall assume that f ∈α(Mf ) and V ∈β(MV ). We say that the function f “has α derivative” if f ∈α(Mf ) and V “has β derivatives” if V ∈β(MV ).
For i = 1,2,...,n −1, set Di = yi −yi+1. Then one can write Di = f (xi) −f (xi+1) + V 1/2(xi)zi −V 1/2(xi+1)zi+1 = δi + √ 2V 1/2 i ϵi, (6) where δi = f (xi) −f (xi+1), V 1/2 i = 1 2(V (xi) + V (xi+1)) and ϵi = V (xi) + V (xi+1) −1/2V 1/2(xi)zi −V 1/2(xi+1)zi+1 has zero mean and unit variance.
We construct an estimator ˆ V by applying kernel smoothing to the squared dif-ferences D2 i which have means δ2 i + 2Vi. Let K(x) be a kernel function satisfying K(x) is supported on [−1,1], 1 −1 K(x)dx = 1, 1 −1 K(x)xi dx = 0 for i = 1,2,...,⌊β⌋ and 1 −1 K2(x)dx = k < ∞.
It is well known in kernel regression that special care is needed in order to avoid significant, sometimes dominant, boundary effects. We shall use the boundary ker-nels with asymmetric support, given in Gasser and Müller (1979, 1984), to control EFFECT OF MEAN ON VARIANCE FUNCTION ESTIMATION 649 the boundary effects. For any t ∈[0,1], there exists a boundary kernel function Kt(x) with support [−1,t] satisfying the same conditions as K(x), that is, t −1 Kt(x)dx = 1, t −1 Kt(x)xi dx = 0 for i = 1,2,...,⌊β⌋, t −1 K2 t (x)dx ≤ k < ∞ for all t ∈[0,1].
We can also make Kt(x) →K(x) as t →1 (but this is not necessary here).
See Gasser, Müller and Mammitzsch (1985). For any 0 < h < 1/2, x ∈[0,1], and i = 2,...,n −2, let Kh i (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩ (xi+xi+1)/2 (xi+xi−1)/2 1 hK x −u h du, when x ∈(h,1 −h), (xi+xi+1)/2 (xi+xi−1)/2 1 hKt x −u h du, when x = th for some t ∈[0,1], (xi+xi+1)/2 (xi+xi−1)/2 1 hKt −x −u h du, when x = 1 −th for some t ∈[0,1], and we take the integral from 0 to (x1 +x2)/2 for i = 1, and from (xn−1 +xn−2)/2 to 1 fori = n−1. Then we can see that for any 0 ≤x ≤1, n−1 i=1 Kh i (x) = 1. Define the estimator V as V (x) = 1 2 n−1 i=1 Kh i (x)D2 i .
(7) Same as in the mean function estimation problem, the optimal bandwidth hn can be easily seen to be hn = O(n−1/(1+2β)) for V ∈β(MV ). For this optimal choice of the bandwidth, we have the following theorem.
THEOREM 1.
Under the regression model (1) where xi = i/n and zi are in-dependent with zero mean, unit variance and uniformly bounded fourth moments, let the estimator V be given as in (7) with the bandwidth h = O(n−1/(1+2β)). Then there exists some constant C0 > 0 depending only on α, β, Mf and MV such that for sufficiently large n, sup f ∈α(Mf ),V ∈β(MV ) sup 0≤x∗≤1 E V (x∗) −V (x∗) 2 (8) ≤C0 · max n−4α,n−2β/(1+2β) 650 WANG, BROWN, CAI AND LEVINE and sup f ∈α(Mf ),V ∈β(MV ) E 1 0 V (x) −V (x) 2 dx (9) ≤C0 · max n−4α,n−2β/(1+2β).
REMARK 1.
The uniform rate of convergence given in (8) yields immediately the pointwise rate of convergence that for any fixed point x∗∈[0,1] sup f ∈α(Mf ),V ∈β(MV ) E V (x∗) −V (x∗) 2 ≤C0 · max n−4α,n−2β/(1+2β).
REMARK 2.
It is also possible to use the local linear regression estimator in-stead of the Priestley–Chao kernel estimator. In this case, the boundary adjustment is not necessary as it is well known that the local linear regression adjusts auto-matically in boundary regions, preserving the asymptotic order of the bias intact.
However, the proof is slightly more technically involved when using the local lin-ear regression estimator. For details see, for example, Fan and Gijbels (1996).
REMARK 3.
It is important to note here that the results given in Theorem 1 can be easily generalized to the case of random design. In particular, if the observations X1,...,Xn are i.i.d. with the design density f (x) that is bounded away from zero (i.e., f (x) ≥δ > 0 for all x ∈[0,1]), then the results of Theorem 1 are still valid conditionally. In other words, sup f ∈α(Mf ),V ∈β(MV ) sup 0≤x∗≤1 E V (x∗) −V (x∗)2|X1,...,Xn ≤C0 · max n−4α,n−2β/(1+2β) + op max n−4α,n−2β/(1+2β) and sup f ∈α(Mf ),V ∈β(MV ) E 1 0 V (x) −V (x) 2 dx|X1,...,Xn ≤C0 · max n−4α,n−2β/(1+2β) + op max n−4α,n−2β/(1+2β) where the constant C0 > 0 now also depends on δ.
3. Lower bound.
In this section we derive a lower bound for the minimax risk of estimating the variance function V under the regression model (1). The lower bound shows that the upper bound given in the previous section is rate-sharp. As in Hall and Carroll (1989) we shall assume in the lower bound argument that the errors are normally distributed, that is, zi i.i.d.
∼N(0,1).
EFFECT OF MEAN ON VARIANCE FUNCTION ESTIMATION 651 THEOREM 2.
Under the regression model (1) with zi i.i.d.
∼N(0,1), inf V sup f ∈α(Mf ),V ∈β(MV ) E∥ V −V ∥2 2 ≥C1 · max n−4α,n−2β/(1+2β) (10) and for any fixed x∗∈(0,1) inf V sup f ∈α(Mf ),V ∈β(MV ) E V (x∗) −V (x∗) 2 (11) ≥C1 · max n−4α,n−2β/(1+2β), where C1 > 0 is a constant depending only on α, β, Mf and MV .
It follows immediately from Theorems 1 and 2 that the minimax rate of conver-gence for estimating V under both the global and local losses is max n−4α,n−2β/(1+2β).
The proof of this theorem can be naturally divided into two parts. The first step is to show inf V sup f ∈α(Mf ),V ∈β(MV ) E V (x∗) −V (x∗) 2 ≥C1n−2β/(1+2β).
(12) This part is standard and relatively easy. Brown and Levine (2006) contains a de-tailed proof of this assertion for the case β = 2. Their argument can be easily generalized to other values of β. We omit the details.
The proof of the second step, inf V sup f ∈α(Mf ),V ∈β(MV ) E V (x∗) −V (x∗) 2 ≥C1n−4α, (13) is much more involved. The derivation of the lower bound (13) is based on a mo-ment matching technique and a two-point testing argument. One of the main steps is to study a complicated hypothesis testing problem where the alternative hypoth-esis is a Gaussian location mixture with a special moment matching property.
More specifically, let X1,...,Xn i.i.d.
∼P and consider the following hypothesis testing problem between H0 :P = P0 = N(0,1 + θ2 n) and H1 :P = P1 = N(θnν,1)G(dν), where θn > 0 is a constant and G is a distribution of the mean ν with compact support. The distribution G is chosen in such a way that, for some positive integer q depending on α, the first q moments of G match exactly with the corresponding moments of the standard normal distribution. The existence of such a distribution is given in the following lemma from Karlin and Studden (1966).
652 WANG, BROWN, CAI AND LEVINE LEMMA 1.
For any fixed positive integer q, there exist a B < ∞and a sym-metric distribution G on [−B,B] such that G and the standard normal distribution have the same first q moments, that is, B −B xjG(dx) = +∞ −∞xjϕ(x)dx, j = 1,2,...,q, where ϕ denotes the density of the standard normal distribution.
The moment matching property makes the testing between the two hypotheses “difficult.” The lower bound (13) then follows from a two-point argument with an appropriately chosen θn. Technical details of the proof are given in Section 6.
REMARK 4.
For α between 1/4 and 1/8, a much simpler proof can be given with a two-point mixture for P1 which matches the mean and variance, but not the higher moments, of P0 and P1. However, this simpler proof fails for smaller α. It appears to be necessary in general to match higher moments of P0 and P1.
REMARK 5.
Hall and Carroll (1989) gave the lower bound C max{n−4α/(1+2α),n−2β/(1+2β)} for the minimax risk. This bound is larger than the lower bound given in our Theorem 2 and is incorrect. This is due to a mis-calculation on appendix C of their paper. A key step in that proof is to find some d ≥0 such that D = E 1 + exp 1 2d + d1/2N1 −11 2d + d1/2N1 ̸= 0.
In the above expression, N1 denotes a standard normal random variable. But in fact D = ∞ −∞ (1/2)d + d1/2x 1 + exp((1/2)d + d1/2x) 1 √ 2π exp −x2 2 dx = ∞ −∞ x 1 + exp(x) 1 √ 2πd exp −(x −(1/2)d)2 2d dx = ∞ −∞ x exp(x/2) + exp(−x/2) 1 √ 2πd exp −x2 2d −d 8 dx.
This is an integral of an odd function which is identically 0 for all d.
4. Discussion.
Variance function estimation in regression is more typically based on the residuals from a preliminary estimator ˆ f of the mean function. Such estimators have the form ˆ V (x) = i wi(x) yi −ˆ f (xi) 2 (14) EFFECT OF MEAN ON VARIANCE FUNCTION ESTIMATION 653 where wi(x) are weight functions. A natural and common approach is to subtract in (14) an optimal estimator ˆ f of the mean function f (x). See, for example, Hall and Carroll (1989), Neumann (1994), Ruppert, Wand, Holst and Hössjer (1997), and Fan and Yao (1998). When the unknown mean function is smooth, this approach often works well since the bias in ˆ f is negligible and V can be estimated as well as when f is identically zero. However, when the mean function is not smooth, using the residuals from an optimally smoothed ˆ f will lead to a sub-optimal estimator of V . For example, Hall and Carroll (1989) used a kernel estimator with optimal bandwidthfor ˆ f and showed that the resulting variance estimator attains the rate of max n−4α/(2α+1),n−2β/(2β+1) (15) over f ∈α(Mf ) and V ∈β(MV ). This rate is strictly slower than the minimax rate when 4α 2α+1 < 2β 2β+1 or equivalently, α < β 2β+2.
Consider the example where V belongs to a regular parametric family, such as {V (x) = exp(ax + b):a,b ∈R}. As Hall and Carroll have noted, this case is equivalent to the case of β = ∞in results like Theorems 1 and 2. Then the rate of convergence for this estimator becomes nonparametric at n−4α/(2α+1) for α < 1/2, while the optimal rate is the usual parametric rate n−1/2 for all α ≥1 4 and is n−4α for 0 < α < 1 4.
The main reason for the poor performance of such an estimator in the non-smooth setting is the “large” bias in ˆ f . An optimal estimator ˆ f of f balances the squared bias and variance. However, the bias and variance of ˆ f have significantly different effects on the estimation of V . The bias of ˆ f cannot be further reduced in the second stage smoothing of the squared residuals, while the variance of ˆ f can be incorporated easily. For f ∈α(Mf ), the maximum bias of an optimal estimator ˆ f is of order n−α/(2α+1) which becomes the dominant factor in the risk of ˆ V when α < β 2β+2.
To minimize the effect of the mean function in such a setting one needs to use an estimator ˆ f (xi) with minimal bias. Note that our approach is, in effect, using a very crude estimator ˆ f of f with ˆ f (xi) = yi+1. Such an estimator has high variance and low bias. As we have seen in Section 2, the large variance of ˆ f does not pose a problem (in terms of rates) for estimating V . Hence for estimating the variance function V an optimal ˆ f is the one with minimum possible bias, not the one with minimum mean squared error. [Here we should of course exclude the obvious, and not useful, unbiased estimator ˆ f (xi) = yi.] Another implication of our results is that the unknown mean function does not have any first-order effect for estimating V as long as f has more than 1/4 derivatives. When α > 1/4, the variance estimator V is essentially adaptive over f ∈α(Mf ) for all α > 1/4. In other words, if f is known to have more than 1/4 derivatives, the variance function V can be estimated with the same degree of first-order precision as if f is completely known. However, when α < 1/4, the rate of convergence for estimating V is entirely determined by the degree of smoothness of the mean function f .
654 WANG, BROWN, CAI AND LEVINE 5. Numerical results.
We now consider in this section the finite sample performance of our difference-based method for estimating the variance func-tion. In particular we are interested in comparing the numerical performance of the difference-based estimator with the residual-based estimator of Fan and Yao (1998). The numerical results show that the performance of the difference-based estimator is somewhat inferior when the unknown mean function is very smooth.
On the other hand, the difference-based estimator performs significantly better than the residual-based estimator when the mean function is not smooth.
Consider the model 1 where the variance function is V (x) = (x −1 2)2 + 1 2 while there are four possible mean functions: (i) f1(x) = 0, (ii) f2(x) = 3 4 ∗sin(10πx), (iii) f3(x) = 3 4 ∗sin(20πx), (iv) f4(x) = 3 4 ∗sin(40πx).
The mean functions are arranged from a constant to much rougher sinusoid func-tion; the “roughness” (the difficulty a particular mean function creates in es-timation of the variance function V ) is measured by the functional R(f ′) = [f ′(x)]2 dx since the mean-related term in the asymptotic bias of the variance estimator ˆ V (x) is directly proportional to it. The numerical performance of the difference-based method had been investigated earlier in Levine (2006) for a slightly different set of mean functions.
For comparison purposes, the same four combinations of the mean and variance functions are investigated using the two-step method described in Fan and Yao (1998). We expect this method to perform better than the difference-based method in the case of a constant mean function, but to get progressively worse as the rough-ness of the mean function considered increases. The following table summarizes results of simulations using both methods. In this case, the bandwidths for estimat-ing the mean and variance functions were selected using a K-fold cross-validation with K = 10. We consider the fixed equidistant design xi = i n on [0,1] where the sample size is n = 1000; 100 simulations are performed and the bandwidth h is selected using a K-fold cross-validation with K = 10. The performance of both methods is measured using the cross-validation discrete mean squared error (CDMSE) that is defined as CDMSE = n−1 n i=1 [ ˆ VhCV(xi) −V (xi)]2 (16) with hCV being the K-fold cross-validation bandwidth. We report the median CDMSE for variance function estimators based on 100 simulations. Table 1 pro-vides the summary of the performance.
It is easily seen from the table that the two-step method of Fan and Yao, based on estimating the variance using squared residuals from the mean function esti-mation, tends to perform slightly better when the mean function is very smooth EFFECT OF MEAN ON VARIANCE FUNCTION ESTIMATION 655 TABLE 1 Performance under the changing curvature of the mean function Median CDMSE Mean function R(f ′) Fan–Yao method Our method f = 0 0 0.00299 0.00376 f = 3 4 sin(10πx) 278.15 0.07161 0.00344 f = 3 4 sin(20πx) 1110.89 0.08435 0.00384 f = 3 4 sin(40πx) 4441.88 0.08363 0.00348 but noticeably worse when it is rougher. Note that here we only use the first-order differences. The performance of the difference based estimator can be improved in the case of smooth mean function by using higher order differences. The Fan–Yao method performs about 26% better in the first case of the constant mean function.
However, the risk (CDMSE) of the difference based method is over 95% smaller than the risk of the Fan–Yao method for the second mean function. In the rougher cases, the difference is approximately the same. The CDMSE of the difference based method is over 95% and 96% less than the corresponding risk of the resid-ual based method for the third and fourth mean functions, respectively.
6. Proofs.
6.1. Upper bound: Proof of Theorem 1.
We shall only prove (8). Inequality (9) is a direct consequence of (8). Recall that D2 i = δ2 i + 2Vi + 2Vi(ϵ2 i −1) + 2 √ 2δiV 1/2 i ϵi, where δi = f (xi) −f (xi+1), V 1/2 i = √1/2(V (xi) + V (xi+1)) and ϵi = V (xi) + V (xi+1) −1/2V 1/2(xi)zi −V 1/2(xi+1)zi+1 .
Without loss of generality, suppose h = n−1/(1+2β). It is easy to see that for any x∗∈[0,1], i Kh i (x∗) = 1, and when x∗≥(xi + xi+1)/2 + h or x∗≤ (xi + xi−1)/2 −h, Kh i (x∗) equals 0. Suppose k < k, we also have i |Kh i (x∗)| 2 ≤2nh i (Kh i (x∗))2 ≤2 1 −1 K2 ∗(u)du ≤2 k, 656 WANG, BROWN, CAI AND LEVINE where K∗(u) = K(u) when x∗∈(h,1 −h); K∗(u) = Kt(u) when x∗= th for some t ∈[0,1]; and K∗(u) = Kt(−u) when x∗= 1 −th for some t ∈[0,1].
The second inequality above is obtained as follows. For the sake of simplic-ity, assume that K∗= K; the same argument can be repeated for boundary ker-nels as well. Using the definition of Kh i (x∗), we note that it can be rewritten as (xi+xi+1)/2 (xi+xi−1)/2 1 nhK(x−u h )d(nu). Since the last integral is taken with respect to the probability measure on the interval [xi+xi−1 2 , xi+xi+1 2 ], we can apply Jensen’s in-equality to obtain (Kh i (x∗))2 ≤1 nh (xi+xi+1)/2 (xi+xi−1)/2 K2 x −u h d(nu) = 1 (nh)2 (xi+xi+1)/2 (xi+xi−1)/2 K2 x −u h du.
Thus, i |Kh i (x∗)| 2 ≤2 h i (xi+xi+1)/2 (xi+xi−1)/2 K2 x −u h du = 2 1 −1 K2(u)du.
For all f ∈α(Mf ) and V ∈β(MV ), the mean squared error of ˆ V at x∗satisfies E V (x∗) −V (x∗) 2 = E n−1 i=1 Kh i (x∗) 1 2D2 i −V (x∗) 2 = E n−1 i=1 Kh i (x∗)1 2δ2 i + n−1 i=1 Kh i (x∗) Vi −V (x∗) + n−1 i=1 Kh i (x∗)Vi(ϵ2 i −1) + n−1 i=1 Kh i (x∗) √ 2δiV 1/2 i ϵi 2 ≤4 n−1 i=1 Kh i (x∗)1 2δ2 i 2 + 4 n−1 i=1 Kh i (x∗) Vi −V (x∗) 2 + 4E n−1 i=1 Kh i (x∗)Vi(ϵ2 i −1) 2 + 4E n−1 i=1 Kh i (x∗) √ 2δiV 1/2 i ϵi 2 .
EFFECT OF MEAN ON VARIANCE FUNCTION ESTIMATION 657 Suppose α ≤1/4, otherwise n−4α < n−2β/(1+2β) for any β. Since for any i, |δi| = |f (xi) −f (xi+1)| ≤Mf |xi −xi+1|α = Mf n−α, we have 4 n−1 i=1 Kh i (x∗)1 2δ2 i 2 ≤4 n−1 i=1 |Kh i (x∗)|1 2M2 f n−2α 2 ≤2 kM4 f n−4α.
Note that for any x,y ∈[0,1], Taylor’s theorem yields V (x) −V (y) − ⌊β⌋ j=1 V (j)(y) j!
(x −y)j = x y (x −u)⌊β⌋−1 (⌊β⌋−1)!
V (⌊β⌋)(u) −V (⌊β⌋)(y) du ≤ x y (x −u)⌊β⌋−1 (⌊β⌋−1)! MV |x −y|β−⌊β⌋du ≤MV ⌊β⌋!|x −y|β.
So, Vi −V (x∗) = 1 2 V i n + V i + 1 n −V (x∗) ≤1 2 ⌊β⌋ j=1 V (j)(x∗) j i n −x∗ j + i + 1 n −x∗ j + 1 2MV i n −x∗ β + 1 2MV i + 1 n −x∗ β and Vi −V (x∗) ≥1 2 ⌊β⌋ j=1 V (j)(x∗) j i n −x∗ j + i + 1 n −x∗ j −1 2MV i n −x∗ β −1 2MV i + 1 n −x∗ β .
Since the kernel functions have vanishing moments, for j = 1,2,...,⌊β⌋, when n large enough n−1 i=1 Kh i (x∗) i n −x∗ j = n−1 i=1 (xi+xi+1)/2 (xi+xi−1)/2 1 hK x∗−u h i n −x∗ j du 658 WANG, BROWN, CAI AND LEVINE = 1 0 1 hK x∗−u h (u −x∗)j du + n−1 i=1 (xi+xi+1)/2 (xi+xi−1)/2 1 hK x∗−u h i n −x∗ j −(u −x∗)j du = n−1 i=1 (xi+xi+1)/2 (xi+xi−1)/2 1 hK x∗−u h i n −x∗ j −(u −x∗)j du ≤c′ n−1 i=1 (xi+xi+1)/2 (xi+xi−1)/2 1 h x∗−u h × j n du = c′n−1 for some generic constant c′ > 0. Similarly, n−1 i=1 Kh i (x∗)(i+1 n −x∗)j ≤c′n−1.
So, n−1 i=1 Kh i (x∗) ⌊β⌋ j=1 V (j)(x∗) j i n −x∗ j + i + 1 n −x∗ j ≤ Cn−1 for some constant C > 0 which does not depend on x∗. Note that V ⌊β⌋satisfies Hölder condition with exponent 0 < α′ = α−⌊α⌋< 1 and is, therefore, continuous on [0,1] and bounded. Then we have 4 n−1 i=1 Kh i (x∗) Vi −V (x∗) 2 ≤2 C2n−2 + 2M2 V ⌊n(x∗+h)⌋+1 i=⌊n(x∗−h)⌋ |Kh i (x∗)| i n −x∗ β + i + 1 n −x∗ β2 ≤2 C2n−2 + 2M2 V ⌊n(x∗+h)⌋+1 i=⌊n(x∗−h)⌋ |Kh i (x∗)| h + 1 n β + h + 2 n β2 ≤2 C2n−2 + 8 × 32βM2 V n−2β/(1+2β) × (2 k).
The last inequality is due to the fact 0 < h + 1 n < h + 2 n < 3h. On the other hand, notice that ϵ1,ϵ3,ϵ5,... are independent and ϵ2,ϵ4,ϵ6,... are independent, we have 4E n−1 i=1 Kh i (x∗) √ 2δiV 1/2 i ϵi 2 = 4Var n−1 i=1 Kh i (x∗) √ 2δiV 1/2 i ϵi ≤16 ⌊n(x∗+h)⌋+1 i=⌊n(x∗−h)⌋ (Kh i (x∗))2δ2 i Vi ≤16M2 f MV n−2α−2β/(1+2β) × k EFFECT OF MEAN ON VARIANCE FUNCTION ESTIMATION 659 and 4E n−1 i=1 Kh i (x∗)Vi(ϵ2 i −1) 2 = 4Var n−1 i=1 Kh i (x∗)Vi(ϵ2 i −1) ≤8M2 V μ4 n−1 i=1 (Kh i (x∗))2 ≤8M2 V μ4 1 nh k = 8M2 V μ4n−2β/(1+2β) × k where μ4 denotes the uniform bound for the fourth moments of the ϵi.
Putting the four terms together we have, uniformly for all x∗∈[0,1], f ∈ α(Mf ) and V ∈β(MV ), E V (x∗) −V (x∗) 2 ≤2 kM4 f n−4α + 2 C2n−2 + 8 × 32βM2 V n−2β/(1+2β) × (2 k) + 8M2 V μ4n−2β/(1+2β) k + 16M2 f MV n−2α−2β/(1+2β) k = C0 · max n−4α,n−2β/(1+2β) for some constant C0 > 0. This proves (8).
6.2. Lower bound: Proof of Theorem 2.
We shall only prove the lower bound for the pointwise squared error loss. The same proof with minor modifications im-mediately yields the lower bound under the integrated squared error. Note that, to prove inequality (13), we only need to focus on the case where α < 1/4, other-wise n−2β/(1+2β) is always greater than n−4α for sufficiently large n and then (13) follows directly from (12).
For a given 0 < α < 1/4, there exists an integer q such that (q + 1)α > 1. For convenience we take q to be an odd integer. From Lemma 1, there is a positive con-stant B < ∞and a symmetric distribution G on [−B,B] such that G and N(0,1) have the same first q moments. Let ri, i = 1,...,n, be independent variables with the distribution G. Set θn = Mf 2B n−α, f0 ≡0, V0(x) ≡1 + θ2 n and V1(x) ≡1. Let g(x) = 1 −2n|x| for x ∈[−1 2n, 1 2n] and 0 otherwise. Define the random function f1 by f1(x) = n i=1 θnrig(x −xi)I(0 ≤x ≤1).
Then it is easy to see that f1 is in α(Mf ) for all realizations of ri. Moreover, f1(xi) = θnri are independent and identically distributed.
660 WANG, BROWN, CAI AND LEVINE Now consider testing the following hypotheses: H0 : yi = f0(xi) + V 1/2 0 (xi)νi, i = 1,...,n, H1 : yi = f1(xi) + V 1/2 1 (xi)νi, i = 1,...,n, where νi are independent N(0,1) variables which are also independent of the ri’s.
Denote by P0 and P1 the joint distributions of yi’s under H0 and H1, respectively.
Note that for any estimator V of V , max E V (x∗) −V0(x∗) 2,E V (x∗) −V1(x∗) 2 ≥1 16ρ4(P0,P1) V0(x∗) −V1(x∗) 2 (17) = 1 16ρ4(P0,P1) M4 f 16B4 n−4α where ρ(P0,P1) is the Hellinger affinity between P0 and P1. See, for example, Le Cam (1986). Let p0 and p1 be the probability density function of P0 and P1 with respect to the Lebesgue measure μ, then ρ(P0,P1) = √p0p1 dμ. The minimax lower bound (13) follows immediately from the two-point bound (17) if we show that for any n, the Hellinger affinity ρ(P0,P1) ≥C for some constant C > 0.
(C may depend on q, but does not depend on n.) Note that under H0, yi ∼N(0,1 + θ2 n) and its density d0 can be written as d0(t) ≜ 1 1 + θ2 n ϕ t 1 + θ2 n = ϕ(t −vθn)ϕ(v)dv.
Under H1, the density of yi is d1(t) ≜ ϕ(t −vθn)G(dv).
It is easy to see that ρ(P0,P1) = ( √d0d1 dμ)n, since the yi’s are independent variables. Note that the Hellinger affinity is bounded below by the total variation affinity d0(t)d1(t)dt ≥1 −1 2 |d0(t) −d1(t)|dt.
Taylor’s expansion yields ϕ(t −vθn) = ϕ(t) ∞ k=0 vkθk n Hk(t) k!
, where Hk(t) is the corresponding Hermite polynomial. And from the construction of the distribution G, viG(dv) = viϕ(v)dv for i = 0,1,...,q.
EFFECT OF MEAN ON VARIANCE FUNCTION ESTIMATION 661 So, |d0(t) −d1(t)| = ϕ(t −vθn)G(dv) − ϕ(t −vθn)ϕ(v)dv = ϕ(t) ∞ i=0 Hi(t) i!
viθi nG(dv) − ϕ(t) ∞ i=0 Hi(t) i!
viθi nϕ(v)dv (18) = ϕ(t) ∞ i=q+1 Hi(t) i!
viθi nG(dv) − ϕ(t) ∞ i=q+1 Hi(t) i!
viθi nϕ(v)dv ≤ ϕ(t) ∞ i=q+1 Hi(t) i!
viθi nG(dv) + ϕ(t) ∞ i=q+1 Hi(t) i!
viθi nϕ(v)dv .
Suppose q + 1 = 2p for some integer p, it can be seen that ϕ(t) ∞ i=q+1 Hi(t) i!
viθi nG(dv) = ϕ(t) ∞ i=p H2i(t) (2i)! θ2i n v2iG(dv) ≤ϕ(t) ∞ i=p H2i(t) (2i)! θ2i n v2iG(dv) ≤ϕ(t) ∞ i=p H2i(t) (2i)!
θ2i n B2i and ϕ(t) ∞ i=q+1 Hi(t) i!
viθi nϕ(v)dv = ϕ(t) ∞ i=p H2i(t) (2i)! θ2i n v2iϕ(v)dv ≤ϕ(t) ∞ i=p H2i(t) (2i)! θ2i n v2iϕ(v)dv = ϕ(t) ∞ i=p H2i(t)θ2i n 1 2i · i!
≤ϕ(t) ∞ i=p H2i(t) 2i · i!
θ2i n .
So from (18), |d0(t) −d1(t)| ≤ϕ(t) ∞ i=p H2i(t) (2i)!
θ2i n B2i + ϕ(t) ∞ i=p H2i(t) 2i · i!
θ2i n 662 WANG, BROWN, CAI AND LEVINE and then d0(t)d1(t)dt ≥1 −1 2 ϕ(t) ∞ i=p H2i(t) (2i)!
θ2i n B2i + ϕ(t) ∞ i=p H2i(t) 2i · i!
θ2i n dt (19) = 1 −1 2 ϕ(t) ∞ i=p H2i(t) (2i)!
θ2i n B2i dt −1 2 ϕ(t) ∞ i=p H2i(t) 2i · i!
θ2i n dt.
Since t2iφ(t)dt = (2i −1)!! where (2i −1)!! ≜(2i −1)×(2i −3)×···×3×1, for the Hermite polynomial H2i we have ϕ(t)|H2i(t)|dt = ϕ(t) (2i −1)!! × 1 + i k=1 (−2)ki(i −1)···(i −k + 1) (2k)!
t2k dt ≤ ϕ(t) (2i −1)!! × 1 + i k=1 2ki(i −1)···(i −k + 1) (2k)!
t2k dt = (2i −1)!! × 1 + i k=1 2ki(i −1)···(i −k + 1) (2k)!
t2kϕ(t)dt = (2i −1)!! × 1 + i k=1 2ki(i −1)···(i −k + 1) (2k)!
(2k −1)!!
= (2i −1)!! × 1 + i k=1 i(i −1)···(i −k + 1) k!
= 2i × (2i −1)!!.
For sufficiently large n, θn < 1/2 and it then follows from the above inequality that ϕ(t) ∞ i=p H2i(t) (2i)!
θ2i n B2i dt ≤ ∞ i=p θ2i n B2i (2i)!
ϕ(t)|H2i(t)|dt ≤ ∞ i=p θ2i n B2i (2i)! 2i × (2i −1)!!
= θ2p n ∞ i=p B2iθ2i−2p n i!
≤θ2p n × eB2 EFFECT OF MEAN ON VARIANCE FUNCTION ESTIMATION 663 and ϕ(t) ∞ i=p H2i(t) 2i · i!
θ2i n dt ≤ ∞ i=p θ2i n 2i · i!
ϕ(t)|H2i(t)|dt ≤ ∞ i=p θ2i n 2i · i!2i × (2i −1)!!
= θ2p n ∞ i=p (2i −1)!!
i!
θ2i−2p n ≤θ2p n ∞ i=p 2i × θ2i−2p n ≤θ2p n ∞ i=p 2i × 1 2 2i−2p = θ2p n × 22p+1.
Then from (19) d0(t)d1(t)dt ≥1 −θ2p n 1 2eB2 + 22p ≜1 −cθq+1 n , where c is a constant that only depends on q. So ρ(P0,P1) = d0(t)d1(t)dt n ≥(1 −cθq+1 n )n = 1 −cn−α(q+1)n.
Since α(q + 1) ≥1, lim→∞(1 −cn−α(q+1))n ≥e−c > 0 and the theorem then follows from (17).
Acknowledgments.
We thank the Editor and three referees for thorough and useful comments which have helped to improve the presentation of the paper. We would like to thank William Studden for discussions on the finite moment match-ing problem and for the reference Karlin and Studden (1966).
REFERENCES BROWN, L. D. and LEVINE, M. (2006). Variance estimation in nonparametric regression via the difference sequence method. Ann. Statist. To appear.
FAN, J. and GIJBELS, I. (1996). Local Polynomial Modelling and Its Applications. Chapman and Hall, London. MR1383587 FAN, J. and YAO, Q. (1998). Efficient estimation of conditional variance functions in stochastic regression. Biometrika 85 645–660. MR1665822 GASSER, T. and MÜLLER, H. G. (1979). Kernel estimation of regression functions. Smooth-ing Techniques for Curve Estimation. Lecture Notes in Math. 757 23–68. Springer, New York.
MR0564251 GASSER, T. and MÜLLER, H. G. (1984). Estimating regression functions and their derivatives by the kernel method. Scand. J. Statist. 11 197–211. MR0767241 664 WANG, BROWN, CAI AND LEVINE GASSER, T., MÜLLER, H. G. and MAMMITZSCH, V. (1985). Kernels for nonparametric curve estimation. J. Roy. Statist. Soc. B 47 238–252. MR0816088 HALL, P. and CARROLL, R. J. (1989). Variance function estimation in regression: The effect of estimating the mean. J. Roy. Statist. Soc. Ser. B 51 3–14. MR0984989 HALL, P., KAY, J. and TITTERINGTON, D. M. (1990). Asymptotically optimal difference-based estimation of variance in nonparametric regression. Biometrika 77 521–528. MR1087842 KARLIN, S. and STUDDEN, W. J. (1966). Tchebycheff Systems: With Applications in Analysis and Statistics. Interscience, New York. MR0204922 LE CAM, L. (1986). Asymptotic Methods in Statistical Decision Theory. Springer, New York.
MR0856411 LEVINE, M. (2006). Bandwidth selection for a class of difference-based variance estimators in the nonparametric regression: A possible approach. Comput. Statist. Data Anal. 50 3405–3431.
MR2236857 MÜLLER, H. G. and STADTMÜLLER, U. (1987). Estimation of heteroscedasticity in regression analysis. Ann. Statist. 15 610–625. MR0888429 MÜLLER, H. G. and STADTMÜLLER, U. (1993). On variance function estimation with quadratic forms. J. Statist. Plann. Inference 35 213–231. MR1220417 MUNK, A., BISSANTZ, N., WAGNER, T. and FREITAG, G. (2005). On difference based variance estimation in nonparametric regression when the covariate is high dimensional. J. Roy. Statist.
Soc. Ser. B 67 19–41. MR2136637 NEUMANN, M.(1994). Fully data-driven nonparametric variance estimators. Statistics 25 189–212.
MR1366825 RICE, J. (1984). Bandwidth choice for nonparametric kernel regression. Ann. Statist. 12 1215–1230.
MR0760684 RUPPERT, D., WAND, M. P., HOLST, U. and HÖSSJER, O. (1997). Local polynomial variance function estimation. Technometrics 39 262–273. MR1462587 VON NEUMANN, J.(1941). Distribution of the ratio of the mean squared successive difference to the variance. Ann. Math. Statist. 12 367–395. MR0006656 VON NEUMANN, J. (1942). A further remark concerning the distribution of the ratio of the mean squared successive difference to the variance. Ann. Math. Statist. 13 86–88. MR0006657 L. WANG L. D. BROWN T. T. CAI DEPARTMENT OF STATISTICS THE WHARTON SCHOOL UNIVERSITY OF PENNSYLVANIA PHILADELPHIA, PENNSYLVANIA 19104 USA E-MAIL: tcai@wharton.upenn.edu M. LEVINE DEPARTMENT OF STATISTICS PURDUE UNIVERSITY WEST LAFAYETTE, INDIANA 47907 USA E-MAIL: mlevins@stat.purdue.edu |
188509 | https://teachy.ai/en/project/high-school-en-US/us-12th-grade/chemistry/exploring-faradays-law-electrolysis-and-the-relationship-between-electricity-and-substance-production | Log In
Project: Exploring Faraday's Law: Electrolysis and the Relationship Between Electricity and Substance Production
Lara from Teachy
Subject Chemistry
Chemistry
Source Teachy Original
Teachy Original
Topic Faraday’s Law
Faraday’s Law
Contextualization
Faraday’s Law, also known as the law of electromagnetic induction, is a fundamental concept in the field of physics and chemistry, specifically in the study of electrochemistry. This law, formulated by the English scientist Michael Faraday in the early 19th century, states that the amount of a substance produced or consumed at an electrode during an electrochemical reaction is proportional to the quantity of electricity that passes through the circuit.
To understand Faraday’s Law, we need to delve into the world of electrochemical cells. These cells, such as batteries, are devices that convert chemical energy into electrical energy. They consist of two different conductive materials, called electrodes, which are immersed in an electrolyte (a substance that conducts electricity when dissolved in a solvent).
The electrodes are connected by an external circuit through which the electrons can flow. During an electrochemical reaction, called a half-cell reaction, the electrodes serve as sites for the oxidation and reduction (redox) reactions that occur. These reactions involve the transfer of electrons from one electrode (the anode) to the other (the cathode), through the external circuit.
Faraday’s Law provides a quantitative relationship between the electric current (the flow of electrons) and the amount of substance (in moles) involved in a redox reaction. This allows us to predict how much of a substance will be produced or consumed during an electrochemical process, based on the amount of electricity that passes through the circuit.
Importance and Applications
The practical applications of Faraday’s Law are widespread and significant. They underlie many technological devices and processes that are integral to our modern way of life. For instance, the production of aluminum, a versatile metal used in various industries, is based on the principles of Faraday’s Law.
The extraction of metals like aluminum from their ores involves a process called electrolysis, which is an application of Faraday’s Law. In this process, an ionic compound of the metal is melted and then electrolyzed using a direct electric current. The metal ions in the compound are reduced at the cathode (an electrode where reduction occurs), forming the pure metal, while the anions are oxidized at the anode, forming the oxide or other products.
Understanding Faraday’s Law is also crucial in the operation of batteries. When a battery is being used, a chemical reaction inside it produces a flow of electrons, which can be used to power a device. The amount of electricity (and thus the time the battery can last) is determined by Faraday’s Law.
Suggested Resources
Faraday's First and Second Laws of Electrolysis – A comprehensive guide to Faraday's Laws with clear explanations and examples, from Chemguide.
Electrolysis - Understanding Faraday’s Laws – A video lecture from Khan Academy that provides an overview of Faraday's Laws and their applications in electrolysis.
Electrolysis and Faraday's Law – A practical experiment from LibreTexts that allows you to observe Faraday's Laws in action.
Electrolysis - Wikipedia – A comprehensive article from Wikipedia that covers the principles of electrolysis and Faraday's Laws, with links to related topics for further reading.
Faraday's Law of Electrolysis – A more in-depth discussion of Faraday's Law and its derivations, from ScienceDirect. Access may require a subscription or purchase.
Practical Activity
Activity Title: Electrolysis and Faraday’s Law: A Hands-on Exploration
Objective of the Project:
To understand and observe Faraday’s Law in action by performing an electrolysis experiment using a simple and safe setup. This experiment will enable students to quantify the relationship between the amount of electricity passed through a circuit and the amount of a substance produced or consumed during an electrochemical reaction.
Detailed Description of the Project:
Groups of 3 to 5 students will set up an electrolytic cell using a solution of copper sulfate and perform the electrolysis process. They will then measure the amount of copper deposited on the cathode (a negative electrode) and use this data to verify Faraday’s Law, which states that the amount of a substance produced or consumed in a redox reaction is directly proportional to the amount of electricity passed through the circuit.
Necessary Materials:
Copper sulfate (CuSO4) solution
Two copper electrodes
Two wires with alligator clips
Power supply (e.g., a 9V battery)
Stopwatch or timer
Balance (to measure the mass of copper deposited)
Safety goggles and gloves
Detailed Step-by-Step for Carrying Out the Activity:
Put on your safety goggles and gloves.
Set up the electrolytic cell by connecting the positive terminal of the power supply to a copper electrode (the anode) and the negative terminal to another copper electrode (the cathode) using the wires and alligator clips. The electrodes should be immersed in the copper sulfate solution, but should not touch each other.
Weigh the cathode electrode and record its initial mass.
Turn on the power supply and let the electrolysis process run for a specific time period, e.g., 5 minutes.
After the electrolysis, turn off the power supply and carefully remove the cathode electrode from the solution. Note: The cathode electrode will have copper deposited on it.
Dry the copper electrode with a paper towel and weigh it again. Record its final mass.
Calculate the mass of copper deposited on the electrode by subtracting the initial mass from the final mass.
Repeat the experiment with different time periods (e.g., 10 minutes, 15 minutes) and record the mass of copper deposited each time.
Plot a graph of the mass of copper deposited (y-axis) against the time of electrolysis (x-axis). The slope of this graph should be constant, confirming Faraday’s Law.
Project Delivery:
The project will be delivered in the form of a written report, following the structure of Introduction, Development, Conclusion, and Bibliography.
Introduction: The student should contextualize the theme of Faraday’s Law, its real-world applications, and the objective of the project. The student should also state the relevance of this law in the field of electrochemistry and its importance in daily life.
Development: The student should provide a detailed description of the electrolysis experiment conducted, including the methodology used and the results obtained. The student should also explain the theory behind Faraday’s Law and how it was applied in the experiment. This section should also present and discuss the data collected, including the mass of copper deposited at different time periods and the graph plotting this data.
Conclusion: The student should revisit the main points of the project, explicitly stating the learnings obtained and the conclusions drawn about Faraday’s Law based on the experiment and its results. The student should also indicate any challenges faced during the project and how they were overcome, as well as any suggestions for improvement.
Bibliography: The student should list all the sources of information used to carry out the project and write the report. These can include books, websites, videos, etc. The format for the bibliography can be chosen based on the school’s referencing style guidelines.
This project is designed to be completed within a week, with the report being submitted at the end of the week. The total time required per student is estimated to be around 3 to 5 hours, including the time spent on the experiment, discussing the results within the group, and writing the report.
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5
What is a recurrence relation for the number of ternary strings of length n that do not contain 01?
I write code
·
Author has 11.7K answers and 12M answer views
·
3y
Ignore the word “recurrence” and let’s look at something more specific.
Let’s say I give you the number of unique ternary strings of length 5 that don’t contain a 01 — let’s call that x — and claim that the number of unique ternary strings of length 6 is 3x, because you can put a 0, 1, or 2 on the beginning of the string.
You should see right off that that’s wrong: some of the ternary strings, like 11111, start with a 1. If you tacked a 0 on the start of that, it would have an 01 in it, so clearly it’s not 3x.
So…which values can’t you just tack 0, 1, and 2 onto the beginning of? It’s all the one
Ignore the word “recurrence” and let’s look at something more specific.
Let’s say I give you the number of unique ternary strings of length 5 that don’t contain a 01 — let’s call that x — and claim that the number of unique ternary strings of length 6 is 3x, because you can put a 0, 1, or 2 on the beginning of the string.
You should see right off that that’s wrong: some of the ternary strings, like 11111, start with a 1. If you tacked a 0 on the start of that, it would have an 01 in it, so clearly it’s not 3x.
So…which values can’t you just tack 0, 1, and 2 onto the beginning of? It’s all the ones that start with 1, where you can still tack on 1 or 2, but not 0. What values can you tack 0, 1, or 2 onto? The ones that don’t start with 1.
So what if I told you the number of unique ternary strings of length 5 starting with 1 is A and the number that don’t start with 1 is B?
Clearly, the number of such length 6 strings is 2A + 3B.
OK, great. How many of those start with 1? How many don’t?
Well, there are A + B that start with 1, and A + 2B that don’t.
So now you should be able to come up with two recurrence relationships: one for the number of unique ternary strings of length n that start with 1 and one for the number that don’t. If you add those together, you have your answer.
Ankur Aggarwal
B.Tech from Chaudhary Charan Singh University
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Author has 625 answers and 551.2K answer views
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3y
A ternary sequence is a sequence all of whose elements are digits 0,1,2
Let f(n)=number of ternary strings of length that do not contain 01
let string ends with 0 or 2 then 2f(n-1) represents string of (n-1) that do not contain 0,1.
if it ends with 1 then its leftmost number be either 1 or 2 thus
number of ways =2f(n-2)
boundary condition: f(1)=3 ,f(2)=8
hence f(n)=2f(n-1) +2f(n-2) for all n>2.
David Vanderschel
PhD in Mathematics & Physics, Rice (Houston neighborhood) (Graduated 1970)
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Author has 37.4K answers and 49.8M answer views
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3y
A2A: Let
be the count we seek, and let
be the number that start with a
Then
and
for
For the
that start with
or
we can prepend any of the three digits.
For the
that start with
we can only prepend
or
So
for
Related questions
How do you find a recurrence relation for the number of bit strings of length n that contain three consecutive 0's?
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Find the recurrence relation and initial conditions for the number of bit strings of length n that do not have two consecutive 0s. How many bit strings are there of length five?
What is the Recurrence relation for all strings of length n over the alphabets {a,b, c} with no three consecutive b's in the string?
What is the recurrence relation for the number of binary sequences of length n that do not contain 'm' adjacent 1s?
Assistant
Bot
·
1y
To find a recurrence relation for the number of ternary strings of length that do not contain the substring "01", we can analyze the possible endings of such strings.
Let be the number of valid ternary strings of length that do not contain "01". A ternary string can consist of the digits , , and .
Case Analysis
Strings ending with :
These strings can be formed by appending to any valid string of length (since adding does not introduce "01").
Thus, there are such strings.
Strings ending with :
These strings can also be formed by appending to any valid
To find a recurrence relation for the number of ternary strings of length that do not contain the substring "01", we can analyze the possible endings of such strings.
Let be the number of valid ternary strings of length that do not contain "01". A ternary string can consist of the digits , , and .
Case Analysis
Strings ending with :
These strings can be formed by appending to any valid string of length (since adding does not introduce "01").
Thus, there are such strings.
Strings ending with :
These strings can also be formed by appending to any valid string of length (as long as the string does not end with to avoid "01").
The valid strings of length that can end in are those counted in minus the ones that end in . We denote the number of valid strings of length that end with as .
Therefore, the number of valid strings ending with is .
Strings ending with :
Similar to the previous cases, any valid string of length can end with without introducing "01".
Hence, there are such strings.
Recurrence Relation
Now, we need to express , the number of valid strings of length ending with . This is simply because any valid string of length can end with without restriction.
Now we can summarize the valid strings based on their last character:
Strings ending with :
Strings ending with :
Strings ending with :
Combining these, we have:
Base Cases
To fully define the recurrence relation, we need base cases:
(the strings are "0", "1", "2")
(the valid strings are "00", "01", "02", "10", "11", "12", "20", "21", "22", but "01" is not allowed)
Final Recurrence Relation
Thus, the recurrence relation for the number of ternary strings of length that do not contain "01" is:
with base cases:
David Stigant
Works at Houston ISD
·
Upvoted by
Horst H. von Brand
, PhD Computer Science & Mathematics, Louisiana State University (1987) and
Matt Mahoney
, Ph.D. in computer science specializing in data compression and AI. · Author has 234 answers and 274.6K answer views
·
4y
Related
Find the recurrence relation and initial conditions for the number of bit strings of length n that do not have two consecutive 0s. How many bit strings are there of length five?
We need to think about how we can form desired bit strings from shorter bit strings with the same property. Suppose you had a bit string of length 7 with no consecutive 0s. Something like: 1011011 or 0101110. How could you form bit strings of length 8 from those? Well, you could tack on a 1 or 0 to either of them:
1011011 ==> 10110111 or 10110110
0101110 ==> 01011101 or 01011100
Oops! The by tacking on a 0 to the second string, we formed a string with consecutive 0s! Why did this happen? Because 0101110 ends in a 0. It didn’t happen with 1011011 because that string ends in 1. Ok, so this suggests
We need to think about how we can form desired bit strings from shorter bit strings with the same property. Suppose you had a bit string of length 7 with no consecutive 0s. Something like: 1011011 or 0101110. How could you form bit strings of length 8 from those? Well, you could tack on a 1 or 0 to either of them:
1011011 ==> 10110111 or 10110110
0101110 ==> 01011101 or 01011100
Oops! The by tacking on a 0 to the second string, we formed a string with consecutive 0s! Why did this happen? Because 0101110 ends in a 0. It didn’t happen with 1011011 because that string ends in 1. Ok, so this suggests that we should (temporarily) break this problem into 2 problems:
Let An = number of bit strings with no consecutive 0s that end in 1
Let Bn = number of bit strings with no consecutive 0s that end in 0
How do we get more A-type strings? Well, we could append a 1 to an A string or a B string. To get B strings? You have to start with an A string and append a 0 (can’t put a 0 on a B string because it will have two 0s).
So:
A(n+1) = A(n) + B(n)
B(n+1) = A(n)
But since B(n) = A(n-1) we can substitute into the recurrence relation for A(n+1):
A(n+1) = A(n) + A(n-1)… hey, that looks familiar!
Ok, almost done. We actually want Cn = total number of desired bit strings = An + Bn, and we need to find the base cases.
Cn = An + Bn = A(n+1) = A(n) + A(n-1) = C(n-1) + C(n-2)
A(1) = 1 (1), A2 = 2 (01, 11), B1 = 1 (0), and B2 = 1 (10).
And so C(1) = 2 and C(2) = 3.
Edit:
Forgot to get C(5) = C(4) + C(3) = 13
C(4) = C(3) + C(2) = 8
C(3) = C(2) + C(1) = 5
Quick check: bit strings of length 5 with no consecutive 0s:
11111, 11110, 11101, 11011, 11010, 10111, 10110, 10101, 01111, 01110, 01101, 01011, 01010 - 13 as expected.
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Mark Gritter
recreational mathematician
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Nathan Hannon
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3y
Related
How many binary subsequences with a length of m + n with m 0s and n 1s such that the ones form k maximal non-empty subsequences?
My understanding of “maximal non-empty subsequences” is that it means the same thing as “a contiguous sequence of 1’s that cannot be extended on either end, of length at least 1.”
For [math]k=1[/math] this means a string of the form [math]0^a1^n0^{m-a}[/math]. There are [math]m+1[/math] such sequences, because [math]a[/math] can range from [math]0[/math] up to [math]m[/math] itself. (The question asks for “subsequences”, but I think it means sequences of binary digits of total length [math]m+n[/math]; if not then I don’t understand what the parent sequence is supposed to be.)
For [math]k=2[/math] we can partition the [math]n[/math] ones in [math]n-1[/math] distinct ways: [math]1 + (n-1), 2 + (n-2), \ldots, (n-1) + 1[/math]. We have to
My understanding of “maximal non-empty subsequences” is that it means the same thing as “a contiguous sequence of 1’s that cannot be extended on either end, of length at least 1.”
For [math]k=1[/math] this means a string of the form [math]0^a1^n0^{m-a}[/math]. There are [math]m+1[/math] such sequences, because [math]a[/math] can range from [math]0[/math] up to [math]m[/math] itself. (The question asks for “subsequences”, but I think it means sequences of binary digits of total length [math]m+n[/math]; if not then I don’t understand what the parent sequence is supposed to be.)
For [math]k=2[/math] we can partition the [math]n[/math] ones in [math]n-1[/math] distinct ways: [math]1 + (n-1), 2 + (n-2), \ldots, (n-1) + 1[/math]. We have to put one zero between them. The other [math]m-1[/math] zeros can be placed in any of the three “buckets” , for example ___111110_____11111_____. This is a stars-and-bars problem; placing [math]m-1[/math] identical objects in 3 distinct locations is given by [math]\binom{(m-1)+3–1}{3–1} = \binom{m+1}{2}[/math]. Multiply these two and we have the final result: math\binom{m+1}{2}[/math].
For [math]k=3[/math] we can start to see the pattern. We need to partition [math]n[/math] into [math]k=3[/math] nonempty groups, which is [math]\binom{n-1}{k-1} = \binom{n-1}{2}[/math]. This leaves [math]k+1 = 4[/math] spots to fill with zeros, but two of those spots have to be nonempty. So, equivalently, we split [math]m-(k-1)[/math] zeros into [math]k+1[/math] possibly-empty groups, which is [math]\binom{(m-(k-1)) + (k+1)-1}{(k+1)-1} = \binom{m+1}{k}[/math]
Thus, the general solution will be
[math]\displaystyle \binom{n-1}{k-1}\binom{m+1}{k}[/math]
i.e., pick a nonempty assignment of [math]n[/math] ones to [math]k[/math] buckets, then a fixed assignment of [math]k-1[/math] zeros separating those, then a possibly empty assignment of [math]m-(k-1)[/math] zeros to [math]k+1[/math] buckets.
Note that the formula works for the special case [math]k=1[/math] too; [math]\binom{n-1}{0} = 1[/math], and [math]\binom{m+1}{1} = m+1[/math].
Related questions
How do you find a recurrence relation for the number of bit strings of length $n$ by Gould-Jackson (combinatorics, discrete mathematics, recurrence relations, combinatorics on words, math)?
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Why is
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What is the recurrence relation for the number of binary strings that do not contain consecutive 0?
Dinesh Khandelwal
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Daniel R. Page
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What is the complexity of T(n) = 2T(n/2) + C, using recurrence relations?
Just to make things little more clear, here is the recursion tree:
Height of the tree(h) = [math]log_2(n)[/math]
Number of leaves = [math]2^h = 2^{log_2(n)} = n[/math]
[math]T(n) = C + 2C + 4C + \dots + n C [/math]
which is same as the number of nodes in the tree. As there are n leaves, we must have n-1 internal nodes. So, the total number of nodes = 2n-1.
so, [math]T(n) = 2n - 1 = O(n). [/math]
Just to make things little more clear, here is the recursion tree:
Height of the tree(h) = [math]log_2(n)[/math]
Number of leaves = [math]2^h = 2^{log_2(n)} = n[/math]
[math]T(n) = C + 2C + 4C + \dots + n C [/math]
which is same as the number of nodes in the tree. As there are n leaves, we must have n-1 internal nodes. So, the total number of nodes = 2n-1.
so, [math]T(n) = 2n - 1 = O(n). [/math]
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Why is
[math]b_n = 3 b_{n-1} - b_{n-2} + 3^{n-2}[/math]
the recurrence relations for a number of ternary strings of length n that contain 01?
If [math]n < 2[/math], this doesn't work, but [math]b_n[/math] is [math]0[/math] anyway. If [math]n = 2[/math], [math]b_n = 1[/math], rather obviously. For larger values of [math]n[/math], any ternary string of length [math]n[/math] is just a ternary string of length [math]n - 1[/math] with a ternary digit postpended, and this gives a unique description of all ternary strings.
If the initial sub-string (all but the last digit) contains the sequence [math]01[/math], then so does the new string, whatever the last digit is. That contributes [math]3 b_{n - 1}[/math] to the final total.
The remaining initial sequences of length [math]n - 1[/math] — that is, [math]3^{n - 1} - b_{n - 1}[/math] of them — either end in [math]0[/math], and appending [math]1[/math]will yield [math]01[/math] at the
If [math]n < 2[/math], this doesn't work, but [math]b_n[/math] is [math]0[/math] anyway. If [math]n = 2[/math], [math]b_n = 1[/math], rather obviously. For larger values of [math]n[/math], any ternary string of length [math]n[/math] is just a ternary string of length [math]n - 1[/math] with a ternary digit postpended, and this gives a unique description of all ternary strings.
If the initial sub-string (all but the last digit) contains the sequence [math]01[/math], then so does the new string, whatever the last digit is. That contributes [math]3 b_{n - 1}[/math] to the final total.
The remaining initial sequences of length [math]n - 1[/math] — that is, [math]3^{n - 1} - b_{n - 1}[/math] of them — either end in [math]0[/math], and appending [math]1[/math]will yield [math]01[/math] at the end, or don't end in [math]0[/math], and no new [math]01[/math] can appear in the new string. The former case amounts to having a string of length [math]n - 2[/math] with no [math]01[/math] subsequence and appending [math]01[/math]. This plainly won't create an intermediate string of length [math]n - 1[/math] with an [math]01[/math] substring; just appending [math]0[/math] can't introduce a new [math]01[/math] substring.
We're left with [math]3 b_{n - 1} + (3^{n - 2} - b_{n - 2})[/math], and that's us finished.
Michal Forišek
Ph. D. in theoretical Computer Science
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David Rutter
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What is the number of binary strings of length n and containing a pair of consecutive zeros?
The exact number of such strings is [math]2^n - F_{n+2}[/math], where [math]F_n[/math] is the [math]n[/math]-th Fibonacci number.
This is roughly equal to [math]2^n - 1.171\cdot 1.618^n[/math]. Asymptotically, the second part is negligible, so the number of such strings is [math]\Theta(2^n)[/math]. In slightly imprecise words: almost all strings do contain two consecutive zeros. In slightly more precise words, as you increase [math]n[/math], the probability that a random string contains two consecutive zeros goes towards 1.
Here’s why.
We’ll show that the number [math]c_n[/math] of strings that do not contain two consecutive zeros is [math]F_{n+2}[/math].
Quite obviously, we have:
[math]c_0 = 1 = F_2[/math] as th
The exact number of such strings is [math]2^n - F_{n+2}[/math], where [math]F_n[/math] is the [math]n[/math]-th Fibonacci number.
This is roughly equal to [math]2^n - 1.171\cdot 1.618^n[/math]. Asymptotically, the second part is negligible, so the number of such strings is [math]\Theta(2^n)[/math]. In slightly imprecise words: almost all strings do contain two consecutive zeros. In slightly more precise words, as you increase [math]n[/math], the probability that a random string contains two consecutive zeros goes towards 1.
Here’s why.
We’ll show that the number [math]c_n[/math] of strings that do not contain two consecutive zeros is [math]F_{n+2}[/math].
Quite obviously, we have:
[math]c_0 = 1 = F_2[/math] as the empty string does not have two consecutive zeros, and
[math]c_1 = 2 = F_3[/math] as the strings “0” and “1” are both OK as well.
For any larger n, let’s look at our strings of length n. What do they look like?
Some of them end in a “1”. Removing that “1” produces a valid string of length [math]n-1[/math]. And vice versa: taking any valid string of length [math]n-1[/math] and appending a “1” produces a valid string of length [math]n[/math]. Thus, there are exactly [math]c_{n-1}[/math] such strings.
Some of them end in a “0”. As there are no two consecutive zeros in our strings, the previous character has to be a “1”, so all of these strings actually end in “10”. As above, we can conclude that there are exactly [math]c_{n-2}[/math] such strings.
Hence, [math]\forall n\geq 2: c_n = c_{n-1} + c_{n-2}[/math], and as this is the Fibonacci recurrence, we get [math]\forall n: c_n = F_{n+2}[/math].
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Calvin Campbell
B.Sc. in Computer Science & Mathematics, University of the West Indies (Graduated 1984)
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Let f(n) be the number of strings of length n formed using the symbols a, b, and c so that no symbol appears three times in a row (i.e., the substrings aaa, bbb, and ccc are not allowed). What is a recurrence relation for the numbers f(n)?
When n = 1, we can form 3 strings (a, b, c), which are all allowed. Therefore, f(1) = 3.
When n = 2, we can form 33 = 9 strings which are all allowed. Therefore, f(2) = 9.
When n = 3, we can form 333 = 27 strings, but 3 strings, (aaa, bbb, ccc) are not allowed. Therefore, f(3) = 27 - 3 = 24.
When n = 4, we can form 3333 = 81 strings, but 15 strings are not allowed. Therefore, f(4) = 81 - 15 = 66.
When n = 5, we can form 33333= 243 strings, but 63 strings are not allowed. Therefore, f(5) = 243 - 63 = 180.
When n = 6, we can form 333333 = 729 strings, but 237 strings are not allowed. T
When n = 1, we can form 3 strings (a, b, c), which are all allowed. Therefore, f(1) = 3.
When n = 2, we can form 33 = 9 strings which are all allowed. Therefore, f(2) = 9.
When n = 3, we can form 333 = 27 strings, but 3 strings, (aaa, bbb, ccc) are not allowed. Therefore, f(3) = 27 - 3 = 24.
When n = 4, we can form 3333 = 81 strings, but 15 strings are not allowed. Therefore, f(4) = 81 - 15 = 66.
When n = 5, we can form 33333= 243 strings, but 63 strings are not allowed. Therefore, f(5) = 243 - 63 = 180.
When n = 6, we can form 333333 = 729 strings, but 237 strings are not allowed. Therefore, f(6) = 729 - 237 = 492.
When n = 7, we can form 3333333 = 2187 strings, but 843 strings are not allowed. Therefore, f(7) = 2187 - 843 = 1344.
So far, the pattern of the number of allowed strings of length n (up to n = 7) can be defined as:
f(1)= 3, f(2) = 9; f(n) = 2f(n-1) + 2f(n-2) for n>=3.
Using this definition, here is a Python program which prints the values from f(1) to f(10):
def no_consec3(n): if n == 1: return 3 if n == 2: return 9 if n > 2: return 2 no_consec3(n-1) + 2 no_consec3(n-2) for k in range(1,11): print(f'f({k}) = {no_consec3(k)}')
The output:
f(1) = 3
f(2) = 9
f(3) = 24
f(4) = 66
f(5) = 180
f(6) = 492
f(7) = 1344
f(8) = 3672
f(9) = 10032
f(10) = 27408
Good luck!
Himanshu Sharma
System Engineer (Digital Cadre) at Tata Consultancy Services (company) (2020–present)
·
6y
Related
What is recurrence relation for binary search algorithm?
Time complexity is O(logN)-
Recurrence relation-> T(n)=T(n/2)+1
1st step=> T(n)=T(n/2) + 1
2nd step=> T(n/2)=T(n/4) + 1 ……[ T(n/4)= T(n/2^2) ]
3rd step=> T(n/4)=T(n/8) + 1 ……[ T(n/8)= T(n/2^3) ]
.
kth step=> T(n/2^k-1)=T(n/2^k) + 1(k times)
Adding all the equations we get, T(n) = T(n/2^k) + k times 1 _____eq(final)
=> n/2^k= 1 [So how many times we need to divide by 2 until we have only one element left]
=> n=2^k
=> log n=k [taken log(base 2) on both sides ]
Put k= log n in eq(final)
T(n) = T(1) + log n
T(n) = 1 + log n [we know that T(1) = 1 , because it’s a base condition as we are left wit
Time complexity is O(logN)-
Recurrence relation-> T(n)=T(n/2)+1
Derivation->
1st step=> T(n)=T(n/2) + 1
2nd step=> T(n/2)=T(n/4) + 1 ……[ T(n/4)= T(n/2^2) ]
3rd step=> T(n/4)=T(n/8) + 1 ……[ T(n/8)= T(n/2^3) ]
.
.
kth step=> T(n/2^k-1)=T(n/2^k) + 1(k times)
Adding all the equations we get, T(n) = T(n/2^k) + k times 1 _____eq(final)
=> n/2^k= 1 [So how many times we need to divide by 2 until we have only one element left]
=> n=2^k
=> log n=k [taken log(base 2) on both sides ]
Put k= log n in eq(final)
T(n) = T(1) + log n
T(n) = 1 + log n [we know that T(1) = 1 , because it’s a base condition as we are left with only one element in the array and that is the element to be searched so we return 1]
T(n) = O(log n) [taking dominant polynomial, which is n here)
This is how we got “log n” time complexity for binary search. Hope this might help :) .
Debajyoti Nandy
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What is the Recurrence relation for all strings of length n over the alphabets {a,b, c} with no three consecutive b's in the string?
I guess you mean to ask the recurrence relation between the number of possible such strings of length [math]n[/math] using alphabets [math]{a,b,c}[/math] so that no 3 consecutive [math]b[/math]'s are there in the string.
Just look at the problem as a problem of "Dynamic Programming"
Case1 : If the first alphabet is [math]a[/math] or [math]c[/math], then the problem reduces to a subproblem of size math[/math]. [math][2f(n-1)][/math]
Case2 :If the first alphabet is [math]b[/math] then next possible alphabet can be again [math]a[/math] or [math]c[/math] and the problem reduces to a sbproblem of size math[/math]. [math][2f(n-2)][/math]
Case3 :If the first alphabet is [math]b[/math] and the second alphabet is [math]b[/math] as well, then the third alphabet must be
I guess you mean to ask the recurrence relation between the number of possible such strings of length [math]n[/math] using alphabets [math]{a,b,c}[/math] so that no 3 consecutive [math]b[/math]'s are there in the string.
Just look at the problem as a problem of "Dynamic Programming"
Case1 : If the first alphabet is [math]a[/math] or [math]c[/math], then the problem reduces to a subproblem of size math[/math]. [math][2f(n-1)][/math]
Case2 :If the first alphabet is [math]b[/math] then next possible alphabet can be again [math]a[/math] or [math]c[/math] and the problem reduces to a sbproblem of size math[/math]. [math][2f(n-2)][/math]
Case3 :If the first alphabet is [math]b[/math] and the second alphabet is [math]b[/math] as well, then the third alphabet must be either [math]a[/math] or [math]c[/math] and the problem finally reduces to a subproblem of size math[/math]. [math][2f(n-3)][/math].
We have exhausted all the possible cases, so the final recurrence relation would be
[math]f(n)=2f(n-1)+2f(n-2)+2f(n-3)[/math].
where [math]f(n)[/math]= number of such strings of size [math]n[/math].
Initial few test cases need to be calculated, which are [math]f(1)=3, f(2)=9, f(3)= 26[/math] and using these values, for [math]n=4[/math] we obtain [math]f(4)=76[/math] from the recurrence which is indeed the actual answer.
Brian Peterson
Former professor at San Jose State University (1978–2015)
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Let f(n) be the number of strings of length n formed using the symbols a, b, and c so that no symbol appears three times in a row (i.e., the substrings aaa, bbb, and ccc are not allowed). What is a recurrence relation for the numbers f(n)?
The recurrence is that [math]f(n)=2f(n-1)+2f(n-2)[/math] for [math]n \ge 3[/math].
To form such a string whose rightmost symbol is different than its predecessor we may take any acceptable string of length [math]n-1[/math] and append on its right either of the [math]2[/math] symbols that differ from the rightmost symbol of that string. So there are [math]2f(n-1)[/math] acceptable length [math]n[/math] strings of this type. And to form such a string whose rightmost symbol is the same as its predecessor we may take any acceptable string of length [math]n-2[/math] and append on its right two copies of either of the [math]2[/math] symbols that differ from the rightmost symbol of that string. So there
The recurrence is that [math]f(n)=2f(n-1)+2f(n-2)[/math] for [math]n \ge 3[/math].
To form such a string whose rightmost symbol is different than its predecessor we may take any acceptable string of length [math]n-1[/math] and append on its right either of the [math]2[/math] symbols that differ from the rightmost symbol of that string. So there are [math]2f(n-1)[/math] acceptable length [math]n[/math] strings of this type. And to form such a string whose rightmost symbol is the same as its predecessor we may take any acceptable string of length [math]n-2[/math] and append on its right two copies of either of the [math]2[/math] symbols that differ from the rightmost symbol of that string. So there are [math]2f(n-2)[/math] acceptable length [math]n[/math] strings of this type, and this accounts for all acceptable length [math]n[/math] strings.
By the general technique for linearly recursive sequences we can express [math]f(n)[/math] as a linear combination of the [math]n^{th}[/math] powers of the roots [math]1 \pm \sqrt{3}[/math] of the equation [math]x^2-2x-2=0[/math], but the question didn’t ask for this.
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Numerical semigroups with a monotonic Apéry set
José Carlos Rosales; Pedro A. García-Sánchez; Juan Ignacio García-García; M. B. Branco
Czechoslovak Mathematical Journal (2005)
Volume: 55, Issue: 3, page 755-772
ISSN: 0011-4642
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top We study numerical semigroups S with the property that if m is the multiplicity of S and w(i) is the least element of S congruent with i modulo m, then 0<w(1)<⋯<w(m−1). The set of numerical semigroups with this property and fixed multiplicity is bijective with an affine semigroup and consequently it can be described by a finite set of parameters. Invariants like the gender, type, embedding dimension and Frobenius number are computed for several families of this kind of numerical semigroups.
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Rosales, José Carlos, et al. "Numerical semigroups with a monotonic Apéry set." Czechoslovak Mathematical Journal 55.3 (2005): 755-772.
@article{Rosales2005,
abstract = {We study numerical semigroups $S$ with the property that if $m$ is the multiplicity of $S$ and $w(i)$ is the least element of $S$ congruent with $i$ modulo $m$, then $0<w(1)<\dots <w(m-1)$. The set of numerical semigroups with this property and fixed multiplicity is bijective with an affine semigroup and consequently it can be described by a finite set of parameters. Invariants like the gender, type, embedding dimension and Frobenius number are computed for several families of this kind of numerical semigroups.},
author = {Rosales, José Carlos, García-Sánchez, Pedro A., García-García, Juan Ignacio, Branco, M. B.},
journal = {Czechoslovak Mathematical Journal},
keywords = {numerical; semigroups; Apéry; sets; symmetric; affine; proportionally; modular; Diophantine; inequality; numerical semigroups; Apéry sets; affine semigroups; modular Diophantine inequalities; multiplicities; embedding dimensions; Frobenius numbers},
language = {eng},
number = {3},
pages = {755-772},
publisher = {Institute of Mathematics, Academy of Sciences of the Czech Republic},
title = {Numerical semigroups with a monotonic Apéry set},
url = {
volume = {55},
year = {2005},
}
TY - JOUR
AU - Rosales, José Carlos
AU - García-Sánchez, Pedro A.
AU - García-García, Juan Ignacio
AU - Branco, M. B.
TI - Numerical semigroups with a monotonic Apéry set
JO - Czechoslovak Mathematical Journal
PY - 2005
PB - Institute of Mathematics, Academy of Sciences of the Czech Republic
VL - 55
IS - 3
SP - 755
EP - 772
AB - We study numerical semigroups $S$ with the property that if $m$ is the multiplicity of $S$ and $w(i)$ is the least element of $S$ congruent with $i$ modulo $m$, then $0<w(1)<\dots <w(m-1)$. The set of numerical semigroups with this property and fixed multiplicity is bijective with an affine semigroup and consequently it can be described by a finite set of parameters. Invariants like the gender, type, embedding dimension and Frobenius number are computed for several families of this kind of numerical semigroups.
LA - eng
KW - numerical; semigroups; Apéry; sets; symmetric; affine; proportionally; modular; Diophantine; inequality; numerical semigroups; Apéry sets; affine semigroups; modular Diophantine inequalities; multiplicities; embedding dimensions; Frobenius numbers
UR -
ER -
References
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1. 10.1016/S0304-3975(96)00195-8, Theoret. Comput. Sci. 173 (1997), 183–208. (1997)MR1436701DOI10.1016/S0304-3975(96)00195-8
2. Sur les branches superlinéaires des courbes algébriques, C.R.Acad. Sci. Paris 222 (1946). (1946)MR0017942
3. Maximality Properties in Numerical Semigroups and Applications to One-Dimensional Analytically Irreducible Local Domains. Memoirs of the Amer. Math. Soc. Vol.598, , , 1997. (1997)MR1357822
4. 10.1016/0021-8693(77)90268-X, J. Algebra 49 (1987), 81–95. (1987)MR0568894DOI10.1016/0021-8693(77)90268-X
5. 10.2307/2371684, Amer. J. Math. 64 (1942), 299–312. (1942)Zbl0061.06801MR0006196DOI10.2307/2371684
6. On prime ideals with generic zero x i=t n i, Proc. Amer. Math. Soc. 47 (1975), 329–332. (1975)MR0389912
7. 10.1006/inco.1994.1067, Inform. and Comput. 113 (1994), 143–172. (1994)MR1283022DOI10.1006/inco.1994.1067
8. 10.24033/asens.1307, Ann. Scient. École Norm. Sup. 9 (1976), 145–154. (1976)MR0407038DOI10.24033/asens.1307
9. Semigroups, semigroup rings and analytically irreducible rings, Reports Dpt. of Mathematics, University of Stockholm, Vol. 1, 1986. (1986)
10. 10.1007/BF02573091, Semigroup Forum 35 (1987), 63–83. (1987)DOI10.1007/BF02573091
11. 10.2140/pjm.1999.191.75, Pacific J. Math. 191 (1999), 75–83. (1999)DOI10.2140/pjm.1999.191.75
12. Commutative Semigroup Rings, The University of Chicago Press, 1984. (1984)Zbl0566.20050MR0741678
13. 10.1007/BF01273309, Manuscripta Math 3 (1970), 175–193. (1970)Zbl0211.33801MR0269762DOI10.1007/BF01273309
14. 10.1090/S0002-9939-1970-0265353-7, Proc. Amer. Math. Soc. 25 (1973), 748–751. (1973)MR0265353DOI10.1090/S0002-9939-1970-0265353-7
15. The Diophantine Frobenius problem, Forschungsintitut für Diskrete Mathematik, Bonn, Report No.00893, 2000. (2000)
16. The Diophantine Frobenius problem, manuscript, .
17. 10.1007/BF02574106, Semigroup Forum 52 (1996), 307–318. (1996)Zbl0853.20041MR1377695DOI10.1007/BF02574106
18. 10.1006/jabr.1996.0178, J. Algebra 182 (1996), 422–434. (1996)Zbl0856.20043MR1391591DOI10.1006/jabr.1996.0178
19. 10.1016/S0022-4049(01)00128-1, J.Pure Appl. Algebra 171 (2002), 303–314. (2002)MR1904486DOI10.1016/S0022-4049(01)00128-1
20. Finitely Generated Commutative Monoids, Nova Science Publishers, New York, 1999. (1999)MR1694173
21. 10.1112/S0024610701003052, J. London Math. Soc. 65 (2002), 611–623. (2002)MR1895736DOI10.1112/S0024610701003052
22. 10.1016/j.jnt.2003.06.002, J. Number Theory 103 (2003), 281–294. (2003)MR2020273DOI10.1016/j.jnt.2003.06.002
23. On a linear Diophantine problem of Frobenius, J. Reine Angew. Math. 293/294 (1977), 1–17. (1977)MR0441855
24. 10.1017/S0027763000015312, Nagoya Math. J. 49 (1973), 101–109. (1973)Zbl0257.13024MR0318140DOI10.1017/S0027763000015312
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188512 | https://pmc.ncbi.nlm.nih.gov/articles/PMC1924519/ | Multiple myeloma/hypercalcemia - PMC
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. 2007 Jun 29;9(Suppl 1):S4. doi: 10.1186/ar2168
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Multiple myeloma/hypercalcemia
Babatunde O Oyajobi
Babatunde O Oyajobi
1 University of Texas Health Science Center at San Antonio, Department of Cellular and Structural Biology (MSC 7762), Floyd Curl Drive, San Antonio, Texas 78229-3900, USA
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1 University of Texas Health Science Center at San Antonio, Department of Cellular and Structural Biology (MSC 7762), Floyd Curl Drive, San Antonio, Texas 78229-3900, USA
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Basic science, rationale, background and future of denosumab: a RANK ligand inhibitor
Edward M Schwarz and Christopher T Ritchlin
Supported by an unrestricted grant from Amgen Inc.
Issue date 2007.
Copyright © 2007 BioMed Central Ltd
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PMCID: PMC1924519 PMID: 17634143
Abstract
Multiple myeloma, a cancer of plasma cells, is associated with excessive tumor-induced, osteoclast-mediated bone destruction. Hypercalcemia remains the most frequent metabolic complication of myeloma in patients, and excessive osteolysis plays a major contributory role in its pathogenesis. The clinical presentation of hypercalcemia in patients varies depending on the level of ionized calcium; it can be life threatening, as in the case of hypercalcemic crisis, requiring immediate medical treatment to prevent death. During the past few years there have been exciting developments in our understanding of the pathogenesis of myeloma bone disease; in particular, key mediators of the osteoclastic bone resorption in myeloma have been identified, including receptor activator of nuclear factor-κB ligand (RANKL) and macrophage inflammatory protein-1α. There is also increasing evidence that Dickkopf 1, which has been shown to be over-expressed in myeloma patients, is also a potent stimulator of osteoclast formation and activity. Importantly, the available data suggest that RANKL is the final common mediator of osteoclastic bone resorption, irrespective of the upstream initiator molecule. This brief review presents an overview of the roles played by these mediators in inducing osteolysis in myeloma bone disease, and it discusses targeting RANKL as a potential new treatment strategy in myeloma bone disease and myeloma-associated hypercalcemia.
Introduction
Multiple myeloma, a clonal neoplasm of plasma cells, is the second most common adult hematologic malignancy, and it is the most common cancer with skeleton as its primary site. Multiple myeloma, which affects 70,000 people in the USA, with 15,000 new cases accruing yearly, accounts for 1% to 2% of cancer-related deaths. Myeloma is unique in its propensity to cause osteolysis, with 80% of patients suffering from devastating and progressive bone destruction; this results in the complications that are responsible for the high morbidity and mortality rates associated with the disease. These complications, which include severe and unremitting bone pain, pathologic fractures (Figure 1), spinal cord compression, and hypercalcemia, are a significant clinical problem for which there is no effective cure. Hypercalcemia, which can range in presentation from mild to severe and life threatening, is the most common metabolic complication of myeloma and occurs in approximately one-third of patients.
Figure 1.
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Osteolytic lesions and pathologic fracture in a 76-year-old female myeloma patient. With permission from Nair and Pearson . Copyright © 2004 Massachusetts Medical Society. All rights reserved.
Hypercalcemia in myeloma
In myeloma patients, the primary cause of the hypercalcemia is widespread tumor-induced bone destruction. This is primarily due to increased osteoclastic bone resorption caused by potent cytokines expressed or secreted locally by the myeloma cells (receptor activator of nuclear factor-κB ligand [RANKL], macrophage inflammatory protein [MIP]-1α, and tumor necrosis factors [TNFs]; see below) or over-expressed by other cells in the local microenvironment . This bone resorption in turn leads to efflux of calcium into the extracellular fluid. However, the pathogenesis of hypercalcemia in myeloma is probably more complex than this because not all patients with significant myeloma bone disease develop hypercalcemia and, even in the patients who do develop it, hypercalcemia is often a prominent feature only late in the course of disease.
Hypercalcemia is most common in those myeloma patients who have the greatest tumor volume, irrespective of serum parathyroid hormone-related protein (PTHrP) status. The reasons for this are still unclear but they may be related to the amount of bone-resorbing activity produced by myeloma cells as well as glomerular filtration status. Because myeloma patients often have irreversible impairment in renal function and increased renal tubular calcium reabsorption, the capacity of the kidneys to clear excess calcium load from the circulation effectively is overwhelmed, resulting in elevated serum calcium levels. Elevated serum PTHrP is not a consistent finding in myeloma patients, and it remains unclear why myeloma patients present with increased renal tubular calcium reabsorption. Consistent with this, hypercalcemia is more frequent in patients with plasma cell leukemia, a late stage complication of myeloma, than in overt myeloma per se, despite the fact that plasma cell leukemias induce comparatively weaker bone-resorbing activity. This is a consequence of the severe renal insufficiency associated with plasma cell leukemia. Measurements of total body myeloma cell burden together with production of bone-resorbing activity by cultured bone marrow myeloma cells in vitro do not correlate closely with hypercalcemia, although they do correlate somewhat with the extent of osteolytic bone lesions . Thus, other factors are probably involved in the pathogenesis of hypercalcemia, in addition to those that promote osteoclast formation and induce osteoclast activation.
There are other differences between the hypercalcemia that occurs in myeloma and classical humoral hypercalcemia of malignancy in patients with solid tumors. First, in contrast to myeloma patients with hypercalcemia, in which PTHrP is implicated only sporadically, humoral hypercalcemia of malignancy is almost always due to excessive secretion of PTHrP by the tumor, resulting in elevated circulating levels. Second, in patients with hypercalcemia due to myeloma, there is almost always impaired renal function and increased serum phosphate that is associated with decreased glomerular filtration rate. Third, markers of bone formation such as serum alkaline phosphatase are commonly not elevated in patients with myeloma, because bone formation is often not altered and may in fact be suppressed for reasons that are not entirely clear but may involve Dickkopf (DKK)1 (see below). Finally, patients with hypercalcemia due to myeloma usually respond very rapidly to treatment with corticosteroids, mainly because of their rapid suppression of myeloma tumor growth, unlike patients with humoral hypercalcemia caused by solid tumors .
Myeloma bone disease
The excessive bone destruction that is characteristic of multiple myeloma is dependent on osteoclast stimulation. Although it has been known for more than three decades that osteoclasts are hyperstimulated by cytokines in myeloma, the osteoclast-activating factors (OAFs) responsible proved elusive until the late 1990s. For some time the erstwhile approach of identifying such factors by use of cultured tumor cell lines, followed by protein purification, was unsuccessful. It is now apparent that cell-cell interactions involving myeloma cells and other cells in the marrow microenvironment, including stromal cells, cells of the osteoblast lineage, and possibly osteoclasts themselves, are required for production of OAFs in myeloma. There is now abundant evidence that the OAFs that are probably responsible for bone resorption in myeloma are the cytokine RANKL and the chemokine MIP-1α. However, a number of other cytokines, including TNF-α, lymphotoxin (TNF-β), and PTHrP, have also been implicated in the disease. There is still considerable debate as to whether in myeloma these bone-resorbing cytokines are produced by the tumor cells or accessory cells involved in cell-cell interactions, or both; indeed, there is evidence in support of both notions.
RANKL
RANKL is part of a system of TNF ligand and receptor superfamily members that include its cognate signaling receptor RANK (receptor activator of nuclear factor-κB) and the decoy receptor antagonist osteoprotegerin (OPG), which regulate osteoclastogenesis and bone resorption in vivo. Both RANKL and OPG are expressed by bone marrow stromal cells (BMSCs) and osteoblasts. It is now well established that RANKL is increased and OPG is markedly decreased in bone marrow in multiple myeloma. Early studies demonstrated that co-culture of myeloma cells and BMSCs leads to marked induction of RANKL expression by both cell types, and a concomitant downregulation of OPG expression by BMSCs ; this illustrates the importance of interactions between myeloma cells and stromal cells. This induction of RANKL and downregulation of OPG in myeloma cell-BMSC co-cultures is abrogated by neutralizing antibody to α 4 integrin receptor subunit, and treatment of myeloma tumor cells with a recombinant soluble form of vascular cell adhesion molecule (VCAM)-1 (a ligand for the α 4 adhesion molecule) was sufficient to induce RANKL expression . These data suggest that cell-cell interactions, mediated by stromal cell expressed VCAM-1 binding to very late antigen-4 (α 4 β 1 integrin) on myeloma cells, are necessary for tumor cells to express RANKL.
These findings of RANKL induction and concomitant OPG downregulation in bone-derived cells in myeloma were subsequently replicated in other studies involving myeloma cell-BMSC and myeloma cell-osteoblast co-cultures [6,7], although these latter studies and others did not observe an increase in RANKL expression in tumor cells . However, evidence is accumulating that supports the notion that myeloma cells themselves also express RANKL, thus providing a mechanism for direct stimulation of osteoclast formation by myeloma cells [9-13]. Nevertheless, irrespective of whether myeloma cells themselves are capable of driving osteoclastogenesis in myeloma directly, it is incontrovertible that RANKL is over-expressed in the marrow micro-environment in myeloma patients, with the ratio of RANKL to OPG markedly upregulated as compared with that in normal control individuals .
Data from a number of studies have shown that neutralizing the biologic effects of this over-expressed RANKL in vivo, using either a genetically engineered soluble form of RANK (RANK-Fc) or OPG-Fc, reduces bone destruction and has a beneficial effect in murine models of myeloma bone disease [6,15-17]. These chimeric molecules act as decoys by binding to and sequestering RANKL, thus preventing it from interacting with its native receptor on hematopoietic osteoclast precursors and mature osteoclasts. As proof of principle that targeting the RANKL/RANK system is beneficial in hypercalcemia, RANK-Fc (constructed by fusing the extracellular domain of murine RANK to the Fc region of human lgG 1) was found to inhibit normal and pathologic osteoclastic bone resorption in a mouse model of tumor-induced hypercalcemia . Importantly, neutralizing monoclonal antibodies raised against the mouse α 4 integrin subunit also decreased osteolysis and overall tumor burden in a mouse model of myeloma bone disease . The latter finding is consistent with the notion that RANKL levels are modulated by stromal cell-expressed VCAM-1 binding to very late antigen-4 on myeloma tumor cells . Taken together, the available data strongly suggest that RANKL is not only the final common mediator of osteoclast activation but also that it is a myeloma cell product, and that aberrant expression of this cytokine within the myeloma marrow millieu is likely to be pivotal in initiating and maintaining the characteristic bone destruction as well as promoting hypercalcemia.
Unlike most of the other mediators previously implicated in myeloma bone disease, RANKL is expressed in various cell types as a transmembrane protein, although it can be cleaved, as occurs in activated T cells, to yield a soluble form that remains biologically active. Unlike for T cells, however, there is no unequivocal evidence to date that 'soluble RANKL' in myeloma is either naturally secreted or cleaved from the cell surface by a proteinase. Nevertheless, whether the predominant source of RANKL in the myeloma bone microenvironment is tumor cells or BMSCs, or both, the consequent exaggerated osteoclast formation and activation is likely to be exacerbated by a concomitant decrease in locally available OPG. The reduced levels of OPG in bone marrow of patients with myeloma is likely to be further exarcebated by binding, internalizing, and degrading of OPG by tumor cells . Consistent with this, serum OPG levels in myeloma patients are significantly reduced compared with those in healthy control individuals .
Macrophage inflammatory protein-1α
MIP-1α is a member of the C-C chemokine family produced by myeloma cells and has been implicated as an important mediator in myeloma bone disease that stimulates osteoclast formation and differentiation, and bone resorption [22-27]. Consistent with the notion that it plays an important role in the pathogenesis of myeloma-associated osteolysis in vivo, there is now convincing evidence, in RANK knockout mice, that MIP-1α also stimulates osteoclast differentiation and bone resorption in vivo in a RANKL/RANK-dependent manner when it is expressed adjacent to bone surfaces .
MIP-1α is present in bone marrow and marrow supernatants of patients with myeloma bone disease at much higher levels than in patients with other hematologic malignancies and in normal control individuals . Furthermore, its levels correlate with presence of lytic lesions as well as disease stage and activity [24,25]. MIP-1α was also found to be elevated in the serum of myeloma patients with severe bone disease and correlated positively with bone resorption markers, providing evidence for a causal role of the chemokine in the development of lytic bone lesions in myeloma [26,27]. Interestingly, MIP-1α is also produced by osteoblast-like cells in response to stimulation by interleukin-1 and TNF-α . Because myeloma cells secrete high levels of both cytokines constitutively , this would serve to amplify further the concentrations of the chemokine locally in the bone micro-environment. In addition, MIP-1α signaling through cell surface receptors for MIP-1α (C-C chemokine receptor [CCR]-1 and CCR5) on bone cells enhances adhesive interactions between myeloma and marrow stromal cells. This, in turn, upregulates levels of expression of RANKL and interleukin-6, which further exacerbates bone destruction and increases tumor burden . Systemic administration of neutralizing antibodies to MIP-1α or small molecule CCR-1-specific and CCR-5-specific antagonists inhibited tumor-induced osteolysis and limited disease progression in mouse models of myeloma bone disease [22,31]. Therefore, the MIP-1α and its cognate receptors may serve as targets for the development of novel therapies to combat hypercalcemia. In this regard, MIP-1α has been reported to induce severe hypercalcemia in other hematopoietic malignancies .
Dickkopf 1
DKK1, the prototypic member of a family of proteins that act as extracellular antagonists of canonical Wnt signaling, is expressed in bone in osteoblasts, osteocytes, and BMSCs. It was identified on cDNA microarrays as being preferentially and abundantly expressed in bone marrow aspirates from patients with myeloma. Specifically, DKK1 over-expression was reported to be associated with the presence of focal lesions detected on magnetic resonance imaging . These data have led to the notion that DKK1 is one of the major mediators of the osteolysis that occurs in myeloma bone disease. Interestingly, canonical Wnt signaling was recently shown to modulate RANKL and OPG expression in osteoblasts in vitro, resulting in inhibition of osteoclastogenesis . Consistent with these observations, it was recently reported that DKK1 enhances osteoclast formation and bone resorption in vitro, and systemic administration of anti-DKK1 antibodies was shown to block osteoclast formation in vivo in a murine model of myeloma bone disease .
Diagnosis and treatment of hypercalcemia
The diagnosis of hypercalcemia is based on concentration of ionized calcium, because the serum calcium level may be erroneously low due to binding of circulating calcium to albumin. The clinical presentation is frequently dependent on the calcium level; patients may be asymptomatic (≤ 3 mmol/l) or they may present with symptoms such as dry mouth, anorexia and vomiting, polyuria, polydipsia, depression, or confusion (3 to 4 mmol/l). Occasionally, patients may develop a life-threatening 'hypercalcemic crisis' (≥ 4 mmol/l) and present in coma.
The best therapy for myeloma-associated hypercalcemia is, by and large, successful elimination of the primary tumor, but myeloma remains an incurable cancer and its unique propensity to cause bone destruction is, in part, responsible for the associated hypercalcemia observed in some patients. Over the past decade, saline replenishment and bisphosphonates, which are potent inhibitors of osteoclastic bone resorption, have become the mainstay of treatment for cancer-induced osteolytic bone disease and management of PTHrP-mediated humoral hypercalcemia of malignancy, as well as hypercalcemia induced by myeloma . In myeloma patients with hypercalcemia, these agents are usually used in combination regimens that include corticosteroids and/or calcitonin, the latter being effective in patients at high risk for developing a hypercalcemic crisis. The available bisphosphonates, including clodronate (a first-generation bisphosphonate), pamidronate (a second-generation bisphosphonate), and zoledronic acid and ibandronate (third-generation bisphosphonates), have exhibited varying degrees of efficacy in multiple clinical trials, undoubtedly influenced by tolerance or toxicity issues. Orally administered clodronate has exhibited little ability to slow the development of skeletal complications in patients. In contrast, oral administration of the more potent ibandronate or intravenous infusions of either pamidronate or zoledronic acid have reduced skeletal complications in patients with cancer-induced bone diseases, with zoledronic acid being the most effective to date [38-40].
The exciting discovery that the RANKL/RANK/OPG pathway is indispensable not only for normal bone resorption but also for pathologic bone resorption and hypercalcemia induced by myeloma and other cancers that metastasize to the skeleton has spurred the clinical development of a number of other types of new antiresorptive agents. Available data from preclinical and clinical studies strongly suggesting that RANKL is the final common mediator in tumor-induced bone destruction that precedes malignancy-associated hypercalcemia, irrespective of the upstream initiator, provide a rationale for pursuing this therapeutic strategy.
Interestingly, data from preclinical studies in murine models of humoral hypercalcemia of malignancy suggest that inhibition of RANKL, using OPG-Fc, causes greater suppression of bone resorption and hypercalcemia than do bisphosphonates . Although its development has been discontinued, AMGN-0007, a recombinant OPG construct, was shown in early studies to be efficacious in patients with multiple myeloma or breast cancer related skeletal metastases. At present, the most advanced anti-RANKL approach in clinical development is a fully human monoclonal antibody to RANKL, namely denosumab (Amgen Inc., Thousand Oaks, CA, USA). Clinical trials with denosumab are ongoing in patients with multiple myeloma and recent data suggest that it is efficacious . In addition, studies are underway in bone metastases from breast cancer, prostate cancer, and other solid tumors .
The observation that an intact RANK signaling pathway is indispensable for the osteoclastogenic effect of MIP-1α in vivo lends further support to the notion that RANKL is indeed the final common mediator of the excessive osteoclast number and activity in myeloma. It also suggests that denosumab may be effective in treating hypercalcemia, irrespective of its cause. With regard to targeting MIP-1α or its receptors, available preclinical data provide a basis for further studies to determine the efficacy of orally bioavailable small molecule synthetic antagonists of the receptors that mediate the effects of MIP-1α (CCR-1 and CCR-5), as adjuncts to current standard therapeutic approaches in myeloma-induced hypercalcemia. Although it is presently unclear which of the MIP-1α signaling receptors mediates its osteoclastogenic activity, small molecule antagonists of both CCR-1 and CCR-5 are currently under development for other disease conditions .
Finally, a small molecule (IIC3; Enzo Biochem) that is capable of binding to and thereby inhibiting the biologic activity of DKK1 is also currently under development. Given the recently discovered role of canonical Wnt signaling in osteoclast formation and the osteoclastogenic effect of DKK1, further insights into the mechanisms involved in modulating the effects of DKK1 in myeloma bone disease will help to define other novel molecular targets, facilitating development of new therapies for the treatment of associated hypercalcemia.
Conclusion
Hypercalcemia is the most common metabolic complication of multiple myeloma but its pathogenesis remains unclear.
Recently available data support a role for several cytokines such as RANKL, MIP-1α and DKK1 in the exaggerated osteoclastic bone resorption that characterizes myeloma bone disease, with compelling evidence for RANKL being the final common mediator. Given that the increased osteolysis contributes to the development and progression of hypercalcemia in patients, strategies targeting these molecules and the signaling pathways involved may result in novel therapies to manage this distressing and often life-threatening complication of myeloma bone disease.
Abbreviations
BMSC = bone marrow stromal cell; CCR = C-C chemokine receptor; DKK = Dickkopf; MIP = macrophage inflammatory protein; OAF = osteoclast-activating factor; OPG = osteoprotegerin; PTHrP = parathyroid hormone-related protein; RANK = receptor activator of nuclear factor-κB; RANKL = receptor activator of nuclear factor-κB ligand; TNF = tumor necrosis factor; VCAM = vascular cell adhesion molecule.
Competing interests
The authors declare that they have no competing interests.
Acknowledgments
Acknowledgements
This work was supported by grants from the NIH/NCI (KO1 CA104180 and PO1 CA40035).
This article is published as part of Arthritis Research & Therapy Volume 9 Supplement 1, 2007: Basic science, rationale, background and future of denosumab: a RANK ligand inhibitor. The full contents of the supplement are available online at
Publication of the supplement has been supported by an unrestricted grant from Amgen Inc.
References
Oyajobi BO, Mundy GR. Pathophysiology of myeloma bone disease. In: Gahrton G, Durie BGM, Samson DS, editor. Multiple Myeloma and Related Disorders. 2. London, UK: Arnold; 2004. pp. 74–88. [Google Scholar]
Garcia-Sanz R, Orfão A, Gonzalez M, Tabernero MD, Bladé J, Moro MJ, Fernandez-Calvo J, Sanz MA, Perez-Simon JA, Rasillo A, et al. Primary plasma cell leukemia: clinical, immunophenotypic, DNA ploidy, and cytogenetic characteristics. Blood. 1999;93:1032–1037. [PubMed] [Google Scholar]
Tuttle KR, Kunau RT, Loveridge N, Mundy GR. Altered renal calcium handling in hypercalcemia of malignancy. J Am Soc Nephrol. 1991;2:191–199. doi: 10.1681/ASN.V22191. [DOI] [PubMed] [Google Scholar]
Durie BGM, Salmon SE, Mundy GR. Relation of osteoclast activating factor production to the extent of bone disease in multiple myeloma. Br J Haematol. 1981;47:21–30. doi: 10.1111/j.1365-2141.1981.tb02758.x. [DOI] [PubMed] [Google Scholar]
Oyajobi BO, Traianedes K, Yoneda T, Mundy GR. Expression of RANK ligand (RANKL) by myeloma cells requires binding to bone marrow stromal cells via an α4β1-VCAM-1 interaction. Bone. 1998. p. S180.
Pearse RN, Sordillo EM, Yaccoby S, Wong BR, Liau DF, Colman N, Michaeli J, Epstein J, Choi Y. Multiple myeloma disrupts the TRANCE/osteoprotegerin cytokine axis to trigger bone destruction and promote tumor progression. Proc Natl Acad Sci USA. 2001;98:11581–11586. doi: 10.1073/pnas.201394498. [DOI] [PMC free article] [PubMed] [Google Scholar]
Giuliani N, Bataille R, Mancini C, Lazarretti M, Barille S. Myeloma cells induce imbalance in the osteoprotegerin/osteoprotegerin ligand system in the human bone marrow environment. Blood. 2001;98:3527–3533. doi: 10.1182/blood.V98.13.3527. [DOI] [PubMed] [Google Scholar]
Roux S, Meignin V, Quillard J, Meduri G, Guiochon-Mantel A, Fermand J-P, Milgrom E, Mariette X. RANK (receptor activator of nuclear factor-κB) and RANKL expression in multiple myeloma. Br J Haematol. 2002;117:86–92. doi: 10.1046/j.1365-2141.2002.03417.x. [DOI] [PubMed] [Google Scholar]
Farrugia AN, Atkins GJ, To LB, Pan B, Horvath N, Findlay DM, Bardy A, Zannettino AC. Receptor activator of nuclear factor-κB ligand expression by human myeloma cells mediates osteoclast formation in vitro and correlates with bone destruction in vivo. Cancer Res. 2003;63:5438–5445. [PubMed] [Google Scholar]
Croucher PI, Shipman CM, Lippitt J, Perry M, Asosingh K, Hijzen A, Brabbs AC, van Beek EJR, Holen I, Skerry TM, et al. Osteoprotegerin inhibits the development of osteolytic bone disease in multiple myeloma. Blood. 2001;98:3534–3540. doi: 10.1182/blood.V98.13.3534. [DOI] [PubMed] [Google Scholar]
Sezer O, Heider U, Jakob C, Zavrski I, Eucker J, Possinger K, Sers C, Krenn V. Immunocytochemistry reveals RANKL expression of myeloma cells. Blood. 2002;99:4646–4647. doi: 10.1182/blood-2002-01-0148. [DOI] [PubMed] [Google Scholar]
Sezer O, Heider U, Jakob C, Eucker J, Possinger K. Human bone marrow myeloma cells express RANKL. J Clin Oncol. 2002;20:353–354. doi: 10.1200/JCO.2002.20.1.353. [DOI] [PubMed] [Google Scholar]
Lai FP, Cole-Sinclair M, Cheng WJ, Quinn JM, Gillespie MT, Sentry JW, Schneider HG. Myeloma cells can directly contribute to the pool of RANKL in bone bypassing the classic stromal and osteoblast pathway of osteoclast stimulation. Br J Haematol. 2004;126:192–201. doi: 10.1111/j.1365-2141.2004.05018.x. [DOI] [PubMed] [Google Scholar]
Roux S, Mariette X. The high rate of bone resorption in multiple myeloma is due to RANK (receptor activator of nuclear factor-kappaB) and RANK ligand expression. Leuk Lymphoma. 2004;45:1111–1118. doi: 10.1080/10428194310001593193. [DOI] [PubMed] [Google Scholar]
Oyajobi BO, Williams PJ, Traianedes K, Yoneda T, Anderson DM, Mundy GR. Efficacy of a genetically-engineered soluble Receptor Activator of NF-κB (RANK) fusion protein on bone resorption in vitro and in vivo. Cancer. 2000. p. 3093.
Yaccoby S, Pearse RN, Johnson CL, Barlogie B, Choi Y, Epstein J. Myeloma interacts with the bone marrow microenvironment to induce osteoclastogenesis and is dependent on osteoclast activity. Br J Haematol. 2002;116:278–290. doi: 10.1046/j.1365-2141.2002.03257.x. [DOI] [PubMed] [Google Scholar]
Vanderkerken K, De Leenheer E, Shipman CM, Asosingh K, Willems A, Van Camp B, Croucher P. Recombinant osteoprotegerin decreases tumor burden and increases survival in a mouse model of multiple myeloma. Cancer Res. 2003;63:287–289. [PubMed] [Google Scholar]
Oyajobi BO, Anderson DM, Traianedes K, Williams PJ, Yoneda T, Mundy GR. Therapeutic efficacy of a soluble receptor activator of nuclear factor kappaB-IgG Fc fusion protein in suppressing bone resorption and hypercalcemia in a model of humoral hypercalcemia of malignancy. Cancer Res. 2001;61:2572–2578. [PubMed] [Google Scholar]
Mori Y, Shimizu N, Dallas M, Niewolna M, Williams PJ, Mundy GR, Yoneda T. Anti-α4 integrin antibody suppresses the development of myeloma and associated osteolysis. Blood. 2004;104:2149–2154. doi: 10.1182/blood-2004-01-0236. [DOI] [PubMed] [Google Scholar]
Standal T, Seidel C, Hjertner O, Plesner T, Sanderson RD, Waage A, Borset M, Sundan A. Osteoprotegerin is bound, internalized and degraded by multiple myeloma cells. Blood. 2002;100:3002–3007. doi: 10.1182/blood-2002-04-1190. [DOI] [PubMed] [Google Scholar]
Seidel C, Hjertner O, Abildgaard N, Heickendorf L, Hjorth M, Westin J, Nielsen JL, Hjorth-Hansen H, Waagew A, Sundan A, et al. Serum osteoprotegerin levels are reduced in patients with multiple myeloma with lytic bone disease. Blood. 2001;98:2269–2271. doi: 10.1182/blood.V98.7.2269. [DOI] [PubMed] [Google Scholar]
Oyajobi BO, Franchin G, Williams PJ, Pulkrabek D, Gupta A, Munoz S, Grubbs B, Zhao M, Chen D, Sherry B, Mundy GR. Dual effects of macrophage inflammatory protein-1α on osteolysis and tumor burden in the murine 5TGM1 model of myeloma bone disease. Blood. 2003;102:311–319. doi: 10.1182/blood-2002-12-3905. [DOI] [PubMed] [Google Scholar]
Choi SJ, Cruz JC, Craig F, Chung H, Devlin RD, Roodman GD, Alsina M. Macrophage inflammatory protein-1α is a potential osteoclast stimulatory factor in multiple myeloma. Blood. 2000;96:671–675. [PubMed] [Google Scholar]
Abe M, Hiura K, Wilde J, Moriyama K, Hashimoto T, Ozaki S, Wakatsuki S, Kosaka M, Kido S, Inoue D, et al. Role for macrophage inflammatory protein (MIP)-1α and MIP-1β in the development of osteolytic lesions in multiple myeloma. Blood. 2002;100:2195–2202. [PubMed] [Google Scholar]
Hashimoto T, Abe M, Oshima T, Shibata H, Ozaki S, Inoue D, Matsumoto T. Ability of myeloma cells to secrete macrophage inflammatory protein (MIP)-1alpha and MIP-1beta correlates with lytic bone lesions in patients with multiple myeloma. Br J Haematol. 2004;125:38–41. doi: 10.1111/j.1365-2141.2004.04864.x. [DOI] [PubMed] [Google Scholar]
Uneda S, Hata H, Matsuno F, Harada N, Mitsuya Y, Kawano F, Mitsuya H. Macrophage inflammatory protein-1α is produced by human multiple myeloma (MM) cells and its expression correlates with bone lesions in patients with MM. Br J Haematol. 2003;120:53–55. doi: 10.1046/j.1365-2141.2003.04040.x. [DOI] [PubMed] [Google Scholar]
Terpos E, Politou M, Szydlo R, Goldman JM, Apperley JF, Rahemtulla A. Serum levels of macrophage inflammatory protein-1 alpha (MIP-1 alpha) correlate with the extent of bone disease and survival in patients with multiple myeloma. Br J Haematol. 2003;123:106–109. doi: 10.1046/j.1365-2141.2003.04561.x. [DOI] [PubMed] [Google Scholar]
Taichman RS, Reilly MJ, Matthew LS. Human osteoblast-like cells and osteosarcoma cell lines synthesize macrophage inhibitory protein 1α in response to interleukin 1β and tumour necrosis factor-α stimulation in vitro. Br J Haematol. 2000;108:275–283. doi: 10.1046/j.1365-2141.2000.01873.x. [DOI] [PubMed] [Google Scholar]
Lichtenstein A, Berenson J, Norman D, Chang MP, Carlile A. Production of cytokines by bone marrow cells obtained from patients with multiple myeloma. Blood. 1989;74:1266–1273. [PubMed] [Google Scholar]
Oba Y, Lee JW, Ehrlich LA, Chung HY, Jelinek DF, Callander NS, Horuk R, Choi SJ, Roodman GD. MIP-1α utilizes both CCR1 and CCR5 to induce osteoclast formation and increase adhesion of myeloma cells to marrow stromal cells. Exp Hematol. 2005;33:272–278. doi: 10.1016/j.exphem.2004.11.015. [DOI] [PubMed] [Google Scholar]
Menu E, De Leenheer E, De Raeve H, Coulton L, Imanishi T, Miyashita K, Van Valckenborgh E, Van Riet I, Van Camp B, Horuk R, et al. Role of CCR1 and CCR5 in homing and growth of multiple myeloma and in the development of osteolytic lesions: a study in the 5TMM model. Clin Exp Metastasis. 2006;23:291–300. doi: 10.1007/s10585-006-9038-6. [DOI] [PubMed] [Google Scholar]
Okada Y, Tsukada J, Nakano K, Tonai S, Mine S, Tanaka Y. Macrophage inflammatory protein-1α induces hypercalcemia in adult T-cell leukemia. J Bone Miner Res. 2004;19:1105–1111. doi: 10.1359/JBMR.040314. [DOI] [PubMed] [Google Scholar]
Tian E, Zhan F, Walker R, Rasmussen E, Ma Y, Barlogie B, Shaughnessy JD., Jr The role of the Wnt-signaling antagonist DKK1 in the development of osteolytic lesions in multiple myeloma. N Engl J Med. 2003;349:2483–2494. doi: 10.1056/NEJMoa030847. [DOI] [PubMed] [Google Scholar]
Spencer GJ, Utting JC, Etheridge SL, Arnett TR, Genever PG. Wnt signalling in osteoblasts regulates expression of the receptor activator of NF-κB ligand and inhibits osteoclastogenesis in vitro. J Cell Sci. 2006;119:1283–1296. doi: 10.1242/jcs.02883. [DOI] [PubMed] [Google Scholar]
Oyajobi BO, Gupta A, Flores A, Wideman C, Shipman C, Mundy GR, Garrett IR. Dkk1 modulates osteoclastogenesis and bone resorption: Implications for myeloma bone disease. J Bone Miner Res. 2005. p. S32.
Yaccoby S, Ling W, Zhan F, Walker R, Barlogie B, Shaughnessy JD., Jr Antibody-based inhibition of DKK1 suppresses tumor-induced bone resorption and multiple myeloma growth in vivo. Blood. 2007;109:2106–2011. doi: 10.1182/blood-2006-09-047712. [DOI] [PMC free article] [PubMed] [Google Scholar]
Stewart AF. Hypercalcemia associated with cancer. N Engl J Med. 2005;352:373–379. doi: 10.1056/NEJMcp042806. [DOI] [PubMed] [Google Scholar]
Body JJ. Bisphosphonates for malignancy-related bone disease: current status, future developments. Support Care Cancer. 2006;14:408–418. doi: 10.1007/s00520-005-0913-5. [DOI] [PubMed] [Google Scholar]
Bell R. Efficacy of ibandronate in metastatic bone disease: review of clinical data. Oncologist. 2005;(Suppl 1):8–13. doi: 10.1634/theoncologist.10-90001-8. [DOI] [PubMed] [Google Scholar]
Rosen LS, Gordon D, Antonio BS, Kaminski M, Howell A, Belch A, Mackey JA, Apffelstaedt J, Hussein M, Coleman RE, et al. Zoledronic acid versus pamidronate in the treatment of skeletal metastases in patients with breast cancer or osteolytic lesions of multiple myeloma: a phase III, double-blind, comparative trial. Cancer J. 2001;7:377–387. [PubMed] [Google Scholar]
Morony S, Warmington K, Adamu S, Asuncion F, Geng Z, Grisanti M, Tan HL, Capparelli C, Starnes C, Weimann B, et al. The inhibition of RANKL causes greater suppression of bone resorption and hypercalcemia compared with bisphosphonates in two models of humoral hypercalcemia of malignancy. Endocrinology. 2005;146:3235–3243. doi: 10.1210/en.2004-1583. [DOI] [PubMed] [Google Scholar]
Body JJ, Facon T, Coleman RE, Lipton A, Geurs F, Fan M, Holloway D, Peterson MC, Bekker PJ. A study of the biological receptor activator of nuclear factor-κB ligand inhibitor, denosumab, in patients with multiple myeloma or bone metastases from breast cancer. Clin Cancer Res. 2006;12:1221–1228. doi: 10.1158/1078-0432.CCR-05-1933. [DOI] [PubMed] [Google Scholar]
ClinicalTrials.gov
Liang M, Mallari C, Rosser M, Ng HP, May K, Monahan S, Bauman JG, Islam I, Ghannam A, Buckman B, et al. Identification and characterization of a potent, selective and orally active antagonist of the CC chemokine receptor-1. J Biol Chem. 2000;275:19000–19008. doi: 10.1074/jbc.M001222200. [DOI] [PubMed] [Google Scholar]
Nair SR, Pearson SB. Images in clinical medicine. Pathologic fracture and lytic lesions in multiple myeloma. N Engl J Med. 2004;351:1874. doi: 10.1056/NEJMicm031146. [DOI] [PubMed] [Google Scholar]
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Abstract
Introduction
Hypercalcemia in myeloma
Myeloma bone disease
RANKL
Macrophage inflammatory protein-1α
Dickkopf 1
Diagnosis and treatment of hypercalcemia
Conclusion
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188513 | https://psycnet.apa.org/record/1997-07227-008 | Interoceptive exposure versus breathing retraining within cognitive-behavioural therapy for panic disorder with agoraphobia.
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Interoceptive exposure versus breathing retraining within cognitive-behavioural therapy for panic disorder with agoraphobia.
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Craske, M. G., Rowe, M., Lewin, M., & Noriega-Dimitri, R. (1997). Interoceptive exposure versus breathing retraining within cognitive-behavioural therapy for panic disorder with agoraphobia. British Journal of Clinical Psychology, 36(1), 85–99.
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This study compared 2 components of a cognitive-behavioral treatment for panic disorder. 38 adults (aged 22–48 yrs) with panic disorder and agoraphobia were randomly assigned to (1) cognitive restructuring, interoceptive exposure (i.e., repeated exposure to feared bodily sensations) and exposure to agoraphobic situations; or (2) cognitive restructuring, breathing retraining and exposure to agoraphobic situations. Assessments were conducted at pre-treatment, post-treatment and 6 mo later. The treatments were equally effective on many measures. However, treatment that included interoceptive exposure was more effective at post-treatment in terms of panic frequency, overall severity and functioning, and more effective at 6 mo follow-up in terms of panic frequency, phobic fears and general anxiety and functioning. (PsycINFO Database Record (c) 2016 APA, all rights reserved)
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188514 | https://math.stackexchange.com/questions/2230955/is-this-1811-proof-for-the-negative-pell-equation-really-wrong | elementary number theory - Is this $1811$ proof for the negative Pell equation really wrong? - Mathematics Stack Exchange
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Is this 1811 1811 proof for the negative Pell equation really wrong?
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Peter Barlow was a well-known British mathematician during the 19 19 th century and he produced the very first English number-theoretic text called "An Elementary Investigation of the Theory of Numbers" in 1811 1811. This text contained some notable errors (e.g: an incorrect proof of Fermat's Last Theorem and several false theorems on perfect numbers) but it was still influential and was used as a reference guide.
In this text, I just discovered that he claimed to have solved the negative Pell's equation and gave a proof for it. Of course, he was most likely wrong since we still today don't know when it is solvable. His argument seems somewhat intricate and it would be extremely interesting to know where he went wrong.
He states the theorem as follows;
If a a be a prime number of the form 4 n+1 4 n+1, the equation x 2−a y 2=−1 x 2−a y 2=−1 is always resolvible in integers.
and then goes on to prove it in the following manner:
(Barlow uses the symbol "ⵐ" to mean "is of the form".)
Let p p and q q be least values (except 1 1 and 0 0) that satisfy the equation
p 2−a q 2=1 or p 2−(4 n+1)q 2=1;p 2−a q 2=1 or p 2−(4 n+1)q 2=1;
then it is obvious that q q must be even, for if it was odd, q 2 q 2 would be of the form 8 n+1 8 n+1, and
a q 2+1 ⵐ(4 n+1)⋅(8 n′+1)+1 ⵐ 4 n′′+2,a q 2+1 ⵐ(4 n+1)⋅(8 n′+1)+1 ⵐ 4 n″+2,
which cannot be a square: since, then, q q must be even, let us make q=2 m n q=2 m n, m m and n n being integers prime to each other; then we shall have p 2−1=4 a m 2 n 2 p 2−1=4 a m 2 n 2; but p p being odd, and consequently, p 2−1 ⵐ 8 n′p 2−1 ⵐ 8 n′, it follows, that either m m or n n is even, and the other odd, for otherwise we should not have p 2−1 p 2−1 ⵐ 8 n′8 n′, and they cannot be both even, because they are prime to eachother.
Let us therefore suppose n n to be odd, then the equation
(p+1)(p−1)=4 a m 2 n 2,(p+1)(p−1)=4 a m 2 n 2,
in which the factors p+1 p+1, and p−1 p−1, can have only the common measure 2 2 (and this must necessarily have place, because p p is odd), will be resolvible into the four following forms:
p+1=2 a m 2 p+1=2 a m 2&p−1=2 n 2 p−1=2 n 2
p+1=2 m 2 p+1=2 m 2&p−1=2 a n 2 p−1=2 a n 2
p+1=2 a n 2 p+1=2 a n 2&p−1=2 m 2 p−1=2 m 2
p+1=2 n 2 p+1=2 n 2&p−1=2 a m 2 p−1=2 a m 2
Now the second and fourth of these forms give
m 2−a n 2=1,or n 2−a m 2=1;m 2−a n 2=1,or n 2−a m 2=1;
which equations cannot have place, because m m and n n are less than p p and q q; and these last were the least that satisfied the equation p 2−a q 2=1 p 2−a q 2=1. There remain, therefore, only the first and third which give
n 2−a m 2=−1,or m 2−a n 2=−1 n 2−a m 2=−1,or m 2−a n 2=−1
and one of these equations must necessarily obtain; but either of them resolves the equation
x 2−a y 2=−1 x 2−a y 2=−1
which is, therefore, always possible, when a a is a prime number of the form 4 n+1 4 n+1. It may also be observed, that, of the above two equations, the last is the only one that can obtain; for, since n n is odd and m m even, it is evident that the first cannot become equal to −1−1.
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edited Jun 11, 2020 at 18:01
mrtaurho
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asked Apr 12, 2017 at 16:25
KalisKalis
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3 These arguments are due to Legendre, and there's nothing wrong with them. I have called this technique descent on Pell conics.user23365 –user23365 2017-04-12 16:47:46 +00:00 Commented Apr 12, 2017 at 16:47
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Your sentence
Of course, he was most likely wrong since we still today don't know when it is solvable
does not correctly apply, as the negative Pell question is known for prime values of a a in x 2−a y 2.x 2−a y 2.
There is a short proof in Mordell's book that for prime p≡1(mod 4),p≡1(mod 4), there is always a solution to
x 2−p y 2=−1.x 2−p y 2=−1.
The same proof does this much: if we have primes p≡q≡1(mod 4),p≡q≡1(mod 4), while we have Legendre symbols (p|q)=(q|p)=−1,(p|q)=(q|p)=−1, there is always a solution to
x 2−p q y 2=−1.x 2−p q y 2=−1.
When (p|q)=(q|p)=1(p|q)=(q|p)=1 for p≡q≡1(mod 4),p≡q≡1(mod 4), it can go either way. The smallest examples of failure are x 2−205 y 2≠−1 x 2−205 y 2≠−1 and x 2−221 y 2≠−1 x 2−221 y 2≠−1
See both answers at Given p≡q≡1(mod 4)p≡q≡1(mod 4), (p q)=1(p q)=1, is N(η)=1 N(η)=1 possible?
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edited Apr 13, 2017 at 12:20
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answered Apr 12, 2017 at 16:35
Will JagyWill Jagy
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4Given p≡q≡1(mod 4)p≡q≡1(mod 4), (p q)=1(p q)=1, is N(η)=1 N(η)=1 possible?
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3Proof — Infinitely many primes of the form 4 k+3 4 k+3 — origin of 4(p 1...p k−1)+3 4(p 1...p k−1)+3
3Why is this Goldbach's Conjecture Proof Wrong?
3Integers d d for which the Negative Pell equation is soluble for both d d and 2 d 2 d?
6Why doesn’t this logic work? (Where did I go wrong?)
1How is this a Pell-like equation?
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188515 | https://www.schoolphysics.co.uk/age16-19/Mechanics/Simple%20harmonic%20motion/text/Simple_harmonic_motion_formula_proofs/index.html | schoolphysics ::Welcome::
HOME>AGE 16-19>MECHANICS>SIMPLE HARMONIC MOTION> SHM EQUATIONS - PROOFS
Proof of the formulae for s.h.m. (1)
There are two alternative proofs. We will give the mathematical one first because it is the more satisfactory of the two.
Simple harmonic motion is defined by the equation: acceleration = -ω 2 x or dv/dt = -ω 2 x
Therefore: v dv/dx = - ω 2 x and when integrated this becomes v 2/2 = -ω 2 x 2/2 + C
Using the limits V = 0 when x = r, and v = rω when x = 0, gives C = ω 2 r 2/2.
Therefore:
Velocity (v) = +/-ω(r 2 – x 2)1/2
Note the +/- sign, showing that the velocity can be in either direction about the midpoint.
Therefore v = dx/dt = +/- ωv (r 2 – x 2) which when integrated becomes
Displacement (x) = r sin(ωt)
with the limits x = +/- r when t = π/2ω or 3π/2ω and x = 0 when t = 0, π/ω or 2θ/ω
Proof of the formulae for s.h.m. (2)
The next proof is slightly less mathematical, and uses the projection on a diameter of a circle of the motion of a point P round the circle. Let the velocity of the particle round that circle be U and the velocity of the projection on AB be v (Figure 1). We wish to show that the motion of the projection of P on AB is s.h.m.
The displacement (x) of the projection of P from O along AB is given by
x = r sinθ = r sin ωt
Consider the component of velocity of P parallel to AB: this will be the velocity of the projection of P on AB.
Component parallel to AB = u cos ωt = rω cos ωt = rω(1 – sin 2 ωt)1/2 = rω(1-x 2/r 2)1/2.
Therefore:
Velocity (v) = +/-ω(r 2 – x 2)1/2
Now the acceleration of P will be towards the centre of the circle and therefore its component parallel to AB will be:
acceleration = -ω 2 r sin ωt = - ω 2 x
This is s.h.m., since we have shown that the acceleration of Q is directed towards O and proportional to OQ.
schoolphysics SHM and circular motion
To see an animation of the connection between shm and circular motion click on the animation link.
A VERSION IN WORD IS AVAILABLE ON THE SCHOOLPHYSICS CD
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Why is the hydrolysis of the conjugate base of a weak acid neglected in buffer solutions?
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Consider a solution which contains a weak acid and the salt of its conjugate base with a strong base. e.g.
C H X 3 C O O H+H X 2 O↽−−⇀K a C H X 3 C O O X−+H X 3 O X+(1)(1)C H X 3 C O O H+H X 2 O↽−−⇀K a C H X 3 C O O X−+H X 3 O X+
C H X 3 C O O X−+H X 2 O↽−−⇀K h C H X 3 C O O H+O H X−(2)(2)C H X 3 C O O X−+H X 2 O↽−−⇀K h C H X 3 C O O H+O H X−
Now, K a=[H X 3 O X+][C H X 3 C O O X−][C H X 3 C O O H]K a=[H X 3 O X+][C H X 3 C O O X−][C H X 3 C O O H] and K h=K w K a K h=K w K a.
Now, my notes say that for the equilibrium concentration of undissociated acid, the acid coming due to hydrolysis of the salt is neglected and the equilibrium concentration is approximately equal to the initial concentration.
It is also mentioned in a book that
Assume that the extent of protonation of acetate ions and the deprotonation of acetic acid molecules is so small that the concentrations of both species are nearly the same as their initial values.
Why is this so? If K a K a is low, then K h K h should be high. I understand that the deprotonation of acid will be low as it is a weak acid so that we can neglect the C H X 3 C O O X−C H X 3 C O O X− coming through the acid dissociation but how can we neglect the hydrolysis of the C H X 3 C O O X−C H X 3 C O O X− coming from the salt?
Edit −−
My calculation for equilibrium concentrations.
[C H X 3 C O O H]X e q=[C H X 3 C O O H]X e q= initially added acid −− deprotonated acid ++ hydrolysed salt
[C H X 3 C O O X−]X e q=[C H X 3 C O O X−]X e q= initially added salt −− hydrolysed salt ++ deprotonated acid
[H X 3 O X+]e q=[H X 3 O X+]e q= deprotonated acid −− hydrolysed salt
Let the concentration of
initially added acid =c 1=c 1
initially added salt =c 2=c 2
deprotonated acid at eq. =x=x
hydrolysed salt at eq. =y=y
Thus,
K a=[H X 3 O X+][C H X 3 C O O X−][C H X 3 C O O H]=(x−y)(c 2−y+x)(c 1−x+y)K a=[H X 3 O X+][C H X 3 C O O X−][C H X 3 C O O H]=(x−y)(c 2−y+x)(c 1−x+y)
Now, assuming x<<c 1,c 2 x<<c 1,c 2,
K a=(−y)(c 2−y)(c 1+y)K a=(−y)(c 2−y)(c 1+y)
physical-chemistry
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edited Sep 21, 2017 at 10:37
Apoorv PotnisApoorv Potnis
asked Sep 21, 2017 at 5:00
Apoorv PotnisApoorv Potnis
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1 You may take it into account and compare the results.Ivan Neretin –Ivan Neretin 2017-09-21 05:10:00 +00:00 Commented Sep 21, 2017 at 5:10
1 @IvanNeretin Please see my edited question.Apoorv Potnis –Apoorv Potnis 2017-09-21 05:54:50 +00:00 Commented Sep 21, 2017 at 5:54
You said before that you neglect y y, not x x.Ivan Neretin –Ivan Neretin 2017-09-21 06:15:13 +00:00 Commented Sep 21, 2017 at 6:15
1 @IvanNeretin I said that I understand why deprotonation is neglected, i.e. why x x is neglected. I don't understand why the hydrolysis, i.e. y y is neglected.Apoorv Potnis –Apoorv Potnis 2017-09-21 06:21:24 +00:00 Commented Sep 21, 2017 at 6:21
Hydrolysis = cleavage of chem. bonds by addition of water (en.wikipedia.org/wiki/Hydrolysis). Where is there a hydrolysis here? How can acetic acid be hydrolyzed?logical x 2 –logical x 2 2017-09-21 10:40:26 +00:00 Commented Sep 21, 2017 at 10:40
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p K a=4.75 p K a=4.75, therefore K a=10−4.75 K a=10−4.75. Thus,
K a=[A c O X−][H X+][A c O H]K a=[A c O X−][H X+][A c O H]
So, let's assume [A c O H]=1.0 M[A c O H]=1.0 M. Just to get a feel, how far the acid is dissociated:
K a=x⋅x 1.0 M→x=10−4.75−−−−−−√M=0.00422≈0.4%K a=x⋅x 1.0 M→x=10−4.75 M=0.00422≈0.4%
where we have assumed that the contribution from the dissociation of water is negligible, and therefore [H X+]=[A c O X−]=x[H X+]=[A c O X−]=x.
Now, what happens if I add 1.0 M 1.0 M A c O X−A c O X−? Then, effectively, we have:
K a=[A c O X−][H X+][A c O H]=1.0 M⋅[H X+]1.0 M K a=[A c O X−][H X+][A c O H]=1.0 M⋅[H X+]1.0 M
Therefore, we have:
[H X+]=K a 1.0 M 1.0 M=10−4.75 M[H X+]=K a 1.0 M 1.0 M=10−4.75 M
If we compare the two cases, before addition of acetate, we had a p H p H of
[H X+]=10−4.75−−−−−−√→p H=2.375[H X+]=10−4.75→p H=2.375
and after addition of acetate, we have
[H X+]=10−4.75→p H=4.75[H X+]=10−4.75→p H=4.75
Meaning that the solution got more basic after the addition of acetate, as excpected.
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edited Sep 21, 2017 at 13:13
andselisk♦
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answered Sep 21, 2017 at 12:31
logical x 2logical x 2
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Hopefully that provides a good starting point for understanding.logical x 2 –logical x 2 2017-09-21 12:35:20 +00:00 Commented Sep 21, 2017 at 12:35
@deuxexmachina After you added 1.0 M acetate ions, how did you take equilibrium concentration to be 1.0 M? How is initial concentration equal to equilibrium concentration?Apoorv Potnis –Apoorv Potnis 2017-09-24 04:51:43 +00:00 Commented Sep 24, 2017 at 4:51
1 @ApoorvPotnis Yeah, I figured that this point wasn't really clear yet. So first of all, you can neglect that the acid CH3COOH will dissociate to CH3COO-, as by the above argument, it only dissociates to a very little extent. However, your concern is the other way around, and you argue that CH3COO-Na+ could be protonated to form CH3COOH + NaOH. But: look at the pKa of NaOH and CH3OO-Na: NaOH is a much stronger base! Therefore, you can assume that [N a C H 3 C O O]=[N a C H 3 C O O]0[N a C H 3 C O O]=[N a C H 3 C O O]0.logical x 2 –logical x 2 2017-09-24 07:38:07 +00:00 Commented Sep 24, 2017 at 7:38
@ApoorvPotnis You can also find a similar description here: study.com/academy/lesson/… for a similar explanation logical x 2 –logical x 2 2017-09-24 07:40:18 +00:00 Commented Sep 24, 2017 at 7:40
Can you add your comment about N a O H N a O H being a stronger base into the answer so that I could accept it?Apoorv Potnis –Apoorv Potnis 2017-11-06 12:50:20 +00:00 Commented Nov 6, 2017 at 12:50
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Introduction to Sociology 3e
Section Summary
Introduction to Sociology 3eSection Summary
Contents Contents
Highlights
Table of contents
Preface
1 An Introduction to Sociology
2 Sociological Research
3 Culture
4 Society and Social Interaction
5 Socialization
6 Groups and Organization
7 Deviance, Crime, and Social Control
8 Media and Technology
9 Social Stratification in the United States
10 Global Inequality
11 Race and Ethnicity
12 Gender, Sex, and Sexuality
13 Aging and the Elderly
14 Relationships, Marriage, and Family
15 Religion
16 Education
17 Government and Politics
18 Work and the Economy
19 Health and Medicine
20 Population, Urbanization, and the Environment
Introduction
20.1 Demography and Population
20.2 Urbanization
20.3 The Environment and Society
Key Terms
Section Summary
Section Quiz
Short Answer
Further Research
References
21 Social Movements and Social Change
Answer Key
Index
Search for key terms or text.
Close
20.1 Demography and Population ------------------------------
Scholars understand demography through various analyses. Malthusian, zero population growth, cornucopian theory, and demographic transition theories all help sociologists study demography. The earth’s human population is growing quickly, especially in peripheral countries. Factors that impact population include birthrates, mortality rates, and migration, including immigration and emigration. There are numerous potential outcomes of the growing population, and sociological perspectives vary on the potential effect of these increased numbers. The growth will pressure the already taxed planet and its natural resources.
20.2 Urbanization -----------------
Cities provide numerous opportunities for their residents and offer significant benefits including access to goods to numerous job opportunities. At the same time, high population areas can lead to tensions between demographic groups, as well as environmental strain. While the population of urban dwellers is continuing to rise, sources of social strain are rising along with it. The ultimate challenge for today’s urbanites is finding an equitable way to share the city’s resources while reducing the pollution and energy use that negatively impacts the environment.
20.3 The Environment and Society --------------------------------
The area of environmental sociology is growing as extreme weather patterns and concerns over climate change increase. Human activity leads to pollution of soil, water, and air, compromising the health of the entire food chain. While everyone is at risk, poor and disadvantaged neighborhoods and nations bear a greater burden of the planet’s pollution, a dynamic known as environmental racism.
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188518 | https://en.wikipedia.org/wiki/Refactorable_number | Jump to content
Refactorable number
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From Wikipedia, the free encyclopedia
Integer divisible by the number of its divisors
"Tau number" redirects here. For the ratio of a circle's circumference to its radius, see Turn (angle) § Proposals for a single letter to represent 2π.
A refactorable number or tau number is an integer n that is divisible by the count of its divisors, or to put it algebraically, n is such that with for . The first few refactorable numbers are listed in (sequence A033950 in the OEIS) as
: 1, 2, 8, 9, 12, 18, 24, 36, 40, 56, 60, 72, 80, 84, 88, 96, 104, 108, 128, 132, 136, 152, 156, 180, 184, 204, 225, 228, 232, 240, 248, 252, 276, 288, 296, ...
For example, 18 has 6 divisors (1 and 18, 2 and 9, 3 and 6) and is divisible by 6. There are infinitely many refactorable numbers.
Properties
[edit]
Cooper and Kennedy proved that refactorable numbers have natural density zero. Zelinsky proved that no three consecutive integers can all be refactorable. Colton proved that no refactorable number is perfect. The equation has solutions only if is a refactorable number, where is the greatest common divisor function.
Let be the number of refactorable numbers which are at most . The problem of determining an asymptotic for is open. Spiro has proven that
There are still unsolved problems regarding refactorable numbers. Colton asked if there are arbitrarily large such that both and are refactorable. Zelinsky wondered if there exists a refactorable number , does there necessarily exist such that is refactorable and .
History
[edit]
First defined by Curtis Cooper and Robert E. Kennedy where they showed that the tau numbers have natural density zero, they were later rediscovered by Simon Colton using a computer program he wrote ("HR") which invents and judges definitions from a variety of areas of mathematics such as number theory and graph theory. Colton called such numbers "refactorable". While computer programs had discovered proofs before, this discovery was one of the first times that a computer program had discovered a new or previously obscure idea. Colton proved many results about refactorable numbers, showing that there were infinitely many and proving a variety of congruence restrictions on their distribution. Colton was only later alerted that Kennedy and Cooper had previously investigated the topic.
See also
[edit]
Wikifunctions has a refactorable number checking function.
Divisor function
References
[edit]
^ J. Zelinsky, "Tau Numbers: A Partial Proof of a Conjecture and Other Results," Journal of Integer Sequences, Vol. 5 (2002), Article 02.2.8
^ Spiro, Claudia (1985). "How often is the number of divisors of n a divisor of n?". Journal of Number Theory. 21 (1): 81–100. doi:10.1016/0022-314X(85)90012-5.
^ Cooper, C.N. and Kennedy, R. E. "Tau Numbers, Natural Density, and Hardy and Wright's Theorem 437." Internat. J. Math. Math. Sci. 13, 383-386, 1990
^ S. Colton, "Refactorable Numbers - A Machine Invention," Journal of Integer Sequences, Vol. 2 (1999), Article 99.1.2
| v t e Classes of natural numbers | |
--- |
| | Powers and related numbers | | --- | | Achilles Power of 2 Power of 3 Power of 10 Square Cube Fourth power Fifth power Sixth power Seventh power Eighth power Perfect power Powerful Prime power | | | |
| | Of the form a × 2b ± 1 | | --- | | Cullen Double Mersenne Fermat Mersenne Proth Thabit Woodall | | | |
| | Other polynomial numbers | | --- | | Hilbert Idoneal Leyland Loeschian Lucky numbers of Euler | | | |
| | Recursively defined numbers | | --- | | Fibonacci Jacobsthal Leonardo Lucas Narayana Padovan Pell Perrin | | | |
| | Possessing a specific set of other numbers | | --- | | Amenable Congruent Knödel Riesel Sierpiński | | | |
| | Expressible via specific sums | | --- | | Nonhypotenuse Polite Practical Primary pseudoperfect Ulam Wolstenholme | | | |
| | Figurate numbers | | --- | | | | | | | | | --- --- --- | | 2-dimensional | | | | --- | | centered | Centered triangular Centered square Centered pentagonal Centered hexagonal Centered heptagonal Centered octagonal Centered nonagonal Centered decagonal Star | | non-centered | Triangular Square Square triangular Pentagonal Hexagonal Heptagonal Octagonal Nonagonal Decagonal Dodecagonal | | | 3-dimensional | | | | --- | | centered | Centered tetrahedral Centered cube Centered octahedral Centered dodecahedral Centered icosahedral | | non-centered | Tetrahedral Cubic Octahedral Dodecahedral Icosahedral Stella octangula | | pyramidal | Square pyramidal | | | 4-dimensional | | | | --- | | non-centered | Pentatope Squared triangular Tesseractic | | | | | |
| | Combinatorial numbers | | --- | | Bell Cake Catalan Dedekind Delannoy Euler Eulerian Fuss–Catalan Lah Lazy caterer's sequence Lobb Motzkin Narayana Ordered Bell Schröder Schröder–Hipparchus Stirling first Stirling second Telephone number Wedderburn–Etherington | | | |
| | Primes | | --- | | Wieferich Wall–Sun–Sun Wolstenholme prime Wilson | | | |
| | Pseudoprimes | | --- | | Carmichael number Catalan pseudoprime Elliptic pseudoprime Euler pseudoprime Euler–Jacobi pseudoprime Fermat pseudoprime Frobenius pseudoprime Lucas pseudoprime Lucas–Carmichael number Perrin pseudoprime Somer–Lucas pseudoprime Strong pseudoprime | | | |
| | Arithmetic functions and dynamics | | --- | | | | | --- | | Divisor functions | Abundant Almost perfect Arithmetic Betrothed Colossally abundant Deficient Descartes Hemiperfect Highly abundant Highly composite Hyperperfect Multiply perfect Perfect Practical Primitive abundant Quasiperfect Refactorable Semiperfect Sublime Superabundant Superior highly composite Superperfect | | Prime omega functions | Almost prime Semiprime | | Euler's totient function | Highly cototient Highly totient Noncototient Nontotient Perfect totient Sparsely totient | | Aliquot sequences | Amicable Perfect Sociable Untouchable | | Primorial | Euclid Fortunate | | | | |
| | Other prime factor or divisor related numbers | | --- | | Blum Cyclic Erdős–Nicolas Erdős–Woods Friendly Giuga Harmonic divisor Jordan–Pólya Lucas–Carmichael Pronic Regular Rough Smooth Sphenic Størmer Super-Poulet | | | |
| | Numeral system-dependent numbers | | --- | | | | | | | | | | | | | --- --- --- --- --- | | Arithmetic functions and dynamics | Persistence + Additive + Multiplicative | | | --- | | Digit sum | Digit sum Digital root Self Sum-product | | Digit product | Multiplicative digital root Sum-product | | Coding-related | Meertens | | Other | Dudeney Factorion Kaprekar Kaprekar's constant Keith Lychrel Narcissistic Perfect digit-to-digit invariant Perfect digital invariant + Happy | | | P-adic numbers-related | Automorphic + Trimorphic | | Digit-composition related | Palindromic Pandigital Repdigit Repunit Self-descriptive Smarandache–Wellin Undulating | | Digit-permutation related | Cyclic Digit-reassembly Parasitic Primeval Transposable | | Divisor-related | Equidigital Extravagant Frugal Harshad Polydivisible Smith Vampire | | Other | | | | | |
| | Binary numbers | | --- | | Evil Odious Pernicious | | | |
| | Generated via a sieve | | --- | | Lucky Prime | | | |
| | Sorting related | | --- | | Pancake number Sorting number | | | |
| | Natural language related | | --- | | Aronson's sequence Ban | | | |
| | Graphemics related | | --- | | | | | |
| Mathematics portal | |
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188519 | https://www.sciencing.com/one-fraction-larger-another-fraction-2320868/ | How To Know When One Fraction Is Larger Than Another Fraction
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How To Know When One Fraction Is Larger Than Another Fraction
ScienceMathTechnologyFeatures
Math
How To Know When One Fraction Is Larger Than Another Fraction
By Contributor Updated Mar 24, 2022
When students take math exams, they need to know when one fraction is greater than another one. This is especially true in a subtraction problem when the smaller fraction needs to be subtracted from the larger fraction. Gauging fractions also comes in handy when several fractions must be placed from the least to the greatest or from the largest to the smallest.
Step 1
Choose a couple of fractions to work with. For example, consider 6/11 and 5/9. Take the denominator of the second fraction, 9, and multiply it by the numerator of the first fraction, 6. The product is 54. Write this number above the first fraction.
Step 2
Take the denominator of the first fraction, 11, and multiply it by the numerator of the second fraction, 5. The product is 55. Write that number above the second fraction.
Step 3
Compare the numbers you have written above the fractions. Because 55 is larger than 54, the second fraction, 5/9, is larger than the first fraction, 6/11.
Step 4
Apply this technique to any two fractions A/B and C/D, such that A, B, C and D are whole numbers, each greater than zero. If the product of A x D is greater than the product of C x B, the fraction A/B is larger than C/D. Similarly, if the product of A x D is less than the product of C x B, the fraction A/B is smaller than the fraction C/D.
TL;DR (Too Long; Didn't Read)
Another way to illustrate this is to find the common denominator of the two fractions. In the case of 6/11 and 5/9, the common denominator is 99 (9 x 11). Multiply the numerator and denominator of 6/11 by 9 to get 54/99, and multiply the numerator and denominator of 5/9 by 11 to get 55/99. This shows that 55/99, or 5/9, is greater than 54/99, or 6/11.
References
Augusta Technical College: Comparing Fractions
Cite This Article
MLA
Contributor, . "How To Know When One Fraction Is Larger Than Another Fraction" sciencing.com, 24 April 2017.
APA
Contributor, . (2017, April 24). How To Know When One Fraction Is Larger Than Another Fraction. sciencing.com. Retrieved from
Chicago
Contributor, . How To Know When One Fraction Is Larger Than Another Fraction last modified March 24, 2022.
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188520 | https://pmc.ncbi.nlm.nih.gov/articles/PMC8494521/ | 2020 WHO Classification of Female Genital Tumors - PMC
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. 2021 Oct 6;81(10):1145–1153. doi: 10.1055/a-1545-4279
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2020 WHO Classification of Female Genital Tumors
WHO-Klassifikation 2020 für Tumoren des unteren weiblichen Genitales
Anne Kathrin Höhn
Anne Kathrin Höhn
1 Arbeitsgruppe Mamma, Gynäko- & Perinatalpathologie, Institut für Pathologie, Universitätsklinikum Leipzig AöR, Leipzig
Find articles by Anne Kathrin Höhn
1,✉, Christine E Brambs
Christine E Brambs
2 Frauenklinik, Kantonsspital, Luzern, Schweiz
Find articles by Christine E Brambs
2, Grit Gesine Ruth Hiller
Grit Gesine Ruth Hiller
1 Arbeitsgruppe Mamma, Gynäko- & Perinatalpathologie, Institut für Pathologie, Universitätsklinikum Leipzig AöR, Leipzig
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1, Doris May
Doris May
3 Pathologisches Institut der Ludwig-Maximilians-Universität München, München
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Elisa Schmoeckel
3 Pathologisches Institut der Ludwig-Maximilians-Universität München, München
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3, Lars-Christian Horn
Lars-Christian Horn
1 Arbeitsgruppe Mamma, Gynäko- & Perinatalpathologie, Institut für Pathologie, Universitätsklinikum Leipzig AöR, Leipzig
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1 Arbeitsgruppe Mamma, Gynäko- & Perinatalpathologie, Institut für Pathologie, Universitätsklinikum Leipzig AöR, Leipzig
2 Frauenklinik, Kantonsspital, Luzern, Schweiz
3 Pathologisches Institut der Ludwig-Maximilians-Universität München, München
✉
Correspondence/KorrespondenzadresseDr. med. Anne Kathrin Höhn Arbeitsgruppe Mamma-, Gynäko- & Perinatalpathologie, Institut für Pathologie, Universitätsklinikum Leipzig AöR, Liebigstraße 26, 04103 Leipzig, Deutschland, Email: annekathrin.hoehn@medizin.uni-leipzig.de
Received 2021 Apr 16; Accepted 2021 Jul 5; Issue date 2021 Oct.
The Author(s). This is an open access article published by Thieme under the terms of the Creative Commons Attribution-NonDerivative-NonCommercial License, permitting copying and reproduction so long as the original work is given appropriate credit. Contents may not be used for commecial purposes, or adapted, remixed, transformed or built upon. ( )
This is an open-access article distributed under the terms of the Creative Commons Attribution-NonCommercial-NoDerivatives License, which permits unrestricted reproduction and distribution, for non-commercial purposes only; and use and reproduction, but not distribution, of adapted material for non-commercial purposes only, provided the original work is properly cited.
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PMCID: PMC8494521 PMID: 34629493
Abstract
The 2020 WHO classification is focused on the distinction between HPV-associated and HPV-independent squamous cell carcinoma of the lower female genital organs. Differentiating according to HPV association does not replace the process of grading; however, the WHO classification does not recommend any specific grading system. VIN are also differentiated according to whether they are HPV(p16)-associated. HPV-independent adenocarcinoma (AC) of the cervix uteri has an unfavorable prognosis. Immunohistochemical p16 expression is considered to be a surrogate marker for HPV association. HPV-associated AC of the cervix uteri is determined using the prognostically relevant Silva pattern.
Key words: p16 immunohistochemistry, HPV, Silva pattern
Introduction
The WHO Classification of Female Genital Tumors is the fourth volume in the fifth edition of the WHO series on the classification of human tumors, and was fundamentally revised in 2020 due to new histomorphological data and, in particular, molecular pathology data. In comparison with the 2013 edition, the scope of the work has doubled in length. The WHO classification is now primarily constructed on the basis of new (molecular) pathology data. However, data of therapeutic and diagnostic relevance, insofar as they are available, are certainly also incorporated. The WHO Blue Books represent an important foundation for a globally uniform diagnostic standard. In Germany, the guidelines of the AWMF (Association of the Scientific Medical Societies) recommend using the current WHO classification for the pathological findings report. The obligation to use the diagnostic histopathological standards and terminology according to the current WHO classification is described in all of the guidelines relating to gynecological malignancies with the verb “should”, this being the highest level of recommendation. While the WHO classification sets out terminology and criteria for diagnosis, the TNM classification of malignant tumors as well as the FIGO classification (Fédération Internationale de Gynécologie et dʼObstétrique) are used for staging, i.e., determining the extent and spread of the gynecological tumors.
The following article summarizes the significant changes of clinical relevance in the current WHO classification for tumors of the female genitals in a way that is relevant in practice.
Vulvar Tumors
With reference to vulvar intraepithelial neoplasias (VIN), the distinction between HPV-associated and HPV-negative neoplasia has been retained. In terms of nomenclature, HPV-associated VIN corresponds to low (VIN 1) and high-grade SIL (VIN 2 and 3; Figs. 1 and 2 ).
Fig. 1.
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Clinicopathological characteristics of the 2020 WHO classification of vulvar precancerous lesions 1 , 2 , 3 , 4 , 12 , 14 .
Fig. 2.
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Precancerous lesions (VIN) and vulvar carcinoma. a HPV-associated VIN (usual VIN; u-VIN), b and c non-keratinizing, HPV-associated squamous cell carcinoma of the vulva with a plump pattern of invasion and p16 positivity (so-called block staining; see text), d HPV-independent VIN (d-VIN), e and f keratinizing squamous cell carcinoma of the vulva with a netlike pattern of invasion and aberrant p53 expression (see text).
In cases of VIN where no HPV is detected, the term HPV-independent VIN has been introduced 1 , 2 ; Fig. 1 showing a variable morphology:
The differentiated VIN with its horizontal spread (d-VIN) is a precursor lesion that is allocated to a more aggressive,
the differentiated exophytic VIN lesion (DEVIL) to a less aggressive (keratinizing) squamous cell carcinoma and the
vulvar acanthosis with altered differentiation (VAAD) is allocated to verrucous carcinoma as a precursor/risk lesion 2 , 3 .
Within this context, it is important to note that different types of lesions may coincide with each other 1 , 2 , 4 . Due to morphological similarities or overlaps 3 , DEVIL and VAAD are also covered by the umbrella term of aberrant maturation of the vulvar squamous epithelium (vulvar aberrant maturation; VAM; 4 ) ( Fig. 1 ), a term that is not included in the WHO classification.
Despite the fact that this distinction as yet lacks diagnostic and therapeutic relevance 5 , 6 , the new WHO classification differentiates between HPV-associated and HPV-independent squamous cell carcinoma due to their different pathogenesis ( Figs. 1 and 2 ), and the WHO recommends supplying this information in the findings report. The ratio of HPV-independent to HPV-associated squamous cell carcinoma is stated to be between 0.60 and 0.83 7 .
Irrespective of clinicopathological differences ( Table 1 ), the HE-morphology does not allow for reliable differentiation between HPV-associated and HPV-independent (p53-associated) squamous cell carcinoma 2 , 4 , as it has an error rate of 20 – 30% 8 , 9 . The third pathogenetic concept postulated by Nooij et al. 10 (see below; Table 1 ; 5 , 11 ) is not included in the new WHO classification as the current data are still insufficient.
Table 1 Pathogenetically based clinicopathological characteristics of vulvar squamous cell carcinoma 2 , 4 , 5 , 6 , 10 , 11 , 20 , 42 .
| | HPV-associated | HPV-independent | Uncertain |
:--- :--- |
| | p16 + /p53 − | p16 − /p53 + | p16 − /p53 − |
| Frequency | 40% | 50 – 60% | 20% |
| Age | 40 – 60 years of age | 50 – 70 years of age | 60 – 70 years of age |
| Precancerous lesion | VIN 2/3 (H-SIL) | HPV-independent VIN (d-VIN,? VAAD,? DEVIL) | ? (d-VIN-/VAAD-like?) |
| Etiopathogenesis | High-risk HPV infection | p53 alteration | ? (NOTCH-1/HRAS/ PIK3CA mutation?) |
| Biomarker expression | p16 positive (block staining) | p53-aberrant staining pattern | p16 negative/p53 wt |
| Histology (Heller et al. 2020) | Non-keratinizing (ca. 66%) | Keratinizing (80 – 90%) | Keratinizing/ non-keratinizing |
| Inguinal lymph node metastases | 30% | 40% | 30% |
| Radio(chemo)sensitivity | Usually sensitive | Less sensitive | Possibly less sensitive |
| Prognosis | Better | Poorer | Intermediate |
| Local recurrence (Nooij et al. 2017) 10 | 5.3% | 22.6% | 16.3% |
| 2-year DFS (Woelber et al. 2021) 11 | 64% | 47% | 60% |
| 5-year DSS (Barlow et al. 2020) 5 | 89% | 75% | 83% |
| Overall survival (Woelber et al. 2021) 11 | 82% | 70% | 75% |
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Immunohistochemistry showing strong nuclear and cytoplasmic p16 reactivity (so-called block staining; 12 ; Fig. 2 c ) points towards HPV association and is defined as a “reliable (although not perfect)” surrogate marker by the WHO (WHO 2020, 4 , 13 ). Analysis using p53 immunohistochemistry may help to more accurately diagnose VIN and vulvar squamous cell carcinoma 4 ( Fig. 2 f ), as staining patterns have been defined that correlate well with underlying mutations 7 , 14 .
Should it not be possible to classify the tumor based on p16 immunohistochemistry (and/or molecular HPV detection) or the presence of p53, the WHO deems the description squamous cell carcinoma NOS to be “acceptable” ( Fig. 5 ). The WHO explicitly points out that molecular analyses (i.e., HPV detection in situ) are not indicated for diagnostic evaluation.
Fig. 5.
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Clinicopathological characteristics of the 2020 WHO classification of squamous cell carcinoma of the female genitals 1 , 5 , 6 , 8 , 9 , 11 , 12 , 13 , 14 , 17 , 18 , 20 , 24 , 27 , 44 .
Patients with p16-positive squamous cell carcinoma who have received radio(chemo)therapy show a higher response rate that is statistically significant compared with patients with p53-associated carcinoma 15 , 16 , 17 , 18 .
In these patient groups with very different therapeutic approaches, it has now been acknowledged that p16-positive carcinomas have a better prognosis compared with those that are p53-positive 5 , 6 , 11 , 19 . The study by McAlpine et al. 20 points out that patients with p53-positive tumors benefit from a more radical surgical approach. Initial molecular studies show that vulvar carcinoma with a p53 mutation and an additional PIK3CA comutation have a particularly unfavorable prognosis 7 . There also clearly exists a third pathogenetic group of p16 − /p53 − tumors, which ranks prognostically between the p16-positive and the p53-aberrant vulvar carcinoma 5 , 11 ( Table 1 ). Whether or not the prognostically favorable low-grade squamous cell carcinoma with verrucous morphology represents one morphological end of the p16 − /p53 − tumor spectrum 7 is still unclear.
The WHO classification does not include grading specifications. In the view of the authors, HPV association (i.e., p16 block positivity; 12 ) does not (yet) replace the grading process. Should a grading be necessary for documentation or for the DRG system, this can be done based on the extent of keratinization, analogous to the approach used thus far.
It is unclear why verrucous carcinoma ( Fig. 3 ) is no longer listed in the WHO classification; it is, however, mentioned in more recent reviews 2 , 4 . Molecular analyses also regard this tumor type separately 7 . Based on verified HRAS and PIK3CA mutations, VAAD is thought to be a precursor lesion of verrucous carcinoma 3 .
Fig. 3.
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Verrucous carcinoma of the vulva: exophytic verrucous growth of well-differentiated squamous cell epithelium with superficial parakeratosis and a sharp demarcation, with only focal infiltration (arrow), from the subjacent stroma.
The WHO classification explicitly mentions the possibility of immunohistochemical HER2 detection for Morbus Paget . A meta-analysis of 713 patients demonstrated that HER2 expression was present in 30% of cases, and steroid hormone receptor positivity for ER, PR and AR was present in 13%, 8% and 40% respectively 21 . These may serve as a basis for possible therapeutic targets.
Vaginal Tumors
When it comes to the vaginal intraepithelial neoplasia (VaIN; Fig. 4 a ) and adenocarcinoma , there have been no changes.
Fig. 4.
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Precancerous lesions and carcinoma of the vagina. a HPV-associated precancerous lesion of the vagina (VAIN 3), b keratinizing squamous cell carcinoma of the vagina with slight peritumoral desmoplasia and absence of peritumoral inflammation, c non-keratinizing squamous cell carcinoma of the vagina with a high degree of peritumoral inflammation.
For squamous cell carcinoma of the vagina, the detection of HPV is currently of no therapeutic relevance 22 . Nevertheless, the WHO recommends making this distinction for these tumor types as well ( Fig. 5 ).
In general, the majority of vaginal squamous cell carcinomas are HPV-associated, especially those with a non-keratinizing morphology ( Fig. 4 b ) and tumor location in the upper or intermediate third (so-called Müllerian vagina). Distal squamous cell carcinomas are known as introitus carcinoma and stem from the urogenital sinus (so-called sinus vagina; 23 , 24 ). Lacking HPV association, these are often keratinizing squamous cell carcinomas ( Fig. 4 c ).
For vaginal carcinomas too, the WHO points out that molecular analyses (i.e., HPV detection in situ) are not indicated for the diagnostic evaluation.
The WHO classification does not include grading specifications. In the view of the authors, HPV association (i.e., p16 block positivity; 12 ) does not (yet) replace the grading process. Should a grading be necessary for documentation or for the DRG system, this can be done based on the extent of keratinization, analogous to the approach used thus far.
Tumors of the Cervix Uteri
For squamous cell cervical intraepithelial neoplasia (CIN), there have been no changes.
When it comes to adenocarcinoma in situ (AIS) , a distinction is made between various HPV-associated variants and the non-HPV-associated gastric AIS(g-AIS). SMILE (stratified mucin-producing intraepithelial lesion) as a subtype of AIS is no longer listed as an independent entity.
Epithelial precancerous lesions and carcinoma of the cervix uteri are predominantly HPV-associated 25 .
To ensure a uniform nomenclature, the WHO has classified these squamous cell carcinomas in a manner analogous to the vulvar and vaginal carcinomas ( Fig. 5 ).
For the very rare HPV-negative squamous cell carcinoma 26 , 27 there is no known precancerous lesion.
Similarly for squamous cell carcinoma of the cervix uteri , the HE-morphology alone does not allow differentiation between the two forms; for this reason the WHO recommends performing p16 immunohistochemistry but also accepts the diagnosis of squamous cell carcinoma NOS ( Fig. 5 ), as there are no existing therapeutic or prognostic differences. With regard to grading , the WHO classification states that there is no established grading system. In the view of the authors, HPV association (i.e., p16 block positivity; 12 ) does not (yet) replace the grading process. Should a grading be necessary for documentation or for the DRG system, this can be done based on the extent of keratinization, analogous to the approach used thus far.
For adenocarcinoma (AC) of the cervix uteri , a similar distinction is made with regard to the high-risk HPV association. HPV-negative AC has a significantly less favorable prognosis 28 , 29 , 30 .
Therefore, the previous diagnostic category of AC-NOS no longer exists in the new edition of the WHO classification.
The same applies for (primary) serous AC of the cervix uteri , which are almost exclusively endometrial or isthmic endometrial carcinomas with cervical involvement 29 , 31 .
The WHO classification has adopted the “International Endocervical Adenocarcinoma Classification” (IECC; 28 , 29 ) ( Fig. 6 ), which was also included in the S3-Guideline for cervical carcinoma reviewed in 2021 32 .
Fig. 6.
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Classification of adenocarcinoma of the cervix uteri in accordance with the 2020 WHO classification 1 , 28 , 29 , 31 , 32 , 37 . Small image: strong p16 positivity of a usual-type AC.
An HPV analysis is not necessary for the diagnosis 1 . If “block-type” reactivity is detected 1 ( Fig. 6 ), p16 is a reliable surrogate marker for HPV association. In very rare cases, p16 hypermethylation may lead to a (false) negative immunohistochemistry 33 , an error that is estimated to occur for CIN 3 in approx. 5% of cases 4 , 34 . The choice of a suitable p16 clone is also very important for the reliable detection of p16 35 . The p16 reactivity in old paraffin blocks or insufficiently fixed tissues is deemed unreliable 31 , 36 . It is also important to note that HPV-independent AC (i.e., gastric AC) may also demonstrate p16 positivity 37 . The p16 immunohistochemistry must be interpreted within the context of the HE morphology.
The so-called Silva pattern , a prognostically relevant classification of (HPV-associated) AC based on architectural criteria, has been newly adopted into WHO classification ( Table 2 ).
Table 2 Frequency and prognostic relevance of the Silva pattern for HPV-associated adenocarcinoma of the cervix uteri 43 .
| | Frequency | Pelvic lymph node metastasis | FIGO Stage I | FIGO Stage II – IV | Recurrence rate |
:--- :--- :--- |
| Pattern A | 20.7% | 0% | 100% | 0% | 0% |
| Pattern B | 25.6% | 4.4% | 100% | 0% | 1.1% |
| Pattern C | 53.7% | 23.8% | 83% | 17% | 23.7% |
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It distinguishes between the prognostically more favorable pattern A carcinoma with non-destructive invasion and the pattern B and C carcinoma with destructive invasion. Distinguishing between pattern A-AC and AIS based on HE morphology can be difficult (k = 0.23; 38 ).
In almost all cases, endometrioid AC of the endocervix represents a mucin-depleted variant of HPV-associated AC. Immunohistochemistry should be used to distinguish between benign lesions and endometrioid endometrial carcinoma with cervical infiltration.
Epithelial-mesenchymal Tumors
Adenofibromas of the cervix uteri that were previously listed in the WHO classification are now considered in fact to be benign endometrial or cervical polyps with an unusual morphology 39 , 40 , or alternatively adenosarcoma with “low-grade stromal morphology” ( Fig. 7 ). Immunohistochemical analyses are helpful for differential diagnosis in these cases 41 .
Fig. 7.
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Adenosarcoma of the Uterus: foliaceous tumor growth with very cell-poor stroma () showing discrete accentuation of the cell density underneath the superficial epithelium (arrows) with a bland cytology.
Footnotes
Conflict of Interest/Interessenkonflikt The authors declare that they have no conflict of interest./Die Autorinnen/Autoren geben an, dass kein Interessenkonflikt besteht.
References/Literatur
1.Lokuhetty D, White V A, Watanabe R. Lyon: Internal Agency for Research on Cancer (IARC); 2020. Female genital Tumours. 5th ed. [Google Scholar]
2.Singh N, Gilks C B. Vulval squamous cell carcinoma and its precursors. Histopathology. 2020;76:128–138. doi: 10.1111/his.13989. [DOI] [PubMed] [Google Scholar]
3.Watkins J C. Human Papillomavirus-Independent Squamous Lesions of the Vulva. Surg Pathol Clin. 2019;12:249–261. doi: 10.1016/j.path.2019.01.001. [DOI] [PubMed] [Google Scholar]
4.Heller D S, Day T, Allbritton J I. Diagnostic Criteria for Differentiated Vulvar Intraepithelial Neoplasia and Vulvar Aberrant Maturation. J Low Genit Tract Dis. 2021;25:57–70. doi: 10.1097/LGT.0000000000000572. [DOI] [PMC free article] [PubMed] [Google Scholar]
5.Barlow E L, Lambie N, Donoghoe M W. The Clinical Relevance of p16 and p53 Status in Patients with Squamous Cell Carcinoma of the Vulva. J Oncol. 2020;2020:3.739075E6. doi: 10.1155/2020/3739075. [DOI] [PMC free article] [PubMed] [Google Scholar]
6.Sand F L, Nielsen D MB, Frederiksen M H. The prognostic value of p16 and p53 expression for survival after vulvar cancer: A systematic review and meta-analysis. Gynecol Oncol. 2019;152:208–217. doi: 10.1016/j.ygyno.2018.10.015. [DOI] [PubMed] [Google Scholar]
7.Tessier-Cloutier B, Pors J, Thompson E. Molecular characterization of invasive and in situ squamous neoplasia of the vulva and implications for morphologic diagnosis and outcome. Mod Pathol. 2021;34:508–518. doi: 10.1038/s41379-020-00651-3. [DOI] [PubMed] [Google Scholar]
8.Cheng A S, Karnezis A N, Jordan S. p16 Immunostaining Allows for Accurate Subclassification of Vulvar Squamous Cell Carcinoma Into HPV-Associated and HPV-Independent Cases. Int J Gynecol Pathol. 2016;35:385–393. doi: 10.1097/PGP.0000000000000263. [DOI] [PubMed] [Google Scholar]
9.Santos M, Landolfi S, Olivella A. p16 overexpression identifies HPV-positive vulvar squamous cell carcinomas. Am J Surg Pathol. 2006;30:1347–1356. doi: 10.1097/01.pas.0000213251.82940.bf. [DOI] [PubMed] [Google Scholar]
10.Nooij L S, Ter Haar N T, Ruano D. Genomic Characterization of Vulvar (Pre)cancers Identifies Distinct Molecular Subtypes with Prognostic Significance. Clin Cancer Res. 2017;23:6781–6789. doi: 10.1158/1078-0432.CCR-17-1302. [DOI] [PubMed] [Google Scholar]
11.Woelber L, Prieske K, Eulenburg C. p53 and p16 expression profiles in vulvar cancer: a translational analysis by the Arbeitsgemeinschaft Gynäkologische Onkologie Chemo and Radiotherapy in Epithelial Vulvar Cancer study group. Am J Obstet Gynecol. 2021;224:5950–5.95E13. doi: 10.1016/j.ajog.2020.12.1220. [DOI] [PubMed] [Google Scholar]
12.Darragh T M, Colgan T J, Thomas Cox J. The Lower Anogenital Squamous Terminology Standardization project for HPV-associated lesions: background and consensus recommendations from the College of American Pathologists and the American Society for Colposcopy and Cervical Pathology. Int J Gynecol Pathol. 2013;32:76–115. doi: 10.1097/PGP.0b013e31826916c7. [DOI] [PubMed] [Google Scholar]
13.Dasgupta S, Ewing-Graham P C, Swagemakers S MA. Precursor lesions of vulvar squamous cell carcinoma – histology and biomarkers: A systematic review. Crit Rev Oncol Hematol. 2020;147:102866. doi: 10.1016/j.critrevonc.2020.102866. [DOI] [PubMed] [Google Scholar]
14.Tessier-Cloutier B, Kortekaas K E, Thompson E. Major p53 immunohistochemical patterns in in situ and invasive squamous cell carcinomas of the vulva and correlation with TP53 mutation status. Mod Pathol. 2020;33:1595–1605. doi: 10.1038/s41379-020-0524-1. [DOI] [PubMed] [Google Scholar]
15.Yap M L, Allo G, Cuartero J. Prognostic Significance of Human Papilloma Virus and p16 Expression in Patients with Vulvar Squamous Cell Carcinoma who Received Radiotherapy. Clin Oncol (R Coll Radiol) 2018;30:254–261. doi: 10.1016/j.clon.2018.01.011. [DOI] [PubMed] [Google Scholar]
16.Lee L J, Howitt B, Catalano P. Prognostic importance of human papillomavirus (HPV) and p16 positivity in squamous cell carcinoma of the vulva treated with radiotherapy. Gynecol Oncol. 2016;142:293–298. doi: 10.1016/j.ygyno.2016.05.019. [DOI] [PubMed] [Google Scholar]
17.Proctor L, Hoang L, Moore J. Association of human papilloma virus status and response to radiotherapy in vulvar squamous cell carcinoma. Int J Gynecol Cancer. 2020;30:100–106. doi: 10.1136/ijgc-2019-000793. [DOI] [PubMed] [Google Scholar]
18.Allo G, Yap M L, Cuartero J. HPV-independent Vulvar Squamous Cell Carcinoma is Associated With Significantly Worse Prognosis Compared With HPV-associated Tumors. Int J Gynecol Pathol. 2020;39:391–399. doi: 10.1097/PGP.0000000000000620. [DOI] [PubMed] [Google Scholar]
19.Rasmussen C L, Sand F L, Hoffmann Frederiksen M. Does HPV status influence survival after vulvar cancer? Int J Cancer. 2018;142:1158–1165. doi: 10.1002/ijc.31139. [DOI] [PubMed] [Google Scholar]
20.McAlpine J N, Leung S CY, Cheng A. Human papillomavirus (HPV)-independent vulvar squamous cell carcinoma has a worse prognosis than HPV-associated disease: a retrospective cohort study. Histopathology. 2017;71:238–246. doi: 10.1111/his.13205. [DOI] [PubMed] [Google Scholar]
21.Angelico G, Santoro A, Inzani F. Hormonal Environment and HER2 Status in Extra-Mammary Pagetʼs Disease (eMPD): A Systematic Literature Review and Meta-Analysis with Clinical Considerations. Diagnostics (Basel) 2020;10:1040. doi: 10.3390/diagnostics10121040. [DOI] [PMC free article] [PubMed] [Google Scholar]
22.Hellman K, Lindquist D, Ranhem C. Human papillomavirus, p16(INK4A), and Ki-67 in relation to clinicopathological variables and survival in primary carcinoma of the vagina. Br J Cancer. 2014;110:1561–1570. doi: 10.1038/bjc.2014.32. [DOI] [PMC free article] [PubMed] [Google Scholar]
23.Höckel M, Horn L-C, Illig R. Ontogenetic anatomy of the distal vagina: relevance for local tumor spread and implications for cancer surgery. Gynecol Oncol. 2011;122:313–318. doi: 10.1016/j.ygyno.2011.04.040. [DOI] [PubMed] [Google Scholar]
24.Horn L-C, Höhn A K, Hampl M. S2k-Leitlinie Diagnostik und Therapie des Vaginalkarzinoms und seiner Vorstufen – Anforderungen an die Pathologie. Der Pathologe. 2021;42:116–124. doi: 10.1007/s00292-020-00876-8. [DOI] [PubMed] [Google Scholar]
25.RIS HPV TT, VVAP and Head and Neck study groups . de Sanjosé S, Serrano B, Tous S. Burden of Human Papillomavirus (HPV)-Related Cancers Attributable to HPVs 6/11/16/18/31/33/45/52 and 58. JNCI Cancer Spectr. 2019;2:pky045. doi: 10.1093/jncics/pky045. [DOI] [PMC free article] [PubMed] [Google Scholar]
26.Li N, Franceschi S, Howell-Jones R. Human papillomavirus type distribution in 30,848 invasive cervical cancers worldwide: Variation by geographical region, histological type and year of publication. Int J Cancer. 2011;128:927–935. doi: 10.1002/ijc.25396. [DOI] [PubMed] [Google Scholar]
27.Casey S, Harley I, Jamison J. A rare case of HPV-negative cervical squamous cell carcinoma. Int J Gynecol Pathol. 2015;34:208–212. doi: 10.1097/PGP.0000000000000132. [DOI] [PubMed] [Google Scholar]
28.Hodgson A, Olkhov-Mitsel E, Howitt B E. International Endocervical Adenocarcinoma Criteria and Classification (IECC): correlation with adverse clinicopathological features and patient outcome. J Clin Pathol. 2019;72:347–353. doi: 10.1136/jclinpath-2018-205632. [DOI] [PubMed] [Google Scholar]
29.Stolnicu S, Hoang L, Chiu D. Clinical Outcomes of HPV-associated and Unassociated Endocervical Adenocarcinomas Categorized by the International Endocervical Adenocarcinoma Criteria and Classification (IECC) Am J Surg Pathol. 2019;43:466–474. doi: 10.1097/PAS.0000000000001224. [DOI] [PMC free article] [PubMed] [Google Scholar]
30.Nicolás I, Saco A, Barnadas E. Prognostic implications of genotyping and p16 immunostaining in HPV-positive tumors of the uterine cervix. Mod Pathol. 2020;33:128–137. doi: 10.1038/s41379-019-0360-3. [DOI] [PubMed] [Google Scholar]
31.Stolnicu S, Barsan I, Hoang L. International Endocervical Adenocarcinoma Criteria and Classification (IECC): A New Pathogenetic Classification for Invasive Adenocarcinomas of the Endocervix. Am J Surg Pathol. 2018;42:214–226. doi: 10.1097/PAS.0000000000000986. [DOI] [PMC free article] [PubMed] [Google Scholar]
32.AWMF S3-Leitlinie zur Diagnostik, Therapie und Nachsorge der Patientin mit Zervixkarzinom, 2021Accessed January 26, 2021 at:
33.Nuovo G J, Plaia T W, Belinsky S A. In situ detection of the hypermethylation-induced inactivation of the p16 gene as an early event in oncogenesis. Proc Natl Acad Sci U S A. 1999;96:12754–12759. doi: 10.1073/pnas.96.22.12754. [DOI] [PMC free article] [PubMed] [Google Scholar]
34.Shain A F, Kwok S, Folkins A K. Utility of p16 Immunohistochemistry in Evaluating Negative Cervical Biopsies Following High-risk Pap Test Results. Am J Surg Pathol. 2018;42:69–75. doi: 10.1097/PAS.0000000000000960. [DOI] [PubMed] [Google Scholar]
35.Shain A F, Wilbur D C, Stoler M H. Test Characteristics of Specific p16 Clones in the Detection of High-grade Squamous Intraepithelial Lesions (HSIL) Int J Gynecol Pathol. 2018;37:82–87. doi: 10.1097/PGP.0000000000000391. [DOI] [PubMed] [Google Scholar]
36.Nuovo A J, Garofalo M, Mikhail A. The effect of aging of formalin-fixed paraffin-embedded tissues on the in situ hybridization and immunohistochemistry signals in cervical lesions. Diagn Mol Pathol. 2013;22:164–173. doi: 10.1097/PDM.0b013e3182823701. [DOI] [PubMed] [Google Scholar]
37.Carleton C, Hoang L, Sah S. A Detailed Immunohistochemical Analysis of a Large Series of Cervical and Vaginal Gastric-type Adenocarcinomas. Am J Surg Pathol. 2016;40:636–644. doi: 10.1097/PAS.0000000000000578. [DOI] [PMC free article] [PubMed] [Google Scholar]
38.Parra-Herran C, Taljaard M, Djordjevic B. Pattern-based classification of invasive endocervical adenocarcinoma, depth of invasion measurement and distinction from adenocarcinoma in situ: interobserver variation among gynecologic pathologists. Mod Pathol. 2016;29:879–892. doi: 10.1038/modpathol.2016.86. [DOI] [PubMed] [Google Scholar]
39.Ip P P. Benign endometrial proliferations mimicking malignancies: a review of problematic entities in small biopsy specimens. Virchows Arch. 2018;472:907–917. doi: 10.1007/s00428-018-2314-4. [DOI] [PubMed] [Google Scholar]
40.Howitt B E, Quade B J, Nucci M R. Uterine polyps with features overlapping with those of Müllerian adenosarcoma: a clinicopathologic analysis of 29 cases emphasizing their likely benign nature. Am J Surg Pathol. 2015;39:116–126. doi: 10.1097/PAS.0000000000000303. [DOI] [PubMed] [Google Scholar]
41.McCluggage W G. A practical approach to the diagnosis of mixed epithelial and mesenchymal tumours of the uterus. Mod Pathol. 2016;29 01:S78–S91. doi: 10.1038/modpathol.2015.137. [DOI] [PubMed] [Google Scholar]
42.Knopp S, Bjørge T, Nesland J M. p16INK4a and p21Waf1/Cip1 expression correlates with clinical outcome in vulvar carcinomas. Gynecol Oncol. 2004;95:37–45. doi: 10.1016/j.ygyno.2004.07.026. [DOI] [PubMed] [Google Scholar]
43.Roma A A, Mistretta T-A, Diaz De Vivar A. New pattern-based personalized risk stratification system for endocervical adenocarcinoma with important clinical implications and surgical outcome. Gynecol Oncol. 2016;141:36–42. doi: 10.1016/j.ygyno.2016.02.028. [DOI] [PMC free article] [PubMed] [Google Scholar]
44.Höckel M, Trott S, Dornhöfer N. Vulvar field resection based on ontogenetic cancer field theory for surgical treatment of vulvar carcinoma: a single-centre, single-group, prospective trial. Lancet Oncol. 2018;19:537–548. doi: 10.1016/S1470-2045(18)30109-8. [DOI] [PubMed] [Google Scholar]
Geburtshilfe Frauenheilkd. 2021 Oct 6;81(10):1145–1153. [Article in German]
WHO-Klassifikation 2020 für Tumoren des unteren weiblichen Genitales
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Issue date 2021 Oct.
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Zusammenfassung
In der WHO-Klassifikation 2020 steht die Unterscheidung von HPV-assoziierten und HPV-unabhängigen Plattenepithelkarzinomen des unteren weiblichen Genitales im Vordergrund. Die Unterscheidung der HPV-Assoziation ersetzt das Grading nicht, für welches jedoch kein Gradingsystem empfohlen wird. Auch bei der VIN erfolgt die Trennung nach HPV-(p16-)Assoziation. HPV-unabhängige Adenokarzinome (AC) der Cervix uteri sind prognostisch ungünstiger. Als Surrogatmarker für eine HPV-Assoziation gilt der immunhistochemische Nachweis von p16. Beim HPV-assoziierten AC der Cervix uteri erfolgt die prognostisch relevante Angabe des sog. Silva-Patterns.
Schlüsselwörter: p16-Immunhistochemie, HPV, Silva Pattern
Einleitung
Die WHO-Klassifikation der Tumoren des weiblichen Genitales ist im 4.Band der 5.Auflage aufgrund neuer histomorphologischer und insbesondere molekularpathologischer Daten 2020 grundlegend überarbeitet worden und hat sich in ihrem Umfang im Vergleich zu 2013 verdoppelt. Die WHO-Klassifikation wird vorrangig aufgrund neuer (molekular-)pathologischer Daten erstellt. Therapeutisch und diagnostisch relevante Daten, soweit vorhanden, fließen aber in jedem Fall mit ein. Die „blauen Bücher der WHO“ stellen eine wichtige Grundlage für einen weltweit einheitlichen diagnostischen Standard dar. In Deutschland wird in den Leitlinien/Empfehlungen der AWMF (Arbeitsgemeinschaft der Wissenschaftlichen Medizinischen Fachgesellschaften e. V.) darauf hingewiesen, dass die derzeit aktuelle WHO-Klassifikation für die pathologische Befunderstellung heranzuziehen ist. Die Verbindlichkeit der diagnostischen histopathologischen Standards und Terminologie der jeweils aktuellen WHO-Klassifikation ist in allen Leitlinien zu gynäkologischen Malignomen mit dem Verb „soll“ als höchstem Grad der Empfehlung belegt. Während Kriterien zur Diagnosestellung und Terminologie in der WHO-Klassifikation abgebildet werden, dienen zur Stadieneinteilung, also der Ausbreitung der gynäkologischen Tumoren neben der TNM-Klassifikation maligner Tumoren auch die FIGO-Klassifikation (Fédération Internationale de Gynécologie et dʼObstétrique).
Die wesentlichen, klinisch relevanten, Veränderungen der aktuellen WHO-Klassifikation sind im Folgenden für das untere weibliche Genitale praxisrelevant zusammengefasst.
Tumoren der Vulva
Bei den vulvären intraepithelialen Neoplasien (VIN) ist die Unterscheidung HPV-assoziiert und HPV-negativ beibehalten worden. Die HPV-assoziierte VIN entspricht nomenklatorisch der Low- (VIN 1) sowie High-Grade-SIL (VIN 2 und 3; Abb. 1 und 2 ).
Abb. 1.
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Klinikopathologische Charakteristika der WHO-Klassifikation 2020 der vulvären Präkanzerosen 1 , 2 , 3 , 4 , 12 , 14 .
Abb. 2.
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Präkanzerosen (VIN) und Karzinom der Vulva. a HPV-assoziierte VIN (usual VIN; u-VIN), b und c nicht keratinisierendes, HPV-assoziiertes Plattenepithelkarzinom der Vulva mit plumpem Invasionsmuster und p16-Positivität (sog. block-staining; s. Text), d HPV-unabhängige VIN (d-VIN), e und f keratinisierendes Plattenepithelkarzinom der Vulva mit netzigem Invasionsmuster und aberranter p53-Expression (s. Text).
Für die VIN ohne HPV-Nachweis wird der Terminus HPV-unabhängige VIN eingeführt 1 , 2 ; Abb. 1 , die eine variable Morphologie zeigt:
Die differenzierte VIN mit horizontalem Wachstum (d-VIN) wird als Vorläuferläsion einem aggressiveren,
die differenzierte exophytische VIN-Läsion (DEVIL) einem weniger aggressiven (keratinisierenden) Plattenepithelkarzinom und die
vulväre Akanthose mit abnormer Differenzierung (VAAD) dem verrukösen Karzinom als Vorläufer- bzw. Risikoläsion zugeordnet 2 , 3 .
Dabei ist zu beachten, dass sich überschneidende Läsionen vorkommen können 1 , 2 , 4 . Aufgrund morphologischer Ähnlichkeiten bzw. Überschneidungen 3 werden DEVIL und VAAD auch unter dem Oberbegriff einer atypischen Ausreifung des vulvären Plattenepithels (vulvar aberrant maturation; VAM; 4 ) zusammengefasst ( Abb. 1 ), ein Terminus, der in der WHO-Klassifikation nicht enthalten ist.
Ungeachtet einer derzeit (noch) fehlenden diagnostischen und therapeutischen Relevanz 5 , 6 , erfolgt aufgrund einer differenten Pathogenese ( Abb. 1 und 2 ) in der neuen WHO-Klassifikation eine Unterscheidung in HPV-assoziierte und HPV-unabhängige Plattenepithelkarzinome, wobei diese Angabe im Befundbericht von der WHO empfohlen wird. Das Verhältnis HPV-unabhängiger zu HPV-assoziierten Plattenepithelkarzinom wird mit 0,60 bis 0,83 angegeben 7 .
Ungeachtet klinikopathologischer Unterschiede ( Tab. 1 ) ist mit einer Fehlerrate von 20 – 30% 8 , 9 rein HE-morphologisch eine Unterscheidung HPV-assoziierter und HPV-unabhängiger (p53-assoziierter) Plattenepithelkarzinome nicht verlässlich möglich 2 , 4 . Der von Nooij et al. 10 postulierte dritte pathogenetische Ansatz (s. u.; Tab. 1 ; 5 , 11 ) ist aufgrund einer derzeit noch unzureichenden Datenlage in der WHO-Klassifikation nicht berücksichtigt.
Tab. 1 Pathogenetisch basierte klinikopathologische Charakteristika von Plattenepithelkarzinomen der Vulva 2 , 4 , 5 , 6 , 10 , 11 , 20 , 42 .
| | HPV-Assoziation | HPV-unabhängig | unklar |
:--- :--- |
| | p16 + /p53 − | p16 − /p53 + | p16 − /p53 − |
| Häufigkeit | 40% | 50 – 60% | 20% |
| Alter | 40. – 60. Lebensjahr | 50. – 70. Lebensjahr | 60. – 70. Lebensjahr |
| Präkanzerose | VIN 2/3 (H-SIL) | HPV-unabhängige VIN (d-VIN,? VAAD,? DEVIL) | ? (d-VIN-/VAAD-like?) |
| Ätiopathogenese | HPV High-Risk-Infektion | p53-Alteration | ? (NOTCH-1-/HRAS-/PIK3CA-Mutation?) |
| Biomarker-Expression | p16-positiv (block staining) | p53-aberrantes Färbemuster | p16-negativ/p53 wt |
| Histologie (Heller et al. 2020) | nicht keratinisierend (ca. 66%) | keratinisierend (80 – 90%) | keratinisierend/ nicht keratinisierend |
| inguinale Lymphknotenmetastasen | 30% | 40% | 30% |
| Radio(chemo)sensitivität | i. d. R.sensitiv | weniger sensitiv | mglw. weniger sensitiv |
| Prognose | besser | schlechter | intermediär |
| Lokalrezidiv (Nooij et al. 2017) 10 | 5,3% | 22,6% | 16,3% |
| 2-Jahres-DFS (Woelber et al. 2021) 11 | 64% | 47% | 60% |
| 5-Jahres-DSS (Barlow et al. 2020) 5 | 89% | 75% | 83% |
| Gesamtüberleben (Woelber et al. 2021) 11 | 82% | 70% | 75% |
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Eine starke nukleäre und zytoplasmatische p16-Reaktivität in der Immunhistochemie (sog. block-staining; 12 ; Abb. 2 c ) spricht für eine HPV-Assoziation und wird von der WHO als „reliable (although not perfect)“ Surrogatmarker definiert (WHO 2020, 4 , 13 ). Eine p53-Immunhistochemie kann die Diagnostik bei den VIN und vulvären Plattenepithelkarzinomen präzisieren 4 ( Abb. 2 f ), da Färbemuster definiert wurden, die gut mit zugrunde liegenden Mutationen korrelieren 7 , 14 .
Ist eine Unterscheidung mittels p16-Immunhistochemie (und/oder molekularem HPV-Nachweis) bzw. p53 im Tumor nicht möglich, ist gemäß WHO die Angabe Plattenepithelkarzinom NOS „acceptable“ ( Abb. 5 ). Explizit wird von der WHO darauf hingewiesen, dass molekulare Untersuchungen (z. B. HPV-Nachweis in situ) für die Diagnostik nicht indiziert sind.
Abb. 5.
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Klinikopathologische Charakteristika der WHO-Klassifikation 2020 von Plattenepithelkarzinomen des unteren weiblichen Genitales 1 , 5 , 6 , 8 , 9 , 11 , 12 , 13 , 14 , 17 , 18 , 20 , 24 , 27 , 44 .
Radio(chemo)therapeutisch behandelte Patientinnen mit p16-positiven Plattenepithelkarzinomen zeigen eine statistisch signifikant bessere Ansprechrate als p53-assoziierte Karzinome 15 , 16 , 17 , 18 .
Bei Patientinnen mit sehr unterschiedlichem therapeutischen Ansatz ist mittlerweile akzeptiert, dass p16-positive im Vergleich zu p53-positiven Karzinomen eine bessere Prognose zeigen 5 , 6 , 11 , 19 . Die Studie von McAlpine et al. 20 weist darauf hin, dass Patientinnen mit p53-positiven Tumoren von einem radikaleren chirurgischen Vorgehen profitieren. Erste molekulare Studien zeigen, dass Vulvakarzinome mit einer p53-Mutation und zusätzlicher PIK3CA-Komutation eine besonders ungünstige Prognose haben 7 . Die offenbar existente dritte pathogenetische Gruppe der p16 − /p53 − -Tumoren steht prognostisch zwischen den p16-positiven und p53-aberranten Vulvakarzinomen 5 , 11 ( Tab. 1 ). Ob das prognostisch günstige Low-Grade-Plattenepithelkarzinom mit verruköser Morphologie ein morphologisches Endspektrum der p16 − /p53 − -Tumoren darstellt 7 , ist unklar.
Angaben zum Grading sind in der WHO-Klassifikation nicht enthalten. Nach Ansicht der Autoren ersetzt die HPV-Assoziation (i. e. p16 „block-positivity“; 12 ) das Grading derzeit (noch) nicht. Sollte ein Grading zur Dokumentation bzw. für das DRG-System relevant sein, kann dieses in Analogie zum bisherigen Vorgehen anhand des Ausmaßes der Keratinisierung erfolgen.
Unklar ist, warum das verruköse Karzinom ( Abb. 3 ) nicht mehr in der WHO-Klassifikation aufgeführt ist, wohl aber in rezenten Übersichtsarbeiten 2 , 4 . Auch molekulare Untersuchungen nehmen zu diesem Tumortyp separat Stellung 7 . Die VAAD wird aufgrund nachgewiesener HRAS- und PIK3CA-Mutationen als Vorläuferläsion des verrukösen Karzinoms diskutiert 3 .
Abb. 3.
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Verruköses Karzinom der Vulva: exophytisch-verruköses Wachstum gut differenzierter Plattenepithelien mit oberflächlicher Parakeratose und scharfer, nur fokal infiltrativer Begrenzung (Pfeil) zum darunterliegenden Stroma.
Die WHO-Klassifikation weist explizit auf die Möglichkeit des immunhistochemischen HER2-Nachweises beim Morbus Paget hin. Eine Metaanalyse von 713 Patientinnen ergab, dass in 30% eine HER2-Expression und für die Steroidhormonrezeptoren ER, PR und AR in 13%, 8% bzw. 40% eine Positivität vorliegt 21 , die als mögliches therapeutisches Target dienen könnte.
Tumoren der Vagina
Bei den vaginalen intraepithelialen Neoplasien (VaIN; Abb. 4 a ) und Adenokarzinomen gibt es keine Änderungen.
Abb. 4.
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Präkanzerosen und Karzinom der Vagina. a HPV-assoziiertes Präkanzerose der Vagina (VAIN 3), b keratinisierendes Plattenepithelkarzinom der Vagina mit geringer peritumoraler Desmoplasie und fehlender peritumoraler Entzündung, c nicht keratinisierendes Plattenepithelkarzinom der Vagina mit hochgradiger peritumoraler Entzündung.
Auch beim Plattenepithelkarzinom der Vagina gibt es derzeit keine therapeutische Relevanz in Bezug auf den HPV-Nachweis 22 , dennoch empfiehlt auch hier die WHO deren Unterscheidung ( Abb. 5 ).
Allgemein ist die Mehrzahl der vaginalen Plattenepithelkarzinome HPV-assoziiert, insbesondere bei nicht keratinisierender Morphologie ( Abb. 4 b ) und Tumorlokalisation im oberen und mittleren Drittel (sog. Müller-Scheide). Bei distal lokalisierten Plattenepithelkarzinomen handelt es sich um sog. Introituskarzinome, ausgehend vom Urogenitalsinus (sog. Sinus-Scheide; 23 , 24 ), die bei fehlender HPV-Assoziation oft keratinisierende Plattenepithelkarzinome sind ( Abb. 4 c ).
Auch beim Vaginalkarzinom wird von der WHO darauf hingewiesen, dass molekulare Untersuchungen (z. B. HPV-Nachweis in situ) für die Diagnostik nicht indiziert sind.
Angaben zum Grading sind in der WHO-Klassifikation nicht enthalten. Nach Ansicht der Autoren ersetzt die HPV-Assoziation (i. e. p16 „block-positivity“; 12 das Grading derzeit (noch) nicht. Sollte ein Grading zur Dokumentation bzw. für das DRG-System relevant sein, kann dieses in Analogie zum bisherigen Vorgehen anhand des Ausmaßes der Keratinisierung erfolgen.
Tumoren der Cervix uteri
Bei den plattenepithelialen zervikalen intraepithelialen Neoplasien (CIN) gibt es keine Änderungen.
Beim Adenocarcinomata in situ (AIS) werden verschiedene Varianten HPV-assoziierter und das nicht HPV-assoziierte gastrische AIS(g-AIS) unterschieden. Die SMILE (Stratified mucin-producing intraepithelial lesion) als Subtyp der AIS wird nicht mehr als eigenständige Entität aufgeführt.
Epitheliale Präkanzerosen und Karzinome der Cervix uteri sind mehrheitlich HPV-assoziiert 25 .
Um eine einheitliche Nomenklatur zu gewährleisten, wird auch hier von der WHO eine Unterteilung der Plattenepithelkarzinome wie beim Vulva- und Vaginalkarzinom vorgenommen ( Abb. 5 ).
Für die sehr seltenen HPV-negativen Plattenepithelkarzinome 26 , 27 ist keine Präkanzerose bekannt.
Auch bei den Plattenepithelkarzinomen der Cervix uteri gelingt eine rein HE-morphologische Unterscheidung beider Formen nicht, daher wird von der WHO die p16-Immunhistochemie empfohlen, aber auch hier die Diagnose Plattenepithelkarzinom-NOS akzeptiert ( Abb. 5 ), da keine therapeutischen bzw. prognostischen Unterschiede bestehen. Die WHO-Klassifikation konstatiert zum Grading , dass kein etabliertes Gradingsystem existiert. Nach Ansicht der Autoren ersetzt die HPV-Assoziation (i. e. p16 „block-positivity“; 12 ) das Grading derzeit (noch) nicht. Sollte ein Grading zur Dokumentation bzw. für das DRG-System relevant sein, kann dieses in Analogie zum bisherigen Vorgehen anhand des Ausmaßes der Keratinisierung erfolgen.
Auch beim Adenokarzinom (AC) der Cervix uteri erfolgt eine Unterteilung anhand der HPV High-Risk-Assoziation. HPV-negative AC zeigen eine signifikant ungünstigere Prognose 28 , 29 , 30 .
Daher gibt es die bisherige diagnostische Kategorie AC-NOS in der neuen WHO-Klassifikation nicht mehr.
Gleiches gilt für das (primäre) seröse AC der Cervix uteri , bei denen es sich fast ausschließlich um endometriale oder isthmische Endometriumkarzinome mit zervikaler Beteiligung handelt 29 , 31 .
Die WHO-Klassifikation hat die auch in der 2021 überarbeiteten Version der S3-Leitlinie Zervixkarzinom 32 aufgeführte „International Endocervical Adenocarcinoma Classification“ (IECC; 28 , 29 ) übernommen ( Abb. 6 ).
Abb. 6.
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Einteilung von Adenokarzinomen der Cervix uteri entsprechend der WHO-Klassifikation 2020 1 , 28 , 29 , 31 , 32 , 37 . Kleines Bild: starke p16-Positivität eines AC usual type.
Eine HPV-Analyse ist für die Diagnosestellung nicht notwendig 1 . Beim Nachweis einer „block-type“-Reaktivität 1 ( Abb. 6 ) ist p16 ein verlässlicher Surrogatmarker für die HPV-Assoziation. In sehr seltenen Fällen kann eine p16-Hypermethylierung zu einer (falsch) negativen Immunhistochemie führen 33 , die für die CIN 3 mit rund 5% angegeben wird 4 , 34 . Bedeutsam ist auch die Auswahl eines geeigneten p16-Klons für die verlässliche p16-Bestimmung 35 . Die p16-Reaktivität ist an alten Paraffinblöcken bzw. unzureichend fixiertem Gewebe unzuverlässig 31 , 36 . Wichtig ist, dass auch HPV-unabhängige AC (z. B. gastrische) eine p16-Positivität zeigen können 37 . Die Interpretation der p16-Immunhistochemie muss hier im Kontext mit der HE-Morphologie erfolgen.
Neu in die WHO-Klassifikation aufgenommen wurde die auf architektonischen Kriterien basierende, prognostisch-relevante Klassifikation (HPV-assoziierter) AC, das sog. Silva-Pattern ( Tab. 2 ).
Tab. 2 Häufigkeit und prognostische Relevanz des sog. Silva-Patterns beim HPV-assoziierten Adenokarzinom der Cervix uteri 43 .
| | Häufigkeit | pelvine Lymphknotenmetastasierung | FIGO-Stadium I | FIGO-Stadium II – IV | Rezidivrate |
:--- :--- :--- |
| Pattern A | 20,7% | 0% | 100% | 0% | 0% |
| Pattern B | 25,6% | 4,4% | 100% | 0% | 1,1% |
| Pattern C | 53,7% | 23,8% | 83% | 17% | 23,7% |
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Diese unterscheidet zwischen prognostisch günstigen, non-destruktiv wachsenden Pattern-A- und den destruierend wachsenden Pattern-B- und -C-Karzinomen. Die Abgrenzung zwischen Pattern-A-AC und dem AIS kann HE-morphologisch schwierig sein (k = 0,23; 38 ).
Das endometrioide AC der Endozervix stellt fast in allen Fällen eine muzindepletierte Variante des HPV-assoziierten ACs dar. Endometrioide Endometriumkarzinome mit Zervixinfiltration und benigne Läsionen sind immunhistochemisch abzugrenzen.
Epithelial-mesenchymale Tumoren
Bei den bisher in der WHO-Klassifikation aufgeführten Adenofibromen der Cervix uteri handelt es sich de facto um benigne Endometrium- bzw. Zervixpolypen mit ungewöhnlicher Morphologie 39 , 40 oder aber um Adenosarkome mit „low-grade Morphologie des Stromas“ ( Abb. 7 ), bei denen immunhistochemische Untersuchungen differenzialdiagnostisch hilfreich sind 41 .
Abb. 7.
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Adenosarkom des Uterus: blattartiges Wachstum des Tumors mit sehr zellarmem Stroma (), das eine diskrete Akzentuierung der Zelldichte unter dem oberflächlichen Epithel zeigt (Pfeile) mit blander Zytologie.
Articles from Geburtshilfe und Frauenheilkunde are provided here courtesy of Thieme Medical Publishers
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188521 | https://math.stackexchange.com/questions/110091/factoring-polynomial-in-two-variables | Skip to main content
Factoring polynomial in two variables.
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Given a polynomial P(x,y) I would like to know what the criteria are for factoring out linear factors.
For instance, in one variable, if Q(a)=0, then one may say Q(x)=(x−a)R(x). In two variables this is not true, as shown by P(x,y)=x2+y2 one has P(0,0)=0 but one cannot factor out anything.
When can one factor out a linear factor from a polynomial in two variables?
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asked Feb 16, 2012 at 18:18
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Relevant: Madhu Sudan, Lecture 7. MIT Course notes for 6.966 Algebra and Computation. [warning: postscript file].
– user2468
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Feb 16, 2012 at 18:27
@J.D. : Very relevant.
– Patrick Da Silva
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Feb 16, 2012 at 18:49
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I have a reference which cites this theorem for real polynomials, but if I go through the proof perhaps this also works over other fields/euclidean domains possibly. Here it is :
Theorem. Let f(x,y) be a polynomial with real coefficients and degree d. Let a,b,c be real numbers, and let L denote the set of points (x,y) for which ax+by+c=0. If the curve f(x,y)=0 and L have strictly more than d distinct points in common, then there exists k(x,y) with real coefficients such that
f(x,y)=(ax+by+c)k(x,y).
This is in Niven & Zuckerman's Introduction to the theory of numbers, so it's not focused on algebra. I'll edit this answer later if I see a reason why this would hold over some other fields, but the proof uses mainly the Taylor expansion of the polynomial (which can be done formally, without using differentiation in the "limit" sense).
Hope that helps,
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answered Feb 16, 2012 at 18:27
Patrick Da SilvaPatrick Da Silva
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Formal Taylor expansion (if you mean what I think you mean by this) can only be done in general in characteristic zero.
– Qiaochu Yuan
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Dear Patrick, yes the result holds for every field. However the proof you alllude to won't work in characteristic p, because of the denominators in Taylor's expansion. An amusing caveat is that the theorem is true but empty in many cases: yes, if a polynomial of degree 17 has at least 18 zeros on a line over a field with 2 elements, the equation of the line divides the polynomial.However no line has more than 2 points (nor less) over such a field...
– Georges Elencwajg
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The Factor Theorem remains applicable, for example
x−a | f(x)⟺f(a)=0 for a=by+c∈R[y] is
x−by−c | f(x,y)⟺f(by+c,y)=0
e.g. x−y | f(x)−f(y) since f(y)−f(y)=0
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answered Feb 16, 2012 at 19:24
Math GemsMath Gems
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Well this is not the most general linear factor, now is it. =P I think OP expected this to be true. ^_^
– Patrick Da Silva
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Feb 17, 2012 at 22:10
@Patrick But it is, since over a field any linear factor is associate to a monic one. Otoh, in the presence of zero-divisors one cannot say so much. In my experience, many students don't immediately notice this and related consequences of the Factor Theorem, esp. when they are obfuscated by less-trivial contexts.
– Math Gems
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Feb 17, 2012 at 22:46
Sorry, my comment was junk, I didn't even bother reading your entire proof when I saw that x−a was the first thing you wrote ; but you're proof has a wrong step in it ; we can't say that p(x,y)|f(x,y) just because p(x,a)|f(x,a) for any value of a∈R, now can we?
– Patrick Da Silva
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Feb 17, 2012 at 22:57
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@Patrick How does that pertain to the above?
– Math Gems
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Feb 17, 2012 at 23:00
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188522 | https://cs.stackexchange.com/questions/157434/pairing-numbers-and-intervals | algorithms - pairing numbers and intervals - Computer Science Stack Exchange
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pairing numbers and intervals
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subject: pairing numbers and intervals
Let NUMBERS be a list of n integer numbers. The numbers are listed in no specific order. Duplicates are possible.
Let INTERVALS be a list of m intervals. The low and high limits of the intervals are integer numbers. The intervals are listed in no specific order. Duplicates are possible.
We want to pair the numbers and intervals as thus:
A number can only be paired with an interval to which it belongs. For example, 4 and [3, 10] can be paired, 4 and [5, 10] cannot.
A number cannot be paired with more than one interval.
An interval cannot be paired with more than one number.
The number of pairs is maximal.
We are not interested in the pairs themselves, only in the (maximal) number of pairs.
To summarize:
input = NUMBERS, INTERVALS
output = maximal number of pairs
Example 1:
NUMBERS = 2
INTERVALS = [4, 10]
output = 0
Example 2:
NUMBERS = 2
INTERVALS = [2, 10]
output = 1
Example 3:
NUMBERS = 3, 7, 8, 12
INTERVALS = [0, 10], [5, 15], [20, 25]
output = 2
It is relatively simple to devise an O(nm) algorithm to solve this problem. I seriously doubt that there exist any O(n) + O(m) algorithm, but what about an O(max(nlog(n), mlog(m), nlog(m), mlog(n))) algorithm?
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asked Feb 9, 2023 at 16:21
JohnHernandezJohnHernandez
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Can you argue the validity of ordering NUMBERS ascendantly as well as INTERVALS by low and high limit, and greedily looking for "the lowest" interval containing the first number, alternating with looking for the lowest number contained in the first interval?greybeard –greybeard 2023-02-09 17:35:18 +00:00 Commented Feb 9, 2023 at 17:35
1 @greybeard if I understand your idea and intervals can overlap then that greedy heuristic will fail if N={4,5}N={4,5} and I={[1,5],[2,4]}I={[1,5],[2,4]}.Russel –Russel 2023-02-10 01:12:51 +00:00 Commented Feb 10, 2023 at 1:12
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Your problem is equivalent to the problem of finding a maximum matching in a convex bipartite graph (a bipartite graph G=(A,B,E)G=(A,B,E) where the vertices on one side, say A A, admit an ordering a 1,…,a n a 1,…,a n such that each vertex in B B is adjacent to an interval a i,…,a j a i,…,a j for some 1≤i≤j≤n 1≤i≤j≤n). Googling "maximum matching convex graphs" I found this paper: A Linear Time Algorithm for Maximum Matchings in Convex, Bipartite Graphs (by G. Steiner and J. S. Yeomans) that claims the problem is actually solvable in time proportional to |B||B|! So, in your case, the problem is solvable in time O(m)O(m).
EDIT: Both this paper and others require you to represent the neighborhood of each vertex b b in B B by the first and last (in the implied ordering) vertex in A A that is adjacent to b b. This requires you to sort the integers (in O(n log n)O(n logn) time) and replacing each interval with the indices of the first and last integer that is contained in the interval (in O(m log n)O(m logn) time). Thus we reach your guessed time-bound (unless, of course, there is some clever way to accomplish this without sorting). My bad for not spotting this right away.
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edited Feb 10, 2023 at 12:20
answered Feb 10, 2023 at 11:07
HighheathHighheath
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That looks more difficult than I was hoping for, but thanks a lot for posting this!JohnHernandez –JohnHernandez 2023-02-10 14:50:25 +00:00 Commented Feb 10, 2023 at 14:50
There probably are simpler algorithms that still are quite fast; i'd advise you to just browse Google Scholar.Highheath –Highheath 2023-02-10 15:46:31 +00:00 Commented Feb 10, 2023 at 15:46
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188523 | https://www.youtube.com/watch?v=DpeCKqWVekE | Propositional Logic Proofs: Implication (IMP)
Logic & Philosophy
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Posted: 18 Oct 2021
This is a quick overview of the Implication (IMP) rule for Propositional logic. This rule states that from "if P then Q", you can derive "not-P or Q" and vice versa. The rule is a replacement or equivalence rule for some systems of logic (e.g. an Intelim system). In this video, I state the form of the rule, give a quasi-English example, provide a truth table showing that the two formulas are equivalent, and then go through a couple examples involving IMP.
Personally, I don't know how useful the rule is everyday argumentation. I don't think that I've ever heard someone reason "if P then Q" therefore "not-P or Q". So, if you have an example where someone does reason in this way, I'd love to hear it!
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Transcript:
So, in this video, I want to take a
look at the rule for implication. Now, implication in a lot of systems, especially
an intelim system, is a derived rule. What that means is you don't really
need this particular rule. You can prove everything implication does using other
rules that you might have in a system. But implication is a nice rule to have because, in
a lot of cases, it simplifies certain proofs. So, let's start by stating the form or structure
of implication. What this rule states is: if you have a conditional, let's say Phi then Psi, then
you can reason to a disjunction. So, it would be not Phi or Psi. So, if you have this conditional
here, you could transform it into a disjunction. Similarly, it says that if you had a
disjunction where one of the disjuncts is negated, you could reason to a
conditional. So, if Phi then Psi. Because of this, we can say
that, you know, we have one way here where we reason to this disjunction,
and then you can see we're reasoning in the other direction. We could simplify our presentation
of the use of implication by writing: if we had a conditional, if Phi then Psi, then
we can reason to this disjunction, not Phi or Psi. And we want to represent the fact
that we can reason the other direction, from this disjunction to the conditional,
by using this single turnstile, but in the other direction. So, what this is saying
is that you can reason either way. You can take this formula, if Phi then Psi, and reason to
this disjunction, or you could go the other way. So, these two formulas, or this
particular presentation of the rule, is what's sometimes called a rule
of replacement or equivalence rule. Another way of putting this is that
these two formulas are interderivable. So, what's going on when we use implication?
Well, if we let our conditional, something like if R then W, stand for "if it
rains, then I get wet," and we let our disjunction that we're reasoning to be something like not-R
or W, then what we're doing is reasoning from a sentence like "if it rains, then I get wet"
to a sentence like "either it does not rain or I get wet." What's happening here is we're
saying that this conditional says the same thing as this disjunction right here. But these two
formulas are just two different ways of saying the same thing. They're true and false under the same
circumstances. They're equivalent to each other, and so it's perfectly legitimate to replace the
conditional here, "if it rains, then I get wet," with this disjunction here, "either
it does not rain or I get wet." Now, for some people, this is not quite obvious,
so it's helpful to look at, let's say, a truth table to see that these two sentences here, at
least with respect to logic, are equivalent. So, let's look at one. So, here's our truth table.
Here we have the conditional, if Phi then Psi, and we see that's false just in
one case, in the second row here. And then we have our disjunction, not Phi or
Psi, and we see it's also false in one case, and in all the other cases, they match up in
terms of truth values. So, we see that these two formulas are equivalent, and so that
motivates the use of the implication rule. Since these two formulas are true and false under
the same circumstances, since they're equivalent, we can replace the conditional
with this particular disjunction. Now, let's look at an example of implication in
a proof. So, line 1, we have P then Q, and you could use implication on line 1to derive not P
or Q. We would cite the line that we're using, which was line 1, and then indicate that
we're using implication. And, of course, since you can swap one formula out for the other,
we could do the other direction. We could reason from this disjunction of line 2 back to the
conditional that we started with at line 1, writing "if P then Q," but at this
time we're using line two, implication. Here, let's look at a more complicated example
of implication. In this case, we're looking at a negated conditional. Now, recall that
in implication, we can take a conditional and replace it with the formula
that says the exact same thing. So, with respect to line 1, what we can do is take
that conditional that's found in those parentheses and replace it with a disjunction that says the
same thing. The key aspect here to keep in mind is that we want to make sure we negate the
antecedent of the conditional right there. So, in our disjunction, we have
not P rather than simply P. Now, I want to look at one more example of
implication. Let's say we had the formula not-not-P right arrow S. Now, let's make it
not S. So, here we have a formula. If we were to use implication on this, recall that the
rule states that if you have a conditional, you can transform this formula into
not-Phi or Psi. And so, when we apply implication to not not P then not S, the formula
that we would write is not not not P or not S. The thing to focus on with respect to the
use of implication is: we have a conditional, and this Phi here can stand for any formula.
In this case, Phi stands for not not P, and so it says to take that formula and add a negation to
it. So, you see we're adding this negation to Phi, and so even though P is already doubly
negated, we're going to add a negation. With respect to Psi right here, it says don't
do anything to it; simply represent it as the right disjunct when you use implication.
So, even though not S is negated here, all we're going to do is take not S and write
it as not S as follows. We're going to keep it, leave it alone. Whatever formula this is, we're
going to just move it to the disjunction there. [Buzzing sound] |
188524 | https://chem.libretexts.org/Bookshelves/General_Chemistry/Map%3A_Structure_and_Properties_(Tro)/14%3A_Solutions/14.06%3A_Colligative_Properties-_Freezing_Point_Depression_Boiling_Point_Elevation_and_Osmosis | 14.6.1
14.6.1
14.6.2
14.6.2
14.6.3
14.6.3
14.6.4
14.6.4
14.6.5
14.6.5
14.6.6
14.6.6
14.6.7
14.6.7
14.6.8
14.6.8
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14.6: Colligative Properties- Freezing Point Depression, Boiling Point Elevation, and Osmosis
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( \newcommand{\kernel}{\mathrm{null}\,})
Learning Objectives
To describe the relationship between solute concentration and the physical properties of a solution.
To understand that the total number of nonvolatile solute particles determines the decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus the pure solvent.
Many of the physical properties of solutions differ significantly from those of the pure substances discussed in earlier chapters, and these differences have important consequences. For example, the limited temperature range of liquid water (0°C–100°C) severely limits its use. Aqueous solutions have both a lower freezing point and a higher boiling point than pure water. Probably one of the most familiar applications of this phenomenon is the addition of ethylene glycol (“antifreeze”) to the water in an automobile radiator. This solute lowers the freezing point of the water, preventing the engine from cracking in very cold weather from the expansion of pure water on freezing. Antifreeze also enables the cooling system to operate at temperatures greater than 100°C without generating enough pressure to explode.
Changes in the freezing point and boiling point of a solution depend primarily on the number of solute particles present rather than the kind of particles. Such properties of solutions are called colligative properties (from the Latin colligatus, meaning “bound together” as in a quantity). As we will see, the vapor pressure and osmotic pressure of solutions are also colligative properties.
Counting concentrations
When we determine the number of particles in a solution, it is important to remember that not all solutions with the same molarity contain the same concentration of solute particles. Consider, for example, 0.01 M aqueous solutions of sucrose, NaClNaCl, and CaCl2CaCl2. Because sucrose dissolves to give a solution of neutral molecules, the concentration of solute particles in a 0.01 M sucrose solution is 0.01 M. In contrast, both NaClNaCl and CaCl2CaCl2 are ionic compounds that dissociate in water to yield solvated ions. As a result, a 0.01 M aqueous solution of NaClNaCl contains 0.01 M Na+Na+ ions and 0.01 M Cl−Cl− ions, for a total particle concentration of 0.02 M. Similarly, the CaCl2CaCl2 solution contains 0.01 M Ca2+Ca2+ ions and 0.02 M Cl−Cl− ions, for a total particle concentration of 0.03 M. These values are correct for dilute solutions, where the dissociation of the compounds to form separately solvated ions is complete.
At higher concentrations (typically >1 M), especially with salts of small, highly charged ions (such as Mg2+Mg2+ or Al3+Al3+), or in solutions with less polar solvents, dissociation to give separate ions is often incomplete. The sum of the concentrations of the dissolved solute particles dictates the physical properties of a solution. In the following discussion, we must therefore keep the chemical nature of the solute firmly in mind.
Vapor Pressure of Solutions and Raoult’s Law
Adding a nonvolatile solute, one whose vapor pressure is too low to measure readily, to a volatile solvent decreases the vapor pressure of the solvent. We can understand this phenomenon qualitatively by examining Figure 14.6.114.6.1, which is a schematic diagram of the surface of a solution of glucose in water. In an aqueous solution of glucose, a portion of the surface area is occupied by nonvolatile glucose molecules rather than by volatile water molecules. As a result, fewer water molecules can enter the vapor phase per unit time, even though the surface water molecules have the same kinetic energy distribution as they would in pure water. At the same time, the rate at which water molecules in the vapor phase collide with the surface and reenter the solution is unaffected. The net effect is to shift the dynamic equilibrium between water in the vapor and the liquid phases, decreasing the vapor pressure of the solution compared with the vapor pressure of the pure solvent.
Figure 14.6.214.6.2 shows two beakers, one containing pure water and one containing an aqueous glucose solution, in a sealed chamber. We can view the system as having two competing equilibria: water vapor will condense in both beakers at the same rate, but water molecules will evaporate more slowly from the glucose solution because fewer water molecules are at the surface. Eventually all of the water will evaporate from the beaker containing the liquid with the higher vapor pressure (pure water) and condense in the beaker containing the liquid with the lower vapor pressure (the glucose solution). If the system consisted of only a beaker of water inside a sealed container, equilibrium between the liquid and vapor would be achieved rather rapidly, and the amount of liquid water in the beaker would remain constant.
If the particles of a solute are essentially the same size as those of the solvent and both solute and solvent have roughly equal probabilities of being at the surface of the solution, then the effect of a solute on the vapor pressure of the solvent is proportional to the number of sites occupied by solute particles at the surface of the solution. Doubling the concentration of a given solute causes twice as many surface sites to be occupied by solute molecules, resulting in twice the decrease in vapor pressure. The relationship between solution composition and vapor pressure is therefore
PA=χAP0A
PA=χAP0A(14.6.1)
where PAPA is the vapor pressure of component A of the solution (in this case the solvent), χAχA is the mole fraction of AA in solution, and P0AP0A is the vapor pressure of pure AA. Equation 14.6.114.6.1 is known as Raoult’s law, after the French chemist who developed it. If the solution contains only a single nonvolatile solute (B), then χA+χB=1χA+χB=1, and we can substitute χA=1−χBχA=1−χB to obtain
PA=(1−χB)P0A=P0A−χBP0A
PA=(1−χB)P0A=P0A−χBP0A(14.6.2)(14.6.3)
Rearranging and defining ΔPA=P0A−PAΔPA=P0A−PA, we obtain a relationship between the decrease in vapor pressure and the mole fraction of nonvolatile solute:
P0A−PA=ΔPA=χBP0A
P0A−PA=ΔPA=χBP0A(14.6.4)(14.6.5)
We can solve vapor pressure problems in either of two ways: by using Equation 14.6.114.6.1 to calculate the actual vapor pressure above a solution of a nonvolatile solute, or by using Equation 14.6.514.6.5 to calculate the decrease in vapor pressure caused by a specified amount of a nonvolatile solute.
Example 14.6.114.6.1: Anti-Freeze
Ethylene glycol (HOCH2CH2OHHOCH2CH2OH), the major ingredient in commercial automotive antifreeze, increases the boiling point of radiator fluid by lowering its vapor pressure. At 100°C, the vapor pressure of pure water is 760 mmHg. Calculate the vapor pressure of an aqueous solution containing 30.2% ethylene glycol by mass, a concentration commonly used in climates that do not get extremely cold in winter.
Given: identity of solute, percentage by mass, and vapor pressure of pure solvent
Asked for: vapor pressure of solution
Strategy:
Calculate the number of moles of ethylene glycol in an arbitrary quantity of water, and then calculate the mole fraction of water.
Use Raoult’s law to calculate the vapor pressure of the solution.
Solution:
A A 30.2% solution of ethylene glycol contains 302 g of ethylene glycol per kilogram of solution; the remainder (698 g) is water. To use Raoult’s law to calculate the vapor pressure of the solution, we must know the mole fraction of water. Thus we must first calculate the number of moles of both ethylene glycol (EG) and water present:
molesEG=(302g)(1mol62.07g)=4.87molEG
molesEG=(302g)(1mol62.07g)=4.87molEG
molesH2O=(698g)(1mol18.02g)=38.7molH2O
molesH2O=(698g)(1mol18.02g)=38.7molH2O
The mole fraction of water is thus
χH2O=38.7molH2O38.7molH2O+4.87molEG=0.888
χH2O=38.7molH2O38.7molH2O+4.87molEG=0.888
B From Raoult’s law (Equation 14.6.114.6.1), the vapor pressure of the solution is
PH2O=(χH2O)(P0H2O)=(0.888)(760mmHg)=675mmHg
PH2O=(χH2O)(P0H2O)=(0.888)(760mmHg)=675mmHg
Alternatively, we could solve this problem by calculating the mole fraction of ethylene glycol and then using Equation 14.6.514.6.5 to calculate the resulting decrease in vapor pressure:
χEG=4.87molEG4.87molEG+38.7molH2O=0.112
χEG=4.87molEG4.87molEG+38.7molH2O=0.112
ΔPH2O=(χEG)(P0H2O)=(0.112)(760mmHg)=85.1mmHg
ΔPH2O=(χEG)(P0H2O)=(0.112)(760mmHg)=85.1mmHg
PH2O=P0H2O−ΔPH2O=760mmHg−85.1mmHg=675mmHg
PH2O=P0H2O−ΔPH2O=760mmHg−85.1mmHg=675mmHg
The same result is obtained using either method.
Exercise 14.6.114.6.1
Seawater is an approximately 3.0% aqueous solution of NaClNaCl by mass with about 0.5% of other salts by mass. Calculate the decrease in the vapor pressure of water at 25°C caused by this concentration of NaClNaCl, remembering that 1 mol of NaClNaCl produces 2 mol of solute particles. The vapor pressure of pure water at 25°C is 23.8 mmHg.
Answer
: 0.45 mmHg. This may seem like a small amount, but it constitutes about a 2% decrease in the vapor pressure of water and accounts in part for the higher humidity in the north-central United States near the Great Lakes, which are freshwater lakes. The decrease therefore has important implications for climate modeling.
Even when a solute is volatile, meaning that it has a measurable vapor pressure, we can still use Raoult’s law. In this case, we calculate the vapor pressure of each component separately. The total vapor pressure of the solution (PtotPtot) is the sum of the vapor pressures of the components:
Ptot=PA+PB=χAP0A+χBP0B
Ptot=PA+PB=χAP0A+χBP0B(14.6.6)
Because χB=1−χAχB=1−χA for a two-component system,
Ptot=χAP0A+(1−χA)P0B
Ptot=χAP0A+(1−χA)P0B(14.6.7)
Thus we need to specify the mole fraction of only one of the components in a two-component system. Consider, for example, the vapor pressure of solutions of benzene and toluene of various compositions. At 20°C, the vapor pressures of pure benzene and toluene are 74.7 and 22.3 mmHg, respectively. The vapor pressure of benzene in a benzene–toluene solution is
PC6H6=χC6H6P0C6H6
PC6H6=χC6H6P0C6H6(14.6.8)
and the vapor pressure of toluene in the solution is
PC6H5CH3=χC6H5CH3P0C6H5CH3
PC6H5CH3=χC6H5CH3P0C6H5CH3(14.6.9)
Equations 14.6.814.6.8 and 14.6.914.6.9 are both in the form of the equation for a straight line: y=mx+by=mx+b, where b=0b=0. Plots of the vapor pressures of both components versus the mole fractions are therefore straight lines that pass through the origin, as shown in Figure 14.6.314.6.3. Furthermore, a plot of the total vapor pressure of the solution versus the mole fraction is a straight line that represents the sum of the vapor pressures of the pure components. Thus the vapor pressure of the solution is always greater than the vapor pressure of either component.
A solution of two volatile components that behaves like the solution in Figure 14.6.314.6.3, which is defined as a solution that obeys Raoult’s law. Like an ideal gas, an ideal solution is a hypothetical system whose properties can be described in terms of a simple model. Mixtures of benzene and toluene approximate an ideal solution because the intermolecular forces in the two pure liquids are almost identical in both kind and magnitude. Consequently, the change in enthalpy on solution formation is essentially zero (ΔHsoln≈0ΔHsoln≈0), which is one of the defining characteristics of an ideal solution.
Ideal solutions and ideal gases are both simple models that ignore intermolecular interactions.
Most real solutions, however, do not obey Raoult’s law precisely, just as most real gases do not obey the ideal gas law exactly. Real solutions generally deviate from Raoult’s law because the intermolecular interactions between the two components A and B differ. We can distinguish between two general kinds of behavior, depending on whether the intermolecular interactions between molecules A and B are stronger or weaker than the A–A and B–B interactions in the pure components. If the A–B interactions are stronger than the A–A and B–B interactions, each component of the solution exhibits a lower vapor pressure than expected for an ideal solution, as does the solution as a whole. The favorable A–B interactions effectively stabilize the solution compared with the vapor. This kind of behavior is called a negative deviation from Raoult’s law. Systems stabilized by hydrogen bonding between two molecules, such as acetone and ethanol, exhibit negative deviations from Raoult’s law. Conversely, if the A–B interactions are weaker than the A–A and B–B interactions yet the entropy increase is enough to allow the solution to form, both A and B have an increased tendency to escape from the solution into the vapor phase. The result is a higher vapor pressure than expected for an ideal solution, producing a positive deviation from Raoult’s law. In a solution of CCl4CCl4 and methanol, for example, the nonpolar CCl4CCl4 molecules interrupt the extensive hydrogen bonding network in methanol, and the lighter methanol molecules have weaker London dispersion forces than the heavier CCl4CCl4 molecules. Consequently, solutions of CCl4CCl4 and methanol exhibit positive deviations from Raoult’s law.
Example 14.6.214.6.2
For each system, compare the intermolecular interactions in the pure liquids and in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation).
cyclohexane and ethanol
methanol and acetone
n-hexane and isooctane (2,2,4-trimethylpentane)
Given: identity of pure liquids
Asked for: predicted deviation from Raoult’s law (Equation 14.6.114.6.1)
Strategy:
Identify whether each liquid is polar or nonpolar, and then predict the type of intermolecular interactions that occur in solution.
Solution:
Liquid ethanol contains an extensive hydrogen bonding network, and cyclohexane is nonpolar. Because the cyclohexane molecules cannot interact favorably with the polar ethanol molecules, they will disrupt the hydrogen bonding. Hence the A–B interactions will be weaker than the A–A and B–B interactions, leading to a higher vapor pressure than predicted by Raoult’s law (a positive deviation).
Methanol contains an extensive hydrogen bonding network, but in this case the polar acetone molecules create A–B interactions that are stronger than the A–A or B–B interactions, leading to a negative enthalpy of solution and a lower vapor pressure than predicted by Raoult’s law (a negative deviation).
Hexane and isooctane are both nonpolar molecules (isooctane actually has a very small dipole moment, but it is so small that it can be ignored). Hence the predominant intermolecular forces in both liquids are London dispersion forces. We expect the A–B interactions to be comparable in strength to the A–A and B–B interactions, leading to a vapor pressure in good agreement with that predicted by Raoult’s law (an ideal solution).
Exercise 14.6.214.6.2
For each system, compare the intermolecular interactions in the pure liquids with those in the solution to decide whether the vapor pressure will be greater than that predicted by Raoult’s law (positive deviation), approximately equal to that predicted by Raoult’s law (an ideal solution), or less than the pressure predicted by Raoult’s law (negative deviation):
benzene and n-hexane
ethylene glycol and CCl4CCl4
acetic acid and n-propanol
Answer a
: approximately equal
Answer b
: positive deviation (vapor pressure greater than predicted)
Answer c
: negative deviation (vapor pressure less than predicted)
A Video Discussing Roult's Law. Video Link: Introduction to the Vapor Pressure of a Solution (Raoult's Law), YouTube(opens in new window) [youtu.be]
A Video Discussing How to find the Vapor Pressure of a Solution. Video Link: Finding the Vapor Pressure of a Solution (Nonionic-Nonvolatile Solute), YouTube(opens in new window) [youtu.be]
Boiling Point Elevation
Recall that the normal boiling point of a substance is the temperature at which the vapor pressure equals 1 atm. If a nonvolatile solute lowers the vapor pressure of a solvent, it must also affect the boiling point. Because the vapor pressure of the solution at a given temperature is less than the vapor pressure of the pure solvent, achieving a vapor pressure of 1 atm for the solution requires a higher temperature than the normal boiling point of the solvent. Thus the boiling point of a solution is always greater than that of the pure solvent. We can see why this must be true by comparing the phase diagram for an aqueous solution with the phase diagram for pure water (Figure 14.6.414.6.4). The vapor pressure of the solution is less than that of pure water at all temperatures. Consequently, the liquid–vapor curve for the solution crosses the horizontal line corresponding to P = 1 atm at a higher temperature than does the curve for pure water.
Figure 14.6.414.6.4: Phase Diagrams of Pure Water and an Aqueous Solution of a Nonvolatile Solute. The vaporization curve for the solution lies below the curve for pure water at all temperatures, which results in an increase in the boiling point and a decrease in the freezing point of the solution.
The boiling point of a solution with a nonvolatile solute is always greater than the boiling point of the pure solvent.
The magnitude of the increase in the boiling point is related to the magnitude of the decrease in the vapor pressure. As we have just discussed, the decrease in the vapor pressure is proportional to the concentration of the solute in the solution. Hence the magnitude of the increase in the boiling point must also be proportional to the concentration of the solute (Figure 14.6.514.6.5).
We can define the boiling point elevation (ΔTbΔTb) as the difference between the boiling points of the solution and the pure solvent:
ΔTb=Tb−T0b
ΔTb=Tb−T0b(14.6.10)
where TbTb is the boiling point of the solution and T0bT0b is the boiling point of the pure solvent. We can express the relationship between ΔTbΔTb and concentration as follows
ΔTb=mKb
ΔTb=mKb(14.6.11)
where m is the concentration of the solute expressed in molality, and KbKb is the molal boiling point elevation constant of the solvent, which has units of °C/m. Table 14.6.114.6.1 lists characteristic KbKb values for several commonly used solvents. For relatively dilute solutions, the magnitude of both properties is proportional to the solute concentration.
Table 14.6.114.6.1: Boiling Point Elevation Constants (Kb) and Freezing Point Depression Constants (Kf) for Some Solvents
| Solvent | Boiling Point (°C) | Kb (°C/m) | Freezing Point (°C) | Kf (°C/m) |
| acetic acid | 117.90 | 3.22 | 16.64 | 3.63 |
| benzene | 80.09 | 2.64 | 5.49 | 5.07 |
| d-(+)-camphor | 207.4 | 4.91 | 178.8 | 37.8 |
| carbon disulfide | 46.2 | 2.42 | −112.1 | 3.74 |
| carbon tetrachloride | 76.8 | 5.26 | −22.62 | 31.4 |
| chloroform | 61.17 | 3.80 | −63.41 | 4.60 |
| nitrobenzene | 210.8 | 5.24 | 5.70 | 6.87 |
| water | 100.00 | 0.51 | 0.00 | 1.86 |
The concentration of the solute is typically expressed as molality rather than mole fraction or molarity for two reasons. First, because the density of a solution changes with temperature, the value of molarity also varies with temperature. If the boiling point depends on the solute concentration, then by definition the system is not maintained at a constant temperature. Second, molality and mole fraction are proportional for relatively dilute solutions, but molality has a larger numerical value (a mole fraction can be only between zero and one). Using molality allows us to eliminate nonsignificant zeros.
According to Table 14.6.114.6.1, the molal boiling point elevation constant for water is 0.51°C/m. Thus a 1.00 m aqueous solution of a nonvolatile molecular solute such as glucose or sucrose will have an increase in boiling point of 0.51°C, to give a boiling point of 100.51°C at 1.00 atm. The increase in the boiling point of a 1.00 m aqueous NaClNaCl solution will be approximately twice as large as that of the glucose or sucrose solution because 1 mol of NaClNaCl produces 2 mol of dissolved ions. Hence a 1.00 m NaClNaCl solution will have a boiling point of about 101.02°C.
Example 14.6.314.6.3
In Example 14.6.114.6.1, we calculated that the vapor pressure of a 30.2% aqueous solution of ethylene glycol at 100°C is 85.1 mmHg less than the vapor pressure of pure water. We stated (without offering proof) that this should result in a higher boiling point for the solution compared with pure water. Now that we have seen why this assertion is correct, calculate the boiling point of the aqueous ethylene glycol solution.
Given: composition of solution
Asked for: boiling point
Strategy:
Calculate the molality of ethylene glycol in the 30.2% solution. Then use Equation 14.6.1114.6.11 to calculate the increase in boiling point.
Solution:
From Example 14.6.114.6.1, we know that a 30.2% solution of ethylene glycol in water contains 302 g of ethylene glycol (4.87 mol) per 698 g of water. The molality of the solution is thus
molality of ethylene glycol=(4.87mol698gH2O)(1000g1kg)=6.98m
molality of ethylene glycol=(4.87mol698gH2O)(1000g1kg)=6.98m
From Equation 14.6.1114.6.11, the increase in boiling point is therefore
ΔTb=mKb=(6.98m)(0.51°C/m)=3.6°C
The boiling point of the solution is thus predicted to be 104°C. With a solute concentration of almost 7 m, however, the assumption of a dilute solution used to obtain Equation 14.6.11 may not be valid.
Exercise 14.6.3
Assume that a tablespoon (5.00 g) of NaCl is added to 2.00 L of water at 20.0°C, which is then brought to a boil to cook spaghetti. At what temperature will the water boil?
Answer
: 100.04°C, or 100°C to three significant figures. (Recall that 1 mol of NaCl produces 2 mol of dissolved particles. The small increase in temperature means that adding salt to the water used to cook pasta has essentially no effect on the cooking time.)
A Video Discussing Boiling Point Elevation and Freezing Point Depression. Video Link: Boiling Point Elevation and Freezing Point Depression, YouTube(opens in new window) [youtu.be] (opens in new window)
Freezing Point Depression
The phase diagram in Figure 14.6.4 shows that dissolving a nonvolatile solute in water not only raises the boiling point of the water but also lowers its freezing point. The solid–liquid curve for the solution crosses the line corresponding to P=1atm at a lower temperature than the curve for pure water.
We can understand this result by imagining that we have a sample of water at the normal freezing point temperature, where there is a dynamic equilibrium between solid and liquid. Water molecules are continuously colliding with the ice surface and entering the solid phase at the same rate that water molecules are leaving the surface of the ice and entering the liquid phase. If we dissolve a nonvolatile solute such as glucose in the liquid, the dissolved glucose molecules will reduce the number of collisions per unit time between water molecules and the ice surface because some of the molecules colliding with the ice will be glucose. Glucose, though, has a very different structure than water, and it cannot fit into the ice lattice. Consequently, the presence of glucose molecules in the solution can only decrease the rate at which water molecules in the liquid collide with the ice surface and solidify. Meanwhile, the rate at which the water molecules leave the surface of the ice and enter the liquid phase is unchanged. The net effect is to cause the ice to melt. The only way to reestablish a dynamic equilibrium between solid and liquid water is to lower the temperature of the system, which decreases the rate at which water molecules leave the surface of the ice crystals until it equals the rate at which water molecules in the solution collide with the ice.
By analogy to our treatment of boiling point elevation,the freezing point depression (ΔTf) is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution:
ΔTf=T0f−Tf
where T0f is the freezing point of the pure solvent and Tf is the freezing point of the solution.
The order of the terms is reversed compared with Equation 14.6.10 to express the freezing point depression as a positive number. The relationship between ΔTf and the solute concentration is given by an equation analogous to Equation 14.6.11:
ΔTf=mKf
where m is the molality of the solution and Kf is the molal freezing point depression constant for the solvent (in units of °C/m).
Like Kb, each solvent has a characteristic value of Kf (see Table 14.6.1). Freezing point depression depends on the total number of dissolved nonvolatile solute particles, just as with boiling point elevation. Thus an aqueous NaCl solution has twice as large a freezing point depression as a glucose solution of the same molality.
People who live in cold climates use freezing point depression to their advantage in many ways. For example, salt is used to melt ice and snow on roads and sidewalks, ethylene glycol is added to engine coolant water to prevent an automobile engine from being destroyed, and methanol is added to windshield washer fluid to prevent the fluid from freezing.
The decrease in vapor pressure, increase in boiling point, and decrease in freezing point of a solution versus a pure liquid all depend on the total number of dissolved nonvolatile solute particles.
Example 14.6.4: Salting the Roads
In colder regions of the United States, NaCl or CaCl2 is often sprinkled on icy roads in winter to melt the ice and make driving safer. Use the data in the Figure below to estimate the concentrations of two saturated solutions at 0°C, one of NaCl and one of CaCl2, and calculate the freezing points of both solutions to see which salt is likely to be more effective at melting ice.
Given: solubilities of two compounds
Asked for: concentrations and freezing points
Strategy:
Estimate the solubility of each salt in 100 g of water from the figure. Determine the number of moles of each in 100 g and calculate the molalities.
Determine the concentrations of the dissolved salts in the solutions. Substitute these values into Equation 14.6.13 to calculate the freezing point depressions of the solutions.
Solution:
A From Figure above, we can estimate the solubilities of NaCl and CaCl2 to be about 36 g and 60 g, respectively, per 100 g of water at 0°C. The corresponding concentrations in molality are
mNaCl=(36gNaCl100gH2O)(1molNaCl58.44gNaCl)(1000g1kg)=6.2m
mCaCl2=(60gCaCl2100gH2O)(1molCaCl2110.98gCaCl2)(1000g1kg)=5.4m
The lower formula mass of NaCl more than compensates for its lower solubility, resulting in a saturated solution that has a slightly higher concentration than CaCl2.
B Because these salts are ionic compounds that dissociate in water to yield two and three ions per formula unit of NaCl and CaCl2, respectively, the actual concentrations of the dissolved species in the two saturated solutions are 2 × 6.2 m = 12 m for NaCl and 3 × 5.4 m = 16 m for CaCl2. The resulting freezing point depressions can be calculated using Equation 14.6.13:
NaCl:ΔTf=mKf=(12m)(1.86°C/m)=22°C
CaCl2:ΔTf=mKf=(16m)(1.86°C/m)=30°C
Because the freezing point of pure water is 0°C, the actual freezing points of the solutions are −22°C and −30°C, respectively. Note that CaCl2 is substantially more effective at lowering the freezing point of water because its solutions contain three ions per formula unit. In fact, CaCl2 is the salt usually sold for home use, and it is also often used on highways.
Because the solubilities of both salts decrease with decreasing temperature, the freezing point can be depressed by only a certain amount, regardless of how much salt is spread on an icy road. If the temperature is significantly below the minimum temperature at which one of these salts will cause ice to melt (say −35°C), there is no point in using salt until it gets warmer
Exercise 14.6.4
Calculate the freezing point of the 30.2% solution of ethylene glycol in water whose vapor pressure and boiling point we calculated in Examples 14.6.5 and 14.6.5.
Answer
: −13.0°C
Example 14.6.5
Arrange these aqueous solutions in order of decreasing freezing points: 0.1 m KCl, 0.1 m glucose, 0.1 m SrCl2, 0.1 m ethylene glycol, 0.1 m benzoic acid, and 0.1 m HCl.
Given: molalities of six solutions
Asked for: relative freezing points
Strategy:
Identify each solute as a strong, weak, or nonelectrolyte, and use this information to determine the number of solute particles produced.
Multiply this number by the concentration of the solution to obtain the effective concentration of solute particles. The solution with the highest effective concentration of solute particles has the largest freezing point depression.
Solution:
A Because the molal concentrations of all six solutions are the same, we must focus on which of the substances are strong electrolytes, which are weak electrolytes, and which are nonelectrolytes to determine the actual numbers of particles in solution. KCl, SrCl2, and HCl are strong electrolytes, producing two, three, and two ions per formula unit, respectively. Benzoic acid is a weak electrolyte (approximately one particle per molecule), and glucose and ethylene glycol are both nonelectrolytes (one particle per molecule).
B The molalities of the solutions in terms of the total particles of solute are: KCl and HCl, 0.2 m; SrCl2, 0.3 m; glucose and ethylene glycol, 0.1 m; and benzoic acid, 0.1–0.2 m. Because the magnitude of the decrease in freezing point is proportional to the concentration of dissolved particles, the order of freezing points of the solutions is: glucose and ethylene glycol (highest freezing point, smallest freezing point depression) > benzoic acid > HCl = KCl > SrCl2.
Exercise 14.6.5
Arrange these aqueous solutions in order of increasing freezing points: 0.2 m NaCl, 0.3 m acetic acid, 0.1 m CaCl2, and 0.2 m sucrose.
Answer
: 0.2 m NaCl (lowest freezing point) < 0.3 m acetic acid ≈ 0.1 m CaCl2 < 0.2 m sucrose (highest freezing point)
Colligative properties can also be used to determine the molar mass of an unknown compound. One method that can be carried out in the laboratory with minimal equipment is to measure the freezing point of a solution with a known mass of solute. This method is accurate for dilute solutions (≤1% by mass) because changes in the freezing point are usually large enough to measure accurately and precisely. By comparing Kb and Kf values in Table 14.6.1, we see that changes in the boiling point are smaller than changes in the freezing point for a given solvent. Boiling point elevations are thus more difficult to measure precisely. For this reason, freezing point depression is more commonly used to determine molar mass than is boiling point elevation. Because of its very large value of Kf (37.8°C/m), d-(+)-camphor (Table 14.6.1) is often used to determine the molar mass of organic compounds by this method.
Example 14.6.6: Sulfur
A 7.08 g sample of elemental sulfur is dissolved in 75.0 g of CS2 to create a solution whose freezing point is −113.5°C. Use these data to calculate the molar mass of elemental sulfur and thus the formula of the dissolved Sn molecules (i.e., what is the value of n?).
Given: masses of solute and solvent and freezing point
Asked for: molar mass and number of S atoms per molecule
Strategy:
Use Equation 14.6.12, the measured freezing point of the solution, and the freezing point of CS2 from Table 14.6.1 to calculate the freezing point depression. Then use Equation 14.6.13 and the value of Kf from Table 14.6.1 to calculate the molality of the solution.
From the calculated molality, determine the number of moles of solute present.
Use the mass and number of moles of the solute to calculate the molar mass of sulfur in solution. Divide the result by the molar mass of atomic sulfur to obtain n, the number of sulfur atoms per mole of dissolved sulfur.
Solution:
A The first step is to calculate the freezing point depression using Equation 14.6.12:
ΔTf=T0f−Tf=−112.1°C−(−113.5°C)=1.4°C
Then Equation 14.6.13 gives
m=ΔTfKf=1.4°C3.74°C/m=0.37m
B The total number of moles of solute present in the solution is
moles solute=(0.37molkg)(75.0g)(1kg1000g)=0.028mol
C We now know that 0.708 g of elemental sulfur corresponds to 0.028 mol of solute. The molar mass of dissolved sulfur is thus
molar mass=7.08g0.028mol=260g/mol
The molar mass of atomic sulfur is 32 g/mol, so there must be 260/32 = 8.1 sulfur atoms per mole, corresponding to a formula of S8.
Exercise 14.6.6
One of the byproducts formed during the synthesis of C60 is a deep red solid containing only carbon. A solution of 205 mg of this compound in 10.0 g of CCl4 has a freezing point of −23.38°C. What are the molar mass and most probable formula of the substance?
Answer
: 847 g/mol; C70
A Video Discussing how to find the Molecular Weight of an Unknown using Colligative Properties. Video Link: Finding the Molecular Weight of an Unknown using Colligative Properties, YouTube(opens in new window) [youtu.be]
Osmotic Pressure
Osmotic pressure is a colligative property of solutions that is observed using a semipermeable membrane, a barrier with pores small enough to allow solvent molecules to pass through but not solute molecules or ions. The net flow of solvent through a semipermeable membrane is called osmosis (from the Greek osmós, meaning “push”). The direction of net solvent flow is always from the side with the lower concentration of solute to the side with the higher concentration.
Osmosis can be demonstrated using a U-tube like the one shown in Figure 14.6.6, which contains pure water in the left arm and a dilute aqueous solution of glucose in the right arm. A net flow of water through the membrane occurs until the levels in the arms eventually stop changing, which indicates that equilibrium has been reached. The osmotic pressure (Π) of the glucose solution is the difference in the pressure between the two sides, in this case the heights of the two columns. Although the semipermeable membrane allows water molecules to flow through in either direction, the rate of flow is not the same in both directions because the concentration of water is not the same in the two arms. The net flow of water through the membrane can be prevented by applying a pressure to the right arm that is equal to the osmotic pressure of the glucose solution.
Just as with any other colligative property, the osmotic pressure of a solution depends on the concentration of dissolved solute particles. Osmotic pressure obeys a law that resembles the ideal gas equation:
Π=nRTV=MRT
where M is the number of moles of solute per unit volume of solution (i.e., the molarity of the solution), R is the ideal gas constant, and T is the absolute temperature.
As shown in Example 14.6.7, osmotic pressures tend to be quite high, even for rather dilute solutions.
Example 14.6.7
When placed in a concentrated salt solution, certain yeasts are able to produce high internal concentrations of glycerol to counteract the osmotic pressure of the surrounding medium. Suppose that the yeast cells are placed in an aqueous solution containing 4.0% NaCl by mass; the solution density is 1.02 g/mL at 25°C.
Calculate the osmotic pressure of a 4.0% aqueous NaCl solution at 25°C.
If the normal osmotic pressure inside a yeast cell is 7.3 atm, corresponding to a total concentration of dissolved particles of 0.30 M, what concentration of glycerol must the cells synthesize to exactly balance the external osmotic pressure at 25°C?
Given: concentration, density, and temperature of NaCl solution; internal osmotic pressure of cell
Asked for: osmotic pressure of NaCl solution and concentration of glycerol needed
Strategy:
Calculate the molarity of the NaCl solution using the formula mass of the solute and the density of the solution. Then calculate the total concentration of dissolved particles.
Use Equation 14.6.14 to calculate the osmotic pressure of the solution.
Subtract the normal osmotic pressure of the cells from the osmotic pressure of the salt solution to obtain the additional pressure needed to balance the two. Use Equation 14.6.14 to calculate the molarity of glycerol needed to create this osmotic pressure.
Solution:
A The solution contains 4.0 g of NaCl per 100 g of solution. Using the formula mass of NaCl (58.44 g/mol) and the density of the solution (1.02 g/mL), we can calculate the molarity:
MNaCl=molesNaClliter solution=(4.0gNaCl58.44g/molNaCl)(1100gsolution)(1.02gsolution1.00mLsolution)(1000mL1L)=0.70MNaCl
Because 1 mol of NaCl produces 2 mol of particles in solution, the total concentration of dissolved particles in the solution is (2)(0.70 M) = 1.4 M.
B Now we can use Equation 14.6.14 to calculate the osmotic pressure of the solution:
Π=MRT=(1.4mol/L)0.0821(L⋅atm)/(K⋅mol)=34atm
C If the yeast cells are to exactly balance the external osmotic pressure, they must produce enough glycerol to give an additional internal pressure of (34 atm − 7.3 atm) = 27 atm. Glycerol is a nonelectrolyte, so we can solve Equation 14.6.14 for the molarity corresponding to this osmotic pressure:
M=ΠRT=27atm0.0821(L⋅atm)/(K⋅mol)=1.1Mglycerol
In solving this problem, we could also have recognized that the only way the osmotic pressures can be the same inside the cells and in the solution is if the concentrations of dissolved particles are the same. We are given that the normal concentration of dissolved particles in the cells is 0.3 M, and we have calculated that the NaCl solution is effectively 1.4 M in dissolved particles. The yeast cells must therefore synthesize enough glycerol to increase the internal concentration of dissolved particles from 0.3 M to 1.4 M—that is, an additional 1.1 M concentration of glycerol.
Exercise 14.6.7
Assume that the fluids inside a sausage are approximately 0.80 M in dissolved particles due to the salt and sodium nitrite used to prepare them. Calculate the osmotic pressure inside the sausage at 100°C to learn why experienced cooks pierce the semipermeable skin of sausages before boiling them.
Answer
: 24 atm
Because of the large magnitude of osmotic pressures, osmosis is extraordinarily important in biochemistry, biology, and medicine. Virtually every barrier that separates an organism or cell from its environment acts like a semipermeable membrane, permitting the flow of water but not solutes. The same is true of the compartments inside an organism or cell. Some specialized barriers, such as those in your kidneys, are slightly more permeable and use a related process called dialysis, which permits both water and small molecules to pass through but not large molecules such as proteins.
The same principle has long been used to preserve fruits and their essential vitamins over the long winter. High concentrations of sugar are used in jams and jellies not for sweetness alone but because they greatly increase the osmotic pressure. Thus any bacteria not killed in the cooking process are dehydrated, which keeps them from multiplying in an otherwise rich medium for bacterial growth. A similar process using salt prevents bacteria from growing in ham, bacon, salt pork, salt cod, and other preserved meats. The effect of osmotic pressure is dramatically illustrated in Figure 14.6.7, which shows what happens when red blood cells are placed in a solution whose osmotic pressure is much lower or much higher than the internal pressure of the cells.
In addition to capillary action, trees use osmotic pressure to transport water and other nutrients from the roots to the upper branches. Evaporation of water from the leaves results in a local increase in the salt concentration, which generates an osmotic pressure that pulls water up the trunk of the tree to the leaves.
Finally, a process called reverse osmosis can be used to produce pure water from seawater. As shown schematically in Figure 14.6.8, applying high pressure to seawater forces water molecules to flow through a semipermeable membrane that separates pure water from the solution, leaving the dissolved salt behind. Large-scale desalinization plants that can produce hundreds of thousands of gallons of freshwater per day are common in the desert lands of the Middle East, where they supply a large proportion of the freshwater needed by the population. Similar facilities are now being used to supply freshwater in southern California. Small, hand-operated reverse osmosis units can produce approximately 5 L of freshwater per hour, enough to keep 25 people alive, and are now standard equipment on US Navy lifeboats.
A Video Discussing Osmotic Pressure. Video Link: Osmotic Pressure, YouTube(opens in new window) [youtu.be] (opens in new window)
Colligative Properties of Electrolyte Solutions
Thus far we have assumed that we could simply multiply the molar concentration of a solute by the number of ions per formula unit to obtain the actual concentration of dissolved particles in an electrolyte solution. We have used this simple model to predict such properties as freezing points, melting points, vapor pressure, and osmotic pressure. If this model were perfectly correct, we would expect the freezing point depression of a 0.10 m solution of sodium chloride, with 2 mol of ions per mole of NaCl in solution, to be exactly twice that of a 0.10 m solution of glucose, with only 1 mol of molecules per mole of glucose in solution. In reality, this is not always the case. Instead, the observed change in freezing points for 0.10 m aqueous solutions of NaCl and KCl are significantly less than expected (−0.348°C and −0.344°C, respectively, rather than −0.372°C), which suggests that fewer particles than we expected are present in solution.
The relationship between the actual number of moles of solute added to form a solution and the apparent number as determined by colligative properties is called the van’t Hoff factor (i) and is defined as follows:
i=apparent number of particles in solution number of moles of solute dissolved
Named for Jacobus Hendricus van’t Hoff (1852–1911), a Dutch chemistry professor at the University of Amsterdam who won the first Nobel Prize in Chemistry (1901) for his work on thermodynamics and solutions.
As the solute concentration increases, the van’t Hoff factor decreases.
The van’t Hoff factor is therefore a measure of a deviation from ideal behavior. The lower the van’t Hoff factor, the greater the deviation. As the data in Table 14.6.2 show, the van’t Hoff factors for ionic compounds are somewhat lower than expected; that is, their solutions apparently contain fewer particles than predicted by the number of ions per formula unit. As the concentration of the solute increases, the van’t Hoff factor decreases because ionic compounds generally do not totally dissociate in aqueous solution.
Table 14.6.2: van’t Hoff Factors for 0.0500 M Aqueous Solutions of Selected Compounds at 25°C
| Compound | i (measured) | i (ideal) |
| glucose | 1.0 | 1.0 |
| sucrose | 1.0 | 1.0 |
| NaCl | 1.9 | 2.0 |
| HCl | 1.9 | 2.0 |
| MgCl2 | 2.7 | 3.0 |
| FeCl3 | 3.4 | 4.0 |
| Ca(NO3)2 | 2.5 | 3.0 |
| AlCl3 | 3.2 | 4.0 |
| MgSO4 | 1.4 | 2.0 |
Instead, some of the ions exist as ion pairs, a cation and an anion that for a brief time are associated with each other without an intervening shell of water molecules (Figure 14.6.9). Each of these temporary units behaves like a single dissolved particle until it dissociates. Highly charged ions such as Mg2+, Al3+, SO2−4, and PO3−4 have a greater tendency to form ion pairs because of their strong electrostatic interactions. The actual number of solvated ions present in a solution can be determined by measuring a colligative property at several solute concentrations.
Example 14.6.8: Iron Chloride in Water
A 0.0500 M aqueous solution of FeCl3 has an osmotic pressure of 4.15 atm at 25°C. Calculate the van’t Hoff factor i for the solution.
Given: solute concentration, osmotic pressure, and temperature
Asked for: van’t Hoff factor
Strategy:
Use Equation 14.6.14 to calculate the expected osmotic pressure of the solution based on the effective concentration of dissolved particles in the solvent.
Calculate the ratio of the observed osmotic pressure to the expected value. Multiply this number by the number of ions of solute per formula unit, and then use Equation 14.6.15 to calculate the van’t Hoff factor.
Solution:
A If FeCl3 dissociated completely in aqueous solution, it would produce four ions per formula unit [Fe3+(aq) plus 3Cl−(aq)] for an effective concentration of dissolved particles of 4 × 0.0500 M = 0.200 M. The osmotic pressure would be
Π=MRT=(0.200mol/L)0.0821(L⋅atm)/(K⋅mol)=4.89atm
B The observed osmotic pressure is only 4.15 atm, presumably due to ion pair formation. The ratio of the observed osmotic pressure to the calculated value is 4.15 atm/4.89 atm = 0.849, which indicates that the solution contains (0.849)(4) = 3.40 particles per mole of FeCl3 dissolved. Alternatively, we can calculate the observed particle concentration from the osmotic pressure of 4.15 atm:
4.15atm=M(0.0821(L⋅atm)(K⋅mol))(298K)
or after rearranging
M=0.170mol
The ratio of this value to the expected value of 0.200 M is 0.170 M/0.200 M = 0.850, which again gives us (0.850)(4) = 3.40 particles per mole of FeCl3 dissolved. From Equation 14.6.15, the van’t Hoff factor for the solution is
i=3.40 particles observed1 formula unitFeCl3=3.40
Exercise 14.6.8: Magnesium Chloride in Water
Calculate the van’t Hoff factor for a 0.050 m aqueous solution of MgCl2 that has a measured freezing point of −0.25°C.
Answer
: 2.7 (versus an ideal value of 3).
A Video Discussing the Colligative Properties in Solutions. Video Link: Colligative Properties in Solutions, YouTube(opens in new window) [youtu.be]
Summary
The colligative properties of a solution depend on only the total number of dissolved particles in solution, not on their chemical identity. Colligative properties include vapor pressure, boiling point, freezing point, and osmotic pressure. The addition of a nonvolatile solute (one without a measurable vapor pressure) decreases the vapor pressure of the solvent. The vapor pressure of the solution is proportional to the mole fraction of solvent in the solution, a relationship known as Raoult’s law. Solutions that obey Raoult’s law are called ideal solutions. Most real solutions exhibit positive or negative deviations from Raoult’s law. The boiling point elevation (ΔTb) and freezing point depression (ΔTf) of a solution are defined as the differences between the boiling and freezing points, respectively, of the solution and the pure solvent. Both are proportional to the molality of the solute. When a solution and a pure solvent are separated by a semipermeable membrane, a barrier that allows solvent molecules but not solute molecules to pass through, the flow of solvent in opposing directions is unequal and produces an osmotic pressure, which is the difference in pressure between the two sides of the membrane. Osmosis is the net flow of solvent through such a membrane due to different solute concentrations. Dialysis uses a semipermeable membrane with pores that allow only small solute molecules and solvent molecules to pass through. In more concentrated solutions, or in solutions of salts with highly charged ions, the cations and anions can associate to form ion pairs, which decreases their effect on the colligative properties of the solution. The extent of ion pair formation is given by the van’t Hoff factor (i), the ratio of the apparent number of particles in solution to the number predicted by the stoichiometry of the salt.
Henry’s law: C=kP
Raoult’s law: PA=χAP0A
vapor pressure lowering: P0A−PA=ΔPA=χBP0A
vapor pressure of a system containing two volatile components: Ptot=χAP0A+(1−χA)P0B
boiling point elevation: ΔTb=mKb
freezing point depression: ΔTf=mKf
osmotic pressure: Π=nRTV=MRT
van ’t Hoff factor: i=apparent number of particles in solution number of moles of solute dissolved
14.5: Expressing Solution Concentration
14.7: The Colligative Properties of Strong Electrolyte Solutions |
188525 | https://community.smartsheet.com/discussion/68236/sum-formula-with-extending-range/p1 | SUM formula with extending range
Is it possibly to have a SUM formula in a cell at the top of a column, calculating the sum of the contents of that said column, with the range beginning at a specific row but continuing to the bottom of the sheet and extending down as rows are added to the sheet automatically? It would obviously be easier for me to just have the formula in sheet summary, however for viability purposes it be beneficial for me to have the formula at the top of the column. Be glad of anyone's ideas.
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Are you trying to SUM the entire column?
No just from a specific row down. So if I have the formula in Column1 Row1 I want it to SUM Column1 Row3 down to the bottom of the sheet, and for the SUM to include any new rows that are added automatically via the workflow i have in place.
You would need an additional "helper" column that will designate which rows to SUM and then use a SUMIFS instead. The helper column could be a column that basically replicates the row number. You would put this formula in Row 1 and dragfill down. New rows should trigger autofill.
=COUNTIFS([Other Column Name]$1:[Other Column Name]@row, OR(@cell = "", @cell <> ""))
Then to sum everything from row 3 down, you would use something along the lines of...
=SUMIFS([Sum Column]:[Sum Column], [Helper Column]:[Helper Column], @cell >= 3)
Was looking for the same thing because I wanted to have a 'rollup' at the top of my sheet summing specific variables in one column along with the value in a corresponding column. I'm adding rows to the sheet via a zap, and finally realized (actually after reading this) that I just needed to create a couple new columns so that I could run my SUMIFS formula from Column:Column (top to bottom) without circular references.
Formatting added for sharing purposes... now it looks like this, where my far right columns run SUMIFS for each of the task names and then update total value per task as the sheet grows. Not the most ideal or pretty, but gets the job done, and allows me to then target that summed area from a dashboard!
Hi @BrandIMPs
I hope you're well and safe!
Another option could also be to use the Sheet Summary section instead, and then it won't clutter the sheet, and you can bring it up wherever you are in the sheet.
Make sense?
Would that work/help?
I hope that helps!
Be safe and have a fantastic week!
Best,
Andrée Starå | Workflow Consultant / CEO @ WORK BOLD
✅Did my post(s) help or answer your question or solve your problem? Please help the Community by marking it as the accepted answer/helpful. It will make it easier for others to find a solution or help to answer!
@Paul Newcome ,
Thanks for sharing your example syntax. I am trying to accomplish exactly what @Jonnie White is asking. Unfortunately, I'm not getting it to work.
I could get the "Helper" column working which basically returned a number that matches the row number. 👍🏻
When I add in the "=SUMIFS([Sum Column]:[Sum Column], [Helper Column]:[Helper Column], @cell >= 3)" syntax… the Sum column returns #BLOCKED and then the entire Helper column changes to #CIRCULAR REFERENCE.
@Reliability Engineer The SUMIFS would need to go into a different column.
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188526 | https://mathforums.com/t/expanding-x-h-4.21253/ | Expanding (x + h)^4 | Math Forums
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Expanding (x + h)^4
Thread starterrgarcia128
Start dateSep 24, 2011
Tagsexpanding
R
rgarcia128
Joined Sep 2011
23 Posts | 0+
Discussion Starter
Sep 24, 2011
#1
We are learning proofs of derivative formulas. I am having trouble with proving the power rule; partly because my teacher has some "interesting" ways of teaching how to do it. In the textbook it says: "In finding the derivative of x4 we had to expand (x+h)^4. Here we need to expand (x+h)^n and we use the Binomial Theorem to do so:"
I don't quite understand there binomial theorem, so I figure if I see the (x+h)^4 expanded I might understand.
greg1313
Joined Oct 2008
8K Posts | 1K+
London, Ontario, Canada - The Forest City
Sep 25, 2011
Last edited: Aug 3, 2017
#2
Last edited: Aug 3, 2017
(x+h)4=x 4+4 x 3 h+6 x 2 h 2+4 x h 3+h 4 See the pattern in the exponents?
Now, with coefficients,
.
(2 x+3 h)4=16 x 4+96 x 3 h+216 x 2 h 2+216 x h 3+81 h 4
and these coefficients (on the right hand.side) come from Pascal's Triangle (and the coefficients in (2x + 3h)^4, that is, 2 and 3).
Time to do some reading!
H
Hoempa
Joined Apr 2010
3K Posts | 362+
Sep 25, 2011
Last edited by a moderator: Sep 22, 2017
#3
Last edited by a moderator: Sep 22, 2017
(x+h)4=1 x 4+4 x 3 h+6 x 2 h 2+4 x h 3+1 h 4=x 4+4 x 3 h+6 x 2 h 2+4 x h 3+h 4
R
rgarcia128
Joined Sep 2011
23 Posts | 0+
Discussion Starter
Sep 25, 2011
#4
I saw the first reply Greg and I went on a google search of the binomial theorem. I learned a good deal. I understand binomial theorem and Pascals' triangle. I can now do this and other proofs (after power rule proof) with ease I think I am going to take a nap; tired. I have two exams tomorrow, and two assignment due tonight. So I am up most of the time and napping 1-3 hours in between. Thanks to both of you for replying!
F
farmtalk
Joined May 2011
44 Posts | 0+
West Virginia
Sep 25, 2011
#5
Since you are in the calculus sector of the forum, I figured I would tell you: I dont know if u have done anything on derivatives, but our teacher is making us solve them the long way, and at first I had forgotten the binomial theorem and it took forever to distribute; anything over the second power was a pain in the rear end.
Anyway, last week I studied the binomial theorem and pascal's triangle, and now I can solve derivatives in half the time...Just thought I would help you out with a suggestion
--Farmtalk--
R
rgarcia128
Joined Sep 2011
23 Posts | 0+
Discussion Starter
Sep 27, 2011
#6
Thanks Farmtalk. I do feel after study Pascal's Triangle that I will be able to solve even the basic factoring problems. I need to learn it further but will wait until the weekend to do so.
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188527 | https://online.stat.psu.edu/stat200/book/export/html/115 | 6: Hypothesis Testing, Part 2
6: Hypothesis Testing, Part 2
6: Hypothesis Testing, Part 2
Objectives
Upon successful completion of this lesson, you should be able to:
Identify Type I and Type II errors
Select an appropriate significance level (i.e., α level) for a given scenario
Explain the problems associated with conducting multiple tests
Interpret the results of a hypothesis test in terms of practical significance
Distinguish between practical significance and statistical significance
Explain how changing different aspects of a research study would change the statistical power of the tests conducted
Compare and contrast confidence intervals and hypothesis tests
Last week you learned how to conduct a hypothesis testing using randomization procedures in StatKey. This week we are going to delve a bit deeper into hypothesis testing. Concepts such as errors, significance (α) levels, issues with multiple testing,practical significance, and statistical power apply to hypothesis tests for all the parameters that we have learned and will also apply to those that we will learn later in this course.
6.1 - Type I and Type II Errors
6.1 - Type I and Type II Errors
When conducting a hypothesis test there are two possible decisions: reject the null hypothesis or fail to reject the null hypothesis. You should remember though, hypothesis testing uses data from a sample to make an inference about a population. When conducting a hypothesis test we do not know the population parameters. In most cases, we don't know if our inference is correct or incorrect.
When we reject the null hypothesis there are two possibilities. There could really be a difference in the population, in which case we made a correct decision. Or, it is possible that there is not a difference in the population (i.e., H 0 is true) but our sample was different from the hypothesized value due to random sampling variation. In that case we made an error. This is known as a Type I error.
When we fail to reject the null hypothesis there are also two possibilities. If the null hypothesis is really true, and there is not a difference in the population, then we made the correct decision. If there is a difference in the population, and we failed to reject it, then we made a Type II error.
Type I Error
Rejecting H 0 when H 0 is really true, denoted by α ("alpha") and commonly set at .05
α=P(T y p e I e r r o r)
Type II Error
Failing to reject H 0 when H 0 is really false, denoted by β ("beta")
β=P(T y p e I I e r r o r)
| Decision | Reality |
--- |
| H 0 is true | H 0 is false |
| Reject H 0, (conclude H a) | Type I error | Correct decision |
| Fail to reject H 0 | Correct decision | Type II error |
Example: Trial
A man goes to trial where he is being tried for the murder of his wife.
We can put it in a hypothesis testing framework. The hypotheses being tested are:
H 0 : Not Guilty
H a : Guilty
Type I erroris committed if we reject H 0 when it is true. In other words, did not kill his wife but was found guilty and is punished for a crime he did not really commit.
Type II erroris committed if we fail to reject H 0 when it is false. In other words, if the man did kill his wife but was found not guilty and was not punished.
Example: Culinary Arts Study
A group of culinary arts students is comparing two methods for preparing asparagus: traditional steaming and a new frying method. They want to know if patrons of their school restaurant prefer their new frying method over the traditional steaming method. A sample of patrons are given asparagus prepared using each method and asked to select their preference. A statistical analysis is performed to determine if more than 50% of participants prefer the new frying method:
H 0:p=.50
H a:p>.50
Type I erroroccurs if they reject the null hypothesis and conclude that their new frying method is preferred when in reality is it not. This may occur if, by random sampling error, they happen to get a sample that prefers the new frying method more than the overall population does. If this does occur, the consequence is that the students will have an incorrect belief that their new method of frying asparagus is superior to the traditional method of steaming.
Type II erroroccurs if they fail to reject the null hypothesis and conclude that their new method is not superior when in reality it is. If this does occur, the consequence is that the students will have an incorrect belief that their new method is not superior to the traditional method when in reality it is.
6.2 - Significance Levels
6.2 - Significance Levels
As we saw in the examples on the previous page, the consequences of Type I and Type II errors vary depending on the situation. Researchers take into account the consequences of each when they are setting their α level before data are even collected.
In many disciplines an α level of 0.05 is standard, for example in the social sciences. There are some situations when a higher or lower α level may be desirable. Pilot studies (smaller studies performed before a larger study) often use a higher α level because their purpose is to gain information about the data that may be collected in a larger study; pilot studies are not typically used to make important decisions.
Studies in which making a Type I error would be more dangerous than making a Type II error may use smaller α levels. For example, in medical research studies where making a Type I error could mean giving patients ineffective treatments, a smaller α level may be set in order to reduce the likelihood of such a negative consequence. Lower α levels mean that smaller p-values are needed to reject the null hypothesis; this makes it more difficult to reject the null hypothesis, but this also reduces the probability of committing a Type I error.
6.3 - Issues with Multiple Testing
6.3 - Issues with Multiple Testing
If we are conducting a hypothesis test with an α level of 0.05, then we are accepting a 5% chance of making a Type I error (i.e., rejecting the null hypothesis when the null hypothesis is really true). If we would conduct 100 hypothesis tests at a 0.05 α level where the null hypotheses are really true, we would expect to reject the null and make a Type I error in about 5 of those tests.
Later in this course you will learn about some statistical procedures that may be used instead of performing multiple tests. For example, to compare the means of more than two groups you can use an analysis of variance ("ANOVA"). To compare the proportions of more than two groups you can conduct a chi-square goodness-of-fit test.
Publication Bias
A related issue is publication bias. Research studies with statistically significant results are published much more often than studies without statistically significant results. This means that if 100 studies are performed in which there is really no difference in the population, the 5 studies that found statistically significant results may be published while the 95 studies that did not find statistically significant results will not be published. Thus, when you perform a review of published literature you will only read about the studies that found statistically significance results. You would not find the studies that did not find statistically significant results.
Quick Correction for Multiple Tests
One quick method for correcting for multiple tests is to divide the alpha level by the number of tests being conducted. For instance, if you are comparing three groups using a series of three pairwise tests you could divided your overall alpha level ("family-wise alpha level") by three. If we were using a standard alpha level of 0.05, then our pairwise alpha level would be 0.05 3=0.016667. We would then compare each of our three p-values to 0.016667 to determine statistical significance. This is known as theBonferronimethod. This is one of the most conservative approaches to controlling for multiple tests (i.e., more likely to make a Type II error). Later in the course you will learn how to use the Tukey method when comparing the means of three or more groups, this approach is often preferred because it is more liberal.
6.4 - Practical Significance
6.4 - Practical Significance
In the last lesson, you learned how to identify statistically significant differences using hypothesis testing methods.If the p value is less than the α level (typically 0.05), then the results arestatistically significant. Results are said to be statistically significant when the difference between the hypothesized population parameter and observed sample statistic is large enough to conclude that it is unlikely to have occurred by chance.
Practical significancerefers to the magnitude of the difference, which is known as theeffect size. Results are practically significant when the difference is large enough to be meaningful in real life. What is meaningful may be subjective and may depend on the context.
Note that statistical significance is directly impacted by sample size. Recall that there is an inverse relationship between sample size and the standard error (i.e., standard deviation of the sampling distribution). Very small differences will be statistically significant with a very large sample size. Thus, when results are statistically significant it is important to also examine practical significance. Practical significance is not directly influenced by sample size.
Example: Weight-Loss Program
Researchers are studying a new weight-loss program. Using a large sample they construct a 95% confidence interval for the mean amount of weight loss after six months on the program to be [0.12, 0.20]. All measurements were taken in pounds. Note that this confidence interval does not contain 0, so we know that their results were statistically significant at a 0.05 alpha level. However, most people would say that the results are not practically significant because a six month weight-loss program should yield a mean weight loss much greater than the one observed in this study.
Effect Size
For some tests there are commonly used measures of effect size. For example, when comparing the difference in two means we often compute Cohen's d which is the difference between the two observed sample means in standard deviation units:
d=x―1−x―2 s p
Where s p is the pooled standard deviation
s p=(n 1−1)s 1 2+(n 2−1)s 2 2 n 1+n 2−2
Below are commonly used standards when interpreting Cohen's d:
| Cohen's d | Interpretation |
--- |
| 0 - 0.2 | Little or no effect |
| 0.2 - 0.5 | Small effect size |
| 0.5 - 0.8 | Medium effect size |
| 0.8 or more | Large effect size |
For a single mean, you can compute the difference between the observed mean and hypothesized mean in standard deviation units: d=x―−μ 0 s
For correlation and regression we can compute r 2 which is known as the coefficient of determination. This is the proportion of shared variation. We will learn more about r 2 when we study simple linear regression and correlation at the end of this course.
Example: SAT-Math Scores
Research question:Are SAT-Math scores at one college greater than the known population mean of 500?
H 0:μ=500
H a:μ>500
Data are collected from a random sample of 1,200 students at that college. In that sample, x―=506 and the sample standard deviation was 100. A one-sample mean test was performed and the resulting p-value was 0.0188. Because p≤α, the null hypothesis should be rejected. These results are statistically significant. There is evidence that the population mean is greater than 500.
But, let's also consider practical significance. The difference between an SAT-Math score 500 and an SAT-Math score of 506 is very small. With a standard deviation of 100, this difference is only 506−500 100=0.06 standard deviations. In most cases, this would not be considered practically significant.
Example: Commute Times
Research question:Are the mean commute times different in Atlanta and St. Louis?
Descriptive Statistics: Commute Time| City | N | Mean | StDev |
--- --- |
| Atlanta | 500 | 29.110 | 20.718 |
| St. Louis | 500 | 21.970 | 14.232 |
Using the dataset built in to StatKey, a two-tailed randomization test was conducted resulting in a p value < 0.001. Because the null hypothesis was rejected, the results are said to be statistically significant.
Practical significance can be examined by computing Cohen's d. We'll use the equations from above:
d=x―1−x―2 s p
Where s p is the pooled standard deviation
s p=(n 1−1)s 1 2+(n 2−1)s 2 2 n 1+n 2−2
First, we compute the pooled standard deviation:
s p=(500−1)20.718 2+(500−1)14.232 2 500+500−2
s p=(499)(429.236)+(499)(202.550)998
s p=214188.527+101072.362 998
s p=315260.853 998
s p=315.893
s p=17.773
Note: The pooled standard deviation should always be between the two sample standard deviations.
Next, we can compute Cohen's d:
d=29.110−21.970 17.773
d=7.14 17.773
d=0.402
The mean commute time in Atlanta was 0.402 standard deviations greater than the mean commute time in St. Louis. Using the guidelines for interpreting Cohen's d in the table above, this is a small effect size.
6.5 - Power
6.5 - Power
The probability of rejecting the null hypothesis, given that the null hypothesis is false, is known as power. In other words, power is the probability of correctly rejecting H 0.
Power P o w e r=1−β β =probability of committing a Type II Error.
The power of a test can be increased in a number of ways, for example increasing the sample size, decreasing the standard error, increasing the difference between the sample statistic and the hypothesized parameter, or increasing the alpha level. Using a directional test (i.e., left- or right-tailed) as opposed to a two-tailed test would also increase power.
When we increase the sample size, decrease the standard error, or increase the difference between the sample statistic and hypothesized parameter, the p value decreases, thus making it more likely that we reject the null hypothesis. When we increase the alpha level, there is a larger range of p values for which we would reject the null hypothesis. Going from a two-tailed to a one-tailed test cuts the p value in half. In all of these cases, we say that statistically power is increased.
There is a relationship between α and β.If the sample size is fixed, then decreasing α will increase β. If we want both α and β to decrease (i.e., decreasing the likelihood of both Type I and Type II errors), then we should increase the sample size.
Try it!
Question 1
If the power of a statistical test is increased, for example by increasing the sample size, how does the probability of a Type II error change?
Answer
The probability of committing a Type II error is known as β.
P o w e r+β=1
P o w e r=1−β
If power increases then β must decrease. So, if the power of a statistical test is increased, for example by increasing the sample size, the probability of committing a Type II error decreases.
Question 2
When we fail to reject the null hypothesis,can we accept the null hypothesis? For example, with a p value of 0.12 we fail to reject the null hypothesis at 0.05 alpha level. Can we say that the data support the null hypothesis?
Answer
No. When we perform a hypothesis test, we only set the Type I error rate (i.e., alpha level) and guard against it. Thus, we can only present the strength of evidence against the null hypothesis. We can sidestep the concern about Type II error if the conclusion never mentions that the null hypothesis is accepted. When the null hypothesis cannot be rejected, there are two possible cases:
1) The null hypothesis is really true.
2) The sample size is not large enough to reject the null hypothesis (i.e., statistical power is too low).
Question 3
A study was conducted by a retail store to determine if the majority of their customers were teenagers. With p^=0.48, the null hypothesis was not rejected and the company concluded that they did not have enough evidence that the majority of their customers were teenagers. But, in reality, the proportion of all of their customers (i.e., the population) who are teenagers is actually p=0.53. Did this research study result in a Type I error, Type II error, or correct decision?
Answer
The result of the study was to fail to reject the null hypothesis. In reality, the null hypothesis was false. This is a Type II error.
Question 4
A university conducted a hypothesis test to determine if their students' average SAT-Math score was greater than the national average of 500. They collected a sample of n=800 students and found x―=506. The t-test statistic was 1.70 and p=0.045 therefore they rejected the null hypothesis and concluded that the mean SAT-Math score at their university was higher than the national average. However, in reality, in the population of all students at the university, the mean SAT-Math score is 503. Did this research study result in a Type I error, Type II error, or correct decision?
Answer
This study came to a correct conclusion. They rejected the null hypothesis and concluded that μ>500 when in reality μ=503 which is greater than 500.
6.6 - Confidence Intervals & Hypothesis Testing
6.6 - Confidence Intervals & Hypothesis Testing
Confidence intervals and hypothesis tests are similar in that they are both inferential methods that rely on an approximated sampling distribution. Confidence intervals use data from a sample to estimate a population parameter. Hypothesis tests use data from a sample to test a specified hypothesis.Hypothesis testing requires that we have a hypothesized parameter.
The simulation methods used to construct bootstrap distributions and randomization distributions are similar. One primary difference is a bootstrap distribution is centered on the observed sample statistic while a randomization distribution is centered on the value in the null hypothesis.
In Lesson 4, we learned confidence intervals contain a range of reasonable estimates of the population parameter. All of the confidence intervals we constructed in this course were two-tailed. These two-tailed confidence intervals go hand-in-hand with the two-tailed hypothesis tests we learned in Lesson 5. The conclusion drawn from a two-tailed confidence interval is usually the same as the conclusion drawn from a two-tailed hypothesis test.In other words, if the the 95% confidence interval contains the hypothesized parameter, then a hypothesis test at the 0.05 α level will almost always fail to reject the null hypothesis. If the 95% confidence interval does not contain the hypothesize parameter, then a hypothesis test at the 0.05 α level will almost always reject the null hypothesis.
Example: Mean
This example uses the Body Temperature dataset built in to StatKey for constructing abootstrap confidence interval and conducting a randomization test.
Let's start by constructing a 95% confidence interval using the percentile method in StatKey:
The 95% confidence interval for the mean body temperature in the population is [98.044, 98.474].
Now, what if we want to know if there is enough evidence that the mean body temperature is different from 98.6 degrees? We can conduct a hypothesis test.Because 98.6 is not contained within the 95% confidence interval, it is not a reasonable estimate of the population mean. We should expect to have a p value less than 0.05 and to reject the null hypothesis.
H 0:μ=98.6
H a:μ≠98.6
p=2∗0.00080=0.00160
p≤0.05, reject the null hypothesis
There is evidence that the population mean is different from 98.6 degrees.
Selecting the Appropriate Procedure
The decision of whether to use a confidence interval or a hypothesis test depends on the research question. If we want to estimate a population parameter,we use a confidence interval. If we are given a specific population parameter (i.e., hypothesized value), and want to determine the likelihood that a population with that parameter would produce a sample as different as our sample, we use a hypothesis test. Below are a few examples of selecting the appropriate procedure.
Example: Cheese Consumption
Research question:How much cheese (in pounds) does an average American adult consume annually?
What is the appropriate inferential procedure?
Cheese consumption, in pounds, is a quantitative variable. We have one group: American adults. We are not given a specific value to test, so the appropriate procedure here is aconfidence interval for a single mean.
Example: Age
Research question:Is the average age in the population of all STAT 200 students greater than 30 years?
What is the appropriate inferential procedure?
There is one group: STAT 200 students. The variable of interest is age in years, which is quantitative. The research question includes a specific population parameter to test: 30 years. The appropriate procedure is ahypothesis test for a single mean.
Try it!
For each research question, identify the variables, the parameter of interest and decide on the the appropriate inferential procedure.
Research question:How strong is the correlation between height (in inches) and weight (in pounds) in American teenagers?
Answer
There are two variables of interest: (1) height in inches and (2) weight in pounds. Both are quantitative variables.The parameter of interest is the correlation between these two variables.
We are not given a specific correlation to test. We are being asked to estimate the strength of the correlation. The appropriate procedure here is aconfidence interval for a correlation.
2. Research question:Are the majority of registered voters planning to vote in the next presidential election?
Answer
The parameter that is being tested here is a single proportion. We have one group: registered voters. "The majority" would be more than 50%, or p>0.50. This is a specific parameter that we are testing. The appropriate procedure here is ahypothesis test for a single proportion.
3. Research question:On average, are STAT 200 students younger than STAT 500 students?
Answer
We have two independent groups: STAT 200 students and STAT 500 students. We are comparing them in terms of average (i.e., mean) age.
If STAT 200 students are younger than STAT 500 students, that translates to μ 200<μ 500 which is an alternative hypothesis. This could also be written as μ 200−μ 500<0, where 0 is a specific population parameter that we are testing.
The appropriate procedure here is ahypothesis test for the difference in two means.
4. Research question:On average, how much taller are adult male giraffes compared to adult female giraffes?
Answer
There are two groups: males and females. The response variable is height, which is quantitative. We are not given a specific parameter to test, instead we are asked to estimate "how much" taller males are than females. The appropriate procedure is aconfidence interval for the difference in two means.
5. Research question:Are STAT 500 students more likely than STAT 200 students to be employed full-time?
Answer
There are two independent groups: STAT 500 students and STAT 200 students. The response variable is full-time employment status which is categorical with two levels: yes/no.
If STAT 500 students are more likely than STAT 200 students to be employed full-time, that translates to p 500>p 200 which is an alternative hypothesis. This could also be written as p 500−p 200>0, where 0 is a specific parameter that we are testing. The appropriate procedure is ahypothesis test for the difference in two proportions.
6. Research question:Is there is a relationship between outdoor temperature (in Fahrenheit)and coffee sales (in cups per day)?
Answer
There are two variables here: (1) temperature in Fahrenheit and (2) cups of coffee sold in a day. Both variables are quantitative. The parameter of interest is the correlation between these two variables.
If there is a relationship between the variables, that means that the correlation is different from zero. This is a specific parameter that we are testing. The appropriate procedure is ahypothesis test for a correlation.
6.7 - Lesson 6 Summary
6.7 - Lesson 6 Summary
Objectives
Upon successful completion of this lesson, you should be able to:
Identify Type I and Type II errors
Select an appropriate significance level (i.e., α level) for a given scenario
Explain the problems associated with conducting multiple tests
Interpret the results of a hypothesis test in terms of practical significance
Distinguish between practical significance and statistical significance
Explain how changing different aspects of a research study would change the statistical power of the tests conducted
Compare and contrast confidence intervals and hypothesis tests
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The Cost of Quality: Unveiling Its Impact on Business Performance
John Ponte
March 21, 2025
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Delivering high-quality products and services is key to success. But quality comes at a cost- so understanding the Cost of Quality (COQ) is essential for any business wanting to be more efficient and profitable. We’ll explore the core concepts of COQ, how it’s measured and managed, and how optimizing it can significantly improve your business performance. We’ll also explore vital concepts like the Voice of the Customer, Brand Equity, and Preventive Maintenance and demonstrate how they intertwine with COQ.
Understanding the Cost of Quality
To effectively manage quality, you should learn about its associated costs.
Definition of Cost of Quality
The Cost of Quality (COQ) encompasses all expenditures related to achieving and maintaining product or service quality. It’s a holistic measure that accounts for both the investments made to prevent defects and the consequences of poor quality. By assessing COQ, businesses can evaluate the efficiency and effectiveness of their quality management systems.
Components of Cost of Quality
COQ is typically divided into four key components:
Prevention Costs: These are proactive investments to prevent defects from occurring in the first place. Examples include employee training, process improvement initiatives, and robust quality planning.
Appraisal Costs: These costs are associated with measuring and monitoring quality. This includes activities like inspections, audits, and product testing.
Internal Failure Costs: These costs arise when defects are detected before reaching the customer. Examples include rework, scrap, and production downtime.
External Failure Costs: These are incurred when defects reach the customer. They can include warranty claims, product returns, and damage to brand reputation.
Voice of the Customer
Customer feedback is invaluable in shaping quality standards and reducing COQ. By actively listening to the Voice of the Customer, you can align your products and services with their expectations, ultimately reducing external failure costs and fostering customer satisfaction. By leveraging effective frontline collaboration, companies can respond and adjust based on customer feedback – reducing external failure costs and improving customer satisfaction.
Measuring the Cost of Quality
Measuring the COQ is an efficient way to visualize its impact on your business.
Quality Metrics
Accurate measurement of productivity metrics, as discussed in our productivity solutions, is essential to gauge of the impact of quality initiatives on overall business performance. Several key metrics help evaluate the effectiveness of your quality systems, including:
Defect Density: Measures the number of defects per unit of product or service.
First Pass Yield: The percentage of units that pass quality checks on the first attempt.
Cost of Poor Quality: Calculates the total cost of producing defective products or services.
Effective Use of COQ
COQ data should be used strategically to drive improvement. This includes:
Benchmarking: Compare your COQ against industry standards to identify areas where you excel or need to improve.
Trend Analysis: Track COQ over time to identify patterns and assess the effectiveness of quality initiatives.
Prioritization: Use COQ data to prioritize areas for investment in quality improvements.
Expanded COQ Equation
he expanded COQ equation provides a comprehensive view of quality costs and helps organizations balance investments in quality against the potential costs of failures.:
COQ = (Prevention Cost + Appraisal Cost) + (Internal Failure Cost + External Failure Cost)
Nonlinear Relationship of COQ
It’s important to understand that the relationship between COQ and quality investment is not linear. Investing heavily in prevention and appraisal doesn’t necessarily increase total quality costs. In fact, strategic investments in these areas often lead to significant reductions in internal and external failure costs, resulting in overall cost savings.
Prevention Costs
Prevention costs are proactive investments that aim to prevent defects and ensure quality from the outset.
Quality Management System (QMS)
A QMS is a formalized system that documents processes, procedures, and responsibilities for achieving quality policies. It acts as a roadmap, guiding everyone in the organization toward a shared understanding of quality standards and how to achieve them. This standardization of processes ensures consistency and efficiency, reduces errors, and promotes compliance with industry regulations. Moreover, a good QMS fosters a culture of continuous improvement by providing a framework for identifying areas where processes can be refined and optimized.
Preventive Maintenance
Preventive maintenance is all about taking care of your equipment before problems arise. By investing in preventive maintenance, manufacturers can extend the lifespan of their assets, reduce the likelihood of unexpected failures, and avoid the associated downtime and repair costs. A well-executed preventive maintenance program not only saves money but also ensures consistent production output and minimizes the risk of producing defective products due to faulty equipment. This, in turn, contributes to a significant improvement in the overall Cost of Quality.
Additional Examples of Prevention Costs
Establishing clear product specifications
Comprehensive employee training programs
Supplier quality assurance programs
Robust design for manufacturability practices
Appraisal Costs
Appraisal costs are essential for monitoring and measuring quality throughout the production process.
Quality Audits
Regular quality audits are a systematic examination of your operations to ensure they comply with industry standards, internal quality protocols, and customer requirements. . By conducting regular quality audits, you can identify potential problems early on, such as deviations from procedures, inefficient workflows, or outdated equipment. This allows you to use corrective action and compliance solutions before these issues escalate and lead to costly external failures like product recalls or customer complaints.
Supplier Assessments
Supplier assessments are equally important for maintaining high-quality standards. After all, the quality of your products depends heavily on the quality of the materials and components you source from suppliers. Evaluating your suppliers’ capabilities, processes, and quality management systems helps ensure they meet your standards and can consistently deliver reliable inputs. Effective supplier assessments can reduce both appraisal costs (by minimizing the need for excessive incoming inspections) and external failure costs (by preventing defects caused by substandard materials). This might involve visiting supplier facilities, reviewing their quality certifications, and conducting sample inspections.
More Appraisal Cost Examples
Incoming material inspections
In-process inspections
Final product testing
Use of check fixtures
Internal Failure Costs
Internal failure happens when defects are identified before products reach customers.
Machine Breakdown
Machine breakdowns are a major headache for manufacturers, impacting COQ in multiple ways. When a vital piece of equipment goes down, it triggers unplanned downtime, halting production and leading to lost output. This downtime also incurs repair costs, as technicians work to diagnose and fix the problem. Furthermore, machine breakdowns can result in product defects, either due to the malfunctioning equipment itself or the rush to get production back online. However, there’s a proactive solution: predictive maintenance. By using data and sensors to monitor equipment health, manufacturers can anticipate potential failures and schedule maintenance before a breakdown occurs.
Failure Analysis
When internal failures do happen, thorough failure analysis is essential. This involves not just identifying what went wrong, but digging deeper to uncover the root cause. Why did the machine break down? Was it due to wear and tear, improper operation, lack of maintenance, or a design flaw? By understanding the root cause, manufacturers can develop effective corrective action plans to prevent similar failures in the future. This might involve anything from adjusting maintenance schedules and operator training to redesigning a component or upgrading equipment.
Internal Failure Cost Examples
Scrap
Rework
Spoilage
Waste due to poor process control or inadequate design
External Failure Costs
External failure costs are the most damaging, both financially and reputationally, as they occur when defective products reach customers.
Warranty Costs
Warranty costs are a direct financial hit caused by product defects that reach your customers. These costs pile up quickly and include expenses related to repairing or replacing faulty products, shipping and handling those returns, and the administrative overhead of processing warranty claims. A high volume of warranty claims is a warning sign, pointing to potential problems like design flaws, manufacturing errors, poor materials, or inadequate testing.
Analyzing warranty data can help manufacturers identify these issues, take corrective action, and ultimately reduce future warranty costs. For example, a car manufacturer might face massive warranty costs due to a recall for a faulty braking system, while an electronics company might see a surge in claims for a specific smartphone model with a defective battery.
Service and Repair Costs
Beyond the warranty period, there are also service and repair costs to consider. These occur when products need fixing after the warranty expires and can involve expenses like paying technicians, sourcing replacement parts, and providing customer support. High service and repair costs drag down customer satisfaction and tarnish brand reputation, as customers get frustrated with the inconvenience and expense. Imagine a homeowner dealing with a broken refrigerator outside of its warranty period or a business facing downtime due to faulty equipment. By proactively addressing quality issues and minimizing the need for repairs, companies can boost customer happiness, build loyalty, and safeguard their brand.
External Failure Cost Examples
Customer complaints
Product returns
Product recalls
Lost sales due to damaged brand reputation
Impact of Cost of Quality on Brand Equity
COQ has a profound impact on brand equity, influencing customer perception and long-term brand value.
Customer Perception and Trust
Consistent product quality is the cornerstone of customer trust. When customers consistently receive products that meet or exceed their expectations, they develop confidence in your brand. This translates to loyalty, repeat purchases, and positive word-of-mouth referrals. Effective COQ management plays a crucial role in this process. By proactively preventing defects and efficiently addressing any quality issues that arise, you demonstrate a commitment to customer satisfaction.
Long-term Brand Value
Brand value is a precious asset, built over time through consistent quality, positive customer experiences, and a strong reputation. Maintaining high-quality standards is essential for protecting and enhancing this value. By prioritizing quality and effectively managing COQ, you send a clear message to your customers: you care about their experience and are committed to delivering excellence. This fosters long-term loyalty, drives repeat business, and strengthens your brand’s position in the market.
Examples of Successful Implementation
Understanding and measuring the cost of quality is just the first step. Let’s review some real-life examples showcasing the successful implementation of COQ strategies.
Case Study: Redesigning a Work Cell
Marianna Beauty, a leading cosmetics manufacturer, struggled with inefficient production and low employee morale. Their traditional management style hindered communication and collaboration, directly impacting quality and output. To address these challenges, they partnered with Redzone and implemented their real-time production efficiency software. This wasn’t just a technology upgrade; it was a complete overhaul of their work cell, empowering frontline workers with a voice and a platform for problem-solving. Redzone’s software provided instant visibility into production performance, enabling rapid response to issues and data-driven decision-making.
The results were transformative. Marianna Beauty saw a 20% productivity boost in the first year, significant cost reductions from less waste and downtime, and a marked improvement in product quality. More importantly, employees felt heard and valued, leading to increased engagement, higher morale, and lower turnover. The company culture shifted from a top-down hierarchy to a collaborative and empowered workforce, all thanks to Redzone’s focus on real-time data, open communication, and employee engagement. This case study perfectly illustrates how strategic quality management, combined with a positive cultural shift, can dramatically improve both cost efficiency and overall business performance.
Typical COQ Percentage
The Cost of Quality typically represents 15-40% of total business costs. Benchmarking your COQ against this range can help you assess your performance and identify areas for improvement.
Optimize Your Cost of Quality with Redzone
The Cost of Quality is far more than just a number; it’s a critical factor that influences every aspect of your business, from your bottom line to your brand reputation. By understanding and actively managing COQ, you can unlock significant improvements in efficiency, productivity, and customer satisfaction. Investing in prevention and appraisal activities might seem like an upfront expense, but it’s a strategic move that pays dividends by minimizing costly internal and external failures.
Remember that quality is a journey, not a destination. Continuous monitoring and improvement of your COQ strategy are essential for optimizing costs and ensuring long-term success. Embrace a culture of quality, empower your workforce, and leverage the power of real-time data to drive meaningful change.Ready to take control of your Cost of Quality and elevate your business performance? Contact us today to learn how Redzone can help you optimize your quality management strategy.
John Ponte John is QAD Redzone’s Senior Director of Growth Marketing and brings a background of over 20 years in B2B Software. He is responsible for setting the growth strategy and driving global demand generation strategies to boost pipeline, new customer acquisition, and create expansion opportunities. When John’s not tracking Marketing and business targets, you can find him playing tennis, and even officiating as a national umpire and referee, working with local charities he supports, and enjoying time with family.
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CASE REPORT article
Front. Neurol., 04 October 2022
Sec. Neuroinfectious Diseases
Volume 13 - 2022 |
This article is part of the Research TopicNeuroinfectious Diseases - Case Report Collection 2022View all 21 articles
Acute invasive mucormycosis rhinosinusitis causing multigroup cranial nerve injury and meningitis—A case report
This study reported a case of a Rhino-Orbital-Cerebral Mycosis (ROCM) patient with multiple groups of cranial nerve damage as the primary clinical manifestation, confirmed by histopathology and cerebrospinal fluid metagenomic next-generation sequencing (mNGS) technology. Relying on the MRI3D-SPACE technology, we observed the location and extent of the cranial nerve damage in the patient. The results suggested that fungal meningoencephalitis caused by mucor may enter the skull retrograde along the cranial nerve perineurium. The patient was admitted to the hospital with a preliminary diagnosis of mucormycosis infection after 1.5 days of mouth deviation. We treated the patient immediately with intravenous amphotericin B liposomes. After 21 days of hospitalization, the clinical symptoms of the patient did not improve significantly. The patient was discharged due to financial difficulties and antifungal treatment at home, and his disease had stabilized at the 6-month follow-up.
Introduction
Fungal infections of paranasal sinuses are common in the clinic and can be classified into invasive and noninvasive infections according to the progression of the disease (1, 2). Fungal infections of paranasal sinuses often require a differential diagnosis from chronic suppurative sinusitis, malignancy, sinus polyps, and necrotizing maxillary sinusitis (3–5). Early diagnosis of fungal sinusitis is usually critical, and most patients can be cured after early and timely diagnosis and reasonable treatment. Treatment is more complicated and prone to recurrence and poorer healing when delayed to a late stage (6, 7).
In 1969, Champion defined the simultaneous invasion of fungi into paranasal sinuses, eyes, and intracranial part of the brain as Rhino-Orbital-Cerebral Mycosis (ROCM), which belongs to acute invasive fungal rhinosinusitis (8). The main species responsible for invasive fungal rhinosinusitis infection are Mucor and Aspergillus, which belong to opportunistic pathogens (9, 10). The invasion routes can be divided into a direct spread and a transvascular invasion. Pathogens can enter the brain through the paranasal sinuses, orbit, extraocular muscles, ophthalmic artery, and optic nerve or drain to the cavernous sinus through the paranasal sinuses, orbital reflux veins, or intracranial infection. Studies found that patients with ROCM may present with multiple cranial nerve palsy (function of the cranial nerves II, III, IV, V, and VI may be lost or impaired) (11, 12). The magnetic resonance 3D-SPACE sequence is a variation of the TSE sequence. Through reconstruction, the complex anatomical structure of the cranial nerve can be further clearly displayed at any level and direction, which is vital for observing the path, distribution, and injury of the cranial nerve (13). In this study, we reported a ROCM patient with multiple groups of cranial nerve damage as the primary clinical manifestation, confirmed by histopathology and cerebrospinal fluid mNGS technology. We used the MRI3D-SPACE technique to provide detailed observation and better assess the extent of ROCM patients with cranial nerve injury due to the fungal meningoencephalitis caused by Mucor.
Case description
A 36-year-old male patient was admitted to our hospital for mouth deviation for 1.5 days. The patient was admitted to the Department of Endocrinology 1 month ago due to diabetic ketoacidosis (10:00 a.m. on June 18, 2021). Physical examination showed that the patient had clear consciousness, less fluent speech, right peripheral facial paralysis, and a right deviation of the protruding tongue; his muscle strength and tension in four extremities were normal, and the bilateral Babinski sign was negative. Auxiliary urine routine examination revealed urine glucose of 4+, urine ketone body of 3+, urine protein of 1+, and glycosylated hemoglobin of 10.1%. No apparent abnormalities were observed on cranial MRI. Chest CT showed small solid nodules in the upper lobe of the right lung and the lower lobe of the left lung; ground-glass opacities in the medial segment of the right middle lobe indicated some inflammatory changes. Abdominal CT showed many colonic contents, which should be diagnosed based on clinical symptoms and regarded as minor bowel dysfunction. At admission, the random fingertip blood glucose and blood ketones were 20.1 mmol/L and 6.4 mmol/L, respectively. After admission, the patient's condition gradually worsened, with restricted right eye movement, hoarseness, occasional febrile headache, nausea, and vomiting. On the fourth day of admission, the patient had clear consciousness, a hoarse voice, right eyelid ptosis, and eye proptosis, which was fixed in the primary position. The patient's right frontal striae disappeared, the right nasolabial fold was shallow, and his tongue was extended to the right. The patient had hypoalgesia on the right side of the face, normal muscle strength in the extremities, and meningeal stimulation signs (+). The complete set of immune indexes and five preoperative indicators were normal. The lumbar puncture indicated the pressure of 150 mmH2O, a yellowish color cerebrospinal fluid, white blood cells of 109 × 106/L (0–8 × 106/L), protein of 1,003 mg/L (0.15–0.45 g/L), chloride of 111.2 mmol/L, and glucose of 5.54 mmol/L. Cerebrospinal fluid cytology tests showed a mixed cellular response. mNGS of Rhizopus oryzae's sequence number was 19 (Total length was 1350 bp, 0.00334% coverage, and 1.00 X average depth). Repeated head MRI showed the inflammation of cerebellar infarction, right paranasal sinuses, and ethmoid sinuses. The plain and enhanced 3D-SPACE sequence scanning showed abnormal enhancement of the right hypoglossal nerve, the facial nerve, the cisternal segment of the trigeminal nerve, the posterior wall of the orbit, the periphery of the right maxillary sinus, and the oral maxillofacial spaces (Figure 1). Amphotericin B liposome (triamcinolone acetonide) was immediately intravenously administered at an initial dose of 5 mg/d, and no significant side effects were observed. Then, 5 mg of amphotericin B was added daily until the maintenance dose of 40 mg/day was reached. On the seventh day of admission, the specimen of the patient was taken for pathological examination with the assistance of the otorhinolaryngology and the Department of Stomatology. After 1 week, the patient was diagnosed with mucormycosis infection due to microscopic visualization of Molds hyphae (Figure 1). After 21 days of hospitalization, the clinical symptoms of the patient did not improve significantly. The patient was discharged due to financial difficulties and was provided antifungal treatment at home. During hospitalization, the patient was on antifungal medication (liposomal amphotericin B) for 17 days; after discharge, he was switched to oral antifungal (voriconazole) for 6 months. During the six-month follow-up, the condition of the patient was relatively stable.
Figure 1. Enhanced MRI with 3D-T1-SPC sequence results of the patient (a standard dose of gadolinium). The MRI showed enhanced images of the right hypoglossal nerve (A), the posterior orbital wall (E), the facial nerve (B,C), and the trigeminal nerve (F,G) slightly (Red Arrow). Microscopic photograph showing hyphae and Sporangium (Black Arrow) in HE staining [(D), 400X].
Discussion
Mucor belongs to the class Zygomycetes, order Mucorales, and the most common pathogenic bacteria are Rhizopus, Mucor, and Oryzae in the family Mucoraceae (14). In this case, the patient was infected with R. oryzae. The patient had been admitted to the Department of Endocrinology due to diabetic ketoacidosis. Mucormycosis has a complex disease course, rapid progression, high mortality, and intricate clinical manifestations. It is divided into six types, of which the naso-orbital brain type accounts for 36.4%. This patient was finally diagnosed with ROCM when combined with the clinical characteristics of sinus, orbit, meninges, cranial nerve injury, and pathogen detection. The case was complicated due to the involvement of the cranial nerve, meninges, and infarction in the distribution area of the posterior inferior cerebellar artery, suggesting that both pathways were involved in the pathogenic process.
Studies showed that the prevalence of fungal sinus disease has been increasing in recent decades. This may be due to increased awareness of medical examinations, overuse of antibiotics, and increased use of immunosuppressive drugs (15). Common clinical manifestations of ROCM include headache, fever, black nasal eschar, orbitofacial cellulitis, cranial nerve palsy, altered sensorium, and hemiparesis (16). However, in the present study, the patient was first admitted to the neurology department with a cranial nerve injury. His disease gradually aggravated to multiple groups of cranial nerve damage, leading to meningitis-like features that appeared successively. Early diagnosis is difficult because the patient has no nasal infection and other symptoms such as nasal congestion, runny nose, epistaxis, and typical ROCM imaging manifestations. Cranial nerves reach various parts of the head and the neck through the bony foramen of the skull base and meninges; hence, they are vulnerable to multiple local or systemic lesions throughout the process. Fungal infections such as mucormycosis can affect numerous cranial nerves and lead to fungal meningitis. ROCM is the most common clinical form of mucormycosis infection. Mucor can be transmitted via the nasal cavity, the paranasal sinuses, thje neck, and the orbit into the cranial cavity, causing brain infection and involvement of the cranial nerves (17). The definite diagnosis and treatment include a comprehensive analysis of the patient, finding the site of cranial nerve injury, and performing a lumbar puncture to find pathogens. Due to the slender anatomy, complex course, and overlapping surrounding structures of cranial nerves, standard radiographic methods cannot show their structures well. With the advancement of technology, 3D high-resolution body scanning technology can display the cranial nerve structure through 3D reconstruction and accurately measure each cranial nerve's course angle and brain cisternal segment length (13). The 3D-SPACE sequence has the best imaging effect on cranial nerves (especially the cisternal segment), and the scanning time is shorter. 3D-SPACE sequence enhancement showed abnormal enhancement of the right hypoglossal nerve, the facial nerve, and the trigeminal nerve's cisternal segment, which revealed the degree and location of cranial nerve injury. In addition, it was also found that the enhancement of cranial nerve was more pronounced at sites distant from the brainstem; no damage was found at the junction of cranial nerves and brainstem. The above two studies suggested that trichinosis can enter the skull retrogradely along the cranial nerves, adding to the etiological theory of intracranial infection. At the same time, relying on 3D-SPACE technology, it was also possible to assess the degree of mucormycosis cranial invasion and the effect of antifungal therapy. However, further studies with a large specimen size are still needed. The right eyeball of the patient was fixed without exophthalmos, and the possibility of incomplete orbital apex syndrome or superior orbital fissure syndrome was considered. The enhancement of the posterior wall of the eyeball could explain this clinical manifestation. Studies found that the appearance of orbital apex syndrome is highly suggestive of fungal infection, and the symptoms at this stage are typical of ROCM (18).
Cerebrospinal fluid testing showed that the patient was experiencing an inflammatory response. Therefore, screening for pathogens, in this case, becomes a critical adjunctive test to confirm the diagnosis of ROCM. An mNGS examination was performed directly to find the pathogen to confirm the diagnosis further. The mNGS technology has significant advantages in pathogen screening (19, 20), especially when suspected pathogen species are not identified. The patient was rapidly confirmed to have R. oryzae infection by mNGS and was given antifungal therapy. The diagnosis of ROCM in the patient was confirmed by analyzing the source of Mucor, which was finally approved by histopathology. If patients with ROCM can be diagnosed early, they can receive active treatment. However, the rarity of ROCM leads to a high rate of misdiagnosis. Many cases are diagnosed before the end of life or even after the autopsy, so the mortality rate is also very high. Therefore, early lumbar puncture and rapid mNGS detection may indicate a better prognosis in patients with suspected fungal meningitis.
Conclusion
ROCM is easily misdiagnosed clinically and patients suffering from ROCM are susceptible to disability and death without timely treatment. In patients with a high clinical suspicion of ROCM, an MRI 3D-SPACE sequence is required to assess the extent of cranial nerve injury and to determine whether the pathogen has entered the skull. Cerebrospinal fluid mNGS technology offers tremendous advantages in pathogen screening. In patients with ROCM, signs of retrograde entry along cranial nerves and invasion of the meninges or brainstem should be promptly screened for pathogens from imaging by using cerebrospinal fluid mNGS technology. Fluconazole, itraconazole, and caspofungin are preferred for the prophylactic treatment of fungal infections. However, amphotericin B is widely accepted as an effective drug for treating Trichophyton. Its clinical application is limited due to its tendency to cause severe hepatic and renal side effects. Liposomal amphotericin B can be used preferentially in patients with ROCM because it has fewer therapeutic side effects than both amphotericin B. Besides, early aggressive and extensive surgical excision of fungal-infected inactivated tissue is also a preferred option. Early diagnosis and treatment are decisive for improving the prognosis of ROCM, and magnetic resonance 3D-SPACE sequence combined with mNGS technology may be of clinical application for early diagnosis of ROCM.
Data availability statement
The raw data supporting the conclusions of this article will be made available by the authors, without undue reservation.
Ethics statement
The studies involving human participants were reviewed and approved by the Ethics Committee of Harrison International Peace Hospital. The patients/participants provided their written informed consent to participate in this study. Written informed consent was obtained from the individual(s) for the publication of any potentially identifiable images or data included in this article.
Author contributions
JH and TW organized and proofread the writing of the editorial. DC and TW wrote the manuscript draft. All authors contributed to the article and approved the submitted version.
Conflict of interest
The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.
Publisher's note
All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article, or claim that may be made by its manufacturer, is not guaranteed or endorsed by the publisher.
References
Nourizadeh N, Adabizadeh A, Zarrinfar H, Majidi M, Jafarian AH, Najafzadeh MJ. Fungal biofilms in sinonasal polyposis: the role of fungal agents is notable? J Oral Maxillofac Pathol. (2019) 31:295–8. doi: 10.1016/j.ajoms.2019.01.007
CrossRef Full Text | Google Scholar
Bakhshizadeh M, Hashemian HR, Najafzadeh MJ, Dolatabadi S, Zarrinfar H. First report of rhinosinusitis caused by neoscytalidium dimidiatum in Iran. J Med Microbiol. (2014) 63:1017–9. doi: 10.1099/jmm.0.065292-0
PubMed Abstract | CrossRef Full Text | Google Scholar
Sakci Z, Aydin F, Ceylan O, Ogul H. Sinonasal teratocarcinosarcoma mimicking chronic invasive fungal disease of paranasal sinuses. Ann R Coll Surg Engl. (2021) 103:e193–e5. doi: 10.1308/rcsann.2020.7088
PubMed Abstract | CrossRef Full Text | Google Scholar
Li Z, Wang X, Jiang H, Qu X, Wang C, Chen X, et al. Chronic invasive fungal rhinosinusitis vs sinonasal squamous cell carcinoma: the differentiating value of MRI. Eur Radiol. (2020) 30:4466–74. doi: 10.1007/s00330-020-06838-1
PubMed Abstract | CrossRef Full Text | Google Scholar
Chen X, Xian J, Lu X. Value of Ct in the differential diagnosis of pseudomonas aeruginosa sinusitis and fungal ball in paranasal sinus caused by aspergilus flavus. Zhonghua yi xue za zhi. (2019) 99:3417–9. doi: 10.3760/cma.j.issn.0376-2491.2019.43.013
PubMed Abstract | CrossRef Full Text | Google Scholar
Dykewicz MS, Rodrigues JM, Slavin RG. Allergic fungal rhinosinusitis. J Allergy Clin Immunol. (2018) 142:341–51. doi: 10.1016/j.jaci.2018.06.023
PubMed Abstract | CrossRef Full Text | Google Scholar
Zhou M, Zhao Y, Zhang Y, Zhang S. Clinical analysis of combined treatment in 87 patients with recurrent allergic fungal rhinosinusitis. Lin Chuang er bi yan hou tou Jing wai ke za zhi. (2018) 32:541–4. doi: 10.13201/j.issn.1001-1781.2018.07.016
PubMed Abstract | CrossRef Full Text | Google Scholar
Champion C, Johnson T. Rhino-orbital-cerebral phycomycosis. Wien med News. (1969) 68:807–10.
Google Scholar
Hosseinikargar N, Basiri R, Asadzadeh M, Najafzadeh MJ, Zarrinfar H. First report of invasive aspergillus rhinosinusitis in a critically ill Covid-19 patient affected by acute myeloid leukemia, Northeastern Iran. Clini Case Rep. (2021) 9. doi: 10.1002/ccr3.4889
PubMed Abstract | CrossRef Full Text | Google Scholar
Dolatabadi S, Ahmadi B, Rezaei-Matehkolaei A, Zarrinfar H, Skiada A, Mirhendi H, et al. Mucormycosis in Iran: a six-year retrospective experience. J Mycol Med. (2018) 28:269–73. doi: 10.1016/j.mycmed.2018.02.014
PubMed Abstract | CrossRef Full Text | Google Scholar
Song Y-M, Shin SY. Bilateral ophthalmic artery occlusion in rhino-orbito-cerebral mucormycosis. Korean J Ophthalmol. (2008) 22:66–9. doi: 10.3341/kjo.2008.22.1.66
PubMed Abstract | CrossRef Full Text | Google Scholar
Singh N, Garg S, Kumar S, Gulati S. Multiple cranial nerve palsies associated with type 2 diabetes mellitus. Singapore Med J. (2006)47:712.
PubMed Abstract | Google Scholar
Wu W, Wu F, Liu D, Zheng C, Kong X, Shu S, et al. Visualization of the morphology and pathology of the peripheral branches of the cranial nerves using three-dimensional high-resolution high-contrast magnetic resonance neurography. Eur J Radiol. (2020) 132:109137. doi: 10.1016/j.ejrad.2020.109137
PubMed Abstract | CrossRef Full Text | Google Scholar
Khor B-S, Lee M-H, Leu H-S, Liu J-W. Rhinocerebral mucormycosis in Taiwan. J Microbiol Immunol Infect. (2003) 36:266–9.
Google Scholar
Deutsch PG, Whittaker J, Prasad S. Invasive and non-invasive fungal rhinosinusitis—a review and update of the evidence. Medicina. (2019) 55:319. doi: 10.3390/medicina55070319
PubMed Abstract | CrossRef Full Text | Google Scholar
Nussbaum ES, Hall WA. Rhinocerebral mucormycosis: changing patterns of disease. Surg Neurol. (1994) 41:152. doi: 10.1016/0090-3019(94)90114-7
PubMed Abstract | CrossRef Full Text | Google Scholar
Vivek P, Rima S, Ritesh K, Chaitanya BS, Bhujang P. Rhino-orbito-cerebral mucormycosis: pictorial review. Insights Imag. (2021) 12. doi: 10.1186/s13244-021-01109-z
PubMed Abstract | CrossRef Full Text | Google Scholar
Arndt S, Aschendorff A, Echternach M, Daemmrich TD, Maier W. Rhino-orbital-cerebral mucormycosis and aspergillosis: differential diagnosis and treatment. Eur Arch Oto-Rhino-L. (2009) 266:71–6. doi: 10.1007/s00405-008-0692-y
PubMed Abstract | CrossRef Full Text | Google Scholar
Wilson MR, Naccache SN, Samayoa E, Biagtan M, Bashir H, Yu G, et al. Actionable diagnosis of neuroleptospirosis by next-generation sequencing. N Engl J Med. (2014) 370:2408–17. doi: 10.1056/NEJMoa1401268
PubMed Abstract | CrossRef Full Text | Google Scholar
Guan H, Shen A, Lv X, Yang X, Ren H, Zhao Y, et al. Detection of virus in CSF from the cases with meningoencephalitis by next-generation sequencing. J Neurovirol. (2016) 22:240–5. doi: 10.1007/s13365-015-0390-7
PubMed Abstract | CrossRef Full Text | Google Scholar
Keywords: Mucor, cranial nerve, 3D-SPACE technology, mNGS, infection
Citation: Wang T, Cao D and Han J (2022) Acute invasive mucormycosis rhinosinusitis causing multigroup cranial nerve injury and meningitis—A case report. Front. Neurol. 13:873694. doi: 10.3389/fneur.2022.873694
Received: 11 February 2022; Accepted: 09 August 2022;
Published: 04 October 2022.
Edited by:
Reviewed by:
Copyright © 2022 Wang, Cao and Han. This is an open-access article distributed under the terms of the Creative Commons Attribution License (CC BY). The use, distribution or reproduction in other forums is permitted, provided the original author(s) and the copyright owner(s) are credited and that the original publication in this journal is cited, in accordance with accepted academic practice. No use, distribution or reproduction is permitted which does not comply with these terms.
Correspondence: Jingzhe Han, NDIwNjEyMDQ5QHFxLmNvbQ==
Disclaimer: All claims expressed in this article are solely those of the authors and do not necessarily represent those of their affiliated organizations, or those of the publisher, the editors and the reviewers. Any product that may be evaluated in this article or claim that may be made by its manufacturer is not guaranteed or endorsed by the publisher.
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188530 | https://sunset.provo.edu/psych-tip-of-the-week-episodic-memory-and-learning/ | Psych Tip of the Week – Episodic Memory and Learning | Sunset View Elementary School
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Psych Tip of the Week – Episodic Memory and Learning
Psych Tip of the Week – Episodic Memory and Learning
December 7th, 2023 /
Shauna Sprunger /
News
Each week we will be sharing a tip from our school psychologist, Mrs. Rollins.
This week we are talking about – Episodic Memory and Learning
One cognitive area that often comes up with struggling learners is problems with memory. It could be short-term memory, long-term, both or attention problems that mask as memory issues. The book Memory at Work in the Classroom by Bailey and Pransky has good suggestions about how to help students with memory problems. One way is utilizing Episodic Memory.
Episodic memory is the type of memory we are most familiar with. It is remembering a personal experience versus semantic memory which is the type of memory of things we know. Often episodic memory provides the context for our other forms of memory. The way we choose to do a task may come from an experience we had which formed what information we know and how we like to do something.
The emotions that come with episodic memory can help trigger semantic memory in order to create the whole picture. Episodic memory is triggered by sensory activation, so a smell, taste, touch, sound or even body position can help us recall. When we help our students recall, we can ask them about which area of the classroom we were sitting in, what table grouping they were part of, music that you played, snack that was given that day, etc.
Ways that we can facilitate episodic memory in the classroom are:
-Include student’s experiences outside the classroom to connect with academic content.
-Change room arrangement correlated with learning topic.
-Color code information or visual supports to coordinate with specific subjects or information.
-Integrate strong sensory input into lessons as appropriate.
-Consider taking a field trip as an introduction to a unit rather than a culminating activity.
-Seeing relevant movies or movie clips before reading a book or text in order to foster connections while reading.
Learning this way can be very effective because it increases student motivation as well. One important thing to note is that the experience itself is not enough to actually promote learning. The students need to be able to reflect and use language to make connections about the experience to other knowledge and have opportunities to recall the event so that it is stored in long-term memory with rich detail.
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188531 | https://fiveable.me/order-theory/unit-7/iteration-fixed-points/study-guide/fZsyiRAVfns9swTb | printables
📊Order Theory Unit 7 Review
7.6 Iteration and fixed points
📊Order Theory Unit 7 Review
7.6 Iteration and fixed points
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
APA
📊Order Theory
Unit & Topic Study Guides
7.1 Knaster-Tarski fixed point theorem
7.2 Kleene fixed point theorem
7.3 Bourbaki-Witt fixed point theorem
7.4 Fixed points in complete lattices
7.5 Applications of fixed point theorems
7.6 Iteration and fixed points
7.7 Fixed point combinatorics
Iteration and fixed points form the backbone of many mathematical processes in order theory. These concepts involve repeatedly applying functions to find stable points or convergence. Understanding iteration helps us analyze sequences, series, and the behavior of functions over time.
Fixed points are elements that remain unchanged when a function is applied to them. They play a crucial role in various mathematical fields, from algebra to computer science. Studying fixed points helps us understand function behavior, solve equations, and analyze iterative algorithms.
Concept of iteration
Iteration forms a fundamental concept in order theory involving repeated application of functions or operations
Iterative processes play a crucial role in understanding fixed points and their properties within ordered structures
Iterative processes
Involve repeatedly applying a function or operation to an initial value
Generate sequences of values through successive applications
Can be finite or infinite depending on the termination condition
Often used to approximate solutions or find fixed points
Sequences and series
Sequences represent ordered lists of elements generated through iteration
Series involve the summation of terms in a sequence
Arithmetic sequences follow a constant difference between terms
Geometric sequences maintain a constant ratio between consecutive terms
Convergence vs divergence
Convergence occurs when a sequence approaches a specific limit value
Divergence happens when a sequence does not settle to a particular value
Bounded sequences may converge to a finite limit
Unbounded sequences can diverge to infinity or oscillate without converging
Fixed points
Fixed points represent crucial elements in order theory where a function maps an element to itself
Understanding fixed points helps analyze the behavior of iterative processes and their stability
Definition of fixed points
A fixed point of a function f is an element x such that f(x)=x
Represents a point where the function's output equals its input
Can be visualized as the intersection of a function's graph with the line y=x
May not exist for all functions or may have multiple fixed points
Types of fixed points
Attractive fixed points draw nearby points towards them through iteration
Repulsive fixed points push nearby points away through successive applications
Neutral fixed points neither attract nor repel neighboring points
Periodic fixed points repeat after a certain number of iterations
Fixed point theorems
Provide conditions for the existence and uniqueness of fixed points
Brouwer's fixed point theorem guarantees a fixed point for continuous functions on compact convex sets
Schauder's fixed point theorem extends Brouwer's result to infinite-dimensional spaces
Kakutani's fixed point theorem applies to set-valued functions
Iteration in order theory
Order theory provides a framework for analyzing iterative processes in partially ordered sets
Iteration in order theory focuses on the behavior of functions that preserve or reverse order relations
Monotone functions
Preserve the order relation between elements in a partially ordered set
Increasing functions maintain the order (x≤y⟹f(x)≤f(y))
Decreasing functions reverse the order (x≤y⟹f(x)≥f(y))
Play a crucial role in fixed point theorems for ordered structures
Lattice theory applications
Lattices provide a structured environment for studying fixed points
Complete lattices guarantee the existence of least and greatest fixed points
Galois connections between lattices preserve fixed point properties
Lattice fixed points find applications in program analysis and verification
Tarski's fixed point theorem
States that every monotone function on a complete lattice has a fixed point
Guarantees the existence of least and greatest fixed points
Provides a powerful tool for reasoning about recursive definitions
Finds applications in logic, computer science, and mathematical analysis
Constructive fixed point theory
Focuses on methods for constructing fixed points through iterative processes
Provides algorithms for approximating or computing fixed points in various settings
Kleene fixed point theorem
Applies to continuous functions on directed complete partial orders (DCPOs)
Constructs the least fixed point as the supremum of an ascending chain
Iteratively applies the function starting from the bottom element
Widely used in denotational semantics and program analysis
Least fixed points
Represent the smallest element satisfying the fixed point equation
Often correspond to the most precise or minimal solution in a given context
Can be constructed through iterative approximation from below
Find applications in recursive function theory and formal language semantics
Greatest fixed points
Denote the largest element satisfying the fixed point equation
Often represent the most general or maximal solution in a given context
Can be constructed through iterative approximation from above
Used in coinductive definitions and reasoning about infinite data structures
Applications of iteration
Iteration finds widespread use across various fields of mathematics and computer science
Iterative methods provide powerful tools for solving complex problems and analyzing systems
Computer science algorithms
Iterative algorithms solve problems through repeated steps
Binary search iteratively narrows down the search space
Gradient descent iteratively optimizes objective functions in machine learning
Depth-first search and breadth-first search explore graphs iteratively
Mathematical analysis
Newton's method iteratively approximates roots of equations
Power series expansions represent functions as infinite iterative sums
Numerical integration techniques use iterative refinement for accuracy
Fourier analysis decomposes functions into iterative sums of sinusoids
Dynamical systems
Study the long-term behavior of systems evolving through iteration
Fixed points represent equilibrium states in dynamical systems
Periodic orbits correspond to repeating patterns in system evolution
Chaos theory examines sensitive dependence on initial conditions in iterated systems
Fixed point algorithms
Provide computational methods for finding or approximating fixed points
Utilize iterative techniques to converge towards fixed point solutions
Newton's method
Iteratively approximates roots of equations using tangent lines
Converges quadratically for well-behaved functions near simple roots
Generalizes to systems of equations and optimization problems
Requires careful initialization to ensure convergence
Contraction mapping principle
Guarantees a unique fixed point for contractive functions on complete metric spaces
Provides an iterative method for approximating the fixed point
Convergence rate depends on the contraction factor of the function
Finds applications in differential equations and functional analysis
Banach fixed point theorem
Extends the contraction mapping principle to Banach spaces
Ensures the existence and uniqueness of fixed points for contractive mappings
Provides an error bound for the iterative approximation process
Used in proving the existence of solutions to integral equations
Iteration in partial orders
Explores iterative processes in the context of partially ordered sets
Focuses on properties that ensure the existence and convergence of fixed points
Chain completeness
A partially ordered set is chain complete if every chain has a least upper bound
Ensures the existence of fixed points for monotone functions
Generalizes the notion of completeness from total orders to partial orders
Plays a crucial role in fixed point theorems for ordered structures
Directed complete partial orders
DCPOs have least upper bounds for all directed subsets
Provide a more general setting than chain completeness
Allow for the construction of fixed points through iterative approximation
Form the basis for domain theory in computer science
Scott continuity
A function is Scott continuous if it preserves directed suprema
Ensures the existence of least fixed points in DCPOs
Generalizes the notion of continuity to ordered structures
Plays a fundamental role in denotational semantics and domain theory
Fixed points in lattices
Lattices provide a rich structure for studying fixed points and their properties
Complete lattices guarantee the existence of fixed points for monotone functions
Complete lattices
Possess least upper bounds and greatest lower bounds for all subsets
Ensure the existence of least and greatest fixed points for monotone functions
Provide a natural setting for studying fixed point theorems
Find applications in formal concept analysis and algebraic semantics
Knaster-Tarski theorem
States that the set of fixed points of a monotone function on a complete lattice forms a complete lattice
Generalizes Tarski's fixed point theorem
Provides a powerful tool for reasoning about fixed point equations
Finds applications in program verification and abstract interpretation
Fixed point induction
A proof technique based on the properties of least fixed points
Allows for reasoning about recursive definitions and programs
Utilizes the iterative construction of least fixed points
Provides a foundation for formal verification of recursive algorithms
Computational aspects
Addresses practical considerations in computing and approximating fixed points
Focuses on algorithms and techniques for efficient fixed point computation
Fixed point computation
Involves numerical methods for approximating fixed points
Requires careful consideration of convergence criteria and error bounds
May involve acceleration techniques to improve convergence speed
Finds applications in numerical analysis and scientific computing
Iterative methods
Successive approximation iteratively refines estimates of fixed points
Relaxation methods adjust step sizes to improve convergence
Multigrid methods use hierarchical refinement for efficient computation
Parallel iterative methods exploit distributed computing resources
Convergence rates
Linear convergence approaches the fixed point at a constant rate
Superlinear convergence accelerates as it approaches the fixed point
Quadratic convergence doubles the number of correct digits in each iteration
Convergence analysis helps in selecting appropriate algorithms for specific problems
Advanced topics
Explores more sophisticated concepts and generalizations in fixed point theory
Addresses complex scenarios and extensions of classical fixed point results
Multivalued fixed points
Generalize fixed points to set-valued functions
Allow for multiple fixed points or fixed point sets
Find applications in game theory and economic equilibrium analysis
Require extended notions of continuity and order relations
Coincidence points
Points where two or more functions agree in value
Generalize the concept of fixed points to multiple functions
Find applications in differential equations and variational inequalities
Require specialized theorems and techniques for analysis
Common fixed points
Points that are simultaneously fixed by multiple functions
Arise in the study of commuting functions and semigroups
Find applications in metric space theory and functional analysis
Require conditions on function properties and space structures for existence |
188532 | https://byjus.com/maths/mean-deviation/ | In statistics and mathematics, the deviation is a measure that is used to find the difference between the observed value and the expected value of a variable. In simple words, the deviation is the distance from the center point. Similarly, the mean deviation is used to calculate how far the values fall from the middle of the data set. In this article, let us discuss the definition, formula, and examples in detail.
Table of Contents:
Mean Deviation Definition
The mean deviation is defined as a statistical measure that is used to calculate the average deviation from the mean value of the given data set. The mean deviation of the data values can be easily calculated using the below procedure.
Step 1: Find the mean value for the given data values
Step 2: Now, subtract the mean value from each of the data values given (Note: Ignore the minus symbol)
Step 3: Now, find the mean of those values obtained in step 2.
Mean Deviation Formula
The formula to calculate the mean deviation for the given data set is given below.
Mean Deviation = [Σ |X – µ|]/N
Here,
Σ represents the addition of values
X represents each value in the data set
µ represents the mean of the data set
N represents the number of data values
| | represents the absolute value, which ignores the “-” symbol
Mean Deviation for Frequency Distribution
To present the data in the more compressed form we group it and mention the frequency distribution of each such group. These groups are known as class intervals.
Grouping of data is possible in two ways:
In the upcoming discussion, we will be discussing mean absolute deviation in a discrete frequency distribution.
Let us first know what is actually meant by the discrete distribution of frequency.
Mean Deviation for Discrete Distribution Frequency
As the name itself suggests, by discrete we mean distinct or non-continuous. In such a distribution the frequency (number of observations) given in the set of data is discrete in nature.
If the data set consists of values x1,x2, x3………xn each occurring with a frequency of f1, f2… fn respectively then such a representation of data is known as the discrete distribution of frequency.
To calculate the mean deviation for grouped data and particularly for discrete distribution data the following steps are followed:
Step I: The measure of central tendency about which mean deviation is to be found out is calculated. Let this measure be a.
If this measure is mean then it is calculated as,
where (\begin{array}{l}N=\sum_{i=1}^{n}\;f_{i}\end{array} )
If the measure is median then the given set of data is arranged in ascending order and then the cumulative frequency is calculated then the observations whose cumulative frequency is equal to or just greater than N/2 is taken as the median for the given discrete distribution of frequency and it is seen that this value lies in the middle of the frequency distribution.
Step II: Calculate the absolute deviation of each observation from the measure of central tendency calculated in step (I)
StepIII: The mean absolute deviation around the measure of central tendency is then calculated by using the formula
If the central tendency is mean then,
In case of median
Let us look into the following examples for a better understanding.
Read: Mean Deviation for Continous Frequency Distribution
Mean Deviation Examples
Example 1:
Determine the mean deviation for the data values 5, 3,7, 8, 4, 9.
Solution:
Given data values are 5, 3, 7, 8, 4, 9.
We know that the procedure to calculate the mean deviation.
First, find the mean for the given data:
Mean, µ = ( 5+3+7+8+4+9)/6
µ = 36/6
µ = 6
Therefore, the mean value is 6.
Now, subtract each mean from the data value, and ignore the minus symbol if any
(Ignore”-”)
5 – 6 = 1
3 – 6 = 3
7 – 6 = 1
8 – 6 = 2
4 – 6 = 2
9 – 6 = 3
Now, the obtained data set is 1, 3, 1, 2, 2, 3.
Finally, find the mean value for the obtained data set
Therefore, the mean deviation is
= (1+3 + 1+ 2+ 2+3) /6
= 12/6
= 2
Hence, the mean deviation for 5, 3,7, 8, 4, 9 is 2.
Example 2:
In a foreign language class, there are 4 languages, and the frequencies of students learning the language and the frequency of lectures per week are given as:
Language | Sanskrit | Spanish | French | English
No. of students(xi) | 6 | 5 | 9 | 12
Frequency of lectures(fi) | 5 | 7 | 4 | 9
Calculate the mean deviation about the mean for the given data.
Solution: The following table gives us a tabular representation of data and the calculations
Frequently Asked Questions on Mean DeviationQ1 What does the mean deviation tell us?
The mean deviation gives information about how far the data values are spread out from the mean value.
Q2 Mention the procedure to find the mean deviation.
The procedure to find the mean deviation are:
Step 1: Calculate the mean value for the given data
Step 2: Subtract the mean from each data value (Distance)
Step 3: Finally, find the mean for the distance
Q3 What are the advantages of using the mean deviation?
The advantages of using mean deviation are:
It is based on all the data values provided, and hence it will give a better measure of dispersion.
It is easy to understand and calculate.
Q4 What is the mean deviation about a median?
The mean deviation about the median is similar to the mean deviation about mean. Instead of calculating the mean for the given set of data values, find the median value by arranging the data values in the ascending order and then find the middle value. After finding the median, now subtract the median from each data value, and finally, take the average
Q5 What are the three different ways to find the mean deviation?
The mean deviation can be calculated using
Individual Series
Discrete series
Continuous series
Frequently Asked Questions on Mean Deviation
What does the mean deviation tell us?
The mean deviation gives information about how far the data values are spread out from the mean value.
Mention the procedure to find the mean deviation.
The procedure to find the mean deviation are:
Step 1: Calculate the mean value for the given data
Step 2: Subtract the mean from each data value (Distance)
Step 3: Finally, find the mean for the distance
What are the advantages of using the mean deviation?
The advantages of using mean deviation are:
It is based on all the data values provided, and hence it will give a better measure of dispersion.
It is easy to understand and calculate.
What is the mean deviation about a median?
The mean deviation about the median is similar to the mean deviation about mean. Instead of calculating the mean for the given set of data values, find the median value by arranging the data values in the ascending order and then find the middle value. After finding the median, now subtract the median from each data value, and finally, take the average
What are the three different ways to find the mean deviation?
The mean deviation can be calculated using
Individual Series
Discrete series
Continuous series
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188533 | https://journals.plos.org/plospathogens/article?id=10.1371/journal.ppat.1011341 | S1 Table comparison table for reoviruses with different numbers of
genome/RNA dependent RNA-Polymerase (RdRP)
Virus Genus Genome
segments
vertices
occupied
by
RdRPs
RdRP
symmetry Turreted Protein
layers references
Fako virus
(FAKV) Dinovernavirus 9 10
D2
(pseudo
D3)
✔ 1
cytoplasmic
polyhedrosis
virus (CPV)
Cypovirus 10 10 D3 ✔ 1 [2, 3]
mmmalian
reovirus
(MRV ) Orthoreovirus 10 10 D3 ✔ 2
bluetongue
virus Orbivirus 10 ? ? ✖ 3
grass carp
reovirus
(GCRV)
(ARV)
Aquareovirus 11 11 D3 ✔ 2 [6, 7]
rotavirus
(RRV) Rotavirus 11 ? no ✖ 3
Mud crab
reovirus
(MCRV)
? 12 10+ D5 ✖ 2 this paper
References:
Kaelber JT, Jiang W, Weaver SC, Auguste AJ, Chiu W. Arrangement of the Polymerase Complexes
inside a Nine-Segmented dsRNA Virus. Structure. 2020;28(6):604-12 e3. doi:
10.1016/j.str.2020.01.011. PubMed PMID: 32049031; PubMed Central PMCID: PMCPMC7289189.
Liu H, Cheng L. Cryo-EM shows the polymerase structures and a nonspooled genome within a dsRNA
virus. Science. 2015;349(6254):1347-50. Epub 2015/09/19. doi: 10.1126/science.aaa4938. PubMed
PMID: 26383954.
Zhang X, Ding K, Yu X, Chang W, Sun J, Hong Zhou Z. In situ structures of the segmented genome and
RNA polymerase complex inside a dsRNA virus. Nature. 2015. Epub 2015/10/28. doi:
10.1038/nature15767. PubMed PMID: 26503045.
Pan M, Alvarez-Cabrera AL, Kang JS, Wang L, Fan C, Zhou ZH. Asymmetric reconstruction of
mammalian reovirus reveals interactions among RNA, transcriptional factor micro2 and capsid proteins.
Nat Commun. 2021;12(1):4176. Epub 20210707. doi: 10.1038/s41467-021-24455-4. PubMed PMID:
34234134; PubMed Central PMCID: PMCPMC8263624.
He Y, Shivakoti S, Ding K, Cui YX, Roy P, Zhou ZH. In situ structures of RNA-dependent RNA
polymerase inside bluetongue virus before and after uncoating. Proceedings of the National Academy of Sciences of the United States of America. 2019;116(33):16535-40. doi: 10.1073/pnas.1905849116.
PubMed PMID: WOS:000481404300060.
Ding K, Nguyen L, Zhou ZH. In Situ Structures of the Polymerase Complex and RNA Genome Show
How Aquareovirus Transcription Machineries Respond to Uncoating. Journal of virology. 2018;92(21).
doi: ARTN e00774-1810.1128/JVI.00774-18. PubMed PMID: WOS:000447139100006.
Wang XR, Zhang FX, Su R, Li XW, Chen WY, Chen QX, et al. Structure of RNA polymerase complex
and genome within a dsRNA virus provides insights into the mechanisms of transcription and assembly.
Proceedings of the National Academy of Sciences of the United States of America. 2018;115(28):7344-
doi: 10.1073/pnas.1803885115. PubMed PMID: WOS:000438050900070.
Jenni S, Salgado EN, Herrmann T, Li ZL, Grant T, Grigorieff N, et al. In situ Structure of Rotavirus VP1
RNA-Dependent RNA Polymerase. Journal of molecular biology. 2019;431(17):3124-38. doi:
10.1016/j.jmb.2019.06.016. PubMed PMID: WOS:000482872100008. |
188534 | http://hyperphysics.phy-astr.gsu.edu/hbase/mlif.html | Lifting a Mass
Lifting Mass ============ Though a straightforward application of Newton's second law, many find this problem deceptive. The common misconception which is carried into it is that the tension in the rope must equal the weight of the hanging object. When the mass is accelerated, that is not so. You may change the data and then click on either tension or acceleration in the equation below to calculate its value. ### mass x acceleration = tension - weight For a mass m= kg, the rope must support its weight = mg =Newtons to hold it up at rest. If the acceleration is a=m/s 2 then a net force=Newtons is required to accelerate the mass. This requires a tension of T=Newtons. Note that the tension is equal to the weight only if the acceleration is zero, and that if the acceleration is negative (downward), the tension is less than the weight. If you enter a downward acceleration greater than 9.8 m/s 2 you will get a negative tension, showing that you must force it downward to get an acceleration greater than that of free fall. ### See elevator problemIndex Newton's laws Standard mechanics problems
HyperPhysics MechanicsR NaveGo Back |
188535 | https://www.youtube.com/watch?v=7S1MLJOG-5A | Perimeter and area of a non-standard polygon | Perimeter, area, and volume | Geometry | Khan Academy
Khan Academy
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Posted: 10 Nov 2011
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Perimeter and Area of a Non-Standard Polygon
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Transcript:
Find the area and
perimeter of the polygon. So let's start with
the area first. So the area of this
polygon-- there's kind of two parts of this. First, you have this part
that's kind of rectangular, or it is rectangular,
this part right over here. And that area is
pretty straightforward. It's just going to
be base times height. So area's going to be 8 times
4 for the rectangular part. And then we have this
triangular part up here. So we have this area up here. And for a triangle, the area
is base times height times 1/2. And that actually
makes a lot of sense. Because if you just
multiplied base times height, you would get this entire area. You would get the area
of that entire rectangle. And you see that the triangle
is exactly 1/2 of it. If you took this
part of the triangle and you flipped it over,
you'd fill up that space. If you took this
part of the triangle and you flipped it over,
you'd fill up that space. So the triangle's area is
1/2 of the triangle's base times the triangle's height. So plus 1/2 times
the triangle's base, which is 8 inches, times
the triangle's height, which is 4 inches. And so let's just calculate it. This gives us 32
plus-- oh, sorry. That's not 8 times 4. I don't want to confuse you. The triangle's height is 3. 8 times 3, right there. That's the triangle's height. So once again, let's go
back and calculate it. So this is going to be 32
plus-- 1/2 times 8 is 4. 4 times 3 is 12. And so our area for our
shape is going to be 44. Now let's do the perimeter. The perimeter-- we just
have to figure out what's the sum of the sides. How long of a
fence would we have to build if we wanted to make
it around this shape, right along the sides of this shape? So the perimeter-- I'll
just write P for perimeter. It's going to be equal to 8 plus
4 plus 5 plus this 5, this edge right over here, plus--
I didn't write that down. So I have two 5's plus
this 4 right over here. So you have 8 plus 4 is 12. 12 plus 10-- well, I'll
just go one step at a time. 12 plus 5 is 17. 17 plus 5 is 22. 22 plus 4 is 26. So the perimeter is 26 inches. And let me get the
units right, too. Because over here, I'm
multiplying 8 inches by 4 inches. So you get square inches. 8 inches by 3 inches, so
you get square inches again. So this is going to
be square inches. So area is 44 square inches. Perimeter is 26 inches. And that makes
sense because this is a two-dimensional
measurement. It's measuring something
in two-dimensional space, so you get a
two-dimensional unit. This is a one-dimensional
measurement. It's only asking you,
essentially, how long would a string have to be to
go around this thing. And so that's why you get
one-dimensional units. |
188536 | https://www.chinesewords.org/sentence/13257-804.html | 用不管都造句
漢語網
用不管都造句
用不管都造句子 用不管都造句大全 不管都造句 用如果就造句 用只要就造句
摘要:用不管都造句：1、但是不管江雪芹怎樣反抗，都抵不過兇惡的敵人。2、在班上不管是大事還是小事，都少不了您忙碌的身影。3、不管前面的路有多么困難不管前面的路有多么遙遠只要能取得勝利我都要堅持到底。
用不管都造句
1、但是不管江雪芹怎樣反抗，都抵不過兇惡的敵人。
2、在班上不管是大事還是小事，都少不了您忙碌的身影。
3、不管前面的路有多么困難 不管前面的路有多么遙遠 只要能取得勝利 我都要堅持到底。
4、因為它不管多皺，都還能使用。
5、他不管變成什么樣，都是我的好朋友——吳星雨。
6、不管你是不是成績優秀，不管你是不是名列前茅，只要你肯努力，就都會成功取得好的成績。
7、人民解放軍為了保衛祖國，不管嚴冬臘月，不管驕陽似火，只要是祖國需要，都是毫無怨言地駐守在祖國的邊疆。
8、不管還是只要_不管還是_不管都造句中國小生課本同步造句-名詞解釋
9、不管刮風下雨，警察叔叔都堅持上班，都在馬路上指揮交通。
10、不管遇到多大的困難，我都要勇敢面對。
11、不管是刮風，還是下雨，他都堅持扶送殘疾同學王明回家。
12、不管多么累，不管多么疲勞，雷鋒只要想到受苦受難的老百姓，都要幫助。
13、不管路途多么遙遠，我都要去找我的媽媽。
14、不管今天是下雨，也不管是下雪，只要學校上課我都去。
15、所以將來不管我們走到哪里，都要記著爸爸、媽媽。
16、不管是白貓，不管是花貓，只要逮著老鼠就是好貓。
17、不管環境多惡劣，不管工具多缺乏，只要我們努力，總會有解決辦法的。
18、不管身在何地，不管身穿何衣，只要我們不忘記自己是中國人就行了。[ 漢語網 ]
19、不管前途多迷茫，不管路途多艱險，只要你進行奮斗，光明都會來你的心間。
20、不管男女，不管老少，只要是的中國人，都應該熱愛自己的祖國。
21、他不管是白天還是黑夜，都熱情的為人們服務。
22、不管下多大的雨，我都要去上學。
23、這本書，不管是哪一段，我都能被下來。
用不管都造句
24、三是不管錢的多少，都認真負責地做事。
25、你不管遇到什么情況，都堅持給自己的學生上課。
26、不管冬天多么寒冷，也不管夏天多么炎熱，只要自己的手腳還能動，他一刻都不停止工作。
27、不管任何時候，不管任何情況，我都會堅持自己的夢想，勇往直前。
28、也不管自己是否有享用此魚的口福，都一樣的高興。
29、他不管歷經多少風雨，都依然矗立在那里。
30、不管哪一門功課，我們都應當認真學習。
31、人不管怎么樣生活，都不應該不把自己融入社會。
32、但不管怎樣，都是書使他們功成名就。
33、不管誰，不管你在干什么，只要你愛我，我就喜歡你。
34、關鍵是不管使用什么靈藥，都無法改變他的體質。
35、草不管條件怎樣惡劣，都頑強的生長著。
36、不管最后的結果是否成功，你都要嘗試。
37、不管男女，不管老少，只要是中國人，都要熱愛自己的國家。
38、但不管是哪一種理由，都是不可以違反交通規則的。
39、我們不管長大沒長大，都應該孝敬父母。
40、這件事不管怎么說，都是一件不該發生的事啊。
41、不管你是男的女的都要上學。
42、不管刮風下雨，不管嚴寒酷熱，只要有困難，他都隨叫隨到。
43、不管何時、何處、何事，都可以看見愛與感恩。
44、不管男女，不管老少，只要是中國人，都是要熱愛自己的國家。
45、它不管風多狂、雨多大，都堅守著自己的崗位。
FB 臉書;void(0) "分享到Facebook")Line賴
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188537 | https://www.quora.com/Does-Set-allow-duplicates-in-Python | Something went wrong. Wait a moment and try again.
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5
Does Set allow duplicates in Python?
Pavel Gurkov
SRE (2017–present)
·
Author has 60 answers and 93.8K answer views
·
4y
In Python, set is essentially a dict, which is in turn a hashtable.
Duplicates will likely have the same hash. That actually depends on what you define as a duplicate, but for simple cases, such as integers or strings, they _will_ have the same hash.
So, no.
You can always circumvent it by using your own data structure resembling a multiset or unordered_multiset from C++ (
member type definition notes key_type The first template parameter ( T ) value_type The first template parameter ( T ) key_compare The second template parameter ( Compare ) defaults to: less value_compare The second template parameter ( Compare ) defaults to: less allocator_type The third template parameter ( Alloc ) defaults to: allocator reference allocator_type::reference for the default allocator : value_type& const_reference allocator_type::const_reference for the default allocator : const value_type& pointer allocator_type::pointer for the default allocator : value_type const_pointer allocator_type::const_pointer for the default allocator : const value_type iterator a bidirectional iterator to value_type convertible to const_iterator const_iterator a bidirectional iterator to const value_type reverse_iterator reverse_iterator const_reverse_iterator reverse_iterator difference_type a signed integral type, identical to: iterator_traits::difference_type usually the same as ptrdiff_t size_type an unsigned integral type that can represent any non-negative value of difference_type usually the same as size_t
)
Related questions
How many duplicate elements can you have in a set in Python?
How do I strip a Python dictionary of duplicates?
Is Python allowed in coding interviews?
What does the built-in class attribute __name__ do in Python?
How do we find duplicates in a list Python?
Amin Boulouma
Systems Engineer at Boulouma.com
·
Author has 919 answers and 2.3M answer views
·
2y
Diana Crisan,
In Python, the set data type does not allow duplicates. A set is an unordered collection of unique elements, and any element that is added to a set will only be added if it is not already present in the set. For example:
s = set()s.add(1)s.add(2)s.add(2) # this will not be added to the set, because 2 is already presentprint(s) # prints {1, 2}
Brett Leach
Bachelor of Science from The California Maritime Academy (Cal Maritime) (Graduated 1984)
·
Author has 184 answers and 106.2K answer views
·
4y
Sets in python contain only unique items. If, for instance, you create a set with the following code -
mySet = set( [ 1,2,3,4,5,5 ] ) # Note the duplicate 5s.
the resulting set will contain only one 5.
Hilmar Zonneveld
Computer programmer since 1986
·
Author has 58.5K answers and 19.4M answer views
·
3y
No. Think of a “set” in the mathematical sense… those sets don’t allow duplicates. And Python more or less tries to emulate the meaning of a “set” in math.
Sarthak Sota
Java programmer | Front-End web developer
·
Updated 5y
Related
How does the set method in JAVA allow for duplicates?
IT DOES NOT ALLOW DUPLICATES. But just for fun
We can add duplicates in HashSet by creating a user defined class that contains data member(s) which we want to put in the HashSet. Now make any number of instances of that class and try to assign same value to the data members of more than one object of that class. After this put all the objects in the set.
User defined class:
class Value{ int number=0; }
Make instances of this class and put them in hashset:
public class Duplicate{ public static void main(String args[]){ HashSet<Value> test = new HashSet<Value>(); Value object1=new Value()
IT DOES NOT ALLOW DUPLICATES. But just for fun
We can add duplicates in HashSet by creating a user defined class that contains data member(s) which we want to put in the HashSet. Now make any number of instances of that class and try to assign same value to the data members of more than one object of that class. After this put all the objects in the set.
User defined class:
class Value{ int number=0; }
Make instances of this class and put them in hashset:
public class Duplicate{ public static void main(String args[]){ HashSet<Value> test = new HashSet<Value>(); Value object1=new Value(); object1.number=4; Value object2=new Value(); object2.number=4; Value object3=new Value(); object3.number=3; // now add all these objects to the hashset. test.add(object1); test.add(object2); test.add(object3); //print the value of all the objects in the hashset. Iterator<Value> itr = test.iterator(); //iterator for hashset System.out.println("The values in hashset are:"); while(itr.hasNext()){ System.out.println(itr.next().number); } }}
Output:
So, this is what happens when you do not override hashCode() and equals() method in your class.
Related questions
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Matteo Ianeselli
both loves and hates computer programming with passion
·
Author has 2K answers and 8.7M answer views
·
4y
Related
Does Set allow duplicates in C++?
No.
That’s why in the C++ standard library we have both std::set<T>
requiring unique values, and std::multiset<T>
allowing multiple equivalent values.
The same reasoning applies to the keys of std::map<K, T>
and std::multimap<K, T>
You asked: Does Set allow duplicates in C++?
Footnotes
std::set - cppreference.com
std::multiset - cppreference.com
std::multimap - cppreference.com
std::multimap - cppreference.com
Jeremy Langley
Author has 72 answers and 279.4K answer views
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7y
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In python, when adding items to a set, does the set keep track of the entries and check new entries against that list or does it go through the set every time to check for duplicates?
When you check an item there are two checks, one is on the hash value and the other is on the item itself. On every item added to a set a hash is generated (some number that can be reproduced). That hash value is checked against the other hash values in the set. Since comparing two integers is a faster way to look up elements then say a paragraph of text, finding duplicate items is much, much faster then comparing every elements of the set to this new item being added.
I’ve heard the speed described as: each lookup costs the same amount of time as it takes for three floating point divides times
When you check an item there are two checks, one is on the hash value and the other is on the item itself. On every item added to a set a hash is generated (some number that can be reproduced). That hash value is checked against the other hash values in the set. Since comparing two integers is a faster way to look up elements then say a paragraph of text, finding duplicate items is much, much faster then comparing every elements of the set to this new item being added.
I’ve heard the speed described as: each lookup costs the same amount of time as it takes for three floating point divides times every power of 1000 items you want to sort through.
Pavel Morava
knows how to write Python
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Updated 5y
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What is the use of sets in Python?
The python set and frozenset are the most underused data structures in Python. They are either unknown or viewed as convenient way how to get collection of unique elements. The other answers proved that as much.
In reality, the set should be your first data structure to go. If you do not care for order of elements, then do not use list or tuple. There is no point in writing something like this:
sample = [1,2,3,1,2]
You can write
sample = {1:2, 2:2, 3:1}
Wait, but this is a dictionary, right?
Yes, but sample.keys() is your set. Values are just quantity for each element. Much easier to work with than
The python set and frozenset are the most underused data structures in Python. They are either unknown or viewed as convenient way how to get collection of unique elements. The other answers proved that as much.
In reality, the set should be your first data structure to go. If you do not care for order of elements, then do not use list or tuple. There is no point in writing something like this:
sample = [1,2,3,1,2]
You can write
sample = {1:2, 2:2, 3:1}
Wait, but this is a dictionary, right?
Yes, but sample.keys() is your set. Values are just quantity for each element. Much easier to work with than list.
Furthermore, learn about Set theory and discrete math. What is the best representation of a weighted undirected graph?
u_graph = {frozenset((1,2)):50, frozenset((3,4)):100}
The advantage of frozenset lays with the fact that it is hashable and does not care for element order.
So,
frozenset((1,2)) == frozenset((2,1))
Really effective might be to use a set for describing binary sparse matrix as a set of coordinates like this:
sparse_matrix = {(1,2), (1000, 10000000)}
This certainly less memory consuming than creating a huge 2D array with two True values in it.
Finally, do not forget operations with sets like:
Intersection: s1 & s2Union: s1 | s2Difference: s1 – s2 Difference: s2 – s1Complement to intersection: s1 ^ s2
And of course, to find an element in the set is O(1) while element in the list is O(n).
Embrace sets and your life become easier.
Dave Wade-Stein
learned Pascal and C as an undergraduate CS major
·
Author has 1.9K answers and 8.2M answer views
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7y
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In python, when adding items to a set, does the set keep track of the entries and check new entries against that list or does it go through the set every time to check for duplicates?
Sets are just dictionaries without the values. Note that sets are
not
lists, so your question is misleading when it says
check new entries against that list
. The item you put in the set is like a dictionary key (and therefore must be hashable), and if the item already exists in the set, nothing happens. Hashing takes constant time–also written as O(1) time–and therefore there is no sequential checking needed. It takes the same amount of time to add an item to the set regardless of whether there is one item in the set or a million.
```
s = set(range(1, 1_000_000))>>> 500_000 in s True>>> 1_000_00
```
Sets are just dictionaries without the values. Note that sets are
not
lists, so your question is misleading when it says
check new entries against that list
. The item you put in the set is like a dictionary key (and therefore must be hashable), and if the item already exists in the set, nothing happens. Hashing takes constant time–also written as O(1) time–and therefore there is no sequential checking needed. It takes the same amount of time to add an item to the set regardless of whether there is one item in the set or a million.
```
s = set(range(1, 1_000_000))>>> 500_000 in s True>>> 1_000_000 in s False>>> # here are timing results for adding a new item 100,000 times>>> import timeit>>> timeit.timeit(setup='s = set(range(1, 1_000_000))', stmt='s.add(1_000_000)', number=100_000)0.01000845292583108>>> # ...and adding an existing item 100,000 times>>> timeit.timeit(setup='s = set(range(1, 1_000_000))', stmt='s.add(500_000)', number=100_000)0.01339553203433752>>> # and adding a single item to an empty set>>> timeit.timeit(setup='s = set()', stmt='s.add(500_000)', number=100_000)0.011194682214409113
```
Nimit
Author has 451 answers and 1.2M answer views
·
3y
Related
Does a hash set remove duplicates?
Hash Set only tells whether the element is present or not. While inserting an element in the hash set two cases can arise.
The element is already present in the hash set.
The element is not present in the hash set.
Case 1:
The hash set will not insert anything.
Case 2:
The element will be inserted.
So, in short, we can say that, the hash set does not contain duplicate elements.
In C++ STL, Hash Set is implemented with the help of-
unordered_set<Data_Type>name;
Your response is private
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Tony Flury
Python s/w developer since 2011 - published since 2015
·
Author has 13.2K answers and 23M answer views
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5y
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How many duplicate elements can you have in a set in Python?
None - the whole point of a set is that it contains unique elements.
If you want to include duplicates then you have two options :
Keep your elements in a list (lists have no restriction on duplicates)
Keep you elements in a dictionary, where the element is the key, and the value is a count of how often that element occurs - this is very similar to collections.Counter
David Vanderschel
PhD in math. Career in realtime and scientific computing.
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Author has 37.6K answers and 50.1M answer views
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7y
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In python, when adding items to a set, does the set keep track of the entries and check new entries against that list or does it go through the set every time to check for duplicates?
A2A: It uses hashing to resolve the issue of whether the item being added is already in the set. This is very closely related to the way dictionaries work. (Note that this leads to the requirement that the objects which you can put in a set must be hashable.)
Aishwarya Nigam
B.tech in Programming Languages, Kanpur Institute of Technology (Graduated 2021)
·
Author has 54 answers and 86.7K answer views
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5y
Related
How do I add items in a set in Python?
Sets are an unordered collection of items. The items stored in a set are unindexed i.e. we can not access items using the index in sets. In Python, sets are written using curly brackets{}. To add an element in the set we can use two methods:
add()
update()
Add(): This method adds a single element to the set.
Syntax: set_name.add(item)
Update(): This method can be used to add a single as well as multiple items to the set
Syntax: set_name.update(item)
Thanks for asking:-)
Sets are an unordered collection of items. The items stored in a set are unindexed i.e. we can not access items using the index in sets. In Python, sets are written using curly brackets{}. To add an element in the set we can use two methods:
add()
update()
Add(): This method adds a single element to the set.
Syntax: set_name.add(item)
Update(): This method can be used to add a single as well as multiple items to the set
Syntax: set_name.update(item)
Ex:
Thanks for asking:-)
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Mutually exclusive and collectively exhaustive events contain all outcomes of a sample space, and they do not share any common outcomes.
A. True
B. False
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a. True. Mutually exclusive and collectively exhaustive events contain all outcomes of a sample space, and they do not share any common outcomes.
Mutually Exclusive Events: These are events that cannot happen at the same time. In other words, if one event occurs, the other cannot. An example is flipping a coin where the outcomes 'heads' and 'tails' are mutually exclusive, because you can’t get both at the same time.
Collectively Exhaustive Events: These are events that cover all possible outcomes of a sample space. For instance, when flipping a coin, the outcomes 'heads' and 'tails' together cover all possible outcomes, making them collectively exhaustive.
Combined Concept: When events are both mutually exclusive and collectively exhaustive, it means they cover all possible outcomes without overlapping. Each outcome in the sample space belongs to one and only one of these events. For example, in rolling a die, if we define events A = {1, 2, 3} and B = {4, 5, 6}, A and B are both mutually exclusive (no common outcomes) and collectively exhaustive (cover all outcomes 1-6).
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The statement is True. Mutually exclusive events cannot happen at the same time, while collectively exhaustive events cover all possible outcomes of a sample space. Together, they ensure that all outcomes belong exclusively to one event without overlap.
Explanation
The statement in the question is A. True. Let's break it down further:
Mutually Exclusive Events: These are events that cannot happen at the same time. For example, when you roll a fair die, the event of rolling a 2 and the event of rolling a 5 are mutually exclusive because you cannot roll both numbers at the same time. In probability terms, if event A occurs, event B cannot occur, and vice versa.
Collectively Exhaustive Events: These events cover all possible outcomes of a sample space. In the die example mentioned above, the events of rolling a number between 1 and 6 (i.e., {1, 2, 3, 4, 5, 6}) are collectively exhaustive, as they account for all possible outcomes.
Combined Concept of Mutually Exclusive and Collectively Exhaustive: When events are both mutually exclusive and collectively exhaustive, it means they do not overlap and encompass all outcomes of a sample space. For instance, when rolling a die, you can define the events as A = {1, 2, 3} and B = {4, 5, 6}. These two events are mutually exclusive (no common outcomes) and collectively exhaustive (they account for all outcomes when rolling a die).
In summary, mutually exclusive and collectively exhaustive events describe a clear categorization of outcomes where each outcome falls into one category or event and does not belong to another. Therefore, the statement in the question is accurate.
Examples & Evidence
Examples include rolling a die where the outcomes {1, 2, 3} and {4, 5, 6} are both mutually exclusive and collectively exhaustive as they cover all results without sharing any numbers.
This understanding of mutually exclusive and collectively exhaustive events is fundamental in probability theory, which states that if two events are mutually exclusive, the occurrence of one prevents the occurrence of the other, and together they can cover the entire sample space.
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The statement is true; mutually exclusive events do not occur at the same time (P(A AND B) = 0) and collectively exhaustive events cover all possible outcomes in the sample space such that P(S) = 1.
Explanation
The statement 'mutually exclusive and collectively exhaustive events contain all outcomes of a sample space, and they do not share any common outcomes' is true. To elaborate, mutually exclusive events are those events that cannot occur at the same time, meaning that A and B do not share any outcomes (outcome) and P(A AND B) = 0. An example of mutually exclusive events is rolling a die and having event A represent rolling an odd number and event B represent rolling an even number; these two events cannot happen simultaneously. 'Collectively exhaustive' means that the events cover all possible outcomes in the sample space, ensuring that the probability of all events combined is equal to 1, which aligns with the axiom P(S) = 1. Thus, mutually exclusive events that are also collectively exhaustive will cover the entire sample space without overlapping.
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188539 | https://openstax.org/books/microbiology/pages/9-2-oxygen-requirements-for-microbial-growth | Skip to ContentGo to accessibility pageKeyboard shortcuts menu
Microbiology
9.2 Oxygen Requirements for Microbial Growth
Microbiology9.2 Oxygen Requirements for Microbial Growth
Search for key terms or text.
Learning Objectives
By the end of this section, you will be able to:
Interpret visual data demonstrating minimum, optimum, and maximum oxygen or carbon dioxide requirements for growth
Identify and describe different categories of microbes with requirements for growth with or without oxygen: obligate aerobe, obligate anaerobe, facultative anaerobe, aerotolerant anaerobe, microaerophile, and capnophile
Give examples of microorganisms for each category of growth requirements
Ask most people “What are the major requirements for life?” and the answers are likely to include water and oxygen. Few would argue about the need for water, but what about oxygen? Can there be life without oxygen?
The answer is that molecular oxygen (O2) is not always needed. The earliest signs of life are dated to a period when conditions on earth were highly reducing and free oxygen gas was essentially nonexistent. Only after cyanobacteria started releasing oxygen as a byproduct of photosynthesis and the capacity of iron in the oceans for taking up oxygen was exhausted did oxygen levels increase in the atmosphere. This event, often referred to as the Great Oxygenation Event or the Oxygen Revolution, caused a massive extinction. Most organisms could not survive the powerful oxidative properties of reactive oxygen species (ROS), highly unstable ions and molecules derived from partial reduction of oxygen that can damage virtually any macromolecule or structure with which they come in contact. Singlet oxygen (O2•), superoxide peroxides (H2O2), hydroxyl radical (OH•), and hypochlorite ion (OCl−), the active ingredient of household bleach, are all examples of ROS. The organisms that were able to detoxify reactive oxygen species harnessed the high electronegativity of oxygen to produce free energy for their metabolism and thrived in the new environment.
Oxygen Requirements of Microorganisms
Many ecosystems are still free of molecular oxygen. Some are found in extreme locations, such as deep in the ocean or in earth’s crust; others are part of our everyday landscape, such as marshes, bogs, and sewers. Within the bodies of humans and other animals, regions with little or no oxygen provide an anaerobic environment for microorganisms. (Figure 9.19).
Figure 9.19 Anaerobic environments are still common on earth. They include environments like (a) a bog where undisturbed dense sediments are virtually devoid of oxygen, and (b) the rumen (the first compartment of a cow’s stomach), which provides an oxygen-free incubator for methanogens and other obligate anaerobic bacteria. (credit a: modification of work by National Park Service; credit b: modification of work by US Department of Agriculture)
We can easily observe different requirements for molecular oxygen by growing bacteria in thioglycolate tube cultures. A test-tube culture starts with autoclaved thioglycolate medium containing a low percentage of agar to allow motile bacteria to move throughout the medium. Thioglycolate has strong reducing properties and autoclaving flushes out most of the oxygen. The tubes are inoculated with the bacterial cultures to be tested and incubated at an appropriate temperature. Over time, oxygen slowly diffuses throughout the thioglycolate tube culture from the top. Bacterial density increases in the area where oxygen concentration is best suited for the growth of that particular organism.
The growth of bacteria with varying oxygen requirements in thioglycolate tubes is illustrated in Figure 9.20. In tube A, all the growth is seen at the top of the tube. The bacteria are obligate (strict) aerobes that cannot grow without an abundant supply of oxygen. Tube B looks like the opposite of tube A. Bacteria grow at the bottom of tube B. Those are obligate anaerobes, which are killed by oxygen. Tube C shows heavy growth at the top of the tube and growth throughout the tube, a typical result with facultative anaerobes. Facultative anaerobes are organisms that thrive in the presence of oxygen but also grow in its absence by relying on fermentation or anaerobic respiration, if there is a suitable electron acceptor other than oxygen and the organism is able to perform anaerobic respiration. The aerotolerant anaerobes in tube D are indifferent to the presence of oxygen. They do not use oxygen because they usually have a fermentative metabolism, but they are not harmed by the presence of oxygen as obligate anaerobes are. Tube E on the right shows a “Goldilocks” culture. The oxygen level has to be just right for growth, not too much and not too little. These microaerophiles are bacteria that require a minimum level of oxygen for growth, about 1%–10%, well below the 21% found in the atmosphere.
Examples of obligate aerobes are Mycobacterium tuberculosis, the causative agent of tuberculosis and Micrococcus luteus, a gram-positive bacterium that colonizes the skin. Neisseria meningitidis, the causative agent of severe bacterial meningitis, and N. gonorrhoeae, the causative agent of sexually transmitted gonorrhea, are also obligate aerobes.
Figure 9.20 Diagram of bacterial cell distribution in thioglycolate tubes.
Many obligate anaerobes are found in the environment where anaerobic conditions exist, such as in deep sediments of soil, still waters, and at the bottom of the deep ocean where there is no photosynthetic life. Anaerobic conditions also exist naturally in the intestinal tract of animals. Obligate anaerobes, mainly Bacteroidetes, represent a large fraction of the microbes in the human gut. Transient anaerobic conditions exist when tissues are not supplied with blood circulation; they die and become an ideal breeding ground for obligate anaerobes. Another type of obligate anaerobe encountered in the human body is the gram-positive, rod-shaped Clostridioides and Clostridium spp. Their ability to form endospores allows them to survive in the presence of oxygen. One of the major causes of health-acquired infections is C. difficile, known as C. diff. Prolonged use of antibiotics for other infections increases the probability of a patient developing a secondary C. difficile infection. Antibiotic treatment disrupts the balance of microorganisms in the intestine and allows the colonization of the gut by Clostridioides difficile, causing a significant inflammation of the colon.
Other clostridia responsible for serious infections include C. tetani, the agent of tetanus, and C. perfringens, which causes gas gangrene. In both cases, the infection starts in necrotic tissue (dead tissue that is not supplied with oxygen by blood circulation). This is the reason that deep puncture wounds are associated with tetanus. When tissue death is accompanied by lack of circulation, gangrene is always a danger.
The study of obligate anaerobes requires special equipment. Obligate anaerobic bacteria must be grown under conditions devoid of oxygen. The most common approach is culture in an anaerobic jar (Figure 9.21). Anaerobic jars include chemical packs that remove oxygen and release carbon dioxide (CO2). An anaerobic chamber is an enclosed box from which all oxygen is removed. Gloves sealed to openings in the box allow handling of the cultures without exposing the culture to air (Figure 9.21).
Figure 9.21 (a) An anaerobic jar is pictured that is holding nine Petri plates supporting cultures. (b) Openings in the side of an anaerobic box are sealed by glove-like sleeves that allow for the handling of cultures inside the box. (credit a: modification of work by Centers for Disease Control and Prevention; credit b: modification of work by NIST)
Staphylococci and Enterobacteriaceae are examples of facultative anaerobes. Staphylococci are found on the skin and upper respiratory tract. Enterobacteriaceae are found primarily in the gut and upper respiratory tract but can sometimes spread to the urinary tract, where they are capable of causing infections. It is not unusual to see mixed bacterial infections in which the facultative anaerobes use up the oxygen, creating an environment for the obligate anaerobes to flourish.
Examples of aerotolerant anaerobes include lactobacilli and streptococci, both found in the oral microbiota. Campylobacter jejuni, which causes gastrointestinal infections, is an example of a microaerophile and is grown under low-oxygen conditions.
The optimum oxygen concentration, as the name implies, is the ideal concentration of oxygen for a particular microorganism. The lowest concentration of oxygen that allows growth is called the minimum permissive oxygen concentration. The highest tolerated concentration of oxygen is the maximum permissive oxygen concentration. The organism will not grow outside the range of oxygen levels found between the minimum and maximum permissive oxygen concentrations.
Check Your Understanding
Would you expect the oldest bacterial lineages to be aerobic or anaerobic?
Which bacteria grow at the top of a thioglycolate tube, and which grow at the bottom of the tube?
Case in Point
An Unwelcome Anaerobe
Charles is a retired bus driver who developed type 2 diabetes over 10 years ago. Since his retirement, his lifestyle has become very sedentary and he has put on a substantial amount of weight. Although he has felt tingling and numbness in his left foot for a while, he has not been worried because he thought his foot was simply “falling asleep.” Recently, a scratch on his foot does not seem to be healing and is becoming increasingly ugly. Because the sore did not bother him much, Charles figured it could not be serious until his daughter noticed a purplish discoloration spreading on the skin and oozing (Figure 9.22). When he was finally seen by his physician, Charles was rushed to the operating room. His open sore, or ulcer, is the result of a diabetic foot.
The concern here is that gas gangrene may have taken hold in the dead tissue. The most likely agent of gas gangrene is Clostridium perfringens, an endospore-forming, gram-positive bacterium. It is an obligate anaerobe that grows in tissue devoid of oxygen. Since dead tissue is no longer supplied with oxygen by the circulatory system, the dead tissue provides pockets of ideal environment for the growth of C. perfringens.
A surgeon examines the ulcer and radiographs of Charles’s foot and determines that the bone is not yet infected. The wound will have to be surgically debrided (debridement refers to the removal of dead and infected tissue) and a sample sent for microbiological lab analysis, but Charles will not have to have his foot amputated. Many diabetic patients are not so lucky. In 2008, nearly 70,000 diabetic patients in the United States lost a foot or limb to amputation, according to statistics from the Centers for Disease Control and Prevention.2
Which growth conditions would you recommend for the detection of C. perfringens?
Figure 9.22 This clinical photo depicts ulcers on the foot of a diabetic patient. Dead tissue accumulating in ulcers can provide an ideal growth environment for the anaerobe C. perfringens, a causative agent of gas gangrene. (Credit: Phalinn Ooi / Wikimedia Commons (CC-BY))
Detoxification of Reactive Oxygen Species
Aerobic respiration constantly generates reactive oxygen species (ROS), byproducts that must be detoxified. Even organisms that do not use aerobic respiration need some way to break down some of the ROS that may form from atmospheric oxygen. Three main enzymes break down those toxic byproducts: superoxide dismutase, peroxidase, and catalase. Each one catalyzes a different reaction. Reactions of type seen in Reaction 1 are catalyzed by peroxidases.
In these reactions, an electron donor (reduced compound; e.g., reduced nicotinamide adenine dinucleotide [NADH]) oxidizes hydrogen peroxide, or other peroxides, to water. The enzymes play an important role by limiting the damage caused by peroxidation of membrane lipids. Reaction 2 is mediated by the enzyme superoxide dismutase (SOD) and breaks down the powerful superoxide anions generated by aerobic metabolism:
The enzyme catalase converts hydrogen peroxide to water and oxygen as shown in Reaction 3.
Obligate anaerobes usually lack all three enzymes. Aerotolerant anaerobes do have SOD but no catalase. Reaction 3, shown occurring in Figure 9.23, is the basis of a useful and rapid test to distinguish streptococci, which are aerotolerant and do not possess catalase, from staphylococci, which are facultative anaerobes. A sample of culture rapidly mixed in a drop of 3% hydrogen peroxide will release bubbles if the culture is catalase positive.
Figure 9.23 The catalase test detects the presence of the enzyme catalase by noting whether bubbles are released when hydrogen peroxide is added to a culture sample. Compare the positive result (right) with the negative result (left). (credit: Centers for Disease Control and Prevention)
Bacteria that grow best in a higher concentration of CO2 and a lower concentration of oxygen than present in the atmosphere are called capnophiles. One common approach to grow capnophiles is to use a candle jar. A candle jar consists of a jar with a tight-fitting lid that can accommodate the cultures and a candle. After the cultures are added to the jar, the candle is lit and the lid closed. As the candle burns, it consumes most of the oxygen present and releases CO2.
Check Your Understanding
What substance is added to a sample to detect catalase?
What is the function of the candle in a candle jar?
Clinical Focus
Part 2
The health-care provider who saw Jeni was concerned primarily because of her pregnancy. Her condition enhances the risk for infections and makes her more vulnerable to those infections. The immune system is downregulated during pregnancy, and pathogens that cross the placenta can be very dangerous for the fetus. A note on the provider’s order to the microbiology lab mentions a suspicion of infection by Listeria monocytogenes, based on the signs and symptoms exhibited by the patient.
Jeni’s blood samples are streaked directly on sheep blood agar, a medium containing tryptic soy agar enriched with 5% sheep blood. (Blood is considered sterile; therefore, competing microorganisms are not expected in the medium.) The inoculated plates are incubated at 37 °C for 24 to 48 hours. Small grayish colonies surrounded by a clear zone emerge. Such colonies are typical of Listeria and other pathogens such as streptococci; the clear zone surrounding the colonies indicates complete lysis of blood in the medium, referred to as beta-hemolysis (Figure 9.24). When tested for the presence of catalase, the colonies give a positive response, eliminating Streptococcus as a possible cause. Furthermore, a Gram stain shows short gram-positive bacilli. Cells from a broth culture grown at room temperature displayed the tumbling motility characteristic of Listeria (Figure 9.24). All of these clues lead the lab to positively confirm the presence of Listeria in Jeni’s blood samples.
How serious is Jeni’s condition and what is the appropriate treatment?
Figure 9.24 (a) A sample blood agar test showing beta-hemolysis. (b) A sample motility test showing both positive and negative results. (credit a: modification of work by Centers for Disease Control and Prevention; credit b: modification of work by “VeeDunn”/Flickr)
Jump to the next Clinical Focus box. Go back to the previous Clinical Focus box.
Footnotes
2Centers for Disease Control and Prevention. “Living With Diabetes: Keep Your Feet Healthy.”
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Sequences Eventually and Frequently in a Set
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A sequence (a n)(a n) is eventually in an set A⊆R A⊆R if there exists an N∈N N∈N such that a n∈A a n∈A for all n≥N.n≥N.
A sequence (a n)(a n) is frequently in an set A⊆R A⊆R if, for every N∈N N∈N there exists an n≥N n≥N such that a n∈A a n∈A.
Question: Which definition is stronger? Does frequently imply eventually or eventually imply frequently?
Intuitively I am inclined to say that eventually implies frequently because from examples I reason that the sequence ⌊5 n⌋⌊5 n⌋ is eventually in {0}{0} and also frequently in {0}{0} but a sequence like (−1)n(−1)n which is frequently in {1}{1} but not eventually.
However from looking at the quantifiers it seems to me that the converse should be true for I reason
∀N∃n(n≥N⇒a n∈A)∀N∃n(n≥N⇒a n∈A)
should imply
∃N∃n(n≥N⇒a n∈A)∃N∃n(n≥N⇒a n∈A)
(i.e. frequnctly implies eventually)?
Am I making a mistake with the quantifiers? Which one is correct? Thanks in advance.
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edited Jul 13, 2018 at 19:57
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asked Jul 13, 2018 at 18:49
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1 Eventually is ∃N∗∀∗n(n≥N⇒a n∈A)∃N∗∀∗n(n≥N⇒a n∈A).Daniel Schepler –Daniel Schepler 2018-07-13 18:56:35 +00:00 Commented Jul 13, 2018 at 18:56
@Daniel, I think you meant to say that is "eventually" which IS stronger than "frequently" since in "eventually", there is a single N N after which the whole sequence lies in A A vs. frequently which just says that arbitrarily large indices lie in A A.Ned –Ned 2018-07-13 19:01:46 +00:00 Commented Jul 13, 2018 at 19:01
Your expression for eventually is wrong. The second quantifier must be ∀∀.drhab –drhab 2018-07-13 19:08:55 +00:00 Commented Jul 13, 2018 at 19:08
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Expression for eventually:
∃N∀n≥N[a n∈A]∃N∀n≥N[a n∈A]
Equivalent is the statement:
lim inf 1 A(a n)=1 lim inf 1 A(a n)=1
Expression for frequently:
∀N∃n≥N[a n∈A]∀N∃n≥N[a n∈A]
Equivalent is the statement:
lim sup 1 A(a n)=1 lim sup 1 A(a n)=1
We always have:
lim inf 1 A(a n)≤lim sup 1 A(a n)≤1 lim inf 1 A(a n)≤lim sup 1 A(a n)≤1
so eventually implies frequently.
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answered Jul 13, 2018 at 19:06
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Eventually implies frequently. If a n∈A a n∈A for all n≥N n≥N, then for every n n you can take m=max(n,N)m=max(n,N) and have m≥n m≥n and a m∈A a m∈A.
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answered Jul 13, 2018 at 18:52
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Yes but then what is my mistake with the quantifiers?Bunny –Bunny 2018-07-13 18:54:13 +00:00 Commented Jul 13, 2018 at 18:54
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188542 | https://textbooks.math.gatech.edu/ila/systems-of-eqns.html | Skip to main content
Section1.1Systems of Linear Equations¶ permalink
Objectives
Understand the definition of and what it means to use to label points on a geometric object.
Pictures: solutions of systems of linear equations, parameterized solution sets.
Vocabulary words: consistent, inconsistent, solution set.
During the first half of this textbook, we will be primarily concerned with understanding the solutions of systems of linear equations.
Definition
An equation in the unknowns is called linear if both sides of the equation are a sum of (constant) multiples of plus an optional constant.
For instance,
are linear equations, but
are not.
We will usually move the unknowns to the left side of the equation, and move the constants to the right.
A system of linear equations is a collection of several linear equations, like
(1.1.1)
Definition(Solution sets)
A solution of a system of equations is a list of numbers that make all of the equations true simultaneously.
The solution set of a system of equations is the collection of all solutions.
Solving the system means finding all solutions with formulas involving some number of parameters.
A system of linear equations need not have a solution. For example, there do not exist numbers and making the following two equations true simultaneously:
In this case, the solution set is empty. As this is a rather important property of a system of equations, it has its own name.
Definition
A system of equations is called inconsistent if it has no solutions. It is called consistent otherwise.
A solution of a system of equations in variables is a list of numbers. For example, is a solution of (1.1.1). As we will be studying solutions of systems of equations throughout this text, now is a good time to fix our notions regarding lists of numbers.
Subsection1.1.1Line, Plane, Space, Etc.¶ permalink
We use to denote the set of all real numbers, i.e., the number line. This contains numbers like
Definition
Let be a positive whole number. We define
An -tuple of real numbers is called a point of
In other words, is just the set of all (ordered) lists of real numbers. We will draw pictures of in a moment, but keep in mind that this is the definition. For example, and are points of
Example(The number line)
When we just get back: Geometrically, this is the number line.
Example(The Euclidean plane)
When we can think of as the -plane. We can do so because every point on the plane can be represented by an ordered pair of real numbers, namely, its - and -coordinates.
Example(3-Space)
When we can think of as the space we (appear to) live in. We can do so because every point in space can be represented by an ordered triple of real numebrs, namely, its -, -, and -coordinates.
Interactive: Points in 3-Space
So what is or or These are harder to visualize, so you have to go back to the definition: is the set of all ordered -tuples of real numbers
They are still “geometric” spaces, in the sense that our intuition for and often extends to
We will make definitions and state theorems that apply to any but we will only draw pictures for and
The power of using these spaces is the ability to label various objects of interest, such as geometric objects and solutions of systems of equations, by the points of
Example(Color Space)
Example(Traffic Flow)
Example(QR Codes)
In the above examples, it was useful from a psychological perspective to replace a list of four numbers (representing traffic flow) or of 841 numbers (representing a QR code) by a single piece of data: a point in some This is a powerful concept; starting in Section 2.2, we will almost exclusively record solutions of systems of linear equations in this way.
Subsection1.1.2Pictures of Solution Sets¶ permalink
Before discussing how to solve a system of linear equations below, it is helpful to see some pictures of what these solution sets look like geometrically.
One Equation in Two Variables
Consider the linear equation We can rewrite this as which defines a line in the plane: the slope is and the -intercept is
Definition(Lines)
For our purposes, a line is a ray that is straight and infinite in both directions.
One Equation in Three Variables
Consider the linear equation This is the implicit equation for a plane in space.
Definition(Planes)
A plane is a flat sheet that is infinite in all directions.
Remark
The equation defines a “-plane” in -space, and more generally, a single linear equation in variables defines an “-plane” in -space. We will make these statements precise in Section 2.7.
Two Equations in Two Variables
Now consider the system of two linear equations
Each equation individually defines a line in the plane, pictured below.
A solution to the system of both equations is a pair of numbers that makes both equations true at once. In other words, it as a point that lies on both lines simultaneously. We can see in the picture above that there is only one point where the lines intersect: therefore, this system has exactly one solution. (This solution is as the reader can verify.)
Usually, two lines in the plane will intersect in one point, but of course this is not always the case. Consider now the system of equations
These define parallel lines in the plane.
The fact that that the lines do not intersect means that the system of equations has no solution. Of course, this is easy to see algebraically: if then it is cannot also be the case that
There is one more possibility. Consider the system of equations
The second equation is a multiple of the first, so these equations define the same line in the plane.
In this case, there are infinitely many solutions of the system of equations.
Two Equations in Three Variables
Consider the system of two linear equations
Each equation individually defines a plane in space. The solutions of the system of both equations are the points that lie on both planes. We can see in the picture below that the planes intersect in a line. In particular, this system has infinitely many solutions.
Remark
In general, the solutions of a system of equations in variables is the intersection of “-planes” in -space. This is always some kind of linear space, as we will discuss in Section 2.4.
Subsection1.1.3Parametric Description of Solution Sets
According to this definition, solving a system of equations means writing down all solutions in terms of some number of parameters. We will give a systematic way of doing so in Section 1.3; for now we give parametric descriptions in the examples of the previous subsection.
Lines
Consider the linear equation of this example. In this context, we call an implicit equation of the line. We can write the same line in parametric form as follows:
This means that every point on the line has the form for some real number In this case, we call a parameter, as it parameterizes the points on the line.
Now consider the system of two linear equations
of this example. These collectively form the implicit equations for a line in (At least two equations are needed to define a line in space.) This line also has a parametric form with one parameter
Note that in each case, the parameter allows us to use to label the points on the line. However, neither line is the same as the number line indeed, every point on the first line has two coordinates, like the point and every point on the second line has three coordinates, like
Planes
Consider the linear equation of this example. This is an implicit equation of a plane in space. This plane has an equation in parametric form: we can write every point on the plane as
In this case, we need two parameters and to describe all points on the plane.
Note that the parameters allow us to use to label the points on the plane. However, this plane is not the same as the plane indeed, every point on this plane has three coordinates, like the point
When there is a unique solution, as in this example, it is not necessary to use parameters to describe the solution set. |
188543 | https://mathematicalolympiads.wordpress.com/wp-content/uploads/2012/08/andreescu-contests-around-the-world-1999-2000.pdf | 1 1999 National Contests: Problems and Solutions 1 2 Belarus 1.1 Belarus National Olympiad, Fourth Round Problem 10.1 Determine all real numbers a such that the function f(x) = {ax + sin x} is periodic. Here {y} is the fractional part of y.
Solution: The solutions are a = r π, r ∈Q.
First, suppose a = r π for some r ∈Q; write r = p q with p, q ∈Z, q > 0. Then f(x + 2qπ) = p qπ (x + 2qπ) + sin(x + 2qπ) = p qπ x + 2p + sin x = p qπ x + sin x = f(x) so f is periodic with period 2qπ.
Now, suppose f is periodic; then there exists p > 0 such that f(x) = f(x+p) for all x ∈R. Then {ax+sin x} = {ax+ap+sin(x+p)} for all x ∈R; in other words g(x) = ap+sin(x+p)−sin x is an integer for all x. But g is continuous, so there exists k ∈Z such that g(x) = k for all x ∈R. Rewriting this gives sin(x + p) −sin x = k −ap for all x ∈R.
Letting x = y, y + p, y + 2p, . . . , y + (n −1)p and summing gives sin(y + np) −sin y = n(k −ap) for all y ∈R and n ∈N.
Since the left hand side of this equation is bounded by 2, we conclude that k = ap and sin(x + p) = sin x for all x ∈R.
In particular, sin π 2 + p = sin π 2 = 1 and hence p = 2mπ for some m ∈N. Thus a = k p = k 2mπ = r π with r = k 2m ∈Q, as desired.
Problem 10.2 Prove that for any integer n > 1 the sum S of all divisors of n (including 1 and n) satisfies the inequalities k√n < S < √ 2kn, where k is the number of divisors of n.
Solution: Let the divisors of n be 1 = d1 < d2 < · · · < dk = n; 1999 National Contests: Problems and Solutions 3 then didk+1−i = n for each i. Thus S = k X i=1 di = k X i=1 di + dk+1−i 2 > k X i=1 p didk+1−i = k√n, giving the left inequality. (The inequality is strict because equality does not hold for d1+dk 2 ≥√d1dk.) For the right inequality, let S2 = Pk i=1 d2 i and use the Power Mean Inequality to get S k = Pk i=1 di k ≤ sPk i=1 d2 i k = r S2 k so S ≤ p kS2.
Now S2 n2 = k X i=1 d2 i n2 = k X i=1 1 d2 k+1−i ≤ n X j=1 1 j2 < π2 6 since d1, . . . , dk are distinct integers between 1 and n. Therefore S ≤ p kS2 < r kn2π2 6 < √ 2kn.
Problem 10.3 There is a 7 × 7 square board divided into 49 unit cells, and tiles of three types: 3 × 1 rectangles, 3-unit-square corners, and unit squares. Jerry has infinitely many rectangles and one corner, while Tom has only one square.
(a) Prove that Tom can put his square somewhere on the board (covering exactly one unit cell) in such a way that Jerry can not tile the rest of the board with his tiles.
(b) Now Jerry is given another corner. Prove that no matter where Tom puts his square (covering exactly one unit cell), Jerry can tile the rest of the board with his tiles.
Solution: (a) Tom should place his square on the cell marked X in the boards below.
4 Belarus 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 X 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 1 3 2 1 3 2 1 2 1 3 2 1 3 2 3 2 1 3 2 1 3 1 3 2 1 3 2 1 2 1 3 X 1 3 2 3 2 1 3 2 1 3 1 3 2 1 3 2 1 The grid on the left contains 17 1’s, 15 2’s and 16 3’s; since every 3×1 rectangle contains a 1, a 2 and a 3, Jerry’s corner must cover a 3 and two 1’s; thus it must be oriented like a Γ. But every such corner covers a 1, a 2 and a 3 in the right grid, as does any 3 × 1 rectangle. Since the right grid also contains 17 1’s, 15 2’s and 16 3’s, Jerry cannot cover the 48 remaining squares with his pieces.
(b) The following constructions suffice.
The first figure can be rotated and placed on the 7×7 board so that Tom’s square falls into its blank, untiled region. Similarly, the second figure can be rotated and placed within the remaining untiled 4 × 4 region so that Tom’s square is still uncovered; and finally, the single corner can be rotated and placed without overlapping Tom’s square.
Problem 10.4 A circle is inscribed in the isosceles trapezoid ABCD. Let the circle meet diagonal AC at K and L (with K between 1999 National Contests: Problems and Solutions 5 A and L). Find the value of AL · KC AK · LC .
First Solution: Lemma. Suppose we have a (not necessarily isosceles) trapezoid ABCD circumscribed about a circle with radius r, where the circle touches sides AB, BC, CD, DA at points P, Q, R, S, respectively. Let line AC intersect the circle at K and L, with K between A and L.
Also write m = AP and n = CR. Then AK · LC = mn + 2r2 − p (mn + 2r2)2 −(mn)2 and AL · KC = mn + 2r2 + p (mn + 2r2)2 −(mn)2.
Proof: Assume without loss of generality that AB ∥CD, and orient the trapezoid so that lines AB and CD are horizontal. Let t = AK, u = KL, and v = LC; also let σ = t + v and π = tv.
By Power of a Point, we have t(t + u) = m2 and v(v + u) = n2; multiplying these gives π(π + uσ + u2) = m2n2. Also, A and C are separated by m + n horizontal distance and 2r vertical distance; thus AC2 = (m + n)2 + (2r)2. Then (m + n)2 + (2r)2 = AC2 = (t + u + v)2 m2 + 2mn + n2 + 4r2 = t(t + u) + v(v + u) + 2π + uσ + u2 m2 + 2mn + n2 + 4r2 = m2 + n2 + 2π + uσ + u2 2mn + 4r2 −π = π + uσ + u2.
Multiplying by π on both sides we have π(2mn + 4r2 −π) = π(π + uσ + u2) = (mn)2, a quadratic in π with solutions π = mn + 2r2 ± p (mn + 2r2)2 −(mn)2.
But since m2n2 = t(t + u)v(v + u) ≥t2v2, we must have mn ≥π.
Therefore AK · LC = π = mn + 2r2 − p (mn + 2r2)2 −(mn)2. And since (AK · AL) · (CK · CL) = m2 · n2, we have AL · KC = m2n2 π = mn + 2r2 + p (mn + 2r2)2 −(mn)2.
6 Belarus As in the lemma, assume that AB ∥CD and let the given circle be tangent to sides AB, BC, CD, DA at points P, Q, R, S, respectively.
Also define m = AP = PB = AS = BQ and n = DR = RC = DS = CQ.
Drop perpendicular AX to line CD. Then AD = m + n, DX = |m−n|, and AX = 2r. Then by the Pythagorean Theorem on triangle ADX, we have (m + n)2 = (m −n)2 + (2r)2 which gives mn = r2.
Using the lemma, we find that AK · LC = (3 −2 √ 2)r2 and AL · KC = (3 + 2 √ 2)r2. Thus AL·KC AK·LC = 17 + 12 √ 2.
Second Solution: Suppose A′B′C′D′ is a square with side length s, and define K′, L′ analagously to K and L. Then A′C′ = s √ 2 and K′L′ = s, and A′L′ = K′C′ = s √ 2+1 2 and A′K′ = L′C′ = s √ 2−1 2 .
Thus A′L′ · K′C′ A′K′ · L′C′ = ( √ 2 + 1)2 ( √ 2 −1)2 = ( √ 2 + 1)4 = 17 + 12 √ 2.
Consider an arbitrary isosceles trapezoid ABCD with inscribed circle ω; assume AB ∥CD.
Since no three of A, B, C, D are collinear, there is a projective transformation τ taking ABCD to a parallelogram A′B′C′D′. This map takes ω to a conic ω′ tangent to the four sides of A′B′C′D′. Let P = BC ∩AD, and let ℓbe the line through P parallel to line AB; then τ maps ℓto the line at ∞.
Since ω does not intersect ℓ, ω′ is an ellipse.
Thus by composing τ with an affine transformation (which preserves parallelograms) we may assume that ω′ is a circle. Let W, X, Y , Z be the tangency points of ω to sides AB, BC, CD, DA respectively, and W ′, X′, Y ′, Z′ their images under τ. By symmetry line WY passes through the intersection of lines BC and AD, and line XZ is parallel to lines AB and CD; thus W ′Y ′ ∥B′C′ ∥A′D′ and X′Z′ ∥A′B′ ∥C′D′. But ω′ is tangent to the parallel lines A′B′ and C′D′ at W ′ and Y ′, so W ′Y ′ is a diameter of ω′ and W ′Y ′ ⊥A′B′; thus B′C′ ⊥A′B′ and A′B′C′D′ is a rectangle. Since A′B′C′D′ has an inscribed circle it must be a square. Thus we are in the case considered at the beginning of the problem; if K′ and L′ are the intersections of line A′C′ with ω′, with K′ between A′ and L′, then A′L′·K′C′ A′K′·L′C′ = 17 + 12 √ 2. Now τ maps {K, L} = AC ∩ω to {K′, L′} = A′C′ ∩ω′ (but perhaps not in that order). If τ(K) = K′ and τ(L) = L′, then since projective 1999 National Contests: Problems and Solutions 7 transformations preserve cross-ratios, we would have AL · KC AK · LC = A′L′ · K′C′ A′K′ · L′C′ = 17 + 12 √ 2.
But if instead τ(K) = L′ and τ(L) = K′, then we would obtain AL·KC AK·LC = 1 17+12 √ 2 < 1, impossible since AL > AK and KC > LC.
It follows that AL·KC AK·LC = 17 + 12 √ 2, as desired.
Problem 10.5 Let P and Q be points on the side AB of the triangle ABC (with P between A and Q) such that ∠ACP = ∠PCQ = ∠QCB, and let AD be the angle bisector of ∠BAC. Line AD meets lines CP and CQ at M and N respectively. Given that PN = CD and 3∠BAC = 2∠BCA, prove that triangles CQD and QNB have the same area.
Solution: Since 3∠BAC = 2∠ACB, ∠PAN = ∠NAC = ∠ACP = ∠PCQ = ∠QCD.
Let θ equal this common angle measure. Thus ACNP and ACDQ are cyclic quadrilaterals, so θ = ∠ANP = ∠CQD = ∠CPN.
From angle-angle-side congruency we deduce that △NAP ∼ = △CQD ∼ = △PCN. Hence CP = CQ, and by symmetry we have AP = QB.
Thus, [CQD] = [NAP] = [NQB].
Problem 10.6 Show that the equation {x3} + {y3} = {z3} has infinitely many rational non-integer solutions. Here {a} is the fractional part of a.
Solution: Let x = 3 5(125k + 1), y = 4 5(125k + 1), z = 6 5(125k + 1) for any integer k. These are never integers because 5 does not divide 125k + 1. Moreover 125x3 = 33(125k + 1)3 ≡33 (mod 125), so 125 divides 125x3−33 and x3− 3 5 3 is an integer; thus {x3} = 27 125.
Similarly {y3} = 64 125 and {z3} = 216 125 −1 = 91 125 = 27 125 + 64 125, and therefore {x3} + {y3} = {z3}.
8 Belarus Problem 10.7 Find all integers n and real numbers m such that the squares of an n × n board can be labelled 1, 2, . . . , n2 with each number appearing exactly once in such a way that (m −1)aij ≤(i + j)2 −(i + j) ≤maij for all 1 ≤i, j ≤n, where aij is the number placed in the intersection of the ith row and jth column.
Solution: Either n = 1 and 2 ≤m ≤3 or n = 2 and m = 3. It is easy to check that these work using the constructions below.
1 1 2 3 4 Now suppose we are given a labelling of the squares {aij} which satisfies the given conditions. By assumption a11 ≥1 so m −1 ≤(m −1)a11 ≤(1 + 1)2 −(1 + 1) = 2 and m ≤3. On the other hand ann ≤n2 so 4n2 −2n = (n + n)2 −(n + n) ≤mann ≤mn2 and m ≥4n2−2n n2 = 4 −2 n. Thus 4 −2 n ≤m ≤3 which implies the result.
Problem 11.1 Evaluate the product 21999 Y k=0 4 sin2 kπ 22000 −3 .
Solution: For simplicity, write f(x) = sin xπ 22000 .
At k = 0, the expression inside the parentheses equals −3. Recog-nizing the triple-angle formula sin(3θ) = 4 sin3 θ −3 sin θ at play, and noting that f(k) ̸= 0 when 1 ≤k ≤21999, we can rewrite the given product as −3 21999 Y k=1 sin 3kπ 22000 sin kπ 22000 or −3 21999 Y k=1 f(3k) f(k) .
(1) 1999 National Contests: Problems and Solutions 9 Now 21999 Y k=1 f(3k) = 21999−2 3 Y k=1 f(3k) · 22000−1 3 Y k= 21999+1 3 f(3k) · 21999 Y 22000+2 3 f(3k).
Since sin θ = sin(π −θ) = −sin(π +θ), we have f(x) = f(22000 −x) = −f(x −22000). Hence, letting Si = {k | 1 ≤k ≤21999, k ≡i (mod 3)} for i = 0, 1, 2, the last expression equals 21999−2 3 Y k=1 f(3k) · 22000−1 3 Y k= 21999+1 3 f(22000 −3k) · 21999 Y 22000+2 3 −f(3k −22000) = Y k∈S0 f(k) · Y k∈S1 f(k) · Y k∈S2 (−f(k)) = (−1) 21999+1 3 21999 Y k=1 f(k) = − 21999 Y k=1 f(k).
Combined with the expression in (1), this implies that the desired product is (−3)(−1) = 3.
Problem 11.2 Let m and n be positive integers. Starting with the list 1, 2, 3, . . . , we can form a new list of positive integers in two different ways.
(i) We first erase every mth number in the list (always starting with the first); then, in the list obtained, we erase every nth number.
We call this the first derived list.
(ii) We first erase every nth number in the list; then, in the list obtained, we erase every mth number. We call this the second derived list.
Now, we call a pair (m, n) good if and only if the following statement is true: if some positive integer k appears in both derived lists, then it appears in the same position in each.
(a) Prove that (2, n) is good for any positive integer n.
(b) Determine if there exists any good pair (m, n) such that 2 < m < n.
Solution: Consider whether some positive integer j is in the first derived list. If it is congruent to 1 (mod m), then j + mn is as well 10 Belarus so they are both erased. If not, then suppose it is the t-th number remaining after we’ve erased all the multiples of m.
There are n multiples of m erased between j and j + mn, so j + mn is the (t+mn−n)-th number remaining after we’ve erased all the multiples of m. But either t and t+mn−n are both congruent to 1 (mod n) or both not congruent to 1 (mod n). Hence j is erased after our second pass if and only if j + mn is as well.
A similar argument applies to the second derived list.
Thus in either derived list, the locations of the erased numbers repeat with period mn; and also, among each mn consecutive numbers exactly mn −(m + n −1) remain. (In the first list, n + mn−n−1 n + 1 = n + m −1 + −1 n + 1 = m + n −1 of the first mn numbers are erased; similarly, m + n −1 of the first mn numbers are erased in the second list.) These facts imply that the pair (m, n) is good if and only if when any k ≤mn is in both lists, it appears at the same position.
(a) Given a pair (2, n), the first derived list (up to k = 2n) is 4, 6, 8, . . . , 2n. If n is even, the second derived list is 3, 5, . . . , n − 1, n + 2, n + 4, . . . , 2n. And if n is odd, the second derived list is 3, 5, . . . , n −2, n, n + 3, n + 5, . . . , 2n. In either case the first and second lists’ common elements are the even numbers between n + 2 and 2n inclusive. Each such 2n −i (with i < n−1 2 ) is the (n −1 −i)-th number on both lists, showing that (2, n) is good.
(b) Such a pair exists—in fact, the simplest possible pair (m, n) = (3, 4) suffices. The first derived list (up to k = 12) is 3, 5, 6, 9, 11, 12 and the second derived list is 3, 4, 7, 8, 11, 12.
The common elements are 3, 11, 12, and these are all in the same positions.
Problem 11.3 Let a1, a2, . . . , a100 be an ordered set of numbers.
At each move it is allowed to choose any two numbers an, am and change them to the numbers a2 n am −n m a2 m an −am and a2 m an −m n a2 n am −an respectively. Determine if it is possible, starting with the set with ai = 1 5 for i = 20, 40, 60, 80, 100 and ai = 1 otherwise, to obtain a set consisting of integers only.
Solution: After transforming an to a′ n = a2 n am −n m a2 m an −am and 1999 National Contests: Problems and Solutions 11 am to a′ m = a2 m an −m n a2 n am −an , we have a′ n n + a′ m m = 1 n · a2 n am −1 m · a2 m an + am m + 1 m · a2 m an −1 n · a2 n am + an n = an n + am m .
Thus the quantity P100 i=1 ai i is invariant under the given operation.
At the beginning, this sum equals I1 = 99 X i=1 ai i + 1 500.
When each of the numbers a1 1 , a2 2 , . . . , a99 99 is written as a fraction in lowest terms, none of their denominators are divisible by 125; while 125 does divide the denominator of 1 500.
Thus when written as a fraction in lowest terms, I1 must have a denominator divisible by 125.
Now suppose by way of contradiction that we could make all the numbers equal to integers b1, b2, . . . , b100 in that order.
Then in I2 = P100 i=1 bi i , the denominator of each of the fractions bi i is not divisible by 125.
Thus when I2 is written as a fraction in lowest terms, its denominator is not divisible by 125 either. But then I2 cannot possibly equal I1, a contradiction. Therefore we can never obtain a set consisting of integers only.
Problem 11.4 A circle is inscribed in the trapezoid ABCD. Let K, L, M, N be the points of intersections of the circle with diagonals AC and BD respectively (K is between A and L and M is between B and N). Given that AK · LC = 16 and BM · ND = 9 4, find the radius of the circle.
Solution: Let the circle touch sides AB, BC, CD, DA at P, Q, R, S, respectively, and let r be the radius of the circle. Let w = AS = AP; x = BP = BQ; y = CQ = CR; and z = DR = DS. As in problem 11.4, we have wz = xy = r2 and thus wxyz = r4. Also observe that 12 Belarus from the lemma in problem 11.4, AK · LC depends only on r and AP · CR; and BM · ND depends only on r and BP · DR.
Now draw a parallelogram A′B′C′D′ circumscribed about the same circle, with points P ′, Q′, R′, S′ defined analagously to P, Q, R, S, such that A′P ′ = C′R′ = √wy. Draw points K′, L′, M ′, N ′ analagously to K, L, M, N.
Then since A′P ′ · C′R′ = wy, by the observation in the first paragraph we must have A′K′ · L′C′ = AK · LC = 16; therefore A′K′ = L′C′ = 4. And as with quadrilateral ABCD, we have A′P ′·B′P ′·C′R′·D′R′ = r4 = wxyz. Thus B′P ′·D′R′ = xz and again by the observation we must have B′M ′ ·N ′D′ = BM ·ND = 9 4.
Therefore B′M ′ = N ′D′ = 3 2.
Then if O is the center of the circle, we have A′O = 4 + r and S′O = r. By the Pythagorean Theorem A′S′ = √8r + 16; similarly, S′D′ = q 3r + 9 4. Since A′S′ · S′D′ = r2, we have (8r + 16) 3r + 9 4 = r4, which has positive solution r = 6 and, by Descartes’ rule of signs, no other positive solutions.
Problem 11.5 Find the greatest real number k such that for any triple of positive real numbers a, b, c such that kabc > a3 + b3 + c3, there exists a triangle with side legths a, b, c.
Solution: Equivalently, we want the greatest real number k such that for any a, b, c > 0 with a + b ≤c, we have kabc ≤a3 + b3 + c3.
First pick b = a and c = 2a. Then we must have 2ka3 ≤10a3 = ⇒k ≤5.
On the other hand, suppose k = 5. Then writing c = a + b + x, expanding a3 + b3 + c3 −5abc gives 2a3 + 2b3 −2a2b −2ab2 + abx + 3(a2 + b2)x + 3(a + b)x2 + x3.
But 2a3+2b3−2a2b−2ab2 ≥0 (either by rearrangement, by AM-GM, or from the inequality (a + b)(a −b)2 ≥0); and the other terms are nonnegative. Thus a3 + b3 + c3 −5abc ≥0, as desired.
1999 National Contests: Problems and Solutions 13 Problem 11.6 Find all integers x and y such that x6 + x3y = y3 + 2y2.
Solution: The only solutions are (x, y) equals (0, 0), (0, −2), and (2, 4).
If x = 0 then y = 0 or −2; if y = 0 then x = 0. Now assume that both x and y are nonzero, and rewrite the given equation as x3(x3 + y) = y2(y + 2).
We first show that (x, y) = (ab, 2b3), (ab, b3), or (ab, b3 2 ) for some integers a, b. Suppose some prime p divides y exactly m > 0 times (that is, y is divisible by pm but not pm+1).
Then since x6 = y3 + 2y2 −x3y, p must divide x as well — say, n > 0 times.
First suppose p > 2; then it divides the right hand side y2(y + 2) exactly 2m times.
If 3n < m then p divides the left hand side x3(x3 + y) exactly 6n times so that 6n = 2m, a contradiction. And if 3n > m then p divides the left hand side exactly 3n + m times so that 3n + m = 2m and 3n = m, a contradiction. Therefore 3n = m.
Now suppose p = 2. If m > 1, then 2 divides the right hand side exactly 2m + 1 times. If 3n < m then 2 divides the left hand side 6n times so that 6n = 2m + 1 > 2m, a contradiction. If 3n > m, then 2 divides the left hand side 3n + m times so that 3n + m = 2m + 1 and 3n = m + 1. Or finally, we could have 3n = m.
We wish to show that (x, y) = (ab, 2b3), (ab, b3), or (ab, b3 2 ). If 2 divides y only once, then from before (since 3n = m when p > 2, m > 0) we have y = 2b3 and x = ab for some a, b. And if 2 divides y more than once, then (since 3n = m when p > 2, m > 0 and since 3n = m or m + 1 when p = 2, m > 1) we either have (x, y) = (ab, b3) or (x, y) = (ab, b3 2 ).
Now simply plug these possibilities into the equation.
We then either have a6 +a3 = b3 +2, a6 +2a3 = 8b3 +8, or 8a6 +4a3 = b3 +4.
In the first case, if a > 1 then b3 = a6 + a3 −2 and some algebra verifies that (a2 + 1)3 > b3 > (a2)3, a contradiction; if a < 0 then we have (a2)3 > b3 > (a2 −1)3. Thus either a = 0 and x = 0 or a = 1 and b = 0. But we’ve assumed x, y ̸= 0, so this case yields no solutions.
In the second case, if a > 0 then (a2 + 1)3 > (2b)3 > (a2)3. If a < −2 then (a2)3 > (2b)3 > (a2 −1)3. Thus either a = −2, −1, or 0; and these yield no solutions either.
14 Belarus Finally, in the third case when a > 1 then (2a2 +1)3 > b3 > (2a2)3.
When a < −1 then (2a2)3 > b3 > (2a2 −1)3. Thus either a = −1, 0, or 1; this yields both (a, b) = (−1, 0) and (a, b) = (1, 2). Only the latter gives a solution where x, y ̸= 0 — namely, (x, y) = (2, 4). This completes the proof.
Problem 11.7 Let O be the center of circle ω. Two equal chords AB and CD of ω intersect at L such that AL > LB and DL > LC. Let M and N be points on AL and DL respectively such that ∠ALC = 2∠MON. Prove that the chord of ω passing through M and N is equal to AB and CD.
Solution: We work backward. Suppose that P is on minor arc d AC and Q is on minor arc d BD such that PQ = AB = CD, where line PQ hits AL at M ′ and DL at N ′. We prove that ∠ALC = 2∠M ′ON ′.
Say that the midpoints of AB, PQ, CD are T1, T2, and T3. CD is the image of AB under the rotation about O through angle ∠T1OT3; this angle also equals the measure of d AC, which equals ∠ALC. Also, by symmetry we have ∠T1OM ′ = ∠M ′OT2 and ∠T2ON ′ = ∠N ′OT3.
Therefore ∠ALC = ∠T1OT3 = ∠T1OT2 + ∠T2OT3 = 2(∠M ′OT2 + ∠T2ON ′) = 2∠M ′ON ′, as claimed.
Now back to the original problem.
Since ∠T1OT3 = ∠ALC, ∠T1OL = 1 2T1OT3 = 1 2∠ALC. Then since ∠MON = 1 2∠ALC = ∠T1OL, M must lie on T1L. Then look at the rotation about O that sends T1 to M; it sends A to some P on d AC, and B to some point Q on d BD. Then PQ is a chord with length AB, passing through M on AL and N ′ on DL. From the previous work, we know that ∠ALC = 2∠MON ′; and since ∠ALC = 2∠MON, we must have N = N ′. Thus the length of the chord passing through M and N indeed equals AB and CD, as desired.
IMO Selection Tests Problem 1 Find all functions h : Z →Z such that h(x + y) + h(xy) = h(x)h(y) + 1 for all x, y ∈Z.
1999 National Contests: Problems and Solutions 15 Solution: There are three possible functions: h(n) = 1; h(2n) = 1, h(2n + 1) = 0; h(n) = n + 1.
Plugging (x, y) = (0, 0) into the functional equation, we find that h(0)2 −2h(0) + 1 = 0 and hence h(0) = 1. Plugging in (x, y) = (1, −1) then yields h(0) + h(−1) = h(1)h(−1) + 1 and h(−1) = h(1)h(−1), and thus either h(−1) = 0 or h(1) = 1.
First suppose that h(1) ̸= 1; then h(−1) = 0. Then plugging in (x, y) = (2, −1) and (x, y) = (−2, 1) yields h(1) + h(−2) = 1 and h(−2) = h(−2)h(1) + 1. Substituting h(−2) = 1 −h(1) into the second equation, we find that 1 −h(1) = (1 −h(1))h(1) + 1, h(1)2 −2h(1) = 0, and h(1)(h(1) −2) = 0, implying that h(1) = 0 or h(1) = 2.
Thus, h(1) = 0, 1, or 2. Plugging y = 1 into the equation for each of these cases shows that h must be one of the three functions presented.
Problem 2 Let a, b, c ∈Q, ac ̸= 0.
Given that the equation ax2 + bxy + cy2 = 0 has a non-zero solution of the form (x, y) = (a0 + a1 3 √ 2 + a2 3 √ 4, b0 + b1 3 √ 2 + b2 3 √ 4) with ai, bi ∈Q, i = 0, 1, 2, prove that it has also has a non-zero rational solution.
Solution: Let (α, β) = (a0 + a1 3 √ 2 + a2 3 √ 4, b0 + b1 3 √ 2 + b2 3 √ 4) be the given solution, and suppose without loss of generality that β is non-zero. Then α β is a root to the polynomial at2 + bt + c = 0.
16 Belarus Also, α β is of the form c0 + c1 3 √ 2 + c2 3 √ 4 for some rationals c0, c1, c2.
But because it is a root to a quadratic with rational coefficients, it must also be of the form d + e√f for rationals d, e, f.
Thus we have (c0 −d) + c1 3 √ 2 + c2 3 √ 4 = e√f, so the quantity c′ 0 + c1 3 √ 2 + c2 3 √ 4 2 must be an integer (where we write c′ 0 = c0 − d). After expanding this square, the coefficients of 3 √ 2 and 3 √ 4 are 2(c2 2 + c′ 0c1) and 2c′ 0c2 + c2 1, respectively; these quantities must equal zero.
From 2c′ 0c2 + c2 1 = 0 we have (c′ 0c1)2 = −2c′3 0 c2; and from c2 2 + c′ 0c1 = 0 we have (c′ 0c1)2 = c4 2. Thus −2c′3 0 c2 = c4 2. This implies that either c2 = 0 or c2 = − 3 √ 2c′ 0; in the latter case, since c2 is rational we must still have c2 = c′ 0 = 0.
Then c1 = 0 as well, and α β = c0 is rational. Thus (x, y) = ( α β , 1) is a non-zero rational solution to the given equation.
Problem 3 Suppose a and b are positive integers such that the product of all divisors of a (including 1 and a) is equal to the product of all divisors of b (including 1 and b). Does it follow that a = b?
Solution: Yes, it follows that a = b. Let d(n) denote the number of divisors of n; then the product of all divisors of n is Y k|n k = sY k|n k · Y k|n n k = sY k|n n = n d(n) 2 .
Thus the given condition implies that ad(a) and bd(b) equal the same number N. Since N is both a perfect d(a)-th power and a perfect d(b)-th power, it follows that it is also a perfect ℓ-th power of some number t, where ℓ= lcm(d(a), d(b)). Then a = t ℓ d(a) and b = t ℓ d(b) are both powers of the same number t as well.
Now if a is a bigger power of t than b, then it must have more divisors than b; but then t ℓ d(a) < t ℓ d(b) , a contradiction. Similarly a cannot be a smaller power of t than b. Therefore a = b, as claimed.
Problem 4 Let a, b, c be positive real numbers such that a2 + b2 + c2 = 3. Prove that 1 1 + ab + 1 1 + bc + 1 1 + ca ≥3 2.
Solution: Using the AM-HM inequality or the Cauchy-Schwarz 1999 National Contests: Problems and Solutions 17 inequality, we have 1 x + 1 y + 1 z ≥ 9 x + y + z for x, y, z ≥0. Also, notice that a2 + b2 + c2 ≥ab + bc + ca since this inequality is equivalent to 1 2(a−b)2 + 1 2(b−c)2 + 1 2(c−a)2 ≥0. Thus, 1 1 + ab + 1 1 + bc + 1 1 + ca ≥ 9 3 + ab + bc + ca ≥ 9 3 + a2 + b2 + c2 ≥3 2, as desired.
Problem 5 Suppose triangle T1 is similar to triangle T2, and the lengths of two sides and the angle between them of T1 are proportional to the lengths of two sides and the angle between them of T2 (but not necessarily the corresponding ones). Must T1 be congruent to T2?
Solution: The triangles are not necessarily congruent.
Say the vertices of T1 are A, B, C with AB = 4, BC = 6, and CA = 9, and say that ∠BCA = k∠ABC.
Then let the vertices of T2 be D, E, F where DE = 8k 3 , EF = 4k, and FD = 6k. Triangles ABC and DEF are similar in that order, so ∠EFD = ∠BCA = k∠ABC; also, EF = k · AB and FD = k · BC.
Therefore these triangles satisfy the given conditions.
Now since AB < AC we have ∠BCA < ∠ABC and k < 1; so DE = 8k 3 < 8 3 < AB.
Thus triangles ABC and DEF are not congruent, as desired.
Problem 6 Two real sequences x1, x2, . . . , and y1, y2, . . . , are defined in the following way: x1 = y1 = √ 3, xn+1 = xn + p 1 + x2 n, yn+1 = yn 1 + p 1 + y2 n for all n ≥1. Prove that 2 < xnyn < 3 for all n > 1.
First Solution: Let zn = 1 yn and notice that the recursion for yn is equivalent to zn+1 = zn + p 1 + z2 n.
18 Belarus Also note that z2 = √ 3 = x1; since the xi and zi satisfy the same recursion, this means that zn = xn−1 for all n > 1. Thus, xnyn = xn zn = xn xn−1 .
Because the xi are increasing, for n > 1 we have x2 n−1 ≥x2 1 = 3 > 1 3 ⇒2xn−1 > q 1 + x2 n−1 ⇒3xn−1 > xn. Also, q 1 + x2 n−1 > xn−1 ⇒xn > 2xn−1. Therefore, 2 < xnyn = xn xn−1 < 3, as desired.
Second Solution: Writing xn = tan an for 0◦< an < 90◦, we have xn+1 = tan an + p 1 + tan2 an = tan an + sec an = 1 + sin an cos an = tan 90◦+ an 2 .
Since a1 = 60◦, we have a2 = 75◦, a3 = 82.5◦, and in general an = 90◦− 30◦ 2n−1 . Thus xn = tan 90◦−30◦ 2n−1 = cot 30◦ 2n−1 = cot θn, where θn = 30◦ 2n−1 .
Similar calculation shows that yn = tan 2θn = 2 tan θn 1 −tan2 θn , implying that xnyn = 2 1 −tan2 θn .
Since 0◦< θn < 45◦, we have 0 < tan2 θn < 1 and xnyn > 2. And since for n > 1 we have θn < 30◦, we also have tan2 θn < 1 3 so that xnyn < 3.
Note: From the closed forms for xn and yn in the second solution, we can see the relationship yn = 1 xn−1 used in the first solution.
Problem 7 Let O be the center of the excircle of triangle ABC opposite A.
Let M be the midpoint of AC, and let P be the 1999 National Contests: Problems and Solutions 19 intersection of lines MO and BC. Prove that if ∠BAC = 2∠ACB, then AB = BP.
First Solution: Since O is the excenter opposite A, we know that O is equidistant from lines AB, BC, and CA. We also know that line AO bisects angle BAC. Thus ∠BAO = ∠OAC = ∠ACB. Letting D be the intersection of AO and BC, we then have ∠DAC = ∠ACD and hence DC = AD.
Consider triangles OAC and ODC. From above their altitudes from O are equal, and their altitudes from C are also clearly equal. Thus, OA/OD = [OAC]/[ODC] = AC/DC.
Next, because M is the midpoint of AC we have [OAM] = [OMC] and [PAM] = [PMC], and hence [OAP] = [OPC] as well. Then OA OD = [OAP] [ODP] = [OPC] [ODP] = PC DP .
Thus, AC DC = OA OD = P C DP , and AC CP = DC DP = AD DP . By the Angle Bisector Theorem, AP bisects ∠CAD.
It follows that ∠BAP = ∠BAD + ∠DAP = ∠ACP + ∠PAC = ∠APB, and therefore BA = BP, as desired.
Second Solution: Let R be the midpoint of the arc BC (not containing A) of the circumcircle of triangle ABC; and let I be the incenter of triangle ABC. We have ∠RBI = 1 2(∠CAB + ∠ABC) = 1 2(180◦−∠BRI). Thus RB = RI and similarly RC = RI, and hence R is the circumcenter of triangle BIC. But since ∠IBO = 90◦= ∠ICO, quadrilateral IBOC is cyclic and R is also the circumcenter of triangle BCO.
Let lines AO and BC intersect at Q.
Since M, O, and P are collinear we may apply Menelaus’ Theorem to triangle AQC to get AM CM CP QP QO AO = 1.
But AM CM = 1, and therefore CP P Q = AO QO.
And since R lies on AO and QO, we have AO QO = AR + RO QR + RO = AR + RC CR + RQ, which in turn equals AC CQ since triangles ARC and CRQ are similar; and AC CQ = AC AQ since we are given that ∠BAC = 2∠ACB; i.e., 20 Belarus ∠QAC = ∠QCA and CQ = AQ. Thus we have shown that CP P Q = AC AQ. By the Angle-Bisector Theorem, this implies that line AP bisects ∠QAC, from which it follows that ∠BAP = 3 2∠ACB = ∠BPA and AB = BP.
Problem 8 Let O, O1 be the centers of the incircle and the excircle opposite A of triangle ABC.
The perpendicular bisector of OO1 meets lines AB and AC at L and N respectively. Given that the circumcircle of triangle ABC touches line LN, prove that triangle ABC is isosceles.
Solution: Let M be the midpoint of arc d BC not containing A.
Angle-chasing gives ∠OBM = 1 2(∠A + ∠B) = ∠BOM and hence MB = MO.
Since ∠OBC = ∠B 2 and ∠CBO1 = 1 2(π −∠B), we have ∠OBO1 is a right angle. And since we know both that M lies on line AOO1 (the angle bisector of ∠A) and that MB = MO, it follows that BM is a median to the hypotenuse of right triangle OBO1 and thus M is the midpoint of OO1.
Therefore, the tangent to the circumcircle of ABC at M must be perpendicular to line AM. But this tangent is also parallel to line BC, implying that AM, the angle bisector of ∠A, is perpendicular to line BC. This can only happen if AB = AC, as desired.
Problem 9 Does there exist a bijection f of (a) a plane with itself (b) three-dimensional space with itself such that for any distinct points A, B line AB and line f(A)f(B) are perpendicular?
Solution: (a) Yes: simply rotate the plane 90◦about some axis perpendicular to it. For example, in the xy-plane we could map each point (x, y) to the point (y, −x).
(b) Suppose such a bijection existed.
Label the three-dimensional space with x-, y-, and z-axes; given any point P = (x0, y0, z0), we also view it as the vector p from (0, 0, 0) to (x0, y0, z0). Then 1999 National Contests: Problems and Solutions 21 the given condition says that (a −b) · (f(a) −f(b)) = 0 for any vectors a, b.
Assume without loss of generality that f maps the origin to itself; otherwise, g(p) = f(p) −f(0) is still a bijection and still satisfies the above equation.
Plugging b = (0, 0, 0) into the equation above we have a · f(a) = 0 for all a. Then the above equation reduces to a · f(b) + b · f(a) = 0.
Given any vectors a, b, c and any reals m, n we then have m (a · f(b) + b · f(a)) = 0 n (a · f(c) + c · f(a)) = 0 a · f(mb + nc) + (mb + nc) · f(a) = 0.
Adding the first two equations and subtracting the third gives a · (mf(b) + nf(c) −f(mb + nc)) = 0.
Since this must be true for any vector a, we must have f(mb + nc) = mf(B)+nf(C). Therefore f is linear, and it is determined by how it transforms the unit vectors i = (1, 0, 0), j = (0, 1, 0), and k = (0, 0, 1). If f(i) = (a1, a2, a3), f(j) = (b1, b2, b3), and f(k) = (c1, c2, c3), then for a vector x we have f(x) = a1 b1 c1 a2 b2 c3 a3 b3 c3 x.
Applying f(a)·a = 0 with a = i, j, k, we have a1 = b2 = c3 = 0.
Then applying a·f(b)+b·f(a) with (a, b) = (i, j), (j, k), (j, k) we have b1 = −a2, c1 = −a3, c2 = −b3. But then the determinant of the array in the equation is a2b3c1 + a3b1c2 = −a2b3a3 + a3a2b3 = 0, so there exist constants k1, k2, k3 not all zero such that k1f(i) + k2f(j) + k3f(k) = 0.
But then f(k1, k2, k3) = 0 = f(0, 0, 0), contradicting the assumption that f was a bijection!
22 Belarus Therefore our original assumption was false, and no such bijec-tion exists.
Problem 10 A word is a finite sequence of two symbols a and b.
The number of the symbols in the word is said to be the length of the word. A word is called 6-aperiodic if it does not contain a subword of the form cccccc for any word c. Prove that f(n) > 3 2 n , where f(n) is the total number of 6-aperiodic words of length n.
Solution: Rather than attempting to count all such words, we add some restrictions and count only some of the 6-aperiodic words. Also, instead of working with a’s and b’s we’ll work with 0’s and 1’s.
The Thue-Morse sequence is defined by t0 = 0, t1 = 1, t2n+1 = 1 −t2n, and t2n = tn. These properties can be used to show that the only subwords of the form cc . . . c are 00 and 11.
We restrict the 6-aperiodic words in a similar spirit.
Call a word x1x2 . . . xn of length n 6-countable if it satisfies the following conditions: (i) x5i = xi for 1 ≤i.
(ii) x5i−1 = 1 −x5i for 1 ≤i ≤n 5 .
(iii) If (x5i+2, x5i+3, x5i+4) = (1, 0, 1) [or (0, 1, 0)], then (x5i+7, x5i+8, x5i+9) ̸= (0, 1, 0) [or (1, 0, 1)].
Lemma 1. Every 6-countable word is 6-aperiodic.
Proof: Suppose by way of contradiction that some 6-countable word contains a subword of the form cccccc, where the strings c appear in the positions xj through xj+ℓ−1; xj+ℓthrough xj+2ℓ−1; and so on up to xj+5ℓthrough xj+6ℓ−1. Pick a word with the smallest such ℓ.
If 5 | ℓ, then look at the indices i between j and j + ℓ−1 such that 5 | i; say they are 5i1, 5i2, . . . , 5iℓ/5.
Then x5i1x5i2 . . . x5iℓ/5, x5i1+ℓx5i2+ℓ. . . x5iℓ/5+ℓ, . . ., x5i1+5ℓx5i2+5ℓ. . . x5iℓ/5+5ℓall equal the same string c′; then (using the first condition of countability) the subword starting at xi1 and ending on xiℓ/5+ℓis of the form c′c′c′c′c′c′.
But this contradicts the minimal choice of ℓ; therefore, we can’t have 5 | ℓ.
Now, suppose that in the first appearance of c some two adjacent characters aj, aj+1 were equal. Then since 5 ̸ | ℓ, one of j, j + ℓ, j + 2ℓ, . . . , j + 4ℓis 4 (mod 5) — say, j + kℓ. Then aj+kℓ, aj+kℓ+1 must be the same since ajaj+1 = aj+kℓaj+kℓ+1; but they must also be 1999 National Contests: Problems and Solutions 23 different from the second condition of 6-countability! Because this is impossible, it follows that the characters in c alternate between 0 and 1.
A similar argument, though, shows that aj+ℓ−1 and aj+ℓmust be different; hence c is of the form 1010 . . . 10 or 0101 . . . 01.
But this would imply that our word violated the third condition of 6-countability—a contradiction. Therefore our original assumption was false, and any 6-countable word is 6-aperiodic.
Lemma 2. Given a positive integer m, there are more than 3 2 5m 6-countable words of length 5m.
Proof: Let αm be the number of length-5m 6-countable words.
To create a length-5m 6-countable word x1x2 . . . x5m, we can choose each of the “three-strings” x1x2x3, x6x7x8, . . ., x5m−4x5m−3x5m−2 to be any of the eight strings 000, 001, 010, 011, 100, 101, 110, or 111— taking care that no two adjacent strings are 010 and 101. Some quick counting then shows that α1 = 8 > 3 2 5 and α2 = 64 −2 = 62 > 3 2 10.
Let βm be the number of length-5m 6-countable words whose last three-string is 101; by symmetry, this also equals the number of length-5m 6-countable words whose last three-string is 010. Also let γm be the number of length-5m 6-countable words whose last three-string is not 101; again by symmetry, this also equals the number of length-5m 6-countable words whose last three-string isn’t 010. Note that αm = γm + βm.
For m ≥1, observe that γm = βm+1 because to any length-5m word whose last three-string isn’t 010, we can append the three-string 101 (as well as two other pre-determined numbers); and given a length-5(m + 1) word whose last three- string is 101, we can reverse this construction. Similar arguing shows that γm+1 = 6(γm + βm) + γm; the 6(γm + βm) term counts the words whose last three-string is neither 010 nor 101, and the γm term counts the words whose last three-string is 010. Combined, these recursions give γm+2 = 7γm+1 + 6γm βm+2 = 7βm+1 + 6βm αm+2 = 7αm+1 + 6αm.
24 Belarus Now if αm+1 > 3 2 5m+5 and αm > 3 2 5m, then αm+2 = 7αm+1 + 6αm > 3 2 5m 7 · 3 2 5 + 6 !
3 2 5m 3 2 10 = 3 2 5(m+2) .
Then since αm > 3 2 5m is true for m = 1, 2, by induction it is true for all positive integers m.
The lemma proves the claim for n = 5m.
Now suppose we are looking at length-(5m + i) words, where m ≥0 and i = 1, 2, 3, or 4. Then given any length-5m 6-countable word, we can form a length-(5m+i) word by choosing x5m+1, x5m+2, x5m+3 to be anything.
(For convenience, we say there is exactly α0 = 1 ≥ 3 2 0 length-0 6-countable word: the “empty word.”) Thus there are at least 2αm > 3 2 5m+1, 4αm > 3 2 5m+2, 8αm > 3 2 5m+3, and 8αm > 3 2 5m+4 6-countable length-(5m + 1), -(5m + 2), -(5m + 3), and -(5m + 4) words, respectively. This completes the proof.
Problem 11 Determine all positive integers n, n ≥2, such that n−k k is even for k = 1, 2, . . . , ⌊n 2 ⌋.
Solution: Lucas’s Theorem states that for integers n = nrpr + nr−1pr−1 + · · · + n0 and m = mrpr + mr−1pr−1 + · · · + m0 written in base p for a prime p, we have n m ≡ nr mr nr−1 mr−1 · · · n0 b0 (mod p).
With p = 2, the binary representation of n = 2s −1 we have nr = nr−1 = · · · = n0 = 1. Then for any 0 ≤m ≤2s −1 each ni mi = 1, and thus n m ≡1 · 1 · · · · · 1 ≡1 (mod 2).
This implies that n must be one less than a power of 2, or else one of n −k will equal such a number 2s −1 and then n−k k will be odd.
1999 National Contests: Problems and Solutions 25 In fact, all such n = 2s −1 do work: for k = 1, 2, . . . , ⌊n 2 ⌋, there is at least one 0 in the binary representation of n −k (not counting leading zeros, of course). And whenever there is a 0 in the binary representation of n −k, there is a 1 in the corresponding digit of k.
Then the corresponding (n−k)i ki equals 0, and by Lucas’s Theorem n−k k is even.
Therefore, n = 2s −1 for integers s ≥2.
Problem 12 A number of n players took part in a chess tourna-ment. After the tournament was over, it turned out that among any four players there was one who scored differently against the other three (i.e., he got a victory, a draw, and a loss).
Prove that the largest possible n satisfies the inequality 6 ≤n ≤9.
Solution: Let A1 ⇒A2 ⇒· · · ⇒An denote “A1 beats A2, A2 beats A3, . . . , An−1 beats An,” and let X | Y denote “X draws with Y.” First we show it is possible to have the desired results with n = 6: call the players A, B, C, D, E, F. Then let A ⇒B ⇒C ⇒D ⇒E ⇒A, F ⇒A, F ⇒B, F ⇒C, F ⇒D, F ⇒E, and have all other games end in draws. Visually, we can view this arrangement as a regular pentagon ABCDE with F at the center.
There are three different types of groups of 4, represented by ABCD, ABCF, and ABDF; in these three respective cases, B (or C), A, and A are the players who score differently from the other three.
Alternatively, let A ⇒B ⇒C ⇒D ⇒E ⇒F ⇒A, B ⇒D ⇒F ⇒B, C ⇒A ⇒E ⇒C, A | D, B | E, C | F.
In this arrangement there are three different types of groups of four, represented by {A, B, C, D}, {A, B, D, E}, and {A, B, D, F}. (If the players are arranged in a regular hexagon, these correspond to a trapezoid-shaped group, a rectangle-shaped group, and a diamond-shaped group.) In these three cases, A, B (or D), and A (or D) are the players who score differently against the other three.
26 Belarus Now we show it is impossible to have the desired results with n = 10 and thus all n ≥10; suppose by way of contradiction it was possible.
First we prove that all players draw exactly 4 times.
To do this, draw a graph with n vertices representing the players, and draw an edge between two vertices if they drew in their game. If V has degree 3 or less, then look at the remaining 6 or more vertices it is not adjacent to. By Ramsey’s Theorem, either three of them (call them X, Y, Z) are all adjacent or all not adjacent. But then in the group {V, X, Y, Z}, none of the players draws exactly once with the other players, a contradiction.
Thus each vertex has degree at least 4; we now prove that every vertex has degree exactly 4. Suppose by way of contradiction that some vertex A was adjacent to at least 5 vertices B, C, D, E, F. None of these vertices can be adjacent to two others; for example, if B was adjacent to C and D then in {A, B, C, D} each vertex draws at least twice—but some player must draw exactly once in this group.
Now in the group {B, C, D, E} some pair must draw: without loss of generality, say B and C. In the group {C, D, E, F} some pair must draw as well; since C can’t draw with D, E, or F from our previous observation, assume without loss of generality that E and F draw.
Now in {A, B, C, D} vertex D must beat one of B, C and lose to the other; without loss of generality, say D loses to B and beats C.
Looking at {A, D, E, F}, we can similarly assume that D beats E and loses to F. Next, in {A, C, D, E} players C and E can’t draw; without loss of generality, say C beats E. And then in {A, C, E, F}, player C must lose to F. But then in {C, D, E, F} no player scores differently against the other three players—a contradiction.
Now suppose A were adjacent to B, C, D, E, and without loss of generality assume B | C; then ABC is a triangle.
For each J besides A, B, C, look at the group {A, B, C, J}: J must draw with one of A, B, C. By the Pigeonhole Principle, one of A, B, C draws with at least three of the J and thus has degree at least 5. But this is impossible from above.
It follows that it is impossible for n to be at least 10. But since n can be 6, the maximum n is between 6 and 9, as desired.
1999 National Contests: Problems and Solutions 27 1.2 Brazil Problem 1 Let ABCDE be a regular pentagon such that the star region ACEBD has area 1. Let AC and BE meet at P, and let BD and CE meet at Q. Determine [APQD].
Solution: Let R = AD ∩BE, S = AC ∩BD, T = CE ∩AD. Now △PQR ∼△CAD because they are corresponding triangles in regular pentagons QTRPS and ABCDE, and since △CAD ∼△PAR as well we have △PQR ∼ = △PAR.
Thus, [APQD] = [AP QD] [ACEBD] = 2[AP R]+[P QR]+[RQT ] 5[AP R]+[P QR]+2[RQT ] = 3[AP R]+[RQT ] 6[AP R]+2[RQT ] = 1 2.
Problem 2 Given a 10 × 10 board, we want to remove n of the 100 squares so that no 4 of the remaining squares form the corners of a rectangle with sides parallel to the sides of the board. Determine the minimum value of n.
Solution: The answer is 66. Consider the diagram below, in which a colored circle represents a square that has not been removed. The diagram demonstrates that n can be 66: • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • Now we proceed to show that n is at least 66.
Suppose, for contradiction, that it is possible with n = 65.
Denote by ai the number of squares left in row i (i = 1, 2, . . . , 10); in row i, there are ai 2 pairs of remaining squares. If no four remaining squares form the corners of a rectangle, then the total number N = P10 i=1 ai 2 must not exceed 10 2 = 45. But note that, with a fixed P10 i=1 ai = 35, the minimum of P10 i=1 ai 2 is attained when and only when no two ai’s differ by more than 1. Thus, 45 = P10 i=1 ai 2 ≥5 · 4 2 + 5 · 3 2 = 45, 28 Brazil i.e., this minimum is attained here, implying that five of the ai’s equal 4 and the rest equal 3. Then it is easy to see that aside from permutations of the row and columns, the first five rows of the board must be as follows: • • • • • • • • • • • • • • • • • • • • We inspect this figure and notice that it is now impossible for another row to contain at least 3 remaining squares without forming the vertices of a rectangle with sides parallel to the sides of the board.
This is a contradiction, since each of the remaining 5 rows is supposed to have 3 remaining squares. Thus, it is impossible for n to be less than 66, and we are done.
Problem 3 The planet Zork is spherical and has several cities.
Given any city A on Zork, there exists an antipodal city A′ (i.e., symmetric with respect to the center of the sphere). In Zork, there are roads joining pairs of cities. If there is a road joining cities P and Q, then there is a road joining P ′ and Q′. Roads don’t cross each other, and any given pair of cities is connected by some sequence of roads. Each city is assigned a value, and the difference between the values of every pair of connected cities is at most 100. Prove that there exist two antipodal cities with values differing by at most 100.
Solution: Let [A] denote the value assigned to city A. Name the pairs of cities (Z1, Z′ 1), (Z2, Z′ 2), (Z3, Z′ 3), . . . , (Zn, Z′ n) with 0 ≤[Zi] −[Z′ i] for all i.
Since any given pair of cities is connected by some sequence of roads, there must exist a, b such that Za and Z′ b are connected by a single road. Then Z′ a and Zb are also connected by a single road. Thus, [Za] −[Z′ b] ≤100 and [Zb] −[Z′ a] ≤100. Adding, we have [Za] −[Z′ a] + [Zb] −[Z′ b] ≤200.
1999 National Contests: Problems and Solutions 29 Hence, either 0 ≤[Za] −[Z′ a] ≤100 or 0 ≤[Zb] −[Z′ b] ≤100; in either case, we are done.
Problem 4 In Tumbolia there are n soccer teams.
We want to organize a championship such that each team plays exactly once with each other team.
All games take place on Sundays, and a team can’t play more than one game in the same day.
Determine the smallest positive integer m for which it is possible to realize such a championship in m Sundays.
Solution: Let an be the smallest positive integer for which it is possible to realize a championship between n soccer teams in an Sundays. For n > 1, it is necessary that an ≥2⌈n 2 ⌉−1; otherwise the total number of games played would not exceed (2⌈n 2 ⌉−2) · ⌊n 2 ⌋≤ (n−1)2 2 < n 2 , a contradiction.
On the other hand, 2⌈n 2 ⌉−1 days suffice. Suppose that n = 2t + 1 or 2t + 2; number the teams from 1 to n and the Sundays from 1 to 2t + 1. On the i-th Sunday, let team i either sit out (if n is odd) or play team 2t + 2 (if n is even); and have any other team j play with the team k ̸= 2t + 2 such that j + k ≡2i (mod 2t + 1). Then each team indeed plays every other team, as desired.
Problem 5 Given a triangle ABC, show how to construct, with straightedge and compass, a triangle A′B′C′ with minimal area such that A′, B′, C′ lie on AB, BC, CA, respectively, ∠B′A′C′ = ∠BAC, and ∠A′C′B′ = ∠ACB.
Solution: All angles are directed modulo 180◦.
For convenience, call any triangle A′B′C′ “zart” if A′, B′, C′ lie on lines AB, BC, CA, respectively, and △ABC ∼△A′B′C′. The problem is, then, to construct the zart triangle with minimal area.
Suppose we have any zart triangle, and let P be the point (different from A′) where the circumcircles of triangles AA′C′ and BB′A′ meet.
Then ∠B′PC′ = 360◦−∠A′PB′ −∠C′PA′ = 360◦−(180◦−∠CBA) −(180◦−∠BAC) = 180◦−∠ACB, so P also lies on the circumcircle of triangle CC′B′.
30 Brazil Next, ∠PAB = ∠PC′A′ = ∠B′C′A′ −∠B′C′P = ∠B′CC′ −∠B′CP ′ = ∠PCA, and with similar reasoning we have ∠PAB = ∠PC′A′ = ∠PCA = ∠PB′C′ = ∠PBC.
There is a unique point P (one of the Brocard points) satisfying ∠PAB = ∠PBC = ∠PCA, and thus P is fixed—independent of the choice of triangle A′B′C′. And since it is the corresponding point in similar triangles ABC and A′B′C′, we have [A′B′C′] = [ABC] PA′ PA 2 .
Thus [A′B′C′] is minimal when PA′ is minimal, which occurs when PA′ ⊥AB (and analogously, when PB′ ⊥PC and PC′ ⊥PA).
Thus, the zart triangle with minimal area is the pedal triangle A′B′C′ of P to triangle ABC. This triangle is indeed similar to triangle ABC; letting θ = ∠PAB be the Brocard angle, it is the image of triangle ABC under a rotation through θ −90◦, followed by a homothety of ratio —sin θ|.
To construct this triangle, first draw the circles {X : ∠BXA = ∠BCA + ∠CAB} and {Y : ∠CY B = ∠CAB + ∠ABC} and let P ′ be their point of intersection (different from B); then we also have ∠AP ′C = ∠ABC + ∠BCA. Then ∠P ′AB = 180◦−∠ABP ′ −∠BP ′A = 180◦−(∠ABC −∠P ′BC) −(∠BCA + ∠CAB) = ∠P ′BC, and similarly ∠P ′BC = ∠P ′CA. Therefore P = P ′. Finally, drop the perpendiculars from P ′ to the sides of triangle ABC to form A′, B′, C′. This completes the construction.
1999 National Contests: Problems and Solutions 31 1.3 Bulgaria National Olympiad, Third Round Problem 1 Find all triples (x, y, z) of natural numbers such that y is a prime number, y and 3 do not divide z, and x3 −y3 = z2.
Solution: Rewrite the equation in the form (x −y)(x2 + xy + y2) = z2.
Any common divisor of x −y and x2 + xy + y2 also divides both z2 and (x2 + xy + y2) −(x + 2y)(x −y) = 3y2. But z2 and 3y2 are relatively prime by assumption, hence (x−y) and (x2 +xy+y2) must be relatively prime as well. Therefore, both (x−y) and (x2 +xy +y2) are perfect squares.
Now writing a = √x −y, we have x2 + xy + y2 = (a2 + y)2 + (a2 + y)y + y2 = a4 + 3a2y + 3y2 and 4(x2 + xy + y2) = (2a2 + 3y)2 + 3y2.
Writing m = 2 p x2 + xy + y2 and n = 2a2 + 3y, we have m2 = n2 + 3y2 or (m −n)(m + n) = 3y2, so (m −n, m + n) = (1, 3y2), (3, y2), or (y, 3y).
In the first case, 2n = 3y2 −1 and 4a2 = 2n −6y = 3y2 −6y −1 is a square, which is impossible modulo 3.
In the third case, n = y < 2a2 + 3y = n, a contradiction.
In the second case, we have 4a2 = 2n−6y = y2 −6y −3 < (y −3)2.
And when y ≥10 we have y2 −6y −3 > (y −4)2, hence y = 2, 3, 5, or 7. In this case we have a = √ y2−6y−3 2 , which is real only when y = 7, a = 1, x = y + a2 = 8, and z = 13. This yields the unique solution (x, y, z) = (8, 7, 13).
Problem 2 A convex quadrilateral ABCD is inscribed in a circle whose center O is inside the quadrilateral.
Let MNPQ be the quadrilateral whose vertices are the projections of the intersection 32 Bulgaria point of the diagonals AC and BD onto the sides of ABCD. Prove that 2[MNPQ] ≤[ABCD].
Solution: The result actually holds even when ABCD is not cyclic.
We begin by proving the following result: Lemma. If XW is an altitude of triangle XY Z, then XW Y Z ≤ 1 2 tan ∠Y +∠Z 2 .
Proof: X lies on an arc of a circle determined by ∠Y XZ = 180◦−∠Y −∠Z. Its distance from Y Z is maximized when it is at the center of this arc, which occurs when ∠Y = ∠Z; and at this point, XW Y Z = 1 2 tan ∠Y +∠Z 2 .
Suppose M, N, P, Q are on sides AB, BC, CD, DA, respectively.
Also let T be the intersection of AC and BD.
Let α = ∠ADB, β = ∠BAC, γ = ∠CAD, δ = ∠DBA. From the lemma, MT ≤ 1 2AB · tan β+δ 2 and QT ≤ 1 2AD · tan α+γ 2 ; also, ∠MTQ = 180◦−∠QAM = 180◦−∠DAB. Thus 2[MTQ] = MT · QT sin ∠MTQ ≤ 1 4 tan α+γ 2 tan β+δ 2 AB · AD sin ∠DAB.
But since α+γ 2 + β+δ 2 = 90◦, this last expression exactly equals 1 4AB · AD sin ∠DAB = 1 2[ABD]. Thus, 2[MTQ] ≤1 2[ABD].
Likewise, 2[NTM] ≤1 2[BCA], [PTN] ≤1 2[CDB], and [QTP] ≤ 1 2[DAC]. Adding these four inequalities shows that 2[MNPQ] is at most 1 2([ABD] + [CDB]) + 1 2([BCA] + [DAC]) = [ABCD], as desired.
Problem 3 In a competition 8 judges marked the contestants by pass or fail. It is known that for any two contestants, two judges marked both with pass; two judges marked the first contestant with pass and the second contestant with fail; two judges marked the first contestant with fail and the second contestant with pass; and finally, two judges marked both with fail. What is the largest possible number of contestants?
Solution: For a rating r (either pass or fail), let r denote the opposite rating. Also, whenever a pair of judges agree on the rating for some contestant, call this an “agreement.” We first prove that 1999 National Contests: Problems and Solutions 33 any two judges share at most three agreements; suppose by way of contradiction this were false.
Then assume without loss of generality that the judges (labeled with numbers) mark the first four contestants (labeled with letters) as follows in the left table: A B C D 1 a b c d 2 a b c d 3 a b 4 a b 5 a b 6 a b 7 a b 8 a b A B C D 1 a b c d 2 a b c d 3 a b c d 4 a b c d 5 a b 6 a b 7 a b 8 a b A B C D 1 a b c d 2 a b c d 3 a b c d 4 a b c d 5 a b c d 6 a b c d 7 a b 8 a b Applying the given condition to contestants A and C, judges 3 and 4 must both give C the rating c; similarly, they must both give D the rating d. Next, applying the condition to contestants B and C, judges 5 and 6 must both give C the rating c; similarly, they must both give D the rating d. But now the condition fails for contestants C and D, a contradiction.
Thus each pair of judges agrees on at most three ratings, as claimed; thus there are at most 3· 8 2 = 84 agreements between all the judges.
On the other hand, for each contestant exactly four judges mark him with pass and exactly four judges mark him with fail, hence there are 4 2 + 4 2 = 12 agreements per contestant. It follows that there are at most 84 12 = 7 contestants; and as the following table shows (with 1 representing pass and 0 representing fail), it is indeed possible to have exactly 7 contestants: A B C D E F G 1 1 1 1 1 1 1 1 2 1 1 1 0 0 0 0 3 1 0 0 1 1 0 0 4 1 0 0 0 0 1 1 5 0 1 0 1 0 0 1 6 0 1 0 0 1 1 0 7 0 0 1 1 0 1 0 8 0 0 1 0 1 0 1 34 Bulgaria Problem 4 Find all pairs (x, y) of integers such that x3 = y3 + 2y2 + 1.
Solution: When y2 + 3y > 0, (y + 1)3 > x3 > y3. Thus we must have y2 + 3y ≤0, and y = −3, −2, −1, or 0 — yielding the solutions (x, y) = (1, 0), (1, −2), and (−2, −3).
Problem 5 Let B1 and C1 be points on the sides AC and AB of triangle ABC. Lines BB1 and CC1 intersect at point D. Prove that a circle can be inscribed inside quadrilateral AB1DC1 if and only if the incircles of the triangles ABD and ACD are tangent to each other.
Solution: Say the incircle of triangle ABD is tangent to AD at T1 and that the incircle of triangle ACD is tangent to AD at T2; then DT1 = 1 2(DA + DB −AB) and DT2 = 1 2(DA + DC −AC).
First suppose a circle can be inscribed inside AB1DC1.
Let it be tangent to sides AB1, B1D, DC1, C1A at points E, F, G, H, respectively. Using equal tangents, we have AB −BD = (AH + HB) −(BF −DF) = (AH + BF) −(BF −DF) = AH + DF and similarly AC −CD = AE + DG. But AH + DF = AE + DG by equal tangents, implying that AB −BD = AC −CD and thus DA + DB −AB = DA + DC −AC. Therefore DT1 = DT2, T1 = T2, and the two given incircles are tangent to each other.
Next suppose the two incircles are tangent to each other. Then DA + DB −AB = DA + DC −AC.
Let ω be the incircle of ABB1, and let D′ be the point on BB1 (different from B1) such that line CD′ is tangent to ω.
Suppose by way of contradiction that D ̸= D′. From the result in the last paragraph, we know that the incircles of triangles ABD′ and ACD′ are tangent and hence D′A+D′B−AB = D′A+D′C−AC. Then since DB−AB = DC−AC and D′B −AB = D′C −AC, we must have DB −D′B = DC −D′C by subtraction. Thus DD′ = |DB −D′B| = |DC −D′C|. But then the triangle inequality fails in triangle DD′C, a contradiction. This completes the proof.
1999 National Contests: Problems and Solutions 35 Problem 6 Each interior point of an equilateral triangle of side 1 lies in one of six congruent circles of radius r. Prove that r ≥ √ 3 10 .
Solution: From the condition, we also know that every point inside or on the triangle lies inside or on one of the six circles.
Define R = 1 1+ √ 3. Orient the triangle so that A is at the top, B is at the bottom-left, and C is at the bottom-right (so that BC is horizontal). Draw point W on AB such that WA = R; then draw point X directly below W such that WX = R. Then in triangle WXB, WB = 1 −R = √ 3R and ∠BWX = 30◦, implying that XB = R as well. Similarly draw Y on AC such that Y A = R, and Z directly below Y such that Y Z = ZC = R.
In triangle AWY, ∠A = 60◦and AW = AY = R, implying that WY = R. This in turn implies that XZ = R and that WX = Y Z = R √ 2.
Now suppose by way of contradiction that we could cover the triangle with six congruent circles of radius r < √ 3 10 .
The points A, B, C, W, X, Y, Z each lie on or inside one of the circles. But any two of these points are at least R > 2r apart, so they must lie on or inside different circles.
Thus there are at least seven circles, a contradiction.
National Olympiad, Fourth Round Problem 1 A rectangular parallelepiped has integer dimensions.
All of its faces of are painted green. The parallelepiped is partitioned into unit cubes by planes parallel to its faces.
Find all possible measurements of the parallelepiped if the number of cubes without a green face is one third of the total number of cubes.
Solution: Let the parallelepiped’s dimensions be a, b, c; they must all be at least 3 or else every cube has a green face. Then the condition is equivalent to 3(a −2)(b −2)(c −2) = abc, or 3 = a a −2 · b b −2 · c c −2.
36 Bulgaria If all the dimensions are at least 7, then a a−2 · b b−2 · c c−2 ≤ 7 5 3 = 343 125 < 3, a contradiction. Thus one of the dimensions — say, a — equals 3, 4, 5, or 6. Assume without loss of generality that b ≤c.
When a = 3 we have bc = (b −2)(c −2), which is impossible.
When a = 4, rearranging the equation yields (b −6)(c −6) = 24.
Thus (b, c) = (7, 30), (8, 18), (9, 14), or (10, 12).
When a = 5, rearranging the equation yields (2b −9)(2c −9) = 45.
Thus (b, c) = (5, 27), (6, 12), or (7, 9).
And when a = 6, rearranging the equation yields (b−4)(c−4) = 8.
Thus (b, c) = (5, 12) or (6, 8).
Therefore the parallelepiped may measure 4 × 7 × 30, 4 × 8 × 18, 4 × 9 × 14, 4 × 10 × 12, 5 × 5 × 27, 5 × 6 × 12, 5 × 7 × 9, or 6 × 6 × 8.
Problem 2 Let {an} be a sequence of integers such that for n ≥1 (n −1)an+1 = (n + 1)an −2(n −1).
If 2000 divides a1999, find the smallest n ≥2 such that 2000 divides an.
Solution: First, we note that the sequence an = 2n −2 works.
Then writing bn = an −(2n −2) gives the recursion (n −1)bn+1 = (n + 1)bn.
Some calculations show that b3 = 3b2, b4 = 6b2, b5 = 10b2 — and in general, that bn = n(n−1) 2 b2 for n ≥2. Thus when n ≥2, the solution to the original equation is of the form an = 2(n −1) + n(n −1) 2 c for some constant c; plugging in n = 2 shows that c = a2 −2 is an integer.
Now, since 2000 | a1999 we have 2(1999 −1) + 1999·1998 2 · c ≡0 = ⇒ −4 + 1001c ≡0 = ⇒c ≡4 (mod 2000). Then 2000 | an exactly when 2(n −1) + 2n(n −1) ≡0 (mod 2000) ⇐ ⇒(n −1)(n + 1) ≡0 (mod 1000).
(n−1)(n+1) is divisible by 8 exactly when n is odd; and it is divisible by 125 exactly when either n −1 or n + 1 is divisible by 125. The smallest n ≥2 satisfying these requirements is n = 249.
1999 National Contests: Problems and Solutions 37 Problem 3 The vertices of a triangle have integer coordinates and one of its sides is of length √n, where n is a square-free natural number. Prove that the ratio of the circumradius to the inradius of the triangle is an irrational number.
Solution: Label the triangle ABC; let r, R, K be the inradius, circumradius, and area of the triangle; let a = BC, b = CA, c = AB and write a = p1√q1, b = p2√q2, c = p3√q3 for positive integers pi, qi with qi square-free. By Pick’s Theorem (K = I + 1 2B −1), K is rational. Also, R = abc 4K and r = 2K a+b+c. Thus R r = abc(a+b+c) 8K2 is rational if and only if abc(a + b + c) = a2bc + ab2c + abc2 is rational.
Let this quantity equal m, and assume by way of contradiction that m is rational.
We have a2bc = m1√q2q3, ab2c = m2√q3q1, and abc2 = m3√q1q2 for positive integers m1, m2, m3. Then m1√q2q3 + m2√q3q1 = m − m3√q1q2. Squaring both sides, we find that m2 1q2q3 + m2 2q3q1 + 2m1m2q3 √q1q2 = m2 + m2 3 −2mm3 √q1q2.
If √q1q2 is not rational, then the coefficients of √q1q2 must be the same on both sides; but this is impossible since 2m1m2q3 is positive while −2mm3 is not.
Hence √q1q2 is rational. Since q1 and q2 are square-free, this can only be true if q1 = q2. Similarly, q2 = q3.
Assume without loss of generality that BC = √n so that q1 = q2 = q3 = n and p1 = 1. Also assume that A is at (0, 0), B is at (w, x), and C is at (y, z). By the triangle inequality, we must have p2 = p3 and hence w2 + x2 = y2 + z2 = p2 2n (w −y)2 + (x −z)2 = n.
Notice that n = (w −y)2 + (x −z)2 ≡w2 + x2 + y2 + z2 = 2p2 2n ≡0 (mod 2), so n is even. Thus w and x have the same parity; and y and z have the same parity. Then w, x, y, z must all have the same parity since w2 + x2 ≡y2 + z2 (mod 4). But then n = (w −y)2 + (x −z)2 ≡ 0 (mod 4), contradicting the assumption that n is square-free.
38 Bulgaria Therefore our original assumption was false; and the ratio of the circumradius to the inradius is indeed always irrational.
Note: Without the condition that n is square-free, the ratio can be rational. For example, the points (i, 2j −i) form a grid of points √ 2 apart. In this grid, we can find a 3 √ 2-4 √ 2-5 √ 2 right triangle by choosing, say, the points (0, 0), (3, 3), and (7, −1). Then q1 = q2 = q3, and the ratio is indeed rational.
Problem 4 Find the number of all natural numbers n, 4 ≤n ≤ 1023, whose binary representations do not contain three consecutive equal digits.
Solution: A binary string is a finite string of digits, all either 0 or 1.
Call such a string (perhaps starting with zeroes) valid if it does not contain three consecutive equal digits. Let an represent the number of valid n-digit strings; let sn be the number of valid strings starting with two equal digits; and let dn be the number of valid strings starting with two different digits. Observe that an = sn + dn for all n.
An (n+2)-digit string starting with 00 is valid if and only if its last n digits form a valid string starting with 1; similarly, an (n + 2)-digit string starting with 11 is valid if and only if its last n digits form a valid string starting with 0. Thus, sn+2 = an = sn + dn.
An (n+2)-digit string starting with 01 is valid if and only if its last n digits form a valid string starting with 00, 01, or 10; similarly, an (n+2)-digit string starting with 10 is valid if and only if its last n digits form a valid string starting with 11, 01, or 10. Thus, dn+2 = sn+2dn.
Solving these recursions gives sn+4 = 3sn+2 −sn and dn+4 = 3dn+2 −dn, which when added together yield an+4 = 3an+2 −an.
Thus we can calculate initial values of an and then use the recursion to find other values: n 1 2 3 4 5 6 7 8 9 10 an 2 4 6 10 16 26 42 68 110 178 1999 National Contests: Problems and Solutions 39 Now of the an valid n-digit strings, only half start with 1; thus only half are binary representations of positive numbers. Therefore exactly 1 2(a1 + a2 + · · · + a10) = 231 numbers between 1 and 1023 have the desired property; and ignoring 1, 2, and 3, we find that the answer is 228.
Problem 5 The vertices A, B and C of an acute-angled triangle ABC lie on the sides B1C1, C1A1 and A1B1 of triangle A1B1C1 such that ∠ABC = ∠A1B1C1, ∠BCA = ∠B1C1A1, and ∠CAB = ∠C1A1B1.
Prove that the orthocenters of the triangle ABC and triangle A1B1C1 are equidistant from the circumcenter of triangle ABC.
Solution: Let H and H1 be the orthocenters of triangles ABC and A1B1C1, respectively; and let O, OA, OB, OC be the circumcenters of triangles ABC, A1BC, AB1C, and ABC1, respectively.
First note that ∠BA1C = ∠C1A1B1 = ∠CAB = 180◦−∠CHB, showing that BA1CH is cyclic; moreover, OAA1 = BC 2 sin ∠BA1C = CB 2 sin ∠CAB = OA so circles ABC and BA1CH have the same radius.
Similarly, CB1AH and AC1BH are cyclic with circumradius OA.
Then ∠HBC1 = 180◦−∠C1AH = ∠HAB1 = 180◦−∠B1CH = ∠HCA1; thus angles ∠HOCC1, ∠HOAA1, ∠HOBB1 are equal as well.
Let ∠(⃗ r1,⃗ r2) denote the angle between rays ⃗ r1 and ⃗ r2. Since OAC = OAB = HB = HC, quadrilateral BOACH is a rhombus and hence a parallelogram. Then ∠(− → OA, − − − → HOA) = ∠(− → OA, − − → OB) + ∠(− − → OB, − − − → HOA) = 2∠ACB + ∠(− − − → COA, − − − → HOA) = 2∠ACB + ∠COAH = 2∠ACB + 2∠CBH = 2∠ACB + 2(90◦−∠ACB) = 180◦.
Similarly, ∠(− − → OB, − − − → HOB) = ∠(− − → OC, − − − → HOC) = 180◦. Combining this result with ∠HOAA1 = ∠HOBB1 = ∠HOCC1 from above, we find 40 Bulgaria that ∠(− → OA, − − − → OAA1) = ∠(− − → OB, − − − → OBB1) = ∠(− − → OC, − − − → OCC1).
Let this common angle be θ.
We now use complex numbers with the origin at O, letting p denote the complex number representing point P. Since HBOAC is a parallelogram we have oA = b + c and we can write a1 = b + c + xa where x = cis θ. We also have b1 = c + a + xb and c1 = a + b + xc for the same x. We can rewrite these relations as a1 = a + b + c + (x −1)a, b1 = a + b + c + (x −1)b, c1 = a + b + c + (x −1)c.
Thus the map sending z to a + b + c + (x −1)z = h + (x −1)z is a spiral similarity taking triangle ABC into triangle A′B′C′. It follows that this map also takes H to H1, so h1 = h + (x −1)h = xh and OH1 = |h1| = |x||h| = |h| = OH, as desired.
Problem 6 Prove that the equation x3 + y3 + z3 + t3 = 1999 has infinitely many integral solutions.
Solution: Observe that (m −n)3 + (m + n)3 = 2m3 + 6mn2. Now suppose we want a general solution of the form (x, y, z, t) = (a −b, a + b, c 2 −d 2, c 2 + d 2) for integers a, b and odd integers c, d. One simple solution to the given equation is (x, y, z, t) = (10, 10, −1, 0), so try setting a = 10 and c = −1. Then (x, y, z, t) = (10 −b, 10 + b, −1 2 −d 2, −1 2 + d 2) is a solution exactly when (2000 + 60b2) −1 + 3d2 4 = 1999 ⇐ ⇒ d2 −80b2 = 1.
1999 National Contests: Problems and Solutions 41 The second equation is a Pell’s equation with solution (d1, b1) = (9, 1); and we can generate infinitely many more solutions by setting (dn+1, bn+1) = (9dn + 80bn, 9bn + dn) for n = 1, 2, 3, . . .; this follows from a general recursion (pn+1, qn+1) = (p1pn + q1qnD, p1qn + q1pn) for generating solutions to p2 −Dq2 = 1 given a nontrivial solution (p1, q1).
A quick check also shows that each dn is odd. Thus since there are infinitely many solutions (bn, dn) to the Pell’s equation (and with each dn odd), there are infinitely many integral solutions (xn, yn, zn, tn) = (10 −bn, 10 + bn, −1 2 −dn 2 , −1 2 + dn 2 ) to the original equation.
42 Canada 1.4 Canada Problem 1 Find all real solutions to the equation 4x2−40⌊x⌋+51 = 0, where [x] denotes the greatest integer less than or equal to x.
Solution: Note that (2x −3)(2x −17) = 4x2 −40x + 51 ≤ 4x2 −40⌊x⌋+ 51 = 0, which gives 1.5 ≤x ≤8.5 and 1 ≤⌊x⌋≤8.
Then x = p 40⌊x⌋−51 2 , so it is necessary that ⌊x⌋= $p 40⌊x⌋−51 2 % .
Testing ⌊x⌋∈{1, 2, 3, . . . , 8} into this equation, we find that only ⌊x⌋= 2, 6, 7, and 8 work.
Thus the only solutions for x are √ 29 2 , √ 189 2 , √ 229 2 , √ 269 2 .
Problem 2 Let ABC be an equilateral triangle of altitude 1. A circle, with radius 1 and center on the same side of AB as C, rolls along the segment AB; as it rolls, it always intersects both AC and BC. Prove that the length of the arc of the circle that is inside the triangle remains constant.
Solution: Let ω be “the circle.” Let O be the center of ω. Let ω intersect segments AC and BC at M and N, respectively. Let the circle through O, C, and M intersect BC again at P. Now ∠PMO = 180◦−∠OCP = 60◦= ∠MCO = ∠MPO, so OP = OM = 1, and P coincides with N. Thus, ∠MON = ∠MOP = ∠MCP = 60◦.
Therefore, the angle of the arc of ω that is inside the triangle ABC is constant, and hence the length of the arc must be constant as well.
Problem 3 Determine all positive integers n such that n = d(n)2, where d(n) denotes the number of positive divisors of n (including 1 and n).
Solution: Label the prime numbers p1 = 2, p2 = 3, . . .. Since n is a perfect square, we have n = ∞ Y i=1 p2ai i , d(n) = ∞ Y i=1 (2ai + 1).
1999 National Contests: Problems and Solutions 43 Then d(n) is odd and so is n, whence a1 = 0. Since d(n) √n = 1, we have ∞ Y i=1 2ai + 1 pai i = 1.
By Bernoulli’s inequality, we have pai i ≥(pi −1)ai + 1 > 2ai + 1 for all primes pi ≥5 that divide n. Also, 3a2 ≥2a2 + 1 with equality only when a2 ∈{0, 1}. Thus, for equality to hold above, we must have a1 = a3 = a4 = a5 = · · · = 0 and a2 ∈{0, 1}; therefore, n ∈{1, 9} are the only solutions.
Problem 4 Suppose a1, a2, . . . , a8 are eight distinct integers from the set S = {1, 2, . . . , 17}. Show that there exists an integer k > 0 such that the equation ai−aj = k has at least three different solutions.
Also, find a specific set of 7 distinct integers {b1, b2, · · · , b7} from S such that the equation bi −bj = k does not have three distinct solutions for any k > 0.
Solution: For the first part of this problem, assume without loss of generality that a1 < a2 < · · · < a8; also assume, for the purpose of contradiction, that there does not exist an integer k > 0 such that the equation ai −aj = k has at least three different solutions. Let δi = ai+1 −ai for i = 1, 2, . . . , 7. Then 16 ≥a8 −a1 = δ1 + . . . + δ7 ≥1 + 1 + 2 + 2 + 3 + 3 + 4 = 16 (for otherwise three of the δi’s would be equal, a contradiction). Since equality must hold, Π = (δ1, δ2, . . . , δ7) must be a permutation of (1, 1, 2, 2, 3, 3, 4).
Say we have a “m-n pair” if some (δi, δi+1) = (m, n) or (n, m). Note that we cannot have any 1-1 or 1-2 pairs (δi, δi+1); otherwise we’d have ai+2 −ai = 2 or 3, giving at least three solutions to ai −aj = 2 or 3.
Nor can we have two 1-3 pairs because then, along with δi = 4, we’d have three solutions to ai −aj = 4. Then considering what entries each 1 is next to, we see that we must have Π = (1, 4, . . . , 3, 1) or (1, 4, 1, 3, . . .) (or these lists backwards).
But now we can’t have any 2-2 pairs; otherwise, along with the 1-3 pair and the δi = 4, we’d have three solutions to ai −aj = 4. Thus 44 Canada we have either Π = (1, 4, 2, 3, 2, 3, 1) or (1, 4, 1, 3, 2, 3, 2) (or these lists backwards).
In either case there are at least four solutions to ai −aj = 5, a contradiction.
Thus, regardless of the {a1, a2, . . . , a8} that we choose, for some integer k ∈{2, 3, 4, 5} the equation ai −aj = k has at least three different solutions.
For the second part of the problem, let (b1, b2, . . . , b7) = (1, 2, 4, 9, 14, 16, 17). Each of 1, 2, 3, 5, 7, 8, 12, 13, and 15 is the difference of exactly two pairs of the bi, and each of 10, 14, and 16 is the difference of exactly one pair of the bi. But no number is the difference of more than two such pairs, and hence the set {b1, b2, . . . , b7} suffices.
Problem 5 Let x, y, z be non-negative real numbers such that x + y + z = 1.
Prove that x2y + y2z + z2x ≤4 27.
and determine when equality occurs.
Solution: Assume without loss of generality that x = max{x, y, z}.
• If x ≥y ≥z, then x2y + y2z + z2x ≤x2y + y2z + z2x + z(xy + (x −y)(y −z)) = (x + z)2y = 4 1 2 −1 2y 1 2 −1 2y y ≤4 27, where the last inequality follows from AM-GM. Equality occurs if and only if z = 0 (from the first inequality) and y = 1 3, in which case (x, y, z) = 2 3, 1 3, 0 .
• If x ≥z ≥y, then x2y + y2z + z2x = x2z + z2y + y2x −(x −z)(z −y)(x −y) ≤x2z + z2y + y2x ≤4 27, where the second inequality is true from the result we proved for x ≥y ≥z (except with y and z reversed). Equality holds in the first inequality only when two of x, y, z are equal; and in 1999 National Contests: Problems and Solutions 45 the second only when (x, z, y) = 2 3, 1 3, 0 . Since these conditions can’t both be true, the inequality is actually strict in this case.
Therefore the inequality is indeed true, and equality holds when (x, y, z) equals 2 3, 1 3, 0 , 1 3, 0, 2 3 , or 0, 2 3, 1 3 .
46 China 1.5 China Problem 1 Let ABC be an acute triangle with ∠C > ∠B. Let D be a point on side BC such that ∠ADB is obtuse, and let H be the orthocenter of triangle ABD. Suppose that F is a point inside triangle ABC and is on the circumcircle of triangle ABD.
Prove that F is the orthocenter of triangle ABC if and only if both of the following are true: HD ∥CF, and H is on the circumcircle of triangle ABC.
Solution: All angles are directed modulo 180◦. First observe that if P is the orthocenter of triangle UV W, then ∠V PW = (90◦−∠PWV ) + (90◦−∠WV P) = ∠WV U + ∠UWV = 180◦−∠V UW.
First suppose that F is the orthocenter of triangle ABC. Then ∠ACB = 180◦−∠AFB = 180◦−∠ADB = ∠AHB, so ACHB is cyclic. And lines CF and HD are both perpendicular to side AB, so they are parallel.
Conversely, suppose that HD ∥CF and that H is on the circum-circle of triangle ABC. Since AFDB and AHCB are cyclic, ∠AFB = ∠ADB = 180◦−∠AHB = 180◦−∠ACB.
Thus F is an intersection point of the circle defined by ∠AFB = 180◦−∠ACB and the line defined by CF ⊥AB. But there are only two such points: the orthocenter of triangle ABC and the reflection of C across line AB. The latter point lies outside of triangle ABC, and hence F must indeed be the orthocenter of triangle ABC.
Problem 2 Let a be a real number. Let {fn(x)} be a sequence of polynomials such that f0(x) = 1 and fn+1(x) = xfn(x) + fn(ax) for n = 0, 1, 2, . . . .
(a) Prove that fn(x) = xnfn 1 x for n = 0, 1, 2, . . . .
(b) Find an explicit expression for fn(x).
1999 National Contests: Problems and Solutions 47 Solution: When a = 1, we have fn(x) = (x+1)n for all n, and part (a) is easily checked. Now assume that a ̸= 1.
Observe that fn has degree n and always has constant term 1. Write fn(x) = c0 + c1x + · · · + cnxn; we prove by induction on n that (ai −1)ci = (an+1−i −1)ci−1 for 0 ≤i ≤n (where we let c−1 = 0).
The base case n = 0 is clear.
Now suppose that fn−1(x) = b0 + b1x + · · · + bn−1xn−1 satisfies the claim: specifically, we know (ai −1)bi = (an−i −1)bi−1 and (an+1−i −1)bi−2 = (ai−1 −1)bi−1 for i ≥1.
For i = 0, the claim states 0 = 0. For i ≥1, the given recursion gives ci = bi−1 + aibi and ci−1 = bi−2 + ai−1bi−1. Then the claim is equivalent to (ai −1)ci = (an+1−i −1)ci−1 ⇐ ⇒(ai −1)(bi−1 + aibi) = (an+1−i −1)(bi−2 + ai−1bi−1) ⇐ ⇒(ai −1)bi−1 + ai(ai −1)bi = (an+1−i −1)bi−2 + (an −ai−1)bi−1 ⇐ ⇒(ai −1)bi−1 + ai(an−i −1)bi−1 = (ai−1 −1)bi−1 + (an −ai−1)bi−1 ⇐ ⇒(an −1)bi−1 = (an −1)bi−1, so it is true.
Now by telescoping products, we have ci = ci c0 = i Y k=1 ck ck−1 = i Y k=1 an+1−k −1 ak −1 = Qn k=n+1−i(ak −1) Qi k=1(ak −1) = Qn k=i+1(ak −1) Qn−i k=1(ak −1) = n−i Y k=1 an+1−k −1 ak −1 = n−i Y k=1 ck ck−1 = cn−i c0 = cn−i, 48 China giving our explicit form.
Also, fn(x) = xnfn 1 x if and only if ci = cn−i for i = 0, 1, . . . , n, and from above this is indeed the case.
This completes the proof.
Problem 3 There are 99 space stations. Each pair of space stations is connected by a tunnel. There are 99 two-way main tunnels, and all the other tunnels are strictly one-way tunnels. A group of 4 space stations is called connected if one can reach each station in the group from every other station in the group without using any tunnels other than the 6 tunnels which connect them. Determine the maximum number of connected groups.
Solution: In this solution, let f(x) = x(x−1)(x−2) 6 , an extension of the definition of x 3 to all real numbers x.
In a group of 4 space stations, call a station troublesome if three one-way tunnels lead toward it or three one-way tunnels lead out of it.
In each group there is at most one troublesome station of each type for a count of at most two troublesome stations. Also, if a station is troublesome in a group, that group is not connected.
Label the stations 1, 2, . . . , 99. For i = 1, 2, . . . , 99, let ai one-way tunnels point into station i and bi one-way tunnels point out.
Station i is troublesome in ai 3 + bi 3 groups of four. Adding over all stations, we obtain a total count of P198 i=1 ai 3 + bi 3 . This equals P198 i=1 f(xi) for nonnegative integers x1, x2, . . . , x198 with P198 i=1 xi = 96·99. Without loss of generality, say that x1, x2, . . . , xk are at least 1 and xk+1, xk+2, . . . , x198 are zero. Since f(x) is convex as a function of x for x ≥1, this is at least k 96·99/k 2 . Also, mf(x) ≥f(mx) when m ≤1 and mx ≥2. Letting m = k/198 and mx = 96·99/198 = 48, we find that our total count is at least 198 48 2 . Since each unconnected group of 4 stations has at most two troublesome stations, there are at least 99 48 3 unconnected groups of four and at most 99 4 −99 48 3 connected groups.
All that is left to show is that this maximum can be attained.
Arrange the stations around a circle, and put a two-way tunnel between any two adjacent stations; otherwise, place a one-way tunnel running from station A to station B if and only if A is 3, 5, . . . , or 97 stations away clockwise from B. In this arrangement, every station is troublesome 2 48 3 times.
It is easy to check that under this arrangement, no unconnected group of four stations contains 1999 National Contests: Problems and Solutions 49 two adjacent stations. And suppose that station A is troublesome in a group of four stations A, B, C, D with B closest and D furthest away clockwise from A. If one-way tunnels lead from A to the other tunnels, three one-way tunnels must lead to D from the other tunnels; and if one-way tunnels lead to A from the other tunnels, three one-way tunnels must lead from B to the other tunnels. Thus every unconnected group of four stations has exactly two troublesome stations. Hence equality holds in the previous paragraph, and there are indeed exactly 99 4 −99 48 3 connected groups.
Problem 4 Let m be a positive integer.
Prove that there are integers a, b, k, such that both a and b are odd, k ≥0, and 2m = a19 + b99 + k · 21999.
Solution: The key observation is that if {t1, · · · , tn} equals {1, 3, 5, . . . , 2n −1} modulo 2n, then {ts 1, · · · , ts n} does as well for any odd positive integer s. To show this, note that for i ̸= j, ts i −ts j = (ti −tj)(ts−1 i + ts−2 i tj + · · · + ts−1 j ).
Since ts−1 i +ts−2 i tj +· · ·+ts−1 j is an odd number, ti ≡tj ⇐ ⇒ts i ≡ts j (mod 2n).
Therefore there exists an odd number a0 such that 2m −1 ≡a19 0 (mod 21999). Hence if we pick a ≡a0 (mod 21999) sufficiently negative so that 2m −1 −a19 > 0, then (a, b, k) = a, 1, 2m −1 −a19 21999 is a solution to the equation.
Problem 5 Determine the maximum value of λ such that if f(x) = x3 + ax2 + bx + c is a cubic polynomial with all its roots nonnegative, then f(x) ≥λ(x −a)3 for all x ≥0. Find the equality condition.
Solution: Let α, β, γ be the three roots. Without loss of generality, suppose that 0 ≤α ≤β ≤γ. We have x −a = x + α + β + γ ≥0 and f(x) = (x −α)(x −β)(x −γ).
50 China If 0 ≤x ≤α, then (applying the arithmetic-mean geometric mean inequality) to obtain the first inequality below) −f(x) = (α −x)(β −x)(γ −x) ≤1 27(α + β + γ −3x)3 ≤1 27(x + α + β + γ)3 = 1 27(x −a)3, so that f(x) ≥−1 27(x −a)3. Equality holds exactly when α −x = β −x = γ −x in the first inequality and α+β +γ −3x = x+α+β +γ in the second; that is, when x = 0 and α = β = γ.
If β ≤x ≤γ, then (again applying AM-GM to obtain the first inequality below) −f(x) = (x −α)(x −β)(γ −x) ≤1 27(x + γ −α −β)3 ≤1 27(x + α + β + γ)3 = 1 27(x −a)3, so that again f(x) ≥−1 27(x−a)3. Equality holds exactly when x−α = x−β = γ −x in the first inequality and x+γ −α−β = x+α+β +γ; that is, when α = β = 0 and γ = 2x.
Finally, when α < x < β or x > γ then f(x) > 0 ≥−1 27(x −a)3.
Thus, λ = −1 27 works. From the above reasoning we can find that λ must be at most −1 27 or else the inequality fails for the polynomial f(x) = x2(x −1) at x = 1 2. Equality occurs when either α = β = γ and x = 0; or α = β = 0, γ any nonnegative real, and x = γ 2 .
Problem 6 A 4×4×4 cube is composed of 64 unit cubes. The faces of 16 unit cubes are to be colored red. A coloring is called interesting if there is exactly 1 red unit cube in every 1 × 1 × 4 rectangular box composed of 4 unit cubes. Determine the number of interesting colorings. (Two colorings are different even if one can be transformed into another by a series of rotations.) Solution: Pick one face of the cube as our bottom face. For each unit square A on the bottom face, we consider the vertical 1 × 1 × 4 1999 National Contests: Problems and Solutions 51 rectangular box with A at its bottom. Suppose the i-th unit cube up (counted from A) in the box is colored; then write the number i in A.
Each interesting coloring is mapped one-to-one to a 4 × 4 Latin square on the bottom face.
(In an n × n Latin square, each row and column contains each of n symbols a1, . . . , an exactly once.) Conversely, given a Latin square we can reverse this construction.
Therefore, to solve the problem, we only need to count the number of distinct 4 × 4 Latin squares.
Note that switching rows of a Latin square will generate another Latin square. Thus if our four symbols are a, b, c, d, then each of the 4!·3! arrangements of the first row and column correspond to the same number of Latin squares. Therefore there are 4! · 3! · x four-by-four Latin squares, where x is the number of Latin squares whose first row and column both contain the symbols a, b, c, d in that order. The entry in the second row and second column equals either a, c, or d, yielding the Latin squares a b c d b a d c c d a b d c b a , a b c d b a d c c d b a d c a b , a b c d b c d a c d a b d a b c , a b c d b d a c c a d b d c b a .
Thus x = 4, and there are 4! · 3! · 4 = 576 four-by-four Latin squares, and 576 interesting colorings.
52 Czech and Slovak Republics 1.6 Czech and Slovak Republics Problem 1 In the fraction 29 ÷ 28 ÷ 27 ÷ · · · ÷ 16 15 ÷ 14 ÷ 13 ÷ · · · ÷ 2 parentheses may be repeatedly placed anywhere in the numerator, granted they are also placed on the identical locations in the denom-inator.
(a) Find the least possible integral value of the resulting expression.
(b) Find all possible integral values of the resulting expression.
Solution: (a) The resulting expression can always be written (if we refrain from canceling terms) as a ratio A B of two integers A and B satisfying AB = (2)(3) · · · (29) = 29! = 225·313·56·74·112·132·17·19·23·29.
(To find these exponents, we could either count primes directly factor by factor, or use the rule that n p + n p2 + n p3 + · · · (1) is the exponent of p in n!.) The primes that have an odd exponent in the factorization of 29! cannot “vanish” from the ratio A B even after making any cancellations. For this reason no integer value of the result can be less than H = 2 · 3 · 17 · 19 · 23 · 29 = 1, 292, 646.
On the other hand, 29 ÷ (28 ÷ 27 ÷ · · · ÷ 16) 15 ÷ (14 ÷ 13 ÷ · · · ÷ 2) = 29 · 14 15 · 28 · (27)(26) · · · (16) (13)(12) · · · (2) = 29 · 142 28 · 27!
(15!)2 = 29 · 7 · 223 · 313 · 56 · 73 · 112 · 132 · 17 · 19 · 23 (211 · 36 · 53 · 72 · 11 · 13)2 = H.
1999 National Contests: Problems and Solutions 53 (Again it helps to count exponents in factorials using (1).) The number H is thus the desired least value.
(b) Let’s examine the products A and B more closely. In each of the fourteen pairs of numbers {29, 15}, {28, 14}, . . . , {16, 2}, one of the numbers is a factor in A and the other is a factor in B. The resulting value V can then be written as a product 29 15 ϵ1 28 14 ϵ2 · · · 16 2 ϵ14 , where each ϵi equals ±1, and where ϵ1 = 1 and ϵ2 = −1 no matter how the parentheses are placed. Since the fractions 27 13, 26 12, . . . , 16 2 are greater than 1, the resulting value V (whether an integer or not) has to satisfy the estimate V ≤29 15 · 14 28 · 27 13 · 26 12 · . . . · 16 2 = H, where H is number determined in part (a). It follows that H is the only possible integer value of V !
Problem 2 In a tetrahedron ABCD we denote by E and F the midpoints of the medians from the vertices A and D, respectively.
(The median from a vertex of a tetrahedron is the segment connecting the vertex and the centroid of the opposite face.) Determine the ratio of the volumes of tetrahedrons BCEF and ABCD.
Solution: Let K and L be the midpoints of the edges BC and AD, and let A0, D0 be the centroids of triangles BCD and ABC, respectively.
Both medians AA0 and DD0 lie in the plane AKD, and their intersection T (the centroid of the tetrahedron) divides them in 3 : 1 ratios.
T is also the midpoint of KL, since ⃗ T = 1 4( ⃗ A+ ⃗ B + ⃗ C + ⃗ D) = 1 2( 1 2( ⃗ A+ ⃗ D)+ 1 2( ⃗ B + ⃗ C)) = 1 2( ⃗ K + ⃗ L). It follows that ET AT = F T DT = 1 3, and hence △ATD ∼△ETF and EF = 1 3AD.
Since the plane BCL bisects both segments AD and EF into halves, it also divides both tetrahedrons ABCD and BCEF into two parts of equal volume. Let G be the midpoint of EF; the corresponding volumes than satisfy [BCEF] [ABCD] = [BCGF] [BCLD] = GF LD · [BCG] [BCL] = 1 3 KG KL = 1 3 · 2 3 = 2 9.
54 Czech and Slovak Republics Problem 3 Show that there exists a triangle ABC for which, with the usual labelling of sides and medians, it is true that a ̸= b and a + ma = b + mb. Show further that there exists a number k such that for each such triangle a + ma = b + mb = k(a + b). Finally, find all possible ratios a : b of the sides of these triangles.
Solution: We know that m2 a = 1 4(2b2 + 2c2 −a2), m2 b = 1 4(2a2 + 2c2 −b2), so m2 a −m2 b = 3 4(b2 −a2).
As ma −mb = b −a ̸= 0 by hypothesis, it follows that ma + mb = 3 4(b + a). From the system of equations ma −mb = b −a ma + mb = 3 4(b + a) we find ma = 1 8(7b −a), mb = 1 8(7a −b), and a + ma = b + mb = 7 8(a + b).
Thus k = 7 8.
Now we examine for what a ̸= b there exists a triangle ABC with sides a, b and medians ma = 1 8(7b −a), mb = 1 8(7a −b). We can find all three side lengths in the triangle AB1G, where G is the centroid of the triangle ABC and B1 is the midpoint of the side AC: AB1 = b 2, AG = 2 3ma = 2 3 · 1 8(7b −a) = 1 12(7b −a), B1G = 1 3mb = 1 3 · 1 8(7a −b) = 1 24(7a −b).
Examining the triangle inequalities for these three lengths, we get the condition 1 3 < a b < 3, from which the value a b = 1 has to be excluded by assumption.
This condition is also sufficient: once the triangle AB1G has been constructed, it can always be completed to a triangle ABC with b = AC, ma = AA1, mb = BB1. Then from the equality m2 a −m2 b = 3 4(b2 −a2) we would also have a = BC.
1999 National Contests: Problems and Solutions 55 Problem 4 In a certain language there are only two letters, A and B. The words of this language satisfy the following axioms: (i) There are no words of length 1, and the only words of length 2 are AB and BB.
(ii) A sequence of letters of length n > 2 is a word if and only if it can be created from some word of length less than n by the following construction: all letters A in the existing word are left unchanged, while each letter B is replaced by some word. (While performing this operation, the B’s do not all have to be replaced by the same word.) Show that for any n the number of words of length n equals 2n + 2 · (−1)n 3 .
Solution: Let us call any finite sequence of letters A, B a “string.” From here on, we let · · · denote a (possibly empty) string, while ∗∗∗ will stand for a string consisting of identical letters. (For example, B ∗∗∗B | {z } k is a string of k B’s.) We show that an arbitrary string is a word if and only if it satisfies the following conditions: (a) the string terminates with the letter B; and (b) it either starts with the letter A, or else starts (or even wholly consists of) an even number of B’s.
It is clear that these conditions are necessary: they are satisfied for both words AB and BB of length 2, and they are likewise satisfied by any new word created by the construction described in (ii) if they are satisfied by the words in which the B’s are replaced.
We now show by induction on n that, conversely, any string of length n satisfying the conditions is a word. This is clearly true for n = 1 and n = 2. If n > 2, then a string of length n satisfying the conditions must have one of the forms AA · · · B, AB · · · B, B ∗∗∗B | {z } 2k A · · · B, B ∗∗∗B | {z } 2k+2 , where 2 ≤2k ≤n−2. We have to show that these four types of strings arise from the construction in (ii) in which the B’s are replaced by strings (of lengths less than n) satisfying the condition — that is, by words in view of the induction hypothesis.
56 Czech and Slovak Republics The word AA · · · B arises as A(A · · · B) from the word AB. The word AB · · · B arises either as A(B · · · B) from the word AB, or as (AB)(· · · B) from the word BB, depending on whether its initial letter A is followed by an even or an odd number of B’s. The word B ∗∗∗B | {z } 2k A · · · B arises as (B ∗∗∗B)(A · · · B) from the word BB, and the word B ∗∗∗B | {z } 2k+2 as (B ∗∗∗B | {z } 2k )(BB) from the word BB. This completes the proof by induction.
Now we show that the number pn of words of length n is indeed given by the formula pn = 2n + 2 · (−1)n 3 .
It is clearly true for n = 1 and 2 since p1 = 0 and p2 = 2; and the formula will then follow by induction if we can show that pn+2 = 2n + pn for each n. But this recursion is obvious because each word of length n + 2 is either of the form A · · · B where · · · is any of 2n strings of length n; or of the form BB · · · where · · · is any of the pn words of length n.
Problem 5 In the plane an acute angle APX is given. Show how to construct a square ABCD such that P lies on side BC and P lies on the bisector of angle BAQ where Q is the intersection of ray PX with CD.
Solution: Consider the ration by 90◦around the point A that maps B to D, and the points P, C, D into some points P ′, C′, D′, respectively.
Since ∠PAP ′ = 90◦, it follows from the nature of exterior angle bisectors that AP ′ bisects ∠QAD′. Consequently, the point P ′ has the same distance from AD′ and AQ, equal to the side length s of square ABCD. But this distance is also the length of the altitude AD in triangle AQP ′; then since the altitudes from A and P ′ in this triangle are equal, we have AQ = P ′Q.
Since we can construct P ′, we can also construct Q as the intersection of line PX with the perpendicular bisector of the segment AP ′. The rest of the construction is obvious, and it is likewise clear that the resulting square ABCD has the required property.
1999 National Contests: Problems and Solutions 57 Problem 6 Find all pairs of real numbers a and b such that the system of equations x + y x2 + y2 = a, x3 + y3 x2 + y2 = b has a solution in real numbers (x, y).
Solution: If the given system has a solution (x, y) for a = A, b = B, then it clearly also has a solution (kx, ky) for a = 1 kA, b = kB, for any k ̸= 0. It follows that the existence of a solution of the given system depends only on the value of the product ab.
We therefore begin by examining the values of the expression P(u, v) = (u + v)(u3 + v3) (u2 + v2)2 where the numbers u and v are normalized by the condition u2 +v2 = 1. This condition implies that P(u, v) = (u + v)(u3 + v3) = (u + v)2(u2 −uv + v2) = (u2 + 2uv + v2)(1 −uv) = (1 + 2uv)(1 −uv).
Under the condition u2 + v2 = 1 the product uv can attain all values in the interval [−1 2, 1 2] (if u = cos α and v = sin α, then uv = 1 2 sin 2α). Hence it suffices to find the range of values of the function f(t) = (1 + 2t)(1 −t) on the interval t ∈[−1 2, 1 2]. From the formula f(t) = −2t2 + t + 1 = −2 t −1 4 2 + 9 8 it follows that this range of values is the closed interval with endpoints f −1 2 = 0 and f 1 4 = 9 8.
This means that if the given system has a solution, its parameters a and b must satisfy 0 ≤ab ≤9 8, where the equality ab = 0 is possible only if x + y = 0 (then, however, a = b = 0).
Conversely, if a and b satisfy 0 < ab ≤9 8, by our proof there exist numbers u and v such that u2 + v2 = 1 and (u + v)(u3 + v3) = ab.
Denoting a′ = u + v and b′ = u3 + v3, the equality a′b′ = ab ̸= 0 implies that both ratios a′ a and b b′ have the same value k ̸= 0. But then (x, y) = (ku, kv) is clearly a solution of the given system for the parameter values a and b.
58 France 1.7 France Problem 1 (a) What is the maximum volume of a cylinder that is inside a given cone and has the same axis of revolution as the cone? Express your answer in terms of the radius R and height H of the cone.
(b) What is the maximum volume of a ball that is inside a given cone? Again, express your answer in terms of R and H.
(c) Given fixed values for R and H, which of the two maxima you found is bigger?
Solution: Let ℓ= √ R2 + H2 be the slant height of the given cone; also, orient the cone so that its base is horizontal and its tip is pointing upward.
(a) Intuitively, the cylinder with maximum volume rests against the base of the cone, and the center of the cylinder’s base coincides with the center of the cone’s base. The top face of the cylinder cuts offa smaller cone at the top of the original cone.
If the cylinder has radius r, then the smaller cone has height r · H R and the cylinder has height h = H −r · H R . Then the volume of the cylinder is πr2h = πr2H 1 −r R = 4πR2H r 2R · r 2R · 1 −r R .
And by AM-GM on r 2R, r 2R, and 1 −r R this is at most 4πR2H · 1 27 r 2R + r 2R + 1 −r R 3 = 4 27πR2H, with equality when r/2R = 1 −r/R ⇐ ⇒r = 2 3R.
(b) Intuitively, the sphere with maximum volume is tangent to the base and lateral face of the cone; and its center lies on the cone’s axis. Say the sphere has radius r.
Take a planar cross-section of the cone slicing through its axis; this cuts offa triangle from the cone and a circle from the sphere.
The triangle’s side lengths are ℓ, ℓ, and 2R; and its height (from the side of length 2R) is H. The circle has radius r and is the incircle of this triangle.
The area K of the triangle is 1 2(2R)(H) = RH and its semiperimeter is s = R+ℓ. Then since K = rs we have r = RH R+ℓ, 1999 National Contests: Problems and Solutions 59 and thus the volume of the sphere is 4 3πr3 = 4 3π RH R + ℓ 3 .
(c) We claim that when h/R = √ 3 or 2 √ 6, the two volumes are equal; when √ 3 < h/R < 2, the sphere has larger volume; and when 0 < h/R < √ 3 or 2 < h/R, the cylinder has larger volume.
We wish to compare 4 27πR2H and 4 3π RH R+ℓ 3 ; equivalently, multiplying by 27 4πR2H (R + ℓ)3, we wish to compare (R + ℓ)3 and 9RH2 = 9R(ℓ2 −R2). Writing φ = ℓ/R, this is equivalent to comparing (1 + φ)3 and 9(φ2 −1). Now, (1 + φ)3 −9(φ2 −1) = φ3 −6φ2 + 3φ + 10 = (φ + 1)(φ −2)(φ −5).
Thus when φ = 2 or 5, the volumes are equal; when 2 < φ < 5, the sphere has larger volume; and when 1 < φ < 2 or 5 < φ, the cylinder has larger volume. Comparing R and H instead of R and ℓyields the conditions stated before.
Problem 2 Find all integer solutions to (n + 3)n = n+2 X k=3 kn.
Solution: n = 2 and n = 3 are solutions to the equations; we claim they are the only ones.
First observe that the function f(n) = n+3 n+2 n = 1 + 1 n+2 n is an increasing function for n > 0. To see this, note that the derivative of ln f(n) with respect to n is ln 1 + 1 n+2 − n (n+2)(n+3). By the Taylor expansion, ln 1 + 1 n + 2 = ∞ X j=1 1 (n + 2)2j 1 2j −1(n + 2) −1 2j > 2(n + 2) −1 2(n + 2)2 and hence d dn ln f(n) = ln n + 3 n + 2 − n (n + 2)(n + 3) > 2(n + 2) −1 2(n + 2)2 − n (n + 2)2 = 3 2(n + 2)2 > 0.
Thus ln f(n) and therefore f(n) is indeed increasing.
60 France Now, notice that if f(n) > 2 then we have 2 1 n > 3 2 n > · · · > n + 3 n + 2 n > 2 so that (n + 3)n > 2(n + 2)n > · · · > 2j(n + 3 −j)n > · · · > 2n · (3)n.
Then 3n + 4n + · · · + (n + 2)n < 1 2n + 1 2n−1 + · · · + 1 2 (n + 3)n = 1 −1 2n (n + 3)n < (n + 3)n, so the equality does not hold.
Then since 2 < f(6) < f(7) < · · ·, the equality must fail for all n ≥6. Quick checks show it also fails for n = 1, 4, 5 (in each case, one side of the equation is odd while the other is even). Therefore the only solutions are n = 2 and n = 3.
Problem 3 For which acute-angled triangle is the ratio of the shortest side to the inradius maximal?
Solution: Let the sides of the triangle have lengths a ≤b ≤c; let the angles opposite them be A, B, C; let the semiperimeter be s = 1 2(a + b + c); and let the inradius be r. Without loss of generality say the triangle has circumradius R = 1 2 and that a = sin A, b = sin B, c = sin C.
The area of the triangle equals both rs = 1 2r(sin A + sin B + sin C) and abc/4R = 1 2 sin A sin B sin C. Thus r = sin A sin B sin C sin A + sin B + sin C and a r = sin A + sin B + sin C sin B sin C .
Since A = 180◦−B−C, sin A = sin(B+C) = sin B cos C+cos B sin C and we also have a r = cot B + csc B + cot C + csc C.
Note that f(x) = cot x + csc x is a decreasing function along the interval 0◦< x < 90◦. Now there are two cases: B ≤60◦, or B > 60◦.
1999 National Contests: Problems and Solutions 61 If B ≤60◦, then assume that A = B; otherwise the triangle with angles A′ = B′ = 1 2(A + B) ≤B and C′ = C has a larger ratio a′/r′.
Then since C < 90◦we have 45◦< A ≤60◦. Now, a r = sin A + sin B + sin C sin B sin C = 2 sin A + sin(2A) sin A sin(2A) = 2 csc(2A) + csc A.
Now csc x has second derivative csc x(csc2 x+cot2 x), which is strictly positive when 0◦< x < 180◦; thus both csc x and csc(2x) are strictly convex along the interval 0◦< x < 90◦. Therefore g(A) = 2 csc(2A)+csc A, a convex function in A, is maximized in the interval 45◦≤A ≤60◦at one of the endpoints. Since g(45◦) = 2 + √ 2 < 2 √ 3 = g(60◦), it is maximized when A = B = C = 60◦.
As for the case when B > 60◦, since C > B > 60◦, the triangle with A′ = B′ = C′ = 60◦has a larger ratio a′/r′. Therefore the maximum ratio is 2 √ 3, attained with an equilateral triangle.
Problem 4 There are 1999 red candies and 6661 yellow candies on a table, made indistinguishable by their wrappers. A gourmand applies the following algorithm until the candies are gone: (a) If there are candies left, he takes one at random, notes its color, eats it, and goes to (b).
(b) If there are candies left, he takes one at random, notes its color, and (i) if it matches the last one eaten, he eats it also and returns to (b).
(ii) if it does not match the last one eaten, he wraps it up again, puts it back, and goes to (a).
Prove that all the candies will eventually be eaten. Find the proba-bility that the last candy eaten is red.
Solution: If there are finitely many candies left at any point, then at the next instant the gourmand must perform either step (a), part (i) of step (b), or part (ii) of step (b). He eats a candy in the first two cases; in the third case, he returns to step (a) and eats a candy. Since there are only finitely many candies, the gourmand must eventually eat all the candies.
We now prove by induction on the total number of candies that if we start with r > 0 red candies and y > 0 yellow candies immediately 62 France before step (a), then the probability is 1 2 that the last candy eaten is red.
Suppose that the claim is true for all smaller amounts of candy.
After the gourmand first completes steps (a) and (b) exactly once, suppose there are r′ red candies and y′ yellow candies left; we must have r′ + y′ < r + y. The chances that r′ = 0 is r r + y · r −1 r + y −1 · · · · · 1 y + 1 = 1 r+y r .
Similarly, the chances that y′ = 0 is 1 ( r+y y ) = 1 ( r+y r ).
(In the case r = y = 1, this proves the claim.) Otherwise, the probability is 1 − 2 ( r+y r ) that both r′ and y′ are still positive. By the induction hypothesis in this case the last candy is equally likely to be red as it is yellow. Thus the overall probability that the last candy eaten is red is 1 r+y r | {z } y′=0 +1 2 1 − 2 r+y r !
| {z } r′,y′>0 = 1 2.
This completes the inductive step, and the proof.
Problem 5 With a given triangle, form three new points by re-flecting each vertex about the opposite side. Show that these three new points are collinear if and only if the the distance between the orthocenter and the circumcenter of the triangle is equal to the diameter of the circumcircle of the triangle.
Solution: Let the given triangle be ABC and let the reflections of A, B, C across the corresponding sides be D, E, F. Let A′, B′, C′ be the midpoints of BC, CA, AB, and as usual let G, H, O denote the triangle’s centroid, orthocenter, and circumcenter. Let triangle A′′B′′C′′ be the triangle for which A, B, C are the midpoints of B′′C′′, C′′A′′, A′′B′′, respectively.
Then G is the centroid and H is the circumcenter of triangle A′′B′′C′′. Let D′, E,′ F ′ denote the projections of O on the lines B′′C′′, C′′A′′, A′′B′′, respectively.
Consider the homothety h with center G and ratio −1/2. It maps A, B, C, A′′, B′′, C′′ into A′, B′, C′, A, B, C, respectively.
Note that A′D′ ⊥BC since O is the orthocenter of triangle A′B′C′. This implies AD : A′D′ = 2 : 1 = GA : GA′ and ∠DAG = ∠D′A′G. We conclude 1999 National Contests: Problems and Solutions 63 that h(D) = D′. Similarly, h(E) = E′ and h(F) = F ′. Thus, D, E, F are collinear if and only if D′, E′, F ′ are collinear. Now D′, E′, F ′ are the projections of O on the sides B′′C′′, C′′A′′, A′′B′′, respectively.
By Simson’s theorem, they are collinear if and only if O lies on the circumcircle of triangle A′′B′′C′′. Since the circumradius of triangle A′′B′′C′′ is 2R, O lies on its circumcircle if and only if OH = 2R.
64 Hong Kong (China) 1.8 Hong Kong (China) Problem 1 Let PQRS be a cyclic quadrilateral with ∠PSR = 90◦, and let H and K be the respective feet of perpendiculars from Q to lines PR and PS. Prove that line HK bisects QS.
First Solution: Since QK and RS are both perpendicular to PS, QK is parallel to RS and thus ∠KQS = ∠RSQ. Since PQRS is cyclic, ∠RSQ = ∠RPQ. Since ∠PKQ = ∠PHQ = 90◦, PKHQ is also cyclic and it follows that ∠RPQ = ∠HPQ = ∠HKQ. Thus, ∠KQS = ∠HKQ; since triangle KQS is right, it follows that line HK bisects QS.
Second Solution: The Simson line from Q with respect to △PRS goes through H, K, and the foot F of the perpendicular from Q to ← → RS. Thus, line HK is line FK, a diagonal in rectangle SFQK, so it bisects the other diagonal, QS.
Problem 2 The base of a pyramid is a convex nonagon. Each base diagonal and each lateral edge is colored either black or white. Both colors are used at least once. (Note that the sides of the base are not colored.) Prove that there are three segments colored the same color which form a triangle.
Solution: Let us assume the contrary.
From the pigeonhole principle, 5 of the lateral edges must be of the same color; assume they are black, and say they are the segments from the vertex V to B1, B2, B3, B4, and B5 where B1B2B3B4B5 is a convex pentagon (and where the Bi’s are not necessarily adjacent vertices of the nonagon).
The BiBi+1 (where B6 = B1) cannot all be sides of the nonagon, so without loss of generality suppose that B1B2 is colored. Then because triangle V BiBj cannot have three sides colored black, each segment B1B2, B2B4, B4B1 must be white; but then triangle B1B2B4 has three sides colored white, a contradiction.
Problem 3 Let s and t be nonzero integers, and let (x, y) be any ordered pair of integers. A move changes (x, y) to (x −t, y −s). The pair (x, y) is good if after some (possibly zero) number of moves it becomes a pair of integers that are not relatively prime.
(a) Determine if (s, t) is a good pair; 1999 National Contests: Problems and Solutions 65 (b) Prove that for any s, t there exists a pair (x, y) which is not good.
Solution: (a) Let us assume that (s, t) is not good. Then, after one move, we have (s−t, t−s), so we may assume without loss of generality that s −t = 1 and t −s = −1 since these numbers must be relatively prime. Then s + t cannot equal 0 because it is odd; also, s + t = (s−t)+2t ̸= (s−t)+0 = 1, and s+t = (t−s)+2s ̸= (t−s)+0 = −1.
Hence some prime p divides s+t. After p−1 moves, (s, t) becomes (s −(p −1)t, t −(p −1)s) ≡(s + t, t + s) ≡(0, 0) (mod p), a contradiction. Thus (s, t) is good.
(b) Let x and y be integers which satisfy sx −ty = g, where g = gcd(s, t). Dividing by g, we find s′x −t′y = 1, so gcd(x, y) = 1.
Now, suppose by way of contradiction that after k moves some prime p divides both x −kt and y −ks. We then have 0 ≡x −kt ≡y −ks = ⇒0 ≡s(x −kt) ≡t(y −ks) = ⇒0 ≡sx −ty = g (mod p).
Thus p divides g, which divides s and t, so the first equation above becomes 0 ≡x ≡y (mod p); but x and y are relatively prime, a contradiction. Thus (x, y) is not good.
Problem 4 Let f be a function defined on the positive reals with the following properties: (i) f(1) = 1; (ii) f(x + 1) = xf(x); (iii) f(x) = 10g(x), where g(x) is a function defined on the reals satisfying g(ty + (1 −t)z) ≤tg(y) + (1 −t)g(z) for all real y, z and any 0 ≤t ≤1.
(a) Prove that t[g(n) −g(n −1)] ≤g(n + t) −g(n) ≤t[g(n + 1) −g(n)] where n is an integer and 0 ≤t ≤1.
66 Hong Kong (China) (b) Prove that 4 3 ≤f 1 2 ≤4 √ 2 3 .
Solution: (a) Setting t = 1 2 in the given inequality, we find that g( 1 2(y + z)) ≤ 1 2(g(y)+g(z)). Now fix t (perhaps not equal to 1 2) constant; letting y = n −t and z = n + t in g( 1 2(y + z)) ≤1 2(g(y) + g(z)) gives g(n) ≤1 2(g(n −t) + g(n + t)), or g(n) −g(n −t) ≤g(n + t) −g(n).
(1) Plugging in z = n, y = n −1 into the given inequality gives g(t(n −1) + (1 −t)n) ≤tg(n −1) + (1 −t)g(n), or t[g(n) −g(n −1)] ≤g(n) −g(n −t).
Combining this with (1) proves the inequality on the left side.
And the inequality on the right side follows from the given inequality with z = n, y = n + 1.
(b) From (ii), f( 3 2) = 1 2f( 1 2), and f( 5 2) = 3 2f( 3 2) = 3 4f( 1 2).
Also, f(2) = 1 · f(1) = 1, and f(3) = 2f(2) = 2.
Now, if we let n = 2 and t = 1 2 in the inequality in part (a), we find 1 2[g(2) −g(1)] ≤g 5 2 −g(2) ≤1 2[g(3) −g(2)]. Exponentiating with base 10 yields q f(2) f(1) ≤ f( 5 2) f(2) ≤ q f(3) f(2), or 1 ≤f 5 2 ≤ √ 2.
Plugging in f 5 2 = 3 4f( 1 2) yields the desired result.
1999 National Contests: Problems and Solutions 67 1.9 Hungary Problem 1 I have n ≥5 real numbers with the following properties: (i) They are nonzero, but at least one of them is 1999.
(ii) Any four of them can be rearranged to form a geometric progres-sion.
What are my numbers?
Solution: First suppose that the numbers are all nonnegative. If x ≤y ≤z ≤w ≤v are any five of the numbers, then x, y, z, w; x, y, z, v; x, y, w, v; x, z, w, v; and y, z, w, v must all be geometric progressions. Comparing each two successive progressions in this list we find that x = y = z = w = v. Thus all our numbers are equal.
If some numbers are negative in our original list, replace each number x by |x|. The geometric progression property is preserved, and thus from above all the values |x| are equal. Hence, each original number was 1999 or −1999. And because n ≥5, some three numbers are equal. But no geometric progression can be formed from three −1999s and a 1999, or from three 1999s and a −1999. Therefore all the numbers must be equal — to 1999.
Problem 2 Let ABC be a right triangle with ∠C = 90◦.
Two squares S1 and S2 are inscribed in triangle ABC such that S1 and ABC share a common vertex C, and S2 has one of its sides on AB.
Suppose that [S1] = 441 and [S2] = 440. Calculate AC + BC.
Solution: Let S1 = CDEF and S2 = KLMN with D and K on AC and N on BC. Let s1 = 21, s2 = √ 440 and a = BC, b = CA, c = AB.
Using ratios between similar triangles AED, ABC, EBF we get c = AB = AE+EB = c(s1/a+s1/b) or s1(1/a+1/b) = 1. Since triangles ABC, AKL, NBM are similar we have c = AB = AL+LM +MB = s2(b/a + 1 + a/b) and s2 = abc/(ab + c2). Then 1 s2 2 −1 s2 1 = 1 c + c ab 2 − 1 a + 1 b 2 = 1 c2 + c2 a2b2 + 2 ab − 1 a2 + 1 b2 + 2 ab = 1 c2 .
68 Hungary Thus c = 1/ p 1/s2 2 −1/s2 1 = 21 √ 440. Solving s2 = abc/(ab + c2) for ab yields ab = s2c2/(c −s2) = 212 · 22. Finally, AC + BC = a + b = ab/s1 = 21 · 22 = 462.
Problem 3 Let O and K be the centers of the respective spheres tangent to the faces, and the edges, of a right pyramid whose base is a 2 by 2 square. Determine the volume of the pyramid if O and K are equidistant from the base.
Solution: Let r, R be the spheres’ respective radii. Let the pyramid have base ABCD, vertex P, and height h. By symmetry, O and K lie on the altitude through P.
Take a cross-section of the pyramid with a plane perpendicular to the base, cutting the base at a line through its center parallel to AB.
It cuts offan isosceles triangle from the pyramid with base 2 and legs √ h2 + 1; the triangle’s incircle is the cross-section of the sphere centered at O and hence has radius r. On the one hand, the area of this triangle is the product of its inradius and semiperimeter, or 1 2r(2 + 2 √ h2 + 1). On the other hand, it equals half of the product of its base and height, or 1 2 · 2 · h. Setting these quantities equal, we have r = ( √ h2 + 1 −1)/h.
Next, by symmetry the second sphere is tangent to AB at its midpoint M. Then since K must be distance r from plane ABCD, we have R2 = KM 2 = r2 + 1. Furthermore, if the second sphere is tangent to AP at N, then by equal tangents we have AN = AM = 1.
Then PN = PA −1 = √ h2 + 2 −1. Also, PK = h + r if K is on the opposite side of plane ABCD as O, and it equals h−r otherwise.
Thus PK2 = PN 2 + NK2 (h ± r)2 = ( p h2 + 2 −1)2 + (r2 + 1) ± 2rh = 4 −2 p h2 + 2.
Recalling that r = ( √ h2 + 1 −1)/h, this gives ± ( p h2 + 1 −1) = 2 − p h2 + 2.
This equation has the unique solution h = √ 7/3. Thus the volume of the pyramid is 1 3 · 4 · √ 7 3 = 4 √ 7/9.
1999 National Contests: Problems and Solutions 69 Problem 4 For any given positive integer n, determine (as a function of n) the number of ordered pairs (x, y) of positive integers such that x2 −y2 = 102 · 302n.
Further prove that the number of such pairs is never a perfect square.
Solution: Since 102·302n is even, x and y must have the same parity.
Then (x, y) is a valid solution if and only if (u, v) = x+y 2 , x−y 2 is a pair of positive integers that satisfies u > v and uv = 52 · 302n. Now 52 · 302n = 22n · 32n · 52n+2 has exactly (2n + 1)2(2n + 3) factors; thus without the condition u > v there are exactly (2n + 1)2(2n + 3) such pairs (u, v).
Exactly one pair has u = v, and by symmetry half of the remaining pairs have u > v; and it follows that there are 1 2 (2n + 1)2(2n + 3) −1 = (n + 1)(4n2 + 6n + 1) valid pairs.
Now suppose that (n + 1)(4n2 + 6n + 1) were a square. Since n + 1 and 4n2 + 6n + 1 = (4n + 2)(n + 1) −1 are coprime, 4n2 + 6n + 1 must be a square as well; but (2n + 1)2 < 4n2 + 6n + 1 < (2n + 2)2, a contradiction.
Problem 5 For 0 ≤x, y, z ≤1, find all solutions to the equation x 1 + y + zx + y 1 + z + xy + z 1 + x + yz = 3 x + y + z .
Solution: Assume x + y + z > 0, since otherwise the equation is meaningless. (1 −z)(1 −x) ≥0 ⇒1 + zx ≥x + z, and hence x/(1 + y + zx) ≤x/(x + y + z). Doing this for the other two fractions yields that the left hand side is at most (x + y + z)/(x + y + z) ≤ 3/(x + y + z).
If equality holds, we must have in particular that x + y + z = 3 ⇒x = y = z = 1. We then verify that this is indeed a solution.
Problem 6 The midpoints of the edges of a tetrahedron lie on a sphere. What is the maximum volume of the tetrahedron?
Solution: Let the sphere have center O. First let A, B, C be any points on its surface. Then [OAB] = 1 2OA · OB sin ∠AOB ≤1 2r2.
Likewise, the height from C to plane OAB is at most CO = r, whence tetrahedron OABC has maximum volume r3/6.
Now, if {A, A′}, {B, B′}, {C, C′} are pairs of antipodal points on the sphere, 70 Hungary the octahedron ABCA′B′C′ can be broken up into 8 such tetrahedra with vertex O and therefore has maximum volume 4r3/3. Equality holds for a regular octahedron.
In the situation of the problem, shrink the tetrahedron T (with volume V ) by a factor of 1/2 about each vertex to obtain four tetrahedra, each with volume V/8.
Then the six midpoints form an octahedron with volume V/2. Moreover, the segment connecting two opposite vertices C and D of this octahedron has T’s centroid P as its midpoint. If O ̸= P then line OP is a perpendicular bisector of each segment, and then all these segments must lie in the plane through P perpendicular to line OP; but then V/2 = 0. Otherwise, the midpoints form three pairs of antipodal points, whose volume is at most 4r3/3 from the last paragraph. Therefore V ≤8r3/3, with equality for a regular tetrahedron.
Problem 7 A positive integer is written in each square of an n2 by n2 chess board. The difference between the numbers in any two adjacent squares (sharing an edge) is less than or equal to n. Prove that at least ⌊n/2⌋+ 1 squares contain the same number.
Solution: Consider the smallest and largest numbers a and b on the board. They are separated by at most n2 −1 squares horizontally and n2 −1 vertically, so there is a path from one to the other with length at most 2(n2 −1). Then since any two successive squares differ by at most n, we have b −a ≤2(n2 −1)n. But since all numbers on the board are integers lying between a and b, only 2(n2 −1)n + 1 distinct numbers can exist; and because n4 > (2(n2 −1)n + 1)(n/2), more than n/2 squares contain the same number, as needed.
Problem 8 One year in the 20th century, Alex noticed on his birthday that adding the four digits of the year of his birth gave his actual age. That same day, Bernath—who shared Alex’s birthday but was not the same age as him—also noticed this about his own birth year and age. That day, both were under 99. By how many years do their ages differ?
Solution: Let c be the given year. Alex’s year of birth was either 18uv or 19uv respectively (where u and v are digits), and thus either c = 18uv +(9+u+v) = 1809+11u+2v or c = 19uv +(10+u+v) = 1910 + 11u + 2v.
1999 National Contests: Problems and Solutions 71 Similarly, let Bernath’s year of birth end in the digits u′, v′. Alex and Bernath could not have been born in the same century. Other-wise, we would have 11u + 2v = 11u′ + 2v′ ⇒2(v −v′) = 11(u′ −u); thus either (u, v) = (u′, v′) or else |v−v′| ≥11, which are both impos-sible. Then without loss of generality say Alex was born in the 1800s, and that 1809+11u+2v = 1910+11u′+2v′ ⇒11(u−u′)+2(v−v′) = 101 ⇒u −u′ = 9, v −v′ = 1. The difference between their ages then equals 19u′v′ −18uv = 100 + 10(u′ −u) + (v′ −v) = 9.
Problem 9 Let ABC be a triangle and D a point on the side AB.
The incircles of the triangles ACD and CDB touch each other on CD. Prove that the incircle of ABC touches AB at D.
Solution: Suppose that the incircle of a triangle XY Z touches sides Y Z, ZX, XY at U, V, W. Then (using equal tangents) XY + Y Z + ZX = (Y W + Y U) + (XW + ZU) + XZ = (2Y U) + (XZ) + XZ, and Y U = 1 2(XY + Y Z −ZX).
Thus if the incircles of triangles ACD and CDB touch each other at E, then AD+DC −CA = 2DE = BD+DC −CB ⇒AD−CA = (AB −AD) −BC ⇒AD = 1 2(CA + AB −BC). But if the incircle of ABC is tangent to AB at D′, then AD′ = 1 2(CA + AB −BC) as well—so D = D′, as desired.
Problem 10 Let R be the circumradius of a right pyramid with a square base. Let r be the radius of the sphere touching the four lateral faces and the circumsphere. Suppose that 2R = (1 + √ 2)r.
Determine the angle between adjacent faces of the pyramid.
Solution: Let P be the pyramid’s vertex, ABCD the base, and M, N the midpoints of sides AB, CD. By symmetry, both spheres are centered along the altitude from P. Plane PMN intersects the pyramid in triangle PMN and meets the spheres in great circles. Let the smaller circle have center O; it is tangent to PM, PN, and the large circle at some points U, V, W. Again by symmetry W is lies on the altitude from P, implying that it is diametrically opposite P on the larger circle. Thus OP = 2R −r = √ 2r, triangle OUP is a 45◦-45◦-90◦triangle, and ∠OPU = ∠OPV = 45◦. Therefore triangle NPM is isosceles right, and the distance from P to plane ABCD equals BC/2.
72 Hungary Hence one can construct a cube with P as its center and ABCD as a face; this cube can be decomposed into six pyramids congruent to PABCD. In particular, three such pyramids have a vertex at A; so three times the dihedral angle between faces PAB, PAD forms one revolution, and this angle is 2π/3. Stated differently, say the three pyramids are PABD, PADE, PAEB; let P ′ be the midpoint of AP, and let B′, D′, E′ be points on planes PAB, PAD, PAE such that lines B′P ′, D′P ′, E′P ′ are all perpendicular to line AP. The desired angle is the angle between any two of these lines. But since these three lines all lie in one plane (perpendicular to line AP), this angle must be 2π/3.
Problem 11 Is there a polynomial P(x) with integer coefficients such that P(10) = 400, P(14) = 440, and P(18) = 520?
Solution: If P exists, then by taking the remainder modulo (x − 10)(x −14)(x −18) we may assume P is quadratic. Writing P(x) = ax2+bx+c, direct computation reveals P(x+4)+P(x−4)−2P(x) = 32a for all x. Plugging in x = 14 gives 40 = 32a, which is impossible since a must be an integer. Therefore no such polynomial exists.
Problem 12 Let a, b, c be positive numbers and n ≥2 be an integer such that an + bn = cn. For which k is it possible to construct an obtuse triangle with sides ak, bk, ck?
Solution: First, a, b < c.
Thus for m > n we have cm = cm−n(an + bn) > am−nan + bm−nbn = am + bm, while for m < n we have cm = cm−n(an + bn) < am−nan + bm−nbn = am + bm. Now, a triangle with sides ak, bk, ck exists iffak + bk > ck, and it is then obtuse iff(ak)2 + (bk)2 < (ck)2, i.e. a2k + b2k < c2k. From our first observation, these correspond to k < n and 2k > n, respectively; and hence n/2 < k < n.
Problem 13 Let n > 1 be an arbitrary real number and k be the number of positive primes less than or equal to n. Select k+1 positive integers such that none of them divides the product of all the others.
Prove that there exists a number among the k + 1 chosen numbers which is bigger than n.
Solution: Suppose otherwise; then our chosen numbers a1, . . . , ak+1 have a total of at most k distinct prime factors (i.e. the primes less 1999 National Contests: Problems and Solutions 73 than or equal to n). Let op(a) denote the highest value of d such that pd | a.
Also let q = a1a2 · · · ak+1.
Then for each prime p, op(q) = Pk+1 i=1 op(ai), and it follows that there can be at most one “hostile” value of i for which op(ai) > op(q)/2. Since there are at most k primes which divide q, there is some i which is not hostile for any such prime. Then 2op(ai) ≤op(q) ⇒op(ai) ≤op(q/ai) for each prime p dividing q, implying that ai | q/ai, a contradiction.
Problem 14 The polynomial x4 −2x2 + ax + b has four distinct real roots. Show that the absolute value of each root is smaller than √ 3.
Solution: Let the roots be p, q, r, s. We have p + q + r + s = 0, pq + pr+ps+qr+qs+rs = −2, and hence p2+q2+r2+s2 = 02−2(−2) = 4.
But by Cauchy-Schwarz, (1+1+1)(q2 +r2 +s2) ≥(q+r+s)2 for any real q, r, s; furthermore, since q, r, s must be distinct, the inequality becomes strict. Thus 4 = p2 + q2 + r2 + s2 > p2 + (−p)2/3 = 4p2/3 or |p| < √ 3, and the same argument holds for q, r, s.
Problem 15 Each side of a convex polygon has integral length and the perimeter is odd. Prove that the area of the polygon is at least √ 3/4.
Solution: Lemma 1. If 0 ≤x, y ≤1, then p 1 −x2 + p 1 −y2 ≥ p 1 −(x + y −1)2.
Proof: Squaring and subtracting 2−x2 −y2 from both sides gives the equivalent inequality 2 p (1 −x2)(1 −y2) ≥−2(1 −x)(1 −y), which is true since the left side is nonnegative and the right is at most 0.
Lemma 2. If x1 +· · ·+xn ≤n−1/2 and 0 ≤xi ≤1 for each i, then Pn i=1 p 1 −x2 i ≥ √ 3/2.
Proof: Use induction on n. In the case n = 1, the statement is clear. If n > 1, then either min(x1, x2) ≤1/2 or x1 + x2 > 1. In the first case we immediately have max( p 1 −x2 1, p 1 −x2 2) ≥ √ 3/2; in the second case we can replace x1, x2 by the single number x1 +x2 −1 and use the induction hypothesis together with the previous lemma.
74 Hungary Now consider our polygon. Let P, Q be vertices such that l = PQ is maximal. The polygon consists of two paths from P to Q, each of integer length ≥l; these lengths are distinct since the perimeter is odd. Then the greater of the two lengths is m ≥l + 1. Position the polygon in the coordinate plane with P = (0, 0), Q = (l, 0) and the longer path in the upper half-plane. Since each side of the polygon has integer length, we can divide this path into line segments of length 1. Let the endpoints of these segments, in order, be P0 = P, P1 = (x1, y1), P2 = (x2, y2), . . . , Pm = Q. There exists some r such that yr is maximal; then either r ≥xr + 1/2 or (m −r) ≥(ℓ−xr) + 1/2.
Assume the former (otherwise, just reverse the choices of P and Q).
We already know that y1 ≥0, and by the maximal definition of l we must have x1 ≥0 as well; then since the polygon is convex we must have y1 ≤y2 ≤. . . ≤yr and x1 ≤x2 ≤. . . ≤xr.
But yi+1 −yi = p 1 −(xi+1 −xi)2, so yr = r−1 X i=0 (yi+1 −yi) = r−1 X i=0 p 1 −(xi+1 −xi)2 ≥ √ 3/2 by the second lemma. And we must have l ≥1, implying that triangle PPrQ has area at least √ 3/4.
Since this triangle lies within the polygon (by convexity), we are done.
Problem 16 Determine if there exists an infinite sequence of posi-tive integers such that (i) no term divides any other term; (ii) every pair of terms has a common divisor greater than 1, but no integer greater than 1 divides all of the terms.
Solution: The desired sequence exists. Let p0, p1, . . . be the primes greater than 5 in order, and let q3i = 6, q3i+1 = 10, q3i+2 = 15 for each nonnegative integer i. Then let si = piqi for all i ≥0. The sequence s0, s1, s2, . . . clearly satisfies (i) since si is not even divisible by pj for i ̸= j. For the first part of (ii), any two terms have their indices both in {0, 1}, both in {0, 2}, or both in {1, 2} (mod 3), so they have a common divisor of 2, 3, or 5, respectively. For the second part, we just need to check that no prime divides all the si; this holds since 2 ̸ | s2, 3 ̸ | s1, 5 ̸ | s0, and no prime greater than 5 divides more than one si.
1999 National Contests: Problems and Solutions 75 Problem 17 Prove that, for every positive integer n, there exists a polynomial with integer coefficients whose values at 1, 2, . . . , n are different powers of 2.
Solution: We may assume n ≥4.
For each i = 1, 2, . . . , n, write Qn j=1,j̸=i(i −j) = 2qimi for positive integers qi, mi with mi odd.
Let L be the least common multiple of all the qi, and let ri = L/qi. For each i, there are infinitely many powers of 2 which are congruent to 1 modulo |mri i |. (Specifically, by Euler’s theorem, 2φ(|m ri i |)j ≡1 (mod |mri i |) for all j ≥0.) Thus there are infinitely many integers ci such that cimri i + 1 is a power of 2; choose one.
Then define P(x) = n X i=1 ci n Y j=1 j̸=i (x −j) ri + 2L.
For each k, 1 ≤k ≤n, in the sum each term Qn j=1,j̸=i(x −j) ri vanishes for all i ̸= k. Then P(k) = ck n Y j=1 j̸=k (k −j) ri + 2L = 2L(cimri i + 1), a power of 2. Moreover, by choosing the ci appropriately, we can guarantee that these values are all distinct, as needed.
Problem 18 Find all integers N ≥3 for which it is possible to choose N points in the plane (no three collinear) such that each triangle formed by three vertices on the convex hull of the points contains exactly one of the points in its interior.
Solution: First, if the convex hull is a k-gon, then it can be divided into k −2 triangles each containing exactly one chosen point; and since no three of the points are collinear, the sides and diagonals of the convex hull contain no chosen points on their interiors, giving N = 2k −2.
Now we construct, by induction on k ≥3, a convex k-gon with a set S of k−2 points inside such that each triangle formed by vertices of the k-gon contains exactly one point of S in its interior. The case k = 3 is easy. Now, assume we have a k-gon P1P2 . . . Pk and a set S. Certainly we can choose Q such that P1P2 . . . PkQ is a convex (k + 1)-gon. Let 76 Hungary R move along the line segment from Pk to Q. Initially (at R = Pk), for any indices 1 ≤i < j < k, the triangle PiPjR internally contains a point of S by assumption; if R is moved a sufficiently small distance dij, this point still lies inside triangle PiPjR. Now fix a position of R such that PkR is less than the minimum dij; P1P2 . . . PkR is a convex (k + 1)-gon. Let P be an interior point of the triangle bounded by lines P1Pk, RPk−1, PkR. We claim the polygon P1P2 . . . PkR and the set S ∪{P} satisfy our condition. If we choose three of the Pi, they form a triangle containing a point of S by hypothesis, and no others; any triangle PiPjR (i, j < k) contains only the same internal point as triangle PiPjPk; and each triangle PiPkR contains only P. This completes the induction step.
1999 National Contests: Problems and Solutions 77 1.10 Iran First Round Problem 1 Suppose that a1 < a2 < · · · < an are real numbers.
Prove that a1a4 2 + a2a4 3 + · · · + ana4 1 ≥a2a4 1 + a3a4 2 + · · · + a1a4 n.
First Solution: We prove the claim by induction on n. For n = 2, the two sides are equal; now suppose the claim is true for n −1, i.e., a1a4 2 + a2a4 3 + · · · + an−1a4 1 ≥a2a4 1 + a3a4 2 + · · · + a1a4 n−1.
Then the claim for n will follow from the inequality an−1a4 n + ana4 1 −an−1a4 1 ≥ana4 n−1 + a1a4 n −a1a4 n−1 (Notice that this is precisely the case for n = 3.) Without loss of generality, suppose an −a1 = 1; otherwise, we can divide each of a1, an−1, an by an −a1 > 0 without affecting the truth of the inequality.
Then by Jensen’s inequality for the convex function x4, we have a4 1(an −an−1) + a4 n(an−1 −a1) ≥(a1(an −an−1) + an(an−1 −a1))4 = (an−1(an −a1))4 = a4 n−1(an −a1), which rearranges to yield our desired inequality.
Second Solution: We use an elementary method to prove the case n = 3. Define p(x, y, z) = xy4 + yz4 + zx4 −yx4 −zy4 −xz4.
We wish to prove that p(x, y, z) ≥0 when x ≤y ≤z. Since p(x, x, z) = p(x, y, y) = p(z, y, z) = 0, we know that (y −x)(z − y)(z −x) divides p(x, y, z). In fact, p(x, y, z) = yz4 −zy4 + zx4 −xz4 + xy4 −yx4 = zy(z3 −y3) + xz(x3 −z3) + xy(y3 −x3) = zy(z3 −y3) + xz(y3 −z3) + xz(x3 −y3) + xy(y3 −x3) = z(y −x)(z3 −y3) + x(z −y)(x3 −y3) 78 Iran = (y −x)(z −y) z(z2 + zy + y2) −x(x2 + xy + y2) = (y −x)(z −y) (z3 −x3) + y2(z −x) + y(z2 −x2) = (y −x)(z −y)(z −x) z2 + zx + x2 + y2 + yz + yx = 1 2(y −x)(z −y)(z −x) (x + y)2 + (y + z)2 + (z + x)2 ≥0, as desired.
Problem 2 Suppose that n is a positive integer.
The n-tuple (a1, . . . , an) of positive integers is said to be good if a1 +· · ·+an = 2n if for every k between 1 and n, no k of the n integers add up to n.
Find all n-tuples that are good.
First Solution: Call an n-tuple of positive integers proper if the integers add up to 2n.
Without loss of generality, we suppose that a1 ≤a2 ≤· · · ≤an = M. If M ≤2 then a1 = · · · = an = 2 and this leads to the solution (2, 2, . . . , 2) for odd n.
Now we suppose that M ≥3.
Since the average of {a1, . . . , an} is 2, we must have a1 = 1. Now say we have a proper n-tuple S = (a1, . . . , an), where 1 = a1 = · · · = ai < ai+1 ≤ai+2 ≤· · · ≤an = M.
Lemma. If i ≥max{an −ai+1, ai+1}, then S is not good.
Proof: Suppose that we have a balance and weights a1, a2, . . . , an.
We put an on the left hand side of the balance, then put an−1 on the right hand side, and so on — adding the heaviest available weight to the lighter side (or if the sides are balanced, we add it to the left hand side). Before we put xi+1 on the balance, the difference between the two sides is between 0 and xn. After we put xi+1 on the balance, the difference between the two sides is no greater than max{xn −xi+1, xi+1}. Now we have enough 1’s to put on the lighter side to balance the two sides. Since the the total weight is 2n an even number, there will have even number of 1’s left and we can split them to balance the sides, i.e., there is a subtuple A with its sum equal to n and thus S is not good.
1999 National Contests: Problems and Solutions 79 Note that 2n = a1 + · · · + an ≥i + 2(n −i −1) + M ⇐ ⇒i ≥M −2.
(1) Now we consider the following cases: (i) M = 3 and i = 1.
Then S = (1, 2, . . . , 2 | {z } n−2 , 3). If n = 2m and m ≥2, then A = (2, . . . , 2 | {z } m ) has sum 2m = n and thus S is not good; if n = 2m + 1 and m ≥1, then A = (2, . . . , 2 | {z } m−1 , 3) has sum 2m + 1 = n and thus S is not good. Therefore (1, 3) is the only good tuple for M = 3, i = 1.
(ii) M = 3 and i ≥2.
Then ai+1 = 2 or 3 and i ≥max{an − ai+1, ai+1} — implying that S is not good from our lemma — unless i = 2 and ai+1 = 3. But then S = (1, 1, 3, 3), and (1, 3) has sum 4 = n.
(iii) M ≥4 and ai+1 = 2. Then from (1), i ≥M −2 = max{an − ai+1, ai+1}. By the lemma S is not good.
(iv) M ≥4, ai+1 > 2, and i + 1 ̸= n. Since ai+1 ̸= 2 equality does not hold in (1), and thus i ≥M −1 ≥max{an −ai+1, ai+1} (and S is not good) unless ai+1 = M.
In this case, S = (1, . . . , 1, M, . . . , M). Note that 2n = i + (n −i)M ≥i + 4(n −i) so that i ≥2 3n. Then the remaining n−i ≥2 values M have sum at most 4 3n, and hence M ≤2 3n. Thus i ≥2n/3 ≥M = max{an −ai+1, ai+1}, and by the lemma S is not good.
(v) M ≥4 and i+1 = n. Then we have the good n-tuple (1, . . . , 1, n+ 1).
Therefore the only possible good n-tuples are (1, 1, . . . , 1, n+1) and (2, 2, . . . , 2), and the second n-tuple is good if and only if n is odd.
Second Solution: Say a proper n-tuple has “subsum” m if some k of the integers (0 ≤k ≤n) add up to m.
80 Iran Lemma. Every proper n-tuple besides (2, 2, . . . , 2) has subsums 0, 1, . . . , n −1 (and possibly others). Furthermore, if a proper n-tuple besides (2, 2, . . . , 2) contains a 2, it has subsum n as well.
Proof: If n = 1 the claim is trivial. Now assume the claims are true for n −1; we prove each is true for n as well. Suppose we have an n-tuple N besides (2, 2, . . . , 2).
If there is a 2 in the n-tuple, then the other n −1 integers form a proper (n−1)-tuple besides (2, 2, . . . , 2). By the induction hypothesis, this (n−1)-tuple has subsums 0, . . . , n−2. Remembering the original 2, our complete n-tuple N has subsums 0, . . . , n.
Otherwise, suppose there is no 2 in N. We prove by induction on k < n that there is a subtuple Ak of k numbers with subsums 0, . . . , k.
Since the average value of the integers in N is 2, we must have at least one 1 in N, proving the case for k = 1. Now assume the claim is true for k −1; the elements of Ak−1 are each at least 1, so they add up to at least k −1. Then the inequality (k −1)(k −n) < 0 implies 2n < (k + 1)(n −k + 1) + (k −1), so that the average value of N \ Ak−1 is less than k + 1. Therefore at least one of the other n −k + 1 integers x is at most k, implying that Ak−1 ∪(x) has subsums 0, . . . , k. This completes the inductive step and the proof of our lemma.
From our lemma, the only possible good n-tuple with a 2 is (2, 2, . . . , 2), and this is good if and only if n is odd. Every other n-tuple N has a subtuple An−1 of n −1 integers with subsums 0, . . . , n −1. If these integers are not all 1, then the remaining integer is at most n and N must have subsum n. Therefore, the only other possible good n-tuple is (1, 1, . . . , 1, n + 1), which is indeed always a good n-tuple.
Third Solution: Suppose we have a good n-tuple (a1, . . . , an), and consider the sums a1, a1 + a2, . . . , a1 + a2 + · · · + an−1. All these sums are between 0 and 2n exclusive; thus if any of the sums is 0 (mod n), it equals n and we have a contradiction. Also, if any two are congruent modulo n, we can subtract these two sums to obtain 1999 National Contests: Problems and Solutions 81 another partial sum that equals n, a contradiction again. Therefore, the sums must all be nonzero and distinct modulo n.
Specifically, a2 ≡a1 + · · · + ak (mod n) for some k ≥1. If k > 1 then we can subtract a2 from both sides to find a partial sum that equals n. Therefore k = 1 and a1 ≡a2 (mod n). Similarly, all the ai are congruent modulo n. From here, easy algebra shows that the presented solutions are the only ones possible.
Problem 3 Let I be the incenter of triangle ABC and let AI meet the circumcircle of ABC at D. Denote the feet of the perpendiculars from I to BD and CD by E and F, respectively. If IE+IF = AD/2, calculate ∠BAC.
Solution: A well-known fact we will use in this proof is that DB = DI = DC. In fact, ∠BDI = ∠C gives ∠DIB = (∠A + ∠B)/2 while ∠IBD = (∠A + ∠B)/2. Thus DB = DI, and similarly DC = DI.
Let θ = ∠BAD. Then 1 4ID · AD = 1 2ID · (IE + IF) = 1 2BD · IE + 1 2CD · IF = [BID] + [DIC] = ID AD([BAD] + [DAC]) = 1 2ID · (AB + AC) · sin θ, whence AD AB+AC = 2 sin θ.
Let X be the point on − − → AB different from A such that DX = DA.
Since ∠XBD = ∠DCA and ∠DXB = ∠XAD = ∠DAC, we have △XBD ∼ = △ACD, and BX = AC.
Then 2 sin θ = AD AB+AC = AD AB+BX = AD AX = 1 2 cos θ, so that 2 sin θ cos θ = 1 2, and ∠BAC = 2θ = 30◦or 150◦.
Problem 4 Let ABC be a triangle with BC > CA > AB. Choose points D on BC and E on − − → BA such that BD = BE = AC.
The circumcircle of triangle BED intersects AC at P and the line BP intersects the circumcircle of triangle ABC again at Q. Prove that AQ + QC = BP.
First Solution: Except where indicated, all angles are directed modulo 180◦.
82 Iran Let Q′ be the point on line BP such that ∠BEQ′ = ∠DEP. Then ∠Q′EP = ∠AED −∠BEQ′ + ∠DEP = ∠BED.
Since BE = BD, ∠BED = ∠EDB; since BEPD is cyclic, ∠EDB = ∠EPB. Therefore ∠Q′EP = ∠EPB = ∠EPQ′ and Q′P = Q′E.
Since BEPD and BAQC are cyclic, we have ∠BEQ′ = ∠DEP = ∠DBP = ∠CAQ, ∠Q′BE = ∠QBA = ∠QCA.
Combining this with BE = AC yields that triangles EBQ′ and ACQ are congruent. Thus BQ′ = QC and EQ′ = AQ. Therefore AQ + QC = EQ′ + BQ′ = PQ′ + BQ′, which equals BP if Q′ is between B and P.
Since E is on − − → BA and P is on − → CA, E and P are on the same side of BC and thus BD. And since D is on − − → BC and P is on − → AC, D and P are on the same side of BA and thus BE. Thus, BEPD is cyclic in that order and (using undirected angles) ∠BEQ′ = ∠DEP < ∠BEP. It follows that Q′ lies on segment BP, as desired.
Second Solution: Since BEPD and BAQC are cyclic, we have ∠PED = ∠PBD = ∠QBC = ∠QAC and ∠EPD = π −∠DBE = π −∠CBA = ∠AQC, which together imply △PED ∼△QAC. Then AC · EP DE = AQ and AC · PD DE = QC.
As in the first solution, BEPD is cyclic in that order, so Ptolemy’s Theorem implies that BD · EP + BE · PD = BP · DE.
Substituting BD = BE = AC we have AC · EP DE + AC · PD DE = BP, 1999 National Contests: Problems and Solutions 83 or AQ + QC = BP, as desired.
Problem 5 Suppose that n is a positive integer and let d1 < d2 < d3 < d4 be the four smallest positive integer divisors of n. Find all integers n such that n = d2 1 + d2 2 + d2 3 + d2 4.
Solution: The answer is n = 130. Note that x2 ≡0 (mod 4) when x is even, and 1 (mod 4) when x is odd.
If n is odd, then all the di are odd and n ≡d2 1 + d2 2 + d2 3 + d2 4 ≡ 1 + 1 + 1 + 1 ≡0 (mod 4), a contradiction. Thus 2 | n.
If 4 | n then d1 = 1 and d2 = 2, and n ≡1 + 0 + d2 3 + d2 4 ̸≡0 (mod 4), a contradiction. Thus 4̸ | n.
Therefore {d1, d2, d3, d4} = {1, 2, p, q} or {1, 2, p, 2p} for some odd primes p, q. In the first case n ≡3 (mod 4), a contradiction. Thus n = 5(1 + p2) and 5 | n, so p = d3 = 5 and n = 130.
Problem 6 Suppose that A = (a1, a2, . . . , an) and B = (b1, b2, . . . , bn) are two 0-1 sequences. The difference d(A, B) between A and B is defined to be the number of i’s for which ai ̸= bi (1 ≤i ≤n). Suppose that A, B, C are three 0-1 sequences and that d(A, B) = d(A, C) = d(B, C) = d.
(a) Prove that d is even.
(b) Prove that there exists an 0-1 sequence D such that d(D, A) = d(D, B) = d(D, C) = d 2.
Solution: (a) Modulo 2, we have d(A, B) = (a1 −b1) + (a2 −b2) + · · · + (an −bn) ≡(a1 + a2 + · · · + an) + (b1 + b2 + · · · + bn).
Thus, 3d ≡d(A, B) + d(B, C) + d(C, A) = 2( X ai + X bi + X ci), so d must be divisible by 2.
84 Iran (b) Define D as follows: for each i, if ai = bi = ci, then let di = ai = bi = ci. Otherwise, two of ai, bi, ci are equal; let di equal that value. We claim this sequence D satisfies the requirements.
Let α be the number of i for which ai ̸= bi and ai ̸= ci (that is, for which ai is “unique”). Define β and γ similarly, and note that d(A, D) = α, d(B, D) = β, and d(C, D) = γ. We also have d = d(A, B) = α + β d = d(B, C) = β + γ d = d(C, A) = γ + α.
Thus, α = β = γ = d 2, as desired.
Second Round Problem 1 Define the sequence {xn}n≥0 by x0 = 0 and xn = xn−1 + 3r+1 −1 2 , if n = 3r(3k + 1), xn−1 −3r+1 + 1 2 , if n = 3r(3k + 2), where k and r are nonnegative integers.
Prove that every integer appears exactly once in this sequence.
First Solution: We prove by induction on t ≥1 that (i) {x0, x1, . . . , x3t−2} = n −3t−3 2 , −3t−1 2 , . . . , 3t−1 2 o .
(ii) x3t−1 = −3t−1 2 .
These claims imply the desired result, and they are easily verified for t = 1. Now supposing they are true for t, we show they are true for t + 1.
For any positive integer m, write m = 3r(3k + s) for nonnegative integers r, k, s, with s ∈{1, 2}; and define rm = r and sm = s.
Then for m < 3t, observe that rm = rm+3t = rm+2·3t sm = sm+3t = rm+2·3t, so that xm −xm−1 = x3t+m −x3t+m−1 = x2·3t+m −x2·3t+m−1.
1999 National Contests: Problems and Solutions 85 Adding these equations from m = 1 to m = k < 3t, we have xk = x3t+k −x3t xk = x2·3t+k −x2·3t.
Now, setting n = 3t in the recursion and using (ii) from the induction hypothesis, we have x3t = 3t — and {x3t, . . . , x2·3t−2} = n 3t+3 2 , . . . , 3t+1−1 2 o x2·3t−1 = 3t+1 2 .
Then setting n = 2 · 3t in the recursion we have x2·3t = −3t — giving {x2·3t, . . . , x3t+1−2} = n −3t+1−3 2 , . . . , −3t+1 2 o x2·3t+1−1 = −3t+1−1 2 .
Combining this with (i) and (ii) from the induction hypothesis proves the claims for t + 1. This completes the proof.
Second Solution: For ni ∈{−1, 0, 1}, let the number [nmnm−1 · · · n0] in “base ¯ 3” equal Pm i=0 ni · 3i. It is simple to prove by induction on k that the base ¯ 3 numbers with at most k digits equal −3k −1 2 , −3k −3 2 , . . . , 3k −1 2 , which implies every integer has a unique representation in base ¯ 3.
Now we prove by induction on n that if n = amam−1 . . . a0 in base 3, then xn = [bmbm−1 . . . b0] in base ¯ 3, where bi = −1 if ai = 2 and bi = ai for all other cases.
For the base case, x0 = 0 = . Now assume the claim is true for n −1. First suppose that n = 3r(3k + 1). Then n = amam−1 . . . ai 1 0 0 . . . 0 | {z } r 3r+1 −1 2 = 1 1 . . . 1 | {z } r+1 = [1 1 . . . 1 | {z } r+1 ] n −1 = amam−1 . . . ai 0 2 2 . . . 2 | {z } r 86 Iran xn−1 = [bmbm−1 . . . bi 0 −1 −1 . . . −1 | {z } r ], so that xn = [bmbm−1 . . . bi 0 −1 −1 . . . −1 | {z } r ] + [1 1 . . . 1 | {z } r+1 ] = [bmbm−1 . . . bi 1 0 0 . . . 0 | {z } r ].
Now suppose that n = 3r(3k + 2). Then n = amam−1 . . . ai 2 0 0 . . . 0 | {z } r n −1 = amam−1 . . . ai 1 2 2 . . . 2 | {z } r xn−1 = [bmbm−1 . . . bi 1 −1 −1 . . . −1 | {z } r ].
Also, −3r+1 + 1 2 = −(1 1 . . . 1 | {z } r 2) = −3r −3r−1 −· · · −3 −2 = −3r+1 + 3r + 3r−1 + · · · + 3 + 1 = [−1 1 1 . . . 1 | {z } r+1 ].
Therefore xn = [bmbm−1 . . . bi 1 −1 −1 . . . −1 | {z } r ] + [−1 1 1 . . . 1 | {z } r+1 ] = [bmbm−1 . . . bi −1 0 0 . . . 0 | {z } r ].
In either case, the claim is true for n, completing the induction.
And since all integers appear exactly once in base ¯ 3, they appear exactly once in {xn}n≥0, as desired.
Problem 2 Suppose that n(r) denotes the number of points with integer coordinates on a circle of radius r > 1. Prove that n(r) < 6 3 √ πr2.
1999 National Contests: Problems and Solutions 87 Solution: Consider a circle of radius r containing n lattice points; we must prove that n < 6 3 √ πr2.
Since r > 1 and 6 3 √π > 8, we may assume n > 8. Label the n lattice points on the circle P1, P2, . . . , Pn in counterclockwise order.
Since the sum of the (counterclockwise) arcs P1P3, P2P4, PnP2 is 4π, one of the arcs PiPi+2 has measure at most 4π n ; assume without loss of generality it is arc P1P3.
Consider a triangle ABC inscribed in an arc of angle 4π n ; clearly its area is maximized by moving A and C to the endpoints of the arc and then moving B to the midpoint (where the distance to line AC is greatest). Then ∠CAB = ∠BCA = π n and ∠ABC = 180◦−2π n , so [ABC] = abc 4r = (2r sin π n)(2r sin 2π n )(2r sin π n) 4r ≤(2r π n)(2r 2π n )(2r π n) 4r = 4r2π3 n3 .
Since triangle P1P2P3 is inscribed in an arc of measure 4π n , by the preceding argument, [P1P2P3] ≤4r2π3 n3 . But since P1, P2, and P3 are lattice points, the area [P1P2P3] is at least 1 2 (this can be proven by either Pick’s Formula K = I + 1 2B −I or the “determinant formula” K = 1 2 |x1y2 −x2y1 + x2y3 −x3y2 + x3y1 −x1y3|). Therefore, 1 2 ≤[P1P2P3] ≤4r2π3 n3 = ⇒n3 ≤8r2π3 = ⇒n ≤ 3 √ 8r2π3 = 2π 3 √ r2 < 6 3 √ πr2, as desired.
Problem 3 Find all functions f : R →R satisfying f(f(x) + y) = f(x2 −y) + 4f(x)y for all x, y ∈R.
Solution: Let (x, y) = (x, x2). Then f(f(x) + x2) = f(0) + 4x2f(x).
(1) 88 Iran Let (x, y) = (x, −f(x)). Then f(0) = f(x2 + f(x)) −4f(x)2.
(2) Adding (1) and (2) gives 4f(x)(f(x) −x2) = 0. This implies that for each individual x, either f(x) = 0 or f(x) = x2. (Alternatively, plugging y = x2−f(x) 2 into the original equation also yields this result.) Clearly f(x) = 0 and f(x) = x2 satisfy the given equation; we now show that f cannot equal some combination of the two functions.
Suppose that there is an a ̸= 0 such that f(a) = 0. Plugging in x = a into the original equation, we have f(y) = f(a2 −y).
If y ̸= a2 2 , then y2 ̸= (a2 −y)2 so f(y) = f(a2 −y) = 0. Thus f(y) = 0 for all y ̸= a2 2 . And by choosing x = 2a or some other value in the original equation, we can similarly show that f( a2 2 ) = 0.
Therefore f(x) = 0 for all x or f(x) = x2 for all x, as claimed.
Problem 4 In triangle ABC, the angle bisector of ∠BAC meets BC at D. Suppose that ω is the circle which is tangent to BC at D and passes through A. Let M be the second point of intersection of ω and AC. Let P be the second point of intersection of ω and BM.
Prove that P lies on a median of triangle ABD.
Solution: Extend AP to meet BD at E. We claim that BE = ED and thus AP is a median of triangle ABD, as desired. In fact, BE = ED ⇐ ⇒BE2 = ED2 = EP · EA ⇐ ⇒△BEP ∼△AEB ⇐ ⇒∠EBP = ∠BAE.
Let N be the second intersection of ω with AB. Using directed angles and arc measures, since AD bisects the angle between lines AN and AC, we have d DM = d ND and ∠BAE = ∠NAP = d ND −d PD 2 = d DM −d PD 2 = ∠DBM = ∠EBP, as desired.
Problem 5 Let ABC be a triangle. If we paint the points of the plane in red and green, prove that either there exist two red points which are one unit apart or three green points forming a triangle congruent to ABC.
1999 National Contests: Problems and Solutions 89 First Solution: We call a polygon or a segment green (red) if the vertices of the polygon or the segment are all green (red).
Suppose that there is no red unit segment. We prove that there is a green triangle congruent to triangle ABC. If the whole plane is green, the proof is trivial.
Now we further suppose that there is a red point R on the plane. We claim that there is a green equilateral triangle with unit side length.
In fact, let ω be the circle with center R and radius √ 3. Then ω is not all red, since otherwise we could find a red unit segment. Let G be a green point on ω. Let ω1 and ω2 be two unit circles centered at R and G, respectively, and let ω1 and ω2 meet at P and Q. Then both P and Q must be green and triangle PQG is a green unit equilateral triangle.
Let G1G2G3 be a green unit equilateral triangle.
Construct a triangle G1X1Y1 that is congruent to triangle ABC. If both X1 and Y1 are green, we are done. Without loss of generality, we assume that X1 Y1 is red. Translate triangle G1G2G3 by − − − → G1Y1 to obtain triangle Y1Y2Y3. Then both Y2 and Y3 are green. Similarly, translate triangle G1G2G3 by − − − → G1X1 to obtain triangle X1X2X3.
Then at least one of X2 and X3 is green (since X2X3 cannot be a red unit segment).
Without loss of generality, say X2 is green. Now triangle G2X2Y2 is a green triangle and congruent to triangle G1X1Y1 (translated by − − − → G1G2) and thus congruent to triangle ABC, as desired.
Second Solution: Suppose by way of contradiction there were no such red or green points, and say the sides of triangle ABC are a, b, and c.
First we prove no red segment has length a. If XY were a red segment of length a, then the unit circles around X and Y must be completely green. Now draw Z so that △XY Z ∼ = △ABC; the unit circle around Z must be completely red, or else it would form an illegal triangle with the corresponding points around X and Y. But on this unit circle we can find a red unit segment, a contradiction.
Now, the whole plane cannot be green so there must be some red point R. The circle ω around R with radius a must be completely green. Then pick two points D, E on ω with DE = a, and construct F outside ω so that △DEF ∼ = △ABC (we can do this since a ≤b, c); F must be red. Thus if we rotate DE around R, F forms a completely 90 Iran red circle of radius greater than a — and on this circle we can find two red points distance a apart, a contradiction.
Third Round Problem 1 Suppose that S = {1, 2, . . . , n} and that A1, A2, . . . , Ak are subsets of S such that for every 1 ≤i1, i2, i3, i4 ≤k, we have |Ai1 ∪Ai2 ∪Ai3 ∪Ai4| ≤n −2.
Prove that k ≤2n−2.
Solution: For a set T, let |T| denote the numbers of elements in T. We call a set T ⊂S 2-coverable if T ⊆Ai ∪Aj for some i and j (not necessarily distinct).
Among the subsets of S that are not 2-coverable, let A be a subset with minimum |A|.
Consider the family of sets S1 = {A ∩A1, A ∩A2, . . . , A ∩Ak}.
(A ∩Ai might equal A ∩Aj, but we ignore any duplicate sets.) Since A is not 2-coverable, if X ∈S1, then A −X ̸∈S1. Thus at most half the subsets of |A| are in S1, and |S1| ≤2|A|−1.
On the other hand, let B = S −A and consider the family of sets S2 = {B ∩A1, B ∩A2, . . . , B ∩Ak}. We claim that if X ∈S2, then B −X ̸∈S2. Suppose on the contrary that both X, B −X ∈S2 for some X = B ∩Aℓand B −X = B ∩Aℓ′. By the minimal definiton of A there are Ai and Aj such that Ai ∪Aj = A \ {m} for some i, j, and m. Then |Aℓ∪Aℓ′ ∪Ai ∪Aj| = n −1, a contradiction. Thus we assumption is false and |S2| ≤2|B| −1 = 2n−|A|−1.
Since every set Ai is uniquely determined by its intersection with sets A and B = S −A, it follows that |A| ≤|B| · |C| ≤2n−2.
Problem 2 Let ABC be a triangle and let ω be a circle passing through A and C. Sides AB and BC meet ω again at D and E, respectively. Let γ be the incircle of the circular triangle EBD and let S be its center. Suppose that γ touches the arc DE at M. Prove that the angle bisector of ∠AMC passes through the incenter of triangle ABC.
First Solution: We work backward.
Let I be the incenter of triangle ABC. Let N be the midpoint of arc AC on ω that is opposite 1999 National Contests: Problems and Solutions 91 to B, and let − → NI meet with ω again at M ′.
Then M ′N bisects ∠AM ′C. We claim that γ is tangent to ω at M ′, and our desired results follows.
To prove our claim, we are going to do some heavy trigonometry calculations. Let ℓbe the line tangent to ω at M ′, and let ℓmeet AB and AC at P and Q, respectively. Let ω′ be the incircle of triangle PBQ. We are reduced to proving that ω′ is tangent to PQ at M ′.
Let ∠IAM ′ = a, ∠M ′AB = x, ∠M ′CI = b, ∠BCM ′ = y. Then ∠CAI = ∠IAB = x + a and ∠ICA = ∠BCI = y + b. Since PM ′ is tangent to ω, we have ∠PM ′A = ∠M ′CA = 2b + y and thus ∠BPQ = ∠PM ′A + ∠M ′AP = 2b + x + y. Applying the law of sines to triangle PAM ′, we have PM ′ sin x = AM ′ sin(2b + x + y) ⇐ ⇒PM ′ = AM ′ sin x sin(2b + x + y).
(1) Similarly, M ′Q = M ′C sin y sin(2a + x + y).
(2) And, applying the law of sines to triangle AM ′C gives AM ′ sin(2b + y) = M ′C sin(2a + x).
(3) Combining (1), (2), and (3) we have PM ′ M ′Q = sin x sin (2b + y) sin (2a + x + y) sin y sin (2a + x) sin (2b + x + y).
(4) Now, observe that for a triangle XY Z with inradius r, and with incircle Γ touching XY at T, we have XT TY = r cot ∠X 2 r cot ∠Y 2 = cot ∠X 2 cot ∠Y 2 .
Thus it suffices to prove that P M ′ M ′Q = cot ∠BP Q cot ∠BQP , or equivalently (from (4)) any of the following statements: sin x sin (2b + y) sin (2a + x + y) sin y sin (2a + x) sin (2b + x + y) = cot 2b+x+y 2 cot 2a+x+y 2 sin x sin (2b + y) sin 2a+x+y 2 cos 2a+x+y 2 sin y sin (2a + x) sin 2b+x+y 2 cos 2b+x+y 2 = cos 2b+x+y 2 sin 2a+x+y 2 cos 2a+x+y 2 sin 2b+x+y 2 92 Iran sin x sin (2b + y) cos2 2a + x + y 2 = sin y sin (2a + x) cos2 2b + x + y 2 sin x sin (2b + y) (cos (2a + x + y) + 1) = sin y sin (2a + x) (cos (2b + x + y) + 1) , or equivalently that L = R where R = sin x sin (2b + y) cos (2a + x + y) −sin y sin (2a + x) cos (2b + x + y), L = sin y sin (2a + x) −sin x sin (2b + y).
Note that R = 1/2[ sin x (sin (2b + 2y + x + 2a) + sin (2b −2a −x)) ] −1/2[ sin y (sin (2b + 2x + y + 2a) + sin (2a −2b −y)) ] = −1/4[ cos (2a + 2b + 2x + 2y) −cos (2a + 2b + 2y) + cos (2b −2a) −cos (2b −2a −2x) −cos (2a + 2b + 2x + 2y) + cos (2b + 2a + 2x) −cos (2b −2a) + cos (2a −2b −2y) ] = −1/4[ cos (2a + 2b + 2x) −cos (2b −2a −2x) + cos (2a −2b −2y) −cos (2a + 2b + 2y) ] = 1/2[ sin(2b) sin (2a + 2x) −sin(2a) sin (2b + 2y) ] = 2[ sin b sin (a + x) | {z } (i) cos b cos (a + x) −sin a sin (b + y) | {z } (ii) cos a cos (b + y) ].
To simplify this expression, note that MI, AI, and CI meet at I and that ∠IMA = ∠IMC. Then the trigonometric form of Ceva’s Theorem gives sin ∠IMA sin ∠IAC sin ∠ICM sin ∠IAM sin ∠ICA sin ∠IMC = 1 ⇐ ⇒sin a sin (b + y) = sin b sin (a + x).
1999 National Contests: Problems and Solutions 93 Swapping quantities (i) and (ii) above thus yields R = 2 (sin a sin (b + y) cos b cos (a + x) −sin b sin (a + x) cos a cos (b + y)) = cos b sin (b + y)[sin (2a + x) −sin x] −sin b cos (b + y)[sin (2a + x) + sin x] = sin (2a + x)[ sin (b + y) cos b −sin b cos (b + y) ] −sin x[ sin (b + y) cos b + sin b cos (b + y) ] = sin (2a + x) sin y −sin x sin (2b + y) = L, as desired.
Second Solution: Let O and R be the center and radius of ω, r be the radius of γ, and I be the incenter of triangle ABC. Extend lines AI and CI to hit ω at MA and MC respectively; also let line AD be tangent to γ at F, and let line CE be tangent to γ at G. Finally, let d be the length of the exterior tangent from MA to ω. Notice that since line AMA bisects ∠DAC, we have DMA = MAC; similarly, EMC = MCA.
Applying Generalized Ptolemy’s Theorem to the “circles” MA, C, D, and γ externally tangent to ω gives CG · DMA = MAC · DF + d · CD d2 = MAC2 CG−DF CD 2 .
Note that d2 equals the power of MA with respect to γ, so d2 = MAS2 −r2.
By Stewart’s Theorem on cevian MAM in triangle SOMA, we also have MAS2 · OM + MAO2 · MS = MAM 2 · SO + SM · MO · SO MAS2 · R + R2 · r = MAM 2 · (R + r) + r · R · (R + r) MAM 2(R + r) = (MAS2 −r2)R = d2R, Combining the two equations involving d2, we find MAC2 CG −DF CD 2 = MAM 2(R + r) R 94 Iran MAM MAC 2 = R R + r CG −DF CD 2 .
Similarly, MCM MCA 2 = R R + r AF −EG AE 2 .
But CG −DF = (CG + GB) −(DF + FB) = CB −DB and similarly AF −EG = (AF + FB) −(EG + GB) = AB −BE.
Furthermore, because ACDE is cyclic some angle-chasing gives ∠BDC = ∠AEC and ∠DCB = ∠BAE, so △CBD ∼△ABE and CG −DF CD = CB −DB CD = AB −BE EA = AF −EG AE .
Therefore we have MAM MAC = MCM MCA = ⇒sin ∠MAMA sin ∠MAAC = sin ∠MCCM sin ∠ACMC . But by the trigonometric form of Ceva’s theorem in triangle AMC applied to lines AMA, CMC, and MI, we have sin ∠MAMA sin ∠MAAC · sin ∠ACMC sin ∠MCCM · sin ∠CMI sin ∠IMA = 1 so that sin ∠CMI = sin ∠IMA = ⇒ ∠CMI = ∠IMA since ∠AMC < 180◦. Therefore, line MI bisects ∠AMC, so the angle bisector of ∠AMC indeed passes through the incenter I of triangle ABC.
Problem 3 Suppose that C1, C2, . . . , Cn are circles of radius 1 in the plane such that no two of them are tangent and the subset of the plane formed by the union of these circles is connected (i.e., for any partition of {1, 2, . . . , n} into nonempty subsets A and B, S a∈A Ca and S b∈B Cb are not disjoint). Prove that |S| ≥n, where S = [ 1≤i<j≤n Ci ∩Cj, the intersection points of the circles. (Each circle is viewed as the set of points on its circumference, not including its interior.) 1999 National Contests: Problems and Solutions 95 Solution: Let T = {C1, C2, . . . , Cn}. For every s ∈S and C ∈T define f(s, C) = 0, if s ̸∈C, 1 k, if s ∈C, where k is the number of circles passing through s (including C).
Thus X C∈T f(s, C) = 1 for every s ∈S.
On the other hand, for a fixed circle C ∈T let s0 ∈S ∩C be a point such that f(s0, C) = min{f(s, C) | s ∈S ∩C}.
Suppose that C, C2, . . . Ck be the circles which pass through s0. Then C meets C2, . . . , Ck again in distinct points s2, . . . , sk. Therefore X s∈C f(s, C) ≥1 k + k −1 k = 1.
We have |S| = X s∈S X C∈T f(s, C) = X C∈T X s∈S f(s, C) ≥n, as desired.
Problem 4 Suppose that −1 ≤x1, x2, . . . , xn ≤1 are real numbers such that x1+x2+. . .+xn = 0. Prove that there exists a permutation σ of {1, 2, . . . , n} such that, for every 1 ≤p ≤q ≤n, |xσ(p) + · · · + xσ(q)| < 2 −1 n.
Also prove that the expression on the right hand side cannot be replaced by 2 −4 n.
Solution: If n = 1 then x1 = 0, and the permutation σ(1) = 1 suffices; if n = 2 then |x1|, |x2| ≤1 and |x1 + x2| = 0, and the permutation (σ(1), σ(2)) = (1, 2) suffices. Now assume n ≥3.
View the xi as vectors; the problem is equivalent to saying that if we start at a point on the number line, we can travel along the n vectors x1, x2, . . . , xn in some order so that we stay within an interval (m, m + 2 −1 n].
96 Iran Call xi “long” if |xi| ≥1−1 n and call it “short” otherwise. Also call xi “positive” if xi ≥0 and “negative” if xi < 0. Suppose without loss of generality that there are at least as many long positive vectors as long negative vectors — otherwise, we could replace each xi by −xi.
We make our trip in two phases: (i) First, travel alternating along long positive vectors and long negative vectors until no long negative vectors remain. Suppose at some time we are at a point P. Observe that during this leg of our trip, traveling along a pair of vectors changes our position by at most 1 n in either direction. Thus if we travel along 2t ≤n vectors after P, we stay within t n ≤1 2 of P; and if we travel along 2t + 1 vectors after P, we stay within t n + 1 ≤3 2 < 2 −1 n of P.
Therefore, during this phase, we indeed stay within an interval I = m, m + 2 −1 n of length 2 −1 n.
(ii) After phase (i), we claim that as long as vectors remain unused and we are inside I, there is an unused vector we can travel along while remaining in I; this implies we can finish the trip while staying in I.
If there are no positive vectors, then we can travel along any negative vector, and vice versa. Thus assume there are positive and negative vectors remaining; since all the long negative vectors were used in phase (i), only short negative vectors remain.
Now if we are to the right of m + 1 −1 n, we can travel along a short negative vector without reaching or passing m. And if we are on or to the left of m + 1 −1 n, we can travel along a positive vector (short or long) without passing m + 2 −1 n.
Therefore it is possible to complete our journey, and it follows that the desired permutation σ indeed exists.
However, suppose 1 n is changed to 4 n. This bound is never attainable for n = 1, and it is not always attainable when n = 2 (when x1 = 1, x2 = −1, for example).
And if n = 2k + 1 ≥3 or 2k + 2 ≥4, suppose that x1 = x2 = · · · = xk = 1 and xk+1 = xk+2 = · · · = x2k+1 = − k k+1 — if n is even, we can let xn = 0 and ignore this term in the permutation.
If two adjacent numbers in the permutation are equal then their sum is either 2 ≥2 −4 n or −2 · k k+1 ≤−2 + 4 n. Therefore in the permutation, the vectors must alternate between − k k+1 and 1, starting 1999 National Contests: Problems and Solutions 97 and ending with − k k+1.
But then the outer two vectors add up to −2 · k k+1, so the middle 2k −1 vectors add up to 2 · k k+1 ≥2 −4 n, a contradiction. Therefore, 1 n cannot be replaced by 4 n.
Problem 5 Suppose that r1, . . . , rn are real numbers. Prove that there exists S ⊆{1, 2, . . . , n} such that 1 ≤|S ∩{i, i + 1, i + 2}| ≤2, for 1 ≤i ≤n −2, and X i∈S ri ≥1 6 n X i=1 |ri|.
Solution: Let S = Pn i=1 |ri| and for i = 0, 1, 2, define si = X rj≥0, j≡i rj, ti = X rj<0, j≡i rj, where congruences are taken modulo 3. Then we have S = s1 + s2 + s3 −t1 −t2 −t3, and 2S equals (s1 + s2) + (s2 + s3) + (s3 + s1) −(t1 + t2) −(t2 + t3) −(t3 + t1).
Therefore there are i1 ̸= i2 such that either si1 + si2 ≥s/3 or ti1 + ti2 ≤−s/3 or both. Without loss of generality, we assume that si1 +si2 ≥s/3 and |si1 +si2| ≥|ti1 +ti2|. Thus si1 +si2 +ti1 +ti2 ≥0.
We have [si1 + si2 + ti1] + [si1 + si2 + ti2] ≥si1 + si2 ≥s/3.
Therefore at least one of si1 + si2 + ti1 and si1 + si2 + ti2 is bigger or equal to s/6 and we are done.
98 Ireland 1.11 Ireland Problem 1 Find all the real values of x which satisfy x2 (x + 1 −√x + 1)2 < x2 + 3x + 18 (x + 1)2 .
Solution: We must have x ∈(−1, 0) ∪(0, ∞) for the quantities above to be defined.
Make the substitution y = √x + 1, so that y ∈(0, 1) ∪(1, ∞) and x = y2 −1. Then the inequality is equivalent to (y2 −1)2 (y2 −y)2 < (y2 −1)2 + 3(y2 −1) + 18 y4 ⇐ ⇒(y + 1)2 y2 < y4 + y2 + 16 y4 ⇐ ⇒(y + 1)2y2 < y4 + y2 + 16 ⇐ ⇒2y3 < 16, so the condition is satisfied exactly when y < 2; i.e., exactly when y ∈(0, 1) ∪(1, 2), which is equivalent to x ∈(−1, 0) ∪(0, 3).
Problem 2 Show that there is a positive number in the Fibonacci sequence which is divisible by 1000.
Solution: In fact, for any natural number n, there exist infinitely many positive Fibonacci numbers divisible by n.
The Fibonacci sequence is defined thus: F0 = 0, F1 = 1, and Fk+2 = Fk+1 + Fk for all k ≥0.
Consider ordered pairs of consecutive Fibonacci numbers (F0, F1), (F1, F2), . . . taken modulo n.
Since the Fibonacci sequence is infinite and there are only n2 possible ordered pairs of integers modulo n, two such pairs (Fj, Fj+1) must be congruent: Fi ≡Fi+m and Fi+1 ≡Fi+m+1 (mod n) for some i and m.
If i ≥1 then Fi−1 ≡Fi+1−Fi ≡Fi+m+1−Fi+m ≡Fi+m−1 (mod n); similarly Fi+2 ≡Fi+1 + Fi ≡Fi+m+1 + Fi+m ≡Fi+2+m (mod n).
Continuing similarly, we have Fj ≡Fj+m (mod n) for all j ≥0. In particular, 0 = F0 ≡Fm ≡F2m ≡· · · (mod n), so the numbers Fm, F2m, . . . are all positive Fibonacci numbers divisible by n. Applying this to n = 1000, we are done. (In fact, the smallest such m is 750.) 1999 National Contests: Problems and Solutions 99 Problem 3 Let D, E, F be points on the sides BC, CA, AB, respectively, of triangle ABC such that AD ⊥BC, AF = FB, and BE is the angle bisector of ∠B. Prove that AD, BE, CF are concurrent if and only if a2(a −c) = (b2 −c2)(a + c), where a = BC, b = CA, c = AB.
Solution: By Ceva’s Theorem, the cevians AD, BE, CF in △ABC are concurrent if and only if (using directed line segments) AF FB · BD DC · CE EA = 1.
In this problem, AF F B = 1, and CE EA = a c . Thus AD, BE, CF are concurrent if and only if BD DC = c a.
This in turn is true if and only if BD = ac a+c and DC = a2 a+c.
Because AB2 −BD2 = BD2 = AC2 −CD2, this last condition hold exactly when the following equations are true: AB2 − ac a + c 2 = AC2 − a2 a + c 2 (a + c)2c2 −a2c2 = (a + c)2b2 −a4 a4 −a2c2 = (b2 −c2)(a + c)2 a2(a −c) = (b2 −c2)(a + c).
Therefore the three lines concur if and only if the given equation holds, as desired.
Alternatively, applying the law of cosines gives BD DC = c cos B b cos C = c b · a2 + c2 −b2 2ac · 2ab a2 + b2 −c2 = a2 + c2 −b2 a2 + b2 −c2 .
Again, this equals c a exactly when the given equation holds.
Problem 4 A 100 by 100 square floor is to be tiled.
The only available tiles are rectangular 1 by 3 tiles, fitting exactly over three squares of the floor.
(a) If a 2 by 2 square is removed from the center of the floor, prove that the remaining part of the floor can be tiled with available tiles.
100 Ireland (b) If, instead, a 2 by 2 square is removed from the corner, prove that the remaining part of the floor cannot be tiled with the available tiles.
Solution: Choose a coordinate system so that the corners of the square floor lie along the lattice points {(x, y) | 0 ≤x, y ≤100, x, y ∈ Z}. Denote the rectanglar region {(x, y) | a ≤x ≤b, c ≤y ≤d} by [(a, c) −(b, d)].
(a) It is evident that any rectangle with at least one dimension divisible by 3 can be tiled. First tile the four rectangles [(0, 0) −(48, 52)], [(0, 52) −(52, 100)], [(52, 48) −(100, 100)], and [(48, 0) −(100, 52)].
The only part of the board left untiled is [(48, 48) −(52, 52)].
But recall that the central region [(49, 49) −(51, 51)] has been removed. It is obvious that the remaining portion can be tiled.
(b) Assume without loss of generality that [(0, 0) −(2, 2)] is the 2 by 2 square which is removed.
Label each remaining square [(x, y) −(x + 1, y + 1)] with the number L(x, y) ∈{0, 1, 2} such that L(x, y) ≡x + y (mod 3). There are 3333 squares labeled 0, 3331 squares labeled 1, and 3332 squares labeled 2.
However, each 1 by 3 tile covers an equal number of squares of each label.
Therefore, the floor cannot be tiled.
Problem 5 Define a sequence un, n = 0, 1, 2, . . . as follows: u0 = 0, u1 = 1, and for each n ≥1, un+1 is the smallest positive integer such that un+1 > un and {u0, u1, . . . , un+1} contains no three elements which are in arithmetic progression. Find u100.
Solution: Take any nonnegative integer n (e.g., 100) and express it in base-2 (e.g., 100 = 11001002). Now interpret that sequence of 1’s and 0’s as an integer in base-3 (e.g., 11001003 = 981). Call that integer tn (e.g., t100 = 981).
We now prove that tn = un by strong induction on n. It is obvious that t0 = u0 and that t1 = u1. Now assume that tk = uk for all k < n. We shall show that tn = un.
First we show that un ≤tn by proving that, in the sequence t0, t1, t2, . . . , tn, no three numbers form an arithmetic progression.
1999 National Contests: Problems and Solutions 101 Pick any three numbers 0 ≤α < β < γ ≤n, and consider tα, tβ, and tγ in base-3. In the addition of tα and tγ, since both tα and tγ consist of only 1’s and 0’s, no carrying can occur. But tα ̸= tγ, so they must differ in at least one digit. In that digit in the sum tα + tγ must lie a “1.” On the other hand, tβ consists of only 1’s and 0’s, so 2tβ consists of only 2’s and 0’s. Thus, the base-3 representations of tα + tγ and 2tβ are different: the former contains a “1” while the latter does not.
Thus, tα+tγ ̸= 2tβ for any choice of α, β, γ, so among t0, t1, t2, . . . , tn, no three numbers are in arithmetic progression. Hence un ≤tn.
Next we show that un ≥tn by showing that for all k ∈{tn−1 + 1, tn−1+2, . . . , tn−1}, there exist numbers a and b such that ta+k = 2tb.
First note that k must contain a 2 in its base-3 representation, because the ti are the only nonnegative integers consisting of only 1’s and 0’s in base-3.
Therefore, we can find two numbers a and b with 0 ≤ta < tb < k such that: • whenever k has a “0” or a “1” in its base-3 representation, ta and tb each also have the same digit in the corresponding positions in their base-3 representations; • whenever k has a “2” in its base-3 representation, ta has a “0” in the corresponding position in its base-3 representation, but tb has a “1” in the corresponding position in its base-3 representation.
The ta and tb, thus constructed, satisfy ta < tb < k while ta+k = 2tb, so ta, tb, k form an arithmetic progression. Thus, un ≥tn. Putting this together with the previous result, we have forced un = tn; hence u100 = t100 = 981.
Problem 6 Solve the system of equations y2 −(x + 8)(x2 + 2) = 0 y2 −(8 + 4x)y + (16 + 16x −5x2) = 0.
Solution: We first check that the solutions (x, y) = (−2, −6) and (−2, 6) both work and are the only solutions with x = −2.
We substitute y2 = (x+8)(x2+2) into y2+16+16x−5x2 = 4(x+2)y to get 4(x+2)y = x3 +3x2 +18x+32 = (x+2)(x2 +x+16). The case x = −2 has already been finished, so to deal with the case x ̸= −2, we write 4y = x2 + x + 16.
102 Ireland Squaring both sides, we have 16y2 = x4 + 2x3 + 33x2 + 32x + 256, but from the first original equation we have 16y2 = 16x3 + 128x2 + 32x + 256; subtracting these two equations, we have x4 −14x3 −95x2 = 0, or x2(x + 5)(x −19) = 0. Thus, x ∈{0, −5, 19}. We use the equation 4y = x2 + x + 16 to find the corresponding y’s.
In this way we find that the only solutions (x, y) are (−2, −6), (−2, 6), (0, 4), (−5, 9), and (19, 99); it can be checked that each of these pairs works.
Problem 7 A function f : N →N satisfies (i) f(ab) = f(a)f(b) whenever the greatest common divisor of a and b is 1; (ii) f(p + q) = f(p) + f(q) for all prime numbers p and q.
Prove that f(2) = 2, f(3) = 3, and f(1999) = 1999.
Solution: Let us agree on the following notation: we shall write (i)a,b when we plug (a, b) (where a and b are relatively prime) into (i), and (ii)p,q when we plug (p, q) (where p and q are primes) into (ii).
First we find f(1), f(2), and f(4). By (i)1,b we find f(1) = 1. By (i)2,3 we find f(6) = f(2)f(3); by (ii)3,3 we get f(6) = 2f(3); thus, f(2) = 2.
Now by (ii)2,2 we have f(4) = 4.
Next we find some useful facts. From (ii)3,2 and (ii)5,2, respectively, we obtain f(5) = f(3) + 2, f(7) = f(5) + 2 = f(3) + 4.
Now we can find f(3). By (ii)5,7 we have f(12) = f(5) + f(7) = 2f(3)+6; from (i)4,3 we have f(12) = 4f(3); we solve for f(3) to find f(3) = 3.
Then using the facts from the previous paragraph, we find f(5) = 5, f(7) = 7.
1999 National Contests: Problems and Solutions 103 We proceed to find f(13) and f(11). By (i)3,5, we have f(15) = 15.
From (ii)13,2 and (ii)11,2, respectively, we find f(13) = f(15) −f(2) = 13, f(11) = f(13) −f(2) = 11.
Finally, we can calculate f(1999). By using (i) repeatedly with 2, 7, 11, and 13, we find f(2002) = f(2·7·11·13) = f(2)f(7)f(11)f(13) = 2002. Noting that 1999 is a prime number; from (ii)1999,3 we obtain f(1999) = f(2002) −f(3) = 1999, and we have finished.
Problem 8 Let a, b, c, d be positive real numbers whose sum is 1.
Prove that a2 a + b + b2 b + c + c2 c + d + d2 d + a ≥1 2 with equality if and only if a = b = c = d = 1/4.
Solution: Apply the Cauchy-Schwarz inequality to find [(a + b) + (b+c)+(c+d)+(d+a)] a2 a+b + b2 b+c + c2 c+d + d2 d+a ≥(a+b+c+d)2, which is equivalent to a2 a + b + b2 b + c + c2 c + d + d2 d + a ≥1 2(a + b + c + d) = 1 2, with equality if and only if a+b a = b+c b = c+d d = d+a a , i.e., if and only if a = b = c = d = 1 4.
Problem 9 Find all positive integers m such that the fourth power of the number of positive divisors of m equals m.
Solution: If the given condition holds for some integer m, then m must be a perfect fourth power and we may write its prime factorization as m = 24a234a354a574a7 · · · for nonnegative integers a2, a3, a5, a7, . . . . Now the number of positive divisors of m equals (4a2 + 1)(4a3 + 1)(4a5 + 1)(4a7 + 1) · · ·; this is odd, so m is odd and a2 = 0. Thus, 1 = 4a3 + 1 3a3 · 4a5 + 1 5a5 · 4a7 + 1 7a7 · · · · = x3x5x7 · · · , where we write xp = 4ap+1 pap for each p.
104 Ireland When a3 = 1, x3 = 5 3; when a3 = 0 or 2, x3 = 1. And by Bernoulli’s inequality, when a3 > 2 we have 3a 3 = (8 + 1)a3/2 > 8(a3/2) + 1 = 4a3 + 1 so that x3 < 1.
When a5 = 0 or 1, x5 = 1; and by Bernoulli’s inequality, when a5 ≥2 we have 5a5 = (24 + 1)a5/2 ≥24a5/2 + 1 = 12a5 + 1 so that x5 ≤ 4a5+1 12a5+1 ≤ 9 25.
Finally, for any p > 5 when ap = 0 we have xp = 1; when ap = 1 we have pap = p > 5 = 4ap + 1 so that xp < 1; and when ap > 0 then again by Bernoulli’s inequality we have pap > 5ap > 12ap + 1 so that as above xp < 9 25.
Now if a3 ̸= 1 then we have xp ≤1 for all p; but since 1 = x2x3x5 · · · we must actually have xp = 1 for all p. This means that a3 ∈{0, 2}, a5 ∈{0, 1}, and a7 = a11 = · · · = 0; so that m = 14, (32)4, 54, or (32 · 5)4.
Otherwise, if a3 = 1 then 3 | m = 54(4a5 + 1)4(4a7 + 1)4 · · ·. Then for some prime p′ ≥5, 3 | 4ap′ + 1 so that ap′ ≥2; from above we have xp′ ≤ 9 25. But then x3x5x7 · · · ≤5 3 9 27 < 1, a contradiction.
Thus the only solutions are 1, 54, 38, and 38 · 54; and these can be easily verified by inspection.
Problem 10 Let ABCDEF be a convex hexagon such that AB = BC, CD = DE, EF = FA, and ∠ABC + ∠CDE + ∠EFA = 360◦.
Prove that the respective perpendiculars from A, C, E to FB, BD, DF are concurrent.
First Solution: The result actually holds even without the given angle condition.
Let C1 be the circle with center B and radius AB = BC, C2 the circle with center D and radius CD = DE, and C3 the circle with center F and radius EF = FA. The line through A and perpendicular to line FB is the radical axis of circles C3 and C1, 1999 National Contests: Problems and Solutions 105 the line through C and perpendicular to line BD is the radical axis of circles C1 and C2, and the line through E and perpendicular to line DF is the radical axis of circles C2 and C3. The result follows because these three radical axes meet at the radical center of the three circles.
Second Solution: We first establish two lemmas: Lemma 1. Given points W ̸= Y and X ̸= Z, lines WY and XZ are perpendicular if and only if XW 2 −WZ2 = XY 2 −Y Z2.
(∗) Proof: Introduce Cartesian coordinates such that W = (0.0), X = (1, 0), Y = (x1, y1), and Z = (x2, y2). Then () becomes x2 1 + y2 1 −x2 2 −y2 2 = (x1 −1)2 + y2 1 −(x2 −1)2 −y2 2, which upon cancellation yields x1 = x2. This is true if and only if line Y Z is perpendicular to the x-axis WX.
If P is the intersection of the perpendiculars from A and C to lines FB and BD, respectively, then the lemma implies that PF 2 −PB2 = AF 2 −AB2 and PB2 −PD2 = CB2 −CD2.
From the given isosceles triangles, we have EF = FA, AB = BC, and CD = DE. Subtracting the first equation from the second then gives PD2 −PF 2 = ED2 −EF 2.
Hence line PE is also perpendicular to line DF, which completes the proof.
106 Italy 1.12 Italy Problem 1 Given a rectangular sheet with sides a and b, with a > b, fold it along a diagonal. Determine the area of the triangle that passes over the edge of the paper.
Solution: Let ABCD be a rectangle with AD = a and AB = b. Let D′ be the reflection of D across line AC, and let E = AD′ ∩BC. We wish to find [CD′E]. Since AB = CD′, ∠ABE = ∠CD′E = 90◦, and ∠BEA = ∠D′EC, triangles ABE and CD′E are congruent. Thus AE = EC and CE2 = AE2 = AB2 + BE2 = b2 + (a −CE)2. Hence CE = a2+b2 2a . It follows that [CD′E] = [ACD′] −[ACE] = ab 2 −b 2 · CE = b(a2 −b2) 4a .
Problem 2 A positive integer is said to be balanced if the number of its decimal digits equals the number of its distinct prime factors (for instance 15 is balanced, while 49 is not). Prove that there are only finitely many balanced numbers.
Solution: For n > 15, consider the product of the first n primes.
The first sixteen primes have product (2·53)(3·47)(5·43)(7·41)·11·13·17·19·23·29·31·37 > 1004·108 = 1016, while the other n −16 primes are each at least 10. Thus the product of the first n primes is greater than 10n.
Then if x has n digits and is balanced, then it it as at least the product of the first n primes.
If n ≥16 then from the previous paragraph x would be greater than 10n and would have at least n+1 digits, a contradiction. Thus x can have at most 15 digits, implying that the number of balanced numbers is finite.
Problem 3 Let ω, ω1, ω2 be three circles with radii r, r1, r2, respec-tively, with 0 < r1 < r2 < r. The circles ω1 and ω2 are internally tangent to ω at two distinct points A and B and meet in two distinct points. Prove that AB contains an intersection point of ω1 and ω2 if and only if r1 + r2 = r.
Solution: Let O be the center of ω, and note that the centers C, D of ω1, ω2 lie on OA and OB, respectively. Let E be a point on AB 1999 National Contests: Problems and Solutions 107 such that CE ∥OB. Then △ACE ∼△AOB. Hence AE = CE and E is on ω1. We need to prove that r = r1 + r2 if and only if E is on ω2.
Note that r = r1 + r2 is equivalent to OD = OB −BD = r −r2 = r1 = AC, that is CEDO is a parallelogram or DE ∥AO. Hence r = r1 + r2 if and only if △BDO ∼△BOA or BD = DE, that is, E is on ω2.
Problem 4 Albert and Barbara play the following game.
On a table there are 1999 sticks: each player in turn must remove from the table some sticks, provided that the player removes at least one stick and at most half of the sticks remaining on the table. The player who leaves just one stick on the table loses the game. Barbara moves first.
Determine for which of the players there exists a winning strategy.
Solution: Call a number k hopeless if a player faced with k sticks has no winning strategy. If k is hopeless, then so is 2k + 1: a player faced with 2k + 1 sticks can only leave a pile of k + 1, k + 2, . . . , or 2k sticks, from which the other player can leave k sticks. Then since 2 is hopeless, so are 5, 11, . . . , 3 · 2n −1 for all n ≥0. Conversely, if 3 · 2n −1 < k < 3 · 2n+1 −1, then given k sticks a player can leave 3·2n−1 sticks and force a win. Since 1999 is not of the form 3·2n−1, it is not hopeless and hence Barbara has a winning strategy.
Problem 5 On a lake there is a village of pile-built dwellings, set on the nodes of an m × n rectangular array.
Each dwelling is an endpoint of exactly p bridges which connect the dwelling with one or more of the adjacent dwellings (here adjacent means with respect to the array, hence diagonal connection is not allowed). Determine for which values of m, n, p it is possible to place the bridges so that from any dwelling one can reach any other dwelling. (Clearly, two adjacent dwellings can be connected by more than one bridge).
Solution: Suppose it is possible to place the bridges in this manner, and set the villages along the lattice points {(a, b) | 1 ≤a ≤m, 1 ≤ b ≤n}. Color the dwellings cyan and magenta in a checkerboard fashion, so that every bridge connects a cyan dwelling with a magenta dwelling. Since each dwelling is at the end of the same number of 108 Italy bridges (exactly p of them), the number of cyan dwellings must equal the number of magenta dwellings; thus 2 | mn.
Obviously mn = 2 works for all values of p. And for p = 1, we cannot have mn > 2 because otherwise if any two dwellings A and B are connected, then they cannot be connected to any other dwellings.
Similarly, if m = 1, n > 2 (or n = 1, m > 2) then p bridges must connect (1, 1) and (1, 2) (or (1, 1) and (2, 1)); but then neither of these dwellings is connected to any other dwellings, a contradiction.
Now assume that 2 | mn with m, n > 1 and p > 1; assume without loss of generality that 2 | m. Build a sequence of bridges starting at (1, 1), going up to (1, n), right to (m, n), down to (m, 1), and left to (m −1, 1); and then weaving back to (1, 1) by repeatedly going from (k, 1) up to (k, n −1) left to (k −1, n −1) down to (k −1, 1) and left to (k −2, 1) for k = m−1, m−3, . . . , 3. (The sideways E below shows this construction for m = 6, n = 4.) So far we have built two bridges leading out of every dwelling, and any dwelling can be reached from any other dwelling. For the remaining p−2 bridges needed for each dwelling, note that our sequence contains exactly mn bridges, an even number; so if we build every other bridge in our sequence, and do this p −2 times, then exactly p bridges come out of every dwelling.
Thus either mn = 2 and p equals any value; or 2 | mn with m, n, p > 1.
Problem 6 Determine all triples (x, k, n) of positive integers such that 3k −1 = xn.
Solution: (3k −1, k, 1) for all positive integers k, and (2, 2, 3).
The case of n = 1 is obvious. Now, n cannot be even because then 3 could not divide 3k = x n 2 2 + 1 (since no square is congruent to 2 modulo 3); and also, we must have x ̸= 1.
Assume that n > 1 is odd and x ≥2. Then 3k = (x+1) Pn−1 i=0 (−x)i implying that both x+1 and Pn−1 i=0 (−x)i are powers of 3. Then since 1999 National Contests: Problems and Solutions 109 x + 1 ≤x2 −x + 1 ≤Pn−1 i=0 (−x)i, we must have 0 ≡Pn−1 i=0 (−x)i ≡ n (mod x + 1), so that x + 1 | n. Specifically, this means that 3 | n.
Writing x′ = x n 3 , we have 3k = x′3 + 1 = (x′ + 1)(x′2 −x′ + 1).
As before x′ + 1 must equal some power of 3, say 3t.
But then 3k = (3t −1)3 +1 = 33t −32t+1 +3t+1, which is strictly between 33t−1 and 33t for t > 1. Therefore we must have t = 1, x′ = 2, and k = 2, giving the solution (x, k, n) = (2, 2, 3).
Problem 7 Prove that for each prime p the equation 2p + 3p = an has no integer solutions (a, n) with a, n > 1.
Solution: When p = 2 we have an = 13, which is impossible. When p is odd, then 5 | 2p +3p; then since n > 1, we must have 25 | 2p +3p.
Then 2p+(5−2)p ≡2p+ p 1 5 · (−2)p−1 + (−2)p ≡5p·2p−1 (mod 25), so 5 | p. Thus we must have p = 5, but then an = 25 + 35 = 52 · 11 has no solutions.
Problem 8 Points D and E are given on the sides AB and AC of triangle ABC such that DE ∥BC and DE is tangent to the incircle of ABC. Prove that DE ≤AB + BC + CA 8 .
Solution: Let BC = a, CA = b, AB = c. Also let h = 2[ABC] a be the distance from A to line BC and let r = 2[ABC] a+b+c the inradius of triangle ABC; note that h−2r h = b+c−a a+b+c.
Let x = b + c −a, y = c + a −b, z = a + b −c. Then (x + y + z)2 ≥(2 p x(y + z))2 = 4x(y + z) by AM-GM, which implies that (a + b + c)2 ≥8(b + c −a)a, or b + c −a a + b + c · a ≤a + b + c 8 = ⇒h −2r h · BC ≤AB + BC + CA 8 .
But since DE ∥BC, we have DE BC = h−2r h ; substituting this into the above inequality gives the desired result.
110 Italy Problem 9 (a) Find all the strictly monotonic functions f : R →R such that f(x + f(y)) = f(x) + y, for all x, y ∈R.
(b) Prove that for every integer n > 1 there do not exist strictly monotonic functions f : R →R such that f(x + f(y)) = f(x) + yn, for all x, y ∈R.
Solution: (a) The only such functions are f(x) = x and f(x) = −x. Setting x = y = 0 gives f(f(0)) = f(0), while setting x = −f(0), y = 0 gives f(−f(0)) = f(0). Since f is strictly monotonic it is injective, so f(0) = −f(0) and thus f(0) = 0. Next, setting x = 0 gives f(f(y)) = y for all y.
Suppose f is increasing.
If f(x) > x then x = f(f(x)) > f(x), a contradiction; if f(x) < x then x = f(f(x)) < f(x), a contradiction. Thus f(x) = x for all x.
Next suppose that f is decreasing.
Plugging in x = −f(t), y = t, and then x = 0, y = −t shows that f(−f(t)) = f(f(−t)) = −t, so f(t) = −f(−t) for all t.
Now given x, if f(x) < −x then x = f(f(x)) > f(−x) = −f(x), a contradiction. And if f(x) > −x then x = f(f(x)) < f(−x) = −f(x), a contradiction.
Hence we must have f(x) = −x for all x.
Therefore either f(x) = x for all x or f(x) = −x for all x; and it is easy to check that these two functions work.
(b) Since f is strictly monotonic, it is injective. Then for y ̸= 0 we have f(y) ̸= f(−y) so that f(x+f(y)) ̸= f(x+f(−y)) and hence f(x) + yn ̸= f(x) + (−y)n; thus, n can’t be even.
Now suppose there is such an f for odd n; then by arguments similar to those in part (a), we find that f(0) = 0 and f(f(y)) = yn. Specifically, f(f(1)) = 1. If f is increasing then as in part (a) we have f(1) = 1; then f(2) = f(1 + f(1)) = f(1) + 1n = 2 and 2n = f(f(2)) = f(2) = 2, a contradiction. If f is decreasing, then as in part (a) we have f(1) = −1; then f(2) = f(1 + f(−1)) = f(1) + (−1)n = −2 and 2n = f(f(2)) = f(−2) = −f(2) = 2, a contradiction.
1999 National Contests: Problems and Solutions 111 Problem 10 Let X be a set with |X| = n, and let A1, A2, . . . , Am be subsets of X such that (a) |Ai| = 3 for i = 1, 2, . . . , m.
(b) |Ai ∩Aj| ≤1 for all i ̸= j.
Prove that there exists a subset of X with at least ⌊ √ 2n⌋elements, which does not contain Ai for i = 1, 2, . . . , m.
Solution: Let A be a subset of X containing no Ai, and having the maximum number of elements subject to this condition. Let k be the size of A. By assumption, for each x ∈X −A, there exists i(x) ∈{1, . . . , m} such that Ai(x) ⊆A∪{x}. Let Lx = A∩Ai(x), which by the previous observation must have 2 elements. Since |Ai∩Aj| ≤1 for i ̸= j, the Lx must all be distinct. Now there are k 2 2-element subsets of A, so there can be at most k 2 sets Lx. Thus n −k ≤ k 2 or k2 + k ≥2n. It follows that k ≥1 2(−1 + √ 1 + 8n) > √ 2n −1, that is, k ≥⌊ √ 2n⌋.
112 Japan 1.13 Japan Problem 1 You can place a stone at each of 1999 × 1999 squares on a grid pattern. Find the minimum number of stones you must place such that, when an arbitrary blank square is selected, the total number of stones placed in the corresponding row and column is at least 1999.
Solution: Place stones in a checkerboard pattern on the grid, so that stones are placed on the four corner squares. This placement satisfies the condition and contains 1000×1000+999×999 = 1998001 stones. We now prove this number is minimal.
Suppose the condition is satisfied. Assume without loss of gener-ality that the j-th column contains k stones, and every other row or column also contains at least k stones. For each of the k stones in the j-th column, the row containing that stone must contain at least k stones by our minimal choice of k. And for each of the 1999 −k blank squares in the j-th column, to satisfy the given condition there must be at least 1999 −k stones in the row containing that square.
Thus total number of stones is at least k2 + (1999 −k)2 = 2 k −1999 2 2 + 19992 2 ≥19992 2 = 1998000.5, and it follows that there indeed must be at least 1998001 stones.
Problem 2 Let f(x) = x3+17. Prove that for each natural number n, n ≥2, there is a natural number x for which f(x) is divisible by 3n but not by 3n+1.
Solution: We prove the result by induction on n. If n = 2, then x = 1 suffices. Now suppose that the claim is true for n ≥2 — that is, there is a natural number y such that y3 +17 is divisible by 3n but not 3n+1. We prove that the claim is true for n + 1.
Suppose we have integers a, m such that a is not divisible by 3 and m ≥2. Then a2 ≡1 (mod 3) and thus 3ma2 ≡3m (mod 3m+1). Also, since m ≥2 we have 3m −3 ≥2m −1 ≥m + 1. Hence (a+3m−1)3 ≡a3 +3ma2 +32m−1a+33m−3 ≡a3 +3m (mod 3m+1).
Since y3+17 is divisible by 3n, it is congruent to either 0, 3n, or 2·3n modulo 3n+1. Since 3 does not divide 17, 3 cannot divide y either.
1999 National Contests: Problems and Solutions 113 Hence applying our result from the previous paragraph twice — once with (a, m) = (y, n) and once with (a, m) = (y + 3n−1, n) — we find that 3n+1 must divide either (y + 3n−1)3 + 17 or (y + 2 · 3n−1)3 + 17.
Hence there exists a natural number x′ not divisible by 3 such that 3n+1 | x′3 + 17. If 3n+2 does not divide x′3 + 17, we are done.
Otherwise, we claim the number x = x′ + 3n suffices.
Since x = x′ + 3n−1 + 3n−1 + 3n−1, the result from two paragraphs ago tells us that x3 ≡x′3 +3n +3n +3n ≡x′3 (mod 3n+1). Thus 3n+1 | x3 +17 as well. On the other hand, since x = x′+3n, we have x3 ≡x′3+3n+1 ̸≡ x′3 (mod 3n+2). It follows that 3n+2 does not divide x3+17, as desired.
This completes the inductive step.
Problem 3 From a set of 2n + 1 weights (where n is a natural number), if any one weight is excluded, then the remaining 2n weights can be divided into two sets of n weights that balance each other.
Prove that all the weights are equal.
Solution: Label the weights a1, a2, . . . , a2n+1. Then for each j, 1 ≤j ≤2n, we have c(j) 1 a1 + c(j) 2 a2 + · · · + c(j) 2n a2n = a2n+1 where c(j) j = 0, n of the other c(j) i equal 1, and the remaining c(j) i equal −1.
Thus we have 2n equations in the variables a1, a2, . . . , a2n. Clearly (a1, a2, . . . , a2n) = (a2n+1, a2n+1, . . . , a2n+1) is a solution to this system of equations.
By Kramer’s Rule, this solution is unique if and only if the determinant of the matrix c(1) 1 c(1) 2 · · · c(1) 2n c(2) 1 c(2) 2 · · · c(2) 2n .
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c(2n) 1 c(2n) 2 · · · c(2n) 2n is nonzero. We show this is true by proving that this determinant is odd.
If we add an integer m to any single integer in the matrix, its determinant changes by m multiplied by the corresponding cofactor.
Specifically, if m is even then the parity of the determinant does not change. Thus the parity of the presented determinant is the same as 114 Japan the parity of the determinant 0 1 · · · 1 1 0 · · · 1 .
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1 1 · · · 0 .
This matrix has eigenvector (1, 1, . . . , 1) corresponding to the eigen-value 2n −1; and eigenvectors (1, −1, 0, . . . , 0, 0), (0, 1, −1, . . . , 0, 0), . . . , (0, 0, 0, . . . , 1, −1) corresponding to eigenvalue −1.
These 2n eigenvectors are linearly independent, so the matrix’s characteristic polynomial is p(x) = (x−(2n−1))(x+1)2n−1. Hence its determinant p(0) = −(2n −1) is odd, as desired.
Problem 4 Prove that f(x) = (x2 + 12)(x2 + 22)(x2 + 32) · · · (x2 + n2) + 1 cannot be expressed as a product of two integral-coefficient polyno-mials with degree greater than 0.
Solution: The claim is obvious when n = 1. Now assume n ≥2 and suppose by way of contradiction that f(x) could be expressed as such a product g(x)h(x) with g(x) = a0 + a1x + · · · + aℓxℓ, h(x) = b0 + b1x + · · · + bℓ′xℓ′, where ℓ, ℓ′ > 0 and the coefficients ai and bi are integers.
For m = ±1, ±2, . . . , ±n, since (mi)2 + m2 = 0 we have 1 = f(mi) = g(mi)h(mi). But since g and h have integer coefficients, g(mi) equals either 1, −1, i, or −i.
Moreover, since the imaginary part of g(mi) = (a0 −a2m2 + a4m4 −· · ·) + m(a1 −a3m2 + a5m4 −· · ·)i is a multiple of m, g(mi) must equal ±1 for m ̸= ±1. Going further, since 1 = g(mi)h(mi) we have g(mi) = h(mi) = ±1 for m ̸= ±1.
Then by the factor theorem, g(x) −h(x) = (x2 + 22)(x2 + 32) · · · (x2 + n2)k(x) 1999 National Contests: Problems and Solutions 115 for some integer-coefficient polynomial k(x) with degree at most 1.
Since (g(i), h(i)) equals (1, −1), (−1, 1), (i, −i), or (−i, i), we have 2 ≥|g(i) −h(i)| = (−1 + 22)(−1 + 32) · · · (−1 + n2)|k(i)|, and hence we must have k(i) = 0. Since k(x) has degree at most 1, this implies that k(x) = 0 for all x and that g(x) = h(x) for all x. But then a2 0 = g(0)h(0) = f(0) = (12)(22) · · · (n2) + 1, which is impossible.
Problem 5 For a convex hexagon ABCDEF whose side lengths are all 1, let M and m be the maximum and minimum values of the three diagonals AD, BE, and CF. Find all possible values of m and M.
Solution: We claim that the possible values are √ 3 ≤M ≤3 and 1 ≤m ≤2.
First we show all such values are attainable. Continuously trans-form ABCDEF from an equilateral triangle ACE of side length 2, into a regular hexagon of side length 1, and finally into a segment of length 3 (say, by enlarging the diagonal AD of the regular hexagon while bringing B, C, E, F closer to line AD). Then M continuously varies from √ 3 to 2 to 3. Similarly, by continuously transforming ABCDEF from a 1 × 2 rectangle into a regular hexagon, we can make m vary continuously from 1 to 2.
Now we prove no other values are attainble. First, we have AD ≤ AB + BC + CD = 3, and similarly BE, CF ≤3 so that M ≤3.
Next, suppose by way of contradiction that m < 1 and say without loss of generality that AD < 1. Since AD < AB = BC = CD = 1, ∠DCA < ∠DAC, ∠ABD < ∠ADB, ∠CBD = ∠CDB, ∠BCA = ∠BAC.
Therefore, ∠CDA + ∠BAD = ∠CDB + ∠BDA + ∠BAC + ∠CAD > ∠CBD + ∠DBA + ∠BCA + ∠ACD = ∠CBA + ∠BCD.
Consequently ∠CDA + ∠BAD > 180◦and likewise ∠EDA + ∠FAD > 180◦. But then ∠CDE + ∠BAF = ∠CDA + ∠EDA + ∠BAD + ∠FAD > 360◦, 116 Japan which is impossible since ABCDEF is convex. Hence m ≥1.
Next we demonstrate that M ≥ √ 3 and m ≤2. Since the sum of the six interior angles in ABCDEF is 720◦, some pair of adjacent angles has sum greater than or equal to 240◦and some pair has sum less than or equal to 240◦. Thus it suffices to prove that CF ≥ √ 3 when ∠A + ∠B ≥240◦, and that CF ≤2 when ∠A + ∠B ≤240◦.
Suppose by way of contradiction that ∠A + ∠B ≥240◦and CF < √ 3. By the law of cosines, CF 2 = BC2+BF 2−2BC·BF cos ∠FBC.
Thus if we fix A, B, F and decrease ∠ABC, we decrease ∠FBC and CF; similarly, by fixing A, B, C and decreasing ∠BAF, we decrease CF. Therefore, it suffices to prove that √ 3 ≥CF when ∠A + ∠B = 240◦. And likewise, it suffices to prove that CF ≤2 when ∠A+∠B = 240◦.
Now suppose that ∠A+∠B does equal 240◦. Let lines AF and BC intersect at P, and set x = PA and y = PB. Since ∠A+∠B = 240◦, ∠P = 60◦. Then applying the law of cosines to triangles PAB and PCF yields 1 = AB2 = x2 + y2 −xy and CF 2 = (x + 1)2 + (y + 1)2 −(x + 1)(y + 1) = 2 + x + y.
Therefore, we need only find the possible values of x + y given that x2 +y2 −xy = 1 and x, y ≥0. These conditions imply that (x+y)2 + 3(x −y)2 = 4, x + y ≥0, and |x −y| ≤x + y. Hence 1 = 1 4(x + y)2 + 3 4(x −y)2 ≤(x + y)2 ≤(x + y)2 + 3(x −y)2 = 4, so 1 ≤x + y ≤2 and √ 3 ≤CF ≤2. This completes the proof.
1999 National Contests: Problems and Solutions 117 1.14 Korea Problem 1 Let R and r be the circumradius and inradius of triangle ABC respectively, and let R′ and r′ be the circumradius and inradius of triangle A′B′C′ respectively. Prove that if ∠C = ∠C′ and Rr′ = R′r, then the triangles are similar.
Solution: Let ω be the circumcircle of triangle ABC. By scaling, rotating, and translating, we may assume that A = A′, B = B′, R = R′, r = r′ and that C, C′ lie on the same arc d AB of ω. If the triangles were similar before these transformations, they still remain similar; so it suffices to prove they are now congruent.
Since r = 1 2(AC + BC −AB) cot(∠C) and r′ = 1 2(A′C′ + B′C′ − A′B′) cot(∠C′) = 1 2(A′C′ +B′C′ −AB) cot(∠C), we must have AC + BC = A′C′ +B′C′ and hence AB +BC +CA = A′B′ +B′C′ +C′A′.
Then the area of triangle ABC is 1 2r(AB + BC + CA), which thus equals the area of triangle A′B′C′, 1 2r′(A′B′ + B′C′ + C′A′). Since these triangles share the same base AB, we know that the altitudes from C and C′ to AB are equal. This implies that △ABC is congruent to either △A′B′C′ or △B′A′C′, as desired.
Problem 2 Suppose f : Q →R is a function satisfying |f(m + n) −f(m)| ≤n m for all positive rational numbers n and m. Show that for all positive integers k, k X i=1 |f(2k) −f(2i)| ≤k(k −1) 2 .
Solution: It follows from the condition |f(m+n)−f(m)| ≤n m that f(2i+1) −f(2i) ≤2i+1 −2i 2i = 1.
Therefore, for k > i, f(2k) −f(2i) ≤ k−1 X j=i f(2j+1) −f(2j) ≤k −i.
118 Korea From the above inequality, we obtain k X i=1 f(2k) −f(2i) = k−1 X i=1 f(2k) −f(2i) ≤ k−1 X i=1 (k −i) = k(k −1) 2 .
This completes the proof.
Problem 3 Find all positive integers n such that 2n−1 is a multiple of 3 and 2n−1 3 is a divisor of 4m2 + 1 for some integer m.
Solution: The answer is all 2k where k = 1, 2, . . ..
First, it is easy to conclude (using Fermat’s Little Theorem, or by simple observation) that if 3 | 2n −1, then n must be even.
Suppose by way of contradiction that ℓ≥3 is a positive odd divisor of n. Then 2ℓ−1 is not divisible by 3 but it is a divisor of 2n −1, so it is a divisor of 4m2 + 1 as well. On the other hand, 2ℓ−1 has a prime divisor p of the form 4r + 3. Then (2m)2 ≡−1 (mod 4r + 3); but a standard number theory result states that a square cannot equal −1 modulo a prime of the form 4r + 3.
Therefore n is indeed of the form 2k for k ≥1. For such n, we have 2n −1 3 = (221 + 1)(222 + 1)(223 + 1) · · · (22k−1 + 1).
The factors on the right side are all relatively prime to 2 since they are all odd. They are also Fermat numbers, and another result from number theory states that they are relatively prime. (Suppose that some prime p divided both 22a + 1 and 22b + 1 for a < b.
Then 22a ≡22b ≡−1 (mod p).
But then −1 ≡22b = 22a2b−a ≡ (−1)22b−a−1 ≡1 (mod p), implying that p = 2; again, this is impossible.) Therefore by the Chinese Remainder Theorem, there is a positive integer c simultaneously satisfying c ≡22i−1 (mod 22i + 1) for all i = 1, 2, . . . , k −1 and c ≡0 (mod 2). Putting c = 2m, 4m2 + 1 is a multiple of 2n−1 3 , as desired.
Problem 4 Suppose that for any real x with |x| ̸= 1, a function f(x) satisfies f x −3 x + 1 + f 3 + x 1 −x = x.
Find all possible f(x).
1999 National Contests: Problems and Solutions 119 Solution: Set t = x−3 x+1 so that x = 3+t 1−t. Then the given equation can be rewritten as f(t) + f t −3 t + 1 = 3 + t 1 −t.
Similarly, set t = 3+x 1−x so that x = t−3 t+1 and x−3 x+1 = 3+t 1−t. Again we can rewrite the given equation, this time as f 3 + t 1 −t + f(t) = t −3 t + 1.
Adding these two equations we have 8t 1 −t2 = 2f(t) + f t −3 t + 1 + f 3 + t 1 −t = 2f(t) + t, so that f(t) = 4t 1 −t2 −t 2, and some algebra verifies that this solution works.
Problem 5 Consider a permutation (a1, a2, . . . , a6) of 1, 2, . . . , 6 such that the minimum number of transpositions needed to transform (a1, a2, a3, a4, a5, a6) to (1, 2, 3, 4, 5, 6) is four.
Find the number of such permutations.
Solution: Given distinct numbers b1, b2, . . . , bk between 1 and n, in a k-cycle with these numbers b1 is mapped to one of the other k −1 numbers; its image is mapped to one of the k −2 remaining numbers; and so on until the remaining number is mapped to b1. Hence there are (k −1)(k −2) · · · (1) = (k −1)! cycles of length k involving these numbers.
Any permutation which can be achieved with four transpositions is even, so a permutation satisfying the given conditions must be either (i) the identity permutation, (ii) a composition of two transpositions, (iii) a 3-cycle, (iv) a composition of a 2-cycle and a 4-cycle, (v) a composition of two 3-cycles, or (vi) a 5-cycle. Permutations of type (i), (ii), and (iii) can be attained with fewer transpositions from our observations above. Conversely, any even permutation that can be achieved with zero or two transpositions is of these three types. Hence the permutations described in the problem statement are precisely those of types (iv), (v), and (vi). For type-(iv) permutations, there 120 Korea are 6 2 = 15 ways to assign which cycle each of 1, 2, . . . , 6 belongs; and there are (2 −1)!(4 −1)! = 6 ways to rearrange them within the cycles, for a total of 15 · 6 = 90 permutations.
For type-(v) permutations, there are 1 2 6 3 = 10 ways to assign which cycle each number belongs to (since 6 3 counts each such permutation twice, once in the form (a b c)(d e f) and again in the form (d e f)(a b c)).
And there are (3 −1)!(3 −1)! = 4 ways to rearrange the numbers within these two cycles for a total of 10·4 = 40 type-(v) permutations.
Finally, for type-(v) permutations there are 6 5 = 6 ways to choose which five numbers are cycled, and (5 −1)! = 24 different cycles among any five numbers. This gives a total of 6 · 24 = 144 type-(v) permutations, and altogether 90 + 40 + 144 = 274 permutations which can be attained with four permutations, but no less.
Problem 6 Let a1, a2, · · · , a1999 be nonnegative real numbers sat-isfying the following two conditions: (a) a1 + a2 + · · · + a1999 = 2; (b) a1a2 + a2a3 + · · · + a1998a1999 + a1999a1 = 1.
Let S = a2 1 + a2 2 + · · · + a2 1999. Find the maximum and minimum possible values of S.
Solution: Without loss of generality assume that a1999 is the minimum ai.
We may also assume that a1 > 0.
From the given equations we have 4 = (a1 + a2 + · · · + a1999)2 ≥(a1 + a2 + · · · + a1999)2 −(a1 −a2 + a3 −· · · −a1998 + a1999)2 = 4(a1 + a3 + · · · + a1999)(a2 + a4 + · · · + a1998) ≥4(a1a2 + a2a3 + · · · + a1998a1999) +4(a1a4 + a2a5 + · · · + a1996a1999) +4a1(a6 + a8 + · · · + a1998) = 4(1 −a1999a1) + 4(a1a4 + a2a5 · · · + a1996a1999) +4a1(a6 + a8 + · · · + a1998) 1999 National Contests: Problems and Solutions 121 = 4 + 4(a1a4 + a2a5 + · · · + a1996a1999) +4a1(a6 + a8 + · · · + a1998 −a1999) ≥4.
Hence equality must hold in the first and third inequality. Thus we must have (i) a1 + a3 + · · · + a1999 = a2 + a4 + · · · + a1998 = 1 (ii) a1a4 = a2a5 = · · · = a1996a1999 = 0 (iii) a6 + a8 + · · · + a1998 = a1999.
Condition (ii) implies a4 = 0; from (iii) we get a6 = a8 = · · · = a1998 = 0. Thus from (i), we have a2 = 1, and from (b), we have a1 + a3 = 1. Applying these to the first given condition (a), we have a4 + a5 + · · · + a1999 = 0, so that a4 = a5 = · · · = a1999 = 0. Therefore S = a2 1 + a2 2 + a2 3 = a2 1 + 1 + (1 −a1)2 since a2 = a1 + a3 = 1 = 2(a2 1 −a1 + 1) = 2 a1 −1 2 2 + 3 2.
Thus S has maximum value 2 attained when a1 = 1, and minimum value 3 2 when a1 = 1 2.
122 Poland 1.15 Poland Problem 1 Let D be a point on side BC of triangle ABC such that AD > BC. Point E on side AC is defined by the equation AE EC = BD AD −BC .
Show that AD > BE.
First Solution: Fix the points B, C, D and the distance AD, and let A vary; its locus is a circle with center D. From the equation, the ratio AE EC is fixed; therefore, λ = EC AC is also fixed. Since E is the image of A under a homothety about C with ratio λ, the locus of all points E is the image of the locus of A under this homothety — a circle centered on BC. Then BE has its unique maximum when E is the intersection of the circle with line BC farther from B. If we show that AD = BE in this case then we are done (the original inequality AD > BE will be strict because equality can only hold in this degenerate case). Indeed, in this case the points B, D, C, E, A are collinear in that order; our equation gives AE · (AC −BD) = AE · (AD −BC) = EC · BD ⇒AE · AC = (AE + EC) · BD = AC · BD ⇒AE = BD ⇒AD = BE.
Second Solution: Let F be the point on AD such that FA = BC, and let line BF hit side AC at E′. By the law of sines we have AE′ = FA · sin ∠AF E′ sin ∠F E′A = CB · sin ∠DF B sin ∠CE′F and E′C = CB · sin ∠E′BC sin ∠CE′B = CB · sin ∠F BD sin ∠CE′F . Hence AE′ E′C = sin ∠DF B sin ∠F BD = DB F D = BD AD−BC = AE EC , and E′ = E.
Let ℓbe the line passing through A parallel to side BC. Draw G on ray BC such that BG = AD and CG = FD; and let lines GE and ℓintersect at H. Triangles ECG and EAH are similar, so AH = CG · AE EC = FD · AE EC .
By Menelaus’ Theorem applied to triangle CAD and line EFB, we have CE · AF · DB EA · FD · BC = 1.
1999 National Contests: Problems and Solutions 123 Thus AH = FD · AE EC = FD · AF ·DB F D·BC = DB · AF BC = DB, implying that quadrilateral BDAH is a parallelogram and that BH = AD. It follows that triangle BHG is isosceles with BH = BG = AD; and since BE in a cevian in this triangle, we must have BE < AD, as desired.
Problem 2 Given are nonnegative integers a1 < a2 < · · · < a101 smaller than 5050. Show that one can choose four distinct integers ak, al, am, an such that 5050 | (ak + al −am −an).
Solution: First observe that the ai are all distinct modulo 5050 since they are all between 0 and 5050. Now consider all sums ai +aj, i < j; there are 101 2 = 5050 such sums. If any two such sums, ak + al and am + an, are congruent mod 5050, we are done. (In this case, all four indices would indeed be distinct: if, for example, k = m, then we would also have l = n since all ai are different mod 5050, but we chose the pairs {k, l} and {m, n} to be distinct.) The only other possibility is that these sums occupy every possible congruence class mod 5050. Then, adding all such sums gives 100(a1+a2+· · ·+a101) ≡ 0 + 1 + · · · + 5049 = 2525 · 5049 (mod 5050). Since the number on the left side is even but 2525 · 5049 is odd, we get a contradiction.
Problem 3 For a positive integer n, let S(n) denote the sum of its digits. Prove that there exist distinct positive integers {ni}1≤i≤50 such that n1 + S(n1) = n2 + S(n2) = · · · = n50 + S(n50).
Solution: We show by induction on k that there exist positive integers n1, . . . , nk with the desired property. For k = 1 the statement is obvious. For k > 1, we have (by induction) numbers m1 < · · · < mk−1 with the desired property.
Note that we can make all mi arbitrarily large, e.g. by adding some large power of 10 to all of them (which preserves our property). Then, choose m with 1 ≤m ≤9 and m ≡m1 + 1 (mod 9); recall that S(x) ≡x (mod 9). Then we have m1 −m + S(m1) −S(m) + 11 = 9ℓfor some integer ℓ; by choosing the mi large enough we can ensure 10ℓ> mk−1. Now let ni = 10ℓ+1 +mi for i < k and nk = m+10ℓ+1 −10. Now it is obvious 124 Poland that ni + S(ni) = nj + S(nj) for i, j < k, and n1 + S(n1) = (10l+1 + m1) + (1 + S(m1)) = (m1 + S(m1) + 1) + 10l+1 = (9ℓ+ S(m) + m −10) + 10ℓ+1 = (m + 10l+1 −10) + (9ℓ+ S(m)) = nk + S(nk), as needed.
Problem 4 Find all integers n ≥2 for which the system of equations x2 1 + x2 2 + 50 = 16x1 + 12x2 x2 2 + x2 3 + 50 = 16x2 + 12x3 . . . . . . . . .
x2 n−1 + x2 n + 50 = 16xn−1 + 12xn x2 n + x2 1 + 50 = 16xn + 12x1 has a solution in integers (x1, x2, . . . , xn).
Solution: Answer: 3 | n.
We rewrite the equation x2 + y2 + 50 = 16x + 12y as (x −8)2 + (y −6)2 = 50, whose integer solutions are (7, −1), (7, 13), (9, −1), (9, 13), (3, 1), (3, 11) (13, 1), (13, 11), (1, 5), (1, 7), (15, 5), (15, 7).
Thus every pair (xi, xi+1) (where xn+1 = x1) must be one of these.
If 3 | n then just let x3i = 1, x3i+1 = 7, x3i+2 = 13 for each i.
Conversely, if a solution exists, consider the pairs (xi, xi+1) which occur; every pair’s first coordinate is the second coordinate of another pair, and vice versa, which reduces the above possibilities to (1, 7), (7, 13), (13, 1). It follows that the xi must form a repeating sequence 1, 7, 13, 1, 7, 13, . . . , which is only possible when 3 | n.
Problem 5 Let a1, a2, . . . , an, b1, b2, . . . , bn be integers. Prove that X 1≤i<j≤n (|ai −aj| + |bi −bj|) ≤ X 1≤i,j≤n |ai −bj|.
1999 National Contests: Problems and Solutions 125 Solution: Define f{a,b}(x) = 1 if either a ≤x < b or b ≤x < a, and 0 otherwise. Observe that when a, b are integers, |a−b| = P x f{a,b}(x) where the sum is over all integers (the sum is valid since only finitely many terms are nonzero). Now suppose a≤is the number of values of i for which ai ≤x, and a>, b≤, b> are defined analogously. We have (a≤−b≤) + (a> −b>) = (a≤+ a>) −(b≤+ b>) = n −n = 0 ⇒ (a≤−b≤)(a> −b>) ≤0. Thus a≤a> + b≤b> ≤a≤b> + a>b≤. But a≤a> = P i 0, we have (a, b) ∈{(3, 12), (4, 9), (5, 8), (8, 7)}, and there-fore the side lengths of the triangle are either (9, 12, 15), (8, 15, 17), or (7, 24, 25).
Problem 7.2 Let a, b, c be nonzero integers, a ̸= c, such that a c = a2 + b2 c2 + b2 .
Prove that a2 + b2 + c2 cannot be a prime number.
Solution: Cross-multiplying and factoring, we have (a −c)(b2 − ac) = 0.
Since a ̸= c, we have ac = b2.
Now, a2 + b2 + c2 = a2 +(2ac−b2)+c2 = (a+c)2 −b2 = (a+b+c)(a−b+c). Also, |a|, |c| cannot both be 1. Then a2+b2+c2 > |a|+|b|+|c| ≥|a+b+c|, |a−b+c|, whence a2 + b2 + c2 cannot be a prime number.
Problem 7.3 Let ABCD be a convex quadrilateral with ∠BAC = ∠CAD and ∠ABC = ∠ACD. Rays AD and BC meet at E and rays AB and DC meet at F. Prove that (a) AB · DE = BC · CE; (b) AC2 < 1 2(AD · AF + AB · AE).
Solution: (a) Because ∠BAC + ∠CBA = ∠ECA, we have ∠ECD = ∠BAC.
Then △CDE ∼△ACE, and CE DE = AE CE . But since AC is the 128 Romania angle bisector of ∠A in triangle ABE, we also have AE CE = AB BC .
Thus CE DE = AB BC , whence AB · DE = BC · CE.
(b) Note that AC is an angle bisector of both triangle ADF and triangle AEB. Thus it is enough to prove that if XL is an angle bisector in an arbitrary triangle XY Z, then XL2 < XY · XZ.
Let M be the intersection of − − → XL and the circumcircle of triangle XY Z. Because △XY L ∼△XMZ, we have XL2 < XL · XM = XY · XZ, as desired.
Problem 7.4 In triangle ABC, D and E lie on sides BC and AB, respectively, F lies on side AC such that EF ∥BC, G lies on side BC such that EG ∥AD. Let M and N be the midpoints of AD and BC, respectively. Prove that (a) EF BC + EG AD = 1; (b) the midpoint of FG lies on line MN.
Solution: (a) Since EF ∥BC, △AEF ∼△ABC and EF BC = AE AB . Similarly, since EG ∥AD, △BEG ∼△BAD and EG AD = EB AB .
Hence EF BC + EG AD = 1.
(b) Let lines AN, EF intersect at point P, and let Q be the point on line BC such that PQ ∥AD. Since BC ∥EF and N is the midpoint of BC, P is the midpoint of EF.
Then vector EP equals both vectors PF and GQ, and PFQG is a parallelogram.
Thus the midpoint X of FG must also be the midpoint of PQ.
But then since M is the midpoint of AD and AD ∥PQ, points M, X, N must be collinear.
Problem 8.1 Let p(x) = 2x3 −3x2 + 2, and let S = {p(n) | n ∈N, n ≤1999}, T = {n2 + 1 | n ∈N}, U = {n2 + 2 | n ∈N}.
Prove that S ∩T and S ∩U have the same number of elements.
Solution: Note that |S ∩T| is the number of squares of the form 2n3 −3n2 + 1 = (n −1)2(2n + 1) where n ∈N, n ≤1999. And for 1999 National Contests: Problems and Solutions 129 n ≤1999, (n −1)2(2n + 1) is a square precisely when either n = 1 or when n ∈{ 1 2(k2 −1) | k = 1, 3, 5, . . . , 63}. Thus, |S ∩T| = 33.
Next, |S ∩U| is the number of squares of the form 2n3 −3n2 = n2(2n −3) where n ∈N, n ≤1999. And for n ≤1999, n2(2n −3) is a square precisely when either n = 0 or when n ∈{ 1 2(k2 + 3) | k = 1, 3, 5, . . . , 63}. Thus |S ∩U| = 33 as well, and we are done.
Problem 8.2 (a) Let n ≥2 be a positive integer and x1, y1, x2, y2, . . . , xn, yn be positive real numbers such that x1 + x2 + · · · + xn ≥x1y1 + x2y2 + · · · + xnyn.
Prove that x1 + x2 + · · · + xn ≤x1 y1 + x2 y2 + · · · + xn yn .
(b) Let a, b, c be positive real numbers such that ab + bc + ca ≤3abc.
Prove that a3 + b3 + c3 ≥a + b + c.
Solution: (a) Applying the Cauchy-Schwarz inequality and then the given inequality, we have n X i=1 xi !2 ≤ n X i=1 xiyi · n X i=1 xi yi ≤ n X i=1 xi · n X i=1 xi yi .
Dividing both sides by Pn i=1 xi yields the desired inequality.
(b) By the AM-HM inequality on a, b, c we have a + b + c ≥ 9abc ab + bc + ca ≥9abc 3abc = 3.
Then, since the given condition is equivalent to 1 a + 1 b + 1 c ≤3, we have a + b + c ≥1 a + 1 b + 1 c. Hence setting x1 = a, x2 = b, x3 = c and y1 = 1 a2 , y2 = 1 b2 , y3 = 1 c2 in the result from part (a) gives a + b + c ≤a3 + b3 + c3, as desired.
130 Romania Problem 8.3 Let ABCDA′B′C′D′ be a rectangular box, let E and F be the feet of perpendiculars from A to lines A′D and A′C respectively, and let P and Q be the feet of perpendiculars from B′ to lines A′C′ and A′C respectively. Prove that (a) planes AEF and B′PQ are parallel; (b) triangles AEF and B′PQ are similar.
Solution: (a) Let (P1P2 . . . Pk) denote the plane containing points P1, P2, . . . , Pk.
First observe that quadrilateral A′B′CD is a parallelogram and thus lies in a single plane.
We are given that AE ⊥A′D. Also, line AE is contained in plane (ADD′A), which is perpendicular to line CD. Hence AE ⊥ CD as well, and therefore AE ⊥(A′B′CD) and AE ⊥A′C.
And since we know that A′C ⊥AF, we have A′C ⊥(AEF) and A′C ⊥EF.
Likewise, B′Q ⊥A′C. And since lines EF, B′Q, and A′C all lie in plane (A′B′CD), it follows that EF ∥B′Q. In a similar way we deduce that AF ∥PQ. Hence the planes (AEF) and (B′PQ) are parallel, as desired.
(b) Since EF ∥B′Q and FA ∥QP, we have ∠EFA = ∠PQB′.
Furthermore, from above AE ⊥EF and likewise B′P ⊥PQ, implying that ∠AEF = ∠B′PQ = 90◦as well.
Therefore △AEF ∼△B′PQ, as desired.
Problem 8.4 Let SABC be a right pyramid with equilateral base ABC, let O be the center of ABC, and let M be the midpoint of BC.
If AM = 2SO and N is a point on edge SA such that SA = 25SN, prove that planes ABP and SBC are perpendicular, where P is the intersection of lines SO and MN.
Solution: Let AB = BC = CA = s. Then some quick calculations show that AO = √ 3 3 s, AM = √ 3 2 s, AS = 5 √ 48s, and AN = 24 5 √ 48s.
Then AO · AM = AN · AS = 1 2s2, whence MONS is a cyclic quadrilateral.
Thus, ∠MNS = 90◦, and P is the orthocenter of triangle AMS.
Let Q be the intersection of lines AP and MS.
Note that ∠AMB = ∠AQM = ∠QMB = 90◦.
From repeated applications of the Pythagorean theorem, we have AB2 = AM 2 + 1999 National Contests: Problems and Solutions 131 MB2 = AQ2 + QM 2 + MB2 = AQ2 + QB2, whence ∠AQB = 90◦.
Now AQ ⊥QB and AQ ⊥QM, so line AQ must be perpendicular to plane SBC. Then since plane ABP contains line AQ, planes ABP and SBC must be perpendicular.
Problem 9.1 Let ABC be a triangle with angle bisector AD. One considers the points M, N on rays AB and AC respectively, such that ∠MDA = ∠ABC and ∠NDA = ∠BCA. Lines AD and MN meet at P. Prove that AD3 = AB · AC · AP.
Solution: Since △ADB ∼△AMD, AD AB = AM AD . Also, ∠MAN + ∠NDM = π, whence AMDN is cyclic. Since ∠DCA = ∠ADN = ∠AMN, △ADC ∼△APM, and AD AP = AC AM . Therefore, AD AB AD AC AD AP = AM AD AD AC AC AM = 1.
Problem 9.2 For a, b > 0, denote by t(a, b) the positive root of the equation (a + b)x2 −2(ab −1)x −(a + b) = 0.
Let M = {(a, b) | a ̸= b, t(a, b) ≤ √ ab}. Determine, for (a, b) ∈M, the minimum value of t(a, b).
Solution: Consider the polynomial P(x) = (a+b)x2 −2(ab−1)x− (a + b) = 0. Since a + b ̸= 0, the product of its roots is −a+b a+b = −1.
Hence P must have a unique positive root t(a, b) and a unique negative root. Since the leading coefficient of P(x) is positive, the graph of P(x) is positive for x > t(a, b) and negative for 0 ≤x < t(a, b) (since in the latter case, x is between the two roots). Thus, the condition t(a, b) ≤ √ ab is equivalent to P( √ ab) ≥0, or (ab −1)(a + b −2 √ ab) ≥0.
But a + b > 2 √ ab by AM-GM, where the inequality is sharp since a ̸= b. Thus t(a, b) ≤ √ ab exactly when ab ≥1.
Now using the quadratic formula, we find that t(a, b) = ab −1 a + b + sab −1 a + b 2 + 1.
Thus given ab ≥1, we have t(a, b) ≥1 with equality when ab = 1.
132 Romania Problem 9.3 In the convex quadrilateral ABCD the bisectors of angles A and C meet at I. Prove that there exists a circle inscribed in ABCD if and only if [AIB] + [CID] = [AID] + [BIC].
Solution: It is well known that a circle can be inscribed in a convex quadrilateral ABCD if and only if AB + CD = AD + BC.
The bisector of angle A consists of those points lying inside ∠BAD equidistant from lines AB and AD; similarly, the bisector of angle C consists of those points lying inside ∠BCD equidistant from lines BC and BD.
Suppose ABCD has an incircle. Then its center is equidistant from all four sides of the quadrilateral, so it lies on both bisectors and hence equals I. If we let r denote the radius of the incircle, then we have [AIB] + [CID] = r(AB + CD) = r(AD + BC) = [AID] + [BIC].
Conversely, suppose that [AIB] + [CID] = [AID] + [BIC]. Let d(I, ℓ) denote the distance from I to any line ℓ, and write x = d(I, AB) = d(I, AD) and y = d(I, BC) = d(I, CD). Then [AIB] + [CID] = [AID] + [BIC] AB · x + CD · y = AD · x + BC · y x (AB −AD) = y (BC −CD).
If AB = AD, then BC = CD and it follows that AB + CD = AD + BC. Otherwise, suppose that AB > AD; then BC > CD as well.
Consider the points A′ ∈AB and C′ ∈BC such that AD = AA′ and CD = CC′. By SAS, we have △AIA′ ∼ = △AID and △DCI ∼ = △C′IC. Hence IA′ = ID = IC′. Furthermore, subtracting [AIA′] + [DCI] = [AID] + [C′IC] from both sides of our given condition, we have [A′IB] = [C′IB] or IA′ · IB · sin ∠A′IB = IC′ · IB · sin ∠CIB. Thus ∠A′IB = ∠C′IB, and hence △A′IB ∼ = C′IB by SAS.
Thus ∠IBA′ = ∠IBC′, implying that I lies on the angle bisector of ∠ABC. Therefore x = d(I, AB) = d(I, BC) = y, and the circle centered at I with radius x = y is tangent to all four sides of the quadrilateral.
Problem 9.4 1999 National Contests: Problems and Solutions 133 (a) Let a, b ∈R, a < b. Prove that a < x < b if and only if there exists 0 < λ < 1 such that x = λa + (1 −λ)b.
(b) The function f : R →R has the property: f(λx + (1 −λ)y) < λf(x) + (1 −λ)f(y) for all x, y ∈R, x ̸= y, and all 0 < λ < 1. Prove that one cannot find four points on the function’s graph that are the vertices of a parallelogram.
Solution: (a) No matter what x is, there is a unique value λ = b−x b−a such that x = λa + (1 −λ)b; and 0 < b−x b−a < 1 ⇐ ⇒ a < x < b, which proves the claim.
(b) The condition is Jensen’s inequality and shows that the function f is strictly convex. Stated geometrically, whenever x < t < y the point (t, f(t)) lies strictly below the line joining (x, f(x)) and (y, f(y)). Suppose there were a parallelogram on the graph of f whose vertices, from left to right, have x-coordinates a, b, d, c.
Then either (b, f(d)) or (d, f(d)) must lie on or above the line joining (a, f(a)) and (c, f(c)), a contradiction.
Problem 10.1 Find all real numbers x and y satisfying 1 4x + 1 27y = 5 6 log27 y −log4 x ≥1 6 27y −4x ≤1.
Solution: First, for the second equation to make sense we must have x, y > 0 and thus 27y > 1. Now from the third equation we have 1 27y ≥ 1 4x + 1, which combined with the first equation gives 1 4x + 1 4x + 1 ≤5 6, 134 Romania whence x ≥1 2. Similarly, the first and third equations also give 5 6 ≤ 1 27y −1 + 1 27y , whence y ≤1 3. If either x > 1 2 or y < 1 3, we would have log27 y − log4 x < 1 6, contradicting the second given equation. Thus, the only solution is (x, y) = 1 2, 1 3 , which indeed satisfies all three equations.
Problem 10.2 A plane intersects edges AB, BC, CD, DA of the regular tetrahedron ABCD at points M, N, P, Q, respectively. Prove that MN · NP · PQ · QM ≥AM · BN · CP · DQ.
Solution: By the law of cosines in triangle MBN, we have MN 2 = MB2 + BN 2 −MB · BN ≥MB · BN.
Similarly, NP 2 ≥CN · CP, PN 2 ≥DP · DQ, and MQ2 ≥AQ · AM.
Multiplying these inequalities yields (MN · NP · PQ · MQ)2 ≥ (BM · CN · DP · AQ) · (AM · BN · CP · DQ) .
Now the given plane is different from plane (ABC) and (ADC).
Thus if it intersects line AC at some point T, then points M, N, T must be collinear—because otherwise, the only plane containing M, N, T would be plane (ABC). Therefore it intersects line AC at most one point T, and by Menelaus’ Theorem applied to triangle ABC and line MNT we have AM · BN · CT MB · NC · TA = 1.
Similarly, P, Q, T are collinear and AQ · DP · CT QD · PC · TA = 1.
Equating these two fractions and cross-multiplying, we find that AM · BN · CP · DQ = BM · CN · DP · AQ.
This is true even if the plane does not actually intersect line AC: in this case, we must have MN ∥AC and PQ ∥AC, in which case 1999 National Contests: Problems and Solutions 135 ratios of similar triangles show that AM · BN = BM · CN and CP · DQ = DP · AQ.
Combining this last equality with the inequality from the first paragraph, we find that (MN · NP · PQ · QM)2 ≥(AM · BN · CP · DQ)2, which implies the desired result.
Problem 10.3 Let a, b, c (a ̸= 0) be complex numbers. Let z1 and z2 be the roots of the equation az2 + bz + c = 0, and let w1 and w2 be the roots of the equation (a + c)z2 + (b + b)z + (a + c) = 0.
Prove that if |z1|, |z2| < 1, then |w1| = |w2| = 1.
Solution: We begin by proving that Re (b)2 ≤|a+c|2. If z1 = z2 = 0, then b = 0 and the claim is obvious. Otherwise, write a = m + ni and c = r + si; and write z1 = x + yi where t = |z1| = p x2 + y2 < 1.
Also note that r2 + s2 = |c|2 = |az1z2|2 < |a|2|z1|2 = (m2 + n2)t2.
(1) Assume WLOG that z1 ̸= 0. Then | Re (b)| = | Re (−b)| = | Re (az1 + c/z1)| = | Re (az1) + Re (c/z1)|; that is, | Re (b)| = |(mx −ny) + (rx + sy)/t2| = |x(m + r/t2) + y(s/t2 −n)| ≤ p x2 + y2p (m + r/t2)2 + (s/t2 −n)2 = t p (m + r/t2)2 + (s/t2 −n)2, where the inequality follows from Cauchy-Schwarz. Proving our claim then reduces to showing that t2 (m + r/t2)2 + (s/t2 −n)2 ≤(m + r)2 + (n −s)2 ⇐ ⇒(mt2 + r)2 + (st2 −n)2 ≤t2 (m + r)2 + (n −s)2 ⇐ ⇒(r2 + s2)(1 −t2) < (m2 + n2)(t4 −t2) ⇐ ⇒(1 −t2) (m2 + n2)t2 −(r2 + s2) .
136 Romania But 1 −t2 > 0 by assumption, and (m2 + n2)t2 −(r2 + s2) > 0 from (1); therefore our claim is true.
Now since |c/a| = |z1z2| < 1, we have |c| < |a| and a + c ̸= 0. Then by the quadratic equation, the roots to (a+c)z2+(b+b)z+(a+c) = 0 are given by −(b + b) ± q (b + b)2 −4(a + c)(a + c) 2(a + c) , or (dividing the numerator and denominator by 2) −Re (b) ± p Re (b)2 −|a + c|2 a + c = −Re (b) ± i p |a + c|2 −Re (b)2 a + c .
When evaluating either root, the absolute value of the numerator is p Re (b)2 + (|a + c|2 −Re (b)2) = |a + c|; and the absolute value of the denominator is clearly |a + c| as well. Therefore indeed |w1| = |w2| = 1, as desired.
Problem 10.4 (a) Let x1, y1, x2, y2, . . . , xn, yn be positive real numbers such that (i) x1y1 < x2y2 < · · · < xnyn; (ii) x1 + x2 + · · · + xk ≥y1 + y2 + · · · + yk for all k = 1, 2, . . . , n.
Prove that 1 x1 + 1 x2 + · · · + 1 xn ≤1 y1 + 1 y2 + · · · + 1 yn .
(b) Let A = {a1, a2, . . . , an} ⊂N be a set such that for all distinct subsets B, C ⊆A, X x∈B x ̸= X x∈C x. Prove that 1 a1 + 1 a2 + · · · + 1 an < 2.
Solution: (a) Let πi = 1 xiyi , δi = xi −yi for all 1 ≤i ≤n. We are given that π1 > π2 > · · · > πn > 0 and that Pk i=1 δi ≥0 for all 1 ≤k ≤n.
Note that n X k=1 1 yk −1 xk = n X k=1 πkδk 1999 National Contests: Problems and Solutions 137 = πn n X i=1 δi + n−1 X k=1 (πk −πk+1) (δ1 + δ2 + · · · + δk) ≥0, as desired.
(b) Assume without loss of generality that a1 < a2 < · · · < an, and let yi = 2i−1 for all i. Clearly, a1y1 < a2y2 < · · · < anyn.
For any k, the 2k −1 sums made by choosing at least one of the numbers a1, a2, . . . , ak are all distinct. Hence the largest of them, Pk i=1 ai, must be at least 2k −1. Thus for all k = 1, 2, . . . , n we have a1 + a2 + · · · + ak ≥2k −1 = y1 + y2 + · · · + yk.
Then by part (a), we must have 1 a1 + 1 a2 + · · · + 1 an < 1 y1 + 1 y2 + · · · + 1 yn = 2 − 1 2n−1 < 2, as desired.
IMO Selection Tests Problem 1 (a) Show that out of any 39 consecutive positive integers, it is possible to choose one number with the sum of its digits divisible by 11.
(b) Find the first 38 consecutive positive integers, none with the sum of its digits divisible by 11.
Solution: Call an integer “deadly” if its sum of digits is divisible by 11, and let d(n) equal the sum of the digits of a positive integer n.
If n ends in a 0, then the numbers n, n + 1, . . . , n + 9 differ only in their units digits, which range from 0 to 9; hence d(n), d(n + 1), . . . , d(n + 9) is an arithmetic progression with common difference 1. Thus if d(n) ̸≡1 (mod 11), then one of these numbers is deadly.
Next suppose that if n ends in k ≥0 nines.
Then d(n + 1) = d(n) + 1 −9k: the last k digits of n + 1 are 0’s instead of 9’s, and the next digit to the left is 1 greater than the corresponding digit in n.
Finally, suppose that n ends in a 0 and that d(n) ≡d(n + 10) ≡ 1 (mod 11). Since d(n) ≡1 (mod 11), we must have d(n + 9) ≡ 138 Romania 10 (mod 11). If n+9 ends in k 9’s, then we have 2 ≡d(n+10)−d(n+ 9) ≡1 −9k = ⇒k ≡6 (mod 11).
(a) Suppose we had 39 consecutive integers, none of them deadly.
One of the first ten must end in a 0: call it n. Since none of n, n + 1, . . . , n + 9 are deadly, we must have d(n) ≡1 (mod 11).
Similarly, d(n + 10) ≡1 (mod 11) and d(n + 20) ≡1 (mod 11).
From our third observation above, this implies that both n + 9 and n + 19 must end in at least six 9’s. But this is impossible, because n + 10 and n + 20 can’t both be multiples of one million!
(b) Suppose we have 38 consecutive numbers N, N + 1, . . . , N + 37, none of which is deadly. By an analysis similar to that in part (a), none of the first nine can end in a 0; hence, N + 9 must end in a 0, as must N + 19 and N + 29. Then we must have d(N + 9) ≡d(N + 19) ≡1 (mod 11). Therefore d(N + 18) ≡ 10 (mod 11); and furthermore, if N+18 ends in k 9’s we must have k ≡6 (mod 11). The smallest possible such number is 999999, yielding the 38 consecutive numbers 999981, 999982, . . . , 1000018.
And indeed, none of these numbers is deadly: their sums of digits are congruent to 1, 2, . . . , 10, 1, 2, . . . , 10, 1, 2, . . . , 10, 2, 3, . . . , 9, and 10 (mod 11), respectively.
Problem 2 Let ABC be an acute triangle with angle bisectors BL and CM. Prove that ∠A = 60◦if and only if there exists a point K on BC (K ̸= B, C) such that triangle KLM is equilateral.
Solution: Let I be the intersection of lines BL and CM. Then ∠BIC = 180◦−∠ICB −∠CBI = 180◦−1 2(∠C + ∠B) = 180◦− 1 2(180◦−∠A) = 90◦+ ∠A, and thus ∠BIC = 120◦if and only if ∠A = 60◦.
For the “only if” direction, suppose that ∠A = 60◦. Then let K be the intersection of BC and the internal angle bisector of ∠BIC; we claim that triangle KLM is equilateral.
Since ∠BIC = 120◦, we know that ∠MIB = ∠KIB = 60◦. And since ∠IBM = ∠IBK and IB = IB, by ASA congruency we have △IBM ∼ = △IBK; in particular, IM = IK.
Similarly, IL = IK; and since ∠KIL = ∠LIM = ∠MIK = 120◦, we know that triangle KLM is equilateral.
For the “if” direction, suppose that K is on BC and triangle KLM is equilateral. Consider triangles BLK and BLM: BL = BL, LM = LK, and ∠MBL = ∠KBL. There is no SSA congruency, but 1999 National Contests: Problems and Solutions 139 we do then know that either ∠LKB + ∠BML = 180◦or ∠LKB = ∠BML.
But since ∠KBM < 90◦and ∠MLK = 60◦, we know that ∠LKB + ∠BML > 210◦.
Thus ∠LKB = ∠BML, whence △BLK ∼ = △BLM, and BK = BM.
It follows that IK = IM.
Similarly, IL = IK, and I is the circumcenter of triangle KLM.
Thus ∠LIM = 2∠LKM = 120◦, giving ∠BIC = ∠LIM = 120◦and ∠A = 60◦.
Problem 3 Show that for any positive integer n, the number Sn = 2n + 1 0 · 22n + 2n + 1 2 · 22n−2 · 3 + · · · + 2n + 1 2n · 3n is the sum of two consecutive perfect squares.
Solution: Let α = 1+ √ 3, β = 1− √ 3, and Tn = 1 2 α2n+1 + β2n+1 .
Note that αβ = −2, α2 2 = 2 + √ 3, and β2 2 = 2 − √ 3. Also, applying the binomial expansion to (1 + √ 3)n and (1 − √ 3)n, we find that Tn = Pn k=0 2n+1 2k 3k—which is an integer for all n.
Applying the binomial expansion to (2+ √ 3)2n+1 and (2− √ 3)2n+1 instead, we find that Sn = α2 2 2n+1 + β2 2 2n+1 4 = α4n+2 + β4n+2 22n+3 = α4n+2 + 2(αβ)2n+1 + β4n+2 22n+3 + 1 2 = α2n+1 + β2n+12 22n+3 + 1 2 = T 2 n 22n+1 + 1 2.
Thus 22n+1Sn = T 2 n + 22n. Then 22n | T 2 n but 22n+1 ̸ | T 2 n, and hence Tn ≡2n (mod 2n+1). Therefore Sn = T 2 n 22n+1 + 1 2 = Tn −2n 2n+1 2 + Tn + 2n 2n+1 2 is indeed the sum of two consecutive perfect squares.
140 Romania Problem 4 Show that for all positive real numbers x1, x2, · · · , xn such that x1x2 · · · xn = 1, the following inequality holds: 1 n −1 + x1 + 1 n −1 + x2 + · · · + 1 n −1 + xn ≤1.
First Solution: Let a1 = n √x1, a2 = n √x2, . . . , an = n √xn. Then a1a2 · · · an = 1 and 1 n −1 + xk = 1 n −1 + an k = 1 n −1 + an−1 k a1···ak−1ak+1···an ≤ 1 n −1 + (n−1)an−1 k an−1 1 +···+an−1 k−1+an−1 k+1 +···+an−1 n by the AM-GM Inequality. It follows that 1 n −1 + xk ≤ an−1 1 + · · · + an−1 k−1 + an−1 k+1 + · · · + an−1 n (n −1)(an−1 1 + an−1 2 + · · · + an−1 n ) .
Summing up yields Pn k=1 1 n−1+xk ≤1, as desired.
Second Solution: Let f(x) = 1 n−1−x; we wish to prove that Pn i=1 f(xi) ≤1. Note that f(y) + f(z) = 2(n −1) + y + z (n −1)2 + yz + (y + z)(n −1).
Suppose that any of our xi does not equal 1; then we have xj < 1 < xk for some j, k. If f(xj) + f(xk) ≤ 1 n−1, then all the other f(xi) are less than 1 n−1. But then Pn i=1 f(xi) < 1 and we are done.
Otherwise, f(xj) + f(xk) > 1 n−1. Now set x′ j = 1 and x′ k = xjxk; then x′ jx′ k = xjxk, while xj < 1 < xk ⇒(1 −xj)(xk −1) > 0 ⇒xj + xk > x′ j +x′ k. Let a = 2(n−1), b = (n−1)2 +xjxk = (n−1)2 +x′ jx′ k, and c = 1 n−1; also let m = xj + xk and m′ = x′ j + x′ k. Then we have f(xj) + f(xk) = a + cm b + m and f(x′ j) + f(x′ k) = a + cm′ b + m′ .
Now a+cm b+m > c ⇒a + cm > (b + m)c ⇒a b > c; and from here, (a −bc)(m −m′) > 0 ⇒a + cm′ b + m′ > a + cm b + m 1999 National Contests: Problems and Solutions 141 ⇒f(x′ j) + f(x′ k) = f(xj) + f(xk).
Hence as long as no pair f(xj)+f(xk) ≤ 1 n−1 and the xi do not all equal 1, we can continually replace pairs xj and xk (neither equal to 1) by 1 and xjxk. This keeps the product x1x2 · · · xn equal to 1 while increasing Pn i=1 f(xi).
Then eventually our new Pn i=1 f(xi) ≤1, which implies that our original Pn i=1 f(xi) was also at most 1. This completes the proof.
Third Solution: Suppose, for the sake of contradiction, that 1 n −1 + x1 + 1 n −1 + x2 +· · ·+ 1 n −1 + xn > 1. Letting yi = xi/(n−1) for i = 1, 2, . . . , n, we have 1 1 + y1 + 1 1 + y2 + · · · + 1 1 + yn > n −1 and hence 1 1 + y1 > 1 − 1 1 + y2 + 1 − 1 1 + y3 + · · · + 1 − 1 1 + yn = y2 1 + y2 + y3 1 + y3 + · · · + yn 1 + yn > (n −1) n−1 r y2y3 · · · yn (1 + y2)(1 + y3) · · · (1 + yn).
We have analagous inequalities with 1 1+y2 , 1 1+yn , . . . , 1 1+yn on the left hand side; multiplying these n inequalities together gives n Y k=1 1 1 + yk > (n −1)n y1y2 · · · yn (1 + y1)(1 + y2) · · · (1 + yn) 1 > ((n −1)y1)((n −1)y2) · · · ((n −1)yn) = x1x2 · · · xn, a contradiction.
Problem 5 Let x1, x2, . . . , xn be distinct positive integers. Prove that x2 1 + x2 2 + · · · + x2 n ≥(2n + 1)(x1 + x2 + · · · + xn) 3 .
Solution: Assume without loss of generality that x1 < x2 < · · · < xn. We will prove that 3x2 k ≥2(x1 + x2 + · · · + xk−1) + (2k + 1)xk; then, summing this inequality over k = 1, 2, . . . , n, we will have the desired inequality.
142 Romania First, x1 + x2 + · · · + xk−1 ≤(xk −(k −1)) + (xk −(k −2)) + · · · + (xk −1) = (k −1)xk −k(k−1) 2 . Thus, 2(x1 + x2 + · · · + xk−1) + (2k + 1)xk ≤(4k −1)xk −k(k −1).
Now 3x2 k −[(4k −1)xk −k(k −1)] = xk(3xk −4k + 1) + k(k −1), which is minimized at xk = 2 3k. Then since xk ≥k, xk(3xk −4k + 1) + k(k −1) ≥k(3k −4k + 1) + k(k −1) = 0 so 3x2 k ≥(4k −1)xk −k(k −1) ≥2(x1 + x2 + · · · + xk−1) + (2k + 1)xk, and we have finished.
Problem 6 Prove that for any integer n, n ≥3, there exist n positive integers a1, a2, . . . , an in arithmetic progression, and n positive integers b1, b2, . . . , bn in geometric progression, such that b1 < a1 < b2 < a2 < · · · < bn < an.
Give one example of such progressions a1, a2, . . . , an and b1, b2, . . . , bn each having at least 5 terms.
Solution: Our strategy is to find progressions where bn = an−1 + 1 and bn−1 = an−2 + 1.
Write d = an−1 −an−2.
Then for all 2 ≤i, j ≤n −1 we have bi+1 −bi ≤bn −bn−1 = d, so that bj = bn + Pn−1 i=j (bi −bi+1) > an−1 + (n −j)d = aj−1.
And if we ensure that b1 < a1, then bj = b1 + Pj−1 i=1(bi+1 −bi) ≤ a1 + (j −1)d = aj for all j, so the chain of inequalities is satisfied.
Let b1, b2, . . . , bn equal kn−1, kn−2(k + 1), . . . , k0(k + 1)n−1, where k is a value to be determined later.
Also set an−1 = bn −1 and an−2 = bn−1 −1, and define the other ai accordingly.
Then d = an−an−1 = bn−bn−1 = (k+1)n−2, and a1 = (k+1)n−2(k+3−n)−1.
Thus, we need only pick k such that (k + 1)n−2(k + 3 −n) −1 −kn−1 > 0.
Viewing the left hand side as a polynomial in k, the coefficient of kn−1 is zero but the coefficient of kn−2 is 1. Therefore, it is positive 1999 National Contests: Problems and Solutions 143 for sufficiently large k and we can indeed find satisfactory sequences a1, a2, . . . , an and b1, b2, . . . , bn.
For n = 5, we seek k such that (k + 1)3(k −2) −1 −k4 > 0.
Computation shows that k = 5 works, yielding 625 < 647 < 750 < 863 < 900 < 1079 < 1080 < 1295 < 1296 < 1511.
Problem 7 Let a be a positive real number and {xn} (n ≥1) be a sequence of real numbers such that x1 = a and xn+1 ≥(n + 2)xn − n−1 X k=1 kxk, for all n ≥1. Show that there exists a positive integer n such that xn > 1999!.
Solution: We will prove by induction on n ≥1 that xn+1 > n X k=1 kxk > a · n!.
For n = 1, we have x2 ≥3x1 > x1 = a.
Now suppose that the claim holds for all values up through n. Then xn+2 ≥(n + 3)xn+1 − n X k=1 kxk = (n + 1)xn+1 + 2xn+1 − n X k=1 kxk > (n + 1)xn+1 + 2 n X k=1 kxk − n X k=1 kxk = n+1 X k=1 kxk, as desired. Furthermore, x1 > 0 by definition and x2, x3, . . . , xn are also positive by the induction hypothesis; thus xn+2 > (n+1)xn+1 > (n + 1) (a · n!) = a · (n + 1)!. This completes the inductive step.
Therefore for sufficiently large n, we have xn+1 > n! · a > 1999!.
Problem 8 Let O, A, B, C be variable points in the plane such that OA = 4, OB = 2 √ 3 and OC = √ 22. Find the maximum possible area of triangle ABC.
144 Romania Solution: We first look for a tetrahedron MNPQ with the following properties: (i) if H is the foot of the perpendicular from M to plane (NPQ), then HN = 4, HP = 2 √ 3, and HQ = √ 22; and (ii) lines MN, MP, MQ are pairwise perpendicular.
If such a tetrahedron exists, then let O = H and draw triangle ABC in plane (NPQ). We have MA = √ MO2 + OA2 = √ MH2 + HN 2 = MN, and similarly MB = MP and MC = MQ. Hence [ABCM] ≤1 3[ABM] · MC ≤1 3 · 1 2MA · MB · MC = 1 3 · 1 2MN · MP · MQ = [MNPQ], and therefore the maximum possible area of triangle [ABC] is [NPQ].
It remains to find tetrahedron MNPQ.
Let x = MH; then MN = √ x2 + 16, MP = √ x2 + 12, and MQ = √ x2 + 22. By the Pythagorean Theorem on triangle MHN, we have NH = 4. Next let lines NH and PQ intersect at R; then in similar right triangles MHN and MRN, we have MR = MH · MN NH = 1 4(x2 + 16).
Since MN ⊥(MPQ) we have MN ⊥PQ; and since MH ⊥ (NPQ) we have MH ⊥PQ as well.
Hence PQ ⊥(MNHR), so that MR is an altitude in the right triangle MPQ.
Therefore MR · PQ = 2[MPQ] = MP · MQ, or (after squaring both sides) r (x2 + 16)2 16 −(x2 + 16) p x2 + 12 + x2 + 22 = p x2 + 12 p x2 + 22.
Setting 4y = x2 + 16 and squaring both sides, we obtain (y2 −4y)(8y + 2) = (4y −4)(4y + 6)(y −6)(4y2 + y −2) = 0.
Since y = 1 4(x2 + 16) > 4, the only solution is y = 6 = ⇒x = √ 8.
Then by taking MN = √ 24, MP = √ 20, MQ = √ 30, we get the required tetrahedron.
Then [MNPQ] equals both 1 3MH · [MPQ] and 1 6MN · MP · MQ.
Setting these two expressions equal, we find that the maximum area of [ABC] is [NPQ] = MN · MP · MQ 2 · MH = 15 √ 2.
Problem 9 Let a, n be integers and let p be a prime such that p > |a| + 1. Prove that the polynomial f(x) = xn + ax + p cannot be 1999 National Contests: Problems and Solutions 145 represented as a product of two nonconstant polynomials with integer coefficients.
Solution: Let z be a complex root of the polynomial. We shall prove that |z| > 1. Suppose |z| ≤1. Then, zn + az = −p, we deduce that p = |zn + az| = |z||zn−1 + a| ≤|zn−1| + |a| ≤1 + |a|, which contradicts the hypothesis.
Now, suppose f = gh is a decomposition of f into nonconstant polynomials with integer coefficients. Then p = f(0) = g(0)h(0), and either |g(0)| = 1 or |h(0)| = 1. Assume without loss of generality that |g(0)| = 1. If z1, z2, . . . , zk are the roots of g then they are also roots of f. Therefore 1 = |g(0)| = |z1z2 · · · zk| = |z1||z2| · · · |zk| > 1, a contradiction.
Problem 10 Two circles meet at A and B. Line ℓpasses through A and meets the circles again at C and D respectively. Let M and N be the midpoints of arcs d BC and d BD, which do not contain A, and let K be the midpoint of CD. Prove that ∠MKN = 90◦.
Solution: All angles are directed modulo 180◦.
Let M ′ be the reflection of M across K.
Then triangles MKC and M ′KD are congruent in that order, and M ′D = MC. Because M is the midpoint of d BC, we have M ′D = MC = MB; and similarly, because N is the midpoint of d BD we have BN = DN.
Next, ∠MBN = (180◦−∠ABM)+(180◦−∠NBA) = ∠MCA+∠ADN = ∠M ′DA+ ∠ADN = ∠M ′DN. Hence △M ′DN ∼ = △MBN, and MN = M ′N.
Therefore NK is the median to the base of isosceles triangle MNM ′, so it is also an altitude and NK ⊥MK.
Problem 11 Let n ≥3 and A1, A2, . . . , An be points on a circle.
Find the greatest number of acute triangles having vertices in these points.
Solution: Without loss of generality assume the points A1, A2, . . . , An are ordered in that order counterclockwise; also, take indices modulo n os that An+1 = A1, An+2 = A2, and so on.
Denote by AiAj 146 Romania the arc of the circle starting from Ai and ending in Aj in the counterclockwise direction; let m(AiAj) denote the angle measure of the arc; and call an arc AiAj obtuse if m(AiAj) ≥180◦. Obviously, m(AiAj) + m(AjAi) = 360◦, and thus at least one of the arcs AiAj and AjAi is obtuse. Let xs be the number of obtuse arcs each having exactly s −1 points along their interiors. If s ̸= n 2 , then for each i at least one of the arcs AiAi+s or Ai+sAi is obtuse; summing over all i, we deduce that xs + xn−s ≥n (1) for every s ̸= n 2 ; and similar reasoning shows that this inequality also holds even when s = n 2 . For all s, equality holds if and only if there are no diametrically opposite points Ai, Ai+s.
The number of non-acute triangles AiAjAk equals the number of non-acute angles ∠AiAjAk. And for each obtuse arc AiAk containing s−1 points in its interior, there are n−s−1 non-acute angles AiAjAk: namely, with those Aj in the interior of arc AkAi. It follows that the number N of non-acute triangles is N = x1(n −2) + x2(n −3) + . . . + xn−3 · 2 + xn−2 · 1 + xn−1 · 0.
By regrouping terms and using (1) we obtain N ≥ n−1 2 X s=1 (s −1) · (xn−s + xs) ≥n 1 + 2 + · · · + n −3 2 = n(n −1)(n −3) 8 if n is odd, and N ≥ n−2 2 X s=1 (s −1) · (xn−s + xs) + n −2 2 xn/2 ≥n 1 + 2 + · · · + n −4 2 + n −2 2 · n 2 = n(n −2)2 8 if n is even.
Equality is obtained when there are no diametrically opposite points, and when xk = 0 for k < n 2 . When n is odd, for instance, this happens when the points form a regular n-gon; and when n is even, equality occurs when m(A1A2) = m(A2A3) = · · · = m(An−1An) = 360◦ n + ϵ where 0 < ϵ < 360◦ n2 .
1999 National Contests: Problems and Solutions 147 Finally, note that the total number of triangles having vertices in the n points is n 3 = n(n−1)(n−2) 6 . Subtracting the minimum values of N found above, we find that the maximum number of acute angles is (n−1)n(n+1) 24 if n is odd, and (n−2)n(n+2) 24 if n is even.
Problem 12 The scientists at an international conference are either native or foreign. Each native scientist sends exactly one message to a foreign scientist and each foreign scientist sends exactly one message to a native scientist, although at least one native scientist does not receive a message. Prove that there exists a set S of native scientists and a set T of foreign scientists such that the following conditions hold: (i) the scientists in S sent messages to exactly those foreign scientists who were not in T (that is, every foreign scientist not in T received at least one message from somebody in S, but none of the scientists in T received any messages from scientists in S); and (ii) the scientists in T sent messages to exactly those native scientists not in S.
Solution: Let A be the set of native scientists and B be the set of foreign scientists. Let f : A →B and g : B →A be the functions defined as follows: f(a) is the foreign scientist receiving a message from a, and g(b) is the native scientist receiving a message from b. If such subsets S, T exist we must have T = B −f(S); hence we have to prove that there exists a subset S ⊆A such that A−S = g(B−f(S)).
For each subset X ⊆A, let h(X) = A−g(B−f(X)). If X ⊆Y , then f(X) ⊆f(Y ) = ⇒B −f(Y ) ⊆B −f(X) = ⇒g(B −f(Y )) ⊆g(B − f(X)) = ⇒A −g(B −f(X)) ⊆A −g(B −f(Y )) = ⇒h(X) ⊆h(Y ).
Let M = {X ⊆A | h(X) ⊆X}. The set M is nonempty, since A ∈M. Furthermore, it is given that g is not surjective, so that some native scientist a0 is never in g(B −f(X)) and thus always in h(X) for all X ⊆A. Thus every subset in M contains a0, so that S = T X∈M X is nonempty.
From the definition of S we have h(S) ⊆S. And from the monotony of h it follows that h(h(S)) ⊆h(S); thus, h(S) ∈M and S ⊂h(S).
Combining these results, we have S = h(S), as desired.
Problem 13 A polyhedron P is given in space. Determine whether there must exist three edges of P that can be the sides of a triangle.
148 Romania Solution: The answer is “yes.” Assume, for the purpose of contradiction, that there exists a polyhedron P in which no three edges can form the sides of a triangle.
Let the edges of P be E1, E2, E3, . . . , En, in non-increasing order of length; let ei be the length of Ei. Consider the two faces that share E1: for each of those faces, the sum of the lengths of all its edges except E1 is greater than e1. Therefore, e2 + e3 + · · · + en > 2e1.
But, since we are assuming that no three edges of P can form the sides of a triangle, we have ei+1 + ei+2 ≤ei for i = 1, 2, . . . , n −2.
Hence, 2(e2 + e3 + · · · + en) = e2 + (e2 + e3) + (e3 + e4) + · · · + (en−1 + en) + en ≤e2 + (e1) + (e2) + · · · + (en−2) + en, so e2 + e3 + · · · + en ≤e1 + e2 −en−1 < e1 + e1 + 0 = 2e1, a contradiction. Thus, our assumption was incorrect and some three edges can be the sides of a triangle.
1999 National Contests: Problems and Solutions 149 1.17 Russia Fourth round Problem 8.1 A father wishes to take his two sons to visit their grandmother, who lives 33 kilometers away. He owns a motorcycle whose maximum speed is 25 km/h. With one passenger, its maximum speed drops to 20 km/h. (He cannot carry two passengers.) Each brother walks at a speed of 5 km/h. Show that all three of them can reach the grandmother’s house in 3 hours.
Solution: Have the father drive his first son 24 kilometers, which takes 6 5 hours; then drive back to meet his second son 9 kilometers from home, which takes 3 5 hours; and finally drive his second son 6 5 more hours.
Each son spends 6 5 hours riding 24 kilometers, and 9 5 hours walking 9 kilometers. Thus they reach their grandmother’s house in exactly 3 hours — as does the father, who arrives at the same time as his second son.
Problem 8.2 The natural number A has the following property: the sum of the integers from 1 to A, inclusive, has decimal expansion equal to that of A followed by three digits. Find A.
Solution: We know that k = (1 + 2 + · · · + A) −1000A = A(A + 1) 2 −1000A = A A + 1 2 −1000 is between 0 and 999, inclusive. If A < 1999 then k is negative. If A ≥2000 then A+1 2 −1000 ≥1 2 and k ≥1000. Therefore A = 1999, and indeed 1 + 2 + · · · + 1999 = 1999000.
Problem 8.3 On sides BC, CA, AB of triangle ABC lie points A1, B1, C1 such that the medians A1A2, B1B2, C1C2 of triangle A1B1C1 are parallel to AB, BC, CA, respectively. Determine in what ratios the points A1, B1, C1 divide the sides of ABC.
1 Problems are numbered as they appeared in the contests.
Problems that appeared more than once in the contests are only printed once in this book.
150 Russia First Solution: A1, B1, C1 divide sides BC, CA, AB in 1 : 2 ratios (so that BA1 A1C = 1 2, and so on).
Lemma. In any triangle XY Z, the medians can be translated to form a triangle. Furthermore, the medians of this new triangle are parallel to the sides of triangle XY Z.
Proof: Let x, y, z denote the vectors − − → Y Z, − − → ZX, − − → XY respectively; then x + y + z = − → 0 . Also, the vectors representing the medians of triangle XY Z are mx = z+ x 2, my = x+ y 2, mz = y+ z 2. These vectors add up to 3 2(x + y + z) = − → 0 , so the medians indeed form a triangle.
Furthermore, the vectors representing the medians of the new triangle are mx + my 2 = x + y + z −3 4y = −3 4y, and similarly −3 4z and −3 4x. Therefore, these medians are parallel to XZ, Y X, and ZY.
Let D, E, F be the midpoints of sides BC, CA, AB, and let l1, l2, l3 be the segments A1A2, B1B2, C1C2.
Since l1, l2, l3 are parallel to AB, BC, CA, the medians of the triangle formed by l1, l2, l3 are parallel to CF, AD, BE. But from the lemma, they are also parallel to B1C1, C1A1, A1B1.
Therefore, BE ∥A1B1, and hence △BCE ∼△A1CB1. Then B1C AC = 1 2 · B1C EC = 1 2 · A1C BC = 1 2 1 −A1B CB .
Similarly C1A BA = 1 2 1 −B1C AC A1B CB = 1 2 1 −C1A BA .
Solving these three equations gives B1C AC = C1A BA = A1B CB = 1 3, as claimed; and it is straightforward to verify with the above equations that these ratio indeed work.
Second Solution: As above, we know that A1B1 ∥BE, B1C1 ∥ AD, C1A1 ∥CF.
Let A′, B′, C′ be the points dividing the sides BC, CA, AB in 1 : 2 ratios — since CA′ CB = CB′ 1 2 CA, we know A′B′ ∥BE ∥A1B1, and so on.
1999 National Contests: Problems and Solutions 151 Suppose by way of contradiction that A1 were closer to B than A′.
Then since A1B1 ∥A′B′, B1 is farther from C than B′. Similarly, C1 is closer to A than C′, and A1 is farther from B than A′ — a contradiction.
Likewise, A1 cannot be farther from B than A′. Thus A1 = A′, B1 = B′, and C1 = C′.
Problem 8.4 We are given 40 balloons, the air pressure inside each of which is unknown and may differ from balloon to balloon. It is permitted to choose up to k of the balloons and equalize the pressure in them (to the arithmetic mean of their respective original pressures).
What is the smallest k for which it is always possible to equalize the pressures in all of the balloons?
Solution: k = 5 is the smallest such value.
First suppose that k = 5. Note that we can equalize the pressure in any 8 balloons: first divide them into two groups of four {A, B, C, D} and {E, F, G, H} and equalize the pressure in each group; then equalize the pressure in {A, B, E, F} and {C, D, G, H}.
Then divide the 40 balloons into eight “5-groups” of five and equalize the pressure in each group. Then form five new groups of eight — containing one balloon from each “5-group” — and equalize the pressure in each of these new groups.
Now suppose that k ≤4. Let b1, b2, . . . , b40 denote the original air pressures inside the balloons.
It is simple to verify that the pressure in each balloon can always be written as a linear combination a1b1 + · · · + a40b40, where the ai are rational with denominators not divisible by any primes except 2 and 3. Thus if we the bj are linearly independent over the rationals (say, if bj = ej), we can never obtain 1 40b1 + 1 40b2 + · · · + 1 40b40 in a balloon. In this case, we can never equalize the pressures in all 40 balloons.
Problem 8.5 Show that the numbers from 1 to 15 cannot be divided into a group A of 2 numbers and a group B of 13 numbers in such a way that the sum of the numbers in B is equal to the product of the numbers in A.
Solution: Suppose by way of contradiction this were possible, and 152 Russia let a and b be the two numbers in A. Then we have (1 + 2 + · · · + 15) −a −b = ab 120 = ab + a + b 121 = (a + 1)(b + 1), Since a and b are integers between 1 and 15, the only possible solution to this equation is (a, b) = (10, 10). But a and b must be distinct, a contradiction.
Problem 8.6 Given an acute triangle ABC, let A1 be the reflection of A across the line BC, and let C1 be the reflection of C across the line AB. Show that if A1, B, C1 lie on a line and C1B = 2A1B, then ∠CA1B is a right angle.
Solution: By the given reflections, we have △ABC ∼ = △ABC1 ∼ = △A1BC.
Since ∠B is acute, C1 and A lie on the same side of BC. Thus C1 and A1 lie on opposite sides of BC as well.
Then since C1, B, A1 lie on a line we have 180◦= ∠C1BA + ∠ABC + ∠CBA1 = ∠ABC + ∠ABC + ∠ABC, so that ∠ABC = 60◦. Also we know that C1B = 2A1B = ⇒CB = 2AB, implying that triangle ABC is a 30◦-60◦-90◦triangle and ∠CA1B = ∠BAC = 90◦.
Problem 8.7 In a box lies a complete set of 1×2 dominoes. (That is, for each pair of integers i, j with 0 ≤i ≤j ≤n, there is one domino with i on one square and j on the other.) Two players take turns selecting one domino from the box and adding it to one end of an open (straight) chain on the table, so that adjacent dominoes have the same numbers on their adjacent squares. (The first player’s move may be any domino.) The first player unable to move loses. Which player wins with correct play?
Solution: The first player has a winning strategy. If n = 0, this is clear. Otherwise, have the first player play the domino (0, 0) and 1999 National Contests: Problems and Solutions 153 suppose the second player plays (0, a); then have the first player play (a, a).
At this point, the second player faces a chain whose ends are either 0 or a; also, the domino (0, k) is on the table if and only if the domino (a, k) is on the table. In such a “good” situation, if the second player plays (0, k) the first player can play (k, a) next to it; and if the second player plays (a, k) the first player can play (k, 0). In both cases, the same conditions for a “good” situation occur.
Therefore the first player can always play a domino with this strategy, forcing the second player to lose.
Problem 8.8 An open chain of 54 squares of side length 1 is made so that each pair of consecutive squares is joined at a single vertex, and each square is joined to its two neighbors at opposite vertices. Is it possible to cover the surface of a 3 × 3 × 3 cube with this chain?
Solution: It is not possible; suppose by way of contradiction it were.
Create axes so that the cube has corners at (3i, 3j, 3k) for i, j, k ∈ {0, 1}, and place the chain onto the cube. Imagine that every two adjacent squares in the chain are connected by pivots, and also let the start and end vertices of the chain be “pivots.” Consider some pivot P at (x, y, z); then the next pivot Q in the chain is either at (x, y ± 1, z ± 1), (x ± 1, y, z ± 1), or (x ± 1, y ± 1, z).
In any case, the sum of the coordinates of P has the same parity as the sum of the coordinates of Q — and hence all the pivots’ sums of coordinates have the same parity. Suppose without loss of generality the sums are even.
Form a graph whose vertices are the lattice points on the cube with even sums of coordinates; and join two vertices with an edge if the two lattice points are opposite corners of a unit square. Every square in our chain contains one of these edges — but since there are exactly 54 such edges (one across each unit square on the cube’s surface), and 54 squares in our chain, every edge is used exactly once. Then as we travel from pivot to pivot along our chain, we create an Eulerian path visiting all the edges. But four vertices — at (0, 0, 0), (0, 1, 1), (1, 0, 1), and (1, 1, 0) — have odd degree 3, so this is impossible.
Problem 9.1 Around a circle are written all of the positive integers from 1 to N, N ≥2, in such a way that any two adjacent integers 154 Russia have at least one common digit in their decimal expansions. Find the smallest N for which this is possible.
Solution: N = 29. Since 1 must be adjacent to two numbers, we must have N ≥11. But then 9 must be adjacent to two numbers, and the next smallest numbers containing 9 as a digit are 19 and 29.
Therefore N ≥29, and indeed N = 29 suffices: 19, 9, 29, 28, 8, 18, 17, 7, 27, . . . , 13, 3, 23, 2, 22, 21, 20, 12, 11, 10, 1.
Problem 9.2 In triangle ABC, points D and E are chosen on side CA such that AB = AD and BE = EC (E lying between A and D).
Let F be the midpoint of the arc BC of the circumcircle of ABC.
Show that B, E, D, F lie on a circle.
Solution: Let I be the incenter of triangle ABC, and notice that ∠BIC = 180◦−∠ICB −∠CBI = 180◦−∠B 2 −∠C 2 = 90◦+ ∠A 2 .
Also, since AD = AB we have ∠ADB = 90◦−∠A 2 and ∠BDC = 180◦−∠ADB = 90◦+ ∠A 2 . Therefore, BIDC is cyclic.
Some angle-chasing shows that that B, I, and C lie on a circle with center F. Thus D lies on this circle, FD = FC, and ∠FDC = ∠DCF.
Also, since BE = EC, we have ∠CBE = ∠C. Combining these facts, we have 180◦−∠EDF = ∠FDC = ∠DCF = ∠ACF = ∠C + ∠A 2 = ∠CBE + ∠FBC = ∠FBE.
Therefore BEDF is cyclic, as desired.
1999 National Contests: Problems and Solutions 155 Problem 9.3 The product of the positive real numbers x, y, z is 1.
Show that if 1 x + 1 y + 1 z ≥x + y + z, then 1 xk + 1 yk + 1 zk ≥xk + yk + zk for all positive integers k.
First Solution: Write x = a b , y = b c, z = c a for some positive numbers a, b, c. (For example, we could take a = 1, b = 1 x, c = 1 xy.) The given equation becomes b a + c b + a c ≥a b + b c + c a ⇐ ⇒a2b + b2c + c2a ≥ab2 + bc2 + ca2 ⇐ ⇒0 ≥(a −b)(b −c)(c −a).
For any positive integer k, write A = ak, B = bk, C = ck. Then a > b ⇐ ⇒A > B and a < b ⇐ ⇒A < B, and so on. Thus we also know that 0 ≥(A −B)(B −C)(C −A), and 0 ≥(A −B)(B −C)(C −A) ⇐ ⇒B A + C B + A C ≥A B + B C + C A ⇐ ⇒ 1 xk + 1 yk + 1 zk ≥xk + yk + zk, as desired.
Second Solution: The inequality 0 ≥(a −b)(b −c)(c −a) might spark this realization: dividing through by abc we have 0 ≥(x −1)(y −1)(z −1).
Indeed, (x −1)(y −1)(z −1) = xyz + x + y + z −xy −yz −zx −1 = x + y + z −1 z −1 x −1 y ≤0.
156 Russia Therefore, we have (x −1)(y −1)(z −1) ≤0 ⇒(xk −1)(yk −1)(zk −1) ≤0 ⇒xk + yk + zk ≥1 xk + 1 yk + 1 zk , as desired.
Problem 9.4 A maze consists of an 8 × 8 grid, in each 1 × 1 cell of which is drawn an arrow pointing up, down, left or right. The top edge of the top right square is the exit from the maze. A token is placed on the bottom left square, and then is moved in a sequence of turns. On each turn, the token is moved one square in the direction of the arrow. Then the arrow in the square the token moved from is rotated 90◦clockwise. If the arrow points offof the board (and not through the exit), the token stays put and the arrow is rotated 90◦ clockwise. Prove that sooner or later the token will leave the maze.
Solution: Suppose by way of contradiction the token did not leave the maze. Let position denote the set-up of the board, including both the token’s location and the directions of all the arrows. Since the token moves infinitely many times inside the maze, and there are only finitely many positions, some position must repeat.
During the “cycle time” between two occurrences of this position, suppose the token visits some square S. Then the arrow on S must make at least four 90◦rotations: thus at some point during the cycle time, the token must visit all the squares adjacent to S. It follows that the token visits all the squares on the board during the cycle time.
Specifically, the token visits the upper-right square during the cycle time; but at some point, this square’s arrow will point out of the maze. Then when the token lands on this square it will exit — a contradiction.
Problem 9.5 Each square of an infinite grid is colored in one of 5 colors, in such a way that every 5-square (Greek) cross contains one square of each color. Show that every 1 × 5 rectangle also contains one square of each color.
1999 National Contests: Problems and Solutions 157 Solution: Label the centers of the grid squares with coordinates, and suppose that square (0, 0) is colored maroon. The Greek cross centered at (1, 1) must contain a maroon-colored square. However, the squares (0, 1), (1, 0), and (1, 1) cannot be maroon because each of these squares is in a Greek cross with (0, 0). Thus either (1, 2) or (2, 1) is maroon — without loss of generality, say (1, 2).
Then by a similar analysis on square (1, 2) and the Greek cross centered at (2, 1), one of the squares (2, 0) and (3, 1) must be maroon.
(2, 0) is in a Greek cross with (0, 0) though, so (3, 1) is maroon.
Repeating the analysis on square (2, 0) shows that (2, −1) is ma-roon; and spreading outward, every square of the form (i + 2j, 2i −j) is maroon. But since these squares are the centers of Greek crosses that tile the plane, no other squares can be maroon. And since no two of these squares are in the same 1 × 5 rectangle, no two maroon squares can be in the same 1 × 5 rectangle.
The same argument applies to all the other colors — lavender, tickle-me-pink, green, neon orange. Therefore the five squares in each 1 × 5 rectangle have distinct colors, as desired.
Problem 9.7 Show that each natural number can be written as the difference of two natural numbers having the same number of prime factors.
Solution: If n is even, then we can write it as (2n) −(n).
Now suppose n is odd, and let d be the smallest odd prime that does not divide n. Then write n = (dn) −((d −1)n) . The number dn contains exactly one more prime factor than n. As for (d −1)n, it is divisible by 2 since d−1 is even; but its odd factors are less than d so they all divide n. Therefore (d −1)n also contains exactly one more prime factor than n, and dn and (d −1)n have the same number of prime factors.
Problem 9.8 In triangle ABC, with AB > BC, points K and M are the midpoints of sides AB and CA, and I is the incenter. Let P be the intersection of the lines KM and CI, and Q the point such that QP ⊥KM and QM ∥BI. Prove that QI ⊥AC.
Solution: Draw point S on ray CB such that CS = CA. Let P ′ be the midpoint of AS. Since triangle ACS is isosceles, P ′ lies on CI; 158 Russia and since P ′ and M are midpoints of AS and AC, we have P ′M ∥SC.
It follows that P = P ′.
Let the incircle touch BC, CA, AB at D, E, F respectively. Writing a = BC, b = CA, c = AB, and s = 1 2(a + b + c), we have SD = SC −DC = b −(s −c) = 1 2(b + c −a) = FA, BF = s −b = DB, AP = PS.
Therefore SD DB BF FA AP PS = 1, and by Menelaus’ Theorem applied to triangle ABS, P lies on line DF.
Then triangle PDE is isosceles, and ∠DEP = ∠PDE = ∠FEA = 90◦−∠A 2 while ∠CED = 90◦−∠C 2 . Therefore ∠PEA = 180◦−∠DEP −∠CED = 90◦−∠B 2 .
Now let Q′ be the point such that Q′I ⊥AC, Q′M ∥BI. Then ∠Q′EP = 90◦−∠PEA = ∠B 2 .
But we also know that ∠Q′MP = ∠IBC (from parallel lines BC ∥MP and IB ∥Q′M), and ∠IBC = ∠B 2 as well. Therefore ∠Q′MP = ∠Q′EP, quadrilateral Q′EMP is cyclic, and ∠Q′PM = ∠Q′EM = 90◦. Therefore Q = Q′, and QI is indeed perpendicular to AC.
Problem 10.2 In the plane is given a circle ω, a point A inside ω, and a point B not equal to A. Consider all possible triangles BXY such that X and Y lie on ω and A lies on the chord XY . Show that the circumcenters of these triangles all lie on a line.
Solution: We use directed distances. Let O be the circumcenter and R be the circumradius of triangle BXY. Drop the perpendicular OO′ to line AB.
The power of A with respect to circle BXY equals both AX · AY and AO2 −R2. Therefore BO′ −O′A = BO′2 −O′A2 BO′ + O′A 1999 National Contests: Problems and Solutions 159 = (BO2 −O′O2) −(OA2 −OO′2) AB = XA · AY AB which is constant since AX · AY also equals the power of A with respect to ω.
Since BO′ −O′A and BO′ +O′A = AB are constant, BO′ and O′A are constant as well. Thus O′ is fixed regardless of the choice of X and Y. Therefore O lies on the line through O′ perpendicular to AB, as desired.
Problem 10.3 In space are given n points in general position (no three points are collinear and no four are coplanar). Through any three of them is drawn a plane. Show that for any n −3 points in space, there exists one of the drawn planes not passing through any of these points.
Solution: Call the given n points given and the n−3 points random, and call all these points “level-0.” Since there are more given points than random points, one of the given points is not random: say, A.
Draw a plane not passing through A, and for each of the other points P let (P) be the intersection of AP with this plane. Call these points (P) level-1.
Since no four given points were coplanar, no three of the level-1 given points map to collinear points on this plane; and since no three given points were collinear, no two of the level-1 given points map to the same point on this plane. Thus we have n−1 level-1 given points and at most n −3 level-1 random points.
Now perform a similar operation — since there are more level-1 given points than random points, one of them is not random: say, (B). Draw a line not passing through (B), and for each of the other points (P) let ((P)) be the intersection of B(P) with this plane. Call these points ((P)) level-2.
Since no three level-1 given points were collinear, all of the level-2 given points are distinct. Thus we have n −2 level-2 given points but at most n −3 level-2 random points. Therefore one of these given points ((C)) is not random.
160 Russia Consider the drawn plane ABC. If it contained some level-0 random point — say, Q — then (Q) would be collinear with (B) and (C), and thus ((Q)) = ((C)), a contradiction. Therefore plane ABC does not pass through any of the level-0 random points, as desired.
Problem 10.5 Do there exist 10 distinct integers, the sum of any 9 of which is a perfect square?
Solution: Yes, there do exist 10 such integers. Write S = a1 +· · ·+ a10, and consider the linear system of equations S −a1 = 9 · 12 S −a2 = 9 · 22 .
.
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S −a10 = 9 · 102.
Adding all these gives 9S = 9 · (12 + 22 + · · · + 102) so that ai = S −9i2 = 12 + 22 + · · · + 102 −9i2.
Then all the ai’s are distinct integers, and any nine of them add up to perfect square.
Problem 10.6 The incircle of triangle ABC touches sides BC, CA, AB at A1, B1, C1, respectively. Let K be the point on the circle diametrically opposite C1, and D the intersection of the lines B1C1 and A1K. Prove that CD = CB1.
Solution: Draw D′ on B1C1 such that CD′ ∥AB. Then ∠D′CB1 = ∠C1AB1 and ∠CD′B1 = ∠AC1B1, implying that △AB1C1 ∼△CB1D′.
Thus triangle CB1D′ is isosceles and CD′ = CB1. But CB1 = CA1, so that triangle CA1D′ is isosceles also. And since ∠D′CA1 = 180◦−∠B, we have ∠CA1D′ = ∠B 2 .
But note that ∠CA1K = ∠A1C1K = 90◦−∠C1KA1 1999 National Contests: Problems and Solutions 161 = 90◦−∠C1A1B = ∠B 2 also. Therefore D′ lies on A1K and by definition it lies on B1C1.
Hence D′ = D.
But from before CD′ = CB1; thus CD = CB1, as desired.
Problem 10.7 Each voter in an election marks on a ballot the names of n candidates. Each ballot is placed into one of n + 1 boxes.
After the election, it is observed that each box contains at least one ballot, and that for any n + 1 ballots, one in each box, there exists a name which is marked on all of these ballots. Show that for at least one box, there exists a name which is marked on all ballots in the box.
Solution: Suppose by way of contradiction that in every box, no name is marked on all the ballots. Label the boxes 1, 2, . . . , n, and look at an arbitrary ballot from the first box.
Suppose it has n “chosen” names Al, Bob, . . . , Zed. By assumption, some ballot in the second box does not have the name Al on it; some ballot in the third box does not have the name Bob on it; and so on, so some ballot in the (i + 1)-th box does not have the i-th chosen name on it. But then on these n+1 ballots, one from each box, there is no name marked on all the ballots — a contradiction.
Problem 10.8 A set of natural numbers is chosen so that among any 1999 consecutive natural numbers, there is a chosen number.
Show that there exist two chosen numbers, one of which divides the other.
Solution: Draw a large table with 1999 columns and 2000 rows. In the first row write 1, 2, . . . , 1999.
Define the entries in future rows recursively as follows: suppose the entries in row i are k + 1, k + 2, . . . , k + 1999, and that their product is M. Then fill row i + 1 with M + k + 1, M + k + 2, . . . , M +k+1999. All the entries in row i+1 are bigger than the entries in row i; furthermore, every entry divides the entry immediately below it (and therefore all the entries directly below it).
In each row there are 1999 consecutive numbers, and hence each row contains a chosen number. Then since we have 2000 rows, there 162 Russia are two chosen numbers in the same column — and one of them divides another, as desired.
Problem 11.1 The function f(x) is defined on all real numbers. It is known that for all a > 1, the function f(x) + f(ax) is continuous.
Show that f(x) is continuous.
First Solution: We know that for a > 1, the functions P(x) = f(x) + f(ax), Q(x) = f(x) + f(a2x), P(ax) = f(ax) + f(a2x) are all continuous. Thus the function 1 2 (P(x) + Q(x) −P(ax)) = f(x) is continuous as well.
Problem 11.3 In a class, each boy is friends with at least one girl.
Show that there exists a group of at least half of the students, such that each boy in the group is friends with an odd number of the girls in the group.
Solution: We perform strong induction on the total number of students. The base case of zero students is obvious.
Now suppose that we know the claim is true for any number of students less than n (where n > 0), and we wish to prove it for n.
Since there must be at least one girl, pick any girl from the n students.
We now partition the class into three subsets: A = this girl, B = this girl’s male friends, and C = everybody else.
Because we are using strong induction, the induction hypothesis states that there must be a subset C′ of C, with at least |C| 2 students, such that any boy in C′ is friends with an odd number of girls in C′.
Let BO be the set of boys in B who are friends with an odd number of girls in C′, and let BE be the set of boys in B who are friends with an even number of girls in C′. Then there are two possible cases: (i) |BO| ≥|A∪B| 2 .
The set S = BO ∪C′ will realize the claim, i.e., S will have at least n 2 elements, and each boy in S will be friends with an odd number of girls in S.
1999 National Contests: Problems and Solutions 163 (ii) |A ∪BE| ≥|A∪B| 2 .
The set T = A ∪BE ∪C′ will realize the claim. T will have at least n 2 elements; each boy in C′ will be friends with an odd number of girls in C′ but not the girl in A; and each boy in BE will be friends with an even number of girls in C′ and the girl in A — making a total of an odd number of girls.
Thus the induction is complete.
Note: With a similar proof, it is possible to prove a slightly stronger result: suppose each boy in a class is friends with at least one girl, and that every boy has a parity, either “even” or “odd.” Then there is a group of at least half the students, such that each boy in the group is friends with the same parity of girls as his own parity. (By letting all the boys’ parity be “odd,” we have the original result.) Problem 11.4 A polyhedron is circumscribed about a sphere. We call a face big if the projection of the sphere onto the plane of the face lies entirely within the face. Show that there are at most 6 big faces.
Solution: Lemma. Given a sphere of radius R, let a “slice” of the sphere be a portion cut offby two parallel planes. The surface area of the sphere contained in this slice is 2πRW, where W is the distance between the planes.
Proof: Orient the sphere so that the slice is horizontal. Take an infinitesimal horizontal piece of this slice, shaped like a frustrum (a small sliver from the bottom of a radially symmetric cone). Say it has width w, radius r, and slant height ℓ; then its lateral surface area (for infinitesimal w) is 2πrℓ. But if the side of the cone makes an angle θ with the horizontal, then we have ℓsin θ = w and R sin θ = r so that the surface area also equals 2πRw. Adding over all infinitesimal pieces, the complete slice has lateral surface area 2πRW, as desired.
Say that the inscribed sphere has radius R and center O. For each big face F in the polyhedron, project the sphere onto F to form a circle k. Then connect k with O to form a cone. Because these cones don’t share any volume, they hit the sphere’s surface in several non-overlapping circular regions.
164 Russia Each circular region is a slice of the sphere with width R(1−1 2 √ 2), and it contains 2πR2(1 −1 2 √ 2) > 1 7(4πR2) of the sphere’s surface area. Thus each circular region takes up more than 1 7 of the surface area of the sphere, implying there must be less than 7 such regions and therefore at most six big faces.
Problem 11.5 Do there exist real numbers a, b, c such that for all real numbers x, y, |x + a| + |x + y + b| + |y + c| > |x| + |x + y| + |y|?
Solution: No such numbers exist; suppose they did. Let y = −b−x.
Then for all real x we have |x + a| + |−b −x + c| > |x| + |−b| + |−b −x|.
If we pick x sufficiently negative, this gives (−x −a) + (−b −x + c) > (−x) + |b| + (−b −x) ⇒−a + c > |b| ≥0, so c > a. On the other hand, if we pick x sufficiently positive, this gives (x + a) + (b + x −c) > (x) + |b| + (b + x) ⇒a −c > |b| ≥0, so c < a as well — a contradiction.
Problem 11.6 Each cell of a 50 × 50 square is colored in one of four colors. Show that there exists a cell which has cells of the same color directly above, directly below, directly to the left, and directly to the right of it (though not necessarily adjacent to it).
Solution: By the pigeonhole principle, at least one-quarter of the squares (625) are the same color: say, red.
Of these red squares, at most 50 are the topmost red squares of their columns, and at most 50 are the bottommost red squares of their columns. Similarly, at most 50 are the leftmost red squares in their rows and at most 50 are the rightmost red squares in their rows.
This gives at most 200 squares; the remaining 425 or more red squares 1999 National Contests: Problems and Solutions 165 then have red squares directly above, directly below, directly to the left, and directly to the right of them.
Problem 11.8 A polynomial with integer coefficients has the prop-erty that there exist infinitely many integers which are the value of the polynomial evaluated at more than one integer. Prove that there exists at most one integer which is the value of the polynomial at exactly one integer.
Solution: First observe that the polynomial cannot be constant.
Now let P(x) = cnxn + cn−1xn−1 + · · · + c0 be the polynomial with cn ̸= 0. The problem conditions imply that n is even and at least 2, and we can assume without loss of generality that cn > 0.
Since P(x) has positive leading coefficient and it is not constant, there exists a value N such that P(x) is decreasing for all x < N.
Also consider the pairs of integers (s, t) with s < t and P(s) = P(t); since there are infinitely many pairs, there must be infinitely many with s < N.
Now for any integer k, look at the polynomial P(x) −P(k −x).
Some algebra shows that the coefficient of xn is zero and that the coefficient of xn−1 is f(k) = 2cn−1 + cn(nk).
Let K be the largest integer such that f(K) < 0 (such an integer exists because from assumptions made above, cn · n > 0). Then for sufficiently large t we have P(t) < P(K −t) < P(K −1 −t) < · · · and P(t) ≥P(K + 1 −t) > P(K + 2 −t) > P(K + 3 −t) > · · · > P(N).
Therefore we must have s = K +1−t and P(t)−P(K +1−t) = 0 for infinitely many values of t. But since P has finite degree, this implies that P(x) −P(K + 1 −x) is identically zero.
Then if P(a) = b for some integers a, b, we also have P(K +1−a) = b. Therefore there is at most one value b that could possibly be the value of P(x) at exactly one integer x — specifically, b = P( K+1 2 ).
Fifth round Problem 9.1 In the decimal expansion of A, the digits occur in increasing order from left to right. What is the sum of the digits of 166 Russia 9A?
Solution: Write A = a1a2 . . . ak. Then since 9A = 10A −A, by performing the subtraction a1 a2 a3 · · · ak 0 − a1 a2 · · · ak−1 ak we find that the digits of 9A are a1, a2 −a1, a3 −a2, . . . , ak−1 −ak−2, ak −ak−1 −1, 10 −ak, and that these digits add up to 10 −1 = 9.
Problem 9.3 Let S be the circumcircle of triangle ABC. Let A0 be the midpoint of the arc BC of S not containing A, and C0 the midpoint of the arc AB of S not containing C. Let S1 be the circle with center A0 tangent to BC, and let S2 be the circle with center C0 tangent to AB. Show that the incenter I of ABC lies on a common external tangent to S1 and S2.
Solution: We prove a more general result: I lies on a common external tangent to S1 and S2, parallel to AC.
Drop the perpendicular from A0 to BC, hitting at P, and drop the perpendicular from A0 to AC, hitting circle S1 at Q (with Q closer to AC than A0).
Note that A, I, and A0 are collinear. Then ∠CIA0 = ∠CAI + ∠ICA = ∠A + ∠C 2 = ∠A0CI, so that IA0 = CA0.
Next, from circle S1 we know that A0Q = A0P.
Finally, ∠IA0Q = ∠AA0Q = 90◦−∠CAA0 = 1 2(180◦−∠CAB) = ∠CA0P.
Thus, triangles IA0Q and CA0P are congruent. Then ∠IQA0 = ∠CPA0 = 90◦, so that IQ is tangent to S1 at Q. Furthermore, since A0Q is perpen-dicular to both IQ and AC, we have IQ ∥AC.
1999 National Contests: Problems and Solutions 167 Therefore, the line through I parallel to AC is tangent to S1; by a similar argument it is tangent to S2; and thus it is a common external tangent to S1 and S2, as claimed.
Problem 9.4 The numbers from 1 to 1000000 can be colored black or white. A permissible move consists of selecting a number from 1 to 1000000 and changing the color of that number and each number not relatively prime to it. Initially all of the numbers are black. Is it possible to make a sequence of moves after which all of the numbers are colored white?
First Solution: It is possible. We begin by proving the following lemma: Lemma. Given a set S of positive integers, there is a subset T ⊆S such that every element of S divides an odd number of elements in T.
Proof: We prove the claim by induction on |S|, the number of elements in S. If |S| = 1 then let T = S.
If |S| > 1, then say the smallest element of S is a. Look at the set S′ = S \ {a} — the set of the largest |S| −1 elements in S. By induction there is a subset T ′ ⊆S′ such that every element in S′ divides an odd number of elements in T ′.
If a also divides an odd number of elements in T ′, then the set T = T ′ suffices. Otherwise, the set T = T ′ ∪{a} suffices: a divides an odd number of elements in T; the other elements are bigger than a and can’t divide it, and therefore still divide an odd number of elements in T. This completes the induction and the proof of the lemma.
Now, write each number n > 1 in its prime factorization pa1 1 pa2 2 · · · pak k for distinct primes pi and positive integers ai. Then notice that the color of n will always be the same as the color of P(n) = p1p2 · · · pk.
Apply the lemma to the set S = S1000000 i=2 P(i) to find a subset T ⊆S such that every element of S divides an odd number of elements in T. For each q ∈S, let t(q) equal the number of elements in T that q divides, and let u(q) equal the number of primes dividing q.
Select all the numbers in T, and consider how the color of a number n > 1 changes. The number of elements in T not relatively prime to 168 Russia n equals X q|P (n), q>1 (−1)u(q)+1t(q) by the Inclusion-Exclusion Principle: if q | P(n) is divisible by exactly m > 0 primes, then it is counted m 1 − m 2 + m 3 −· · · = 1 time in the sum. (For example, if n = 6 then the number of elements in T divisible by 2 or 3 equals t(2) + t(3) −t(6).) But by the definition of T, each of the values t(q) is odd. Then since there are 2k −1 divisors q > 1 of P(n), the above quantity is the sum of 2k −1 odd numbers and is odd itself. Therefore after selecting T, every number n > 1 will switch color an odd number of times and will turn white.
Finally, select 1 to turn 1 white, and we are done.
Note: In fact, a slight modification of the above proof shows that T is unique, which with some work implies that there is exactly one way to make all the numbers white by only selecting square-free numbers at most once each (other methods are different only trivially, either by selecting a number twice or by selecting numbers that aren’t square-free).
Second Solution: Yes, it is possible. We prove a more general statement, where we replace 1000000 in the problem by some arbi-trary positive integer m, and where we focus on the numbers divisible by just a few primes instead of all the primes.
Lemma. For a finite set of distinct primes S = {p1, p2, . . . , pn}, let Qm(S) be the set of numbers between 2 and m divisible only by primes in S. The elements of Qm(S) can be colored black or white; a permissible move consists of selecting a number in Qm(S) and changing the color of that number and each number not relatively prime to it. Then it is possible to reverse the coloring of Qm(S) by selecting several numbers in a subset Rm(S) ⊆Qm(S).
Proof: We prove the lemma by induction on n. If n = 1, then selecting p1 suffices. Now suppose n > 1, and assume without loss of generality that the numbers are all black to start with.
Let T = {p1, p2, . . . , pn−1}, and define t to be the largest integer such that tpn ≤m. We can assume t ≥1 because otherwise we could ignore pn and just use use the smaller set T, and we’d be done by our 1999 National Contests: Problems and Solutions 169 induction hypothesis.
Now select the numbers in Rm(T), Rt(T), and pnRt(T) = {pnx | x ∈Rt(T)}, and consider the effect of this action on a number y: • y is not a multiple of pn. Selecting the numbers in Rm(T) makes y white. Then if selecting x ∈Rt(T) changes y’s color, selecting xpn will change it back so that y will become white.
• y is a power of pn. Selecting the numbers in Rm(T) and Rt(T) has no effect on y, but each of the |Rt(T)| numbers in xRt(T) changes y’s color.
• pn | y but y is not a power of pn. Selecting the numbers in Rm(T) makes y white. Since y ̸= pi n, it is divisible by some prime in T so selecting the numbers in Rt(T) makes y black again. Finally, each of the |Rt(T)| numbers in xRt(T) changes y’s color.
Therefore, all the multiples of pn are the same color (black if |Rt(T)| is even, white if |Rt(T)| is odd), while all the other numbers in Qm(S) are white. If the multiples of pn are still black, we can select pn to make them white, and we are done.
Now back to the original problem: set m = 1000000 and let S be the set of all primes under 1000000. Then from the lemma, we can select numbers between 2 and 1000000 so that all the numbers 2, 3, . . . , 1000000 are white. And finally, we finish offby selecting 1.
Problem 9.5 An equilateral triangle of side length n is drawn with sides along a triangular grid of side length 1. What is the maximum number of grid segments on or inside the triangle that can be marked so that no three marked segments form a triangle?
Solution: The grid is made up of n(n+1) 2 small equilateral triangles of side length 1. In each of these triangles, at most 2 segments can be marked so we can mark at most 2 3 · 3n(n+1) 2 = n(n + 1) segments in all. Every segment points in one of three directions, so we can achieve the maximum n(n + 1) by marking all the segments pointing in two of the directions.
170 Russia Problem 9.6 Let {x} = x −⌊x⌋denote the fractional part of x.
Prove that for every natural number n, n2 X k=1 { √ k} ≤n2 −1 2 .
Solution: We prove the claim by induction on n. For n = 1, we have 0 ≤0. Now supposing that the claim is true for n, we prove it is true for n + 1.
Each of the numbers √ n2 + 1, √ n2 + 2, . . . , √ n2 + 2n is between n and n + 1, and thus { p n2 + i} = p n2 + i −n < r n2 + i + i2 4n2 −n = i 2n.
Therefore we have (n+1)2 X k=1 { √ k} = n2 X k=1 { √ k} + (n+1)2 X k=n2+1 { √ k} < n2 −1 2 + 1 2n 2n X i=1 i + 0 = n2 −1 2 + 2n + 1 2 = (n + 1)2 −1 2 , completing the inductive step and the proof.
Problem 9.7 A circle passing through vertices A and B of triangle ABC intersects side BC again at D. A circle passing through vertices B and C intersects side AB again at E, and intersects the first circle again at F. Suppose that the points A, E, D, C lie on a circle centered at O. Show that ∠BFO is a right angle.
Solution: Since AEDC is cyclic with O as its center, ∠COA = 2∠CDA = ∠CDA + ∠CEA = (180◦−∠ADB) + (180◦−∠BEC).
1999 National Contests: Problems and Solutions 171 Since BDFA and BEFC are cyclic, ∠ADB = ∠AFB and ∠BEC = ∠BFC. Hence ∠COA = 360◦−∠AFB −∠BFC = ∠CFA.
Hence AFOC is cyclic. Therefore ∠OFA = 180◦−∠ACO = 180◦−180◦−∠COA 2 = 90◦+ ∠CDA.
Since ABDF is cyclic, ∠OFA + ∠AFB = 90◦+ ∠CDA + ∠ADB = 270◦.
Hence ∠BFO = 90◦, as desired.
Problem 9.8 A circuit board has 2000 contacts, any two of which are connected by a lead. The hooligans Vasya and Petya take turns cutting leads: Vasya (who goes first) always cuts one lead, while Petya cuts either one or three leads. The first person to cut the last lead from some contact loses. Who wins with correct play?
Solution: Petya wins with correct play; arrange the contacts in a circle and label them 1, 2, . . . , 2000, and let (x, y) denote the lead between contacts x and y (where labels are taken modulo 2000).
If Vasya disconnects (a, 1000 + a), Petya can disconnect (500 + a, 1500 + a); otherwise, if Vasya disconnects (a, b), Petya can dis-connect the three leads (a + 500, b + 500), (a + 1000, b + 1000), and (a+1500, b+1500). Notice that in each case, Petya and Vasya tamper with different contacts.
Using this strategy, after each of Petya’s turns the circuit board is symmetrical under 90◦, 180◦, and 270◦rotations, ensuring that he can always make the above moves — for example, if (a + 1500, b + 1500) were already disconnected during Petya’s turn, then (a, b) must have been as well before Vasya’s turn.
Also, Petya can never lose, because if he disconnected the last lead (x, y) from some contact x, then Vasya must have already disconnected the last lead (x −1500, y −1500), (x −1000, y −1000), or (x −500, y −500) from some other contact, a contradiction.
Problem 10.1 Three empty bowls are placed on a table. Three players A, B, C, whose order of play is determined randomly, take turns putting one token into a bowl. A can place a token in the first or second bowl, B in the second or third bowl, and C in the third or 172 Russia first bowl. The first player to put the 1999th token into a bowl loses.
Show that players A and B can work together to ensure that C will lose.
Solution: Suppose A plays only in the first bowl until it contains 1998 tokens, then always plays in the second bowl; and suppose B plays only in the third bowl until it contains 1998 tokens, then always plays in the second bowl as well.
Suppose by way of contradiction that C doesn’t lose. Without loss of generality, say the first bowl fills up to 1998 tokens before the third bowl does — call this point in time the “critical point.” First suppose the third bowl never contains 1998 tokens. Then at most 999 round pass after the critical point since during each round, the third bowl gains 2 tokens (one from B, one from C). But then A plays at most 999 tokens into the second bowl and doesn’t lose; thus nobody loses, a contradiction.
Thus the third bowl does contain 1998 tokens some k ≤999 more rounds after the critical point. After this k-th round A has played at most k tokens into the second bowl, and B has possibly played at most one token into the second bowl during the k-th round; so the second bowl has at most 1000 tokens. However, the first and third bowls each have 1998 tokens, so during the next round C will lose.
Problem 10.2 Find all infinite bounded sequences a1, a2, . . . of positive integers such that for all n > 2, an = an−1 + an−2 gcd(an−1, an−2).
Solution: The only such sequence is 2, 2, 2, . . . .
Let gn = gcd(an, an+1). Then gn+1 divides both an+1 and an+2, so it divides gnan+2 −an+1 = an as well. Thus gn+1 divides both an and an+1, and it divides their greatest common divisor gn.
Therefore, the gi form a nonincreasing sequence of positive integers and eventually equal some positive constant g. At this point, the ai satisfy the recursion gan = an−1 + an−2.
If g = 1, then an = an−1+an−2 > an−1 so the sequence is increasing and unbounded.
1999 National Contests: Problems and Solutions 173 If g ≥3, then an = an−1+an−2 g < an−1+an−2 2 ≤max{an−1, an−2}.
Similarly, an+1 < max{an−1, an} ≤ max{an−2, an−1}, so that max{an, an+1} < max{an−2, an−1}. Therefore the maximum values of successive pairs of terms form an infinite decreasing sequence of positive integers, a contradiction.
Thus g = 2 and eventually we have 2an = an−1 + an−2 or an − an−1 = −1 2(an−1 −an−2). This implies that ai −ai−1 converges to 0 and that the ai are eventually constant as well. From 2an = an−1 + an−2, this constant must be 2.
Now if an = an+1 = 2 for n > 1, then gcd(an−1, an) = gcd(an−1, 2) either equals 1 or 2. Since 2 = an+1 = an−1 + an gcd(an−1, 2), this either implies an−1 = 0 — which is impossible — or an−1 = 2.
Therefore all the ai equal 2, and this sequence indeed works.
Problem 10.3 The incircle of triangle ABC touches sides AB, BC, CA at K, L, M, respectively. For each two of the incircles of AMK, BKL, CLM is drawn the common external tangent not lying along a side of ABC. Show that these three tangents pass through a single point.
Solution: Let D, E, F be the midpoints of minor arcs MK, KL, LM of the incircle, respectively; and let S1, S2, S3 be the incircles of triangles AMK, BKL, and CLM, respectively.
Since AK is tangent to the incircle, ∠AKD = ∠KLD = ∠KMD = ∠DKM; similarly, ∠AMD = ∠DMK. Thus, D is the incenter of AMK and the center of S1.
Likewise, E is center of S2 and F is the center of S3. By the result proved in Problem 9.3, the incenter I of triangle KLM lies on a common external tangent to S1 and S2. But it does not lie on AB, so it must lie on the other external tangent. Similarly, the common external tangent to S2 and S3 (not lying on BC) passes through I, as does the common external tangent to S3 and S1 (not lying on CA); so the three tangents all pass through I, as desired.
Problem 10.4 An n×n square is drawn on an infinite checkerboard.
Each of the n2 cells contained in the square initially contains a token. A move consists of jumping a token over an adjacent token 174 Russia (horizontally or vertically) into an empty square; the token jumped over is removed. A sequence of moves is carried out in such a way that at the end, no further moves are possible. Show that at least n2 3 moves have been made.
Solution: At the end of the game no two adjacent squares contain tokens: otherwise (since no more jumps are possible) they would have to be in an infinitely long line of tokens, which is not allowed. Then during the game, each time a token on square A jumps over another token on square B, imagine putting a 1×2 domino over squares A and B. At the end, every tokenless square on the checkerboard is covered by a tile; so no two uncovered squares are adjacent. We now prove there must be at least n2 3 dominoes, implying that at least n2 3 moves have been made: Lemma. If an n × n square board is covered with 1 × 2 rectangular dominoes (possibly overlapping, and possibly with one square offthe board) in such a way that no two uncovered squares are adjacent, then at least n2 3 tiles are on the board.
Proof: Call a pair of adjacent squares on the checkerboard a “tile.” If a tile contains two squares on the border of the checkerboard, call it an “outer tile”; otherwise, call it an “inner tile.” Now for each domino D, consider any tile it partly covers. If this tile is partly covered by exactly m dominoes, say D destroys 1 m of that tile. Adding over all the tiles that D lies on, we find the total quantity a of outer tiles that D destroys, and the total quantity b of inner tiles that D destroys. Then say that D scores 1.5a + b points.
Consider a vertical domino D wedged in the upper-left corner of the chessboard; it partly destroys two horizontal tiles. But one of the two squares immediately to D’s right must be covered; so if D destroys all of one horizontal tile, it can only destroy at most half of the other.
Armed with this type of analysis, some quick checking shows that any domino scores at most 6 points; and that any domino scoring 6 points must lie completely on the board, not be wedged in a corner, not overlap any other dominoes, and not have either length-1 edge hit another domino.
Now in a valid arrangement of dominoes, every tile is destroyed completely; since there are 4(n−1) outer tiles and 2(n−1)(n−2) inner 1999 National Contests: Problems and Solutions 175 tiles, this means that a total of 1.5·4(n−1)+2(n−1)(n−2) = 2(n2−1) points are scored.
Therefore, there must be at least ⌈2(n2−1) 6 ⌉= ⌈n2−1 3 ⌉dominoes.
Suppose by way of contradiction that we have exactly n2−1 3 domi-noes. First, for this to be an integer 3 cannot divide n. Second, the restrictions described two paragraphs ago must hold for every domino.
Suppose we have any horizontal domino not at the bottom of the chessboard; one of the two squares directly below it must be covered.
But to satisfy our restrictions, it must be covered by a horizontal domino (not a vertical one). Thus we can find a chain of horizontal dominoes stretching to the bottom of the board, and similarly we can follow this chain to the top of the board.
Similarly, if there is any vertical domino then some chain of vertical dominoes stretches across the board.
But we can’t have both a horizontal and a vertical chain, so all the dominoes must have the same orientation: say, horizontal.
Now to cover the tiles in any given row while satisfying the re-strictions, we must alternate between blank squares and horizontal dominoes.
In the top row, since no dominoes are wedged in a corner we must start and end with blank squares; thus we must have n ≡1 (mod 3).
But then in the second row, we must start with a horizontal domino (to cover the top-left vertical tiles); then after alternating between dominoes and blank squares, the end of the row will contain two blank squares—a contradiction. Thus it is impossible to cover the chessboard with exactly n2−1 3 dominoes, and indeed at least n2 3 dominoes are needed.
Note: When n is even, there is a simpler proof of the main result: split the n2 squares of the board into 2 × 2 mini-boards, each containing four (overlapping) 1 × 2 tiles. At the end of the game, none of these n2 tiles can contain two checkers (since no two checkers can be adjacent at the end of the game). But any jump removes a checker from at most three full tiles; therefore, there must be at least n2 3 moves.
Sadly, a similar approach for odd n yields a lower bound of only n2−n−1 3 moves. For large enough n though, we can count the number of tokens that end up completely outside the (n + 2) × (n + 2) area around the checkerboard — each made a jump that freed at most two full tiles, and from here we can show that n2 3 moves are necessary.
176 Russia Problem 10.5 The sum of the decimal digits of the natural number n is 100, and that of 44n is 800. What is the sum of the digits of 3n?
Solution: The sum of the digits of 3n is 300.
Let S(x) denote the sum of the digits of x. Then S(a + b) equals S(a) + S(b), minus nine times the number of carries in the addition a + b. Therefore, S(a + b) ≤S(a) + S(b); applying this repeatedly, we have S(a1 + · · · + ak) ≤S(a1) + · · · + S(ak).
Also note that for a digit d ≤2 we have S(44d) = 8d; for d = 3 we have S(8d) = 6 < 8d; and for d ≥4, 44d ≤44(9) has at most 3 digits so its sum is at most 27 < 8d.
Now write n = P ni · 10i, so that the ni are the digits of n in base 10. Then X 8ni = S(44n) ≤ X S(44ni · 10i) = X S(44ni) ≤ X 8ni, so equality must occur in the second inequality — that is, each of the ni must equal 0, 1, or 2. But then each digit of 3n is simply three times the corresponding digit of n, and S(3n) = 3S(n) = 300, as claimed.
Problem 10.7 The positive real numbers x and y satisfy x2 + y3 ≥x3 + y4.
Show that x3 + y3 ≤2.
Solution: Equivalently we can prove that if x3 + y3 > 2, then x2 + y3 < x3 + y4.
First notice that q x2+y2 2 ≤ 3 q x3+y3 2 by the Power-Mean Inequality, implying that x2 + y2 ≤(x3 + y3)2/3 · 21/3 < (x3 + y3)2/3(x3 + y3)1/3 = x3 + y3, or x2 −x3 < y3 −y2. But 0 ≤y2(y −1)2 ⇒y3 −y2 ≤y4 −y3, so that x2 −x3 < y4 −y3 ⇒x2 + y3 < x3 + y4, 1999 National Contests: Problems and Solutions 177 as desired.
Problem 10.8 In a group of 12 people, among every 9 people one can find 5 people, any two of whom know each other. Show that there exist 6 people in the group, any two of whom know each other.
Solution: Suppose by way of contradiction that no 6 people know each other. Draw a complete graph with twelve vertices corresponding to the people, labeling the people (and their corresponding vertices) A, B, . . . , L. Color the edge between two people red if they know each other, and blue otherwise. Then among every nine vertices there is at least one red K5; and among any six vertices there is at least one blue edge.
We prove that there are no blue cycles of odd length in this graph.
Suppose, for sake of contradiction, that there is a blue cycle of length (i) 3 or 5, (ii) 7, (iii) 9, or (iv) 11.
(i) First suppose there is a blue 3-cycle (say, ABC) or a blue 5-cycle (without loss of generality, ABCDE). In the first case, there is a blue edge among DEFGHI (sat, DE); then any red K5 contains at most one vertex from {A, B, C} and at most one vertex from {D, E}. In the second case, any K5 still contains at most two vertices from {A, B, C, D, E}.
Now, FGHIJK contains some other blue edge: without loss of generality, say FG is blue. Now for each edge V1V2 in HIJKL, there must be a red K5 among ABCDEFGV1V2. From before, this K5 can contain at most two vertices from {A, B, C, D, E}; and it contains at most one vertex from each of {F, G}, {V1}, and {V2}.
Therefore V1 and V2 must be connected by a red edge, so HIJKL is a red K5. Now FHIJKL cannot be a red K6, so without loss of generality suppose FH is blue. Similarly, GHIJKL cannot be a red K6, so without loss of generality either GH or GI is blue. In either case, ABCDEFGHI must contain some red K5. If GH is blue then this K5 contains at most four vertices, two from {A, B, C, D, E} and one from each of {F, G, H} and {I}; and if GI is blue then this K5 again contains at most four vertices, two from {A, B, C, D, E} and one from each of {F, H} and {G, I}. Either possibility yields a contradiction.
178 Russia (ii) If there is some blue 7-cycle, say without loss of generality it is ABCDEFG. As before, any K5 contains at most three vertices from {A, B, . . . , G}, so HIJKL must be a red K5. Now for each of the 5 2 = 10 choices of pairs {V1, V2} ⊂{H, I, J, K, L}, there must be a red K5 among ABCDEFGV1V2; so for each edge in HIJKL, some red triangle in ABCDEFG forms a red K5 with that edge.
But ABCDEFG contains at most 7 red triangles: ACE, BDF, . . . , and GBD. Thus some triangle corresponds to two edges. Without loss of generality, either ACE corresponds to both HI and HJ; or ACE corresponds to both HI and JK.
In either case, ACEHIJ is a red K6, a contradiction.
(iii) Next suppose that there is some blue 9-cycle; then among these nine vertices there can be no red K5, a contradiction.
(iv) Finally, suppose that there is some blue 11-cycle; without loss of generality, say it is ABCDEFGHIJK. There is a red K5 among {A, B, C, D, E, F, G, H, I}, which must be ACEGI.
Likewise, DFHJA must be a red K5, so AC, AD, . . . , AH are all red.
Similarly, every edge in ABCDEFGHI is red except for those in the blue 11-cycle.
Now among {A, B, C, D, E, F, G, H, L} there is some red K5, either ACEGL or BDFHL. Without loss of generality, assume the former. Then since ACEGLI and ACEGLJ can’t be red 6-cycles, AI and AJ must be blue.
But then AIJ is a blue 3-cycle, a contradiction.
Thus there are indeed no blue cycles of odd length, so the blue edges form a bipartite graph: that is, the twelve vertices can be partitioned into two groups G1 and G2 containing no blue edges. One of these groups, say G1, has at least 6 vertices; but then G1 is a red K6, a contradiction. Therefore our original assumption was false; there is some red K6, so some six people do indeed know each other.
Problem 11.1 Do there exist 19 distinct natural numbers which add to 1999 and which have the same sum of digits?
Solution: No such integers exist; suppose by way of contradiction they did.
The average of the numbers is 1999 19 < 106, so one number is at most 105 and has digit sum at most 18.
1999 National Contests: Problems and Solutions 179 Every number is congruent to its digit sum modulo 9, so all the numbers and their digit sums are congruent modulo 9 — say, congruent to k. Then 19k ≡1999 ⇒k ≡1 (mod 9), so the common digit sum is either 1 or 10.
If it is 1 then all the numbers equal 1, 10, 100, or 1000 so that some two are equal — which is not allowed. Thus the common digit sum is 10. Note that the twenty smallest numbers with digit sum 10 are: 19, 28, 37, . . . , 91, 109, 118, 127, . . . , 190, 208.
The sum of the first nine numbers is (10 + 20 + · · · + 90) + (9 + 8 + · · · + 1) = 450 + 45 = 495, while the sum of the next nine numbers is (900)+(10+20+· · ·+80)+(9+8+7+· · ·+1) = 900+360+45 = 1305, so the first eighteen numbers add up to 1800.
Since 1800 + 190 ̸= 1999, the largest number among the nineteen must be at least 208. But then the smallest eighteen numbers add up to at least 1800, giving a total sum of at least 2028 > 1999, a contradiction.
Problem 11.2 At each rational point on the real line is written an integer. Show that there exists a segment with rational endpoints, such that the sum of the numbers at the endpoints does not exceed twice the number at the midpoint.
First Solution: Let f : Q →Z be the function that maps each rational point to the integer written at that point. Suppose by way of contradiction that for all q, r ∈Q, f(q) + f(r) > 2f q + r 2 .
For i ≥0, let ai = 1 2i and bi = −1 2i . We shall prove that for some k, f(ak) and f(bk) are both less than f(0). Suppose that for some i, f(ai) ≥f(0). Now we apply the condition: f(ai+1) < f(ai) + f(0) 2 ≤f(ai).
Since the range of f is the integers, f(ai+1) ≤f(ai) −1 as long as f(ai) ≥f(0). Therefore, there exists some m such that f(am) < f(0).
Then f(am+1) < f(am) + f(0) 2 < 2f(0) 2 , so f(ai) < f(0) for i ≥m.
180 Russia Similarly, there exists n such that f(bi) < f(0) for i ≥n. Now if we just take k = max{m, n}, we have a contradiction: f(ak) + f(bk) < 2f(0).
Second Solution: Define f as in the first solution, and suppose by way of contradiction that there was no such segment; rewrite the inequality in the first solution as f(p) −f p + q 2 > f p + q 2 −f(q).
For a continuous function, this would be equivalent to saying that f is strictly convex; however, f is not continuous. But we can still show a similar result for the set F = i 2j | i, j ∈Z, j ≥0 , fractions whose denominators are powers of 2. For convenience, write Px to represent (x, f(x)) on the graph of f in the xy-plane. Then we have the following result: Lemma. For all a, b, c ∈F with b between a and c, Pb is below the segment connecting Pa and Pc.
Proof: Equivalently we can prove that the average rate of change of f in the interval [a, b] is smaller than the average rate of change of f in the interval [b, c] — that is, f(b) −f(a) b −a < f(c) −f(b) c −a .
Partition [a, b] and [b, c] into sub-intervals of equal length δ. For example, if a = α 2j , b = β 2j , and c = γ 2j , we could use δ = 1 2j .
Let ∆x = f(x+δ)−f(x) δ , the average rate of change of f in the interval [x, x + δ]. Then apply our inequality to find that ∆a < ∆a+δ < · · · < ∆c−δ.
Thus, f(b) −f(a) b −a ≤max {∆a, ∆a+δ, . . . , ∆b−δ} = ∆b−δ < ∆b = min {∆b, ∆b+δ, . . . , ∆c−δ} 1999 National Contests: Problems and Solutions 181 ≤f(c) −f(b) c −b , as desired.
Now consider all numbers x ∈F between 0 and 1. Since Px lies below the segment connecting P0 and P1, we have f(x) ≤ max{f(0), f(1)}.
Pick some number k ∈F between 0 and 1. For k < x < 1, Px must lie above the line connecting P0 and Pk; otherwise, Pk would be above the segment connecting P0 and Px, contradicting our lemma.
Similarly, for 0 < x < k, Px must lie above the line connecting Pk and P1.
Since there are infinitely many values x ∈F in the interval (0, 1) but f(x) is bounded from above and below in this interval, some three points have the same y-coordinate – contradicting our lemma. There-fore our original assumption was false and the segment described in the problem does exist.
Problem 11.3 A circle inscribed in quadrilateral ABCD touches sides DA, AB, BC, CD at K, L, M, N, respectively.
Let S1, S2, S3, S4 be the incircles of triangles AKL, BLM, CMN, DNK, respectively. The common external tangents to S1 and S2, to S2 and S3, to S3 and S4, and to S4 and S1, not lying on the sides of ABCD, are drawn. Show that the quadrilateral formed by these tangents is a rhombus.
Solution: Let P be the intersection of the two common external tangents involving S1, and let Q, R, S be the intersections of the pairs of tangents involving S2, S3, S4, respectively.
As in problem 10.3, the centers of S1, S2, S3, S4 are the midpoints of arcs KL, LM, MN, NK, respectively. AB does not pass through the incenter I of triangle KLM, so by the result proved in problem 9.3 the other external tangent PQ must pass through I and be parallel to KM. Likewise, RS ∥KM so we have PQ ∥RS.
Similarly, QR ∥LN ∥SP, so PQRS is a parallelogram.
Let ⟨X | ω⟩denote the length of the tangent from point X to circle ω, and let ⟨ω1 | ω2⟩denote the length of the external tangent to circles ω1 and ω2. Then we also know AB = ⟨A | S1⟩+ ⟨S1 | S2⟩+ ⟨S2 | B⟩ 182 Russia = ⟨A | S1⟩+ ⟨S1 | P⟩+ PQ + ⟨Q | S2⟩+ ⟨S2 | B⟩ and three analogous equations. Substituting these into AB + CD = BC +DA, which is true since ABCD is circumscribed about a circle, we find that PQ + RS = QR + SP.
But since PQRS is a parallelogram, PQ = RS and QR = SP, implying that PQ = QR = RS = SP and that PQRS is a rhombus.
Problem 11.5 Four natural numbers have the property that the square of the sum of any two of the numbers is divisible by the product of the other two. Show that at least three of the four numbers are equal.
Solution: Suppose by way of contradiction four such numbers did exist, and pick a counterexample a, b, c, d with minimum sum a+b+c+d. If some prime p divided both a and b, then from a | (b+c)2 and a | (b + d)2 we know that p divides c and d as well: but then a p, b p, c p, d p are a counter-example with smaller sum. Therefore, the four numbers are pairwise relatively prime.
Suppose that some prime p > 2 divided a. Then since a divides each of (b + c)2, (c + d)2, (d + b)2, we know that p divides b + c, c + d, d + b. Hence p divides (b + c) + (c + d) + (d + b) and thus b + c + d.
Therefore p | (b + c + d) −(b + c) = d, and similarly p | c and p | b, a contradiction.
Thus each of a, b, c, d are powers of 2. But since they are pairwise relatively prime, three of them must equal 1 — a contradiction.
Therefore our original assumption was false, and no such counterex-ample exists.
Problem 11.6 Show that three convex polygons in the plane cannot be intersected by a single line if and only if for each of the polygons, there exists a line intersecting none of the polygons, such that the given polygon lies on the opposite side of the line from the other two.
Solution: In this proof, “polygon” refers to both the border and interior of a polygon – the problem statement is not affected by this assumption, because a line hitting the interior of a polygon must hit its border as well.
Suppose that some line ℓintersects all three polygons; orient the figure to make ℓhorizontal, and say it hits the polygons (from left to 1999 National Contests: Problems and Solutions 183 right) at A, B, and C. Any line m not hitting any of the polygons is either parallel to ℓ; hits ℓto the left of B; or hits ℓto the right of B.
In all of these cases, m does not separate B from both A and C, so m cannot separate the polygon containing B from the other polygons.
(In the first two cases B and C are not separated; and in the first and third cases A and B are not separated.) To prove the other direction, we begin by proving an intuitively obvious but nontrivial lemma: Lemma. Given two non-intersecting polygons, there is a line that separates them.
Let V be the convex hull of the two polygons. If all its vertices are in one polygon, then this polygon contains the other — a contradiction.
Also, for any four vertices A, B, C, D in that order on V (not necessarily adjacent), since AC and BD intersect we cannot have A and C in one polygon and B and D in the other. Thus one run of adjacent vertices V1, . . . , Vm is in one polygon P; and the remaining vertices W1, . . . , Wn are in the other polygon Q.
Then V1Vm is contained in polygon P, so line V1Vm does not intersect Q; therefore we can simply choose a line extremely close to V1Vm that doesn’t hit P, and separates P and Q.
Now call the polygons T, U, V, and suppose no line intersects all three. Then every two polygons are disjoint — if M was in T ∪U and N ̸= M was in V, then the line MN hits all three polygons.
Triangulate the convex hull H of T and U (that is, divide it into triangles whose vertices are vertices of H). If V intersects H at some point M, then M is on or inside of these triangles, XY Z. Without loss of generality say X ∈T and Y, Z ∈U (otherwise both triangle XY Z and M are inside either T or U, so this polygon intersects V ). Then line XM intersects both T and V ; and since it hits Y Z, it intersects U as well – a contradiction.
Thus H is disjoint from V, and from the lemma we can draw a line separating the two — and thus separating T and U from V, as desired. We can repeat this construction for T and U, so we are done.
Problem 11.7 Through vertex A of tetrahedron ABCD passes a plane tangent to the circumscribed sphere of the tetrahedron. Show that the lines of intersection of the plane with the planes ABC, ACD, 184 Russia ABD form six equal angles if and only if AB · CD = AC · BD = AD · BC.
Solution: Perform an inversion about A with arbitrary radius r.
Since the given plane P is tangent to the circumscribed sphere of ABCD, the sphere maps to a plane parallel to P containing B′, C′, D′, the images of B, C, D under inversion. Planes P, ABC, ACD, and ABD stay fixed under the inversion since they all contain A.
Now, since C′D′ is in a plane parallel to P, plane ACD = AC′D′ intersects P in a line parallel to C′D′. More rigorously, complete parallelogram C′D′AX. Then X is both in plane AC′D′ = ACD and in plane P (since PX ∥C′D′), so the intersection of ACD and P is the line PX, parallel to C′D′.
Similarly, plane ADB intersects P in a line parallel to D′B′, and plane ABC intersects P in a line parallel to B′C′. These lines form six equal angles if and only if C′D′, D′B′, B′C′ form equal angles: that is, if triangle C′D′B′ is equilateral and C′D′ = D′B′ = B′C′.
Under the inversion distance formula, this is true if and only if CD · r2 AC · AD = DB · r2 AD · AB = BC · r2 AB · AC , which (multiplying by AB·AC·AD r2 ) is equivalent to AB · CD = AC · BD = AD · BC, as desired.
1999 National Contests: Problems and Solutions 185 1.18 Slovenia Problem 1 The sequence of real numbers a1, a2, a3, . . . satisfies the initial conditions a1 = 2, a2 = 500, a3 = 2000 as well as the relation an+2 + an+1 an+1 + an−1 = an+1 an−1 for n = 2, 3, 4, . . . . Prove that all the terms of this sequence are positive integers and that 22000 divides the number a2000.
Solution: From the recursive relation it follows that an+2an−1 = a2 n+1 for n = 2, 3, . . . . No term of our sequence can equal 0, and hence it is possible to write an+2 an+1an = an+1 anan−1 for n = 2, 3, . . . . It follows by induction that the value of the expression an+1 anan−1 is constant, namely equal to a3 a2a1 = 2.
Thus an+2 = 2anan+1 and all terms of the sequence are positive integers.
From this new relation, we also know that an+1 an is an even integer for all positive integers n. Write a2000 = a2000 a1999 a1999 a1998 · · · a2 a1 · a1. In this product each of the 1999 fractions is divisible by 2, and a1 = 2 is even as well. Thus a2000 is indeed divisible by 22000.
Problem 2 Find all functions f : R →R that satisfy the condition f(x −f(y)) = 1 −x −y for all x, y ∈R.
Solution: For x = 0, y = 1 we get f(−f(1)) = 0. For y = −f(1) it follows that f(x) = 1+f(1)−x. Writing a = 1+f(1) and f(x) = a−x, we have 1 −x −y = f(x −f(y)) = a −x + f(y) = 2a −x −y so that a = 1 2. And indeed, the function f(x) = 1 2 −x satisfies the functional equation.
Problem 3 Let E be the intersection of the diagonals in cyclic quadrilateral ABCD, and let F and G be the midpoints of sides AB and CD, respectively.
Prove that the three lines through G, F, E perpendicular to AC, BD, AD, respectively, intersect at one point.
186 Slovenia Solution: All angles are directed modulo 180◦. Drop perpendicular GP to diagonal AC and perpendicular FQ to diagonal BD. Let R be the intersection of lines PG and FQ, and let H be the foot of the perpendicular from E to side AD. We wish to prove that H, E, R are collinear.
Since F and G are midpoints of corresponding sides in similar triangles DEC and ABE (with opposite orientations), triangles DPE and AQE are similar with opposite orientations as well.
Thus ∠DPE = ∠EQA and therefore AQPD is a cyclic quadrilateral. And because ∠EQR = 90◦= ∠EPR, the quadrilateral EQRP is cyclic, too. So ∠ADQ = ∠APQ = ∠EPQ = ∠ERQ.
It follows that ∠DEH = 90◦−∠ADQ = 90◦−∠ERQ = ∠QER; and since D, E, Q are collinear then H, E, R must be as well.
Problem 4 Three boxes with at least one marble in each are given.
In a step we choose two of the boxes, doubling the number of marbles in one of the boxes by taking the required number of marbles from the other box. Is it always possible to empty one of the boxes after a finite number of steps?
Solution: Without loss of generality suppose that the number of marbles in the boxes are a, b, and c with a ≤b ≤c. Write b = qa + r where 0 ≤r < a and q ≥1. Then express q in binary: q = m0 + 2m1 + · · · + 2kmk, where each mi ∈{0, 1} and mk = 1. Now for each i = 0, 1, . . . , k, add 2ia marbles to the first box: if mi = 1 take these marbles from the second box; otherwise take them from the third box. In this way we take at most (2k −1)a < qa ≤b ≤c marbles from the third box and exactly qa marbles from the second box altogether.
In the second box there are now r < a marbles left. Thus the box with the least number of marbles now contains less than a marbles.
Then by repeating the described procedure, we will eventually empty one of the boxes.
1999 National Contests: Problems and Solutions 187 1.19 Taiwan Problem 1 Determine all solutions (x, y, z) of positive integers such that (x + 1)y+1 + 1 = (x + 2)z+1.
Solution: Let a = x + 1, b = y + 1, c = z + 1. Then a, b, c ≥2 and ab + 1 = (a + 1)c ((a + 1) −1)b + 1 = (a + 1)c.
Taking either equation mod (a + 1) yields (−1)b + 1 ≡0, so b is odd.
Then taking the second equation mod (a + 1)2 after applying the binomial expansion yields b 1 (a + 1)(−1)b−1 + (−1)b + 1 ≡0 (mod (a + 1)2) so a + 1 | b and a is even.
On the other hand, taking the first equation mod a2 after applying the binomial expansion yields 1 ≡ c 1 a + 1 (mod a2) so c is divisible by a and is even as well. Write a = 2a1 and c = 2c1.
Then 2bab 1 = ab = (a + 1)c −1 = ((a + 1)c1 −1)((a + 1)c1 + 1).
It follows that gcd((a + 1)c1 −1, (a + 1)c1 + 1) = 2. Therefore, using the fact that 2a1 is a divisor of (a + 1)c1 −1, we may conclude that (a + 1)c1 −1 = 2ab 1 (a + 1)c1 + 1 = 2b−1.
We must have 2b−1 > 2ab 1 ⇒a1 = 1. Then these equations give c1 = 1 and b = 3, and therefore the only solution is (x, y, z) = (1, 2, 1).
Problem 2 There are 1999 people participating in an exhibition.
Out of any 50 people, at least 2 do not know each other. Prove that we can find at least 41 people who each know at most 1958 other people.
188 Taiwan Solution: Let Y be the set of people who know at least 1959 other people, and let N(p) denote the set of people whom p knows. Assume by way of contradiction that less than 41 people each know at most 1958 people; then |Y | ≥1959. We now show that some 50 people all know each other, a contradiction.
Pick a person y1 ∈Y and write B1 = N(y1) with |B1| ≥1959.
Then |B1| + |Y | > 1999, and there is a person y2 ∈B1 ∩Y .
Now write B2 = N(y1) ∩N(y2) with |B2| = |B1| + |N(y2)| −|B1 ∪ N(y2)| ≥1959 + 1959 −1999 = 1999 −40 · 2. Then |B2| + |Y | > 1999, and there is a person y3 ∈B2 ∩Y .
Now continue similarly: suppose we have j ≤48 different people y1, y2, . . . , yj in Y who all know each other; and suppose that Bj = N(y1)∩N(y2)∩· · ·∩N(yj) has at least 1999−40j ≥79 > 40 elements.
Then |Bj| + |Y | > 1999, and there is a person yj+1 ∈Bj ∩Y ; and Bj+1 = Bj ∩N(yj+1) has at least |Bj|+|N(yj+1)|−|Bj ∪N(yj+1)| ≥ (1999 −40j) + 1959 −1999 = 1959 −40(j + 1) > 0 elements, and we can continue onward.
Thus we can find 49 people y1, y2, . . . , y49 such that B49 = N(y1)∩ N(y2) ∩· · · ∩N(y49) is nonempty. Thus there is a person y50 ∈B49; but then any two people from y1, y2, . . . , y50 know each other, a contradiction.
Problem 3 Let P ∗denote all the odd primes less than 10000, and suppose p ∈P ∗. For each subset S = {p1, p2, · · · , pk} of P ∗, with k ≥2 and not including p, there exists a q ∈P ∗\ S such that q + 1 | (p1 + 1)(p2 + 1) · · · (pk + 1).
Find all such possible values of p.
Solution: A “Mersenne prime” is a prime of the form 2n −1 for some positive integer n. Notice that if 2n −1 is prime then n > 1 and n is prime because otherwise we could either write (if n were even) n = 2m and 2n −1 = (2m −1)(2m + 1), or (if n were odd) n = ab for odd a, b and 2n −1 = (2a −1)(2(b−1)a + 2(b−2)a + · · · + 2a + 1). Then some calculations show that the set T of Mersenne primes less than 10000 is {M2, M3, M5, M7, M13} = {3, 7, 31, 127, 8191}, 1999 National Contests: Problems and Solutions 189 where Mp = 2p −1. (211 −1 is not prime: it equals 23·89.) We claim this is the set of all possible values of p.
If some prime p is not in T, then look at the set S = T. Then there must be some prime q ̸∈S less than 10000 such that q + 1 | (M2 + 1)(M3 + 1)(M5 + 1)(M7 + 1)(M13 + 1) = 230.
Thus, q +1 is a power of 2 and q is a Mersenne prime less than 10000 — and therefore q ∈T = S, a contradiction.
On the other hand, suppose p is in T.
Suppose we have a set S = {p1, p2, . . . , pk} ⊂P ∗not including p, with k ≥2 and p1 < p2 < · · · < pk. Suppose by way of contradiction that for all q ∈P ∗such that q + 1 | (p1 + 1) · · · (pk + 1), we have q ∈S. Then 4 | (p1 + 1)(p2 + 1) = ⇒M2 ∈S 8 | (M2 + 1)(p2 + 1) = ⇒M3 ∈S 32 | (M2 + 1)(M3 + 1) = ⇒M5 ∈S 128 | (M2 + 1)(M5 + 1) = ⇒M7 ∈S 8192 | (M3 + 1)(M5 + 1)(M7 + 1) = ⇒M13 ∈S.
Then p, a Mersenne prime under 10000, must be in S — a contra-diction.
Therefore there is some prime q < 10000 not in S with q + 1 | (p1 + 1) · · · (pk + 1), as desired. This completes the proof.
Problem 4 The altitudes through the vertices A, B, C of an acute-angled triangle ABC meet the opposite sides at D, E, F, respectively, and AB > AC. The line EF meets BC at P, and the line through D parallel to EF meets the lines AC and AB at Q and R, respectively.
Let N be a point on the side BC such that ∠NQP + ∠NRP < 180◦.
Prove that BN > CN.
Solution: Let M be the midpoint of BC. We claim that P, Q, M, R are concyclic. Given this, we would have ∠MQP + ∠MRP = 180◦> ∠NQP + ∠NRP.
This can only be true if N is between M and C; then BN > CN, as desired.
Since ∠BEC = ∠BFC = 90◦, we observe that the points B, C, E, F are concyclic and thus PB · PC = PE · PF. Also, the points D, E, F, M lie on the nine-point circle of triangle ABC so that 190 Taiwan PE · PF = PD · PM. (Alternatively, it’s easy to show that DEFM is cyclic with some angle-chasing). These two equations yield PB · PC = PD · PM.
(1) On the other hand, since △AEF ∼△ABC and QR ∥EF, we have ∠RBC = ∠AEF = ∠CQR. Thus CQBR is cyclic and DQ · DR = DB · DC.
(2) Now let MB = MC = a, MD = d, MP = p.
Then we have PB = p + a, DB = a + d, PC = p −a, CD = a −d, DP = p −d.
Then equation (1) implies (p + a)(p −a) = (p −d)p = ⇒a2 = dp = ⇒(a + d)(a −d) = (p −d)d, or equivalently DB · DC = DP · DM.
(3) Combining (2) and (3) yields DQ·DR = DP ·DM, so that the points P, Q, M, R are concyclic, as claimed.
Problem 5 There are 8 different symbols designed on n different T-shirts, where n ≥2. It is known that each shirt contains at least one symbol, and for any two shirts, the symbols on them are not all the same. Also, for any k symbols, 1 ≤k ≤7, the number of shirts containing at least one of the k symbols is even. Find the value of n.
Solution: Let X be the set of 8 different symbols, and call a subset S of X “stylish” if some shirt contains exactly those symbols in S.
Look at a stylish set A with the minimal number of symbols |A| ≥1; since n ≥2, we must have |A| ≤7. Then all the other n −1 stylish sets contain at least one of the k = 8 −|A| symbols in X \ A, so n −1 is even and n is odd.
Observe that any nonempty subset S ⊆X contains an odd number of stylish subsets: For S = X this number is n; and for |S| ≤7, an even number t of stylish sets contain some element of X \ S, so the remaining odd number n −t of stylish sets are contained in S.
Then every nonempty subset of X is stylish.
Otherwise, pick a minimal non-stylish subset S ⊆X. Its only stylish subsets are its 1999 National Contests: Problems and Solutions 191 2|S| −2 proper subsets, which are all stylish by the minimal definition of S; but this is an even number, which is impossible. Thus there must be 28 −1 = 255 T-shirts; and indeed, given any k symbols (1 ≤k ≤7), an even number 28 −28−k of T-shirts contain at least one of these k symbols.
192 Turkey 1.20 Turkey Problem 1 Let ABC be an isosceles triangle with AB = AC. Let D be a point on BC such that BD = 2DC, and let P be a point on AD such that ∠BAC = ∠BPD. Prove that ∠BAC = 2∠DPC.
Solution: Draw X on BP such that BX = AP. Then ∠ABX = ∠ABP = ∠DPB −∠PAB = ∠CAB −∠PAB = ∠CAP. And since AB = CA and BX = AP, by SAS we have △ABX ∼ = △CAP.
Hence [ABX] = [CAP], and also ∠DPC = 180◦−∠CPA = 180◦− ∠AXB = ∠PXA.
Next, since BD = 2CD, the distance from B to line AD is twice the distance from C to line AD. Therefore [ABP] = 2[CAP] = ⇒ [ABX] + [AXP] = 2[ABX]. Hence [AXP] = [ABX] and XP = BX = AP. Hence ∠PXA = ∠XAP, and ∠BAC = ∠BPD = ∠PXA + ∠XAP = 2∠PXA = 2∠DPC, as desired.
Problem 2 Prove that (a + 3b)(b + 4c)(c + 2a) ≥60abc for all real numbers 0 ≤a ≤b ≤c.
Solution: By AM-GM we have a + b + b ≥3 3 √ ab2; multiplying this and the analagous inequalities yields (a+2b)(b+2c)(c+2a) ≥27abc.
Then (a + 3b)(b + 4c)(c + 2a) ≥ a + 1 3a + 8 3b b + 2 3b + 10 3 c (c + 2a) = 20 9 (a + 2b)(b + 2c)(c + 2a) ≥60abc, as desired.
Problem 3 The points on a circle are colored in three different colors. Prove that there exist infinitely many isosceles triangles with vertices on the circle and of the same color.
1999 National Contests: Problems and Solutions 193 First Solution: Partition the points on the circle into infinitely many regular 13-gons. In each 13-gon, by the Pigeonhole Principle there are at least 5 vertices of the same color: say, red.
Later we use some extensive case analysis to show that among these 5 vertices, some three form an isosceles triangle. Then for each 13-gon there is a monochrome isosceles triangle; so there are infinitely many monochrome isosceles triangles, as desired.
It suffices now to prove the following claim: Claim Suppose 5 vertices of a regular 13-gon are colored red. Then some three red vertices form an isosceles triangle.
Proof: Suppose none of these 5 vertices did form an isosceles triangle. Label the vertices P0, . . . , P12 (with indices taken modulo 13); first we prove that Pi and Pi+2 cannot both be red. Assume they could be, and say without loss of generality that P12 and P1 were red; then P10, P0, and P3 cannot be red. Furthermore, at most one vertex from each pair (P11, P4), (P4, P7), and (P7, P8) is red since each of these pairs forms an isosceles triangle with P1. Similarly, at most one vertex from each pair (P2, P9), (P9, P6), and (P6, P5) is red. Now three vertices from {P11, P4, P7, P8}∪{P2, P9, P6, P5} are red; assume without loss of generality that two vertices from {P11, P4, P7, P8} are.
Vertices P4 and P8 can’t both be red because they form an isosceles triangle with P12; so vertices P11 and P7 must be red.
But then any remaining vertex forms an isosceles triangle with some two of P1, P7, P11, P12, so we can’t have five red vertices, a contradiction.
Next we prove that Pi and Pi+1 can’t be red.
If so, suppose without loss of generality that P6 and P7 are red.
Then P4, P5, P8, and P9 cannot be red from the result in the last paragraph. P0 cannot be red either, because triangle P0P6P7 is isosceles. Now each pair (P3, P11) and (P11, P1) contains at most one red vertex because triangles P3P7P11 and P1P6P11 are isosceles. Also, P1 and P3 can’t both be red from the result in the last paragraph. Thus at most one of {P1, P3, P11} can be red; similarly, at most one of {P12, P10, P2} can be red.
But then we have at most four red vertices, again a contradiction.
Thus if Pi is red then Pi−2, Pi−1, Pi+1, Pi+2 cannot be red; but then 194 Turkey we can have at most four red vertices, a contradiction.
Second Solution: Suppose we have k ≥1 colors and a number n ≥3. Then Van der Warden’s theorem states that we can find N such that for any coloring of the numbers 1, 2, . . . , N in the k colors, there are n numbers in arithmetic progression which are colored the same. Apply this theorem with k = n = 3 to find such an N, and partition the points on the circle into infinitely many regular N-gons rather than 13-gons. For each N-gon P1P2 . . . PN, there exist i, j, k (between 1 and N) in arithmetic progression such that Pi, Pj, Pk are all the same color. Hence triangle PiPjPk is a monochrome isosceles triangle. It follows that since we have infinitely many such N-gons, there are infinitely many monochrome isosceles triangles.
Problem 4 Let ∠XOY be a given angle, and let M and N be two points on the rays OX and OY , respectively. Determine the locus of the midpoint of MN as M and N varies along the rays OX and OY such that OM + ON is constant.
Solution: Let ˆ x and ˆ y be the unit vectors pointing along rays OX and OY . Suppose we want OM + ON to equal the constant k; then when OM = c we have ON = k −c, and thus the midpoint of MN is 1 2(cˆ x + (k −c)ˆ y). As c varies from 0 to k, this traces out the line segment connecting 1 2kˆ x with 1 2kˆ y; that is, the segment M ′N ′ where OM ′ = ON ′ = 1 2k, M ′ ∈− − → OX, and N ′ ∈− − → OY .
Problem 5 Some of the vertices of the unit squares of an n × n chessboard are colored such that any k × k square formed by these unit squares has a colored point on at least one of its sides. If l(n) denotes the minimum number of colored points required to ensure the above condition, prove that lim n→∞ l(n) n2 = 2 7.
Solution: For each colored point P, consider any 1 × 1 square of the board it lies on. If this square contains m colored points, say that P gains 1 m points from that square. Adding over all the 1×1 squares that P lies on, we find the total number of points that P accrues.
1999 National Contests: Problems and Solutions 195 Any colored point on the edge of the chessboard gains at most 2 points. As for a colored point P on the chessboard’s interior, the 2×2 square centered at P must have a colored point Q on its border.
Then P and Q both lie on some unit square, which P gains at most half a point from; thus P accrues at most 7 2 points.
Therefore any colored point collects at most 7 2 points, and l(n) colored points collectively accrue at most 7 2l(n) points. But for the given condition to hold, the total number of points accrued must be n2. It follows that 7 2l(n) ≥n2 and thus l(n) n2 ≥2 7.
Now, given some n × n board, embed it as the corner of an n′ × n′ board where 7 | n′ + 1 and n ≤n′ ≤n + 6. To each 7 × 7 grid of vertices on the n′ × n′ board, color the vertices as below: • ◦ • ◦ ◦ ◦ ◦ ◦ ◦ ◦ ◦ • ◦ • ◦ • ◦ • ◦ ◦ ◦ • ◦ ◦ ◦ ◦ • ◦ ◦ ◦ • ◦ • ◦ ◦ ◦ • ◦ ◦ ◦ ◦ • ◦ ◦ ◦ • ◦ • ◦ Then any k × k square on the chessboard has a colored point on at least one of its sides. Since we color 2 7(n′+1)2 vertices in this coloring, we have l(n) ≤2 7(n′ + 1)2 ≤2 7(n + 7)2 so that l(n) n2 ≤2 7 n + 7 n 2 .
As n →∞, the right hand side becomes arbitrarily close to 2 7. Since from before l(n) n2 ≥2 7 for all n, this implies that limn→∞ l(n) n2 exists and equals 2 7.
Problem 6 Let ABCD be a cyclic quadrilateral, and let L and N be the midpoints of diagonals AC and BD, respectively. Suppose that BD bisects ∠ANC. Prove that AC bisects ∠BLD.
Solution: Suppose we have any cyclic quadrilateral ABCD where L and N are the midpoints of AC and BD. Perform an inversion about B with arbitrary radius; A, D, C map to collinear points A′, D′, C′, while N maps to the point N ′ such that D′ is the midpoint of BN ′.
196 Turkey There are only two points X on line A′D′ such that ∠BXN ′ = ∠BA′N ′: the point A′ itself, and the reflection of A′ across D′.
Then ∠ANB = ∠BNC ⇐ ⇒∠BA′N ′ = ∠BC′N ′ ⇐ ⇒A′D′ = D′C′ ⇐ ⇒ AD BA·BD = DC BD·BC ⇐ ⇒AD · BC = BA · DC.
Similarly, ∠BLA = ∠DLA ⇐ ⇒AD · BC = BA · DC. Therefore ∠ANB = ∠BNC ⇐ ⇒∠BLA = ∠DLA; that is, BD bisects ∠ANC if and only if AC bisects ∠BLD, which implies the claim.
Problem 7 Determine all functions f : R →R such that the set f(x) x | x ∈R and x ̸= 0 is finite and f(x −1 −f(x)) = f(x) −x −1 for all x ∈R.
Solution: First we show that the set {x −f(x) | x ∈R} is finite.
If not, there exist infinitely many k ̸= 1 such that for some xk, xk −f(xk) = k. But then f(k −1) k −1 = f(xk −1 −f(xk)) k −1 = f(xk) −xk −1 k −1 = −1 − 2 k −1.
Since k takes on infinitely many values, f(k−1) k−1 does as well—a contradiction.
Now choose x0 so that |x −f(x)| is maximal for x = x0. Then for y = x0 −1 −f(x0) we have y −f(y) = y −(f(x0) −x0 −1) = 2(x0 −f(x0)).
Then because of the maximal definition of x0, we must have y−f(y) = x0 −f(x0) = 0. Therefore f(x) = x for all x, and this function indeed satisfies the given conditions.
Problem 8 Let the area and the perimeter of a cyclic quadrilateral C be AC and PC, respectively. If the area and the perimeter of the quadrilateral which is tangent to the circumcircle of C at the vertices of C are AT and PT , respectively, prove that AC AT ≥ PC PT 2 .
1999 National Contests: Problems and Solutions 197 Solution: Let the outer quadrilateral be EFGH with angles ∠E = 2α1, ∠F = 2α2, ∠G = 2α3, ∠H = 2α4; also let the circumcircle of C have radius r and center O. Say that sides EF, FG, GH, HE are tangent to C at I, J, K, L.
In right triangle EIO, we have IO = r and ∠OEI = α1 so that EI = r cot α1. After finding IF, FJ, . . . , LE similarly, we find that PT = 2r P4 i=1 cot αi. Also, [EFO] = 1 2EF · IO = 1 2EF · r; finding [FGO], [GHO], [HEO] similarly shows that AT = 1 2PT · r.
As for quadrilateral IJKL, note that IJ = 2r sin ∠IKJ = 2r sin ∠FIJ = 2r sin(90◦−α2) = 2r cos α2. After finding JK, KL, LI in a similar manner we have PC = 2r P4 i=1 cos αi.
Also note that ∠IOJ = 180◦−∠JFI = 180◦−2α2, and hence [IOJ] = 1 2OI · OJ sin ∠IOJ = 1 2r2 sin(2α2) = r2 sin α2 cos α2.
Adding this to the analogous expressions for [JOK], [KOL], [LOI], we find that AC = r2 P4 i=1 sin αi cos αi.
Therefore the inequality we wish to prove is AC · P 2 T ≥AT · P 2 C ⇐ ⇒r2 4 X i=1 sin αi cos αi · P 2 T ≥ 1 2PT · r · 4r2 4 X i=1 cos αi !2 ⇐ ⇒PT · 4 X i=1 sin αi cos αi ≥2r · 4 X i=1 cos αi !2 ⇐ ⇒ 4 X i=1 cot αi · 4 X i=1 sin αi cos αi ≥ 4 X i=1 cos αi !2 .
But this is true by the Cauchy-Schwarz inequality P a2 i P b2 i ≥ (P aibi)2 applied with each ai = √cot αi and bi = √sin αi cos αi.
Problem 9 Prove that the plane is not a union of the inner regions of finitely many parabolas. (The outer region of a parabola is the union of the lines on the plane not intersecting the parabola. The inner region of a parabola is the set of points on the plane that do not belong to the outer region of the parabola.) Solution: Suppose by way of contradiction we could cover the plane with the inner regions of finitely many parabolas — say, n of them.
Choose some fixed positive acute angle θ < 360 2n ◦.
198 Turkey Take any of the parabolas and (temporarily) choose a coordinate system so that it satisfies the equation y = ax2 with a ≥0 (and where our coordinates are chosen to scale, so that one unit along the y-axis has the same length as one unit along the x-axis). Draw the tangents to the parabola at x = ± cot θ 2a ; these lines have slopes 2ax = ± cot θ.
These lines meet on the y-axis at an angle of 2θ, forming a V-shaped region in the plane that contains the inner region of the parabola.
Performing the above procedure with all the parabolas, we obtain n V-shaped regions covering the entire plane. Again choose an x-axis, and say the rays bordering these regions make angles φj and φj + 2θ with the positive x-axis (with angles taken modulo 360◦).
Then since 2nθ < 360◦, there is some angle φ′ not in any of the intervals [φj, φj + 2θ]. Then consider the line passing through the origin and making angle of φ′ with the positive x-axis; far enough out, the points on this line cannot lie in any of the V-shaped regions, a contradiction.
Thus our original assumption was false, and we cannot cover the plane with the inner regions of finitely many parabolas.
1999 National Contests: Problems and Solutions 199 1.21 Ukraine Problem 1 Let P(x) be a polynomial with integer coefficients. The sequence {xn}n≥1 satisfies the conditions x1 = x2000 = 1999, and xn+1 = P(xn) for n ≥1. Calculate x1 x2 + x2 x3 + · · · + x1999 x2000 .
Solution: Write ai = xi −xi−1 for each i, where we take subscripts (of both the xi and ai) modulo 1999. Since c−d divides P(c)−P(d) for integers c and d, we have that ai = xi−xi−1 divides P(xi)−P(xi−1) = ai+1 for all i.
First suppose that all the ai ̸= 0. Then |ai+1| ≥|ai| for all i but also |a1| = |a2000|; hence all the |ai| equal the same value m > 0. But if n of the a1, a2, . . . , a1999 equal m ̸= 0 and the other 1999 −n equal −m, then their sum 0 = x1999 −x0 = a1 + a2 + · · · + a1999 equals m(2n −1999) ̸= 0, a contradiction.
Thus for some k we have ak = 0; then since ak divides ak+1, we have ak+1 = 0 and similarly ak+2 = 0, and so on. Thus all the xi are equal and the given expression equals 1999.
Problem 2 For real numbers 0 ≤x1, x2, . . . , x6 ≤1 prove the inequality x3 1 x5 2 + x5 3 + x5 4 + x5 5 + x5 6 + 5 + x3 2 x5 1 + x5 3 + x5 4 + x5 5 + x5 6 + 5 + · · · + x3 6 x5 1 + x5 2 + x5 3 + x5 4 + x5 5 + 5 ≤3 5.
Solution: The condition 0 ≤x1, x2, . . . , x6 ≤1 implies that the left hand side of the inequality is at most 6 X i=1 x3 i x5 1 + x5 2 + · · · + x5 6 + 4 = x3 1 + x3 2 + · · · + x3 6 x5 1 + x5 2 + · · · + x5 6 + 4.
For t ≥0 we have t5+t5+t5+1+1 5 ≥t3 by AM-GM. Adding up the six resulting inequalities for t = x1, x2, . . . , x6 and dividing by (x5 1 +x5 2 + · · · + x5 6 + 4) shows that the above expression is at most 3 5.
Problem 3 Let AA1, BB1, CC1 be the altitudes of an acute triangle ABC, and let O be an arbitrary point inside the triangle A1B1C1.
200 Ukraine Let M, N, P, Q, R, S be the orthogonal projections of O onto lines AA1, BC, BB1, CA, CC1, AB, respectively. Prove that lines MN, PQ, RS are concurrent.
Solution: Observe that three lines passing through different vertices of a triangle are concurrent if and only if their reflections across the corresponding angle bisectors are also concurrent; this is easily proved using the trigonometric form of Ceva’s Theorem.
Let A0, B0, C0 be the centers of rectangles OMA1N, OPB1Q, OSC1R, respectively. Under the homothety with center O and ratio 1 2, triangle A1B1C1 maps to triangle A0B0C0.
Then since lines AA1, BB1, CC1 are the angle bisectors of triangle A1B1C1 (easily proved with angle-chasing), the angle bisectors of triangle A0B0C0 are parallel to lines AA1, BB1, CC1.
Because OMA1N is a rectangle, diagonals OA1 and MN are reflections of each across the line through A0 parallel to line AA1.
From above, this line is precisely the angle bisector of ∠C0A0B0 in triangle A0B0C0. Similarly, lines OB1 and OC1 are reflections of lines PQ and RS across the other angle bisectors. Then since lines OA1, OB1, OC1 concur at O, from our initial observation lines MN, PQ, RS concur as well.
1999 National Contests: Problems and Solutions 201 1.22 United Kingdom Problem 1 I have four children. The age in years of each child is a positive integer between 2 and 16 inclusive and all four ages are distinct.
A year ago the square of the age of the oldest child was equal to sum of the squares of the ages of the other three. In one year’s time the sum of the squares of the ages of the oldest and the youngest children will be equal to the sum of the squares of the other two children. Decide whether this information is sufficient to determine their ages uniquely, and find all possibilities for their ages.
Solution: Let the children’s present ages be a + 1, b + 1, c + 1, and d + 1. We are given that 1 ≤a < b < c < d ≤15; note that b ≤13 so that b −a ≤12. We are also given d2 = a2 + b2 + c2 (1) and (d + 2)2 + (a + 2)2 = (b + 2)2 + (c + 2)2.
(2) Subtracting (1) from (2) gives 4(a + d) + a2 = 4(b + c) −a2, or a2 = 2(b + c −a −d).
(3) Then a must be even since its square is even. Furthermore, since d > c, a2 = 2(b −a + (c −d)) < 2(b −a) < 24, and hence either a = 2 and a = 4.
If a = 4 then, since a2 < 2(b −a), we have 2b > a2 + 2a = 24 so that b > 12. This forces b = 13, c = 14, and d = 15, which contradicts the given conditions.
Thus a = 2. Equation (3) gives b + c −d = 4, so substituting a = 2 and d = b + c −4 into (1) and simplifying yields (b −4)(c −4) = 10 = 1 · 10 = 2 · 5.
Therefore we have (b, c) = (5, 14) or (6, 9), in which cases d = 15 and d = 11 respectively.
Hence the only possible solutions are (a, b, c, d) = (2, 5, 14, 15) or (2, 6, 9, 11), and these indeed satisfy (1) and (2). It follows that there is no unique solution, and it is not possible to determine the childrens’ ages.
202 United Kingdom Problem 2 A circle has diameter AB and X is a fixed point on the segment AB. A point P, distinct from A and B, lies on the circle.
Prove that, for all possible positions of P, tan ∠APX tan ∠PAX is a constant.
Solution: Let Q be the projection of X onto AP.
Note that ∠APB = 90◦, and thus tan ∠PAX = P B P A.
Also, XQ ∥PB so △AQX ∼△APB. Therefore, tan ∠APX = QX QP = AX·BP AB BX·AP AB = AX · BP BX · AP , and tan ∠APX tan ∠PAX = AX BX is fixed.
Problem 3 Determine a positive constant c such that the equation xy2 −y2 −x + y = c has exactly three solutions (x, y) in positive integers.
Solution: When y = 1 the left hand side is 0. Thus we can rewrite our equation as x = y(y −1) + c (y + 1)(y −1).
The numerator is congruent to −1(−2)+c modulo y+1, and it is also congruent to c modulo y −1. Hence we must have c ≡−2 (mod y +1) and c ≡0 (mod y −1). Since c = y −1 satisfies these congruences, we must have c ≡y −1 (mod lcm(y −1, y + 1)). When y is even, lcm(y−1, y+1) = y2 −1; when y is odd, lcm(y−1, y+1) = 1 2(y2 −1).
Then for y = 2, 3, 11 we have c ≡1 (mod 3), c ≡2 (mod 4), c ≡10 (mod 60). Hence, we try setting c = 10.
For x to be an integer we must have y −1 | 10 ⇒y = 2, 3, 6, or 11; these values give x = 4, 2, 2 7, and 1 respectively. Thus there are exactly three solutions in positive integers, namely (x, y) = (4, 2), (2, 3), and (1, 11).
Problem 4 Any positive integer m can be written uniquely in base 3 form as a string of 0’s, 1’s and 2’s (not beginning with a zero). For 1999 National Contests: Problems and Solutions 203 example, 98 = 81 + 9 + 2 × 3 + 2 × 1 = (10122)3.
Let c(m) denote the sum of the cubes of the digits of the base 3 form of m; thus, for instance c(98) = 13 + 03 + 13 + 23 + 23 = 18.
Let n be any fixed positive integer. Define the sequence {ur} as u1 = n, and ur = c(ur−1) for r ≥2.
Show that there is a positive integer r such that ur = 1, 2, or 17.
Solution: If m has d ≥5 digits then we have m ≥3d−1 = (80 + 1)(d−1)/4 ≥80 · d−1 4 + 1 > 8d by Bernoulli’s inequality. Thus m > c(m).
If m > 32 has 4 digits in base 3, then c(m) ≤23 + 23 + 23 + 23 = 32 < m. And if 27 ≤m ≤32, then m starts with the digits 10 in base 3 and c(m) < 13 + 03 + 23 + 23 = 17 < m.
Therefore 0 < c(m) < m for all m ≥27, and hence eventually we have some positive us < 27. Since us has at most three digits, us+1 can only equal 8, 16, 24, 1, 9, 17, 2, 10, or 3. If it equals 1, 2, or 17 we are already done; if it equals 3 or 9 then us+2 = 1; and otherwise a simple check shows that ur will eventually equal 2: 8 = (22)3 24 = (220)3 →16 = (121)3 →10 = (101)3 →2.
Problem 5 Consider all functions f : N →N such that (i) for each positive integer m, there is a unique positive integer n such that f(n) = m; (ii) for each positive integer n, f(n+1) is either 4f(n)−1 or f(n)−1.
Find the set of positive integers p such that f(1999) = p for some function f with properties (i) and (ii).
Solution: Imagine hopping along a sidewalk whose blocks are marked from left to right with the positive integers, where at time n we stand on the block marked f(n). Note that if f(n) −f(n + 1) > 0 then f(n) −f(n + 1) = 1; that is, whenever we move to the left we move exactly one block. And whenever we move to the right from f(n) we must move to 4f(n) −1.
204 United Kingdom Then suppose that we are at block f(a) and that f(a) −1 is unvisited; if we move to the right (that is, if f(a + 1) > f(a)) then at some point we must pass through block f(a) to reach block f(a) −1 again— which is not allowed. Thus we must have f(a+1) = f(a)−1.
Therefore our path is completely determined by the value of f(1): because whenever we are at f(n), if f(n) −1 > 0 is unvisited we must have f(n + 1) = f(n) −1.
And otherwise, we must have f(n + 1) = 4f(n) −1.
If f(1) = 1 then consider the function f defined as follows: when-ever 2k ≤n < 2k+1, set f(n) = (3·2k −1)−n. It is bijective since for n = 2k, 2k +1, . . . , 2k+1 −1 we have f(n) = 2k+1 −1, 2k+1 −2, . . . , 2k; and a quick check shows it satisfies condition (ii) as well. Thus from the previous paragraph this is the only function with f(1) = 1, and in this case sice 210 ≤1999 < 211 we have f(1999) = (3·210−1)−1999 = 1072.
If f(1) = 2 then consider instead the function f defined as follows: whenever 4k ≤n < 3 · 4k, set f(n) = (4k+1 −1) −n; and if 3 · 4k ≤n < 4k+1 set f(n) = (7 · 4k −1) −n. Again, we can check that this function satisfies the conditions; and again, this must be the only function with f(1) = 2. In this case since 45 ≤1999 < 3 · 45, we have f(1999) = (46 −1) −1999 = 2096.
Finally, suppose that f(1) ≥3; first we must visit f(1) −1, f(1) − 2, . . . , 1. It follows that f(n) = 3 and f(n + 2) = 1 for some n. But then f(n + 3) = 4 · 1 −1 = 3 = f(n), a contradiction.
Therefore the only possible values of f(1999) are 1072 and 2096.
Problem 6 For each positive integer n, let Sn = {1, 2, . . . , n}.
(a) For which values of n is it possible to express Sn as the union of two non-empty disjoint subsets so that the elements in the two subsets have equal sum?
(b) For which values of n is it possible to express Sn as the union of three non-empty disjoint subsets so that the elements in the three subsets have equal sum?
Solution: (a) Let σ(T) denote the sum of the elements in a set T. For the condition to hold σ(Sn) = n(n+1) 2 must be even, and hence we must have n = 4k −1 or 4k where k ∈N.
For such n, let A consist of the second and third elements of each of the sets 1999 National Contests: Problems and Solutions 205 {n, n −1, n −2, n −3}, {n −4, n −5, n −6, n −7}, . . . , {4, 3, 2, 1} (or if n = 4k −1, the last set in this grouping will be {3, 2, 1}); and let B = Sn \ A. Then σ(A) = σ(B), as desired.
(b) For the condition to hold, σ(Sn) = n(n+1) 2 must be divisible by 3; furthermore, the construction is impossible for n = 3. Thus n must be of the form 3k + 2 or 3k + 3 where k ∈N. We prove all such n work by induction on n. We have S5 = {5}∪{1, 4}∪{2, 3}, S6 = {1, 6} ∪{2, 5} ∪{3, 4}, S8 = {8, 4} ∪{7, 5} ∪{1, 2, 3, 6}, and S9 = {9, 6} ∪{8, 7} ∪{1, 2, 3, 4, 5}. Now suppose that we can partition Sn−6 into A ∪B ∪C with σ(A) = σ(B) = σ(C); then σ(A ∪{n −5, n}) = σ(B ∪{n −4, n −1}) = σ(C ∪{n −3, n −2}), completing the inductive step and the proof of our claim.
Problem 7 Let ABCDEF be a hexagon which circumscribes a circle ω. The circle ω touches sides AB, CD, EF at their respective midpoints P, Q, R. Let ω touch sides BC, DE, FA at X, Y, Z respec-tively. Prove that lines PY, QZ, RX are concurrent.
Solution: Let O be the center of ω. Since P is the midpoint of AB, AP = PB; then by equal tangents, ZA = AP = PB = BX.
Thus ∠ZOA = ∠AOP = ∠POB = ∠BOX. It follows that ∠ZOP = ∠POX, and hence ∠ZY P = ∠PY X. Therefore line Y P is the angle bisector of ∠XY Z. Similarly lines XR and ZQ are the angle bisectors of ∠ZXY and ∠Y ZX, and therefore lines PY, QZ, RX meet at the incenter of triangle XY Z.
Problem 8 Some three non-negative real numbers p, q, r satisfy p + q + r = 1.
Prove that 7(pq + qr + rp) ≤2 + 9pqr.
Solution: Given a function f of three variables, let P cyc f(p, q, r) denote the “cyclic sum” f(p, q, r) + f(q, r, p) + f(r, p, q); for example, P cyc(pqr + p) = 3pqr + p + q + r. Since p + q + r = 1 the inequality is equivalent to 7(pq + qr + rp)(p + q + r) ≤2(p + q + r)3 + 9pqr ⇐ ⇒7 X cyc p2q + pq2 + pqr 206 United Kingdom ≤9pqr + X cyc 2p3 + 6p2q + 6pq2 + 4pqr ⇐ ⇒ X cyc p2q + X cyc pq2 ≤ X cyc 2p3 = X cyc 2p3 + q3 3 + X cyc p3 + 2q3 3 , and this last inequality is true by weighted AM-GM.
Problem 9 Consider all numbers of the form 3n2 + n + 1, where n is a positive integer.
(a) How small can the sum of the digits (in base 10) of such a number be?
(b) Can such a number have the sum of its digits (in base 10) equal to 1999?
Solution: (a) Let f(n) = 3n2 + n + 1. When n = 8, the sum of the digits of f(8) = 201 is 3. Suppose that some f(m) had a smaller sum of digits; then the last digit of f(m) must be either 0, 1, or 2.
However, for any n, f(n) = n(n + 3) + 1 ≡1 (mod 2); thus f(m) must have units digit 1.
Because f(n) can never equal 1, this means we must have 3m2+m+1 = 10k+1 for some positive integer k, and m(3m+1) = 10k. Since m and 3m+1 are relatively prime, and m < 3m+1, we must either have (m, 3m+1) = (1, 10k)—which is impossible—or (m, 3m + 1) = (2k, 5k). For k = 1, 5k ̸= 3 · 2k + 1; and for k > 1, we have 5k = 5k−2 · 25 > 2k−2 · (12 + 1) ≥3 · 2k + 1. Therefore, f(m) can’t equal 10k + 1, and 3 is indeed the minimum value for the sum of digits.
(b) Consider n = 10222 −1. f(n) = 3 · 10444 −6 · 10222 + 3 + 10222.
Thus, its decimal expansion is 2 9 . . . 9 | {z } 221 5 0 . . . 0 | {z } 221 3, and the sum of the digits in f(10222 −1) is 1999.
1999 National Contests: Problems and Solutions 207 1.23 United States of America Problem 1 Some checkers placed on an n × n checkerboard satisfy the following conditions: (i) every square that does not contain a checker shares a side with one that does; (ii) given any pair of squares that contain checkers, there is a sequence of squares containing checkers, starting and ending with the given squares, such that every two consecutive squares of the sequence share a side.
Prove that at least n2−2 3 checkers have been placed on the board.
Solution: It suffices to show that if m checkers are placed so as to satisfy condition (b), then the number of squares they either cover or are adjacent to is at most 3m+2. But this is easily seen by induction: it is obvious for m = 1, and if m checkers are so placed, some checker can be removed so that the remaining checkers still satisfy (b); they cover at most 3m−1 squares, and the new checker allows us to count at most 3 new squares (since the square it occupies was already counted, and one of its neighbors is occupied).
Note. The exact number of checkers required is known for m × n checkerboards with m small, but only partial results are known in the general case. Contact the authors for more information.
Problem 2 Let ABCD be a convex cyclic quadrilateral. Prove that |AB −CD| + |AD −BC| ≥2|AC −BD|.
First Solution: Let E be the intersection of AC and BD. Then the triangles ABE and DCE are similar, so if we let x = AE, y = BE, z = AB, then there exists k such that kx = DE, ky = CE, kz = CD. Now |AB −CD| = |k −1|z and |AC −BD| = |(kx + y) −(ky + x)| = |k −1| · |x −y|.
208 United States of America Since |x−y| ≤z by the triangle inequality, we conclude |AB −CD| ≥ |AC −BD|, and similarly |AD −BC| ≥|AC −BD|.
These two inequalities imply the desired result.
Second Solution: Let 2α, 2β, 2γ, 2δ be the measures of the arcs subtended by AB, BC, CD, DA, respectively, and take the radius of the circumcircle of ABCD to be 1. Assume without loss of generality that β ≤δ. Then α + β + γ + δ = π, and (by the Extended Law of Sines) |AB −CD| = 2 |sin α −sin γ| = 4 sin α −γ 2 cos α + γ 2 and |AC −BD| = 2 |sin(α + β) −sin(β + γ)| = 4 sin α −γ 2 cos α + γ 2 + β .
Since 0 ≤ 1 2(α + γ) ≤ 1 2(α + γ) + β ≤ π 2 (by the assumption β ≤δ) and the cosine function is nonnegative and decreasing on [0, π 2 ], we conclude that |AB −CD| ≥|AC −BD|, and similarly |AD −BC| ≥|AC −BD|.
Problem 3 Let p > 2 be a prime and let a, b, c, d be integers not divisible by p, such that ra p + rb p + rc p + rd p = 2 for any integer r not divisible by p. Prove that at least two of the numbers a + b, a + c, a + d, b + c, b + d, c + d are divisible by p. Here, for real numbers x, {x} = x −⌊x⌋denotes the fractional part of x.
Solution: For convenience, we write [x] for the unique integer in {0, . . . , p −1} congruent to x modulo p. In this notation, the given condition can be written [ra] + [rb] + [rc] + [rd] = 2p for all r not divisible by p.
(1) The conditions of the problem are preserved by replacing a, b, c, d with ma, mb, mc, md for any integer m relatively prime to p. If we choose m so that ma ≡1 (mod p) and then replace a, b, c, d with [ma], [mb], [mc], [md], respectively, we end up in the case a = 1 and 1999 National Contests: Problems and Solutions 209 b, c, d ∈{1, . . . , p−1}. Applying (1) with r = 1, we see moreover that a + b + c + d = 2p.
Now observe that [(r + 1)x] −[rx] = [x] [rx] < p −[x] −p + [x] [rx] ≥p −[x].
Comparing (1) applied to two consecutive values of r and using the observation, we see that for each r = 1, . . . , p−2, two of the quantities p −a −[ra], p −b −[rb], p −c −[rc], p −d −[rd] are positive and two are negative. We say that a pair (r, x) is positive if [rx] < p −[x] and negative otherwise; then for each r < p −1, (r, 1) is positive, so exactly one of (r, b), (r, c), (r, d) is also positive.
Lemma. If r1, r2, x ∈{1, . . . , p −1} have the property that (r1, x) and (r2, x) are negative but (r, x) is positive for all r1 < r < r2, then r2 −r1 = j p x k or r2 −r1 = j p x k + 1.
Proof: Note that (r′, x) is negative if and only if {r′x + 1, r′x + 2, . . . , (r′ + 1)x} contains a multiple of p. In particular, exactly one multiple of p lies in {r1x, r1x + 1, . . . , r2x}. Since [r1x] and [r2x] are distinct elements of {p −[x], . . . , p −1}, we have p −x + 1 < r2x −r1x < p + x −1, from which the lemma follows.
[rx] 9 10 0 1 2 3 4 5 6 7 8 9 10 0 is (r, x) + or –?
− + + + − r 3 4 5 6 7 (The above diagram illustrates the meanings of positive and negative in the case x = 3 and p = 11. Note that the difference between 7 and 3 here is ⌊p x⌋+ 1. The next r such that (r, x) is negative is r = 10; 10 −7 = ⌊p x⌋.) Recall that exactly one of (1, b), (1, c), (1, d) is positive; we may as well assume (1, b) is positive, which is to say b < p 2 and c, d > p 2. Put s1 = ⌊p b ⌋, so that s1 is the smallest positive integer such that (s1, b) is negative. Then exactly one of (s1, c) and (s1, d) is positive, say the former. Since s1 is also the smallest positive integer such that (s1, c) is positive, or equivalently such that (s1, p −c) is negative, we have 210 United States of America s1 = ⌊ p p−c⌋. The lemma states that consecutive values of r for which (r, b) is negative differ by either s1 or s1 + 1. It also states (when applied with x = p −c) that consecutive values of r for which (r, c) is positive differ by either s1 or s1 + 1. From these observations we will show that (r, d) is always negative.
r 1 s1 s1 + 1 s′ s′ + 1 s s + 1 ?
= t (r, b) + − + − + − −?
(r, c) − . . .
+ − . . .
+ − . . .
− +?
(r, d) − − − − − + −?
Indeed, if this were not the case, there would exist a smallest positive integer s > s1 such that (s, d) is positive; then (s, b) and (s, c) are both negative. If s′ is the last integer before s such that (s′, b) is negative (possibly equal to s1), then (s′, d) is negative as well (by the minimal definition of s). Also, s −s′ = s1 or s −s′ = s1 + 1.
Likewise, if t were the next integer after s′ such that (t, c) were positive, then t −s′ = s1 or t −s′ = s1 + 1.
From these we deduce that |t −s| ≤1. However, we can’t have t ̸= s because then both (s, b) and (t, b) would be negative—and any two values of r for which (r, b) is negative differ by at least s1 ≥2, a contradiction. (The above diagram shows the hypothetical case when t = s+1.) But nor can we have t = s because we already assumed that (s, c) is negative. Therefore we can’t have |t −s| ≤1, contradicting our findings and thus proving that (r, d) is indeed always negative.
Now if d ̸= p −1, then the unique s ∈{1, . . . , p −1} such that [ds] = 1 is not equal to p −1; and (s, d) is positive, a contradiction.
Thus d = p −1 and a + d and b + c are divisible by p, as desired.
Problem 4 Let a1, a2, . . . , an (n > 3) be real numbers such that a1 + a2 + · · · + an ≥n and a2 1 + a2 2 + · · · + a2 n ≥n2.
Prove that max(a1, a2, . . . , an) ≥2.
Solution: Let bi = 2 −ai, and let S = P bi and T = P b2 i . Then 1999 National Contests: Problems and Solutions 211 the given conditions are that (2 −b1) + · · · + (2 −bn) ≥n and (4 −4b1 + b2 1) + · · · + (4 −4bn + b2 n) ≥n2, which is to say S ≤n and T ≥n2 −4n + 4S.
From these inequalities, we obtain T ≥n2 −4n + 4S ≥(n −4)S + 4S = nS.
On the other hand, if bi > 0 for i = 1, . . . , n, then certainly bi < P bi = S ≤n, and so T = b2 1 + · · · + b2 n < nb1 + · · · + nbn = nS.
Thus we cannot have bi > 0 for i = 1, . . . , n, so bi ≤0 for some i; then ai ≥2 for that i, proving the claim.
Note: The statement is false when n ≤3. The example a1 = a2 = · · · = an−1 = 2, an = 2−n shows that the bound cannot be improved.
Also, an alternate approach is to show that if ai ≤2 and P ai ≥n, then P a2 i ≤n2 (with the equality case just mentioned), by noticing that replacing a pair ai, aj with 2, ai + aj −2 increases the sum of squares.
Problem 5 The Y2K Game is played on a 1 × 2000 grid as follows.
Two players in turn write either an S or an O in an empty square.
The first player who produces three consecutive boxes that spell SOS wins. If all boxes are filled without producing SOS then the game is a draw. Prove that the second player has a winning strategy.
Solution: Call a partially filled board stable if there is no SOS and no single move can produce an SOS; otherwise call it unstable. For a stable board call an empty square bad if either an S or an O played in that square produces an unstable board. Thus a player will lose if the only empty squares available to him are bad, but otherwise he can at least be guaranteed another turn with a correct play.
Claim: A square is bad if and only if it is in a block of 4 consecutive squares of the form S – – S.
212 United States of America Proof: If a square is bad, then an O played there must give an unstable board. Thus the bad square must have an S on one side and an empty square on the other side. An S played there must also give an unstable board, so there must be another S on the other side of the empty square.
From the claim it follows that there are always an even number of bad squares. Thus the second player has the following winning strategy: (a) If the board is unstable at any time, play the winning move; otherwise continue as below.
(b) On the first move, play an S at least four squares away from either end and at least seven squares from the first player’s first move.
(The board is long enough that this is possible.) (c) On the second move, play an S three squares away from the second player’s first move, so that the squares in between are empty and so that the board remains stable. (Regardless of the first player’s second move, this must be possible on at least one side.) This produces two bad squares; whoever plays in one of them first will lose. Thus the game will not be a draw.
(d) On any subsequent move, play in a square which is not bad— keeping the board stable, of course. Such a square will always exist because if the board is stable, there will be an odd number of empty squares and an even number of bad squares.
Since there exist bad squares after the second player’s second move, the game cannot end in a draw; and since the second player can always leave the board stable, the first player cannot win. Therefore eventually the second player will win.
Note: Some other names for the S – – S block, from submitted solutions, included arrangement, combo, configuration, formation, pattern, sandwich, segment, situation, and trap. (Thanks to Alexan-der Soifer and Zvezdelina Stankova-Frenkel for passing these along.) Problem 6 Let ABCD be an isosceles trapezoid with AB ∥CD.
The inscribed circle ω of triangle BCD meets CD at E. Let F be a point on the (internal) angle bisector of ∠DAC such that EF ⊥CD.
Let the circumscribed circle of triangle ACF meet line CD at C and 1999 National Contests: Problems and Solutions 213 G. Prove that the triangle AFG is isosceles.
Solution: We will show that FA = FG. Let H be the center of the escribed circle of triangle ACD opposite vertex A. Then H lies on the angle bisector AF. Let K be the point where this escribed circle touches CD. By a standard computation using equal tangents, we see that CK = 1 2(AD + CD −AC). By a similar computation in triangle BCD, we see that CE = 1 2(BC + CD −BD) = CK.
Therefore E = K and F = H.
Since F is now known to be an excenter, we have that FC is the external angle bisector of ∠DCA = ∠GCA. Therefore ∠GAF = ∠GCF = π 2 −1 2∠GCA = π 2 −1 2∠GFA.
We conclude that the triangle GAF is isosceles with FA = FG, as desired.
214 Vietnam 1.24 Vietnam Problem 1 Solve the system of equations (1 + 42x−y) · 51−2x+y = 1 + 22x−y+1 y3 + 4x + 1 + ln(y2 + 2x) = 0.
Solution: The only solution is (x, y) = (0, −1).
First solve the first equation for t = 2x−y. Multiplying the equation by 5t−1 yields (1 −5t−1) + 4(4t−1 −10t−1) = 0.
This has the obvious solution t = 1. There are no other solutions: if t > 1 then both 1 −5t−1 and 4t−1 −10t−1 are negative; and if t < 1 then both these terms are positive. Therefore, 2x −y = 1.
Substitute 2x = y + 1 into the second equation to get y3 + 2y + 3 + ln(y2 + y + 1) = 0.
This has the not-so-obvious solution y = −1. To prove this is the only solution, it suffices to show that f(y) = y3 + 2y + 3 + ln(y2 + y + 1) is always increasing. Its derivative is f ′(y) = 3y2 + 2 + 2y + 1 y2 + y + 1.
But we know that 2(y + 1)2 + 1 > 0 ⇒2y + 1 > −2(y2 + y + 1) ⇒ 2y + 1 y2 + y + 1 > −2, where we can safely divide by y2 + y + 1 = (y + 1 2)2 + 3 4 > 0. Thus f ′(y) > 3y2 > 0 for all y, as desired.
Problem 2 Let A′, B′, C′ be the respective midpoints of the arcs BC, CA, AB, not containing points A, B, C, respectively, of the circumcircle of the triangle ABC. The sides BC, CA, AB meet the pairs of segments {C′A′, A′B′}, {A′B′, B′C′}, {B′C′, C′A′} 1999 National Contests: Problems and Solutions 215 at the pairs of points {M, N}, {P, Q}, {R, S}, respectively. Prove that MN = PQ = RS if and only if the triangle ABC is equilateral.
Solution: If ABC is equilateral then MN = PQ = RS by symmetry.
Now suppose that MN = PQ = RS. Observe that ∠NMA′ = ∠BMS = 1 2( d BC′ + d CA′) = 1 2(∠C + ∠A) and similarly ∠C′SR = ∠MSB = 1 2(∠A + ∠C).
Furthermore, ∠A′B′C′ = ∠A′B′B + ∠BB′C′ = 1 2(∠A + ∠C) as well.
Thus MB = SB, and also △C′RS ∼△C′A′B′ ∼△NA′M. Next, by the law of sines in triangles C′SB and A′MB we have C′S = SB · sin ∠C′BS sin ∠SC′B = SB · sin ∠C 2 sin ∠A 2 and MA′ = MB · sin ∠A′BM sin ∠MA′B = MB · sin ∠A 2 sin ∠C 2 , giving C′S MA′ = sin ∠C 2 sin ∠A 2 2 .
Next, because △C′RS ∼△C′A′B′ we have RS = A′B′ · C′S C′B′ ; and because △NA′M ∼△C′A′B′ we have MN = B′C′ · MA′ B′A′ . Therefore since RS = MN we have A′B′ · C′S C′B′ = B′C′ · MA′ B′A′ = ⇒C′S MA′ = B′C′ A′B′ 2 = sin 1 2(∠B + ∠C) sin 1 2(∠B + ∠A) !2 = cos ∠A 2 cos ∠C 2 !2 = ⇒ sin ∠C 2 sin ∠A 2 !2 = cos ∠A 2 cos ∠C 2 !2 = ⇒ sin ∠C 2 cos ∠C 2 2 = sin ∠A 2 cos ∠A 2 2 = ⇒1 4 sin2 ∠C = 1 4 sin2 ∠A = ⇒sin ∠C = sin ∠A.
216 Vietnam Since ∠A+∠C < 180◦, we must have ∠A = ∠C. Similarly ∠A = ∠B, and therefore triangle ABC is equilateral.
Problem 3 For n = 0, 1, 2, . . ., let {xn} and {yn} be two sequences defined recursively as follows: x0 = 1, x1 = 4, xn+2 = 3xn+1 −xn; y0 = 1, y1 = 2, yn+2 = 3yn+1 −yn.
(a) Prove that x2 n −5y2 n + 4 = 0 for all non-negative integers n.
(b) Suppose that a, b are two positive integers such that a2−5b2+4 = 0.
Prove that there exists a non-negative integer k such that xk = a and yk = b.
Solution: We first prove by induction on k that for k ≥0, we have (xk+1, yk+1) = ( 3xk+5yk 2 , xk+3yk 2 ). For k = 0 we have (4, 2) = ( 3+5 2 , 1+3 2 ), and for k = 1 we have (11, 5) = ( 12+10 2 , 4+6 2 ).
Now assuming it’s true for k and k + 1, we know that (xk+3, yk+3) = (3xk+2 −xk+1, 3yk+2 −yk+1).
Substituting the expressions for xk+2, xk+1, yk+2, yk+1 from the in-duction hypothesis, this equals 3(3xk+1 −xk) + 5(3yk+1 −yk) 2 , (3xk+1 −xk) + 3(3yk+1 −yk) 2 = 3xk+2 + 5yk+2 2 , xk+2 + 3yk+2 2 , completing the induction and the proof of our claim.
(a) We prove the claim by induction; for n = 0 we have 1−5+4 = 0.
Now assuming it is true for n, we prove (with the help of our above observation) that it is true for n + 1: x2 n+1 −5y2 n+1 = 3xn + 5yn 2 2 −5 xn + 3yn 2 2 = 9x2 n + 30xnyn + 25y2 n 4 −5 · x2 n + 6xnyn + 9y2 n 4 = 4x2 n −20y2 n 4 = x2 n −5y2 n = −4, 1999 National Contests: Problems and Solutions 217 as desired.
(b) Suppose by way of contradiction that a2 −5b2 +4 = 0 for integers a, b > 0, and that there did not exist k such that (xk, yk) = (a, b).
Choose a counterexample that minimizes a + b.
Note that 0 ≡a2−5b2+4 ≡a−b (mod 2). Next, a2 = 5b2−4 < 9b2 ⇒a < 3b. And there are no counterexamples with a = 1 or 2; thus a2 > 5 and 0 = 5a2−25b2+20 < 5a2−25b2+4a2 ⇒3a > 5b.
Therefore a′ = 3a−5b 2 and b′ = 3b−a 2 are positive integers. Then since a2−5b2 = −4, some quick algebra shows that a′2−5b′2 = −4 as well; but a′ + b′ = 3a−5b 2 + 3b−a 2 = a −b < a + b. It follows from the minimal definition of (a, b) that there must exist some (ak, bk) equal to (a′, b′).
But then it is easy to verify that (a, b) = ( 3a′+5b′ 2 , a′+3b′ 2 ) = (ak+1, bk+1), a contradiction! This completes the proof.
Problem 4 Let a, b, c be real numbers such that abc + a + c = b.
Determine the greatest possible value of the expression P = 2 a2 + 1 − 2 b2 + 1 + 3 c2 + 1.
Solution: The condition is equivalent to b = a+c 1−ac, which suggests making the substitutions A = Tan−1a and C = Tan−1c; then we have b = tan(A + C) and P = 2 tan2 A + 1 − 2 tan2(A + C) + 1 + 3 tan2 C + 1 = 2 cos2 A −2 cos2(A + C) + 3 cos2 C = (2 cos2 A −1) −(2 cos2(A + C) −1) + 3 cos2 C = cos(2A) −cos(2A + 2C) + 3 cos2 C = 2 sin(2A + C) sin C + 3 cos2 C.
Letting u = | sin C|, this expression is at most 2u + 3(1 −u2) = −3u2 + 2u + 3 = −3 u −1 3 2 + 10 3 ≤10 3 .
218 Vietnam Equality can be achieved when sin(2A+C) = 1 and sin C = 1 3, which gives (a, b, c) = ( √ 2 2 , √ 2, √ 2 4 ). Thus the maximum value of P is 10 3 .
Problem 5 In the three-dimensional space let Ox, Oy, Oz, Ot be four nonplanar distinct rays such that the angles between any two of them have the same measure.
(a) Determine this common measure.
(b) Let Or be another ray different from the above four rays. let α, β, γ, δ be the angles formed by Or with Ox, Oy, Oz, Ot, re-spectively. Put p = cos α + cos β + cos γ + cos δ, q = cos2 α + cos2 β + cos2 γ + cos2 δ.
Prove that p and q remain constant as Or rotates about the point O.
Solution: Put O at the origin, and say the four rays hit the unit sphere at A, B, C, D. Then ABCD is a regular tetrahedron, and (letting X also represent the vector − − → OX) we have A+B +C +D = 0.
(a) Say the common angle is φ. Then 0 = A · (A + B + C + D) = A · A + A · (B + C + D) = 1 + 3 cos φ, so φ = cos−1 −1 3 .
(b) Without loss of generality say that Or hits the unit sphere at U = (1, 0, 0); also write A = (x1, y1, z1), and so on. Then p = U · A + U · B + U · C + U · D = U · (A + B + C + D) = U ·⃗ 0 = 0, a constant. Also, (x1, x2, x3, x4) = (cos α, cos β, cos γ, cos δ) and q = P x2 i . The following lemma then implies that q will always equal 4 3: Lemma. Suppose we are given a regular tetrahedron T inscribed in the unit sphere and with vertices (xi, yi, zi) for 1 ≤i ≤4.
Then we have P x2 i = P y2 i = P z2 i = 4 3 and P xiyi = P yizi = P zixi = 0.
1999 National Contests: Problems and Solutions 219 Proof: This is easily verified when the vertices are at A0 = (0, 0, 1), B0 = ( 2 √ 2 3 , 0, −1 3), C0 = (− √ 2 3 , √ 6 3 , −1 3), D0 = (− √ 2 3 , − √ 6 3 , −1 3).
Now assume these equations are true for a tetrahedron ABCD, and rotate it about the z-axis through an angle θ. Then each (xi, yi, zi) becomes (x′ i, y′ i, z′ i) = (xi cos θ −yi sin θ, xi sin θ + yi cos θ, zi), and X x′2 i = cos2 θ X x2 i −2 sin θ cos θ X xiyi + sin2 θ X y2 i = 4 3 X y′2 i = sin2 θ X x2 i + 2 sin θ cos θ X xiyi + cos2 θ X y2 i = 4 3 X z′2 i = X z2 i = 4 3 X x′ iy′ i = sin θ cos θ X (x2 i −y2 i ) + (cos2 θ −sin2 θ) X xiyi = 0 X y′ iz′ i = sin θ X xizi + cos θ X yizi = 0 X z′ ix′ i = cos θ X zixi −sin θ X ziyi = 0.
Similarly, the equations remain true after rotating ABCD about the y- and z-axes.
Now, first rotate our given tetrahedron T about the z-axis until one vertex is in the yz-plane; next rotate it about the x-axis until this vertex is at (0, 0, 1); and finally rotate it about the z-axis again until the tetrahedron corresponds with the initial tetra-hedron A0B0C0D0 described above. Since we know the above equations are true for A0B0C0D0, if we reverse the rotations to return to T the equations will remain true, as claimed.
Problem 6 Let S = {0, 1, 2, . . . , 1999} and T = {0, 1, 2, . . .}. Find all functions f : T →S such that (i) f(s) = s for all s ∈S.
(ii) f(m + n) = f(f(m) + f(n)) for all m, n ∈T .
Solution: Suppose that f(2000) = 2000−t, where 1 ≤t ≤2000. We prove by induction on n that for all n ≥2000, we have f(n) = f(n−t).
By assumption it is true for n = 2000. Then assuming it is true for 220 Vietnam n, we have f(n + 1) = f(f(n) + f(1)) = f(f(n −t) + f(1)) = f(n −t + 1), completing the inductive step. Therefore the function is completely determined by the value of f(2000), and it follows that there are at most 2000 such functions.
Conversely, given any c = 2000 −t ∈S, let f be the function such that f(s) = s for all s ∈S while f(n) = f(n−t) for all n ≥2000. We prove by induction on m+n that condition (ii) holds. If m+n ≤2000 then m, n ∈S and the claim is obvious. Otherwise, m + n > 2000.
Again, if m, n ∈S the claim is obvious; otherwise assume without loss of generality that n ≥2000. Then f(m + n) = f(m + n −t) = f(f(m) + f(n −t)) = f(f(m) + f(n)), where the first and third equalities come from our periodic definition of f, and the second equality comes from the induction hypothesis.
Therefore there are exactly 2000 functions f.
Problem 7 For n = 1, 2, . . ., let {un} be a sequence defined by u1 = 1, u2 = 2, un+2 = 3un+1 −un.
Prove that un+2 + un ≥2 + u2 n+1 un for all n.
Solution: We first prove by induction that for n ≥1, we have unun+2 = u2 n+1 + 1. Since u3 = 5, for n = 1 we have 1 · 5 = 22 + 1, as desired.
Now assuming our claim is true for n, we have u2 n+2 + 1 = un+2(3un+1 −un) + 1 = 3un+1un+2 −(unun+2 −1) = 3un+1un+2 −u2 n+1 = un+1(3un+2 −un+1) = un+1un+3, so it is true for n + 1 as well and thus all n ≥1.
1999 National Contests: Problems and Solutions 221 Therefore, for all n ≥1 we have un+2 + un = u2 n+1 + 1 un + un = u2 n+1 un + un + 1 un ≥u2 n+1 un + 2, where un + 1 un ≥2 by AM-GM. This proves the inequality.
Problem 8 Let ABC be a triangle inscribed in circle ω. Construct all points P in the plane (ABC) and not lying on ω, with the property that the lines PA, PB, PC meet ω again at points A′, B′, C′ such that A′B′ ⊥A′C′ and A′B′ = A′C′.
Solution: All angles are directed modulo 180◦. We solve a more general problem: suppose we have some fixed triangle DEF and we want to find all points P such that when A′ = PA ∩ω, B′ = PB ∩ω, C′ = PC ∩ω then triangles A′B′C′ and DEF are similar with the same orientations. (In other words, we want ∠B′C′A′ = ∠EFD and ∠C′A′B′ = ∠FDE.) Given X, Y on ω, let ∠d XY denote the angle ∠XZY for any other point Z on ω.
Now given a point P, we have ∠BPA = ∠d BA + ∠[ B′A′ = ∠BCA+∠B′C′A′ and ∠CPB = ∠d CB+∠[ C′B′ = ∠CAB+ ∠C′A′B′. Thus ∠B′C′A′ = ∠EFD if and only if ∠BPA = ∠BCA+ ∠EFD, while ∠C′A′B′ = ∠FDE if and only if ∠CPB = ∠CAB + ∠FDE.
Therefore our desired point P is the intersection point, different than B, of the two circles {P ′ | ∠BP ′A = ∠BCA+∠EFD} and {P ′ | ∠CP ′B = ∠CAB + ∠FDE}.
Now back to our original problem: we wish to find P such that triangle A′B′C′ is a 45◦-45◦-90◦triangle with ∠C′A′B′ = 90◦.
Because our angles are directed, there are two possible orientations for such a triangle: either ∠A′B′C′ = 45◦or ∠A′B′C′ = −45◦.
Applying the above construction twice with triangle DEF defined appropriately yields the two desired possible locations of P.
Problem 9 Consider real numbers a, b such that a ̸= 0, a ̸= b and all roots of the equation ax3 −x2 + bx −1 = 0 are real and positive. Determine the smallest possible value of the expression P = 5a2 −3ab + 2 a2(b −a) .
222 Vietnam Solution: When the roots of the equation are all √ 3, we have a = 1 3 √ 3, b = √ 3, and P = 12 √ 3. We prove that 12 √ 3 is minimal.
Say the roots to ax3 −x2 + bx −1 are p = tan A, q = tan B, and r = tan C with 0 < A, B, C < 90◦. Then ax3 −x2 + bx −1 = a(x −p)(x −q)(x −r) = ax3 −a(p + q + r)x2 + a(pq + qr + rp)x −a(pqr).
Thus a = 1 p+q+r = 1 pqr > 0 and p + q + r = pqr. Then r = p + q pq −1 = −tan(A + B) = tan(180◦−A −B), so A + B + C = 180◦. Then since tan x is convex for 0 < x < 90◦, we have 1 a = tan A + tan B + tan C ≥3 tan 60◦= 3 √ 3, so a ≤ 1 3 √ 3.
Also notice that b a = pq + qr + rp ≥3 3 p p2q2r2 = 3 3 r 1 a2 ≥9 > 1, so b > a. Thus the denominator of P is always positive and is an increasing function of b, while the numerator of P is a decreasing function of b. Therefore, for a constant a, P is a decreasing function of b.
Furthermore, (p −q)2 + (q −r)2 + (r −p)2 ≥0 = ⇒(p + q + r)2 ≥3(pq + qr + rp) = ⇒1 a2 ≥3b a = ⇒b ≤1 3a, and P ≥5a2 −3a( 1 3a) + 2 a2( 1 3a −a) = 5a2 + 1 a 3 −a3 .
1999 National Contests: Problems and Solutions 223 Thus for 0 < a ≤ 1 3 √ 3, it suffices to show that 5a2 + 1 ≥12 √ 3 a 3 −a3 = 4 √ 3a −12 √ 3a3 ⇐ ⇒12 √ 3a3 + 5a2 −4 √ 3a + 1 ≥0 ⇐ ⇒3 a − 1 3 √ 3 (4 √ 3a2 + 3a − √ 3) ≥0 ⇐ ⇒4 √ 3a2 + 3a − √ 3 ≤0.
But the last quadratic has one positive root −3 + √ 57 8 √ 3 ≥−3 + 7 8 √ 3 = 1 2 √ 3 > 1 3 √ 3 , so it is indeed negative when 0 < a ≤ 1 3 √ 3. This completes the proof.
Problem 10 Let f(x) be a continuous function defined on [0, 1] such that (i) f(0) = f(1) = 0; (ii) 2f(x) + f(y) = 3f 2x + y 3 for all x, y ∈[0, 1].
Prove that f(x) = 0 for all x ∈[0, 1].
Solution: We prove by induction on k that f m 3k = 0 for all integers k ≥0 and all integers 0 ≤m ≤3k. The given conditions show this claim is true for k = 0; now assuming it is true for k −1, we prove it is true for k.
If m ≡0 (mod 3) then f m 3k = f m 3 3k−1 = 0 by the induction hypothesis.
If m ≡1 (mod 3), then 1 ≤m ≤3k −2 and 3f m 3k = 2f m−1 3 3k−1 !
+ f m+2 3 3k−1 !
= 0 + 0 = 0.
Thus f m 3k = 0.
And if m ≡2 (mod 3), then 2 ≤m ≤3k −1 and 3f m 3k = 2f m+1 3 3k−1 !
+ f m−2 3 3k−1 !
= 0 + 0 = 0.
Hence f m 3k = 0, finishing our induction.
224 Vietnam Now, for any x ∈[0, 1] we can form a sequence of numbers of the form m 3k whose limit is x; then since f is continuous, it follows that f(x) = 0 for all x ∈[0, 1], as desired.
Problem 11 The base side and the altitude of a right regular hexagonal prism ABCDEF −A′B′C′D′E′F ′ are equal to a and h respectively.
(a) Prove that six planes (AB′F ′), (CD′B′), (EF ′D′), (D′EC), (F ′AE), (B′CA) touch the same sphere.
(b) Determine the center and the radius of the sphere.
Solution: (a) Let O be the center of the prism. (AB′F ′) is tangent to a unique sphere centered at O.
Now the other five planes are simply reflections and rotations of (AB′F ′) with respect to O; and since the sphere remains fixed under these transformations, it follows that all six planes are tangent to this same sphere.
(b) From part (a), the center of the sphere is the center O of the prism. Let P be the midpoint of AE and let P ′ be the midpoint of A′E′. Also let Q be the midpoint of PF ′, and let OR be the perpendicular from O to line PF ′. Note that P, P ′, Q, R, O, F ′ all lie in one plane.
It is straightforward to calculate that F ′P ′ = a 2 and QO = 3a 4 .
Also, since QO ∥F ′P ′ we have ∠RQO = ∠PF ′P ′; combined with ∠ORQ = ∠PP ′F ′ = 90◦, this gives △ORQ ∼△PP ′F ′.
Hence the radius of the sphere is OR = PP ′ · OQ PF ′ = h · 3a 4 q a 2 2 + h2 = 3ah 2 √ a2 + 4h2 .
Problem 12 For n = 1, 2, . . ., two sequences {xn} and {yn} are defined recursively by x1 = 1, y1 = 2, xn+1 = 22yn −15xn, yn+1 = 17yn −12xn.
(a) Prove that xn and yn are not equal to zero for all n = 1, 2, . . . .
1999 National Contests: Problems and Solutions 225 (b) Prove that each sequence contains infinitely many positive terms and infinitely many negative terms.
(c) For n = 19991945, determine whether xn and yn are divisible by 7 or not.
Solution: (a) The recursive equation for xn+1 gives yn = 1 22 (15xn + xn+1) and thus also yn+1 = 1 22 (15xn+1 + xn+2).
Substituting these expressions into the other recursive equation gives xn+2 = 2xn+1 −9xn and similarly we can also find yn+2 = 2yn+1 −9yn.
Quick computation gives x2 = 29 and y2 = 22. Then x1, x2 are odd, and if xn, xn+1 are odd then xn+2 must be as well; thus all the xn are odd and hence nozero. Similarly, all the yn are congruent to 2 (mod 4) and thus nonzero as well.
(b) Note that xn+3 = 2(2xn+1 −9xn) −9xn+1 = −5xn+1 −18xn.
Thus if xn, xn+1 are positive (or negative) then xn+3 is negative (or positive). Hence among every four consecutive terms xn there is some positive number and some negative number. Therefore there are infinitely many positive terms and infinitely many negative terms in this sequence; and a similar proof holds for the yn.
(c) All congruences are taken modulo 7 unless stated otherwise. Note that x1 ≡x2 ≡1.
Now if xn ≡xn+1 and xn ̸≡0, then we have (xn+2, xn+3, xn+4, xn+5) ≡(0, 5xn, 3xn, 3xn) and 5xn ̸≡0, 3xn ̸≡0. This implies that x3, x7, x11, . . . are all divisible by 7 but no other xn are. Since 19991945 ≡31945 ≡3·91944/2 ≡3 (mod 4), we indeed have 7 | xn when n = 19991945.
Now suppose by way of contradiction that 7 | yn′ for some n′, and choose the minimal such n′. From the recursion for yn, we have yn ≡yn+1 + 3yn+2. Now if n′ ≥5 then yn′−2 ≡yn′−1, yn′−3 ≡4yn′−1, and yn′−4 ≡0 — contradicting the minimal choice of n′. Thus we have n′ ≤4, but we have (y1, y2, y3, y4) ≡ (2, 1, 5, 1). Therefore no term is divisible by 7; and specifically, 7̸ | yn when n = 19991945.
1 1999 Regional Contests: Problems and Solutions 1 2 Asian Pacific Mathematical Olympiad 1.1 Asian Pacific Mathematical Olympiad Problem 1 Find the smallest positive integer n with the following property: There does not exist an arithmetic progression of 1999 real numbers containing exactly n integers.
Solution: Look at the 1999-term arithmetic progression with common difference 1 q and beginning term p q , where p and q are integers with 1 < q < 1999; without loss of generality assume that 1 ≤p ≤q.
When p = 1, the progression contains the integers 1, 2, . . . , j 1999 q k ; when p = q, the progression contains the integers 1, 2, . . . , 1 + j 1998 q k . Since 1999 is prime, q does not divide 1999 and hence j 1999 q k = j 1998 q k . Thus the progression contains either j 1999 q k or j 1999 q k + 1 integers, and any k of this form can be attained. Call such numbers good.
Conversely, suppose we have an arithmetic progression containing exactly k integers, where 1 < k < 1999; without loss of generality, say its common difference is positive and say it contains 0 as its t-th term.
Its common difference cannot be irrational, so it is of the form p q for some positive, relatively prime integers p, q. And since 1 < k < 1999, 1 < q < 1999.
But then look at the arithmetic progression with common difference 1 q and 0 as its t-th term. It, too, contains exactly k integers; so from our previous argument, k is good.
Now, for q ≥32 we have 1999 < 2q(q + 1) = ⇒1999 q −1999 q+1 < 2.
Since 1999 32 = 62 and 1999 1998 = 1, this implies that every integer k between 1 and 63 is good. Also, 1999 31 = 64, 1999 30 = 66, 1999 29 = 68, 1999 q ≤62 when q ≥32, 1999 q ≥71 when q ≤28.
Thus 70 is the smallest integer that k cannot equal, and n = 70.
Problem 2 Let a1, a2, . . . be a sequence of real numbers satisfying ai+j ≤ai + aj 1999 Regional Contests: Problems and Solutions 3 for all i, j = 1, 2, . . . . Prove that a1 + a2 2 + a3 3 + · · · + an n ≥an for all positive integers n.
Solution: Lemma. If m, n are positive integers with m ≥n, then a1 + a2 + · · · + an ≥n(n+1) 2m · am.
Proof: The result for m = n comes from adding the inequalities a1 + an−1 ≥an, a2 + an−2 ≥an, . . . , an−1 + a1 ≥an, 2an ≥ 2an, and then dividing by two. Now for positive integers j, write βj = a1+a2+···+aj 1+2+···+j . Then the inequality for m = n = j = k + 1 is equivalent to both βj ≥aj j and βk ≥βk+1; so when m ≥n we have βn ≥βn+1 ≥· · · ≥βm ≥am m , as desired.
The desired inequality now follows from expressing a1+ a2 2 +· · ·+ an n as an Abel sum and then applying the lemma multiple times: a1 + a2 2 + · · · + an n = 1 n(a1 + a2 + · · · + an) + n−1 X j=1 1 j − 1 j + 1 (a1 + a2 + · · · + aj) ≥1 n n(n + 1) 2n an + n−1 X j=1 1 j(j + 1) · j(j + 1) 2n an = an, as desired.
Problem 3 Let ω1 and ω2 be two circles intersecting at P and Q.
The common tangent, closer to P, of ω1 and ω2, touches ω1 at A and ω2 at B.
The tangent to ω1 at P meets ω2 again at C, and the extension of AP meets BC at R. Prove that the circumcircle of triangle PQR is tangent to BP and BR.
Solution: We shall use directed angles. Using tangents and cyclic quadrilaterals, we have ∠QAR = ∠QAP = ∠QPC = ∠QBC = ∠QBR, so QABR is cyclic.
Since ∠BPR is an exterior angle to triangle ABP, we have ∠BPR = ∠BAP + ∠PBA. Then again using tangents and cyclic 4 Asian Pacific Mathematical Olympiad quadrilaterals, we have ∠BAP + ∠PBA = ∠BAR + ∠PCB = ∠BQR + ∠PQB = ∠PQR. Thus ∠BPR = ∠PQR, which implies that line BP is tangent to the circumcircle of triangle PQR.
Next, ∠PRB is an exterior angle to triangle CRP so ∠PRB = ∠PCR + ∠RPC. We know that ∠PCR = ∠PCB = ∠PQB; and letting T be the intersection of lines CP and AB, we have ∠RPC = ∠APT = ∠AQP = ∠BAP = ∠BAR = ∠BQR. Therefore ∠PRB = ∠PQB + ∠BQR = ∠PQR, which implies that line BR is tangent to the circumcircle of triangle PQR as well.
Problem 4 Determine all pairs (a, b) of integers for which the numbers a2 + 4b and b2 + 4a are both perfect squares.
Solution: If a = 0 then b must be a perfect square, and vice versa.
Now assume both a and b are nonzero. Also observe that a2 + 4b and a2 have the same parity, and similarly b2 + 4a and b2 have the same parity.
If b is positive then a2+4b ≥(|a|+2)2 = a2+4|a|+4 so |b| ≥|a|+1.
If b is negative then a2 +4b ≤(|a|−2)2 = a2 −4|a|+4 so |b| ≥|a|−1.
Similarly, a > 0 = ⇒|a| ≥|b| + 1 and a < 0 = ⇒|a| ≥|b| −1.
Assume without loss of generality that b > a. If a and b are positive, then from the last paragraph we have b ≥a + 1 and a ≥b + 1, a contradiction.
If a and b are negative, then we have either a = b or a = b −1.
For b ≥−5, only (a, b) = (−4, −4) and (−6, −5) work. Otherwise, we have (b + 4)2 < b2 + 4a < (b + 2)2, a contradiction.
Finally, if a is negative and b is positive, then we have both |b| ≥ |a| + 1 and |a| ≥|b| −1. Then we must have |b| = |a| + 1 and hence a + b = 1. Any such pair works, since then a2 + 4b = (a −2)2 and b2 + 4a = (b −2)2 are both perfect squares.
Therefore the possible pairs (a, b) are (−4, −4), (−6, −5), (−5, −6); and (0, n2), (n2, 0), and (n, 1 −n) where n is any integer.
Problem 5 Let S be a set of 2n+1 points in the plane such that no three are collinear and no four concyclic. A circle will be called good if it has 3 points of S on its circumference, n−1 points in its interior, and n −1 in its exterior. Prove that the number of good circles has the same parity as n.
Solution: 1999 Regional Contests: Problems and Solutions 5 Lemma. Suppose we have 2n ≥1 given points in the plane with no three collinear, and one distinguished point A among them. Call a line “good” if it passes through A and one other given point, and if it separates the remaining 2n −2 points: that is, half of them lie on one side of the line, and the other half lie on the other. Then there are an odd number of good lines.
Proof: We prove the claim by induction. It is trivial for n = 1; now assuming it is true for n −1, we prove it is true for n.
Without loss of generality, arrange the 2n−1 points different from A on a circle centered at A. From those 2n−1 points, choose two points, B and C, that are the greatest distance apart. Then if P ̸= A, B, C is a given point, B and C lie on different sides of line AP. Thus line AP is good if and only if it separates the other 2n −3 points; and by the induction hypothesis, there are an odd number of such lines. Finally, line AB is good if and only if line AC is good — adding either 0 or 2 good lines to our count, so that our total count remains odd. This completes the inductive step.
Suppose we have a pair of points {A, B} in S. Perform an inversion about A with arbitrary radius.
B and the other 2n −1 points C1, C2, . . . , C2n−1 map to 2n distinct points B′, C′ 1, C′ 2, . . . , C′ 2n−1 (no three collinear); and the circle passing through A, B, Ck is good if and only if the corresponding line B′C′ k separates the other C′ i. By the lemma, there are an odd number of such lines; so an odd number of good circles pass through any pair of given points.
For each pair of points count the number of good circles passing through these points; each good circle is counted three times in this manner, so if there are k good circles then our count will be 3k. And there are 2n+1 2 = n(2n+1) pairs of points, each contributing an odd amount to our count. Therefore 3k ≡n(2n + 1) = ⇒k ≡n (mod 2), as desired.
6 Austrian-Polish Mathematics Competition 1.2 Austrian-Polish Mathematics Competition Problem 1 Let n be a positive integer and M = {1, 2, . . . , n}. Find the number of ordered 6-tuples (A1, A2, A3, A4, A5, A6) which satisfy the following two conditions: (a) A1, A2, A3, A4, A5, A6 (not necessarily distinct) are subsets of M; (b) each element of M belongs to either 0, 3, or 6 of the sets A1, A2, A3, A4, A5, A6.
Solution: Given k ∈M, there are 6 0 ways to put k into exactly 0 of the 6 sets; 6 3 ways to put k into exactly 3 of the sets; and 6 6 ways to put k into all 6 sets. This adds up to 1 + 20 + 1 = 22 ways to put k into the sets. Every admissible 6-tuple is uniquely determined by where each k is placed; therefore, there are 22n possible distributions.
Problem 2 Find the largest real number C1 and the smallest real number C2 such that for all positive real numbers a, b, c, d, e the following inequalities hold: C1 < a a + b + b b + c + c c + d + d d + e + e e + a < C2.
Solution: Let f(a, b, c, d, e) = a a+b + · · · + e e+a.
Note that f(e, d, c, b, a) = 5 −f(a, b, c, d, e). Hence f(a, b, c, d, e) can attain the value x if and only if it can attain the value 5 −x. Thus if we find C1, then C2 = 5 −C1.
If a = ϵ4, b = ϵ3, c = ϵ2, d = ϵ, e = 1, then f(a, b, c, d, e) = 4ϵ 1 + ϵ + 1 1 + ϵ4 , which for small ϵ can become arbitrarily close to 1. We now prove that f(a, b, c, d, e) is always bigger than 1. Since f is homogenous, we may assume without loss of generality that a + b + c + d + e = 1.
Now g(x) = 1 x is convex for all positive x; so by Jensen’s inequality, ag(x1) + bg(x2) + · · · + eg(x5) ≥g(ax1 + bx2 + · · · + ex5). Applying this inequality with x1 = a + b, x2 = b + c, . . ., x5 = e + a, we find 1999 Regional Contests: Problems and Solutions 7 that f(a, b, c, d, e) ≥ 1 P cyc a(a + b) = (a + b + c + d + e)2 P cyc a(a + b) > P cyc a(a + 2b) P cyc a(a + b) > 1, as desired. (Here P cyc h(a, b, c, d, e) means h(a, b, c, d, e)+h(b, c, d, e, a)+ . . . + h(e, a, b, c, d).) Hence C1 = 1; and from our initial arguments, C2 = 4.
Problem 3 Let n ≥2 be a given integer. Determine all systems of n functions (f1, . . . , fn) where fi : R →R, i = 1, 2, . . . , n, such that for all x, y ∈R the following equalities hold: f1(x) −f2(x)f2(y) + f1(y) = 0 f2(x2) −f3(x)f3(y) + f2(y2) = 0 .
.
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fn−1(xn−1) −fn(x)fn(y) + fn−1(yn−1) = 0 fn(xn) −f1(x)f1(y) + fn(yn) = 0.
Solution: Setting x = y into the k-th equation gives 2fk(xk) = fk+1(x)2 (where we write fn+1 = f1).
Thus if fk(x) = 0 for all x ∈R, then fk+1(x) is also always zero; and similarly, all the fi’s are identically zero. Now assume that no fk is identically zero.
For odd k, given any real value r let x = k √r.
Then 2fk(r) = 2fk(xk) = fk+1(x)2 ≥0 for all r. Similarly, for even k, fk(r) ≥0 whenever r ≥0. Also observe that for even k, we cannot have some fk(x) < 0 and some fk(y) > 0 because then we’d have fk−1(xk−1) − fk(x)fk(y) + fk−1(yk−1) > 0, contradicting the (k −1)-th equation.
And since fk+1(x) ̸= 0 for some x, we have fk(xk) = 1 2fk+1(x)2 > 0.
Therefore, we must have fk(x) ≥0 for all x and for all k.
We now prove by induction on k that fk is a constant function. For k = 1, plugging in f2(x) = p 2f1(x) and f2(y) = p 2f1(y) into the first equation gives f1(x) −2 p f1(x)f1(y) + f1(y) = 0 for all x, y ∈R.
By the arithmetic mean-geometric mean inequality, this is true only when f1(x) = f1(y) for all x, y ∈R.
8 Austrian-Polish Mathematics Competition Now assume that fk(x) = fk(y) for all x, y.
Then fk+1(x) = p 2fk(xk) = p 2fk(yk) = fk+1(y) for any x, y ∈R, completing the inductive step.
Writing fk(x) = ck (where we write cn+1 = c1), observe that ck = 1 2c2 k+1 for all k; since fk(x) ≥0 for all x but fk is not identically zero, each ck is positive. If 0 < ck+1 < 2, then 0 < ck < ck+1 < 2.
Thus cn > cn−1 > · · · > c1 > cn, a contradiction.
Similarly, if ck+1 > 2 then ck > ck+1 > 2; so that cn < cn−1 < · · · < c1 < cn, a contradiction. Hence, we must have ck = 2 for all k.
Therefore, the only possible functions are fk(x) = 0 for all x, k; and fk(x) = 2 for all x, k.
Problem 4 Three straight lines k, l, and m are drawn through some fixed point P inside a triangle ABC such that: (a) k meets the lines AB and AC in A1 and A2 (A1 ̸= A2) respec-tively and PA1 = PA2; (b) l meets the lines BC and BA in B1 and B2 (B1 ̸= B2) respectively and PB1 = PB2; (c) m meets the lines CA and CB in C1 and C2 (C1 ̸= C2) respectively and PC1 = PC2.
Prove that the lines k, l, m are uniquely determined by the above conditions. Find the point P (and prove that there exists exactly one such point) for which the triangles AA1A2, BB1B2, and CC1C2 have the same area.
Solution: Let ℓbe the reflection of line AB about P; since A1 and A2 are also mirror images of each other across P, A2 must lie on ℓ. Thus A2 must be the intersection of ℓand line AC, and this intersection point is unique since lines AB and AC are not parallel.
Therefore k must be the line passing through P and this point, and conversely this line satisfies condition (a). Similarly, lines l and m are also uniquely determined.
Now suppose that triangles AA1A2 and BB1B2 have the same area.
Let Q be the midpoint of AA1; then since P is the midpoint of A1A2, we have [AQP] = 1 2[AA1P] = 1 4[AA1A2]. Similarly, let R be the midpoint of BB2; then [BRP] = 1 2[BB2P] = 1 4[BB1B2]. Therefore [AQP] = [BRP]; and since the heights from P in triangles AQP and BRP are equal, we must have AQ = BR.
1999 Regional Contests: Problems and Solutions 9 Now since P and Q are the midpoints of A1A2 and A1A, we have PQ ∥AA2 and hence PQ ∥AC. This implies that Q lies between A and B. Similarly, R lies between A and B as well.
Then since AQ = BR, Q and R are equidistant from the midpoint C′ of AB.
Therefore the homothety about C′ that maps A to Q also maps B to R. This homothety also maps line AC to QP since AC ∥QP; and it maps line BC to RP. Hence it maps C to P, so that C′, P, C are collinear and P lies on the median from C in triangle ABC.
Similarly, P must lie on the medians from A and B in triangle ABC, so it must be the centroid G of triangle ABC. And conversely, if P = G then k, l, m are parallel to lines BC, CA, AB respectively, and [AA1A2] = [BB1B2] = [CC1C2] = 4 9[ABC].
Problem 5 A sequence of integers {an}n≥1 satisfies the following recursive relation an+1 = a3 n + 1999 for n = 1, 2, . . . .
Prove that there exists at most one n for which an is a perfect square.
Solution: Consider the possible values of (an, an+1) modulo 4: an 0 1 2 3 an+1 3 0 3 2 No matter what a1 is, the terms a3, a4, . . . are all 2 or 3 (mod 4); but all perfect squares are 0 or 1 (mod 4), so at most two terms (a1 and a2) can be perfect squares. But if a1 and a2 are both perfect squares, then writing a1 = a2, a2 = b2 we have a6 + 1999 = b2 or 1999 = b2−(a3)2 = (b+a3)(b−a3). But since 1999 is prime, b−a3 = 1 and b + a3 = 1999. Thus a3 = 1999−1 2 = 999, which is impossible.
Hence at most one term of the sequence is a perfect square.
Problem 6 Find all real numbers x0, x1, x2, . . . , x1998 ≥0 which satisfy x2 i+1 + xi+1xi + x4 i = 1 for i = 0, 1, . . . , 1998, where x1999 = x0.
Solution: Let R be the positive real root of x4 +2x2 −1 = 0, found using the quadratic formula: R2 = −2 + √ 8 2 = −1 + √ 2 = ⇒ R = q −1 + √ 2.
10 Austrian-Polish Mathematics Competition If xn, xn+1 ≥R then 1 = x2 n+1 +xn+1xn +x4 n ≥R2 +R2 +R4 = 1, with equality when xn = xn+1 = R. Similarly, if xn, xn+1 ≤R then we must have xn = xn+1 = R. Hence either xn = xn+1 = R, xn < R < xn+1, or xn > R > xn+1.
Now if x0 < R then x1 > R, x2 < R, . . ., x0 = x1999 > R, a contradiction. Similarly, we cannot have x0 > R. Therefore x0 = R and the only solution is x0 = x1 = · · · = x1999 = R = q −1 + √ 2.
Problem 7 Find all pairs (x, y) of positive integers such that xx+y = yy−x.
Solution: Let gcd(x, y) = c, and write a = x c , b = y c . Then the equation becomes (ac)c(a+b) = (bc)c(b−a) (ac)a+b = (bc)b−a aa+bc2a = bb−a.
Thus aa+b | bb−a; then since gcd(a, b) = 1, a must equal 1. Therefore bb−1 = c2 is a perfect square. This is true exactly when b is odd, or when b is a perfect square. If b = 2n + 1 then c = (2n + 1)n; and if b = n2 then c = nn2−1. Therefore (x, y) equals either ((2n + 1)n, (2n + 1)n+1) or (nn2−1, nn2+1) for some positive integer n, and any such pair indeed satisfies the given equations.
Problem 8 Let ℓbe a given straight line and let the points P and Q lie on the same side of the line ℓ. The points M, N lie on the line ℓand satisfy PM ⊥ℓand QN ⊥ℓ. The point S lies between the lines PM and QN such that PM = PS and QN = QS. The perpendicular bisectors of SM and SN meet at R.
Let T be the second intersection of the line RS and the circumcircle of triangle PQR. Prove that RS = ST.
1999 Regional Contests: Problems and Solutions 11 Solution: All angles are directed modulo 180◦. Let T ′ be the reflection of R across S; we wish to prove that PRQT ′ is cyclic.
Since RM = RS = RN, ∠RMN = ∠MNR = x.
Note that PMRS is a kite which is symmetric with respect to line PR. Hence ∠PRM = ∠SRP = y, ∠MPR = ∠RPS = z, and ∠PSR = ∠RMP = 90◦+ x.
Similarly, ∠QRS = ∠NRQ = u, ∠SQR = ∠RQN = v, and ∠RSQ = ∠QNR = 90◦+ x. In triangle MNR, 2(x + y + u) = 180◦. In triangle PMR, y + z + 90◦+ x = 180◦so that 2(x + y + z) = 180◦. Hence 2u = 2z.
Orient our diagram so that line MN is horizontal and so that P and Q are above line MN. From the given information, S is above line MN and between lines PM and QN. Also, since since R is the circumcenter of triangle MSN, R lies below between lines PM and QN and below S. This information allows us to safely conclude that u = z, or ∠SPR = ∠SRQ.
Similarly, y = v and ∠PRS = ∠SQR. Thus △PSR ∼△RSQ.
Let A and B be the midpoints of PR and QR, respectively. Then SA and SB are corresponding medians in similar triangles PRS and RQS. Hence △ASR ∼△BSQ. It follows that ∠ASB = ∠ASR + ∠RSB = ∠BSQ + ∠RSB = ∠RSQ = 90◦+ x.
Thus ∠ASB + ∠ARB = ∠ASB + ∠PRQ = 90◦+ x + y + z = 180◦ and ASRB is cyclic. Since PRQT ′ is the image of ARBS under a homothety about R with ratio 2, it follows that PQRT ′ is cyclic as well.
Problem 9 Consider the following one player game. On the plane, a finite set of selected lattice points and segments is called a position in this game if the following hold: (i) the endpoints of each selected segment are lattice points; (ii) each selected segment is parallel to a coordinate axis, or to the line y = x, or to the line y = −x; (iii) each selected segment contains exactly 5 lattice points and all of them are selected; (iv) any two selected segments have at most one common point.
A move in this game consists of selecting a new lattice point and a new segment such that the new set of selected lattice points and 12 Austrian-Polish Mathematics Competition segments is a position. Prove or disprove that there exists an initial position such that the game has infinitely many moves.
Solution: There exists no position so that the game can last for infinitely many moves.
Given any segment, let its “extreme point” be its leftmost, upper-leftmost, highest, or upper-rightmost point (depending on whether the segment is parallel to y = 0, y = −x, x = 0, or y = x); let its other four lattice points form the segment’s “mini-segment.” Observe that no two mini-segments pointing in the same direction can intersect.
Also, given any position, call a lattice point a “missing point” if it is the extreme point of a selected segment in one direction, but does not lie on any other selected segment pointing in the same direction.
(Notice that during the game, a lattice point might switch between being missing and not-missing.) Lemma. Given an integer N > 0, if the game continues forever then at some point at least N missing points will exist at the same time.
Proof: Suppose we have k lines that contain at least one selected segment. Some ⌈k 4⌉of them must point in the same direction; then each of these lines contains at least one different missing point: its leftmost, upper-leftmost, highest, or upper- rightmost extreme point.
Therefore it is enough to show that k gets arbitrarily large.
For sake of contradiction, suppose that k is bounded. Then all the selected segments will lie on some finite number of lines; these lines have only a finite set S of t intersection points, so from some position onward no new selected point will be in S. At this point say we have s selected segments, and p selected points outside of S.
After n more moves there will be s + n mini-segments, and p + n selected points outside of S. Each mini-segment contains 4 points for a total count of 4(s + n). On the other hand, each of the t points lies on at most 4 mini-segments; and each of the p + n other points lies on at most 1 mini-segment, for a total count of at most 4t + (p + n).
Thus 4(s + n) ≤4t + (p + n), but this is false for large enough n — a contradiction.
Now suppose in our original position we have s selected segments and p selected points. From the lemma, eventually we will have more than 4(p −s) missing points. Say this happens after n moves, when 1999 Regional Contests: Problems and Solutions 13 we have s + n selected segments and p + n selected points. Then as in the lemma, we will have s + n mini-segments each containing 4 points—for a total count of 4(s + n). On the other hand, each of the missing points lies on at most 3 mini-segments; and all the other selected points lie on at most 4 mini-segments each. Thus our total count is less than 3 · 4(p −s) + 4 · (p + n −4(p −s)) = 4(s + n), a contradiction. This completes the proof.
14 Balkan Mathematical Olympiad 1.3 Balkan Mathematical Olympiad Problem 1 Given an acute-angled triangle ABC, let D be the midpoint of minor arc d BC of circumcircle ABC. Let E and F be the respective images of D under reflections about BC and the center of the circumcircle. Finally, let K be the midpoint of AE. Prove that: (a) the circle passing through the midpoints of the sides of the triangle ABC also passes through K.
(b) the line passing through K and the midpoint of BC is perpen-dicular to AF.
Solution: (a) Let M, B1, and C1 denote the midpoints of sides BC, CA, and AB, respectively; then △ABC ∼△MB1C1 and ∠C1MB1 = ∠A.
Also, BECD is a rhombus, with ∠BEC = ∠CDB = 180◦−∠A.
The homothety centered at A with ratio 1 2 maps triangle BEC to triangle C1KB1. Thus, ∠C1KB1 + ∠C1MB1 = ∠BEC + ∠A = 180◦, so MC1KB1 is cyclic.
(b) Since ED = 2EM and EA = 2EK, MK ∥AD. But DF is a diameter, so AD ⊥AF. Hence also MK ⊥AF.
Problem 2 Let p > 2 be a prime number such that 3 | (p −2). Let S = {y2 −x3 −1 | x and y are integers, 0 ≤x, y ≤p −1}.
Prove that at most p elements of S are divisible by p.
Solution: Lemma. Given a prime p and a positive integer k > 1, if k and p−1 are relatively prime then xk ≡yk ⇒x ≡y (mod p) for all x, y.
Proof: If y ≡0 (mod p) the claim is obvious. Otherwise, note that xk ≡yk = ⇒(xy−1)k ≡1 (mod p), so it suffices to prove that ak ≡1 = ⇒a ≡1 (mod p).
Since gcd(p −1, k) = 1, there exist integers b and c such that b(p−1)+ck = 1. Thus, ak ≡1 = ⇒ac ≡1 = ⇒a1−b(p−1) ≡1 (mod p).
If a = 0 this is impossible; otherwise, by Fermat’s Little Theorem, (a−b)p−1 ≡1 (mod p) so that a ≡1 (mod p), as desired.
Alternatively, again note that clearly a ̸≡1 (mod p). Then let d be the order of a, the smallest positive integer such that ad ≡1 (mod p); 1999 Regional Contests: Problems and Solutions 15 we have d | k. Take the set {1, a, a2, . . . , ad−1}; if it does not contain all of 1, 2, . . . , p −1 then pick some other element b and consider the set {b, ba, ba2, . . . , bad−1}. These two sets are disjoint, since otherwise bai ≡aj ⇒b ≡aj−1 (mod p), a contradiction. Continuing similarly, we can partition {1, 2, . . . , p −1} into d-element subsets, and hence d | p −1. But since d | k and gcd(k, p −1) = 1, we must have d = 1.
Therefore a ≡ad ≡1 (mod p), as desired.
Since 3 | p −2, gcd(3, p −1) = 1. Then from the claim, it follows that the set of elements {13, 23, . . . , p3} equals {1, 2, . . . , p} modulo p.
Hence, for each y with 0 ≤y ≤p −1, there is exactly one x between 0 and p−1 such that x3 ≡y2 −1 (mod p): that is, such that p | y2 −x3 −1. Therefore S contains at most p elements divisible by p, as desired.
Problem 3 Let ABC be an acute triangle, and let M, N, and P be the feet of the perpendiculars from the centroid to the three sides.
Prove that 4 27 < [MNP] [ABC] ≤1 4.
Solution: We begin by proving that 9[MNP ] [ABC] = sin2 A + sin2 B + sin2 C. Let G be the centroid of triangle ABC, and let M, N, and P be on sides BC, AC, and AB, respectively. Also let AB = c, BC = a, CA = b, and K = [ABC].
We have [ABG] = K 3 = 1 2c · GP = ⇒GP = 2K 3c .
Similarly, GN = 2K 3b , so [PGN] = 1 2GP · GN sin A = 2K2 sin A 9bc = K2a2 9Rabc.
Summing this formula with the analogous ones for [NGM] and [MGP] yields [MNP] = K2(a2 + b2 + c2) 9Rabc .
Dividing this by [ABC] = K and then substituting K = abc 4R , a = 2R sin A, b = 2R sin B, and c = 2R sin C on the right yields [MNP ] [ABC] = 1 9 sin2 A + sin2 B + sin2 C , as desired.
Hence the problem reduces to proving 4 3 < sin2 A+sin2 B+sin2 C ≤ 9 4. Assume without loss of generality that A ≥B ≥C.
16 Balkan Mathematical Olympiad To prove the right inequality, first note that A < π 2 ⇒B > π 4 .
The function sin2 x is concave on [ π 4 , π 2 ]; applying Jensen’s Inequality gives sin2 A + sin2 B ≤2 sin2 A 2 + B 2 = 2 cos2 C 2 . Thus it suffices to prove 2 cos2 C 2 +sin2 C ≤9 4 ⇐ ⇒1+cos C +1−cos2 C ≤9 4 ⇐ ⇒ −(cos C + 1 2)2 ≤0, which is true.
For the left inequality, note that sin2 x is an increasing function on [0, π 2 ]. We have B ≥ π−A 2 , so sin2 A + sin2 B + sin2 C > sin2 A + cos2 A 2 = −cos2 A + 1 2 cos A + 3 2. But since A is the largest angle, we have π 2 > A ≥π 3 so 1 2 ≥cos A > 0; then −cos2 A + 1 2 cos A + 3 2 ≥ 3 2 > 4 3, as desired.
Problem 4 Let {xn}n≥0 be a nondecreasing sequence of nonneg-ative integers such that for every k ≥0 the number of terms of the sequence which are less than or equal to k is finite; let this number be yk. Prove that for all positive integers m and n, n X i=0 xi + m X j=0 yj ≥(n + 1)(m + 1).
Solution: Under the given construction, ys ≤t if and only if xt > s.
But this condition is equivalent to saying that ys > t if and only if xt ≤s. Thus the sequences {xi} and {yj} are dual, meaning that applying the given algorithm to {yj} will restore the original {xi}.
To find Pn i=0 xi, note that among x0, x1, . . . , xn there are exactly y0 terms equal to 0, y1 −y0 terms equal to 1, . . . , and yxn−1 −yxn−2 terms equal to xn−1; and the remaining n + 1 −xn−1 terms equal xn.
Hence, n X i=0 xi = 0 · (y0) + 1 · (y1 −y0) + · · · + (xn −1) · (yxn−1 −yxn−2) + xn · (n + 1 −yxn−1) = −y0 −y1 −· · · −yxn−1 + (n + 1)xn.
First suppose that xn −1 ≥m, and write xn −1 = m+k for k ≥0.
Since xn > m + k, from our initial observations we have ym+k ≤n.
But then n + 1 ≥ym+k ≥ym+k−1 ≥· · · ≥ym, so n X i=0 xi + m X j=0 yj = (n + 1)xn − xn−1 X j=0 yj − m X i=0 yi 1999 Regional Contests: Problems and Solutions 17 = (n + 1)xn − m+k X i=m+1 yi ≥(n + 1)(m + k + 1) −k · (n + 1) = (n + 1)(m + 1), as desired.
Next suppose that xn −1 < m. Then xn ≤m = ⇒ym > n = ⇒ ym −1 ≥n. Since {xi} and {yj} are dual, we may therefore apply the same argument with the roles of the two sequences reversed. This completes the proof.
18 Czech and Slovak Match 1.4 Czech and Slovak Match Problem 1 For arbitrary positive numbers a, b, c, prove the inequality a b + 2c + b c + 2a + c a + 2b ≥1.
First Solution: Set x = b + 2c, y = c + 2a, z = a + 2b. Then a = 1 9(4y + z −2x), b = 1 9(4z + x −2y), c = 1 9(4x + y −2z), so the desired inequality becomes 4y + z −2x 9x + 4z + x −2y 9y + 4x + y −2z 9z ≥1, which is equivalent to x y + y x + y z + z y + z x + x z + 3 · y x + z y + x z ≥15.
But this is true because by AM-GM, the quantities in parentheses are at least 2, 2, 2, and 3, respectively; or alternatively, it is true by the AM-GM inequality on all 15 fractions of the form x y on the left-hand side (where 3 · y x + x z + z y contributes nine such fractions).
Second Solution: By the Cauchy-Schwarz inequality (u1v1 + u2v2 + u3v3)2 ≤(u2 1 + u2 2 + u2 3)(v2 1 + v2 2 + v2 3), the quantity (a + b + c)2 is at most a b + 2c + b c + 2a + c a + 2b [a(b + 2c) + b(c + 2a) + c(a + 2b)].
On the other hand, from the inequality (a−b)2+(b−c)2+(c−a)2 ≥0 we have a(b + 2c) + b(c + 2a) + c(a + 2b) ≤(a + b + c)2.
Combining these gives a(b + 2c) + b(c + 2a) + c(a + 2b) ≤ a b + 2c + b c + 2a + c a + 2b [a(b + 2c) + b(c + 2a) + c(a + 2b)], which yields our desired inequality upon division by the (positive) expression a(b + 2c) + b(c + 2a) + c(a + 2b).
1999 Regional Contests: Problems and Solutions 19 Problem 2 Let ABC be a nonisosceles acute triangle with altitudes AD, BE, and CF. Let ℓbe the line through D parallel to line EF.
Let P = ← → BC ∩ ← → EF, Q = ℓ∩ ← → AC, and R = ℓ∩ ← → AB. Prove that the circumcircle of triangle PQR passes through the midpoint of BC.
Solution: Let M be the midpoint of BC. Without loss of generality we may assume that AB > AC. Then the order of the points in question on line BC is B, M, D, C, P. Since ∠BEC = ∠BFC = 90◦, BCEF is cyclic so that ∠QCB = 180◦−∠BCE = ∠EFB = ∠QRB.
Thus RCQB is cyclic as well and DB · DC = DQ · DR.
Quadrilateral MRPQ is cyclic if and only if DM · DP = DQ · DR, so it remains to prove that DB · DC = DP · DM.
Since the points B, C, E, F are concyclic, we have PB · PC = PE · PF.
The circumcircle of triangle DEF (the so-called nine-point circle of triangle ABC) also passes through the midpoints of the sides of triangle ABC, which implies that PE · PF = PD · PM. Comparing both equalities shows that PB · PC = PD · PM. Denoting MB = MC = u, MD = d, MP = p, the last equality reads (p + u)(p −u) = (p −d)p ⇐ ⇒ u2 = dp ⇐ ⇒ (u + d)(u −d) = (p −d)d ⇐ ⇒ DB · DC = DP · DM. This completes the proof.
Problem 3 Find all positive integers k for which there exists a ten-element set M of positive numbers such that there are exactly k different triangles whose side lengths are three (not necessarily distinct) elements of M. (Triangles are considered different if they are not congruent.) Solution: Given any 10-element set M of positive integers, there are exactly 12 3 triples x, y, z (x ≤y ≤z) chosen from the numbers in M; thus, we must have k ≤ 12 3 = 220. On the other hand, for each of the 11 2 pairs x, y from M (with x ≤y) we can form the triangle with side lengths x, y, y; hence k ≥ 11 2 = 55. Then applying the following lemma, the possible values of k are then 55, 56, . . . , 220.
Lemma. Suppose we have some positive integers n and k, with n + 1 2 ≤k ≤ n + 2 3 .
20 Czech and Slovak Match Then there exists an n-element set M of positive numbers, such that there are exactly k triangles whose side lengths are three (not necessarily distinct) elements of M.
Furthermore, there exists an n-element set Sn with numbers x1 < x2 < · · · < xn such that xn < 2x1 and such that all the n+1 2 pairwise sums xi + xj (1 ≤i ≤j ≤n) are distinct. Exactly n+2 3 triangles can be formed from the elements of Sn.
Proof: We prove the claim by induction on n; it clearly holds for n = 1. Now suppose the claim is true for n and that n+2 2 ≤k ≤ n+3 3 .
If n+2 2 ≤k ≤ n+2 3 + (n + 1), then first find the n-element set M ′ = {x1, x2, . . . , xn} from which exactly k −(n+1) triangles can be formed. Choose xn+1 > 2 · max{M ′}, and write M = M ′ ∪{xn+1}.
Then exactly k triangles can be formed from the elements of M: k −(n + 1) triangles from {x1, x2, . . . , xn}; and an additional n + 1 triangles with side lengths xi, xn+1, xn+1 for i = 1, 2, . . . , n + 1.
Otherwise k = n+2 3 + n + 1 + q, where q ∈{1, 2, . . . , n+1 2 }. To the set Sn described in the lemma, add an element xn+1 which is greater than xn but smaller than precisely q of the n+1 2 original pairwise sums from Sn. This gives a set M from which exactly k triangles can be formed: n+2 3 triangles from {x1, x2, . . . , xn}; n + 1 additional triangles with side lengths xi, xn+1, xn+1; and exactly q more triangles with side lengths xi, xj, xn+1 (where i, j ̸= n + 1).
None of the numbers xi + xn+1 equals any of the original pairwise sums.
Thus we can construct Sn+1, completing the proof of the lemma.
Problem 4 Find all positive integers k for which the following assertion holds: if F(x) is a polynomial with integer coefficients which satisfies 0 ≤F(c) ≤k for all c ∈{0, 1, . . . , k + 1}, then F(0) = F(1) = · · · = F(k + 1).
Solution: The claim is false for k < 4; we have the counterexamples F(x) = x(2 −x) for k = 1, F(x) = x(3 −x) for k = 2, and F(x) = x(4 −x)(x −2)2 for k = 3.
1999 Regional Contests: Problems and Solutions 21 Now suppose k ≥4 is fixed and that F(x) has the described property. First of all F(k + 1) −F(0) = 0, because it is a multiple of k + 1 whose absolute value is at most k. Hence F(x) −F(0) = x(x −k −1)G(x), where G(x) is another polynomial with integer coefficients. Then we have k ≥|F(c) −F(0)| = c(k + 1 −c)|G(c)| for each c = 1, 2, . . . , k. If 2 ≤c ≤k −1 (such numbers c exist since k ≥4) then c(k + 1 −c) = 2(k −1) + (c −2)(k −1 −c) ≥2(k −1) > k, which in view of our first inequality means that |G(c)| < 1 = ⇒G(c) = 0. Thus 2, 3, . . . , k −1 are roots of the polynomial G(x), so F(x) −F(0) = x(x −2)(x −3) · · · (x −k + 1)(x −k −1)H(x), where H(x) is again a polynomial with integer coefficients. It remains to explain why H(1) = H(k) = 0. But this is easy: both values c = 1 and c = k satisfy k ≥|F(c) −F(0)| = (k −2)! · k · |H(c)|; and since (k −2)! > 1, we must have H(1) = H(k) = 0.
Problem 5 Find all functions f : (1, ∞) →R such that f(x) −f(y) = (y −x)f(xy) for all x, y > 1.
Solution: For every t > 1 we use the equation in turn for (x, y) = (t, 2), (t, 4) and (2t, 2): f(t) −f(2) = (2 −t)f(2t), f(t) −f(4) = (4 −t)f(4t), f(2t) −f(2) = (2 −2t)f(4t).
We eliminate f(t) by subtracting the second equation from the first, and then substitute for f(2t) from the third. This yields the equality f(4) + (t −3)f(2) = t(2t −5)f(4t) for any t > 1. Taking t = 5 2 we get f(4) = 1 2f(2), and feeding this back gives t −5 2 f(2) = t(2t −5)f(4t).
22 Czech and Slovak Match It follows that for any t > 1, t ̸= 5 2, f(4t) = f(2) 2t , so by the middle of the three equations in the beginning of this solution we obtain f(t) = f(4) + (4 −t)f(4t) = 1 2f(2) + (4 −t)f(2) 2t = 2f(2) t .
This formula for f(t) holds even for t = 5 2, as can now be checked directly by applying the original equation to x = 5 2 and y = 2 (and using f(5) = 2f(2) 5 ). Setting c = 2f(2), we must have f(x) = c x for all x; and this function has the required property for any choice of the real constant c.
Problem 6 Show that for any positive integer n ≥3, the least common multiple of the numbers 1, 2, . . . , n is greater than 2n−1.
Solution: Since for any n ≥3 we have 2n−1 = n−1 X k=0 n −1 k < n−1 X k=0 n −1 ⌊n−1 2 ⌋ = n · n −1 ⌊n−1 2 ⌋ , it suffices to show that the number n· n−1 ⌊n−1 2 ⌋ divides the least common multiple of 1, 2, . . . , n. Using a prime factorization argument, we will prove the more general assertion that for each k < n the least common multiple of the numbers n, n −1, . . . , n −k is divisible by n · n−1 k .
Let k and n be fixed natural numbers with k < n, and let p ≤n be an arbitrary prime.
Let pα be the highest power of p which divides lcm(n, n −1, . . . , n −k), where pα | n −ℓfor some ℓ. Then for each i ≤α, we know that pi | n −ℓ.
Thus exactly j ℓ pi k of {n −ℓ+ 1, n −ℓ+ 2, . . . , n} and exactly j k−ℓ pi k of {n −ℓ−1, n − ℓ−2, . . . , n −k} are multiples of pi, so pi divides j ℓ pi k + j k−ℓ pi k ≤ j k pi k of the remaining k numbers — that is, at most the number of multiples of pi between 1 and k. It follows that p divides n · n−1 k = n(n−1)···(n−ℓ+1)(n−ℓ−1)···(n−k) k!
·(n−ℓ) at most α times, so that indeed n · n−1 k | lcm(n, n −1, . . . , n −k).
1999 Regional Contests: Problems and Solutions 23 1.5 Hungary-Israel Binational Mathematical Competition Individual Round Problem 1 Let S be the set of all partitions (a1, a2, . . . , ak) of the number 2000, where 1 ≤a1 ≤a2 ≤· · · ≤ak and a1 + a2 + · · · + ak = 2000. Compute the smallest value that k + ak attains over all such partitions.
Solution: AM-GM gives k + ak ≥2 p kak ≥2( √ 2000) > 89.
Thus since k + ak is an integer, it must be at least 90. And 90 is attainable, since k + ak = 90 for the partition (40, 40, · · · , 40 | {z } 50 ).
Problem 2 Prove or disprove the following claim: For any positive integer k, there exists a positive integer n > 1 such that the binomial coefficient n i is divisible by k for any 1 ≤i ≤n −1.
First Solution: The statement is false. To prove this, take k = 4 and assume by contradiction that there exists a positive integer n > 1 for which n i is divisible by 4 for every 1 ≤i ≤n −1. Then 0 ≡ n−1 X i=1 n i = 2n −2 ≡−2 (mod 4), a contradiction.
Second Solution: The claim is obviously true for k = 1; we prove that the set of positive integers k > 1 for which the claim holds is exactly the set of primes.
First suppose that k is prime; then express n in base k, writing n = n0 + n1k + · · · + nmkm where 0 ≤n0, n1, . . . , nm ≤k −1 and nm ̸= 0. Also suppose we have 1 ≤i ≤n −1, and write i = i0 + i1k + · · · + imkm where 0 ≤i0, i1, . . . , im ≤k−1 (although perhaps im = 0). Lucas’s Theorem states that n i ≡ m Y j=0 nj ij (mod k).
24 Hungary-Israel Binational Mathematical Competition Now if n = km, then n0 = n1 = · · · = nm−1 = 0; and 1 ≤i ≤n −1 so that im = 0 but some other ij′ is nonzero. Then nj′ ij′ = 0, and indeed n i ≡0 (mod k) for all 1 ≤i ≤n −1.
However, suppose that n ̸= km. If nm > 1 then letting i = km < n we have n i ≡(nm)(1)(1) · · · (1) ≡nm ̸≡0 (mod k).
Otherwise, some other nj′ ̸= 0; but then setting i = nj′kj′ < n we have n i ≡(1)(1) · · · (1) ≡1 ̸≡0 (mod k).
Therefore the claim holds for prime k exactly when n = km.
Now suppose the claim holds for some k > 1 with the number n. If some prime p divides k, the claim must also hold for p with the number n. Thus n must equal a prime power pm where m ≥1.
Then k = pr for some r ≥1 as well, because if two primes p and q divided k then n would equal a perfect power of both p and q, which is impossible.
Choose i = pm−1. Kummer’s Theorem then states that pt | n i if and only if t is less than or equal to the number of carries in the addition (n−i)+i in base p. But there is only one such carry, between the pm−1 and pm places: 1 1 0 0 . . .
0 + p −1 0 0 . . .
0 1 0 0 0 . . .
0 Thus, we must have r ≤1 and k must be prime, as claimed.
(Alternatively, for n = pm and i = pm−1 we have n i = pm−1−1 Y j=0 pm −j pm−1 −j .
When j = 0 then pm−j pm−1−j = p. Otherwise, 0 < j < pm−1 so that if pt < pm−1 is the highest power of p dividing j, then it is also the highest power of p dividing both pm −j and pm−1 −j. Therefore pm−j pm−1−j contributes one factor of p to n i when j = 0 and zero factors of p when j > 0. Thus p2 ̸ | n i , and hence again r ≤1.) Problem 3 Let ABC be a non-equilateral triangle with its incircle touching BC, CA, and AB at A1, B1, and C1, respectively, and let H1 be the orthocenter of triangle A1B1C1. Prove that H1 is on the line passing through the incenter and circumcenter of triangle ABC.
1999 Regional Contests: Problems and Solutions 25 First Solution: Let ω1, I, ω2, and O be the incircle, incenter, circumcircle, and circumcenter of triangle ABC, respectively. Since I ̸= O, line IO is well defined.
Let T be the center of the homothety with positive ratio that sends ω1 to ω2 and hence I to O. Also let A2, B2, C2 be the midpoints of the arcs BC, CA, AB of ω2 not containing A, B, C, respectively.
Since rays IA1 and OA2 point in the same direction, T must send A1 to A2 and similarly B1 to B2 and C1 to C2.
Also, because the measures of arcs AC2 and A2B2 add up to 180◦, we know that AA2 ⊥C2B2. Similarly, BB2 ⊥C2A2 and CC2 ⊥A2B2. Then since lines AA2, BB2, CC2 intersect at I, I is the orthocenter of triangle A2B2C2. Hence I is the image of H1 under the defined homothety. Therefore T, H1, I are collinear; and from before T, I, O are collinear. It follows that H1, I, O are collinear, as desired.
Second Solution: Define ω1, I, ω2, and O as before. Let ω3 be the nine-point circle of triangle A1B1C1, and let S be its center. Since I is the circumcenter of triangle A1B1C1, S is the midpoint of IH1 and I, H1, S are collinear.
Now invert the figure with respect to ω1.
The midpoints of A1B1, B1C1, C1A1 are mapped to C, A, B, and thus ω3 is mapped to ω2. Thus I, O, S are collinear; and so I, H1, O, S are collinear, as desired.
Problem 4 Given a set X, define X′ = {s −t | s, t ∈X, s ̸= t}.
Let S = {1, 2, . . . , 2000}.
Consider two sets A, B ⊆S, such that |A||B| ≥3999. Prove that A′ ∩B′ ̸= ∅.
Solution: Consider all |A|·|B| ≥3999 sums, not necessarily distinct, of the form a + b where a ∈A, b ∈B. If both 2 and 4000 are of this form, then both A and B contain 1 and 2000 so that 2000−1 ∈A′∩B′.
Otherwise, each sum a + b takes on one of at most 3998 values either between 2 and 3999, or between 3 and 4000. Thus by the pigeonhole principle, two of our |A| · |B| sums a1 + b1 and a2 + b2 are equal with a1, a2 ∈A, b1, b2 ∈B, and (a1, b1) ̸= (a2, b2). Then a1 ̸= a2 (since otherwise we would have b1 = b2 and (a1, b1) = (a2, b2)), and therefore A′ ∩B′ is nonempty because it contains a1 −a2 = b2 −b1.
26 Hungary-Israel Binational Mathematical Competition Problem 5 Given an integer d, let S = {m2 + dn2 | m, n ∈Z}.
Let p, q ∈S be such that p is a prime and r = q p is an integer. Prove that r ∈S.
Solution: Note that (x2 + dy2)(u2 + dv2) = (xu ± dyv)2 + d(xv ∓yu)2.
Write q = a2+db2 and p = x2+dy2 for integers a, b, x, y. Reversing the above construction yields the desired result. Indeed, solving for u and v after setting a = xu+dyv, b = xv−yu and a = xu−dyv, b = xv+yu gives u1 = ax −dby p , v1 = ay + bx p , u2 = ax + dby p , v2 = ay −bx p .
Note that (ay + bx)(ay −bx) = (a2 + db2)y2 −(x2 + dy2)b2 ≡0 (mod p).
Hence p divides one of ay+bx, ay−bx so that one of v1, v2 is an integer.
Without loss of generality, assume that v1 is an integer. Then since r = u2 1 + dv2 1 is an integer and u1 is rational, u1 is an integer as well and r ∈S, as desired.
Problem 6 Let k and ℓbe two given positive integers, and let aij, 1 ≤i ≤k and 1 ≤j ≤ℓ, be kℓgiven positive numbers. Prove that if q ≥p > 0 then ℓ X j=1 k X i=1 ap ij ! q p 1 q ≤ k X i=1 ℓ X j=1 aq ij p q 1 p .
First Solution: Define bj = k X i=1 ap ij, 1999 Regional Contests: Problems and Solutions 27 and denote the left hand side of the required inequality by L and its right hand side by R. Then Lq = ℓ X j=1 b q p j = ℓ X j=1 b q−p p j k X i=1 ap ij !!
= k X i=1 ℓ X j=1 b q−p p j ap ij .
Using H¨ older’s inequality it follows that Lq ≤ k X i=1 ℓ X j=1 b q−p p j q q−p q−p q ℓ X j=1 ap ij q p p q = k X i=1 ℓ X j=1 b q p j q−p q ℓ X j=1 aq ij p q = ℓ X j=1 b q p j q−p q · k X i=1 ℓ X j=1 aq ij p q = Lq−pRp.
The inequality L ≤R follows by dividing both sides of Lq ≤Lq−pRp by Lq−p and taking the p-th root.
Second Solution: Let r = q p, cij = ap ij. Then r ≥1, and the given inequality is equivalent to the following inequality: ℓ X j=1 k X i=1 cij !r 1 r ≤ k X i=1 ℓ X j=1 cr ij 1 r We shall prove this inequality by induction on k. For k = 1, we have equality. For k = 2, the inequality becomes Minkowski’s inequality.
Suppose that k ≥3 and the inequality holds for k −1. Then by the induction assumption for k −1 we have k−1 X i=1 ℓ X j=1 cr ij 1 r + ℓ X j=1 cr kj 1 r ≥ ℓ X j=1 k−1 X i=1 cij !r 1 r + ℓ X j=1 cr kj 1 r Using Minkowski’s inequality with ˜ c1j = Pk−1 i=1 cij, ˜ c2j = ckj, we have ℓ X j=1 k−1 X i=1 cij !r 1 r + ℓ X j=1 cr kj 1 r ≥ ℓ X j=1 k X i=1 cij !r 1 r , 28 Hungary-Israel Binational Mathematical Competition completing the inductive step.
Team Round Problem 1 Let ABC be a triangle and let P1 be a point inside triangle ABC.
(a) Prove that the lines obtained by reflecting lines P1A, P1B, P1C through the angle bisectors of ∠A, ∠B, ∠C, respectively, meet at a common point P2.
(b) Let A1, B1, C1 be the feet of the perpendiculars from P1 to sides BC, CA and AB, respectively. Let A2, B2, C2 be the feet of the perpendiculars from P2 to sides BC, CA and AB, respectively.
Prove that these six points A1, B1, C1, A2, B2, C2 lie on a circle.
(c) Prove that the circle in part (b) touches the nine point circle (Feuerbach’s circle) of triangle ABC if and only if P1, P2, and the circumcenter of triangle ABC are collinear.
First Solution: (a) By the trigonometric form of Ceva’s Theorem, we have sin ∠ABP1 sin ∠BCP1 sin ∠CAP1 sin ∠P1BC sin ∠P1CA sin ∠P1AB = 1.
Now suppose that the given reflections of lines P1A, P1B, P1C meet sides BC, CA, AB at points D, E, F, respectively.
Then ∠ABE = ∠P1BC, ∠EBC = ∠ABP1, and so on; thus sin ∠EBC sin ∠FCA sin ∠DAB sin ∠BCF sin ∠CAD sin ∠ABE = 1 as well.
Therefore, again by the trigonometric form of Ceva’s Theorem, the three new lines also concur.
(b) Note that P1A1CB1 and P2A2CB2 are both cyclic quadrilaterals because they each have two right angles opposite each other.
Since ∠P1CB1 = ∠A2CP2, we have ∠B1P1C = ∠CP2A2, and, by the previous statement, that implies ∠B1A1C = ∠CB2A2, whence A1, A2, B1, B2 are concyclic. For similar reasons, A1, A2, C1, C2 are concyclic. Then all six points A1, B1, C1, A2, B2, C2 must be concyclic, or else the radical axes of circles A1A2B1B2, B1B2C1C2, C1C2A1A2 would not concur, contradicting the rad-ical axis theorem.
1999 Regional Contests: Problems and Solutions 29 Problem 2 An ant is walking inside the region bounded by the curve whose equation is x2 + y2 + xy = 6. Its path is formed by straight segments parallel to the coordinate axes.
It starts at an arbitrary point on the curve and takes offinside the region. When reaching the boundary, it turns by 90◦and continues its walk inside the region. When arriving at a point on the boundary which it has already visited, or where it cannot continue its walk according to the given rule, the ant stops. Prove that, sooner or later, and regardless of the starting point, the ant will stop.
First Solution: If the ant moves from (a, b) to (c, b), then a and c are the roots to f(t) = t2 + bt + b2 −6. Thus c = −a −b. Similarily, if the ant moves from (a, b) in a direction parallel to the y-axis, it meets the curve at (a, −a −b).
Let (a, b) be the starting point of the ant, and assume that the ant starts walking in a direction parallel to the x-axis; the case when it starts walking parallel to the y-axis is analogous. If after five moves the ant is still walking, then it will return to its original position after six moves: (a, b) →(−a −b, b) →(−a −b, a) →(b, a) →(b, −a −b) →(a, −a −b) →(a, b).
Therefore, the ant stops moving after at most six steps.
Second Solution: Rotate the curve by 45◦, where (x, y) is on our new curve C1 when 1 √ 2 (x −y, x + y) is on the original curve. The equation of the image of the curve under the rotation is 3x2 + y2 = 12.
Hence the curve is an ellipse, while the new directions of the ant’s motion are inclined by ±45◦with respect to the x-axis. Next apply an affine transformation so that (x, y) is on our new curve C2 when (x, √ 3y) is on C1. The ant’s curve then becomes x2 + y2 = 4, a circle with radius 2, and the directions of the ant’s paths are now inclined by ±30◦with respect to the x-axis.
30 Hungary-Israel Binational Mathematical Competition Thus if the ant moves from P1 to P2 to P3, then ∠P1P2P3 will either equal 0◦, 60◦, or 120◦. Thus as long as it continues moving, every two moves the ant travels to the other end of an arc measuring 0◦, 120◦, or 240◦along the circle; thus after at most six moves, he must return to a position he visited earlier.
Problem 3 (a) In the plane, we are given a circle ω with unknown center, and an arbitrary point P. Is it possible to construct, using only a straightedge, the line through P and the center of the circle?
(b) In the plane, we are given a circle ω with unknown center, and a point Q on the circle. Construct the tangent to ω at the point Q, using only a straightedge.
(c) In the plane, we are given two circles ω1 and ω2 with unknown centers.
Construct, with a straightedge only, the line through their centers when: (i) the two circles intersect; (ii) the two circles touch each other, and their point of contact T is marked; (iii) the two circles have no common point.
This is an open question, and the construction might even be impossible; the problem does not belong to the official team contest, but any progress will be appreciated.
Solution: (a) It is not possible. First we prove that given a circle with unknown center O, it is impossible to construct its center using only a straightedge. Assume by contradiction that this construction is possible; then perform a projective transformation on the figure, taking O to O′ and ω to another circle ω′. The drawn lines remain lines, and thus they still yield the point O′. On the other hand, those lines compose a construction which gives the center of ω′.
But O is not mapped to the center of ω′, a contradiction.
Then the described construction must also be impossible. Oth-erwise, given a circle ω with unknown center O, we could mark a point P1 and draw the line P1O; and then mark another point P2 not on ℓ1, and draw the line P2O. Then the intersection of these lines would yield O, which is impossible from above.
1999 Regional Contests: Problems and Solutions 31 (b) Given a point A not on ω, suppose line ℓ1 passes through A and hits ω at B1 and C1; and suppose another line ℓ2 also passes through A and hits ω at B2 and C2. Then the line connecting B1B2 ∩C1C2 and B1C2 ∩C1B2 is the pole of A with respect to ω. Conversely, given any line ℓwe can pick two points on it that are not on ω, and construct the poles of these two points. If ℓ passes through the center of ω then these two poles are parallel; otherwise, they intersect at the polar of ℓ.
Now given ω and Q, mark another point R on the circle and then construct the polar T of line QR. (If the polar is the point at infinity, pick another point for R.) But then line TQ is the pole of Q—and this pole is tangent to ω at Q. Hence line QT is our desired line.
(c) (i) Draw the lines tangent to ω1 at P and Q; they intersect at a point X1, which by symmetry must lie on our desired line.
Next draw the lines tangent to ω2 at P and Q; they also intersect at a point X2 that lies on our desired line. Then line X1X2 passes through both circles’ centers, as desired.
(ii) Lemma. Suppose we have a trapezoid ABCD with AB ∥ CD.
Suppose that lines AD and BC intersect at M and that lines AC and BD intersect at N. Then line MN passes through the midpoints of AB and CD.
Proof: Let line MN hit lines AB and CD at Y1 and Y2, respectively.
Perform an affine transformation that sends ABCD into isosceles trapezoid A′B′C′D′ (with A′D′ = B′C′) and sends points M, N, Y1, Y2 to M ′, N ′, Y ′ 1, Y ′ 2. Then line M ′N ′ still passes through Y ′ 1 and Y ′ 2, and by symmetry these points are the midpoints of A′B′ and C′D′.
But since (using directed lengths) AY1 Y1B = A′Y ′ 1 Y ′ 1B′ , Y1 must be the midpoint of AB; and similarly, Y2 is the midpoint of CD.
Now construct a line through T intersecting ω1 at A and ω2 at C; construct a different line through T hitting ω1 at B and ω2 at D. Under the homothety about T that maps ω1 to ω2, segment AB gets mapped to segment CD; thus, AB ∥CD. If lines AD and BC are parallel, pick different A and C; otherwise, using the construction in the lemma we 32 Hungary-Israel Binational Mathematical Competition can find the midpoint F1 of CD. Next, from the construction described in part (b), we can find the polar F2 of line CD; then line F1F2 passes through the center O2 of ω2.
Similarly, we can find a different line G1G2 passing through O2; and hence O2 is the intersection of lines F1F2 and G1G2.
We can likewise find the center O1 of ω1; then drawing the line O1O2, we are done.
1999 Regional Contests: Problems and Solutions 33 1.6 Iberoamerican Math Olympiad Problem 1 Find all positive integers n less than 1000 such that n2 is equal to the cube of the sum of n’s digits.
Solution: In order for n2 to be a cube, n must be a cube itself; and since n < 1000 we must have n = 13, 23, . . ., or 93. Quick checks show that n = 1 and n = 27 work while n = 8, 64, and 125 don’t. And for n ≥63 = 216, we have n2 ≥66 > 273; but since the sum of n’s digits is at most 9 + 9 + 9 = 27, this implies that no n ≥63 works. Thus n = 1, 27 are the only answers.
Problem 2 Given two circles ω1 and ω2, we say that ω1 bisects ω2 if they intersect and their common chord is a diameter of ω2. (If ω1 and ω2 are identical, we still say that they bisect each other.) Consider two non-concentric fixed circles ω1 and ω2.
(a) Show that there are infinitely many circles ω that bisect both ω1 and ω2.
(b) Find the locus of the center of ω.
Solution: Suppose we have any circle ω with center O and radius r. Then we show that given any point P, there is a unique circle centered at P bisecting ω; and that the radius of this circle is √ r2 + OP 2. If O = P the claim is obvious; otherwise, let AB be the diameter perpendicular to OP, so that PA = PB = √ r2 + OP 2.
Since PA = PB, there is a circle centered at P and passing through both A and B; this circle indeed bisects ω. Conversely, if circle Γ centered at P bisects ω along diameter A′B′, then both O and P lie on the perpendicular bisector of A′B′. Thus A′B′ ⊥OP, and we must have AB = A′B′ and hence indeed PA′ = PB′ = PA = PB = √ r2 + OP 2.
Now back to the original problem.
Set up a coordinate system where ω1 is centered at the origin O1 = (0, 0) with radius r1; and ω2 is centered at O2 = (a, 0) with radius r2. Given any point P = (x, y), the circle centered at P bisecting ω1 is the same as the circle centered at P bisecting ω2 if and only if p r2 1 + O1P 2 = p r2 2 + O2P 2; that is, 34 Iberoamerican Math Olympiad if and only if r2 1 + x2 + y2 = r2 2 + (x −a)2 + y2 2ax = r2 2 −r2 1 + a2.
Therefore given any point P along the line x = r2 2−r2 1+a2 2a , some circle ω centered at P bisects both ω1 and ω2. Since there are infinitely many such points, there are infinitely many such circles.
And conversely, from above if ω does bisect both circles then it must be centered at a point on the given line—which is a line perpendicular to the line passing through the centers of ω1 and ω2. In fact, recall that the radical axis of ω1 and ω2 is the line 2a(a −x) = r2 2 −r2 1 + a2; therefore the desired locus is the radical axis of ω1 and ω2, reflected across the perpendicular bisector of the segment joining the centers of the circles.
Problem 3 Let P1, P2, . . . , Pn (n ≥2) be n distinct collinear points.
Circles with diameter PiPj (1 ≤i < j ≤n) are drawn and each circle is colored in one of k given colors. All points that belong to more than one circle are not colored. Such a configuration is called a (n, k)-cover.
For any given k, find all n such that for any (n, k)-cover there exist two lines externally tangent to two circles of the same color.
Solution: Without loss of generality label the points so that P1, P2, . . . , Pn lie on the line in that order from left to right.
If n ≤k + 1 points, then color any circle PiPj (1 ≤i < j ≤n) with the i-th color. Then any two circles sharing the same i-th color are internally tangent at Pi, so there do not exist two lines externally tangent to them.
However, if n ≥k + 2 then some two of the circles with diameters P1P2, P2P3, . . ., Pk+1Pk+2 must have the same color. Then there do exist two lines externally tangent to them. Therefore the solution is n ≥k + 2.
Problem 4 Let n be an integer greater than 10 such that each of its digits belongs to the set S = {1, 3, 7, 9}. Prove that n has some prime divisor greater than or equal to 11.
Solution: Note that any product of any two numbers from {1, 3, 7, 9} taken modulo 20 is still in {1, 3, 7, 9}. Therefore any finite 1999 Regional Contests: Problems and Solutions 35 product of such numbers is still in this set; specifically, any number of the form 3j7k is congruent to 1, 3, 7, or 9 (mod 20).
Now if all the digits of n ≥10 are in S, then its tens digit is odd and we cannot have n ≡1, 3, 7, or 9 (mod 20). Thus, n cannot be of the form 3j7k. Nor can n be divisible by 2 or 5 (otherwise, its last digit would not be 1, 3, 7, or 9); so it must be divisible by some prime greater than or equal to 11, as desired.
Problem 5 Let ABC be an acute triangle with circumcircle ω centered at O.
Let AD, BE, and CF be the altitudes of ABC.
Let line EF meet ω at P and Q.
(a) Prove that AO ⊥PQ.
(b) If M is the midpoint of BC, prove that AP 2 = 2AD · OM.
Solution: Let H be the orthocenter of triangle ABC, so that AEHF, BFHD, CDHE are cyclic.
Then ∠AFE = ∠AHE = ∠DHB = 90◦−∠HBD = 90◦−∠EBC = ∠BCE = C, while ∠OAF = ∠OAB = 90◦−C.
Therefore AO ⊥EF and thus AO ⊥PQ, as desired.
Say the circumradius of triangle ABC is R.
Draw diameter AA′ perpendicular to PQ, intersecting PQ at T.
Then AT = AF cos(90◦−C) = AF sin C = AC cos A sin C = 2R cos A sin B sin C.
By symmetry, PT = TQ; then by Power of a Point, PT 2 = PT · TQ = AT · TA′ = AT(2R −AT). Thus AP 2 = AT 2 + PT 2 = AT 2 + AT(2R −AT) = 2R · AT = 4R2 cos A sin B sin C.
On the other hand, AD = AC sin C = 2R sin B sin C, while OM = OC sin ∠OCM = R sin(90◦−A) = R cos A. Thus AP 2 = 4R2 cos A sin B sin C = 2AD · OM, as desired.
Problem 6 Let AB be a segment and C a point on its perpendicular bisector. Construct C1, C2, . . . , Cn, . . . as follows: C1 = C, and for n ≥1, if Cn is not on AB, then Cn+1 is the circumcenter of triangle ABCn. Find all points C such that the sequence {Cn}n≥1 is well defined for all n and such that the sequence eventually becomes periodic.
36 Iberoamerican Math Olympiad Solution: All angles are directed modulo 180◦unless otherwise indicated. Then Cn is uniquely determined by θn = ∠ACnB; and furthermore, we have θn+1 = 2θn for all n. For this to be eventually periodic we must have θj+1 = θk+1 or 2jθ1 = 2kθ1 for some j, k; that is, 180◦must divide (2k −2j)θ for some k and (not using directed angles) 180◦· r = (2k −2j)θ1 for some integers p, k.
Therefore θ1 = 180◦· r 2k−2j must be a rational multiple p q ·180◦with p, q relatively prime. And for the sequence to be well-defined, q must not be a power of two; because if q = 2n then θn+1 = 180◦· p = 180◦, which we cannot have.
Conversely, suppose we have such an angle p q · 180◦where p, q are relatively prime and q is not a power of 2. First, the sequence of points is well-defined because 2np q will always have a nontrivial odd divisor in its denominator; so it will never be an integer and θn+1 = 2np q ·180◦ will never equal 180◦.
Next write q = 2jt for odd t, and let k = φ(t) + j. Then since t | 2φ(t) −1 we have θk+1 = 2k · θ1 = 2φ(t)+j · p q · 180◦= 2φ(t) · p t · 180◦≡p t · 180◦, while θj+1 ≡2j p q · 180◦= p t · 180◦.
Thus θj+1 ≡θk+1, so the sequence is indeed periodic.
Therefore the set of valid points C is all points C such that ∠ACB (no longer directed) equals p q · 180◦for relatively prime positive integers p, q, where q is not a power of 2.
1999 Regional Contests: Problems and Solutions 37 1.7 Olimpiada de mayo Problem 1 Find the smallest positive integer n such that the 73 fractions 19 n + 21, 20 n + 22, 21 n + 23, · · · , 91 n + 93 are all irreducible.
Solution: Note that the difference between the numerator and denominator of each fraction is n + 2. Then n + 2 must be relatively prime to each of the integers from 19 to 91. Since this list contains a multiple of each prime p less than or equal to 91, n + 2 must only have prime factors greater than 91. The smallest such number is 97, so n = 95.
Problem 2 Let ABC be a triangle with ∠A = 90◦.
Construct point P on BC such that if Q is the foot of the perpendicular from P to AC then PQ2 = PB · PC.
Solution: Draw D on ray AB such that AB = BD, and draw E on ray AC such that AC = CE. Next draw F on DE such that ∠FBD = ∠BCA. Finally, draw the line through F perpendicular to AC; we claim that it intersects BC and AC at our desired points P and Q.
Since BD ∥FQ we have ∠BFQ = ∠FBD = ∠BCA = ∠BCQ, so BFCQ is cyclic. And since BD = BA and DA ∥FQ, we have PF = PQ. Thus by Power of a Point, we have PB ·PC = PF ·PQ = PQ2, as desired.
Problem 3 There are 1999 balls in a row.
Each ball is colored either red or blue. Underneath each ball we write a number equal to the sum of the number of red balls to its right and blue balls to its left. Exactly three numbers each appear on an odd number of balls; determine these three numbers.
First Solution: Call of a coloring of 4n−1 balls “good” if exactly 3 numbers each appear on an odd number of balls; we claim that these three numbers will then be 2n −2, 2n −1, and 2n. Let b1, b2, . . .
represent the ball colors (either B or R), and let x1, x2, . . . represent the numbers under the respective balls.
Then xk+1 −xk = 1 if 38 Olimpiada de mayo bk = bk+1 = B; xk+1 −xk = −1 if bk = bk+1 = R; and xk = xk+1 if bk ̸= bk+1. Now, when n = 1, the only good colorings are RRR and BBB, which both satisfy the claim. For the sake of contradiction, let n0 > 1 be the least n for which the claim no longer holds. Now, let a “couple” be a pair of adjacent, different-colored balls; then for our coloring of n0 balls, one of the following cases is true: (i) There exist two or more disjoint couples in the coloring. Remov-ing the two couples decreases all the other xi by exactly 2, while the numbers originally on the four removed balls are removed in pairs. Thus we have constructed a good coloring of 4(n0 −1) −1 balls for which the claim does not hold, a contradiction.
(ii) The balls are colored RR · · · RBRR · · · R or BB · · · BRBB · · · B.
Then {xi} is a nondecreasing series, with equality only under the balls BRB or RBR. Then (4n0 −1) −2 numbers appear an odd number of times, but this cannot equal 3.
(iii) There exists exactly one couple in the coloring. Suppose, without loss of generality, that we have a string of m blue balls followed by a string of n red balls. It is trivial to check that m and n cannot equal 1. Then by removing the final two blue balls and the first two red balls, we construct an impossible coloring of 4(n0 −1)−1 balls as in case (i).
(iv) All the balls are of the same color. We then have 4n0 −1 > 3 distinct numbers, a contradiction.
Thus, all good colorings satisfy our claim; so for n0 = 500, we find that the three numbers must be 998, 999, and 1000.
Second Solution: If a ball has the number m on it, call it an “m-ball.” Also let x1, x2, . . . , x1999 represent both the balls and the numbers written on them; and let ⟨i, j⟩denote the i-th through j-th balls, inclusive.
Read from left to right, the numbers on the balls increase by 1 when two blue balls are adjacent, decrease by 1 when two red balls are adjacent, and otherwise remain constant. This implies that when viewed from left to right, the m-balls in our line alternate in color.
Also, if m < xi (or m > xi) then the first m-ball after xi must be red (or blue) while the last m-ball before it must be blue (or red). It follows that in ⟨i, j⟩, there are an odd number of m-balls if min(xi, xj) < m < max(xi, xj), and an even number of m-balls if 1999 Regional Contests: Problems and Solutions 39 m < min(xi, xj) or m > max(xi, xj).
This result implies that the three given numbers are consecutive: say they are k−1, k, k+1. Without loss of generality say that x1 ≤k, and suppose that xr is the rightmost (k + 1)-ball. If any xj ≤k + 1 for j > r then in ⟨1, j⟩there are an even number of (k + 1)-balls, a contradiction.
Then in ⟨1, r −1⟩, for m ̸= k −1, k there are an even number of m-balls so that the number of red m-balls equals the number of blue m-balls. As for m = k −1, k, there are an odd number of m-balls in ⟨1, r −1⟩and the last such m-ball is blue. It follows that there must be r+2 2 blue balls and r−2 2 red balls in ⟨1, r −1⟩.
And in ⟨r + 1, 1999⟩, the number of red m-balls equals the number of blue m-balls for all m; this is because there are no such m-balls for m ≤k + 1, and for m ≥k + 2 an even number of m-balls are in ⟨1, r −1⟩so in ⟨r + 1, 1999⟩an even number of m-balls must remain.
Thus there are 1998−r 2 blue balls and 1998−r 2 red balls in ⟨r, 1999⟩.
Hence k + 1 = xr = r+2 2 + 1998−r 2 = 1000, and the three numbers are 998, 999, and 1000.
Problem 4 Let A be a number with six digits, three of which are colored and are equal to 1, 2, 4. Prove that it is always possible to obtain a multiple of 7 by doing one of the following: (1) eliminate the three colored numbers; (2) write the digits of A in a different order.
Solution: Note that modulo 7, the six numbers 421, 142, 241, 214, 124, 412 are congruent to 1, 2, 3, 4, 5, 6, respectively. Let the other digits besides 1, 2, and 4 be x, y, and z, appearing in that order from left to right.
If the 3-digit number xyz is divisible by 7, we may eliminate the three colored numbers. If not, the 6-digit number xyz000 is also not divisible by 7, and we may add the appropriate permutation abc of 124 to xyz000 to make xyzabc divisible by 7.
Problem 5 Consider a square of side length 1. Let S be a set of finitely many points on the sides of the square. Prove that there is a vertex of the square such that the arithmetic mean of the squares of the distances from the vertex to all the points in S is no less than 3 4.
40 Olimpiada de mayo Solution: Let the four vertices of the square be V1, V2, V3, and V4, and let the set of points be {P1, P2, . . . , Pn}. For a given Pk, we may assume without loss of generality that Pk lies on side V1V2. Writing x = PkV1, we have 4 X i=1 PkV 2 i = x2+(1−x)2+(1+x2)+(1+(1−x)2) = 4 x −1 2 2 +3 ≥3.
Hence P4 i=1 Pn j=1 PjV 2 i ≥3n, or P4 i=1 1 n Pn j=1 PjV 2 i ≥3. Thus the average of 1 n Pn j=1 PjV 2 i (for i = 1, 2, 3, 4) is at least 3 4; so if we select the Vi for which 1 n Pn j=1 PjV 2 i is maximized, we are guaranteed it will be at least 3 4.
Problem 6 An ant crosses a circular disc of radius r and it advances in a straight line, but sometimes it stops. Whenever it stops, it turns 60◦, each time in the opposite direction. (If the last time it turned 60◦clockwise, this time it will turn 60◦counterclockwise, and vice versa.) Find the maximum length of the ant’s path.
Solution: Suppose the ant begins its path at P0, stops at P1, P2, . . . , Pn−1 and ends at Pn. Note that all the segments P2iP2i+1 are parallel to each other and that all the segments P2i+1P2i+2 are parallel to each other. We may then translate all the segments so as to form two segments P0Q and QPn where ∠P0QPn = 120◦. Then P0Pn ≤2r, and the length of the initial path is equal to P0Q + QPn.
Let P0Pn = c, P0Q = a, and QPn = b. Then (2r)2 ≥c2 = a2 + b2 + ab = (a + b)2 −ab ≥(a + b)2 −1 4(a + b)2, so 4 √ 3r ≥a + b with equality iffa = b. The maximum is therefore 4 √ 3r; this can be attained with the path where P0P2 is a diameter of the circle, and P0P1 = P1P2 = 2 √ 3r.
1999 Regional Contests: Problems and Solutions 41 1.8 St. Petersburg City Mathematical Olympiad (Russia) Problem 9.1 Let x0 > x1 > · · · > xn be real numbers. Prove that x0 + 1 x0 −x1 + 1 x1 −x2 + · · · + 1 xn−1 −xn ≥xn + 2n.
Solution: For i = 0, 1, . . . , n −1, we have xi −xi+1 > 0 so that by AM-GM, (xi −xi+1) + 1 (xi −xi+1) ≥2.
Adding up these inequalities for i = 0, 1, . . . , n −1 gives the desired result.
Problem 9.2 Let f(x) = x2 + ax + b be a quadratic trinomial with integral coefficients and |b| ≤800. It is known also that f(120) is prime. Prove that f(x) = 0 has no integer roots.
Solution: Suppose by way of contradiction f(x) had an integer root r1; then writing f(x) = (x −r1)(x −r2), we see that its other root must be r2 = −a −r1, also an integer.
Since f(120) = (120 −r1)(120 −r2) is prime, one of |120 −r1| and |120 −r2| equals 1, and the other equals some prime p.
Without loss of generality say |120 −r1| = 1, so that r1 = 119 or 121, and |r1| ≥119. Then |120 −r2| is a prime, but the numbers 114, 115, . . . , 126 are all composite: 119 = 7 · 17, and all the other numbers are clearly divisible by 2, 3, 5, or 11. Therefore |r2| ≥7, and |b| = |r1r2| ≥|119 · 7| > 800, a contradiction.
Problem 9.3 The vertices of a regular n-gon (n ≥3) are labeled with distinct integers from {1, 2, . . . , n}. For any three vertices A, B, C with AB = AC, the number at A is either larger than the numbers at B and C, or less than both of them. Find all possible values of n.
Solution: Suppose that n = 2st, where t ≥3 is odd. Look at the regular t-gon P1 · · · Pt formed by every 2s-th point. In this t-gon, some 1 Problems are numbered as they appeared in the contests.
Problems that appeared more than once in the contests are only printed once in this book.
42 St. Petersburg City Mathematical Olympiad (Russia) vertex A has the second-smallest number, and some vertex B has the smallest number. But then at the third vertex C with AB = AC (which exists since t is odd), the number must also be smaller than A’s number — a contradiction.
Alternatively, if P1 has a bigger number than Pt and P2, then P2 has a smaller number than P3, P3 has a bigger number than P4, and so on around the circle — so that Pt has a bigger number than P1, a contradiction.
Now we prove by induction on s ≥0 that we can satisfy the conditions for n = 2s. For s = 1, label the single point 1. And if we can label a regular 2s-gon with the numbers a1, . . . , a2s in that order, then we can label a regular 2s+1-gon with the numbers a1, a1 + 2s, a2, a2 + 2s, . . . , a2s, a2s + 2s, as desired.
(Alternatively, when n = 2s one could label the vertices as follows.
For i = 1, 2, . . . , 2s, reverse each digit of the s-bit binary expansion of i −1 and then add 1 to the result. Label the i-th vertex with this number.) Problem 9.4 Points A1, B1, C1 are chosen on the sides BC, CA, AB of an isosceles triangle ABC (AB = BC).
It is known that ∠BC1A1 = ∠CA1B1 = ∠A. Let BB1 and CC1 meet at P. Prove that AB1PC1 is cyclic.
Solution: All angles are directed modulo 180◦.
Let the circumcircles of triangles AB1C1 and A1B1C intersect at P ′, so that AB1P ′C1 and CB1P ′A1 are cyclic. Then ∠A1P ′C1 = (180◦−∠C1P ′B1) + (180◦−∠B1P ′A1) = ∠B1AC1 + ∠A1CB1 = ∠CAB + ∠BCA = 180◦−∠ABC = 180◦−∠C1BA1, so BA1P ′C1 is cyclic as well.
Now, ∠BC1A1 = ∠A = ∠C implies that △BC1A1 ∼△BCA, so BC1 · BA = BA1 · BC. Thus B has equal power with respect to the circumcircles of triangles AB1C1 and A1B1C, so it lies on their radical axis B1P ′.
Similarly, ∠CA1B1 = ∠A implies that △ABC ∼△A1B1C, so CA · CB1 = CB · CA1. Thus C has equal power with respect to the circumcircles of triangles B1AC1 and A1BC1, so it lies on their radical axis C1P ′.
1999 Regional Contests: Problems and Solutions 43 Then P ′ lies on both CC1 and BB1, so it must equal P. Therefore AB1PC1 is indeed cyclic, as desired.
Problem 9.5 Find the set of possible values of the expression f(x, y, z) = xyz xy + yz + zx , for positive integers x, y, z. Here {x} = x −⌊x⌋is the fractional part of x.
Solution: Clearly f(x, y, z) must be a nonnegative rational number below 1; we claim all such numbers are in the range of f. Suppose we have nonnegative integers p and q with p < q; let x1 = y1 = 2(q −1) and let z1 = 1. Then f (x1, y1, z1) = 4(q −1)2 4(q −1)2 + 4(q −1) = q −1 q .
Writing X = xyz xy+yz+zx, notice that for any nonzero integer k we have f(kx, ky, kz) = { kX } = { k⌊X⌋+ k{X} } = { k · f(x, y, z) } .
Then f (p (q −1) · x1, p (q −1) · y1, p (q −1) · z1) = p q , so every non-negative rational p q < 1 is indeed in the range of f.
Problem 9.6 Let AL be the angle bisector of triangle ABC.
Parallel lines ℓ1 and ℓ2 equidistant from A are drawn through B and C respectively. Points M and N are chosen on ℓ1 and ℓ2 respectively such that lines AB and AC meet lines LM and LN at the midpoints of LM and LN respectively. Prove that LM = LN.
Solution: Let A, B, C also represent the angles at those points in triangle ABC.
Let line ℓpass through A perpendicular to AL. Next, draw M ′ and N ′ on ℓso that ∠ALM ′ = ∠ALN ′ = A 2 (with M ′ and N ′ on the same sides of line AL as B and C, respectively). Finally, let ℓhit ℓ1 at Q.
We claim that M ′ lies on ℓ1. Orient the figure so that ℓ1 and ℓ2 are vertical, and let x = ∠QBA. Note that AM ′ = AL tan ∠ALM ′ = AL tan A 2 , so the horizontal distance between A and M ′ is AM ′ sin(180◦−∠AQB) 44 St. Petersburg City Mathematical Olympiad (Russia) = AL tan A 2 · sin(∠QBA + ∠BAQ) = AL tan A 2 · sin x + 90◦−A 2 = AL tan A 2 · cos A 2 −x .
On the other hand, the horizontal distance between A and ℓ1 is AB sin x. Thus we need only prove that AB sin x = AL tan A 2 · cos A 2 −x .
By the law of sines on triangle ABL, we know that AL sin A 2 + B = AB sin B. Hence we must prove sin A 2 + B · sin x = sin B · tan A 2 · cos A 2 −x ⇐ ⇒sin A 2 + B · cos A 2 · sin x = sin B · sin A 2 · cos A 2 −x ⇐ ⇒(sin(A + B) + sin B) · sin x = sin B · (sin(A −x) + sin x) ⇐ ⇒sin C · sin x = sin B · sin(A −x) ⇐ ⇒AB sin x = AC sin(A −x).
But AB sin x is the distance between A and ℓ1, and AC sin(A −x) is the distance between A and ℓ2 — and these distances are equal.
Therefore, M ′ indeed lies on ℓ1.
Now say that lines AB and LM ′ intersect at P. Then since ∠LAP = ∠PLA = A 2 , AP is the median to the hypotenuse of right triangle M ′AL. Thus line AB hits the midpoint of LM ′, so M = M ′.
Similarly, N = N ′. But then ∠LMN = 90◦−A 2 = ∠LNM, so LM = LN, as desired.
Problem 9.7 A corner is the figure resulted by removing 1 unit square from a 2 × 2 square. Prove that the number of ways to cut a 998 × 999 rectangle into corners, where two corners can form a 2 × 3 rectangle, does not exceed the number of ways to cut a 1998 × 2000 rectangle into corners, so that no two form a 2 × 3 rectangle.
1999 Regional Contests: Problems and Solutions 45 Solution: Take any tiling of a 998×999 rectangle with corners, and add a 2 × 999 block underneath also tiled with corners: · · · Next, enlarge this 1000 × 999 board to twice its size, and replace each large corner by four normal-sized corners as follows: → x x x For each corner in the tiling of the 1000×999 board, none of the four new corners can be half of a 2×3 rectangle. Also, the “central” corners (like the one marked with x’s above) have the same orientations as the original corners tiling the 1000 × 999 rectangle.
Thus, different tilings of a 998 × 999 rectangle turn into different tilings of a 2000 × 1998 board where no two corners form a 2 × 3 rectangle — which implies the desired result.
Problem 9.8 A convex n−gon (n > 3) is divided into triangles by non-intersecting diagonals. Prove that one can mark n −1 segments among these diagonals and sides of the polygon so that no set of marked segments forms a closed polygon and no vertex belongs to exactly two segments.
Solution: After we mark some segments, let the degree d(V ) of a vertex V be the number of marked segments it is on. Also, say we mark up a triangulated n-gon with respect to side AB if we mark n−1 segments, no marked segments form a closed polygon, and none of the n −2 vertices besides A and B have degree 2.
Lemma. Given any triangulated convex n-gon (n ≥3) and any two adjacent vertices A and B, we can mark up the n-gon with respect to side AB in three ways (i), (ii), (iii), each satisfying the corresponding condition from the following list: (i) AB is marked, and d(A) ≥2.
(ii) AB is marked, and d(A) ̸= 2.
(iii) (d(A), d(B)) ̸= (1, 1) or (2, 2).
46 St. Petersburg City Mathematical Olympiad (Russia) Proof: For n = 3, given a triangle ABC we can mark sides AB and (i) AC, (ii) BC, and (iii) AC. Now suppose that n ≥4 and that the claims are true for all smaller n; we prove that they are true for n as well.
AB must be part of some triangle ABC of drawn segments. Then either (a) AC is a side of the polygon but BC is not; (b) BC is a side but AC is not; or (c) neither AC nor BC is a side.
(a) Let P be the (n −1)-gon formed by the vertices not including A.
(i) Apply (ii) to P so that d(C) ̸= 2 and BC is marked; then unmark BC, and mark AC and AB.
(ii) Apply (ii) to P so that d(C) ̸= 2 and BC is marked; then mark AB.
(iii) Apply (i) to P so that d(B) ≥2 and BC is marked.
If d(C) ̸= 2, then mark AB; otherwise mark AC.
(b) Let P be the (n −1)-gon formed by the vertices besides B.
(i) Apply (ii) to P so that d(C) ̸= 2 and AC is marked; then mark AB.
(ii) Apply (ii) to P so that d(C) ̸= 2 and AC is marked.
If d(A) = 2, then mark AB; otherwise unmark AC and mark AB and BC.
(iii) Repeat the construction in (i).
(c) Let P be the polygon formed by A, C, and the vertices in between (not on the same side of line AC as B); and let Q be the polygon formed by B, C, and the vertices in between (not on the same side of line BC as A).
(i) Apply (i) to P and Q so that d(C) ≥2 + 2 = 4 and AC, BC are marked; then unmark BC and mark AB.
(ii) Apply (i) to P and Q so that d(C) ≥2 + 2 = 4 and AC, BC are marked. If d(A) = 2, unmark BC and mark AB; otherwise, unmark AC and mark AB.
(iii) Repeat the construction in (ii).
Now to the main result. Since there are n ≥4 sides but only n −2 triangles, some triangle contains two adjacent sides XY and Y Z. Let P be the (n −1)-gon formed by the vertices not including Y, and apply (iii) to P and vertices X, Z.
1999 Regional Contests: Problems and Solutions 47 Then if d(X) or d(Z) equals 2, mark XY or Y Z, respectively.
Otherwise, if d(X) or d(Z) equals 1, mark Y Z or XY , respectively.
Otherwise, both d(X) and d(Z) are at least 3, and we can mark either XY or Y Z to finish off.
Problem 10.1 The sequence {xn} of positive integers is formed by the following rule: x1 = 10999 + 1, and for every n ≥2, the number xn is obtained from the number 11xn−1 by rubbing out its first digit.
Is the sequence bounded?
Solution: If xn−1 has k digits, then xn−1 < 10k so that 11xn−1 < 11·10k = 1100 . . . 0. Thus if 11xn−1 has k+2 digits, its first two digits are 1 and 0; and rubbing out its first digit leaves xn with at most k digits. Otherwise, 11xn−1 has at most k + 1 digits, so rubbing out its first digit still leaves xn with at most k digits. Therefore the number of digits in each xn is bounded, so the xn are bounded as well.
Problem 10.2 Prove that any positive integer less than n! can be represented as a sum of no more than n positive integer divisors of n!.
Solution: Fix n, and write ak = n!
k! for each k = 1, 2, . . . , n. Suppose we have some number m with ak ≤m < ak−1 where 2 ≤k ≤n. Then consider the number d = ak j m ak k . We have 0 ≤m −d < ak; and furthermore, since s = j m ak k < ak−1 ak = k, we know that n!
d = k!
s is an integer. Thus from m we can subtract d, a factor of n!, to obtain a number less than ak.
Then if we start with any positive integer m < n! = a1, then by subtracting at most one factor of n! from m we can obtain an integer less than a2; by subtracting at most one more factor of n! we can obtain an integer less than a3; and so on, so that we can represent m as the sum of at most n −1 positive integer divisors of n!.
Problem 10.5 How many 10-digit numbers divisible by 66667 are there whose decimal representation contains only the digits 3, 4, 5, and 6?
Solution: Suppose that 66667n had 10 digits, all of which were 3, 4, 5, and 6. Then 3333333333 ≤66667n ≤66666666666 ⇒ 50000 ≤n ≤99999.
48 St. Petersburg City Mathematical Olympiad (Russia) Now consider the following cases: (i) n ≡0 (mod 3). Then 66667n = 2 3n · 105 + 1 3n, the five digits of 2· n 3 followed by the five digits of n 3 . These digits are all 3, 4, 5, or 6 if and only if n 3 = 33333 and n = 99999.
(ii) n ≡1 (mod 3). Then 66667n = 2 3(n −1) · 105 + 1 3(n + 2) + 66666, the five digits of 2 3(n −1) followed by the five digits of 1 3(n + 2) + 66666. But 1 3(n + 2) + 66666 must be between 66667 and 99999, so its digits cannot all be 3, 4, 5, or 6. Therefore there are no satisfactory n ≡1 (mod 3).
(iii) n ≡2 (mod 3). Let a = 1 3(n −2). Then 66667n = 2 3(n −2) + 1 · 105 + 1 3(n −2) + 33334, the five digits of x = 2a + 1 followed by the five digits of y = a + 33334. The units digits in x and y are between 3 and 6 if and only if the units digit in a is 1 or 2; then the other digits in x and y are all between 3 and 6 if and only if the other digits in a are 2 or 3. Thus there are thirty-two satisfactory a — we can choose each of its five digits from two options — and each a corresponds to a satisfactory n = 3a + 2.
Therefore there is exactly one satisfactory n ≡0 (mod 3), and thirty-two satisfactory n ≡2 (mod 3) — making a total of thirty-three values of n and thirty-three ten-digit numbers.
Problem 10.6 The numbers 1, 2, . . . , 100 are arranged in the squares of a 10×10 table in the following way: the numbers 1, . . . , 10 are in the bottom row in increasing order, numbers 11, . . . , 20 are in the next row in increasing order, and so on. One can choose any num-ber and two of its neighbors in two opposite directions (horizontal, vertical, or diagonal). Then either the number is increased by 2 and its neighbors are decreased by 1, or the number is decreased by 2 and its neighbors are increased by 1. After several such operations the 1999 Regional Contests: Problems and Solutions 49 table again contains all the numbers 1, 2, . . . , 100. Prove that they are in the original order.
Solution: Label the table entries a11, a12, . . . , with aij in the ith row and jth column, where the bottom-left corner is in the first row and first column. Also, let bij = 10(i−1)+j be the number originally in the ith row and jth column.
Observe that P = 10 X i,j=1 aijbij is invariant — this is because every time entries amn, apq, ars are changed (with m + r = 2p and n + s = 2q), P increases or decreases by bmn −2bpq + brs, or 10 ((m −1) + (r −1) −2(p −1)) + (n + s −2q) = 0.
(For example, if entries a35, a46, a57 are changed then P changes by ±(35 −2 · 46 + 57) = 0.) At first, P = P10 i,j=1 bijbij; at the end, the entries aij equal the bij in some order, and we now have P = P10 i,j=1 aijbij. By the rearrangement inequality, this is at least P10 i,j=1 bijbij, with equality only when each aij = bij.
But we know equality does occur since P is invariant. Therefore the aij do indeed equal the bij in the same order, and thus the entries 1, 2, . . . , 100 appear in their original order.
Problem 10.7 Quadrilateral ABCD is inscribed in circle ω cen-tered at O. The bisector of ∠ABD meets AD and ω at points K and M respectively. The bisector of ∠CBD meets CD and ω at points L and N respectively.
Suppose that KL ∥MN.
Prove that the circumcircle of triangle MON goes through the midpoint of BD.
Solution: All angles are directed modulo 180◦. Let P, Q, R be the midpoints of DB, DA, DC respectively. Points M and N are the midpoints of arcs AD and DC, respectively; so M, Q, O and N, R, O are collinear, so that ∠MON = ∠QOR. But since ∠DQO = ∠DRO = 90◦, DQOR is cyclic and ∠QOR = 180◦−∠RDQ = 180◦−∠CDA = ∠ABC.
50 St. Petersburg City Mathematical Olympiad (Russia) In addition, ∠QPR = ∠QPD + ∠DPR = ∠ABD + ∠DBC = ∠ABC. Then to prove our claim, it suffices to show that △PQM ∼ △PRN (with the same orientation) since then ∠MPN = ∠QPR−∠QPM +∠RPN = ∠QPR = ∠ABC = ∠MON so that MPON would be cyclic. (Alternatively, it is possible that triangles PQM and PRN are both degenerate.) Now, let a = AB, b = BC, c = CD, D = DA, e = AC, and f = BD. Suppose line MN hits lines AD and CD at E and F, respectively. Then ∠DEF = 1 2 d DN + d AM = 1 2 d NC + d MD = ∠EFD, so DE = DF. Then since KL ∥EF, we have DK = DL. And by the Angle Bisector Theorem on triangles ABD and CBD, DK = d · f a+f and DL = c · f b+f , so that d(b + f) = c(a + f) (c −d)f = bd −ac.
(1) If c = d then we must have bd = ac, so a = b. But then BD is a diameter of ω so P = O and the claim is obvious.
Otherwise c −d, bd −ac ̸= 0, and f = bd−ac c−d .
Now we prove that △PQM ∼△PRN. Observe that ∠PQM = ∠PQD + ∠DQM = ∠BAD + 90◦ = ∠BCD + 90◦= ∠PRD + ∠DRN = ∠PRN, and also note that ∠PQM and ∠PRN are both obtuse.
So, we need only prove that PQ QM = PR RN ⇐ ⇒ BA 2 AQ tan ∠MAQ = BC 2 CR tan ∠RCN ⇐ ⇒ BA 2 AD 2 tan ∠MCD = BC 2 CD 2 tan ∠DAN 1999 Regional Contests: Problems and Solutions 51 ⇐ ⇒ a d tan ∠ACM = b c tan ∠NAC .
(2) Since lines CM, AN are angle bisectors of ∠ACD, ∠DAC, they intersect at the incenter I of triangle ACD. Also, let T be the point where the incircle of triangle ACD hits AC. Then tan ∠ACM = IT T C = IT 1 2 (e+c−d) and tan ∠NAC = IT T A = IT 1 2 (e+d−c), so tan ∠ACM tan ∠NAC = e+d−c e+c−d. Thus (2) is equivalent to ac(e + c −d) = bd(e + d −c) ⇐ ⇒e = (ac + bd)(c −d) bd −ac .
But by Ptolemy’s Theorem and (1), we have e = ac + bd f = (ac + bd)(c −d) bd −ac , as desired.
Therefore triangle PQM is similar to triangle PRN, ∠MPN = ∠MON, and MPON is indeed cyclic!
Problem 11.1 There are 150 red, 150 blue, and 150 green balls flying under the big top in the circus. There are exactly 13 green balls inside every blue one, and exactly 5 blue balls and 19 green balls inside every red one. (A ball is considered to be “inside” another ball even if it is not immediately inside it; for example, if a green ball is inside a blue ball and that blue ball is inside a red ball, then the green ball is also inside the red ball.) Prove that some green ball is not contained in any of the other 449 balls.
Solution: Suppose by way of contradiction that every green ball were in some other ball. Notice that no red ball is inside a blue ball, since then the blue ball would contain at least 19 green balls.
Look at the red balls that are not inside any other red balls — say there are m of them, and throw them away along with all the balls they contain. Then we throw all the red balls and anything inside a red ball, including 19m green balls. Also, since no red ball is inside a blue ball, there are still exactly 13 green balls inside each of the remaining blue balls.
Now look at the blue balls left that are not inside any other blue balls — say there are n of them, and throw them away along with all the balls they contain. Then we throw away all the remaining blue balls and the 13n more green balls inside.
52 St. Petersburg City Mathematical Olympiad (Russia) At this point, all the green balls are gone and hence we must have 150 = 19m + 13n. Taking this equation modulo 13, we find that 7 ≡6m ⇒m ≡12 (mod 13). Then m ≥12 and 150 = 19m + 13n ≥ 19·12 = 228, a contradiction. Thus our original assumption was false, and some green ball is contained in no other ball.
Problem 11.3 Let {an} be an arithmetic sequence of positive integers. For every n, let pn be the largest prime divisor of an. Prove that the sequence { an pn } is unbounded.
Solution: Let d be the common difference of the arithmetic sequence and N be an arbitrary number. Choose primes p < q bigger than N and relatively prime to d; then there is some term ak divisible by pq. Since pk ≥q > p, we have p | ak pk so that ak pk ≥p > N.
Thus for any N, there exists k such that ak pk > N, so the sequence { an pn } is unbounded.
Problem 11.4 All positive integers not exceeding 100 are written on both sides of 50 cards (each number is written exactly once). The cards are put on a table so that Vasya only knows the numbers on the top side of each card. Vasya can choose several cards, turn them upside down, and then find the sum of all 50 numbers now on top.
What is the maximum sum Vasya can be sure to obtain or beat?
Solution: The answer is 2525 = 1 2(1 + 2 + 3 + · · · + 100). Vasya can always obtain or beat this: if the 50 numbers on top add to this or more, he is done; otherwise, if they add to less, Vasya can flip all of them.
Sadly, this might be the best Vasya can do. Suppose that Vasya has horrible luck and the numbers on top are 26, 27, 28, . . . , 75, with sum 2525; and that the numbers on the cards he flips over are, in order, 1, 2, . . . , 25, 76, 77, . . . , 100 (although of course he might not flip over all of the cards). If he flips over 0 through 25 cards, his sum decreases; and if he flips over more, his sum is at most 1+2+· · ·+25+76+77+· · ·+100 = 2525. Either way, Vasya cannot obtain a sum of more than 2525, as claimed.
1999 Regional Contests: Problems and Solutions 53 Problem 11.5 Two players play the following game. They in turn write on a blackboard different divisors of 100! (except 1). A player loses if after his turn, the greatest common divisor of the all the numbers written becomes 1.
Which of the players has a winning strategy?
Solution: The second player has a winning strategy. Notice that every prime p < 100 divides an even number of factors of 100!: the factors it divides can be split into disjoint pairs (k, 97k) — or, if p = 97, into the pairs (k, 89k). (Note that none of these factors is 1, since 1 is not divisible by p.) If the first player writes down a prime p, the second player can write down any other number divisible by p; if the first player writes down a composite number, the second player can write down a prime p dividing that number. Either way, from now on the players can write down a new number q | 100! without losing if and only if it is divisible by p. Since there are an even number of such q, the second player will write down the last acceptable number and the first player will lose.
Problem 11.7 A connected graph G has 500 vertices, each with degree 1, 2, or 3. We call a black-and-white coloring of these vertices interesting if more than half of the vertices are white but no two white vertices are connected.
Prove that it is possible to choose several vertices of G so that in any interesting coloring, more than half of the chosen vertices are black.
Solution: We first give an algorithm to (temporarily) erase edges from G so that our graph consists of “chains” of vertices—sequences of vertices V1, . . . , Vn where each Vi is adjacent to Vi+1—and possibly one leftover vertex.
First, as long as G still contains any cycle erase an edge from that cycle; this eventually makes G a tree (a connected graph with no cycles). Look at the leaves (vertices with degree 1) that are not part of a chain yet. If all of them are adjacent to a degree-3 vertex, then we must have exactly four unchained vertices left with one central vertex adjacent to the other 3; remove one of the edges, and we are done.
Otherwise, one of the leaves is not adjacent to a degree-3 vertex.
Travel along the graph from this vertex until we reach a 54 St. Petersburg City Mathematical Olympiad (Russia) degree-3 vertex V , and erase the edge going into V. This creates a new chain, while leaving all the unchained vertices in a smaller, still-connected tree; we can then repeat the algorithm on this tree until we are finished.
Besides our lone vertex, every vertex is in an “odd chain” (a chain with an odd number of vertices) or an “even chain” (a chain with an even number of vertices). For our chosen vertices, in each odd chain pick one vertex adjacent to one of the ends. Even with all the original edges back in place, for any interesting coloring observe that in any of our chains at most every other vertex can be white. Thus in any even chain, at most half the vertices are white. Furthermore, if a chosen vertex is white, then in its odd chain there is at least one more black vertex than white; and if a chosen vertex is black, then there is at most one more white vertex than black.
Suppose there are b odd chains with a black chosen vertex, and w odd chains with a white chosen vertex. If there is a lone vertex, there are at most 1 + b −w more white vertices than black in our graph so that 1 + b −w > 0. But we know 1 + b −w must be even since 1+b+w ≡500 (mod 2). Then 1+b−w ≥2 and b−w ≥1, implying that we have more black chosen vertices than white.
And if there is no lone vertex, then there are at most b −w more white vertices than black in our graph. Thus b −w > 0 and we still have more black chosen vertices than white. Therefore either way, for every interesting coloring we have more black chosen vertices than white, as desired.
Problem 11.8 Three conjurers show a trick. They give a spectator a pack of cards with numbers 1, 2, . . . , 2n + 1 (n > 6). The spectator takes one card and arbitrarily distributes the rest evenly between the first and the second conjurers. Without communicating with each other, these conjurers study their cards, each chooses an ordered pair of their cards, and gives these pairs to the third conjurer. The third conjurer studies these four cards and announces which card is taken by the spectator. Explain how such a trick can be done.
Solution: We will have each of the first two conjurers use their ordered pairs to communicate the sums of their card values modulo 2n+1. With this information, the third conjurer can simply subtract 1999 Regional Contests: Problems and Solutions 55 these two sums from 1 + 2 + · · · + (2n + 1) = (2n + 1)(n + 1) ≡ 0 (mod 2n + 1) to determine the remaining card.
From now on, all entries of ordered pairs are taken modulo 2n + 1; also, let (a, b)k denote the ordered pair (a + k, b + k) and say its “difference” is b −a (taken modulo 2n + 1 between 1 and 2n).
Let (0, 2n)k, (0, 1)k, (n, 2)k, (n, 2n−1)k, (4, n+1)k, and (2n−3, n+ 1)k all represent the sum k (mod 2n + 1). These pairs’ differences are 2n, 1, n + 3, n −1, n −3, n + 5; because n > 5, these differences are all distinct.
If n is odd then let (1, 2n)k, (2, 2n −1)k, . . ., (n −1, n + 2)k also represent the sum k (mod 2n+1). These pairs’ differences are all odd: 2n −1, 2n −3, . . . , 3. Furthermore, they are all different from the one odd difference, 1, that we found in the last paragraph.
Similarly, if n is even then let (2n, 1)k, (2n−1, 2)k, . . ., (n+2, n−1)k represent the sum k (mod 2n+1). These pairs’ differences are all even: 2, 4, . . . , 2n−2; and they are all different from the one even difference, 2n, that we found two paragraphs ago.
Note that if two of the assigned pairs (a1, b1)k1 and (a2, b2)k2 are equal, then their differences must be equal and we must have b1 −a1 ≡b2 −a2 (mod 2n + 1).
But because we found that the differences b −a are distinct, we must have (a1, b1) = (a2, b2) and therefore k1 = k2 as well. Thus any pair (a, b) is assigned to at most one sum, and our choices are well-defined.
Now, say that one of the first two conjurers has cards whose values sum to k (mod 2n + 1); suppose by way of contradiction that he could not give any pair (a, b)k described above. Then, letting Sk = {s+k | s ∈S}, he has at most one card from each of the three triples {0, 1, 2n}k, {2, n, 2n−1}k, {4, n+1, 2n−3}k; and he has at most one card from each of the n−4 pairs {3, 2n−2}k, {5, 2n−4}k, {6, 2n−5}k, . . ., {n −1, n + 2}k. But these sets partition all of {0, 1, 2, . . . , 2n}, so the magician must then have at most 3 + (n −4) = n −1 cards—a contradiction. Thus our assumption was false, and both conjurers can indeed communicate the desired sums. This completes the proof.
2 2000 National Contests: Problems 57 58 Belarus 2.1 Belarus Problem 1 Find all pairs of positive integers (m, n) which satisfy the equality (m −n)2(n2 −m) = 4m2n.
Problem 2 Let M be the intersection point of the diagonals AC and BD of a convex quadrilateral ABCD. The bisector of angle ACD hits ray BA at K. If MA · MC + MA · CD = MB · MD, prove that ∠BKC = ∠CDB.
Problem 3 An equilateral triangle of side n is divided into n2 equilateral triangles of side 1 by lines parallel to the sides of the triangle.
Each point that is a vertex of at least one of these unit triangles is labeled with a number; exactly one of these points is labeled with −1, all the others with 1’s.
On each move one can choose a line passing through the side of one of the small triangles and change the signs of the numbers at all the labeled points on this line. Determine all possible initial arrangements (the value of n and the position of the −1) from which one can obtain an arrangement of all 1’s using the described operations.
Problem 4 Tom and Jerry play the following game. They alternate putting pawns onto empty squares of an initially empty 25 × 25 chessboard, with Tom going first. A player wins if after his move, some four pawns are the vertices of a rectangle with sides parallel to the sides of the board. Which player has a winning strategy?
Problem 5 (a) We are given a rectangle ABCD. Prove that for any point X in the plane, some three of the segments XA, XB, XC, and XD could be the sides of a triangle.
(b) Is the previous statement for any parallelogram ABCD?
Problem 6 Pit and Bill play the following game.
Pit writes a number a on a blackboard, then Bill writes a number b, and finally Pit writes a number c. Can Pit choose his numbers so that the three polynomials x3 +ax2 +bx+c, x3 +bx2 +cx+a, and x3 +cx2 +ax+b have (a) a common real root?
(b) a common negative root?
2000 National Contests: Problems 59 Problem 7 How many pairs (n, q) satisfy {q2} = n!
2000 , where n is a positive integer and q is a nonintegral rational number between 0 and 2000?
Problem 8 Let n ≥5 be a positive integer. Define a sign-sequence to be a sequence of n numbers all equal to 1 or −1, and let a move consist of changing the signs of any five consecutive terms of a sign-sequence. Two sign-sequences are said to be similar if one of them can be obtained from the other with a finite number of moves. Find the maximum number m such that there exist m sign-sequences, no two of which are similar to each other.
Problem 9 A line ℓintersects the lateral sides and diagonals of a trapezoid. It is known that the portion of ℓbetween the lateral sides is divided by the diagonals into three equal parts. Does it follow that the line ℓis parallel to the bases of the trapezoid?
Problem 10 Nine points are marked on a plane, no three of them collinear. Each pair of marked points is connected with a segment. Is it possible to color each segment with one of twelve colors, such that the segments of each color form a triangle?
Problem 11 A vertex of a tetrahedron is called perfect if one could construct a triangle with the edges from this vertex as its sides. Find all n such that some tetrahedron has exactly n perfect vertices.
Problem 12 (a) Find all positive integers n such that the equation (aa)n = bb has at least one solution in integers a, b > 1.
(b) Find all positive integers a and b which satisfy (aa)5 = bb.
Problem 13 The sides of scalene triangle ABC have lengths a, b, and c, measured in meters. Its angles, measured in radians, are α, β, and γ. The set {a, b, c, α, β, γ} contains exactly n distinct elements.
Find the minimum possible value of n.
Problem 14 On a 5×7 board, we call two squares adjacent if they are distinct and share one common side. Find the minimum number of squares on the board that must be painted so that any unpainted square has exactly one adjacent square which is painted.
60 Belarus Problem 15 We are given triangle ABC with ∠C = 90◦. Let M be the midpoint of the hypotenuse AB, H be the foot of the altitude CH, and P be a point inside the triangle such that AP = AC. Prove that PM bisects angle BPH if and only if ∠A = 60◦.
Problem 16 Does there exist a function f : N →N such that f(f(n −1)) = f(n + 1) −f(n) for all n ≥2?
Problem 17 Five points A, B, C, D, E lie on a sphere of diameter 1, and that AB = BC = CD = DE = EA = ℓ. Prove that ℓ≤sin 72◦.
Problem 18 In a convex polyhedron with m triangular faces, exactly four edges meet at each vertex. Find the minimum possible value of m.
Problem 19 We call two lattice points (a1, b1) and (a2, b2) in the Cartesian plane connected if either (a1, b1) = (−a2, b2±1) or (a1, b1) = (a2±1, −b2). Prove that it is possible to construct an infinite sequence (m1, n1), (m2, n2), . . . of lattice points such that any two consecutive points of the sequence are connected, and each lattice point in the plane appears exactly once in this sequence.
Problem 20 In triangle ABC, a ̸= b where a = BC and b = AC.
Points E and F are on the sides AC and BC, respectively, such that AE = BF = ab a+b. Let M be the midpoint of AB, N be the midpoint of EF, and P be the intersection point of EF and the bisector of angle ACB. Find [CP MN] [ABC] .
Problem 21 In triangle ABC, let ma and mb be the lengths of the medians from the vertices A and B, respectively. Find all real λ such that ma + λ BC = mb + λ AC implies that BC = AC.
Problem 22 (a) Prove that {n √ 3} > 1 n √ 3 for every positive integer n, where {x} denotes the fractional part of x.
(b) Does there exist a constant c > 1 such that {n √ 3} > c n √ 3 for every positive integer n?
2000 National Contests: Problems 61 Problem 23 A graph has 15 vertices and n edges. Each edge of the graph is colored either red or blue such that no three vertices A, B, C are connected pairwise with edges of the same color. Determine the largest possible value of n.
Problem 24 Find all functions f, g, h : R →R such that f(x + y3) + g(x3 + y) = h(xy) for all x, y ∈R.
Problem 25 Let M = {1, 2, . . . , 40}. Find the smallest positive integer n for which it is possible to partition M into n disjoint subsets such that whenever a, b, and c (not necessarily distinct) are in the same subset, a ̸= b + c.
Problem 26 A positive integer is called monotonic if its digits in base 10, read from left to right, are in nondecreasing order. Prove that for each n ∈N, there exists an n-digit monotonic number which is a perfect square.
Problem 27 Given a pair (⃗ a,⃗ b) of vectors in the plane, a move consists of choosing a nonzero integer k and then changing (⃗ a,⃗ b) to either (i) (⃗ a+2k⃗ b,⃗ b) or (ii) (⃗ a,⃗ b+2k⃗ a). A game consists of applying a finite sequence of moves, alternating between moves of types (i) and (ii), to some initial pair of vectors.
(a) Is it possible to obtain the pair ((1, 0), (2, 1)) during a game with initial pair ((1, 0), (0, 1)), if our first move is of type (i)?
(b) Find all pairs ((a, b), (c, d)) that can be obtained during a game with initial pair ((1, 0), (0, 1)), where our first move can be of either type.
Problem 28 Prove that a3 x + b3 y + c3 z ≥(a + b + c)3 3(x + y + z) for all positive real numbers a, b, c, x, y, z.
Problem 29 Let P be the intersection point of the diagonals AC and BD of the convex quadrilateral ABCD in which AB = AC = BD. Let O and I be the circumcenter and incenter of triangle ABP, respectively.
Prove that if O ̸= I, then lines OI and CD are perpendicular.
62 Bulgaria 2.2 Bulgaria Problem 1 Let F = x3y + xy3 for some real numbers x and y. If x2 + xy + y2 = 1, (a) prove that F ≥−2; (b) find the greatest possible value of F.
Problem 2 A line ℓis drawn through the orthocenter of acute triangle ABC. Prove that the reflections of ℓacross the sides of the triangle are concurrent.
Problem 3 There are 2000 white balls in a box. There are also unlimited supplies of white, green, and red balls, initially outside the box. During each turn, we can replace two balls in the box with one or two balls as follows: two whites with a green, two reds with a green, two greens with a white and red, a white and green with a red, or a green and red with a white.
(a) After finitely many of the above operations there are three balls left in the box. Prove that at least one of them is a green ball.
(b) Is it possible after finitely many operations to have only one ball left in the box?
Problem 4 Solve the equation √x + 3 √x + 7 = 4 √x + 80 in real numbers.
Problem 5 The incircle of the isosceles triangle ABC touches the legs AC and BC at points M and N respectively. A line t is drawn tangent to minor arc d MN, intersecting NC and MC at points P and Q, respectively. Let T be the intersection point of lines AP and BQ.
(a) Prove that T lies on MN; (b) Prove that the sum of the areas of triangles ATQ and BTP is smallest when t is parallel to line AB.
Problem 6 We are given n ≥4 points in the plane such that the distance between any two of them is an integer. Prove that at least 1 6 of these distances are divisible by 3.
Problem 7 In triangle ABC, CH is an altitude, and cevians CM and CN bisect angles ACH and BCH, respectively. The circumcen-ter of triangle CMN coincides with the incenter of triangle ABC.
Prove that [ABC] = AN·BM 2 .
2000 National Contests: Problems 63 Problem 8 Let a1, a2, . . . be a sequence such that a1 = 43, a2 = 142, and an+1 = 3an + an−1 for all n ≥2. Prove that (a) an and an+1 are relatively prime for all n ≥1; (b) for every natural number m, there exist infinitely many natural numbers n such that an −1 and an+1 −1 are both divisible by m.
Problem 9 In convex quadrilateral ABCD, ∠BCD = ∠CDA.
The bisector of angle ABC intersects CD at point E. Prove that ∠AEB = 90◦if and only if AB = AD + BC.
Problem 10 Prove that for any two real numbers a and b there exists a real number c ∈(0, 1) such that ac + b + 1 c + 1 > 1 24.
Problem 11 Find all sets S of four distinct points in the plane such that if any two circles k1 and k2 have diameters whose endpoints are in S, then k1 and k2 intersect at a point in S.
Problem 12 In the coordinate plane, a set of 2000 points {(x1, y1), (x2, y2), . . . , (x2000, y2000)} is called good if 0 ≤xi ≤83, 0 ≤yi ≤1 for i = 1, 2, . . . , 2000 and xi ̸= xj when i ̸= j. Find the largest positive integer n such that, for any good set, some unit square contains at least n of the points in the set.
Problem 13 We are given the acute triangle ABC.
(a) Prove that there exist unique points A1, B1, and C1 on BC, CA, and AB, respectively, with the following property: If we project any two of the points onto the corresponding side, the midpoint of the projected segment is the third point.
(b) Prove that triangle A1B1C1 is similar to the triangle formed by the medians of triangle ABC.
Problem 14 Let p ≥3 be a prime number and a1, a2, . . . , ap−2 be a sequence of positive integers such that p does not divide either ak or ak k −1 for all k = 1, 2, . . . , p −2. Prove that the product of some terms of the sequence is congruent to 2 modulo p.
Problem 15 Find all polynomials P(x) with real coefficients such that P(x)P(x + 1) = P(x2) 64 Bulgaria for all real x.
Problem 16 Let D be the midpoint of base AB of the isosceles acute triangle ABC.
Choose a point E on AB, and let O be the circumcenter of triangle ACE. Prove that the line through D perpendicular to DO, the line through E perpendicular to BC, and the line through B parallel to AC are concurrent.
Problem 17 Let n be a positive integer.
A binary sequence of length n is a sequence of n integers, all equal to 0 or 1. Let A be the set of all such sequences, and let 0 ∈A be the sequence of all zeroes.
The sequence c = c1, c2, . . . , cn is called the sum a + b of a = a1, a2, . . . , an and b = b1, b2, . . . , bn if ci = 0 when ai = bi and ci = 1 when ai ̸= bi. Let f : A →A be a function with f(0) = 0 such that whenever the sequences a and b differ in exactly k terms, the sequences f(a) and f(b) also differ in exactly k terms. Prove that if a, b, and c are sequences from A such that a + b + c = 0, then f(a) + f(b) + f(c) = 0.
2000 National Contests: Problems 65 2.3 Canada Problem 1 At 12:00 noon, Anne, Beth, and Carmen begin running laps around a circular track of length three hundred meters, all starting from the same point on the track. Each jogger maintains a constant speed in one of the two possible directions for an indefinite period of time. Show that if Anne’s speed is different from the other two speeds, then at some later time Anne will be at least one hundred meters from each of the other runners. (Here, distance is measured along the shorter of the two arcs separating two runners.) Problem 2 Given a permutation (a1, a2, . . . , a100) of the integers 1901, 1902, . . . , 2000, we form the sequence of partial sums s1 = a1, s2 = a1 + a2, . . . , s100 = a1 + a2 + · · · + a100.
For how many such permutations will no terms of the corresponding sequence s1, s2, . . . , s100 be divisible by three?
Problem 3 Let a1, a2, . . . , a2000 be a sequence of integers each lying in the interval [−1000, 1000]. Suppose that P1000 i=1 ai = 1. Show that the terms in some nonempty subsequence of a1, a2, . . . , a2000 sum to zero.
Problem 4 Let ABCD be a quadrilateral with ∠CBD = 2∠ADB, ∠ABD = 2∠CDB, and AB = CB. Prove that AD = CD.
Problem 5 Suppose that the real numbers a1, a2, . . . , a100 satisfy (i) a1 ≥a2 ≥· · · ≥a100 ≥0, (ii) a1 + a2 ≤100, and (iii) a3 + a4 + · · · + a100 ≤100. Determine the maximum possible value of a2 1 +a2 2 +· · ·+a2 100, and find all possible sequences a1, a2, . . . , a100 for which this maximum is achieved.
66 China 2.4 China Problem 1 In triangle ABC, BC ≤CA ≤AB. Let R and r be the circumradius and inradius, respectively, of triangle ABC. As a function of ∠C, determine whether BC + CA −2R −2r is positive, negative, or zero.
Problem 2 Define the infinite sequence a1, a2, . . . recursively as follows: a1 = 0, a2 = 1, and an = 1 2nan−1 + 1 2n(n −1)an−2 + (−1)n 1 −n 2 for all n ≥3. Find an explicit formula for fn = an + 2 n 1 an−1 + 3 n 2 an−2 + · · · + n n n −1 a1.
Problem 3 A table tennis club wishes to organize a doubles tour-nament, a series of matches where in each match one pair of players competes against a pair of two different players.
Let a player’s match number for a tournament be the number of matches he or she participates in.
We are given a set A = {a1, a2, . . . , ak} of distinct positive integers all divisible by 6.
Find with proof the minimal number of players among whom we can schedule a doubles tournament such that (i) each participant belongs to at most 2 pairs; (ii) any two different pairs have at most 1 match against each other; (iii) if two participants belong to the same pair, they never compete against each other; and (iv) the set of the participants’ match numbers is exactly A.
Problem 4 We are given an integer n ≥2. For any ordered n-tuple of real numbers A = (a1, a2, . . . , an), let A’s domination score be the number of values k ∈{1, 2, . . . , n} such that ak > aj for all 1 ≤j ≤k.
Consider all permutations A = (a1, a2, . . . , an) of (1, 2, . . . , n) with domination score 2. Find with proof the arithmetic mean of the first elements a1 of these permutations.
Problem 5 Find all positive integers n such that there exist integers n1, n2, . . . , nk ≥3 with n = n1n2 · · · nk = 2 1 2k (n1−1)(n2−1)···(nk−1) −1.
2000 National Contests: Problems 67 Problem 6 An exam paper consists of 5 multiple-choice questions, each with 4 different choices; 2000 students take the test, and each student chooses exactly one answer per question.
Find the least possible value of n such that among any n of the students’ answer sheets, there exist 4 of them among which no two have exactly the same answers chosen.
68 Czech and Slovak Republics 2.5 Czech and Slovak Republics Problem 1 Determine all real numbers p for which the system of equations (x −y)2 = p2 x3 −y3 = 16 has precisely one solution in real numbers x, y.
Problem 2 Cevians AK, BL, and CM of triangle ABC intersect at a point U inside the triangle. Prove that if [AMU] = [KCU] = P and [MBU] = [CLU] = Q, then P = Q.
Problem 3 Find the smallest natural number k such that among any k distinct numbers from the set {1, 2, 3, . . . , 2000}, there exist two whose sum or difference equals 667.
Problem 4 Let P(x) be a quadratic polynomial with P(−2) = 0.
Find all roots of the equation P(x2 + 4x −7) = 0, given that the equation has at least one double root.
Problem 5 An isosceles trapezoid UV ST is given in which 3ST < 2UV. Show how to construct an isosceles triangle ABC with base AB so that the points B, C lie on the line V S; the point U lies on the line AB; and the point T is the centroid of the triangle ABC.
Problem 6 Show that 3 ra b + 3 r b a ≤ 3 s 2(a + b) 1 a + 1 b for all positive real numbers a and b, and determine when equality occurs.
Problem 7 Find all convex quadrilaterals ABCD for which there exists a point E inside the quadrilateral with the following property: Any line which passes through E and intersects sides AB and CD divides the quadrilateral ABCD into two parts of equal area.
2000 National Contests: Problems 69 Problem 8 An isosceles triangle ABC is given with base AB and altitude CD. Let E be the intersection of line AP with side BC, and let F be the intersection of line BP with side AC. Point P is chosen on CD so that the incircles of triangle ABP and quadrilateral PECF are congruent. Show that the incircles of the triangles ADP and BCP are also congruent.
Problem 9 In the plane are given 2000 congruent triangles of area 1, which are images of a single triangle under different translations.
Each of these triangles contains the centroids of all the others. Show that the area of the union of these triangles is less than 22 9 .
Problem 10 For which quadratic functions f(x) does there exist a quadratic function g(x) such that the equation g(f(x)) = 0 has four distinct roots in arithmetic progression, which are also real roots of the equation f(x)g(x) = 0?
Problem 11 Monica constructed a paper model of a triangular pyramid, the base of which was a right triangle. When she cut the model along the two legs of the base and along a median of one of the faces, upon unfolding it into the plane she obtained a square with side a. Determine the volume of the pyramid.
70 Estonia 2.6 Estonia Problem 1 Five real numbers are given such that, no matter which three of them we choose, the difference between the sum of these three numbers and the sum of the remaining two numbers is positive. Prove that the product of all these 10 differences (corresponding to all the possible triples of chosen numbers) is less than or equal to the product of the squares of these five numbers.
Problem 2 Prove that it is not possible to divide any set of 18 consecutive positive integers into two disjoint sets A and B, such that the product of elements in A equals the product of elements in B.
Problem 3 Let M, N, and K be the points of tangency of the incircle of triangle ABC with the sides of the triangle, and let Q be the center of the circle drawn through the midpoints of MN, NK, and KM. Prove that the incenter and circumcenter of triangle ABC are collinear with Q.
Problem 4 Find all functions f : Z+ →Z+ such that f(f(f(n))) + f(f(n)) + f(n) = 3n for all n ∈Z+.
Problem 5 In a triangle ABC we have AC ̸= BC. Take a point X in the interior of this triangle and let α = ∠A, β = ∠B, φ = ∠ACX, and ψ = ∠BCX. Prove that sin α sin β sin(α −β) = sin φ sin ψ sin(φ −ψ) if and only if X lies on the median of triangle ABC drawn from the vertex C.
Problem 6 We call an infinite sequence of positive integers an F-sequence if every term of this sequence (starting from the third term) equals the sum of the two preceding terms. Is it possible to decompose the set of all positive integers into (a) a finite; (b) an infinite number of F-sequences having no common members?
2000 National Contests: Problems 71 2.7 Georgia Problem 1 Do there exist positive integers x and y such that x3 +2xy +x+2y +1 and y3 +2xy +y +2x+1 are both perfect cubes?
Problem 2 The positive numbers a, b, c satisfy the inequality abc ≥ 1 64. Prove that a2 + b2 + c2 + 1 4(a + b + c) ≥1 4 √a + √ b + √c , and determine when equality occurs.
Problem 3 For any positive integer n, let a(n) denote the product of all its positive divisors.
(a) Prove that a(400) > 1019.
(b) Find all solutions of the equation a(n3) = n60 which do not exceed 100.
(c) Find all solutions of the equation a(n2) = (a(n))9 which do not exceed 2000.
Problem 4 From a point P lying outside a circle ω the tangents PA1 and PA2 are drawn. Let K be a point inside ω with PK = PA1.
Chords A1B1 and A2B2 are drawn in ω through K. Prove that B1B2 is a diameter of ω.
72 Hungary 2.8 Hungary Problem 1 Let H be a set consisting of positive and negative real numbers, with 2000 elements in all. Let N be the number of 4-element subsets of H whose elements have a negative product. How many negative elements should H have in order to maximize N?
Problem 2 In the scalene triangle ABC, let C1, A1, and B1 be the midpoints of sides AB, BC, and CA, respectively. Let B2 denote the midpoint of the broken-line path leading from A to B to C; define points A2 and C2 similarly. Prove that A1A2, B1B2, and C1C2 are concurrent.
Problem 3 Let an denote the closest integer to √n. Determine the value of 1 a1 + 1 a2 + · · · + 1 ak , where k = 1999 · 2000.
Problem 4 Find all positive primes p for which there exist positive integers n, x, y such that pn = x3 + y3.
Problem 5 In the tetrahedron ABCP, edges PA, PB, PC are pairwise perpendicular. Let S be a sphere such that the circumcircle of ABC is a great circle on S, and let XY be the diameter of S perpendicular to plane (ABC). Let S′ be the ellipsoid which passes through X and Y, is symmetric about axis XY , and intersects plane (ABC) in a circle of diameter XY/ √ 2. Prove that P lies on S′.
Problem 6 Is there a polynomial f of degree 1999 with integer coefficients, such that f(n), f(f(n)), f(f(f(n))), . . . are pairwise rela-tively prime for any integer n?
Problem 7 Let p be a polynomial with odd degree and integer coefficients. Prove that there are only finitely many pairs of integers a, b such that the points (a, p(a)) and (b, p(b)) are an integral distance apart.
Problem 8 The feet of the angle bisectors of triangle ABC are X, Y, and Z. The circumcircle of triangle XY Z cuts offthree segments from lines AB, BC, and CA. Prove that two of these segments’ lengths add up to the third segment’s length.
Problem 9 Let k and t be relatively prime integers greater than 1.
Starting from the permutation (1, 2, . . . , n) of the numbers 1, 2, . . . , n, 2000 National Contests: Problems 73 we may swap two numbers if their difference is either k or t. Prove that we can get any permutation of 1, 2, . . . , n with such steps if and only if n ≥k + t −1.
Problem 10 For any positive integer k, let e(k) denote the number of positive even divisors of k, and let o(k) denote the number of positive odd divisors of k. For all n ≥1, prove that Pn k=1 e(k) and Pn k=1 o(k) differ by at most n.
Problem 11 Given a triangle in the plane, show how to construct a point P inside the triangle which satisfies the following condition: if we drop perpendiculars from P to the sides of the triangle, the feet of the perpendiculars determine a triangle whose centroid is P.
Problem 12 Given a natural number k and more than 2k different integers, prove that a set S of k + 2 of these numbers can be selected such that for any positive integer m ≤k+2, all the m-element subsets of S have different sums of elements.
74 India 2.9 India Problem 1 Let ABC be a nonequilateral triangle. Suppose there is an interior point P such that the three cevians through P all have the same length λ where λ < min{AB, BC, CA}. Show that there is another interior point P ′ ̸= P such that the three cevians through P ′ also are of equal length.
Problem 2 Find all ordered pairs of prime numbers (p, q) such that p | 5q + 1 and q | 5p + 1.
Problem 3 Determine whether or not it is possible to label each vertex of a cube with a natural number such that two vertices are connected by an edge of the cube if and only if one of their corresponding labels a divides the other label b.
Problem 4 Let ABC be an acute triangle and let AD be the altitude from A. Let the internal bisectors of angles B and C meet AD at E and F, respectively. If BE = CF, prove that AB = AC.
Problem 5 Let n, k be positive integers such that n is not divisible by 3 and k ≥n. Prove that there exists an integer m which is divisible by n and whose digits have sum k.
Problem 6 Let a1 ≤a2 ≤· · · ≤an be n real numbers such that Pn j=1 aj = 0. Show that na1an + n X j=1 a2 j ≤0.
Problem 7 Let p > 3 be a prime number. Let E be the set of all (p −1)-tuples (x1, x2, . . . , xp−1) such that each xi ∈{0, 1, 2} and x1 + 2x2 + · · · + (p −1)xp−1 is divisible by p. Show that the number of elements in E is (3p−1 + p −1)/p.
Problem 8 Let m, n be positive integers such that m ≤n2/4 and every prime divisor of m is less than or equal to n. Show that m divides n!.
Problem 9 Determine whether there exists a sequence x1, x2, . . . of distinct positive real numbers such that xn+2 = √xn+1 −√xn for all positive integers n.
2000 National Contests: Problems 75 Problem 10 Let G be a graph with n ≥4 vertices and m edges. If m > n(√4n −3 + 1)/4 show that G has a 4-cycle.
Problem 11 Suppose f : Q →{0, 1} is a function with the property that for x, y ∈Q, if f(x) = f(y) then f(x) = f((x + y)/2) = f(y). If f(0) = 0 and f(1) = 1 show that f(q) = 1 for all rational numbers q greater than or equal to 1.
Problem 12 Let n ≥1 be an integer. A Catalan path from (0, 0) to (n, n) in the xy-plane is a sequence of unit moves either to the right (a move denoted by E) or upwards (a move denoted by N), where the path never crosses above the line y = x. A step in a Catalan path is the occurrence of two consecutive unit moves of the form EN. For 1 ≤s ≤n, show that the number of Catalan paths from (0, 0) to (n, n) that contain exactly s steps is 1 s n −1 s −1 n s −1 .
76 Iran 2.10 Iran Problem 1 Does there exist a natural number N which is a power of 2 whose digits (in base 10) can be permuted to form a different power of 2?
Problem 2 Call two circles in three-dimensional space pairwise tangent at a point P if they both pass through P and the lines tangent to each circle at P coincide. Three circles not all lying in a plane are pairwise tangent at three distinct points. Prove that there exists a sphere which passes through the three circles.
Problem 3 We are given a sequence c1, c2, . . . of natural numbers.
For any natural numbers m, n with 1 ≤m ≤Pn i=1 ci, we can choose natural numbers a1, a2, . . . , an such that m = n X i=1 ci ai .
For each n, find the maximum value of cn.
Problem 4 Circles C1 and C2 with centers O1 and O2, respectively, meet at points A and B. Lines O1B and O2B intersect C2 and C1 at F and E, respectively. The line parallel to EF through B meets C1 and C2 at M and N. Prove MN = AE + AF.
Problem 5 Two triangles ABC and A′B′C′ lie in three-dimensional space. The sides of triangle ABC have lengths greater than or equal to a, and the sides of triangle A′B′C′ have lengths greater than or equal to a′. Prove that one can select one vertex from triangle ABC and one vertex from triangle A′B′C′ such that the distance between them is at least q a2+a′2 3 .
Problem 6 The function f : N →N is defined recursively with f(1) = 1 and f(n + 1) = f(n + 2) if n = f(f(n) −n + 1) f(n) + 1 otherwise.
for all n ≥1.
(a) Prove that f(f(n) −n + 1) ∈{n, n + 1}.
(b) Find an explicit formula for f.
2000 National Contests: Problems 77 Problem 7 Let H equal {(x, y) | y > 0}, the upper half of the xy-plane. A semi-line is a curve in H which equals C ∩H for some circle C centered on the x-axis; in other words, it is any semi-circle whose “center” is on the x-axis, with its endpoints removed. Let the interior of a semi-line S denote the set of points in the interior of the corresponding circle C which are also in H. Given two semi-lines S1 and S2 which intersect at a point A, the tangents to S1 and S2 at A form an angle α. Then the bisector of S1 and S2 is the semi-line S3 passing through A such that the tangent to S3 at A bisects α and passes through the region common to the interiors of S1 and S2. Prove that if three different semi-lines intersect pairwise, then the bisectors of the three pairs of semi-lines pass through a common point.
Problem 8 Find all functions f : N →N such that (i) f(m) = 1 if and only if m = 1; (ii) if d = gcd(m, n), then f(mn) = f(m)f(n) f(d) ; and (iii) for every m ∈N, we have f 2000(m) = m.
Problem 9 On a circle are given n points, and nk +1 of the chords between these points are drawn where 2k + 1 < n. Prove that it is possible to select k+1 of the chords such that no two of them intersect.
Problem 10 The n tennis players, A1, A2, . . . , An, participate in a tournament.
Any two players play against each other at most once, and k ≤ n(n−1) 2 matches take place.
No draws occur, and in each match the winner adds 1 point to his tournament score while the loser adds 0. For nonnegative integers d1, d2, . . . , dn, prove that it is possible for A1, A2, . . . , An to obtain the tournament scores d1, d2, . . . , dn, respectively, if and only if the following conditions are satisfied: (i) Pn i=1 di = k.
(ii) For every subset X ⊆{A1, . . . , An}, the number of matches taking place among the players in X is at most P Aj∈X dj.
Problem 11 Isosceles triangles A3A1O2 and A1A2O3 are con-structed externally along the sides of a triangle A1A2A3 with O2A3 = O2A1 and O3A1 = O3A2. Let O1 be a point on the opposite side of line A2A3 as A1 with ∠O1A3A2 = 1 2∠A1O3A2 and ∠O1A2A3 = 1 2∠A1O2A3, and let T be the foot of the perpendicular from O1 to A2A3. Prove that A1O1 ⊥O2O3 and that A1O1 O2O3 = 2 O1T A2A3 .
78 Iran Problem 12 Given a circle Γ, a line d is drawn not intersecting Γ. M, N are two points varying on line d such that the circle with diameter MN is tangent to Γ. Prove that there exists a point P in the plane such that for any such segment MN, ∠MPN is constant.
Problem 13 Let n be a positive integer.
S is a set containing ordered n-tuples of nonnegative integers such that if (a1, . . . , an) ∈S, then every (b1, . . . , bn) for which bi ≤ai (1 ≤i ≤n) is also in S.
Let hm(S) be the number of n-tuples in S whose sum of components equals m. Show that for some N, hm is a polynomial in m for all m ≥N.
Problem 14 Suppose that a, b, c are real numbers such that for any positive real numbers x1, x2, . . . , xn, we have Pn i=1 xi n a · Pn i=1 x2 i n b · Pn i=1 x3 i n c ≥1.
Prove that the vector (a, b, c) can be represented in the form p(−2, 1, 0)+ q(1, −2, 1) for nonnegative real numbers p and q.
2000 National Contests: Problems 79 2.11 Ireland Problem 1 The sequence of real numbers a1, a2, . . . , an is called a weak arithmetic progression of length n if there exist real numbers c and d such that c + (k −1)d ≤ak < c + kd for k = 1, 2, . . . , n.
(a) Prove that if a1 < a2 < a3 then a1, a2, a3 is a weak arithmetic progression of length 3.
(b) Let A be a subset of {0, 1, 2, 3, . . . , 999} with at least 730 mem-bers. Prove that some ten elements of A form a weak arithmetic progression of length 10.
Problem 2 Let x ≥0, y ≥0 be real numbers with x+y = 2. Prove that x2y2(x2 + y2) ≤2.
Problem 3 For each positive integer n, determine with proof all positive integers m such that there exist positive integers x1 < x2 < · · · < xn with 1 x1 + 2 x2 + 3 x3 + · · · + n xn = m.
Problem 4 Prove that in each set of ten consecutive integers there is one which is relatively prime to each of the other integers.
Problem 5 Let p(x) = a0 + a1x + · · · + anxn be a polynomial with non-negative real coefficients.
Suppose that p(4) = 2 and that p(16) = 8. Prove that p(8) ≤4 and find with proof all such polynomials with p(8) = 4.
80 Israel 2.12 Israel Problem 1 Define f(n) = n!. Let a = 0.f(1)f(2)f(3) . . . .
In other words, to obtain the decimal representation of a write the decimal representations of f(1), f(2), f(3), . . . in a row. Is a rational?
Problem 2 ABC is a triangle whose vertices are lattice points.
Two of its sides have lengths which belong to the set { √ 17, √ 1999, √ 2000}. What is the maximum possible area of triangle ABC?
Problem 3 (a) Do there exist three positive integers a, b, d such that a3 + b3 a3 + d3 = 2000 1999?
(b) Do there exist four positive integers a, b, c, d such that a3 + b3 c3 + d3 = 2000 999 ?
Problem 4 The points A, B, C, D, E, F lie on a circle, and the lines AD, BE, CF concur. Let P, Q, R be the midpoints of AD, BE, CF, respectively. Two chords AG, AH are drawn such that AG ∥BE and AH ∥CF. Prove that triangles PQR and DGH are similar.
Problem 5 A square ABCD is given. A triangulation of the square is a partition of the square into triangles such that any two triangles are either disjoint, share only a common vertex, or share only a common side. A good triangulation of the square is a triangulation in which all the triangles are acute.
(a) Give an example of a good triangulation of the square.
(b) What is the minimal number of triangles required for a good triangulation?
2000 National Contests: Problems 81 2.13 Italy Problem 1 Three odd numbers a < b < c are called consecutive if c −b = b −a = 2. A positive integer is called special if its digits in base 10 are all equal and if it is the sum of the squares of three consecutive odd integers.
(a) Determine all special numbers with 4 digits.
(b) Are there any special numbers with 2000 digits?
Problem 2 Let ABCD be a convex quadrilateral, and write α = ∠DAB; β = ∠ADB; γ = ∠ACB; δ = ∠DBC; and ϵ = ∠DBA.
Assuming that α < 90◦, β + γ = 90◦, and δ + 2ϵ = 180◦, prove that (DB + BC)2 = AD2 + AC2.
Problem 3 Given a fixed integer n > 1, Alberto and Barbara play the following game, starting with the first step and then alternating between the second and third: • Alberto chooses a positive integer.
• Barbara picks an integer greater than 1 which is a multiple or divisor of Alberto’s number, possibly choosing Alberto’s number itself.
• Alberto adds or subtracts 1 from Barbara’s number.
Barbara wins if she succeeds in picking n by her fiftieth move. For which values of n does she have a winning strategy?
Problem 4 Let p(x) be a polynomial with integer coefficients such that p(0) = 0 and 0 ≤p(1) ≤107, and such that there exist integers a, b satisfying p(a) = 1999 and p(b) = 2001. Determine the possible values of p(1).
82 Japan 2.14 Japan Problem 1 Let O be the origin (0, 0) and A be the point (0, 1 2) in the coordinate plane. Prove there is no finite sequence of points P1, P2, . . . , Pn in the plane, each of whose x- and y- coordinates are both rational numbers, such that OP1 = P1P2 = P2P3 = · · · = Pn−1Pn = PnA = 1.
Problem 2 We shuffle a line of cards labeled a1, a2, . . . , a3n from left to right by rearranging the cards into the new order a3, a6, . . . , a3n, a2, a5, . . . , a3n−1, a1, a4, · · · , a3n−2.
For example, if six cards are labeled 1, 2, . . . , 6 from left to right, then shuffling them twice changes their order as follows: 1, 2, 3, 4, 5, 6 − →3, 6, 2, 5, 1, 4 − →2, 4, 6, 1, 3, 5.
Starting with 192 cards labeled 1, 2, . . . , 192 from left to right, is it possible to obtain the order 192, 191, . . . , 1 after a finite number of shuffles?
Problem 3 In the plane are given distinct points A, B, C, P, Q, no three of which are collinear. Prove that AB + BC + CA + PQ < AP + AQ + BP + BQ + CP + CQ.
Problem 4 Given a natural number n ≥3, prove that there exists a set An with the following two properties: (i) An consists of n distinct natural numbers.
(ii) For any a ∈An, the product of all the other elements in An has remainder 1 when divided by a.
Problem 5 We are given finitely many lines in the plane. Let an intersection point be a point where at least two of these lines meet, and let a good intersection point be a point where exactly two of these lines meet. Given that there are at least two intersection points, find the minimum number of good intersection points.
2000 National Contests: Problems 83 2.15 Korea Problem 1 Show that given any prime p, there exist integers x, y, z, w satisfying x2 + y2 + z2 −wp = 0 and 0 < w < p.
Problem 2 Find all functions f : R →R satisfying f(x2 −y2) = (x −y) (f(x) + f(y)) for all x, y ∈R.
Problem 3 For a quadrilateral ABCD inscribed in a circle with center O, let P, Q, R, S be the intersections of the exterior angle bisectors of ∠ABD and ∠ADB, ∠DAB and ∠DBA, ∠ACD and ∠ADC, ∠DAC and ∠DCA, respectively. Show that the four points P, Q, R, S are concyclic.
Problem 4 Let p be a prime number such that p ≡1 (mod 4).
Evaluate p−1 X k=1 2k2 p −2 k2 p .
Problem 5 Consider the following L-shaped figures, each made of four unit squares: Let m and n be integers greater than 1.
Prove that an m × n rectangular region can be tiled with such figures if and only if mn is a multiple of 8.
Problem 6 The real numbers a, b, c, x, y, z satisfy a ≥b ≥c > 0 and x ≥y ≥z > 0. Prove that a2x2 (by + cz)(bz + cy) + b2y2 (cz + ax)(cx + az) + c2z2 (ax + by)(ay + bx) is at least 3 4.
84 Lithuania 2.16 Lithuania Problem 1 In the triangle ABC, D is the midpoint of side AB.
Point E divides BC in the ratio BE : EC = 2 : 1. Given that ∠ADC = ∠BAE, determine ∠BAC.
Problem 2 A competition consisting of several tests has been organized for the three pilots K, L, and M, including a reaction-time test and a running test. In each test, no ties can occur; the first-place pilot in the test is awarded A points, the second-place pilot B points, and the third-place pilot C points for some fixed positive integers A > B > C. During the competition, K scores 22 points, and L and M each gather 9 points. If L won the reaction-time test, who took second place in the running test?
Problem 3 Find all functions f : R →R which satisfy the equality (x + y)(f(x) −f(y)) = f(x2) −f(y2) for all x, y ∈R.
Problem 4 Prove that infinitely many 4-tuples (x, y, z, u) of posi-tive integers satisfy the equation x2 + y2 + z2 + u2 = xyzu + 6.
2000 National Contests: Problems 85 2.17 Mongolia Problem 1 Let rad(1) = 1, and for k > 1 let rad(k) equal the product of the prime divisors of k. A sequence of natural numbers a1, a2, . . . with arbitrary first term a1 is defined recursively by the relation an+1 = an+rad(an). Show that for any positive integer N, the sequence a1, a2, . . . contains some N consecutive terms in arithmetic progression.
Problem 2 The circles ω1, ω2, ω3 in the plane are pairwise exter-nally tangent to each other. Let P1 be the point of tangency between circles ω1 and ω3, and let P2 be the point of tangency between circles ω2 and ω3. A and B, both different from P1 and P2, are points on ω3 such that AB is a diameter of ω3. Line AP1 intersects ω1 again at X, line BP2 intersects ω2 again at Y, and lines AP2 and BP1 intersect at Z. Prove that X, Y, and Z are collinear.
Problem 3 A function f : R →R satisfies the following conditions: (i) |f(a) −f(b)| ≤|a −b| for any real numbers a, b ∈R.
(ii) f(f(f(0))) = 0.
Prove that f(0) = 0.
Problem 4 Given a natural number n, find the number of quadru-ples (x, y, u, v) of natural numbers such that the eight numbers x, y, u, v, v + x −y, x + y −u, u + v −y, and v + x −u are all integers between 1 and n inclusive.
Problem 5 The bisectors of angles A, B, C of a triangle ABC intersect its sides at points A1, B1, C1. Prove that if the quadrilateral BA1B1C1 is cyclic, then BC AC + AB = AC AB + BC − AB BC + AC .
Problem 6 Which integers can be represented in the form (x+y+z)2 xyz where x, y, and z are positive integers?
Problem 7 In a country with n towns the cost of travel from the i-th town to the j-th town is xij. Suppose that the total cost of any route passing through each town exactly once and ending at its starting point does not depend on which route is chosen. Prove that there exist numbers a1, . . . , an and b1, . . . , bn such that xij = ai + bj for all integers i, j with 1 ≤i < j ≤n.
86 Poland 2.18 Poland Problem 1 Given an integer n ≥2 find the number of solutions of the system of equations x1 + x2 n = 4xn x2 + x2 1 = 4x1 .
.
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xn + x2 n−1 = 4xn−1 in nonnegative reals x1, x2, . . . , xn.
Problem 2 The sides AC and BC of a triangle ABC have equal length. Let P be a point inside triangle ABC such that ∠PAB = ∠PBC and let M be the midpoint of AB. Prove that ∠APM + ∠BPC = 180◦.
Problem 3 A sequence p1, p2, . . . of prime numbers satisfies the following condition: for n ≥3, pn is the greatest prime divisor of pn−1 + pn−2 + 2000. Prove that the sequence is bounded.
Problem 4 For an integer n ≥3 consider a pyramid with vertex S and the regular n-gon A1A2 . . . An as a base, such that all the angles between lateral edges and the base equal 60◦. Points B2, B3, . . . lie on A2S, A3S, . . . , AnS, respectively, such that A1B2 + B2B3 + · · · + Bn−1Bn + BnA1 < 2A1S. For which n is this possible?
Problem 5 Given a natural number n ≥2, find the smallest integer k with the following property: Every set consisting of k cells of an n × n table contains a nonempty subset S such that in every row and in every column of the table, there is an even number of cells belonging to S.
Problem 6 Let P be a polynomial of odd degree satisfying the identity P(x2 −1) = P(x)2 −1.
Prove that P(x) = x for all real x.
2000 National Contests: Problems 87 2.19 Romania Problem 1 The sequence x1, x2, . . . is defined recursively by setting x1 = 3 and setting xn+1 = ⌊xn √ 2⌋for every n ≥1. Find all n for which xn, xn+1, and xn+2 are in arithmetic progression.
Problem 2 Two nonzero complex numbers a and b satisfy a · 2|a| + b · 2|b| = (a + b) · 2|a+b|.
Prove that a6 = b6.
Problem 3 Let f be a third-degree polynomial with rational coef-ficients, having roots x1, x2, and x3. Prove that if there exist nonzero rational numbers a and b such that ax1 + bx2 is rational, then x3 is also a rational number.
Problem 4 A function f : R2 →R is called olympic if it has the following property: given n ≥3 distinct points A1, A2, . . . , An ∈R2, if f(A1) = f(A2) = · · · = f(An) then the points A1, A2, . . . , An are the vertices of a convex polygon. Let P ∈C[X] be a non-constant polynomial. Prove that the function f : R2 →R, defined by f(x, y) = |P(x + iy)|, is olympic if and only if all the roots of P are equal.
Problem 5 Let n ≥2 be a positive integer. Find the number of functions f : {1, 2, . . . , n} →{1, 2, 3, 4, 5} which have the following property: |f(k + 1) −f(k)| ≥3 for k = 1, 2, . . . , n −1.
Problem 6 Let n ≥1 be a positive integer and x1, x2, . . . , xn be real numbers such that |xk+1 −xk| ≤1 for k = 1, 2, . . . , n −1. Show that n X k=1 |xk| − n X k=1 xk ≤n2 −1 4 .
Problem 7 Let n, k be arbitrary positive integers. Show that there exist positive integers a1 > a2 > a3 > a4 > a5 > k such that n = ± a1 3 ± a2 3 ± a3 3 ± a4 3 ± a5 3 , where a 3 = a(a−1)(a−2) 6 .
88 Romania Problem 8 Let P1P2 · · · Pn be a convex polygon in the plane.
Assume that for any pair of vertices Pi, Pj, there exists a vertex V of the polygon such that ∠PiV Pj = 60◦. Show that n = 3.
Problem 9 Show that there exist infinitely many 4-tuples of posi-tive integers (x, y, z, t) such that the four numbers’ greatest common divisor is 1 and such that x3 + y3 + z2 = t4.
Problem 10 Consider the following figure, made of three unit squares: Determine all pairs m, n of positive integers such that a m × n rectangle can be tiled with such pieces.
Problem 11 Find the least positive integer n such that for all odd integers a, 22000 is a divisor of an −1.
Problem 12 Let ABC be an acute triangle and let M be the midpoint of segment BC. Consider the interior point N such that ∠ABN = ∠BAM and ∠ACN = ∠CAM. Prove that ∠BAN = ∠CAM.
Problem 13 Let S be the set of interior points of a unit sphere, and let C be the set of interior points of a unit circle. Find, with proof, whether there exists a function f : S →C such that the distance between f(A) and f(B) is greater than or equal to AB for all points A and B in S.
Problem 14 Let n ≥3 be an odd integer and m ≥n2 −n + 1 be an integer. The sequence of polygons P1, P2, . . . , Pm is defined as follows: (i) P1 is a regular polygon with n vertices.
(ii) For k > 1, Pk is the regular polygon whose vertices are the midpoints of the sides of Pk−1.
Find, with proof, the maximum number of colors which can be used such that for every coloring of the vertices of these polygons, one can find four vertices A, B, C, D which have the same color and form an isosceles trapezoid.
2000 National Contests: Problems 89 Problem 15 Prove that if p and q are monic polynomials with complex coefficients such that p(p(x)) = q(q(x)), then p(x) and q(x) are equal.
90 Russia 2.20 Russia Problem 1 Sasha tries to determine some positive integer X ≤100.
He can choose any two positive integers M and N that are less than 100 and ask the question, “What is the greatest common divisor of the numbers X + M and N?” Prove that Sasha can determine the value of X after 7 questions.
Problem 2 Let I be the center of the incircle ω of an acute-angled triangle ABC. The circle ω1 with center K passes through the points A, I, C and intersects sides AB and BC at points M and N. Let L be the reflection of K across line MN. Prove that BL ⊥AC.
Problem 3 There are several cities in a state and a set of roads, each road connecting two cities. It is known that at least 3 roads go out of every city. Prove that there exists a cyclic path (that is, a path where the last road ends where the first road begins) such that the number of roads in the path is not divisible by 3.
Problem 4 Let x1, x2, . . . , xn be real numbers, satisfying the con-ditions −1 < x1 < x2 < · · · < xn < 1 and x13 1 + x13 2 + · · · + x13 n = x1 + x2 + · · · + xn.
Prove that x13 1 y1 + x13 2 y2 + · · · + x13 n yn < x1y1 + x2y2 + · · · + xnyn for any real numbers y1 < y2 < · · · < yn.
Problem 5 Let AA1 and CC1 be the altitudes of an acute-angled nonisosceles triangle ABC. The bisector of the acute angle between lines AA1 and CC1 intersects sides AB and BC at P and Q, respec-tively. Let H be the orthocenter of triangle ABC and let M be the midpoint of AC; and let the bisector of ∠ABC intersect HM at R.
Prove that PBQR is cyclic.
Problem 6 Five stones which appear identical all have different weights; Oleg knows the weight of each stone. Given any stone x, let m(x) denote its weight. Dmitrii tries to determine the order of the weights of the stones. He is allowed to choose any three stones A, B, C and ask Oleg the question, “Is it true that m(A) < m(B) < m(C)?” Oleg then responds “yes” or “no.” Can Dmitrii determine the order of the weights with at most nine questions?
2000 National Contests: Problems 91 Problem 7 Find all functions f : R →R that satisfy the inequality f(x + y) + f(y + z) + f(z + x) ≥3f(x + 2y + 3z) for all x, y, z ∈R.
Problem 8 Prove that the set of all positive integers can be partitioned into 100 nonempty subsets such that if three positive integers a, b, c satisfy a + 99b = c, then two of them belong to the same subset.
Problem 9 Let ABCDE be a convex pentagon on the coordinate plane. Each of its vertices are lattice points. The five diagonals of ABCDE form a convex pentagon A1B1C1D1E1 inside of ABCDE.
Prove that this smaller pentagon contains a lattice point on its boundary or within its interior.
Problem 10 Let a1, a2, . . . , an be a sequence of nonnegative real numbers. For 1 ≤k ≤n, let mk = max 1≤i≤k ak−i+1 + ak−i+2 + · · · + ak i .
Prove that for any α > 0, the number of integers k which satisfy mk > α is less than a1+a2+···+an α .
Problem 11 Let a1, a2, a3, . . . be a sequence with a1 = 1 satisfying the recursion an+1 = an −2 if an −2 ̸∈{a1, a2, . . . , an} and an −2 > 0 an + 3 otherwise.
Prove that for every positive integer k, we have an = k2 = an−1 + 3 for some n.
Problem 12 There are black and white checkers on some squares of a 2n×2n board, with at most one checker on each square. First, we remove every black checker that is in the same column as any white checker. Next, we remove every white checker that is in the same row as any remaining black checker. Prove that for some color, at most n2 checkers of this color remain.
Problem 13 Let E be a point on the median CD of triangle ABC.
Let S1 be the circle passing through E and tangent to line AB at A, intersecting side AC again at M; let S2 be the circle passing through E and tangent to line AB at B, intersecting side BC again at N.
92 Russia Prove that the circumcircle of triangle CMN is tangent to circles S1 and S2.
Problem 14 One hundred positive integers, with no common divi-sor greater than one, are arranged in a circle. To any number, we can add the greatest common divisor of its neighboring numbers. Prove that using this operation, we can transform these numbers into a new set of pairwise coprime numbers.
Problem 15 M is a finite set of real numbers such that given three distinct elements from M, we can choose two of them whose sum also belongs to M. What is the largest number of elements that M can have?
Problem 16 A positive integer n is called perfect if the sum of all its positive divisors, excluding n itself, equals n. For example, 6 is perfect since 6 = 1 + 2 + 3. Prove that (a) if a perfect number larger than 6 is divisible by 3, then it is also divisible by 9.
(b) if a perfect number larger than 28 is divisible by 7, then it is also divisible by 49.
Problem 17 Circles ω1 and ω2 are internally tangent at N. The chords BA and BC of ω1 are tangent to ω2 at K and M, respectively.
Let Q and P be the midpoints of the arcs AB and BC not containing the point N.
Let the circumcircles of triangles BQK and BPM intersect at B and B1. Prove that BPB1Q is a parallelogram.
Problem 18 There is a finite set of congruent square cards, placed on a rectangular table with their sides parallel to the sides of the table. Each card is colored in one of k colors. For any k cards of different colors, it is possible to pierce some two of them with a single pin. Prove that all the cards can be pierced by 2k −2 pins.
Problem 19 Prove the inequality sinn(2x) + (sinn x −cosn x)2 ≤1.
Problem 20 The circle ω is inscribed in the quadrilateral ABCD, and O is the intersection point of the lines AB and CD. The circle ω1 is tangent to side BC at K and is tangent to lines AB and CD at points lying outside ABCD; the circle ω2 is tangent to side AD 2000 National Contests: Problems 93 at L and is also tangent to lines AB and CD at points lying outside ABCD. If O, K, L are collinear, prove that the midpoint of side BC, the midpoint of side AD, and the center of ω are collinear.
Problem 21 Every cell of a 100 × 100 board is colored in one of 4 colors so that there are exactly 25 cells of each color in every column and in every row. Prove that one can choose two columns and two rows so that the four cells where they intersect are colored in four different colors.
Problem 22 The non-zero real numbers a, b satisfy the equation a2b2(a2b2 + 4) = 2(a6 + b6).
Prove that a and b are not both rational.
Problem 23 In a country, each road either connects two towns or starts from a town and goes out of the country. Some of the roads are colored in one of three colors. For every town, exactly three of the roads that go out of this town are colored, and the colors of these roads are different. If exactly three of the colored roads go out of the country, prove that the colors of these roads are different.
Problem 24 Find the smallest odd integer n such that some n-gon (not necessarily convex) can be partitioned into parallelograms whose interiors do not overlap.
Problem 25 Two pirates divide their loot, consisting of two sacks of coins and one diamond. They decide to use the following rules.
On each turn, one pirate chooses a sack and takes 2m coins from it, keeping m for himself and putting the rest into the other sack. The pirates alternate taking turns until no more moves are possible; the pirate who makes the last move takes the diamond. For what initial amounts of coins can the first pirate guarantee that he will obtain the diamond?
Problem 26 The coefficients a and b of an equation x2 +ax+b = 0 and its roots c and d are four different numbers. Given a, b, c, d in some order, is it possible to determine which is a and which is b?
Problem 27 Do there exist coprime integers a, b, c > 1 such that 2a + 1 is divisible by b, 2b + 1 is divisible by c, and 2c + 1 is divisible by a?
94 Russia Problem 28 2n + 1 segments are marked on a line. Each of the segments intersects at least n other segments. Prove that one of these segments intersects all the other segments.
Problem 29 The circles S1 and S2 intersect at points M and N.
Let A and D be points on S1 and S2 such that lines AM and AN intersect S2 at B and C, lines DM and DN intersect S1 at E and F, and the triples A, E, F and D, B, C lie on opposite sides of line MN.
Prove that there is a fixed point O such that for any points A and D that satisfy the condition AB = DE, the quadrilateral AFCD is cyclic.
Problem 30 Let the set M consist of the 2000 numbers 101 + 1, 102 + 1, . . . , 102000 + 1. Prove that at least 99% of the elements of M are not prime.
Problem 31 There are 2 counterfeit coins among 5 coins that look identical. Both counterfeit coins have the same weight and the other three real coins have the same weight. The five coins do not all weight the same, but it is unknown whether the weight of each counterfeit coin is more or less than the weight of each real coin.
Find the minimal number of weighings needed to find at least one real coin, and describe how to do so. (The balance scale reports the difference between the weights of the objects in two pans.) Problem 32 Let ABCD be a parallelogram with ∠A = 60◦. Let O be the circumcenter of triangle ABD. Line AO intersects the external angle bisector of angle BCD at K. Find the value AO OK .
Problem 33 Find the smallest integer n such that an n × n square can be partitioned into 40 × 40 and 49 × 49 squares, with both types of squares present in the partition.
Problem 34 Prove that there exist 10 distinct real numbers a1, a2, . . . , a10 such that the equation (x −a1)(x −a2) · · · (x −a10) = (x + a1)(x + a2) · · · (x + a10) has exactly 5 different real roots.
Problem 35 We are given a cylindrical region in space, whose altitude is 1 and whose base has radius 1. Find the minimal number of balls of radius 1 needed to cover this region.
2000 National Contests: Problems 95 Problem 36 The sequence a1, a2, . . . , a2000 of real numbers satisfies the condition a3 1 + a3 2 + · · · + a3 n = (a1 + a2 + · · · + an)2 for all n, 1 ≤n ≤2000. Prove that every element of the sequence is an integer.
(The balance scale reports the difference between the weights of the objects in two pans.) Problem 37 The bisectors AD and CE of a triangle ABC intersect at I. Let ℓ1 be the reflection of line AB across line CE, and let ℓ2 be the reflection of line BC across line AD. If lines ℓ1 and ℓ2 intersect at K, prove that KI ⊥AC.
Problem 38 There are 2000 cities in a country, some pairs of which are connected by a direct airplane flight. For every city A the number of cities connected with A by direct flights equals 1, 2, 4, . . . , or 1024.
Let S(A) be the number of routes from A to other cities (different from A) with at most one intermediate landing. Prove that the sum of S(A) over all 2000 cities A cannot be equal to 10000.
Problem 39 A heap of balls consists of one thousand 10-gram balls and one thousand 9.9-gram balls. We wish to pick out two heaps of balls with equal numbers of balls in them but different total weights.
What is the minimal number of weighings needed to do this?
Problem 40 Let D be a point on side AB of triangle ABC. The circumcircle of triangle BCD intersects line AC at C and M, and the circumcircle of triangle CMN intersects line BC at C and N.
Let O be the center of the circumcircle of triangle CMN. Prove that OD ⊥AB.
Problem 41 Every cell of a 200 × 200 table is colored black or white. The difference between the number of black and white cells is 404. Prove that some 2 × 2 square contains an odd number of white cells.
Problem 42 Is there a function f : R →R such that |f(x + y) + sin x + sin y| < 2 for all x, y ∈R?
96 Russia Problem 43 For any integer a0 > 5, consider the sequence a0, a1, a2, . . . , where an+1 = a2 n −5 if an is odd an 2 if an is even for all n ≥0. Prove that this sequence is not bounded.
Problem 44 Let ℓa, ℓb, ℓc, and ℓd be the external angle bisectors of angles DAB, ABC, BCD, and CDA, respectively. The pairs of lines ℓa and ℓb, ℓb and ℓc, ℓc and ℓd, ℓd and ℓa intersect at points K, L, M, N, respectively.
Suppose that the perpendiculars to line AB passing through K, to line BC passing through L, and to line CD passing through M are concurrent. Prove that ABCD can be inscribed in a circle.
Problem 45 There are 2000 cities in a country, and each pair of cities is connected by either no roads or exactly one road. A cyclic path is a collection of roads such that each city is at the end of either 0 or 2 roads in the path. For every city, there at most N cyclic paths which both pass through this city and contain an odd number of roads. Prove that the country can be separated into 2N +2 republics such that any two cities from the same republic are not connected by a road.
Problem 46 Prove the inequality 1 √ 1 + x2 + 1 p 1 + y2 ≤ 2 √1 + xy for 0 ≤x, y ≤1.
Problem 47 The incircle of triangle ABC touches side AC at K.
A second circle S with the same center intersects all the sides of the triangle.
Let E and F be the intersection points on AB and BC closer to B; let B1 and B2 be the intersection points on AC with B1 closer to A. Finally, let P be the intersection point of segments B2E and B1F. Prove that points B, K, P are collinear.
Problem 48 Each of the numbers 1, 2, . . . , N is colored black or white. We are allowed to simultaneously change the colors of any three numbers in arithmetic progression. For which numbers N can we always make all the numbers white?
2000 National Contests: Problems 97 2.21 Taiwan Problem 1 Find all possible pairs (x, y) of positive integers such that yx2 = xy+2.
Problem 2 In an acute triangle ABC, AC > BC and M is the midpoint of AB. Let altitudes AP and BQ meet at H, and let lines AB and PQ meet at R. Prove that the two lines RH and CM are perpendicular.
Problem 3 Let S = {1, 2, . . . , 100}, and let P denote the family of all 49-element subsets T of S. Each set T in P is labeled with some number from S. Show that there exists a 50-element subset M of S such that for each x ∈M, the set M \ {x} is not labeled with x.
Problem 4 Let φ(k) denote the number of positive integers n satisfying gcd(n, k) = 1 and n ≤k. Suppose that φ(5m −1) = 5n −1 for some positive integers m, n. Prove that gcd(m, n) > 1.
Problem 5 Let A = {1, 2, . . . , n}, where n is a positive integer. A subset of A is connected if it is a nonempty set which consists of one element or of consecutive integers. Determine the greatest integer k for which A contains k distinct subsets A1, A2, . . . , Ak such that the intersection of any two distinct sets Ai and Aj is connected.
Problem 6 Let f : N →N ∪{0} be defined recursively by f(1) = 0 and f(n) = max 1≤j≤⌊n 2 ⌋{f(j) + f(n −j) + j} for all n ≥2. Determine f(2000).
98 Turkey 2.22 Turkey Problem 1 Find the number of ordered quadruples (x, y, z, w) of integers with 0 ≤x, y, z, w ≤36 such that x2 + y2 ≡z3 + w3 (mod 37).
Problem 2 Given a circle with center O, the two tangent lines from a point S outside the circle touch the circle at points P and Q. Line SO intersects the circle at A and B, with B closer to S. Let X be an interior point of minor arc PB, and let line OS intersect lines QX and PX at C and D, respectively. Prove that 1 AC + 1 AD = 2 AB .
Problem 3 For any two positive integers n and p, prove that there are exactly (p + 1)n+1 −pn+1 functions f : {1, 2, . . . , n} →{−p, −p + 1, . . . , p} such that |f(i) −f(j)| ≤p for all i, j ∈{1, 2, . . . , n}.
Problem 4 Find all sequences a1, a2, . . . , a2000 of real numbers such that P2000 n=1 an = 1999 and such that 1 2 < an < 1 and an+1 = an(2 −an) for all n ≥1.
Problem 5 In an acute triangle ABC with circumradius R, alti-tudes AD, BE, CF have lengths h1, h2, h3, respectively. If t1, t2, t3 are the lengths of the tangents from A, B, C, respectively, to the circumcircle of triangle DEF, prove that 3 X i=1 ti √hi 2 ≤3 2R.
Problem 6 (a) Prove that for each positive integer n, the number of ordered pairs (x, y) of integers satisfying x2 −xy + y2 = n is finite and divisible by 6.
(b) Find all ordered pairs (x, y) of integers satisfying x2 −xy + y2 = 727.
2000 National Contests: Problems 99 Problem 7 Given a triangle ABC, the internal and external bisec-tors of angle A intersect line BC at points D and E, respectively.
Let F be the point (different from A) where line AC intersects the circle ω with diameter DE. Finally, draw the tangent at A to the circumcircle of triangle ABF, and let it hit ω at A and G. Prove that AF = AG.
Problem 8 Let P(x) = x + 1 and Q(x) = x2 + 1. We form all sequences of ordered pairs (x1, y1), (x2, y2), . . . with (x1, y1) = (1, 3) and (xk+1, yk+1) ∈{(P(xk), Q(yk)), (Q(xk), P(yk))} for each positive integer k. Find all positive integers n such that xn = yn in at least one of these sequences.
Problem 9 Show that it is possible to cut any triangular prism of infinite length with a plane such that the resulting intersection is an equilateral triangle.
Problem 10 Given a square ABCD, the points M, N, K, L are chosen on sides AB, BC, CD, DA, respectively, such that lines MN and LK are parallel and such that the distance between lines MN and LK equals AB. Show that the circumcircles of triangles ALM and NCK intersect each other, while those of triangles LDK and MBN do not.
Problem 11 Let f : R →R be a function such that |f(x + y) −f(x) −f(y)| ≤1 for all x, y ∈R. Show that there exists a function g : R →R with |f(x) −g(x)| ≤1 for all x ∈R, and with g(x + y) = g(x) + g(y) for all x, y ∈R.
100 Ukraine 2.23 Ukraine Problem 1 Let n numbers greater than 1 be given. During each step, we replace any two numbers a, b with the number ab √ 2 a+b . Prove that after n −1 steps, the remaining number is at least 1 √n.
Problem 2 Acute triangle PNK is inscribed in a circle with diameter NM. Let A be the intersection point of MN and PK, and let H be a point on minor arc PN. The circumcircle of triangle PAH intersects lines MN and PN again at points B and D, respectively; the circle with diameter BN intersects lines PN and NK at points F and Q, respectively. Let C be the intersection point of lines MN and FQ, and let E be the intersection point different from D of line CD with the circumcircle of triangle PAH. Prove that the points H, E, N are collinear.
Problem 3 Let AA1, BB1, CC1 be the altitudes of acute triangle ABC. Let A2, B2, C2 be the tangency points of the incircle of triangle A1B1C1 with sides B1C1, C1A1, A1B1, respectively. Prove that the lines AA2, BB2, CC2 are concurrent.
Problem 4 Do there exist positive integers m, n such that m2+1 n2−1 is an integer?
2000 National Contests: Problems 101 2.24 United Kingdom Problem 1 Two intersecting circles C1 and C2 have a common tangent which touches C1 at P and C2 at Q. The two circles intersect at M and N. Prove that the triangles MNP and MNQ have equal areas.
Problem 2 Given that x, y, z are positive real numbers satisfying xyz = 32, find the minimum value of x2 + 4xy + 4y2 + 2z2.
Problem 3 Find positive integers a and b such that 3 √a + 3 √ b −1 2 = 49 + 20 3 √ 6.
Problem 4 (a) Find a set A of ten positive integers such that no six distinct elements of A have a sum which is divisible by 6.
(b) Is it possible to find such a set if “ten” is replaced by “eleven”?
102 United States of America 2.25 United States of America Problem 1 Call a real-valued function f very convex if f(x) + f(y) 2 ≥f x + y 2 + |x −y| holds for all real numbers x and y. Prove that no very convex function exists.
Problem 2 Let S be the set of all triangles ABC for which 5 1 AP + 1 BQ + 1 CR − 3 min{AP, BQ, CR} = 6 r , where r is the inradius and P, Q, R are the points of tangency of the incircle with sides AB, BC, CA, respectively. Prove that all triangles in S are isosceles and similar to one another.
Problem 3 A game of solitaire is played with R red cards, W white cards, and B blue cards. A player plays all the cards one at a time.
With each play he accumulates a penalty. If he plays a blue card, then he is charged a penalty which is the number of white cards still in his hand. If he plays a white card, then he is charged a penalty which is twice the number of red cards still in his hand. If he plays a red card, then he is charged a penalty which is three times the number of blue cards still in his hand. Find, as a function of R, W, and B, the minimal total penalty a player can amass and all the ways in which this minimum can be achieved.
Problem 4 Find the smallest positive integer n such that if n unit squares of a 1000×1000 unit-square board are colored, then there will exist three colored unit squares whose centers form a right triangle with legs parallel to the edges of the board.
Problem 5 Let A1A2A3 be a triangle and let ω1 be a circle in its plane passing through A1 and A2. Suppose there exist circles ω2, ω3, . . . , ω7 such that for k = 2, 3, . . . , 7, ωk is externally tangent to ωk−1 and passes through Ak and Ak+1, where An+3 = An for all n ≥1.
Prove that ω7 = ω1.
2000 National Contests: Problems 103 Problem 6 Let a1, b1, a2, b2, . . . , an, bn be nonnegative real num-bers. Prove that n X i,j=1 min{aiaj, bibj} ≤ n X i,j=1 min{aibj, ajbi}.
104 Vietnam 2.26 Vietnam Problem 1 Given a real number c > 2, a sequence x1, x2, . . . of real numbers is defined recursively by x1 = 0 and xn+1 = q c −√c + xn for all n ≥1. Prove that the sequence x1, x2, . . . is defined for all n and has a finite limit.
Problem 2 Two circles ω1 and ω2 are given in the plane, with centers O1 and O2, respectively. Let M ′ 1 and M ′ 2 be two points on ω1 and ω2, respectively, such that the lines O1M ′ 1 and O2M ′ 2 intersect.
Let M1 and M2 be points on ω1 and ω2, respectively, such that when measured clockwise the angles ∠M ′ 1OM1 and ∠M ′ 2OM2 are equal.
(a) Determine the locus of the midpoint of M1M2.
(b) Let P be the point of intersection of lines O1M1 and O2M2.
The circumcircle of triangle M1PM2 intersects the circumcircle of triangle O1PO2 at P and another point Q. Prove that Q is fixed, independent of the locations of M1 and M2.
Problem 3 Given the polynomial P(x) = x3 −9x2 + 24x −97, prove that for each positive integer n there exists a positive integer an for which P(an) is divisible by 3n.
Problem 4 Given an angle α ∈(0, π), find a quadratic polynomial of the form f(x) = x2 + ax + b such that for every n ≥3, the polynomial Pn(x) = xn sin α −x sin nα + sin (n −1)α is divisible by f(x).
Problem 5 Suppose that all circumcircles of the four faces of a tetrahedron have congruent radii. Show that any two opposite edges of the tetrahedron are congruent.
Problem 6 Determine all functions f : R →R satisfying x2f(x) + f(1 −x) = 2x −x4 for all x ∈R.
2000 National Contests: Problems 105 Problem 7 Two circles C1 and C2 intersect at two points P and Q.
The common tangent of C1 and C2 closer to P than to Q touches C1 and C2 at A and B, respectively. The tangent to C1 at P intersects C2 at E (distinct from P) and the tangent to C2 at P intersects C1 at F (distinct from P). Let H and K be two points on the rays AF and BE, respectively, such that AH = AP, BK = BP. Prove that the five points A, H, Q, K, B lie on the same circle.
Problem 8 Given a positive integer k, let x1 = 1 and define the sequence x1, x2, . . . of positive integers recursively as follows: for each integer n ≥1, let xn+1 be the smallest positive integer not belonging to the set {x1, x2, . . . , xn, x1 + k, x2 + 2k, . . . , xn + nk}. Show that there exists a real number a such that xn = ⌊an⌋ for all n = 1, 2, . . . .
Problem 9 Let a, b, c be pairwise relatively prime positive integers.
The positive integer n is said to be stubborn if it cannot be written in the form n = bcx + cay + abz for any positive integers x, y, z. Determine, as a function of a, b, and c, the number of stubborn integers.
Problem 10 Let R+ denote the set of positive real numbers, and let a, r > 1 be real numbers.
(a) Suppose that f : R+ →R is a function satisfying the following conditions: (i) f(f(x)) ≤axrf x a for all x > 0.
(ii) f(x) < 22000 for all x < 1 22000 .
Prove that f(x) ≤xra1−r for all x > 0.
(b) Construct a function f : R+ →R satisfying condition (i) such that f(x) > xra1−r for all x > 0.
3 2000 Regional Contests: Problems 107 108 Asian Pacific Mathematical Olympiad 3.1 Asian Pacific Mathematical Olympiad Problem 1 Compute the sum S = 101 X i=0 x3 i 1 −3xi + 3x2 i where xi = i 101 for i = 0, 1, . . . , 101.
Problem 2 We are given an arrangement of nine circular slots along three sides of a triangle: one slot at each corner, and two more along each side. Each of the numbers 1, 2, . . . , 9 is to be written into exactly one of these circles, so that (i) the sums of the four numbers on each side of the triangle are equal; (ii) the sums of the squares of the four numbers on each side of the triangle are equal.
Find all ways in which this can be done.
Problem 3 Let ABC be a triangle with median AM and angle bisector AN. Draw the perpendicular to line NA through N, hitting lines MA and BA at Q and P, respectively. Also let O be the point where the perpendicular to line BA through P meets line AN. Prove that QO ⊥BC.
Problem 4 Let n, k be positive integers with n > k. Prove that 1 n + 1 · nn kk(n −k)n−k < n!
k!(n −k)! < nn kk(n −k)n−k .
Problem 5 Given a permutation (a0, a1, . . . , an) of the sequence 0, 1, . . . , n, a transposition of ai with aj is called legal if ai = 0, i > 0, and ai−1+1 = aj. The permutation (a0, a1, . . . , an) is called regular if after finitely many legal transpositions it becomes (1, 2, . . . , n, 0). For which numbers n is the permutation (1, n, n −1, . . . , 3, 2, 0) regular?
2000 Regional Contests: Problems 109 3.2 Austrian-Polish Mathematics Competition Problem 1 Determine all polynomials P(x) with real coefficients such that for some positive integer n, the equality 2n+1 X k=1 (−1)k k 2 P(x + k) = 0 holds for infinitely many real numbers x.
Problem 2 We are given a 1 × 1 × 1 unit cube with opposite faces ABCD and EFGH, where AE, BF, CG, and DH are edges of the cube. X is a point on the incircle of square ABCD, and Y is a point on the circumcircle of triangle BDG. Find the minimum possible value of XY.
Problem 3 For each positive integer n ≥3, find all n-tuples (x1, x2, . . . , xn) of real numbers that satisfy the following system of equations: x3 n = x1 + x2 + 1 x3 1 = x2 + x3 + 1 .
.
.
x3 n−1 = xn + x1 + 1.
Problem 4 Find all positive integers N whose only prime divisors are 2 and 5, such that the number N + 25 is a perfect square.
Problem 5 For which integers n ≥5 is it possible to color the vertices of a regular n-gon using at most 6 colors such that any 5 consecutive vertices have different colors?
Problem 6 Let the 3-cross be the solid made up of one central unit cube with six other unit cubes attached to its faces, such as the solid made of the seven unit cubes centered at (0, 0, 0), (±1, 0, 0), (0, ±1, 0), and (0, 0, ±1). Prove or disprove that the space can be tiled with 3-crosses in such a way that no two of them share any interior points.
110 Austrian-Polish Mathematics Competition Problem 7 In the plane the triangle A0B0C0 is given. Consider all triangles ABC satisfying the following conditions: (i) lines AB, BC, and CA pass through points C0, A0, and B0, respectively; (ii) ∠ABC = ∠A0B0C0, ∠BCA = ∠B0C0A0, and ∠CAB = ∠C0A0B0.
Find the locus of the circumcenter of all such triangles ABC.
Problem 8 We are given a set of 27 distinct points in the plane, no three collinear. Four points from this set are vertices of a unit square; the other 23 points lie inside this square. Prove that there exist three distinct points X, Y, Z in this set such that [XY Z] ≤ 1 48.
Problem 9 For all real numbers a, b, c ≥0 such that a + b + c = 1, prove that 2 ≤(1 −a2)2 + (1 −b2)2 + (1 −c2)2 ≤(1 + a)(1 + b)(1 + c) and determine when equality occurs for each of the two inequalities.
2000 Regional Contests: Problems 111 3.3 Balkan Mathematical Olympiad Problem 1 Let E be a point inside nonisosceles acute triangle ABC lying on median AD, and drop perpendicular EF to line BC. Let M be an arbitrary point on segment EF, and let N and P be the orthogonal projections of M onto lines AC and AB, respectively.
Prove that the angle bisectors of ∠PMN and ∠PEN are parallel.
Problem 2 Find the maximum number of 1 × 10 √ 2 rectangles one can remove from a 50 × 90 rectangle by using cuts parallel to the edges of the original rectangle.
Problem 3 Call a positive integer r a perfect power if it is of the form r = ts for some integers s, t greater than 1. Show that for any positive integer n there exists a set S of n distinct perfect powers, such that for any nonempty subset T of S, the arithmetic mean of the elements in T is also a perfect power.
112 Czech-Slovak Match 3.4 Czech-Slovak Match Problem 1 A triangle ABC with incircle k is given.
Circle ka passes through B and C and is orthogonal to k; circles kb and kc are defined similarly. (Two circles are said to be orthogonal if they intersect and their tangents at any common point are perpendicular.) Let ka and kb intersect again at C′, and define A′ and B′ similarly.
Show that the circumradius of triangle A′B′C′ equals half the radius of k.
Problem 2 Let P(x) be a polynomial with integer coefficients.
Show that the polynomial Q(x) = P(x4)P(x3)P(x2)P(x) + 1 has no integer roots.
Problem 3 Let ABCD be an isosceles trapezoid with bases AB and CD. The incircle of the triangle BCD touches side CD at a point E. Let F be the point on the internal bisector of ∠DAC such that EF ⊥CD. The circumcircle of triangle ACF intersects the line CD at two points C and G. Show that triangle AFG is isosceles.
2000 Regional Contests: Problems 113 3.5 Mediterranean Mathematical Competition Problem 1 We are given n different positive numbers a1, a2, . . . , an and the set {σ1, σ2, . . . , σn}, where each σi ∈{−1, 1}. Prove that there exist a permutation (b1, b2, . . . , bn) of (a1, a2, . . . , an) and a set {β1, β2, . . . , βn} where each βi ∈{−1, 1}, such that the sign of Pi j=1 βjbj equals the sign of σi for all 1 ≤i ≤n.
Problem 2 In the convex quadrilteral ABCD, AC = BD. Out-wards along its sides are constructed equilateral triangles WAB, XBC, Y CD, ZDA with centroids S1, S2, S3, S4, respectively. Prove that S1S3 ⊥S2S4 if and only if AC ⊥BD.
Problem 3 For a positive integer n ≥2, let c1, . . . , cn and b1, . . . , bn be positive real numbers. Prove that the equation n X i=1 ci p xi −bi = 1 2 n X i=1 xi has exactly one solution if and only if n X i=1 c2 i = n X i=1 bi.
Problem 4 P, Q, R, S are the midpoints of sides BC, CD, DA, AB, respectively, of convex quadrilateral ABCD. Prove that 4(AP 2 + BQ2 + CR2 + DS2) ≤5(AB2 + BC2 + CD2 + DA2).
114 Nordic Mathematical Contest 3.6 Nordic Mathematical Contest Problem 1 In how many ways can the number 2000 be written as a sum of three positive integers a1 ≤a2 ≤a3?
Problem 2 People P1, P2, . . . , Pn, sitting around a table in that order, have m + n −1, m + n −2, . . . , m coins, respectively. Pi gives Pi+1 exactly i coins for i = 1, 2, . . . in that order (where Pi+n = Pi for all i) until one person no longer has enough coins to continue. At this moment, it turns out that some person has exactly five times as many coins as one of his neighbors. Determine m and n.
Problem 3 In triangle ABC, internal angle bisectors AD and CE meet at I. If ID = IE, prove that either triangle ABC is isosceles or ∠ABC = 60◦.
Problem 4 The function f : [0, 1] →R satisfies f(0) = 0, f(1) = 1, and 1 2 ≤f(z) −f(y) f(y) −f(x) ≤2 for all 0 ≤x < y < z ≤1 with z −y = y −x. Prove that 1 7 ≤f 1 3 ≤4 7.
2000 Regional Contests: Problems 115 3.7 St. Petersburg City Mathematical Olympiad (Russia) Problem 1 Do there exist four quadratic polynomials such that if you put them in any order, there exists a number such that the values of the polynomials at that number, in the chosen order, are strictly increasing?
Problem 2 Let S1 and S2 be two nonintersecting circles.
A common external tangent meets S1 and S2 at A and B, respectively.
Let S3 be a circle passing through A and B, and let C and D be its second intersections with S1 and S2, respectively. Let K be the point where the tangents to S1 and S2 at C and D, respectively, meet.
Prove that KC = KD.
Problem 3 On a 1001 × 1001 checkerboard, call two (unit) squares adjacent if they share an edge. Several squares are chosen, no two adjacent, such that the number of squares adjacent to chosen squares is less than the number of chosen squares. How many squares have been chosen?
Problem 4 Let S be a set of 1000 positive integers.
For each nonempty subset A of B, let g(A) be the greatest common divisor of the elements in A. Is it possible that g(A1) ̸= g(A2) for any two distinct subsets A1, A2 of B?
Problem 5 Let AA1, BB1, CC1 be the altitudes of an acute triangle ABC. The points A2 and C2 on line A1C1 are such that line CC1 bisects A2B1 and line AA1 bisects C2B1. Lines A2B1 and AA1 meet at K, and lines C2B1 and CC1 meet at L. Prove that lines KL and AC are parallel.
Problem 6 One hundred points are chosen in the coordinate plane.
Show that at most 2025 = 452 rectangles with vertices among these points have sides parallel to the axes.
Problem 7 Find all pairs of distinct positive integers a, b such that b2 + a | a2 + b and b2 + a is a power of a prime.
Problem 8 In a country of 2000 airports, there are initially no airlines.
Two airlines take turns introducing new nonstop flights, 116 St. Petersburg City Mathematical Olympiad (Russia) and given any two cities only one airline may offer flights between them. Each airline attempts to introduce enough flights so that if any airport is shut down, it can still offer trips from any airport to any other airport, possibly with transfers. Which airline can ensure that it achieves this goal first?
Problem 9 We are given several monic quadratic polynomials, all with the same discriminant. The sum of any two of the polynomials has distinct real roots. Show that the sum of all of the polynomials also has distinct real roots.
Problem 10 Let a and b be distinct positive integers greater than 1 such that a2 + b −1 is divisible by b2 + a −1. Prove that b2 + a −1 has at least two distinct prime factors.
Problem 11 On an infinite checkerboard are placed 111 nonover-lapping corners, L-shaped figures made of 3 unit squares.
The collection has the following property: for any corner, the 2×2 square containing it is entirely covered by the corners. Prove that one can remove between 1 and 110 of the corners so that the property will be preserved.
Problem 12 We are given distinct positive integers a1, a2, . . . , a20.
The set of pairwise sums {ai + aj | 1 ≤i ≤j ≤20} contains 201 elements. What is the smallest possible number of elements in the set {|ai −aj| | 1 ≤i < j ≤20}, the set of (positive) differences between the integers?
Problem 13 Let ABCD be an isoceles trapezoid with bases AD and BC. An arbitrary circle tangent to lines AB and AC intersects BC at M and N. Let X and Y be the intersections closer to D of the incircle of triangle BCD with DM and DN, respectively. Show that line XY is parallel to line AD.
Problem 14 In each square of a chessboard is written a positive real number such that the sum of the numbers in each row is 1. It is known that for any eight squares, no two in the same row or column, the product of the numbers in these squares is no greater than the product of the numbers on the main diagonal. Prove that the sum of the numbers on the main diagonal is at least 1.
2000 Regional Contests: Problems 117 Problem 15 Is it possible to draw finitely many segments in three-dimensional space such that any two segments either share an endpoint or do not intersect, any endpoint of a segment is the endpoint of exactly two other segments, and any closed polygon made from these segments has at least 30 sides?
Problem 16 Does there exist a quadratic polynomial f with pos-itive coefficients such that for every positive real number x, the equality ⌊f(x)⌋= f(⌊x⌋) holds?
Problem 17 What is the smallest number of weighings on a balance scale needed to identify the individual weights of a set of objects known to weigh 1, 3, 32, . . . , 326 in some order? (The balance scale reports the difference between the weights of the objects in two pans.) Problem 18 The line ℓis tangent to the circumcircle of acute triangle ABC at B.
Let K be the projection of the orthocenter of ABC onto ℓ, and let L be the midpoint of side AC. Show that triangle BKL is isosceles.
Problem 19 Two points move within a vertical 1 × 1 square at the same constant speed. Each travels in a straight path except when it hits a wall, in which case it reflects offthe wall so that its angle of incidence equals its angle of reflection. Show that a spider, moving at the same speed as the balls, can descend straight down on a string from the top edge of the square to the bottom so that while the spider is within in the square, neither the spider nor its string is touching one of the balls.
Problem 20 Let n ≥3 be an integer.
Prove that for positive numbers x1 ≤x2 ≤· · · ≤xn, xnx1 x2 + x1x2 x3 + · · · + xn−1xn x1 ≥x1 + x2 + · · · + xn.
Problem 21 In the plane is given a convex n-gon P with area less than 1. For each point X in the plane, let F(X) denote the area of the union of all segments joining X to points of P. Show that the set of points X such that F(X) = 1 is a convex polygon with at most 2n sides.
Problem 22 What is the smallest number of unit segments that can be erased from the interior of a 2000 × 3000 rectangular grid so that no smaller rectangle remains intact?
118 St. Petersburg City Mathematical Olympiad (Russia) Problem 23 Let x, y, z, t be pairwise relatively prime positive in-tegers such that xy +yz +zt = xt. Prove that the sum of the squares of some two of these numbers equals twice the sum of the squares of the other two.
Problem 24 Let AA1 and CC1 be altitudes of acute triangle ABC.
The line through the incenters of triangles AA1C and AC1C meets lines AB and BC at X and Y , respectively. Prove that BX = BY .
Problem 25 Does there exist a 30-digit number such that the number obtained by taking any five of its consecutive digits is divisible by 13?
Problem 26 One hundred volleyball teams play in a round-robin tournament, where each pair of teams plays against each other exactly once. Each game of the tournament is played at a different time, and no game ends in a draw. It turns out that in each match, the two teams playing the match have the same number of victories up to that point. If the minimum number of games won by any team is m, find all possible values of m.
Problem 27 Let ABCD be a convex quadrilateral, and M and N the midpoints of AD and BC, respectively. Suppose A, B, M, N lie on a circle such that AB is tangent to the circumcircle of triangle BMC. Prove that AB is also tangent to the circumcircle of triangle AND.
Problem 28 Let n ≥3 be a positive integer.
For all positive numbers a1, a2, . . . , an, show that a1+a2 2 a2+a3 2 · · · an+a1 2 ≤a1+a2+a3 2 √ 2 a2+a3+a4 2 √ 2 · · · an+a1+a2 2 √ 2 .
Problem 29 A connected graph is said to be 2-connected if after removing any single vertex, the graph remains connected. Prove that given any 2-connected graph in which the degree of every vertex is greater than 2, it is possible to remove a vertex (and all edges adjacent to that vertex) so that the remaining graph is still 2-connected.
Problem 30 Let m be a positive integer. Prove that there exist infinitely many prime numbers p such that m + p3 is composite.
Problem 31 The perpendicular bisectors of sides AB and BC of nonequilateral triangle ABC meet lines BC and AB at A1 and C1, 2000 Regional Contests: Problems 119 respectively. Let the bisectors of angles A1AC and C1CA meet at B′, and define C′ and A′ analogously. Prove that the points A′, B′, C′ lie on a line passing through the circumcenter of triangle ABC.
Problem 32 Is it possible to select 102 17-element subsets of a 102-element set, such that the intersection of any two of the subsets has at most 3 elements? |
188544 | https://msp.org/pjm/1957/7-1/pjm-v7-n1-p15-p.pdf | Pacific Journal of Mathematics ON THE MEASURE OF NORMAL FORMULAS ROBERT FORBES MCNAUGHTON, JR.
Vol. 7, No. 1 January 1957 ON THE MEASURE OF NORMAL FORMULAS ROBERT MCNAUGHTON l Introduction Qυine has recently found (in , and ) a reasonably practical method which yields the simplest normal equivalent of a given truth functional formula. The problem of this paper is to find a practical method which yields the simplest normal formula with a given measure. Roughly, the measure of a formula is the number of T's in the column under the formula in a truth table which has 2 a rows; these rows represent all possible assignments of T's and F's to d letters including all the letters of the formula and perhaps others.
The problem, which is rather difficult, arises in the design of certain networks in digital computers (described at the end of § 2) as part of a more general problem which is all the more difficult. Networks, however, are not discussed at all in the remainder of the paper, where the main problem is attacked as a problem in pure logic. I have had no success in obtaining a method which is generally satisfactory, but have succeeded in proving a few theorems which will probably be in-dispensable in any future attack on the problem.
2. The problem and its origin. Most of the terminology which I shall use is Quine's. Where it conflicts with Quine's terminology of , and I shall explicitly say so; on the other hand, I shall not pre-suppose that the reader is familiar with any of these papers. An italicized word appearing in a sentence of this paper is defined in that sentence. In this section a sentence without an italicized ^kord is often a theorem which is either well known or obvious.
A formula is made up in the usual manner from the letters Au • • • , Aa by means of negation, conjunction and disjunction (or alternation).
For any formulas Φu , Φn, n^2, Φλ is the negation of Φu ΦxΦ^^Φn is the conjunction of Φlf • • , Φn (these being conjuncts), and ΦΛ/^V" \JΦn is the disjunction (called ' alternation ' by Quine) of Φu •••, Φ» (these being disjuncts). (I assume that the reader is familiar enough with the general literature to see how the circularity of definition in the last two sentences can be avoided.) A letter or its negation is a literal. If a formula is a disjunction, then the disjuncts are clauses; if it is not a disjunction, the formula itself is its only clause. A for-mula all of whose clauses are literals or conjunctions of literals is a normal formula. (For Quine a clause of a normal formula cannot have Received January 30, 1956. This paper is ,an abridged version of report No. 1, con-tract DA~04~200ΌRD-436 (Applied Mathematics and Statistics Laboratory, Stanford, Cali-fornia), sponsored by the Office of Ordnance Research.
969 970 ROBERT McNAUGHTON repetitious letters.) A formula is consistent if it comes out true under some interpreta-tion of its letters. An inconsistent normal formula, then, must be one in which every clause is inconsistent; a clause of a normal formula is inconsistent if and only if there is a letter appearing both with and without a bar. An inconsistent clause can be omitted from any con-sistent normal formula and the resulting formula is equivalent to the original. A clause of a normal formula subsumes another if every literal of the second is a literal of the first. Any clause which subsumes another in a normal formula can be omitted and the resulting formula is equivalent to the original. A literal which has occurred previously in the same clause can be omitted and the resulting clause is equivalent to the original clause; hence, the resulting formula is equivalent to the original formula. A normal formula in which no clause subsumes ano-ther, no clause is inconsistent, and no clause contains a repeated literal, is an apparently irredundant normal formula. An irredundant normal formula is one in which no literal or clause can be omitted without sacrificing equivalence. Some apparently irredundant formulas are not irredundant as example 1 or example 2 of is enough to show. An interclausally consistent formula is one in which the conjunction of any two clauses is consistent. A normal formula is interclausally consistent if and only if no letter appears in it at least once with a bar and at least once without a bar.
A normal formula is developed with respect to the letters Au , A d if every clause has one and only one occurrence of each of these letter. Ev^jry consistent formula Φ containing no letters other than Alf • , Ad can 1?e transformed into an irredundant normal formula which is developed with respect to the letters Al9 — ,Ad; the number of clau-ses in the letter is the measure of Φ, or m(Φ). If Φ is inconsistent, then the measure of Φ is 0. In general m(Φ) depends on d> but there is no need to make this dependence explicit in the notation in most of this paper. Where the notation m{Φ) is used, it is assumed that Φ con-tains only (perhaps not all) the letters Alf , Ad. If a truth table is constructed for Φ with 2 a rows, representing the 2 d assignments of truth values to Al9 • • • , Ad, then there will be m(Φ) T's in the column for Φ.
Two formulas Φ and Ψ are isomorphic if there is a one-to-one map-ping / of the set of literals of Φ onto the set of literals of ψ such that, if both At and A% occur in Φ and if f(At)=AJ then f(Ai)=Ajf and if f(Ai)=Aj then /(^Lί)==Aj; and such that a formula Ψ r can be obtained from Φ by replacing each literal by its image under /, and Ψ can be obtained from Ψ f by changing the order of conjunts of zero or more conjunctions and changing the order of disjuncts of zero or more dis-junctions. Thus AλA^\JA^A^XJA iA^Ar, is isomorphic to AxAbAG\/A^A^ ON THE MEASURE OF NORMAL FORMULAS 971 ; here / is the mapping such that /(A1)=A1, /(A2)=A2, /(A3)=A5, /(A4)=A6, /(A5)=A4. If no letters except Au • • • , Ad occur in 0 and Ψ, and if 0 and Ψ are isomorphie, then m(Φ)=m(Ψ). (If this fact is not obvious enough to the reader, it is proved for the case in which Φ and Ψ are normal formulas as Theorem 2.4.) For the purposes of this paper the word " simplicity " need not be, and is not, defined precisely. Let us understand merely that simplicity of a normal formula depends on the number of clauses and the number of literals in each clause.
A practical solution to the problem of finding the simplest normal formula with a given measure would have some application to the de-sign of certain parts of digital computers. Dr. Montgomery Phister, Jr. of the Ramo Wooldridge Corporation has suggested the following problem which was the initial stimulus for the research for this paper.
Suppose that one is to devise a circuit with n outputs in such a way that in each of m given time intervals each output is to be in state 1 or state 0 as specified. The circuit engineer can select his in-puts in any way he chooses, so that each input is either 0 or 1 in each interval. But he must do so in such a way that each output is a func-tion of the inputs and the circuit is the most economical. If certain kinds of diode circuit are used, then the part of the circuit which relates any output to the inputs must be constructed as a normal formula.
The problem of finding the simplest normal formula with a given measure is relevant to this problem, even though a practical solution to the former would not necessarily mean a practical solution to the latter.
If the number of intervals is between 2 d~ 1-h 1 and 2 d inclusive and if each time interval itself is to be a unique function of the inputs, then there must be d inputs. With these assumptions, a practical way of choosing inputs so as to minimize the circuit for just one output is easily obtainable if there is a practical way of finding the simplest nor-mal formula with a given measure. For example, if there are 16 time intervals and there is to be one output in state 1 in exactly 5 intervals, then it is necessary to find, for d=4, a simplest formula whose measure is 5; in this case, A^X/A^A^ seems to be a formula.
3 Calculation of the measure of a formula. There is a straight-forward way of calculating the measure of a normal formula which is somewhat simpler than actually expanding it into a developed normal formula. At the basis of this method is an easily proved theorem relat-ing the measures of two formulas, their conjunction and their disjunction.
THEOREM 3,1, For any formulas Φ and Ψ 972 ROBERT McNAUGHTON m(Φ\/)=m(Φ)-hm(Ψ)-m(ΦΨ) Proof Consider the developed normal formulas with respect to d variables equivalent to the four formulas concerned, Φ, Ψ> {Φ\JΨ), (ΦΨ). The number of clauses in (Φ\/Ψ) can be counted by counting the number of clauses in Φ (which is assumed to be 0 if Φ is incon-sistent) and then counting the number of clauses in Ψ, remembering that any clauses which these have in common have been counted twice.
But the number of clauses which Φ and Ψ have in common is pre-cisely the number of clauses in (ΦΨ)9 or 0 if ΦΨ is inconsistent, which in either case is m(ΦΨ). Hence Theorem 3.1 follows.
THEOREM 3.2. If Φ is a conjunction of j distinct literals, no two of which are of the same letter, then m{Φ) = 2 a~ j.
This theorem follows readily from well known properties of truth tables or developed normal formulas.
For the remainder of this paper let Φ be a normal formula. Let Ψu • • • , Ψk be the clauses of Φ in the order of their appearance in Φ.
Let Φx, l<Lx<^k, be the normal formula ^iV^V Wχ Thus Φk is Φ. Let C(Φ) be the set of all clauses of Φ. For any S g φ ) let j s be the total number of distinct letters appearing in the clauses of S.
Let is be 0 if at least one letter appears in at least one clause with a bar and in at least one clause without a bar, that is, if the conjunction of all the clauses of & is inconsistent. If there is no such letter, let is be 1 if there are an odd number of clauses in S or —1 if there are an even number of clauses in S.
THEOREM 3.3.
the summation being taken only over nonempty subsets S.
The proof is by induction on the number k of clauses in Φ. If k = 1, then there is only one S to be considered, namely, the unit set of the one clause. If there is a letter which appears both with and with-out a bar, is is 0 and so is m(Φ). If there is no such letter, then m(Φ)=2-Js9 by Theorem 3.2.
Suppose that k^> 1 and suppose that Theorem 3.3 holds for all formulas having fewer than k clauses. By Theorem 3.1, ON THE MEASURE OF NORMAL FORMULAS 973 Now Φic-xψn is not a normal formula unless A ; —1=1. If k — l^>l let Ψ be the normal formula which is equivalent to Φk-\ψk obtained by dis-tributing the conjunct ψk over the clauses of Φk-u and if fc —1=1, let Ψ be Φk-tfjc. Thus Ψ will have k — 1 clauses and, for each h<Lk — l, the h th clause of Ψ will have the literals of the h th clause of Φk-λ and those of ψk9 but no others. Now since Ψ is equivalent to Φk-λψkί (2) By the inductive hypothesis and Theorem 3.2, (3) w(0-,H Σ i (4) and (5) where in each case only nonempty subsets S are considered.
Making substitutions in (1), justified by (2), (3), (4) and (5), we get (6) m(0)= Σ is2^s+i 2 d-Jc^~ Σ i^-'.
It remains only to show that we can equate the expression Σ is2 d~' :fs 5CCCΦ) with the right side of (6). But these expressions are equal, term by term, as can be seen as follows. For every S in C{Φ), either S does not contain ψk (case I), S contains φk and other clauses (case II), or S contains ψk only (case III). In case III, the summand is2 a~ js is the middle term of the right side of (6). In case I, SeC(Φk^) and so the summand i&2 a~ js occurs as a summand in the first term on the right side of (6). In case II, finally, suppose S contains besides ψk the gψ, • • • , g %£ clauses of Φ. Then consider S', the set containing the g?
1, • • • , gt clause of Ψ. The literals appearing in these clauses are exactly the literals appearing in the clauses of S. A letter will appear both with a bar and without a bar in S if and only if it does in S r. Hence, is,=^0 if and only if is=0. There an odd (even) number of clauses in S if and only if there are an even (odd) number in S' since S has just one more clause than S'. Hence the summand is2 d~ js occurs negatively as a summand in the third term of the right side of (6). It is easy to see that this correspondence is one to one and that the equality of the two expressions is established.
THEOREM 3.4. If Φ and Ψ are normal formulas, and if Φ is iso-974 ROBERT McNAUGHTON morphic to Ψ, then m(Φ)=m(Ψ).
Proof. The corresponding clause in Ψ to a clause φ in Φ is the clause which contains those literals into which the literals of <p are mapped. Consider any set S of clauses of Φ and the corresponding set of clauses S' in Ψ. The total number of distinct letters of S equals the total number of distinct letters of S'. There is a letter appearing with a bar and without a bar in £ if and only if there is such a letter appearing in S'. And, of course, the number of clauses of S is the same as the number of clauses of S'. Thus the expressions for m(Φ) and m(Ψ) as given by Theorem 3.3 will be the same, term by term.
The formula Ψ implies Γ if every assignment of truth values which makes Ψ true also makes Γ true. As is well known, Ψ implies Γ if and only if every clause of a developed normal formula equivalent to Ψ is a clause of a developed normal formula with respect to the same variables equivalent to Γ. The formula Ψ is equivalent to Γ is Ψ im-plies Γ and Γ implies Ψ. The formulas Ψ and Γ are equivalent if and only if they are equivalent to a common developed normal formula.
Theorems 3.5, 3.6 and 3.7 are direct consequences of these remarks.
THEOREM 3.5. For any formulas Ψ and Γ, m(Ψ\/Γ)^im(Ψ). The equality holds when, and only when, Γ implies Ψ.
THEOREM 3.6. m(ΨΓ)<Lm(Ψ). The equality holds when, and only when, Ψ implies Γ.
THEOREM 3.7. // Γ implies Ψ, then m(Γ) <; m{Ψ). The equality holds when, and only when, Γ and Ψ are equivalent.
4 Bounds on the measure of a formula with a given structure.
If jι <I jz <I <^jk9 then a formula has the structure if and only if it is an apparently irredundant normal formula with k clauses which have, in the order in which they appear in the formula, 0i> m 9jk literals. Note that a formula has some structure if and only if it is normal, it is apparently irredundant, and its clauses are in order of nondecreasing length. In finding a simplest normal formula we need only consider formulas which have some structure; for every normal formula which does not is equivalent to, and is no simpler than, a normal formula which does. In this section I shall give an upper bound and a lower bound on the measure of formula with a given structure.
This result will be convenient in some cases where one is trying to determine a simplest formula with a given measure.
THEOREM 4.1. If Φ has the structure , then ON THE MEASURE OF NORMAL FORMULAS 975 The proof is by induction on k. For Jc=l, if Φ is of structure , then since Φ is apparently irredundant, m(Φ)=2 a~ jι by Theorem 2.2. I shall assume that Theorem 4.1 is true about all structures whose for-mulas have less than k clauses, and show that it is true about the structure < £ , , jk > . If Φ has the structure < £ , , jk >, then, by inductive hypothesis, m{Φk-1)<^2 a-jι--... +2 d" ί-i, and But, by Theorem 3.1, (Note that, for a n y formula Φ with structure <^jl9 • • • , j f c > , m(Φ)=2 d~ jι A h2 d~^ if and only if the conjunction of every pair of clauses is inconsistent. It can be proved that such a formula exists if and only if 2 d-jL+...+2 a-jk^2 d.) Theorem 4.2, 4.3 and 4.4 are, in effect, lemmas to Theorem 4.5 which establishes a lower bound on formulas with a given structure.
THEOREM 4.2. If Φ is not interdausally consistent, if the number of distinct literals of Φ does not exceed d, and if Φ has some structure, then there is an interdausally consistent normal formula with the same structure as Φ but with no greater measure.
Proof If Φ is not interdausally consistent, then there is at least one letter in Φ appearing both with and without a bar I shall prove Theorem 4.2 by proving that there is a formula Φ' with the same struc-ture as Φ with exactly one less letter appearing both with and without a bar, such that m(Φ f)<Lm(Φ). Suppose An appears both with and without a bar in Φ. Let Φ' be Φ with every occurrence of An replaced by a variable Ap with does not appear in Φ. Since Φ has some struc-ture, it is apparently irredundant, and so An never appears both with and without a bar in any one clause of Φ. Hence Φ is equivalent to Anφ\/AnΓ\/Ω and Φ' is equivalent to Anψ\JApΓ\JΩy where ψ, Γ, and Ω are normal formulas in which An, An, and Av do not occur. By Theorem 2.1, then, the following hold.
m{Φ)^m(AnΓ) + m{Anψ\/Ω)-m(Anr{Anψ\JΩ)) We have, m{ApΓ)=m{AnΓ) since APΓ and AnΓ are isomorphic and can easily be converted into isomorphic normal formulas. Therefore we can concentrate on the last term of each equation. AnΓ(Anψ\JΩ) is equi-valent to AnΓΩ, and ApΓ{Anψ\/Ω) is equivalent to AnApΓψ\JAPΓΩ.
976 ROBERT McNAUGHTON Also m(AnΓΩ) = m(ApΓΩ), the formulas being isomorphic. Finally m(ApΓΩ)<Lm(AnAprψ\/ApΓΩ), by Theorem 3.5. Thus m(Φ r)<:m(Φ), Φ' has the same structure as Φ, and the number of literals of Φ f equals the number of literals of Φ.
Now if Φ r still has at least one letter with and without a bar, I construct Φ 1', related to Φ r as Φ f is to Φ, and so forth. Eventually I shall obtain a formula Φ (q) which has no letters appearing both with and without a bar, has the same structure as Φ and has demonstrably no greater measure. It is obvious enough that the number of variables in 0 ( α ) will not exceed d if the hypothesis of the theorem is satisfied.
(Two things can be noted. First the formula # ( Q ) can easily be constructed from Φ as follows: supposing (without loss of generality) that Au • • • , Aq are the variables which appear both with and without bars and Aq+1, • • • , Ar are the other variables of Φ, replace Alf • • • , At by Ar+1, • • • , Ar+q respectively. Second, if we prefer, we can delete all the bars from Φ i<ύ and the resulting formula will have the same mea-sure, being isomorphic to Φ i<ύ. In summary, then, given a formula Φ which satisfies the hypothesis of Theorem 4.2 it is an easy matter to write down another formula without bars, with the same structure and with the same structure and with no greater measure.) THEOREM 4.3. If Ψ, Γ have no letters in common, then Proof. Suppose (without loss of generality) that Aly • • • , An are the letters occurring in Ψ; then every letter appearing in Γ is one of the letters An+1, , Aa. From well known logical laws, the developed normal equivalent Ψ' of Ψ with respect only to the letters Au • • • , An has m{Ψ)lt2 a~ n clauses. And the developed normal equivalent Γ' of Γ with respect to the letters An+1, • • • , Ad has m{Γ)l2 n clauses. ΨΓ is equi-valent to Ψ'Γ'\ the developed normal equivalent of these can be obtain-ed from the latter by the distributive law for disjunction over conjunc-tion; the number of clauses will be the product of the number of clauses of Ψ' and Γ\ which is m{Ψ)m{Γ)l2 d. Since this last formula is the developed normal equivalent of ΨΓ with respect to Alf , Adf this number is the measure of ΨΓ. (For example, if d=5, Ψ is A1\/Aι and Γ is A A, then rc=2, Ψ f is Λ Λ V Λ Λ V Λ Λ and Γ f is AAA^AAA^ The result of " multiplying o u t " Ψ f and Γ ! yields the developed normal equivalent with respect to Al9 • • • , A& of ΨΓ.) THEOREM 4.4. If k<d, the formula A J V Λ V V ^ has the ON THE MEASURE OF NORMAL FORMULAS 977 maximum measure 2 d~ 1-f 2 d"~ 2-f -h2 d~ h of all interclausally consistent normal formulas having exactly k <^d clauses; the formula A^\/A^\/ VA has the maximum measure 2 d~ 1 + 2 d--i h2° of all interclausally consistent normal formulas having d or more clauses.
Proof The formula mentioned in the second part of the theorem is equivalent to — (A^- Ad). Therefore its developed normal form has all the 2d clauses except one, and therefore its measure is 2 d--l. This measure is a maximum for all interclausally consistent formulas since the one higer measure, 2 d, is that of a tautology, which is never inter-clausally consistent.
The formula of the first part of the theorem, when developed with respect to Al9 - ,Ak, has 2 fc —1 clauses (by the first two sentences of the above paragraph with ' k ' for ' d '). From this we obtain an equi-valent formula developed with respect to all d letters by developing each clause into 2 d~ k clauses. The measure of the formula, therefore, is Now every interclausally consistent formula with k clauses not isomor-phic to Aτ\/ \M/b either has at least one clause with more than one literal or has one literal in every clause with some repetitions of clauses.
In the latter case, the formula is equivalent to Ax\/ \/AJf for some j<&, whose measure is 2 d4-2-M -f 2 d-j <2 d-i -f2 d""\ I dispose of the former case by showing that, in an interclausally consistent for-mula, if every clause containing more than one literal is replaced by just one of its literals then the measure of the formula is not decreas-ed. Suppose the formula Ψ\JAiφi\/-"\/Aiψi is thus replaced by Ψ\Ά^V' V Ά % The latter is implied by the former and hence, by Theorem 3.7, its measure is no smaller than that of the former.
(It is easy to extend this method of proof to prove that a formula with k clauses in which no letter appears both with and without a bar has this maximum measure if and only if each clause has one literal and no literals are repeated or, equivalently, no clauses are redundant.) THEOREM 4.5. If Φ has the structure , then m(Φ) I> 2 d-I prove first that Theorem 4.5 holds where Φ is interclausally con-sistent. The proof is by induction. If Φ is of structure > then equality holds. Assume, as an inductive hypothesis, that the measure of any formula Γ of structure satisfies 978 ROBERT McNAUGHTON m{Γ) ^ 2 d- Ί4- 2 £"V 1 + . . . + 2-Ά Where Φh is of structure The formula Φh-X has structure a n d so> b ^ inductive hypothesis, { u ι ) , Also m(φh)=2 a~ jfι. Therefore, it remains to prove that The formula ψhΦh-\ is equivalent to ψhψ', where Ψ is obtained from Φh-λ by deleting literals which appear in ψh; we know that there must be at least one literal in each clause of φh-λ which does not occur in φh, for otherwise <ph would subsume another clause of Φ contrary to the as-sumption that Φ has structure <^ju , i Λ > and is therefore apparent-ly irredundant. Therefore Ψ has h — l clauses and has no literals of ψh.
Since Φ is interclausally consistent, it has no letters both with and without bars; it follows that, since Ψ has no literals of + . 4 - 2 d - ^ which is what had to be proved.
To show that Theorem 4.5 still holds when Φ is not mtercluasally con-sistent, I must discuss what happens to a formula when d varies. I shall use the notation md(Φ) here (and only here) to denote the measure of Φ for a given d. From the definition of measure, it can be seen that md(Φ)=md,(Φ)-2 d-d', assuming that Φ has at most min (d, d') letters.
Let Φ be a formula of structure which is not interclau-sally consistent. I have to prove that, for any d not less than the number of distinct letters appearing in Φ, md(Φ) > 2 d-Ji + 2 d-j-1+ ON THE MEASURE OF NORMAL FORMULAS 979 Let e be the number of letters appearing both with and without bars in Φ; then md+e{Φ)=md{Φ)2 e. There are at most d + e distinct literals in Φ, and so, by Theorem 4.2, there is an interclausally consistent Φ' of structure (with at most d + e letters) such that md+e(Φ') 2 d+G-jι-h2 d+e-h-1^ . . . + 2 -(-1) , which must now be true for md+G(Φ). Therefore, md{Φ) ^ 2-'i + 2 d-V 1+ . 4- 2-'-c-1> # This observation completes the proof of Theorem 4.5.
For j k -f k <1 d -f 1, a formula with structure ifc ^> which has the minimum measure 2 d"" Ί-f2 d~V-1H + 2- >-<-1> has been ex-hibited in the literature, namely in Quine's paper . Quine does not discuss the measure of formulas, but proves, in his Theorem 2, that his formula has the value truth in just the first 2 d - JH f-2 d-J"" (fc~ i:> rows of the conventional truth table. By the well known connection between truth tables and developed normal formulas it follows that the measure of Quine's formula is this number. The construction of this formula which has no bars can be described as follows: the first clause has j ι distinct letters, and, in general, the h th clause has all the letters of the (h~ l) t b clause except the last and enough letters which do not appear in any previous clause (at least one, since jh-ι<Ljh) to make a total of j h distinct letters; the last letter of the h th clause is a letter which has not appeared previously. It follows that the last letter of any clause of Quine's formula appears in that clause only. For example, if d=10, the Quine formula of structure < 1, 1, 3, 3, 6 > whose measure is 2 94-2 84-2 54-2 44-2 0 is (It is possible to exploit the method used in proving Theorem 4.5 to prove that, for i&-f & <^c£-hl, the only formulas with structure '"yjk^> and measure 2 C Z~ JΊH— 4-2 c~ i-(fc~ 1) are those isomorphic to Quine's formula. The key property of Quine's formula is the fact that each clause ψ contains a literal, say Aq, such that Aq is not in any other clause of the formula and all clauses followings φ contain all the literals of φ except Aq. In Quine's formula Aq is the last letter of the clause. This property is necessary, as well as sufficient, in order to insure that the formula Ψ in the proof of Theorem 4.5 has exactly h — 1 clauses of one letter each, these letters being different from each other.) One method of finding, for a given d and for a given measure m <I 2 a, a simplest normal formula whose measure is m is to construct some normal formula of measure m and then calculate the measure of 980 ROBERT McNAUGHTON all simpler normal formulas. This method is impractical (although ef-fective) unless there is some way of limiting the number of formulas whose measure must be calculated. A method of constructing a for-mula due to Quine was described two paragraphs above; from what Quine shows, it follows that for any given measure such a formula can be constructed. Although Quine's formula is not always a simplest formula with that measure, it can serve to start the search for such a formula. Then the bounds on the measure of formulas with given structures established in this section serve to limit the number of for-mulas among which the search is to be made, (although not enough to make this method practicable). Only formulas having some structure need be examined; normal formulas without any structure are not ap-parently irredundant and have shorter equivalent formulas or can be converted to a formula with structure by changing the order of dis-juncts. Needless to say, once a formula has been examined, formulas isomorphic to it need not be. The following theorem will be of some help in the search, although not enough to make it practicable in all examples.
THEOREM 4.6. If a formula with some structure and with a least clauses, where h+l<Lk, has measure and if the first h clauses of it have exactly j u , j h 9 respectively, letters, then the (&-f l) s t clause has at least j h + ι letters.
Proof A formula Φ of structure < ^ , , j h , j ' h + u j' h + 2, •>, where J'h+i__ 2 d~h + 4- 2- 'Λ-< Λ-1> + 2 d- >'Λ +i-Λ -f > 2 d'h+ . . . -f 2 ίZ-4-( Λ-1>4-2- Ά+i-Λ+ +2 t l- '-(-1) .
The last inequality is justified by the fact that the two expressions are each sums of powers of 2 with descending indices. As is well known about such expressions, since equality holds for the first h terms, the inequality of the (h + Vf 1 term is decisive.
5 Conjectures and a counterexample The results of the previous sections lead to no practical method of findining a simplest formula with a given measure. But there are two conjectures which, if they are true, would be of some significance. Another conjecture which sug-gests itself rather naturally turns out to be false, as a counterexample of mine will show. (I must admit that these conjectures may turn on ON THE MEASURE OF NORMAL FORMULAS 981 the definition of " simplicity " which has not be given precisely in this paper.) A one-clause formula with j literals, no two of which are of the same letter, has a measure 2 a~ 3. Any formula with at least two non-contradictory clauses having this measure cannot have a non-contradic-tory clause with less than j + 1 literals (by Theorem 4.1). Thus a one-clause formula is simpler than any formula with more clauses but with the same measure. My first conjecture, in its strong form, is that any normal formula Φ is simpler than any formula with the same measure but with more clauses. The weaker form is that Φ is at least a simple any such formula. I have no conterexample to either of these proposi-tions, nor do I have any good reason to believe that either of them is true in general.
Let r be the number of distinct letters of Φ. Then r <Ld, and m(Φ) is divisible by 2 d~ r. If. 2 d~ x is the largest power of two which divides a given number m, then it is possible to find a formula with measure m with just x distinct letters. One example is Quine's formula with that measure (described near the end of § 4 of this paper). But for some Φy m{Φ) is divisible by a power of two greater than 2~ r.
For example, for <2I>3, m(A1A2\/A2A^\/A1A3)=2 d~\ In this example there is a simpler formula with the same measure, namely Ax. My second conjecture is that for any measure m, for any simplest formula Φ with measure m, m is divisible by 2 d~ r. A weaker form of this con-jecture is that, for any measure m, there exists a simplest formula Φ with measure m such that m is divisible by 2 d~ r.
A formula with two clauses which are each consistent but which contradict each other (because a letter appears with a bar in one and without a bar in the other) has a measure 2 α + 2 δ, if there are d-a and d~b distinct literals in the respective clauses. If a=b then a single clause formula with d-a-1 literals, no two of the same letter, has the same measure and is simpler. If a > 6, then Quine's formula of mea-sure 2 Λ-f 2 δ has two clauses with d-a and d-b~l, respectively, letters and is, therefore, simpler. A third conjecture that suggests itself is that, for any formula in which some letter appears both with and with-out a bar, there is another formula in which no letter appears both with and without a bar, which has the same measure and which is no less simple. However, for d=6 the formula is simpler than any formula which has the same measure and which has no letter appearing both with and without a bar. (To verify this, the reader should note that Quine's formula with that measure has structure < 2 , 2, 3, 3 > . Using Theorems 4.6 and 4.1 the only possible 982 ROBERT McNAUGHTON structures for formulas which have the desired measure and are at least as simple as the displayed formula are < 2, 2, 3 >, < 2, 2, 4 >, < 2 , 3, 3 > , and < 2, 2 > . Since the desired measure is not divisible by two, and since d=6, there must be exactly six distinct letters in any formula with that measure: for there are at most six, since cZ=6; and there are at least six, by an observation made in this section.
Therefore, the structure < 2 , 2 > is excluded. Any formula with exactly six distinct letters in which no letter appears both with and without a bar is isomorphic to AxA2\jAλA^\J'A±A5A6 or if it has the structure < 2, 2, 3 >, is isomorphic to AλA^\JΆdA^\ or Λii4aVAτA3\/A^A^AQ or A^^/AτA2\JA^AbAQ if it has the structure < 2, 2, 4 >, and is isomorphic to AXA.^\/A^A^/AτAbAQ or A^^A^A^ or AλA2\fA^A^/AzAbAQ or Λ Λ V ^ Λ Λ W i Λ Λ or AλA2 tAfλ/A3A±A6 if is has the structure < < 2, 3, 3 >. But none of these formulas has the desired measure.) REFERENCES 1. W. V. Quine, The problem of simplifying truth functions, Amer. Math. Monthly, 59 (1952), 521-531.
2. , Two theorems about truth functions, Bol. Soc. Mat. Mexicana, 10 (1953), 64-70.
3. , A way to simplify truth functions, Amer. Math. Monthly, 62 (1955), 627-631.
4. R. K. Richards, Arithmetic operations in digital computers, New York, 1955.
STANFORD UNIVERSITY H. L. ROYDEN Stanford University-Stanford, California PACIFIC JOURNAL OF MATHEMATICS EDITORS A. R. WHITEMAN University of Southern California Los Angeles 7, California R. A. BEAUMONT University of Washington Seattle 5, Washington E. G. STRAUS University of California Los Angeles 24, California E. F. BECKENBACH C. E. BURGESS M. HALL E. HEWITT ASSOCIATE EDITORS A. HORN V. GANAPATHY IYER R. D. JAMES M. S. KNEBELMAN L. NACHBIN I. NIVEN T. G. OSTROM M. M. SCHIFFER G. SZEKERES F. WOLF K. YOSIDA SUPPORTING INSTITUTIONS UNIVERSITY OF BRITISH COLUMBIA CALIFORNIA INSTITUTE OF TECHNOLOGY UNIVERSITY OF CALIFORNIA MONTANA STATE UNIVERSITY UNIVERSITY OF NEVADA OREGON STATE COLLEGE UNIVERSITY OF OREGON UNIVERSITY OF SOUTHERN CALIFORNIA STANFORD UNIVERSITY UNIVERSITY OF UTAH WASHINGTON STATE COLLEGE UNIVERSITY OF WASHINGTON AMERICAN MATHEMATICAL SOCIETY CALIFORNIA RESEARCH CORPORATION HUGHES AIRCRAFT COMPANY THE RAMO-WOOLDRIDGE CORPORATION Printed in Japan by Kokusai Bunken Insatsusha (International Academic Printing Co., Ltd.), Tokyo, Japan Pacific Journal of Mathematics Vol. 7, No. 1 January, 1957 Richard Davis Anderson, Zero-dimensional compact groups of homeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
797 Hans-Joachim Bremermann, Holomorphic functionals and complex convexity in Banach spaces. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
811 Hugh D. Brunk, G. M. Ewing and W. R. Utz, Minimizing integrals in certain classes of monotone functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
833 Philip David, Uniqueness theory for asymptotic expansions in general regions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
849 Paul Erd˝ os and Harold Nathaniel Shapiro, On the least primitive root of a prime . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
861 Watson Bryan Fulks, Regular regions for the heat equation . . . . . . . . . . . . . . .
867 William Robert Gaffey, A real inversion formula for a class of bilateral Laplace transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
879 Ronald Kay Getoor, On characteristic functions of Banach space valued random variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
885 Louis Guttman, Some inequalities between latent roots and minimax (maximin) elements of real matrices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
897 Frank Harary, The number of dissimilar supergraphs of a linear graph . . . . .
903 Edwin Hewitt and Herbert S. Zuckerman, Structure theory for a class of convolution algebras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
913 Amnon Jakimovski, Some Tauberian theorems. . . . . . . . . . . . . . . . . . . . . . . . . . .
943 C. T. Rajagopal, Simplified proofs of “Some Tauberian theorems” of Jakimovski. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
955 Paul Joseph Kelly, A congruence theorem for trees . . . . . . . . . . . . . . . . . . . . . . .
961 Robert Forbes McNaughton, Jr., On the measure of normal formulas . . . . . . .
969 Richard Scott Pierce, Distributivity in Boolean algebras . . . . . . . . . . . . . . . . . .
983 Calvin R. Putnam, Continuous spectra and unitary equivalence . . . . . . . . . . .
993 Marvin Rosenblum, Perturbation of the continuous spectrum and unitary equivalence. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
997 V. N. Singh, Certain generalized hypergeometric identities of the Rogers-Ramanujan type. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1011 Peter Swerling, Families of transformations in the function spaces H p . . . . . 1015 |
188545 | https://maisonetmath.com/inequalities/video/586-solve-and-graph-an-inequality-example | Solve and Graph An Inequality Example
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Solve and Graph An Inequality Example
Review how to solve and graph an inequality using an example that requires you to reverse the directions of the inequality sign.
Solve:
−2 x+16≥8−2 x+16≥8
Step 1: Subtract 16 16 from both sides to isolate the term with the variable:
−2 x+16−16≥8−16−2 x+16−16≥8−16
This simplifies to:
−2 x≥−8−2 x≥−8
Step 2: Divide both sides by −2−2. Remember, when dividing or multiplying both sides of an inequality by a negative number, you must reverse the direction of the inequality sign.
−2 x−2≤−8−2−2 x−2≤−8−2
This simplifies to:
x≤4 x≤4
So, the solution to the inequality is x≤4 x≤4.
Explanation of the sign reversal:
When we multiplied both sides of the inequality by −2−2 in step 2, we needed to flip the inequality sign from ≥≥ to ≤≤. This is because when you multiply or divide both sides of an inequality by a negative number, the direction of the inequality sign flips.
This principle can be understood intuitively. Consider the inequality −2 x≥8−2 x≥8. If we divide both sides by −2−2 without flipping the sign, we would get x≤−4 x≤−4, which would imply that all numbers less than or equal to −4−4 satisfy the inequality. However, if you check the original inequality with a number greater than −4−4, say x=0 x=0, you'll find that it satisfies the inequality (−2(0)+16≥8−2(0)+16≥8), which contradicts the original statement. Therefore, we must flip the sign to maintain the correct relationship between the numbers.
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188546 | https://en.wikipedia.org/wiki/Cold_agglutinin_disease | Jump to content
Cold agglutinin disease
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From Wikipedia, the free encyclopedia
Blood disorder in which cold sensitive antibodies target red blood cells
Medical condition
| Cold agglutinin disease | |
--- |
| Specialty | Hematology |
Cold agglutinin disease (CAD) is a rare autoimmune disease characterized by the presence of high concentrations of circulating cold sensitive antibodies, usually IgM and autoantibodies that are also active at temperatures below 30 °C (86 °F), directed against red blood cells, causing them to agglutinate and undergo lysis. It is a form of autoimmune hemolytic anemia, specifically one in which antibodies bind red blood cells only at low body temperatures, typically 28–31 °C.
When affected people's blood is exposed to cold temperatures (32 °F (0 °C; 273 K) to 50 °F (10 °C; 283 K)), certain proteins that normally attack bacteria (IgM antibodies) attach themselves to red blood cells and bind them together into clumps (agglutination). This eventually causes red blood cells to be prematurely destroyed (hemolysis) leading to anemia and other associated signs and symptoms.
Cold agglutinin disease can be primary (unknown cause) or secondary, due to an underlying condition such as an infection, another autoimmune disease, or certain cancers. Treatment depends on many factors including the severity of the condition, the signs and symptoms present in each person, and the underlying cause.
Cold agglutinin disease was first described in 1957.
Signs and symptoms
[edit]
Symptoms of cold agglutinin disease (CAD) are often triggered or made worse by cold temperatures or a viral infection. Therefore, symptoms generally are worse during winter months. Symptoms may arise suddenly leading to abrupt onset of severe anemia and hemoglobinuria or develop more gradually and insidiously in the background without patient's consciousness and precaution.
Most people with CAD have symptoms of hemolytic anemia (destruction of red blood cells, causing low levels of red blood cells). However, the number of symptoms and severity of symptoms may depend on how severe the anemia is. Signs and symptoms of hemolytic anemia may include:
Tiredness (fatigue)
Dizziness
Headaches
Cold hands and feet
Pale skin
Dark urine
Jaundice
Chest pain
Pain in the back or legs
Vomiting or diarrhea
Heart problems such as an irregular heartbeat (arrhythmia), a heart murmur, an enlarged heart, or heart failure. These may occur because the heart has to work harder to make sure the body gets enough healthy red blood cells.
Many people with CAD also experience pain and bluish coloring of the hands and feet (acrocyanosis) or Raynaud's disease. These symptoms result from slow or poor circulation and can range from mild to disabling.
Other signs and symptoms of CAD may include enlargement of the spleen (splenomegaly) and mottled discoloration of the skin (livedo reticularis).
In people with secondary CAD (associated with another underlying condition), there may be additional signs and symptoms depending on the condition present. For example: Mycoplasma pneumoniae infection (the most common cause of secondary CAD) may cause respiratory symptoms. Various infections or cancers may cause enlarged or swollen lymph nodes.
80–99% of people have these symptoms
Arthralgia
Autoimmunity
Fatigue
Hemolytic anemia
Muscular weakness
Pallor
5–29% of people have these symptoms
Abnormal urine color
Back pain
Diarrhea
Headache
Hepatomegaly
Lymphadenopathy
Nausea and vomiting
Splenomegaly
Cause
[edit]
Cold agglutinin disease can be either primary (unknown cause) or secondary (a result of another pathology/ caused by an underlying condition).
Primary cold agglutinin disease
[edit]
The primary form is caused by excessive cell proliferation of B lymphocytes, characterized by clonal lymphoproliferative disorder.
Primary cold agglutinin disease occurs after the fifth decade of life and peaks prevalence in a person's 70s and 80s.
Secondary cold agglutinin disease
[edit]
Secondary cold agglutinin syndrome occurs when autoantibodies bind to red blood cells, rendering them subject to attack by the complement system. It is a result of an underlying condition potentially associated with either monoclonal cold-reacting autoantibodies or polyclonal cold-reacting autoantibodies predominantly caused by infection or lymphoproliferative disorders. In adults, this is typically due to:
Bacterial infections such as mycoplasma, Legionnaires' disease, syphilis, listeriosis, or E. coli.
Viral infections such Epstein-Barr virus, cytomegalovirus, mumps, varicella, rubella, adenovirus, HIV, influenza, or hepatitis C.
Parasitic infections such as malaria or trypanosomiasis.
Other autoimmune diseases such as systemic lupus erythematosus.
Certain types of cancers such as lymphoma, chronic lymphocytic leukemia, Waldenström macroglobulinemia, multiple myeloma, and Kaposi sarcoma.
In children, cold agglutinin disease is often secondary to an infection, such as Mycoplasma pneumonia, mononucleosis, and HIV.
Inheritance
[edit]
Cold agglutinin disease is not an inherited condition. It is designated as either primary (unknown cause) or secondary (associated with or caused by another condition). In some cases, cold agglutinin may be multifactorial which means that multiple environmental factors and genes likely interact to predispose a person to developing the condition. However, no disease-causing genes have been identified and no familial cases have been reported.
Pathophysiology
[edit]
All individuals have circulating antibodies directed against red blood cells, but their concentrations are often too low to trigger disease (titers under 64 at 4 °C). In individuals with cold agglutinin disease, these antibodies are in much higher concentrations (titers over 1000 at 4 °C).[citation needed]
At body temperatures of 28–31 °C, such as those encountered during winter months, and occasionally at body temperatures of 37 °C, antibodies (generally IgM) bind to the polysaccharide region of glycoproteins on the surface of red blood cells (typically the I antigen or Pr antigen). Binding of antibodies to red blood cells activates the classical pathway of the complement system. If the complement response is sufficient, red blood cells are damaged by the membrane attack complex, an effector of the complement cascade. In the formation of the membrane attack complex, several complement proteins are inserted into the red blood cell membrane, forming pores that lead to membrane instability and intravascular hemolysis (destruction of the red blood cell within the blood vessels).
If the complement response is insufficient to form membrane attack complexes, then extravascular lysis will be favored over intravascular red blood cell lysis. In lieu of the membrane attack complex, complement proteins (particularly C3b and C4b) are deposited on red blood cells. This opsonization enhances the clearance of red blood cell by phagocytes in the liver, spleen, and lungs, a process termed extravascular hemolysis.[citation needed]
Individuals with cold agglutinin disease present with signs and symptoms of hemolytic anemia. Those with secondary agglutinin disease may also present with an underlying disease, often autoimmune.
Diagnosis
[edit]
Detection of antibodies (cold or warm) and /or complement system on RBC from the patient is a direct Coombs antiglobulin test. Detection of antibodies in serum of the patient (still circulating in the blood, that have not yet formed any complexes with RBC) is an indirect Coombs antiglobulin test.[citation needed]
A diagnosis of cold agglutinin disease may be made after several types of tests are performed by a health care provider. In some cases, the diagnosis is first suspected by chance if a routine complete blood count (CBC) detects abnormal clumping (agglutination) of the red blood cells. In most cases, the diagnosis is based on evidence of hemolytic anemia (from symptoms and/or blood tests). A person may also be physically examined for spleen or liver enlargement. An antiglobulin test (called the Coombs test) may be performed to determine the presence of a specific type of antibody. In people with cold agglutinin disease, the Coombs test is almost always positive for immunoglobulin M (IgM).
Treatment
[edit]
Avoid cold weather.
Treat the underlying lymphoma.
No cold drinks; all drinks should be at room temperature (or above).
Requires heater to maintain temperature in cold places.
Treatment with rituximab has been described.
The treatment of cold agglutinin disease depends on many factors including the severity of the condition, the signs and symptoms present in each person, and the underlying cause. For example, in those affected by secondary cold agglutinin disease, it is important to diagnose and treat the underlying condition which may include certain types of cancer; bacterial, viral, or parasitic infections; and/or other autoimmune disease. People with few symptoms and/or mild anemia may not require any specific treatment. These cases are often managed by simply avoiding exposure to the cold.
In severe cases of hemolysis, medical interventions may be necessary. Rituximab (an antibody that selectively reduces specific types of immune cells) is effective in about 60% of cases of severe cold agglutinin disease. Medical researchers have found the response to rituximab is seen on average within 1 to 2 months of treatment and the effect of the treatment lasts for about 1 to 2 years. Rituximab may be used after a second and even a third relapse, however the success rate is less. Combined treatment with rituximab and fludarabine has resulted in higher response rates (76% of cases) and longer periods of remissions (on average, 6.5 years). However the combined treatment may include serious side effects so is presently only recommended when rituximab has not worked alone. Finally, plasmapheresis, which involves filtering blood to remove antibodies, may be useful in acute hemolytic crisis and before surgery requiring hypothermia, however its effect is only short term. Removing the spleen is not recommended for cold agglutinin disease. In addition, because severe cold agglutinin disease requires very high doses of corticosteroids (levels not considered safe), corticosteroid treatment is no longer a recommended treatment in severe cases.
Several possible therapies have been reported in a few case reports to have successfully treated people who are not responding to the treatments listed above.
However more studies need to be performed before the safety and effectiveness of these therapies can be determined.
Sutimlimab (Enjaymo) was approved for medical use in the United States in February 2022.
Prognosis
[edit]
The long-term outlook (prognosis) for people with cold agglutinin disease varies based on many factors including the severity of the condition, the signs and symptoms present in each person and the underlying cause. For example, people with cold agglutinin disease caused by bacterial or viral infections tend to have an excellent prognosis; in these cases, the symptoms typically disappear within 6 months after the infection has resolved. Mild to moderate primary (unknown cause) cold agglutinin disease can also be associated with a good prognosis if excessive exposure to the cold is avoided. Those with cold agglutinin disease caused by HIV infection or certain types of cancer generally have a poor prognosis due to the nature of the underlying condition.
Epidemiology
[edit]
Cold agglutinin disease most commonly affects adults who are of middle age and older. Some studies also report a slight bias in favor of females in the incidence of cold agglutinin disease, particularly in older populations. People with infectious mononucleosis, lymphoproliferative diseases, or mycoplasma pneumonia are more susceptible to this condition. Cold agglutinin disease represents an estimated 16–32% of autoimmune hemolytic anemia, whose annual incidence is estimated to be between 1/35,000-1/80,000 in North America and Western Europe. In patients with infectious mononucleosis, more than 60% of whom develop cold agglutinins disease.
Primary cold agglutinin disease
[edit]
In single-center series, primary CAD has been found to account for 13–15% of the cases of AIHA. In a population-based clinical study of primary CAD in Norway, the prevalence was found to be 16 per million in habitants and the incidence rate 1 per million inhabitants per year. Little is known about possible geographic variations. Median age of CAD patients was 76 years and median age at onset of symptoms was approximately 67 years. The male/female ratio has been reported to be 0.5–0.6 which is not very different from a male/female ratio of 0.72 in an age-matched general population. The frequency of auto-immune disorders other than CAD does probably not differ from what is to be expected in an elderly population with some female predominance. as about 12.5 years from diagnosis and median age at death was 82 years, which implies a life expectancy in these patients similar to that of an age-matched general population. More than 90% of patients with primary CAD have Cold-induced circulatory symptoms ranging from moderate acrocyanosis to severe Raynaud phenomena precipitated even by very slight cold exposure.
History
[edit]
Cold hemagglutination was first reported by Landsteiner in 1903 and found to occur in human beings in 1918. The association of cold hemagglutination with hemolysis was described in 1937 by Rosenthal and Corten. During the 1960s, Dacie and Schubothe published systematic descriptions of 16 CAD patients each. The auto antibodies responsible for hemagglutination at low temperatures, cold agglutinins (CA), may be found in the sera of healthy subjects as well as in patients with AIHA of the cold reactive types. CA bind to erythrocyte surface antigens at a temperature optimum of 0–4 °C. In contrast to polyclonal CA in healthy individuals, monoclonal CA often have a high-thermal amplitude, which contributes to their pathogenicity at temperatures approaching 37 °C.
Binding of CA causes agglutination of erythrocytes and the antigen–antibody complex induces complement (C) activation and hemolysis. Essential clinical manifestations of primary CAD are hemolytic anemia and cold-induced circulatory symptoms. Exact estimates of the severity of anemia and the frequency of cold-induced symptoms, however, have not been provided until recent years.
See also
[edit]
Cold shock response
Donath–Landsteiner hemolytic anemia
List of hematologic conditions
Paroxysmal cold hemoglobinuria
Warm antibody autoimmune hemolytic anemia
References
[edit]
^ Jump up to: a b c "Cold agglutinin disease". Orphanet. Archived from the original on 2015-10-06. Retrieved 2019-02-09.
^ Cold Agglutinin Disease at eMedicine
^ Jump up to: a b c "Anemia, Hemolytic, Cold Antibody". NORD (National Organization for Rare Disorders). 2004-10-28. Archived from the original on 2017-02-21. Retrieved 2019-02-09.
^ Jump up to: a b c d e f g h i j k l m n o p q r s t u v w x y z aa ab "Cold agglutinin disease". Genetic and Rare Diseases Information Center (GARD) – an NCATS Program. 2019-01-24. Archived from the original on 2018-11-23. Retrieved 2019-02-09. This article incorporates text from this source, which is in the public domain.
^ Gertz, Moric A (2006). "Cold agglutinin disease". Haematologica. 91 (4): 439–41. PMID 16585009.
^ Dacie, J. V.; Crookston, J. H.; Christenson, W. N. (1957). "'Incomplete' Cold Antibodies: Role of Complement in Sensitization to Antiglobulin Serum by Potentially Haemolytic Antibodies". British Journal of Haematology. 3 (1): 77–87. doi:10.1111/j.1365-2141.1957.tb05773.x. PMID 13413095. S2CID 38159450.
^ Jump up to: a b "Orphanet: Cold agglutinin disease". Orphanet. 2019-02-08. Archived from the original on 2015-10-06. Retrieved 2019-02-09.
^ Jump up to: a b c Berentsen, Sigbjørn; Randen, Ulla; Tjønnfjord, Geir E. (2015). "Cold Agglutinin-Mediated Autoimmune Hemolytic Anemia". Hematology/Oncology Clinics of North America. 29 (3). Elsevier BV: 455–471. doi:10.1016/j.hoc.2015.01.002. ISSN 0889-8588. PMID 26043385.
^ Jump up to: a b "Cold Agglutinin Disease: Practice Essentials, Pathophysiology, Etiology". Medscape Reference. 2019-02-02. Retrieved 2019-02-09.
^ Jump up to: a b c Swiecicki, Paul L.; Hegerova, Livia T.; Gertz, Morie A. (2013-08-15). "Cold agglutinin disease". Blood. 122 (7): 1114–1121. doi:10.1182/blood-2013-02-474437. ISSN 0006-4971. PMID 23757733.
^ Berentsen, Sigbjørn; Beiske, Klaus; Tjønnfjord, Geir E. (2007-07-21). "Primary chronic cold agglutinin disease: An update on pathogenesis, clinical features and therapy". Hematology. 12 (5): 361–370. doi:10.1080/10245330701445392. ISSN 1607-8454. PMC 2409172. PMID 17891600.
^ Berentsen, Sigbjørn; Tjønnfjord, Geir E. (2012). "Diagnosis and treatment of cold agglutinin mediated autoimmune hemolytic anemia". Blood Reviews. 26 (3): 107–15. doi:10.1016/j.blre.2012.01.002. PMID 22330255.
^ Berentsen, Sigbjørn; Randen, Ulla; Tjønnfjord, Geir E. (2015). "Cold Agglutinin-Mediated Autoimmune Hemolytic Anemia". Hematology/Oncology Clinics of North America. 29 (3): 455–71. doi:10.1016/j.hoc.2015.01.002. PMID 26043385.
^ Swiecicki, P. L.; Hegerova, L. T.; Gertz, M. A. (2013-06-11). "Cold agglutinin disease". Blood. 122 (7). American Society of Hematology: 1114–1121. doi:10.1182/blood-2013-02-474437. ISSN 0006-4971. PMID 23757733.
^ Ma ecka, A.; Troen, G.; Tierens, A.; Ostlie, I.; Ma ecki, J.; Randen, U.; Berentsen, S.; Tjonnfjord, G. E.; Delabie, J. M. A. (2016-05-19). "Immunoglobulin heavy and light chain gene features are correlated with primary cold agglutinin disease onset and activity". Haematologica. 101 (9). Ferrata Storti Foundation (Haematologica): e361 – e364. doi:10.3324/haematol.2016.146126. ISSN 0390-6078. PMC 5060031. PMID 27198717.
^ Jump up to: a b c d e "Cold Agglutinin Disease Workup: Approach Considerations, Complete Blood Cell Count and Peripheral Smear, Reticulocytes and Spherocytes". Medscape Reference. 2019-02-02. Archived from the original on 2017-10-06. Retrieved 2019-02-09.
^ Baines, Andrea C.; Brodsky, Robert A. (2017). "Complementopathies". Blood Reviews. 31 (4): 213–223. doi:10.1016/j.blre.2017.02.003. ISSN 0268-960X. PMC 5513767. PMID 28215731.
^ Reference, Genetics Home (2019-02-05). "What are complex or multifactorial disorders?". Genetics Home Reference. Retrieved 2019-02-10.
^ Jump up to: a b Jaffe, C J; Atkinson, J P; Frank, M M (1976-10-01). "The role of complement in the clearance of cold agglutinin-sensitized erythrocytes in man". The Journal of Clinical Investigation. 58 (4). American Society for Clinical Investigation: 942–949. doi:10.1172/jci108547. ISSN 0021-9738. PMC 333257. PMID 965497.
^ Jump up to: a b Dacie J. The auto-immune haemolytic anaemias: Introduction. In: Dacie J, editor. The haemolytic anaemias. vol. 3. London: Churchill Livingstone; 1992. pp. 1–5.
^ Berentsen, Sigbjørn; Ulvestad, Elling; Gjertsen, Bjørn Tore; Hjorth-Hansen, Henrik; Langholm, Ruth; Knutsen, Håvar; Ghanima, Waleed; Shammas, Fuad Victor; Tjønnfjord, Geir E. (2004). "Rituximab for primary chronic cold agglutinin disease: a prospective study of 37 courses of therapy in 27 patients". Blood. 103 (8): 2925–8. doi:10.1182/blood-2003-10-3597. PMID 15070665.
^ Jump up to: a b c Barcellini, Wilma (2015-07-31). "Current treatment strategies in autoimmune hemolytic disorders". Expert Review of Hematology. 8 (5). Informa UK Limited: 681–691. doi:10.1586/17474086.2015.1073105. ISSN 1747-4086. PMID 26343892. S2CID 27741019.
^ Jump up to: a b c "clinical-features-and-treatment-of-autoimmune-hemolytic-anemia-cold-agglutinins". UpToDate. Retrieved 2019-02-09.
^ "FDA approves treatment for adults with rare type of anemia". U.S. Food and Drug Administration. 4 February 2022. Archived from the original on February 6, 2022. Retrieved 6 February 2022. This article incorporates text from this source, which is in the public domain.
^ Salman Abdullah Aljubran (2019-02-02). "Cold Agglutinin Disease: Practice Essentials, Pathophysiology, Etiology". Medscape Reference. Retrieved 2019-02-11.
^ Jump up to: a b c d e f g h i j k l m Berentsen, Sigbjørn; Beiske, Klaus; Tjønnfjord, Geir E. (2007-07-21). "Primary chronic cold agglutinin disease: An update on pathogenesis, clinical features and therapy". Hematology (Amsterdam, Netherlands). 12 (5). Informa UK Limited: 361–370. doi:10.1080/10245330701445392. ISSN 1607-8454. PMC 2409172. PMID 17891600.
^ Genty, I; Michel, M; Hermine, O; Schaeffer, A; Godeau, B; Rochant, H (2002). "[Characteristics of autoimmune hemolytic anemia in adults: retrospective analysis of 83 cases]". La Revue de Médecine Interne (in French). 23 (11): 901–9. doi:10.1016/S0248-8663(02)00688-4. ISSN 0248-8663. PMID 12481390.
^ Jump up to: a b c d e Berentsen, S; Ulvestad, E; Langholm, R; Beiske, K; Hjorth-Hansen, H; Ghanima, W; Sørbø, JH; Tjønnfjord, GE (2006). "Primary chronic cold agglutinin disease: a population based clinical study of 86 patients". Haematologica. 91 (4): 460–6. ISSN 0390-6078. PMID 16585012.
^ Jump up to: a b c Ulvestad, E; Berentsen, S; Bø, K; Shammas, FV (1999), "Clinical immunology of chronic cold agglutinin disease.", European Journal of Haematology, 63 (4): 259–66, doi:10.1111/j.1600-0609.1999.tb01887.x, ISSN 0902-4441, PMID 10530415, S2CID 34579070
^ Landsteiner K. Über Beziehungen zwischen dem Blutserum und den Körperzellen. Münchener medizinische Wochenschrift. 1903;50:1812–1814.
^ Clough MC, Richter IM. A study of an auto-agglutinin occurring in human serum. Johns Hopkins Hosp Bull. 1918;29:86–93.
^ Rosenthal F, Corten M. Über das Phänomen der Auto-hämagglutination und über die Eigenscaften der Kältehämagglutinine. Folia Haematol (Leipzig) 1937;58:64–90.
^ Jump up to: a b c d e Dacie, J. (1992). "Auto-immune haemolytic anaemia (AIHA): Cold antibody syndromes I: Idiopathic types: Clinical presentation and haematological and serological findings.". In Dacie, J. (ed.). The Haemolytic Anaemias. Vol. 3 (3rd ed.). London, UK: Churchill Livingstone. pp. 1–5. ISBN 978-0-443-03502-9.
^ Jump up to: a b c d Schubothe, H (1966). "The cold hemagglutinin disease". Seminars in Hematology. 3 (1): 27–47. ISSN 0037-1963. PMID 5323366.
^ Gertz, M. A. (2006-01-01). "Cold Hemolytic Syndrome". Hematology. American Society of Hematology. Education Program. 2006 (1). American Society of Hematology: 19–23. doi:10.1182/asheducation-2006.1.19. ISSN 1520-4391. PMID 17124034.
^ Jump up to: a b Olesen, H. (1966). "Thermodynamics of the Cold Agglutinin Reaction". Scandinavian Journal of Clinical and Laboratory Investigation. 18 (1). Informa UK Limited: 1–15. doi:10.3109/00365516609065601. ISSN 0036-5513. PMID 5918670.
^ Zilow, G; Kirschfink, M; Roelcke, D (1994). "Red cell destruction in cold agglutinin disease". Infusionstherapie und Transfusionsmedizin. 21 (6): 410–5. doi:10.1159/000223021. ISSN 1019-8466. PMID 7873920.
^ RØRVIK, K (1954). "The syndrome of high-titre cold haemagglutination; a survey and a case report". Acta Medica Scandinavica. 148 (4): 299–308. doi:10.1111/j.0954-6820.1954.tb01722.x. ISSN 0001-6101. PMID 13157944.
^ Kirschfink, M; Knoblauch, K; Roelcke, D (1994). "Activation of complement by cold agglutinins". Infusionstherapie und Transfusionsmedizin. 21 (6): 405–9. doi:10.1159/000223020. ISSN 1019-8466. PMID 7873919.
^ Nydegger, UE; Kazatchkine, MD; Miescher, PA (1991). "Immunopathologic and clinical features of hemolytic anemia due to cold agglutinins". Seminars in Hematology. 28 (1): 66–77. ISSN 0037-1963. PMID 1708169.
^ Gertz, MA (2006), "Cold agglutinin disease.", Haematologica, 91 (4): 439–41, ISSN 0390-6078, PMID 16585009
External links
[edit]
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| Classification | D ICD-10: D59.1 ICD-9-CM: 283.0 MeSH: D000744 DiseasesDB: 2949 |
| External resources | eMedicine: med/408 |
| Diseases of red blood cells | |
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| ↑ | | | | --- | | Polycythemia | Polycythemia vera | |
| ↓ | | | | | | | | | | | | | | | | | | --- --- --- --- --- --- --- --- | | Anemia | | | | --- | | Nutritional | Micro-: Iron-deficiency anemia + Plummer–Vinson syndrome Macro-: Megaloblastic anemia + Pernicious anemia | | Hemolytic (mostly normo-) | | | | --- | | Hereditary | enzymopathy: Glucose-6-phosphate dehydrogenase deficiency glycolysis + pyruvate kinase deficiency + triosephosphate isomerase deficiency + hexokinase deficiency hemoglobinopathy: Thalassemia + alpha + beta + delta Sickle cell disease/trait Hemoglobin C disease membrane: Hereditary spherocytosis + Minkowski–Chauffard syndrome Hereditary elliptocytosis + Southeast Asian ovalocytosis Hereditary stomatocytosis | | Acquired | | | | --- | | AIHA | Warm antibody autoimmune hemolytic anemia Cold agglutinin disease Donath–Landsteiner hemolytic anemia + Paroxysmal cold hemoglobinuria Mixed autoimmune hemolytic anemia | membrane + paroxysmal nocturnal hemoglobinuria Microangiopathic hemolytic anemia Thrombotic microangiopathy + Hemolytic–uremic syndrome Drug-induced autoimmune Drug-induced nonautoimmune Hemolytic disease of the newborn | | | Aplastic (mostly normo-) | - Hereditary: Fanconi anemia - Diamond–Blackfan anemia Acquired: Pure red cell aplasia Sideroblastic anemia Myelophthisic | | Blood tests | Mean corpuscular volume + normocytic + microcytic + macrocytic Mean corpuscular hemoglobin concentration + normochromic + hypochromic | | | Other | Methemoglobinemia Sulfhemoglobinemia Reticulocytopenia Hereditary persistence of fetal hemoglobin | |
| | |
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| Authority control databases: National | United States Israel |
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188547 | https://www.maths.tcd.ie/pub/Maths/Courseware/NumberTheory/ch13.pdf | Chapter 13 Quadratic fields and quadratic number rings 12.1 Quadratic number fields Definition 12.1. A quadratic number field is a number field of degree 2.
The integer d ∈Z is said to be square-free if it has no square factor, ie a2 | d = ⇒a = ±1.
Thus the square-free integers are ±1, ±2, ±3, ±5, . . . .
Proposition 12.1. Suppose d ̸= 1 is square-free. Then the numbers x + y √ d (x, y ∈Q) form a quadratic number field Q( √ d.
Moreover, every quadratic number field is of this form; and different square-free integers d, d′ ̸= 1 give rise to different quadratic number fields.
Proof. Recall the classic proof that √ d is irrational; √ d = m n = ⇒n2d = m2, and if any prime factor p | d divides the left hand side to an odd power, and the right to an even power.
It is trivial to see that the numbers x + y √ d form a commutative ring, while 1 x + y √ d = x −y √ d (x −y √ d)(x + y √ d) = x −y √ d x2 −dy2 , where x2 −dy2 ̸= 0 since √ d / ∈Q.
12–1 It follows that these numbers form a field; and the degree of the field is 2 since 1, √ d form a basis for the vector space.
Conversely, suppose F is a quadratic number field. Let 1, θ be a basis for the vector space. Then 1, θ, θ2 are linearly independent, ie θ satisfies a quadratic equation aθ2 + bθ + c = 0 (a, b, c ∈Q).
Since F is of degree 2, a ̸= 0, and we can take a = 1. Thus θ = −b ± √ D 2 , with D = b2 −4c.
Now D = a2d, where d is a square-free integer (with a ∈Q). It follows easily that F = Q( √ d).
Finally if d ̸= d′ then Q( √ d) ̸= Q( √ d′). For otherwise √ d′ = x + y √ d for some x, y ∈Q; and so, on squaring, d′2 = x2 + dy2 + 2xy √ d.
But this implies that √ d ∈Q if xy ̸= 0; while y = 0 = ⇒ √ d = x ∈Q, and x = 0 = ⇒d′ = dy2, which is easily seen to be incompatible with d, d′ being square-free.
12.2 Conjugacy We suppose in the rest of the Chapter that we are working in a specific quadratic number field Q( √ d).
Definition 12.2. We define the conjugate of z = x + y √ d to be ¯ z = x −y √ d If d < 0 then this coincides with the complex conjugate; but if d > 0 then both z and ¯ z are real; and z = ¯ z ⇐ ⇒z ∈Q.
12–2 Proposition 12.2. The map z 7→¯ z : Q( √ d) →Q( √ d) is an automorphism of Q( √ d). In fact it is the only such automorphism apart from the trivial map z 7→z.
The proof is identical to that we gave for gaussian numbers.
Definition 12.3. The norm of z = x + y √ d ∈Q( √ d) is N(z) = z¯ z = x2 −dy2.
Proposition 12.3.
1. N(z) ∈Q; 2. N(z) = 0 ⇐ ⇒z = 0; 3. N(zw) = N(z)N(w); 4. If a ∈Q then N(a) = a2; Again, the proof is identical to that we gave for the corresponding result for gaussian numbers.
12.3 Quadratic number rings We want to determine the number ring A = Q( √ d) ∩¯ Z associated to the number field Q( √ d), ie we want to find which numbers x + y √ d are algebraic integers.
Theorem 12.1. Suppose z = x + y √ d ∈Q( √ d).
Then 1. If d ̸≡1 mod 4 z ∈¯ Z ⇐ ⇒z = m + n √ d, where m, n ∈Z.
2. If d ≡1 mod 4 then z ∈¯ Z ⇐ ⇒z = m + n √ d 2 , where m, n ∈Z and m ≡n mod 2.
12–3 Proof. If z = x + y √ d ∈¯ Z then ¯ z = x ∈y √ d ∈¯ Z since z and ¯ z satisfy the same polynomials over Q. Hence z + ¯ z = 2x ∈¯ Z ∩Q = Z.
Also N(z) = z¯ z = x2 −dy2 ∈Z.
It follows that 4dy2 = d(2y)2 ∈Z = ⇒2y ∈Z since d is square-free. (For suppose 2y = m/n, where gcd(m, n) = 1. Then dm2/n2 ∈Z. If the prime p | n then p2 | dm2 = ⇒p2 | d, which is impossible since d is square-free.) Thus z = m + n √ d 2 , where m, n ∈Z. Now N(z) = m2 −dn2 4 ∈Z, ie m2 ≡dn2 mod 4.
If n is even then so is m; and if m is even then so is n, since 4 ∤d. On the other hand if m, n are both odd then m2 ≡n2 ≡1 mod 4.
It follows that d ≡1 mod 4.
In other words, if d ̸≡1 mod 4 then m, n are even, and so z = a + b √ d, with a, b ∈Z.
On the other hand, if d ≡1 mod 4 then m, n are both even or both odd.
It only remains to show that if d ≡1 mod 4 and m, n are both odd then z = m + n √ d 2 ∈¯ Z, 12–4 It is sufficient to show that θ = 1 + √ d 2 ∈¯ Z, since z = (a + b √ d) + θ, where a = (m −1)/2, b = (n −1)/2 ∈Z.
But (θ −1/2)2 = d/4, ie θ2 −θ + (1 −d)/4.
But (1 −d)/4 ∈Z if d ≡1 mod 4. Hence θ ∈¯ Z.
12.4 Units I: Imaginary quadratic fields Suppose F is a number field, with associated number ring A (the algebraic integers in F). By ‘abuse of language’, as the French say, we shall speak of the units of F when we are really referring to the units in A.
Proposition 12.4. Suppose z ∈Q( √ d) is an algebraic integer. Then z is a unit ⇐ ⇒N(z) = ±1.
Proof. Suppose z is a unit, say zw = 1, where w is also an integer. Then N(zw) = N(z)N(w) = N(1) = 12 = 1.
Since N(z), N(w) ∈Z it follows that N(z) = N(w) = ±1.
On the other hand, if N(z) = z¯ z = ±1 then z−1 = ±¯ z ∈¯ Z.
12–5 Theorem 12.2. Suppose d is square-free and d < 0. Then the group of units is finite. More precisely, 1. If d = −1 there are 4 units: ±1, ±i; 2. if d = −3 there are 6 units: ±1, ±ω, ±ω2, where ω = (1 + √−3)/2; 3. in all other cases, there are just 2 units: ±1.
Proof. Suppose ϵ is a unit.
If d ̸≡1 mod 4 then ϵ = m + n √ d (m, n ∈Z).
Thus N(ϵ) = m2 + dn2 = 1, If d < −1 then it follows that m = ±1, n = 0. If d = −1 then there are the additional solutions m = 0, n = ±1, as we know.
If d ≡1 mod 4 then ϵ = m + n √ d 2 , where m, n ∈Z with m ≡n mod 2. In this case, N(ϵ) = m2 −dn2 4 = 1, ie m2 −dn2 = 4.
If d ≤−7 then this implies that m = ±1, n = 0. This only leaves the case d = −3, where m2 + 3n2 = 4.
This has 6 solutions: m = ±2, n = 0, giving ϵ = ±1; and m = ±1, n = ±1, giving ϵ = ±ω, ±ω2.
Units in real quadratic fields (where d > 0) have a very different character, requiring a completely new idea from the theory of diophantine approxima-tion; we leave this to another Chapter.
12–6 |
188548 | https://vallance.chem.ox.ac.uk/pdfs/Equipartition.pdf | The Equipartition Theorem The equipartition theorem states that energy is shared equally amongst all energetically accessible degrees of freedom of a system. This is not a particularly surprising result, and can be thought of as another way of saying that a system will generally try to maximise its entropy (i.e. how ‘spread out’ the energy is in the system) by distributing the available energy evenly amongst all the accessible modes of motion. To give a rather contrived example, consider a container in which we have placed a number of ping-pong balls. Initially the balls are stationary. Imagine we now throw some energy randomly into our box, which will be shared out amongst the ping-pong balls in some way such that they begin to move about. While you might not realise it, intuitively you know what this motion will look like. For example, you would be very surprised if the particle motion looked like this: You would probably predict something more like this: i.e. completely random motion of the ping-pong balls. This is exactly the same result as predicted by the equipartition theorem – the energy is shared out evenly amongst the x, y, and z translational degrees of freedom. The equipartition theorem can go further than simply predicting that the available energy will be shared evenly amongst the accessible modes of motion, and can make quantitative predictions about how much energy will appear in each degree of freedom. Specifically, it states that each quadratic degree of freedom will, on average, possess an energy ½kT. A ‘quadratic degree of freedom’ is one for which the energy depends on the square of some property. Consider the kinetic and potential energies associated with translational, rotational and vibrational energy. Translational degrees of freedom K = ½mv2 Rotational degrees of freedom K = ½Iω2 Vibrational degrees of freedom K = ½mv2 V = ½kx2 These three types of degrees of freedom all have a quadratic dependence on the velocity (or angular velocity in the case of rotation) and therefore all follow the equipartition theorem. Note that when considering vibration in a harmonic oscillator potential (V, above), we consider both the kinetic energy and the potential energy i.e the potential energy counts as an additional degree of freedom. Another point about vibrations is that vibrational motion in molecules is highly quantised, and at room temperature most molecules are in their ground vibrational state and higher levels are not thermally accessible. As a consequence, equipartition contributions from vibrational degrees of freedom need only usually be considered at very high temperatures. Conversely, at room temperature many rotational and translational states are occupied, and they can be treated classically (i.e. as if their energy levels were not quantised) to a very good approximation. A simple derivation of the equipartition result for translational motion We can use the Maxwell-Boltzmann distribution of molecular speeds to determine the average kinetic energy of a particle in a gas, and show that it agrees with the equipartition result. The Maxwell-Boltzmann distribution of molecular speeds is: f(v) = 4π ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ m 2πkT 3/2 v2 exp⎝ ⎜ ⎛ ⎠ ⎟ ⎞ -mv2 2kT The average kinetic energy of a particle in the gas is then < K > = < 1 2 mv2 > = ⌡ ⌠ 0 ∞ 1 2 mv2 f(v) dv Substituting for f(v) and taking all constant terms outside the integral gives < K > = 1 2 m 4π ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ m 2πkT 3/2 ⌡ ⎮ ⌠ 0 ∞ v4 exp⎝ ⎜ ⎛ ⎠ ⎟ ⎞ -mv2 2kT dv We can evaluate the integral by using the general result that ⌡ ⌠ 0 ∞ x2s exp(-ax2) = (2s-1)!!
2s+1 as ⎝ ⎛ ⎠ ⎞ π a 1/2 where n!! indicates a double factorial, n(n-2)(n-4)... etc. Identifying x = v and a = m/2kT in our integral above gives ⌡ ⎮ ⌠ 0 ∞ v4 exp⎝ ⎜ ⎛ ⎠ ⎟ ⎞ -mv2 2kT dv = 3!!
8(m/2kT)2 ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ 2πkT m 1/2 = 3 2 ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ kT m 2 ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ 2πkT m 1/2 Substituting back into our expression for < K > gives < K > = 1 2 m 4π ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ m 2πkT 3/2 3 2 ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ kT m 2 ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ 2πkT m 1/2 = 3 2 kT The average translational kinetic energy of a particle in a gas is therefore 3 2 kT, or ½ kT per translational degree of freedom, in agreement with the equipartition theorem. A more general derivation of the equipartition theorem A general derivation of the equipartition theorem requires statistical mechanics, which is covered in the second year of the M.Chem. course. I will try and provide enough explanation here so that those of you in the first year can also follow what’s going on. The partition function in statistical mechanics tells us the number of quantum states of a system that are thermally accessible at a given temperature. It is defined as: q = ∑ i exp(-Ei/kT) where Ei are the energies of the quantum states i. Once we know the partition function, we can calculate many of the macroscopic properties of our system using standard equations from statistical mechanics. Of particular relevance here, in deriving the equipartition theorem we will use the partition function to calculate the internal energy U associated with a single degree of freedom of the system. Before we do this though, we need to consider the difference between a quantum and a classical system. If we are treating the particle motions classically (which we should do if we’re expecting to derive the classical equipartition theorem), then it doesn’t make much sense to express the partition function as a sum of discrete terms as we have above. Classically, the position and momentum of a particle can vary continuously, and the ‘energy levels’ are also continuous. As a result, the classical partition function takes the form of an integral rather than a sum. q = ⌡ ⌠ exp⎝ ⎛ ⎠ ⎞ -E(x1,x2...,p1,p2...) kT dx1 dx2... dp1 dp2... where the energy E can be a function of the particle positions xi and momenta pi. If we assume that we can write the energy as a sum of contributions from each degree of freedom, then the exponential functional dependence on the energy means that we can separate the integral into the product of integrals over each degree of freedom. i.e. E(x1,x2,...p1,p2...) = E(x1) + E(x2) + ... + E(p1) + E(p2) + ... so exp⎝ ⎛ ⎠ ⎞ -E(x1,x2,...p1,p2...) kT = exp⎝ ⎛ ⎠ ⎞ -E(x1 ) - E(x2) - ... - E(p1 ) - E(p2) ...
kT = exp⎝ ⎛ ⎠ ⎞ -E(x1) kT exp⎝ ⎛ ⎠ ⎞ -E(x2) kT ... exp⎝ ⎛ ⎠ ⎞ -E(p1) kT exp⎝ ⎛ ⎠ ⎞ -E(p2) kT ... and the integral may be written q = ⌡ ⌠ exp⎝ ⎛ ⎠ ⎞ -E(x1) kT dx1 ⌡ ⌠ exp⎝ ⎛ ⎠ ⎞ -E(x2) kT dx2 ... ⌡ ⌠ exp⎝ ⎛ ⎠ ⎞ -E(p1) kT dp1 ⌡ ⌠ exp⎝ ⎛ ⎠ ⎞ -E(p2) kT dp2 ... = q(x1) q(x2) ... q(p1) q(p2) ... The consequence of this is that we have separated the partition function into the product of partition functions for each degree of freedom. In general, we may write the partition function for a single degree of freedom in which the energy depends quadratically on the coordinate x (i.e. E(x) = cx2 with c a constant) as q(x) = ⌡ ⌠ -∞ ∞ exp⎝ ⎛ ⎠ ⎞ -E(x) kT dx = ⌡ ⌠ -∞ ∞ exp⎝ ⎛ ⎠ ⎞ -cx2 kT dx = ⎝ ⎜ ⎛ ⎠ ⎟ ⎞ π kT c 1/2 where we have used the standard integral ⌡ ⌠ -∞ ∞ exp(-ax2) dx = ⎝ ⎛ ⎠ ⎞ π a 1/2 Once we know the partition function, we can calculate the internal energy of the system according to the standard result from statistical mechanics: U = kT2 d(lnq) dT Substituting in our partition function, the internal energy associated with one degree of freedom is therefore U = kT2 d dT ln⎝ ⎜ ⎛ ⎠ ⎟ ⎞ πkT c 1/2 = 1 2 kT2 d dT ln⎝ ⎜ ⎛ ⎠ ⎟ ⎞ πkT c (since ln xn = n ln x) = kT2 2 c πkT πk c (since dln y dx = 1 y dy dx ) = 1 2 kT The energy associated with each quadratic degree of freedom is therefore ½kT, and we have proved the equipartition theorem. |
188549 | https://artofproblemsolving.com/wiki/index.php/1987_AIME_Problems/Problem_14?srsltid=AfmBOopZ7JX7KxQyjvZMDmOBlbBZT0WzgoZmJzrt2WdepsumozDAnoI4 | Art of Problem Solving
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1987 AIME Problems/Problem 14
Contents
[hide]
1 Problem
2 Solution 1 (Sophie Germain Identity)
3 Solution 2 (Completing the Square and Difference of Squares)
4 Solution 3 (Complex Numbers)
4.1 Solution 3.1 (Polar Form)
4.2 Solution 3.2 (Rectangular Form)
5 Solution 4
6 Video Solution by OmegaLearn
7 Video Solution
8 See also
Problem
Compute
Solution 1 (Sophie Germain Identity)
The Sophie Germain Identity states that can be factored as Each of the terms is in the form of Using Sophie Germain, we get that so the original expression becomes which simplifies to Almost all of the terms cancel out! We are left with
~Azjps (Solution)
~MRENTHUSIASM (Minor Reformatting)
Solution 2 (Completing the Square and Difference of Squares)
In both the numerator and the denominator, each factor is of the form for some positive integer
We factor by completing the square, then applying the difference of squares: The original expression now becomes ~MRENTHUSIASM
Solution 3 (Complex Numbers)
In both the numerator and the denominator, each factor is of the form for some positive integer
We factor by solving the equation or
Two solutions follow from here:
Solution 3.1 (Polar Form)
We rewrite to the polar form where is the magnitude of such that and is the argument of such that
By De Moivre's Theorem, we have from which
so
so
By the Factor Theorem, we get We continue with the last paragraph of Solution 2 to get the answer
~MRENTHUSIASM
Solution 3.2 (Rectangular Form)
We rewrite to the rectangular form for some real numbers and
Note that so there are two cases:
We have We need from which or
We have We need from which or
By the Factor Theorem, we get We continue with the last paragraph of Solution 2 to get the answer
~MRENTHUSIASM
Solution 4
We use Sophie Germain's Identity to rewrite the first couple of multiplicands in the numerator and denominator. By Sophie Germain's: If we only had these terms, then the fraction would rewrite to . However, we notice most of the terms cancel, leaving us only with the largest term in the numerator and the smallest term in the denominator (). We hypothesize that this will happen with the fraction as a whole. Then we will only be left with the largest term in the numerator, which is . The fraction simplifies to .
~ cxsmi
Video Solution by OmegaLearn
~ pi_is_3.14
Video Solution
~Michael Penn
See also
1987 AIME (Problems • Answer Key • Resources)
Preceded by
Problem 13Followed by
Problem 15
1•2•3•4•5•6•7•8•9•10•11•12•13•14•15
All AIME Problems and Solutions
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188550 | https://en.wikipedia.org/wiki/Periodic_points_of_complex_quadratic_mappings | Periodic points of complex quadratic mappings - Wikipedia
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1 Definitions
2 Number of periodic points
3 Stability of periodic points (orbit) - multiplier
4 Period-1 points (fixed points)Toggle Period-1 points (fixed points) subsection
4.1 Finite fixed points
4.1.1 Complex dynamics
4.1.2 Special cases
4.1.3 Only one fixed point
4.2 Infinite fixed point
5 Period-2 cyclesToggle Period-2 cycles subsection
5.1 First method of factorization
5.2 Second method of factorization
5.2.1 Special cases
6 Cycles for period greater than 2
7 References
8 Further reading
9 External links
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From Wikipedia, the free encyclopedia
This article describes periodic points of some complex quadratic maps. A map is a formula for computing a value of a variable based on its own previous value or values; a quadratic map is one that involves the previous value raised to the powers one and two; and a complex map is one in which the variable and the parameters are complex numbers. A periodic point of a map is a value of the variable that occurs repeatedly after intervals of a fixed length.
These periodic points play a role in the theories of Fatou and Julia sets.
Definitions
[edit]
Let
f c(z)=z 2+c{\displaystyle f_{c}(z)=z^{2}+c\,}
be the complex quadratic mapping, where z{\displaystyle z} and c{\displaystyle c} are complex numbers.
Notationally, f c(k)(z){\displaystyle f_{c}^{(k)}(z)} is the k{\displaystyle k}-fold composition of f c{\displaystyle f_{c}} with itself (not to be confused with the k{\displaystyle k}th derivative of f c{\displaystyle f_{c}})—that is, the value after the k-th iteration of the functionf c.{\displaystyle f_{c}.} Thus
f c(k)(z)=f c(f c(k−1)(z)).{\displaystyle f_{c}^{(k)}(z)=f_{c}(f_{c}^{(k-1)}(z)).}
Periodic points of a complex quadratic mapping of periodp{\displaystyle p} are points z{\displaystyle z} of the dynamical plane such that
f c(p)(z)=z,{\displaystyle f_{c}^{(p)}(z)=z,}
where p{\displaystyle p} is the smallest positive integer for which the equation holds at that z.
We can introduce a new function:
F p(z,f)=f c(p)(z)−z,{\displaystyle F_{p}(z,f)=f_{c}^{(p)}(z)-z,}
so periodic points are zeros of function F p(z,f){\displaystyle F_{p}(z,f)}: points z satisfying
F p(z,f)=0,{\displaystyle F_{p}(z,f)=0,}
which is a polynomial of degree2 p.{\displaystyle 2^{p}.}
Number of periodic points
[edit]
The degree of the polynomial F p(z,f){\displaystyle F_{p}(z,f)} describing periodic points is d=2 p{\displaystyle d=2^{p}} so it has exactlyd=2 p{\displaystyle d=2^{p}} complex roots (= periodic points), counted with multiplicity.
Stability of periodic points (orbit) - multiplier
[edit]
Stability index of periodic points along horizontal axis
boundaries of regions of parameter plane with attracting orbit of periods 1-6
Critical orbit of discrete dynamical system based on complex quadratic polynomial. It tends to weakly attractingfixed point with abs(multiplier) = 0.99993612384259
The multiplier (or eigenvalue, derivative) m(f p,z 0)=λ{\displaystyle m(f^{p},z_{0})=\lambda } of a rational map f{\displaystyle f} iterated p{\displaystyle p} times at cyclic point z 0{\displaystyle z_{0}} is defined as:
m(f p,z 0)=λ={f p′(z 0),if z 0≠∞1 f p′(z 0),if z 0=∞{\displaystyle m(f^{p},z_{0})=\lambda ={\begin{cases}f^{p\prime }(z_{0}),&{\mbox{if }}z_{0}\neq \infty \{\frac {1}{f^{p\prime }(z_{0})}},&{\mbox{if }}z_{0}=\infty \end{cases}}}
where f p′(z 0){\displaystyle f^{p\prime }(z_{0})} is the first derivative of f p{\displaystyle f^{p}} with respect to z{\displaystyle z} at z 0{\displaystyle z_{0}}.
Because the multiplier is the same at all periodic points on a given orbit, it is called a multiplier of the periodic orbit.
The multiplier is:
a complex number;
invariant under conjugation of any rational map at its fixed point;
used to check stability of periodic (also fixed) points with stability indexa b s(λ).{\displaystyle abs(\lambda ).\,}
A periodic point is
attracting when a b s(λ)<1;{\displaystyle abs(\lambda )<1;}
super-attracting when a b s(λ)=0;{\displaystyle abs(\lambda )=0;}
attracting but not super-attracting when 0<a b s(λ)<1;{\displaystyle 0<abs(\lambda )<1;}
indifferent when a b s(λ)=1;{\displaystyle abs(\lambda )=1;}
rationally indifferent or parabolic if λ{\displaystyle \lambda } is a root of unity;
irrationally indifferent if a b s(λ)=1{\displaystyle abs(\lambda )=1} but multiplier is not a root of unity;
repelling when a b s(λ)>1.{\displaystyle abs(\lambda )>1.}
Periodic points
that are attracting are always in the Fatou set;
that are repelling are in the Julia set;
that are indifferent fixed points may be in one or the other. A parabolic periodic point is in the Julia set.
Period-1 points (fixed points)
[edit]
Finite fixed points
[edit]
Let us begin by finding all finite points left unchanged by one application of f{\displaystyle f}. These are the points that satisfy f c(z)=z{\displaystyle f_{c}(z)=z}. That is, we wish to solve
z 2+c=z,{\displaystyle z^{2}+c=z,\,}
which can be rewritten as
z 2−z+c=0.{\displaystyle \ z^{2}-z+c=0.}
Since this is an ordinary quadratic equation in one unknown, we can apply the standard quadratic solution formula:
α 1=1−1−4 c 2{\displaystyle \alpha {1}={\frac {1-{\sqrt {1-4c}}}{2}}} and α 2=1+1−4 c 2.{\displaystyle \alpha {2}={\frac {1+{\sqrt {1-4c}}}{2}}.}
So for c∈C∖{1/4}{\displaystyle c\in \mathbb {C} \setminus {1/4}} we have two finite fixed points α 1{\displaystyle \alpha {1}} and α 2{\displaystyle \alpha {2}}.
Since
α 1=1 2−m{\displaystyle \alpha {1}={\frac {1}{2}}-m} and α 2=1 2+m{\displaystyle \alpha {2}={\frac {1}{2}}+m} where m=1−4 c 2,{\displaystyle m={\frac {\sqrt {1-4c}}{2}},}
we have α 1+α 2=1{\displaystyle \alpha {1}+\alpha {2}=1}.
Thus fixed points are symmetrical about z=1/2{\displaystyle z=1/2}.
This image shows fixed points (both repelling)
Complex dynamics
[edit]
Fixed points for c along horizontal axis
Fatou set for F(z) = zz with marked fixed point
Here different notation is commonly used:
α c=1−1−4 c 2{\displaystyle \alpha {c}={\frac {1-{\sqrt {1-4c}}}{2}}} with multiplier λ α c=1−1−4 c{\displaystyle \lambda {\alpha _{c}}=1-{\sqrt {1-4c}}}
and
β c=1+1−4 c 2{\displaystyle \beta {c}={\frac {1+{\sqrt {1-4c}}}{2}}} with multiplier λ β c=1+1−4 c.{\displaystyle \lambda {\beta _{c}}=1+{\sqrt {1-4c}}.}
Again we have
α c+β c=1.{\displaystyle \alpha {c}+\beta {c}=1.}
Since the derivative with respect to z is
P c′(z)=d d z P c(z)=2 z,{\displaystyle P_{c}'(z)={\frac {d}{dz}}P_{c}(z)=2z,}
we have
P c′(α c)+P c′(β c)=2 α c+2 β c=2(α c+β c)=2.{\displaystyle P_{c}'(\alpha {c})+P{c}'(\beta {c})=2\alpha {c}+2\beta {c}=2(\alpha {c}+\beta {c})=2.}
This implies that P c{\displaystyle P{c}} can have at most one attractive fixed point.
These points are distinguished by the facts that:
β c{\displaystyle \beta _{c}} is:
the landing point of the external ray for angle=0 for c∈M∖{1/4}{\displaystyle c\in M\setminus \left{1/4\right}}
the most repelling fixed point of the Julia set
the one on the right (whenever fixed point are not symmetrical around the real axis), it is the extreme right point for connected Julia sets (except for cauliflower).
α c{\displaystyle \alpha _{c}} is:
the landing point of several rays
attracting when c{\displaystyle c} is in the main cardioid of the Mandelbrot set, in which case it is in the interior of a filled-in Julia set, and therefore belongs to the Fatou set (strictly to the basin of attraction of finite fixed point)
parabolic at the root point of the limb of the Mandelbrot set
repelling for other values of c{\displaystyle c}
Special cases
[edit]
An important case of the quadratic mapping is c=0{\displaystyle c=0}. In this case, we get α 1=0{\displaystyle \alpha {1}=0} and α 2=1{\displaystyle \alpha {2}=1}. In this case, 0 is a superattractive fixed point, and 1 belongs to the Julia set.
Only one fixed point
[edit]
We have α 1=α 2{\displaystyle \alpha {1}=\alpha {2}} exactly when 1−4 c=0.{\displaystyle 1-4c=0.} This equation has one solution, c=1/4,{\displaystyle c=1/4,} in which case α 1=α 2=1/2{\displaystyle \alpha {1}=\alpha {2}=1/2}. In fact c=1/4{\displaystyle c=1/4} is the largest positive, purely real value for which a finite attractor exists.
Infinite fixed point
[edit]
We can extend the complex planeC{\displaystyle \mathbb {C} } to the Riemann sphere (extended complex plane)C^{\displaystyle \mathbb {\hat {C}} } by adding infinity:
C^=C∪{∞}{\displaystyle \mathbb {\hat {C}} =\mathbb {C} \cup {\infty }}
and extend f c{\displaystyle f_{c}} such that f c(∞)=∞.{\displaystyle f_{c}(\infty )=\infty .}
Then infinity is:
superattracting
a fixed point of f c{\displaystyle f_{c}}:f c(∞)=∞=f c−1(∞).{\displaystyle f_{c}(\infty )=\infty =f_{c}^{-1}(\infty ).}
Period-2 cycles
[edit]
Bifurcation from period 1 to 2 for complex quadratic map
Bifurcation of periodic points from period 1 to 2 for fc(z)=zz +c
Period-2 cycles are two distinct points β 1{\displaystyle \beta {1}} and β 2{\displaystyle \beta {2}} such that f c(β 1)=β 2{\displaystyle f_{c}(\beta {1})=\beta {2}} and f c(β 2)=β 1{\displaystyle f_{c}(\beta {2})=\beta {1}}, and hence
f c(f c(β n))=β n{\displaystyle f_{c}(f_{c}(\beta {n}))=\beta {n}}
for n∈{1,2}{\displaystyle n\in {1,2}}:
f c(f c(z))=(z 2+c)2+c=z 4+2 c z 2+c 2+c.{\displaystyle f_{c}(f_{c}(z))=(z^{2}+c)^{2}+c=z^{4}+2cz^{2}+c^{2}+c.}
Equating this to z, we obtain
z 4+2 c z 2−z+c 2+c=0.{\displaystyle z^{4}+2cz^{2}-z+c^{2}+c=0.}
This equation is a polynomial of degree 4, and so has four (possibly non-distinct) solutions. However, we already know two of the solutions. They are α 1{\displaystyle \alpha {1}} and α 2{\displaystyle \alpha {2}}, computed above, since if these points are left unchanged by one application of f{\displaystyle f}, then clearly they will be unchanged by more than one application of f{\displaystyle f}.
Our 4th-order polynomial can therefore be factored in 2 ways:
First method of factorization
[edit]
(z−α 1)(z−α 2)(z−β 1)(z−β 2)=0.{\displaystyle (z-\alpha {1})(z-\alpha {2})(z-\beta {1})(z-\beta {2})=0.\,}
This expands directly as x 4−A x 3+B x 2−C x+D=0{\displaystyle x^{4}-Ax^{3}+Bx^{2}-Cx+D=0} (note the alternating signs), where
D=α 1 α 2 β 1 β 2,{\displaystyle D=\alpha {1}\alpha {2}\beta {1}\beta {2},\,}C=α 1 α 2 β 1+α 1 α 2 β 2+α 1 β 1 β 2+α 2 β 1 β 2,{\displaystyle C=\alpha {1}\alpha {2}\beta {1}+\alpha {1}\alpha {2}\beta {2}+\alpha {1}\beta {1}\beta {2}+\alpha {2}\beta {1}\beta {2},\,}B=α 1 α 2+α 1 β 1+α 1 β 2+α 2 β 1+α 2 β 2+β 1 β 2,{\displaystyle B=\alpha {1}\alpha {2}+\alpha {1}\beta {1}+\alpha {1}\beta {2}+\alpha {2}\beta {1}+\alpha {2}\beta {2}+\beta {1}\beta {2},\,}A=α 1+α 2+β 1+β 2.{\displaystyle A=\alpha {1}+\alpha {2}+\beta {1}+\beta {2}.\,}
We already have two solutions, and only need the other two. Hence the problem is equivalent to solving a quadratic polynomial. In particular, note that
α 1+α 2=1−1−4 c 2+1+1−4 c 2=1+1 2=1{\displaystyle \alpha {1}+\alpha {2}={\frac {1-{\sqrt {1-4c}}}{2}}+{\frac {1+{\sqrt {1-4c}}}{2}}={\frac {1+1}{2}}=1}
and
α 1 α 2=(1−1−4 c)(1+1−4 c)4=1 2−(1−4 c)2 4=1−1+4 c 4=4 c 4=c.{\displaystyle \alpha {1}\alpha {2}={\frac {(1-{\sqrt {1-4c}})(1+{\sqrt {1-4c}})}{4}}={\frac {1^{2}-({\sqrt {1-4c}})^{2}}{4}}={\frac {1-1+4c}{4}}={\frac {4c}{4}}=c.}
Adding these to the above, we get D=c β 1 β 2{\displaystyle D=c\beta {1}\beta {2}} and A=1+β 1+β 2{\displaystyle A=1+\beta {1}+\beta {2}}. Matching these against the coefficients from expanding f{\displaystyle f}, we get
D=c β 1 β 2=c 2+c{\displaystyle D=c\beta {1}\beta {2}=c^{2}+c} and A=1+β 1+β 2=0.{\displaystyle A=1+\beta {1}+\beta {2}=0.}
From this, we easily get
β 1 β 2=c+1{\displaystyle \beta {1}\beta {2}=c+1} and β 1+β 2=−1{\displaystyle \beta {1}+\beta {2}=-1}.
From here, we construct a quadratic equation with A′=1,B=1,C=c+1{\displaystyle A'=1,B=1,C=c+1} and apply the standard solution formula to get
β 1=−1−−3−4 c 2{\displaystyle \beta {1}={\frac {-1-{\sqrt {-3-4c}}}{2}}} and β 2=−1+−3−4 c 2.{\displaystyle \beta {2}={\frac {-1+{\sqrt {-3-4c}}}{2}}.}
Closer examination shows that:
f c(β 1)=β 2{\displaystyle f_{c}(\beta {1})=\beta {2}} and f c(β 2)=β 1,{\displaystyle f_{c}(\beta {2})=\beta {1},}
meaning these two points are the two points on a single period-2 cycle.
Second method of factorization
[edit]
We can factor the quartic by using polynomial long division to divide out the factors (z−α 1){\displaystyle (z-\alpha {1})} and (z−α 2),{\displaystyle (z-\alpha {2}),} which account for the two fixed points α 1{\displaystyle \alpha {1}} and α 2{\displaystyle \alpha {2}} (whose values were given earlier and which still remain at the fixed point after two iterations):
(z 2+c)2+c−z=(z 2+c−z)(z 2+z+c+1).{\displaystyle (z^{2}+c)^{2}+c-z=(z^{2}+c-z)(z^{2}+z+c+1).\,}
The roots of the first factor are the two fixed points. They are repelling outside the main cardioid.
The second factor has the two roots
−1±−3−4 c 2.{\displaystyle {\frac {-1\pm {\sqrt {-3-4c}}}{2}}.\,}
These two roots, which are the same as those found by the first method, form the period-2 orbit.
Special cases
[edit]
Again, let us look at c=0{\displaystyle c=0}. Then
β 1=−1−i 3 2{\displaystyle \beta {1}={\frac {-1-i{\sqrt {3}}}{2}}} and β 2=−1+i 3 2,{\displaystyle \beta {2}={\frac {-1+i{\sqrt {3}}}{2}},}
both of which are complex numbers. We have |β 1|=|β 2|=1{\displaystyle |\beta {1}|=|\beta {2}|=1}. Thus, both these points are "hiding" in the Julia set. Another special case is c=−1{\displaystyle c=-1}, which gives β 1=0{\displaystyle \beta {1}=0} and β 2=−1{\displaystyle \beta {2}=-1}. This gives the well-known superattractive cycle found in the largest period-2 lobe of the quadratic Mandelbrot set.
Cycles for period greater than 2
[edit]
Periodic points of f(z) = zz−0.75 for period =6 as intersections of 2 implicit curves
The degree of the equation f(n)(z)=z{\displaystyle f^{(n)}(z)=z} is 2 n; thus for example, to find the points on a 3-cycle we would need to solve an equation of degree 8. After factoring out the factors giving the two fixed points, we would have a sixth degree equation.
There is no general solution in radicals to polynomial equations of degree five or higher, so the points on a cycle of period greater than 2 must in general be computed using numerical methods. However, in the specific case of period 4 the cyclical points have lengthy expressions in radicals.
In the case c = –2, trigonometric solutions exist for the periodic points of all periods. The case z n+1=z n 2−2{\displaystyle z_{n+1}=z_{n}^{2}-2} is equivalent to the logistic map case r = 4: x n+1=4 x n(1−x n).{\displaystyle x_{n+1}=4x_{n}(1-x_{n}).} Here the equivalence is given by z=2−4 x.{\displaystyle z=2-4x.} One of the k-cycles of the logistic variable x (all of which cycles are repelling) is
sin 2(2 π 2 k−1),sin 2(2⋅2 π 2 k−1),sin 2(2 2⋅2 π 2 k−1),sin 2(2 3⋅2 π 2 k−1),…,sin 2(2 k−1 2 π 2 k−1).{\displaystyle \sin ^{2}\left({\frac {2\pi }{2^{k}-1}}\right),\,\sin ^{2}\left(2\cdot {\frac {2\pi }{2^{k}-1}}\right),\,\sin ^{2}\left(2^{2}\cdot {\frac {2\pi }{2^{k}-1}}\right),\,\sin ^{2}\left(2^{3}\cdot {\frac {2\pi }{2^{k}-1}}\right),\dots ,\sin ^{2}\left(2^{k-1}{\frac {2\pi }{2^{k}-1}}\right).}
References
[edit]
^Alan F. Beardon, Iteration of Rational Functions, Springer 1991, ISBN0-387-95151-2, p. 41
^Alan F. Beardon, Iteration of Rational Functions, Springer 1991, ISBN0-387-95151-2, page 99
^Some Julia sets by Michael Becker
^On the regular leaf space of the cauliflower by Tomoki Kawahira Source: Kodai Math. J. Volume 26, Number 2 (2003), 167-178.Archived 2011-07-17 at the Wayback Machine
^Periodic attractor by Evgeny DemidovArchived 2008-05-11 at the Wayback Machine
^R L Devaney, L Keen (Editor): Chaos and Fractals: The Mathematics Behind the Computer Graphics. Publisher: Amer Mathematical Society July 1989, ISBN0-8218-0137-6 , ISBN978-0-8218-0137-6
^Period 2 orbit by Evgeny DemidovArchived 2008-05-11 at the Wayback Machine
^Gvozden Rukavina: Quadratic recurrence equations - exact explicit solution of period four fixed points functions in bifurcation diagram
Further reading
[edit]
Geometrical properties of polynomial roots
Alan F. Beardon, Iteration of Rational Functions, Springer 1991, ISBN0-387-95151-2
Michael F. Barnsley (Author), Stephen G. Demko (Editor), Chaotic Dynamics and Fractals (Notes and Reports in Mathematics in Science and Engineering Series) Academic Pr (April 1986), ISBN0-12-079060-2
Wolf Jung: Homeomorphisms on Edges of the Mandelbrot Set. Ph.D. thesis of 2002
The permutations of periodic points in quadratic polynominials by J Leahy
External links
[edit]
Wikibooks has a book on the topic of: Fractals
Algebraic solution of Mandelbrot orbital boundaries by Donald D. Cross
Brown Method by Robert P. Munafo
arXiv:hep-th/0501235v2 V.Dolotin, A.Morozov: Algebraic Geometry of Discrete Dynamics. The case of one variable.
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The Claus Sulfur Recovery Process
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The Claus Sulfur Recovery Process
Sulfur recovery refers to the conversion of hydrogen sulfide (H2S) to elemental sulfur. Hydrogen sulfide is a by-product of processing natural gas and refining high-sulfur crude oils. There are many sulfur recovery technologies available for different applications, however the most common conversion method used is the Claus process. Approximately 90 to 95 percent of recovered sulfur is produced by the Claus process. The Claus process typically recovers 95 to 98 percent of the hydrogen sulfide feedstream. Only the Claus process will be described here, however please contact SRE for information related to any sulfur recovery process.
HYDROGEN SULFIDE (H2S)
SYNONYMS: SOUR GAS OR ACID GAS
‘Sour Gas’ is commonly present in natural gas and oil formations.
At low concentrations, it smells like rotten eggs
EXTREMELY dangerous as it is heavier than air, impairs our ability to smell/detect with increasing concentration, and exposure to >500 ppm can be fatal!
With respect to safety and product specifications, it must be removed from natural gas and petroleum products.
Thus, how do we safely rid ourselves of H2S and make another potentially valuable product?
THE SULFUR RECOVERY UNIT (SRU)
The Sulfur Recovery Unit (SRU) is based on the modified Claus Process, which is a two-step reaction scheme overall. A portion of the total H2S is burned in the Reaction Furnace to form SO2. Then, the H2S and SO2 react, at an optimal 2:1 ratio, to form elemental Sulfur (Sx) across the Claus Reactors. After each catalytic stage, liquid sulfur is recovered in the Claus Condensers. The remaining unreacted H2S and SO2 then proceed to the next stage, where the equilibrium-limited Claus reaction continues in the presence of Claus catalyst.
MEETING SULFUR RECOVERY REQUIREMENTS
In general, gas processing plants and oil refineries are required to recover between 95 and 99.99 percent of the total sulfur introduced to the SRU. A conventional 2-stage SRU, with two Claus Reactors, is typically expected to recover 96+ percent. For a 3-stage SRU, the general expectation is 98+ percent. If necessary, a Tail Gas Cleanup Unit (TGCU) is required to recover the remainder.
Emissions regulations are only becoming more stringent each and every year! That’s where SRE can help out. Our specialized testing services ensure that the operation of your SRU is optimized for long-term performance and reliability purposes!
Performance Evaluation Services
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Featured
The Fundamentals of Sulfur Recovery Units (SRUs)
This video covers fundamentals of Claus process, focusing on process chemistry, design principles, and typical configurations. We'll discuss how to optimize conversion and recovery efficiency, manage key operating parameters like the H2S ratio, and implement best operating practices.
This video covers fundamentals of Claus process, focusing on process chemistry, design principles, and typical configurations. We'll discuss how to optimize conversion and recovery efficiency, manage key operating parameters like the H2S ratio, and implement best operating practices.
Book a 1-1 Meeting with SRE
SRU Performance Evaluation Case Study
The SRU Optimization series, which describes how to optimize the performance of each individual process unit, and covers 2 case studies from past testing experiences at SRE.
This is the final case study, where we’ll look at a testing job from about ten years ago at a refinery, and this is still a big client of ours to this day. So upon initial sampling we found that both feed streams had above threshold amounts of higher hydrocarbons (C2-C6), indicating over-circulation in the amine unit, and inadequate skimming operation of the SWS.
The low quality feed streams (& high contaminants) caused low thermal conversions.
High h/c content caused high CS2 formation. Poor hydrolysis, leading to 1.3% combined losses. And then their ADA signal was a little wonky from the inaccurate TGA readings. They were getting just under 96% recovery efficiency, with pretty much each stage underperforming.
SRE engineers then built a simulation model of the plant on that first night of testing and were able to provide some good guidance the following day. We increased the RF co-firing for better destruction temp, we increased the converter 1 temp, while reducing converter 3’s temp. And finally we lowered the final condenser temperature to 130°C. The increased RF temperature lowered the ammonia and BTEX breakthrough and jumped the thermal conversion efficiency to 50%. Increased the converter 1 hydrolysis by nearly 10%. But, with the increased co-firing rates came elevated gas mass velocities in the condensers, this increases the risk of liquid sulfur carryover to the next stages. Therefore, it was recommended that all the tubesheet blanks be removed to bring the velocities back down to normal. And, since the CS2 formation was still high, we recommended they review the circulation rates in the amine unit and the skimming operation of the SWS unit to attempt to reduce hydrocarbons and BTEX to the SRU.
So when we returned for the next set of quarterly testing, we found a much more accurate TGA, both feed streams were more optimized with less hydrocarbons, and the mass velocities dropped down into the normal range because they removed the tube blanks. This performance test, along with an incinerator optimization study, boosted the client’s recovery efficiency to well above the license, it improved the upstream unit’s operation, and the catalyst life was increased. This granted them huge savings in maintenance and fines, and greatly reduced their greenhouse gas emissions. It ended up being one of many examples of why testing and optimizing an SRU is often much more beneficial than some people might think.
And that brings us to the end of this webinar. Thank you for attending, and if you liked this presentation – consider looking into our on-site training, where we can send an engineer to your location to provide a wide range of courses regarding SRU’s and amine sweetening units. There is just one cost per course, not per person. So now I will take the time to answer any of your questions.
Troubleshooting the SRU Front End
We will start this section by discussing the common issues associated with SRU feed streams. Since the SRU is only as good as what it receives, there are many potential areas for troubleshooting with regards to upsets with the feed stream units.
All SRUs have an Amine Acid Gas feed stream, and many refineries have an additional Sour Water Stripper Acid Gas feed stream. The first indication that there may be an issue with your feed streams is swings within your air demand signal. Large and uncontrollable swings in the combustion air demand signal occur due to fluctuations in the acid gas compositions and flow rates. Typically, when these fluctuate by more than 10%, the Air Demand Signal will fall outside of the optimal range and result in an immediate loss in recovery efficiency.
While plant swings can be unpredictable, they can be prepared for, new gas streams should be introduced slowly for a smoother transition. This can, for example, give the ADA time to adjust to the changing H2S content in the feed stream. Plant upsets can never be fully eliminated, so the SRU instrumentation must be regularly calibrated in order to be prepared for feed stream swings.
Full performance testing of the upstream amine and SWS units is the best way to optimize performance and minimize swings in the feed stream compositions and flows.
Another common issue with SRU feed streams is when they have high hydrocarbon contents. Hydrocarbons require much more oxygen to oxidize in the reaction furnace, which throws off the air demand. Also, more hydrocarbons means more CS2 production, which will hurt the recovery efficiency if it isn’t hydrolyzed. If sufficient amounts of hydrocarbon make it into the converters, catalyst poisoning will occur and deactivate the catalyst.
Excessive accumulation of HCs in the feed streams can be due to:
• Over circulation of amine, or amine temperatures falling below inlet gas temperatures, which results in HC condensation.
• Insufficient flash tank residence time or skimming operation can also increase hydrocarbons in the amine or sws acid gas. The same goes for the reflux drum.
• Sometimes the Reflux Drum purge rate is too low, or sometimes the pump does not start, resulting in LPG carryover to the RF.
The best way to minimize hydrocarbons in the feed streams is to perform regular maintenance, testing, and optimization of the amine and sour water units.
It is very important to maintain the SWS acid gas feed stream above 180F in order to keep this line from plugging. If the temperature is allowed to drop below this value then NH3 salts may begin to precipitate out and can eventually plug the line off entirely. The piping needs to be adequately heat traced and insulated all the way to the Reaction Furnace
The next common troubleshooting item we will discuss is high pressure drops in the SRU. Because SRUs operate at such low pressures, typically below 5 psi, it is not always easy to identify an abnormal drop in pressure across any given vessel. Trending the front-end pressure at the Reaction Furnace is often the best way to observe any abnormal pressures. In the worst case, the high pressure drop will limit the amount of feed gas that can be processed in the unit; thus, limiting the capacity of the train.
Pressure drops most commonly occur in converters and condensers, and they can be caused from the following:
• Sooting off the top of the first Converter while conducting a start-up or shutdown procedure (meaning a sub-stoichiometric natural gas burn has taken place).
• Amine carryover to the SRU can result in the fouling of the first Converter and cause a pressure drop.
• A flooded condenser due to plugged rundown lines/seal legs can cause small pressure drops in the SRU, but can go unnoticed (so its important to check rundown flows multiple times per day).
• Incomplete destruction of Ammonia in RF can result in the formation of ammonia salts, which can plug off condenser tubes.
• Incomplete combustion of acid gas feed stream contaminants can result in the formation of carbonaceous deposits, which will discolor the sulfur and deactivate and plug off the Converter.
• And finally, sulfur laydown is a common cause of pressure drops in SRU converters, and can be caused by operating at sub-dewpoint temperatures, or from liquid sulfur entrainment from the upstream condensers.
Troubleshooting SRU Burners (RF & Reheaters)
Troubleshooting SRU Burners (RF & Reheaters)
From there we will move onto some of the specific issues that can happen with the reaction furnace, starting with the instrumentation.
• Failures of the flame scanners is often due to the lens being coated with sulfur, soot, condensation, or corrosion products. All of this can be avoided with a properly set up purge system. If the purge systems do not have a rotameter for measuring flow then the purge air flow must be proven via another means. Flame scanners are a crucial safety element for the BMS. And also igniters are critical for safe start-ups and shutdowns, so there should be extras kept on-site.
• Fuel gas lines are not used often, so the system should be completely checked out before attempting any start-up or shutdown. Typical problems include plugged lines from moisture and corrosion. Having the fuel metered is highly recommended for safety and for historical records
• Failures in the optical pyrometers is normally due to the lens being coated, which can be rectified by ensuring that there is adequate flow of the purge air followed by a cleaning of the lens
• Ceramic encased thermowells must have a N2 purge to ensure that the corrosive gases from the RF do not penetrate the inner ceramic. This purge must be 3 to 5 psig higher than the pressure in the RF
• Proper air purges must be conducted before each attempt at a startup. If this is not performed, an explosion can occur within the RF- which can damage the main burner and checker wall
• Main Burner damage can be caused by burn-back due to low acid gas or fuel gas flow rates, corrosion, vibration, and plugging. All BMS logic should contain low flow shutdowns for both fuel and acid gas. Hydrocarbon carryover from the amine unit can cause fouling of the vanes and jets which can lead to plugging and vibration problems
• It is important that the Reaction Furnace rain shield is intact so that corrosion does not occur to the carbon steel shell
And this is what severe burner damage looks like, we can see that not only has the refractory lining eroded but also the burner shape itself has deformed. Operating the reaction furnace with a burner like this will create uneven mixing within the chamber and result in cold spots.
And that will bring us to reheaters
• Indirect steam reheaters can suffer from tube leaks, which can be indicated by unusual behavior with respect to process temperatures in the Reheater outlet and the corresponding Converter temperature profile.
• Liquid sulfur entrainment from the upstream Condenser can result in plugging and serious corrosion problems at or near the downstream reheater, this also causes a reduction in the Claus reaction in the converter.
• For Direct-fired reheaters, the burner management system, or BMS, must be reviewed regularly to ensure the proper burn stoichiometry is still being achieved, more specifically this means having the correct air to fuel gas ratio, or acid gas if it’s burning that. A burn ratio above 100 percent will result in excess air going to the converter, which can cause sulfation and risks a sulfur fire starting. Burn ratios that are too low, like below 70% in general, can result in soot deposition onto the catalyst. Our performance testing includes a DFR evaluation with downstream oxygen measurements to verify the correct burn stoich is being achieved.
• And then lastly, like any burner, having fully functional and calibrated instruments is key. We must know that the air and fuel gas flow rates are correct, and that the temperature indicator and control loop is functioning properly. Otherwise the BMS wont be working at its full potential, and we could have false converter inlet temperatures.
Troubleshooting WHB and Claus Condensers
The SRU Troubleshooting series, which describes how to troubleshoot the common issues with each individual process unit, and covers 3 case studies from past testing experiences at SRE.
Now onto the wasteheat boiler and condensers, and seal legs.
• Thermal shocks to the WHB can result in a tubesheet leak so it is very important to make sure that the RF temperatures are controlled when warming up or cooling down. The recommended heating and cooling rate is 50°C per hour.
• Low levels in the WHB are particularly dangerous since exposed tubes are subject to extremely high temperatures from the Reaction Furnace, and it is therefore critical that the level column be reporting accurate boiler feed water levels
Condenser tubes can be fouled by Sulcrete, NH3 salts, or just solid Sulfur
• Like the WHB tube sheet failures are caused by thermal shock and corrosion
• Tubesheet failures can result in freezing/solidifying the sulfur which can plug off the rundowns and seal legs. A tube or tube sheet failure can be identified by steam coming from the lookbox or from seeing slug flows of the sulfur. In the worst case, the solid sulfur ‘bridging’ in the outlet end of the condenser completely plugs off and allows the condenser to flood entirely with liquid sulfur, which is then carried over to the downstream Reheater/Converter
• Failures in a Converter bed screening/grating can result in catalyst falling into the outlet nozzle of the Converter and into the Condenser, finally ending up in the bottom of the seal leg. Depending on the amount of catalyst that has migrated into the seal legs, the legs can be either pulled and cleaned or isolated, then 50 psig steam or compressed air can be used to blow the catalyst out of the seal leg.
• Rodding the rundowns, seal legs, and up into the outlet section of the Condenser may allow for continued operations until the train can be brought down for repairs
The most important factor in designing sulfur condensers is the mass velocity of the process stream. If velocities are too low, sulfur fogging can occur, if velocities are too high, sulfur re-entrainment can occur, both leading to sulfur carryover, and a reduced recovery. Liquid sulfur entrainment is especially detrimental when it occurs in the final condenser, where losses can be upwards of 1.5% off the recovery efficiency.
Capacity evaluations by way of VMG simulation software can determine the optimal mass velocity through the condenser process tubes. These capacity evals should be up to date with the process parameters and are essential when operating at a high turndown, or expecting to in the future.
Here are some pictures of the common condenser problems, in the top left we see condenser tubesheet that’s been pulled and water tested, all the water coming out of the tubes would suggest that it failed this leak test miserably. The top right picture is what sulfur concrete, or ‘sulcrete’ looks like in a rundown, and sulcrete can be formed when there are impurities in the liquid sulfur such as catalyst dust or hydrocarbon soot. And we will typically see a buildup of sulcrete when the rundowns haven’t been cleaned before a start-up and there's still dust and contaminants in there. The bottom left picture is a hole in a multi-pass condenser and we’ll actually cover a case study about that later on. Then the bottom right picture is of sulfur that has built up and partially plugged the outlet of a condenser which could have been due to long continuous sulfur entrainment, or from a tubesheet leak.
Troubleshooting Claus Converters
The SRU Troubleshooting series, which describes how to troubleshoot the common issues with each individual process unit, and covers 3 case studies from past testing experiences at SRE
Converters, where the catalytic reaction takes place.
• If the temperature rise across a converter is observed to be lessening, this indicates that deactivation is occurring. This phenomena may be accompanied by an increase in the temperature rise in the downstream Converter since it now has to do ‘more work’. We also need to be aware of an increase in the RF pressure, which may indicate that one of the Converters has gone ‘sub-dewpoint’ and is plugging off due to liquid sulfur.
There are a number of mechanisms that can cause catalyst deactivation. They include BTEX and methanol poisoning; sulfation; carbon fouling, hydrothermal ageing; and normal ageing.
BTEX poisoning occurs when the RF is unable to completely destroy BTEX components in the acid gas feed stream(s) to the SRU. The resulting effect is a ‘cracking’ or ‘polymerizing’ of these components on the Claus catalyst.
Methanol poisoning is normally due to an SRU with an acid gas by-pass that allows methanol to by-pass the RF.
Both of these poisoning mechanisms are permanent.
Soot deposition and liquid sulfur deposition on the top of the catalyst, results in plugging of the converter beds, but these can be reversible with a heat soak.
Sulfation of catalyst occurs when excessive free oxygen is carried over from either the RF or direct-fired reheaters;
Hydrothermal ageing results when the catalyst is exposed to excessive amounts of water vapor over a long period of time. Although the actual physical mechanism is still not completely understood, it can occur when either excessive steam is introduced into the process and may also occur due to serious tube or tubesheet leaks (BFW being on the shell side) from the Wasteheat boiler or Condensers.
Thermal ageing is caused by ‘thermal excursions’ or ‘sulfur fires’ in the catalyst beds. Temperatures above 1300 F, which are all too easy to obtain during a serious sulfur fire, can result in ‘fusing’ of the catalyst into large solid pieces.
And then more quick notes;
• Exposure to large amounts of condensed water will result in immediate destruction of the catalyst pellets.
• And If there is a sudden step-change reduction in the temperature rise across the first Converter, this would indicate that a severe carryover of contaminants such as hydrocarbons and/or amine has occurred
• A sudden increase in temperature in the Converters will be due to a ‘sulfur fire’. This means that there is free oxygen getting to the Converters either from the RF and/or the Reheaters. If the problem is not corrected quickly, it will be necessary to introduce ‘snuffing steam’ or an inert gas to stop the fire. Do not allow the temperatures to go above 450C in the Converters or there will be severe damage done to the internals.
• Damage in the Converter internals normally shows up as a failure in the catalyst supporting mesh screen and grating, which will result in catalyst falling through and ending up in the Condensers, Rundowns, and Seal Legs (often plugging is the result). In the worst case scenario the carbon steel support beams will bow downward or collapse.
So to quickly sum up the troubleshooting process for catalyst deactivation, once the specific mechanism has been identified, we need to eliminate the cause, whether it be air, soot, or poison going onto the catalyst, or if its just normal aging that we need to minimize with smoother start-up and shutdowns. Once the cause has been taken care of, we need to assess the performance of the converter, is the temperature profile still ok, is the conversion reasonable? Then we can take the necessary planning steps to either do a heat soak to try a reversible, if applicable. Or to plan a catalyst changeout for the next shutdown.
Troubleshooting Sulfur Degassing Pits Webinar
The SRU Troubleshooting series, which describes how to troubleshoot the common issues with each individual process unit, and covers 3 case studies from past testing experiences at SRE.
So now onto the sulfur pits, or degassing pits. The two key elements for degassing are agitation and gas sweeping, where the agitation can be from continuous circulation, from dropping through tray columns, or from an actual agitator which shakes the liquid sulfur. These processes break the H2S out of solution, so then the gas sweeping aspect can move the H2S out of the pits and to the incinerator. Some degassing pits also use a chemical catalyst to improve the process of breaking the H2S out of solution.
So when the degassing becomes less effective and too much H2S is remaining in the sulfur, it could be due to a number of issues in the process. Sometimes the agitation equipment malfunctions, like a broken pump or agitator. And sometimes the air blowers will stop working. The reason for poor degassing can be difficult to identify so often times professional sampling of the liquid sulfur must be done. This process would include sampling each different stage of the degassing process as well as the sulfur pit inlets and outlets. By sampling incrementally and moving upstream, we can quickly identify the problem area and take action to bring the sulfur product back up to spec.
And that brings us to our next case study, which is about the sulfur pits. Last summer we were brought in to a gas plant to investigate increasingly severe glycol losses from an unknown location in the SRU. This plant has 3 SRU trains, with a common MCRC tail gas unit. All 3 condensers used glycol, and all the rundowns and the lines to the pit were heated with glycol, so there were quite a few locations to look for a possible leak.
With a 60:40 glycol to water mixing ratio, the amount of total diluted glycol lost was estimated to be over 1,500 L/day. So we had to quickly come up with method to test for the source of the leak without having to add any tracers to the process, which may impact the quality of the sulfur product.
SRE devised a plan to sample from the pit outlets first, then move backwards (or upstream) until the exact source of the glycol was found. We first tested the pit outlet lines and found glycol in 2 of those 3 outlets. Next we tested each stage within those 2 pits, and we found that the entire #1 pit had glycol contamination, but only the last 2 sections of the #2 Pit. So this led us to conclude that there had to be a leak somewhere between the first and second sections of Pit 2. Then we had to check all the Condenser rundowns that led to Pit #1 to find that leak, which ended up being in the second Condenser’s tubesheet, and was found to be much more severe of a leak. Now that the leaks were identified, they could be repaired during a fairly quick shutdown.
So our team stayed on site to assist with a smooth shutdown of those 2 trains, and we were even able to see how bad the condenser leak really was. That, and the pit tracing leak were patched up, and the SRU was started back up 24 hours later.
In conclusion, both leaks were identified and then repaired in a timely manner. We didn’t have to use tracers, and we analyzed all those liquid sulfur samples on site, with fast results. And we saved them a bunch of money by stopping the glycol losses, and cleaning up the sulfur product back to normal.
Troubleshooting the Incinerator and TGA Webinar
The SRU Troubleshooting series, which describes how to troubleshoot the common issues with each individual process unit, and covers 3 case studies from past testing experiences at SRE.
So now well move onto the Thermal Incinerator, and well start by reviewing some of the possible reasons for seeing an increase of emissions at the Continuous Stack Emissions Monitor (CSEM) which is typically located on the incinerator stack. So increased emissions could be due to:
• An upset in the Claus plant, which results in off-ratio conditions and a reduced overall sulfur recovery
• It could be from Liquid sulfur entrainment in the final Condenser, or from
• Flooding of the final Condenser due to plugged off rundowns or seal legs, which could be from a tube sheet leak, or sulcrete forming in the rundowns, or just sulfur solidifying there.
• Also Deactivation in one or more of the Converters will obviously increase emissions
• Damage or failure of the Reaction Furnace’s main burner would lower the thermal conversion and increase emissions
• And then a sulfur fire in a piece of equipment or a vessel will also create more emissions
And now we’ll talk about common issues that happen with the incinerator itself.
The first is operating with an incorrect burn stoichiometry, which typically leads to too much excess oxygen in the stack. If there is 2 to 4 percent O2 then the stoich is good, but often times the O2 analyzer is unreliable due to the acidic nature of the stack gas, and more specifically the sulfuric acid formation taking place. These analyzers need to calibrated and maintained often, just like all instrumentation. When the O2 levels are too high then that means fuel gas usage is likely too high as well, and the more FG burned, then the more CO2 will be produced. This is why knowing the normal flow rates and temperatures in the stack is key, we must recognize when these conditions are deviating from the optimal levels.
The next common issue with the incinerator is operating at an elevated temperature, which is often associated with the excessive oxygen condition. But it can also be due to design flaws or the need for an updated operating design (or BMS). The threshold temperature in the stack is 650C. SRE performs Stack Top Temperature Reduction studies to determine the minimum allowable temperature that will still minimize the Total Reduced Sulfur. Operating at higher than optimal temperatures can also be caused by inaccurate thermocouple readings, or inaccurate flow measurements of the fuel gas or oxygen. Then lastly, when the Claus plant is performing poorly, then more TRS will be sent to the incinerator. And this can make it difficult for the BMS to maintain the correct stoich and temperature. So similar to the streams feeding the Reaction furnace, the Incinerator is only as good as what it receives in terms of Claus tail gas.
Historically, tail gas analyzer sample lines have been the cause of the bulk of this instrument’s operating failure. The simple reason is that the sample loop needs to be extremely well heated to ensure that the line does not plug with liquid or solid sulfur. The preferred method to use here is electrical heat tracing as it is easily controlled and it’s not affected by process upsets or steam trap failures. The critical area is where the sampling system connects to the process piping, since electrically heated valves are not practical, steam tracing is used at this connecting juncture. This is also why is it highly recommended to have a sample probe here that goes right into the process tail gas stream.
The second most common problem for tail gas analyzers is the loss of sample line flow. This may be due to a failure in the aspiration system or the possibly excessive liquid sulfur in the tail gas line due to liquid sulfur entrainment from the final condenser.
A reduction in the sample line flow rate will increase the lag time of the feedback signal to the trim air controller and can be detected by long and ‘looping’ air demand signal trends on the DCS. Keep in mind that sample lines should be as short as possible to minimize the response time of the sampling system.
By far the most common method for metering the acid gas feed streams and main/trim air flow rate is with orifice meters. These meters are accurate if they are set up properly and have a wide range for flow, which can be particularly useful during startups and shutdowns
Troubleshooting a Low Recovery Efficiency (Case Study)
The SRU Troubleshooting series, which describes how to troubleshoot the common issues with each individual process unit, and covers 3 case studies from past testing experiences at SRE.
So now we will move on to our next case study which is troubleshooting a low recovery efficiency. This was an existing client whom SRE visited several times in the past, they brought us in to troubleshoot their recently unexplained decrease in recovery efficiency.
Upon initial compositional analysis of the process gas, SRE found that the conversion across Converter 2 was low, other than that the SRU was being operated well and nothing else had changed since the previous test period. But the prilling station downstream of the SRU was also reporting high H2S levels in the product, suggesting poor degassing operation, but well get to that on the next slide.
Then the simulation was completed which reconfirmed the reduced activity observed in the second converter bed. The sudden reduction here made no sense, as the catalyst was just replaced 6 months prior, and virtually nothing else had changed which might deactivate the catalyst. Upon further analysis SRE discovered that the converter inlet temperature was above the normal recommendation, so we then installed a sample probe at the inlet of converter 2 and tested there. This analysis revealed a different composition here than at the Converter 1 outlet, when they should have been identical. This analysis, combined with much troubleshooting experience, led SRE to conclude that there must be hole in the multi-pass condenser, which was allowing some process gas from the first pass to leak through into the second pass, and therefore contaminating the Converter 2 inlet stream with RF outlet gas. This would fully explain the reduced conversion across that converter, as well as the increased inlet temperature to it. Now the degassing operation just needed reviewing.
Before the client shut down to repair the condenser leak, SRE wanted to evaluate the degassing pits to determine why 200 ppm of H2S was still in the liquid sulfur product. So we got the operator to make incremental adjustments, first to the agitator, then to the circulation rates from the pumps, and we sampled after each adjustment to observe the effect it would have on the H2S levels. We quickly found that the circulation rate was too low for adequate degassing and that it could not really be increased, it was then discovered that one of the two sulfur pumps was barely moving any sulfur and needed repairs. A pump repair was then planned, along with the condenser repair during the shutdown of the SRU. SRE assisted with the shutdown and then returned to site once the maintenance was completed, we assisted with a smooth start-up, and then performed another test to verify the results of the repairs. The overall recovery efficiency jumped back up above the license of 98%, and the liquid sulfur was back to the normal H2S content of about 50 ppm.
And here is a picture of that hole in the multi-pass condenser, and also a picture of some sulfur pumps on top of the degassing pit. To reiterate the conclusion of this case study, once SRE tested the SRU and degassing pits, the troubleshooting process was quick and prevented any further damage to the SRU, and prevented a sizeable amount of contaminated product from going to the prilling station. So it always pays off to catch the problem early and repair it before things get worse, and many of these SRU issues can only be identified with formal testing to see whats going on under the hood.
And that brings us to the end of this webinar. Thank you for attending, and if you liked this presentation – consider looking into our on-site training, where we can send an engineer to your location to provide a wide range of courses regarding SRU’s and amine sweetening units. There is just one cost per course, not per person.
Cost Savings Full Webinar with Q&A
This video is the full recording of the SRU Cost Savings Webinar from July 14, 2016. The video describes the costs that are associated with the SRU, and methods for reducing them. It also covers 3 case studies from past testing experiences at SRE.
1) Introduction
2) About SRE (2:46)
3) SRU Associated Costs (5:57)
4) Methods of Cost Savings (16:45)
5) Payback Period Examples (30:18)
6) Wrap-up (40:56)
7) Q&A (42:03)
Sulfur Wash Full Webinar
Is plugging getting you down? Did you have a hiccup during your last SRU start-up?
This is the full recording of SRE's Sulfur Wash Webinar
Is plugging getting you down? Did you have a hiccup during your last SRU start-up? SRE has routinely helped our clients through this simple and effective method, avoiding the need of a costly emergency SRU shutdown and a catalyst skimming operation.
This video teaches how to identify the key indicators of a problem and the steps involved in performing a successful sulfur wash. Data compiled from a recent sulfur wash operation is presented.
1) Introduction
2) What is a sulfur wash? (6:07)
3) Identifying the problem (7:11)
4) Steps of a sulfur wash (21:22)
5) Recommended follow-up (31:21)
6) Wrap-up (35:54)
7) Question & Answer (38:35)
SRU Optimization Case Studies
This video explains the process, and the benefits, of performing a complete Incinerator Optimization Study with Sulfur Recovery Engineering (SRE). |
188552 | https://brainly.com/question/4280121 | [FREE] Three circles with centers A, B, and C have respective radii of 50, 30, and 20 inches and are tangent to - brainly.com
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Three circles with centers A, B, and C have respective radii of 50, 30, and 20 inches and are tangent to each other externally. Find the area (in [i n 2]) of the curvilinear triangle formed by the three circles.
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Refer to the figure below which illustrates the given problem. The yellow
shaded area is the curvilinear triangle.
First, we should find angles A, B, and C of triangle ABC.
a = 50 in, b = 70 in, c = 80 in.
From the Law of Cosines: c² = a² + b² - 2ab cosC
cos C = (a² + b² - c²)/(2ab) = (50²+70²-80²)/(2 50 70) = 0.1429
C = arcos(0.1429) = 81.79°.
From the Law of Sines,
sin(B)/b = sin(C)/c
sin(B) = (b/c) sin(C) = (70/80)sin(81.79°) = 0.866
B = arcsin(0.866) = 60°
Because the sum of angles in a triangle is 180°, therefore
A = 180 - (60 + 81.79) = 38.21°.
Calculate the area of ΔABC from the Law of Sines.
Area = (1/2)ab sinC = (1/2)50 70sin(81.79°) = 1732.065 in²
Calculate the areas of arc segments created by circle centers A, B, and C.
Aa = (38.2/360)π(50²) = 833.39 in²
Ab = (60/360)π (30²) = 471.2 in²
Ac = (81.79/360)π(20²) = 285.5 in²
The area of the curvilinear triangle is
1732.065 - (833.39 + 471.2 + 285.5) = 141.975 in²
Answer: 142.0 in² (nearest tenth)
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The area of the curvilinear triangle formed by the three tangent circles is approximately 142.0 square inches. This was determined by calculating the angles of triangle ABC using the Law of Cosines, then finding the area of the triangle, and subtracting the areas of the circular segments from it. The radii of the circles were crucial in these calculations.
Explanation
To find the area of the curvilinear triangle formed by three externally tangent circles with centers at A, B, and C, we will go through a few steps.
Identify the Radii and Distance Between Centers
The circles have the following radii:
Circle A: 50 inches
Circle B: 30 inches
Circle C: 20 inches
The distances between the centers of the circles (AB, BC, and CA) can be calculated as the sum of their respective radii:
Distance AB = Radius A + Radius B = 50 + 30 = 80 inches
Distance BC = Radius B + Radius C = 30 + 20 = 50 inches
Distance CA = Radius C + Radius A = 20 + 50 = 70 inches
Use the Law of Cosines to Find Angles
Now we find angles at each vertex of triangle ABC using the Law of Cosines:
For angle C:
c 2=a 2+b 2−2 ab⋅cos(C)
Let,
a = 50 inches (distance CA)
b = 80 inches (distance AB)
c = 70 inches (distance BC)
We can calculate C:
cos(C)=2 ab a 2+b 2−c 2
cos(C)=2⋅50⋅80 5 0 2+8 0 2−7 0 2=8000 2500+6400−4900=8000 4000=0.5
Thus, C=arccos(0.5)=6 0∘.
For angle A:
A=180−(B+C)=180−(60+38.21)=81.7 9∘
And angle B can similarly be calculated.
Calculate the Area of Triangle ABC
Using the area formula:
Area=2 1ab⋅sin(C)
Substituting in from above:
Area=2 1⋅50⋅80⋅sin(60)=2 1⋅50⋅80⋅2 3≈1732.1 in 2
Calculate Areas of Circular Segments
The area of each circular segment can be calculated using the formula:
Area se g m e n t=360 θ⋅π r 2
Where θ is in degrees:
For Circle A (angle 60°): Area A=360 60⋅π(5 0 2)
For Circle B (angle 80°): Area B=360 80⋅π(3 0 2)
For Circle C (angle 90°): Area C=360 90⋅π(2 0 2)
Final Area of the Curvilinear Triangle
Now, subtract the sum of the circular segment areas from the area of triangle ABC:
Curvilinear area=Area A BC−(Area A+Area B+Area C)≈142.0 in 2
Thus, the area of the curvilinear triangle is approximately 142.0 square inches.
Examples & Evidence
For instance, using the radii of the circles, the distance between centers can be calculated and then angles in triangle ABC determined. Each circle's segment area is calculated based on its radius and the angles we found, which contribute to finding the curvilinear area.
The use of basic geometric principles such as the Law of Cosines for finding angles and circle segment area formulas are well-established methods in mathematics.
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To find the area of the curvilinear triangle formed by the three circles, we can start by drawing the circles and connecting their centers to form an equilateral triangle. The area of the curvilinear triangle is 2500sqrt(3) square inches.
Explanation
To find the area of the curvilinear triangle formed by the three circles, we can start by drawing the circles and connecting their centers to form an equilateral triangle.
The side length of this triangle is equal to the sum of the radii of the circles: 50 + 30 + 20 = 100 inches.
Since the triangle is equilateral, the height can be found using the Pythagorean theorem: height = sqrt(100^2 - 50^2) = sqrt(7500) = 50sqrt(3) inches.
The area of an equilateral triangle can be found using the formula: area = (sqrt(3)/4) side_length^2.
Plugging in the values we found, the area of the curvilinear triangle is (sqrt(3)/4) 100^2 = 2500sqrt(3) square inches.
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Grade 6 Math: Absolute Value
Absolute Value
The absolute value of any number is defined as the distance from zero to that number, without regard to the direction.
The size of a number itself is independent of the absolute value of the number. As positive numbers get larger, the absolute value of the number gets larger. As negative numbers get smaller, the absolute value of the number gets larger. The positive and negative values of any number have the same absolute value.
These properties of positive and negative numbers are illustrated on a number line. In the problems that follow, students are given two numbers and must determine which is the larger or smaller number and which has the larger or smaller absolute value.
The opposite of a number is defined as the number that has the same absolute value as the original number, but the opposite sign. If the original number is positive, the opposite is negative. If the original number is negative, the opposite is positive.
This is illustrated on a number line as a ray whose length represents the absolute value that is rotated about the zero point.
In the problems in this lesson, students are given positive and negative numbers and they must give the opposite of that number.
« Back to Math Topics by Grade Level
Grade 6 Topics
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Percents, Part 2
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Absolute Value
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188554 | https://pubmed.ncbi.nlm.nih.gov/7521884/ | C5a-induced expression of P-selectin in endothelial cells - PubMed
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. 1994 Sep;94(3):1147-55.
doi: 10.1172/JCI117430.
C5a-induced expression of P-selectin in endothelial cells
K E Foreman1,A A Vaporciyan,B K Bonish,M L Jones,K J Johnson,M M Glovsky,S M Eddy,P A Ward
Affiliations Expand
Affiliation
1 Parke-Davis Research Division of Warner-Lambert Company, Ann Arbor, Michigan 48105.
PMID: 7521884
PMCID: PMC295185
DOI: 10.1172/JCI117430
Item in Clipboard
C5a-induced expression of P-selectin in endothelial cells
K E Foreman et al. J Clin Invest.1994 Sep.
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J Clin Invest
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. 1994 Sep;94(3):1147-55.
doi: 10.1172/JCI117430.
Authors
K E Foreman1,A A Vaporciyan,B K Bonish,M L Jones,K J Johnson,M M Glovsky,S M Eddy,P A Ward
Affiliation
1 Parke-Davis Research Division of Warner-Lambert Company, Ann Arbor, Michigan 48105.
PMID: 7521884
PMCID: PMC295185
DOI: 10.1172/JCI117430
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Abstract
Human umbilical vein endothelial cells have recently been shown to respond to C5a with increases in intracellular Ca2+, production of D-myo-inositol 1,4,5-triphosphate and superoxide anion generation. In the current studies, C5a had been found to cause in a time- and dose-dependent manner rapid expression of endothelial P-selectin, secretion of von Willebrand factor, and adhesiveness for human neutrophils. The effects of C5a in P-selectin expression and adhesiveness of neutrophils were similar to the effects of histamine and thrombin on endothelial cells. The adhesiveness of C5a-stimulated endothelium for neutrophils was blocked by anti-P-selectin, but not by antibodies to intercellular adhesion molecule 1, E-selectin, or CD18. A cell-based ELISA technique has confirmed upregulation of P-selectin in endothelial cells exposed to C5a. Binding of C5a to endothelial cells has been demonstrated, with molecules bound being approximately 10% of those binding to neutrophils. By a reverse transcriptase-PCR technique, endothelial cells have been shown to contain mRNA for the C5a receptor. These data suggest that C5a may be an important inflammatory mediator for the early adhesive interactions between neutrophils and endothelial cells in the acute inflammatory response.
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References
J Clin Invest. 1988 Jul;82(1):360-3 - PubMed
Science. 1993 Dec 3;262(5139):1580-2 - PubMed
J Biol Chem. 1989 Jan 25;264(3):1760-6 - PubMed
Blood. 1989 Apr;73(5):1109-12 - PubMed
J Biol Chem. 1989 May 15;264(14):7768-71 - PubMed
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188555 | https://arxiv.org/abs/1811.05032 | [1811.05032] Gelfand-Fuchs cohomology in algebraic geometry and factorization algebras
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arXiv:1811.05032 (math)
[Submitted on 12 Nov 2018 (v1), last revised 25 Jan 2022 (this version, v2)]
Title:Gelfand-Fuchs cohomology in algebraic geometry and factorization algebras
Authors:Benjamin Hennion, Mikhail Kapranov
View a PDF of the paper titled Gelfand-Fuchs cohomology in algebraic geometry and factorization algebras, by Benjamin Hennion and 1 other authors
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Abstract:Let X be a smooth affine variety over a field k of characteristic 0 and T(X) be the Lie algebra of regular vector fields on X. We compute the Lie algebra cohomology of T(X) with coefficients in k. The answer is given in topological terms relative to any embedding of k into complex numbers and is analogous to the classical Gelfand-Fuks computation for smooth vector fields on a C-infinity manifold. Unlike the C-infinity case, our setup is purely algebraic: no topology on T(X) is present. The proof is based on the techniques of factorization algebras, both in algebro-geometric and topological contexts.
Comments:Revised version, 101 pages
Subjects:Algebraic Geometry (math.AG)
Cite as:arXiv:1811.05032 [math.AG]
(or arXiv:1811.05032v2 [math.AG] for this version)
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arXiv-issued DOI via DataCite
Submission history
From: Mikhail Kapranov [view email]
[v1] Mon, 12 Nov 2018 22:56:03 UTC (66 KB)
[v2] Tue, 25 Jan 2022 09:21:11 UTC (98 KB)
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188556 | https://artofproblemsolving.com/wiki/index.php/Shoelace_Theorem?srsltid=AfmBOopUY2azk5H1S2C9Vr72qG3wCalXKKETNCKIMx0pZpQvWOi_LnoV | Art of Problem Solving
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Shoelace Theorem
The Shoelace Theorem is a nifty formula for finding the area of a simple polygon given the coordinates of its vertices.
Contents
[hide]
1 Theorem
2 Other Forms
3 Proof 1
3.1 Proof of claim 1:
3.2 Proof:
4 Proof 2
5 Proof 3
6 Problems
6.1 Introductory
6.2 Exploratory
7 External Links
Theorem
Suppose the polygon has vertices , , ... , , listed in clockwise order. Then the area () of is
You can also go counterclockwise order, as long as you find the absolute value of the answer.
The Shoelace Theorem gets its name because if one lists the coordinates in a column, and marks the pairs of coordinates to be multiplied, the resulting image looks like laced-up shoes.
Other Forms
This can also be written in form of a summation or in terms of determinants as which is useful in the variant of the Shoelace theorem. Note here that and .
The formula may also be considered a special case of Green's Theorem
where and so .
Proof 1
Claim 1: The area of a triangle with coordinates , , and is .
Proof of claim 1:
Writing the coordinates in 3D and translating so that we get the new coordinates , , and . Now if we let and then by definition of the cross product .
)
Proof:
We will proceed with induction.
By claim 1, the shoelace theorem holds for any triangle. We will show that if it is true for some polygon then it is also true for .
We cut into two polygons, and . Let the coordinates of point be . Then, applying the shoelace theorem on and we get
Hence
as claimed.
~ShreyJ
Proof 2
Let be the set of points belonging to the polygon. We have that where . The volume form is an exact form since , where Using this substitution, we have Next, we use the Theorem of Stokes to obtain We can write , where is the line segment from to . With this notation, we may write If we substitute for , we obtain If we parameterize, we get Performing the integration, we get More algebra yields the result
Proof 3
This is a very nice approach that directly helps in understanding the sum as terms which are areas of trapezoids.
The proof is in this book:
(The only thing that needs to be slightly modified is that one must shift the entire polygon up by k, until all the y coordinates are positive, but this term gets canceled in the resulting sum.)
Problems
Introductory
In right triangle , we have , , and . Medians and are drawn to sides and , respectively. and intersect at point . Find the area of .
Exploratory
Observe that is the area of a triangle with vertices and is the volume of a tetrahedron with vertices . Does a similar formula hold for Dimensional triangles for any ? If so how can we use this to derive the D Shoelace Formula?
External Links
A good explanation and exploration into why the theorem works by James Tanton:
Nice geometric approach and discussion for proving the 3D Shoelace Theorem by Nicholas Patrick and Nadya Pramita:
Nice integral approach for proving the 3D Shoelace Theorem (ignoring sign of volume) by @george2079: AOPS
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188557 | https://patents.google.com/patent/US3365510A/en | US3365510A - Low temperature conversion of acetylene to pure benzene - Google Patents
US3365510A - Low temperature conversion of acetylene to pure benzene - Google Patents
Low temperature conversion of acetylene to pure benzene Download PDF
Info
Publication number US3365510A US3365510A US424804A US42480465A US3365510A US 3365510 A US3365510 A US 3365510A US 424804 A US424804 A US 424804A US 42480465 A US42480465 A US 42480465A US 3365510 A US3365510 A US 3365510A Authority US United States Prior art keywords benzene acetylene catalyst oxide catalysts Prior art date 1965-01-11 Legal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)Expired - Lifetime Application number US424804A Inventor John E Noakes Current Assignee (The listed assignees may be inaccurate. Google has not performed a legal analysis and makes no representation or warranty as to the accuracy of the list.) JOHN E NOAKES Original Assignee John E. Noakes Priority date (The priority date is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the date listed.)1965-01-11 Filing date 1965-01-11 Publication date 1968-01-23 1965-01-11 Application filed by John E. Noakes filed Critical John E. Noakes 1965-01-11 Priority to US424804A priority Critical patent/US3365510A/en 1968-01-23 Application granted granted Critical 1968-01-23 Publication of US3365510A publication Critical patent/US3365510A/en 1985-01-23 Anticipated expiration legal-status Critical Status Expired - Lifetime legal-status Critical Current
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UHOVQNZJYSORNB-UHFFFAOYSA-N Benzene Chemical compound C1=CC=CC=C1 UHOVQNZJYSORNB-UHFFFAOYSA-N 0.000 title description 191
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PQQKPALAQIIWST-UHFFFAOYSA-N oxomolybdenum Chemical compound[Mo]=O PQQKPALAQIIWST-UHFFFAOYSA-N 0.000 description 1
239000002245 particle Substances 0.000 description 1
239000012071 phase Substances 0.000 description 1
238000006116 polymerization reaction Methods 0.000 description 1
238000010791 quenching Methods 0.000 description 1
230000000171 quenching effect Effects 0.000 description 1
230000001105 regulatory effect Effects 0.000 description 1
230000004044 response Effects 0.000 description 1
230000035945 sensitivity Effects 0.000 description 1
239000000741 silica gel Substances 0.000 description 1
229910002027 silica gel Inorganic materials 0.000 description 1
239000007787 solid Substances 0.000 description 1
238000004611 spectroscopical analysis Methods 0.000 description 1
Classifications
C—CHEMISTRY; METALLURGY
C07—ORGANIC CHEMISTRY
C07C—ACYCLIC OR CARBOCYCLIC COMPOUNDS
C07C2/00—Preparation of hydrocarbons from hydrocarbons containing a smaller number of carbon atoms
C07C2/02—Preparation of hydrocarbons from hydrocarbons containing a smaller number of carbon atoms by addition between unsaturated hydrocarbons
C07C2/42—Preparation of hydrocarbons from hydrocarbons containing a smaller number of carbon atoms by addition between unsaturated hydrocarbons homo- or co-oligomerisation with ring formation, not being a Diels-Alder conversion
C07C2/48—Preparation of hydrocarbons from hydrocarbons containing a smaller number of carbon atoms by addition between unsaturated hydrocarbons homo- or co-oligomerisation with ring formation, not being a Diels-Alder conversion of only hydrocarbons containing a carbon-to-carbon triple bond
C—CHEMISTRY; METALLURGY
C07—ORGANIC CHEMISTRY
C07C—ACYCLIC OR CARBOCYCLIC COMPOUNDS
C07C2523/00—Catalysts comprising metals or metal oxides or hydroxides, not provided for in group C07C2521/00
C07C2523/16—Catalysts comprising metals or metal oxides or hydroxides, not provided for in group C07C2521/00 of arsenic, antimony, bismuth, vanadium, niobium, tantalum, polonium, chromium, molybdenum, tungsten, manganese, technetium or rhenium
C07C2523/20—Vanadium, niobium or tantalum
C—CHEMISTRY; METALLURGY
C07—ORGANIC CHEMISTRY
C07C—ACYCLIC OR CARBOCYCLIC COMPOUNDS
C07C2523/00—Catalysts comprising metals or metal oxides or hydroxides, not provided for in group C07C2521/00
C07C2523/16—Catalysts comprising metals or metal oxides or hydroxides, not provided for in group C07C2521/00 of arsenic, antimony, bismuth, vanadium, niobium, tantalum, polonium, chromium, molybdenum, tungsten, manganese, technetium or rhenium
C07C2523/24—Chromium, molybdenum or tungsten
Y—GENERAL TAGGING OF NEW TECHNOLOGICAL DEVELOPMENTS; GENERAL TAGGING OF CROSS-SECTIONAL TECHNOLOGIES SPANNING OVER SEVERAL SECTIONS OF THE IPC; TECHNICAL SUBJECTS COVERED BY FORMER USPC CROSS-REFERENCE ART COLLECTIONS [XRACs] AND DIGESTS
Y10—TECHNICAL SUBJECTS COVERED BY FORMER USPC
Y10S—TECHNICAL SUBJECTS COVERED BY FORMER USPC CROSS-REFERENCE ART COLLECTIONS [XRACs] AND DIGESTS
Y10S585/00—Chemistry of hydrocarbon compounds
Y10S585/929—Special chemical considerations
Y10S585/942—Production of carbonium ion or hydrocarbon free-radical
Y—GENERAL TAGGING OF NEW TECHNOLOGICAL DEVELOPMENTS; GENERAL TAGGING OF CROSS-SECTIONAL TECHNOLOGIES SPANNING OVER SEVERAL SECTIONS OF THE IPC; TECHNICAL SUBJECTS COVERED BY FORMER USPC CROSS-REFERENCE ART COLLECTIONS [XRACs] AND DIGESTS
Y10—TECHNICAL SUBJECTS COVERED BY FORMER USPC
Y10S—TECHNICAL SUBJECTS COVERED BY FORMER USPC CROSS-REFERENCE ART COLLECTIONS [XRACs] AND DIGESTS
Y10S585/00—Chemistry of hydrocarbon compounds
Y10S585/929—Special chemical considerations
Y10S585/943—Synthesis from methane or inorganic carbon source, e.g. coal
Definitions
This invention in one of its aspects pertains to the synthesis of benzene.
the invention in a more specific aspect pertains to the preparation of benzene of such high purity that it can be usedin radiocarbon dating.
an ambient temperature process which converts acetylene to benzene in a manner ideally suited for liquid scintillation radiocarbon dating. Impure benzene cannot be tolerated for liquid scintillation counting and radiocarbon dating. Purity is therefore of primary importance, and of secondary importance is the greater chance in known methods for carbon isotope fractionation.
a process for synthesizing pure benzene from acetylene by passing the acetylene over the surface of an alumina based promoter at a temperature below 150 C., and as low as room temperature (25 C.)
the promoter in part is activated alumina, that is, aluminum oxide which has been calcined, or otherwise heated with steam or air, etc., to raise its surface area above 29 say, 30 to 400 square meters per gram or higher, preferably to 350.
This promoter also contains at least 3, and preferably 5 to 25 weight percent based on the total, of the metals oxide possessing strong Lewis acid properties.
Such oxides are those oxides in group V-B and VIB of the Periodic Table having valences greater than four.
the valence V of the metal is 4 V 7.
Such metal oxides are V205, Nb205, T3205, CrO M003, W03 and M0205. Normally there is no advantage to employing more than 30 weight percent metal oxide based on the catalyst.
a postulated mechanism for the functioning of these catalysts is that the presence of cations of +5 or greater valence on large surface area give maximum exposure of strong Lewis acid sites for attachments of acetylene molecules. Acetylene molecules in the gaseous state coming in the vicinity of the +5 cation site are then polarized. In addition, induced polarization is set up in the attached acetylene molecule. As a result additional acetylene molecules are attracted to this attached acetylene molecule. When three molecules of acetylene are thus attached, angular rotation is possible and a ring structure of benzene is thereby formed. At higher temperatures, say above 100 C.
the ring formation sometimes does not take place until after more than three acetylene molecules become so attached to a catalyst site, resulting in ethyl benzene being synthesized, thereby decreasing the benzene yield.
a temperature of 25 C. to 65 C.is preferred, in the absence of reaction chamber cooling to compensate for the exotherm.
the acetylene is passed through the catalyst mass.
the catalyst thus contacted contains an alumina base material making available a large surface area.
vanadium oxide on alumina has been found to give excellent results with alumina having a surface area of 214 square meters per gram, but poor results using the same amount of vanadium oxide on alumina having a surface area of one square meter per gram.
the rate of acetylene gas introduction over the preheated catalyst is related to both the temperature and the benzene synthesis yield. Investigation of the reaction shows that the catalyst reaction temperature and the acetylene flow rate are directly related, whereas the benzene yield and the acetylene flow rate are inversely related. The exact flow rate consequently depends upon reaction conditions, but normally it will be in the range of 5 cc. per min. to 25 cc. per min.
the catalyst involved is heated at an elevated temperature (300 C. under vacuum) to extract water vapor.
elevated temperature 300 C. under vacuum
the dehydration step increases benzene synthesis and enables pure benzene to be extracted devoid of any water.
the catalyst is then cooled and stored under nitrogen gas until ready for use.
benzene can be extracted from the catalyst by placing the reaction column in an oven heated to about 100 C. An applied vacuum is pulled on the column to extract the benzene, which is collected in a liquid nitrogen freeze-out trap. An extraction of 100 C. is used so that if any higher molecular material is synthesized its boiling point will keep it in the column.
benzene can be made at ambient temperatures
another advantage of this invention is that the benzene produced by this process is extremely pure. Benzene extracted from the reaction column was checked for purity by infra-red and gas chromatography. The only product found other than benzene has been ethyl benzene and only in trace amounts. In addition the ethyl benzene was found to occur only at high temperatures when a large acetylene fiow rate was used. As the yield increases only pure benzene is obtained. In addition, mass spectrometric analysis of benzene produced shows that no carbon isotope fractionation was present in either low or high yield benzene reactions.
the catalysts were first dehydrated. This pretreatment was limited to the elimination of associated water so that no water would be present in the formed benzene, the catalyst being heated for two hours under vacuum at 200 C.
the acetylene catalysis was accomplished by passing acetylene gas through the catalyst mass d at a controlled rate as given in the following tables.
the acetylene was obtained from tank acetylene and the flow rate regulated by a lv.athenson Gas Proportioner with an accuracy of i2%.
the catalytic acetylene reaction was exothermic and was accompanied by a darkening of the catalyst material. To insure that no excessive acetylene pressures would build up during the reaction an exhaust valve and displacement trap were connected to the reaction column to allow gas of greater than one atmosphere to escape and to be coll cted.
Feed stream Acetylene.
Table I also shows that catalysts are improved by cobait modification. One to five percent is usually employed based on the catalyst.
additional cobalt molybdate catalysts were purchased and examined. For the purposes of comparison with catalysts VII and VIII of Table I, several additional catalysts were selected. The compositions of these catalysts were as follows:
Feed stream Acetylene.
Infra-red analyses were made with a Beckman IR-lO model using a CaF cell with a 200 length light path.
the range of wave numbers scanned for each analysis was from 1000 to 4000.
the presence of acetone and trace amounts of water and ethyl benzene was confirmed only in the benzene samples of low yield.
the ratio of atomic mass units (79/78) of benzene were determined for seven benzene samples, ranging from low to high yields. No variation in the C/C mass ratios was observed for low or high yield samples. The average C/C value for the seven benzene samples analyzed was 0.0655. The calculated percent C abundances for acetylene and benzene to be 1.07 percent. The percent C abundance calculations using the mass ratio data is reliable for three significant figures.
a low temperature method for the quantitative synthesis of a pure benzene which has no energy quenching products present and which does not require purifying distillation procedures.
the benzene thus produced is ideally suited as a scintillation solvent.
a process for synthesizing substantially pure benzene by cyclic trimerization of acetylene which comprises trimerizing the acetylene in the gaseous phase at a temperature of 20 to 150 C. by bringing it into contact at said temperature with an oxide of one of the metals of Group VB and Group VIB of the Pen'odic Table having a valence V of 4 V 7, the oxide being supported on an activated alumina carrier, the metal being in said valence state in the oxide to confer Lewis acid properties on the combination, said carrier containing 3 to 30 weight percent of said oxide based on the total.
the alumina has a surface area of 100 to 350 square meters per gram, wherein the Group VB-Group VI-B metal oxide is employed in an amount of 5 to 25 weight percent based on the total, and wherein said oxide is present in combination with 1 to 5 weight percent based on the total of C00.
the process of claim 3 wherein the oxide conferring Lewis acid properties is V 0 '7.
the process of claim 3 wherein the oxide conferring Lewis acid properties is Ta O 8.
the process of claim 3 wherein the oxide conferring Lewis acid properties is CrO 9.
the process of claim 3 wherein the oxide conferring Lewis acid properties is M00 10.
the process of claim 3 wherein the oxide conferring Lewis acid properties is W0 References Cited UNITED STATES PATENTS 2,846,490 12/1953 Witt 260673 2,990,434 6/1961 Smith 260673 PAUL M. COUGHLAN, ]R., Primary Examiner.
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Chemical & Material Sciences (AREA)
Organic Chemistry (AREA)
Organic Low-Molecular-Weight Compounds And Preparation Thereof (AREA)
Description
United States Patent 3,365,510 LOW TEMPERATURE CONVERSEON 0F ACETYLENE TO PURE BENZENE John E. Noakes, P.0. Box 117,
Oak Ridge, Tenn. 37830 No Drawing. Filed Jan. 11, 1965, Ser. No. 424,804 Claims. (Cl. 260673) ABSTRACT OF THE DISCLOSURE High purity benzene, particularly useful in radiocarbon dating because of its purity, is made using a low temperature, C. to 100 C., pure gas phase system. Under these reaction conditions a metal ion having a valence of +5 or +6 induces polarization and initiates the trimerization of acetylene gas to benzene. Oxides of metals of Group V-B and VIB of the Periodic Table are employed on an activated alumina carrier.
This invention in one of its aspects pertains to the synthesis of benzene. In a more specific aspect the invention relates to the preparation of benzene of such high purity that it can be usedin radiocarbon dating.
When cosmic rays enter the earths atmosphere they collide with various atoms in the air to form neutrons,
mesons, protons and other particles. Some of the resulting neutrons strike nitrogen atoms causing them to disintegrate, giving olf a proton. The result is carbon 14 or radiocarbon. Living things constantly absorb carbon 14 from the atmosphere. This discovery of carbon 14 in nature led to its development as an archaeological and geological calendar.
In dating organic matter by its carbon 14 content, the
successfully in 1954. The high sensitivity and precision in radiocarbon dating by use of liquid scintillation are due to the large amount of carbon which can be incorporated into liquid samples and counted with high efficiency. The conversion of carbon samples to benzene is preferred because of benzenes excellent liquid scintillation counting properties (high energy transmitting property and low optical density) and its high carbon content (92 percent). With recent improvements in liquid scintillation spectrometry, such as summation counting, and the development of wide response quartz-faced phototubes (3000 A.- 4500 A.), counting efiiciencies approaching 90 percent haveb'een realized with benzene samples. Nevertheless, in the past the advantages of liquid scintillation have been offset by the diificulties involved in the complex synthesis required for liquid sample preparation. Thus, benzene has been partially formed by the cyclic trimerization of acetylene at 500 C. However, this process has the disadvantage that the high temperature required at this stage enhances the formation of compounds other than benzene which are poor scintillating solvents and also increases the possibility of carbon isotope fractionation. A process is preferred whereby benzene can be synthesized from acetylene gas at near ambient temperature which will yield only pure benzene and no carbon isotope fractionation.
The conversion of acetylene to benzene at a temperature below 500 C. has been described in J. Amer. Chem. Soc., 79, 3294, 1957. This method involves passing the acetylene gas over a catalyst of silica gel, reacted, after removal of absorbed water, with diborane. The fact that boron hydrides have a catalytic eifect on the polymerization of acetylenes poses the question of whether the attached boron, the aluminum, or the silicon atoms are the sites of attachment for acetylene molecule. This invention is a result of the study of this reaction. While the affinity of boron for electrons would seem to be sufficient to initiate attachment and to induce polarization and cyclization of acetylene to benzene, it appears that the close proximity of the boron hydride group to the aluminum and silicon atoms shifts their pi cloud to the boron, thereby increasing the Lewis acid character of the +4 silicon atoms. Based on these studies it has been found that outstanding alumina based promoters or catalysts for benzene synthesis can be made by combining alumina with a metal oxide which possesses strong Lewis acid properties. By this invention, no hazardous diborane gas or other activating gas is needed to enhance the Lewis acid character of the promoter.
Thus, in accordance with this invention an ambient temperature process is provided which converts acetylene to benzene in a manner ideally suited for liquid scintillation radiocarbon dating. Impure benzene cannot be tolerated for liquid scintillation counting and radiocarbon dating. Purity is therefore of primary importance, and of secondary importance is the greater chance in known methods for carbon isotope fractionation. According to the practice of the invention a process is provided for synthesizing pure benzene from acetylene by passing the acetylene over the surface of an alumina based promoter at a temperature below 150 C., and as low as room temperature (25 C.) The promoter, in part is activated alumina, that is, aluminum oxide which has been calcined, or otherwise heated with steam or air, etc., to raise its surface area above 29 say, 30 to 400 square meters per gram or higher, preferably to 350. This promoter also contains at least 3, and preferably 5 to 25 weight percent based on the total, of the metals oxide possessing strong Lewis acid properties. Such oxides are those oxides in group V-B and VIB of the Periodic Table having valences greater than four. Thus in the oxide employed herein the valence V of the metal is 4 V 7. Such metal oxides are V205, Nb205, T3205, CrO M003, W03 and M0205. Normally there is no advantage to employing more than 30 weight percent metal oxide based on the catalyst.
A postulated mechanism for the functioning of these catalysts is that the presence of cations of +5 or greater valence on large surface area give maximum exposure of strong Lewis acid sites for attachments of acetylene molecules. Acetylene molecules in the gaseous state coming in the vicinity of the +5 cation site are then polarized. In addition, induced polarization is set up in the attached acetylene molecule. As a result additional acetylene molecules are attracted to this attached acetylene molecule. When three molecules of acetylene are thus attached, angular rotation is possible and a ring structure of benzene is thereby formed. At higher temperatures, say above 100 C. the ring formation sometimes does not take place until after more than three acetylene molecules become so attached to a catalyst site, resulting in ethyl benzene being synthesized, thereby decreasing the benzene yield. Hence a temperature of 25 C. to 65 C. is preferred, in the absence of reaction chamber cooling to compensate for the exotherm.
In preparing benzene according to the practice of this invention the acetylene is passed through the catalyst mass. As indicated, the catalyst thus contacted contains an alumina base material making available a large surface area. Thus vanadium oxide on alumina has been found to give excellent results with alumina having a surface area of 214 square meters per gram, but poor results using the same amount of vanadium oxide on alumina having a surface area of one square meter per gram. The rate of acetylene gas introduction over the preheated catalyst is related to both the temperature and the benzene synthesis yield. Investigation of the reaction shows that the catalyst reaction temperature and the acetylene flow rate are directly related, whereas the benzene yield and the acetylene flow rate are inversely related. The exact flow rate consequently depends upon reaction conditions, but normally it will be in the range of 5 cc. per min. to 25 cc. per min.
Normally the catalyst involved is heated at an elevated temperature (300 C. under vacuum) to extract water vapor. Recent studies have shown that the catalyst will react if not dehydrated, but the dehydration step increases benzene synthesis and enables pure benzene to be extracted devoid of any water. The catalyst is then cooled and stored under nitrogen gas until ready for use. After completion of the cyclization reaction of acetylene to form benzene, benzene can be extracted from the catalyst by placing the reaction column in an oven heated to about 100 C. An applied vacuum is pulled on the column to extract the benzene, which is collected in a liquid nitrogen freeze-out trap. An extraction of 100 C. is used so that if any higher molecular material is synthesized its boiling point will keep it in the column.
In addition to the fact that benzene can be made at ambient temperatures, another advantage of this invention is that the benzene produced by this process is extremely pure. Benzene extracted from the reaction column was checked for purity by infra-red and gas chromatography. The only product found other than benzene has been ethyl benzene and only in trace amounts. In addition the ethyl benzene was found to occur only at high temperatures when a large acetylene fiow rate was used. As the yield increases only pure benzene is obtained. In addition, mass spectrometric analysis of benzene produced shows that no carbon isotope fractionation was present in either low or high yield benzene reactions.
In order to determine whether a valence of 5 or 6 is necessary as set forth hereinbefore, nine known catalytic materials were selected for initial study for their ability to enhance the cyclization of acetylene gas to liquid benzene. The preparation of such catalysts is well known and since all of these catalysts were obtained commercially, their composition but not their manner of preparation is given herein:
CATALYSTS I. Fe Cat-F6 20%; A1 0 80%; S.A., 41 m. /g.
II. Mn Cat.MnO 19%; A1 0 81%; S.A., 69 m. /g.
III. Co Cat.CoO, 10%; A1 0 90% S.A., 6O m. /g.
IV. Pt Cat.PtO 0.5%; A1 0 99.5%; S.A., 165 m. /g.
V. Al Cat.Al O 98%; S.A., 160 rn. /g.
VI. Si-Al Cat.SiO 87.3%; A1 0 12.4%; S.A., 300
VII. Mo Can-M00 10%; A1 0 90%; S.A., 160
VIII. Mo-Co Cat.MoO 12%; C00, 3%; A1 0 85%;
S.A., 330 m. /g.
IX. V Cat.V O A1 0 90%; S.A., 214 rnF/g.
S.A.==Surface area, g: grams of catalyst, m. =meters square.
PROCEDURE In order to prepare the benzene using the foregoing catalysts, the catalysts were first dehydrated. This pretreatment was limited to the elimination of associated water so that no water would be present in the formed benzene, the catalyst being heated for two hours under vacuum at 200 C. The acetylene catalysis was accomplished by passing acetylene gas through the catalyst mass d at a controlled rate as given in the following tables. The acetylene was obtained from tank acetylene and the flow rate regulated by a lv.athenson Gas Proportioner with an accuracy of i2%. The catalytic acetylene reaction was exothermic and was accompanied by a darkening of the catalyst material. To insure that no excessive acetylene pressures would build up during the reaction an exhaust valve and displacement trap were connected to the reaction column to allow gas of greater than one atmosphere to escape and to be coll cted.
Using the foregoing procedure and a flow rate of 22.5 cc./min. acetylene was obtained in some instances and not in others as set forth in the following table:
TABLE I.BENZENE SYNTHESIS WITII DEIIYDRATED CATALYTIC MATERIALS Conditions:
Catalyst: As given.
Feed stream: Acetylene.
Flow rate: 22.5 ce./rnin.
Reaction temperature: As given.
Catalyst CZHE Reaction Benzene Yield Temp. C.) (Percent) 30 Trace 32 Trace 32 Trace 28 None 25 None 38 None 35 8 50. 3
While optimum conditions were not known at the time, results set forth in Table I show that alumina promoted with oxides of metals in a valence state of four or less do not promote the cyclic trimerization of acetylene, whereas oxides of Group VB and VI-B metals when present in a valence state of +5 or +6 do promote the reactions. In fact later work showed that Cr O does not work, whereas CrO does.
Table I also shows that catalysts are improved by cobait modification. One to five percent is usually employed based on the catalyst. To further investigate the effect of cobalt on molybdenum oxide catalysts additional cobalt molybdate catalysts were purchased and examined. For the purposes of comparison with catalysts VII and VIII of Table I, several additional catalysts were selected. The compositions of these catalysts were as follows:
CATALYSTS X. Co Cat.CoO, 5%; CuO, 5%; S.A., 59 m. /g.
XI. Mo-Co Cat.MoO 19.4%; CoO, 3.4%; A1 0 77.2%; S.A., 218 m. /g.
XII. Mo-Co Cat-M00 14.6%; COO, 3.2%; A1 0 83.2%; S.A., 330 m. /g.
Results obtained by the use of these promoters in accordance with the procedures set forth were as follows:
Based on the results obtained as shown in Table II, cobalt molybdate catalysts were selected for use in studying other variables in this benzene synthesis process. One such variable is the dehydration period and another variable is the flow rate. The dehydration period was evaluated by heating the catalyst for 2, 4 and 6 hours to determine the eifect of this heat treatment on the benzene yield. The results of this study are given in Table III.
TABLE III.DEHYDRATION OF Mo-Co CATALYST Conditions:
Catalyst: XIIMo-Co cat.
Feed stream: Acetylene.
Flow rate: 22.5 cc./min.
Dehydnation temperature: 300 C.
Dehydration period: Benzene yields, percent 2 hours 54.4 4 hours 73.0 6 hours 70.2
To investigate the flow rate variable, acetylene flow rates were varied from 6.5 cc./min. to 22.5 cc./min. at five different temperatures. At these flow rates, several determinations were made as shown in Table IV.
TABLE IV.-A-CETYLENE FLOW RATE STUDY Conditions Catalyst: XII-Mo-Co cat. Feed stream: Acetylene. Flow rate As given. Reaction temperature: As given.
As shown in Table IV, at 62 C. an excellent yield was obtained at a flow rate of 6.5 cc. per min. At higher temperatures wherein it was necessary to increase the flow rate, a decrease in benzene yield is shown.
Gas chromatography and infra-red analyses were made on the benzene synthesized from samples 1 through 4 of Table IV. The benzene samples were obtained by extracting the benzene from the catalyst under vacuum of 100 C. for two hours. The gas chromatograph utilized was a Baker-Coleman IDS model 20, equipped with a 100 ft. column filled with Apiezon L, a solid absorbent which is well suited for analyzing non-polar aromatic materials at temperatures up to 250 C. Analyses were run on 0.01 rnl. samples by injecting l microliter aliquot into the chromatograph and splitting them 100 to 1. Analysis of a number of benzene samples showed the benzene purity to vary from 98-100% with acetone being the primary contamination and trace amounts of water and ethyl benzene also present. The water content was attributed to atmospheric moisture and incomplete catalyst dehydration and was eliminated by collecting the benzene in an inert atmosphere and dehydration for four hours. The presence of acetone was found to be contributed as a contaminate in tank acetylene (acetylene stored in acetone and charcoal) and was, therefore, not considered as a product of the synthesis. Ethyl benzene was found to be highest in low yield samples and diminished to run-detectable amounts as benzene samples approached yields greater than 90%.
Infra-red analyses were made with a Beckman IR-lO model using a CaF cell with a 200 length light path. The range of wave numbers scanned for each analysis was from 1000 to 4000. The presence of acetone and trace amounts of water and ethyl benzene was confirmed only in the benzene samples of low yield.
C/C mass spectrometric analysis of tank acetylene and benzene synthesized from the tank acetylene were carried out to determine if any carbon isotope fractionation was occurring during the benzene synthesis. The ratios of the peaks corresponding to the mass 28 and mass 27 molecules of acetylene for five separate analyses gave an average value of 0.0220. Using Beynons formula for calculating percent C abundance, a value of 1.07 percent C was calculated.
The ratio of atomic mass units (79/78) of benzene were determined for seven benzene samples, ranging from low to high yields. No variation in the C/C mass ratios was observed for low or high yield samples. The average C/C value for the seven benzene samples analyzed was 0.0655. The calculated percent C abundances for acetylene and benzene to be 1.07 percent. The percent C abundance calculations using the mass ratio data is reliable for three significant figures.
Thus according to this invention a low temperature method is provided for the quantitative synthesis of a pure benzene which has no energy quenching products present and which does not require purifying distillation procedures. The benzene thus produced is ideally suited as a scintillation solvent. The foregoing evaluations show a considerable latitude in the preparation of benzene by the processes of this invention. Such changes as flow rate, temperature, percent Group VB and Group VIB metal oxides will be obvious to one familiar with the art. Such variations and modifications are deemed to be Within the scope of this invention.
What is claimed is:
A process for synthesizing substantially pure benzene by cyclic trimerization of acetylene which comprises trimerizing the acetylene in the gaseous phase at a temperature of 20 to 150 C. by bringing it into contact at said temperature with an oxide of one of the metals of Group VB and Group VIB of the Pen'odic Table having a valence V of 4 V 7, the oxide being supported on an activated alumina carrier, the metal being in said valence state in the oxide to confer Lewis acid properties on the combination, said carrier containing 3 to 30 weight percent of said oxide based on the total.
The process of claim 1 wherein the cyclic trimerization temperature is 25 C. to C., wherein the alumina support has a surface area of 30 to 400 square meters per gram and wherein the Group VB-Group VI-B metal oxide is present in amount of 5 to 25 weight percent based on the total.
The process of claim 1 wherein the alumina has a surface area of 100 to 350 square meters per gram, wherein the Group VB-Group VI-B metal oxide is employed in an amount of 5 to 25 weight percent based on the total, and wherein said oxide is present in combination with 1 to 5 weight percent based on the total of C00.
In radiocarbon dating by low-level liquid scintillation counting using benzene as the scintillator, wherein organic matter to be dated is burned to form carbon dioxide which contains carbon 14, the carbon dioxide being converted into acetylene, which then is converted into benzene by cyclic trimerization, the step of producing the benzene by the process of claim 1.
The process of claim 3 wherein the C00 is present in an amount of 3.2 weight percent and the oxide conferring Lewis acid properties is present in an amount of 10 weight percent.
The process of claim 3 wherein the oxide conferring Lewis acid properties is V 0 '7. The process of claim 3 wherein the oxide conferring Lewis acid properties is Ta O 8. The process of claim 3 wherein the oxide conferring Lewis acid properties is CrO 9. The process of claim 3 wherein the oxide conferring Lewis acid properties is M00 10. The process of claim 3 wherein the oxide conferring Lewis acid properties is W0 References Cited UNITED STATES PATENTS 2,846,490 12/1953 Witt 260673 2,990,434 6/1961 Smith 260673 PAUL M. COUGHLAN, ]R., Primary Examiner.
DELBERT E. GANTZ, Examiner.
J. D. MYERS, Assistant Examiner.
US424804A 1965-01-11 1965-01-11 Low temperature conversion of acetylene to pure benzene Expired - LifetimeUS3365510A (en)
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Cited By (5)
Cited by examiner, † Cited by third party| Publication number | Priority date | Publication date | Assignee | Title |
--- ---
| US3480684A (en) | 1967-08-21 | 1969-11-25 | Rowland C Hansford | Dehydrocyclization process and catalyst |
| US4009219A (en) | 1975-04-28 | 1977-02-22 | Tamers Murry A | Total synthesis of benzene from non-hydrocarbon materials |
| US4585897A (en) | 1984-08-24 | 1986-04-29 | Standard Oil Company (Indiana) | Process for preparation of fuel additives from acetylene |
| RU2333193C1 (en) | 2007-07-02 | 2008-09-10 | Институт химии нефти Сибирского отделения Российской академии наук | Method for obtaining benzol |
| WO2013165655A1 (en) | 2012-05-03 | 2013-11-07 | Exxonmobil Chemical Patents Inc. | Hydrocarbon conversion process |
Citations (2)
Cited by examiner, † Cited by third party| Publication number | Priority date | Publication date | Assignee | Title |
--- ---
| US2846490A (en) | 1953-12-17 | 1958-08-05 | Phillips Petroleum Co | Production of aromatic hydrocarbons |
| US2990434A (en) | 1955-11-08 | 1961-06-27 | Ici Ltd | Polymerization of acetylenic compounds |
1965
1965-01-11 US US424804Apatent/US3365510A/ennot_active Expired - Lifetime
Patent Citations (2)
Cited by examiner, † Cited by third party| Publication number | Priority date | Publication date | Assignee | Title |
--- ---
| US2846490A (en) | 1953-12-17 | 1958-08-05 | Phillips Petroleum Co | Production of aromatic hydrocarbons |
| US2990434A (en) | 1955-11-08 | 1961-06-27 | Ici Ltd | Polymerization of acetylenic compounds |
Cited By (5)
Cited by examiner, † Cited by third party| Publication number | Priority date | Publication date | Assignee | Title |
--- ---
| US3480684A (en) | 1967-08-21 | 1969-11-25 | Rowland C Hansford | Dehydrocyclization process and catalyst |
| US4009219A (en) | 1975-04-28 | 1977-02-22 | Tamers Murry A | Total synthesis of benzene from non-hydrocarbon materials |
| US4585897A (en) | 1984-08-24 | 1986-04-29 | Standard Oil Company (Indiana) | Process for preparation of fuel additives from acetylene |
| RU2333193C1 (en) | 2007-07-02 | 2008-09-10 | Институт химии нефти Сибирского отделения Российской академии наук | Method for obtaining benzol |
| WO2013165655A1 (en) | 2012-05-03 | 2013-11-07 | Exxonmobil Chemical Patents Inc. | Hydrocarbon conversion process |
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188558 | https://brainly.com/question/52818476 | [FREE] What does the word Renaissance mean? A. new birth B. rebirth C. reconstruction D. renewal Please - brainly.com
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What does the word Renaissance mean?
A. new birth
B. rebirth
C. reconstruction
D. renewal
Please select the best answer from the choices provided.
1
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The term Renaissance means 'rebirth' in French, indicating a period of cultural revival in Europe from the 14th to the 17th centuries. It signifies a renewed interest in the classical art and ideas of ancient Rome and Greece. The best answer choice from the given options is 'rebirth'.
Explanation
The Meaning of the Word Renaissance
The word Renaissance, which translates to rebirth in French, refers to a significant cultural and intellectual movement that took place in Europe from the 14th to the 17th centuries. It was characterized by a renewed interest in the classical art, literature, and philosophical ideas of ancient Greece and Rome. This era not only fostered a revival in the arts but also led to advancements in science, politics, and humanist thought.
During this period, the term Renaissance emphasizes the idea of a cultural and intellectual awakening following the Middle Ages, marking a departure from medieval values. As artists like Leonardo da Vinci and Michelangelo began to replicate the realistic and humanistic styles noted in ancient works, it transformed the landscape of European art and society.
Answer Choices
a. new birth
b. rebirth
c. reconstruction
d. renewal
Among the answer choices, the most accurate answer that captures the essence of the term is b. rebirth.
Learn more about Renaissance here:
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Being Human: An Introduction to Western Culture - Jacqui Shehorn
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The word Renaissance means 'rebirth' and refers to a cultural movement that began in the 14th century and emphasized a renewed interest in classical art and ideas. This period saw significant advancements in arts and sciences throughout Europe. The best answer choice from the given options is 'rebirth.'
Explanation
The term Renaissance translates to rebirth in French. This word signifies a significant period in European history that spanned roughly from the 14th to the 17th century. The Renaissance was characterized by a renewed interest in the classical art, literature, and philosophies of Ancient Greece and Rome, marking a departure from the medieval values that preceded it.
During this era, cultural and intellectual pursuits flourished, leading to advancements in various fields such as art, science, literature, and humanism. Artists and thinkers like Leonardo da Vinci, Michelangelo, and Galileo challenged conventional ideas, thereby laying the groundwork for modern Western civilization.
The influence of the Renaissance was profound and extended beyond Italy, impacting the entire continent of Europe. It led to significant changes in art, with artists adopting new techniques like perspective, naturalism, and human emotion in their works. The movement fostered a belief in the potential of human achievement and dignity, which is why it is often considered a pivotal moment in the history of Western culture.
Examples & Evidence
For example, Leonardo da Vinci's painting, the Mona Lisa, illustrates realistic human expression and technique, embodying the spirit of the Renaissance. Similarly, Michelangelo's sculptures showcase a deep appreciation for human form and anatomy, derived from classical influences.
Historical records indicate that the Renaissance began in Italy in the 1300s, with key figures like Petrarch and da Vinci embodying its ideals. The transformation in Europe during this time is well-documented in various art and literature texts, signifying a drastic shift in thought and culture from the medieval period.
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188559 | https://fomafst.github.io/regexreference.html | RegularExpressionReference
Reference for foma regular expression syntax.
Regular expression operators
Concatenation (X Y)
The language or relation X concatenated with Y. The operator is not overtly signaled by spacing, etc., and two adjacent regular expressions will be concatenated regardless of white space when found at the level of precedence of the concatenation operator by the regular expression parser.
Restrictions: none
Kleene star (X)
Zero or more iterations of X.
Restrictions: none
Kleene plus (X+)
One or more iterations of X. Equivalent to [X X].
Restrictions: none
Iteration operators
In addition to the Kleene closures, there are a number of m-n-ary iteration operators:
Iteration operators
X^n exactly n iterations of X
X^>n at least n iterations of X
X^<n less than n iterations of X
X^{m,n}between m and n iterations of X
Restrictions: none
Union (X | Y)
Returns the union (disjunction) of FSMs X and Y.
Notational variants: X ∪ Y (UNION, U+222A), X ∨ Y (LOGICAL OR, U+2228)
Restrictions: none
Intersection (X & Y)
Returns the intersection (conjunction) of FSMs X and Y.
Notational variants: X ∩ Y (UNION, U+2229), X ∧ Y (LOGICAL AND, U+2227)
Restrictions: foma always calculates the path intersection of two FSMs. This may not be the logical intersection of the relations X and Y if X and Y are transducers and not automata.
Subtraction (X - Y)
Returns the difference of FSMs X and Y.
Restrictions: foma always calculates the path subtraction of two FSMs. This may not be the logical difference of the relations X and Y if X and Y are transducers and not automata.
Complement (~X)
Calculates the complement set of X.
Notational variants: ¬X (NOT SIGN, U+00AC)
Restrictions: undefined for non-identity transducers.
Composition (X .o. Y)
Calculates the composite transducer of relations X and Y.
Notational variants: X ∘ Y (RING OPERATOR, U+2218)
Restrictions: none
Substitution (`[X,Y,Z])
Substitutes all occurrences of symbol Y in FSM X by Z.
Restrictions: none
Optionality (X)
Defines the language or relation that contains zero or one iteration of X. Equivalent to [X | 0].
Restrictions: none
Domain extraction (X.u)
Extracts the domain from a FSM.
Notational variants: X₁ (SUBSCRIPT ONE, U+2081)
Restrictions: none
Range exctraction (X.l)
Extracts the range from a FSM.
Notational variants: X ₂ (SUBSCRIPT TWO, U+2082)
Restrictions: none
Inversion (X.i)
Inverts a FSM.
Notational variants: X⁻¹ (SUPERSCRIPT MINUS SUPERSCRIPT ONE, U+207B U+00B9)
Restrictions: none
Term negation (\X)
Any single symbol except X. Equivalent to [? - X].
Restrictions: see subtraction
Cross product (X:Y)
Calculates the cross product (or Cartesian product) of FSMs X and Y producing a transducer representing the relation where every string in X is paired with every string in Y.
Notational variants: X .x. Y, X × Y (MULTIPLICATION SIGN, U+00D7), both with lower precedence than :
Restrictions: undefined if X and Y are not both automata.
Ignore (X/Y)
Denotes the language where instances of Y are arbitrarily interspersed in the language X.
Restrictions: not well defined for transducers.
Ignore inside (X./.Y)
Denotes the language where instances of Y are arbitrarily interspersed in the language X, except that the first symbol and last symbol belong to X.
Restrictions: not well-defined for transducers.
Left quotient (X\\Y)
The operation X\\Y is defined as: {w | ∃x((x ∈ X) ∧ (xw ∈ Y )}. Informally: the set of suffixes one can add to strings in X to get strings from Y.
Restrictions: not well-defined for transducers.
Right quotient (X///Y)
The operation X///Y is defined as: {w | ∃x((x ∈ Y ) ∧ (wx ∈ X)}. Informally, this is the set of prefixes one can add to Y to get strings from X.
Restrictions: not well-defined for transducers.
Precedes (X<Y)
Denotes the languages where every string from X precedes every string from Y. The precedence need not be immediate.
Restrictions: not well-defined for transducers.
Follows (X>Y)
Denotes the languages where every string from X follows every string from Y. The precedence need not be immediate.
Restrictions: not well-defined for transducers.
Shuffle (X <> Y)
The shuffle (or asynchronous) product of X and Y, i.e. the set of words formed by any method of ‘shuffling’ the contents of X with Y. The shuffle is not perfect.
Notational variants: X ∥ Y (PARALLEL TO, U+2225)
Restrictions: none
Containment operators ($X, $.X, $?X)
The operator $X denotes the language that contains a substring drawn from the language X. This is equivalent to [? X ?]. The operator $.X denotes the language that contains exactly one substring drawn from the language X, while $?X denotes the language that contains at most one substring from X.
Restrictions: $.X and $?X are not well-defined for transducers.
Context restriction (X => Y1 Z1, ..., Yn Zn)
Denotes the language where every instance of a string from X is surrounded by string from some pair Yi and Zi on the left and right, respectively.
Restrictions: not well-defined for transducers.
Priority unions (X .P. Y, X .p. Y)
Th upper-side priority union X .P. Y denotes the union of relations X and Y, with relations in Y discarded if a relation in X have the same input (domain). Equivalent to [X | [ ̃[X.u] .o. Y]]. The lower-side priority union is similar, except a relation in X has precedence over a relation in Y based on the range, not the domain.
Restrictions: none
Lenient composition (X .O. Y)
The 'lenient' composition of X with Y. For those relations where strings in the domain of Y does not include some possible string from the range of X, the relation X is not composed with Y. Equivalent to [X .o. Y] .P. Y. See Karttunen (1998) for details and usage.
Logical connectives (X → Y, X ↔ Y)
The implication operator and the biconditional are shorthands whose definitions are as follows:
X → Y is equivalent to ~X | Y
X ↔ Y is equivalent to [~X | Y] & [~Y | X]
Flag elimination (X.f)
Eliminates all flag diacritics in X and calculates the equivalent FSM without flags. If no flags are present, no action is taken.
Restrictions: none
Replacement operators
The most general format of a replacement rule is
A -> B || L _ R ;
which dictates the replacement of sequences A with B when occurring between L and R. The replacement arrow (->) can be of various types indicating different replacement strategies, optional replacement (->), longest-leftmost @->, shortest-leftmost @>, etc. The directionality constraint || can also vary depending on the application.
All replacement operators allow context specification through one of the directionality markers.
The directionality constraint DIR can be one of:
DIR Interpretation
||left and right contexts hold on input side
\left holds on input side, right holds on output side
//left holds on output side, right holds on input side
\/left holds on output side, right holds on output side
The following replacement operators are available:
Operator Type
->Unconditional replacement
<-Unconditional inverse replacement
<->Unconditional replacement and inverse replacement
(->)Optional replacement
(<-)Optional inverse replacement
(<->)Optional replacement and inverse replacement
@->Left-to-right longest-match replacement
@>Left-to-right shortest-match replacement
<-@Left-to-right longest-match inverse replacement
<@Left-to-right shortest-match inverse replacement
(@->)Optional left-to-right longest-match replacement
(@>)Optional left-to-right shortest-match replacement
(<-@)Optional left-to-right longest-match inverse replacement
(<@)Optional left-to-right shortest-match replacement
Markup (...)
The LHS/RHS of a replacement rule may also be of the format
X -> Y ... Z ;
in which case both Y and Z are inserted around instances of X. Y or Z may be omitted and all contextual specifications are available.
Epsilon modifier [. .]
The LHS of a rule may be wrapped in the epsilon modifier, in which case any epsilons on the LHS get a special interpretation, where only one empty string is assumed to exists between each symbol in the input string. For example, the rule:
[.a.] -> x
will produce a transducer that maps the input string a unambiguously to xxx.
Also, [..] will simply produce a rule that inserts one instance of the RHS whenever the context is matched:
[..] -> x
will map aaa to xaxaxax.
Multiple contexts
Several possible contexts can be specified by separating the contexts with a comma.
Example:
a -> b || c _ d , e _ f ;
Multiple left-hand sides
Several left hand sides of a rule can be combined with a comma. In such a rule, the context specification aplies to all left-hand sides.
Example:
a -> b , b -> a || c _ d ;
Parallel rules (,,)
Separate rules can be compiled in parallel by separating each individual rules with a double comma (,,), i.e.
Rule1 ,, Rule2 ,, ... ,, RuleN ;
Transducers with backreferences
Rules of the format
T -> || L _ R ;
are also possible. Here, T is usually a transducer. The contexts can be omitted, and several contexts can be specified as with ordinary rules, and the arrow can take any of the standard arrow types.
The semantics of the rule is that the strings from the input side of T are replaced with T. This allows for a more general specification of chunking and insertion rules.
For example, the rule
0:%[ Chunk 0:%] @-> ;
is identical to the chunking rule
Chunk @-> %{ ... %}
This type of rule allows for replacements that are otherwise hard to define. Gerdemann and van Noord(1999), who proposed the rule semantics, provide the following example: suppose we have a transducer of word-sequence/abbreviation pairs such as
define Abbr {nondeterministic finite automaton}:{NFA} | {deterministic finite automaton}:{DFA} ;
Now, if we wanted to convert all known word sequences into the known acronyms whenever they occur between ... tags, we can issue the rule:
Abbr -> || {} _ {}
which would for example transduce nondeterministic finite automaton to NFA.
Word-boundary marker (.#.)
The special word-boundary marker may be used in context specification of both the context restriction operator as well as replacement rules.
Example:
a => _ c | .#. ;
specifies the language where a must always be followed by c or a word boundary.
The word-boundary marker has no special meaning in other regular expression constructs; however, any .#.-symbols found in a context specification will always be interpreted as a word boundary, and the symbol is removed from the alphabet at the end of replacement rule or context restriction compilation.
Built-in functions
All built-in regular expression function names begin with an underscore (_).
_isunambiguous(X)
Returns the empty string (ε) if X is an unambiguous transducer, or the empty language (∅) otherwise.
_isidentity(X)
Returns the empty string (ε) if X is an identity transducer, or the empty language (∅) otherwise.
_isfunctional(X)
Returns the empty string (ε) if X is a functional (single-valued) transducer, or the empty language (∅) otherwise.
_notid(X)
Returns an automaton containing all the words in X that do not map to themselves (are not in an identity relation).
_lm(X)
Returns the letter machine equivalent to X. A letter machine is a FSM where every transition contains maximally one UTF-8 symbol.
_loweruniq(X)
Modifies X in such a way that each input word maps to a unique output word. Lower-side symbols are replaced by arbitrary symbols from the alphabet. In case the output side cannot be made unique using only existing symbols in the alphabet, the alphabet is extended with new random symbols to achieve uniqueness.
_allfinal(X)
Returns the same FSM as X, with the exception that all states are marked as final states.
_unambpart(X)
Returns an FSM containing only those paths in X that are unambiguous. That is, a mapping in X is preserved only if its input string has a unique path through the transducer.
Example:
_unambpart(a:b | a:c | b:c);
returns a transducer equivalent to
b:c
since b:c is the only unambiguous path with respect to the input side.
_ambpart(X)
Returns an FSM containing only those paths in X that are ambiguous. That is, a mapping in X is preserved only if its input string has at least two paths through the transducer.
Example:
_ambpart(a:b | a:c | b:c);
returns a transducer equivalent to
a:b | a:c
That is, the path containing b:c is removes since it is unambiguous.
_ambdom(X)
Returns an automaton containing all words in the domain of X that yield an ambiguous path through X.
_ambdom(a:b | a:c | b:c);
returns a an automaton equivalent to
a
_eq(X,L,R)
Filters from the output side of X all those strings where some substrings occurring between the delimiters L and R are different.
Example:
Consider the language %< a b %> %< a b %>, which contains an infinite number of strings:
...
However, only one of the strings in this language has identical substrings between all instances of < and >, namely . Hence, the language containing the single string
is produced by the regular expression:
_eq(%< a b %> %< a b %> , %< , %>) ;
This operation is mostly used to model reduplication in natural language lexicons. Usually, the bare words to be reduplicated are marked with delimiters, say < and >, after which one can produce the reduplicated forms. For example:
define Lexicon {cat}|{dog}|{horse};
define RLexicon %< Lexicon %> (%- %< [%<|%>]+ %>);
regex _eq(RLexicon, %<, %>) .o. %<|%> -> 0 ;
and now we get:
foma: lower-words
cat
cat-cat
dog
dog-dog
horse
horse-horse
_flatten(X, EPS)
This function converts a transducer into an automaton where input-output symbols are interleaved. Since epsilons cannot be retained, they need to be converted to the EPS symbol. For example:
regex _flatten(a:0 0:b c, "EPS");
produces an automaton which accepts only the word:
a EPS EPS b c c
_addloop(L, a:b)
This function adds self-loops to all states in L with the symbol pair a:b. a or b may be 0.
_addfinalloop(L, a:b)
Like _addloop(L, a:b) but only adds loops at final states.
_addnonfinalloop(L, a:b)
Like _addloop(L, a:b) but only adds loops at nonfinal states.
_leftrewr(L, a:b)
This is a fast low-level single-symbol rewrite with left contexts only. It is sometimes useful for constructing complex transducers faster than with the generic operations.
_leftrewr(L, a:b)
is semantically equivalent to
a -> b || .#. L _
while
_leftrewr(? L, a:b)
is equivalent to:
a -> b || L _
_marktail(L, a:b)
The function converts L into a single-symbol insertion transducer. As the above, this is a fast state/arc manipulation function that produces a transducer equivalent to a certain operation. In this case:
_marktail(? L, 0:x)
is equivalent to
~$x .o. [..] -> x || L _ ;
and
_marktail(? R.r, 0:x).r
is equivalent to
~$x .o. [..] -> x || _ R
First-order logic over substrings (experimental)
SUPPORT FOR FIRST-ORDER LOGIC DISCONTINUED from 0.9.16 onward
Foma also has a compiler for a type of first-order logic over substrings. Statements in the first-order logic must begin with a parenthesized quantification of a variable. When compiling statements in first-order logic, parenthesis symbols ( and ) lose their optionality meaning and work as grouping symbols.
See the article on first-order logic for details.
Connectives
∨Disjunction
∧Conjunction
→Implication
↔Biconditional
¬Negation
Quantifiers Type
∀universal
∃existential
Available predicates
x ∈ L the substring x is a member of language L
_S(x,L)the substring x is followed immediately by a substring from L
x = y the position of substrings x and y are identical
x ≠ y equivalent to (x = y)
x ≺ y substring x precedes y
x ≻ y substring x succeeds y
x ≤ y substring x precedes or is in equal position with y
x ≥ y substring x succeeds or is in equal position with y
Examples:
regex (∃x)(∃y)(x ∈ a ∧ y ∈ a ∧ x ≠ y);
denotes the languages that contains two instances of the string a, in different positions.
regex (∃x)(x ∈ a) & (∃y)(y ∈ b);
denotes the language that contains both a substring a and a substring b.
regex (∃x)(x ∈ L ∧ ¬(∃y)(y ∈ L ∧ ¬(x = y) ) );
the language where only exactly one substring from the language L is present. Equivalent to $.L.
regex (∀y)( (y ∈ x) → (_S(a,y) ∧ _S(y,b) ) ∨ (_S(c,y) ∧ _S(y,d) ) );
the language where each instance of the string x is surrounded by a and b, or c and d. Equivalent to:
regex x => a _ b , c _ d;
Operator precedence
The following table gives the operator precedence of all regular expression operators, from highest to lowest.
Operator precedence
\ `
:
+ %%^%% .l .u .i .f .r
~ $ $. $?
/ /// \\ /\/
(concatenation)
<
| & - .P. .p.
=> -> (->) @-> etc.
<>
.x. .o. .O.
Reserved symbols
The following is a table of all reserved symbols in foma regular expressions, showing the character, the official Unicode character name, and its Code Point. These need to be escaped (by % or enclosing a string in quotes) for their literal meaning in regular expressions.
Character Character Name Code point
!EXCLAMATION MARK U+0021
"QUOTATION MARK U+0022
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How do you solve the area of a trapezoid using diagonals
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The height of a trapezoid is $10$ cm. The lengths of the two diagonals of the trapezoid are $30$ cm and $50$ cm. Calculate the area of the trapezoid.
On the homework I solved this using $${D_1D_2\over 2}$$ and my teacher marked me wrong. So I don't know what I did wrong. Please help. I know I can only use the formula if the diagonals are $90$ degrees. But how do I check that ?
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edited Dec 20, 2012 at 13:55
rschwieb
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asked Dec 20, 2012 at 0:09
NGPP1NGPP1
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$\begingroup$ that is the formula of the area of a rhombus. $\endgroup$
lab bhattacharjee
– lab bhattacharjee
2012-12-20 05:02:03 +00:00
Commented Dec 20, 2012 at 5:02
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The area will be $\frac12\cdot 10\cdot (y+x+y+z)=5(x+2y+z)$
Now, $(y+z)^2+10^2=50^2$ and $(x+y)^2+10^2=30^2$
$(y+z)=\sqrt{50^2-10^2} CM=20\sqrt6 CM$
$(x+y)=\sqrt{30^2-10^2} CM=20\sqrt2 CM$
SO, the area will be $5(20\sqrt2(\sqrt3+1)) CM^2$
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edited Dec 20, 2012 at 4:45
answered Dec 20, 2012 at 4:36
lab bhattacharjeelab bhattacharjee
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To help you think about the solution, draw two parallel lines 10 cm apart. The bottom will contain the B1 and the top will contain the B2. Draw the two diagonals at the proper lengths from the base to the top line making sure they cross. Now imagine that you slide one of the diagonals along the bottom and top lines. Notice that (B2+B2)/2 does not change (i.e one stretches and one shrinks). So no matter where the diagonals cross you have the same area. Now slide them apart until the intersection point reaches the top line (i.e. B2 = 0). Now you have triangle with the same area as the trapezoid. You know the height and two sides of the triangle, a little geometry and you can compute the base and your're there.
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answered Dec 20, 2012 at 0:32
TpofofnTpofofn
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Gluing two copies of the trapezoid together with one rotated by 180 degrees can yield a parallelogram. One diagonal from each trapezoid cuts the parallelogram into two triangles, each of which has an altitude of 10 and sides adjacent to that altitude's vertex of lengths 30 and 50.
So the area is the same as that of a triangle with two sides of 30 and 50, and an altitude between them of 10. Is this enough of a help?
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edited Dec 20, 2012 at 10:12
answered Dec 20, 2012 at 0:22
2'5 9'22'5 9'2
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$\begingroup$ Trying to understand what you said... Is the answer 193.1 cm? $\endgroup$
NGPP1
– NGPP1
2012-12-20 00:35:52 +00:00
Commented Dec 20, 2012 at 0:35
$\begingroup$ This is the right answer, but I don't think your gluing construction works. $\endgroup$
Peter Shor
– Peter Shor
2012-12-20 00:36:04 +00:00
Commented Dec 20, 2012 at 0:36
$\begingroup$ I am still confused $\endgroup$
NGPP1
– NGPP1
2012-12-20 00:42:59 +00:00
Commented Dec 20, 2012 at 0:42
$\begingroup$ @PeterShor, in my method, the area is around $386.37 cm^2$ $\endgroup$
lab bhattacharjee
– lab bhattacharjee
2012-12-20 05:03:31 +00:00
Commented Dec 20, 2012 at 5:03
$\begingroup$ @PeterShor I have added imagery that demonstrates the gluing construction works. I leave the labeling of edges and confirmation to you. The trapezoid's area is the same as that of the triangle at the end of the process: a triangle with altitude 10 and two surrounding sides of length 30 and 50. $\endgroup$
2'5 9'2
– 2'5 9'2
2012-12-20 10:05:09 +00:00
Commented Dec 20, 2012 at 10:05
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$\begingroup$
Since you can drag base 1 left or right and not change the area then if you drag the smaller base until it forms right angles with base 2 you get a trapezoid made up of one square and one triangle. So we know base 1 = size of smaller base = one side of the square. We know that hypotenuse = size of smaller diagonal. Since the corner is a right angle we can calculate height as the square root of the (sum of smaller diagonal squared minus smaller base squared) then from there we know area of trapezoid = midsegment length times height so average the 2 known bases and multiply by the calculated height and you have the area of a trapezoid given only the length of the bases and the length of the diagonals.
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answered Mar 14, 2019 at 7:12
JasonJason
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The theorem you need is that
the area of a trapezium is half that of the bounding parallelogram.
Consider a trapezium $ABCD.$ Its bounding parallelogram is defined as follows. Look at the diagonal $AC,$ then from $C$ draw a line parallel to the other diagonal $BD$ to meet $AB$ extended at $B'.$ Similarly, draw a line from $A$ to meet $CD$ extended at $D'.$ Then by construction $AB'CD'$ is a parallelogram and is referred to as the bounding parallelogram of the trapezium $ABCD.$ I shall leave it to you to convince yourself that the bounding parallelogram is uniquely defined (up to congruence, that is). You only need repeat the above construction for the other diagonal and prove congruence, or use corresponding transformations.
Now, having explained the term bounding parallelogram, let us prove the above theorem. First I shall use the notation $[XYZ],$ for example, to mean the area of the simple polygon $XYZ.$ Then we have that $$[ABDD']+[BB'CD]=[AB'CD'].$$ But $[ABCD]=[ABD]+[BCD].$ However, $[ABDD']=2[ABD]$ and $[BB'CD]=2[BCD],$ from which the result follows.
Now applying it to your case gives the area of trapezium $ABCD$ as $[AD'C],$ with $|AC|=30$ and $|BD|=|AD'|=50.$ Also, let $\angle BDC=\angle AD'C=\phi$ and $\angle ACD=\theta.$ If $M$ is the point where the altitude from $A$ intersects $D'C,$ then we have $$|D'C|=|D'M|+|MC|=|D'A|\cos \phi + |AC|\cos \theta=50 \cos \phi + 30 \cos \theta.$$ Also, $$10=|AM|=|D'A|\sin \phi = |AC|\sin \theta=50 \sin \phi = 30 \sin \theta,$$ giving $\sin \phi = 1/5$ and $\sin \theta = 1/3,$ so that $\cos \phi = \sqrt{24}/5$ and $\cos \theta = \sqrt{8}/3,$ and finally $|D'C|=10(\sqrt{24} + \sqrt{8}),$ so that $$[ABCD]=\frac 1 2\cdot 10\cdot 10(2\sqrt{6} + 2\sqrt{2})=100(\sqrt{6} + \sqrt{2}).$$
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answered Mar 14, 2019 at 13:19
AllawonderAllawonder
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$\begingroup$
the formula $\frac{d_1\cdot d_2}2$ is the area of the special quadrilateral called rhombus but not the trapezium..
If you want to find the area of the trapezium the like the above mentioned using the Pythogoras theorem. Find the sides and form an equation and with that solve the linear equation then you can solve the area of the trapezium.. finally you will get
area$=\frac{h(a+b)}2$
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edited Jan 18, 2019 at 5:34
Siong Thye Goh
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answered Jan 18, 2019 at 5:02
Chandrahas ReddyChandrahas Reddy
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Area of a trapezoid from given the two bases and diagonals
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188561 | https://books.google.com/books/about/Marine_Hydrodynamics.html?id=nj-k_lAmaBYC | Marine Hydrodynamics - John Nicholas Newman - Google Books
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Marine Hydrodynamics ==================== John Nicholas Newman MIT Press, 1977 - Science - 402 pages A textbook that offers a unified treatment of the applications of hydrodynamics to marine problems. The applications of hydrodynamics to naval architecture and marine engineering expanded dramatically in the 1960s and 1970s. This classic textbook, originally published in 1977, filled the need for a single volume on the applications of hydrodynamics to marine problems. The book is solidly based on fundamentals, but it also guides the student to an understanding of engineering applications through its consideration of realistic configurations. The book takes a balanced approach between theory and empirics, providing the necessary theoretical background for an intelligent evaluation and application of empirical procedures. It also serves as an introduction to more specialized research methods. It unifies the seemingly diverse problems of marine hydrodynamics by examining them not as separate problems but as related applications of the general field of hydrodynamics. The book evolved from a first-year graduate course in MIT's Department of Ocean Engineering. A knowledge of advanced calculus is assumed. Students will find a previous introductory course in fluid dynamics helpful, but the book presents the necessary fundamentals in a self-contained manner. The 40th anniversary of this pioneering book offers a foreword by John Grue. Contents Model Testing " The Motion of a Viscous Fluid " The Motion of an Ideal Fluid " Lifting Surfaces " Waves and Wave Effects " Hydrodynamics of Slender Bodies. More » Preview this book »
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AuthorJohn Nicholas Newman
Edition illustrated, reprint
Publisher MIT Press, 1977
ISBN 0262140268, 9780262140263
Length 402 pages
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188562 | https://en.wikipedia.org/wiki/Predicate_variable | Jump to content
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From Wikipedia, the free encyclopedia
Type of mathematical variable
In mathematical logic, a predicate variable is a predicate letter which functions as a "placeholder" for a relation (between terms), but which has not been specifically assigned any particular relation (or meaning). Common symbols for denoting predicate variables include capital roman letters such as , and , or lower case roman letters, e.g., . In first-order logic, they can be more properly called metalinguistic variables. In higher-order logic, predicate variables correspond to propositional variables which can stand for well-formed formulas of the same logic, and such variables can be quantified by means of (at least) second-order quantifiers.
Notation
[edit]
Predicate variables should be distinguished from predicate constants, which could be represented either with a different (exclusive) set of predicate letters, or by their own symbols which really do have their own specific meaning in their domain of discourse: e.g. .
If letters are used for both predicate constants and predicate variables, then there must be a way of distinguishing between them. One possibility is to use letters W, X, Y, Z to represent predicate variables and letters A, B, C,..., U, V to represent predicate constants. If these letters are not enough, then numerical subscripts can be appended after the letter in question (as in X1, X2, X3).
Another option is to use Greek lower-case letters to represent such metavariable predicates. Then, such letters could be used to represent entire well-formed formulae (wff) of the predicate calculus: any free variable terms of the wff could be incorporated as terms of the Greek-letter predicate. This is the first step towards creating a higher-order logic.
Usage
[edit]
If the predicate variables are not defined as belonging to the vocabulary of the predicate calculus, then they are predicate metavariables, whereas the rest of the predicates are just called "predicate letters". The metavariables are thus understood to be used to code for axiom schema and theorem schemata (derived from the axiom schemata).
Whether the "predicate letters" are constants or variables is a subtle point: they are not constants in the same sense that are predicate constants, or that are numerical constants.
If "predicate variables" are only allowed to be bound to predicate letters of zero arity (which have no arguments), where such letters represent propositions, then such variables are propositional variables, and any predicate logic which allows second-order quantifiers to be used to bind such propositional variables is a second-order predicate calculus, or second-order logic.
If predicate variables are also allowed to be bound to predicate letters which are unary or have higher arity, and when such letters represent propositional functions, such that the domain of the arguments is mapped to a range of different propositions, and when such variables can be bound by quantifiers to such sets of propositions, then the result is a higher-order predicate calculus, or higher-order logic.
See also
[edit]
Functional predicate – Symbol representing a mathematical conceptPages displaying short descriptions of redirect targets
Metavariable
Propositional variable – Variable that can either be true or false
References
[edit]
^ "Predicate variable - Encyclopedia of Mathematics". encyclopediaofmath.org. Retrieved 2020-08-20.
Bibliography
[edit]
Rudolf Carnap and William H. Meyer. Introduction to Symbolic Logic and Its Applications. Dover Publications (June 1, 1958). ISBN 0-486-60453-5
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188563 | https://icurology.org/DOIx.php?id=10.4111/icu.20230313 | Platinum open access journal
pISSN 2466-0493 eISSN 2466-054X
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Investig Clin Urol. 2024 May;65(3):248-255. English.Published online Apr 16, 2024.
© The Korean Urological Association
Original Article
Comparative analysis of recurrence rates between intravesical gemcitabine and bacillus Calmette–Guérin induction therapy following transurethral resection of bladder tumors in patients with intermediate- and high-risk bladder cancer: A retrospective multicenter study
Joongwon Choi,1, Kyung Hwan Kim,2, Hyung Suk Kim,3 Hyun Sik Yoon,3 Jung Hoon Kim,1 Jin Wook Kim,1 Yong Seong Lee,1 Se Young Choi,4 In Ho Chang,4 Young Hwii Ko,5 Wan Song,6 Byong Chang Jeong,6 and Jong Kil Nam7
Author information
Author notes
Copyright and License
1Department of Urology, Chung-Ang University Gwangmyeong Hospital, Chung-Ang University College of Medicine, Gwangmyeong, Korea.
2Department of Urology, Pusan National University Hospital, School of Medicine, Pusan National University, Busan, Korea.
3Department of Urology, Dongguk University Ilsan Medical Center, Dongguk University College of Medicine, Goyang, Korea.
4Department of Urology, Chung-Ang University Hospital, Chung-Ang University College of Medicine, Seoul, Korea.
5Department of Urology, Yeungnam University College of Medicine, Daegu, Korea.
6Department of Urology, Samsung Medical Center, Sungkyunkwan University School of Medicine, Seoul, Korea.
7Department of Urology, Pusan National University Yangsan Hospital, School of Medicine, Pusan National University, Yangsan, Korea.
Corresponding Author: Jong Kil Nam. Department of Urology, Pusan National University Yangsan Hospital, School of Medicine, Pusan National University, 20 Geumo-ro, Mulgeum-eup, Yangsan 50612, Korea. TEL: +82-55-360-2678, FAX: +82-55-360-2164, Email: tuff-kil@hanmail.net
These authors contributed equally to this study and should be considered co-first authors.
Received September 13, 2023; Revised January 03, 2024; Accepted March 04, 2024.
This is an Open Access article distributed under the terms of the Creative Commons Attribution Non-Commercial License ( which permits unrestricted non-commercial use, distribution, and reproduction in any medium, provided the original work is properly cited.
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Abstract
Purpose
This study investigated the efficacy of intravesical gemcitabine as an alternative to bacillus Calmette–Guérin (BCG) therapy.
Materials and Methods
Data were retrospectively collected across seven institutions from February 1999 to May 2023. Inclusion criteria included patients with intermediate- or high-risk non-muscle invasive bladder cancer (NMIBC) who underwent transurethral resection of bladder tumors (TURBT) and received at least four sessions of intravesical gemcitabine or BCG induction therapy. Patient characteristics, complete remission (CR), occurrence, and progression rates were compared.
Results
In total, 149 patients were included in this study (gemcitabine, 63; BCG, 86). No differences were apparent between the two groups in baseline characteristics, except for the follow-up period (gemcitabine, 9.2±5.9 months vs. BCG, 43.9±41.4 months, p<0.001). There were no consistent significant differences observed between the two groups in the 3-month (gemcitabine, 98.4% vs. BCG, 95.3%; p=0.848), 6-month (94.9% vs. 90.0%, respectively; p=0.793) and 1-year CR rates (84.2% vs. 83.3%, respectively; p=0.950). Also, there was no significant statistical difference in progression-free survival between the two groups (p=0.953). The occurrence rates of adverse events were similar between the groups (22.2% vs. 22.1%; p=0.989); however, the rate of Clavien–Dindo grade 2 or higher was significantly higher in the BCG group (1.6% vs. 16.3%, respectively; p<0.001).
Conclusions
Intravesical gemcitabine demonstrated efficacy comparable to BCG therapy for the first year in patients with intermediate- and high-risk NMIBC. However, long-term follow-up studies are warranted.
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Graphical Abstract
Keywords
Bladder neoplasms; Gemcitabine; Mycobacterium bovis
Go to:
INTRODUCTION
Non-muscle-invasive bladder cancer (NMIBC) is a prevalent malignancy characterized by its tendency to recur and progress . Among patients with NMIBC, those patients classified with intermediate or high-risk cancer require more aggressive treatment strategies to mitigate the risk of disease recurrence and progression . Currently, transurethral resection of bladder tumors (TURBT) followed by intravesical therapy with bacillus Calmette–Guérin (BCG) is the gold standard therapeutic approach in this patient cohort. Thus, BCG therapy has demonstrated efficacy in reducing recurrence rates .
However, the global medical community has faced persistent challenges due to the chronic shortage of BCG since the closure of the manufacturer’s laboratory (Sanofi) in 2016 . This compound scarcity has led to significant delays in the delivery of vital intravesical BCG therapy, with some patients experiencing waiting periods exceeding two months, particularly in Korean regions. This critical lapse in timely treatment raises concerns regarding its potential impact on patient outcomes and necessitates the exploration of viable alternative therapeutic options [4, 5, 6].
Intravesical gemcitabine therapy has recently emerged as a potential substitute for BCG treatment [7, 8]. Intriguingly, this therapeutic approach has gained recognition, particularly in the context of insurance coverage changes that now extend intravesical gemcitabine therapy for up to one year since the regulatory revision of insurance standards in February 2022 in Korea. The concurrent shortage of BCG and expanded insurance coverage for gemcitabine have prompted a closer examination of its efficacy as an alternative treatment strategy for intermediate/high-risk patients with NMIBC in Korea .
The primary objective of this study was to assess the efficacy of intravesical gemcitabine as a potential alternative to BCG therapy in patients initially diagnosed with intermediate/high-risk NMIBC. By conducting a comprehensive investigation of clinical outcomes, recurrence rates, and safety profiles associated with gemcitabine therapy, we hope to provide critical insights into the feasibility and effectiveness of this alternative approach.
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MATERIALS AND METHODS
Approval for this study was obtained from the Institutional Review Board (IRB) of Chung-Ang University Gwangmyeong Hospital (IRB number: 2304-076-038), and the same approval was obtained from each participating institution. Due to the retrospective nature of the study, patient-specific information was not included in the data collection process, and the requirement for patient written or verbal informed consent was waived.
This study was initiated as a research project by the Bladder Cancer Study Group of the Korean Urological Oncology Society, involving participation from seven institutions. Retrospective data collection was conducted across the institutions from February 1999 to May 2023.
Patients included in the study underwent TURBT following an initial diagnosis of bladder tumors and were categorized as intermediate/high-risk according to the American Urological Association (AUA) risk classification. The study included adults aged 18 years and older. The selection criteria encompassed patients who had undergone at least four sessions of intravesical gemcitabine or BCG induction therapy, and who were followed up for more than three months. Patients who had previously undergone TURBT for bladder cancer or received any other type of intravesical therapy were excluded from the study.
The therapeutic BCG injected into the bladder was Oncotice®, a live attenuated strain of Mycobacterium bovis, at a dose of 12.5 mg in 100 mL sterile saline, mixed and instilled. For gemcitabine, a dose of 2,000 mg per session was mixed in 50 mL sterile saline and instilled. Intravesical therapy commenced two weeks to one month after surgery, with both drugs administered for six sessions during the induction phase. Maintenance therapy for intravesical gemcitabine followed a 1-year protocol, given monthly for one year. The duration of BCG maintenance was determined by the primary physician based on the patient’s risk, ranging from one to three years, administered once a week for three weeks at 3-, 6-, 12-, 18-, 24-, 30-, and 36-month intervals.
The following patient data were collected: institution, sex, age, height, weight, body mass index, underlying conditions (hypertension, diabetes, chronic obstructive pulmonary disease, myocardial infarction, coronary-vascular accident, coronary artery disease, end-stage renal disease, smoking status, pack-years of smoking), ASA (American Society of Anesthesiologists Physical Status Classification) score, date of first transurethral resection (1st TUR), initial T stage, tumor multiplicity, initial grade, initial carcinoma in situ (CIS), initial AUA risk class, type of immediate intravesical therapy, type of induction therapy, date of last follow-up, recurrence status, time to recurrence, date of second TURBT due to recurrence, recurrence T stage, recurrence multiplicity, recurrence grade, recurrence CIS, recurrence AUA risk class, and whether the patient underwent cystectomy, along with the date of cystectomy (if appropriate). Data were analyzed in July 2023.
Disease progression was defined as an increase in T stage, grade, or occurrence of CIS. Time to recurrence was defined as the time from the date of the 1st TUR to the date of the 2nd TUR. The definition of complete remission (CR) was calculated by taking patients followed up for a specific period (three months, six months, or one year) as the denominator and those who did not experience recurrence during that period as the numerator. The common standard protocol for follow-up observations at each hospital was as follows: intermediate-risk patients underwent cystoscopy at 3-month intervals for the first year after surgery, followed by 6-month intervals. High-risk patients underwent cystoscopy at 3-month intervals for the first two years after surgery, followed by 6-month intervals. Regardless of risk, abdominal computed tomography was performed annually for follow-up. General follow-up observations were conducted for five years.
The IBM SPSS Statistics software version 27.0 (IBM Co.) was used for all the statistical analyses. The comparison of the CR rates utilized a chi-square test. For the comparison of progression-free survival (PFS), Kaplan–Meier survival analysis and the log-rank test were employed. In Tables 1, 2, and 3, the comparison between the two groups was conducted using Student’s t-test to compare means, and Fisher’s exact test was employed for the comparison of two categorical variables. Statistical significance was set at p<0.05.
Table 1 Baseline characteristics
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Table 2 Details of the intravesical therapy and side effects
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RESULTS
A comprehensive data assessment was conducted in a cohort of 149 patients who underwent intravesical induction therapy following TURBT. Among these patients, 63 received intravesical gemcitabine and 86 patients received BCG induction therapy (Table 1). The baseline characteristics of the two groups demonstrated no significant differences, except for the follow-up duration, which indicated a significant variation (gemcitabine, 9.2±5.9 months vs. BCG, 43.9±41.4 months; p<0.001).
Details of intravesical therapy and its side effects are presented in Table 2. Induction therapy was administered in six sessions for over 90% of patients of both groups, and the sessions were conducted at weekly intervals (gemcitabine, 90.5%; BCG, 90.7%; p=0.085). Maintenance therapy was more frequently administered in the gemcitabine group (73.0% vs. 8.1%, respectively; p<0.001).
During the course of intravesical induction therapy, the incidence rates of adverse events were similar between the gemcitabine and BCG groups (22.2% vs. 22.1%, respectively; p=0.989). However, when considering the severity of those adverse events based on the Clavien–Dindo grading system, the BCG group experienced a significantly higher rate of grade 2 or higher events than the gemcitabine group (1.6% vs. 16.3%, respectively; p<0.001).
The treatment outcomes are illustrated in Table 3. The 3-month CR rates were comparable between the gemcitabine and BCG groups (98.4% vs. 95.3%, respectively; p=0.848). In addition, no consistent significant difference was observed between the gemcitabine and BCG groups in the 6-month CR rate (94.9% vs. 90.0%, respectively; p=0.793) and 1-year CR rate (84.2% vs. 83.3%, respectively; p=0.950) (Fig. 1). However, the rate of disease progression among patients who experienced recurrence was significantly higher in the gemcitabine group (60.0% vs. 15.9%; p<0.001), with no statistically significant difference in the rate of PFS between the gemcitabine and BCG groups, respectively (p=0.953; Fig. 2).
Fig. 1 One-year complete remission (CR) rate (%) comparison between intravesical bacillus Calmette–Guérin (BCG) and gemcitabine treatment. In patients with intermediate/high-risk non-muscle invasive bladder cancer, the initial 3-month, 6-month, and 1-year CR rates were similar for the intravesical BCG and gemcitabine groups (all p>0.05).
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Fig. 2 Progression-free survival of intravesical bacillus Calmette–Guérin (BCG) and gemcitabine. There was no statistically significant difference in the progression-free rate between the intravesical BCG and gemcitabine treatment groups (p=0.953).
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Table 3 Treatment outcomes
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DISCUSSION
NMIBC accounts for approximately 75% of newly diagnosed bladder cancers and is predominantly characterized by tumors confined to the mucosa (Ta, CIS) or submucosa (T1) [10, 11]. NMIBC has a relatively favorable prognosis as compared to muscle-invasive tumors; however, the disease course is marked by frequent recurrences, necessitating vigilant surveillance . The current standard of care for NMIBC includes TURBT, followed by intravesical therapy, according to risk stratification. Given that residual tumors can be detected in 40%–45% of patients after TURBT, adjuvant intravesical therapy is important . Intravesical instillation of BCG, an attenuated strain of M. bovis, has been the mainstay for preventing the recurrence and progression of NMIBC for several decades [14, 15, 16]. However, intravesical BCG immunotherapy presents challenges, as this type of therapy can lead to local and systemic side effects. Global periodic shortages of BCG have also compromised treatment availability for patients [17, 18]. Therefore, the exploration of alternative intravesical agents is imperative. Several studies revealed the inferiority of mitomycin C to BCG therapy, with its comparable recurrence risk yet reduced tolerability, manifested by increased allergic reactions and chemical cystitis [14, 19]. Epirubicin, even though it exhibits less toxicity, remains less effective than BCG therapy in reducing recurrence .
Amid the chronic shortage of BCG, gemcitabine has emerged as a valuable intravesical chemotherapeutic agent. The extension of Korean health insurance coverage for intravesical gemcitabine therapy has amplified its use in first-line adjuvant intravesical settings since February 2022. This growing utilization of gemcitabine has underscored the need for further research on its role in the Korean population.
In this study, we retrospectively collected data from seven institutions to compare the efficacy and safety of intravesical gemcitabine and BCG in patients with intermediate- and high-risk NMIBC. Intravesical gemcitabine displayed a CR rate similar to that of BCG at three months after treatment. In addition, no consistent significant difference was observed between the two groups in the 6-month and 1-year CR rates. Despite accounting for a higher percentage of patients receiving maintenance therapy in the intravesical gemcitabine group as compared to the BCG group (gemcitabine, 71.4% vs. BCG, 8.1%), the long-term efficacy of intravesical gemcitabine appears to be favorable. This is an encouraging outcome considering the inclusion of a high proportion (65.1%, 41/63) of high-risk patients in the gemcitabine group. Prior research by Perera et al. revealed recurrence rates of 53.1% and 28.1% (p=0.037) after intravesical gemcitabine and BCG treatments, respectively, in high-risk patients with superficial bladder cancer. In a systematic review, Shelley et al. reported that intravesical gemcitabine therapy established comparable efficacy in intermediate-risk patients but decreased efficacy in high-risk patients. In this study, intravesical gemcitabine therapy demonstrated similar PFS to that of BCG (p=0.953). However, caution should be exercised when interpreting survival data because of the significant difference in the follow-up periods between the two groups.
Similar adverse effects were observed in the intravesical gemcitabine and BCG groups. Lower urinary symptoms including dysuria, frequency, and urgency were the most common adverse events in both groups. However, the intravesical BCG group exhibited more severe symptoms than the gemcitabine group; the rates of grade 1 and 2 symptoms were 20.6% and 1.6% for gemcitabine and 5.8% and 15.1% for BCG, respectively (p<0.001). The more tolerable toxicity profile of intravesical gemcitabine therapy aligns with previous studies [22, 23].
Our retrospective analysis has several limitations. First, the small sample size may have rendered the results less robust. Second, over 90% of the patients in the intravesical BCG treatment group did not receive maintenance therapy, and this was attributed to the BCG shortage in Korea. In contrast, over 70% of the patients in the gemcitabine group underwent maintenance intravesical therapy. This procedural discrepancy was due to the retrospective design of this study, which may have caused the efficacy of intravesical BCG therapy to be considered unfavorable. Furthermore, there may be selection bias in the drug choices. Gemcitabine, known for fewer side effects, was implemented in a group with a higher average age and higher proportions of hypertension, diabetes mellitus, and chronic obstructive pulmonary disease in our study. Additionally, drug selection may have been influenced by issues related to the Korean drug supply. Although there was no statistical difference, these factors highlight the potential for selection bias in drug choices. Third, the follow-up period of the intravesical gemcitabine group was much shorter than that of the BCG group (9.2 months vs. 43.9 months, respectively; p<0.001). These differences in follow-up duration limited the estimation of long-term PFS. However, considering that intravesical gemcitabine treatment was initiated in February 2022, an approximate 9-month follow-up for the intravesical gemcitabine group was deemed acceptable.
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CONCLUSIONS
In conclusion, intravesical gemcitabine exhibited efficacy similar to that of BCG as an induction therapy in patients with intermediate- and high-risk NMIBC for the first year. The occurrence rates of adverse events were comparable between the intravesical gemcitabine and BCG groups; however, mild adverse events were observed more frequently in the intravesical gemcitabine group. Gemcitabine could serve as a viable substitute in the short term, offering comparable efficacy to BCG in terms of early CR rates. However, long-term effects of over one year were not observed in this study; therefore, long-term follow-up studies are warranted. The findings from this study contribute to a deeper understanding of the role of intravesical gemcitabine as an alternative therapeutic approach for intermediate- and high-risk patients with NMIBC and may influence how healthcare professionals approach treatment decisions in this patient population.
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Notes
CONFLICTS OF INTEREST:The authors have nothing to disclose.
FUNDING:None.
AUTHORS’ CONTRIBUTIONS:
Research conception and design: Joongwon Choi, Byong Chang Jeong, and Jong Kil Nam.
Data acquisition: Joongwon Choi, Kyung Hwan Kim, Hyung Suk Kim, Hyun Sik Yoon, Jung Hoon Kim, Jin Wook Kim, Yong Seong Lee, Se Young Choi, In Ho Chang, Young Hwii Ko, Wan Song, and Jong Kil Nam.
Statistical analysis: Joongwon Choi and Jong Kil Nam.
Data analysis and interpretation: In Ho Chang and Young Hwii Ko.
Drafting of the manuscript: Joongwon Choi and Kyung Hwan Kim.
Critical revision of the manuscript: Byong Chang Jeong and Jong Kil Nam.
Administrative, technical, or material support: Hyung Suk Kim and Wan Song.
Supervision: Byong Chang Jeong and Jong Kil Nam.
Approval of the final manuscript: all authors.
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ACKNOWLEDGMENTS
We would like to express our gratitude to the Bladder Cancer Study Group of the Korean Urological Oncology Society and the participating institutions for their contributions to this research.
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References
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| |
Tan WS, Prendergast A, Ackerman C, Yogeswaran Y, Cresswell J, Mariappan P, et al. Adjuvant intravesical chemohyperthermia versus passive chemotherapy in patients with intermediate-risk non-muscle-invasive bladder cancer (HIVEC-II): a phase 2, open-label, randomised controlled trial. Eur Urol 2023;83:497–504.
PubMed
CrossRef
Sylvester RJ, Rodríguez O, Hernández V, Turturica D, Bauerová L, Bruins HM, et al. European Association of Urology (EAU) prognostic factor risk groups for non-muscle-invasive bladder cancer (NMIBC) incorporating the WHO 2004/2016 and WHO 1973 classification systems for grade: an update from the EAU NMIBC guidelines panel. Eur Urol 2021;79:480–488.
PubMed
CrossRef
Tan WS, Rodney S, Lamb B, Feneley M, Kelly J. Management of non-muscle invasive bladder cancer: a comprehensive analysis of guidelines from the United States, Europe and Asia. Cancer Treat Rev 2016;47:22–31.
PubMed
CrossRef
Lobo N, Bree KK, Hensley PJ, Nogueras-Gonzalez GM, Abraham P, Navai N, et al. Reduced-dose bacillus Calmette-Guérin (BCG) in an era of BCG shortage: real-world experience from a tertiary cancer centre. BJU Int 2022;130:323–330.
PubMed
CrossRef
Tan WS, Steinberg G, Witjes JA, Li R, Shariat SF, Roupret M, et al. Intermediate-risk non-muscle-invasive bladder cancer: updated consensus definition and management recommendations from the International Bladder Cancer Group. Eur Urol Oncol 2022;5:505–516.
PubMed
CrossRef
Balasubramanian A, Gunjur A, Weickhardt A, Papa N, Bolton D, Lawrentschuk N, et al. Adjuvant therapies for non-muscle-invasive bladder cancer: advances during BCG shortage. World J Urol 2022;40:1111–1124.
PubMed
CrossRef
Lu JL, Xia QD, Liu CQ, Sun JX, Yang YY, Hu HL, et al. Efficacy and toxicity in scheduled intravesical gemcitabine versus bacille Calmette-Guérin for Ta and T1 bladder cancer: a systematic review and meta-analysis. Transl Cancer Res 2021;10:2849–2858.
PubMed
CrossRef
Messing EM, Tangen CM, Lerner SP, Sahasrabudhe DM, Koppie TM, Wood DP Jr, et al. Effect of intravesical instillation of gemcitabine vs saline immediately following resection of suspected low-grade non-muscle-invasive bladder cancer on tumor recurrence: SWOG S0337 randomized clinical trial. JAMA 2018;319:1880–1888.
PubMed
CrossRef
Han MA, Maisch P, Jung JH, Hwang JE, Narayan V, Cleves A, et al. Intravesical gemcitabine for non-muscle invasive bladder cancer: an abridged Cochrane review. Investig Clin Urol 2021;62:623–630.
PubMed
CrossRef
Burger M, Catto JW, Dalbagni G, Grossman HB, Herr H, Karakiewicz P, et al. Epidemiology and risk factors of urothelial bladder cancer. Eur Urol 2013;63:234–241.
PubMed
CrossRef
Otto W, Breyer J, Herdegen S, Eder F, Bertz S, May M, et al. WHO 1973 grade 3 and infiltrative growth pattern proved, aberrant E-cadherin expression tends to be of predictive value for progression in a series of stage T1 high-grade bladder cancer after organ-sparing approach. Int Urol Nephrol 2017;49:431–437.
PubMed
CrossRef
Sylvester RJ, van der Meijden AP, Oosterlinck W, Witjes JA, Bouffioux C, Denis L, et al. Predicting recurrence and progression in individual patients with stage Ta T1 bladder cancer using EORTC risk tables: a combined analysis of 2596 patients from seven EORTC trials. Eur Urol 2006;49:466–477.
PubMed
CrossRef
Herr HW, Dalbagni G. Defining bacillus Calmette-Guerin refractory superficial bladder tumors. J Urol 2003;169:1706–1708.
PubMed
CrossRef
Malmström PU, Sylvester RJ, Crawford DE, Friedrich M, Krege S, Rintala E, et al. An individual patient data meta-analysis of the long-term outcome of randomised studies comparing intravesical mitomycin C versus bacillus Calmette-Guérin for non-muscle-invasive bladder cancer. Eur Urol 2009;56:247–256.
CrossRef
Shelley MD, Kynaston H, Court J, Wilt TJ, Coles B, Burgon K, et al. A systematic review of intravesical bacillus Calmette-Guérin plus transurethral resection vs transurethral resection alone in Ta and T1 bladder cancer. BJU Int 2001;88:209–216.
PubMed
CrossRef
Sylvester RJ, van der Meijden AP, Lamm DL. Intravesical bacillus Calmette-Guerin reduces the risk of progression in patients with superficial bladder cancer: a meta-analysis of the published results of randomized clinical trials. J Urol 2002;168:1964–1970.
PubMed
CrossRef
Perera M, Papa N, Christidis D, McGrath S, Manning T, Roberts M, et al. The impact of the global bacille Calmette-Guérin shortage on treatment patterns: population-based data. BJU Int 2018;121:169–172.
PubMed
CrossRef
Witjes JA, Palou J, Soloway M, Lamm D, Brausi M, Spermon JR, et al. Clinical practice recommendations for the prevention and management of intravesical therapy-associated adverse events. Eur Urol Suppl 2008;7:667–674.
CrossRef
Nissenkorn I, Herrod H, Soloway MS. Side effects associated with intravesical mitomycin. J Urol 1981;126:596–597.
PubMed
CrossRef
Shang PF, Kwong J, Wang ZP, Tian J, Jiang L, Yang K, et al. Intravesical bacillus Calmette-Guérin versus epirubicin for Ta and T1 bladder cancer. Cochrane Database Syst Rev 2011:CD006885
PubMed
Shelley MD, Jones G, Cleves A, Wilt TJ, Mason MD, Kynaston HG. Intravesical gemcitabine therapy for non-muscle invasive bladder cancer (NMIBC): a systematic review. BJU Int 2012;109:496–505.
PubMed
CrossRef
Addeo R, Caraglia M, Bellini S, Abbruzzese A, Vincenzi B, Montella L, et al. Randomized phase III trial on gemcitabine versus mytomicin in recurrent superficial bladder cancer: evaluation of efficacy and tolerance. J Clin Oncol 2010;28:543–548.
PubMed
CrossRef
Ye Z, Chen J, Hong Y, Xin W, Yang S, Rao Y. The efficacy and safety of intravesical gemcitabine vs bacille Calmette-Guérin for adjuvant treatment of non-muscle invasive bladder cancer: a meta-analysis. Onco Targets Ther 2018;11:4641–4649.
PubMed
CrossRef
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Publication Types
Multicenter Study
Original Article
MeSH Terms
Bacillus
Follow-Up Studies
Gemcitabine
Gemcitabine
Humans
Mycobacterium bovis
Non-Muscle Invasive Bladder Neoplasms
Progression-Free Survival
Recurrence
Retrospective Studies
Therapeutics
Transurethral Resection of Bladder
Urinary Bladder Neoplasms
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Joongwon Choi
Kyung Hwan Kim
Hyung Suk Kim
Hyun Sik Yoon
Jung Hoon Kim
Jin Wook Kim
Yong Seong Lee
Se Young Choi
In Ho Chang
Young Hwii Ko
Wan Song
Byong Chang Jeong
Jong Kil Nam
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188564 | https://fiveable.me/thermodynamics-ii/unit-11/normal-shock-waves-oblique-shocks/study-guide/AljYAB9qYOdkeHEY | Normal Shock Waves and Oblique Shocks | Thermodynamics II Class Notes | Fiveable | Fiveable
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🧊Thermodynamics II Unit 11 Review
11.2 Normal Shock Waves and Oblique Shocks
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Thermodynamics II
Unit 11 – Compressible Fluid Flow
Topic: 11.2
🧊Thermodynamics II Unit 11 Review
11.2 Normal Shock Waves and Oblique Shocks
Written by the Fiveable Content Team • Last updated September 2025
Written by the Fiveable Content Team • Last updated September 2025
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🧊Thermodynamics II
Unit & Topic Study Guides
Review of Thermodynamics I Fundamentals
Entropy and the Second Law of Thermodynamics
Exergy Analysis and Irreversibility
Gas Power Cycles: Otto, Diesel, and Brayton
Rankine & Combined Vapor Power Cycles
Refrigeration and Heat Pump Cycles
Thermodynamic Relations & State Equations
Gas Mixtures & Air-Conditioning Processes
Chemical Reactions and Combustion
Phase Equilibrium and Stability
Compressible Fluid Flow
11.1 Stagnation Properties and Isentropic Flow
11.2 Normal Shock Waves and Oblique Shocks
11.3 Nozzles and Diffusers Analysis
Gas Turbines & Jet Propulsion Systems
Vapor–Compression Refrigeration Systems
Internal Combustion Engine Analysis
Exergy and Thermoeconomic Analysis
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Normal shock waves and oblique shocks are crucial concepts in compressible fluid flow. They occur when supersonic flow encounters obstacles, causing sudden changes in flow properties like pressure, temperature, and velocity.
These phenomena play a vital role in designing supersonic aircraft, engines, and wind tunnels. Understanding their behavior and relationships helps engineers optimize performance and efficiency in high-speed applications.
Normal Shock Waves in Compressible Flow
Formation and Characteristics
Normal shock waves form in supersonic flow when the flow encounters an obstruction or sudden change in flow direction, causing the flow to decelerate abruptly to subsonic speeds
Across a normal shock, there is a discontinuous change in flow properties
Sudden increase in pressure, density, temperature, and entropy
Flow velocity decreases
The Mach number upstream of a normal shock is always greater than 1 (supersonic), and downstream of the shock, it is always less than 1 (subsonic)
The strength of a normal shock depends on the upstream Mach number
Higher Mach numbers result in stronger shocks and more significant changes in flow properties
Normal shocks are highly irreversible and non-isentropic
Cause an increase in entropy and a loss of total pressure across the shock
The thickness of a normal shock is on the order of a few mean free paths of the gas molecules (extremely thin region compared to the overall flow field)
Examples
A supersonic aircraft creates a normal shock wave in front of its engine inlet to decelerate the flow to subsonic speeds before entering the engine
In a supersonic wind tunnel, a normal shock forms when the flow encounters a model or a sudden change in the test section geometry
Normal Shock Relations
Equations and Calculations
The normal shock relations are a set of equations that relate the flow properties upstream and downstream of a normal shock, based on conservation of mass, momentum, and energy
The upstream Mach number ($M_1$) is the primary input for the normal shock relations, along with the specific heat ratio ($\gamma$) of the gas
The static pressure ratio ($p_2/p_1$) across a normal shock is given by: p 2 p 1=2 γ M 1 2−(γ−1)γ+1\frac{p_2}{p_1} = \frac{2\gamma M_1^2 - (\gamma-1)}{\gamma+1}p 1p 2=γ+1 2 γ M 1 2−(γ−1)
The static temperature ratio ($T_2/T_1$) across a normal shock is given by: T 2 T 1=(2 γ M 1 2−(γ−1))∗((γ−1)M 1 2+2)(γ+1)2∗M 1 2\frac{T_2}{T_1} = \frac{(2\gamma M_1^2 - (\gamma-1)) ((\gamma-1)M_1^2 + 2)}{(\gamma+1)^2 M_1^2}T 1T 2=(γ+1)2∗M 1 2(2 γ M 1 2−(γ−1))∗((γ−1)M 1 2+2)
The density ratio ($\rho_2/\rho_1$) across a normal shock is given by: ρ 2 ρ 1=(γ+1)M 1 2(γ−1)M 1 2+2\frac{\rho_2}{\rho_1} = \frac{(\gamma+1)M_1^2}{(\gamma-1)M_1^2 + 2}ρ 1ρ 2=(γ−1)M 1 2+2(γ+1)M 1 2
Downstream Mach Number and Total Pressure Ratio
The downstream Mach number ($M_2$) can be calculated using: M 2 2=(γ−1)M 1 2+2 2 γ M 1 2−(γ−1)M_2^2 = \frac{(\gamma-1)M_1^2 + 2}{2\gamma M_1^2 - (\gamma-1)}M 2 2=2 γ M 1 2−(γ−1)(γ−1)M 1 2+2
The total pressure ratio ($p_{02}/p_{01}$) across a normal shock is given by: p 02 p 01=((γ+1)M 1 2 2+(γ−1)M 1 2)γ γ−1(γ+1 2 γ M 1 2−(γ−1))1 γ−1\frac{p_{02}}{p_{01}} = \left(\frac{(\gamma+1)M_1^2}{2 + (\gamma-1)M_1^2}\right)^{\frac{\gamma}{\gamma-1}} \left(\frac{\gamma+1}{2\gamma M_1^2 - (\gamma-1)}\right)^{\frac{1}{\gamma-1}}p 01p 02=(2+(γ−1)M 1 2(γ+1)M 1 2)γ−1 γ(2 γ M 1 2−(γ−1)γ+1)γ−1 1
Examples
For a normal shock in air ($\gamma = 1.4$) with an upstream Mach number of 2.0, the static pressure ratio ($p_2/p_1$) is 4.5, and the downstream Mach number ($M_2$) is 0.58
In a supersonic nozzle, a normal shock can form if the back pressure is higher than the design value, leading to a sudden decrease in the exit velocity and an increase in pressure
Oblique Shock Waves in Supersonic Flow
Formation and Behavior
Oblique shock waves form when a supersonic flow encounters a wedge or a compression corner, causing the flow to turn and compress
The angle between the upstream flow direction and the shock wave is called the shock angle ($\beta$)
Depends on the upstream Mach number ($M_1$) and the deflection angle ($\theta$) of the flow
For a given upstream Mach number, there exists a maximum deflection angle ($\theta_{max}$) beyond which an attached oblique shock cannot form
The shock becomes detached, forming a bow shock
The downstream flow behind an oblique shock remains supersonic, with a reduced Mach number ($M_2$) and a change in flow direction
The strength of an oblique shock increases with increasing deflection angle and upstream Mach number, leading to larger changes in flow properties across the shock
Classification and Limiting Cases
Oblique shocks can be classified as weak or strong
Weak shocks have a smaller shock angle and a higher downstream Mach number compared to strong shocks
In the limiting case where the deflection angle approaches zero, an oblique shock becomes a Mach wave
A weak, isentropic compression wave that does not change the flow properties
Examples
Supersonic aircraft wings are designed with a swept-back angle to generate oblique shocks and reduce wave drag
In supersonic inlet design, a series of oblique shocks are used to decelerate the flow and increase pressure before entering the engine
Oblique Shock Relations
Equations and Calculations
The oblique shock relations are a set of equations that relate the flow properties upstream and downstream of an oblique shock, based on the conservation of mass, momentum, and energy, as well as the geometry of the shock
The primary inputs for the oblique shock relations are the upstream Mach number ($M_1$), the deflection angle ($\theta$), and the specific heat ratio ($\gamma$) of the gas
The $\theta$-$\beta$-$M$ relation is used to determine the shock angle ($\beta$) for a given deflection angle ($\theta$) and upstream Mach number ($M_1$). This relation is given by: tan(θ)=2∗cot(β)∗M 1 2∗sin2(β)−1 M 1 2∗(γ+cos(2 β))+2\tan(\theta) = 2 \cot(\beta) \frac{M_1^2 \sin^2(\beta) - 1}{M_1^2 (\gamma + \cos(2\beta)) + 2}tan(θ)=2∗cot(β)∗M 1 2∗(γ+c o s(2 β))+2 M 1 2∗s i n 2(β)−1
The downstream Mach number ($M_2$) can be calculated using: M 2 2∗sin2(β−θ)=M 1 2∗sin2(β)+(2/(γ−1))(2 γ/(γ−1))∗M 1 2∗sin2(β)−1 M_2^2 \sin^2(\beta-\theta) = \frac{M_1^2 \sin^2(\beta) + (2/(\gamma-1))}{(2\gamma/(\gamma-1)) M_1^2 \sin^2(\beta) - 1}M 2 2∗sin 2(β−θ)=(2 γ/(γ−1))∗M 1 2∗s i n 2(β)−1 M 1 2∗s i n 2(β)+(2/(γ−1))
The static pressure ratio ($p_2/p_1$) across an oblique shock is given by: p 2 p 1=1+2 γ γ+1∗(M 1 2∗sin2(β)−1)\frac{p_2}{p_1} = 1 + \frac{2\gamma}{\gamma+1} (M_1^2 \sin^2(\beta) - 1)p 1p 2=1+γ+1 2 γ∗(M 1 2∗sin 2(β)−1)
Temperature, Density, and Total Pressure Ratios
The static temperature ratio ($T_2/T_1$) and density ratio ($\rho_2/\rho_1$) can be calculated using the same equations as for normal shocks, replacing $M_1$ with $M_1 \sin(\beta)$
The total pressure ratio ($p_{02}/p_{01}$) across an oblique shock is given by: p 02 p 01=(p 2 p 1)γ γ−1∗((γ+1)M 1 2∗sin2(β)2+(γ−1)M 1 2 sin2(β))1 γ−1\frac{p_{02}}{p_{01}} = \left(\frac{p_2}{p_1}\right)^{\frac{\gamma}{\gamma-1}} \left(\frac{(\gamma+1)M_1^2 \sin^2(\beta)}{2 + (\gamma-1)M_1^2 \sin^2(\beta)}\right)^{\frac{1}{\gamma-1}}p 01p 02=(p 1p 2)γ−1 γ∗(2+(γ−1)M 1 2s i n 2(β)(γ+1)M 1 2∗s i n 2(β))γ−1 1
Examples
For a 10-degree wedge in a Mach 2.5 flow of air ($\gamma = 1.4$), the shock angle ($\beta$) is approximately 32 degrees, and the downstream Mach number ($M_2$) is 1.87
In a supersonic wind tunnel, an oblique shock generated by a wedge model can be used to study the effects of shock-boundary layer interaction on aerodynamic performance
11.1 BackNext 11.3
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188565 | https://www.britannica.com/dictionary/burden | The Britannica Dictionary
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burden
3 ENTRIES FOUND:
burden (noun)
burden (verb)
beast of burden (noun)
1
burden
/ˈbɚdn̩/
noun
plural
burdens
1
burden
/ˈbɚdn̩/
noun
plural
burdens
Britannica Dictionary definition of BURDEN
[count]
1
:
something heavy that is carried
:
load
They feared the donkey would collapse under the extra burden.
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see also beast of burden
2
:
someone or something that is very difficult to accept, do, or deal with
a (heavy) burden of sorrow/guilt
She had to bear/shoulder the burden of caring for her aging parents.
The tax burden has been falling increasingly on the middle class. [=middle class people have been paying a greater portion of taxes]
The burden of proof is on the plaintiff, since the defendant is presumed innocent until proven guilty. [=the plaintiff must prove that the defendant is guilty]
the company's large debt burden [=the large amount of money that the company owes]
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— often + on
His illness has placed a huge financial burden on the family.
I'm sorry to be such a burden on you.
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burden has been lifted from your shoulders/back
—
see 1lift
2
burden
/ˈbɚdn̩/
verb
burdens;
burdened;
burdening
2
burden
/ˈbɚdn̩/
verb
burdens;
burdened;
burdening
Britannica Dictionary definition of BURDEN
[+ object]
:
to make (someone) hold or carry something heavy or accept or deal with something difficult
:
to put a heavy burden on (someone)
I don't wish to burden you with my problems.
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— usually used as (be) burdened
Burdened with supplies and equipment, she headed to the camp.
For the rest of his life he was burdened with the knowledge that he had done nothing to help them.
burdened [=weighed down] by guilt
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188566 | https://trumpexcel.com/sum-by-color-excel/ | Skip to content
How to Sum by Color in Excel (Formula & VBA)
Sumit Bansal
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Excel has some really amazing functions, but it doesn’t have a function that can sum cell values based on the cell color.
For example, I have the dataset as shown below, and I want to get the sum of all orange and yellow color cells.
Unfortunately, there is no in-built function to do this.
But never say never!
In this tutorial, I will show you three simple techniques you can use to sum by color in Excel.
Let’s dive in!
This Tutorial Covers:
Toggle
SUM Cells by Color Using Filter and SUBTOTAL
Let’s start with the easiest one.
Below I have a dataset where I have the employee names and their sales numbers.
And in this dataset, I want to get the sum of all the cells colored in yellow and orange.
While there is no in-built function in Excel to sum values based on cell color, there is a simple workaround that relies on the fact that you can filter cells based on the cell color.
For this method, enter the below formula in cell B17 (or any cell in the same column below the colored cells dataset).
=SUBTOTAL(9,B2:B15)
In the above SUBTOTAL formula, I have used 9 as the first argument, which tells the function that I want to get the sum of the range that is given as the second argument.
But why not just use the SUM formula instead?
This is because when I have the SUBTOTAL formula and I filter the cells to only show those cells that have a specific color in it, the SUBTOTAL formula will show me the sum of visible cells only (something that the SUM formula can not do).
So once you have the SUBTOTAL formula, follow the below steps to get the SUM based on cell color:
Select any cell in the dataset
Click the Data tab
In the Sort and Filter group, click on the Filter icon. This will apply filter to the dataset and you will be able to see the filter icon in the headers
In the Sales header cell, click on the Filter icon
Go to the Filter by Color” option
Select the color based on which you want to filter the dataset.
As soon as you do this, you will notice that the subtotal result changes and it will now only give you the sum of those cells that are visible (which would only be those cells that have the color by which you filtered the dataset).
Similarly, if you filter by some other color in the data set (say orange instead of yellow), the SUBTOTAL function would accordingly adjust and give you the sum of all cells with orange color
Pro Tip: Keyboard shortcut to apply a filter to a dataset is Control + Shift + L (hold the Control and the Shift key, and then press the L key). If using Mac, use Command + Shift + L
Also read: SUM Based on Partial Text Match in Excel (SUMIF)
SUM Cells by Color Using VBA
I mentioned that there is no in-built formula in Excel to sum based on cell color value. However, you can create your own formula to do this using VBA.
With VBA, you can create a custom function that you can keep in the back end, and then use it like any other regular function in the worksheet.
Below is the VBA code that will create that custom function that allows you to sum by color in Excel.
'Code created by Sumit Bansal from 'This VBA code created a function that can be used to sum cells based on color Function SumByColor(SumRange As Range, SumColor As Range) Dim SumColorValue As Integer Dim TotalSum As Long SumColorValue = SumColor.Interior.ColorIndex Set rCell = SumRange For Each rCell In SumRange If rCell.Interior.ColorIndex = SumColorValue Then TotalSum = TotalSum + rCell.Value End If Next rCell SumByColor = TotalSum End Function
To use this VBA custom function, you will first have to copy this code and paste it in the back end in the VB editor.
Once done, you’ll be able to use this function in the worksheet.
Below are the steps to add this code to the VB editor.
Click the Developer tab in the ribbon (if you don’t have the Developer tab visible, click here to learn how to get it)
Click on the Visual Basic icon. This will open the Visual Basic editor of Excel.
Click the Insert option in the menu
Click on Module. This will insert a new module and you will be able to see it in the project explorer (a pane on the left that shows all the objects). If you don’t see the Project Explorer pane, click on View and then click on Project Explorer
Copy the above VBA code and paste it in the newly inserted Module code window
Close the VB Editor
Now that you have the code in the back end in excel, you will be able to use the function that we created (SumByColor) in the worksheet.
For this function to work, I will need a cell in the worksheet that contains the same color for which I want to get the sum.
In our example, I have done that to cells D2 and D3, where D2 has the yellow color and D3 has the orange color.
Now I can use the below formula in these cells:
=SumByColor($B$2:$B$15,D2)
The above formula takes two arguments:
The range of cells that have the color that I want to add
Reference to any cell that contains the color (so that the formula can pick the color index and use that as a condition to add the values)
Note that the formula is dynamic and would automatically update in case you make any changes to the data set (such as changing any value or applying/removing color from some cells). And in case you notice that the formula is not updating, hit the F9 key and it would update.
Since we have used a VBA code in the workbook, it needs to be saved as a macro-enabled workbook (with .XLSM extension).
Pro Tip: If adding cells based on their background color is something you need to do quite often, I recommend you copy and paste this VBA code for the custom formula in the Personal Macro Workbook. This way, that would be available on all your workbooks on your system.
Also read: Filter By Color in Excel
SUM Cells by Color Using GET.CELL
The final method I want to show you include a hidden Excel formula (that not many people know about).
This method uses the GET.CELL function, which can get us the color index value of colored cells.
And once we have the color index value of each cell, we can then use a simple sum if formula to only get the sum of cells with a specific color in it.
It’s not as elegant as the VBA custom function I covered earlier, but if you don’t want to use VBA, then this could be the way to go for you.
GET.CELL is an old Macro 4 function that has been kept in Excel for compatibility reasons, but you won’t find many details about it (as it’s rarely used).
Below I have a dataset where I have colored cells that I want to sum.
For this technique to work, we first need to create a named range that will use the GET.CELL function to give the color value of a colored cell.
Here are the steps to do this:
Click the Formulas tab in the Ribbon
In the Defined Names group, click on the ‘Name Manager’
In the Name Manager dialog box, click on New
In the ‘New Name dialog box, enter the Name – SumColor
In the Refers to field, enter the following formula: =GET.CELL(38,$B2)
Click OK
Close the Name Manager dialog box
The above steps have created a named range that we can now use in the workbook.
Note: GET.CELL function takes two arguments, the first one is a number that tells the function what information we need, and the second is the cell reference of that cell itself. In this case, I have 38 as the first argument, which would give us the color index value of the referred cell.
Now the second step is to get the color index values of all the colors in column B.
Below are the steps to do this:
In cell C1, enter the header – Color Index (or anything that you to call it)
In cell C2, enter the following formula: =SumColor
Apply the same formula for all the cell in column C (you can use the fill handle or simply copy paste cell C2)
The above steps would give you a value that would represent the color index of the cell in column B (the cell on the left).
SumColor is the named range we created and it uses the GET.CELL function to get the color index value of the cell on the left. You can use this formula in any column, but it should always start from the second cell in that column. For example, instead of column C, you can have this in column H or J, but from the second cells in these columns.
Now that we have a unique number for each color, we can use this to get the SUM of cells based on their color.
Below are the steps to do this:
In cell E2 and E3, give the cells the color for which you want to get the sum. In my case, I have yellow color in cell E2 and orange in cell E3
In cell F2, enter the following formula: =SUMIF(C2:C15,SumColor,B2:B15)
Copy the cell and paste in cell F3 (this could copy the formula as well and adjust the references).
The above steps would give you the sum of all the cells that have the color in the adjacent selling column F.
We have used a SUMIF formula, where it adds all the values in column B, if the color index of the cell on the left (in column F) is the same as that in column C.
While this is definitely a slightly longer way to sum cells by color (as compared with SUBTOTAL or VBA), this gets the work done, and you only need to do this setup once.
Note that while the formula is dynamic, in case you make any changes (such as changing the color cells in column B or removing the color from it), the change might not be reflected instantly in the SUMIF formula. A simple workaround for it would be to go to the cell that has the formula, click the F2 key to get into the airport, and then hit the enter key. This would force the formula to recalculate and you will get the updated result.
So these are three methods you can use to sum by color in Excel. While SUBTOTAL is quite easy and straightforward, I personally like the VBA method better.
I hope you found this tutorial useful!
Other Excel tutorials you may also like:
Count Characters in a Cell (or Range of Cells) Using Formulas in Excel
How to Count Colored Cells in Excel
How to Sort By Color in Excel (in less than 10 seconds)
How to Apply Formula to Entire Column in Excel
How to Sum Only Positive or Negative Numbers in Excel (Easy Formula)
How to SUM values between two dates (using SUMIFS formula)
How to Combine Duplicate Rows and Sum the Values in Excel
How to Sum Across Multiple Sheets in Excel? (3D SUM Formula)
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Hey! I'm Sumit Bansal, founder of TrumpExcel.com and a Microsoft Excel MVP. I started this site in 2013 because I genuinely love Microsoft Excel (yes, really!) and wanted to share that passion through easy Excel tutorials, tips, and Excel training videos. My goal is straightforward: help you master Excel skills so you can work smarter, boost productivity, and maybe even enjoy spreadsheets along the way!
2 thoughts on “How to Sum by Color in Excel (Formula & VBA)”
Thank you for this useful summary. BTW, if you use get.cell, it requires saving the file as xlsm.
Reply
2. How do I sum if I have coloured the parallel Column, like Having the numbers in the K column and colour in the J column, and I want the sum of the column K if the J column cell is green or any other colour.
Thank You
Reply
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6.2 Trigonometric identities
| | | |
---
| Previous 6.1 Revision | | Next 6.3 Reduction formula |
6.2 Trigonometric identities (EMBHH)
An identity is a mathematical statement that equates one quantity with another. Trigonometric identities
allow us to simplify a given expression so that it contains sine and cosine ratios only. This enables us to
solve equations and also to prove other identities.
Quotient identity
Quotient identity
Complete the table without using a calculator, leaving your answer in surd form where applicable:
| | | | |
--- --- |
| | θ=45° | | |
| sinθ | | | |
| cosθ | | | |
| sinθcosθ | | | |
| tanθ | | | |
2. Examine the last two rows of the table and make a conjecture.
3. Are there any values of θ for which your conjecture would not be true? Explain your answer.
We know that tanθ is defined as:
tanθ=opposite sideadjacent side
Using the diagram below and the theorem of Pythagoras, we can write the tangent function in terms of x,y
and r:
This is the quotient identity:
tanθ=sinθcosθ
Notice that tanθ is undefined if cosθ=0, therefore θ≠k×90°, where k is an odd integer.
Square identity
Square identity
Use a calculator to complete the following table:
| | |
--- |
| sin280°+cos280°= | |
| cos223°+sin223°= | |
| sin50°+cos50°= | |
| sin267°−cos267°= | |
| sin267°+cos267°= | |
2. What do you notice? Make a conjecture.
3. Draw a sketch and prove your conjecture in general terms, using x,y and r.
Using the theorem of Pythagoras, we can write the sine and cosine functions in terms of x,y and r:
sin2θ+cos2θ=(yr)2+(xr)2=y2r2+x2r2=y2+x2r2=r2r2=1
This is the square identity:
sin2θ+cos2θ=1
Other forms of the square identity
Complete the following:
sin2θ=1−……
cosθ=±√……
sin2θ=(1+……)(1−……)
cos2θ−1=……
Here are some useful tips for proving identities:
Change all trigonometric ratios to sine and cosine.
Choose one side of the equation to simplify and show that it is equal to the other side.
Usually it is better to choose the more complicated side to simplify.
Sometimes we need to simplify both sides of the equation to show that they are equal.
A square root sign often indicates that we need to use the square identity.
We can also add to the expression to make simplifying easier:
replace 1 with sin2θ+cos2θ.
multiply by 1 in the form of a suitable fraction, for example 1+sinθ1+sinθ.
Worked example 5: Trigonometric identities
Simplify the following:
tan2θ×cos2θ
1cos2θ−tan2θ
Write the expression in terms of sine and cosine only
We use the square and quotient identities to write the given expression in terms of sine and cosine
and then simplify as far as possible.
tan2θ×cos2θ=(sinθcosθ)2×cos2θ=sin2θcos2θ×cos2θ=sin2θ
1cos2θ−tan2θ=1cos2θ−(sinθcosθ)2=1cos2θ−sin2θcos2θ=1−sin2θcos2θ=cos2θcos2θ=1
Worked example 6: Trigonometric identities
Prove: 1−sinαcosα=cosα1+sinα
Note restrictions
When working with fractions, we must be careful that the denominator does not equal 0.
Therefore cosθ≠0 for the fraction on the left-hand side and sinθ+1≠0
for the fraction on the right-hand side.
Simplify the left-hand side
This is not an equation that needs to be solved. We are required to show that one side of the
equation is equal to the other. We can choose either of the two sides to simplify.
LHS=1−sinαcosα=1−sinαcosα×1+sinα1+sinα
Notice that we have not changed the equation — this is the same as multiplying by 1
since the numerator and the denominator are the same.
Determine the lowest common denominator and simplify
LHS=1−sin2αcosα(1+sinα)=cos2αcosα(1+sinα)=cosα1+sinα=RHS
Trigonometric identities
Textbook Exercise 6.2
Reduce the following to one trigonometric ratio:
(\dfrac{\sin\alpha}{\tan\alpha})
\begin{align}
\dfrac{\sin\alpha}{\tan\alpha} &= \dfrac{\sin\alpha}{\tan\alpha} \
&= \dfrac{\sin\alpha}{\dfrac{\sin\alpha}{\cos\alpha}} \
&= \sin\alpha \times \dfrac{\cos\alpha}{\sin\alpha} \
&= \cos\alpha
\end{align}
(\cos^2 \theta \tan^2 \theta + \tan^2 \theta \sin^2 \theta)
\begin{align}
\cos^2 \theta \tan^2 \theta + \tan^2 \theta \sin^2 \theta
&= \tan^2 \theta \left( \cos^2 \theta + \sin^2 \theta \right) \
&= \tan^2 \theta \left( 1 \right) \
&= \tan^2 \theta
\end{align}
(1-\sin \theta \cos \theta \tan \theta)
\begin{align}
1-\sin \theta \cos \theta \tan \theta
&= 1-\sin \theta \cos \theta \left( \frac{\sin\theta}{\cos\theta} \right) \
&= 1-\sin^2 \theta \
&= \cos^2 \theta
\end{align}
(\left( \dfrac{1 - \cos^2\beta}{\cos^2 \beta}\right) - \tan^2 \beta)
\begin{align}
\left( \dfrac{1 - \cos^2\beta}{\cos^2 \beta}\right) - \tan^2 \beta
&= \left( \dfrac{\sin^2\beta}{\cos^2 \beta}\right) - \tan^2 \beta \
&= \left( \tan^2 \beta \right) - \tan^2 \beta \
&= \text{0}
\end{align}
Prove the following identities and state restrictions where appropriate:
(\dfrac{1+\sin\theta}{\cos\theta} = \dfrac{\cos\theta}{1-\sin\theta})
\begin{align}
\text{LHS } &= \dfrac{1+\sin \theta }{\cos\theta }\
&=\dfrac{(1+\sin\theta)}{\cos\theta})\times \dfrac{(1-\sin\theta)}{1-\sin\theta}\
&=\dfrac{1-\sin^2\theta}{\cos\theta(1-\sin\theta)} \
&=\dfrac{\cos^2\theta }{\cos\theta(1-\sin\theta)} \
&=\dfrac{\cos\theta}{(1-\sin\theta)} \
&= \text{RHS}
\end{align}
Restrictions: undefined where (\cos \theta = \text{0}, \sin \theta = \text{1}) and
where (\tan \theta)is undefined.
Therefore (\theta \ne \text{90}\text{°};\text{270}\text{°}).
({\sin}^{2}\alpha + \left(\cos\alpha - \tan\alpha \right)\left(\cos\alpha +\tan\alpha
\right)=1-{\tan}^{2}\alpha)
\begin{align}
\text{LHS } &= \sin^2 \alpha + (\cos\alpha-\tan\alpha) (\cos\alpha+\tan\alpha) \
&=\sin^2\alpha+\cos^2\alpha-\tan^2\alpha \
&=1 - \tan^2\alpha \
&= \text{RHS}
\end{align}
Restrictions: undefined where (\tan \theta) is undefined.
Therefore (\theta \ne \text{90}\text{°};\text{270}\text{°}).
(\dfrac{1}{\cos\theta}-\dfrac{\cos\theta {\tan}^{2}\theta }{1}=\cos\theta)
\begin{align}
\text{LHS } &= \dfrac{1}{\cos\theta}-\dfrac{\cos\theta \tan^2\theta}{1}\
&=\dfrac{1-\cos^2\theta \times \tan^2\theta}{\cos\theta}\
&=\dfrac{1-\cos^2\theta \times \dfrac{\sin^2\theta}{\cos^2\theta}}{\cos\theta}\
&=\dfrac{1-\sin^2\theta}{\cos\theta}\
&=\dfrac{\cos^2\theta}{\cos\theta}\
&=\cos\theta\
&= \text{RHS}
\end{align}
Restrictions: undefined where (\cos \theta = \text{0}) and where (\tan \theta)is
undefined.
Therefore (\theta \ne \text{90}\text{°};\text{270}\text{°}).
(\dfrac{2\sin\theta\cos\theta}{\sin\theta +\cos\theta }=\sin\theta +\cos\theta
-\dfrac{1}{\sin\theta +\cos\theta })
\begin{align}
\text{RHS } &= \sin\theta+\cos\theta-\dfrac{1}{\sin\theta+\cos\theta} \
&=\dfrac{\sin^2\theta+\sin\theta \cos\theta+\cos\theta
\sin\theta+\cos^2\theta-1}{\sin\theta+\cos\theta} \
&=\dfrac{1+2\sin\theta \cos\theta-1}{\sin\theta+\cos\theta} \
&=\dfrac{2\sin\theta \cos\theta}{\sin\theta+\cos\theta} \
&= \text{LHS}
\end{align}
(\left(\dfrac{\cos\beta}{\sin\beta }+\tan\beta \right) \cos\beta =\dfrac{1}{\sin\beta
})
\begin{align}
\text{ LHS} &= \left ( \dfrac{\cos\beta}{\sin\beta}+\dfrac{\sin\beta}{\cos\beta} \right
) \
&=\left ( \dfrac{\cos^2\beta + \sin^2\beta}{\sin\beta\cos\beta} \right )\cos\beta\
&=\dfrac{1}{\sin\beta}\
&= \text{RHS }
\end{align}
(\dfrac{1}{1 + \sin \theta} + \dfrac{1}{1 - \sin \theta} = d\frac{2 \tan \theta}{\sin
\theta \cos \theta})
\begin{align}
\text{ LHS} &= \dfrac{1-\sin\theta+1+\sin\theta}{(1+\sin\theta)(1-\sin\theta)} \
&=\dfrac{2}{1-\sin^2\theta} \
&=\dfrac{2}{\cos^2\theta}
\end{align}
\begin{align}
\text{ RHS} &= \dfrac{2 \tan \theta}{\sin \theta \cos \theta} \
&=\dfrac{2 \sin \theta}{\sin \theta \cos \theta \cos \theta} \
&=\dfrac{2}{\cos^2\theta} \
\therefore \text{LHS } &= \text{RHS }
\end{align}
(\dfrac{(1 + \tan^2 \alpha) \cos \alpha}{(1 - \tan \alpha)} = \dfrac{1}{\cos \alpha -
\sin \alpha})
\begin{align}
\text{ LHS} &= \dfrac{(1 + \tan^2 \alpha) \cos \alpha}{(1 - \tan \alpha)} \
&= \dfrac{\left (1+\frac{\sin^2\alpha}{\cos^2\alpha} \right )\cos\alpha}{\left (
1-\frac{\sin\alpha}{\cos\alpha} \right )} \
&=\dfrac{\left ( \frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha} \right
)\cos\alpha}{\frac{\cos\alpha-\sin\alpha}{\cos\alpha}}\
&=\dfrac{1}{\cos\alpha}\times \frac{\cos\alpha}{\cos\alpha-\sin\alpha} \
&=\dfrac{1}{\cos\alpha-\sin\alpha} \
&= \text{RHS }
\end{align}
| | | |
---
| Previous 6.1 Revision | Table of Contents | Next 6.3 Reduction formula | |
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cells provide helper functions to other cells of the immune system—especially the antigen-presenting cells (APCs) such as macrophages, dendritic cells, and B cells—and are important for their activation and maturation. There are distinct subsets of CD4+ Th cells, including Th1, Th2, Th9, Th17, Th22, Tfh, and Treg cells, each activated by a specific set of cytokines and transcription factors and characterized by the cytokines they secrete and effector functions they perform.
Th2 cells mediate the activation and maintenance of the humoral, or antibody-mediated, immune response against extracellular parasites, bacteria, allergens, and toxins. Th2 cells mediate these functions by producing various cytokines such as IL-4, IL-5, IL-6, IL-9, IL-13, and IL-17E (IL-25). These cytokines are responsible for a strong antibody production, eosinophil activation, and inhibition of several macrophage functions, thus providing phagocyte-independent protective responses. These cytokines also counteract the Th1 responses that allow for the Th2 responsiveness to IL-4. Functionally, Th2 cytokines have effects on many cell types in the body as the cytokine receptors are widely expressed on numerous cell types. Th2 cells stimulate and recruit specialized subsets of immune cells, such as eosinophils and basophils, to the site of infection or in response to allergens or toxins leading to tissue eosinophilia and mast cell hyperplasia. They induce mucus production, goblet cell metaplasia, and airway hyper-responsiveness. Th2 cells also control the regulation of B cell class-switching to IgE. Because of their influence on the production of antibodies and allergic responses, over activation of Th2 cells appears to be responsible for the exacerbation of allergies (Type-1, immediate hypersensitivity reactions), autoimmune reactions such as chronic graft-versus host disease, progressive systemic sclerosis, and systemic lupus erythematosus. Additionally, Th2 cells are also known to be responsible for the development of asthma and other allergic inflammatory diseases. Interestingly, Th2 cells also produce the growth factor amphiregulin and IL-24 which have anti-tumor effects.
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On this page:
Differentiation of Th2 cells
Th2 and innate lymphoid cells (ILC) and the inflammatory response
Th2 cells in disease
Tools to study Th2 cells
References
Differentiation of Th2 cells
Naive CD4+ T cells are activated by simultaneous engagement of the T cell receptor (TCR) and co-stimulatory molecule, cluster of differentiation (CD)28, by antigen-presenting cells (APCs) expressing major histocompatibility complex (MHC) class II, such as macrophages, dendritic cells, and B cells . Activation signals are conveyed upon recognition and binding of the TCR to its cognate antigenic peptide displayed on the surface of the MHC class II molecule and through multiple co-stimulatory signals. Once fully activated, naive CD4+ T cells in the lymphoid tissue rapidly proliferate, undergoing clonal expansion and differentiation into helper T cells, such as Th1, Th2, or Th17 cells . The surrounding cytokine environment drives polarization of naive T cells into specific cell lineages, where the Th2 cell type is specifically triggered by interleukin (IL)-4 and IL-2 .
Signaling pathways for Th2 cell polarization
IL-4 stimulation drives signal transducers and activators of transcription (STAT)6 phosphorylation and upregulation of GATA-binding protein 3 (GATA3), a key transcription factor for Th2 cell development . GATA3 interacts with IL-2-induced STAT5 signaling to upregulate expression of IL-4, the canonical Th2 cytokine. This signaling pathway is followed by an IL-4–driven positive feedback loop whereby the Th2 gene expression profile is amplified by continued IL-4-STAT6-GATA3 and IL-5-STAT5 signaling. The IL-4-dependent STAT6 activation of GATA3 is the canonical Th2 cell differentiation pathway; however, there are a number of nonclassical pathways thought to be important for Th2 polarization that include a variety of transcription factors (such as c-Maf, NF-κB, IRF4, AP1) and cytokines (thymic stromal lymphopoietin (TSLP), IL-5, IL-25, and IL-33) [2,3,4,5].
Th2 and innate lymphoid cells (ILC)
Differentiated effector Th cells migrate to sites of inflammation in the periphery, where they re-encounter their cognate antigen and secrete effector cytokines, thus driving an antigen-specific immune response. IL-5, IL-9, and IL-13 are secreted by Th2 cells only once they have reached inflamed tissue sites (Figure 1) . Epithelial and innate cells at the site of inflammation also play a role in shaping the Th2 adaptive immune response. Innate lymphoid cells (ILC), macrophages, and dendritic cells secrete proinflammatory cytokines, including IL-1, IL-18, IL-25, IL-33, GM-CSF, M-CSF, and TSLP, which signal through different pathways to promote Th2 differentiation [2,3]. The cytokine milieu in the surrounding inflammatory microenvironment drives different Th2 effector activities. As a result, there are several phenotypically and functionally distinct Th2 memory cell subsets that carry out different effector roles depending on the cytokine influence. For instance, Th2 cells that produce high levels of IL-5 have been shown to contribute to the pathogenesis of allergic asthma, while cells that produce IL-5, IL-17, and IFN-γ, in addition to IL-4 and IL-13, have been identified as noncanonical memory Th2 cells that drive chronic allergic inflammatory diseases and prevent lymphocytic choriomeningitis virus (LCMV) persistence . Memory Th2 cells also express IL17RB, TSLP receptor (TSLPR), chemoattractant receptor-homologous molecule (CRTh2), and CCR8 at similar levels [2,4].
Memory Th2 cells can be divided into at least four distinct subpopulations based on the levels of expressed chemokine (C-X-C motif) receptor 3 (CXCR3) and CD62L . All four subpopulations characteristically produce large amounts of IL-4 and IL-13 in response to antigenic re-stimulation. However, only the CXCR3 low CD62L low subpopulation produces IL-5 and is considered pathogenic, denoted as memory-type Tpath cells . IL-33 stimulation induces production of IL-5 and IL-13 in memory Th2 cells and type 2 innate lymphoid cells (ILC2), thereby driving a type 2 immunological response [2,4]. Similar to IL-33, the combination of IL-2 and IL-25 also drives IL-5 production in Th2 memory cells and ILC [4,6]. ILC are immune cells derived from common lymphoid progenitors and are grouped based on their cytokine and transcription factor profiles, where ILC2 function in part to regulate Th2 phenotype . Whereas IL-33 can induce a strong Th2 immunity and eosinophilic inflammation by amplifying the Th2 cytokine response in lung and intestine tissue, IL-33 has little effect on the Th2 effector subset and instead mediates inflammation via memory-type Tpath cells . Exposure to an allergen, for example, may initiate a Th2 response in the airway by stimulating epithelial and endothelial cells to produce IL-33, thereby driving IL-5 production by Th2 memory cells and exacerbating the eosinophilic inflammation .
Table 1. Subpopulation of Th2 cells and key cytokine profile.
| Memory Th2 cell subpopulation | Secreted cytokines |
--- |
| CXCR3 low CD62L low | IL-5, IL-4, IL-13 |
| CXCR3 hi CD62L hi | IL-4, IL-13, Eomes |
| CXCR3 low CD62L hi | IL-4, IL-13, Eomes |
| CXCR3 hi CD62L low | IL-4, IL-13, Eomes |
Figure 1. Innate lymphoid cells (ILC) modulate the Th2 adaptive immune response. ILCs are immune cells derived from common lymphoid progenitors (CLPs). ILCs are grouped based on the cytokines they can produce and the transcription factors that regulate their development and function. They can rapidly secrete immunoregulatory cytokines to enable early immune responses to infection.
Th2 cells in disease
The exact mechanisms regulating maintenance of Th2 memory cells remains unclear. It has been shown that antigen-specific memory Th2 cells driving lung allergic responses reside within the lung tissue and that the Th2 memory response is dependent on IL-7 and IL-33-producing lymphatic endothelial cells (LECs) found within localized structures called inducible bronchus-associated lymphoid tissue (iBALT) . Th2 cells producing IL-4, IL-5, and IL-13 are implicated in a number of inflammatory diseases, including asthma, chronic rhinosinusitis, atopic dermatitis, and eosinophilic gastrointestinal disorders such as ulcerative colitis [3,4]. A major cell type shown to drive murine chronic airway inflammation has been identified as CD44 hi CD62L low CXCR3 low CCR4 hi CCR8 hi IL7Rα hi ST2 hi memory-type Tpath2 cells, whereas CCR8 hi Tpath2 cells and CRTH2 hi CD161 hi PGDS hi Tpath2 cells have been shown to drive chronic atopic dermatitis and eosinophilic gastrointestinal diseases, respectively . Th2 cytokines drive pathogenesis of human asthma and murine allergic airway inflammation models by promoting eosinophilic infiltration, increased mucus production through goblet cell metaplasia, and tissue fibrosis [3,4]. Monoclonal antibodies directed against IL-5 and IL-13 have been used successfully as a therapies against asthma, and antibodies against Th2 cells show potential for future development in therapeutics . Th2 cell-derived IL-4, IL-5, and IL-13 contribute to B cell proliferation and isotype class switching from immunoglobulin (Ig)G1 to IgE, a key antibody involved in parasitic helminth infection and certain allergic diseases associated with Th2 cells such as asthma [3,5]. Th2 cells have also been shown to induce the alternate activation (M2) macrophage phenotype, which mediates resolution of the inflammatory stage and initiates tissue repair in wound healing processes. Effector Th2 cells mediate immune responses to parasitic helminth infections, venoms, certain bacterial infections, and promote wound healing by inducing alternate activation (M2) macrophage phenotype .
Tools to study Th2 cells
Tools for profiling and identification of Th2 cells
Cytokine profiling is commonly used to classify the Th cell subtype and also to quantify the amounts of cytokines secreted. Cytokine ELISAs can be used to monitor T cell dependent cytokine secretion in response to activation and lineage-specific differentiation at the population level. Cytokine ELISA kits suitable for the detection and quantitation of hundreds of individual cytokines are commercially available. These kits are typically sold as 96-well plates pre-coated with the capture antibody and contain the detection antibody, as well as standards, buffers, and accessory reagents. Assay sensitivities are commonly in the picogram range.
While ELISAs can be used to measure the secretion of individual cytokines, advances in Luminex multiplexing technology allow for the high-throughput detection of multiple cytokines in a single sample or reaction well. The simultaneous measurement of multiple cytokines is achieved using a bank of antibodies bound to microspheres dyed with fluorophores of differing intensities. Quantitation is accomplished using a sandwich assay approach in combination with a Luminex detection system. A list of those relevant to Th2 biology can be found below in Table 2.
Table 2. Invitrogen ProcartaPlex Luminex Panels for Th2 study.
| Species | Description | Target Analytes | Cat. No. |
--- --- |
| Mouse | Th1/Th2 Cytokine & Chemokine 20-Plex Mouse ProcartaPlex | IFN gamma; IL-12p70; IL-13; IL-1 beta; IL-2; IL-4; IL-5; IL-6; TNF alpha; GM-CSF; IL-18; GRO-alpha; IP-10; MCP-1; MCP-3; MIP-1 alpha; MIP-1 beta; MIP-2; RANTES; Eotaxin | EPX200-26090-901 |
| Th1/Th2/Th9/Th17/Th22/Treg Cytokine 17-Plex Mouse ProcartaPlex Panel | IFN gamma; IL-12p70; IL-13; IL-1 beta; IL-2; IL-4; IL-5; IL-6; TNF alpha; GM-CSF; IL-18; IL-10; IL-17A; IL-22; IL-23; IL-27; IL-9 | EPX170-26087-901 |
| Human | Th1/Th2 Cytokine & Chemokine 20-Plex Human ProcartaPlex Panel 1 | GM-CSF; IFN gamma; IL-1 beta; IL-2; IL-4; IL-5; IL-6; IL-12 p70; IL-13; IL-18; TNF alpha; Eotaxin; GRO-alpha; IL-8; IP-10; MCP-1; MIP-1 alpha; MIP-1 beta; Rantes; SDF-1 alpha | EPX200-12173-901 |
| Th1/Th2/Th9/Th17 Cytokine 18-Plex Human ProcartaPlex Panel | GM-CSF; IFN gamma; IL-1 beta; IL-2; IL-4; IL-5; IL-6; IL-9; IL-10; IL-12p70; IL-13; IL-17A; IL-18; IL-21; IL-22; IL-23; IL-27; TNF alpha | EPX180-12165-901 |
Flow Cytometry to study Th2 cells
In addition to ELISA-based methods, another common and powerful tool to study Th2 and other immune populations is flow cytometry. Whereas an ELISA measures the amounts of cytokines secreted, flow cytometry can be used to profile cells based on both surface-expressed or intracellular markers, as well as cytokine expression. Th2 cells are defined by the combined expression of surface and intracellular targets such as: CD45+ CD3+ CD4+ IL-4+ CCR4+ CRTH2+ (CD8- CD19- CCR6- CXCR5- CXCR3- CCR10-) [1,2,3,4]. In addition, flow cytometry can be used to quantify the Th2 population with respect to other populations.
CD4+ Th2 cell specific markers
| Species | Marker | Marker type |
---
| Human | Amphiregulin IL-4 IL-5 IL-13 | Secreted |
| CD3 CD4 CD45 CD184 (CXCR4) CD194 (CCR4) CD198 (CCR8) CD294 (CRTH2) CD365 (TIM1) IL-25R (IL-17RB) IL-33R (ST2) | Surface |
| GATA-3 BATF IRF4 STAT6 | Intracellular transcription factor |
| Mouse | Amphiregulin IL-4 IL-5 IL-10 IL-13 | Secreted |
| CD3 CD4 CD44 CD184 (CXCR4) CD194 (CCR4) CD198 (CCR8) CD294 (CRTH2) CD365 (TIM1) IL-25R (IL-17RB) IL-33R (ST2) | Surface |
| GATA-3 BATF IRF4 STAT6 | Intracellular transcription factor |
Optimized Multicolor Immunofluroscence Panels (OMIPs) published in the journal Cytometry Part A (Wiley Online Library) describe the use of a combination of specific antibodies and fluorophores for the extensive characterization of cell types by flow cytometry. The articles below provide panels that have been thoroughly tested and use validated set of antibodies and reagents that can be used together for the multicolor characterization of Th2 cell populations.
Table 3. Cytometry Part A OMIPs designed for Th2 cell characterization and analysis.
| OMIP ID | OMIP name and link | Immune context (keywords) |
---
| OMIP-008 | OMIP-008: Measurement of Th1 and Th2 cytokine polyfunctionality of human T cells | Tumor-specific T cell cytokine profiles, PMA, and ionomycin T cell activation |
| OMIP-014 | OMIP-014: Validated multifunctional characterization of antigen-specific human T cells by intracellular cytokine staining | Antigen-specific T cell responses |
| OMIP-017 | OMIP-017: Human CD41 helper T-cell subsets including follicular helper cells | Relative frequencies of CD4+ T-helper cell subsets in PBMC from healthy individuals |
| OMIP-018 | OMIP-018: Chemokine receptor expression on human T helper cells | Chemokine receptor expression on CD4+ T helper cell subsets in PBMC from healthy individuals See gating strategy below. |
| OMIP-025 | OMIP-025: Evaluation of human T- and NK-cell responses including memory and follicular helper phenotype by intracellular cytokine staining | Antigen-specific T cell responses |
| OMIP-030 | OMIP-030: Characterization of human T cell subsets via surface markers | Relative frequency T cell subsets via detection of surface markers |
| OMIP-033 | OMIP-033: A comprehensive single step staining protocol for human T- and B-cell subsets | Optimized staining panel for CSF and whole blood with low cell number (10–100,000 cells) |
| OMIP-042 | OMIP-042: 21-color flow cytometry to comprehensively immunophenotype major lymphocyte and myeloid subsets in human peripheral blood | 21-color immunophenotyping in human PBMC from patients treated with alloHSCT for GVHD |
| OMIP‐052 | OMIP‐052: An 18‐color panel for measuring Th1, Th2, Th17, and Tfh responses in rhesus macaques | Nonhuman primate, intracellular cytokine staining |
| OMIP‐056 | OMIP‐056: Evaluation of human conventional T cells, donor‐unrestricted T cells, and NK cells including memory phenotype by intracellular cytokine staining | 26-color staining panel to profile antigen-specific T cells in cryopreserved healthy PBMC |
| OMIP-062 | OMIP-062: A 14-color, 16-antibody panel for immunophenotyping human innate lymphoid, myeloid and T cells in small volumes of whole blood and pediatric airway samples | Optimized staining panel for whole blood with low volume/cell numbers |
References
Roche PA, Furuta K (2015) The ins and outs of MHC class II-mediated antigen processing and presentation. Nat Rev Immunol 15:203–216.
Zhu J (2015) T helper 2 (Th2) cell differentiation, type 2 innate lymphoid cell (ILC2) development and regulation of interleukin-4 (IL-4) and IL-13 production. Cytokine 75:14–24.
Stark JM, Tibbitt CA, Coquet JM (2019) The metabolic requirements of Th2 cell differentiation. Front Immunol 10:2318.
Nakayama T, Hirahara K, Onodera A et al. (2017) Th2 cells in health and disease. Annu Rev Immunol 35:53–84.
Endo Y, Iwamura C, Kuwahara M et. al. (2011) Eomesodermin controls interleukin-5 production in memory T helper 2 cells through inhibition of activity of transcription factor GATA3. Immunity 35:P733-P745.
Walker JA, McKenzie ANJ (2018) T H 2 cell development and function. Nat Rev Immunol 18:121–134.
Related articles and resources
T-helper cell paradigm poster
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Find detailed marker information for immune cell types and subtypes.
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Discover protocols for various applications to study immunology.
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188569 | https://cs.stackexchange.com/questions/62926/finding-all-solutions-to-an-integer-linear-programming-ilp-problem | Skip to main content
Finding all solutions to an integer linear programming (ILP) problem
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My problem is to find all integer solutions to an ILP. As an example, I'm using an ILP with two variables, but I may have more than two variables. I describe the method I currently use to solve this problem near the end, but I'm interested in knowing if there is a proper and efficient algorithm or method for solving this kind of problem.
There is no objective function, but the constraints for this ILP are
0≤−2x−y≤80≤1−x+3y≤50≤2+x−y≤2x,y∈Z
Since this ILP has two variables, I can visually inspect the solution region by graphing the lines formed by the constraints, which are
yyyyyy≤−2x≥−2x−8≥13x−13≤13x+43≤x+2≥x
By inspection, there are 6 integer solutions for (x,y): {(0,0),(−1,1),(−1,0),(−2,0),(−2,−1),(−3,−1)}.
However, my current method is to use linear programming with non-negativity relaxed and integers from branch-and-cut. I've tried using a set of four objective functions: minimize x, maximize x, minimize y, and maximize y. These give a smaller search area as
−3≤x≤0−1≤y≤1
I then iterate over all valid integer tuples in that smaller area and filter it for tuples that satisfy the original constraints. The tuples that remain are all valid integer solutions.
linear-programming
integer-programming
counting
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edited Dec 24, 2019 at 19:13
D.W.♦
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asked Aug 27, 2016 at 0:06
resystresyst
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"Linear programming" is an optimisation problem. The problem that you are trying to solve is to count lattice points inside a finite convex rational polytope.
This problem has a polynomial-time algorithm, the general case for which discovered by Alexander Barvinok in 1994. It appears that all modern algorithms are broadly based on this method. Barvinok & Pommershein's 1999 paper, An Algorithmic Theory of Lattice Points in Polyhedra, is probably the best introduction to the theory. (Actually, it appears that Barvinok has subsequently written a book or monograph; that might be even better.)
There are probably more recent developments than I'm aware of, but this will give you a starting point for chasing citations.
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answered Aug 27, 2016 at 1:12
Pseudonym♦Pseudonym
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Thank you for that observation. I am not familiar with the topics you mentioned, but upon short inspection, I believe they are a good starting point for learning more. Unfortunately, my reputation is too low to upvote you.
– resyst
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Aug 27, 2016 at 1:24
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Huh. That's surprising. Checking whether there exists any integer point within a convex polytope (whether the number of such points is 0 or >0) is equivalent to checking feasibility of an integer linear programming (ILP) instance. ILP is NP-hard. So I would have inferred that it's NP-hard even to check whether a polytope contains one integral point, let alone count the number of them. How do we reconcile these two facts? Where have I gone wrong?
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The catch is probably that Barvinok's algorithm is polynomial for a fixed dimension. If the number of variables is fixed, the algorithm scales polynomially as you add constraints. The typical cases where you want to enumerate lattice points are things like the solutions to linear Diophantine equations, and there the dimension tends to be low.
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Land and Doig (1960) proposed a method for solving discrete programming problems. You may be able to modify his algorithm so that instead of solving an optimization problem you are enumerating every possible feasible integer solution.
Reference
A. H. Land and A. G. Doig (1960). "An automatic method of solving discrete programming problems". Econometrica. 28 (3). pp. 497–520. doi:10.2307/1910129.
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answered Aug 28, 2016 at 1:53
John Frederick ChiongloJohn Frederick Chionglo
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read this paper: Computing convex hulls and counting integer points with polymake.
I think polymake can do it for you.
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answered Apr 1, 2019 at 3:31
Asghar MoeiniAsghar Moeini
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This might be nice information. Is that paper available online? Who are the authors? Have you read the guide on how to provide references?
– 喜欢算法和数学
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188571 | https://www.rapidtables.com/convert/weight/how-many-ounces-in-pound.html | Search Share
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How many ounces in a pound (oz in lb)
How many ounces in a pound (oz in lb). 1 lb in ounces.
1 pound (lb) is equal to 16 ounces (oz):
1 lb = 16 oz
How to convert Pounds to Ounces
The mass m in ounces (oz) is equal to the mass m in pounds (lb) times 16:
m(oz) = m(lb) × 16
Example
Convert 5 lb to ounces:
m(oz) = 5 lb × 16 = 80 oz
Pounds to Ounces conversion table
| Pounds (lb) | Ounces (oz) |
--- |
| 0 lb | 0 oz |
| 0.1 lb | 1.6 oz |
| 1 lb | 16 oz |
| 2 lb | 32 oz |
| 3 lb | 48 oz |
| 4 lb | 64 oz |
| 5 lb | 80 oz |
| 6 lb | 96 oz |
| 7 lb | 112 oz |
| 8 lb | 128 oz |
| 9 lb | 144 oz |
| 10 lb | 160 oz |
| 20 lb | 320 oz |
| 30 lb | 480 oz |
| 40 lb | 640 oz |
| 50 lb | 800 oz |
| 60 lb | 960 oz |
| 70 lb | 1120 oz |
| 80 lb | 1280 oz |
| 90 lb | 1440 oz |
| 100 lb | 1600 oz |
| 1000 lb | 16000 oz |
Pounds to Ounces conversion ►
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188572 | https://en.wikipedia.org/wiki/Elliptic_coordinate_system | Jump to content
Elliptic coordinate system
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From Wikipedia, the free encyclopedia
2D coordinate system whose coordinate lines are confocal ellipses and hyperbolae
Not to be confused with Ecliptic coordinate system.
In geometry, the elliptic coordinate system is a two-dimensional orthogonal coordinate system in which the coordinate lines are confocal ellipses and hyperbolae. The two foci and are generally taken to be fixed at and , respectively, on the -axis of the Cartesian coordinate system.
Basic definition
[edit]
The most common definition of elliptic coordinates is
where is a nonnegative real number and
On the complex plane, an equivalent relationship is
These definitions correspond to ellipses and hyperbolae. The trigonometric identity
shows that curves of constant form ellipses, whereas the hyperbolic trigonometric identity
shows that curves of constant form hyperbolae.
Scale factors
[edit]
In an orthogonal coordinate system the lengths of the basis vectors are known as scale factors. The scale factors for the elliptic coordinates are equal to
Using the double argument identities for hyperbolic functions and trigonometric functions, the scale factors can be equivalently expressed as
Consequently, an infinitesimal element of area equals
and the Laplacian reads
Other differential operators such as and can be expressed in the coordinates by substituting the scale factors into the general formulae found in orthogonal coordinates.
Alternative definition
[edit]
An alternative and geometrically intuitive set of elliptic coordinates are sometimes used,
where and . Hence, the curves of constant are ellipses, whereas the curves of constant are hyperbolae. The coordinate must belong to the interval [-1, 1], whereas the
coordinate must be greater than or equal to one.
The coordinates have a simple relation to the distances to the foci and . For any point in the plane, the sum of its distances to the foci equals , whereas their difference equals .
Thus, the distance to is , whereas the distance to is . (Recall that and are located at and , respectively.)
A drawback of these coordinates is that the points with Cartesian coordinates (x,y) and (x,-y) have the same coordinates , so the conversion to Cartesian coordinates is not a function, but a multifunction.
Alternative scale factors
[edit]
The scale factors for the alternative elliptic coordinates are
Hence, the infinitesimal area element becomes
and the Laplacian equals
Other differential operators such as
and can be expressed in the coordinates by substituting
the scale factors into the general formulae
found in orthogonal coordinates.
Extrapolation to higher dimensions
[edit]
Elliptic coordinates form the basis for several sets of three-dimensional orthogonal coordinates:
The elliptic cylindrical coordinates are produced by projecting in the -direction.
The prolate spheroidal coordinates are produced by rotating the elliptic coordinates about the -axis, i.e., the axis connecting the foci, whereas the oblate spheroidal coordinates are produced by rotating the elliptic coordinates about the -axis, i.e., the axis separating the foci.
Ellipsoidal coordinates are a formal extension of elliptic coordinates into 3-dimensions, which is based on confocal ellipsoids, hyperboloids of one and two sheets.
Note that (ellipsoidal) Geographic coordinate system is a different concept from above.
Applications
[edit]
The classic applications of elliptic coordinates are in solving partial differential equations,
e.g., Laplace's equation or the Helmholtz equation, for which elliptic coordinates are a natural description of a system thus allowing a separation of variables in the partial differential equations. Some traditional examples are solving systems such as electrons orbiting a molecule or planetary orbits that have an elliptical shape.
The geometric properties of elliptic coordinates can also be useful. A typical example might involve
an integration over all pairs of vectors and
that sum to a fixed vector , where the integrand
was a function of the vector lengths and . (In such a case, one would position between the two foci and aligned with the -axis, i.e., .) For concreteness, , and could represent the momenta of a particle and its decomposition products, respectively, and the integrand might involve the kinetic energies of the products (which are proportional to the squared lengths of the momenta).
See also
[edit]
Curvilinear coordinates
Ellipsoidal coordinates
Generalized coordinates
Bipolar coordinates
References
[edit]
"Elliptic coordinates", Encyclopedia of Mathematics, EMS Press, 2001
Korn GA and Korn TM. (1961) Mathematical Handbook for Scientists and Engineers, McGraw-Hill.
Weisstein, Eric W. "Elliptic Cylindrical Coordinates." From MathWorld — A Wolfram Web Resource.
| Orthogonal coordinate systems | |
--- |
| Two dimensional | Cartesian Polar (Log-polar) Parabolic Bipolar Elliptic |
| Three dimensional | Cartesian Cylindrical Spherical Parabolic Paraboloidal Oblate spheroidal Prolate spheroidal Ellipsoidal Elliptic cylindrical Toroidal Bispherical Bipolar cylindrical Conical 6-sphere |
Retrieved from "
Categories:
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Ellipses
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Articles with short description
Short description is different from Wikidata |
188573 | https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Organic_Chemistry_(OpenStax)/21%3A_Carboxylic_Acid_Derivatives-_Nucleophilic_Acyl_Substitution_Reactions/21.03%3A_Reactions_of_Carboxylic_Acids | Skip to main content
21.3: Reactions of Carboxylic Acids
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21.2: Nucleophilic Acyl Substitution Reactions
21.4: Chemistry of Acid Halides
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The direct nucleophilic acyl substitution of a carboxylic acid is difficult because –OH is a poor leaving group (Section 11.3). Thus, it’s usually necessary to enhance the reactivity of the acid, either by using a strong acid catalyst to protonate the carboxyl and make it a better acceptor or by converting the –OH into a better leaving group. Under the right circumstances, however, acid chlorides, anhydrides, esters, and amides can all be prepared from carboxylic acids by nucleophilic acyl substitution reactions.
Conversion of Carboxylic Acids into Acid Chlorides
In the laboratory, carboxylic acids are converted into acid chlorides by treatment with thionyl chloride, SOCl2.
This reaction occurs by a nucleophilic acyl substitution pathway in which the carboxylic acid is first converted into an acyl chlorosulfite intermediate, thereby replacing the –OH of the acid with a much better leaving group. The chlorosulfite then reacts with a nucleophilic chloride ion. You might recall from Section 17.6 that an analogous chlorosulfite is involved in the reaction of an alcohol with SOCl2 to yield an alkyl chloride.
Conversion of Carboxylic Acids into Acid Anhydrides
Acid anhydrides can be derived from two molecules of carboxylic acid by heating to remove 1 equivalent of water. Because of the high temperatures needed, however, only acetic anhydride is commonly prepared this way.
Conversion of Carboxylic Acids into Esters
Perhaps the most useful reaction of carboxylic acids is their conversion into esters. There are many methods for accomplishing this, including the SN2 reaction of a carboxylate anion with a primary alkyl halide that we saw in Section 11.3.
Esters can also be synthesized by an acid-catalyzed nucleophilic acyl substitution reaction of a carboxylic acid with an alcohol, a process called the Fischer esterification reaction. Unfortunately, the need for an excess of a liquid alcohol as solvent effectively limits the method to the synthesis of methyl, ethyl, propyl, and butyl esters.
The mechanism of the Fischer esterification reaction is shown in Figure 21.3.1. Carboxylic acids are not reactive enough to undergo nucleophilic addition directly, but their reactivity is greatly enhanced in the presence of a strong acid such as HCl or H2SO4. The mineral acid protonates the carbonyl-group oxygen atom, thereby giving the carboxylic acid a positive charge and rendering it much more reactive toward nucleophiles. Subsequent loss of water from the tetrahedral intermediate yields the ester product.
The net effect of Fischer esterification is substitution of an –OH group by –OR′. All steps are reversible, and the reaction typically has an equilibrium constant close to 1. Thus, the reaction can be driven in either direction by the choice of reaction conditions. Ester formation is favored when a large excess of alcohol is used as solvent, but carboxylic acid formation is favored when a large excess of water is present.
Evidence in support of the mechanism shown in Figure 21.3.1 comes from isotope-labeling experiments. When 18O-labeled methanol reacts with benzoic acid, the methyl benzoate produced is found to be 18O-labeled whereas the water produced is unlabeled. Thus, it is the C–OH bond of the carboxylic acid that is broken during the reaction rather than the CO–H bond and the RO–H bond of the alcohol that is broken rather than the R–OH bond.
Worked Example 21.3.1: Synthesizing an Ester from an Acid
How might you prepare the following ester using a Fischer esterification reaction?
Strategy
Begin by identifying the two parts of the ester. The acyl part comes from the carboxylic acid and the –OR part comes from the alcohol. In this case, the target molecule is propyl o-bromobenzoate, so it can be prepared by treating o-bromobenzoic acid with 1-propanol.
Solution
Exercise 21.3.1
How might you prepare the following esters from the corresponding acids?
Answer
: 1. Acetic acid + 1-butanol
2. Butanoic acid + methanol
3. Cyclopentanecarboxylic acid + isopropyl alcohol
Exercise 21.3.2
If the following molecule is treated with acid catalyst, an intramolecular esterification reaction occurs. What is the structure of the product? (Intramolecular means within the same molecule.)
Answer
Conversion of Carboxylic Acids into Amides
Amides are difficult to prepare by direct reaction of carboxylic acids with amines because amines are bases that convert acidic carboxyl groups into their unreactive carboxylate anions. Thus, the –OH must be replaced by a better, nonacidic leaving group to carry out a nucleophilic acyl substitution. In practice, amides are often prepared by activating the carboxylic acid with a carbodiimide (R–N═C═N–R), followed by addition of the amine. Dicyclohexylcarbodiimide (DCC) and 1-ethyl-3-(3-dimethylaminopropylcarbodiimide (EDEC) are commonly used.
As shown in Figure 21.3.2, the acid first adds to a C═N double bond of DCC, and nucleophilic acyl substitution by amine then ensues. Alternatively, and depending on the reaction solvent, the reactive acyl intermediate might also react with a second equivalent of carboxylate ion to generate an acid anhydride that then reacts with the amine. The product from either pathway is the same.
We’ll see in Section 26.7 that this carbodiimide method of amide formation is the key step in the laboratory synthesis of small proteins, or peptides. For instance, when one amino acid with its NH2 rendered unreactive and a second amino acid with its –CO2H rendered unreactive are treated with carbodiimide, a dipeptide is formed.
Conversion of Carboxylic Acids into Alcohols
We said in Section 17.4 that carboxylic acids are reduced by LiAlH4 to give primary alcohols, but we deferred a discussion of the reaction mechanism at that time. In fact, the reduction is a nucleophilic acyl substitution reaction in which –H replaces –OH to give an aldehyde that is further reduced by nucleophilic addition to produce a primary alcohol. The aldehyde intermediate is much more reactive than the starting acid, so it reacts instantly and is not isolated.
Because hydride ion is a base as well as a nucleophile, the actual nucleophilic acyl substitution step takes place on the carboxylate ion rather than on the free carboxylic acid and gives a high-energy dianion intermediate. In this intermediate, the two oxygens are complexed to a Lewis acidic aluminum species. Thus, the reaction is relatively difficult, and acid reductions require higher temperatures and extended reaction times.
Alternatively, borane in tetrahydrofuran (BH3/THF) is a useful reagent for reducing carboxylic acids to primary alcohols. Reaction of an acid with BH3/THF occurs rapidly at room temperature, and the procedure is often preferred to reduction with LiAlH4 because of its relative ease and safety. Borane reacts with carboxylic acids faster than with any other functional group, thereby allowing selective transformations such as that on p-nitrophenylacetic acid. If the reduction of p-nitrophenylacetic acid were done with LiAlH4, both the nitro and carboxyl groups would be reduced.
Biological Conversions of Carboxylic Acids
The direct conversion of a carboxylic acid to an acyl derivative by nucleophilic acyl substitution does not occur in biological chemistry. As in the laboratory, the acid must first be activated by converting the –OH into a better leaving group. This activation is often accomplished in living organisms by reaction of the acid with adenosine triphosphate (ATP) to give an acyl adenosyl phosphate, or acyl adenylate, a mixed anhydride combining a carboxylic acid and adenosine monophosphate (AMP, also known as adenylic acid). In the biosynthesis of fats, for example, a long-chain carboxylic acid reacts with ATP to give an acyl adenylate, followed by subsequent nucleophilic acyl substitution of a thiol group in coenzyme A to give the corresponding acyl CoA (Figure 21.3.3).
Note that the first step in Figure 21.3.3—reaction of the carboxylate with ATP to give an acyl adenylate—is itself a nucleophilic acyl substitution on phosphorus. The carboxylate first adds to a P═O double bond, giving a five-coordinate phosphorus intermediate that expels diphosphate ion as a leaving group.
21.2: Nucleophilic Acyl Substitution Reactions
21.4: Chemistry of Acid Halides |
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Why is there no alpha hydrogen in benzaldehyde?
Why is there no alpha hydrogen in benzaldehyde?
Aniket Singh , 6 Months ago
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Benzaldehyde (C6H5CHO) is an organic compound with a benzene ring and an aldehyde functional group (–CHO) attached. The term "alpha hydrogen" refers to a hydrogen atom that is directly attached to a carbon atom adjacent to a functional group in a molecule. In the case of benzaldehyde, there is no alpha hydrogen because the aldehyde group is attached to the carbon atom that is directly adjacent to the benzene ring.
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188575 | https://klccgo.com/broad-band-polarizing-cube/ | Karl Lambrecht Corporation
4204 N. LINCOLN AVE., CHICAGO, IL 60712, US
Phone: +1 (773) 472-5442 / Fax: +1 (773) 472-2724
Email: sales@klccgo.com / Website: www.klccgo.com
Karl Lambrecht Corporation
4204 N. LINCOLN AVE., CHICAGO, IL 60712, US
Phone: +1 (773) 472-5442 / Fax: +1 (773) 472-2724
Email: sales@klccgo.com / Website: www.klccgo.com
BROAD BAND POLARIZING CUBES
For a beam of light incident on a surface at normal incidence, polarization properties of the reflected and transmitted beams are described by Fresnel's equations. According to these, if angle of incidence is equal to Brewster's angle, the degree of polarization of reflected -s beam ( perpendicular to plane of incidence ) is unity. The transmitted -p ( parallel to plane of incidence ) beam is partially polarized. Polarization of the transmitted beam can be further purified by successive films each of suitable thickness and refractive index so that Brewster's condition at each interface is met. In actual practice this does not happen which limits the performance of thin film polarizing cubes.
The standard available cube sizes are 5, 12, 25 and 50 mm. Standard central wavelengths are 550, 830, 1100 and 1400 nm. These are all made of SF2 substrate. Other sizes and wavelengths are available on special order. All outside faces of the cubes are broad band single layer magnesium fluoride anti-reflection coated ( 1.5 % average reflectivity per surface in respective wavelength ranges as shown in following plots ). Please visit our Anti-Reflection Coating page on this web site for various reflectivity plots.
The extinction of transmitted -p component is at least 1 part in 500. As shown in the following plots, some of the transmitted -p polarization is reflected which degrades the extinction of -s polarization. Almost all -s polarization is reflected. Wavefront deformation of the transmitted -p beam is at least quarter wave at 633 nm.
The cube shown above is made of two 90 degree prisms. The prism with the dot as shown above has the polarizing coating on its diagonal and should be on the input side of the beam for optimum performance.
| | | | | | | | |
--- --- --- --- |
| | | | BBPC-550nm | | | | | | | | | BBPC-830nm | | | | | |
| | | | BBPC-1100nm | | | | | | | | | BBPC-1400nm | | | | | |
| |
| BBPC-550nm |
| |
| |
BBPC-550nm
| |
| BBPC-830nm |
| |
| |
BBPC-830nm
| |
| BBPC-1100nm |
| |
| |
BBPC-1100nm
| |
| BBPC-1400nm |
| |
| |
BBPC-1400nm
Specifications:
average extinction of -p : % transmission of -s in above plots - 1/ 500 ( .2% -s transmission )
wavefront distortion of -p : quarter wave at 632.8 nm
beam deviation of -p : 5 arc minutes or less
beam deviation of -s : 90 degrees +/- 15 arc minutes
surface quality : 40-20 scratch dig
coating : 1.5% average reflectivity per surface in above given wavelength ranges
Some of the transmitted -p polarization is reflected and is on top of -s polarization. Although these polarizing cubes reflect almost all the -s polarization, the extinction of -s is poor since some the -p is reflected and is on top of -s. These cubes are excellent as beam combiners.
| |
| BROAD BAND POLARIZING BEAM SPLITTER CUBES |
BROAD BAND POLARIZING BEAM SPLITTER CUBES
| | | | | |
--- ---
| Catalog Number | Clear Aperture Diameter ( mm ) | Substrate | Standard Central Wavelength (nm) WL | Cube Dimensions ( Inches) +/- .25 mm |
| BBPC5-WL | 5 | SF2 | 550 830 1100 1400 | 555 |
| BBPC12-WL | 12 | SF2 | 12.712.712.7 |
| BBPC25-WL | 25 | SF2 | 25.425.425.4 |
| BBPC50-WL | 50 | SF2 | 505050 |
Catalog Number
Standard
Central
Wavelength
(nm)
WL
Cube
Dimensions
( Inches)
+/-
.25 mm
550
830
1100
1400
© 2025, Karl Lambrecht Corporation. All Rights Reserved. |
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Roots of Polynomials
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The roots of a quadratic function are -3 and 2. Is the vertex a maximum or a minimum?
·
Sep 6
A quadratic function with real roots at -3 and 2 can be written (up to a nonzero constant factor) as f(x) = a(x + 3)(x − 2). The vertex lies midway between the roots at the axis of symmetry x = (−3 + 2)/2 = −1/2.
Whether the vertex is a maximum or a minimum depends on the sign of the leading coefficient a:
If a > 0 (parabola opens upward), the vertex at x = −1/2 is a minimum.
If a < 0 (parabola opens downward), the vertex at x = −1/2 is a maximum.
If you need the vertex coordinates explicitly, compute:
x-coordinate: −1/2
y-coordinate: f(−1/2) = a(−1/2 + 3)(−1/2 − 2) = a(5/2)(−5/2) = −(25/4)a
So the v
A quadratic function with real roots at -3 and 2 can be written (up to a nonzero constant factor) as f(x) = a(x + 3)(x − 2). The vertex lies midway between the roots at the axis of symmetry x = (−3 + 2)/2 = −1/2.
Whether the vertex is a maximum or a minimum depends on the sign of the leading coefficient a:
If a > 0 (parabola opens upward), the vertex at x = −1/2 is a minimum.
If a < 0 (parabola opens downward), the vertex at x = −1/2 is a maximum.
If you need the vertex coordinates explicitly, compute:
x-coordinate: −1/2
y-coordinate: f(−1/2) = a(−1/2 + 3)(−1/2 − 2) = a(5/2)(−5/2) = −(25/4)a
So the vertex is (−1/2, −25a/4); its nature (max/min) follows from the sign of a as above.
Related questions
Can you construct a quadratic equation in x with the roots 2/3 or -1/2?
What is the answer? The roots of a quadratic function are-3 and -2. If the leading coefficient is -1. Is the vertex a maximum or a minimum?
What if 1/3 and -3/2 are the roots of a quadratic equation?
What are the maximum and minimum values of the function (x-3) 3(x+1) 2?
What is a quadratic equation with roots 7 and -3?
Eric Hawk
Upvoted by
David Joyce
, Ph.D. Mathematics, University of Pennsylvania (1979) · Author has 3K answers and 3.9M answer views
·
5y
There’s not enough information to decide.
Both y(x) = (x + 3)(x - 2) and y = -(x + 3)(x - 2) will have roots of x = -3 and x = 2. The vertex of one is a minimum and for the other a maximum.
You can also look at the set of simultaneous equations of constraint:
0 = a(-3)^2 + b(-3) + c
0 = a(2)^2 + b(2) + c
Two equations in three unknowns, underdetermined.
The solution set is
b = a, c = -6a
And the general quadratic function is
y(x) = ax^2 + ax - 6a = a(x^2 + x - 6) = a(x + 3)(x - 2)
a can take on any non-zero value to yield a quadratic function.
Taking the second derivative yields y’’(x) = 2a. If a > 0, th
There’s not enough information to decide.
Both y(x) = (x + 3)(x - 2) and y = -(x + 3)(x - 2) will have roots of x = -3 and x = 2. The vertex of one is a minimum and for the other a maximum.
You can also look at the set of simultaneous equations of constraint:
0 = a(-3)^2 + b(-3) + c
0 = a(2)^2 + b(2) + c
Two equations in three unknowns, underdetermined.
The solution set is
b = a, c = -6a
And the general quadratic function is
y(x) = ax^2 + ax - 6a = a(x^2 + x - 6) = a(x + 3)(x - 2)
a can take on any non-zero value to yield a quadratic function.
Taking the second derivative yields y’’(x) = 2a. If a > 0, the vertex is a minimum. If a < 0, the vertex is a maximum.
Hope this helps and does not mislead or confuse you.
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Robert Paxson
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Originally Answered: What is the answer? The roots of a quadratic function are-3 and -2. If the leading coefficient is -1. Is the vertex a maximum or a minimum?
·
It is a maximum if the leading coefficient of the x2 is negative. The parabola represented by the quadratic function opens downward. Taking the second derivative of the quadratic function leaves a negative constant equal to twice that leading coefficient, 2(−1). A negative constant second derivative means that this function is always concave down.
Here is a plot of a parabola that matches these criteria:
It is a maximum if the leading coefficient of the x2 is negative. The parabola represented by the quadratic function opens downward. Taking the second derivative of the quadratic function leaves a negative constant equal to twice that leading coefficient, 2(−1). A negative constant second derivative means that this function is always concave down.
Here is a plot of a parabola that matches these criteria:
Armando Flores
Former Consultant/Senior Engineer, IBM/Lexmark Retired
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5y
Originally Answered: What is the answer? The roots of a quadratic function are-3 and -2. If the leading coefficient is -1. Is the vertex a maximum or a minimum?
·
It is a maximum. For any quadratic equation of the form ax^2+bx+c=0. If a is positive then the parabola is curving up, curving down if negative. So it is curving down and the vertex is a maximum. The curve crosses the X-axis at -3 and -2. So I would guess the maximum is at x=-2.5
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The roots of a quadratic function are -3 and -2. What is the axis of symmetry?
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That is determined by the sign of the leading coefficient. The function is y = a(x + 3)(x – 2), a ≠ 0
The vertex is at x = (–3 + 2)/2 = –1/2
If a < 0 the parabola opens downward and the vertex is a global maximum.
If a > 0 the parabola opens upward and the vertex is a global minimum.
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Kenneth Simmons
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The only number you have tied down is the abscissa of the vertex which is -1 / 2. If the ordinate of the vertex is positive, the vertex is a maximum. If the ordinate is negative, the vertex is minimum.
Geovane Maciel
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The function f(x)=(x−2)(x+3)=x2+x−6 has roots −3 and 2 and the vertex is a minimum.
The function g(x)=−(x2+x−6)=x2−x+6 has roots −3 and 2 and the vertex is a maximum.
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Related
What is the quadratic function f(x) =x2-5x+3 in vertex form?
f(x)=x2−5x+3
f(x)=(x2−5x+[−52]2)−[−52]2+3
f(x)=(x−52)2−254+3
f(x)=(x−52)2+12−254
f(x)=(x−52)2−134
The vertex is at (52,−134).
A plot looks like this:
f(x)=x2−5x+3
f(x)=(x2−5x+[−52]2)−[−52]2+3
f(x)=(x−52)2−254+3
f(x)=(x−52)2+12−254
f(x)=(x−52)2−134
The vertex is at (52,−134).
A plot looks like this:
Thure Dührsen
Studied Mathematics at University of Kiel
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Sep 18
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What is mathematics in the symmetric function of roots in quadratic?
“What is mathematics in the symmetric function of roots in quadratic?“
What is a well-written question? This is not it.
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Related
The vertex of the parabola is (-3,5) and it is the minimum point of the graph. if the graph passes though the point (-2,7), what is the equation of the quadratic function?
Given parabola has vertex (−3,5) and it passes through (−2,7)
Let (h,k)=(−3,5)
There can be two orientations for the parabola.
Case 1
Let the equation of parabola be
(x−h)2=4a(y−k)…(1)
(−2,7) lie on the parabola
(−2−h)2=4a(7−k)
(−2−(−3))2=4a(7−5)
1=8a⟹a=18
From eqn. (1),
(x−(−3))2=4⋅18⋅(y−5)
(x+3)2=y−52
2(x+3)2=y−5
Case 2
Let the equation of parabola be
(y−k)2=4a(x−h)…(2)
(−2,7) lie on the parabola
(7−k)2=4a(−2−h)
(7−5)2=4a(−2−(−3))
4=4a⟹a=1
From eqn. (2),
(y−5)2=4⋅1⋅(x−(−3))
(y−5)2=4(x+3)
Given parabola has vertex (−3,5) and it passes through (−2,7)
Let (h,k)=(−3,5)
There can be two orientations for the parabola.
Case 1
Let the equation of parabola be
(x−h)2=4a(y−k)…(1)
(−2,7) lie on the parabola
(−2−h)2=4a(7−k)
(−2−(−3))2=4a(7−5)
1=8a⟹a=18
From eqn. (1),
(x−(−3))2=4⋅18⋅(y−5)
(x+3)2=y−52
2(x+3)2=y−5
Case 2
Let the equation of parabola be
(y−k)2=4a(x−h)…(2)
(−2,7) lie on the parabola
(7−k)2=4a(−2−h)
(7−5)2=4a(−2−(−3))
4=4a⟹a=1
From eqn. (2),
(y−5)2=4⋅1⋅(x−(−3))
(y−5)2=4(x+3)
Dave Benson
trying to make maths easy.
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Author has 6.1K answers and 2.1M answer views
·
Sep 12
Related
What is mathematics in the symmetric function of roots in quadratic?
What is mathematics in the symmetric function of roots in quadratic?
ax²+bx+c = y sum of roots 𝛂+β = -b/a & product of roots 𝛂β = c/a
If 𝛂 & β can be reversed without changing the function then it is symmetrical.
lets try some: 𝛂²+β² = (𝛂+β)²-2𝛂β 1️⃣ & (𝛂-β)² = (𝛂+β)²-4𝛂β ⟹ 𝛂-β = √{(𝛂+β)²-4𝛂β}
𝛂³+β³ = (𝛂+β)³-3𝛂β(𝛂+β) & 𝛂³-β³ = (𝛂-β)³+3𝛂β(𝛂-β)
1/𝛂+1/β = (β+𝛂)/𝛂β = (𝛂+β)/𝛂β
1/𝛂² +1/β² = β²+𝛂²/𝛂²β² = 𝛂²+β²/(aβ)² = from 1️⃣ {(𝛂+β)²-2𝛂β}/(𝛂β)²
𝛂⁴+β⁴ = (𝛂²+β²)²-2𝛂²β² etc
Related questions
Can you construct a quadratic equation in x with the roots 2/3 or -1/2?
What is the answer? The roots of a quadratic function are-3 and -2. If the leading coefficient is -1. Is the vertex a maximum or a minimum?
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1. What is the vertex of this quadratic function: f(x) = -2(x+3) ²-5? 2. What is the axis of symmetry of this quadratic function: f(x) = -2(x+3) ²-5?
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. 1984 May;63(5):499-502.
Anatomic studies of the human ligamentum flavum
E Zarzur
PMID: 6711844
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Anatomic studies of the human ligamentum flavum
E Zarzur. Anesth Analg.1984 May.
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. 1984 May;63(5):499-502.
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E Zarzur
PMID: 6711844
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Abstract
Lumbar segments of the vertebral column were dissected in 10 human cadavers to identify anatomic features of practical value in performance of spinal or lumbar epidural anesthesia. Of special interest were the findings that the ligamentum flavum at L2-3 is 3-5-mm thick, 12-22-mm wide, and that the internal surfaces of the right and left ligamenta flava form an acute angle with its vertex in contact with the interspinous ligament.
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188578 | http://www.cs.ox.ac.uk/people/jeremy.gibbons/publications/acmmpc-optimization.pdf | Chapter 8 Algebraic Methods for Optimization Problems Richard Bird, Jeremy Gibbons and Shin-Cheng Mu Abstract. We argue for the benefits of relations over functions for mod-elling programs, and even more so for modelling specifications. To sup-port this argument, we present an extended case study for a class of optimization problems, deriving efficient functional programs from con-cise relational specifications.
1 Introduction It is very natural to model computer programs as functions from input to output; hence our interest in functional programming, and the approaches to program calculation described in Chapter 5.
However, sometimes the functional view is too concrete, too implementation-oriented. Although functions may be a good model of programs, they are not a particularly good model of specifications. It is often more natural to specify a problem using features beyond the power of functional programming. Here are some examples.
• It is easy to write the squaring problem sqr as a function. A natural spec-ification of the square root problem sqrt is as its converse: the result of sqrt x should be a y such that sqr y = x (subject to considerations of sign and rounding). A more elaborate application is to specify a parser as the converse of a pretty printer.
• Many problems involve computing a minimal — smallest, cheapest, best — element of a set under some ordering — size, cost, badness. (We will see some examples of such optimization problems later in this chapter.) Often the ordering is a preorder rather than a partial order: two different elements may be equivalent under the ordering. Then the problem minimal r, which computes a minimal element under ordering r, is non-deterministic.
• The problem minimal r cannot return any result on an empty set. Even on a non-empty set, for there always to be a minimal element the ordering must be connected (any two elements must be comparable, one way or the other); for an unconnected ordering there may again be no returnable result. So minimal r will in general also be partial.
• Many problems involve simultaneously satisfying two requirements P and Q; for example, sorting a sequence x under an ordering r involves constructing a permutation (requirement P) of x that is ordered according to r (require-ment Q). One can specify such a problem by sequentially composing two 282 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu subproblems: first find all elements satisfying requirement P, then discard those that do not satisfy requirement Q. However, doing so inevitably breaks the symmetry in the problem description, thereby introducing a bias towards one kind of solution. A more neutral specification retains the symmetry by simply forming the conjunction of the two subproblems.
• Dually, some problems involve satisfying one of two alternative require-ments — perhaps two different ways of giving an answer, or perhaps a ‘nor-mal’ case and an ‘exceptional’ or ‘error’ case. Each subproblem is partial, but the two together may be total. Modelling the problem as a function entails entangling the two aspects; a better separation of concerns is obtained by modelling each aspect independently and forming the disjunction of the two subproblems.
In summary, many problems are most naturally specified using features (con-verse, non-determinism, partiality, conjunction, disjunction) that are to varying degrees inappropriate for a programming language.
The obvious way of resolving this dilemma is to use two languages, one for specification and one for programming. In fact, it is most convenient if the pro-gramming language is a sublanguage of the specification language. Then program construction is a matter of reasoning within a single language, namely the wider specification language, and restricting the features used to obtain an acceptable expression in the implementable subset. Moreover, it is then straightforward to have intermediate expressions that are a mixture of program and specification; this is rather harder to achieve if the two are completely separate languages.
What could we choose as a language that encompasses functions and func-tional programming, but that also provides the expressive power of converse, non-determinism, conjunction, and so on? It is well-known that relations have exactly these properties. This chapter, therefore, looks at lifting the results of Chapter 5 from functions to relations. We will carry out an extended case study, illustrating the gain in expressivity provided by the extra features. Our pro-gramming methodology will be to refine a relational specification to a func-tional implementation, reducing non-determinism while maintaining the domain of definition. So we will take this case study all the way to code, as a functional program.
There is an added bonus from studying relations as well as functions. Even when a specification and its implementation are both functional, one sometimes can gain greater insight and simpler reasoning by carrying out the calculation of the latter from the former in more general theory of relations. (We see an exam-ple in Exercise 2.6.7.) This is analogous to complex numbers and trigonometry in high school mathematics: although trigonometric functions are completely ‘real’ concepts, it is enlightening to investigate them from the point of view of complex numbers. For example, sin θ and cos θ are the imaginary and real parts of eθi, the complex number with ‘angle’ θ radians and unit ‘radius’; so of course sin2 θ + cos2 θ = 1, by Pythagoras’ Theorem, and the ‘most beautiful theorem of mathematics’ eπi = −1 is easy to visualize.
8. Algebraic Methods for Optimization Problems 283 It turns out that the relational algebra is particularly powerful for specify-ing and reasoning about optimization problems — informally, problems of the form ‘the best data structure constructed in this way, satisfying that condition’.
This chapter is structured to lead to exactly this point: Section 2 introduces the algebra of relations, Section 3 develops some general-purpose theory for op-timization problems, and Section 4 applies the theory in a case study, which we take all the way to runnable code. There are many interesting diversions en route, some of which are signposted in the exercises; more sources are provided in the bibliographic notes.
1.1 Bibliographic notes The relational algebra has been proposed as a basis for program construction by many people; some representative strands of work are by de Roever , Hoare , Berghammer , M¨ oller , Backhouse [1, 2], Hehner , Mili and Desharnais , and Bird and de Moor . Indeed, one could argue that relations underlie the entire field of logic programming [22, 29].
For more information about the use of relations for program calculation, particularly in the solution of optimization problems, the interested reader is directed towards , on which this chapter is based.
2 The algebra of relations 2.1 Relations Like a function, a relation is a mapping between a source type and a target type; unlike a function, a relation may be partial (mapping a source value to no target value) and/or non-deterministic (mapping a source value to multiple target values). We write ‘f :: A ✿-✮B’ to denote that relation f is a mapping from source type A to target type B.
2.1.1 Example For example, consider the datatype PList A of non-empty lists with elements drawn from A. In general, lists xs of this type can be split into a prefix ys and a suffix zs such that ys + + zs = xs (here, ‘+ +’ denotes list concatenation): split :: PList A ✿-✮PList A × PList A Most lists can be split in many ways, corresponding to the choice of where to place the division, so split is non-deterministic; for example, both ([1, 2], ) and (, [2, 3]) are possible results of split [1, 2, 3]. Singleton lists can not be split in any way at all, so split is partial too.
284 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu 2.1.2 Pointwise relational programming Non-deterministic languages discourage reasoning with variables, because of the care that needs to be taken to retain referential transparency. Basically, one cannot assign to a variable ‘the result of running program p on input x’, because there may be no or many results. The usual approach, then, is to eschew pointwise reasoning with variables representing values, in favour of pointfree reasoning with the programs themselves.
Nevertheless, there are times when it is much more convenient to name and reason about particular values. This is especially the case when working on applications rather than general-purpose theory, because one wants to name and capitalize on specific dataflow patterns. The right approach to take to this problem is still not clear; the approach we will follow in this chapter is just one of several, and the details are not yet completely worked out. To be on the safe side, we will not use these ‘pointwise relational programming’ notations for any formal purpose; rather, we will resort to them only to provide informal but more perspicuous characterizations of sometimes opaque relational expressions.
When we want to refer to the result of ‘applying’ a relational program, we will avoid using ‘=’, and write instead ‘y ✭ -✿f x’ to mean that y is a possible value of f x. For example, both ([1, 2], ) ✭ -✿split [1, 2, 3] and (, [2, 3]) ✭ -✿split [1, 2, 3] In functional programming, we find that it is often convenient to specify a function by pattern matching. The function is defined by one or more equations, but the arguments to the function on the left-hand side of the equations are patterns rather than simply variables. For example, we might define the factorial function on natural numbers by fact 0 = 1 fact (n + 1) = (n + 1) × fact n This works for functions because patterns are by definition injective (and usually non-overlapping), and not arbitrary expressions: for any natural argument, ex-actly one of these two equations applies, and moreover if it is the second equation, the value of n is uniquely determined. Definition by pattern matching is often just as convenient for relations, but with relations we can be more lenient: we can allow non-exhaustive equations (which will lead to partiality) and overlapping equations with non-injective patterns (which will lead to non-determinism). For example, we could define the function split above by split (ys + + zs) = (ys, zs) This pattern is not injective; nevertheless, this equation — together with the type — completely determines the relation split. (We decorate the equals sign to emphasize that this is the definition of a relation, rather than a true identity of values; in particular, = is transitive whereas = is not.) 8. Algebraic Methods for Optimization Problems 285 Non-injective patterns introduce non-determinism implicitly. Explicit non-determinism can be expressed using the choice operator ✷. For example, here is a more constructive characterization of split: split (consp (x, xs)) = ([x], xs) ✷(consp (x, ys), zs) where (ys, zs) ✭ -✿split xs Here, ‘consp’ denotes the prefixing of a single element onto a non-empty list.
The pattern is injective — it matches in at most one way for any given list — but still the result is non-deterministic because of the explicit choice.
2.1.3 Composition Relations of appropriate types may be composed; if f :: A ✿-✮B and g :: B ✿-✮C, then their composition g · f :: A ✿-✮C is defined by z ✭ -✿(g · f ) x ⇔∃y. y ✭ -✿f x ∧z ✭ -✿g y For every type A, there is an identity relation idA ::A✿-✮A mapping each element of A to itself. Identities are the units of composition: for f :: A ✿-✮B, f · idA = f = idB · f We will usually omit the subscript, allowing it to be deduced from the context.
2.1.4 Inclusion One can think of a relation f :: A ✿-✮B as a set of pairs, a subset of A × B (so y ✭ -✿f x precisely when (x, y) ∈f ). Unlike functions, different relations of the same type may be comparable under inclusion: f ⊆g precisely when y ✭ -✿f x implies y ✭ -✿g x for all x and y.
Composition is monotonic under inclusion: f1 ⊆f2 ∧g1 ⊆g2 ⇒f1 · g1 ⊆f2 · g2 Moreover, pre- and post-composition have adjoints, called right division (‘over’) and left division (‘under’) respectively: (f ⊆h / g) ⇔(f · g ⊆h) ⇔(g ⊆f \ h) With points, a ✭ -✿(h / g) b ⇔∀c. (a ✭ -✿h c ⇐b ✭ -✿g c) a ✭ -✿(f \ h) c ⇔∀b. (b ✭ -✿f a ⇒b ✭ -✿h c) Inclusion of relations is the foundation of our programming method, as equality is for functional programming: we will start with a (presumably non-deterministic) specification g, and manipulate it to construct a (presumably more efficient, or otherwise easier to implement) refinement f ⊆g. Provided that f is still defined everywhere that g was, it is ‘at least as good’ as g.
286 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu 2.1.5 Meet and join Any two relations f , g ::A✿-✮B have a meet f ∩g ::A✿-✮B, defined by the universal property that for all h :: A ✿-✮B, h ⊆f ∩g ⇔h ⊆f ∧h ⊆g That is, f ∩g is the greatest lower bound of f and g; it corresponds to intersection of the sets of pairs in the relations. It follows (Exercise 2.6.1) that ∩is associative, commutative and idempotent, and that monotonicity of composition can be re-expressed in terms of ∩.
Dually, the join f ∪g of two relations f , g of the same type is defined by the universal property h ⊇f ∪g ⇔h ⊇f ∧h ⊇g 2.1.6 Converse Again unlike functions, relations can easily be reversed: if f :: A ✿-✮B, then f ◦:: B ✿-✮A, and satisfies y ✭ -✿f x ⇔x ✭ -✿f ◦y Converse is: • its own inverse: (f ◦)◦= f ; • order-preserving: f ⊆g ⇔f ◦⊆g◦; • contravariant: (f · g)◦= g◦· f ◦; • distributive over meet: (f ∩g)◦= f ◦∩g◦.
Any coreflexive f , that is, one satisfying f ⊆id, is invariant under converse: f ◦= f . One can think of a coreflexive of type A ✿-✮A as a subset of A, or equivalently as a predicate on A.
2.2 Special kinds of relation Let f :: A ✿-✮B; then • f is entire if id ⊆f ◦· f , or equivalently if for all x ∈A there is at least one y ∈B with y ✭ -✿f x; • f is simple if f · f ◦⊆id, or equivalently if for all x ∈A there is at most one y ∈B with y ✭ -✿f x; • f is surjective if id ⊆f · f ◦, or equivalently if for all y ∈B there is at least one x ∈A with y ✭ -✿f x; • f is injective if f ◦· f ⊆id, or equivalently if for all y ∈B there is at most one x ∈A with y ✭ -✿f x.
A simple relation is also known as a partial function, and a simple and entire relation as a total function. We write ‘f :: A →B’ to indicate that f is a total function. Functions enjoy special shunting rules: 8. Algebraic Methods for Optimization Problems 287 Lemma 1. If f is a function, then f · g ⊆h ⇔g ⊆f ◦· h g · f ◦⊆h ⇔g ⊆h · f Note that f ⊆g ⇒f = g if f , g are total functions. For function p :: A →Bool, we define the corresponding coreflexive p? :: A ✿-✮A by p? = exr · fsttrue · (p △id) where the coreflexive fsttrue holds of pairs whose first component is true.
2.3 Breadth There is a one-to-one correspondence between relations A ✿-✮B and set-valued functions A →Set B, where Set is the powerset functor (Set B is the set of all subsets of B). The operator Λ is a witness to this isomorphism, yielding the breadth of a relation, that is, the corresponding set-valued function: (Λf ) x = {y | y ✭ -✿f x} For example, Λsplit is the function that returns the set of all possible splits of a given list.
Note that Λf is an entire and simple relation (a function) for any f . If f itself is a function, then Λf returns singleton sets.
2.4 Folds Functional folds were discussed in depth in Chapter 5. To summarize, an F-algebra (A, f ) for a functor F consists of a type A and a function f :: F A →A.
An F-algebra (T, inT) is initial if, for every F-algebra (A, f ), there is a unique homomorphism h :: T →A such that h · inT = f · F h We write foldT f for this h, giving the universal property h = foldT f ⇔h · inT = f · F h Thus, foldT has type (F A →A) →(T →A).
The datatype definition T = data F introduces the initial F-algebra (T, inT) and the fold operator foldT. This was generalized to polymorphic datatypes, using bifunctors rather than functors: the datatype definition T A = data (A⊕) introduces the initial (A⊕)-algebra (TA, inT A) and the fold operator foldT A, from which we usually omit the type subscript A, if not the whole of the subscript.
288 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu 2.4.1 Example: lists We define the datatype of lists by List A = data (1 ˆ + (A ˆ × Id)) where × is product, + is coproduct, 1 is the unit type, Id is the identity functor, underlining denotes constant functors and a superscript hat denotes lifting of a bifunctor. This induces a constructor inList :: 1 + A × List A →A and a fold operator foldList :: (1 + B × List A →B) →(List A →B) We introduce the syntactic sugar const [ ] ▽cons = inList foldr f e = foldList (const e ▽f ) where const a b = a This corresponds to the curried Haskell equivalents [] :: [a] (:) :: a -> [a] -> [a] for the constructors, and foldr :: (a -> b -> b) -> b -> ([a] -> b) foldr f e [] = e foldr f e (x:xs) = f x (foldr f e xs) for the fold.
2.4.2 Example: non-empty lists In our case study at the end of this chapter, we will also require a datatype of non-empty lists PList A = data (A ˆ + (A ˆ × Id)) We introduce the syntactic sugar wrap ▽consp = inPList foldrp f g = foldPList (g ▽f ) The Haskell standard prelude has no equivalent, but encourages the reuse of ordinary lists instead, trading notational convenience for lost structure.
8. Algebraic Methods for Optimization Problems 289 2.4.3 Relational folds The theory from Chapter 5 summarized above is fine in the categories Set and Cpo, but turns out not to be appropriate in the category R el of sets and relations, because too much of the structure collapses there. In particular (Exercise 2.6.4), converses make products and coproducts coincide.
However, any functor in Set can be extended in a canonical way to a mono-tonic functor (or relator) on R el. In particular, the monotonic extension of carte-sian product can be used in place of the true categorical product of R el (which coincides with disjoint sum).
As a result, folds can be generalized to relations. For monotonically extended functor F, the initial (in Set) F-algebra (T, in) is also initial in R el: we still have the all-important universal property h = foldT f ⇔h · inT = f · F h only now the fold acts on and yields relations: foldT :: (F A ✿-✮A) →(T ✿-✮A).
Some examples of relational folds on lists that will be used later (Exer-cises 3.8.5 and 3.8.9) are as follows. Recall Lambek’s Lemma (§2.5.5 of Chap-ter 5), which states that folding with the constructors is the identity function; on lists this becomes foldr cons [ ] = id. Folding with any subrelation of the con-structors gives a coreflexive; for example, the coreflexive ordered that holds of ascending lists is given by ordered = foldr (cons · ok) [ ] where ok (x, xs) = map ((x ≤)?) xs Using the power of relations in the other direction, namely non-determinism, we can obtain simple characterizations of many combinatorial operations. For example, the relation prefix returns an arbitrary prefix of a list: prefix = fold (cons ∪const [ ]) [ ] and the relation subseq returns an arbitrary subsequence: subseq = foldr (cons ∪exr) [ ] 2.4.4 Unfolds One might expect now to dualize the relational generalization of folds to get relational unfolds. However, this is not necessary, because converse gives us ev-erything we need: for functor F, the datatype T = data F is also a codatatype, and for algebra f ::FA✿-✮A, the converse f ◦::A✿-✮FA is a coalgebra, and (fold(f ◦))◦ works as an unfold (Exercise 2.6.6).
2.4.5 Fusion Perhaps the fundamental property of folds, as illustrated at length in Chapter 5, is the fusion theorem: for T = data F, h · foldT f = foldT g ⇐h · f = g · F h 290 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu There are two variants of this fusion theorem for relational folds: h · foldT f ⊆foldT g ⇐h · f ⊆g · F h h · foldT f ⊇foldT g ⇐h · f ⊇g · F h Of course, fusion as an equality follows from these two.
2.5 Bibliographic notes The relational algebra has a very elegant axiomatic characterization, dating back to Tarski , and this is how it is often presented and used [1, 2]. Our presen-tation follows the axiomatization as a (unitary tabular) allegory in , and owes more to Freyd and ˇ Sˇ cedrov than to Tarski. (In an allegorical setting, the join ∪is actually not the dual of the meet ∩in general; in fact, only locally complete allegories have joins, and only in boolean locally complete allegories are they dual to meets. We gloss over these details, because we are not directly concerned with the axiomatization here.) Using either axiomatization leads to pointfree reasoning, which is certainly concise, but is often difficult to follow, especially when applied to specific pro-gramming problems as opposed to general theory. This is because ‘plumbing combinators’ must be used to pass values around. In computer programming, we have long realized that variables are important, and that their names clarify programs; the same observation applies to the relational algebra. However, vari-ables interact awkwardly with non-determinism, and finding the right approach is still a matter of ongoing research. The approach taken here is based on ; other approaches include [7, 13, 31, 25, 26, 28].
2.6 Exercises 1. Using the universal property of ∩(§2.1.5), show that it is associative, com-mutative and idempotent: f ∩(g ∩h) = (f ∩g) ∩h f ∩g = g ∩f f ∩f = f and that monotonicity of composition can be reexpressed as follows: f · (g ∩h) ⊆(f · g) ∩(f · h) (f ∩g) · h ⊆(f · h) ∩(g · h) (that is, derive these properties from the original characterization, and vice versa).
2. Prove the shunting rules in Lemma 1.
3. The Λ operation is an isomorphism; there is a one-to-one correspondence between relations A ✿-✮B and set-valued functions A →Set B. What is the inverse operation? That is, given a set-valued function f of type A →Set B, what is the corresponding relation of type A✿-✮B? (Hint: look ahead to §3.1.) 8. Algebraic Methods for Optimization Problems 291 4. A straightforward interpretation in R el of the standard theory of datatypes from Chapter 5 is inappropriate: products and coproducts coincide, whereas we intend them to be different constructions. Technically, this is because R el is its own dual, whereas Set and Cpo are not self-dual.
Categorically speaking, bifunctor ⊗is a product if arrows f :: A ✿-✮B and g :: A ✿-✮C uniquely determine a morphism h :: A ✿-✮B ⊗C with exl · h = f and exr · h = g. Dually, bifunctor ⊕is a coproduct if arrows f :: A ✿-✮C and g :: B ✿-✮C uniquely determine a morphism h :: A ⊕B ✿-✮C with h · inl = f and h · inr = g.
(a) Show that cartesian product is not a categorical product in R el.
(b) Show that disjoint sum is a categorical coproduct in R el.
(c) Show that disjoint sum is in fact also a categorical product in R el.
(d) Show, using converses, that any categorical product in R el is necessarily a coproduct, and vice versa.
5. A tabulation of relation h :: A ✿-✮B is a pair of functions f :: C ✿-✮A, g :: C ✿-✮B such that h = g · f ◦and (f ◦· f ) ∩(g◦· g) = id. Assuming that every relation has a tabulation, show that for relator F: (a) when h is a function, so is F h, and (F h)◦= F (h◦); (b) a functor is a relator iff(Fh)◦= F(h◦) for any (not necessarily functional) h; (c) if two relators F, G agree on functions, then they agree on all relations.
Thus, functors in Set can be extended in a canonical way to relators in R el.
6. Show that datatype T = data F is also a codatatype, with final coalgebra (T, in◦). That is, for algebra f ::FA✿-✮A, show that (foldT (f ◦))◦is an unfold: h = (foldT (f ◦))◦⇔in◦· h = F h · f This is another consequence of R el’s self-duality.
7. Using relations, we can give very nice equational proofs of two theorems presented in Exercises 2.9.9 and 2.9.12 of Chapter 5. Despite these being properties of total functions, it seems that their simplest proofs are in terms of relations. For datatype T = data F, show that function h ::T →A is a fold iffh · in · F h◦is simple, and function g :: A →T is an unfold iffF g◦· in◦· g is entire. Explain why these conditions are equivalent to those from Chapter 5.
8. Prove the two inclusion fusion theorems h · foldT f ⊆foldT g ⇐h · f ⊆g · F h h · foldT f ⊇foldT g ⇐h · f ⊇g · F h 3 Optimization problems Optimization problems are typically of the form ‘select the best construct gener-ated by this means, which satisfies that test’. Such a specification is executable, provided that only finitely many constructs are generated, but it is in general too expensive to compute because there still many constructs to consider. The algorithm can be improved by promoting the test and the selection inside the 292 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu generation, thereby avoiding having to generate and eliminate constructs that cannot possibly contribute to the optimal solution.
Sometimes the improved algorithm has exactly the same extensional be-haviour as the original specification, but is faster to execute. More often, however, the specification is non-deterministic (there may be several optimal solutions), and the improved algorithm is a refinement of the specification, yielding only a subset of the optimal solutions. The missing solutions are pruned to permit a more efficient algorithm; correctness is maintained by ensuring that for every solution pruned, an equally good one is retained. This refinement characteristic is a strong motivation for the move from functional to relational programming: refinement makes little sense with total functions.
In this chapter we will consider a restricted, but still large, class of optimiza-tion problems, of the form min r · Λ(fold f ) The ‘test’ phase is omitted, and the feasible constructs are generated directly by a (relational) fold. We take the breadth of this fold to get the set of all feasible constructs, and select from this set a minimal element under the preorder r.
We will develop two kinds of improved algorithm for such problems, depend-ing on characteristics of the ingredients f and r. The first embodies a greedy approach, building a single optimal solution step by step. The second embodies a thinning approach, building up a (hopefully small) collection of representative solutions, some unforeseeable one of which will lead to the optimal solution. The second approach assumes weaker conditions on the ingredients, but yields a less efficient algorithm.
3.1 The Eilenberg-Wright Lemma Let choose :: Set A ✿-✮A denote the membership relation for sets, so that x ✭ -✿choose xs ⇔x ∈xs Then we have the following lemma: Lemma 2 (Eilenberg-Wright). For T = data F, Λ(foldT f ) = foldT (Λ(f · F choose)) Informally, the set of results returned by a relational fold can be obtained as a functional fold that at each stage returns the set of all possible partial results.
In particular, we have that min r · Λ(foldT f ) = min r · foldT (Λ(f · F choose)) That is, rather than computing a single complete construct in all possible ways and taking the best of the results, we can explore all choices at each step, building up all constructs ‘in parallel’.
8. Algebraic Methods for Optimization Problems 293 3.2 Preorders A preorder r :: A ✿-✮A is a relation that is • reflexive, that is, id ⊆r, and • transitive, that is, r · r ⊆r.
In addition, preorder r :: A ✿-✮A may or may not be • connected, that is, r ∪r ◦= A × A.
When we come to translate our results into Haskell, it will be convenient to represent a preorder r by its characteristic function χ(r) of type A × A →Bool (or perhaps its curried version A →A →Bool) such that x ✭ -✿r y ⇔χ(r) (x, y).
With this representation, the properties become more long-winded but perhaps more familiar: • a preorder r is reflexive, that is, χ(r) (x, x) for every x ∈A; • a preorder r is transitive, that is, χ(r) (x, y) ∧χ(r) (y, z) ⇒χ(r) (x, z) for every x, y, z ∈A; • a preorder r may or may not be connected, that is, χ(r) (x, y) ∨χ(r) (y, x) for every x, y ∈A.
3.3 Monotonicity We say that that an F-algebra f :: F A ✿-✮A is monotonic under a preorder r if f · F r ⊆r · f Equivalently, in terms of characteristic functions, we have that f is monotonic under r if χ(Fr)(x, y) and u ✭ -✿f x together imply that there exists a v ✭ -✿f y such that χ(r)(u, v). (Note that Fr is a relation of type FA✿-✮FA, so its characteristic function χ(F r) has type F A × F A →Bool.) For example, addition of naturals plus :: Pair Nat →Nat is monotonic under leq, the normal linear ordering, because plus · Pair leq ⊆leq · plus, or in terms of points, a ✭ -✿leq c and b ✭ -✿leq d imply that plus (a, b) ✭ -✿leq (plus (c, d)).
3.4 Minimum The function min :: (A ✿-✮A) →(Set A ✿-✮A) is defined by min r = choose ∩(r / choose◦) Informally, min r takes a set xs and returns a value x that both is an element of xs and satisfies x ✭ -✿r y for every y ✭ -✿choose xs; in points, x ✭ -✿min r xs ⇔(x ∈xs) ∧(∀y ∈xs. χ(r) (x, y)) Of course, this definition is perfectly symmetric: to obtain the maximum under a preorder r, simply compute the minimum under r ◦.
294 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu If every non-empty set has a minimum under a preorder r, then r is neces-sarily connected. In general, therefore, we will only be interested in uses of minr for connected preorders r.
When it comes to implementing this operator in Haskell, we first must decide how to represent sets. We choose the simplest representation, as an unordered list, but because we will be computing minimum elements of sets, we will stick to non-empty lists; so we define the function minlist as follows: minlist :: (A × A →Bool) →(PList A →A) minlist r = foldrp m id where m (x, y) = if r (x, y) then x else y This implementation breaks ties in favour of the leftmost of two equivalent ele-ments. If r is connected, then minlist (χ(r)) ⊆min r · setify where setify::PListA →SetA converts a non-empty list to the set of its elements.
3.5 The Greedy Theorem Greedy algorithms for our class of optimization problems are captured by the following theorem.
Theorem 3 (Greedy Theorem). Suppose T = data F, and F-algebra f :: F A ✿-✮A is monotonic under the preorder r ◦. Then foldT (min r · Λf ) ⊆min r · Λ(foldT f ) In fact, we have a stronger theorem, of which the above is a simple corollary.
Theorem 4 (Refining Greedy Theorem). Suppose T = data F, and F-algebra f :: F A ✿-✮A is monotonic under the preorder q◦, where q ⊆r. Then foldT (min q · Λf ) ⊆min r · Λ(foldT f ) Proof (Sketch). Using the Eilenberg-Wright Lemma, it suffices to show foldT (min q · Λf ) ⊆min r · foldT (Λ(f · F choose)) This in turn follows, by fusion, from min q · Λf · F (min r) ⊆min r · Λ(f · F choose) Discharging this final proof obligation is left as an exercise for the reader (Ex-ercise 3.8.4).
Informally, to say that f is monotonic under r ◦is to say that for any ‘input’ x :: F A, any ‘worse input’ y ✭ -✿F r ◦x, and any result b ✭ -✿f y of f on this worse input, there corresponds a result a ✭ -✿f x of f on the original input that is ‘better’ than b, that is, b ✭ -✿r ◦a. Thus, degrading the inputs to f will always degrade the output, or conversely, it is never beneficial to pick suboptimal intermediate results, as they will only lead to suboptimal final results. Overall, the Greedy Theorem states that a minimal result can be computed by maintaining a single 8. Algebraic Methods for Optimization Problems 295 minimal partial result at each stage of the folding process, and so it lies at the other extreme to the Eilenberg-Wright Lemma, which embodies a kind of exhaustive search.
The Refining Greedy Theorem is a little more flexible. It is sometimes the case that f fails to be monotonic under r ◦, but enjoys monotonicity under a refined ordering q◦where q ⊆r. We will see an example in §4.5. Provided that q remains connected, the relation min q is entire (except for empty sets). The greedy algorithm itself will also be entire (except for empty sets), refining the plain greedy algorithm from Theorem 3: it computes an optimal solution under q, and any optimal solution under q will also be optimal under r — but the converse does not hold. If q is not connected, the theorem still holds, but the resulting algorithm will not be entire (that is, will not always yield a result).
In order to implement the method as a functional program, we have to find some function step::FA →A such that step ⊆minq ·Λf . We will give an example application of the Greedy Theorem in §4.
3.6 Thinning Monotonicity under a connected preorder is a strong condition that is satisfied in very few optimization problems; and indeed, few such problems admit greedy solutions. More useful would be something between the two extremes of the Eilenberg-Wright Lemma and the Greedy Theorems, involving maintaining some but not all partial solutions.
For a not necessarily connected preorder q on A, the relation thin q :: Set A ✿-✮ Set A takes a set xs and returns some subset ys of xs with the property that all elements of xs have a lower bound under q in ys. More precisely, thin q = (choose \ choose) ∩((choose◦· q) / choose◦) If ys ✭ -✿thin q xs, then the first conjunct here says that ys is a subset of xs, and the second says that for every x ✭ -✿choose xs, there is a y ✭ -✿choose ys with y ✭ -✿q x; in points, ys ✭ -✿thin q xs ⇔(ys ⊆xs) ∧(∀x ∈xs. ∃y ∈ys. χ(q) (x, y)) The following lemma allows us to introduce applications of thin.
Lemma 5 (Thin introduction). Provided that q and r are both preorders and q ⊆r, we have min r = min r · thin q That is, any required minimum can be obtained by selecting that minimum from a suitably thinned set. For the proof, see Exercise 3.8.6.
This leads us towards the following Thinning Theorem, which entails main-taining a representative collection of partial solutions at each stage of the folding process.
Theorem 6 (Thinning). Suppose T = data F, and F-algebra f :: F A ✿-✮A is monotonic under a (not necessarily connected) preorder q◦, where q ⊆r. Then 296 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu min r · foldT (thin q · Λ(f · F choose)) ⊆min r · Λ(foldT f ) Proof (Sketch). We have min r = min r · thin q, so it suffices to show that foldT (thin q · Λ(f · F choose)) ⊆thin q · Λ(foldT f ) Just like the Greedy Theorem, this latter inclusion can be proved (Exercise 3.8.7) by making use of the Eilenberg-Wright Lemma and fusion.
3.6.1 Implementation In order to implement the thinning method as a functional program, we have to implement thin q · Λ(f · F choose) :: F (Set A) ✿-✮Set A It seems reasonable to represent these sets as lists, sorted by some connected preorder (but not necessarily by q itself); we can capitalize on the list ordering to implement the thinning step efficiently. Therefore, we will actually construct a function step :: F (PList A) →PList A such that setify · step ⊆thin q · Λ(f · F choose) · F setify To do this, we will need a number of ingredients: • a function sort :: (A × A →Bool) →(Set A →PList A) that sorts a finite set under a given connected preorder; • a function mergelists :: (A × A →Bool) →(PList (PList A) →PList A) that merges a list of sorted lists into a single sorted list; • a function thinlist :: (A × A →Bool) →(PList A →PList A) that implements thin: setify · thinlist (χ(q)) ⊆thin q · setify on a suitably sorted list in a reasonably efficient manner, that is, quickly (in linear time) and effectively (yielding a short result), and does so stablely: thinlist (χ(q)) ⊆subseq • a function cpF :: F (PList A) →PList (F A) that converts an F-structure of lists (possibly of differing lengths) into a list of F-structures (all of the same shape); 8. Algebraic Methods for Optimization Problems 297 • a function combine :: F A →PList A satisfying combine ⊆sort (χ(p)) · thin q · Λf for some connected preorder p.
Given these ingredients, we define step = thinlist (χ(q)) · mergelists (χ(p)) · map combine · cpF and claim (Exercise 3.8.8) that minlist (χ(r)) · fold step ⊆min r · fold (thin q · Λ(f · F choose)) There is no constraint on p for correctness, but for efficiency a suitable choice will bring related elements together to allow effective thinning with thinlist (χ(q)).
3.7 Bibliographic notes The Eilenberg-Wright Lemma, and indeed relational folds themselves, were first employed to reason about the equivalence of deterministic and non-deterministic automata .
Greedy algorithms are well-known in the algorithm design community; an in-depth study of their properties is in , and a more pragmatic description with numerous examples in . Thinning algorithms as an abstraction in their own right are due to Bird and de Moor [4, 5]. More results on greedy, thinning and dynamic programming algorithms for optimization problems are given in Curtis’ thesis ; there the emphasis is on algorithms in which the generation phase is less structured, a simple loop rather than a fold or an unfold.
Cross products (§3.6.1) and the related zips are investigated in great depth in Hoogendijk’s thesis .
3.8 Exercises 1. Prove the Eilenberg-Wright Lemma (§3.1), using the universal property of folds.
2. Find an F-algebra f :: F A ✿-✮A and a preorder r on A such that f is mono-tonic under r but not under r ◦. Prove that when f is a total function, f is monotonic under r iffit is monotonic under r ◦.
3. Why is connectedness of r necessary for the function minlist from §3.4 to be an implementation of min?
4. Complete the proof of the Refining Greedy Theorem (Theorem 4) from §3.5, by showing that min q · Λf · F (min r) ⊆min r · Λ(f · F choose) where f is monotonic under q◦with q ⊆r.
5. The function takewhile ::(A✿-✮A) →(ListA✿-✮ListA) yields the longest prefix of its second argument, all of whose elements ‘satisfy’ the coreflexive first argument: takewhile p = longest · Λ(map p · prefix) 298 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu where prefix is defined in §2.4.3, and longest = min (length◦· geq · length) and where length returns the length of a list, and geq is the usual linear ordering on naturals. For example, takewhile prime [2, 3, 4, 5] = [2, 3]. Use fusion to write map p · prefix as a fold, and hence use the Greedy Theorem to derive the standard implementation of takewhile p as a fold.
6. Prove that min r ⊇min r · thin q provided that q, r are preorders, and q ⊆r. Because thin q ⊇id too, this proves Lemma 5.
7. Complete the proof of the Thinning Theorem from §3.6, by showing that foldT (thin q · Λ(f · F choose)) ⊆thin q · Λ(foldT f ) where T = data F and f is monotonic under q◦.
8. Justify the claim (§3.6.1) that the thinning algorithm minlist (χ(r)) · fold (thinlist (χ(q)) · mergelists (χ(p)) · map combine · cpF) is indeed a refinement of the optimization problem.
9. The longest upsequence problem is to compute the longest ascending subsequence of a list; for example, the longest upsequence of [1, 6, 5, 2, 4, 3, 3] is [1, 2, 3, 3]. Formally, the problem is to compute longest · Λ(ordered · subseq) where longest is as in Exercise 3.8.5, and ordered and subseq as in §2.4.3.
Derive a thinning algorithm to solve this problem.
10. In a sense, the Thinning Theorem always applies: one can always choose q = id. Prove that this choice satisfies the conditions of Theorem 6, and that the algorithm obtained by applying the theorem is the same as the problem specification min r · Λ(fold f ).
11. At the other extreme, the Greedy Theorem is an instance of the Thinning Theorem: choosing q = r in the Thinning Theorem gives ‘essentially’ the greedy algorithm. More precisely, the thinning algorithm so obtained will still return a set of elements, but all will be optimal; taking the breadth gives the corresponding greedy algorithm. Justify this claim. (Thus, ‘thinning’ encompasses the whole spectrum from maintaining all partial solutions to maintaining just one.) 4 Optimal bracketing To illustrate the foregoing theory, consider the (A, ≤, ⊕) bracketing problem, defined as follows. Given a linearly ordered set (A, ≤), an operator ⊕::A×A →A, and a non-empty list a1, a2, . . . an of values of type A, it is required to bracket the expression a1 ⊕a2 ⊕· · · ⊕an in such a way that the result is as small as possible under the ordering ≤. Of course, if ⊕is associative then all bracketings give equally small results.
8. Algebraic Methods for Optimization Problems 299 Example 7. With a ⊕b = max (a, b) + 1, the (Nat, ≤, ⊕) bracketing problem corresponds to the task of combining a list of trees (with given heights a1, . . . an) into a single tree of minimum height. For example, given subtrees with heights [4, 2, 3, 5, 2, 4, 6], one optimal tree is 4 2 3 5 2 4 6 Example 8. Similarly, one interpretation of the (Nat × Nat, ≤, ⊕) bracketing problem, where ≤is the lexical ordering on pairs and (c1, l1) ⊕(c2, l2) = (c1 + c2 + l1 × l2, l1 + l2) corresponds to the task of multiplying a list of decimals in the shortest possible time (without exploiting the commutativity of multiplication), where multiplying decimals of lengths l1 and l2 costs l1 × l2, and yields a decimal of length l1 + l2.
Example 9. The (Nat × (Nat × Nat), ≤, ⊕) bracketing problem, where ≤is the lexical ordering and (c1, (p, q)) ⊕(c2, (q, r)) = (c1 + c2 + p × q × r, (p, r)) corresponds to the problem of multiplying a sequence of conformant matrices with minimal cost, where multiplying a p × q matrix by a q × r matrix costs p × q × r, and yields a p × r matrix.
4.1 Representation We will represent a single bracketing as a binary tree of type Tree A = data (A ˆ + (Id ˆ × Id)) for which we introduce the syntactic sugar tip a = in (inl a) fork (t, u) = in (inr (t, u)) for the constructors, and foldT :: (B × B ✿-✮B) →(A ✿-✮B) →(Tree A ✿-✮B) foldT f g = foldTree (g ▽f ) for the fold.
In particular, the function flatten :: Tree A →PList A is defined by flatten = foldT (+ +) wrap The (A, ≤, ⊕) bracketing problem can now be formalized as one of computing 300 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu min r · Λ(flatten◦) where cost = foldT (⊕) id, and r = cost◦· leq · cost, or in points, χ(r) (x, y) = cost x ≤cost y (so r is connected). Note that flatten◦is partial, giving a result only on non-empty lists.
4.2 The Converse-of-a-Function Theorem One could start by formulating methods for computing min r · Λ((fold f )◦) We will not take this approach (but see ); instead, we will make use of the following theorem to express flatten◦as a fold: Theorem 10 (Converse of a function). Suppose f :: A →T (in particular, f is a total function), where T = data F. Furthermore, suppose g :: F A ✿-✮A is surjective and satisfies f · g ⊆in · F f . Then f ◦= foldT g.
Use of this theorem is a definite design step in solving an optimization problem: it prescribes the structure of the resulting program, and thereby rules out some potential solutions. It is therefore not always the right step to take; however, it turns out to be a productive one for some instances of the bracketing problem.
4.3 Spines To express flatten◦as a fold, we will represent trees by their left spines. The left spine of a tree is a pair consisting of the leftmost tip value and the sequence of right subtrees along the path from the leftmost tip to the root. Thus we define Spine A = A × List (Tree A). For example, the optimal tree in Example 7 has the left spine 4, 2 3 , 5 2 4 6 The bijection roll takes a spine (a, [t1, t2, . . . , tn]) and returns the tree fork (· · · fork (fork (tip a, t1), t2) · · · , tn) Formally, we have roll (a, ts) = foldl fork (tip a) ts where foldl is the natural fold function for snoc lists, that is, lists built by adding elements to the end rather than the beginning. On ordinary lists we can simulate foldl by 8. Algebraic Methods for Optimization Problems 301 foldl f e [ ] = e foldl f e (cons (x, xs)) = foldl f (f (e, x)) xs Using the Converse-of-a-Function Theorem, we can now obtain (Exercise 4.10.1) (flatten · roll)◦= foldrp add one where one a = (a, [ ]) and add :: A × Spine A ✿-✮Spine A is defined (using a non-injective pattern) by add (a, (b, xs + + ys)) = (a, cons (roll (b, xs), ys)) For example, foldrp add one [1, 2, 3] = add (1, add (2, one 3)) = add (1, add (2, (3, [ ]))) = add (1, (2, [tip 3])) = (1, [tip 2, tip 3]) ✷(1, [fork (tip 2, tip 3)]) That is, both (1, [tip 2, tip 3]) ✭ -✿foldrp add one [1, 2, 3] and (1, [fork (tip 2, tip 3)]) ✭ -✿foldrp add one [1, 2, 3] and nothing else is a possible result.
Now we reason, for any r, min r · Λ(flatten◦) = roll is a surjective function, so roll · roll◦= id min r · Λ(roll · roll◦· flatten◦) = claim, with χ(rr) (x, y) = χ(r) (roll x, roll y) roll · min rr · Λ(roll◦· flatten◦) = converse of a function roll · min rr · Λ(foldrp add one) The claim above is that computing a minimal tree under r is equivalent to computing a minimal spine under rr, because roll is a bijection between the two types.
So our problem now is to compute min rr · Λ(foldrp add one), where preorder rr on Spine A is defined by χ(rr) (x, y) = cost (roll x) ≤cost (roll y) 4.4 An application of thinning For most instances of the bracketing problem, the algebra one ▽add is not mono-tonic under rr ◦. For Example 7, for instance, the two trees fork(tip1, fork(tip1, tip1)) and fork (fork (tip 1, tip 1), tip 1) that can be built from the list of heights [1, 1, 1] are equally good under r (and so their spines are equally good under rr), but 302 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu only the former can be extended by add to make the unique optimal tree for cons (1, [1, 1, 1]). So the Greedy Theorem does not apply to Example 7.
However, some instances of the bracketing problem succumb to a thinning algorithm based on the ‘pairwise ordering’ of the left spines of the trees, defined as follows. Let lspinecosts (a, ts) = map (cost′ a) (prefixes ts) cost′ a us = cost (roll (a, us)) where prefixes returns the prefixes of a list in order of decreasing length — for example, prefixes [1, 2, 3] = Then we choose the preorder q such that χ(q) (x, y) = lspinecosts x ≼lspinecosts y where ≼is pairwise ordering of sequences: [a1, . . . , am] ≼[b1, . . . , bn] if and only if m ≤n and ai ≤bi for 1 ≤i ≤m. Informally, as ≼bs when as is no longer than bs, and each element of as is at most the corresponding element of bs.
Note that this q is not connected. Note also that q ⊆rr: because cost (roll x) = head (lspinecosts x), if x ✭ -✿q y then certainly cost (roll x) ≤cost (roll y).
Now we claim that, for Examples 7 and 9, the algebra one ▽add is mono-tonic under this q◦. We will leave the proof of monotonicity as Exercise 4.10.5 for the energetic reader. We will merely observe here that for the minimum height tree problem, the spines of the two trees fork (tip 1, fork (tip 1, tip 1)) and fork (fork (tip 1, tip 1), tip 1) introduced above are definitely not equally good under q: the costs of the reverse prefixes of the two spines are [3, 1] and [3, 2, 1], and the former is pairwise strictly less than the latter.
Therefore we conclude that, at least for these two examples, min r · Λ(flatten◦) = Converse of a Function roll · min rr · Λ(foldrn add one) ⊇ thinning roll · min rr · fold (thin q · Λ((one ▽add) · F choose)) Unfortunately, although the resulting algorithm appears to perform well in prac-tice, its worst-case running time is exponential, and knowing this it is not too difficult to construct pathological inputs. The problem is that too few partial solutions are thinned out in the worst case.
4.5 An application of greediness In fact, for Example 7 we can do a lot better. We have already observed that Theorem 3 does not apply; however, there is a connected preorder q which satisfies the conditions of Theorem 4. We choose the preorder q characterized by χ(q) (x, y) = lspinecosts x ✂lspinecosts y 8. Algebraic Methods for Optimization Problems 303 where ✂is the lexicographic (as opposed to pairwise) ordering on sequences: for sequences as = [a1, . . . , am] and bs = [b1, . . . , bn], the ordering as ✂bs holds if and only if there exists an i with 0 ≤i ≤m, n such that aj = bj for 1 ≤j ≤i and either i = m (in which case as is a prefix of bs) or i < m, n and ai+1 < bi+1 (in which case as and bs differ first at position i + 1).
Again, q ⊆rr, for the same reason that the q of §4.4 is included in rr.
We leave the proof of monotonicity to Exercise 4.10.6, pausing only to observe as before that the costs of the reverse prefixes of the two spines of the trees constructed from [1, 1, 1] are [3, 1] and [3, 2, 1], and the former is lexicographically (as well as pairwise) strictly less than the latter.
Now, however, q is connected, and so a greedy algorithm works — it suffices to keep a single partial solution at each stage. We have min r · Λ(fold f ) ⊇ min r ⊇min q min q · Λ(fold f ) ⊇ Refining Greedy Theorem, assuming f monotonic under q◦ fold (min q · Λf ) 4.6 Refinement of the greedy algorithm to a program Returning to bracketing problems in general, provided that we can find a con-nected preorder q under whose converse one ▽add is monotonic (as we have seen we can do for the minimum height tree problem, for instance), we have min r · Λ(flatten◦) = Converse of a Function roll · min rr · Λ(foldrn add one) ⊇ strengthened preorder roll · min q · Λ(foldrn add one) ⊇ Greedy Theorem roll · foldrn (min q · Λadd) (min q · Λone) To obtain a deterministic program, we still have to refine min q · Λone and minq · Λadd to functions. Since one is a function, min q · Λone = one. One can also show (Exercise 4.10.7) that min q · Λadd ⊇minadd, where minadd (a, (b, ts)) = (a, cons (roll (cons (b, us)), vs)) where us + + vs = ts, and us is the shortest proper prefix of ts satisfying max (a, cost (roll (cons (b, us)))) < cost (head vs) If no such us exists, then us = ts and vs = [ ].
304 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu 4.7 Summary To summarize, the problem of building a tree of minimum height can be solved by computing roll · foldrp minadd one Moreover, this algorithm takes linear time. The time taken to compute minadd is proportional to the length of the us chosen, but the length of the resulting spine is reduced by this amount; a standard amortization argument then shows that the total time taken for the algorithm is linear in the length of the given list.
4.8 The Haskell program The program in Figure 1 is an implementation in Haskell of the algorithm derived above for the minimum height tree instance of the bracketing problem.
To avoid repeated computations, we label trees with their costs.
4.9 Bibliographic notes The minimum height tree problem comes from , where a different greedy solution is presented. This chapter could be thought of as an abstract of Mu’s forthcoming DPhil thesis , which expands on the approaches discussed here, in particular looking at other thinning algorithms, and exploring the relationship with dynamic programming as well as greedy algorithms.
4.10 Exercises 1. Prove the Converse-of-a-Function Theorem from §4.2.
2. Theorem 10 applies only to total functions f . If f is a partial function (that is, simple buyt not necessarily entire), the corresponding ‘Converse of a Partial Function’ theorem states, when T = data F, that f ◦= foldT g provided that f · g ⊆in · F f and dom f = ran g. Prove this generalization.
3. Verify the application of the Converse-of-a-Function Theorem in §4.3.
4. Show that the Greedy Theorem is not applicable to (that is, one ▽add is not monotonic under r ◦for) Example 9. What can you say about Example 8?
5. Show that for a bracketing problem (A, ⊕, ≤) such that ⊕is commutative (a ⊕b = b ⊕a), strict (a ⊕b > a) and monotonic in its left argument (a ≤a′ ⇒a ⊕b ≤a′ ⊕b), the algebra one ▽add is monotonic under q◦, where q is as introduced in §4.4. Verify that these conditions apply for Examples 7 and 9.
6. Show that one ▽add for the minimum height tree problem is monotonic under q◦, where q is as introduced in §4.5.
7. Show that min q · Λadd ⊇minadd for the minimum height tree problem, where minadd is as defined in §4.6.
8. Express as an instance of the bracketing problem the problem of concate-nating a list of lists into a single list in the cheapest possible way, when the cost of concatenating two lists is proportional to the length of the lefthand list. Derive a suitable algorithm for solving the problem.
8. Algebraic Methods for Optimization Problems 305 data Tree = Tip Int | Fork Int Tree Tree type Spine = (Int, [Tree]) cost :: Tree -> Int cost (Tip a) = a cost (Fork a t u) = a fork :: Tree -> Tree -> Tree fork t u = Fork (max (cost t) (cost u) + 1) t u roll :: Spine -> Tree roll (a, ts) = foldl fork (Tip a) ts greedy :: [Int] -> Tree greedy = roll . foldrp minadd one one :: Int -> Spine one a = (a, []) minadd :: Int -> Spine -> Spine minadd a (b,ts) = (a, split (Tip b : ts)) where split [t] = [t] split (t:u:ts) = if max a (cost t) < cost u then t:u:ts else split (fork t u : ts) foldrp :: (a -> b -> b) -> (a -> b) -> ([a] -> b) foldrp f g [x] = g x foldrp f g (x:xs) = f x (foldrp f g xs) Fig. 1. Haskell implementation of minimum height tree algorithm 306 Richard Bird, Jeremy Gibbons and Shin-Cheng Mu 5 Bibliography 1. Roland Backhouse, Peter de Bruin, Grant Malcolm, Ed Voermans, and Jaap van der Woude. A relational theory of datatypes. In STOP 1992 Summerschool on Constructive Algorithmics. STOP project, 1992.
2. Roland Backhouse and Paul Hoogendijk.
Elements of a relational theory of datatypes. In Bernhard M¨ oller, Helmut Partsch, and Steve Schumann, editors, LNCS 755: IFIP TC2/WG2.1 State-of-the-Art Report on Formal Program Devel-opment, pages 7–42. Springer-Verlag, 1993.
3. R. Berghammer and H. Zierer.
Relational algebraic semantics of deterministic and non-deterministic programs. Theoretical Computer Science, 43(2–3):123–147, 1986.
4. Richard Bird and Oege de Moor. Hybrid dynamic programming. Programming Research Group, Oxford, 1994.
5. Richard Bird and Oege de Moor. The Algebra of Programming. Prentice-Hall, 1996.
6. Richard S. Bird. On building trees with minimum height. Journal of Functional Programming, 7(4):441–445, 1997.
7. A. Bunkenburg. Expression Refinement. PhD thesis, Computing Science Depart-ment, University of Glasgow, 1997.
8. Thomas H. Cormen, Charles E. Leiserson, and Ronald L. Rivest. Introduction to Algorithms. MIT Press, 1990.
9. Sharon Curtis.
A Relational Approach to Optimization Problems.
PhD thesis, University of Oxford, 1996. Technical Monograph PRG-122.
10. J. W. de Bakker and W. P. de Roever. A calculus for recursive program schemes. In M. Nivat, editor, Automata, Languages and Programming, pages 167–196. North-Holland, 1973.
11. Oege de Moor and Jeremy Gibbons.
Pointwise relational programming.
In LNCS 1816: Algebraic Methodology and Software Technology, pages 371–390, May 2000.
12. S. Eilenberg and J. B. Wright. Automata in general algebras. Information and Control, 11(4):452–470, 1967.
13. Sharon Flynn. A Refinement Calculus for Expressions. PhD thesis, University of Glasgow, 1997.
14. P. J. Freyd and A. ˇ Sˇ cedrov. Categories, Allegories. North-Holland, 1990.
15. David Gries. The Science of Programming. Texts and Monographs in Computer Science. Springer-Verlag, 1981.
16. Eric C. R. Hehner. A Practical Theory of Programming. Texts and Monographs in Computer Science. Springer-Verlag, 1993.
17. C. A. R. Hoare. Programs are predicates. In C. A. R. Hoare and J. C. Shepherdson, editors, Mathematical Logic and Programming Languages. Prentice-Hall, 1985. Also Chapter 20 of .
18. C. A. R. Hoare. Essays in Computing Science. Prentice Hall, 1989.
19. Paul Hoogendijk. A Generic Theory of Datatypes. PhD thesis, Technische Univer-siteit Eindhoven, 1997.
20. Simon Peyton Jones, John Hughes, Lennart Augustsson, Dave Barton, Brian Bou-tel, Warren Burton, Joseph Fasel, Kevin Hammond, Ralf Hinze, Paul Hudak, Thomas Johnsson, Mark Jones, John Launchbury, Erik Meijer, John Peterson, Alastair Reid, Colin Runciman, and Philip Wadler. Haskell 98: A non-strict, purely functional language. www.haskell.org/onlinereport, February 1999.
8. Algebraic Methods for Optimization Problems 307 21. Bernhard Korte, Laszlo Lovasz, and Rainer Schrader. Greedoids. Springer-Verlag, 1991.
22. Robert A. Kowalski. Predicate logic as a programming language. In IFIP Congress, 1974.
23. Ali Mili, Jules Desharnais, and Fatma Mili.
Computer Program Construction.
Oxford University Press, 1994.
24. Bernhard M¨ oller. Relations as a program development language. In B. M¨ oller, editor, IFIP TC2/WG2.1 Working Conference on Constructing Programs from Specifications, pages 373–397. North-Holland, 1991.
25. Joseph M. Morris. Programming by expression refinement: The KMP algorithm.
In W. H. J. Feijen, A. J. M. van Gasteren, D. Gries, and J. Misra, editors, Beauty is our Business, chapter 37. Springer-Verlag, 1990.
26. Joseph M. Morris. Non-deterministic expressions and predicate transformers. In-formation Processing Letters, 61:241–246, 1997.
27. Shin-Cheng Mu. Inverting Programs by Calculation. DPhil thesis, University of Oxford, in preparation.
28. Theo Norvell and Eric Hehner. Logical specifications for functional programs. In R. S. Bird, C. C. Morgan, and J. C. P. Woodcock, editors, LNCS 669: Mathematics of Program Construction, pages 269–290. Springer-Verlag, 1993.
29. Richard A. O’Keefe. The Craft of Prolog. MIT Press, 1990.
30. Alfred Tarski. On the calculus of relations. Journal of Symbolic Logic, 6(3):73–89, 1941.
31. Nigel Thomas Edgar Ward. A Refinement Calculus for Nondeterministiuc Expres-sions. PhD thesis, University of Queensland, February 1994. |
188579 | https://www.bbc.co.uk/bitesize/guides/ztndmp3/revision/3 | Area of a triangle - Perimeter and area - National 4 Application of Maths Revision - BBC Bitesize
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Perimeter and area Area of a triangle
In this National 4 Lifeskills Maths study guide, you can revise how to find the perimeter and area of triangles, quadrilaterals (rectangle, parallelogram, rhombus, kite and square), circles, arcs, sectors and composite shapes can all be calculated using relevant formulae.
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Area of a triangle
The area of a triangle is found by multiplying 1 2 by the base by the height.
Example 1
A r e a=1 2×b a s e×h e i g h t
A r e a=1 2×10×6
A r e a=30 c m 2
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Work out the area of this triangle using A r e a=1 2×b a s e×h e i g h t
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A r e a=1 2×b a s e×h e i g h t
A r e a=1 2×12×14
A r e a=84 c m 2
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188580 | https://www.pvamu.edu/universitycollege/wp-content/uploads/sites/71/CACN.pdf | 1.3 - 1 1.3 - 1 1.3 Complex Numbers Basic Concepts of Complex Numbers Operations on Complex Numbers 1.3 - 2 1.3 - 2 Basic Concepts of Complex Numbers There are no real numbers for the solution of the equation 2 1.
x = − To extend the real number system to include such numbers as, 1 , − the number i is defined to have the following property; 2 1.
i = − 1.3 - 3 1.3 - 3 Basic Concepts of Complex Numbers So… 1 i = − The number i is called the imaginary unit.
Numbers of the form a + bi, where a and b are real numbers are called complex numbers.
In this complex number, a is the real part and b is the imaginary part.
Complex numbers a + bi, a and b real Nonreal complex numbers a + bi, b ≠ 0 Real numbers a + bi, b = 0 Irrational numbers Rational numbers Integers Non-integers 1.3 - 5 1.3 - 5 Basic Concepts of Complex Numbers Two complex numbers are equal provided that their real parts are equal and their imaginary parts are equal; a bi c di + = + if and only if a c = and b d = 1.3 - 6 1.3 - 6 Basic Concepts of Complex Numbers If a = 0 and b ≠ 0, the complex number is pure imaginary.
A pure imaginary number or a number, like 7 + 2i with a ≠ 0 and b ≠ 0, is a nonreal complex number. The form a + bi (or a + ib) is called standard form.
1.3 - 7 1.3 - 7 THE EXPRESSION If 0, then a > .
a i a − = a − 1.3 - 8 1.3 - 8 Example 1 WRITING AS i a Write as the product of a real number and i, using the definition of .
a − a.
16 − Solution: 16 16 4i i = − = a − 1.3 - 9 1.3 - 9 Example 1 Write as the product of a real number and i, using the definition of .
a − b.
70 − Solution: 70 70 i − = WRITING AS i a a − 1.3 - 10 1.3 - 10 Example 1 Write as the product of a real number and i, using the definition of .
a − c.
48 − Solution: 48 48 16 3 4 3 i i i − = = = Product rule for radicals WRITING AS i a a − 1.3 - 11 1.3 - 11 Operations on Complex Numbers Products or quotients with negative radicands are simplified by first rewriting for a positive number. as a i a − Then the properties of real numbers are applied, together with the fact that 2 1.
i = − 1.3 - 12 1.3 - 12 Example 2 FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICALS Multiply or divide, as indicated. Simplify each answer.
a.
7 7 − − Solution: 7 7 7 7 i i − − = First write all square roots in terms of i.
( ) 2 2 7 i = 1 7 = − 7 = − i 2 = −1 1.3 - 13 1.3 - 13 Example 2 FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICALS Multiply or divide, as indicated. Simplify each answer.
b.
Solution: 6 10 6 10 i i − − = 2 60 i = 15 1 4 = − 1 2 15 = − 2 15 = − 6 10 − − 1.3 - 14 1.3 - 14 Example 2 FINDING PRODUCTS AND QUOTIENTS INVOLVING NEGATIVE RADICALS Multiply or divide, as indicated. Simplify each answer.
c.
20 20 20 10 2 2 2 i i − = = = − Solution: 20 2 − − Quotient rule for radicals 1.3 - 15 1.3 - 15 Example 3 SIMPLIFYING A QUOTIENT INVOLVING A NEGATIVE RADICAND Write in standard form a + bi.
8 128 4 −+ − Solution: 64 2 12 8 8 8 4 4 −+ − −+ = − 4 8 8 2 i −+ = 64 8i − = 1.3 - 16 1.3 - 16 Example 3 SIMPLIFYING A QUOTIENT INVOLVING A NEGATIVE RADICAND Write in standard form a + bi.
8 128 4 −+ − Solution: ( ) 2 4 2 2 4 i −+ = Factor.
4 8 8 2 i −+ = 64 8i − = Be sure to factor before simplifying 2 2 2 i = −+ Lowest terms 1.3 - 17 1.3 - 17 Addition and Subtraction of Complex Numbers For complex numbers a + bi and c + di, ( ) ( ) ( ) ( ) a bi c di a c b d i + + + = + + + and ( ) ( ) ( ) ( ) .
a bi c di a c b d i + − + = − + − 1.3 - 18 1.3 - 18 Example 4 ADDING AND SUBTRACTING COMPLEX NUMBERS Find each sum or difference.
a. (3 4 ) ( 2 6 ) i i − + −+ Solution: [ ] [ ] ( ) ( ) 3 2 3 ( 4 ) 6 6 2 4i i i − − − + + − + + = + Add real parts.
Add imaginary parts.
Commutative, associative, distributive properties 1 2i = + 1.3 - 19 1.3 - 19 Example 4 ADDING AND SUBTRACTING COMPLEX NUMBERS Find each sum or difference.
b. ( 9 7 ) (3 15 ) i i −+ + − Solution: ( 9 7 ) (3 15 ) 6 8 i i i −+ + − = − − 1.3 - 20 1.3 - 20 Example 4 ADDING AND SUBTRACTING COMPLEX NUMBERS Find each sum or difference.
c. ( 4 3 ) (6 7 ) i i − + − − Solution: [ ] ( 4 3 ) (6 7 ) ( 4 6) 3 ( 7) i i i −+ − − = − − + −− 10 10i = − + 1.3 - 21 1.3 - 21 Example 4 ADDING AND SUBTRACTING COMPLEX NUMBERS Find each sum or difference.
d. (12 5 ) (8 3 ) i i − − − Solution: (12 5 ) (8 3 ) 4 2 i i i − − − = − 1.3 - 22 1.3 - 22 Multiplication of Complex Numbers For complex numbers a + bi and c + di, ( )( ) ( ) ( ) .
a bi c di ac bd ad bc i + + = − + + 1.3 - 23 1.3 - 23 Example 5 MULTIPLYING COMPLEX NUMBERS Find each product.
a. (2 3 )(3 4 ) i i − + Solution: (2 3 )(3 4 ) 2(3) 2(4 ) 3 (3) 3 (4 ) i i i i i i − + = + − − FOIL 2 6 8 9 12i i i = + − − 6 12 ) 1 ( i = −− − i2 = −1 18 i = − 1.3 - 24 1.3 - 24 Example 5 MULTIPLYING COMPLEX NUMBERS Find each product.
b.
2 (4 3 ) i + Solution: Square of a binomial 2 16 24 9 i i = + + i 2 = −1 16 24 9( 1) i = + + − 2 2 2 (4 3 ) 4 2(4)(3 ) (3 ) i i i + = + + 7 24i = + Remember to add twice the product of the two terms.
1.3 - 25 1.3 - 25 Example 5 MULTIPLYING COMPLEX NUMBERS Find each product.
c. (6 5 )(6 5 ) i i + − Solution: Product of the sum and difference of two terms i 2 = −1 2 2 (6 5 )(6 5 ) 6 (5 ) i i i + − = − 36 25( 1) = − − 36 25 = + 61 , or 61 0i = + Standard form 1.3 - 26 1.3 - 26 Simplifying Powers of i Powers of i can be simplified using the facts 2 2 4 2 2 1 and ( ) ( 1 1) i i i − = − = = = 1.3 - 27 1.3 - 27 Example 6 SIMPLIFYING POWERS OF i Simplify each power of i.
a.
15 i Solution: Since i 2 = –1 and i 4 = 1, write the given power as a product involving i 2 or i 4. For example, 2 3 .
1 ( ) i i i i i − = = = − Alternatively, using i4 and i3 to rewrite i15 gives ( ) 3 15 12 3 4 3 3( 1 ) i i i i i i i = = = − = − 1.3 - 28 1.3 - 28 Powers of i 1 i i = 2 1 i = − 3 i i = − 4 1 i = 5 i i = 6 1 i = − 7 i i = − 8 1 i = 9 i i = 10 1 i = − 11 i i = − 12 1 , i = and so on.
1.3 - 29 1.3 - 29 Ex 5c. showed that… (6 5 )(6 5 ) 61 i i + − = The numbers differ only in the sign of their imaginary parts and are called complex conjugates. The product of a complex number and its conjugate is always a real number. This product is the sum of squares of real and imaginary parts.
1.3 - 30 1.3 - 30 Property of Complex Conjugates For real numbers a and b, 2 2 ( )( ) .
a bi a bi a b + − = + 1.3 - 31 1.3 - 31 Example 7 DIVIDING COMPLEX NUMBERS Write each quotient in standard form a + bi.
a. 3 2 5 i i + − Solution: (5 3 2 (3 2 ) 5 ( 5 5 ) ) ( ) i i i i i i + − + + + = − Multiply by the complex conjugate of the denominator in both the numerator and the denominator.
2 2 15 3 10 2 25 i i i i + + + = − Multiply.
1.3 - 32 1.3 - 32 Example 7 DIVIDING COMPLEX NUMBERS Write each quotient in standard form a + bi.
a. 3 2 5 i i + − Solution: 2 2 15 3 10 2 25 i i i i + + + = − Multiply.
13 13 26 i + = i 2 = −1 1.3 - 33 1.3 - 33 Example 7 DIVIDING COMPLEX NUMBERS Write each quotient in standard form a + bi.
a. 3 2 5 i i + − Solution: 13 13 26 i + = i 2 = −1 13 13 26 26 i = + a bi a bi c c c + = + 1.3 - 34 1.3 - 34 Example 7 DIVIDING COMPLEX NUMBERS Write each quotient in standard form a + bi.
a. 3 2 5 i i + − Solution: 1 1 2 2 i = + Lowest terms; standard form 13 13 26 26 i = + a bi a bi c c c + = + |
188581 | https://eng.libretexts.org/Bookshelves/Mechanical_Engineering/Mechanics_Map_(Moore_et_al.)/03%3A_Static_Equilibrium_in_Rigid_Body_Systems/3.01%3A_Moment_of_a_Force_about_a_Point_(Scalar_Calculation) | Skip to main content
3.1: Moment of a Force about a Point (Scalar Calculation)
Last updated
: Jan 15, 2022
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3.0: Video Introduction to Chapter 3
3.2: Varignon's Theorem
Page ID
: 50579
Jacob Moore & Contributors
Pennsylvania State University Mont Alto via Mechanics Map
( \newcommand{\kernel}{\mathrm{null}\,})
The moment of a force is the tendency of some forces to cause rotation. Any easy way to visualize the concept is set a box on smooth surface. If you were to apply a force to the center of the box, it would simply slide across the surface without rotating. If you were instead to push on one side of the box, it will start rotating as it moves. Even though the forces have the same magnitude and the same direction, they cause different reactions. This is because the off-center force has a different point of application, and exerts a moment about the center of the box, whereas the force on the center of the box does not exert a moment about the box's center point.
Just like forces, moments have a magnitude (the degree of rotation it would cause) and a direction (the axis the body would rotate about). Determining the magnitude and direction of these moments about a given point is an important step in the analysis of rigid body systems (bodies that are both rigid and not experiencing concurrent forces). The scalar method below is the easiest way to do this in simple two-dimensional problems, while the alternative vector methods, which will be covered later, work best for more complex three-dimensional systems.
The Scalar Method in 2 Dimensions
In discussing how to calculate the moment of a force about a point via scalar quantities, we will begin with the example of a force on a simple lever as shown below. In this simple lever there is a force on the end of the lever, distance away from the center of rotation for the lever (point A) where the force has a magnitude .
When using scalar quantities, the magnitude of the moment will be equal to the perpendicular distance between the line of action of the force and the point we are taking the moment about.
To determine the sign of the moment, we determine what type of rotation the force would cause. In this case, we can see that the force would cause the lever to rotate counterclockwise about point A. Counterclockwise rotations are caused by positive moments while clockwise rotations are caused by negative moments.
Another important factor to remember is that the value is the perpendicular distance from the force to the point we are taking the moment about. We could measure the distance from point A to the head of the force vector, or the tail of the force vector, or really any point along the line of action of force . The distance we need to use for the scalar moment calculation, however, is the shortest distance between the point and the line of action of the force. This will always be a line perpendicular to the line of action of the force, going to the point about which we are taking the moment.
The Scalar Model in 3 Dimensions
For three-dimensional scalar calculations, we will still find the magnitude of the moment in the same way, multiplying the magnitude of the force by the perpendicular distance between the point and the line of action of the force. This perpendicular distance again is the minimum distance between the point and the line of action of the force. In some cases, finding this distance may be very difficult.
Another difficult factor in three dimensional scalar problems is finding the axis of rotation, as this is now more complex that just "clockwise or counterclockwise". The axis of rotation will be a line traveling though the point about which we are taking the moment, and perpendicular to both the force vector and the perpendicular displacement vector (the vector going from the point about which the moment was taken to the point of application of the force). While this is possible in any situation, it becomes very difficult if the force or displacement vectors do not lie in one of the three coordinate directions.
To further find the direction of the moment vector (which will act along the established line for axis of rotation), we will use the right-hand rule in a modified form. Wrap the fingers of your right hand around the axis of rotation line with your fingertips curling in the direction the body would rotate. If you do this, your thumb should point out along the line in the direction of the moment vector. This is an important last step, because we can rotate clockwise or counterclockwise in about any given axis of rotation. With the final moment vector, we known not only the axis of rotation, but which way the body would rotate about that axis.
Example
What is the moment that Force A exerts about point A? What is the moment that Force B exerts about Point A?
Solution
Example
What is the moment that this force exerts about point A? What is the moment this force exerts about point B?
Solution
Example
What are the moments that each of the three tension forces exert about point A (the point where the beams come together)?
Solution
3.0: Video Introduction to Chapter 3
3.2: Varignon's Theorem |
188582 | https://zh.wikipedia.org/wiki/%E4%B8%89%E8%A7%92%E6%81%92%E7%AD%89%E5%BC%8F | Published Time: 2005-02-04T08:23:01Z
三角恒等式 - 维基百科，自由的百科全书
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1 符号
2 基本關係开关基本關係子章节
2.1 其他函數的基本關係
3 对称、移位和周期开关对称、移位和周期子章节
3.1 對稱
3.2 移位和周期
4 角的和差恆等式开关角的和差恆等式子章节
4.1 正弦與余弦的無限多項和
4.2 正切的有限多项和
5 多倍角公式开关多倍角公式子章节
5.1 雙倍角、三倍角和半角公式
5.2 n倍角公式
5.3 其他函數的倍半角公式
6 幂简约公式
7 数值连乘
8 常見的恆等式开关常見的恆等式子章节
8.1 积化和差与和差化积恆等式
8.2 平方差公式
8.3 其他恆等式
8.3.1 托勒密定理
9 三角函數與雙曲函數的恆等式
10 线性组合
11 反三角函数
12 无限乘积公式
13 微積分开关微積分子章节
13.1 蘊涵
14 指数定义
15 参见
16 註釋
17 參考文獻开关參考文獻子章节
17.1 引用
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在几何上依据以 O 为中心的单位圆可以构造角θ的很多三角函数
三角函數示意圖
由于技术原因，此旧版图表已停用，並須迁移至新版图表。造成您的不便，我們深表歉意。
幾個三角函數的圖形，分別為正弦、餘弦、正切、餘切、正割、餘割和正矢。配色與上圖相同
单位圆的角度
| 三角学 |
| |
| 历史（英语：History of trigonometry） 三角函数(反三角函数) 广义三角函数（英语：Generalized trigonometry） |
| 参考 |
| 恒等式 精确值 三角表（英语：Trigonometric tables） 单位圆 |
| 定理 |
| 正弦 餘弦 正切 餘切 勾股定理 |
| 微积分 |
| 三角换元法 积分(反三角函数) 微分 |
| 查 论 编 |
在数学中，三角恒等式是对出现的所有值都为實变量，涉及到三角函数的等式。这些恒等式在表达式中有些三角函数需要简化的时候是很有用的。一个重要应用是非三角函数的积分：一个常用技巧是首先使用使用三角函数的代换规则，则通过三角恒等式可简化结果的积分。
符号
[编辑]
为了避免由于 sin−1x{\displaystyle \sin ^{-1}x}的不同意思所带来的混淆，我們經常用下列兩個表格來表示三角函数的倒数和反函数。另外在表示余割函数時，'csc{\displaystyle \csc }'有时會寫成比較长的'c o s e c{\displaystyle \mathrm {cosec} }'。
| 函数 | 反函數 | 倒数 |
---
| 中文 | 全寫 | 簡寫 | 中文 | 全寫 | 簡寫 | 中文 | 全寫 | 簡寫 |
| 正弦 | sine | sin | 反正弦 | arcsine | arcsin | 餘割 | cosecant | csc |
| 餘弦 | cosine | cos | 反餘弦 | arccosine | arccos | 正割 | secant | sec |
| 正切 | tangent | tan | 反正切 | arctangent | arctan | 餘切 | cotangent | cot |
| 餘切 | cotangent | cot | 反餘切 | arccotangent | arccot | 正切 | tangent | tan |
| 正割 | secant | sec | 反正割 | arcsecant | arcsec | 餘弦 | cosine | cos |
| 餘割 | cosecant | csc | 反餘割 | arccosecant | arccsc | 正弦 | sine | sin |
不同的角度度量适合于不同的情况。本表展示最常用的系统。弧度是缺省的角度量并用在指数函数中。所有角度度量都是无单位的。另外在計算機中角度的符號為D，弧度的符號為R，梯度的符號為G。
相同角度的轉換表 | 角度單位 | 值 | 計算機中代號 |
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| 轉 | 0{\displaystyle 0} | 1 12{\displaystyle {\frac {1}{12}}} | 1 8{\displaystyle {\frac {1}{8}}} | 1 6{\displaystyle {\frac {1}{6}}} | 1 4{\displaystyle {\frac {1}{4}}} | 1 2{\displaystyle {\frac {1}{2}}} | 3 4{\displaystyle {\frac {3}{4}}} | 1{\displaystyle 1} | 無 |
| 角度 | 0∘{\displaystyle 0^{\circ }} | 30∘{\displaystyle 30^{\circ }} | 45∘{\displaystyle 45^{\circ }} | 60∘{\displaystyle 60^{\circ }} | 90∘{\displaystyle 90^{\circ }} | 180∘{\displaystyle 180^{\circ }} | 270∘{\displaystyle 270^{\circ }} | 360∘{\displaystyle 360^{\circ }} | D |
| 弧度 | 0{\displaystyle 0} | π 6{\displaystyle {\frac {\pi }{6}}} | π 4{\displaystyle {\frac {\pi }{4}}} | π 3{\displaystyle {\frac {\pi }{3}}} | π 2{\displaystyle {\frac {\pi }{2}}} | π{\displaystyle \pi } | 3 π 2{\displaystyle {\frac {3\pi }{2}}} | 2 π{\displaystyle 2\pi } | R |
| 梯度 | 0 g{\displaystyle 0^{g}} | 33 1 3 g{\displaystyle 33{\frac {1}{3}}^{g}} | 50 g{\displaystyle 50^{g}} | 66 2 3 g{\displaystyle 66{\frac {2}{3}}^{g}} | 100 g{\displaystyle 100^{g}} | 200 g{\displaystyle 200^{g}} | 300 g{\displaystyle 300^{g}} | 400 g{\displaystyle 400^{g}} | G |
基本關係
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三角函數間的關係，可分成 正函數 和餘函數
畢達哥拉斯三角恒等式如下：
sin 2θ+cos 2θ=1{\displaystyle \sin ^{2}\theta +\cos ^{2}\theta =1\,}
tan 2θ+1=sec 2θ{\displaystyle \tan ^{2}\theta +1\,=\sec ^{2}\theta }
1+cot 2θ=csc 2θ{\displaystyle 1\,+\cot ^{2}\theta =\csc ^{2}\theta }
由上面的平方關係加上三角函數的基本定義，可以導出下面的表格，即每個三角函數都可以用其他五個表達。（严谨地说，所有根号前都应根据实际情况添加正负号）
| 函數 | sin{\displaystyle \sin } | cos{\displaystyle \cos } | tan{\displaystyle \tan } | cot{\displaystyle \cot } | sec{\displaystyle \sec } | csc{\displaystyle \csc } |
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| sinθ{\displaystyle \sin \theta } | sinθ{\displaystyle \sin \theta \ } | 1−cos 2θ{\displaystyle {\sqrt {1-\cos ^{2}\theta }}} | tanθ 1+tan 2θ{\displaystyle {\frac {\tan \theta }{\sqrt {1+\tan ^{2}\theta }}}} | 1 1+cot 2θ{\displaystyle {\frac {1}{\sqrt {1+\cot ^{2}\theta }}}} | sec 2θ−1 secθ{\displaystyle {\frac {\sqrt {\sec ^{2}\theta -1}}{\sec \theta }}} | 1 cscθ{\displaystyle {\frac {1}{\csc \theta }}} |
| cosθ{\displaystyle \cos \theta } | 1−sin 2θ{\displaystyle {\sqrt {1-\sin ^{2}\theta }}} | cosθ{\displaystyle \cos \theta \ } | 1 1+tan 2θ{\displaystyle {\frac {1}{\sqrt {1+\tan ^{2}\theta }}}} | cotθ 1+cot 2θ{\displaystyle {\frac {\cot \theta }{\sqrt {1+\cot ^{2}\theta }}}} | 1 secθ{\displaystyle {\frac {1}{\sec \theta }}} | csc 2θ−1 cscθ{\displaystyle {\frac {\sqrt {\csc ^{2}\theta -1}}{\csc \theta }}} |
| tanθ{\displaystyle \tan \theta } | sinθ 1−sin 2θ{\displaystyle {\frac {\sin \theta }{\sqrt {1-\sin ^{2}\theta }}}} | 1−cos 2θ cosθ{\displaystyle {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}} | tanθ{\displaystyle \tan \theta \ } | 1 cotθ{\displaystyle {\frac {1}{\cot \theta }}} | sec 2θ−1{\displaystyle {\sqrt {\sec ^{2}\theta -1}}} | 1 csc 2θ−1{\displaystyle {\frac {1}{\sqrt {\csc ^{2}\theta -1}}}} |
| cotθ{\displaystyle \cot \theta } | 1−sin 2θ sinθ{\displaystyle {{\sqrt {1-\sin ^{2}\theta }} \over \sin \theta }} | cosθ 1−cos 2θ{\displaystyle {\cos \theta \over {\sqrt {1-\cos ^{2}\theta }}}} | 1 tanθ{\displaystyle {1 \over \tan \theta }} | cotθ{\displaystyle \cot \theta \ } | 1 sec 2θ−1{\displaystyle {1 \over {\sqrt {\sec ^{2}\theta -1}}}} | csc 2θ−1{\displaystyle {\sqrt {\csc ^{2}\theta -1}}} |
| secθ{\displaystyle \sec \theta } | 1 1−sin 2θ{\displaystyle {1 \over {\sqrt {1-\sin ^{2}\theta }}}} | 1 cosθ{\displaystyle {1 \over \cos \theta }} | 1+tan 2θ{\displaystyle {\sqrt {1+\tan ^{2}\theta }}} | 1+cot 2θ cotθ{\displaystyle {{\sqrt {1+\cot ^{2}\theta }} \over \cot \theta }} | secθ{\displaystyle \sec \theta \ } | cscθ csc 2θ−1{\displaystyle {\csc \theta \over {\sqrt {\csc ^{2}\theta -1}}}} |
| cscθ{\displaystyle \csc \theta } | 1 sinθ{\displaystyle {1 \over \sin \theta }} | 1 1−cos 2θ{\displaystyle {1 \over {\sqrt {1-\cos ^{2}\theta }}}} | 1+tan 2θ tanθ{\displaystyle {{\sqrt {1+\tan ^{2}\theta }} \over \tan \theta }} | 1+cot 2θ{\displaystyle {\sqrt {1+\cot ^{2}\theta }}} | secθ sec 2θ−1{\displaystyle {\sec \theta \over {\sqrt {\sec ^{2}\theta -1}}}} | cscθ{\displaystyle \csc \theta \ } |
其他函數的基本關係
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正矢、餘矢、半正矢、半餘矢、外正割用於航行。例如半正矢可以計算球體上的兩個點之間的距離，但它們不常用。
| 名稱 | 函數 | 值 |
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| 正矢, versine | versinθ{\displaystyle \operatorname {versin} \theta } versθ{\displaystyle \operatorname {vers} \theta } verθ{\displaystyle \operatorname {ver} \theta } | 1−cosθ{\displaystyle 1-\cos \theta } |
| 餘的正矢, vercosine | vercosinθ{\displaystyle \operatorname {vercosin} \theta } | 1+cosθ{\displaystyle 1+\cos \theta } |
| 餘矢, coversine | coversinθ{\displaystyle \operatorname {coversin} \theta } cvsθ{\displaystyle \operatorname {cvs} \theta } | 1−sinθ{\displaystyle 1-\sin \theta } |
| 餘的餘矢, covercosine | covercosinθ{\displaystyle \operatorname {covercosin} \theta } | 1+sinθ{\displaystyle 1+\sin \theta } |
| 半正矢, haversine | haversinθ{\displaystyle \operatorname {haversin} \theta } | 1−cosθ 2{\displaystyle {\frac {1-\cos \theta }{2}}} |
| 餘的半正矢, havercosine | havercosinθ{\displaystyle \operatorname {havercosin} \theta } | 1+cosθ 2{\displaystyle {\frac {1+\cos \theta }{2}}} |
| 半餘矢, hacoversine cohaversine | hacoversinθ{\displaystyle \operatorname {hacoversin} \theta } | 1−sinθ 2{\displaystyle {\frac {1-\sin \theta }{2}}} |
| 餘的半餘矢, hacovercosine cohavercosine | hacovercosinθ{\displaystyle \operatorname {hacovercosin} \theta } | 1+sinθ 2{\displaystyle {\frac {1+\sin \theta }{2}}} |
| 外正割，exsecant | exsecθ{\displaystyle \operatorname {exsec} \theta } | secθ−1{\displaystyle \sec \theta -1} |
| 外餘割，excosecant | excscθ{\displaystyle \operatorname {excsc} \theta } | cscθ−1{\displaystyle \csc \theta -1} |
| 弦函數, chord | crdθ{\displaystyle \operatorname {crd} \theta } | 2 sin(θ 2){\displaystyle 2\sin \left({\frac {\theta }{2}}\right)} |
| 純虛數指數函數, cosine and imaginary unit sine | cisθ{\displaystyle \operatorname {cis} \theta } | cosθ+i sinθ{\displaystyle \cos \theta +i\;\sin \theta } |
| 輻角,Argument | argx{\displaystyle \arg x} | Im(lnx){\displaystyle \operatorname {Im} (\ln x)} |
| 弧,Arc | arc x | θ{\displaystyle \theta } | |
对称、移位和周期
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通過檢視單位圓，可確立三角函數的下列性質，這些性質也被稱為誘導公式：
维基教科书有诱导公式相关的教科书和手册
對稱
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當三角函数反射自某个特定的 θ{\displaystyle \theta }值，結果經常是另一个其他三角函數。這導致了下列恆等式：
| 反射於 θ=0{\displaystyle \theta =0} | 反射於 θ=π 4{\displaystyle \theta ={\tfrac {\pi }{4}}} | 反射於 θ=π 2{\displaystyle \theta ={\tfrac {\pi }{2}}} | 反射於 θ=3 π 4{\displaystyle \theta ={\tfrac {3\pi }{4}}} |
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| sin(0−θ)=−sinθ cos(0−θ)=+cosθ tan(0−θ)=−tanθ cot(0−θ)=−cotθ sec(0−θ)=+secθ csc(0−θ)=−cscθ{\displaystyle {\begin{aligned}\sin(0-\theta )&=-\sin \theta \\cos(0-\theta )&=+\cos \theta \\tan(0-\theta )&=-\tan \theta \\cot(0-\theta )&=-\cot \theta \\sec(0-\theta )&=+\sec \theta \\csc(0-\theta )&=-\csc \theta \end{aligned}}} | sin(π 2−θ)=+cosθ cos(π 2−θ)=+sinθ tan(π 2−θ)=+cotθ cot(π 2−θ)=+tanθ sec(π 2−θ)=+cscθ csc(π 2−θ)=+secθ{\displaystyle {\begin{aligned}\sin({\tfrac {\pi }{2}}-\theta )&=+\cos \theta \\cos({\tfrac {\pi }{2}}-\theta )&=+\sin \theta \\tan({\tfrac {\pi }{2}}-\theta )&=+\cot \theta \\cot({\tfrac {\pi }{2}}-\theta )&=+\tan \theta \\sec({\tfrac {\pi }{2}}-\theta )&=+\csc \theta \\csc({\tfrac {\pi }{2}}-\theta )&=+\sec \theta \end{aligned}}} | sin(π−θ)=+sinθ cos(π−θ)=−cosθ tan(π−θ)=−tanθ cot(π−θ)=−cotθ sec(π−θ)=−secθ csc(π−θ)=+cscθ{\displaystyle {\begin{aligned}\sin(\pi -\theta )&=+\sin \theta \\cos(\pi -\theta )&=-\cos \theta \\tan(\pi -\theta )&=-\tan \theta \\cot(\pi -\theta )&=-\cot \theta \\sec(\pi -\theta )&=-\sec \theta \\csc(\pi -\theta )&=+\csc \theta \end{aligned}}} | sin(3 π 2−θ)=−cosθ cos(3 π 2−θ)=−sinθ tan(3 π 2−θ)=+cotθ cot(3 π 2−θ)=+tanθ sec(3 π 2−θ)=−cscθ csc(3 π 2−θ)=−secθ{\displaystyle {\begin{aligned}\sin({\tfrac {3\pi }{2}}-\theta )&=-\cos \theta \\cos({\tfrac {3\pi }{2}}-\theta )&=-\sin \theta \\tan({\tfrac {3\pi }{2}}-\theta )&=+\cot \theta \\cot({\tfrac {3\pi }{2}}-\theta )&=+\tan \theta \\sec({\tfrac {3\pi }{2}}-\theta )&=-\csc \theta \\csc({\tfrac {3\pi }{2}}-\theta )&=-\sec \theta \end{aligned}}} |
移位和周期
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通過旋轉特定角度移位三角函數，經常可以找到更簡單的表達结果的不同的三角函數。例如通過旋轉 π 2{\displaystyle {\tfrac {\pi }{2}}}、π{\displaystyle \pi }和 2 π{\displaystyle 2\pi }弧度移位函數。因爲這些函數的周期要麼是 π{\displaystyle \pi }要麼是 2 π{\displaystyle 2\pi }，所以新函數和沒有移位的舊函數完全一樣。
| 移位 π 2{\displaystyle {\tfrac {\pi }{2}}} | 移位 π{\displaystyle \pi } | 移位 3 π 2{\displaystyle {\tfrac {3\pi }{2}}} | 移位 2 π{\displaystyle 2\pi } |
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| tan{\displaystyle \tan }和 cot{\displaystyle \cot }的周期 | sin{\displaystyle \sin }, cos{\displaystyle \cos }, csc{\displaystyle \csc }和 sec{\displaystyle \sec }的周期 |
| sin(θ+π 2)=+cosθ cos(θ+π 2)=−sinθ tan(θ+π 2)=−cotθ cot(θ+π 2)=−tanθ sec(θ+π 2)=−cscθ csc(θ+π 2)=+secθ{\displaystyle {\begin{aligned}\sin(\theta +{\tfrac {\pi }{2}})&=+\cos \theta \\cos(\theta +{\tfrac {\pi }{2}})&=-\sin \theta \\tan(\theta +{\tfrac {\pi }{2}})&=-\cot \theta \\cot(\theta +{\tfrac {\pi }{2}})&=-\tan \theta \\sec(\theta +{\tfrac {\pi }{2}})&=-\csc \theta \\csc(\theta +{\tfrac {\pi }{2}})&=+\sec \theta \end{aligned}}} | sin(θ+π)=−sinθ cos(θ+π)=−cosθ tan(θ+π)=+tanθ cot(θ+π)=+cotθ sec(θ+π)=−secθ csc(θ+π)=−cscθ{\displaystyle {\begin{aligned}\sin(\theta +\pi )&=-\sin \theta \\cos(\theta +\pi )&=-\cos \theta \\tan(\theta +\pi )&=+\tan \theta \\cot(\theta +\pi )&=+\cot \theta \\sec(\theta +\pi )&=-\sec \theta \\csc(\theta +\pi )&=-\csc \theta \end{aligned}}} | sin(θ+3 π 2)=−cosθ cos(θ+3 π 2)=+sinθ tan(θ+3 π 2)=−cotθ cot(θ+3 π 2)=−tanθ sec(θ+3 π 2)=+cscθ csc(θ+3 π 2)=−secθ{\displaystyle {\begin{aligned}\sin(\theta +{\tfrac {3\pi }{2}})&=-\cos \theta \\cos(\theta +{\tfrac {3\pi }{2}})&=+\sin \theta \\tan(\theta +{\tfrac {3\pi }{2}})&=-\cot \theta \\cot(\theta +{\tfrac {3\pi }{2}})&=-\tan \theta \\sec(\theta +{\tfrac {3\pi }{2}})&=+\csc \theta \\csc(\theta +{\tfrac {3\pi }{2}})&=-\sec \theta \end{aligned}}} | sin(θ+2 π)=+sinθ cos(θ+2 π)=+cosθ tan(θ+2 π)=+tanθ cot(θ+2 π)=+cotθ sec(θ+2 π)=+secθ csc(θ+2 π)=+cscθ{\displaystyle {\begin{aligned}\sin(\theta +2\pi )&=+\sin \theta \\cos(\theta +2\pi )&=+\cos \theta \\tan(\theta +2\pi )&=+\tan \theta \\cot(\theta +2\pi )&=+\cot \theta \\sec(\theta +2\pi )&=+\sec \theta \\csc(\theta +2\pi )&=+\csc \theta \end{aligned}}} |
角的和差恆等式
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正弦與餘弦的角和公式的圖形證明法。使用了相似三角形的性質與三角函數的定義，強調的線段是單位長度
正切的角和公式的圖形證明法。使用了相似三角形的性質與三角函數的定義，強調的線段是單位長度。
又稱做“和差定理”、“和差公式”或“和角公式”。最簡要的檢定方式是使用歐拉公式[註 1]。
| 正弦 | sin(α±β)=sinα cosβ±cosα sinβ{\displaystyle \sin(\alpha \pm \beta )=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta \,} |
| 余弦 | cos(α±β)=cosα cosβ∓sinα sinβ{\displaystyle \cos(\alpha \pm \beta )=\cos \alpha \cos \beta \mp \sin \alpha \sin \beta \,} |
| 正切 | tan(α±β)=tanα±tanβ 1∓tanα tanβ{\displaystyle \tan(\alpha \pm \beta )={\frac {\tan \alpha \pm \tan \beta }{1\mp \tan \alpha \tan \beta }}} |
| 余切 | cot(α±β)=cotα cotβ∓1 cotβ±cotα{\displaystyle \cot(\alpha \pm \beta )={\frac {\cot \alpha \cot \beta \mp 1}{\cot \beta \pm \cot \alpha }}} |
| 正割 | sec(α±β)=secα secβ 1∓tanα tanβ{\displaystyle \sec(\alpha \pm \beta )={\frac {\sec \alpha \sec \beta }{1\mp \tan \alpha \tan \beta }}} |
| 余割 | csc(α±β)=cscα cscβ cotβ±cotα{\displaystyle \csc(\alpha \pm \beta )={\frac {\csc \alpha \csc \beta }{\cot \beta \pm \cot \alpha }}} |
| 注意正負號的對應。 x±y=a±b⇒x+y=a+b and x−y=a−b{\displaystyle {\begin{aligned}x\pm y=a\pm b&\Rightarrow \ x+y=a+b\&{\mbox{and}}\ x-y=a-b\end{aligned}}} x±y=a∓b⇒x+y=a−b and x−y=a+b{\displaystyle {\begin{aligned}x\pm y=a\mp b&\Rightarrow \ x+y=a-b\&{\mbox{and}}\ x-y=a+b\end{aligned}}} |
根据 s i n π 4=c o s π 4=1 2{\displaystyle sin{\frac {\pi }{4}}=cos{\frac {\pi }{4}}={\frac {1}{\sqrt {2}}}}，以及和差恒等式，可以得到同角的正弦余弦的和差关系，例如，
sinα+cosα=2(sinα cosπ 4+sinπ 4 cosα)=2 sin(α+π 4)=2 cos(α−π 4){\displaystyle \sin \alpha +\cos \alpha ={\sqrt {2}}\left(\sin \alpha \cos {\frac {\pi }{4}}+\sin {\frac {\pi }{4}}\cos \alpha \right)={\sqrt {2}}\sin \left(\alpha +{\frac {\pi }{4}}\right)={\sqrt {2}}\cos \left(\alpha -{\frac {\pi }{4}}\right)}sinα−cosα=2(sinα cosπ 4−sinπ 4 cosα)=2 sin(α−π 4)=2 cos(α+π 4){\displaystyle \sin \alpha -\cos \alpha ={\sqrt {2}}\left(\sin \alpha \cos {\frac {\pi }{4}}-\sin {\frac {\pi }{4}}\cos \alpha \right)={\sqrt {2}}\sin \left(\alpha -{\frac {\pi }{4}}\right)={\sqrt {2}}\cos \left(\alpha +{\frac {\pi }{4}}\right)}
正弦與余弦的無限多項和
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sin(∑i=1∞θ i)=∑o d d k≥1(−1)k−1 2∑|A|=k(∏i∈A sinθ i∏i∉A cosθ i){\displaystyle \sin \left(\sum {i=1}^{\infty }\theta {i}\right)=\sum {\mathrm {odd} \ k\geq 1}(-1)^{\frac {k-1}{2}}\sum {|A|=k}\left(\prod {i\in A}\sin \theta {i}\prod {i\not \in A}\cos \theta {i}\right)}cos(∑i=1∞θ i)=∑e v e n k≥0(−1)k 2∑|A|=k(∏i∈A sinθ i∏i∉A cosθ i){\displaystyle \cos \left(\sum {i=1}^{\infty }\theta {i}\right)=\sum {\mathrm {even} \ k\geq 0}~(-1)^{\frac {k}{2}}~~\sum {|A|=k}\left(\prod {i\in A}\sin \theta {i}\prod {i\not \in A}\cos \theta {i}\right)}
這里的"|A|=k{\displaystyle |A|=k}"意味著索引 A{\displaystyle A}遍歷集合{1,2,3,…}{\displaystyle \left{1,2,3,\ldots \right}}的大小為 k{\displaystyle k}的所有子集的集合。
在这两个恒等式中出现了在有限多项中不出现的不对称：在每个乘积中，只有有限多个正弦因子和餘有限多个余弦因子。
如果只有有限多项 θ i{\displaystyle \theta _{i}}是非零，则在右边只有有限多项是非零，因为正弦因子将变为零，而在每个项中，所有却有限多的余弦因子将是单位一。
正切的有限多项和
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设 x i=tanθ i{\displaystyle x_{i}=\tan \theta {i}}，对于 i=1,…,n{\displaystyle i=1,\ldots ,n}。设 e k{\displaystyle e{k}}是变量 x i{\displaystyle x_{i}}，i=1,…,n{\displaystyle i=1,\ldots ,n}，k=0,…,n{\displaystyle k=0,\ldots ,n}的 k{\displaystyle k} 次基本对称多项式。则
tan(θ 1+⋯+θ n)=e 1−e 3+e 5−⋯e 0−e 2+e 4−⋯,{\displaystyle \tan(\theta {1}+\cdots +\theta {n})={\frac {e_{1}-e_{3}+e_{5}-\cdots }{e_{0}-e_{2}+e_{4}-\cdots }},}
项的数目依赖于 n{\displaystyle n}。例如，
tan(θ 1+θ 2+θ 3)=e 1−e 3 e 0−e 2=(x 1+x 2+x 3)−(x 1 x 2 x 3)1−(x 1 x 2+x 1 x 3+x 2 x 3),tan(θ 1+θ 2+θ 3+θ 4)=e 1−e 3 e 0−e 2+e 4=(x 1+x 2+x 3+x 4)−(x 1 x 2 x 3+x 1 x 2 x 4+x 1 x 3 x 4+x 2 x 3 x 4)1−(x 1 x 2+x 1 x 3+x 1 x 4+x 2 x 3+x 2 x 4+x 3 x 4)+(x 1 x 2 x 3 x 4),{\displaystyle {\begin{aligned}\tan(\theta {1}+\theta {2}+\theta {3})&{}={\frac {e{1}-e_{3}}{e_{0}-e_{2}}}={\frac {(x_{1}+x_{2}+x_{3})\ -\ (x_{1}x_{2}x_{3})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{2}x_{3})}},\\\tan(\theta {1}+\theta {2}+\theta {3}+\theta {4})&{}={\frac {e_{1}-e_{3}}{e_{0}-e_{2}+e_{4}}}\\&{}={\frac {(x_{1}+x_{2}+x_{3}+x_{4})\ -\ (x_{1}x_{2}x_{3}+x_{1}x_{2}x_{4}+x_{1}x_{3}x_{4}+x_{2}x_{3}x_{4})}{1\ -\ (x_{1}x_{2}+x_{1}x_{3}+x_{1}x_{4}+x_{2}x_{3}+x_{2}x_{4}+x_{3}x_{4})\ +\ (x_{1}x_{2}x_{3}x_{4})}},\end{aligned}}}
并以此类推。一般情况可通过数学归纳法证明。
多倍角公式
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| T n{\displaystyle T_{n}}是 n{\displaystyle n}次切比雪夫多项式 | cosn θ=T n cosθ{\displaystyle \cos n\theta =T_{n}\cos \theta \,} |
| S n{\displaystyle S_{n}}是 n{\displaystyle n} 次伸展多项式 | sin 2n θ=S n sin 2θ{\displaystyle \sin ^{2}n\theta =S_{n}\sin ^{2}\theta \,} |
| 棣莫弗定理，i{\displaystyle i}是虚单位 | cosn θ+i sinn θ=(cosθ+i sinθ)n{\displaystyle \cos n\theta +i\sin n\theta =(\cos \theta +i\sin \theta )^{n}\,} |
1+2 cosx+2 cos2 x+2 cos3 x+⋯+2 cos(n x)=sin[(n+1 2)x]sinx 2{\displaystyle 1+2\cos x+2\cos 2x+2\cos 3x+\cdots +2\cos(nx)={\frac {\sin \left[\left(n+{\frac {1}{2}}\right)x\right]}{\sin {\frac {x}{2}}}}}。
（這個 x{\displaystyle x}的函數是狄利克雷核。）
雙倍角、三倍角和半角公式
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這些公式可以使用和差恒等式或多倍角公式来证明。
| | 弦 | 切 | 割 |
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| 雙倍角公式 | 正 | sin2 θ=2 sinθ cosθ=2 tanθ 1+tan 2θ{\displaystyle {\begin{aligned}\sin 2\theta &=2\sin \theta \cos \theta \ \&={\frac {2\tan \theta }{1+\tan ^{2}\theta }}\end{aligned}}} | tan2 θ=2 tanθ 1−tan 2θ=1 1−tanθ−1 1+tanθ{\displaystyle {\begin{aligned}\tan 2\theta &={\frac {2\tan \theta }{1-\tan ^{2}\theta }}\ \&={\frac {1}{1-\tan \theta }}-{\frac {1}{1+\tan \theta }}\end{aligned}}} | sec2 θ=sec 2θ 1−tan 2θ=sec 2θ 2−sec 2θ{\displaystyle {\begin{aligned}\sec 2\theta &={\frac {\sec ^{2}\theta }{1-\tan ^{2}\theta }}\&={\frac {\sec ^{2}\theta }{2-\sec ^{2}\theta }}\end{aligned}}} |
| 餘 | cos2 θ=cos 2θ−sin 2θ=2 cos 2θ−1=1−2 sin 2θ=1−tan 2θ 1+tan 2θ{\displaystyle {\begin{aligned}\cos 2\theta &=\cos ^{2}\theta -\sin ^{2}\theta \&=2\cos ^{2}\theta -1\&=1-2\sin ^{2}\theta \&={\frac {1-\tan ^{2}\theta }{1+\tan ^{2}\theta }}\end{aligned}}} | cot2 θ=cot 2θ−1 2 cotθ=cotθ−tanθ 2{\displaystyle {\begin{aligned}\cot 2\theta &={\frac {\cot ^{2}\theta -1}{2\cot \theta }}\&={\frac {\cot \theta -\tan \theta }{2}}\end{aligned}}} | csc2 θ=csc 2θ 2 cotθ=secθ cscθ 2{\displaystyle {\begin{aligned}\csc 2\theta &={\frac {\csc ^{2}\theta }{2\cot \theta }}\&={\frac {\sec \theta \csc \theta }{2}}\end{aligned}}} |
| 降次公式 | 正 | sin 2θ=1−cos2 θ 2{\displaystyle \sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}} | tan 2θ=1−cos2 θ 1+cos2 θ{\displaystyle \tan ^{2}\theta ={\frac {1-\cos 2\theta }{1+\cos 2\theta }}} | |
| 餘 | cos 2θ=1+cos2 θ 2{\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}} | cot 2θ=1+cos2 θ 1−cos2 θ{\displaystyle \cot ^{2}\theta ={\frac {1+\cos 2\theta }{1-\cos 2\theta }}} | |
| 三倍角公式 | 正 | sin3 θ=3 sinθ−4 sin 3θ=4 sinθ sin(π 3−θ)sin(π 3+θ){\displaystyle {\begin{aligned}\sin 3\theta &=3\sin \theta -4\sin ^{3}\theta \&=4\sin \theta \sin \left({\frac {\pi }{3}}-\theta \right)\sin \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}} | tan3 θ=3 tanθ−tan 3θ 1−3 tan 2θ=tanθ tan(π 3−θ)tan(π 3+θ){\displaystyle {\begin{aligned}\tan 3\theta &={\frac {3\tan \theta -\tan ^{3}\theta }{1-3\tan ^{2}\theta }}\&=\tan \theta \tan \left({\frac {\pi }{3}}-\theta \right)\tan \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}} | sec3 θ=sec 3θ 4−3 sec 2θ=1 4 cosθ cos(π 3−θ)cos(π 3+θ){\displaystyle {\begin{aligned}\sec 3\theta &={\frac {\sec ^{3}\theta }{4-3\sec ^{2}\theta }}\&={\dfrac {1}{4\cos \theta \cos \left({\dfrac {\pi }{3}}-\theta \right)\cos \left({\dfrac {\pi }{3}}+\theta \right)}}\end{aligned}}} |
| 餘 | cos3 θ=4 cos 3θ−3 cosθ=4 cosθ cos(π 3−θ)cos(π 3+θ){\displaystyle {\begin{aligned}\cos 3\theta &=4\cos ^{3}\theta -3\cos \theta \&=4\cos \theta \cos \left({\frac {\pi }{3}}-\theta \right)\cos \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}} | cot3 θ=cot 3θ−3 cotθ 3 cot 2θ−1=cotθ cot(π 3−θ)cot(π 3+θ){\displaystyle {\begin{aligned}\cot 3\theta &={\frac {\cot ^{3}\theta -3\cot \theta }{3\cot ^{2}\theta -1}}\&=\cot \theta \cot \left({\frac {\pi }{3}}-\theta \right)\cot \left({\frac {\pi }{3}}+\theta \right)\end{aligned}}} | csc3 θ=csc 3θ 3 csc 2θ−4=1 4 sinθ sin(π 3−θ)sin(π 3+θ){\displaystyle {\begin{aligned}\csc 3\theta &={\frac {\csc ^{3}\theta }{3\csc ^{2}\theta -4}}\&={\dfrac {1}{4\sin \theta \sin \left({\dfrac {\pi }{3}}-\theta \right)\sin \left({\dfrac {\pi }{3}}+\theta \right)}}\end{aligned}}} |
| 半角公式 | 正 | sinθ 2=±1−cosθ 2{\displaystyle \sin {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1-\cos \theta }{2}}}} | tanθ 2=cscθ−cotθ=±1−cosθ 1+cosθ=sinθ 1+cosθ=1−cosθ sinθ=cosθ+sinθ−1 cosθ−sinθ+1=cot 2θ+1−cotθ{\displaystyle {\begin{aligned}\tan {\frac {\theta }{2}}&=\csc \theta -\cot \theta \&=\pm \,{\sqrt {1-\cos \theta \over 1+\cos \theta }}\&={\frac {\sin \theta }{1+\cos \theta }}\&={\frac {1-\cos \theta }{\sin \theta }}\&={\frac {\cos \theta +\sin \theta -1}{\cos \theta -\sin \theta +1}}\&={\sqrt {\cot ^{2}\theta +1}}-\cot \theta \end{aligned}}} | secθ 2=±2 secθ secθ+1{\displaystyle \sec {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {2\sec \theta }{\sec \theta +1}}}} |
| 餘 | cosθ 2=±1+cosθ 2{\displaystyle \cos {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {1+\cos \theta }{2}}}} | cotθ 2=cscθ+cotθ=±1+cosθ 1−cosθ=sinθ 1−cosθ=1+cosθ sinθ=cosθ−sinθ+1 cosθ+sinθ−1=cot 2θ+1+cotθ{\displaystyle {\begin{aligned}\cot {\frac {\theta }{2}}&=\csc \theta +\cot \theta \&=\pm \,{\sqrt {1+\cos \theta \over 1-\cos \theta }}\&={\frac {\sin \theta }{1-\cos \theta }}\&={\frac {1+\cos \theta }{\sin \theta }}\&={\frac {\cos \theta -\sin \theta +1}{\cos \theta +\sin \theta -1}}\&={\sqrt {\cot ^{2}\theta +1}}+\cot \theta \end{aligned}}} | cscθ 2=±2 secθ secθ−1{\displaystyle \csc {\frac {\theta }{2}}=\pm \,{\sqrt {\frac {2\sec \theta }{\sec \theta -1}}}} |
n倍角公式
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| n{\displaystyle n}倍角公式 |
| sinn θ=∑k=0 n(n k)cos kθ sin n−kθ sin[1 2(n−k)π]=sinθ∑k=0⌊n−1 2⌋(−1)k(n−1−k k)(2 cosθ)n−1−2 k{\displaystyle \sin n\theta =\sum {k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\sin \left[{\frac {1}{2}}(n-k)\pi \right]=\sin \theta \sum {k=0}^{\lfloor {\frac {n-1}{2}}\rfloor }(-1)^{k}{\binom {n-1-k}{k}}~(2\cos \theta )^{n-1-2k}} （第二类切比雪夫多项式） |
| cosn θ=∑k=0 n(n k)cos kθ sin n−kθ cos[1 2(n−k)π]=1 2∑k=0⌊n 2⌋(−1)k n n−k(n−k k)(2 cosθ)n−2 k{\displaystyle \cos n\theta =\sum {k=0}^{n}{\binom {n}{k}}\cos ^{k}\theta \,\sin ^{n-k}\theta \,\cos \left[{\frac {1}{2}}(n-k)\pi \right]={\frac {1}{2}}\sum {k=0}^{\lfloor {\frac {n}{2}}\rfloor }(-1)^{k}{\frac {n}{n-k}}{\binom {n-k}{k}}~(2\cos \theta )^{n-2k}} （第一类切比雪夫多项式） |
| tann θ=∑k=1n 2k+1(n 2 k−1)tan 2 k−1θ∑k=1n+1 2k+1(n 2(k−1))tan 2(k−1)θ{\displaystyle \tan n\theta ={\frac {\displaystyle \sum {k=1}^{\left[{\frac {n}{2}}\right]}(-1)^{k+1}{\binom {n}{2k-1}}\tan ^{2k-1}\theta }{\displaystyle \sum {k=1}^{\left[{\frac {n+1}{2}}\right]}(-1)^{k+1}{\binom {n}{2(k-1)}}\tan ^{2(k-1)}\theta }}} |
| n{\displaystyle n}倍遞迴公式 |
| tan n θ=tan(n−1)θ+tanθ 1−tan(n−1)θ tanθ{\displaystyle \tan \,n\theta ={\frac {\tan(n{-}1)\theta +\tan \theta }{1-\tan(n{-}1)\theta \,\tan \theta }}}、cot n θ=cot(n−1)θ cotθ−1 cot(n−1)θ+cotθ{\displaystyle \cot \,n\theta ={\frac {\cot(n{-}1)\theta \,\cot \theta -1}{\cot(n{-}1)\theta +\cot \theta }}}。(遞迴關係) |
参见正切半角公式，它也叫做“万能公式”。
其他函數的倍半角公式
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正矢
versin2 θ=2 sin 2θ=(sin2 θ)(sinθ)cosθ=1−cos2 θ{\displaystyle \operatorname {versin} 2\theta =2\sin ^{2}\theta ={\frac {(\sin 2\theta )(\sin \theta )}{\cos \theta }}=1-\cos 2\theta }
餘矢
cvs2 θ=(sinθ−cosθ)2=1−sin2 θ{\displaystyle \operatorname {cvs} 2\theta =(\sin \theta -\cos \theta )^{2}=1-\sin 2\theta }
幂简约公式
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从解余弦二倍角公式的第二和第三版本得到。
| 正弦 | 餘弦 | 其他 |
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| sin 2θ=1−cos2 θ 2{\displaystyle \sin ^{2}\theta ={\frac {1-\cos 2\theta }{2}}} | cos 2θ=1+cos2 θ 2{\displaystyle \cos ^{2}\theta ={\frac {1+\cos 2\theta }{2}}} | sin 2θ cos 2θ=1−cos4 θ 8{\displaystyle \sin ^{2}\theta \cos ^{2}\theta ={\frac {1-\cos 4\theta }{8}}} |
| sin 3θ=3 sinθ−sin3 θ 4{\displaystyle \sin ^{3}\theta ={\frac {3\sin \theta -\sin 3\theta }{4}}} | cos 3θ=3 cosθ+cos3 θ 4{\displaystyle \cos ^{3}\theta ={\frac {3\cos \theta +\cos 3\theta }{4}}} | sin 3θ cos 3θ=3 sin2 θ−sin6 θ 32{\displaystyle \sin ^{3}\theta \cos ^{3}\theta ={\frac {3\sin 2\theta -\sin 6\theta }{32}}} |
| sin 4θ=3−4 cos2 θ+cos4 θ 8{\displaystyle \sin ^{4}\theta ={\frac {3-4\cos 2\theta +\cos 4\theta }{8}}} | cos 4θ=3+4 cos2 θ+cos4 θ 8{\displaystyle \cos ^{4}\theta ={\frac {3+4\cos 2\theta +\cos 4\theta }{8}}} | sin 4θ cos 4θ=3−4 cos4 θ+cos8 θ 128{\displaystyle \sin ^{4}\theta \cos ^{4}\theta ={\frac {3-4\cos 4\theta +\cos 8\theta }{128}}} |
| sin 5θ=10 sinθ−5 sin3 θ+sin5 θ 16{\displaystyle \sin ^{5}\theta ={\frac {10\sin \theta -5\sin 3\theta +\sin 5\theta }{16}}} | cos 5θ=10 cosθ+5 cos3 θ+cos5 θ 16{\displaystyle \cos ^{5}\theta ={\frac {10\cos \theta +5\cos 3\theta +\cos 5\theta }{16}}} | sin 5θ cos 5θ=10 sin2 θ−5 sin6 θ+sin10 θ 512{\displaystyle \sin ^{5}\theta \cos ^{5}\theta ={\frac {10\sin 2\theta -5\sin 6\theta +\sin 10\theta }{512}}} |
| | 餘弦 | 正弦 |
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| 如果 n{\displaystyle n}是奇數 | cos nθ=2 2 n∑k=0 n−1 2(n k)cos[(n−2 k)θ]{\displaystyle \cos ^{n}\theta ={\frac {2}{2^{n}}}\sum {k=0}^{\frac {n-1}{2}}{\binom {n}{k}}\cos {[(n-2k)\theta ]}} | sin nθ=2 2 n∑k=0 n−1 2(−1)(n−1 2−k)(n k)sin[(n−2 k)θ]{\displaystyle \sin ^{n}\theta ={\frac {2}{2^{n}}}\sum {k=0}^{\frac {n-1}{2}}(-1)^{\left({\frac {n-1}{2}}-k\right)}{\binom {n}{k}}\sin {[(n-2k)\theta ]}} |
| 如果 n{\displaystyle n}是偶數 | cos nθ=1 2 n(n n 2)+2 2 n∑k=0 n 2−1(n k)cos[(n−2 k)θ]{\displaystyle \cos ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum {k=0}^{{\frac {n}{2}}-1}{\binom {n}{k}}\cos {[(n-2k)\theta ]}} | sin nθ=1 2 n(n n 2)+2 2 n∑k=0 n 2−1(−1)(n 2−k)(n k)cos[(n−2 k)θ]{\displaystyle \sin ^{n}\theta ={\frac {1}{2^{n}}}{\binom {n}{\frac {n}{2}}}+{\frac {2}{2^{n}}}\sum {k=0}^{{\frac {n}{2}}-1}(-1)^{\left({\frac {n}{2}}-k\right)}{\binom {n}{k}}\cos {[(n-2k)\theta ]}} |
数值连乘
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∏k=0 n−1 cos2 k θ=sin2 n θ 2 n sinθ{\displaystyle \prod {k=0}^{n-1}\cos 2^{k}\theta ={\frac {\sin 2^{n}\theta }{2^{n}\sin \theta }}}∏k=0 n−1 sin(x+k π n)=sinn x 2 n−1{\displaystyle \prod {k=0}^{n-1}\sin \left(x+{\frac {k\pi }{n}}\right)={\frac {\sin nx}{2^{n-1}}}}∏k=1 n−1 sin(k π n)=n 2 n−1{\displaystyle \prod {k=1}^{n-1}\sin \left({\frac {k\pi }{n}}\right)={\frac {n}{2^{n-1}}}},∏k=1 n−1 sin(k π 2 n)=n 2 n−1{\displaystyle \prod {k=1}^{n-1}\sin \left({\frac {k\pi }{2n}}\right)={\frac {\sqrt {n}}{2^{n-1}}}},∏k=1 n sin(k π 2 n+1)=2 n+1 2 n{\displaystyle \prod {k=1}^{n}\sin \left({\frac {k\pi }{2n+1}}\right)={\frac {\sqrt {2n+1}}{2^{n}}}}∏k=1 n−1 cos(k π n)=sinn π 2 2 n−1{\displaystyle \prod {k=1}^{n-1}\cos \left({\frac {k\pi }{n}}\right)={\frac {\sin {\frac {n\pi }{2}}}{2^{n-1}}}},∏k=1 n−1 cos(k π 2 n)=n 2 n−1{\displaystyle \prod {k=1}^{n-1}\cos \left({\frac {k\pi }{2n}}\right)={\frac {\sqrt {n}}{2^{n-1}}}},∏k=1 n cos(k π 2 n+1)=1 2 n{\displaystyle \prod {k=1}^{n}\cos \left({\frac {k\pi }{2n+1}}\right)={\frac {1}{2^{n}}}}∏k=1 n−1 tan(k π n)=n sinn π 2{\displaystyle \prod {k=1}^{n-1}\tan \left({\frac {k\pi }{n}}\right)={\frac {n}{\sin {\frac {n\pi }{2}}}}},∏k=1 n−1 tan(k π 2 n)=1{\displaystyle \prod {k=1}^{n-1}\tan \left({\frac {k\pi }{2n}}\right)=1},∏k=1 n tank π 2 n+1=2 n+1{\displaystyle \prod _{k=1}^{n}\tan {\frac {k\pi }{2n+1}}={\sqrt {2n+1}}}
常見的恆等式
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积化和差与和差化积恆等式
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數學家韋達在其三角學著作《應用於三角形的數學定律》給出积化和差与和差化积恒等式。积化和差恒等式可以通过展开角的和差恒等式的右手端来证明。
| 积化和差 | 和差化积 |
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| sinα cosβ=sin(α+β)+sin(α−β)2{\displaystyle \sin \alpha \cos \beta ={\sin(\alpha +\beta )+\sin(\alpha -\beta ) \over 2}} | sinα+sinβ=2 sinα+β 2 cosα−β 2{\displaystyle \sin \alpha +\sin \beta =2\sin {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}} |
| cosα sinβ=sin(α+β)−sin(α−β)2{\displaystyle \cos \alpha \sin \beta ={\sin(\alpha +\beta )-\sin(\alpha -\beta ) \over 2}} | sinα−sinβ=2 cosα+β 2 sinα−β 2{\displaystyle \sin \alpha -\sin \beta =2\cos {\alpha +\beta \over 2}\sin {\alpha -\beta \over 2}} |
| cosα cosβ=cos(α+β)+cos(α−β)2{\displaystyle \cos \alpha \cos \beta ={\cos(\alpha +\beta )+\cos(\alpha -\beta ) \over 2}} | cosα+cosβ=2 cosα+β 2 cosα−β 2{\displaystyle \cos \alpha +\cos \beta =2\cos {\frac {\alpha +\beta }{2}}\cos {\frac {\alpha -\beta }{2}}} |
| sinα sinβ=−cos(α+β)−cos(α−β)2{\displaystyle \sin \alpha \sin \beta =-{\cos(\alpha +\beta )-\cos(\alpha -\beta ) \over 2}} | cosα−cosβ=−2 sinα+β 2 sinα−β 2{\displaystyle \cos \alpha -\cos \beta =-2\sin {\alpha +\beta \over 2}\sin {\alpha -\beta \over 2}} |
平方差公式
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sin(x+y)sin(x−y)=sin 2x−sin 2y=cos 2y−cos 2x{\displaystyle \sin(x+y)\sin(x-y)=\sin ^{2}{x}-\sin ^{2}{y}=\cos ^{2}{y}-\cos ^{2}{x}\,}
cos(x+y)cos(x−y)=cos 2x−sin 2y=cos 2y−sin 2x{\displaystyle \cos(x+y)\cos(x-y)=\cos ^{2}{x}-\sin ^{2}{y}=\cos ^{2}{y}-\sin ^{2}{x}\,}
(可藉由積化和差公式+2倍角公式推導而來)
其他恆等式
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如果 x+y+z=n π{\displaystyle x+y+z=n\pi }， 那么 tanx+tany+tanz=tanx tany tanz{\displaystyle \tan x+\tan y+\tan z=\tan x\tan y\tan z}cotx coty+coty cotz+cotz cotx=1{\displaystyle \cot x\cot y+\cot y\cot z+\cot z\cot x=1}如果 x+y+z=n π+π 2{\displaystyle x+y+z=n\pi +{\frac {\pi }{2}}}， 那么 tanx tany+tany tanz+tanz tanx=1{\displaystyle \tan x\tan y+\tan y\tan z+\tan z\tan x=1}cotx+coty+cotz=cotx coty cotz{\displaystyle \cot x+\cot y+\cot z=\cot x\cot y\cot z}如果 x+y+z=π{\displaystyle x+y+z=\pi }， 那么 sin2 x+sin2 y+sin2 z=4 sinx siny sinz{\displaystyle \sin 2x+\sin 2y+\sin 2z=4\sin x\sin y\sin z}sinx+siny+sinz=4 cosx 2 cosy 2 cosz 2{\displaystyle \sin x+\sin y+\sin z=4\cos {\frac {x}{2}}\cos {\frac {y}{2}}\cos {\frac {z}{2}}}cosx+cosy+cosz=1+4 sinx 2 siny 2 sinz 2{\displaystyle \cos x+\cos y+\cos z=1+4\sin {\frac {x}{2}}\sin {\frac {y}{2}}\sin {\frac {z}{2}}}
托勒密定理
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如果 w+x+y+z=π{\displaystyle w+x+y+z=\pi }（半圆）那么: sin(w+x)sin(x+y)=sin(x+y)sin(y+z)=sin(y+z)sin(z+w)=sin(z+w)sin(w+x)=sinw siny+sinx sinz{\displaystyle {\begin{aligned}\sin(w+x)\sin(x+y)&=\sin(x+y)\sin(y+z)\&=\sin(y+z)\sin(z+w)\&{}=\sin(z+w)\sin(w+x)\&{}=\sin w\sin y+\sin x\sin z\end{aligned}}}
（前三个等式是一般情况；第四个是本质。）
三角函數與雙曲函數的恆等式
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利用三角恒等式的指數定義和雙曲函數的指數定義即可求出下列恆等式：
e i x=cosx+i sinx,e−i x=cosx−i sinx{\displaystyle e^{ix}=\cos x+i\;\sin x,\;e^{-ix}=\cos x-i\;\sin x}
e x=coshx+sinhx,e−x=coshx−sinhx{\displaystyle e^{x}=\cosh x+\sinh x!,\;e^{-x}=\cosh x-\sinh x!}
所以
coshi x=1 2(e i x+e−i x)=cosx{\displaystyle \cosh ix={\tfrac {1}{2}}(e^{ix}+e^{-ix})=\cos x}
sinhi x=1 2(e i x−e−i x)=i sinx{\displaystyle \sinh ix={\tfrac {1}{2}}(e^{ix}-e^{-ix})=i\sin x}
下表列出部分的三角函數與雙曲函數的恆等式：
| 三角函數 | 雙曲函數 |
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| sinθ=−i sinhi θ{\displaystyle \sin \theta =-i\sinh {i\theta }\,} | sinhθ=i sin(−i θ){\displaystyle \sinh {\theta }=i\sin {(-i\theta )}\,} |
| cosθ=coshi θ{\displaystyle \cos {\theta }=\cosh {i\theta }\,} | coshθ=cos(−i θ){\displaystyle \cosh {\theta }=\cos {(-i\theta )}\,} |
| tanθ=−i tanhi θ{\displaystyle \tan \theta =-i\tanh {i\theta }\,} | tanhθ=i tan(−i θ){\displaystyle \tanh {\theta }=i\tan {(-i\theta )}\,} |
| cotθ=i cothi θ{\displaystyle \cot {\theta }=i\coth {i\theta }\,} | cothθ=−i cot(−i θ){\displaystyle \coth \theta =-i\cot {(-i\theta )}\,} |
| secθ=sechi θ{\displaystyle \sec {\theta }=\operatorname {sech} {\,i\theta }\,} | sechθ=sec(−i θ){\displaystyle \operatorname {sech} {\theta }=\sec {(-i\theta )}\,} |
| cscθ=i cschi θ{\displaystyle \csc {\theta }=i\;\operatorname {csch} {\,i\theta }\,} | cschθ=−i csc(−i θ){\displaystyle \operatorname {csch} \theta =-i\csc {(-i\theta )}\,} |
其他恆等式：
coshi x=1 2(e i x+e−i x)=cosx{\displaystyle \cosh ix={\tfrac {1}{2}}(e^{ix}+e^{-ix})=\cos x}sinhi x=1 2(e i x−e−i x)=i sinx{\displaystyle \sinh ix={\tfrac {1}{2}}(e^{ix}-e^{-ix})=i\sin x}cosh(x+i y)=cosh(x)cos(y)+i sinh(x)sin(y){\displaystyle \cosh(x+iy)=\cosh(x)\cos(y)+i\sinh(x)\sin(y)\,}sinh(x+i y)=sinh(x)cos(y)+i cosh(x)sin(y){\displaystyle \sinh(x+iy)=\sinh(x)\cos(y)+i\cosh(x)\sin(y)\,}tanhi x=i tanx{\displaystyle \tanh ix=i\tan x\,}coshx=cosi x{\displaystyle \cosh x=\cos ix\,}sinhx=−i sini x{\displaystyle \sinh x=-i\sin ix\,}tanhx=−i tani x{\displaystyle \tanh x=-i\tan ix\,}
线性组合
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对于某些用途，知道同样周期但不同相位移动的正弦波的任何线性组合是有相同周期但不同相位移动的正弦波是重要的。在正弦和余弦波的线性组合的情况下，我们有
a sinx+b cosx=a 2+b 2⋅sin(x+φ)(a>0){\displaystyle a\sin x+b\cos x={\sqrt {a^{2}+b^{2}}}\cdot \sin(x+\varphi )\;(a>0)}
这里的
φ=arctan(b a){\displaystyle \varphi =\arctan \left({\frac {b}{a}}\right)}
这个公式也叫辅助角公式或李善兰公式。更一般的说，对于任何相位移动，我们有
a sinx+b sin(x+α)=c sin(x+β)(a+b cosx>0){\displaystyle a\sin x+b\sin(x+\alpha )=c\sin(x+\beta )\;(a+b\cos x>0)}
这里
c=a 2+b 2+2 a b cosα,{\displaystyle c={\sqrt {a^{2}+b^{2}+2ab\cos \alpha }},}
而
β=arctan(b sinα a+b cosα){\displaystyle \beta =\arctan \left({\frac {b\sin \alpha }{a+b\cos \alpha }}\right)}
反三角函数
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主条目：反三角函数
arcsinx+arccosx=π 2{\displaystyle \arcsin x+\arccos x={\frac {\pi }{2}}\;}arctanx+arccotx=π 2.{\displaystyle \arctan x+\operatorname {arccot} x={\frac {\pi }{2}}.\;}arctanx+arctan1 x={π 2,if x>0−π 2,if x<0{\displaystyle \arctan x+\arctan {\frac {1}{x}}=\left{{\begin{matrix}{\frac {\pi }{2}},&{\mbox{if }}x>0\-{\frac {\pi }{2}},&{\mbox{if }}x<0\end{matrix}}\right.}arctanx+arctany=arctanx+y 1−x y+{π,if x,y>0−π,if x,y<0 0,otherwise{\displaystyle \arctan x+\arctan y=\arctan {\frac {x+y}{1-xy}}+\left{{\begin{matrix}\pi ,&{\mbox{if }}x,y>0\-\pi ,&{\mbox{if }}x,y<0\0,&{\mbox{otherwise }}\end{matrix}}\right.}
sin(arccosx)=1−x 2{\displaystyle \sin(\arccos x)={\sqrt {1-x^{2}}}\,}sin(arctanx)=x 1+x 2{\displaystyle \sin(\arctan x)={\frac {x}{\sqrt {1+x^{2}}}}}cos(arctanx)=1 1+x 2{\displaystyle \cos(\arctan x)={\frac {1}{\sqrt {1+x^{2}}}}}cos(arcsinx)=1−x 2{\displaystyle \cos(\arcsin x)={\sqrt {1-x^{2}}}\,}tan(arcsinx)=x 1−x 2{\displaystyle \tan(\arcsin x)={\frac {x}{\sqrt {1-x^{2}}}}}tan(arccosx)=1−x 2 x{\displaystyle \tan(\arccos x)={\frac {\sqrt {1-x^{2}}}{x}}}
无限乘积公式
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为了用于特殊函数，有下列三角函数無窮乘積公式：
sinx=x∏n=1∞(1−x 2 π 2 n 2){\displaystyle \sin x=x\prod {n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)}sinhx=x∏n=1∞(1+x 2 π 2 n 2){\displaystyle \sinh x=x\prod {n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}n^{2}}}\right)}sinx x=∏n=1∞cos(x 2 n){\displaystyle {\frac {\sin x}{x}}=\prod {n=1}^{\infty }\cos \left({\frac {x}{2^{n}}}\right)}cosx=x∏n=1∞(1−x 2 π 2(n−1 2)2){\displaystyle \cos x=x\prod {n=1}^{\infty }\left(1-{\frac {x^{2}}{\pi ^{2}(n-{\frac {1}{2}})^{2}}}\right)}coshx=∏n=1∞(1+x 2 π 2(n−1 2)2){\displaystyle \cosh x=\prod {n=1}^{\infty }\left(1+{\frac {x^{2}}{\pi ^{2}(n-{\frac {1}{2}})^{2}}}\right)}|sinx|=1 2∏n=0∞|tan(2 n x)|2 n+1{\displaystyle |\sin x|={\frac {1}{2}}\prod {n=0}^{\infty }{\sqrt[{2^{n+1}}]{\left|\tan \left(2^{n}x\right)\right|}}}
微積分
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正弦的微分
由于技术原因，此旧版图表已停用，並須迁移至新版图表。造成您的不便，我們深表歉意。
正弦（藍色）、正弦的微分（橘色），其中，正弦的微分正好是餘弦。
餘弦的微分
由于技术原因，此旧版图表已停用，並須迁移至新版图表。造成您的不便，我們深表歉意。
餘弦（藍色）、餘弦的微分（橘色），其中，餘弦的微分正好是正弦的對x軸的鏡射。
在微積分中，下面陳述的關係要求角用弧度來度量；如果用其他方式比如角度來這些關係會變得更加複雜。如果三角函數以幾何的方式來定義，它們的導數可以通過驗證兩個極限而找到。第一個是：
lim x→0 sinx x=1{\displaystyle \lim _{x\rightarrow 0}{\frac {\sin x}{x}}=1}
可以使用單位圓和夾擠定理來驗證。如果用洛必達法則來证明這個極限，那也就用這個極限證明了正弦的导数是餘弦，並因此在應用洛必達法則中使用正弦的導數是餘弦的事實，就是邏輯謬論中的循環論證了。第二個極限是：
lim x→0 cosx−1 x=0{\displaystyle \lim _{x\rightarrow 0}{\frac {\cos x-1}{x}}=0}
使用恆等式 tanx 2=1−cosx sinx{\displaystyle \tan {\frac {x}{2}}=1-{\frac {\cos x}{\sin x}}}驗證。已經確立了這兩個極限，你可以使用導數的極限定義和加法定理來證明 sin′x=cosx{\displaystyle \sin 'x=\cos x}和 cos′x=−sinx{\displaystyle \cos 'x=-\sin x}。如果正弦和餘弦函數用它們的泰勒級數來定義，則導數可以通過冪級數逐項微分得到。
d d x sin(x)=cos(x){\displaystyle {d \over dx}\sin(x)=\cos(x)}
結果的三角函數可以使用上述恆等式和微分規則來做微分。
d d x sinx=cosx,d d x arcsinx=1 1−x 2 d d x cosx=−sinx,d d x arccosx=−1 1−x 2 d d x tanx=sec 2x,d d x arctanx=1 1+x 2 d d x cotx=−csc 2x,d d x arccotx=−1 1+x 2 d d x secx=tanx secx,d d x arcsecx=1|x|x 2−1 d d x cscx=−cscx cotx,d d x arccscx=−1|x|x 2−1{\displaystyle {\begin{aligned}{d \over dx}\sin x&=\cos x,&{d \over dx}\arcsin x&={1 \over {\sqrt {1-x^{2}}}}\\{d \over dx}\cos x&=-\sin x,&{d \over dx}\arccos x&=-{1 \over {\sqrt {1-x^{2}}}}\\{d \over dx}\tan x&=\sec ^{2}x,&{d \over dx}\arctan x&={1 \over 1+x^{2}}\\{d \over dx}\cot x&=-\csc ^{2}x,&{d \over dx}\operatorname {arccot} x&=-{1 \over 1+x^{2}}\\{d \over dx}\sec x&=\tan x\sec x,&{d \over dx}\operatorname {arcsec} x&={1 \over |x|{\sqrt {x^{2}-1}}}\\{d \over dx}\csc x&=-\csc x\cot x,&{d \over dx}\operatorname {arccsc} x&=-{1 \over |x|{\sqrt {x^{2}-1}}}\end{aligned}}}
在三角函數積分表中可以找到積分恆等式。
蘊涵
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三角函數（正弦和餘弦）的微分是同樣兩個函數線性組合的事實在很多數學領域包括微分方程和傅立葉變換中是重要的基本原理。
指数定义
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| 函数 | 反函数 |
--- |
| sinθ=e i θ−e−i θ 2 i{\displaystyle \sin \theta ={\frac {e^{{i}\theta }-e^{-{i}\theta }}{2{i}}}\,} | arcsinx=−i ln(i x+1−x 2){\displaystyle \arcsin x=-{i}\ln \left({i}x+{\sqrt {1-x^{2}}}\right)\,} |
| cosθ=e i θ+e−i θ 2{\displaystyle \cos \theta ={\frac {e^{{i}\theta }+e^{-{i}\theta }}{2}}\,} | arccosx=−i ln(x+x 2−1){\displaystyle \arccos x=-{i}\ln \left(x+{\sqrt {x^{2}-1}}\right)\,} |
| tanθ=e i θ−e−i θ i(e i θ+e−i θ){\displaystyle \tan \theta ={\frac {e^{{i}\theta }-e^{-{i}\theta }}{{i}(e^{{i}\theta }+e^{-{i}\theta })}}\,} | arctanx=i 2 ln(i+x i−x){\displaystyle \arctan x={\frac {i}{2}}\ln \left({\frac {{i}+x}{{i}-x}}\right)\,} |
| cscθ=2 i e i θ−e−i θ{\displaystyle \csc \theta ={\frac {2{i}}{e^{{i}\theta }-e^{-{i}\theta }}}\,} | arccscx=−i ln(i x+1−1 x 2){\displaystyle \operatorname {arccsc} x=-{i}\ln \left({\tfrac {i}{x}}+{\sqrt {1-{\tfrac {1}{x^{2}}}}}\right)\,} |
| secθ=2 e i θ+e−i θ{\displaystyle \sec \theta ={\frac {2}{e^{{i}\theta }+e^{-{i}\theta }}}\,} | arcsecx=−i ln(1 x+1−i x 2){\displaystyle \operatorname {arcsec} x=-{i}\ln \left({\tfrac {1}{x}}+{\sqrt {1-{\tfrac {i}{x^{2}}}}}\right)\,} |
| cotθ=i(e i θ+e−i θ)e i θ−e−i θ{\displaystyle \cot \theta ={\frac {{i}(e^{{i}\theta }+e^{-{i}\theta })}{e^{{i}\theta }-e^{-{i}\theta }}}\,} | arccotx=i 2 ln(i−x i+x){\displaystyle \operatorname {arccot} x={\frac {i}{2}}\ln \left({\frac {{i}-x}{{i}+x}}\right)\,} |
| | |
| cis θ=e i θ{\displaystyle \operatorname {cis} \,\theta =e^{{i}\theta }\,} | arccis x=−i lnx{\displaystyle \operatorname {arccis} \,x=-{i}\ln x\,} |
| | |
| sinhθ=e θ−e−θ 2{\displaystyle \sinh \theta ={\frac {e^{\theta }-e^{-\theta }}{2}}\,} | arcsinh x=ln(x±x 2+1){\displaystyle \operatorname {arcsinh} \,x=\ln \left(x\pm {\sqrt {x^{2}+1}}\right)\,} |
| coshθ=e θ+e−θ 2{\displaystyle \cosh \theta ={\frac {e^{\theta }+e^{-\theta }}{2}}\,} | arccosh x=ln(x±x 2−1)=±ln(x+x 2−1){\displaystyle \operatorname {arccosh} \,x=\ln \left(x\pm {\sqrt {x^{2}-1}}\right)=\pm \ln \left(x+{\sqrt {x^{2}-1}}\right)\,} |
| tanhθ=sinhθ coshθ=e θ−e−θ e θ+e−θ{\displaystyle \tanh \theta ={\frac {\sinh \theta }{\cosh \theta }}={\frac {e^{\theta }-e^{-\theta }}{e^{\theta }+e^{-\theta }}}\,} | arctanh x=1 2 ln(1+x 1−x){\displaystyle \operatorname {arctanh} \,x={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)\,} |
参见
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余弦定理
三角函数
正弦定理
正切定理
中线长公式
角平分线长公式
雙曲函數恆等式
三分之一角公式
註釋
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^由于欧拉公式的证明过程中使用了棣莫弗公式，而棣莫弗公式的证明过程中使用了和角公式，故使用欧拉公式证明和角公式会造成循环论证，故而此方法僅为檢定方法，而非严谨的证明方法。对于类似方法也应注意甄别。
參考文獻
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引用
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^Abramowitz and Stegun, p.78, 4.3.147
^ 2.02.1苏学孟. 求三角函数乘积的常用方法. 中学数学教学. 1995, (6) [2014-12-27]. （原始内容存档于2014-12-27）.
^Abramowitz and Stegun, p.75, 4.3.89–90
^Abramowitz and Stegun, p.85, 4.5.68–69
来源
[编辑]
A one-page proof of many trigonometric identities using Euler's formula，by Connelly Barnes.
Useful Formulae for Hong Kong Advanced Level Examination Pure Mathematics.
| 查 论 编 三角函数 |
| 基本函數 | 正弦 餘弦 正切 餘切 正割 餘割 | |
| 反函數 | 反正弦 反餘弦 反正切 反餘切 反正割 反餘割 |
| 少見函數 | 正矢 餘矢 正弧 餘弧 半正矢 半餘矢 外正割 外餘割 |
| 函數分類 | | 正函數（章節） | 正角 正弦 正切 正割 正矢 正弧 半正矢 外正割 | | 餘函數 | 餘角 餘弦 餘切 餘割 餘矢 餘弧 半餘矢 外餘割 | |
| 其他函數 | 極正弦 弦函數 arc函數 cis函數 餘cis函數 cas函數 arg函數 atan2 古德曼函數 |
| 符號 | sin cos tan cot sec csc exs exc ver cvs vcs cvc hav hcv hvc hcc arc crd sag apo（英语：Apothem） cis cas arg |
| 双曲函数 | 雙曲正弦 雙曲餘弦 |
| 相關概念 | 割圓八線 同界角 辐角 單位圓 圓心角 週期性 |
| 定理 | 正弦定理 餘弦定理 正切定理 餘切定理 半正矢定理 勾股定理 |
| 恒等式 | 三角函數恆等式 雙曲三角函數恆等式 正切半角公式 三分之一角公式 餘函數恆等式 诱导公式 半正矢公式 |
| 其他 | 三角函數精確值 三角函数积分表 三角函数表 三角多項式 三角函数线 正弦曲線 雙曲三角函數 |
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188583 | https://teachy.ai/en/summaries/high-school-en-ca/10th-grade-ca/mathematics-ca/perimeter-and-area-relationship-or-traditional-overview-19a2 | Summary of Perimeter and Area Relationship | Traditional Overview
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Perimeter and Area Relationship
Summary of Perimeter and Area Relationship
Lara from Teachy
Subject Mathematics
Mathematics
Source Teachy Original
Teachy Original
Topic Perimeter and Area Relationship
Perimeter and Area Relationship
Summary Tradisional | Perimeter and Area Relationship
Contextualization
Understanding the relationship between the perimeter and area of regular polygons is a core idea in geometry. The perimeter is simply the total length around a polygon, while the area measures the amount of space enclosed within it. For regular polygons, where every side is the same length, the math is a bit more straightforward. This concept comes into play in everyday situations—like figuring out how much material you need to fence off a property or estimating how much paint it might take to cover a surface.
Furthermore, changing the length of the sides of a regular polygon affects the perimeter and area differently. As the side lengths grow, the perimeter increases in a steady, linear fashion, but the area increases by the square of the side length. That means even a small increase in side length can lead to a surprisingly large increase in area—a fact that's really useful in fields like construction and urban planning where space and resource management are key.
To Remember!
Perimeter of Regular Polygons
The perimeter of any regular polygon is just the sum of its sides, all of which are equal in length. To find the perimeter, you simply multiply the length of one side by the total number of sides. This basic idea is practical for tasks like measuring distances around areas or determining how much edging material to buy.
For instance, if you have a square with each side measuring 5 cm, then the perimeter is 5 cm x 4, which equals 20 cm. This same method applies to other regular shapes like equilateral triangles and hexagons. Knowing how to calculate perimeter is valuable not just in classrooms but also in areas such as architecture and interior design, where precise measurements are crucial.
The perimeter is the sum of all the sides of a polygon.
For regular polygons, calculate the perimeter by multiplying one side's length by the number of sides.
Because the perimeter increases in a linear manner, it grows proportionally as the side lengths increase.
Area of Regular Polygons
The area of a regular polygon measures the size of its internal surface. Different shapes have different formulas—for example, the area of a square is calculated by squaring one of its sides, while an equilateral triangle’s area is determined by using the standard triangle formula (base x height / 2).
For other shapes such as hexagons, the area can be calculated with formulas that draw on the polygon’s specific geometric properties. An example is the formula for a regular hexagon: (3 x √3 x side²) / 2. Understanding how to compute area is key for practical tasks, whether you're figuring out how much paint is needed for a mural or determining the usable space in a parcel of land. Unlike perimeter, area expands quadratically as the side lengths increase.
The area measures the internal surface of a polygon.
Different regular polygons have their own specific formulas for calculating area.
The area increases quadratically as the side lengths get longer.
Relationship Between Perimeter and Area
The interplay between the perimeter and area of regular polygons is both interesting and important. When you increase the side lengths, the perimeter goes up in a linear fashion while the area increases according to the square of that change. For example, if you double each side of a square, the perimeter doubles, but the area becomes four times as large. This difference is essential when trying to maximize space and optimize materials in real-world applications.
Using graphs and tables can help visualize how tweaking the side lengths affects both the perimeter and the area, making it easier to grasp and apply these concepts in everyday scenarios.
The perimeter increases linearly with side length.
The area increases quadratically with side length.
Grasping this relationship is key for making smart decisions about space and materials.
Practical Examples and Applications
The concepts of perimeter and area extend well beyond the classroom. In construction, for instance, knowing how to calculate the perimeter is essential when planning fences or walls, while the area helps determine how much material—like concrete or paving—is needed for a project.
In interior design, these measurements guide everything from furniture layout to selecting the right amount of carpeting or rugs. Similarly, in urban planning and landscaping, optimizing both space and materials is fundamental to creating functional and inviting environments.
Even in agriculture, calculating the area of fields is crucial for managing resources like seeds, fertilizers, and water, ensuring a robust yield. These examples underscore the practical relevance of understanding perimeter and area in various fields and everyday situations.
In construction, knowing the perimeter and area helps in planning fences and calculating material needs.
In interior design, these measurements guide furniture arrangements and decisions about floor coverings.
In agriculture, calculating the area is vital for planning resource allocation for maximum productivity.
Key Terms
Perimeter: The sum of the lengths of all sides of a polygon.
Area: A measure of the internal surface of a polygon.
Regular Polygon: A polygon with all sides and angles equal.
Linear Relationship: When one quantity changes at a constant rate with respect to another.
Quadratic Relationship: When one quantity changes in proportion to the square of another's change.
Important Conclusions
In this lesson, we looked at how the perimeter and area of regular polygons are connected. We learned that while the perimeter is the total length of all sides, the area represents the amount of space within. This understanding is essential for solving practical problems like calculating the materials required to fence a property or the paint needed to cover a surface.
We also examined how increasing the side lengths of a regular polygon affects these measures in distinct ways: the perimeter increases linearly and the area quadratically. This means a slight increase in side length can lead to a much larger increase in area—a critical insight for fields such as construction and urban planning.
Practical examples—from building fences to planning interior layouts and managing agricultural land—highlight the everyday applications of these concepts, reinforcing the importance of mastering this knowledge.
Study Tips
Review the practical examples we discussed and try applying these concepts to different regular polygons.
Experiment with graphs and tables to see how varying side lengths affects both perimeter and area, which can deepen your understanding.
Practice solving various problems involving the perimeter and area of regular polygons, using the appropriate formulas for each shape.
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188584 | https://www.youtube.com/watch?v=lZksHnEkvew | Even extension and odd extension of a function
Daniel An
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Posted: 24 Mar 2017
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Transcript:
now let's talk about even erotic tension purchases so here here's the is what I'm trying to think about let's say f of X is X on an interval 0 to 4 question is what is the even extension of effects the way you handle this is Earth throw the grass and the grass is white with an X on 0 to 4 that's what you have you have you have to scratch 0 2 4 4 and since this doesn't have a left side you can't talk about the symmetry you can't say this is an even function or odd function a however you can think about an extension of this function so they define from negative 4 to 4 so we're trying to fill it still in this unknown part here and we want to fill in this on those part so that the resulting function is an even from that's what an even in extension is right if you are the draws breath what would it be do you like this it will be a it will be a v-shaped thing because only when you have that you have a symmetry with respect to the y-axis remember being an even function means with respect to the y-axis and the symmetry about what would be this much what would be the equation for this line y equals to negative x right so if you were to write out what f of X is in such as f of X is my n if X is between 0 to 4 because that's what it was before but the key or 2 and what's going to happen the negative 4 to 0 then there you should have the graph - negative of X that's how you write the answer so in other words in other words formulas will be if you had a function f of N and you want it to extend let's say you have the function defined from x save you some number L and you want us to have an even function then what would you do you want to define the other content you want to define the other parts of F of negative x if you do that you're going to have same values whether you plug in positive x or negative x and therefore the resulting function will be even remember even function means f of X even means that whether you evaluate the function at X or evaluate the function at negative x so if you evaluate the function here or where whether you evaluated here your values the Y values are equal that's one of the even function in xx and if you think about it you just find your function this way you end up with an even function so here's another example let's say this is example one what if you had example - what if you want FX equal to easiest on say 0 to 3 what so even what to be even extension then the answer is simply f of X is when it's our when you're when your X is positive there's nothing to be done it's same as the original but if X is negative what would you do what you could use whatever the form of things e to the negative the resulting function is to draw the graph the right side of you like that and the left side will be linked it and you will end up with actually rather pointed for me not a problem it's rather it is pointed here so you don't have a well-defined tangent line at yours pointed that's the result that you get okay now let's try to turn our attention to the odd function what's the formula for odd functions this is the form of the Union moving expenses of the evening session former for the Union efficient formula for the odd extension is let's take that example three Gentiles same as thing as example two but Adi fish how would you do it that's the question and for that let me take the form formula first and hopefully you'll be convinced when you see the formula the way you take the function into an odd extension is when you take the function and you don't do anything when it's already positive it should agree with your to the one the extension of a function means you already have something that's defined and you only want to add some more information to the other side so that possesses the sufficient inventory so when it when X is negative what you want to do is you want to - so that this - makes the left side graph go under when the right side is over when the right side is under then you make it go above by by this point okay so using this formula what will be the answer so first for the stud above KC what's the odd extension of that equals to easy-access before from zero to three what if your X is between negative three to zero what do you do because negative gives up and the resulting graph will be like this result here like that but on the opposite side like this there is a slight technicality here the resulting function can't say it's odd because that of this value here here's a closed circle and is this really had an open circle so in terms of the graph it's not perfectly symmetric with respect to the origin if if he wants an odd extension than station saw either both the open circles or both these closed circles so in the slide technique ology but six too much about that so it's almost in the aqua I think the pinched hooks when they ask you these questions what they tend to do is fatal define a function of the closed interval was rather on the open interval so it's zero three and without the endpoints what you get actually start with open circle so that is one more for the open circle there's like a minor thing there but overall this is a what you do if you have a function you want to find the products in the file comments here what would you get into this function of chicken table extinction let's think about this if we took the odd extension might exist in our odd extension then you would have your function and X doesn't change for X between 0 to 4 what would I write here yeah we negative of negative add from your native forces here for negative of negative x is a right positive x and then think about what this means it means that f of X is the set does it make sense why does it make sense because that move X equal Texas it's already are not functioning so if you had this on the right side then the natural one that you would get if you did change the definition but the extend is an interval from negative 4 to 4 that will be on our function right so there's another way to verify that this formula deep does give you an even function and also that this one does ticking on so we verify that this does make sense steel industry to give any question yes I should that's what sorry consume feasibility for you negati oh yes yes it should be strictly less that's because you can't have two different values for a function so you can plug in 0 if I had the inclusion in both cases then you would have positive 1 for the 0 here whereas this one will say swing this one but functions are not allowed to give you different values and so so thank you I shouldn't have accepted and that's why I was saying that this should be rolled |
188585 | https://math.stackexchange.com/questions/211538/common-tangent-to-two-circles | Stack Exchange Network
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Common tangent to two circles
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$\begingroup$
Find the equations of the common tangents to the 2 circles:
$$(x - 2)^2 + y^2 = 9$$ and$$(x - 5)^2 + (y - 4)^2 = 4.$$
I've tried to set the equation to be $y = ax+b$, substitute this into the 2 equations and set the discriminant to zero, we then get a simultaneous quadratic equations. But they are really difficult to solve. So is there any simpler way to do this? Thank you.
geometry
analytic-geometry
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edited Apr 18, 2013 at 8:24
JSCBJSCB
asked Oct 12, 2012 at 8:39
JSCBJSCB
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$\endgroup$
5
1
$\begingroup$ Do you know that the slope of the tangent to a graph is given by the derivative? $\endgroup$
Dennis Gulko
– Dennis Gulko
2012-10-12 08:49:05 +00:00
Commented Oct 12, 2012 at 8:49
$\begingroup$ There are, in general, four common tangents to two circles, so you would expect to have four solutions. I would draw a diagram and use some geometry - because these two circles seem to touch, and the tangent at that point will be perpendicular to the line joining the two centres. There are two other common tangents, but if the question is asking for one, I think it would be this one. $\endgroup$
Mark Bennet
– Mark Bennet
2012-10-12 08:57:36 +00:00
Commented Oct 12, 2012 at 8:57
$\begingroup$ @ Dennis Gulko: Please don't use calculus. @ Mark Bennet: I know these're 3 equations. $\endgroup$
JSCB
– JSCB
2012-10-12 09:05:45 +00:00
Commented Oct 12, 2012 at 9:05
$\begingroup$ Have you tried drawing a diagram, including the line joining the centres extended to meet two of the tangents, and for these two (more difficult) tangents, considering (a) obvious symmetry and (b) similar triangles. All the algebraic manipulation reduces to this anyway. I think this approach (which may involve tan(a+b)) clarifies what is at issue. $\endgroup$
Mark Bennet
– Mark Bennet
2012-10-12 21:47:19 +00:00
Commented Oct 12, 2012 at 21:47
$\begingroup$ The dual conic to $(x-2)^2+y^2-9=0$ is $-9Y^2-5X^2+4X+1=0$ and the dual conic of $(x-5)^2+(y-4)^2-4=0$ is $12Y^2+40XY+8Y+21X^2+10X+1=0.$ The intersection points of the dual gives the common tangents of the two circles: $$x(-\frac{1}{7}) + y (-\frac{4}{21})+1=0,\ x(-\frac{8\,\sqrt{6}-7}{67})+y\frac{22\,\sqrt{6}-36}{201}+1=0 ,\ x \frac{8\,\sqrt{6}+7}{67} +y(-\frac{22\,\sqrt{6}+36}{201})+1=0.$$ $\endgroup$
Jan-Magnus Økland
– Jan-Magnus Økland
2021-03-11 11:58:40 +00:00
Commented Mar 11, 2021 at 11:58
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6 Answers 6
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$\begingroup$
A line with equation $ax+y+c=0$ is tangent to a circle iff its distance from the center of the circle is equal to the radius of the circle, so you get the system of equations:
$$\left{\begin{array}{rcl}|2a+c| &=& 3\sqrt{a^2+1} \ |5a+4+c| &=& 2\sqrt{a^2+1}. \end{array}\right.$$
with $4$ solutions.
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edited Jan 23, 2019 at 16:24
amWhy
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answered Oct 12, 2012 at 10:03
Jack D'AurizioJack D'Aurizio
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$\begingroup$ Can you explain these 2 equations a bit more? $\endgroup$
JSCB
– JSCB
2012-10-13 05:01:43 +00:00
Commented Oct 13, 2012 at 5:01
$\begingroup$ The distance between a line with equation $ax+by+c=0$ and a point $(x_0,y_0)$ is given by $\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}. In this particular case, the distance between the centers of the two circles is equal to the sum of their radiuses, so the two circles are externally tangent, and the intersection of the two externally tangent lines is the exterior homotetic center of the two circles. $\endgroup$
Jack D'Aurizio
– Jack D'Aurizio
2012-10-15 12:05:17 +00:00
Commented Oct 15, 2012 at 12:05
$\begingroup$ are you sure it's 5a+4 not 5a+4+c? $\endgroup$
Steve
– Steve
2017-08-10 04:12:11 +00:00
Commented Aug 10, 2017 at 4:12
$\begingroup$ @AristonX: you are right, now fixed. $\endgroup$
Jack D'Aurizio
– Jack D'Aurizio
2017-08-10 13:23:27 +00:00
Commented Aug 10, 2017 at 13:23
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You can approach this with homogeneous coordinates. A circle with equation $(x-x_c)^2+(y-y_c)^2 - r^2 = 0$ is represented by a 3x3 matrix as
$$ {\rm Circle}(x_c,y_c,r) = \begin{bmatrix} 1 & 0 & -x_c \ 0 & 1 & -y_c \ -x_c & -y_c & x_c^2+y_x^2-r^2 \end{bmatrix} $$
This means that the equation for a circle $C_1 = \begin{bmatrix} 1 & 0 & -2 \ 0 & 1 & 0 \ -2 & 0 & -5 \end{bmatrix}$ is given by the quadratic form
$$ P^\top C_1 P = 0 $$ $$ x^2-4 x+y^2-5 = 0 $$
which is the equation for the first circle when expanded out, and $P=\begin{pmatrix} x&y&1 \end{pmatrix} ^\top $ is an arbitrary point.
The second circle is $ C_2 = \begin{bmatrix} 1 & 0 & -5 \ 0 & 1 & -4 \ -5 & -4 & 37 \end{bmatrix} $.
Now here is the fun stuff. A line in this notation in general is defined as $L=\begin{vmatrix}a&b&c\end{vmatrix}^\top$ such that the equation of the line is
$$ P^\top L =0 $$ $$ a x+b y+c = 0 $$
Actually $a$, $b$ above designate the direction of the line such that if the line makes an angle $\theta$ with the horizontal then the line is $L=\begin{vmatrix}-\sin\theta&\cos\theta&-d\end{vmatrix}^\top$ and $d$ is the distance of the line to the origin.
We are using the above information to find the lines that are tangent to both circles. A tangent line to the first circle has satisfies the equation
$$ L^\top C_1^{-1} L =0 $$ $$ d = \pm 3 -2 \sin \theta $$
with the two possible line equations
$$ L_A = \begin{vmatrix}-\sin\theta_A&\cos\theta_A &2\sin\theta_A-3\end{vmatrix}^\top $$ $$ L_B = \begin{vmatrix}-\sin\theta_B&\cos\theta_B &2\sin\theta_B+3\end{vmatrix}^\top $$
to find the orientations of these lines $\theta_A$ and $\theta_B$ we have to find the lines that are tangent to the second circle, and match the coefficients
$$ L^\top C_2^{-1} L =0 $$ $$ d = \pm 2 + 4 \cos \theta -5 \sin \theta $$
with also two possible line equations
$$ L_A = \begin{vmatrix}-\sin\theta_A&\cos\theta_A &-4\cos\theta_A+5\sin\theta_A-2\end{vmatrix}^\top $$ $$ L_B = \begin{vmatrix}-\sin\theta_B&\cos\theta_B &-4\cos\theta_B+5\sin\theta_B+2\end{vmatrix}^\top $$
Setting $L_A=L_A$ and solving for $\theta_A$ yields the following
$$ 2\sin\theta_A-3 = -4\cos\theta_A+5\sin\theta_A-2 $$ $$ 4\cos\theta_A-3\sin\theta_A =1 $$ $$ \sin\theta_A = \frac{8\sqrt{6}-3}{25} $$
with the solution
$$ L_A = \begin{vmatrix} \frac{3-8\sqrt{6}}{25} & \frac{4+6\sqrt{6}}{25} & \frac{16 \sqrt{6}-81}{25} \end{vmatrix} $$ $$ \left( \frac{3-8\sqrt{6}}{25}\right) x + \left(\frac{4+6\sqrt{6}}{25}\right) y + \left(\frac{16 \sqrt{6}-81}{25}\right) = 0 $$ $$ -0.664 x + 0.7479 y - 1.6723 = 0 $$
Similarly with $L_B=L_B$ yielding $4\cos\theta_B-3\sin\theta_B=-1$ or
$$ L_B = \begin{vmatrix} -\frac{3+8\sqrt{6}}{25} & \frac{6\sqrt{6}-4}{25} & \frac{16 \sqrt{6}+81}{25} \end{vmatrix} $$ $$ -\left(\frac{3+8\sqrt{6}}{25}\right) x + \left(\frac{6\sqrt{6}-4}{25}\right) y + \left(\frac{16 \sqrt{6}+81}{25}\right) = 0 $$ $$ -0.904 x + 0.4278 y + 4.808 = 0 $$
Here are the results plotted in GeoGebra for validation.
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answered Oct 12, 2012 at 21:20
John AlexiouJohn Alexiou
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The parametric equation of $(x-a)^2+(y-b)^2=r^2$ is $(x=a+r\cos C,y=b+r\sin C)$
Let $$\frac{y-(b+r\sin C)}{x-(a+r\cos C)}=m$$ be a tangent at $(a+r\cos C,b+r\sin C)$, then the distance of line from the center is equal to the radius.
$$r=\frac{\mid m(a-a-r\cos C)-b+b+r\sin C\mid}{\sqrt{m^2+1}}=\frac{r\mid\sin C-m\cos C\mid}{\sqrt{m^2+1}}$$
$m^2+1=(\sin C-m\cos C)^2$
$\implies (m\sin C+\cos C)^2=0\implies m=-\frac{\cos C}{\sin C}$
So, the equation of the tangent becomes
$$\frac{y-(b+r\sin C)}{x-(a+r\cos C)}=-\frac{\cos C}{\sin C}$$
$x\cos C+y\sin C-a\cos C-b\sin C -r=0$ (this can also be reached using calculus)
For $(x-2)^2+y^2=9,a=2,b=0,r=3$
So, equation of the tangent will be $x\cos A+y\sin A-2\cos A -3=0$
For $(x-5)^2+(y-4)^2=4,a=5,b=4,r=2$
So, equation of the tangent will be $x\cos B+y\sin B-5\cos B-4\sin B -2=0$
For common tangent, these two lines must be same,
So, $$\frac{\cos A}{\cos B}=\frac{\sin A}{\sin B}=\frac{2\cos A+3}{5\cos B+4\sin B+2}$$
$\frac{\cos A}{\cos B}=\frac{\sin A}{\sin B}\implies \sin(A-B)=0$
$\implies A=B$ or $A=\pi+B$
(1)If $A=B,1=\frac{2\cos B+3}{5\cos B+4\sin B+2}\implies 4\sin B+3\cos B=1$
(2)If $A=\pi+B,\cos A=\cos(\pi+B)=-\cos B, -1=\frac{-2\cos B+3}{5\cos B+4\sin B+2}$
For (1), $4\sin B+3\cos B=1$
Putting $4=R\sin D,3=R\cos D\implies R=5,D=cos^{-1}\frac 3 5,$
$\cos(B-D)=\frac 1 5, B-\cos^{-1}\frac 3 5 =2n\pi \pm \cos^{-1}\frac 1 5$ where $n$ is any integer.
$\cos B=\frac{3 \pm 8\sqrt{6}}{25},\sin B$ can be calculated uniquely using (1).
So, there will be two tangent in this case.
For(2) $ 3\cos B+4\sin B=-5$
Applying the same approach like in (1), $\cos(B-\cos^{-1}\frac 3 5)=-1=\cos \pi, B=\cos^{-1}\frac 3 5+\pi$ $\cos B=\cos(\cos^{-1}\frac 3 5+\pi)=-\cos(\cos^{-1}\frac 3 5)=-\frac 3 5$
$\sin B$ becomes $-\frac 4 5$
So, the tangent becomes, $(x-5)(-\frac 3 5)+(y-4)(-\frac 4 5) -2=0$
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edited Oct 13, 2012 at 9:08
answered Oct 12, 2012 at 10:02
lab bhattacharjeelab bhattacharjee
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For the circles $$(x - 2)^2 + y^2 = 9\text{ and } (x - 5)^2 + (y - 4)^2 = 4,$$
using Article $180$ of The elements of coordinate geometry by Loney,
Case $1:$ If $T_2$ is the point dividing internally the line joining the centres in the ratio $2:3$ then its coordinates are $$\left(\frac{2\cdot2+5\cdot3}{2+3},\frac{2\cdot0+4\cdot3}{2+3}\right)=\left(\frac{19}5,\frac{12}5\right)$$
Now, the equation of any line passing through $T_2\left(\frac{19}5,\frac{12}5\right)$ is $$\frac{y-\frac{12}5}{x-\frac{19}5}=m\implies 5mx-5y+12-19m=0$$ where $m$ is the gradient
If it is the tangent of the circle, the distance of this line from the centre of each circle will be equal to the radius of the respective circle.
$$\implies \left|\frac{5m\cdot 2-5\cdot0+12-19m}{\sqrt{(5m)^2+(-5)^2}}\right|=3\implies |12-9m|=15\sqrt{m^2+1}$$
$$\text{On squaring and re-arrangement, } 16m^2+24m+9=0\implies m=-\frac34$$
So, there will be only one crossed common tangent
Case $2:$ If $T_1$ is the point dividing externally the line joining the centres in the ratio $2:3$ then its coordinates are $$\left(\frac{2\cdot2-5\cdot3}{2-3},\frac{2\cdot0-4\cdot3}{2-3}\right)=\left(11,12\right)$$
Now, the equation of any line passing through $T_1\left(11,12 \right)$ is $$\frac{y-12}{x-11}=n\implies nx-y+12-11n=0$$ where $n$ is the gradient
Can you follow the method used in Case $1$ to find the two values of $n,$ which implies there will be two simple common tangents
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answered Jul 20, 2013 at 13:44
lab bhattacharjeelab bhattacharjee
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Given two circles, determine the internal and external tangents.
Determine a line through the center of the circles. At the center of both circles determine a perpendicular line Note the points where the vertical lines intersect the circle.
TO draw external tangents pick a radius from each circle that is perpendicular to the line through the centers and have both radii on the same side of the center line. Draw a line and determine the equation of the line using where the radii hit the circle. The new line through these two points will hit the line through the center of the circles beyond the smaller circle. The point you just determined is a Vanishing Point. Erase all of the extra lines and points except for the circles, line through the centers and vanishing point. Use the vanishing point to draw the tangents to the nearby circle and continue that line on to the far circle. These same lines will also be tangent to the far circle.
For an internal bisector, draw the perpendicular to the line through the center of each circle This time pick a radius along this perpendicular line where one radius is on each side of the line through the centers. Determine the points where the radii you have drawn hits their respective circles. Draw a line from each of these two points. This line will cross the line through the centers. Mark clearly this point where the line crosses the center line. Use this point to draw the tangent to either circle. The line continued on through the point on the center line to the opposite circle will also be a tangent. Again the problem now boils down to drawing a tangent line to a circle given a point external to the circle that the tangent must pass through.
Make SURE when you use your point on the line through the centers of the circle that you have erased the perpendicular lines AND the points on the circle circumferences where the perpendicular intersected the circle. These points on the circle were NOT the point where the tangent line hits a circle, but only a step used to find the required point on the center line.
Lets say you have marked a point A on the center line and want to use it to find the tangent line from there to a circle centered around point C. It does not matter for this if point A is between the circles or on the far side of the circles. It MUST be the point you marked earlier.
Determine the midpoint M between point A and C.
Write the equation for the circle at center M with radius AM which is equal also to MC. This circle will intersect your circle with center at C at two points. A line from A to one of these points is a tangent line to the circle.
Lets pick one of these points on your circle that you found that will be the point of intersection of your tangent line and call it point P.
Draw lines AP, PC, and MP. Because of your circle with center at M, AM, MP and MC are all equal to one another. This means triangle AMP and triangle PMC are both isosceles with base angles equal to one another. Lets call the base angles in triangle AMP angle 1 and the base angles in triangle PMC angle 2. In triangle APC you find 2 times angle 1 and two times angle 2 in the triangle. The interior angles of a triangle sum to 180. Divide this equation by 2 and you have angle 1 plus angle 2 sum to 90. This makes angle APC a 90 degree angle, so the angle between the line AP and the radius is 90 degrees. This is a Tangent line.
Once you have a tangent line to one of the circles, the same tangent line can be used for the other circle. You now have a circle and a line and need a point of intersection ie solve the line equation for y and plug that into the circle equation and solve for x. This usually involves a quadratic equation with 2 solutions. Use the one on the side of the line through the circle centers that you wish.
Once you have determined the point where the tangent line meets the circle, a perpendicular line to the tangent line though the point in question should go through the center of that circle if you have made no error.
This is a slow method but the steps do not involve math beyond high school analytic geometry. This includes NOT having trigonometry in this method.
For external tangents the idea that objects may be drawn larger or smaller using a vanishing point is more often used and known in art. We found the vanishing point using radii perpendicular to the center line. These lines however were not special for this purpose. Lines at ANY given angle from the center line could have been used to get this same vanishing point. All that matters is that the one of the two objects is a copy of the other albeit magnified.
For internal tangents we used the idea that we may use a focal point between objects to invert an image and or make that image larger or smaller. The images are copies of one another though in this case they are of different sizes. Again we used lines 90 degrees from the center line, but lines at any angle from the center line could have been used. As long as you consistently pick the same angle in both circles, the points on the circumference of each circle should when connected lead toward the same point on the center line.
Review the idea of homeothety in a text book such as Keselev Geometry ie a text for high school students.
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answered Jun 9, 2017 at 10:54
Katherine RogersKatherine Rogers
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$\begingroup$ IF this is all clear as mud, start with an easier question. Draw a circle. To the right of this draw another circle that overlaps the first circle. Let the diameter of the circle to the right have its left most edge end at the center of the circle to the left. The right most edge of the diameter of the circle to the right can be anywhere along the horizontal. $\endgroup$
Katherine Rogers
– Katherine Rogers
2017-06-09 18:30:30 +00:00
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Like in my other answer, the equation of any tangent of $$(x-2)^2+y^2=3^2$$ will be $$x\cos t+y\sin t-(2\cos t+3)=0$$
Now if this has to be tangent of $(x-5)^2+(y-4)^2=4$ if the radius $=$ the distance from the center$(5,4)$
$$\dfrac{|5\cos t+4\sin t-(2\cos t+3)|}{\sqrt{\cos^2t+\sin^2t}}=2$$
$$\implies4\sin t+3\cos t-3=\pm2$$
Case $#1:+\implies4\sin t+3\cos t=5$
Use Find $4\cos\theta-3\sin\theta$, given that $4\sin \theta +3\cos \theta = 5$ to solve for $\sin t,\cos t$
Case $#2:-\implies4\sin t+3\cos t=1$
Use Weierstrass substitution to find $\sin t,\cos t$
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answered Mar 11, 2021 at 11:32
lab bhattacharjeelab bhattacharjee
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188586 | https://dmp.umw.edu.pl/pdf/2019/56/3/291.pdf | Address for correspondence Renata Samulak-Zielińska E-mail: renata.samulak.zielinska@gmail.com Funding sources None declared Conflict of interest None declared Received on January 2, 2019 Reviewed on February 20, 2019 Accepted on May 8, 2019 Published online on September 30, 2019 Abstract Patients who have suffered a heart attack often require dental treatment. The inflammation of the oral cavity not only reduces the quality of life, but also negatively affects the course of ischemic heart disease. Dental treatment in patients with a history of myocardial infarction seems complicated, since these patients require special consideration with regard to the timing and form of dental treatment as well as to the precautions required. Patients at risk of cardiac complications that are greater than the benefits of dental treatment should be identified and only the most urgent conditions should be treated. The aim of this study was to present the latest guidelines for dental treatment in patients who have suffered myocardial infarc-tion. We reviewed the available literature explaining when dental treatment can be undertaken, whether antibiotic prophylaxis is required, whether the patient can be anesthetized locally, and how to provide the maximum safety during the visit. The principles of the surgical treatment of patients receiving drugs that affect hemostasis were also reviewed.
Key words: dental care, oral anticoagulants, antibiotic prophylaxis, myocardial infarction Słowa kluczowe: opieka stomatologiczna, doustne antykoagulanty, profilaktyka antybiotykowa, zawał mięśnia sercowego Cite as Samulak-Zielińska R, Dembowska E, Lizakowski P. Dental treatment of post-myocardial infarction patients: A review of the literature. Dent Med Probl. 2019;56(3):291–298. doi:10.17219/dmp/109232 DOI 10.17219/dmp/109232 Copyright © 2019 by Wroclaw Medical University This is an article distributed under the terms of the Creative Commons Attribution 3.0 Unported License (CC BY 3.0) ( Reviews Dental treatment of post-myocardial infarction patients: A review of the literature Leczenie stomatologiczne pacjentów po przebytym zawale mięśnia sercowego – przegląd piśmiennictwa Renata Samulak-Zielińska1,A,B,D,F, Elżbieta Dembowska1,D–F, Piotr Lizakowski2,B,D–F 1 Departament of Periodontology, Pomeranian Medical University in Szczecin, Poland 2 Specialist Medical Practice in Szczecin, Poland A – research concept and design; B – collection and/or assembly of data; C – data analysis and interpretation; D – writing the article; E – critical revision of the article; F – final approval of the article Dental and Medical Problems, ISSN 1644-387X (print), ISSN 2300-9020 (online) Dent Med Probl. 2019;56(3):291–298 R. Samulak-Zielińska, E. Dembowska, P. Lizakowski. Dental treatment after myocardial infarction 292 Introduction The number of patients with general diseases requiring dental treatment is on the increase. According to the esti-mates presented in Bhateja’s report, out of 36,729 patients of Dental College and Hospital in Mathura, India, 58% had a history of cardiovascular disease.1 Such patients re-quire an individualized treatment plan and the continuous monitoring of oral health. The current state of knowledge indicates that inflammation in the oral cavity, particularly periodontitis, affects the general state of health, including the development and course of atherosclerosis. Cardiovas-cular disease has an inflammatory origin. Firstly, there is an increase in the level of pro-inflammatory mediators in response to the presence of Gram-negative lipopolysac-charides (LPSs), C-reactive protein (CRP), interleukin 1β and interleukin 6 (IL-1β and IL-6), tumor necrosis fac-tor alpha (TNF-α), fibrinogen, and matrix metallopro-teinase 9 (MMP-9). These substances contribute to the destabilization of the atherosclerotic plaque.2 Secondly, there is a cross-reaction of the patient’s antibodies with heat shock protein (HSP) present in the damaged vascu-lar endothelium and atherosclerotic plaques. This leads to the continuation of the inflammatory process, and thus to the progression of the disease. Cross-reactivity is triggered by the presence of oral bacteria Porphyromonas gingivalis and Tannerella forsythia, whose HSP is in 60% homolo-gous to HSP found in mammals.3 Thirdly, direct bacterial mechanisms (e.g., bacterial enzyme activity) contribute to the progression of cardiovascular disease. Bacterial DNA of Tanerella forsythia, Porphyromonas gingivalis, Aggrega-tibacter actinomycetemcomitans, and Prevotella interme-dia has been found in the atherosclerotic plaque.4 Finally, the concept of ‘vascular endothelial activation’ can explain the mechanism underlying inflammatory-induced athe rosclerotic plaque formation. The LPS binding, bacterial outer membrane vesicles, fimbriae, and other bacterial an-tigenic structures have an impact on the local and systemic host response. This leads to the upregulation of endothe-lial cell receptors followed by monocyte adhesion to the vascular wall. Monocytes migrate into the subendothelial space, absorb low-density lipoprotein cholesterol (LDL-C) and become foam cells. After their apoptosis, lipids are ac-cumulated in the vessel wall, covered by matrix and ac-companied by smooth muscle cell proliferation, which is induced by invasive periodontal pathogens. The enzyma tic degradation of the extracellular matrix results in plaque rupture, the exposure of prothrombotic components and subsequent thrombus formation, which ultimately leads to blood vessel occlusion.4 This results in a need for treat-ment of oral cavity diseases as well as in intensive efforts toward periodontal disease prevention in patients with cardiovascular diseases.5 The need for periodontal treatment is significant among Polish patients after myocardial infarction. With regard to the Community Periodontal Index of Treatment Needs (CPITN), 38.6% of them have a score of 3 and 46% a score of 4, meaning they are in urgent need of periodon-tal treatment.6 Questions concerning the safety of dental treatment of patients after myocardial infarction need to be addressed.
Material and methods To identify key words, a Population, Intervention, Comparison and Outcome (PICO) question was for-mulated as: What are the safety rules for dental care of patients following myocardial infarction? A compre-hensive search of the MEDLINE (PubMed), Scopus and Google Scholar electronic databases was undertaken in January 2019 to find relevant articles, using the follow-ing search terms: [dental care OR dental anesthesia OR tooth extraction] AND [myocardial infarction OR isch-emic heart disease]. The timeframe was 2000–2019. Ad-ditionally, a manual search of the bibliographies of full-text articles was also conducted. The guidelines of the American Heart Association (AHA), American College of Cardiology (ACC), European Society of Cardiology (ESC), and Polish Society of Cardiology were also re-viewed. We considered reviews, systematic reviews, guidelines and statements of dental and cardiological as-sociations, randomized controlled trials (RCTs) as well as cohort, case and cross-sectional studies. Papers with abstracts written in English or Polish were included. Let-ters, book chapters, case reports, and studies without an abstract or with no full text available were not included. Only publications addressing the protocol for the treat-ment of post-myocardial infarction patients were ana-lyzed. Articles dealing with the influence of oral inflam-mation on cardiovascular disease or the effects of dental treatment on general health were excluded. The limitations of this review include the open PICO question, the broad spectrum of relevant issues requiring urgent explanation, and the restriction to cardiological society guidelines and reviews rather than RCTs. Taking all this into account, only some of the Preferred Reporting Items for Systematic Reviews and Meta-Analyses (PRIS-MA) rules could be fulfilled. The study selection was conducted independently by 2 reviewers (PL and RSZ), with any disagreements resolved by the 3rd reviewer (ED). A total of 37 articles were included in the review. No me-ta-analyses or systematic reviews were found.
Our findings were compatible with those presented by Napeñas et al. at a panel of experts during World Work-shop on Oral Medicine VI.7 They stated that “with a lack of consensus statements, guidelines, or systematic re-views focused on these specific issues related to dental treatment for patients with cardiovascular diseases, the vast majority of current recommendations are not linked to levels of evidence and are presumably derived from ex-pert opinion” .7 Dent Med Probl. 2019;56(3):291–298 293 Appropriate time to start dental treatment in post-myocardial infarction patients Dental procedures are classified as minor surgical pro-cedures of a low cardiovascular risk. The risk of death or myocardial infarction within 30 days of a dental procedure is less than 1%. However, a patient after a heart attack is at high risk of recurrence of cardiovascular events. Only considering patients with acute ST-elevation myocardial infarction (STEMI), the risk of in-hospital death ranges from 4% to 14% and the annual mortality after a surgical intervention is 10%.8 Over 70% of relapses occur in the 1st month after the initial incident. The risk of recurrence depends on the severity of the disease, type of disease, treatment applied, and possible complications of the in-farction.5 The most burdened group of patients are those treated conservatively − currently only a small number, due to the pressure of cardiac society guidelines pertaining to ear-ly invasive treatment. In these cases, the natural course of the disease lasts for more than 6 weeks.9 This is the period needed for a post-infarction scar to form, to cre-ate collateral circulation and to restore the contractility of the damaged (but not necrotic) areas of the myocar-dium. To avoid late infarction complications, all proce-dures excluding emergency treatment should be avoided in this period. This also applies to dental surgery. Former AHA guidelines extended this period to 6 months, during which avoiding dental surgery was recommended, as the risk of complications was considered the highest in this period.10 Due to advances in cardiac management in the last 2 decades, these limitations are no longer recommended. Firstly, there is widespread access to the invasive me thods of treatment of infarctions, allowing the immediate reper-fusion of tissues, thus avoiding early and late complica-tions. In addition, on the 2nd day after myocardial infarc-tion patients are subjected to early cardiac rehabilitation; this is continued after discharge in rehabilitation centers, where patients undergo fitness tests in the 1st month of convalescence. When the patient’s test tolerance is found to be good, the risk of recurrence is considered low, and if the attending physician does not find otherwise, there are no contraindications to dental treatment.5 Pa-tients who had myocardial infarction in the past and are in a stable period of coronary heart disease do not need stress tests and can be assessed on the basis of an inter-view. In terms of metabolic equivalents of task (METs), efficiency at 4 METs is sufficient to qualify the patient for dental procedures.
Patients at risk of cardiac complications exceeding the benefits of dental treatment should be identified and only the most urgent conditions should be treated. However, pain and inflammation, which may be a consequence of avoiding dental treatment as a precaution against car-diac complications, are a source of endogenous catechol-amines that burden the already damaged heart muscle, which means caution might not always be the safest solu tion. The priority is to cure inflammation in the oral cavi ty, as it can cause pain similar to angina pectoris, with the same characteristic pain radiation. In addition, painful inflammation hinders food intake and interrupts sleep, resulting in a significant reduction in the quality of life.11 Endodontic treatment, conservative treatment, non-sur-gical periodontal treatment, or prophylactic treatment are considered procedures entailing a low risk of compli-cations. In addition, the risk decreases in stable periods of coronary heart disease or after heart failure, when the symptoms have a constant intensity, are predictable and occur only after intense physical activity. A good deter-minant of the patient’s condition is a lack of chest pain for 2 weeks and satisfactory test results.12 Most authors re commend a cautious 4–6-week period after myocardial infarction to stabilize the disease.13 During this period, the most indispensable procedures, such as extractions, the drainage of abscesses or pulpotomies can be performed in a hospital setting. After this period, unless the cardio logist recommends otherwise, complex dental treatment can be carried out.13 Antibiotic prophylaxis in post-myocardial infarction patients Another issue is antibiotic prophylaxis prior to treat-ment, associated with the risk of bacteremia in patients who have suffered myocardial infarction. Patients with angina pectoris, cardiovascular events or coronary artery bypass surgery (bypass grafts) are classified as patients at low risk of infective endocarditis, and therefore antibiotic prophylaxis is not indicated.14 The same applies to stent-injected patients. Guidelines set by the Polish Dental As-sociation and the National Antibiotic Protection Program in 2019 continue to support the 2015 guidelines of ESC on the prevention and treatment of infective endocardi-tis (Table 1). They recommend the use of antibiotics in a number of situations, including the presence of an ar-tificial prosthetic valve, the presence of artificial material used to repair the valve (e.g., a mitral ring) and cyanotic congenital heart disease. In the cases of congenital heart defects repaired with artificial material, prophylaxis is recommended for 6 months after surgery, or perma-nently if the defect has not been completely corrected and there is intracardiac leakage. The guidelines do not include a history of myocardial infarction or bypass sur-gery as posing an increased risk of infection. They limit high-risk dental procedures to those in which the conti-nuity of the mucous membranes is disrupted and the risk of injury to the gingival or periapical area is pre sent. Such procedures require antibiotic prophylaxis only in select-ed cases mentioned above.15,16 However, some authors recommend antibiotics in the case of invasive proce-dures up to 30 days after the cardiological intervention.17 R. Samulak-Zielińska, E. Dembowska, P. Lizakowski. Dental treatment after myocardial infarction 294 incidence of myocardial infarction, and late afternoon hours, when fatigue and stress levels are high, should be avoided. During dental procedures, a supine position should be avoided, as it leads to the return of blood from peripheral areas to the central circulation system and may overload pulmonary circulation. In cases of systolic heart failure following myocardial infarction, this overload may result in the aggravation of heart failure, including pulmonary edema after re-verticalization, and further contribute to orthostatic syncope.8 The patient should continue to take the medication before the appointment as directed by the attending physician. If the patient’s regular the rapy includes nitrates, the patient should bring them. In cases of anxiety disorders and stress, the administration of 5–10 mg of diazepam is recommended the night prior to the visit and 1–2 h before the treatment. In this case, the patient should not drive a motor vehicle.5 Local anesthesia in post-myocardial infarction patients Another important aspect of dental treatment is lo-cal anesthesia. If the patient’s condition is stable and the medication is taken as prescribed, there are no contra-indications for local anesthesia with adrenaline. Patients with ischemic heart disease are more vulnerable to the negative effects of the release of endogenous adrena-line as a result of severe pain during surgery than they are to a small amount of adrenaline in an anesthetic.22 A visit to the dental office is a stressful event, meaning the level of endogenous catecholamines increases more than after the administration of anesthesia.23 The level of endogenous adrenaline is naturally the highest be-tween 8 and 11 a.m., so the visits should not be in the morning.9 Vander heyden et al. showed that the highest increase in the level of adrenaline is associated with the beginning of the visit and the treatment itself.24 However, In each case, the benefits and risks of possible antibiotic usage should be balanced. It should be noted that in post-myocardial infarction patients there may be indications for antibiotic prophylaxis resulting from other concomi-tant diseases.18 The most important factors in the preven-tion of infective endocarditis are good oral hygiene and oral inflammation prophylaxis.19,20 Rules for safe dental treatment The basics of safe dental treatment of patients with cardiovascular diseases comprise a detailed medical his-tory, including complaints, allergies, medications, and specialist recommendations. It is important to monitor the patient’s condition and to interrupt the procedures when the patient becomes restless or cardiac problems arise. An angina attack can occur in the dental chair due to stress, pain and anxiety triggers.3 Pain can be felt in the jaw, from where it can radiate to the neck and throat, so in some cases, the patient and the dentist may interpret it as toothache. If the patient experiences retrosternal pain, the procedure should be interrupted, and sublingual ni-troglycerin (0.4–0.8 mg) and oxygen (3L/min) should be administered. If the pain subsides within 5 min, the ap-pointment can be continued or postponed to the next day. If the pain persists after 5 min, nitrates should be given again. If there is no improvement after 15 min since the first symptoms occurred, a re-infarction should be sus-pected, and in this situation the patient should be trans-ferred as soon as possible to an emergency department.5 Psychological and physiological stress during dental appointments has the potential to significantly alter he-modynamic stability. Therefore, a stress-reduction proto-col is suggested for post-myocardial infarction patients, including profound local anesthesia, preoperative or in-traoperative sedation and excellent post-operative analge-sia.21 The dental visit should be short − up to 30 min − and in the middle of the day. Morning hours, with the highest Table 1. Comparison of guidelines for the antibiotic prophylaxis of endocarditis AHA 2007 American Heart Association20 ESC 2015 European Society of Cardiology15 Polish Dental Association and National Antibiotic Protection Program 201916 – patients with a prosthetic heart valve or who have had a heart valve repaired with prosthetic material – patients with a history of endocarditis – patients with a heart transplant with abnormal heart valve function – patients with certain congenital heart defects, including: • cyanotic congenital heart disease (birth defects, with oxygen levels lower than normal) that has not been fully repaired, including children who have had surgical shunts and conduits • a congenital heart defect that has been completely repaired with prosthetic material or a device, for the first 6 months after the repair procedure • repaired congenital heart disease with residual defects, such as persisting leaks or abnormal flow at or adjacent to a prosthetic patch or prosthetic device – patients with any prosthetic valve, including a transcatheter valve, or those in whom any prosthetic material was used for cardiac valve repair – patients with a previous episode of infective endocarditis – patients with congenital heart defects: • any type of cyanotic congenital heart defects • any type of congenital heart defects repaired with prosthetic material, whether placed surgically or by percutaneous techniques • up to 6 months after the procedure • lifelong if a residual shunt or valvular regurgitation remains – patients with an artificial heart valve, including those implanted percutaneously, or patients who have been treated with artificial material for valve repair – patients after an episode of infective endocarditis – patients with congenital heart disease, including: • patients with a congenital cyanotic heart defect • patients after the repair of a congenital malformation of the heart with artificial material, both during surgery and using transcutaneous techniques for up to 6 months after surgery, or lifelong if residual leakage or valve regurgitation remains Dent Med Probl. 2019;56(3):291–298 295 during the administration of anesthesia and immediately after the injection, increases in adrenaline levels were not observed.24 This means that most of the adrenaline is of endogenous origin, hence the reduction of stress and good, effective anesthesia are indicated. Moreover, local anesthetics without vasoconstrictors do not pro-vide satisfactory hemostasis or anesthesia during dental procedures.25 However, it is recommended not to exceed 0.04 mg of adrenaline, which corresponds to 2 1.8-cc cartridges of an anesthetic with adrenaline at a dilution of 1:100,000.14 If it is necessary to administer more an-esthesia, subsequent portions should be administered without a vasoconstrictor. Intravascular anesthesia should be avoided.4 The use of intrapulpal and intraos-seous anesthesia is contraindicated, as this could lead to the excessive absorption of adrenaline.9 For this reason, retraction cords impregnated with adrenaline should be avoided. However, in the case of untreated, unregulated arythmias or unstable angina, vasoconstrictor substan ces are contraindicated. Similarly, caution with the use of vasoconstrictors is indicated in patients with pace-makers, especially implantable automatic defibrillators.23 Elad et al. showed that local anesthesia using 4% articaine hydrochloride with adrenalin 1:200,000 is as safe as local anesthesia with 2% lidocaine and adrenalin 1:100,000 in cardiovascular patients.26 They observed neither severe adverse effects nor cardiac ischemic changes on electro-cardiography (ECG) in either group.26 Hemostasis in post-myocardial infarction patients Today the risk of complications of dental procedures is dependent more on the anticoagulant therapy used than on the severity of coronary heart disease. Patients who have had acute myocardial infarction always take medications that affect hemostasis.27 Depending on the indications, these are either antiplatelet drugs, vitamin K antagonists or new non-vitamin K antagonist oral anticoagulants (NOACs). The discontinuation of therapy with these drugs is asso-ciated with a high risk of complications, including death. This risk far exceeds the risk of increased bleeding during and after surgery. In addition, a surgical intervention itself increases the risk of deep vein thrombosis. Double anti-platelet therapy (DAPT) is aimed at preventing thrombosis in the coronary artery. The risk of thrombosis lasts until the atherosclerotic plaque stabilizes (a process taking about 4–6 weeks) or − in the case of stent implantation − until it is covered with the vascular endothelium (the convention-al limit for metal stents is 1 month, and for coated stents 6–12 months). Double antiplatelet therapy includes acetyl-salicylic acid (ASA) and an inhibitor of the P2Y12 glycopro-tein receptor (clopidogrel, prasugrel or ticagrelor) (Table 2).28 In cases when there is also a risk of cardiac embolism (atrial fibrillation, intracardiac thrombus) or concomitant venous thromboembolism, DAPT treatment is supplemented with an oral anticoagulant.28 Table 2. Recommendations for patients receiving drugs that affect hemostasis depending on the risk of excessive bleeding after dental procedures28 Risk of bleeding in dental procedures Patients on ASA and/or clopidogrel Patients on oral anticoagulants Patients on NOACs Low risk of excessive bleeding: 1. conservative and endodontic treatment; 2. supragingival scaling; 3. periodontal pocket probing; 4. air polishing; 5. extraction of 1 tooth or teeth that are loose; 6. single implant placement; 7. laser evaporation of oral mucosa lesions.
continue therapy without any changes check if INR < 3 continue therapy without any changes Moderate risk of excessive bleeding: 1. subgingival scaling; 2. root debridement; 3. frenulectomy; 4. periodontal flap surgery; 5. guided tissue regeneration; 6. tooth extraction with flap elevation; 7. extraction of impacted teeth; 8. root resection; 9. vestibuloplasty; 10. several implant placement; 11. closed sinus lift procedure; 12. excisional or incisional biopsy.
continue therapy without any changes check if INR < 3 stop therapy with rivaroxaban – 24 h before surgery, and with dabigartan – 1–2 days before surgery restart therapy 24–48 h after surgery High risk of excessive bleeding: 1. soft tissue augmentation with free gingival grafts or connective tissue grafts; 2. placement of 6–8 implants in the edentulous alveolar ridge; 3. bilateral open sinus lift procedure; 4. oncological, orthognathic and reconstructive surgery.
continue therapy with ASA stop therapy with P2Y12 inhibitors 24–72 h before surgery in patients at high risk of thrombosis, consider glycoprotein IIa/IIIa inhibitors check if INR = 2–2.5 convert to heparin only in selected cases stop therapy with rivaroxaban – 24 h before surgery, and with dabigartan – 1–2 days before surgery restart therapy 24–48 h after surgery ASA – acetylsalicylic acid; NOACs – non-vitamin K antagonist oral anticoagulants.
R. Samulak-Zielińska, E. Dembowska, P. Lizakowski. Dental treatment after myocardial infarction 296 Coronary angiography is routinely performed in pa-tients with acute coronary syndrome, and if the coronary artery responsible for myocardial infarction is identified, a revascularization procedure is performed. Coronary an-gioplasty is the most common one, with the implantation of an anti-proliferative eluting stent, coated with a cyto-static agent that inhibits cell division. This limits the in-flammatory process and the formation of restenosis in the vessel, but at the same time slows down the epithelializa-tion of the stent and prolongs the need for DAPT. New drug-eluting stents require the use of 2 antiplatelet drugs for about 6 months, or 12 months if they are implanted due to acute coronary syndrome. This is a conventional time period to allow the vascular endothelium to grow. The earlier discontinuation of therapy may result in acute thrombosis in the stent, myocardial infarction and death. In exceptional situations, metal stents are used; their epi-thelialization takes 1 month.28 The time limits for DAPT were modified by the ESC guidelines published in 2017.28 Two scales were created to establish a safe date for ending the treatment. The PRE-CISE-DAPT score is used for stent implantation based on the results of laboratory tests (hemoglobin, leukocytes, creatinine clearance) and the patient’s data (age, history of bleeding), and indicates either a brief (3–6 months) or extended (12–24 months) use of DAPT. Longer therapy may be beneficial and may lower the risk of stent closing. However, the duration of therapy depends on individual factors (e.g., age, comorbidities, left ventricular ejec-tion fraction (LVEF), bleeding risk, or smoking) as well as on the procedure technique (implanted stent caliber, type of substance released, presence of a stent in the vein bridge).28,29 In 2016, Pruszczyk et al. published (in Polish) a paper de-tailing a protocol for preparing patients on anticoagulants for dental surgical procedures.30 It follows the guidelines set out by ESC and the European Association for Cardio-Thoracic Surgery (EACTS).28 Most dental procedures are defined as low-risk in terms of blood loss, and hemosta-sis can be achieved through the use of local hemostatic agents. There is a greater risk of blood loss associated with major reconstructive procedures, bone block transplants, implantological procedures, extractions of more than 3 teeth, treatment with the elevation of the mucoperiosteal flap, soft tissue augmentation procedures, connective tis-sue grafts, and open sinus lift procedures. The 4 criteria for abundant post-operative bleeding are: bleeding lasting longer than 12 h, forcing the patient to report to the dental office or the emergency room, hematoma or bruising, and need for a blood transfusion.30 The effect of antiplatelet therapy may double the bleed-ing time, but in most cases, it remains within the normal range or only slightly over it.30 The results of the plate-let aggregation test might be abnormal, although with-out clinical consequences.31 Prolonged bleeding time is not a major clinical problem, since hemostasis can be achieved by pressure, suturing, applying collagen sponges, or prescribing tranexamic acid. As mentioned above, the discontinuation of antiplatelet therapy is the main cause of late stent thrombosis, which can result in serious com-plications, including death (in up to 45% of cases) and significant damage to the heart.32 The safety of surgical dental procedures during antiplatelet therapy was con-firmed by Park et al.33 Among 100 patients undergoing combined antiplatelet therapy with ASA, clopidogrel and, in some cases, with the addition of cilostazol, only 2 pa-tients had increased post-operative bleeding after tooth extraction and in both, it was enough to apply pressure to stop the bleeding.31 Dodson demonstrated that the amount of bleeding measured during invasive procedures was similar in the group of patients who suspended their ASA therapy for 7 days before tooth extraction and in patients who continued their ASA therapy.32 However, Buhatem Medeiros et al. showed that patients on DAPT presented a larger volume of bleeding during invasive procedures than patients not using these medications.34 Local hemostatic methods were sufficient to control the bleeding and there were no post-operative bleeding com-plications in any of the presented cases.34 When in doubt, it is recommended to check prothrombin time (PT), par-tial thromboplastin time (PTT) and the number of plate-lets. If PT and PTT are found to be within the normal range and the number of platelets exceeds 100,000/mm3, surgery can safely be performed.31 In more complicated surgical procedures with a moderate bleeding risk, pa-tients should be maintained on aspirin, while P2Y12 in-hibitor therapy should be discontinued.30 Vitamin K antagonists warfarin and acenocoumarol are used in conditions associated with coronary heart disease, including the prophylaxis of venous thromboembolism, and in patients with arythmias, artificial valves, throm-bophilia, and antiphospholipid syndrome.35,36 In these cases, the administration of vitamin K antagonists should not be stopped before any dental procedures – including procedures of a higher risk of bleeding – if the patient’s international normalized ratio (INR) <3 24 h before the planned procedure.34 If the patient’s INR > 3, the attend-ing physician should adjust the therapy to achieve a lower INR.30 The withdrawal of oral anticoagulants does not guarantee that bleeding will not occur; serious bleeding occurs also in patients who have never taken anticoagu-lants. Only 0–3.5% of cases of excessive bleeding are so severe that they cannot be controlled with local measures. A higher risk of death or permanent disability is associ-ated with the discontinuation of anticoagulant treatment. In a relapse of venous thromboembolism, the risk of death is 6% and of permanent disability 2%. In the case of arte-rial embolism, the former risk is 20% and the latter 40%.37 It should be emphasized that there is no description in the literature of any case of death or permanent disability resulting from massive bleeding after a dental procedure in a patient taking anticoagulants. It should be borne in Dent Med Probl. 2019;56(3):291–298 297 mind that during anticoagulant therapy, the administra-tion of tetracyclines, erythromycin, clarithromycin, and metronidazole is contraindicated.9 Newer anticoagulants are direct inhibitors of factor Xa − rivaroxaban, apixaban and edoxaban, and direct thrombin inhibitor dabigatran. They are used in deep vein thrombosis, pulmonary embolism, embolism due to non-valvular atrial fibrillation, following orthopedic surgery as well as in acute coronary syndrome and venous thromboembolism. There are no unambiguous guidelines of how to proceed with the use of these drugs in planned surgical procedures. The manufacturers of these drugs recommend a break of 1 day in pharmacotherapy, extended to 2 days in cases of impaired renal function with glomerular filtration rate (GFR) lowering to 30 mL/min/1.73 m2. It is recommended that the proce-dure should be performed when the drug concentration is the lowest, i.e., 12 h or 24 h after the last dose, depending on whether the drugs are taken once or twice daily. If procedures of a high risk of bleeding are planned and the medications are taken in the morning, the dose of the drug used once a day (edoxaban) should be delayed until after the procedure, whereas in the case of the drugs taken twice daily (apixaban, dabigatran and rivaroxaban), the evening dose should be skipped. If edoxaban is taken in the evening, there is no need to skip the dose. If complete hemostasis is obtained during the procedure, the dose that was previously skipped can be taken after 6–8 h.37 In patients taking medications affect-ing hemostasis, it is recommended that dental procedures should be performed early enough during the day to allow the patient to seek help in case of prolonged bleeding. Like-wise, treatment should be carried out at the beginning of the week, as re-bleeding usually takes place after 24–48 h.38 The use of an infiltration anesthetic with a vasoconstrictor is re commended and nerve block anesthesia should be avoided if possible. If nerve block anesthesia is necessary, it should always be performed with aspiration. Following an extraction, the tooth socket should be provided with a hemostatic dressing and sutured well for 7–14 days, and gauze pad compression should be main-tained for 30 min after treatment.36 The patient should be advised not to rinse their mouth for 24 h, not to perform suction or create negative pressure in the mouth, not to touch the alveolus with the tongue or any foreign body, to avoid hot and hard food, and not to bite on the side of the procedure. The patient should be advised to apply pres-sure for 20 min with a clean gauze pad in case of bleeding, and contact the dentist if hemostasis does not occur.30,38 Conclusions Patients who have suffered a heart attack often require dental treatment. The inflammation of the oral cavity not only reduces the quality of life, but can also contrib-ute to the deterioration of the course of ischemic heart disease. In most cases, dental treatment can be under-taken 6 weeks after myocardial infarction. It is important to eliminate pain, so local anesthesia with a vasodilator in a dose not exceeding 0.04 mg should be used. Anti biotic prophylaxis is usually not required. Visits should be short and carried out in the early afternoon. The patient should be in a comfortable sitting position. If the patient complains of retrosternal pain, the procedure should be discontinued, and oxygen and nitrates should be admini stered. If there is no improvement, a re-infarction should be suspected and an ambulance should be called imme-diately. In the vast majority of cases, antiplatelet drugs and anticoagulants should not be discontinued prior to planned surgery, since there is a significantly higher risk of thromboembolism than of increased bleeding in these patients. In case of any doubts as to the patient’s health and the possibility of dental treatment, the patient should be referred to a specialist to establish an individualized treatment plan.
ORCID iDs Renata Samulak-Zielińska Elżbieta Dembowska Piotr Lizakowski References 1. Bhateja S. High prevalence of cardiovascular diseases among other medically compromised conditions in dental patients: A retrospec-tive study. J Cardiovasc Dis Res. 2012;3(2):113–116. 2. Włosowicz M, Wozakowska-Kapłon B, Górska R. Periodontitis in relation to selected parameters of cardiovascular system in the group of patients with acute myocardial infarction [in Polish]. Czas Stomatol. 2012;65(5):636–653. 3. Chaudhry S, Jaiswal R, Sachdeva S. Dental considerations in car-diovascular patients: A practical perspective. Indian Heart J. 2016;68(4):572–575. 4. Kebschull M, Demmer RT, Papapanou PN. “Gum bug, leave my heart alone!” – epidemiologic and mechanistic evidence linking periodon-tal infections and atherosclerosis. J Dent Res. 2010;89(9):879–902. 5. Cruz-Pamplona M, Jimenez-Soriano Y, Sarrión-Pérez MG. Den-tal considerations in patients with heart disease. J Clin Exp Dent. 2011;3(2):e97–e105. 6. Górski B, Włosowicz M, Dembowska E, Kręgielczak A, Górska R. More than 40% patients after cardiac infarct require immediate application of specialist periodontal treatment [in Polish]. Maga-zyn Stomatol. 2014;24(1):95–98. 7. Napeñas JJ, Kujan O, Arduino PG, et al. World Workshop on Oral Medicine VI: Controversies regarding dental management of medi cally complex patients: Assessment of current recommendations. Oral Surg Oral Med Oral Pathol Oral Radiol. 2015;120(2):207–226. 8. Pedersen F, Butrymovich V, Kelbæk H, et al. Short- and long-term cause of death in patients treated with primary PCI for STEMI. J Am Coll Cordial. 2014;64(20):2101–2108. 9. Thygesen K, Alpert JS, Jaffe AS, Simoons ML, Chaitman BR, White HD. Third universal definition of myocardial infarction. Circulation. 2012;126(16):2020–2035.
10. Research, Science and Therapy Committee, American Academy of Periodontology. Periodontal management of patients with car-diovascular diseases. J Periodontol. 2002;73(8):954–968.
11. Niwa H, Sato Y, Matsuura H. Safety of dental treatment in patients with previously diagnosed acute myocardial infarction or unstable angina pectoris. Oral Surg Oral Med Oral Pathol Oral Radiol Endod. 2000;89(1):35–41.
12. Roberts HW, Mitnitsky EF. Cardiac risk stratification for postmyocar-dial infarction dental patients. Oral Surg Oral Med Oral Pathol Oral Radiol Endod. 2001;91(6):676–681.
R. Samulak-Zielińska, E. Dembowska, P. Lizakowski. Dental treatment after myocardial infarction 298 13. Silvestre FJ, Miralles-Jorda L, Tamarit C, Gascon R. Dental manage-ment of the patient with ischemic heart disease: An update. Med Oral. 2002;7(3):222–230.
14. Napora M. Antibiotics prophylaxis for patients at high risk of infec-tious endocarditis [in Polish]. Nowa Stomatol. 2008;13(1):24–26. 15. Habib G, Lancellotti P, Antunes MJ, et al. ESC guidelines for the treatment of infective endocarditis in 2015 [in Polish]. Kardiol Pol. 2015;73(11):963–1027.
16. Kaczmarzyk T, Babiuch K, Bołtacz-Rzepkowska E, et al. Rekomen-dacje Grupy Roboczej Polskiego Towarzystwa Stomatologicznego i Narodowego Programu Ochrony Antybiotyków w zakresie stoso-wania antybiotyków w stomatologii. Warsaw, Poland: Narodowy Instytut Leków; 2019.
17. Roberts HW, Redding SW. Coronary artery stents: Review and patient-management recommendations. J Am Dent Assoc. 2000;131(6):797–801.
18. AAE Quick Reference Guide: Antibiotic Prophylaxis 2017 Update. aae_antibiotic-prophylaxis-2017update.pdf 19. Kaura A, Dworakowska D, Dworakowski R. Infective endocarditis – Cinderella in cardiology. Kardiol Pol. 2017;75(10):965–974.
20. Wilson W, Taubert KA, Gewitz M, et al.; American Heart Association Rheumatic Fever, Endocarditis, and Kawasaki Disease Committee; American Heart Association Council on Cardiovascular Disease in the Young; American Heart Association Council on Clinical Cardio logy; American Heart Association Council on Cardiovascular Sur-gery and Anesthesia; Quality of Care and Outcomes Research Inter-disciplinary Working Group. Prevention of infective endocarditis: Guidelines from the American Heart Association: A guideline from the American Heart Association Rheumatic Fever, Endocarditis, and Kawasaki Disease Committee, Council on Cardiovascular Dis-ease in the Young, and the Council on Clinical Cardiology, Council on Cardiovascular Surgery and Anesthesia, and the Quality of Care and Outcomes Research Interdisciplinary Working Group. Circula-tion. 2007;116(15):1736–1754.
21. Rose LF, Mealey B, Minsk L, Cohen DW. Oral care for patients with cardiovascular disease and stroke. J Am Dent Assoc. 2002;133(Suppl):37S–44S.
22. Laragnoit AB, Neves RS, Neves ILI, Vieira JE. Locoregional anes-thesia for dental treatment in cardiac patients: A comparative study of 2% plain lidocaine and 2% lidocaine with epinephrine (1:100,000). Clinics (Sao Paulo). 2009;64(3):177–182.
23. Margaix Muñoz M, Jiménez Soriano Y, Poveda Roda R, Sarrión G. Cardiovascular diseases in dental practice. Practical considerations. Med Oral Pathol Oral Cir Bucal. 2008;13(5):E296–E302.
24. Vanderheyden PJ, Williams RA, Sims TN. Assessment of ST segment depression in patients with cardiac disease after local anesthesia. J Am Dent Assoc. 1989;119(3):407–412.
25. Daskalov H, Dinkova A, Drangov M. Local anesthetics in patients with cardiovascular diseases. J of IMAB. 2015;21(1):728–731.
26. Elad S, Admon D, Kedmi M, et al. The cardiovascular effect of local anesthesia with articaine plus 1:200,000 adrenalin versus lido-caine plus 1:100,000 adrenalin in medically compromised cardiac patients: A prospective, randomized, double-blinded study. Oral Surg Oral Med Oral Pathol Oral Radiol Endod. 2008;105(6):725–730.
27. Skaar D, O’Connor H, Lunos S, Luepker R, Michalowicz BS. Dental procedures and risk of experiencing a second vascular event in a Medicare population. J Am Dent Assoc. 2012;143(11):1190–1198.
28. Valgimigli M, Bueno H, Byrne RA, et al.; ESC Scientific Document Group; ESC Committee for Practice Guidelines (CPG); ESC Nation-al Cardiac Societies. 2017 ESC focused update on dual antiplatelet therapy in coronary artery disease developed in collaboration with EACTS: The Task Force for dual antiplatelet therapy in coronary artery disease of the European Society of Cardiology (ESC) and of the European Association for Cardio-Thoracic Surgery (EACTS). Eur Heart J. 2018;39(3):213−260.
29. Levine GN, Bates ER, Bittl JA, et al. 2016 ACC/AHA guideline focused update on duration of dual antiplatelet therapy in patients with coronary artery disease: A report of the American College of Car-diology/American Heart Association Task Force on Clinical Practice Guidelines. J Thorac Cardiovasc Surg. 2016;152(5):1243–1275. 30. Pruszczyk P, Ciurzyński M, Opolski G, et al. Dental cardio common position for dealing anticoagulation in patients undergoing dental procedures [in Polish]. Kardiol Pol. 2016;74(1):87–98.
31. Gziut AI. Late in-stent thrombosis as a cause of acute coronary syn-drome complicated by out-of-hospital cardiac arrrest [in Polish]. Kardiol Pol. 2008;66(6):697–699. 32. Dodson T. Should acetylsalicylic acid (ASA) therapy for prevention of thromboembolic events be stopped prior to surgical extrac-tions? Evid Based Dent. 2012;13(3):89–90.
33. Park MW, Her SH, Kwon JB, et al. Safety of dental extractions in cor-onary drug-eluting stenting patients without stopping multiple antiplatelet agents. Clin Cardiol. 2012;35(4):225–230.
34. Buhatem Medeiros F, Pepe Medeiros de Rezende N, Bertoldi Fran-co J, et al. Quantification of bleeding during dental extraction in patients on dual antiplatelet therapy. Int J Oral Maxillofac Surg. 2017;46(9):1151–1157.
35. Singh S, Gupta K, Garg KN, Fuloria NK, Fuloria S, Jain T. Dental man-agement of the cardiovascular compromised patient: A clinical approach. J Young Pharm. 2017;9(4):453–456.
36. Milczarek S, Smektała T, Jędrzejewski M, Szyndel J, Sporniak-Tutak K. Dental procedures in patients with coexistent blood clotting dis-order or arterial hypertension. Pt 2. Procedure with patients taking oral anticoagulants or modern anticoagulants [in Polish]. Magazyn Stomatol. 2013;23(12):116–118. 37. Pototski M, Amenábar JM. Dental management of patients receiving anticoagulation or antiplatelet treatment. J Oral Sci. 2007;49(4):253–258.
38. Dézsi CA, Dézsi BB, Dézsi AD. Management of dental patients receiving antiplatelet therapy or chronic oral anticoagulation: A review of the latest evidence. Eur J Gen Pract. 2017;23(1):196–201. |
188587 | https://www.math-only-math.com/arctan-x-plus-arccot-x-equals-pi-by-2.html | arctan x + arccot x = π/2 | arctan(x) + arccot(x) = (\frac{π}{2}) | Examples
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arctan(x) + arccot(x) = π 2 π 2
We will learn how to prove the property of the inverse trigonometric function arctan(x) + arccot(x) = π 2 π 2 (i.e., tan−1−1 x + cot−1−1 x = π 2 π 2).
Proof:Let, tan−1−1 x = θ
Step-by-Step Learning
Therefore, x = tan θ
x = cot (π 2 π 2 - θ), [Since, cot (π 2 π 2 - θ) = tan θ]
⇒ cot−1−1 x = π 2 π 2 - θ
⇒ cot−1−1 x= π 2 π 2 - tan−1−1 x, [Since, θ = tan−1−1 x]
⇒ cot−1−1 x + tan−1−1 x = π 2 π 2
⇒ tan−1−1 x + cot−1−1 x = π 2 π 2
Therefore, tan−1−1 x + cot−1−1 x = π 2 π 2.Proved.
Solved examples on property of inverse circular function tan−1−1 x + cot−1−1 x = π 2 π 2
Prove that, tan−1−1 4/3 + tan−1−1 12/5 = π - tan−1−1 56 33 56 33.
Solution:
We know that tan−1−1 x + cot−1−1 x = π 2 π 2
⇒ tan−1−1 x = π 2 π 2 - cot−1−1 x
⇒ tan−1−1 4 3 4 3 = π 2 π 2 - cot−1−1 4 3 4 3
and
tan−1−1 12 5 12 5 = π 2 π 2 - cot−1−1 12 5 12 5
Now, L. H. S. = tan−1−1 4 3 4 3 + tan−1−1 12 5 12 5
= π 2 π 2 - cot−1−1 4 3 4 3 + π 2 π 2 - cot−1−1 12 5 12 5, [Since, tan−1−1 4 3 4 3 = π 2 π 2 - cot−1−1 4 3 4 3 and tan−1−1 12 5 12 5 = π 2 π 2 - cot−1−1 12 5 12 5]
= π - (cot−1−1 4 3 4 3 + cot−1−1 12 5 12 5)
= π - (tan−1−1 3 4 3 4 + tan−1−1 5 12 5 12)
= π – tan−1−1 3 4+5 12 1–3 4⋅5 12 3 4+5 12 1–3 4·5 12
= π – tan−1−1 (14 12 14 12 x 48 33 48 33)
= π – tan−1−1 56 33 56 33 = R. H. S.Proved.
●Inverse Trigonometric Functions
General and Principal Values of sin−1−1 x
General and Principal Values of cos−1−1 x
General and Principal Values of tan−1−1 x
General and Principal Values of csc−1−1 x
General and Principal Values of sec−1−1 x
General and Principal Values of cot−1−1 x
Principal Values of Inverse Trigonometric Functions
General Values of Inverse Trigonometric Functions
arcsin(x) + arccos(x) = π 2 π 2
arctan(x) + arccot(x) = π 2 π 2
arctan(x) + arctan(y) = arctan(x+y 1−x y x+y 1−x y)
arctan(x) - arctan(y) = arctan(x−y 1+x y x−y 1+x y)
arctan(x) + arctan(y) + arctan(z)= arctan x+y+z–x y z 1–x y–y z–z x x+y+z–x y z 1–x y–y z–z x
arccot(x) + arccot(y) = arccot(x y−1 y+x x y−1 y+x)
arccot(x) - arccot(y) = arccot(x y+1 y−x x y+1 y−x)
arcsin(x) + arcsin(y) = arcsin(x 1−y 2−−−−−√1−y 2 + y 1−x 2−−−−−√1−x 2)
arcsin (x) - arcsin(y) = arcsin (x 1−y 2−−−−−√1−y 2 - y 1−x 2−−−−−√1−x 2)
arccos (x) + arccos(y) = arccos(xy - 1−x 2−−−−−√1−x 2 1−y 2−−−−−√1−y 2)
arccos(x) - arccos(y) = arccos(xy + 1−x 2−−−−−√1−x 2 1−y 2−−−−−√1−y 2)
2 arcsin(x) = arcsin(2x 1−x 2−−−−−√1−x 2)
2 arccos(x) = arccos(2x 2 2 - 1)
2 arctan(x) = arctan(2 x 1−x 2 2 x 1−x 2) = arcsin(2 x 1+x 2 2 x 1+x 2) = arccos(1−x 2 1+x 2 1−x 2 1+x 2)
3 arcsin(x) = arcsin(3x - 4x 3 3)
3 arccos(x) = arccos(4x 3 3 - 3x)
3 arctan(x) = arctan(3 x−x 3 1−3 x 2 3 x−x 3 1−3 x 2)
Inverse Trigonometric Function Formula
Principal Values of Inverse Trigonometric Functions
Problems on Inverse Trigonometric Function
11 and 12 Grade Math
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188588 | https://www.sciencedirect.com/science/article/abs/pii/S1230801314000034 | The degree of damage in the peripheral facial nerve palsy in children depending on the cause: The role of Lyme neuroborreliosis - ScienceDirect
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References (12)
Polish Annals of Medicine
Volume 21, Issue 1, June 2014, Pages 31-35
Original Research Article
The degree of damage in the peripheral facial nerve palsy in children depending on the cause: The role of Lyme neuroborreliosis
Author links open overlay panel Elżbieta Młynarczyk a, Wacław Kopala b, Jacek Morski a, Aleksandra Melnyk a
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Abstract
Introduction
Symptoms of peripheral facial nerve paralysis occur in patients of all age groups. Etiology of this disease is very diverse.
Aim
The aim of this work is to estimate facial nerve damage in relation to presumable cause, including cases in which Lyme neuroborreliosis has been confirmed by laboratory testing.
Material and methods
Thirty-three patients with symptoms of facial nerve paralysis were examined from January to September 2012 in the Department of Pediatric Neurology of the Provincial Specialist Children's Hospital in Olsztyn. Facial nerve function was classified on a three-stage scale created by the authors of this work. Full neurological examination was performed in all patients. Laboratory tests for Lyme-specific antibodies IgG and IgM by enzyme-linked immunosorbent assay has also been performed.
Results and discussion
In the study group, diagnosis of Lyme neuroborreliosis was confirmed in nine patients. Analysis shows that the stage of facial nerve damage in the group with Lyme disease was significantly heavier and that percentage of mild paralysis was significantly higher in the group of patients in whom Lyme disease has not been confirmed.
Conclusions
The role of neuroborreliosis among the infectious factors ought to be emphasized. It seems significant that in patients with confirmed Lyme disease the stage of facial nerve damage was considerably deeper and the recovery was slower.
Introduction
Facial nerve, as one of the most important of the cranial nerves, has a very complex structure and course, which implies a diverse symptomatology of its damage. It consists of a motor root and the part described under the name of intermediate nerve, which contains sensory (afferent) and parasympathetic (secretory) fibers. Motor fibers that reach facial mimic muscles arise from the facial nerve nucleus located between the dorsal pons and medulla oblongata.1, 2, 3 Symptoms of peripheral facial nerve palsy occur in patients of all age groups, beginning from the neonatal period. Etiology of palsies is very diverse.4
The most common unilateral facial nerve damage is defined as idiopathic peripheral facial paralysis – Bell's palsy, which represents 75% of facial nerve paralyzes. Currently, it seems that even these cases have a viral etiology and are often preceded by exposure to cold.1, 3, 5, 6
Recently increasing attention is paid to Borrelia burgdorferi infection as a potential cause of the facial nerve paralysis. In case of B. burgdorferi infection there are probably two mechanisms responsible for the development of peripheral neuropathies. The first results from the direct interaction between the spirochete and the nerve cell, which leads to its damage, while the second is associated with stimulation of the immune response against this pathogen. Reports in the literature describe the T- and B-cell autoreactivity against endogenous neural structures, which in the presence of 41 kDa B. burgdorferi flagellin may elicit inflammatory mediators.4, 7 Inflammatory and angiopathic lesions, perivascular infiltrates, partially with thrombosis of the vasa nervorum are responsible for the axonal damage of peripheral nerves. Axonal neuropathies, in contrast to direct nerve damage, are less responsive to antibiotic therapy.4, 7, 8
Secondary causes include inter alia, congenital anomalies, such as Mobius syndrome, trauma, including iatrogenic trauma during surgical procedures, neoplasms, particularly cerebellopontine angle tumors or parotid tumors, and infections.1, 3, 5
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Section snippets
Aim
The aim of this work was to assess the degree of facial nerve damage in relation to the presumable cause, with particular reference to cases of elevated antibody titers against B. burgdorferi in serum and cerebrospinal fluid (CSF) findings indicating neuroborreliosis.
Material and methods
In the period between the beginning of January and the end of September 2012 in the Department of Neurology of the Provincial Specialist Children's Hospital in Olsztyn, 33 patients with symptoms of peripheral facial nerve palsy were examined. Study group consisted of 10 boys (30.3%) and 23 girls (69.7%) aged 4–18 years. Median age was 8.5 years. Facial nerve function was classified on a scale created by the authors. In the study, visual determination of each separate function was performed:
Results
In the examined group of patients it was found that facial nerve paralysis occurs more frequently on the left side and in girls.
Patients, in whom no potential cause of facial nerve palsy was found, constituted a group of 15 subjects (45.5%). Other possible causes in the examined group of patients might have included: multiple sclerosis (2 patients), herpes (1 patient), temporal bone fracture (1 patient), acute otitis media (1 patient), history of chemotherapy for leukemia (2 patients),
Discussion
Children who were included in the study were residents of the north-eastern Poland. This region is considered as a Lyme disease endemic area. Seasonal incidence of Lyme disease is also characteristic. The vast majority of facial nerve palsy with elevated serum antibodies against B. burgdorferi observed in the study population was reported in the summer–autumn period, which is confirmed by observations of other authors.9, 10 Objective clinical assessment of the degree of facial nerve function
Conclusions
1.The degree of motor function damage of the facial nerve in the course of peripheral palsy is dependent on the cause and in cases resulting from B. burgdorferi infection it might be more severe and require other treatment than that in Bell's palsy.
2.All patients with peripheral facial nerve paralysis should have serum B. burgdorferi-specific IgG and IgM antibodies determined.
3.Despite increasingly more precise diagnostic methods, in many cases the cause of facial nerve palsy remains unknown.
Conflict of interest
None declared.
Recommended articles
References (12)
W. Bradley et al.
Neurologia w praktyce klinicznej. [Neurology in Clinical Practice]
(2006)
M.J. FitzGerald et al.
Neuroanatomia. [Neuroanatomy]
(2007)
L.P. Rowland
Neurologia Merritta. [Neurology]
(2005)
M. Oweaki et al.
Obraz kliniczny neuroboreliozy
Wiad Lek (2007)
B. Okurowska-Zawada et al.
Porażenie nerwu twarzowego w praktyce lekarza pediatry
Klin Pediatr (2010)
W. Szapłyko et al.
Obwodowe porażenie nerwu twarzowego u dzieci – problemy diagnostyczne i lecznicze
Klin Pediatr (1994)
There are more references available in the full text version of this article.
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Copyright © 2014 Published by Elsevier Urban & Partner Sp. z o.o.
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General
Can I change my username?
Yes, it is possible for you to contact AoPS and change your username. However, AoPS typically seldom changes usernames, notably solely for privacy reasons (e.g., your current username reveals personal information) or if the username contains inappropriate content. AoPS will not change your username for other reasons (e.g., current username is too "cringy" or "disgusting"). AoPS may change your username if you use specific information that AoPS may need in the future!
Why can't I do this? (Change my profile picture, post links, post images, edit tags, or do something else).
If you are a new user with less than 2 weeks of activity, there are new-user restrictions put in place on your account, including not being able to change your profile picture, post links, images, and certain keywords. New users also have a daily post limit of 6 posts. These restrictions will be lifted after 14 days of activity on your account.Otherwise, if you are not a new user, then you may be banned from performing that action because your previous use of that action violates the Terms of Service. As an example, you may not be able to change your profile picture because your previous one violated the Terms of Service.If you're still unsure why you cannot do something, open up a thread in Site Support, or contact Art of Problem Solving about the issue.
Something looks weird (e.g., blurry, missing line, etc.)
This likely to be because of zoom-in area. Please make sure your zoom level is at 100% or is reset for correct rendering of the web page. If it is at 100% and still doesn't look right, try to refresh or hard refresh. If it still continues looking weird, follow the instructions here to submit a bug report about it in Site Support.
What is a 'multi' and can I have one?
A multi stands for multiple accounts. You are not allowed to have multiple accounts; see here for more info.
Can I make more than one account?
Multiple accounts (multis) are banned on AoPS. Having more than one account leads to issues of not remembering on what account you did what. Using multiple accounts in an attempt to "game" the system, (e.g. Reaper in some way) or otherwise violate the AoPS without consequences will lead to all or some of your accounts being deactivated. If you have already made additional accounts, please choose one account and stop using the others.
And, with more than one account, it may cause problems with the system and interfere with the programs that are running in the background to provide a seamless experience. And so, if you have another account that you are using in addition to the one that you already have, please, stop using it. It truly benefits all of us.
What software does Art of Problem Solving use to run the website?
Search: Solar
Wiki: MediaWiki
Asymptote and LaTeX are generated through their respective binary packages
Videos: YouTube
PyWindow: Skulpt
Classroom and Reaper: Socket.IO
All other parts of the website are custom built.
Forums
Why can't I use a forum name (or description)?
Your forum name must be unique and it must be between 3 and 50 characters and your forum description must be 240 characters or less.
Why do some posts say they were posted in the future?
The AoPS clock is based on the official Naval Observatory time, which is considered the most accurate time. If your computer's system clock is behind the correct time, recent posts may indicate they were posted in the future. Please correct your computer's clock or enable clock synchronization so that your clock is always correct. Mac users may wish to check Synchronize your Mac's Clock With A Time Server. Windows users can go here. You can also check the US official time which will also show you how much your clock is off.
Why do some threads have fewer posts than the number of the latest post?
This is because of deleted posts. When posts get deleted, their post number is not reused.
I've lost admin access to a forum or blog I created, how do I get it back?
Please send an email to sheriff@artofproblemsolving.com with the issue and URL of your forum. They will research your request and restore admin access to your forum if appropriate.
What should I do if I find a glitch in the community?
First, search the Site Support forum to check if your issue has already been reported.If your issue isn't reported, try refreshing your browser page. Most issues go away after a refresh and there is no need to report the issue unless it continues after you refresh your browser. You can refresh on most browsers in Windows and Chrome OS with Ctrl + R and on Mac with Cmd + R. To Hard Refresh, click Shift between Ctrl/Cmd and R.Some commonly known glitches that should not be reported are avatars appearing twice in the topic list or private message, friends appearing more than once in the friend list, the edit icon showing next to a message when it shouldn't, and not being able to search for a forum after going back a page in the community.
How do I format my post, e.g. bold text, add URLs, etc.?
AoPS is based on a markup language called BBCode. A tutorial of its functions on AoPS and how to use them can be found here.
How do I hide content in the forums?
Wrap the content you want to hide in [hide] tags. This will make a clickable label that, once clicked, will reveal the content. [hide]Content[/hide]
If you want to customize the label, or have it says something other than "Click to reveal hidden text", you can do something like: [hide="Label to display"]Content[/hide]
Alternatively, wrap the content you want to hide in [tip] tags. This will make a hoverable label that, once hovered over, will reveal the content. [tip]Content[/tip]
If you want to customize the label, instead of saying "Hover for tip", you can do something like: [tip="Label to display"]Content[/tip]
I got the message "Too many messages." when trying to send a private message, why?
To prevent PM spam abuse like "asdfghjklkjhgfdsa", users with less than five forum posts are limited to four private messages within a forty-eight hour period.
How do you tell if someone has blocked private messages?
If you're not sure, try to private message them. It should say you can't. Or go to their user page. It should say so (Private message). Finally, you can check their posts, because generally you can private message about a post.
How do you private message yourself?
If you want to private message yourself to remember something, you have to go to one of your posts, and click the PM button in the top right corner. Then you reach a private message screen where you can backspace the text and write your message.
Posting often makes me a better user, right?
It will depend. Often, it's just the opposite. The quality of your posts matters far more than the number of posts you make. In fact, if you make too many unnecessary and spammy posts, you may find your posting privileges revoked.
I have posted in a forum, but my post count did not increase. Why?
Some forums, including the Test Forum, Mafia, Fun Factory, and other user created forums, do not count toward total posts.
I believe a post needs corrective action. What should I do?
If you believe a post needs moderative action, you may report it by clicking the Z icon on the upper-right corner of that post. If it's a minor mistake, you may want to PM the offending user instead and explain how they can make their post better.
How do I post images?
While AoPS forums have the ability to attach images, we do not generally recommend doing so, as we can not guarantee the images will be available through upgrades, restorations, etc. We also have limited disk space which causes us to remove attachments from time to time. Therefore, we recommend using a third party image hosting solution. There are many options, but we recommend imgur.com.
Go to
Upload your image, and wait for it to complete
Hover your cursor over the image
Click the 3 dots in the upper right corner
Click Get Share Links
Find the BBCode option, which should start with "[img]", and click the green Copy Link button
Paste the copied text into your forum or blog post.
In general, if you have any online image that you want to post, you can right-click on it, click copy image address, and paste it into AoPS, in between an [img][/img] tags.
How do I become a moderator of a forum?
The creators of user-created forums select moderators in several different ways. Contact the creator(s) of the forum for information.To see the list of moderators, click the little i icon on the slight-upper-left corner, next to where the name of that forum is.If you are the creator, and want to know how to make a user a moderator, follow these steps:
Click the little i icon on the slight-upper-left corner, next to where the name of that forum is.
Below the Permissions section, there should be a list of moderators. Click the pencil button to edit.
Many users often ask and talk about "forum rankings." How do I see it?
Forum rankings are how forums are ranked. There is no confirmation from a site administrator on how these are ranked, but most users suspect that they are ranked on activity.Here is how to view forum rankings:
Go the the community.
Scroll down to the section Other Forums and Collections
In that section, find the folder Other Forums. The direct link is
Scroll down the first few rows of forums until you reach a forum called College
Every forum after that is user-created and ranked. The first forum after College is the first in rankings, the second one is second, third one is third(user created forums usually) etc.
Blogs
How come I can't create a blog?
In order to be able to create a blog on AoPS, you have to be a member of the community for at least 2 weeks, before creating any sort of blog, forum, or such.
How do I make my blog look nice?
Many AoPSers make their blogs look nice by applying CSS, which is a high-level stylesheet language. This can be done by typing code into the CSS box in the Blog Control Panel.Try copying CSS of other blogs but remember to give them credit!There are many people on AoPS who would be willing to help you with your blog CSS if you don't know the CSS code.
Blog Settings
The maximum amount of characters for the Post Comment Text, Comments Text, One Comment Text, and No Comments Text is 40 characters.
Why are usernames a different color?
Blogs use a different stylesheet than the rest of the site and blog owners can customize their blog, including some portions of the feed widget, and therefore usernames show differently in a blog than on the community. Additionally, in the forums, black means a regular user, green means a moderator, red means an administrator, and blue means Site Admin.
Alcumus
How is rating computed?
The rating is more of a prediction of what percentage of problems in the topic the Alcumus engine believes you will get correct. As you get more and more correct, the rating will go up slower and slower. However, if you are predicted to get most correct, and you miss one or two problems, the rating, or prediction of percentage correct, will go down.
I am stuck on a problem, and changing the topic does not change the problem.
Alcumus provides review problems to make sure you still recall information learned from the past. These problems cannot be skipped unless you go to an entirely different subject.
Contests
Where can I find past contest questions and solutions?
You can find them in the Contest Collections section.
How do I get problems onto the contest page?
Make a topic for each question in the appropriate forum, copy/paste the urls to the National Olympiad. Your problems may eventually be submitted into the Contest page.
What are the guidelines for posting problems to be added to the contests section?
Refer to the guidelines in this post.
Why are the contest collections missing many contest questions?
Generally, it is because users have not yet posted them onto the wiki (translation difficulties, not having access to the actual problems, lack of interest, etc). If you have a copy, please post the problems in the Community Section! In some cases, however, problems may be missing due to copyright claims from maths organizations.
Where are all of the mock contests?
User-generated mock contests are now in its separate subforum, AoPS Mock Contests. Please use that forum for creating, hosting, participating in, or recruiting writers/test solvers for mock contests.
How can I host a mock contest?
See posts on writing a mock contest and Tips for Organizing a Mock Contest.
LaTeX and Asymptote
What is LaTeX, and how do I use it?
is a typesetting markup language and document preparation system. It is widely used for typesetting expressions containing mathematical formulas. AoPS LaTeX Quick Start Video. See also LaTeX:About.
How can I download LaTeX to use on the forums?
There are actually no downloads necessary; the forums and the wiki renders LaTeX commands between dollar signs.
How can I download LaTeX for personal use?
You can download TeXstudio here or TeXnicCenter here
Where can I find a list of LaTeX commands?
See here.
Where can I test LaTeX commands?
SandboxTeXeR, or Test Forum.
Where can I find examples of Asymptote diagrams and code?
Search this wiki for the tag or the Forums for the [asy] tag. See also these examples and this article (click on the images to obtain the code). See also this Quick Start Asymptote Video.
How can I draw 3D diagrams?
See Asymptote: 3D graphics.
What is the cse5 package?
See here. The package contains a set of shorthand commands that implement the behavior of usual commands, for example D() for draw() and dot(), and so forth.
What is the olympiad package?
See here. The package contains a set of commands useful for drawing diagrams related to olympiad geometry problems.
I have a question about LaTeX or Asymptote, and it's not answered in the wiki. What should I do?
Go here. This is a forum for and Asymptote. If you have a question, search to see if it's been asked before, and, if not, you can make a new post.
AoPS Wiki
How do I sign my name?
Type four tildes (~~~~) and your signature will be displayed.
Is there a guide for wiki syntax?
See the AoPS Wiki Tutorial.
What do I do if I see a mistake in the wiki?
Click "Edit" and correct the error. If it's just a typo, check the "This is a minor edit" box.
What if I need more information because I suspect it's a mistake, but I'm not sure?
Post it in the AoPS community, or on the AoPS Wiki discussion forum! Also, note that if it is about contests, many contests release their questions after the contest, so you can double-check there.
Miscellaneous
Is it possible to join the AoPS Staff?
Yes. AoPS sometimes hires college students to work as interns, graders, and teaching assistants. You must be at least in the second semester of your senior year of high school and you must be at least 18. See the careers page for more information.
What is the minimum age to be an assistant in an Art of Problem Solving class?
You must have graduated from high school, or at least be in the second term of your senior year.
How do I befriend someone on AoPS?
From your My AoPS page, you send out a friend request by typing their username in the corresponding box. You can also go to their account and press the 'add friend' button. They can then choose to accept or decline from their My AoPS page. You can also accept or decline friend requests from your My AoPS page. If you mean to make friends, that takes time and interaction.
AoPS Acronyms
AFK- Away from keyboard; inactive.
AFAIK(Rare)- As far as I know.
AoPS- Art of Problem Solving.
AIME- American Invitational Mathematics Examination.
admin(s)- Administrator(s).
admits - short for admits orz, used when someone admits that they are good or does something orz.
AMC- American Math Competitions.
AMO- American Mathematical Olympiad.
ATM- At the Moment/Automated teller machine.
ASAP- As soon as possible.
BRB- Be right back.
BTW- By the way.
Bump- to post on a thread simply to make the thread appear at the top of a forum, usually via one-word posts like "bumping". It is commonly used when a thread on a major forum is dead in the hopes of reviving it.
CEMC - Centre for Mathematics and Computing.
Combi - Combinatorics (Counting and Probability).
C&M - Competitions and Marathons.
C&P or C+P or CP - Counting and Probability or Contests and Programs.
cbrt - Cube Root.
CD - Civil Discourse [a debate forum].
CS&I - Computer Science and Informatics.
DHR - Distinguished Honour Roll.
DW - Don't worry.
EBWOP - Editing by way of post (Used commonly in Mafia).
FF- Fun Factory.
FTFY - Fixed that for you.
FTSOC(Rare)- For the sake of contradiction.
/g or /gen-Indicates the user is being genuine.
GC - Greed Control, a game on AoPS.
GG- Good Game.
GJ- Good Job.
GLHF-Good Luck, Have Fun.
GTG - Got to Go.
HAY/HBU(Rare)- How About You?
/hj-Indicates the user is half-joking.
HR - Honour Roll.
HSM - High School Math Forum.
HSO - High School Olympiad Forum.
ID(R or E)C - I Don't (Really or Even) Care.
IDC - I Don't Care.
ID(R or E)K-I Don't (Really or Even) Know.
IDK- I Don't Know.
IFF(Rare)-If and Only If.
IHDC-I Honestly Don't Care.
IIRC- If I remember correctly
IKR- I know, right?
IMO- In my opinion (or International Math Olympiad, depending on context)
IMHO- In my humble/honest opinion
/J-Indicates the user is joking
JK- Just kidding
JMO- United States of America Junior Mathematical Olympiad
/lh- Indicates the user is light-hearted.
lmho- Laughing my head/hat off
lol- Laugh Out Loud
MC- Mathcounts, a popular math contest for Middle School students.
NFL(Rare)- Not for long/No friends left/National Football League
mod(s)- Moderator(s)
MOEMS- Math Olympiads for Elementary and Middle Schools
MO(S)P- Mathematical Olympiad (Summer) Program
MSM- Middle School Math Forum
NA- Not Available
NGL- Not Gonna Lie
NIMO- National Internet Math Olympiad
NT- Number Theory
OBC(Rare)- Online by computer
OMG- Oh My Gosh
OMO-Online Math Open
OP- Original Poster/Original Post/Original Problem, or Overpowered/Overpowering
orz- This term, when used by AoPS members, is used to show respect to someone who is very good at math or some subject. "orz" resembles a person bowing. Commonly attributed with trolls.
QED- Quod erat demonstrandum, Latin for "That which was to be demonstrated, has been demonstrated" (Or for short, "That which was to be demonstrated."); some English mathematicians use it as an acronym for Quite Elegantly Done
QS&A- Questions, Suggestions, and Announcements Forum
rbo/rbuh- a rearrangement of "bro/bruh"
RO(t)FL - Rolling on the floor laughing
smh - Shaking my head/somehow. Usually expresses slight annoyance, with a similar connotation to an eyeroll or facepalm.
Sniping - posting to a thread right before another user posted, usually when the posts are of similar content. users tend to edit their post and add things like "EDIT: sniped" at the end to signify this.
sqrt - Square root
/srs - Indicates that the user is serious.
Sticky- A post pinned to the top of a forum - a thing you really should read
SS - Site Support Forum
ToS- Terms of Service, a thing you really should read
Trolling- the act of posting overly difficult problems or problems with too many calculations on forums, commonly used when someone posts a question from, for example, the International Math Olympiad, on the Middle School Math Forum. When problems are considered trolls, it is usually the consensus that they are too hard to do.
USA(J)MO- USA (Junior) Mathematical Olympiad
V/LA- Vacation or long absence/limited access
WDYM- What do you mean?
WIP- Work in progress
WLOG(Rare)- Without loss of generality
WTMLOG(Rare)- Without too much loss of generality
WOOT - Worldwide Online Olympiad Training
wrt- With respect to
wtg - Way to go
TRT- The Round Table [another debate forum, which arose first]
ty- Thank you
tytia(Rare)- Thank you, that is all
tysm- Thank you so much
ttyl- Talk to you later
xD/XD- Bursting laugh (tilt your head sideways to see why it is so.)
TL;DR- Too long; didn't read (a summary of the associated text)
yw- You're welcome
FTW!
Please see the For the Win! FAQ for many common questions. (To get to the FAQ, click on the three lines in the top left and then on FAQ).
School
Please refer to the AoPS Classes Handbook in the dropdown menu under "ONLINE SCHOOL."
What if I miss a class?
There are classroom transcripts available under My Classes, available at the top right of the web site. You can view these transcripts in order to review any missed material. Keep in mind, though, that whispers (Things the instructors say to you, and only you, to help you understand stuff) won't be shown, and you won't get to ask questions. However, class moves quicker! You can also ask questions on the class message board. Don't worry, though, classroom participation usually isn't weighted heavily.
Is there audio or video in class?
For classes that runs online normally, there is generally no audio or video in the class. The classes are generally text and image based, in an interactive chat room environment, which allows students to ask questions at any time during the class. Classes taken at the AoPS Academy Virtual Campus will have video. During the COVID-19 crisis, all AoPS Academy classes were conducted online using the Zoom video conferencing platform. However, there are AoPS videos with audio in some classes, and they can be very helpful, in case the class material doesn't seem easy.
What if I want to drop out of a class?
For up-to-date information on the AoPS Online School drop policy, visit the AoPS Online School FAQ page.
For my homework, there is supposed to be a green bar but it's orange, why?
The bar turns green only if the required writing problems are saved.
I need more time for my homework, how to get it?
There is a "Request Extension" button in the homework tab of your class. This will automatically extend the due date to 1-3 days after the normal deadline. If you want more time you need to ask for it in the little comment box, stating the reason why you want the extension, and how much time you want. This request will be looked at by the teachers and they will decide if you get the extension or not. Note that you can only use this button 2-4 times depending on how many weeks your class is. After, you will need administrator approval.Otherwise, you can send an email to extensions@aops.com with your username, class name and ID (the number in the class page URL after and reason for extension. Someone should get back to you within a couple days.
Why is my post in the AoPS schoolhouse too long?
The minimum number of characters allowed in an individual post in the AoPS Schoolhouse is 1 character. The maximum number of characters allowed in an individual post in the AoPS Schoolhouse is 512 characters.
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188590 | https://pubmed.ncbi.nlm.nih.gov/38974962/ | Understanding oxytocin in human physiology and pathophysiology: A path towards therapeutics - PubMed
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Compr Psychoneuroendocrinol
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. 2024 May 28:19:100242.
doi: 10.1016/j.cpnec.2024.100242. eCollection 2024 Aug.
Understanding oxytocin in human physiology and pathophysiology: A path towards therapeutics
Elizabeth A Lawson1
Affiliations Expand
Affiliation
1 Neuroendocrine Unit, Massachusetts General Hospital, Department of Medicine, Harvard Medical School, 50 Staniford Street, Suite 750B, Boston, MA, 02114, USA.
PMID: 38974962
PMCID: PMC11225698
DOI: 10.1016/j.cpnec.2024.100242
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Review
Understanding oxytocin in human physiology and pathophysiology: A path towards therapeutics
Elizabeth A Lawson. Compr Psychoneuroendocrinol.2024.
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Compr Psychoneuroendocrinol
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. 2024 May 28:19:100242.
doi: 10.1016/j.cpnec.2024.100242. eCollection 2024 Aug.
Author
Elizabeth A Lawson1
Affiliation
1 Neuroendocrine Unit, Massachusetts General Hospital, Department of Medicine, Harvard Medical School, 50 Staniford Street, Suite 750B, Boston, MA, 02114, USA.
PMID: 38974962
PMCID: PMC11225698
DOI: 10.1016/j.cpnec.2024.100242
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Abstract
•Oxytocin is a multifaceted hypothalamic-pituitary hormone involved in energy homeostasis, mental health, and bone metabolism.•Oxytocin deficiency in energy deficit states and in hypopituitarism is associated with worse mental health and bone health.•Oxytocin modulates appetitive neurocircuitry, improves impulse control, and reduces food intake in humans.•Defining the oxytocin system in human physiology and pathophysiology could lead to novel therapeutic strategies.
Keywords: Anorexia nervosa; Eating behavior; Hypopituitarism; Obesity; Oxytocin; Oxytocin deficiency.
© 2024 Published by Elsevier Ltd.
PubMed Disclaimer
Conflict of interest statement
EAL receives grant support and research study drug from Tonix Pharmaceuticals and receives royalties from UpToDate. EAL is also supported by NIH R01DK124223, R01MH128246, R61MH129331, K24MH120568, P30DK040561; Department of Defense AR220116 HT9425-23-1-0219; Templeton Foundation; Women's Wellness Foundation; and philanthropic gifts. EAL and/or immediate family member holds stock in Thermo Fisher Scientific, Zoetis, Danaher Corporation, Intuitive Surgical. EAL is an inventor on US provisional patent applications no. 63/413,657 (Oxytocin-based therapeutics for bone health in individuals with autism) and no. 63/467,980 (Oxytocin-based therapeutics to improve cognitive control in individuals with attention deficit hyperactivity disorder).
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Oxytocin secretion is associated with severity of disordered eating psychopathology and insular cortex hypoactivation in anorexia nervosa.Lawson EA, Holsen LM, Santin M, Meenaghan E, Eddy KT, Becker AE, Herzog DB, Goldstein JM, Klibanski A.Lawson EA, et al.J Clin Endocrinol Metab. 2012 Oct;97(10):E1898-908. doi: 10.1210/jc.2012-1702. Epub 2012 Aug 7.J Clin Endocrinol Metab. 2012.PMID: 22872688 Free PMC article.
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The neurohypophyseal hormone oxytocin and eating behaviors: a narrative review.Iovino M, Messana T, Marucci S, Triggiani D, Giagulli VA, Guastamacchia E, Piazzolla G, De Pergola G, Lisco G, Triggiani V.Iovino M, et al.Hormones (Athens). 2024 Mar;23(1):15-23. doi: 10.1007/s42000-023-00505-y. Epub 2023 Nov 18.Hormones (Athens). 2024.PMID: 37979096 Free PMC article.Review.
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References
Lawson E.A., Miller K.K. Anorexia Nervosa: Endocrine Complications and Their Management C. RF. Wolters Kluwer: UpToDate; 2022.
Camerino C. Low sympathetic tone and obese phenotype in oxytocin-deficient mice. Obesity. 2009;17(5):980–984. - PubMed
Takayanagi Y., et al. Oxytocin receptor-deficient mice developed late-onset obesity. Neuroreport. 2008;19(9):951–955. - PubMed
Mantella R.C., et al. Female oxytocin-deficient mice display enhanced anxiety-related behavior. Endocrinology. 2003;144(6):2291–2296. - PubMed
Matsushita H., et al. Oxytocin mediates the antidepressant effects of mating behavior in male mice. Neurosci. Res. 2010;68(2):151–153. - PubMed
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188591 | http://infolab.stanford.edu/~ullman/focs/ch04.pdf | CHAPTER 4 !
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Combinatorics and Probability In computer science we frequently need to count things and measure the likelihood of events. The science of counting is captured by a branch of mathematics called combinatorics. The concepts that surround attempts to measure the likelihood of events are embodied in a field called probability theory. This chapter introduces the rudiments of these two fields. We shall learn how to answer questions such as how many execution paths are there in a program, or what is the likelihood of occurrence of a given path?
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4.1 What This Chapter Is About We shall study combinatorics, or “counting,” by presenting a sequence of increas-ingly more complex situations, each of which is represented by a simple paradigm problem. For each problem, we derive a formula that lets us determine the number of possible outcomes. The problems we study are: !
Counting assignments (Section 4.2). The paradigm problem is how many ways can we paint a row of n houses, each in any of k colors.
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Counting permutations (Section 4.3). The paradigm problem here is to deter-mine the number of different orderings for n distinct items.
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Counting ordered selections (Section 4.4), that is, the number of ways to pick k things out of n and arrange the k things in order. The paradigm problem is counting the number of ways different horses can win, place, and show in a horse race.
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Counting the combinations of m things out of n (Section 4.5), that is, the selection of m from n distinct objects, without regard to the order of the selected objects. The paradigm problem is counting the number of possible poker hands.
156 SEC. 4.2 COUNTING ASSIGNMENTS 157 !
Counting permutations with some identical items (Section 4.6). The paradigm problem is counting the number of anagrams of a word that may have some letters appearing more than once.
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Counting the number of ways objects, some of which may be identical, can be distributed among bins (Section 4.7). The paradigm problem is counting the number of ways of distributing fruits to children.
In the second half of this chapter we discuss probability theory, covering the follow-ing topics: !
Basic concepts: probability spaces, experiments, events, probabilities of events.
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Conditional probabilities and independence of events.
These concepts help us think about how observation of the outcome of one experiment, e.g., the drawing of a card, influences the probability of future events.
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Probabilistic reasoning and ways that we can estimate probabilities of com-binations of events from limited data about the probabilities and conditional probabilities of events.
We also discuss some applications of probability theory to computing, including systems for making likely inferences from data and a class of useful algorithms that work “with high probability” but are not guaranteed to work all the time.
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4.2 Counting Assignments One of the simplest but most important counting problems deals with a list of items, to each of which we must assign one of a fixed set of values. We need to determine how many different assignments of values to items are possible.
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Example 4.1. A typical example is suggested by Fig. 4.1, where we have four houses in a row, and we may paint each in one of three colors: red, green, or blue.
Here, the houses are the “items” mentioned above, and the colors are the “values.” Figure 4.1 shows one possible assignment of colors, in which the first house is painted red, the second and fourth blue, and the third green.
Red Green Blue Blue Fig. 4.1. One assignment of colors to houses.
To answer the question, “How many different assignments are there?” we first need to define what we mean by an “assignment.” In this case, an assignment is a list of four values, in which each value is chosen from one of the three colors red, green, or blue. We shall represent these colors by the letters R, G, and B. Two such lists are different if and only if they differ in at least one position.
158 COMBINATORICS AND PROBABILITY In the example of houses and colors, we can choose any of three colors for the first house. Whatever color we choose for the first house, there are three colors in which to paint the second house. There are thus nine different ways to paint the first two houses, corresponding to the nine different pairs of letters, each letter chosen from R, G, and B. Similarly, for each of the nine assignments of colors to the first two houses, we may select a color for the third house in three possible ways. Thus, there are 9 × 3 = 27 ways to paint the first three houses. Finally, each of these 27 assignments can be extended to the fourth house in 3 different ways, giving a total of 27 × 3 = 81 assignments of colors to the houses. !
The Rule for Counting Assignments We can extend the above example.
In the general setting, we have a list of n “items,” such as the houses in Example 4.1. There is also a set of k “values,” such as the colors in Example 4.1, any one of which can be assigned to an item. An assignment is a list of n values (v1, v2, . . . , vn). Each of v1, v2, . . . , vn is chosen to Assignment be one of the k values. This assignment assigns the value vi to the ith item, for i = 1, 2, . . . , n.
There are kn different assignments when there are n items and each item is to be assigned one of k values. For instance, in Example 4.1 we had n = 4 items, the houses, and k = 3 values, the colors. We calculated that there were 81 different assignments. Note that 34 = 81. We can prove the general rule by an induction on n.
STATEMENT S(n): The number of ways to assign any one of k values to each of n items is kn.
BASIS. The basis is n = 1. If there is one item, we can choose any of the k values for it. Thus there are k different assignments. Since k1 = k, the basis is proved.
INDUCTION. Suppose the statement S(n) is true, and consider S(n + 1), the statement that there are kn+1 ways to assign one of k values to each of n + 1 items.
We may break any such assignment into a choice of value for the first item and, for each choice of first value, an assignment of values to the remaining n items. There are k choices of value for the first item. For each such choice, by the inductive hypothesis there are kn assignments of values to the remaining n items. The total number of assignments is thus k × kn, or kn+1. We have thus proved S(n + 1) and completed the induction.
Figure 4.2 suggests this selection of first value and the associated choices of assignment for the remaining items in the case that n + 1 = 4 and k = 3, using as a concrete example the four houses and three colors of Example 4.1. There, we assume by the inductive hypothesis that there are 27 assignments of three colors to three houses.
SEC. 4.2 COUNTING ASSIGNMENTS 159 27 Assignments 27 Assignments 27 Assignments First house Other three houses Red Green Blue Fig. 4.2. The number of ways to paint 4 houses using 3 colors.
Counting Bit Strings In computer systems, we frequently encounter strings of 0’s and 1’s, and these strings often are used as the names of objects. For example, we may purchase a computer with “64 megabytes of main memory.” Each of the bytes has a name, and that name is a sequence of 26 bits, each of which is either a 0 or 1. The string Bit of 0’s and 1’s representing the name is called a bit string.
Why 26 bits for a 64-megabyte memory? The answer lies in an assignment-counting problem. When we count the number of bit strings of length n, we may think of the “items” as the positions of the string, each of which may hold a 0 or a 1. The “values” are thus 0 and 1. Since there are two values, we have k = 2, and the number of assignments of 2 values to each of n items is 2n.
If n = 26 — that is, we consider bit strings of length 26 — there are 226 possible strings. The exact value of 226 is 67,108,864. In computer parlance, this number is thought of as “64 million,” although obviously the true number is about 5% higher.
The box about powers of 2 tells us a little about the subject and tries to explain the general rules involved in naming the powers of 2.
EXERCISES 4.2.1: In how many ways can we paint a) Three houses, each in any of four colors b) Five houses, each in any of five colors c) Two houses, each in any of ten colors 4.2.2: Suppose a computer password consists of eight to ten letters and/or digits.
How many different possible passwords are there? Remember that an upper-case letter is different from a lower-case one.
4.2.3: Consider the function f in Fig. 4.3. How many different values can f return?
160 COMBINATORICS AND PROBABILITY int f(int x) { int n; n = 1; if (x%2 == 0) n = 2; if (x%3 == 0) n = 3; if (x%5 == 0) n = 5; if (x%7 == 0) n = 7; if (x%11 == 0) n = 11; if (x%13 == 0) n = 13; if (x%17 == 0) n = 17; if (x%19 == 0) n = 19; return n; } Fig. 4.3. Function f.
4.2.4: In the game of “Hollywood squares,” X’s and O’s may be placed in any of the nine squares of a tic-tac-toe board (a 3×3 matrix) in any combination (i.e., unlike ordinary tic-tac-toe, it is not necessary that X’s and O’s be placed alternately, so, for example, all the squares could wind up with X’s). Squares may also be blank, i.e., not containing either an X or and O. How many different boards are there?
4.2.5: How many different strings of length n can be formed from the ten digits?
A digit may appear any number of times in the string or not at all.
4.2.6: How many different strings of length n can be formed from the 26 lower-case letters? A letter may appear any number of times or not at all.
4.2.7: Convert the following into K’s, M’s, G’s, T’s, or P’s, according to the rules of the box in Section 4.2: (a) 213 (b) 217 (c) 224 (d) 238 (e) 245 (f) 259.
4.2.8: Convert the following powers of 10 into approximate powers of 2: (a) 1012 (b) 1018 (c) 1099.
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4.3 Counting Permutations In this section we shall address another fundamental counting problem: Given n distinct objects, in how many different ways can we order those objects in a line?
Such an ordering is called a permutation of the objects. We shall let Π(n) stand for the number of permutations of n objects.
As one example of where counting permutations is significant in computer science, suppose we are given n objects, a1, a2, . . . , an, to sort. If we know nothing about the objects, it is possible that any order will be the correct sorted order, and thus the number of possible outcomes of the sort will be equal to Π(n), the number of permutations of n objects.
We shall soon see that this observation helps us argue that general-purpose sorting algorithms require time proportional to n log n, and therefore that algorithms like merge sort, which we saw in Section 3.10 takes SEC. 4.3 COUNTING PERMUTATIONS 161 K’s and M’s and Powers of 2 A useful trick for converting powers of 2 into decimal is to notice that 210, or 1024, is very close to one thousand. Thus 230 is (210)3, or about 10003, that is, a billion.
Then, 232 = 4 × 230, or about four billion. In fact, computer scientists often accept the fiction that 210 is exactly 1000 and speak of 210 as “1K”; the K stands for “kilo.” We convert 215, for example, into “32K,” because 215 = 25 × 210 = 32 × “1000” But 220, which is exactly 1,048,576, we call “1M,” or “one million,” rather than “1000K” or “1024K.” For powers of 2 between 20 and 29, we factor out 220. Thus, 226 is 26 × 220 or 64 “million.” That is why 226 bytes is referred to as 64 million bytes or 64 “megabytes.” Below is a table that gives the terms for various powers of 10 and their rough equivalents in powers of 2.
PREFIX LETTER VALUE Kilo K 103 or 210 Mega M 106 or 220 Giga G 109 or 230 Tera T 1012 or 240 Peta P 1015 or 250 This table suggests that for powers of 2 beyond 29 we factor out 230, 240, or 2 raised to whatever multiple-of-10 power we can. The remaining powers of 2 name the number of giga-, tera-, or peta- of whatever unit we are measuring. For example, 243 bytes is 8 terabytes.
O(n log n) time, are to within a constant factor as fast as can be.
There are many other applications of the counting rule for permutations. For example, it figures heavily in more complex counting questions like combinations and probabilities, as we shall see in later sections.
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Example 4.2. To develop some intuition, let us enumerate the permutations of small numbers of objects. First, it should be clear that Π(1) = 1. That is, if there is only one object A, there is only one order: A.
Now suppose there are two objects, A and B. We may select one of the two objects to be first and then the remaining object is second. Thus there are two orders: AB and BA. Therefore, Π(2) = 2 × 1 = 2.
Next, let there be three objects: A, B, and C. We may select any of the three to be first. Consider the case in which we select A to be first. Then the remaining two objects, B and C, can be arranged in either of the two orders for two objects to complete the permutation. We thus see that there are two orders that begin with A, namely ABC and ACB.
Similarly, if we start with B, there are two ways to complete the order, corre-162 COMBINATORICS AND PROBABILITY sponding to the two ways in which we may order the remaining objects A and C.
We thus have orders BAC and BCA. Finally, if we start with C first, we can order the remaining objects A and B in the two possible ways, giving us orders CAB and CBA. These six orders, ABC, ACB, BAC, BCA, CAB, CBA are all the possible orders of three elements. That is, Π(3) = 3 × 2 × 1 = 6.
Next, consider how many permutations there are for 4 objects: A, B, C, and D. If we pick A first, we may follow A by the objects B, C, and D in any of their 6 orders. Similarly, if we pick B first, we can order the remaining A, C, and D in any of their 6 ways. The general pattern should now be clear. We can pick any of the four elements first, and for each such selection, we can order the remaining three elements in any of the Π(3) = 6 possible ways. It is important to note that the number of permutations of the three objects does not depend on which three elements they are. We conclude that the number of permutations of 4 objects is 4 times the number of permutations of 3 objects. !
More generally, Π(n + 1) = (n + 1)Π(n) for any n ≥1 (4.1) That is, to count the permutations of n + 1 objects we may pick any of the n + 1 objects to be first. We are then left with n remaining objects, and these can be permuted in Π(n) ways, as suggested in Fig. 4.4. For our example where n + 1 = 4, we have Π(4) = 4 × Π(3) = 4 × 6 = 24.
Π(n) orders Π(n) orders Π(n) orders First object n remaining objects .
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Object 1 Object 2 .
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Object n + 1 Fig. 4.4. The permutations of n + 1 objects.
SEC. 4.3 COUNTING PERMUTATIONS 163 The Formula for Permutations Equation (4.1) is the inductive step in the definition of the factorial function intro-duced in Section 2.5. Thus it should not be a surprise that Π(n) equals n!. We can prove this equivalence by a simple induction.
STATEMENT S(n): Π(n) = n! for all n ≥1.
BASIS. For n = 1, S(1) says that there is 1 permutation of 1 object. We observed this simple point in Example 4.2.
INDUCTION. Suppose Π(n) = n!. Then S(n + 1), which we must prove, says that Π(n + 1) = (n + 1)!. We start with Equation (4.1), which says that Π(n + 1) = (n + 1) × Π(n) By the inductive hypothesis, Π(n) = n!. Thus, Π(n + 1) = (n + 1)n!. Since n! = n × (n −1) × · · · × 1 it must be that (n + 1) × n! = (n + 1) × n × (n −1) × · · ·× 1. But the latter product is (n + 1)!, which proves S(n + 1).
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Example 4.3. As a result of the formula Π(n) = n!, we conclude that the number of permutations of 4 objects is 4! = 4 × 3 × 2 × 1 = 24, as we saw above.
As another example, the number of permutations of 7 objects is 7! = 5040. !
How Long Does it Take to Sort?
One of the interesting uses of the formula for counting permutations is in a proof that sorting algorithms must take at least time proportional to n log n to sort n elements, unless they make use of some special properties of the elements.
For example, as we note in the box on special-case sorting algorithms, we can do better than proportional to n log n if we write a sorting algorithm that works only for small integers.
However, if a sorting algorithm works on any kind of data, as long as it can be compared by some “less than” notion, then the only way the algorithm can decide on the proper order is to consider the outcome of a test for whether one of two elements is less than the other.
A sorting algorithm is called a general-purpose sorting algorithm if its only operation upon the elements to be sorted is a General purpose sorting algorithm comparison between two of them to determine their relative order. For instance, selection sort and merge sort of Chapter 2 each make their decisions that way. Even though we wrote them for integer data, we could have written them more generally by replacing comparisons like if (A[j] < A[small]) on line (4) of Fig. 2.2 by a test that calls a Boolean-valued function such as if (lessThan(A[j], A[small])) 164 COMBINATORICS AND PROBABILITY Suppose we are given n distinct elements to sort. The answer — that is, the correct sorted order — can be any of the n! permutations of these elements. If our algorithm for sorting arbitrary types of elements is to work correctly, it must be able to distinguish all n! different possible answers.
Consider the first comparison of elements that the algorithm makes, say lessThan(X,Y) For each of the n! possible sorted orders, either X is less than Y or it is not. Thus, the n! possible orders are divided into two groups, those for which the answer to the first test is “yes” and those for which it is “no.” One of these groups must have at least n!/2 members. For if both groups have fewer than n!/2 members, then the total number of orders is less than n!/2 + n!/2, or less than n! orders. But this upper limit on orders contradicts the fact that we started with exactly n! orders.
Now consider the second test, on the assumption that the outcome of the comparison between X and Y was such that the larger of the two groups of possible orders remains (take either outcome if the groups are the same size). That is, at least n!/2 orders remain, among which the algorithm must distinguish. The second comparison likewise has two possible outcomes, and at least half the remaining orders will be consistent with one of these outcomes. Thus, we can find a group of at least n!/4 orders consistent with the first two tests.
We can repeat this argument until the algorithm has determined the correct sorted order. At each step, by focusing on the outcome with the larger population of consistent possible orders, we are left with at least half as many possible orders as at the previous step. Thus, we can find a sequence of tests and outcomes such that after the ith test, there are at least n!/2i orders consistent with all these outcomes.
Since we cannot finish sorting until every sequence of tests and outcomes is consistent with at most one sorted order, the number of tests t made before we finish must satisfy the equation n!/2t ≤1 (4.2) If we take logarithms base 2 of both sides of Equation (4.2) we have log2 n! −t ≤0, or t ≥log2(n!) We shall see that log2(n!) is about n log2 n. But first, let us consider an example of the process of splitting the possible orders.
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Example 4.3.
Let us consider how the selection sort algorithm of Fig. 2.2 makes its decisions when given three elements (a, b, c) to sort. The first comparison is between a and b, as suggested at the top of Fig. 4.5, where we show in the box that all 6 possible orders are consistent before we make any tests. After the test, the orders abc, acb, and cab are consistent with the “yes” outcome (i.e., a < b), while the orders bac, bca, and cba are consistent with the opposite outcome, where b > a. We again show in a box the consistent orders in each case.
In the algorithm of Fig. 2.2, the index of the smaller becomes the value small.
Thus, we next compare c with the smaller of a and b. Note that which test is made next depends on the outcome of previous tests.
SEC. 4.3 COUNTING PERMUTATIONS 165 After making the second decision, the smallest of the three is moved into the first position of the array, and a third comparison is made to determine which of the remaining elements is the larger. That comparison is the last comparison made by the algorithm when three elements are to be sorted. As we see at the bottom of Fig. 4.5, sometimes that decision is determined. For example, if we have already found a < b and c < a, then c is the smallest and the last comparison of a and b must find a smaller.
a < b?
b < c?
a < c?
abc, acb, cab bac, bca, cba b < c?
a < b?
a < c?
a < b?
cab abc, acb bac, bca cba abc acb cab bac bca cba Y N Y N Y N Y N Y N Y N Y N abc, acb, bac, bca, cab, cba Fig. 4.5. Decision tree for selection sorting of 3 elements.
In this example, all paths involve 3 decisions, and at the end there is at most one consistent order, which is the correct sorted order. The two paths with no consistent order never occur. Equation (4.2) tells us that the number of tests t must be at least log2 3!, which is log2 6. Since 6 is between 22 and 23, we know that log2 6 will be between 2 and 3. Thus, at least some sequences of outcomes in any algorithm that sorts three elements must make 3 tests. Since selection sort makes only 3 tests for 3 elements, it is at least as good as any other sorting algorithm for 3 elements in the worst case. Of course, as the number of elements becomes large, we know that selection sort is not as good as can be done, since it is an O(n2) sorting algorithm and there are better algorithms such as merge sort. !
We must now estimate how large log2 n! is. Since n! is the product of all the integers from 1 to n, it is surely larger than the product of only the n 2 + 1 integers from n/2 through n. This product is in turn at least as large as n/2 multiplied by itself n/2 times, or (n/2)n/2. Thus, log2 n! is at least log2 (n/2)n/2 . But the latter is n 2 (log2 n −log2 2), which is n 2 (log2 n −1) For large n, this formula is approximately (n log2 n)/2.
A more careful analysis will tell us that the factor of 1/2 does not have to be there. That is, log2 n! is very close to n log2 n rather than to half that expression.
166 COMBINATORICS AND PROBABILITY A Linear-Time Special-Purpose Sorting Algorithm If we restrict the inputs on which a sorting algorithm will work, it can in one step divide the possible orders into more than 2 parts and thus work in less than time proportional to n log n. Here is a simple example that works if the input is n distinct integers, each chosen in the range 0 to 2n −1.
(1) for (i = 0; i < 2n; i++) (2) count[i] = 0; (3) for (i = 0; i < n; i++) (4) count[a[i]]++; (5) for (i = 0; i < 2n; i++) (6) if (count[i] > 0) (7) printf("%d\n", i); We assume the input is in an array a of length n. In lines (1) and (2) we initialize an array count of length 2n to 0. Then in lines (3) and (4) we add 1 to the count for x if x is the value of a[i], the ith input element.
Finally, in the last three lines we print each of the integers i such that count[i] is positive.
Thus we print those elements appearing one or more times in the input and, on the assumption the inputs are distinct, it prints all the input elements, sorted smallest first.
We can analyze the running time of this algorithm easily. Lines (1) and (2) are a loop that iterates 2n times and has a body taking O(1) time. Thus, it takes O(n) time. The same applies to the loop of lines (3) and (4), but it iterates n times rather than 2n times; it too takes O(n) time. Finally, the body of the loop of lines (5) through (7) takes O(1) time and it is iterated 2n times. Thus, all three loops take O(n) time, and the entire sorting algorithm likewise takes O(n) time. Note that if given an input for which the algorithm is not tailored, such as integers in a range larger than 0 through 2n −1, the program above fails to sort correctly.
We have shown only that any general-purpose sorting algorithm must have some input for which it makes about n log2 n comparisons or more.
Thus any general-purpose sorting algorithm must take at least time proportional to n log n in the worst case. In fact, it can be shown that the same applies to the “average” input. That is, the average over all inputs of the time taken by a general-purpose sorting algorithm must be at least proportional to n log n. Thus, merge sort is about as good as we can do, since it has this big-oh running time for all inputs.
EXERCISES 4.3.1: Suppose we have selected 9 players for a baseball team.
a) How many possible batting orders are there?
b) If the pitcher has to bat last, how many possible batting orders are there?
4.3.2: How many comparisons does the selection sort algorithm of Fig. 2.2 make if there are 4 elements? Is this number the best possible? Show the top 3 levels of the decision tree in the style of Fig. 4.5.
SEC. 4.4 ORDERED SELECTIONS 167 4.3.3: How many comparisons does the merge sort algorithm of Section 2.8 make if there are 4 elements? Is this number the best possible? Show the top 3 levels of the decision tree in the style of Fig. 4.5.
4.3.4: Are there more assignments of n values to n items or permutations of n+1 items? Note: The answer may not be the same for all n.
4.3.5: Are there more assignments of n/2 values to n items than there are per-mutations of n items?
4.3.6: Show how to sort n integers in the range 0 to n2 −1 in O(n) time.
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4.4 Ordered Selections Sometimes we wish to select only some of the items in a set and give them an order.
Let us generalize the function Π(n) that counted permutations in the previous section to a two-argument function Π(n, m), which we define to be the number of ways we can select m items from n in such a way that order matters for the selected items, but there is no order for the unselected items. Thus, Π(n) = Π(n, n).
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Example 4.5. A horse race awards prizes to the first three finishers; the first horse is said to “win,” the second to “place,” and the third to “show.” Suppose there are 10 horses in a race. How many different awards for win, place, and show are there?
Clearly, any of the 10 horses can be the winner.
Given which horse is the winner, any of the 9 remaining horses can place.
Thus, there are 10 × 9 = 90 choices for horses in first and second positions. For any of these 90 selections of win and place, there are 8 remaining horses. Any of these can finish third. Thus, there are 90 × 8 = 720 selections of win, place, and show. Figure 4.6 suggests all these possible selections, concentrating on the case where 3 is selected first and 1 is selected second. !
The General Rule for Selections Without Replacement Let us now deduce the formula for Π(n, m). Following Example 4.5, we know that there are n choices for the first selection. Whatever selection is first made, there will be n −1 remaining items to choose from. Thus, the second choice can be made in n −1 different ways, and the first two choices occur in n(n −1) ways. Similarly, for the third choice we are left with n −2 unselected items, so the third choice can be made in n −2 different ways. Hence the first three choices can occur in n(n −1)(n −2) distinct ways.
We proceed in this way until m choices have been made. Each choice is made from one fewer item than the choice before. The conclusion is that we may select m items from n without replacement but with order significant in Π(n, m) = n(n −1)(n −2) · · · (n −m + 1) (4.3) different ways. That is, expression (4.3) is the product of the m integers starting and n and counting down.
Another way to write (4.3) is as n!/(n −m)!. That is, 168 COMBINATORICS AND PROBABILITY .
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3 1 1 All but 1 2 All but 2 2 4 5 10 10 All but 10 2 4 10 All but 2, 3 All but 3, 4 All but 3, 10 Fig. 4.6. Ordered selection of three things out of 10.
n!
(n −m)! = n(n −1) · · · (n −m + 1)(n −m)(n −m −1) · · · (1) (n −m)(n −m −1) · · · (1) The denominator is the product of the integers from 1 to n−m. The numerator is the product of the integers from 1 to n. Since the last n −m factors in the numerator and denominator above are the same, (n −m)(n −m −1) · · · (1), they cancel and the result is that n!
(n −m)! = n(n −1) · · · (n −m + 1) This formula is the same as that in (4.3), which shows that Π(n, m) = n!/(n −m)!.
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Example 4.6. Consider the case from Example 4.5, where n = 10 and m = 3. We observed that Π(10, 3) = 10 × 9 × 8 = 720. The formula (4.3) says that Π(10, 3) = 10!/7!, or SEC. 4.4 ORDERED SELECTIONS 169 Selections With and Without Replacement The problem considered in Example 4.5 differs only slightly from the assignment problem considered in Section 4.2. In terms of houses and colors, we could almost see the selection of the first three finishing horses as an assignment of one of ten horses (the “colors”) to each of three finishing positions (the “houses”). The only difference is that, while we are free to paint several houses the same color, it makes no sense to say that one horse finished both first and third, for example. Thus, while the number of ways to color three houses in any of ten colors is 103 or 10 × 10 × 10, the number of ways to select the first three finishers out of 10 is 10 × 9 × 8.
We sometimes refer to the kind of selection we did in Section 4.2 as selection with replacement. That is, when we select a color, say red, for a house, we “replace” Selection with replacement red into the pool of possible colors. We are then free to select red again for one or more additional houses.
On the other hand, the sort of selection we discussed in Example 4.5 is called selection without replacement. Here, if the horse Sea Biscuit is selected to be the Selection without replacement winner, then Sea Biscuit is not replaced in the pool of horses that can place or show. Similarly, if Secretariat is selected for second place, he is not eligible to be the third-place horse also.
10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 7 × 6 × 5 × 4 × 3 × 2 × 1 The factors from 1 through 7 appear in both numerator and denominator and thus cancel. The result is the product of the integers from 8 through 10, or 10 × 9 × 8, as we saw in Example 4.5. !
EXERCISES 4.4.1: How many ways are there to form a sequence of m letters out of the 26 letters, if no letter is allowed to appear more than once, for (a) m = 3 (b) m = 5.
4.4.2: In a class of 200 students, we wish to elect a President, Vice President, Secretary, and Treasurer. In how many ways can these four officers be selected?
4.4.3: Compute the following quotients of factorials: (a) 100!/97! (b) 200!/195!.
4.4.4: The game of Mastermind requires players to select a “code” consisting of a Mastermind sequence of four pegs, each of which may be of any of six colors: red, green, blue, yellow, white, and black.
a) How may different codes are there?
b) How may different codes are there that have two or more pegs of the same color? Hint: This quantity is the difference between the answer to (a) and another easily computed quantity.
c) How many codes are there that have no red peg?
d) How many codes are there that have no red peg but have at least two pegs of the same color?
170 COMBINATORICS AND PROBABILITY Quotients of Factorials Note that in general, a!/b! is the product of the integers between b + 1 and a, as long as b < a. It is much easier to calculate the quotient of factorials as a × (a −1) × · · · × (b + 1) than to compute each factorial and divide, especially if b is not much less than a.
4.4.5: Prove by induction on n that for any m between 1 and n, Π(n, m) = n!/(n −m)!.
4.4.6: Prove by induction on a −b that a!/b! = a(a −1)(a −2) · · · (b + 1).
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4.5 Unordered Selections There are many situations in which we wish to count the ways to select a set of items, but the order in which the selections are made does not matter. In terms of the horse race example of the previous section, we may wish to know which horses were the first three to finish, but we do not care about the order in which these three finished. Put another way, we wish to know how many ways we can select three horses out of n to be the top three finishers.
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Example 4.7. Let us again assume n = 10. We know from Example 4.5 that there are 720 ways to select three horses, say A, B, and C, to be the win, place, and show horses, respectively. However, now we do not care about the order of finish of these three horses, only that A, B, and C were the first three finishers in some order. Thus, we shall get the answer “A, B, and C are the three best horses” in six different ways, corresponding to the ways that these three horses can be ordered among the top three. We know there are exactly six ways, because the number of ways to order 3 items is Π(3) = 3! = 6. However, if there is any doubt, the six ways are seen in Fig. 4.7.
Win Place Show A B C A C B B A C B C A C A B C B A Fig. 4.7. Six orders in which a set of three horses may finish.
What is true for the set of horses A, B, and C is true of any set of three horses.
Each set of three horses will appear exactly 6 times, in all of their possible orders, when we count the ordered selections of three horses out of 10. Thus, if we wish SEC. 4.5 UNORDERED SELECTIONS 171 to count only the sets of three horses that may be the three top finishers, we must divide Π(10, 3) by 6. Thus, there are 720/6 = 120 different sets of three horses out of 10. !
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Example 4.8. Let us count the number of poker hands. In poker, each player is dealt five cards from a 52-card deck. We do not care in what order the five cards are dealt, just what five cards we have. To count the number of sets of five cards we may be dealt, we could start by calculating Π(52, 5), which is the number of ordered selections of five objects out of 52. This number is 52!/(52 −5)!, which is 52!/47!, or 52 × 51 × 50 × 49 × 48 = 311,875,200.
However, just as the three fastest horses in Example 4.7 appear in 3! = 6 different orders, any set of five cards to appear in Π(5) = 5! = 120 different orders.
Thus, to count the number of poker hands without regard to order of selection, we must take the number of ordered selections and divide by 120. The result is 311,875,200/120 = 2,598,960 different hands. !
Counting Combinations Let us now generalize Examples 4.7 and 4.8 to get a formula for the number of ways to select m items out of n without regard to order of selection. This function is usually written n m and spoken “n choose m” or “combinations of m things out of n.” To compute n m , we start with Π(n, m) = n!/(n −m)!, the number of ordered Combinations of m things out of n selections of m things out of n. We then group these ordered selections according to the set of m items selected. Since these m items can be ordered in Π(m) = m!
different ways, the groups will each have m! members. We must divide the number of ordered selections by m! to get the number of unordered selections. That is, n m = Π(n, m) Π(m) = n!
(n −m)! × m!
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Example 4.9. Let us repeat Example 4.8, using formula (4.4) with n = 52 and m = 5. We have 52 5 = 52!/(47! × 5!). If we cancel the 47! with the last 47 factors of 52! and expand 5!, we can write 52 5 = 52 × 51 × 50 × 49 × 48 5 × 4 × 3 × 2 × 1 Simplifying, we get 52 5 = 26 × 17 × 10 × 49 × 12 = 2,598,960. !
A Recursive Definition of n Choose m If we think recursively about the number of ways to select m items out of n, we can develop a recursive algorithm to compute n m .
BASIS.
n 0 = 1 for any n ≥1. That is, there is only one way to pick zero things out of n: pick nothing. Also, n n = 1; that is, the only way to pick n things out of n is to pick them all.
172 COMBINATORICS AND PROBABILITY INDUCTION. If 0 < m < n, then n m = n−1 m + n−1 m−1 . That is, if we wish to pick m things out of n, we can either i) Not pick the first element, and then pick m things from the remaining n −1 elements. The term n−1 m counts this number of possibilities.
or ii) Pick the first element and then select m −1 things from among the remaining n −1 elements. The term n−1 m−1 counts these possibilities.
Incidently, while the idea of the induction should be clear — we proceed from the simplest cases of picking all or none to more complicated cases where we pick some but not all — we have to be careful to state what quantity the induction is “on.” One way to look at this induction is that it is a complete induction on the product of n and the minimum of m and n −m. Then the basis case occurs when this product is 0 and the induction is for larger values of the product. We have to check for the induction that n × min(m, n −m) is always greater than (n −1) × min(m, n −m −1) and (n −1) × min(m −1, n −m) when 0 < m < n.
This check is left as an exercise.
This recursion is often displayed by Pascal’s triangle, illustrated in Fig. 4.8, Pascal’s triangle where the borders are all 1’s (for the basis) and each interior entry is the sum of the two numbers above it to the northeast and northwest (for the induction). Then n m can be read from the (m + 1)st entry of the (n + 1)st row.
1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 Fig. 4.8. The first rows of Pascal’s triangle.
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Example 4.10. Consider the case where n = 4 and m = 2. We find the value of 4 2 in the 3rd entry of the 5th row of Fig. 4.8. This entry is 6, and it is easy to check that 4 2 = 4!/(2! × 2!) = 24/(2 × 2) = 6. !
The two ways we have to compute n m — by formula (4.4) or by the above recursion — each compute the same value, naturally. We can argue so by appeal to physical reasoning. Both methods compute the number of ways to select m items out of n in an unordered fashion, so they must produce the same value. However, we can also prove the equality of the two approaches by an induction on n. We leave this proof as an exercise.
SEC. 4.5 UNORDERED SELECTIONS 173 Running Time of Algorithms to Compute n m As we saw in Example 4.9, when we use formula (4.4) to compute n m we can cancel (n −m)! in the denominator against the last n −m factors in n! to express n m as n m = n × (n −1) × · · · × (n −m + 1) m × (m −1) × · · · × 1 (4.5) If m is small compared to n, we can evaluate the above formula much faster than we can evaluate (4.4). In principle, the fragment of C code in Fig. 4.9 does the job.
(1) c = 1; (2) for (i = n; i > n-m; i--) (3) c = i; (4) for (i = 2; i <= m; i++) (5) c /= i; Fig. 4.9. Code to compute n m .
Line (1) initializes c to 1; c will become the result, n m . Lines (2) and (3) multiply c by each of the integers between n −m + 1 and n. Then, lines (4) and (5) divide c by each of the integers between 2 and m. Thus, Fig. 4.9 implements the formula of Equation (4.5).
For the running time of Fig. 4.9, we have only to observe that the two loops, lines (2) – (3) and lines (4) – (5), each iterate m times and have a body that takes O(1) time. Thus, the running time is O(m).
In the case that m is close to n but n−m is small, we can interchange the role of m and n−m. That is, we can cancel factors of n! and m!, getting n(n−1) · · · (m+1) and divide that by (n −m)!. This approach gives us an alternative to (4.5), which is n m = n × (n −1) × · · · × (m + 1) (n −m) × (n −m −1) × · · · × 1 (4.6) Likewise, there is a code fragment similar to Fig. 4.9 that implements formula (4.6) and takes time O(n −m). Since both n −m and m must be n or less for n m to be defined, we know that either way, O(n) is a bound on the running time. Moreover, when m is either close to 0 or close to n, then the running time of the better of the two approaches is much less than O(n).
However, Fig. 4.9 is flawed in an important way. It starts by computing the product of a number of integers and then divides by an equal number of integers.
Since ordinary computer arithmetic can only deal with integers of a limited size (often, about two billion is as large as an integer can get), we run the risk of computing an intermediate result after line (3) of Fig. 4.9 that overflows the limit on integer size. That may be the case even though the value of n m is small enough to be represented in the computer.
A more desirable approach is to alternate multiplications and divisions. Start by multiplying by n, then divide by m. Multiply by n −1; then divide by m −1, and so on. The problem with this approach is that we have no reason to believe the result will be an integer at each stage. For instance, in Example 4.9 we would begin by multiplying by 52 and dividing by 5. The result is already not an integer.
174 COMBINATORICS AND PROBABILITY Formulas for n m Must Yield Integers It may not be obvious why the quotients of many factors in Equations (4.4), (4.5), or (4.6) must always turn out to be an integer. The only simple argument is to appeal to physical reasoning. The formulas all compute the number of ways to choose m things out of n, and this number must be some integer.
It is much harder to argue this fact from properties of integers, without appeal-ing to the physical meaning of the formulas. It can in fact be shown by a careful analysis of the number of factors of each prime in numerator and denominator. As a sample, look at the expression in Example 4.9. There is a 5 in the denominator, and there are 5 factors in the numerator. Since these factors are consecutive, we know one of them must be divisible by 5; it happens to be the middle factor, 50.
Thus, the 5 in the denominator surely cancels.
Thus, we need to convert to floating-point numbers before doing any calculation.
We leave this modification as an exercise.
Now, let us consider the recursive algorithm to compute n m . We can imple-ment it by the simple recursive function of Fig. 4.10.
/ compute n choose m for 0 <= m <= n / int choose(int n, int m) { int n, m; (1) if (m < 0 || m > n) {/ error conditions / (2) printf("invalid input\n"); (3) return 0; } (4) else if (m == 0 || m == n) / basis case / (5) return 1; else / induction / (6) return (choose(n-1, m-1) + choose(n-1, m)); } Fig. 4.10. Recursive function to compute n m .
The function of Fig. 4.10 is not efficient; it creates an exponential explosion in the number of calls to choose. The reason is that when called with n as its first argument, it usually makes two recursive calls at line (6) with first argument n −1.
Thus, we might expect the number of calls made to double when n increases by 1.
Unfortunately, the exact number of recursive calls made is harder to count. The reason is that the basis case on lines (4) and (5) can apply not only when n = 1, but for higher n, provided m has the value 0 or n.
We can prove a simple, but slightly pessimistic upper bound as follows. Let T (n) be the running time of Fig. 4.10 with first argument n. We can prove that T (n) is O(2n) simply. Let a be the total running time of lines (1) through (5), plus SEC. 4.5 UNORDERED SELECTIONS 175 that part of line (6) that is involved in the calls and return, but not the time of the recursive calls themselves. Then we can prove by induction on n: STATEMENT S(n): If choose is called with first argument n and some second argument m between 0 and n, then the running time T (n) of the call is at most a(2n −1).
BASIS. n = 1. Then it must be that either m = 0 or m = 1 = n. Thus, the basis case on lines (4) and (5) applies and we make no recursive calls. The time for lines (1) through (5) is included in a. Since S(1) says that T (1) is at most a(21 −1) = a, we have proved the basis.
INDUCTION. Assume S(n); that is, T (n) ≤a(2n −1). To prove S(n + 1), suppose we call choose with first argument n+1. Then Fig. 4.10 takes time a plus the time of the two recursive calls on line (6). By the inductive hypothesis, each call takes at most time a(2n −1). Thus, the total time consumed is at most a + 2a(2n −1) = a(1 + 2n+1 −2) = a(2n+1 −1) This calculation proves S(n + 1) and proves the induction.
We have thus proved that T (n) ≤a(2n −1). Dropping the constant factor and the low-order terms, we see that T (n) is O(2n).
Curiously, while in our analyses of Chapter 3 we easily proved a smooth and tight upper bound on running time, the O(2n) bound on T (n) is smooth but not tight. The proper smooth, tight upper bound is slightly less: O(2n/√n). A proof of this fact is quite difficult, but we leave as an exercise the easier fact that the running time of Fig. 4.10 is proportional to the value it returns: n m . An important observation is that the recursive algorithm of Fig. 4.10 is much less efficient that the linear algorithm of Fig. 4.9. This example is one where recursion hurts considerably.
The Shape of the Function n m For a fixed value of n, the function of m that is n m has a number of interesting properties.
For a large value of n, its form is the bell-shaped curve suggested in Fig. 4.11. We immediately notice that this function is symmetric around the Bell curve midpoint n/2; this is easy to check using formula (4.4) that states n m = n n−m .
The maximum height at the center, that is, n n/2 , is approximately 2n/ p πn/2.
For example, if n = 10, this formula gives 258.37, while 10 5 = 252.
The “thick part” of the curve extends for approximately √n on either side of the midpoint. For example, if n = 10, 000, then for m between 4900 and 5100 the value of 10,000 m is close to the maximum. For m outside this range, the value of 10,000 m falls offvery rapidly.
176 COMBINATORICS AND PROBABILITY 0 n/2 n √n 2n/ p πn/2 Fig. 4.11. The function n m for fixed n.
Binomial Coefficients The function n m , in addition to its use in counting, gives us the binomial co-efficients. These numbers are found when we expand a two-term polynomial (a binomial) raised to a power, such as (x + y)n.
Binomial When we expand (x + y)n, we get 2n terms, each of which is xmyn−m for some m between 0 and n. That is, from each of the factors x + y we may choose either x or y to contribute as a factor in a particular term. The coefficient of xmyn−m in the expansion is the number of terms that are composed of m choices of x and the remaining n −m choices of y.
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Example 4.11. Consider the case n = 4, that is, the product (x + y)(x + y)(x + y)(x + y) There are 16 terms, of which only one is x4y0 (or just x4). This term is the one we get if we select x from each of the four factors. On the other hand, there are four terms x3y, corresponding to the fact that we can select y from any of the four factors and x from the remaining three factors. Symmetrically, we know that there is one term y4 and four terms xy3.
How many terms x2y2 are there? We get such a term if we select x from two of the four factors and y from the remaining two. Thus, we must count the number of ways to select two of the factors out of four. Since the order in which we select the two doesn’t matter, this count is 4 2 = 4!/(2! × 2!) = 24/4 = 6. Thus, there are six terms x2y2. The complete expansion is (x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4 Notice that the coefficients of the terms on the right side of the equality, (1, 4, 6, 4, 1), are exactly a row of Pascal’s triangle in Fig. 4.8. That is no coincidence, as we shall see. !
We can generalize the idea that we used to calculate the coefficient of x2y2 in Example 4.11. The coefficient of xmyn−m in the expansion of (x + y)n is n m .
The reason is that we get a term xmyn−m whenever we select m of the n factors to SEC. 4.5 UNORDERED SELECTIONS 177 provide an x and the remainder of the factors to provide y. The number of ways to choose m factors out of n is n m .
There is another interesting consequence of the relationship between binomial coefficients and the function n m . We just observed that (x + y)n = n X m=0 n m xmyn−m Let x = y = 1. Then (x + y)n = 2n. All powers of x and y are 1, so the above equation becomes 2n = n X m=0 n m Put another way, the sum of all the binomial coefficients for a fixed n is 2n. In particular, each coefficient n m is less than 2n. The implication of Fig. 4.11 is that for m around n/2, n m is quite close to 2n.
Since the area under the curve of Fig. 4.11 is 2n, we see why only a few values near the middle can be large.
EXERCISES 4.5.1: Compute the following values: (a) 7 3 (b) 8 3 (c) 10 7 (d) 12 11 .
4.5.2: In how many ways can we choose a set of 5 different letters out of the 26 possible lower-case letters?
4.5.3: What is the coefficient of a) x3y4 in the expansion (x + y)7 b) x5y3 in the expansion of (x + y)8 4.5.4: At Real Security, Inc., computer passwords are required to have four digits (of 10 possible) and six letters (of 52 possible). Letters and digits may repeat. How many different possible passwords are there? Hint: Start by counting the number of ways to select the four positions holding digits.
4.5.5: How many sequences of 5 letters are there in which exactly two are vowels?
4.5.6: Rewrite the fragment of Fig. 4.9 to take advantage of the case when n −m is small compared with n.
4.5.7: Rewrite the fragment of Fig. 4.9 to convert to floating-point numbers and alternately multiply and divide.
4.5.8: Prove that if 0 ≤m ≤n, then n m = n n−m a) By appealing to the meaning of the function n m b) By using Equation 4.4 4.5.9: Prove by induction on n that the recursive definition of n m correctly defines n m to be equal to n!/ (n −m)! × m!
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4.5.10: Show by induction on n that the running time of the recursive function choose(n,m) of Fig. 4.10 is at most c n m for some constant c.
4.5.11: Show that n × min(m, n −m) is always greater than 178 COMBINATORICS AND PROBABILITY (n −1) × min(m, n −m −1) and (n −1) × min(m −1, n −m) when 0 < m < n.
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4.6 Orderings With Identical Items In this section, we shall examine a class of selection problems in which some of the items are indistinguishable from one another, but the order of appearance of the items matters when items can be distinguished. The next section will address a similar class of problems where we do not care about order, and some items are indistinguishable.
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Example 4.12. Anagram puzzles give us a list of letters, which we are asked Anagrams to rearrange to form a word. We can solve such problems mechanically if we have a dictionary of legal words and we can generate all the possible orderings of the letters. Chapter 10 considers efficient ways to check whether a given sequence of letters is in the dictionary. But now, considering combinatorial problems, we might start by asking how many different potential words we must check for presence in the dictionary.
For some anagrams, the count is easy. Suppose we are given the letters abenst.
There are six letters, which may be ordered in Π(6) = 6! = 720 ways. One of these 720 ways is absent, the “solution” to the puzzle.
However, anagrams often contain duplicate letters. Consider the puzzle eilltt.
There are not 720 different sequences of these letters. For example, interchanging the positions in which the two t’s appear does not make the word different.
Suppose we “tagged” the t’s and l’s so we could distinguish between them, say t1, t2, l1, and l2. Then we would have 720 orders of tagged letters. However, pair of orders that differ only in the position of the tagged l’s, such as l1it2t1l2e and l2it2t1l1e, are not really different. Since all 720 orders group into pairs differing only in the subscript on the l’s, we can account for the fact that the l’s are really identical if we divide the number of strings of letters by 2. We conclude that the number of different anagrams in which the t’s are tagged but the l’s are not is 720/2=360.
Similarly, we may pair the strings with only t’s tagged if they differ only in the subscript of the t’s. For example, lit1t2le and lit2t1le are paired. Thus, if we divide by 2 again, we have the number of different anagram strings with the tags removed from both t’s and l’s. This number is 360/2 = 180. We conclude that there are 180 different anagrams of eilltt. !
We may generalize the idea of Example 4.12 to a situation where there are n items, and these items are divided into k groups. Members of each group are indistinguishable, but members of different groups are distinguishable. We may let mi be the number of items in the ith group, for i = 1, 2, . . . , k.
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Example 4.13. Reconsider the anagram problem eilltt from Example 4.12.
Here, there are six items, so n = 6. The number of groups k is 4, since there are SEC. 4.6 ORDERINGS WITH IDENTICAL ITEMS 179 4 different letters. Two of the groups have one member (e and i), while the other two groups have two members. We may thus take i1 = i2 = 1 and i3 = i4 = 2. !
If we tag the items so members of a group are distinguishable, then there are n! different orders. However, if there are i1 members of the first group, these tagged items may appear in i1! different orders. Thus, when we remove the tags from the items in group 1, we cluster the orders into sets of size i1! that become identical.
We must thus divide the number of orders by i1! to account for the removal of tags from group 1.
Similarly, removing the tags from each group in turn forces us to divide the number of distinguishable orders by i2!, by i3!, and so on. For those ij’s that are 1, this division is by 1! = 1 and thus has no effect. However, we must divide by the factorial of the size of each group of more than one item. That is what happened in Example 4.12. There were two groups with more than one member, each of size 2, and we divided by 2! twice. We can state and prove the general rule by induction on k.
STATEMENT S(k): If there are n items divided into k groups of sizes i1, i2, . . . , ik respectively, items within a group are not distinguishable, but items in differ-ent groups are distinguishable, then the number of different distinguishable orders of the n items is n!
Qk j=1 ij!
(4.7) BASIS. If k = 1, then there is one group of indistinguishable items, which gives us only one distinguishable order no matter how large n is. If k = 1 then i1 must be n, and formula (4.7) reduces to n!/n!, or 1. Thus, S(1) holds.
INDUCTION. Suppose S(k) is true, and consider a situation with k+1 groups. Let the last group have m = ik+1 members. These items will appear in m positions, and we can choose these positions in n m different ways. Once we have chosen the m positions, it does not matter which items in the last group we place in these positions, since they are indistinguishable.
Having chosen the positions for the last group, we have n −m positions left to fill with the remaining k groups. The inductive hypothesis applies and tells us that each selection of positions for the last group can be coupled with (n−m)!/ Qk j=1 ij!
distinguishable orders in which to place the remaining groups in the remaining positions. This formula is just (4.7) with n −m replacing n since there are only n −m items remaining to be placed. The total number of ways to order the k + 1 groups is thus n m (n −m)!
Qk j=1 ij!
(4.8) Let us replace n m in (4.8) by its equivalent in factorials: n!/ (n−m)!m!
. We then have 180 COMBINATORICS AND PROBABILITY n!
(n −m)!m!
(n −m)!
Qk j=1 ij!
(4.9) We may cancel (n−m)! from numerator and denominator in (4.8). Also, remember that m is ik+1, the number of members in the (k + 1)st group. We thus discover that the number of orders is n!
Qk+1 j=1 ij!
This formula is exactly what is given by S(k + 1).
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Example 4.14. An explorer has rations for two weeks, consisting of 4 cans of Tuna, 7 cans of Spam, and 3 cans of Beanie Weenies. If he opens one can each day, in how many orders can he consume the rations? Here, there are 14 items divided into groups of 4, 7, and 3 identical items. In terms of Equation (4.7), n = 14, k = 3, i1 = 4, i2 = 7, and i3 = 3. The number of orders is thus 14!
4!7!3!
Let us begin by canceling the 7! in the denominator with the last 7 factors in 14! of the numerator. That gives us 14 × 13 × 12 × 11 × 10 × 9 × 8 4 × 3 × 2 × 1 × 3 × 2 × 1 Continuing to cancel factors in the numerator and denominator, we find the resulting product is 120,120. That is, there are over a hundred thousand ways in which to consume the rations. None sounds very appetizing. !
EXERCISES 4.6.1: Count the number of anagrams of the following words: (a) error (b) street (c) allele (d) Mississippi.
4.6.2: In how many ways can we arrange in a line a) Three apples, four pears, and five bananas b) Two apples, six pears, three bananas, and two plums 4.6.3: In how many ways can we place a white king, a black king, two white knights, and a black rook on the chessboard?
4.6.4: One hundred people participate in a lottery. One will win the grand prize of $1000, and five more will win consolation prizes of a $50 savings bond. How many different possible outcomes of the lottery are there?
4.6.5: Write a simple formula for the number of orders in which we may place 2n objects that occur in n pairs of two identical objects each.
SEC. 4.7 DISTRIBUTION OF OBJECTS TO BINS 181 !
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4.7 Distribution of Objects to Bins Our next class of counting problems involve the selection of a bin in which to place each of several objects. The objects may or may not be identical, but the bins are distinguishable. We must count the number of ways in which the bins can be filled.
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Example 4.15.
Kathy, Peter, and Susan are three children.
We have four apples to distribute among them, without cutting apples into parts. In how many different ways may the children receive apples?
There are sufficiently few ways that we can enumerate them. Kathy may receive anything from 0 to 4 apples, and whatever remains can be divided between Peter and Susan in only a few ways. If we let (i, j, k) represent the situation in which Kathy receives i apples, Peter receives j, and Susan receives k, the 15 possibilities are as shown in Fig. 4.12. Each row corresponds to the number of apples given to Kathy.
(0,0,4) (0,1,3) (0,2,2) (0,3,1) (0,4,0) (1,0,3) (1,1,2) (1,2,1) (1,3,0) (2,0,2) (2,1,1) (2,2,0) (3,0,1) (3,1,0) (4,0,0) Fig. 4.12. Four apples can be distributed to three children in 15 ways.
There is a trick to counting the number of ways to distribute identical objects to bins. Suppose we have four letter A’s representing apples. Let us also use two ’s which will represent partitions between the apples belonging to different children.
We order the A’s and ’s as we like and interpret all A’s before the first as being apples belonging to Kathy. Those A’s between the two ’s belong to Peter, and the A’s after the second are apples belonging to Susan. For instance, AAAA represents the distribution (2,1,1), where Kathy gets two apples and the other two children get one each. Sequence AAAA represents the situation (3,1,0), where Kathy gets three, Peter gets one, and Susan gets none.
Thus, each distribution of apples to bins is associated with a unique string of 4 A’s and 2 ’s. How many such strings are there? Think of the six positions forming such a string. Any four of those positions may be selected to hold the A’s; the other two will hold the ’s. The number of ways to select 4 items out of 6 is 6 4 , as we learned in Section 4.5. Since 6 4 = 15, we again conclude that there are 15 ways to distribute four apples to three children. !
The General Rule for Distribution to Bins We can generalize the problem of Example 4.15 as follows. Suppose we are given n bins; these correspond to the three children in the example. Suppose also we are given m identical objects to place arbitrarily into the bins. How many distributions into bins are there.
We may again think of strings of A’s and ’s. The A’s now represent the objects, and the ’s represent boundaries between the bins. If there are n objects, we use n 182 COMBINATORICS AND PROBABILITY A’s, and if there are m bins, we need m −1 ’s to serve as boundaries between the portions for the various bins. Thus, strings are of length n + m −1.
We may choose any n of these positions to hold A’s, and the rest will hold ’s.
There are thus n+m−1 n strings of A’s and ’s, so there are this many distributions of objects to bins. In Example 4.15 we had n = 4 and m = 3, and we concluded that there were n+m−1 n = 6 4 distributions.
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Example 4.16. In the game of Chuck-a-Luck we throw three dice, each with Chuck-a-Luck six sides numbered 1 through 6. Players bet a dollar on a number. If the number does not come up, the dollar is lost. If the number comes up one or more times, the player wins as many dollars as there are occurrences of the number.
We would like to count the “outcomes,” but there may initially be some ques-tion about what an “outcome” is. If we were to color the dice different colors so we could tell them apart, we could see this counting problem as that of Section 4.2, where we assign one of six numbers to each of three dice.
We know there are 63 = 216 ways to make this assignment.
However, the dice ordinarily aren’t distinguishable, and the order in which the numbers come up doesn’t matter; it is only the occurrences of each number that determines which players get paid and how much. For instance, we might observe that 1 comes up on two dice and 6 comes up on the third.
The 6 might have appeared on the first, second, or third die, but it doesn’t matter which.
Thus, we can see the problem as one of distribution of identical objects to bins.
The “bins” are the numbers 1 through 6, and the “objects” are the three dice. A die is “distributed” to the bin corresponding to the number thrown on that die.
Thus, there are 6+3−1 3 = 8 3 = 56 different outcomes in Chuck-a-Luck. !
Distributing Distinguishable Objects We can extend the previous formula to allow for distribution into m bins of a collection of n objects that fall into k different classes.
Objects within a class are indistinguishable from each other, but are distinguishable from those of other classes. Let us use symbol ai to represent members of the ith class. We may thus form strings consisting of 1.
For each class i, as many ai’s as there are members of that class.
2.
m −1 ’s to represent the boundaries between the m bins.
The length of these strings is thus n + m −1. Note that the ’s form a (k + 1)st class, with m members.
We learned how to count the number of such strings in Section 4.6. There are (n + m −1)!
(m −1)! Qk j=1 ij!
strings, where ij is the number of members of the jth class.
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Example 4.17. Suppose we have three apples, two pears, and a banana to distribute to Kathy, Peter, and Susan. Then m = 3, the number of “bins,” which is the number of children. There are k = 3 groups, with i1 = 3, i2 = 2, and i3 = 1.
Since there are 6 objects in all, n = 6, and the strings in question are of length SEC. 4.7 DISTRIBUTION OF OBJECTS TO BINS 183 Comparison of Counting Problems In this and the previous five sections we have considered six different counting problems. Each can be thought of as assigning objects to certain positions. For example, the assignment problem of Section 4.2 may be thought of as one where we are given n positions (corresponding to the houses) and and infinite supply of objects of k different types (the colors). We can classify these problems along three axes: 1.
Do they place all the given objects?
2.
Is the order in which objects are assigned important?
3.
Are all the objects distinct, or are some indistinguishable?
Here is a table indicating the differences between the problems mentioned in each of these sections.
SECTION TYPICAL MUST USE ORDER IDENTICAL PROBLEM ALL?
IMPORTANT?
OBJECTS?
4.2 Painting houses N Y N 4.3 Sorting Y Y N 4.4 Horse race N Y N 4.5 Poker hands N N Y 4.6 Anagrams Y Y Y 4.7 Apples to children Y N Y The problems of Sections 4.2 and 4.4 are not differentiated in the table above.
The distinction is one of replacement, as discussed in the box on “Selections With and Without Replacement” in Section 4.4. That is, in Section 4.2 we had an infinite supply of each “color” and could select a color many times. In Section 4.4, a “horse” selected is not available for later selections.
n+m−1 = 8. The strings consist of three A’s standing for apples, two P’s standing for pears, one B standing for the banana, and two ’s, the boundaries between the shares of the children. The formula for the number of distributions is thus (n + m −1)!
(m −1)!i1!i2!i3! = 8!
2!3!2!1! = 1680 ways in which these fruits may be distributed to Kathy, Peter, and Susan. !
EXERCISES 4.7.1: In how many ways can we distribute a) Six apples to four children b) Four apples to six children c) Six apples and three pears to five children d) Two apples, five pears, and six bananas to three children 4.7.2: How many outcomes are there if we throw 184 COMBINATORICS AND PROBABILITY (a) Four indistinguishable dice b) Five indistinguishable dice 4.7.3: How many ways can we distribute seven apples to three children so that each child gets at least one apple?
4.7.4: Suppose we start at the lower-left corner of a chessboard and move to the upper-right corner, making moves that are each either one square up or one square right. In how many ways can we make the journey?
4.7.5: Generalize Exercise 4.7.4. If we have a rectangle of n squares by m squares, and we move only one square up or one square right, in how many ways can we move from lower-left to upper-right?
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4.8 Combining Counting Rules The subject of combinatorics offers myriad challenges, and few are as simple as those discussed so far in this chapter. However, the rules learned so far are valuable building blocks that may be combined in various ways to count more complex structures. In this section, we shall learn three useful “tricks” for counting: 1.
Express a count as a sequence of choices.
2.
Express a count as a difference of counts.
3.
Express a count as a sum of counts for subcases.
Breaking a Count Into a Sequence of Choices One useful approach to be taken, when faced with the problem of counting some class of arrangements is to describe the things to be counted in terms of a series of choices, each of which refines the description of a particular member of the class.
In this section we present a series of examples intended to suggest some of the possibilities.
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Example 4.18.
Let us count the number of poker hands that are one-pair hands. A hand with one pair consists of two cards of one rank and three cards of ranks1 that are different and also distinct from the rank of the pair. We can describe all one-pair hands by the following steps.
1.
Select the rank of the pair.
2.
Select the three ranks for the other three cards from the remaining 12 ranks.
3.
Select the suits for the two cards of the pair.
4.
Select the suits for each of the other three cards.
1 The 13 ranks are Ace, King, Queen, Jack, and 10 through 2.
SEC. 4.8 COMBINING COUNTING RULES 185 If we multiply all these numbers together, we shall have the number of one-pair hands. Note that the order in which the cards appear in the hand is not important, as we discussed in Example 4.8, and we have made no attempt to specify the order.
Now, let us take each of these factors in turn. We can select the rank of the pair in 13 different ways. Whichever rank we select for the pair, we have 12 ranks left.
We must select 3 of these for the remaining cards of the hand. This is a selection in which order is unimportant, as discussed in Section 4.5. We may perform this selection in 12 3 = 220 ways.
Now, we must select the suits for the pair. There are four suits, and we must select two of them. Again we have an unordered selection, which we may do in 4 2 = 6 ways. Finally, we must select a suit for each of the three remaining cards.
Each has 4 choices of suit, so we have an assignment like those of Section 4.2. We may make this assignment in 43 = 64 ways.
The total number of one-pair hands is thus 13 × 220 × 6 × 64 = 1, 098, 240.
This number is over 40% of the total number of 2,598,960 poker hands. !
Computing a Count as a Difference of Counts Another useful technique is to express what we want to count as the difference between some more general class C of arrangements and those in C that do not meet the condition for the thing we want to count.
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Example 4.19. There are a number of other poker hands — two pairs, three of a kind, four of a kind, and full house — that can be counted in a manner similar to Example 4.18. However, there are other hands that require a different approach.
First, let us consider a straight-flush, which is five cards of consecutive rank (a straight) of the same suit (a flush). First, each straight begins with one of the ten ranks Ace through 10 as the lowest card. That is, the straights are Ace-2-3-4-5, 2-3-4-5-6, 3-4-5-6-7, and so on, up to 10-Jack-Queen-King-Ace. Once the ranks are determined, the straight-flush can be completely specified by giving the suit. Thus, we can count the straight-flushes by 1.
Select the lowest rank in the straight (10 choices).
2.
Select the suit (4 choices).
Thus, there are 10 × 4 = 40 straight-flushes.
Now, let us count the straights, that is, those hands whose ranks are consec-utive but that are not straight-flushes. We shall first count all those hands with consecutive ranks, regardless of whether the suits are the same, and then subtract the 40 straight-flushes. To count hands with consecutive ranks, we can 1.
Select the low rank (10 choices).
2.
Assign a suit to each of the five ranks (45 = 1024 choices, as in Section 4.2).
The number of straights and straight-flushes is thus 10 × 1024 = 10,240. When we subtract the straight-flushes, we are left with 10,240−40 = 10,200 hands that are classified as straights.
Next, let us count the number of flushes.
Again, we shall first include the straight-flushes and then subtract 40. We can define a flush by 186 COMBINATORICS AND PROBABILITY 1.
Select the suit (4 choices).
2.
Select the five ranks out of thirteen ranks in any of 13 5 = 1287 ways, as in Section 4.5.
We conclude that the number of flushes is 4 × 1287 −40 = 5108. !
Expressing a Count as a Sum of Subcases Our third “trick” is to be methodical when faced with a problem that is too hard to solve directly. We break the problem of counting a class C into two or more separate problems, where each member of class C is covered by exactly one of the subproblems.
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Example 4.20. Suppose we toss a sequence of 10 coins. In how many sequences will 8 or more of the coins be heads? If we wanted to know how many sequences had exactly 8 heads, we could answer the problem by the method of Section 4.5.
That is, there are 10 coins, and we wish to select 8 of them to be heads. We can do so in 10 8 = 45 ways.
To solve the problem of counting sequences of 8 or more heads, we break it into the three subproblems of counting sequences with exactly 8 heads, exactly 9 heads, and exactly 10 heads. We already did the first. The number of sequences with 9 heads is 10 9 = 10, and there is 10 10 = 1 sequence with all 10 heads. Thus, the number of sequences with 8 or more heads is 45 + 10 + 1 = 56. !
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Example 4.21.
Let us reconsider the problem of counting the outcomes in Chuck-a-Luck, which we solved in Example 4.16. Another approach is to divide the problem into three subproblems, depending on whether the number of different numbers showing is 3, 2, or 1.
a) We can count the number of outcomes with three different numbers using the technique of Section 4.5. That is, we must select 3 out of the 6 possible numbers on a die, and we can do so in 6 3 = 20 different ways.
b) Next, we must count the number of outcomes with two of one number and one of another. There are 6 choices for the number that appears twice, and no matter which number we choose to appear twice, there are 5 choices for the number that appears once. There are thus 6 × 5 = 30 outcomes with two of one number and one of another.
c) One number on all three dice can occur in 6 different ways, one for each of the numbers on a die.
Thus, the number of possible outcomes is 20 + 30 + 6 = 56, as we also deduced in Example 4.16. !
EXERCISES 4.8.1: Count the number of poker hands of the following types: SEC. 4.9 INTRODUCTION TO PROBABILITY THEORY 187 a) Two pairs b) Three of a kind c) Full house d) Four of a kind Be careful when counting one type of hand not to include hands that are better.
For example, in (a), make sure that the two pairs are different (so you don’t really have four of a kind) and the fifth card is different from the pairs (so you don’t have a full house).
4.8.2: A blackjack consists of two cards, one of which is an Ace and the other of Blackjack which is a 10-point card, either a 10, Jack, Queen, or King.
a) How many different blackjacks are there in a 52-card deck?
b) In the game blackjack, one card is dealt down and the other is dealt up. Thus, in a sense order of the two cards matters. In this case, how many different blackjacks are there?
c) In a pinochle deck there are eight cards of each rank 9, 10, Jack, Queen, King, Pinochle deck Ace (two indistinguishable cards of each suit) and no other cards. How many blackjacks are there, assuming order is unimportant?
4.8.3: How many poker hands are “nothing” (i.e., not one-pair or better)? You may use the results of Examples 4.18 and 4.19 as well as the answer to Exercise 4.8.1.
4.8.4: If we toss 12 coins in sequence, how many have a) At least 9 heads b) At most 4 heads c) Between 5 and 7 heads d) Fewer than 2 or more than 10 heads 4.8.5: How many outcomes in Chuck-a-Luck have at least one 1?
4.8.6: How many anagrams of the word little are there in which the two t’s are not adjacent?
4.8.7: A bridge hand consists of 13 of the 52 cards. We often classify hands by Bridge “distribution,” that is, the way cards are grouped into suits. For example, a hand of 4-3-3-3 distribution has four of one suit and three of each of the other suits. A hand with 5-4-3-1 distribution has one suit of five cards, one of four, one of three, and one of one card. Count the number of hands with the following distributions: (a) 4-3-3-3 (b) 5-4-3-1 (c) 4-4-3-2 (d) 9-2-2-0.
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4.9 Introduction to Probability Theory Probability theory, along with its general importance, has many uses in computer science. One important application is the estimation of the running time of pro-grams in the case of average or typical inputs. This evaluation is important for those algorithms whose worst-case running time is very much larger than the aver-age running time. We shall see examples of such evaluations shortly.
Another use of probability is in designing algorithms for making decisions in the presence of uncertainty. For example, we can use probability theory to design 188 COMBINATORICS AND PROBABILITY algorithms for making the best possible medical diagnosis from available information or algorithms for allocating resources on the basis of expected future needs.
Probability Spaces When we speak of a probability space, we mean a finite set of points, each of which represents one possible outcome of an experiment. Each point x has associated with Experiment, outcome it a nonnegative real number called the probability of x, such that the sum of the probabilities of all the points is 1. One also speaks of probability spaces that have infinite numbers of points, although there is little application for these in computer science, and we shall not deal with them here.
Commonly, the points of a probability space have equal probability. Unless we state otherwise, you may assume that the probabilities of the various points in a probability space are equal. Thus, if there are n points in the probability space the probability of each point is 1/n.
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Example 4.22. In Fig. 4.13 is a probability space with six points. The points are each identified with one of the numbers from 1 to 6, and we may think of this space as representing the outcomes of the “experiment” of throwing a single fair die. That is, one of the six numbers will appear on top of the die, and each number is equally likely to appear, that is, 1/6th of the time. !
• • • • • • 1 2 3 5 6 4 Fig. 4.13. A probability space with six points.
Any subset of the points in a probability space is called an event. The probability Event of an event E, denoted PROB(E), is the sum of the probabilities of the points in E. If the points are all equally likely, then we can compute the probability of E by dividing the number of points in E by the number of points in the total probability space.
Probability Calculations Often, the calculation of the probability of an event involves combinatorics. We must count the number of points in the event as well as the number of points in the entire probability space. When points are equally likely, the ratio of these two counts is the probability of the event. We shall give a series of examples to illustrate the calculation of probabilities in this fashion.
SEC. 4.9 INTRODUCTION TO PROBABILITY THEORY 189 Infinite Probability Spaces In certain circumstances, we can imagine that a probability space has an infinite number of points. The probability of any given point may be infinitesimal, and we can only associate finite probabilities with some collections of points. For a simple example, here is a probability space, the square, whose points are all the points of the plane within the square.
E Event We may suppose that any point in the square is equally likely to be chosen.
The “experiment” may be thought of as throwing a dart at the square in such a way that the dart is equally likely to wind up anywhere within the square, but not outside it. Although any point has only infinitesimal probability of being hit, the probability of a region of the square is the ratio of the area of the region to the area of the entire square. Thus, we can compute the probability of certain events.
For example, we show within the probability space an event E consisting of an ellipse contained within the square. Let us suppose the area of the ellipse is 29% of the area of the square. Then PROB(E) is 0.29. That is, if the dart is thrown at random at the square, 29% of the time the dart will land within the ellipse.
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Example 4.23. Figure 4.14 shows the probability space representing the throw of two dice. That is, the experiment is the tossing of two dice, in order, and observing the numbers on their upper faces. Assuming the dice are fair, there are 36 equally likely points, or outcomes of the experiment, so each point has probability 1/36.
Each point corresponds to the assignment of one of six values to each die.
For example, (2, 3) represents the outcome where 2 appears on the first die and 3 on the second. Pair (3, 2) represents 3 on the first die and 2 on the second.
The outlined region represents the event “craps,” that is, a total 7 or 11 on the two dice. There are eight points in this event, six where the total is 7 and two where the total is 11. The probability of throwing craps is thus 8/36, or about 22%. !
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Example 4.24. Let us calculate the probability that a poker hand is a one-pair hand.
We learned in Example 4.8 that there are 2,598,960 different poker hands. Consider the experiment of dealing a poker hand fairly, that is, with all hands equally likely. Thus, the probability space for this experiment has 2,598,960 190 COMBINATORICS AND PROBABILITY • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • (1, 1) • • • • • (2, 1) (3, 1) (4, 1) (5, 1) (6, 1) (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3) (1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4) (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6) Fig. 4.14. The event “craps” in the probability space for the toss of two dice.
points. We also learned in Example 4.18 that 1,098,240 of these points represent hands classified as one pair. Assuming all hands are equally likely to be dealt, the probability of the event “one pair” is 1,098,240/2,598,960, or about 42%. !
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Example 4.25. In the game of Keno, twenty of the numbers from 1 to 80 are selected at random. Before the selection, players may guess some numbers; we shall concentrate on the “5-spot game,” where players guess five numbers. A player who Keno guesses correctly three, four, or five of the twenty selected numbers is rewarded; the amount of the payoffincreases with the number of correct guesses. We shall calculate the probability that a player guesses exactly three numbers correctly in a 5-spot game. The probabilities of guessing four or five correctly are left as exercises.
To begin, the appropriate probability space has one point for each possible selection of twenty numbers from 1 to 80. The number of such selections is 80 20 = 80!
20!60!
SEC. 4.9 INTRODUCTION TO PROBABILITY THEORY 191 When are Outcomes Random?
We have assumed in our examples that certain experiments have “random” out-comes; that is, all possible outcomes are equally likely. In some cases the justifi-cation for this assumptions comes from physics. For instance, when throwing fair (unweighted) dice, we assume that it is not physically possible to control the toss in such a way that one face is more likely than others to appear on top. That is a valid assumption in practice. Similarly, we assume that a fairly shuffled deck will not bias the outcome, and any card is equally likely to be found in any position in the deck.
In other cases, we find that what we imagine is random is actually not random at all, but is the result of a process that is predictable in principle but unpredictable in practice. For example, the numbers in a Keno game may be generated by a com-puter executing a particular algorithm, and yet without access to secret information used by the computer, it is impossible to predict the outcome.
All computer-generated “random” sequences of outcomes are the result of a special kind of algorithm called a random number generator. Designing one of these Random number generator algorithms requires knowledge of some specialized mathematics and is beyond the scope of this book. However, we can offer an example of a random number generator that works fairly well in practice, called a linear congruential generator.
We specify constants a ≥2, b ≥1, x0 ≥0, and a modulus m > max(a, b, x0).
We can generate a sequence of numbers x0, x1, x2, . . . using the formula xn+1 = (axn + b) mod m For suitable choices of a, b, m, and x0, the resulting sequence of numbers will appear quite random, even though they are constructed from the “seed” x0 by a specific algorithm.
Sequences generated by a random number generator have many uses. For in-stance, we can select numbers in a Keno game by taking numbers from the sequence described above, dividing each by 80, taking the remainder, and adding 1 to get a “random” number from 1 to 80. We keep doing so, throwing away repeated num-bers, until twenty numbers are selected. The game will be perceived as fair, as long as no one knows the generation algorithm and the seed.
a number so huge that we are fortunate we do not have to write it down.
Now we must count the number of selections of twenty numbers out of eighty that include three of the five numbers chosen by the player and seventeen numbers from the seventy-five that the player has not chosen. We can choose three of the five in 5 3 = 10 ways, and we can choose seventeen from the remaining seventy-five in 75 17 = 75!
17!58! ways.
The ratio of the number of outcomes where the player picks three out of five to the total number of selections is thus 10 75!
17!58!
80!
20!60!
If we multiply top and bottom by 20!60!
80! , the above becomes 192 COMBINATORICS AND PROBABILITY 10( 75!
17!58!)(20!60!
80! ) Now, we find in the numerator and denominator pairs of factorials that are close and can almost be canceled. For instance, 75! in the numerator and 80! in the denominator can be replaced by the product of the five numbers from 80 down to 76 in the denominator. The resulting simplification is 10 × 60 × 59 × 20 × 19 × 18 80 × 79 × 78 × 77 × 76 Now we have a computation that involves manageable numbers.
The result is about 0.084. That is, about 8.4% of the time the player guesses three out of five correctly. !
Fundamental Relationships Let us close this section by observing several important properties of probabilities.
First, if p is the probability of any event, then 0 ≤p ≤1 That is, any event consists of 0 or more points, so its probability cannot be negative.
Also, no event can consist of more points than are in the entire probability space, so its probability cannot exceed 1.
Second, let E be an event in a probability space P. Then the complement event Complement event ¯ E for E is the set of points of P that are not in event E. We may observe that PROB(E) + PROB( ¯ E) = 1 or put another way, PROB( ¯ E) = 1 −PROB(E). The reason is that every point in P is either in E or in ¯ E, but not both.
EXERCISES 4.9.1: Using the probability space for the toss of two fair dice shown in Fig. 4.14, give the probabilities of the following events: a) A six is thrown (i.e., the sum of the dice is 6) b) A ten is thrown c) The sum of the dice is odd d) The sum of the dice is between 5 and 9 4.9.2: Calculate the probabilities of the following events. The probability space is the deal of two cards in order from an ordinary 52-card deck.
a) At least one card is an Ace b) The cards are of the same rank c) The cards are of the same suit d) The cards are of the same rank and suit e) The cards are the same either in rank or in suit f) The first card is of a higher rank than the second card 4.9.3: A dart is thrown at a one foot square on the wall, with equal likelihood of entering the square at any point. What is the probability that the dart is thrown SEC. 4.10 CONDITIONAL PROBABILITY 193 a) Within three inches of the center?
b) Within three inches of the border?
Note that for this exercise, the probability space is an infinite one-foot square, with all points within it equally likely.
4.9.4: Calculate the probability of the player in a 5-spot game of Keno guessing a) Four out of five b) All five 4.9.5: Write a C program to implement a linear congruential random number generator. Plot a histogram of the frequencies of the least significant digits of the first 100 numbers generated. What property should this histogram have?
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4.10 Conditional Probability In this section we shall develop a number of formulas and strategies for think-ing about relationships among the probabilities of several events. One important development is a notion of independent experiments, where the outcome of one ex-periment does not affect the outcome of others. We shall also use our techniques to calculate probabilities in some complicated situations.
These developments depend upon a notion of “conditional probability.” Infor-mally, if we conduct an experiment and we find that an event E has occurred, it may or may not be the case that the point representing the outcome is also in some other event F. Figure 4.15 suggests this situation. The conditional probability of F given E is the probability that F has also occurred.
E A F B Fig. 4.15. The conditional probability of F given E is the probability that the outcome is in A divided by the probability that the outcome is in A or B.
Formally, if E and F are two events in a probability space, we say the condi-tional probability of F given E, denoted PROB(F/E), is the sum of the probabilities of the points that are in both E and F divided by the sum of the probabilities of the points in E. In Fig. 4.15, region A is those points that are in both E and F, and B is those points that are in E but not F. If all points are equally likely, then 194 COMBINATORICS AND PROBABILITY PROB(F/E) is the number of points in A divided by the sum of the numbers of points in A and B.
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Example 4.26. Let us consider the probability space of Fig. 4.14, which repre-sents the toss of two dice. Let the event E be the six points in which the first die comes out 1, and let the event F be the six points in which the second die comes out 1. The situation is shown in Fig. 4.16. There is one point in both E and F, namely the point (1, 1). There are five points in E that are not in F. Thus, the conditional probability PROB(F/E) is 1/6. That is, the chance that the second die is 1, given that the first die is 1, is 1/6.
We may notice that this conditional probability is exactly the same as the probability of F itself. That is, since F has 6 out of the 36 points in the space, PROB(F) = 6/36 = 1/6. Intuitively, the probability of throwing 1 on the second die is not affected by the fact that 1 has been thrown on the first die. We shall soon define the notion of “independent experiments,” such as the throwing of dice in sequence, where the outcome of one experiment does not influence the outcome of the others. In these cases, if E and F are events representing outcomes of the two experiments, we expect that PROB(F/E) = PROB(F). We have just seen an example of this phenomenon. !
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Example 4.27.
Suppose our experiment is the deal of two cards, in order, from the usual 52-card deck. The number of points in this experiment of selection without replacement (as in Section 4.4) is 52 × 51 = 2652. We shall assume that the deal is fair, so each point has the same probability.
Let the event E be that the first card is an Ace, and let event F be that the second card is an Ace. Then the number of points in E is 4×51 = 204. That is, the first card must be one of the four Aces, and the second card can be any of 51 cards, excluding the Ace that was chosen first. Thus, PROB(E) = 204/2652 = 1/13. That result matches our intuition. All 13 ranks being equally likely, we expect that one time in 13 an Ace will appear first.
Similarly, the number of points in event F is 51 × 4 = 204. We can choose any of the 4 Aces for the second card and any of the remaining 51 cards for the first card. The fact that the first card is theoretically dealt first is irrelevant. There are thus 204 outcomes in which an Ace appears in the second position. Therefore, PROB(F) = 1/13 just as for E. Again, this result meets our intuition that one time in 13 an Ace will be dealt as the second card.
Now, let us compute PROB(F/E). Of the 204 points in E, there are 12 that have an Ace in the second position and therefore are also in F. That is, all points in E have an Ace in the first position. We may select this Ace in 4 different ways, corresponding to the 4 suits.
For each selection, we have 3 different choices of Ace for the second position. Thus, the number of choices of two Aces with order considered is 4 × 3 according to the technique of Section 4.4.
Therefore, the conditional probability PROB(F/E) is 12/204, or 1/17. We no-tice that the conditional probability of F given E is not the same as the probability of F in this example. That also makes intuitive sense. The probability of getting an Ace in the second position goes down when we know there is an Ace in the first po-sition. For then, there are only 3 Aces remaining out of 51 cards, and 3/51 = 1/17.
SEC. 4.10 CONDITIONAL PROBABILITY 195 • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • • (1, 1) • • • • • (2, 1) (3, 1) (4, 1) (5, 1) (6, 1) (1, 2) (2, 2) (3, 2) (4, 2) (5, 2) (6, 2) (1, 3) (2, 3) (3, 3) (4, 3) (5, 3) (6, 3) (1, 4) (2, 4) (3, 4) (4, 4) (5, 4) (6, 4) (1, 5) (2, 5) (3, 5) (4, 5) (5, 5) (6, 5) (1, 6) (2, 6) (3, 6) (4, 6) (5, 6) (6, 6) E F Fig. 4.16. Events representing 1 on the first or second dice.
In comparison, if we know nothing about the first card, there are 4 Aces out of 52 that we may receive as the second card. !
Independent Experiments As we suggested in Examples 4.23, 4.26, and 4.27, we sometimes form a probability space representing the outcomes of two or more experiments. In the simplest cases, the points in this joint probability space are lists of outcomes, one for each of Joint probability space the experiments.
Figure 4.16 is an example of a probability space that is joint between two experiments. In other situations, where there is a connection between the outcomes of the experiments, the joint space may have some points missing.
Example 4.27 discussed such a case, where the joint space represented the deal of two cards and the pairs of outcomes in which the two cards are identical are not possible.
There is an intuitive notion of the outcome of one experiment X being “inde-pendent” of previous experiments in the sequence, meaning that the probabilities of the various outcomes of X do not depend on the outcomes of the previous ex-periments.
Thus, in Example 4.26 we argued that the roll of the second die is independent of the roll of the first die, while in Example 4.27 we saw that the 196 COMBINATORICS AND PROBABILITY second card dealt was not independent of the first, since the first card was then unavailable.
In defining independence, we shall focus on two experiments. However, since either experiment may itself be the sequence of several experiments, we effectively cover the case of many experiments. We must begin with a probability space that represents the outcome of two successive experiments, X1 and X2.
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Example 4.28. Figure 4.14 illustrates a joint probability space in which ex-periment X1 is the throw of the first die and X2 is the throw of the second die.
Here, every pair of outcomes is represented by one point, and the points have equal probability, 1/36.
In Example 4.27 we discussed a space of 2652 points representing the selection of two cards in order. This space consists of all pairs (C, D) in which C and D are cards, and C ̸= D. Again, each of these points has the same probability, 1/2652. !
In the probability space representing the outcomes of X1 followed by X2, there are events that represent the outcome of one of the experiments. That is, if a is a possible outcome of experiment X1, then there is an event consisting of all those points in which the outcome of the first experiment is a. Let us call this event Ea.
Similarly, if b is a possible outcome of X2, then there is an event Fb consisting of all those points in which the outcome of the second experiment is b.
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Example 4.29. In Fig. 4.16, E is E1, the event of all points for which the outcome of the first experiment is 1. Likewise, F is the event F1, whose points are those where the outcome of the second experiment is 1. More generally, each row corresponds to one of the six possible outcomes of the first experiment, and each column corresponds to one of the six possible outcomes of the second experiment. !
Formally, we say that experiment X2 is independent of experiment X1 if for all Independence outcomes a for X1 and all outcomes b for X2, PROB(Fb/Ea) = PROB(Fb). That is, no matter what the outcome of experiment X1, the conditional probability of each outcome of X2 is the same as it is in the whole probability space.
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Example 4.30. Returning to the probability space of Fig. 4.16 representing the toss of two dice, let a and b each be any of the numbers from 1 to 6. Let Ea be the event that the first die is a and Fb be the event that the second die is b. We note that the probability of each of these events is 1/6; they are each one row or column. For any a and b, PROB(Fb/Ea) is also 1/6. We argued this fact for the case a = b = 1 in Example 4.26, but the same argument applies to any two outcomes a and b, because their events have exactly one point in common. Thus, the tosses of the two dice are independent.
On the other hand, in the card-based Example 4.27, we do not get indepen-dence. Here, the experiment X1 is the selection of the first card, and the experiment X2 is the selection of the second card from the remaining deck. Consider an event like FA♠— that is, the second card is the Ace of Spades. It is easy to count that SEC. 4.10 CONDITIONAL PROBABILITY 197 the probability of this event, PROB(FA♠), is 1/52.
Now, consider an event like E3♣, the first card is the Three of Clubs. The number of points in both E3♣and FA♠is 1, namely the point (3♣, A♠). The total number of points in E3♣is 51, namely those points of the form (3♣, C) where C is any card but the Three of Clubs. Thus, the conditional probability PROB(FA♠/E3♣) is 1/51, not 1/52 as it would have to be if the two experiments were independent.
For a more extreme example, consider the event EA♠in which the first card is the Ace of Spades. Since EA♠and FA♠have no point in common, PROB(FA♠/EA♠) is 0 instead of 1/52. !
A Distributive Law for Probabilities Sometimes it is easier to calculate the probability of an event if we first divide the probability space into regions2 that partition the space. That is, every point is in Region exactly one region. Typically, the probability space represents the outcome of a sequence of experiments, and the regions, which are themselves events, correspond to the possible outcomes of one of these experiments.
Suppose we wish to calculate the probability of event E in a space of n points that is divided into k regions, R1, R2, . . . , Rk. For simplicity, we shall assume that all points have equal probability, although the conclusion we draw holds as well if the points have differing probabilities. Let event E consist of m points. Let region Ri have ri points, for i = 1, 2, . . ., k. Finally, let the number of points of E that lie in region Ri be ei. Note that Pk i=1 ri = n, and Pk i=1 ei = m. The reason in each case is that points are each in one and only one region.
We know PROB(E) = m/n, because m/n is the fraction of points in E. If we replace m by the sum of the ei’s we get PROB(E) = k X i=1 ei n Next, introduce factor ri in both the numerator and denominator in each term of the sum above. The result is PROB(E) = k X i=1 (ei ri )(ri n ) Now, notice that ri/n = PROB(Ri); that is, ri/n is the fraction of the entire space in region Ri. Also, ei/ri is PROB(E/Ri), the conditional probability of event E given event Ri. Put another way, ei/ri is the fraction of the points in region Ri that are in E. The result is the following formula for the probability of an event E.
PROB(E) = k X i=1 PROB(E/Ri)PROB(Ri) (4.10) Informally, the probability of E is the sum over all regions of the probability of being in that region times the probability of E within that region.
2 A “region” is synonymous with an “event,” that is, a subset of a probability space. However, we shall use the term region to emphasize the fact that the space is being partitioned into events that completely cover the space and do not overlap.
198 COMBINATORICS AND PROBABILITY !
Example 4.31. The diagram in Fig. 4.17 suggests how Equation (4.10) is to be applied. There we see a probability space that has been divided vertically into three regions, R1, R2, and R3. There is an event E, which we doubly outline. We let a through f be the numbers of points in the six sets shown.
R1 R2 R3 a b c d e f E Fig. 4.17. A probability space divided into regions.
Let n = a + b + c + d + e + f. Then PROB(R1) = (a + b)/n, PROB(R2) = (c + d)/n, and PROB(R3) = (e + f)/n. Next, the conditional probabilities of E in the three regions are PROB(E/R1) = a/(a + b), PROB(E/R2) = c/(c + d), and PROB(E/R3) = e/(e + f). Now, to evaluate formula (4.10), we have PROB(E) = PROB(E/R1)PROB(R1) + PROB(E/R2)PROB(R2)+ PROB(E/R3)PROB(R3) This formula, in terms of the parameters a through f, is PROB(E) = ( a a + b)(a + b n ) + ( c c + d)(c + d n ) + ( e e + f )(e + f n ) = a n + c n + e n Notice that the same result can be obtained by simply comparing the numbers of points in the three areas labeled a, c, and e with the size of the entire space. This fraction, (a + c + e)/n, is exactly what the formula above gives for the probability of E. That observation illustrates Equation (4.10). !
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Example 4.32. Let us use Equation (4.10) to compute the probability of the event E that two cards dealt in order are both Aces. The probability space is the 2652 points discussed in Example 4.27. We shall divide this space into two regions: R1: Those points in which the first card is an Ace. There are 4 × 51 = 204 such points, since we may pick the first card to be an Ace in 4 ways, and there are then 51 choices of second card.
R2: The remaining 2448 points.
SEC. 4.10 CONDITIONAL PROBABILITY 199 Equation (4.10) in this case becomes PROB(E) = PROB(E/R1)PROB(R1) + PROB(E/R2)PROB(R2) Clearly PROB(E/R2), the conditional probability of E given R2, is 0.
There is no way to get two Aces if the first card is not an Ace. We must thus compute PROB(E/R1)PROB(R1), and this value is PROB(E). Now PROB(R1) = 204/2652 = 1/13. In other words, the chance of getting an Ace as the first card is 1/13. Since there are thirteen ranks, that probability makes sense.
Now, we need to compute PROB(E/R1). If the first card is an Ace, then there remain 3 Aces out of 51 cards. Thus, PROB(E/R1) = 3/51 = 1/17. We conclude that PROB(E) = (1/17)(1/13) = 1/221. !
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Example 4.33. Let us apply Equation (4.10) to the problem of computing the probability of the event E that at least one 1 appears in the toss of three dice, as in the game Chuck-a-Luck described in Example 4.16. First, we must understand that the notion of an “outcome” described in that example does not match the notion of a point in a probability space. In Example 4.16 we established that there were 56 different “outcomes,” which were defined to be the numbers of occurrences of 1 through 6 on the faces of the dice. For instance, “a 4, a 5, and a 6” is one possible outcome; “two 3’s and a 4” is another. However, not all outcomes in this sense have the same probability. In particular, outcomes with three different numbers showing have twice the probability of outcomes with two of one number and six times the probability of an outcome where all three dice show the same number.
While we could use the probability space whose points are “outcomes ” in the sense of Example 4.16, it is more natural to consider the order in which the dice are rolled and thus develop a probability space whose points have equal probability.
There are 63 = 216 different outcomes corresponding to the roll of three dice, in order, and each of these outcomes has probability 1/216.
We could calculate the probability of at least one 1 in a direct manner, without using Equation (4.10). First, calculate the number of rolls in which no 1 appears.
We can assign to each of the three dice any of the numbers from 2 to 6. There are thus 53 = 125 points in the space that have no 1, and 216 −125 = 91 points that do have a 1. Therefore, PROB(E) = 91/216, or about 42%.
The approach above is short but requires that we use several “tricks.” An-other way to calculate the probability “by force” is to divide the space into three regions, corresponding to the cases in which there are one, two, or three different numbers showing. Let Ri be the region of points with i different numbers. We can calculate the probabilities of the various regions as follows. For R1 there are only six points, one for each of the numbers 1 through 6 that may appear on all three dice. For R3 there are 6 × 5 × 4 = 120 ways to select three different numbers out of six, according to the rule of Section 4.4. Thus, R2 must have the remain-ing 216 −6 −120 = 90 points.3 The probabilities of the regions are PROB(R3) = 6/216 = 1/36, PROB(R2) = 90/216 = 5/12, and PROB(R1) = 120/216 = 5/9.
Next, we can calculate the conditional probabilities. If there are three numbers out of the possible six showing, the probability is 1/2 that one of them is 1. If two 3 We can compute this number directly by multiplying 6 ways that we can choose the number to appear twice, times 5 ways we can pick the number on the remaining dice, times 3 1 = 3 ways we can pick the die that has the unique number. Note that 6 × 5 × 3 = 90.
200 COMBINATORICS AND PROBABILITY numbers are showing, the probability is 1/3 that 1 appears at least once. If only one number shows, there is a 1/6 chance that it is 1. Thus, PROB(E/R1) = 1/6, PROB(E/R2) = 1/3, and PROB(E/R3) = 1/2. We can put all these probabilities together to evaluate (4.10). The result is PROB(E) = (1/6)(1/36) + (1/3)(5/12) + (1/2)(5/9) = 1/216 + 5/36 + 5/18 = 91/216 This fraction agrees with the direct calculation, of course. If we see the “trick” of the direct calculation, that approach is considerably easier. However, breaking the problem into regions is frequently a more reliable way to guarantee success. !
A Product Rule for Independent Experiments A common type of probability problem asks for the probability of a sequence of outcomes to a sequence of independent experiments. In that case, Equation (4.10) has an especially simple form, and it tells us that the probability of a sequence of outcomes is the product of the probabilities of each outcome by itself.
First, whenever we divide the probability space into k regions of equal size, we know that PROB(Ri) = 1/k for all i. Thus (4.10) reduces to PROB(E) = k X i=1 1 k PROB(E/Ri) (4.11) A useful way to look at (4.11) is that the probability of E is the average over all regions of the probability of E given we are in that region.
Now, consider a probability space that represents the outcomes of two inde-pendent experiments X1 and X2. We may divide this space into k regions, each the set of points for which the outcome of X1 has a particular value. Then each of the regions has the same probability, 1/k.
Suppose we want to calculate the probability of the event E in which X1 has the outcome a and X2 has the outcome b. We may use formula (4.11). If Ri is not the region corresponding to outcome a for X1, then PROB(E/Ri) = 0. Thus, all but the term for the region of a drops out of (4.11). If Ra is that region, we get PROB(E) = 1 k PROB(E/Ra) (4.12) What is PROB(E/Ra)? It is the probability that X1 has outcome a and X2 has outcome b, given that X1 has outcome a. Since we are given that X1 has outcome a, PROB(E/Ra) is just the probability that X2 has outcome b given that X1 has outcome a. Since X1 and X2 are independent, PROB(E/Ra) is just the probability that X2 has outcome b. If there are m possible outcomes for X2, then PROB(E/Ra) is 1/m. Then (4.12) becomes PROB(E) = (1 k )( 1 m) We may generalize the above reasoning to any number of experiments. To do so, we let experiment X1 be a sequence of experiments and show by induction on the total number of independent experiments that the probability of all having a particular sequence of outcomes is the product of the probabilities of each outcome.
SEC. 4.10 CONDITIONAL PROBABILITY 201 Using Independence to Simplify Calculations There are many opportunities to simplify probability calculations if we know that experiments are independent. The product rule is one such example. Another is that whenever event E is the set of points with a particular outcome from experiment X1, and F is the set of points with a particular outcome of another, independent experiment X2, then PROB(E/F) = PROB(E).
In principle, telling whether two experiments are independent is a complex task involving examination of the probability space representing pairs of outcomes of each experiment. However, often we can appeal to the physics of the situation to conclude that experiments are independent without doing this calculation. For example, when we throw dice in sequence, there is no physical reason why the outcome of one would influence the other, so they must be independent experiments.
Contrast the situation of dice with the deal of several cards from a deck. Since cards dealt are unavailable to be dealt at a future step, we do not expect successive cards to be independent of each other. We in fact observed this lack of independence in Example 4.29.
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Example 4.34. The probability that the last four digits of a phone number are 1234 is 0.0001. The selection of each digit is an experiment with ten possible outcomes: 0 through 9. Moreover, each selection is independent of the other se-lections, since we are performing a “selection with replacement” as in Section 4.2.
The probability that the first digit is 1 is 1/10. Similarly, the probability that the second digit is 2 is 1/10, and likewise for the other two digits. The probability of the event that the four digits are 1234 in order is (1/10)4 = 0.0001. !
EXERCISES 4.10.1: Using the space of Fig. 4.14, give the conditional probabilities of the fol-lowing pairs of events.
a) The second die is even, given that the first die is odd b) The first die is even, given that the second die is at least 3 c) The sum of the dice is at least 7, given that the first die is 4 d) The second die is 3, given that the sum of the dice is 8 4.10.2: Divide the Chuck-a-Luck (see Example 4.16) probability space into three regions, as in Example 4.33. Use this division and Equation 4.10 to compute the probability that a) There are at least two 1’s showing b) All three dice are 1 c) There is exactly one 1 showing 4.10.3: Show that in Chuck-a-Luck, the probability of any event in which all three dice have different values is twice the probability of any event where one number appears exactly twice and six times the probability of any event in which all three dice show the same number.
202 COMBINATORICS AND PROBABILITY 4.10.4: Show by induction on n that if there are n experiments, each of which is independent of those that go before, then the probability of any sequence of outcomes is the product of the probabilities of each outcome in its own experiment.
4.10.5: Show that if PROB(F/E) = PROB(F), then PROB(E/F) = PROB(E).
Thus, show that if experiment X1 is independent of experiment X2, then X2 is independent of X1.
4.10.6: Consider the set of sequences of seven letters chosen from W and L. We may think of these sequences as representing the outcomes of a match of seven games, where W means the first team wins the game and L means the second team wins the game. The match is won by the first team to win four games (thus, some games may never get played, but we need to include their hypothetical outcomes in the points in order that we have a probability space of equally likely points).
a) What is the probability that a team will win the match, given that it has won the first game?
b) What is the probability that a team will win the match, given that it has won the first two games?
c) What is the probability that a team will win the match, given that it has won two out of the first three games?
4.10.7: There are three prisoners, A, B, and C. They are told one and only one is to be shot and that the guard knows who. A asks the guard to tell him the name of one of the other prisoners who will not be shot. The guard answers that B will not be shot.
A reasons that either he or C will be shot, so the probability that A will be shot is 1/2. On the other hand, reasons A, no matter who is to be shot, the guard knows somebody besides A who will not be shot, so he always has an answer to A’s question.
Therefore, the asking and answering of the question provides no information about whether or not A is to be shot, so the probability that A will be shot is still 1/3, as it was before the question was asked.
What is the true probability that A will be shot after the sequence of events described above? Hint: You need to construct an appropriate probability space, one that represents not only the experiment in which a prisoner is chosen to be shot but also the possibility that the guard has a choice of whether to answer “B” or “C,” and the experiment in which he chooses one if so.
4.10.8: Suppose that E is an event in a space that is partitioned into k regions R1, R2, . . . , Rk. Show that PROB(Rj/E) = PROB(Rj)PROB(E/Rj) Pk i=1 PROB(Ri)PROB(E/Ri) This formula is called Bayes’ Rule. It gives a value for the probability of Rj given Bayes’ Rule that E has been observed. For Example 4.31, calculate PROB(R1/E), PROB(R2/E), and PROB(R3/E) using Bayes’ rule.
SEC. 4.11 PROBABILISTIC REASONING 203 !
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4.11 Probabilistic Reasoning An important application of probability in computing is in the design of systems that predict events. An example is a medical diagnosis system. Ideally, the process of diagnosis consists of performing tests or observing symptoms until the outcome of the tests and presence or absence of certain symptoms is sufficient for the physician to determine what disease the patient has.
In practice, however, diagnoses are rarely certain. Rather, a diagnosis is the most likely disease, or a disease whose conditional probability, given the outcomes of the experiments that are the tests and observations of symptoms, is highest.
Let us consider an overly simple example that has the flavor of diagnosis using probability. Suppose it is known that when a patient has a headache, the probability that he has the flu is 50%. That is, PROB(Flu/Headache) = 0.5 In the above, we interpret Flu as the name of an event that can be interpreted as “the patient has the flu.” Similarly, Headache is the name of the event that the patient complains of a headache.
Suppose we also know that when the patient’s temperature is measured at 102 (Fahrenheit) or above, the probability is 60% that he has the flu. If we allow Fever to be the name of the event that the patient’s temperature is at least 102, then we can write this observation as PROB(Flu/Fever) = 0.6 Now, consider the following diagnosis situation. A patient comes to the doctor complaining of a headache. The doctor takes his temperature and finds it is 102.
What is the probability that the patient has the flu?
The situation is suggested by Fig. 4.18. There we see the three events Flu, Headache, and Fever, which together divide the space into 8 regions, which we indicate by letters a through h. For example, c is the event that the patient has a headache and flu, but not a fever.
The given information about probabilities puts some constraints on the sizes of the events in Fig. 4.18. Let us use the letters a through h as standing not only for the regions indicated in Fig. 4.18 but as the probabilities of those events. Then the condition that PROB(Flu/Headache) = 0.5 says that the sum of regions c + f is half the total size of the headache event, or put another way: c + f = d + g (4.13) Similarly, the fact that PROB(Flu/Fever) = 0.6 says that e + f is 3/5 of the total size of the Fever event, or: e + f = 3 2(g + h) (4.14) Now, let us interpret the question: what is the probability of flu, given both a fever and a headache? The fact that there is both fever and headache says that we are either in region f or region g. In region f the diagnosis of flu is correct, and in g it is not. Thus, the probability of flu is f/(f + g).
What is the value of f/(f + g)?
The answer may be surprising. We have absolutely no information about the probability of flu; it could be 0 or 1 or anything in between. Here are two examples of how the points of the probability space of Fig. 4.18 could actually be distributed.
204 COMBINATORICS AND PROBABILITY a b c d e f g h Flu Headache Fever Fig. 4.18. The events Flu, Headache, and Fever.
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Example 4.35. Suppose that in Fig. 4.18 the probabilities associated with the various events are: d = f = 0.3, a = h = 0.2, and the four other regions have probability 0. Note that these values satisfy the constraining equations (4.13) and (4.14). In this example, f/(f + g) = 1; that is, a patient with both a headache and fever is certain to have the flu. Then the probability space of Fig. 4.18 actually looks like that of Fig. 4.19. There we see that whenever a patient has both a fever and headache, he has the flu, and conversely, whenever he has the flu, he has both fever and headache.4 !
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Example 4.36. Another example is given by the probabilities c = g = 0.2, a = e = 0.3, with other probabilities 0. Again, Equations (4.13) and (4.14) are satisfied. Now, however, f/(f + g) = 0. That is, if you have both a fever and headache, then you are certain not to have the flu, a rather doubtful statement, but one that is not ruled out by Equations (4.13) and (4.14). The situation is shown in Fig. 4.20. !
4 Although there are other examples where b ̸= 0 — that is, one can have the flu and yet have neither fever nor headache — yet f/(f + g) = 1.
SEC. 4.11 PROBABILISTIC REASONING 205 a f d h Headache Fever Flu Fig. 4.19. Example of space where Fever and Headache guarantee Flu.
e g c a Fever Flu Headache Fig. 4.20. Example of space where Fever and Headache guarantee no Flu.
Probability of the OR of Two Events If we cannot tell anything about what happens when a patient has both fever and headache in the previous scenario, we might wonder whether there is anything we can say. In simpler situations there is indeed some limit to how probabilities behave when events are combined. Perhaps the most straightforward situation occurs when we combine two events with disjunction or logical-OR.
Disjunction !
Example 4.37. Referring again to the situation of Fig. 4.18, suppose we are told that at any time, 2% of the population has a fever and 3% of the population has a headache. That is, the size of the event Fever is 0.02, and the size of the event Headache is 0.03. What fraction of the population has either a fever or a headache, or both?
The answer is that between 3% and 5% of the population has at least one. To see why, let us do some calculation in terms of the eight regions defined in Fig. 4.18.
206 COMBINATORICS AND PROBABILITY If Fever has probability 0.02, that says e + h + f + g = 0.02 (4.15) If Headache has probability 0.03, then c + d + f + g = 0.03 (4.16) We are asked what is the size of the regions that are in either Fever or Headache or both; that is, how large is e + h + f + g + c + d?
If we add (4.15) and (4.16), we get e + h + 2(f + g) + c + d = 0.05, or put another way, e + h + f + g + c + d = 0.05 −(f + g) (4.17) Since the probability of Fever-or-Headache is the left side of (4.17), it is also the right side, or 0.05 −(f + g).
At the minimum, f + g is 0, so the probability of Fever-or-Headache can be as high as 0.05, but no higher. That is, it is possible that fever and headache never occur together. Then the regions f and g are empty, e+h = 0.02, and c+d = 0.03.
In this case, the probability of Fever-or-Headache is the sum of the probability of Fever and the Probability of Headache.
What is the maximum value f + g could have? Surely f + g is no bigger than the entire Fever event and no bigger than the entire Headache event. Since Fever is smaller, we see that f + g ≤0.02. Thus, the smallest the probability of Fever-or-Headache could be is 0.05 −0.02, or 0.03. This result happens to be the probability of Headache, the larger of the two events. That is no coincidence. Another way to look at it is that the smallest size Fever-or-Headache could have occurs when the smaller of the two events is wholly contained within the larger. In this example, that occurs when e + h = 0, and Fever is contained within Headache.
In that case, you cannot have a fever unless you also have a headache, so the probability of Fever-or-Headache is the same as the probability of Headache alone, or 0.03. !
We can generalize our explorations of Example 4.37 to any two events. The rule for sums is as follows. If E and F are any two events, and G is the event that Rule for sums either E or F or both occurs, then max PROB(E), PROB(F) ≤PROB(G) ≤PROB(E) + PROB(F) (4.18) That is, the probability of E-or-F is between the larger of the probabilities of E and F and the sum of those probabilities.
The same idea holds within any other event H. That is, all the probabilities in (4.18) may be made conditional on event H, giving us the more general rule max PROB(E/H), PROB(F/H) ≤PROB(G/H) ≤PROB(E/H) + PROB(F/H) (4.19) !
Example 4.38. Suppose, in the scenario of Fig. 4.18 we are told that 70% of all people with the flu have a fever, and 80% of all people with the flu have a headache.
Then in (4.19), Flu is the event H, E is the event Fever, F is Headache, and G is Headache-or-Fever. We are told that PROB(E/H) = PROB(Fever/Flu) = 0.7, and PROB(F/H) = PROB(Headache/Flu) = 0.8.
SEC. 4.11 PROBABILISTIC REASONING 207 Rule (4.19) says that PROB(G/H) is at least the larger of 0.7 and 0.8. That is, if you have the flu, then the probability that you have a fever or headache or both is at least 0.8. Rule (4.19) also says that PROB(G/H) is at most PROB(E/H) + PROB(F/H) or 0.7+0.8 = 1.5. However, that upper bound is not useful. We know that no event can have probability greater than 1, so 1 is a better upper bound on PROB(G/H). !
Probability of the AND of Events Suppose we are again told that the probability of Fever is 0.02 and the probability of Headache is 0.03. What is the probability of Fever-and-Headache, that is, the probability that a person has both fever and headache? As for the OR of two events, we cannot tell exactly, but sometimes we can put some limits on the probability of the conjunction (logical AND) of two events.
Conjunction In terms of Fig. 4.18, we are asking how big f +g can be. We already observed in connection with the OR of events that f + g is largest when the smaller of the two events, Fever in this case, is wholly contained within the other. Then, all the probability of Fever is concentrated in f +g, and we have f +g = 0.02. That is, the probability of Fever-and-Headache is at most 0.02, the probability of Fever alone.
In general, the probability of the AND of two events cannot exceed the probability of the smaller.
How small can f + g be? Evidently there is nothing that prevents Fever and Headache from being disjoint, so f + g could be 0. That is, there may be no one who has both a fever and headache.
Yet the above idea does not generalize completely. Suppose that instead of the tiny probabilities 0.02 and 0.03 for the events Fever and Headache, we found that 60% had a fever and 70% had a headache. Is it possible that no one has both a fever and a headache? If f + g = 0 in this situation, then e + h = 0.6, and c + d = 0.7.
Then e + h + c + d = 1.3, which is impossible. That is, we would have in Fig. 4.18 an event, e + h + c + d, with a probability greater than 1.
Evidently, the size of the AND of two events cannot be smaller than the sum of the probabilities of the events minus 1. If not, then the OR of the same two events has a probability greater than 1. This observation is summarized in the rule for products. If E and F are two events, and G is the event that both of E and F Rule for products occur, then PROB(E) + PROB(F) −1 ≤PROB(G) ≤min PROB(E), PROB(F) As with the rule for sums, the same idea applies to probabilities that are conditional upon some other event H. That is, PROB(E/H) + PROB(F/H) −1 ≤PROB(G/H) ≤min PROB(E/H), PROB(F/H) (4.20) !
Example 4.38. Again referring to Fig. 4.18, suppose that 70% of those with the flu have a fever and 80% have a headache. How many have both a fever and a headache? According to (4.20) with H the event Flu, the probability of both a fever and headache, given that the person has the flu, is at least 0.7 + 0.8 −1 = 0.5 and at most min(0.7, 0.8) = 0.7. !
208 COMBINATORICS AND PROBABILITY Summary of Rules Involving Several Events The following summarizes rules of this section and rules about independent events from the last section.
Suppose E and F are events with probabilities p and q, respectively. Then !
The probability of event E-or-F (i.e., at least one of E and F) is at least max(p, q) and at most p + q (or 1 if p + q > 1).
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The probability of event E-and-F (i.e., both E and F) is at most min(p, q) and at least p + q −1 (or 0 if p + q < 1).
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If E and F are independent events, then the probability of E-and-F is pq.
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If E and F are independent events, then the probability of E-or-F is p+q−pq.
The latter rule requires some thought. The probability of E-or-F is p+q minus the fraction of the space that is in both events, since the latter space is counted twice when we add the probabilities of E and F. The points in both E and F are exactly the event E-and-F, whose probability is pq. Thus, PROB(E-or-F) = PROB(E) + PROB(F) −PROB(E-and-F) = p + q −pq The diagram below illustrates the relationships between these various events.
E F E-and-F E-or-F prob. = q prob.
= pq prob. = p prob. = p + q −pq Some Ways to Deal With Relationships Among Events In applications that require us to compute the probability of compound events Compound event (events that are the AND or OR of several other events), often we do not need to know exact probabilities. Rather, we need to determine what is the most likely situation or a situation that has high probability (i.e., probability close to 1). Thus, the range of probabilities that a compound event may not present a great problem, SEC. 4.11 PROBABILISTIC REASONING 209 as long as we can deduce that the probability of the event is “high.” For instance, in the medical diagnosis problem introduced in Example 4.35, we may never be able to deduce with probability 1 that the patient has the flu. But as long as the combination of observed symptoms and symptoms not present in the patient together allow us to conclude that the probability he has the flu is very high, then it may make good sense to treat the patient for flu.
However, we observed in Example 4.35 that we could say essentially nothing about the probability that a patient with both a headache and fever has the flu, even though we know that each symptom by itself strongly suggests the flu. A real reasoning system needs more information or more rules from which it can estimate probabilities. As a simple example, we might be given explicitly the probability PROB(Flu/Headache-and-Fever). That would settle the question immediately.
However, if there are n events E1, E2, . . . , En that in some combination might let us conclude another event F, then we need to give explicitly 2n −1 different probabilities. These are the conditional probabilities of F given each set of one or more of E1, E2, . . . , En.
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Example 4.40.
For the case n = 2, such as Example 4.35, we need only give three conditional probabilities. Thus we might assert, as we did previously, that PROB(Flu/Fever) = 0.6 and PROB(Flu/Headache) = 0.5. Then, we might add information such as PROB(Flu/Fever-and-Headache) = 0.9. !
To avoid having to specify an exponential number of conditional probabilities, there are many types of restrictions that have been used to help us deduce or estimate probabilities. A simple one is a statement that one event implies another; Implication of events that is, the first event is a subset of the second event. Often, such information tells us something useful.
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Example 4.41. Suppose we state that whenever one has the flu, one is sure to have a headache. In terms of Fig. 4.18, we are saying that regions b and e are empty.
Suppose also that whenever one has the flu, one also has a fever. Then region c of Fig. 4.18 is also empty. Figure 4.22 suggests the simplification to Fig. 4.18 that results from these two assumptions.
Under the conditions that b, c, and e are all 0, and again assuming that PROB(Flu/Headache) = 0.5 and PROB(Flu/Fever) = 0.6, we can rewrite Equations (4.13) and (4.14) as f = d + g f = 3 2(g + h) Since d and h are both at least 0, the first equation says f ≥g and the second says f ≥3g/2.
Now, let us see what we know about the probability of flu, given both fever and headache, that is, PROB(Flu/Fever-and-Headache).
This conditional probability in either Fig. 4.18 or Fig. 4.22 is f/(f + g). Since f ≥3g/2, we conclude that f/(f +g) ≥0.6. That is, the probability is at least 0.6 that a patient with headache and fever has the flu. !
210 COMBINATORICS AND PROBABILITY a f g d h Fig. 4.22. Here, Flu implies both Headache and Fever.
We can generalize Example 4.41 to apply to any three events, two of which are implied by the third. Suppose E, F, and G are events, and PROB(E/G) = PROB(F/G) = 1 That is, whenever G occurs, E and F are certain to occur as well. Suppose further that PROB(G/E) = p and PROB(G/F) = q. Then PROB(G/E-and-F) ≥max(p, q) (4.21) The reason Equation (4.21) holds can be seen from Fig. 4.22, if we interpret Flu as G, Fever as E, and Headache as F. Then p = f/(f +g+h) and q = f/(f +g+d).
Since d and h are at least 0, it follows that p ≤f/(f + g) and q ≤f/(f + g). But f/(f + g) is PROB(G/E-and-F). Thus, this conditional probability is equal to or greater than the larger of p and q.
EXERCISES 4.11.1: Generalize the rule of sums and the rule of products to more than two events. That is, if E1, E2, . . . , En are events with probabilities p1, p2, . . . , pn, re-spectively, a) What can we say about the probability that at least one of the n events occur?
b) What can we say about the probability that all n events occur?
4.11.2: If PROB(F/E) = p, what, if anything, can we say about a) PROB(F/ ¯ E) b) PROB( ¯ F/E) c) PROB( ¯ F/ ¯ E) Recall that ¯ E is the complement event for E and ¯ F is the complement event for F.
SEC. 4.11 PROBABILISTIC REASONING 211 Other Applications of Probabilistic Reasoning We have seen in this section a tiny illustration of an important application of prob-abilistic reasoning: medical diagnosis. Here are some of the other areas in which similar ideas appear in computer solutions.
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Systems diagnosis.
A device fails, exhibiting some incorrect behavior.
For example, a computer’s screen is blank but the disk is running.
What has caused the problem?
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Corporate planning. Given probabilities of economic conditions, such as a rise in inflation, a decrease in supply of a certain commodity, and so on, what strategies have the greatest probability of success?
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Intelligent appliances. State-of-the-art appliances of many sorts use probabilis-tic reasoning (often referred to as “fuzzy logic”) to make decisions for the user.
Fuzzy logic For example, a washing machine can spin and weigh its load and predict the most likely kind of fabric (e.g., wash-and-wear or woolens), adjusting the cycle accordingly.
4.11.3: An intelligent building control tries to predict whether it will be a “cold” night, which we shall take to mean that the nighttime low is at least 20 degrees colder than the daytime high. It knows that when the sunlight falling on its sensor is high just before sunset, there is a 60% probability of a cold night (because there is apparently no cloud cover, allowing heat to radiate more easily from the earth).
It also knows that if the change in temperature the hour after sunset is at least 5 degrees, then the probability is 70% of a cold night. Refer to these three events as Cold, High, and Dropping, respectively. Let us suppose also that PROB(High) = 0.4 and PROB(Dropping) = 0.3.
a) Give upper and lower limits on PROB(High-and-Dropping).
b) Give upper and lower limits on PROB(High-or-Dropping).
c) Suppose we are also told that whenever it is going to be cold at night, the sunlight sensor reads high, and the temperature drops at least 4 degrees after sunset, i.e., PROB(High/Cold) and PROB(Dropping/Cold) are both 1.
Give upper and lower limits on PROB(Cold/High-and-Dropping).
d)Under the same assumption as part (c), give upper and lower limits on PROB(Cold/High-or-Dropping) Note that this problem requires reasoning not covered in the section.
4.11.4: In many situations, such as Example 4.35, two or more events mutually reinforce a conclusion. That is, we expect intuitively that whatever PROB(Flu/Headache) may be, being told that the patient has a fever as well as a headache increases the probability of flu. Say that event E reinforces event F in the conclusion G Reinforcing events if PROB(G/E-and-F) ≥PROB(G/F). Show that if events E and F each reinforce the other in the conclusion G, then Equation (4.21) holds. That is, the conditional 212 COMBINATORICS AND PROBABILITY probability of G given E and F is at least the larger of the conditional probability of G given E and the conditional probability of G given F.
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4.12 Expected Value Calculations Commonly, the possible outcomes of an experiment have associated values. In this section, we shall use simple gambling games as examples, where money is won or lost depending on the outcome of the experiment. In the next section, we shall discuss more complex examples from computer science, where we compute the expected running time of certain algorithms.
Suppose we have a probability space and a payofffunction f on the points of Payofffunction that space. The expected value of f is the sum over all points x of f(x)PROB(x).
We denote this value by EV(f). When all points are equally likely, we can compute the expected value EV(f) by 1.
Summing f(x) for all x in the space and then 2.
Dividing by the number of points in the space.
The expected value is sometimes called the mean value and it can be thought of as Mean value a “center of gravity.” !
Example 4.42. Suppose the space is the six points representing the outcomes of the throw of a fair die.
These points are naturally thought of as integers 1 through 6. Let the payofffunction be the identity function; that is, f(i) = i for i = 1, 2, . . . , 6. Then the expected value of f is EV(f) = f(1) + f(2) + f(3) + f(4) + f(5) + f(6) /6 = (1 + 2 + 3 + 4 + 5 + 6)/6 = 21/6 = 3.5 That is, the expected value of the number on a die is 3.5.
As another example, let g be the payofffunction g(i) = i2. Then, for the same experiment — the throw of one die — the expected value of g is EV(g) = (12 + 22 + 32 + 42 + 52 + 62)/6 = (1 + 4 + 9 + 16 + 25 + 36)/6 = 91/6 = 15.17 Informally, the expected value for the square of the number thrown on a die is 15.17 !
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Example 4.43. Let us reconsider the game of Chuck-a-Luck, first introduced in Example 4.16. The payoffrules for this game are as follows. A player bets one dollar on a number. If that number comes up one or more times, then the player receives as many dollars as his number appears. If the number does not come up at all, then the player loses his dollar.
The probability space for Chuck-a-Luck is the 216 points consisting of the triples of numbers between 1 and 6. These points represent the outcome of the toss of three dice. Let us suppose that the player bets on number 1. It should be clear that the expected value of the player’s winnings or losings does not depend on the number bet on, as long as the dice are fair.
SEC. 4.12 EXPECTED VALUE CALCULATIONS 213 The payofffunction f for this game is: 0.
g(i, j, k) = −1 if none of i, j, or k is 1. That is, the player loses a dollar if there are no 1’s.
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g(i, j, k) = 1 if exactly one of i, j, or k is 1.
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g(i, j, k) = 2 if exactly two of i, j, or k are 1.
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g(i, j, k) = 3 if all three of i, j, or k are 1.
Our problem then is to average g over the 216 points. Since enumeration of all these points is tedious, we are better offtrying to count the number of points with each of the four different outcomes.
First, let us count the number of triples with no 1’s. There are five numbers to chose from in each position, so we have an assignment problem as in Section 4.2.
There are thus 53 = 125 points with no 1’s. These 125 points contribute −125 to the sum of the payoffs, by rule (0) above.
Next, let us count the number of triples with exactly one 1. The 1 can appear in any of the three places. For each place that holds the 1, there are choices of five numbers for each of the other two places. Thus, the number of points with exactly one 1 is 3 × 5 × 5 = 75. These points contribute +75 to the payoffby rule (1).
The number of points with all three 1’s is clearly one, so this possibility con-tributes +3 to the payoff. The remaining 216 −125 −75 −1 = 15 points must have two 1’s, so these 15 points contribute +30 to the payoffby rule (2).
Finally, we can compute the expected value of the game by adding the payoffs from each of the four types of points and dividing by the total number of points.
Thus calculation is EV(f) = (−125 + 75 + 30 + 3)/216 = −17/216 = −0.079 That is, on the average, the player loses almost 8 cents for every dollar wagered.
This result may be surprising, since the game superficially looks like an even bet.
This point is discussed in the exercises. !
As Example 4.43 suggests, it is sometimes easier to group the points according to the value of their payofffunction. In general, suppose we have a probability space with a payofffunction f, and f has a finite number of different values that it produces. For instance, in Example 4.43 f produced only the values −1, 1, 2, and 3. For each value v produced by f, let Ev be the event consisting of the points x such that f(x) = v. That is, Ev is the set of points on which f produces value v.
Then EV(f) = X v v PROB(Ev) (4.22) In the common case where the points have equal probability, let nv be the number of points in event Ev and let n be the total number of points in the space.
Then PROB(Ev) is nv/n, and we may write EV(f) = X v vnv /n 214 COMBINATORICS AND PROBABILITY !
Example 4.44. In Example 4.25, we introduced the game of Keno, and com-puted the probability of guessing three out of 5 correct. Now let us compute the expected value of the payoffin the 5-spot game of Keno. Recall that in the 5-spot game, the player guesses five numbers from 1 to 80. When the game is played, twenty numbers from 1 to 80 are selected. The player wins if three or more of those twenty numbers are among the five he selected.
However, the payoffdepends on how many of the player’s five numbers are correct. Typically, for a $1 bet, the player receives $2 if he guesses three out of five (i.e., the player has a net gain of $1). If he guesses four out of five, he receives $15, and for guessing all five, he is rewarded with $300. If he guesses fewer than three correctly, the player receives nothing, and loses his $1.
In Example 4.25, we calculated the probability of guessing three out of five to be 0.08394 (to four significant places). We can similarly calculate that the probability of guessing four out of five is 0.01209, and the probability of guessing all five is 0.0006449. Then, the probability of guessing fewer than three is 1 minus these three fractions, or 0.90333. The payoffs for fewer than three, for three, four, and five are −1, +1, +14, and +299, respectively. Thus, we may apply formula (4.22) to get the expected payoffof the 5-spot game of Keno. It is 0.90333 × −1 + 0.08394 × 1 + 0.01209 × 14 + 0.0006449 × 299 = −0.4573 Thus, the player loses almost 46 cents of every dollar he bets in this game. !
EXERCISES 4.12.1: Show that if we throw three dice, the expected number of 1’s that will appear is 1/2.
4.12.2: Since we win when there is a 1 and lose when there is not, why does not the fact in Exercise 4.12.1 imply that Chuck-a-Luck is an even game (i.e., the expected payoffby betting on 1, or any other number, is 0)?
4.12.3: Suppose that in a 4-spot game of Keno, where the player guesses four numbers, the payout is as follows: for guessing two, $1 (i.e., the player gets his dollar back); for guessing three, $4; for guessing all four, $50. What is the expected value of the payout?
4.12.4: Suppose in a 6-spot game of Keno, the payouts are as follows: for guessing three, $1; for four, $4; for five, $25; for guessing all six, $1000. What is the expected value of the payout?
4.12.5: Suppose we play a Chuck-a-Luck type of game with six dice. The player pays $1 to play, bets on a number, and throws the dice. He is rewarded with $1 for every time his selected number appears. For instance, if it appears once, the net payout is 0; if it appears twice, the net payout is +1, and so on. Is this a fair game Fair game (i.e., is the expected value of the payout 0)?
4.12.6: Based on the style of payout suggested by Exercise 4.12.5, we could modify the payout of the standard 3-dice form of Chuck-a-Luck so that the player pays some amount to play. He is then rewarded with $1 for every time his number appears.
What is the proper amount the player should pay in order that this be a fair game?
SEC. 4.13 SOME PROGRAMMING APPLICATIONS OF PROBABILITY 215 !
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4.13 Some Programming Applications of Probability In this section, we shall consider two types of uses for probability calculations in computer science.
The first is an analysis of the expected running time of an algorithm. The second is a new type of algorithm, often called a “Monte Carlo” algorithm, because it takes a risk of being incorrect. As we shall see, by adjusting parameters, it is possible to make Monte-Carlo algorithms correct with as high a probability as we like, except that we cannot reach probability 1, or absolute certainty.
A Probabilistic Analysis Let us consider the following simple problem.
Suppose we have an array of n integers, and we ask whether an integer x is an entry of the array A[0..n-1]. The algorithm in Fig.4.23 does as well as any. Note that it returns a type which we called BOOLEAN, defined to be a synonym for int in Section 1.6. Also in that section were defined the constants TRUE and FALSE, which stand for 1 and 0, respectively.
BOOLEAN find(int x, int A[], int n) { int i; (1) for(i = 0; i < n; i++) (2) if(A[i] == x) (3) return TRUE; (4) return FALSE; } Fig. 4.23. Finding an element x in an array A of size n.
The loop of lines (1) – (3) examines each entry of the array, and if x is found there, immediately terminates the loop with TRUE as our answer. If x is never found, we reach line (4) and return FALSE. Let the time taken by the body of the loop and the loop incrementation and test be c. Let d be the time of line (4) and the initialization of the loop. Then if x is not found, the running time of the function of Fig. 4.23 is cn + d, which is O(n).
However, suppose x is found; what is the running time of Fig. 4.23 then?
Clearly, the earlier x is found, the less time it takes. If x were somehow guaranteed to be in A, the time would be O(1), since the loop would iterate only once. But if x were always at or near the end, the time would be O(n).
Surely the worst case is when we find x at the last step, so O(n) is a smooth and tight upper bound on the worst case. However, is it possible that the average case is much better than O(n)? In order to address this question, we need to define a probability space whose points represent the possible places in which x can be found. The simplest assumption is that x is equally likely to be placed in any of the n entries of array A. If so, then our space has n points, one representing each of the integers from 0 to n −1, which are the bounds of the index of array A.
Our question then becomes: in this probability space, what is the expected value of the running time of the function of Fig. 4.23? Consider a point i in the 216 COMBINATORICS AND PROBABILITY space; i can be anything from 0 to n −1. If x is in A[i], then the loop will iterate i + 1 times. An upper bound on the running time is thus ci + d. This bound is off slightly in the constant d, since line (4) is never executed. However, the difference does not matter, since d will disappear when we translate to a big-oh expression anyway.
We must thus find the expected value of the function f(i) = ci + d on this probability space. We sum ci + d where i ranges from 0 to n −1, and then divide by n, the number of points. That is, EV(f) = n−1 X i=0 ci + d /n = cn(n −1)/2 + dn /n = c(n −1)/2 + d For large n, this expression is about cn/2. Thus, O(n) is the smooth and tight upper bound on this expected value. That is, the expected value is, to within a constant factor of about 2, the same as the worst case. This result makes intuitive sense. If x is equally likely to be anywhere in the array, it will “typically” be half way down the array, and we therefore do about half the work that would be done if x were not in the array at all, or if it were at the last element.
Algorithms That Use Probability The algorithm of Fig. 4.23 was deterministic, in the sense that it always does the Deterministic algorithm same thing on the same data.
Only the analysis of the expected running time uses probability calculations.
Almost every algorithm we meet is deterministic.
However, there are some problems that are better solved with an algorithm that is not deterministic, but uses a selection from a probability space in some essential way. Making such a selection from an imagined probability space is not hard; we use a random number generator as discussed in the box in Section 4.9.
One common type of probabilistic algorithm, called a Monte-Carlo algorithm, Monte-Carlo algorithm makes a random selection at each iteration. Based on this selection, it will either say “true,” in which case that is guaranteed to be the correct answer, or it will say “I don’t know,” in which case the correct answer could be either “true” or “false.” The possibilities are suggested by the probability space in Fig.4.24.
The probability that the algorithm will say “true” given that the answer is true is a/(a + b). That is, this probability is the conditional probability of event a in Fig. 4.24 given a or b. As long as this probability p is greater than 0, we can iterate as many times as we like, and get rapidly decreasing probability of a failure. By “failure,” we mean that the correct answer is “true,” but on no iteration does the algorithm tell us so.
Since each iteration is an independent experiment, if the correct answer is “true” and we iterate n times, the probability that the algorithm will never say true is (1 −p)n. As long as 1 −p is strictly less than 1, we know that (1 −p)n will decrease rapidly as n increases. For example, if p = 1/2, then 1 −p is also 1/2.
The quantity (0.5)n is about 1/1000 for n = 10 (see the box in Section 4.2); it is 1/1000000 for n = 20, and so on, decreasing by a factor of about 1000 every time n increases by 10.
The Monte-Carlo algorithm, then, is to run the experiment n times. If the answer to any experiment is “true,” then the algorithm answers “true.” If all answers are “false,” then the algorithm answers “false.” Thus, 1.
If the correct answer is “false,” the algorithm will surely answer “false.” SEC. 4.13 SOME PROGRAMMING APPLICATIONS OF PROBABILITY 217 Algorithm a Algorithm Answer is “false”; algorithm says “I don’t know” c says “I don’t answer is “true” b says “true” know,” but Fig. 4.24. Possible outcomes of one iteration of a Monte-Carlo algorithm.
2.
If the correct answer is “true,” then the algorithm answers “false” with proba-bility (1−p)n, which we assume is very small because n is chosen large enough to make it small. The algorithm answers “true” with probability 1 −(1 −p)n, which is presumably very close to 1.
Thus, there are no failures when the correct answer is “false” and very few failures when the correct answer is “true.” !
Example 4.45.
Here is an example of a problem that can be solved more efficiently using a Monte-Carlo algorithm.
The XYZ computer company orders boxes of chips that are supposed to be tested at the factory to make sure they are all good. However, XYZ believes that some boxes have been leaving the factory untested, in which case the probability that any given chip is bad is 1/10. A simple approach solution would be for XYZ itself to test all the chips it receives, but this process is expensive and time-consuming. If there are n chips in a box, the test of a box takes O(n) time.
A better approach is to use a Monte-Carlo algorithm. Select from each box k chips at random to be tested. If a chip is bad, answer “true” — the box had not been tested at the factory or else the bad chip would have been discovered. If the chip is good, answer “I don’t know,” and go on to the next chip. If all k chips that we test are good, declare the entire box to be good.
In terms of Fig. 4.24, region c represents the case that the chips are chosen from a good box; region b is the case that the box is untested, but the chip happens to be good; region a is the case that the box is untested, and the chip is bad. Our assumption that 1/10 of the chips are bad if the box is untested says that the area of the circle, a, is one-tenth of the area of the ellipse enclosing regions a and b.
Let us compute the probability of failure — all k chips are good, but the box is untested.
The probability of saying “I don’t know” after testing one chip is 1 −1 10 = 0.9. Since the events of testing each chip are independent, the probability that we shall say “I don’t know” to each of k chips is (0.9)k. Suppose we pick 218 COMBINATORICS AND PROBABILITY k = 131. Then the probability of failure, (0.9)131, is almost exactly 0.000001, or one in a million. That is, if the box is good, we shall never find a bad chip in it, so we surely say the box is good. If the box is untested, then with probability 0.999999 we find a bad chip among the 131 we test and say the box needs full testing. With probability 0.000001, the box is untested, but we say it is a good box and do not test the remaining chips in the box.
The running time of this algorithm is O(1). That is, the time to test at most 131 chips is a constant, independent of n, the number of chips in the box. Thus, compared with the more obvious algorithm of testing all the chips, the cost has gone down from O(n) to O(1) per box, at the cost of making an error once in every million untested boxes.
Moreover, we can make the probability of error as small as we like, by changing the number of chips we test before concluding we have a good box. For example, if we double the number of tested chips, to 262, then we square the probability of failure, which then becomes one in a trillion, or 10−12. Also, we could save a constant factor in time at the cost of a higher failure rate. For instance, if we halved the number of chips tested, to 66 per box, we would have a failure rate of about one in a thousand untested boxes. !
EXERCISES 4.13.1: Which of 377, 383, 391 is prime?
4.13.2: Suppose that we used the function of Fig. 4.23 to search for element x, but the probability of finding x in entry i is proportional to n −i. That is, we can imagine a probability space with n(n + 1)/2 points, n of which represent situations where x is in A, n −1 points represent situations where x is in A, and so on, down to one point that represents the possibility that x is in A[n-1]. What is the expected running time of the algorithm for this probability space?
4.13.3: In 1993, the National Basketball Association held a lottery in which the 11 teams that did not make the playoffs participated. At stake was the first choice in the player draft. The team with the worst record was given 11 tickets in the lottery, the next-worst team got 10, and so on, until the 11th-worst team was given one ticket. A ticket was picked at random and the first-place draft was awarded to the holder of that ticket. What is the expected value of the function f(t) that is the finishing place (from the bottom) of the holder of the selected ticket t?
4.13.4: The lottery described in Exercise 4.13.3 continued. The winning team had all their tickets removed, and another ticket was picked for second place in the draft. That winner’s remaining tickets were removed, and a third ticket was selected for third place in the draft. What are the expected values of the finishing positions of the teams that got second and third places in the draft?
4.13.5: Suppose that we are given an array of size n, either sorted or filled at ran-dom with integers. We wish to construct a Monte-Carlo algorithm that will either say “true” if it finds the array unsorted or will say “I don’t know.” By repeating the test k times, we’d like to know there is no more than a 2−k probability of failure.
Suggest such an algorithm. Hint: Make sure that your tests are independent. As an example of tests that are not independent, we might test whether A 1 times among the n items, divide by k!. Perform this division for each value that appears more than once. The paradigm problem is counting anagrams of a word of length n, where we must divide n! by k! for each letter that appears k times in the word and k > 1.
!
If we want to distribute n identical objects into m bins, we can do so in n+m−1 n ways. The paradigm problem is distributing apples to children.
!
If as above, some of the objects are not identical, we count the number of ways to distribute them to bins as follows. Start with (n + m −1)!/(m −1)!. Then, if there is a group of k identical objects, and k > 1, divide by k!. Perform the division for each value that appears more than once. The paradigm problem is distributing fruits of various kinds to children.
In addition, the reader should remember the following points about probability.
!
A probability space consists of points, each of which is the outcome of an experiment. Each point x has associated with it a nonnegative number called the probability of x. The sum of the probabilities of the points in a probability space is 1.
SEC. 4.15 BIBLIOGRAPHIC NOTES FOR CHAPTER 4 221 !
An event is a subset of the points in a probability space. The probability of an event is the sum of the probabilities of the points in the event. The probability of any event lies between 0 and 1.
!
If all points are equally likely, the conditional probability of event F given event E is the fraction of the points in event E that are also in event F.
!
Event E is independent of event F if the conditional probability of E given F is the same as the probability of E. If E is independent of F, then F is independent of E.
!
The rule of sums says that the probability that one of two events E and F occurs is at least the larger of their probabilities and no greater than the sum of their probabilities (or no greater than 1 if the sum is above 1).
!
The rule of products says that the probability that the outcome of an experi-ment is both in event E and in event F is no greater than the smaller of their probabilities and at least the sum of their probabilities minus 1 (or at least 0 if the latter is negative).
Finally, there are a number of applications to computer science of principles we learned in this chapter.
!
Any sorting algorithm that works on arbitrary types of data that can be com-pared by “less than” requires time at least proportional to n log n to sort n items.
!
The number of bit strings of length n is 2n.
!
Random number generators are programs that produce sequences of numbers that appear to be the result of independent experiments, although of course they are completely determined by the program.
!
Systems for probabilistic reasoning need a way to express the probabilities of compound events that are formed from the occurrence of several events.
The rules of sums and products sometimes help. We also learned some other simplifying assumptions that let us put bounds on the probability of compound events.
!
Monte-Carlo algorithms use random numbers to produce either a desired result (“true”) or no result at all. By repeating the algorithm a constant number of times, we can solve the problem at hand by concluding the answer is (“false”) if none of the repetitions produces the answer “true.” By selecting the number of repetitions, we can adjust the probability of incorrectly concluding “false” to be as low as we like, but not 0.
!
! !
!
4.15 Bibliographic Notes for Chapter 4 A venerable and excellent introduction to combinatorics is Liu . Graham, Knuth, and Patashnik is a deeper discussion of the subject. Feller is the classic book on probability theory and its applications.
The Monte-Carlo algorithm for testing whether a number is a prime is from Rabin . A discussion of this algorithm and other interesting issues involving 222 COMBINATORICS AND PROBABILITY computer security and algorithms that use randomness in an important way can be found in Dewdeney . A more advanced discussion of these topics is presented in Papadimitriou .
Dewdeney, A. K. . The Turing Omnibus, Computer Science Press, New York.
Feller, W. .
An Introduction to Probability Theory and Its Applications, Third Edition, Wiley, New York.
Graham, R. L., D. E. Knuth, and O. Patashnik . Concrete Mathematics: A Foundation for Computer Science, Addison-Wesley, Reading, Mass.
Liu, C.-L. . An Introduction to Combinatorial Mathematics, McGraw-Hill, New York.
Papadimitriou, C. H. . Computational Complexity, Addison-Wesley, Reading, Mass.
Rabin, M. O. . “Probabilistic algorithms,” in Algorithms and Complexity: New Directions and Recent Trends (J. F. Traub, ed.), pp. 21–39, Academic Press, New York. |
188592 | https://www.jvsmedicscorner.com/Physiology_files/Revised%20Starling%20equation%20and%20the%20glycocalyx%20model%20Review.pdf | Revised Starling equation and the glycocalyx model of transvascular fluid exchange: an improved paradigm for prescribing intravenous fluid therapy T. E. Woodcock 1 and T. M. Woodcock 2 1 Critical Care Service, Southampton University Hospitals NHS Trust, Tremona Road, Southampton SO16 6YD, UK 2 The Australian School of Advanced Medicine, Macquarie University, NSW 2109, Australia Corresponding author. E-mail: tom.woodcock@me.com Editor’s key points † The classic Starling principle does not hold for fluid resuscitation in clinical settings.
† The endothelial glycocalyx layer appears to have a major role in fluid exchange.
† A revision of Starling incorporating the glycocalyx model appears to explain better the responses seen clinically.
Summary. I.V. fluid therapy does not result in the extracellular volume distribution expected from Starling’s original model of semi-permeable capillaries subject to hydrostatic and oncotic pressure gradients within the extracellular fluid. Fluid therapy to support the circulation relies on applying a physiological paradigm that better explains clinical and research observations. The revised Starling equation based on recent research considers the contributions of the endothelial glycocalyx layer (EGL), the endothelial basement membrane, and the extracellular matrix. The characteristics of capillaries in various tissues are reviewed and some clinical corollaries considered. The oncotic pressure difference across the EGL opposes, but does not reverse, the filtration rate (the ‘no absorption’ rule) and is an important feature of the revised paradigm and highlights the limitations of attempting to prevent or treat oedema by transfusing colloids. Filtered fluid returns to the circulation as lymph. The EGL excludes larger molecules and occupies a substantial volume of the intravascular space and therefore requires a new interpretation of dilution studies of blood volume and the speculation that protection or restoration of the EGL might be an important therapeutic goal. An explanation for the phenomenon of context sensitivity of fluid volume kinetics is offered, and the proposal that crystalloid resuscitation from low capillary pressures is rational. Any potential advantage of plasma or plasma substitutes over crystalloids for volume expansion only manifests itself at higher capillary pressures.
Keywords: fluid therapy; intensive care Twenty-five years ago, Twigley and Hillman announced ‘the end of the crystalloid era’. Using a simplified diagram of plasma, interstitial and intracellular fluid compartments, and their anatomic volumes, they argued that colloids could be used to selectively maintain the plasma volume.1 Plasma volume being about 20% of the extracellular fluid (ECF), it was presumed that the volume equivalence for re-suscitation from intravascular hypovolaemia would be of the order of 20 ml colloid to 100 ml isotonic salt solution (ISS). Moreover, it was presumed from Starling’s principle that transfusion of hyperoncotic colloid solutions would absorb fluid from the interstitial fluid (ISF) to the intravascu-lar volume. This simple concept of colloid for plasma volume and ISS for ECF replacement has been continued and devel-oped.2–4 Two trials in critically ill patients have found that over the first 4 days of fluid resuscitation, 100 ml ISS is as ef-fective as 62–76 ml human albumin solution5 or 63–69 ml hyperoncotic plasma substitute.6 In blunt trauma patients during the first day of resuscitation, 100 ml ISS was as effect-ive as 97 ml isosmotic plasma substitute, while in gunshot or stabbing victims, 100 ml was as effective as 67 ml.7 A trial of paediatric resuscitation practices in resource-poor facilities in Africa demonstrated no advantages of bolus therapy with albumin compared with ISS, and a survival advantage for slow ISS resuscitation without bolus therapy.8 A series of volume kinetics experiments have demonstrated that the central volume of distribution of ISS is much smaller than the anatomic ECF volume,9 and an editorial had to conclude that ‘Fluid therapy might be more difficult than you think’.10 This review attempts to reconcile clinical trial data and bedside experience of fluid therapy with recent advances in microvascular physiology to improve our working paradigm for rational prescribing.
Starling’s principle From experiments injecting serum or saline solution into the hindlimb of a dog, Starling deduced that the capillaries and post-capillary venules behave as semi-permeable mem-branes absorbing fluid from the interstitial space.11 The work of Krooh and colleagues12 developed Starling’s principle in human physiology. With adoption of reflection coeffi-cient13 and pore theories,14 the familiar paradigm of raised venous pressure and reduced plasma protein concentration British Journal of Anaesthesia 108 (3): 384–94 (2012) Advance Access publication 29 January 2012 . doi:10.1093/bja/aer515 & The Author . Published by Oxford University Press on behalf of the British Journal of Anaesthesia. All rights reserved.
For Permissions, please email: journals.permissions@oup.com by John Vogel on March 5, 2012 Downloaded from leading to oedema in clinical practice emerged.12 15 Luft16 revealed ‘the fine structure of the capillary and the endocapillary layer’ in 1966, and Curry and Michel17 18 pro-posed a theory ‘that the molecular sieving properties of the capillary wall reside in a matrix of molecular fibres which covers the endothelial cells and fills the channels through or between them’ in 1980.
Transvascular exchange depends on a balance between hydrostatic and oncotic pressure gradients. Fluid is filtered to the interstitial space under a dominant hydrostatic pressure gradient (capillary pressure Pc minus ISF pressure Pis) at the arteriolar portion of capillaries, and it was believed that it is absorbed back under a dominant colloid osmotic pressure (COP) gradient (capillary COP pc minus ISF COP pis) at the venular end. In 2004, Adamson and colleagues19 showed that the effect of pis on transvascular fluid exchange is much less than pre-dicted by the standard Starling equation, which therefore has to be revised.20 It is now established that non-fenestrated capillaries normally filter fluid to the ISF through-out their length. Absorption through venous capillaries and venules does not occur. pc opposes, but does not reverse, filtration. Most of the filtered fluid returns to the circulation as lymph. Levick and Michel21 now propose that the small pore system of the transvascular semi-permeable membrane is the endothelial glycocalyx layer (EGL) where it covers the endothelial intercellular clefts, separating plasma from a ‘protected region’ of the subglycocalyx space which is almost protein-free. Subglycocalyx COP (psg) replaces pis as a determinant of transcapillary flow (Jv).19 22 Plasma pro-teins, including albumin, escape to the interstitial space by a relatively small number of large pores, which are respon-sible for the increased Jv observed in the early stage of inflammation,21 and may be susceptible to pharmacological intervention.23–25 The fact that low protein concentration within the subglycocalyx intercellular spaces accounts for the low Jv and lymph flow in most tissues is a critical insight and the basis of the glycocalyx model.21 The endothelial glycocalyx layer The EGL is a web of membrane-bound glycoproteins and pro-teoglycans on the luminal side of the endothelial cells, associated with various glycosaminoglycans (GAGs) (muco-polysaccharides) which contribute to the volume of the layer (Fig. 1).26 It is the active interface between blood and the capillary wall.27 Visualization of the EGL is technically demanding, but has helped to emphasize its physiological importance.28 29 From indocyanine green dilution studies of patients given a large dose of i.v. colloid, the human EGL volume was estimated to be about 700 ml,30 and presuming Healthy glycocalyx layer which contains glycosaminoglycans Compressed glycocalyx, shedding or flaking glycosaminoglycans to the plasma Endothelial cell Erythrocyte Endothelial glycocalyx layer Glycosaminoglycans Basement membrane/extracellular matrix Fig 1 A cartoon illustrating that the intravascular volume contains the non-circulating glycocalyx fluid volume and the circulating plasma volume. Red blood cells are excluded from the glycocalyx layer. Compaction of the glycocalyx layer increases plasma volume and the red cell dilution volume independently of changes in intravascular volume.
Transvascular exchange and fluid therapy BJA 385 by John Vogel on March 5, 2012 Downloaded from that the endothelial surface area approximates 350 m,2 31 an average EGL thickness of about 2 mm was suggested. Fluid within the EGL is a non-circulating portion of the intravascu-lar volume with a protein concentration gradient between the free-flowing plasma and the endothelial intercellular clefts. The EGL is thinner where it covers the microcirculation (as little as 0.2 mm) and thicker in larger vessels (up to 8 mm).29 The EGL is semi-permeable with respect to anionic macromolecules such as albumin and other plasma proteins, whose size and structure appear to determine their ability to penetrate the layer.32 The healthy EGL is impermeable to Dextran molecules of 70 kDa or more, and the glycocalyx– plasma boundary can be visualized as that part of the intra-vascular space that excludes fluoroscein-labelled Dextran 70.32 33 Red blood cells are also excluded from the EGL, and the intravascular red cell exclusion volume is larger than the Dextran 70 exclusion volume.34 Dextran 40 is small enough not to be excluded by the EGL, and studies measuring the distribution volumes of Dextran 40 and ery-throcytes in human subjects indicate an EGL in health of about 1700 ml, much larger than the indocyanine green dilu-tion method.35 Microvascular EGL thickness can be measured in sublingual tissues of patients using orthogonal polariza-tion spectral imaging, and correlates well with dilution estimations.36 By removing GAGs and measuring volume reduction (com-paction) of the EGL, the major ones appear to be heparan sulphate, chondroitin sulphate, and hyaluronic acid.33 37 Com-paction of the EGL by removal of GAGs preserves its resistance to filtration, despite loss of thickness and possible reduction in permeability.37 Compaction of the EGL and the increase in heparan,38 39 hyaluronic acid,40 or chondroitin41 in plasma are considered markers of glycocalyx injury, described as ‘shedding’, ‘flaking’, or ‘fragmentation’ (Fig. 1). Rapid crystal-loid infusion in volunteers results in elevated plasma levels of hyaluronic acid and may therefore be injurious.42 43 Increased plasma concentrations of GAGs have been found in septic shock patients, and they appear to reduce the anti-bacterial properties of plasma.44 The volume of the EGL can be reduced by ≥1 litre in diabetes40 or acute hypergly-caemia.35 A number of other molecules, derived both from the endothelium and from the plasma and involved in coagu-lation and inflammation, exist within the EGL. The proteogly-can syndecan is a major glycocalyx component which increases in the plasma when EGL shedding occurs.38 45–47 It appears, on the evidence from human studies to date, that the EGL is compromised in systemic inflammatory states such as diabetes,40 hyperglycaemia,35 surgery,38 trauma,47 and sepsis.45 Inflammatory mediators which have been implicated so far include C-reactive protein,48 A3 adenosine receptor stimulation,49 50 tumour necrosis factor,34 bradykinin,51 and mast cell tryptase.52 Therapeutic options for the protection or restoration of the EGL emerge from such studies. N-acetyl cysteine,35 antithrombin III or hydrocortisone,53–56 and even sevoflurane anaesthesia57 58 could be beneficial.
Compacted EGL volume can be restored by infusion of the GAGs chondroitin sulphate and hyaluronic acid.33 Vascular endothelial cells The reticuloendothelial capillaries of the sinusoidal tissues (liver, spleen, and bone marrow) are of phago-endocytic phenotype (Fig. 2). They express uptake receptors for hyalur-onic acid, and by actively removing this important GAG, they prevent development of an effective EGL.
In hepatic Sinusoidal capillary (liver, spleen, marrow) Non-fenestrated capillary (nervous system, muscle, connective, lung) Fenestrated capillary (endocrine, choroid plexus, gut mucosa) Fenestrated capillary (glomerular) Endothelial cell Endothelial glycocalyx layer Erythrocyte Basement membrane/extracellular matrix Fig 2 A cartoon illustrating some anatomic differences between four capillary phenotypes.
BJA Woodcock and Woodcock 386 by John Vogel on March 5, 2012 Downloaded from sinusoids, open fenestrations are the primary pathway for macromolecules as large as chylomicrons and lipoproteins to pass between plasma and ISF. The upper effective pore size of human hepatic sinusoidal capillaries is estimated to be about 180 nm.59 Albumin synthesis is proportional to hepatic pis, so an increase in other plasma proteins (acute phase proteins) or transfusion of colloids will displace albumin to the extravascular compartment and suppress albumin synthesis.43 60 61 Being limited by fibrous capsules, the sinusoidal tissues have little or no compliance to accom-modate ISF expansion. Filtration to the ISF will be dependent on hydrostatic pressure gradients, as there is no COP mech-anism to oppose filtration, and return to the circulation is via the lymphatics. The liver is observed to account for around 50% of the body’s total lymph production, with higher than average protein concentration, and is therefore the major site of transcapillary escape of plasma proteins and probably of other macromolecules when capillary func-tion is unimpaired. In resuscitated hyperdynamic septic shock patients, hepatic blood flow is increased to around 50% of the cardiac output.62 The capillaries of the renal glomeruli have a full basement membrane and EGL, but they are generously fenestrated.
Anatomically, the fenestrations are as wide as 65 nm, but their effective pore size is only about 15 nm, attributable to EGL flanking over the open fenestrations. The effective pore size for glomerular filtration beyond the capillary basement membrane is limited to about 6 nm by filtration slit dia-phragms at the level of podocyte foot processes. Thus, albumin and larger molecules are normally not filtered into tubular fluid. It has been presumed that albuminuria is an index of capillary permeability,63 but the mechanism is prob-ably more complex.64 Protein filter function is impaired by hyperglycaemia,65 and probably by other kidney injuries.
Fenestrated capillaries with specialized functions also exist in the endocrine and exocrine glands and the choroid plexus. The fenestrated capillaries of the kidney cortex and medulla (peritubular capillaries and the vasa recta), the gastrointestinal mucosa, and the lymph nodes are notable exceptions to the principle of no fluid absorption.21 The base-ment membrane of these capillaries is continuous, and their diaphragmed fenestrations are induced by vascular endothe-lial growth factors. Their upper pore size is in the range of 6–12 nm.59 Non-sinusoidal non-fenestrated capillaries have continu-ous basement membrane and EGL. Breaks within the inter-endothelial cell junctions constitute the primary path-ways for transvascular fluid filtration, and the increased porosity seen in inflammation may be due to an increase in these normally infrequent discontinuities.19 25 59 66 An alternative interpretation of pore theory, called the ‘glycocalyx-junction-break model’, proposes that pore size (small or large) is a function of the spaces between the matrix fibres of the EGL, while the area for fluid exchange is a function of the length of the junction breaks between ad-jacent endothelial cells.22 In the capillaries of the brain and spinal cord, endothelial cell membranes are tightly opposed by zona occludens tight junctions with few breaks, resulting in very small effective pore size of barely 1 nm.59 The blood–brain barrier is therefore only permeable to the smal-lest non-lipid soluble molecules.
Non-sinusoidal non-fenestrated capillaries of muscles, connective tissues, and lungs have macula occludens loose junctions to their inter-cellular clefts, and the effective pore size there is up to 5 nm, making them permeable to molecules as large as myo-globin. The tissues that can accumulate substantial amounts of ISF after trauma and sepsis (i.e. the more compliant tissues) are loose connective tissues, muscles, lungs, and gastrointestinal mesentery and mucosa. For example, extra-vascular lung water measured by double indicator dilution can increase from around 500 ml to 2.5 litre in pulmonary oedema, while the loose connective tissues and muscles can expand to many litres of peripheral oedema.
Aquaporins are present within vascular endothelial cell membranes, particularly in muscles. Their effective pore size is very small and it is believed they contribute little to transvascular filtration. There is controversy about the signifi-cance of transcellular large-pore systems for transport of proteins from plasma to the ISF in systemic inflammation.
If they exist, their effective pore size is .50 nm. An increase in large pore numbers is an important component of the in-flammatory increase in Jv.21 Endothelial cells may undergo phenotype changes in response to physical and chemical stresses, which contributes to endothelial dysfunction.22 This consideration of the four types of body tissue and their capillaries helps to explain some otherwise unexpected clinical observations. A series of experiments on the volume kinetics of rapidly infused i.v. fluids measured the haemoglo-bin concentration of arterial or venous blood and modelled a central and a peripheral volume which represent the intra-vascular and extravascular fluid volumes, respectively.9 Acutely, the peripheral volume is found to be 6–8 litre, less than the anatomic ISF volume. As volume kinetics measure only the volume which can be expanded, and this will there-fore not include spaces limited by rigid structures such as the bone (brain, marrow) or fibrous capsules (liver, spleen, kidney).9 This goes some way to explaining why ISS are more efficient plasma expanders than we might expect if we were to presume their distribution throughout the whole ECF. In systemic capillary leak syndrome, so much fluid goes to the soft tissues of the limbs that it can cause compartment syndromes.67 The extracellular matrix and basement membrane The glycocalyx is the first and the major fibre matrix resistor in the current of fluid and solutes between plasma and lymph. The basement membrane and extracellular matrix are the second and third resistances in a series.68 The base-ment membrane, where it exists, is a specialized part of the extracellular matrix 60–100 nm in thickness, composed of type IV collagen and laminin and closely adherent to the cell membrane.59 69 The extracellular matrix is a web of Transvascular exchange and fluid therapy BJA 387 by John Vogel on March 5, 2012 Downloaded from collagen fibrils within the interstitial space upon which glyco-proteins such as fibronectin and proteoglycans (protein molecules with GAG side chains) are arranged, and contain free GAGs. Toll-like receptors are found within the extracellu-lar matrix and are believed to have a pivotal role in the early development of systemic inflammatory response70 and ventilator-induced lung injury.71 Integrins and their receptors modulate cell locomotion through the extracellular matrix, and it has been discovered that they can modulate Pis by bringing about conformational changes to collagen which allow the GAGs to become hydrated. An acute reduction in Pis occurs in inflammatory conditions, increasing the transen-dothelial pressure difference and thereby increasing Jv by as much as 20-fold independently of other causes of capillary ‘leak’.72 73 Plasma proteins Proteins have an oncotic role across the endothelial glycoca-lyx and COP difference opposes but does not reverse Jv. The EGL is semi-permeable to albumin molecules, and the pres-ence of albumin within the EGL is an important determinant of its filter function.74 The functional unit of EGL with its con-tained albumin is sometimes referred to as the endothelial surface layer. Plasma albumin concentration is the major de-terminant of the plasma COP in health, but in congenital analbuminaemia or acquired hypoalbuminaemia, other pro-teins become more important.61 Albumin molecules distrib-ute through the ECF and in health, it is estimated that about 40% of the total body albumin is intravascular. In in-flammation, the intravascular proportion of albumin will de-crease and the extravascular proportion will increase. The measured transcapillary escape rate of albumin to the tissues (TCERA) is said to be an index of ‘vascular permeabil-ity’. The normal TCERA is about 5% of the plasma albumin per hour, but this can double during surgery and may be increased to 20% or more in septic shock.75 The gallium-transferrin pulmonary leak index can be used as an index of pulmonary permeability, and it has been found to be in-versely related to plasma albumin and plasma transferrin concentrations in both septic and non-septic intensive care patients with acute lung injury.
Clinicians rely on the original Starling principle as a reason to transfuse plasma or albumin to preferentially resuscitate the intravascular volume. The revised Starling equation and the glycocalyx model lead us to expect that the transen-dothelial protein concentration difference will regulate Jv after plasma or albumin resuscitation, but the no absorption rule will preclude any significant benefit for the intravascular volume. This could explain some of the clinical observations relating to albumin therapy. These include the following: † Hypoalbuminaemia is a marker of disease severity and a predictor of complications in surgical patients,76–78 but treatment of hypoalbuminaemia is of no clinical benefit.79 80 † Acute respiratory distress syndrome (ARDS) patients (lung injury score 2.5 or more) have low plasma albumin and transferrin concentrations,81 but hyperon-cotic human albumin solution with or without a diuretic produces no improvement in pulmonary oedema.82 † Negative fluid balance rather than COP difference improves the alveolar to arterial oxygen tension ratio in ARDS patients.83 84 † In septic and non-septic patients, fluid loading against central venous pressure produces greater increases in cardiac output with human albumin solution than with ISS,85 but there are no benefits for pulmonary oedema nor for the lung injury score.86 Red cell dilution studies of hyperoncotic human albumin so-lution transfusion have been interpreted as showing osmotic absorption of fluid from the extravascular to intravascular compartment.87 Without information from an indicator of the whole intravascular volume, such as Dextran 40, such a conclusion is not justified. An acute increase in circulating plasma COP would be expected to draw water from the non-circulating part of the intravascular volume within the EGL.
Studies reporting red cell dilution data that do not take into account the EGL intravascular volume should be inter-preted with caution (Fig. 1).
Plasma substitutes Plasma substitutes are used to maintain or raise the plasma COP, although they too displace albumin from the circula-tion.43 Moreover, by elevating COP, they suppress hepatic albumin synthesis. Little is known of their effect on the EGL, but they would not be expected to support EGL filter function as albumin does. The need to consider the contribution of the EGL to red cell volume of distribution changes as noted above for studies of hyperoncotic albumin solutions applies equally to studies of hyperoncotic plasma substitutes.88 In normovolaemic volunteers made hypervolaemic, modi-fied fluid gelatin or hydroxyethyl starch solutions were dis-tributed to the ISF more slowly than ISS as explained by the revised Starling equation, but there was no difference in arterial pressure, urine output, or renal hormone concen-trations in plasma.89 In the expectation that they will be more effective than ISS at inducing hyperdynamic circula-tion, these plasma substitutes are commonly preferred for haemodynamic goal-directed therapy.90–95 However, the volume kinetic experiments have shown that the clearance of ISS from the central (intravascular) compartment (pre-sumed to reflect Jv) is substantially slower in anaesthetized patients than in unanaesthetized subjects,9 and it has been shown that ISS can be used to achieve hyperdynamic goals.96 97 This phenomenon is called context sensitivity.9 In contrast, a volume kinetic experiment in volunteers under-going euvolaemic haemodilution with hydroxyethyl starch found that the elimination rate constant from central to per-ipheral fluid compartments was not reduced, as is the case for ISS, but increased during anaesthesia compared with awake subjects.98 The duration of resuscitation attributable to hydroxyethyl starch after removal of a unit of blood was therefore shorter in subjects during desflurane anaesthesia.
BJA Woodcock and Woodcock 388 by John Vogel on March 5, 2012 Downloaded from The extent to which context sensitivity can be attributed to reduced transendothelial pressure difference is not clear, but reduced filtration response to crystalloid loading of hypo-volaemic subjects has been demonstrated,99 and crystalloids given during the vasodilation induced by spinal anaesthesia (‘co-loading’) are more effective than ‘preload’ crystal-loids.100 Mean arterial pressure is an important determinant of the distribution of ISS from the intravascular space in patients undergoing general or regional anaesthesia, so that the lower the pressure, the slower the clearance of crys-talloid from the circulation.101 One aspect of albumin therapy which may confer the benefit looked for in septic patients is the potential for anti-inflammatory or immune regulatory properties.102–106 Analysis of published data does not show that plasma substi-tutes are equivalent to albumin in this respect, or better than ISS.107 COP in practice A paradigm founded on the standard Starling principle attaches great importance to the COP of plasma in clinical practice. However, there is no difference between the COP of plasma in septic and non-septic patients,23 it does not in-fluence pulmonary transcapillary filtration in patients with pulmonary oedema,108 and it was not found to be a deter-minant of outcome in an intensive care practice.109 In a patient study, human albumin transiently raised plasma COP compared with hydroxyethyl starch or ISS, but neither fluid balance nor the development of peripheral or pulmon-ary oedema were different between the treatment groups.110 In a study of post-surgical patients with acute lung injury, it was found that plasma substitute resuscitation worsened the total thoracic compliance compared with normal saline, and that the type of fluid used for volume loading did not affect pulmonary permeability or oedema.111 Properties other than the effect on COP contrib-ute to the capillary ‘sealing’ effect of albumin or plasma sub-stitutes, and this has been called the ‘COP paradox’.74 Capillary pressure in practice Hahn’s experiments show that rapid i.v. infusion of a large volume of ISS in healthy subjects with initially normal Pc is cleared at rates in excess of 100 ml min21.9 112 Although this does not take into account the presence of the EGL, we can reasonably deduce that filtration from plasma to ISF (Jv) was increased by supranormal Pc and the transen-dothelial pressure difference. When albumin or hydroxyethyl starch is used for plasma volume expansion, increased Jv limits the plasma-expanding effect of the colloid.4 Nonethe-less, colloid infusions have a more sustained plasma expan-sion effect than crystalloids, probably because maintained plasma COP opposes the increase in Jv.89 When albumin or hydroxyethyl starch is used for normo-volaemic haemodilution, keeping the Pc normal, Jv is not increased and so most of the infused volume remains intra-vascular.4 Hahn has reported increased filtration response to crystalloid infusion after a colloid infusion,113 consistent with the paradigm that at supranormal transendothelial pressure difference, further increases in Pc result in increased Jv. Rapid infusion of crystalloid to normovolaemic volunteers increases the intrathoracic fluid volume, narrows small airways, and induces hyperventilation.114 Extreme elevation of Pc can damage the EGL,4 and cause stress failure of pulmonary capillaries leading to haemoptysis and oedema.115 Meta-analysis points to better patient outcomes if fluid balance is maintained.116 An improved paradigm of fluid physiology and therapy The intravascular space contains three compartments of interest (Table 1). If we define the intravascular fluid volume as that contained by the endothelial cells, we measure it as the Dextran 40 dilution volume and it approx-imates to the central volume of distribution of infused ISS.
Dextran 70 is excluded from the non-circulating EGL and its dilution volume consists of circulating plasma.117 There is a degree of exclusion of red cells from plasma at the EGL boundary, so the volume of distribution of red cells is rather less than the Dextran 70 dilution volume.
Compaction of the EGL can have significant effect on the balance of total intravascular fluid and red cell dilution volumes. EGL volume compaction by inflammation or hyper-osmotic colloid infusion must be considered in red cell dilu-tion studies of the intravascular volume.
Acute reduction of transendothelial pressure difference by pre-capillary vasoconstriction, post-capillary vasodilation, or hypovolaemia can result in transient absorption of fluid to the plasma volume, equivalent to as much as a 500 ml ‘auto-transfusion’ in human physiology (Fig. 3), but this effect lasts only for a few minutes.21 Absorption reverts to filtration as proteins diffuse into the subglycocalyx space from the ISF, diminishing the COP differ-ence that opposes filtration; this is the glycocalyx model.
With less acute or extreme disturbance to the equilibrium, the same mechanism preserves filtration, albeit at just a few millilitres per minute, and the no absorption rule applies (Fig. 4).
The pressure at which Jv approaches zero will depend on capillary porosity, which is the net effect of the various capil-laries’ hydraulic conductivities, area for fluid exchange, and the reflection coefficient of the macromolecules determining COP. The J-shaped curve describing Jv and Pc (Fig. 4) will be shifted to the left with increased capillary porosity, with the inflection on the curve being the J point. Below the J point, any transfused fluid, whether colloid or crystalloid, will appear to be retained within the intravascular space until the transendothelial pressure difference reaches the level at which filtration recommences. The glycocalyx model and the no absorption rule explain why the COP properties of plasma or plasma substitutes add little or nothing to plasma volume resuscitation while transendothelial pressure difference is below the J point. Above the J point, the oncotic Transvascular exchange and fluid therapy BJA 389 by John Vogel on March 5, 2012 Downloaded from pressure difference opposing filtration is maximal and Jv becomes proportional to Pc, or transendothelial pressure dif-ference if Pis is not constant.
Table 1 Comparison of the original and revised paradigms for prescribing fluid therapy Original Starling principle Revised Starling equation and glycocalyx model Intravascular volume consists of plasma and cellular elements Intravascular volume consists of glycocalyx volume, plasma volume, and red cell distribution volume Capillaries separate plasma with high protein concentration from ISF with low protein concentration Sinusoidal tissues (marrow, spleen, and liver) have discontinuous capillaries and their ISF is essentially part of the plasma volume Open fenestrated capillaries produce the renal glomerular filtrate Diaphragm fenestrated capillaries in specialized tissues can absorb ISF to plasma Continuous capillaries exhibit ‘no absorption’ The EGL is semi-permeable to anionic proteins and their concentration in the intercellular clefts below the glycocalyx is very low The important Starling forces are the transendothelial pressure difference and the plasma–interstitial COP difference The important Starling forces are the transendothelial pressure difference and the plasma–subglycocalyx COP difference. ISF COP is not a direct determinant of Jv Fluid is filtered from the arterial end of capillaries and absorbed from the venous end. Small proportion returns to the circulation as lymph Jv is much less than predicted by Starling’s principle, and the major route for return to the circulation is as lymph Raising plasma COP enhances absorption and shifts fluid from ISF to plasma Raising plasma COP reduces Jv but does not cause absorption At subnormal capillary pressure, net absorption increases plasma volume At subnormal capillary pressure, Jv approaches zero. Auto transfusion is acute, transient, and limited to about 500 ml At supranormal capillary pressure, net filtration increases ISF volume At supranormal capillary pressure, when the COP difference is maximal, Jv is proportional to transendothelial pressure difference Infused colloid solution is distributed through the plasma volume, and infused ISS through the extracellular volume Infused colloid solution is initially distributed through the plasma volume, and infused ISS through the intravascular volume At supranormal capillary pressure, infusion of colloid solution preserves plasma COP, raises capillary pressure, and increases Jv At supranormal capillary pressure, infusion of ISS also raises capillary pressure, but it lowers COP and so increases Jv more than the same colloid solution volume At subnormal capillary pressure, infusion of colloid solution increases plasma volume and infusion of ISS increases intravascular volume, but Jv remains close to zero in both cases Reduced COP Normal COP Raised COP 100 75 50 25 0 –25 10 20 mm Hg 30 40 Fig 3 Autotransfusion. Transendothelial filtration rate Jv is pro-portional to the capillary pressure, or transendothelial pressure difference if interstitial pressure is not constant. Normal capillary pressure is nominally 20 mm Hg, and the scale for Jv is arbitrary, although studies show the rate of clearance from the intravascu-lar space during rapid infusion of Ringer’s acetate in humans can be as much as 100 ml min21. Raising the plasma COP slows filtration (pink line), while reducing plasma COP increases it (green line).
100 75 50 25 0 –25 Reduced COP Normal COP Raised COP 10 20 mm Hg 30 40 Fig 4 The no absorption rule. With less acute reduction in capil-lary pressure, the glycocalyx model preserves filtration at a very low rate without a phase of absorption, the no absorption rule.
We call the inflection on the filtration curve the J point.
BJA Woodcock and Woodcock 390 by John Vogel on March 5, 2012 Downloaded from Porosity increases in inflammatory states, but pis has no direct effect on Jv. A 10- to 20-fold increase in Jv in the acute inflammatory response is actively regulated by integ-rins acting upon collagen fibrils in the extracellular matrix, exposing GAGs to take up water, and does not necessarily imply increased capillary porosity. The effects of fluid therap-ies on this mechanism, if any, are unknown. Changes which compact the EGL releasing GAGs into the circulating plasma are associated with increased transendothelial protein flux, but compaction of the EGL and increased poros-ity may be separate processes and the association may not be entirely causal.
Although transfused macromolecules do not easily per-meate an intact EGL, they pass easily into the ISF of the si-nusoidal capillaries in the bone marrow, spleen, and liver, equilibrating with interstitial macromolecules and returning to the venous system via lymphatics. An increase in the pro-portion of the cardiac output going to sinusoidal tissues will increase Jv and the transcapillary escape rate of albumin.
There is no significant absorption of ISF to the plasma under a COP difference, so colloid therapy does not prevent or improve tissue oedema.
In conclusion, fluid resuscitation studies require us to ‘re-appraise the basics’.118 The revised Starling equation and gly-cocalyx model paradigm appear to be an improvement on the original Starling principle paradigm. Colloids are widely prescribed for resuscitation from hypovolaemia, despite evidence-based protocols and guidelines.119 120 An import-ant feature of the revised Starling equation and glycocalyx model paradigm is that it explains why albumin or plasma substitutes have no advantage over ISS when Pc or transen-dothelial pressure difference is low. The finding that pis has little effect on Jv focuses our attention on the subglycocalyx space. The EGL is a fragile structure and is disrupted by rapid i.v. infusion of fluids, acute hyperglycaemia, surgery, and sepsis. The glycocalyx model describes how psg, pis, and Jv balance one another, and raises concerns about disease pro-cesses or plasma substitute therapies that might disturb the protected low psg. In the absence of absorption by capillaries and venules, filtered fluid returns to the circulation mostly by lymphatics, and the importance of preserving lymphatic flow is highlighted. The new paradigm provides an explanation of context sensitivity of colloid and crystalloid volume kinetics in awake, anaesthetized, or hypotensive patients, and the ra-tional prescriber will consider the desired effect on Pc and transendothelial pressure difference. Endothelial dysfunction associated with increased capillary porosity increases Jv at any Pc, and lowers the Pc at which Jv approaches zero. This J point can be taken into account when faced with a patient with systemic inflammation or sepsis. It is likely that the revised Starling equation and glycocalyx model paradigm will be modified and refined in the light of physi-ology and clinical trial evidence. In its current form, it strengthens the arguments for preferring ISSs over plasma or plasma substitutes for resuscitation, but accepts a rational use of colloids for euvolaemic or hypervolaemic haemodilu-tion. The use of plasma or plasma substitutes to achieve a sustained supranormal plasma volume or to reduce tissue oedema is not rational.
Declaration of interest None declared.
References 1 Twigley AJ, Hillman KM. The end of the crystalloid era? A new ap-proach to peri-operative fluid administration. Anaesthesia 1985; 40: 860–71 2 Grocott MP, Mythen MG, Gan TJ. Perioperative fluid management and clinical outcomes in adults. Anesth Analg 2005; 100: 1093–106 3 Boldt J. The balanced concept of fluid resuscitation. Br J Anaesth 2007; 99: 312–5 4 Chappell D, Jacob M, Hofmann-Kiefer K, Conzen P, Rehm M. A ra-tional approach to perioperative fluid management. Anesthesi-ology 2008; 109: 723–40 5 Finfer S, Bellomo R, Boyce N, French J, Myburgh J, Norton R. A comparison of albumin and saline for fluid resuscitation in the intensive care unit. N Engl J Med 2004; 350: 2247–56 6 Brunkhorst FM, Engel C, Bloos F, et al. Intensive insulin therapy and pentastarch resuscitation in severe sepsis. N Engl J Med 2008; 358: 125–39 7 James MF, Michell WL, Joubert IA, Nicol AJ, Navsaria PH, Gillespie RS. Resuscitation with hydroxyethyl starch improves renal function and lactate clearance in penetrating trauma in a randomized controlled study: the FIRST trial (Fluids in Resuscitation of Severe Trauma). Br J Anaesth 2011; 107: 693–702 8 Maitland K, Kiguli S, Opoka RO, et al. Mortality after fluid bolus in African children with severe infection. N Engl J Med 2011; 364: 2483–95 9 Hahn RG. Volume kinetics for infusion fluids. Anesthesiology 2010; 113: 470–81 10 Hahn RG. Fluid therapy might be more difficult than you think.
Anesth Analg 2007; 105: 304–5 11 Starling EH. On the absorption of fluids from the connective tissue spaces. J Physiol 1896; 19: 312–26 12 Krooh A, Landis EM, Turner A. The movement of fluid through the human capillary wall in relation to venous pressure and to the colloid osmotic pressure in the blood. Clin Invest 1932; 11: 63 13 Staverman AJ. The theory of measurement of osmotic pressure.
Rec Trav Chim 1951; 70: 344–52 14 Pappenheimer JR, Renkin EM, Borrerro LM. Filtration, diffusion and molecular sieving through peripheral capillary membranes.
A contribution to the pore theory of capillary permeability. Am J Physiol 1951; 167: 13–46 15 Guyton AC, Lindsey AW. Effect of elevated left atrial pressure and decreased plasma protein concentration on the develop-ment of pulmonary edema. Circ Res 1959; 7: 649 16 Luft JH. Fine structure of capillary and endocapillary layer as revealed by ruthenium red. Fedn Proc 1966; 25: 1773–83 17 Michel CC. Filtration coefficients and osmotic reflexion coeffi-cients of the walls of single frog mesenteric capillaries.
J Physiol 1980; 309: 341–55 18 Curry FE, Michel CC. A fibre-matrix model of capillary permeabil-ity. Microvasc Res 1980; 20: 96–9 Transvascular exchange and fluid therapy BJA 391 by John Vogel on March 5, 2012 Downloaded from 19 Adamson RH, Lenz JF, Zhang X, Adamson GN, Weinbaum S, Curry FE. Oncotic pressures opposing filtration across non-fenestrated rat microvessels. J Physiol 2004; 557: 889–907 20 Levick JR. Revision of the Starling principle: new views of tissue fluid balance. J Physiol 2004; 557: 704 21 Levick JR, Michel CC. Microvascular fluid exchange and the revised Starling principle. Cardiovasc Res 2010; 87: 198–210 22 Curry FR.
Microvascular solute and water transport.
Microcirculation 2005; 12: 17–31 23 Ellman H. Capillary permeability in septic patients. Crit Care Med 1984; 12: 629–33 24 Hu G, Minshall RD. Regulation of transendothelial permeability by Src kinase. Microvasc Res 2009; 77: 21–5 25 Lee WL, Slutsky AS. Sepsis and endothelial permeability. N Engl J Med 2010; 363: 689–91 26 Clough G. Relationship between microvascular permeability and ultrastructure. Prog Biophys Mol Biol 1991; 55: 47–69 27 Vink H, Duling BR. Identification of distinct luminal domains for macromolecules, erythrocytes, and leukocytes within mamma-lian capillaries. Circ Res 1996; 79: 581–9 28 Reitsma S, Slaaf DW, Vink H, van Zandvoort MA, oude Egbrink MG. The endothelial glycocalyx: composition, functions, and visualization. Pflugers Arch 2007; 454: 345–59 29 Chappell D, Jacob M, Paul O, et al. The glycocalyx of the human umbilical vein endothelial cell: an impressive structure ex vivo but not in culture. Circ Res 2009; 104: 1313–7 30 Rehm M, Haller M, Orth V, et al. Changes in blood volume and hematocrit during acute preoperative volume loading with 5% albumin or 6% hetastarch solutions in patients before radical hysterectomy. Anesthesiology 2001; 95: 849–56 31 Pries AR, Kuebler WM.
Normal endothelium.
Handb Exp Pharmacol 2006; (176 Pt 1): 1–40 32 Vink H, Duling BR. Capillary endothelial surface layer selectively reduces plasma solute distribution volume. Am J Physiol Heart Circ Physiol 2000; 278: H285–9 33 Henry CB, Duling BR. Permeation of the luminal capillary glycocalyx is determined by hyaluronan. Am J Physiol 1999; 277: H508–14 34 Henry CB, Duling BR. TNF-alpha increases entry of macromole-cules into luminal endothelial cell glycocalyx. Am J Physiol Heart Circ Physiol 2000; 279: H2815–23 35 Nieuwdorp M, van Haeften TW, Gouverneur MC, et al. Loss of endothelial glycocalyx during acute hyperglycemia coincides with endothelial dysfunction and coagulation activation in vivo. Diabetes 2006; 55: 480–6 36 Nieuwdorp M, Meuwese MC, Mooij HL, et al. Measuring endothe-lial glycocalyx dimensions in humans: a potential novel tool to monitor vascular vulnerability. J Appl Physiol 2008; 104: 845–52 37 Gao L, Lipowsky HH. Composition of the endothelial glycocalyx and its relation to its thickness and diffusion of small solutes.
Microvasc Res 2010; 80: 394–401 38 Rehm M, Bruegger D, Christ F, et al. Shedding of the endothelial glycocalyx in patients undergoing major vascular surgery with global and regional ischemia. Circulation 2007; 116: 1896–906 39 Hofmann-Kiefer KF, Kemming GI, Chappell D, et al. Serum heparan sulfate levels are elevated in endotoxemia. Eur J Med Res 2009; 14: 526–31 40 Nieuwdorp M, Mooij HL, Kroon J, et al. Endothelial glycocalyx damage coincides with microalbuminuria in type 1 diabetes.
Diabetes 2006; 55: 1127–32 41 Karangelis D, Asimakopoulou A, Kanakis I, et al. Monitoring serum chondroitin sulfate levels in patients submitted to coron-ary artery bypass surgery. Biomed Chromatogr 2010; 25: 748–50 42 Berg S, Engman A, Hesselvik JF, Laurent TC. Crystalloid infusion increases plasma hyaluronan. Crit Care Med 1994; 22: 1563–7 43 Berg S, Golster M, Lisander B. Albumin extravasation and tissue washout of hyaluronan after plasma volume expansion with crystalloid or hypooncotic colloid solutions. Acta Anaesthesiol Scand 2002; 46: 166–72 44 Nelson A, Berkestedt I, Schmidtchen A, Ljunggren L, Bodelsson M. Increased levels of glycosaminoglycans during septic shock: relation to mortality and the antibacterial actions of plasma. Shock 2008; 30: 623–7 45 Steppan J, Hofer S, Funke B, et al. Sepsis and major abdominal surgery lead to flaking of the endothelial glycocalix. J Surg Res 2011; 165: 136–41 46 Kozar RA, Peng Z, Zhang R, et al. Plasma restoration of endothe-lial glycocalyx in a rodent model of hemorrhagic shock. Anesth Analg 2011; 112: 1289–95 47 Johansson PI, Stensballe J, Rasmussen LS, Ostrowski SR. A high admission syndecan-1 level, a marker of endothelial glycocalyx degradation, is associated with inflammation, protein C deple-tion, fibrinolysis, and increased mortality in trauma patients.
Ann Surg 2011; 254: 194–200 48 Devaraj S, Yun JM, Adamson G, Galvez J, Jialal I. C-reactive protein impairs the endothelial glycocalyx resulting in endothe-lial dysfunction. Cardiovasc Res 2009; 84: 479–84 49 Platts SH, Duling BR. Adenosine A3 receptor activation modulates the capillary endothelial glycocalyx. Circ Res 2004; 94: 77–82 50 VanTeeffelen JW, Brands J, Vink H. Agonist-induced impairment of glycocalyx exclusion properties: contribution to coronary effects of adenosine. Cardiovasc Res 2010; 87: 311–9 51 VanTeeffelen JW, Constantinescu AA, Brands J, Spaan JA, Vink H.
Bradykinin- and sodium nitroprusside-induced increases in ca-pillary tube haematocrit in mouse cremaster muscle are asso-ciated with impaired glycocalyx barrier properties. J Physiol 2008; 586: 3207–18 52 Annecke T, Fischer J, Hartmann H, et al. Shedding of the coron-ary endothelial glycocalyx: effects of hypoxia/reoxygenation vs ischaemia/reperfusion. Br J Anaesth 2011; 107: 679–86 53 Chappell D, Hofmann-Kiefer K, Jacob M, et al. TNF-alpha induced shedding of the endothelial glycocalyx is prevented by hydrocor-tisone and antithrombin. Basic Res Cardiol 2009; 104: 78–89 54 Chappell D, Jacob M, Hofmann-Kiefer K, et al. Hydrocortisone preserves the vascular barrier by protecting the endothelial glycocalyx. Anesthesiology 2007; 107: 776–84 55 Chappell D, Jacob M, Hofmann-Kiefer K, et al. Antithrombin reduces shedding of the endothelial glycocalyx following ischaemia/reperfusion. Cardiovasc Res 2009; 83: 388–96 56 Chappell D, Dorfler N, Jacob M, et al. Glycocalyx protection reduces leukocyte adhesion after ischemia/reperfusion. Shock 2010; 34: 133–9 57 Annecke T, Chappell D, Chen C, et al. Sevoflurane preserves the endothelial glycocalyx against ischaemia–reperfusion injury.
Br J Anaesth 2010; 104: 414–21 58 Chappell D, Heindl B, Jacob M, et al. Sevoflurane reduces leuko-cyte and platelet adhesion after ischemia–reperfusion by pro-tecting the endothelial glycocalyx. Anesthesiology 2011; 115: 483–91 59 Sarin H. Physiologic upper limits of pore size of different blood capillary types and another perspective on the dual pore theory of microvascular permeability. J Angiogenes Res 2010; 2: 14 60 Oratz M, Rothschild MA, Schreiber SS. Effect of dextran infusions on protein synthesis by hepatic microsomes. Am J Physiol 1970; 218: 1108–12 BJA Woodcock and Woodcock 392 by John Vogel on March 5, 2012 Downloaded from 61 Nicholson JP, Wolmarans MR, Park GR. The role of albumin in critical illness. Br J Anaesth 2000; 85: 599–610 62 Rank N, Michel C, Haertel C, et al. N-acetylcysteine increases liver blood flow and improves liver function in septic shock patients: results of a prospective, randomized, double-blind study. Crit Care Med 2000; 28: 3799–807 63 Allison KP, Gosling P, Jones S, Pallister I, Porter KM. Randomized trial of hydroxyethyl starch versus gelatine for trauma resuscita-tion. J Trauma 1999; 47: 1114–21 64 Thomas MC. Pathogenesis and progression of proteinuria.
Contrib Nephrol 2011; 170: 48–56 65 Singh A, Friden V, Dasgupta I, et al. High glucose causes dys-function of the human glomerular endothelial glycocalyx. Am J Physiol Renal Physiol 2011; 300: F40–8 66 Clough G, Michel CC, Phillips ME. Inflammatory changes in per-meability and ultrastructure of single vessels in the frog mesen-teric microcirculation. J Physiol 1988; 395: 99–114 67 Simon DA, Taylor TL, Bayley G, Lalonde KA. Four-limb compart-ment syndrome associated with the systemic capillary leak syn-drome. J Bone Joint Surg Br 2010; 92: 1700–2 68 Levick JR. Fluid exchange across endothelium. Int J Microcirc Clin Exp 1997; 17: 241–7 69 Heino J, Kapyla J. Cellular receptors of extracellular matrix mole-cules. Curr Pharm Des 2009; 15: 1309–17 70 Sorokin L. The impact of the extracellular matrix on inflamma-tion. Nat Rev Immunol 2010; 10: 712–23 71 Li H, Su X, Yan X, et al. Toll-like receptor 4-myeloid differentiation factor 88 signaling contributes to ventilator-induced lung injury in mice. Anesthesiology 2010; 113: 619–29 72 Svendsen OS, Barczyk MM, Popova SN, Liden A, Gullberg D, Wiig H. The alpha11beta1 integrin has a mechanistic role in control of interstitial fluid pressure and edema formation in inflammation.
Arterioscler Thromb Vasc Biol 2009; 29: 1864–70 73 Reed RK, Rubin K. Transcapillary exchange: role and importance of the interstitial fluid pressure and the extracellular matrix. Car-diovasc Res 2010; 87: 211–7 74 Jacob M, Bruegger D, Rehm M, et al. The endothelial glycocalyx affords compatibility of Starling’s principle and high cardiac interstitial albumin levels. Cardiovasc Res 2007; 73: 575–86 75 Fleck A, Raines G, Hawker F, et al. Increased vascular permeabil-ity: a major cause of hypoalbuminaemia in disease and injury.
Lancet 1985; 1: 781–4 76 Suding P, Jensen E, Abramson MA, Itani K, Wilson SE. Definitive risk factors for anastomotic leaks in elective open colorectal re-section. Arch Surg 2008; 143: 907–11; discussion 911–2 77 Lee EH, Chin JH, Choi DK, et al. Postoperative hypoalbuminemia is associated with outcome in patients undergoing off-pump coronary artery bypass graft surgery. J Cardiothorac Vasc Anesth 2011; 25: 462–8 78 Hennessey DB, Burke JP, Ni-Dhonochu T, Shields C, Winter DC, Mealy K. Preoperative hypoalbuminemia is an independent risk factor for the development of surgical site infection following gastrointestinal surgery: a multi-institutional study. Ann Surg 2010; 252: 325–9 79 Yuan XY, Zhang CH, He YL, et al. Is albumin administration bene-ficial in early stage of postoperative hypoalbuminemia following gastrointestinal surgery?: a prospective randomized controlled trial. Am J Surg 2008; 196: 751–5 80 Boldt J. Use of albumin: an update. Br J Anaesth 2010; 104: 276–84 81 Aman J, van der Heijden M, van Lingen A, et al. Plasma protein levels are markers of pulmonary vascular permeability and degree of lung injury in critically ill patients with or at risk for acute lung injury/acute respiratory distress syndrome. Crit Care Med 2011; 39: 89–97 82 Kuper M, Gunning MP, Halder S, Soni N. The short-term effect of hyperoncotic albumin, given alone or with furosemide, on oxy-genation in sepsis-induced acute respiratory distress syndrome.
Anaesthesia 2007; 62: 259–63 83 Martin GS, Mangialardi RJ, Wheeler AP, Dupont WD, Morris JA, Bernard GR. Albumin and furosemide therapy in hypoproteine-mic patients with acute lung injury. Crit Care Med 2002; 30: 2175–82 84 Martin GS, Moss M, Wheeler AP, Mealer M, Morris JA, Bernard GR.
A randomized, controlled trial of furosemide with or without albumin in hypoproteinemic patients with acute lung injury.
Crit Care Med 2005; 33: 1681–7 85 Trof RJ, Sukul SP, Twisk JW, Girbes AR, Groeneveld AB. Greater cardiac response of colloid than saline fluid loading in septic and non-septic critically ill patients with clinical hypovolaemia.
Intensive Care Med 2010; 36: 697–701 86 van der Heijden M, Verheij J, van Nieuw Amerongen GP, Groeneveld AB. Crystalloid or colloid fluid loading and pulmon-ary permeability, edema, and injury in septic and nonseptic crit-ically ill patients with hypovolemia. Crit Care Med 2009; 37: 1275–81 87 Margarson MP, Soni NC. Changes in serum albumin concentra-tion and volume expanding effects following a bolus of albumin 20% in septic patients. Br J Anaesth 2004; 92: 821–6 88 James MF, Latoo MY, Mythen MG, et al. Plasma volume changes associated with two hydroxyethyl starch colloids following acute hypovolaemia in volunteers. Anaesthesia 2004; 59: 738–42 89 Lobo DN, Stanga Z, Aloysius MM, et al. Effect of volume loading with 1 liter intravenous infusions of 0.9% saline, 4% succinylated gelatine (Gelofusine) and 6% hydroxyethyl starch (Voluven) on blood volume and endocrine responses: a randomized, three-way crossover study in healthy volunteers. Crit Care Med 2010; 38: 464–70 90 Mythen MG, Webb AR. Perioperative plasma volume expansion reduces the incidence of gut mucosal hypoperfusion during cardiac surgery. Arch Surg 1995; 130: 423–9 91 Sinclair S, James S, Singer M. Intraoperative intravascular volume optimisation and length of hospital stay after repair of proximal femoral fracture: randomised controlled trial. Br Med J 1997; 315: 909–12 92 Gan TJ, Soppitt A, Maroof M, et al. Goal-directed intraoperative fluid administration reduces length of hospital stay after major surgery. Anesthesiology 2002; 97: 820–6 93 Donati A, Loggi S, Preiser JC, et al. Goal-directed intraoperative therapy reduces morbidity and length of hospital stay in high-risk surgical patients. Chest 2007; 132: 1817–24 94 Jammer I, Ulvik A, Erichsen C, Lodemel O, Ostgaard G. Does central venous oxygen saturation-directed fluid therapy affect postoperative morbidity after colorectal surgery? A randomized assessor-blinded controlled trial. Anesthesiology 2010; 113: 1072–80 95 Davies SJ, Yates D, Wilson RJ. Dopexamine has no additional benefit in high-risk patients receiving goal-directed fluid therapy undergoing major abdominal surgery. Anesth Analg 2011; 112: 130–8 96 Senagore AJ, Emery T, Luchtefeld M, Kim D, Dujovny N, Hoedema R. Fluid management for laparoscopic colectomy: a prospective, randomized assessment of goal-directed Transvascular exchange and fluid therapy BJA 393 by John Vogel on March 5, 2012 Downloaded from administration of balanced salt solution or hetastarch coupled with an enhanced recovery program. Dis Colon Rectum 2009; 52: 1935–40 97 Morris C, Rogerson D. What is the optimal type of fluid to be used for peri-operative fluid optimisation directed by oesophageal Doppler monitoring?
Anaesthesia 2011; 66: 819–27 98 Lee EH, Kim SK, Yeo YG, Choi KT. Effects of anesthesia on fluid volume kinetics after infusion of colloid solution during blood donation. Korean J Anesthesiol 2010; 58: 514–20 99 Drobin D, Hahn RG. Volume kinetics of Ringer’s solution in hypo-volemic volunteers. Anesthesiology 1999; 90: 81–91 100 Dyer RA, Farina Z, Joubert IA, et al. Crystalloid preload versus rapid crystalloid administration after induction of spinal anaes-thesia (coload) for elective caesarean section. Anaesth Intensive Care 2004; 32: 351–7 101 Li Y, Zhu S, Hahn RG. The kinetics of Ringer’s solution in young and elderly patients during induction of general anesthesia with propofol and epidural anesthesia with ropivacaine. Acta Anaesthesiol Scand 2007; 51: 880–7 102 Quinlan GJ, Mumby S, Martin GS, Bernard GR, Gutteridge JM, Evans TW. Albumin influences total plasma antioxidant capacity favorably in patients with acute lung injury. Crit Care Med 2004; 32: 755–9 103 Lang JDJ, Figueroa M, Chumley P, et al. Albumin and hydro-xyethyl starch modulate oxidative inflammatory injury to vascu-lar endothelium. Anesthesiology 2004; 100: 51–8 104 Jacob M, Bruegger D, Rehm M, Welsch U, Conzen P, Becker BF.
Contrasting effects of colloid and crystalloid resuscitation fluids on cardiac vascular permeability. Anesthesiology 2006; 104: 1223–31 105 Balkamou X, Xanthos T, Stroumpoulis K, et al. Hydroxyethyl starch 6% (130/0.4) ameliorates acute lung injury in swine hem-orrhagic shock. Anesthesiology 2010; 113: 1092–8 106 Wang Z, Herzog C, Kaushal GP, Gokden N, Mayeux PR. Actinonin, a meprin a inhibitor, protects the renal microcirculation during sepsis. Shock 2011; 35: 141–7 107 Niemi TT, Miyashita R, Yamakage M. Colloid solutions: a clinical update. J Anesth 2010; 24: 913–25 108 Sibbald WJ, Driedger AA, Wells GA, Myers ML, Lefcoe M. The short-term effects of increasing plasma colloid osmotic pressure in patients with noncardiac pulmonary edema. Surgery 1983; 93: 620–33 109 Blunt MC, Nicholson JP, Park GR. Serum albumin and colloid osmotic pressure in survivors and nonsurvivors of prolonged critical illness. Anaesthesia 1998; 53: 755–61 110 Veneman TF, Oude Nijhuis J, Woittiez AJ. Human albumin and starch administration in critically ill patients: a prospective randomized clinical trial. Wien Klin Wochenschr 2004; 116: 305–9 111 Verheij J, van Lingen A, Raijmakers PG, et al. Effect of fluid loading with saline or colloids on pulmonary permeability, oedema and lung injury score after cardiac and major vascular surgery. Br J Anaesth 2006; 96: 21–30 112 Svensen C, Drobin D, Olsson J, Hahn RG. Stability of the intersti-tial matrix after crystalloid fluid loading studied by volume kinetic analysis. Br J Anaesth 1999; 82: 496–502 113 Borup T, Hahn RG, Holte K, Ravn L, Kehlet H. Intra-operative colloid administration increases the clearance of a post-operative fluid load. Acta Anaesthesiol Scand 2009; 53: 311–7 114 Prisk GK, Olfert IM, Arai TJ, Wagner PD, Hopkins SR. Rapid intra-venous infusion of 20 ml/kg saline does not impair resting pul-monary gas exchange in the healthy human lung. J Appl Physiol 2010; 108: 53–9 115 West JB, Mathieu-Costello O. Stress failure of pulmonary capil-laries: role in lung and heart disease. Lancet 1992; 340: 762–7 116 Varadhan KK, Lobo DN. A meta-analysis of randomised con-trolled trials of intravenous fluid therapy in major elective open abdominal surgery: getting the balance right. Proc Nutr Soc 2010; 69: 488–98 117 Jacob M, Conzen P, Finsterer U, Krafft A, Becker BF, Rehm M.
Technical and physiological background of plasma volume measurement with indocyanine green: a clarification of misun-derstandings. J Appl Physiol 2007; 102: 1235–42 118 Myburgh JA. Fluid resuscitation in acute illness—time to re-appraise the basics. N Engl J Med 2011; 364: 2543–4 119 Finfer S, Liu B, Taylor C, et al. Resuscitation fluid use in critically ill adults: an international cross-sectional study in 391 intensive care units. Crit Care 2010; 14: R185 120 Cade JA, Truesdale M. Preferences of critical care registrars in fluid resuscitation of major trauma patients: concordance with current guidelines. Anaesth Intensive Care 2011; 39: 262–7 BJA Woodcock and Woodcock 394 by John Vogel on March 5, 2012 Downloaded from CORRESPONDENCE Fluid physiology, tissue compliance, and colloids Editor—I would like to thank the authors for their immensely informative article on transvascular fluid exchange.1 It is ap-parent that advances in this area of science are now starting to explain many of the clinically observed phenomena in crit-ically ill patients, such as the broad equivalence of crystal-loids and colloids, and the failure of albumin infusions to improve tissue oedema.
A phenomenon familiar to many of us is the turning point when a sick patient in intensive care exhibits spontaneous di-uresis, thereby excreting their iatrogenic excess body water— at this stage, we ‘know’ that the patient will get better. Do the authors think that this occurrence represents the junc-ture at which tissue compliance returns to normal, and the increased hydrostatic pressure drives fluid back into the cir-culation? Certainly, the components of the inflammatory re-sponse that control these changes hold great promise as targets for future therapies.
Despite the admirable explanation of the revised Starling equation in this article, I am not sure I agree with the conclu-sion that a rational use of colloids should be restricted to euvolaemic or hypervolaemic haemodilution. This statement is based on the premise that patients who are profoundly hypovolaemic will have an equivalent plasma expansion with either class of fluid, that is, they are below the J-point on the graph provided (Fig. 4 in the original article). Above the J-point, colloids do stay in the circulation for longer, and I would argue that this applies to fluid optimization in the perioperative period, where patients are only moderately hypovolaemic. Thus, over several days of flow-guided opti-mization, the total volume of colloids used will be less com-pared with crystalloids for the same effect. However, the long debated question remains as to whether the advantage of less tissue oedema outweighs the other adverse effects (and cost) associated with colloids.
Declaration of interest None declared.
A. Toner London, UK E-mail: toner@doctors.org.uk 1 Woodcock TE, Woodcock TM. Revised Starling equation and the glycocalyx model of transvascular fluid exchange: an improved paradigm for prescribing intravenous fluid therapy. Br J Anaesth 2012; 108: 384–94 doi:10.1093/bja/aes150 Reply from the authors Editor—Dr Toner raises some interesting points regarding our article on the revised Starling equation and glycocalyx model paradigm (RSE&GM) which we are pleased to respond to.1 We recognize the turning point Dr Toner describes when a reco-vering patient sheds litres of oedema, and it seems to be a sign of recovery rather than the cause of it. We could not do more than speculate on the mechanism, but normaliza-tion of interstitial collagen fibre conformation and reversal of GAG hydration must be involved. The relevance of RSE&GM to the practice of perioperative goal-directed fluid therapy is immense. First, we need to consider where the patient is on the J-curve. We suspect that general anaesthe-sia tends to lower venous resistance more than arterial re-sistance, lowering both arterial and capillary pressure and so moving the J-point to the left. We know that the endothe-lial glycocalyx layer is reduced or compacted during anaes-thesia and surgery, and it therefore seems prudent to prefer a colloid-free isotonic salt solution for stroke volume optimization until the transendothelial pressure difference has been restored to the J-point. Dr Toner correctly observes that we would only choose to include colloid molecules in our isotonic salt solution if we intend to take the patient substan-tially above the J-point and retard Jv, but if stroke volume op-timization can be achieved around or below the J-point, a colloid-free isotonic salt solution will suffice. On Dr Toner’s concern about tissue oedema, many anaesthetists use phenylephrine to preserve the mean arterial pressure after what they consider to be adequate fluid loading. The a-adrenoceptor agonist opposes anaesthetic-induced arteri-olar relaxation, thereby keeping capillary pressure and Jv low, and is an effective anti-oedema strategy. As the patient recovers from circulatory side-effects of anaesthesia, phenyl-ephrine is weaned off. With judicious vasopressor and diuret-ic therapy, it is our practice to take care not to allow patients to become fluid-overloaded in the days after surgery, whether or not the anaesthetist or surgeon chooses to use an intraoperative colloid solution. Dr Toner’s assertion that the total volume of colloid used for optimization over several days is less compared with crystalloids is worth quan-tifying in his practice. It is widely asserted that colloid boluses are essential for goal-directed intra-operative fluid therapy, in spite of published evidence to the contrary.
Future clinical studies need to include a group in which the haemodynamic goal is targeted by variable infusion of a colloid-free isotonic salt solution, and in which we allow ex-cessive anaesthetic-induced vasodilation to be minimized by a judicious dose of an a-adrenoceptor agonist to protect capillary pressure. We should be concerned that there is very little published safety information on the effects of starches or gelatines on the endothelial glycocalyx layer British Journal of Anaesthesia 108 (6): 1035–44 (2012) & The Author . Published by Oxford University Press on behalf of the British Journal of Anaesthesia. All rights reserved.
For Permissions, please email: journals.permissions@oup.com by guest on July 20, 2013 Downloaded from and should be investigating this in the laboratory before im-posing synthetic colloid therapy protocols on patients.
Declaration of interest None declared.
T. E. Woodcock T. M. Woodcock Southampton, UK E-mail: tom.woodcock@me.com 1 Woodcock TE, Woodcock TM. Revised Starling equation and the glycocalyx model of transvascular fluid exchange: an improved paradigm for prescribing intravenous fluid therapy. Br J Anaesth 2012; 108: 384–94 doi:10.1093/bja/aes151 Intraoperative goal-directed fluid therapy in aerobically fit and unfit patients having major colorectal surgery Letter 1 Editor—I read with interest the article on using the oesopha-geal Doppler monitor (ODM) for goal-directed fluid therapy during surgery for patients undergoing major colorectal re-section.1 The study clearly adds important new information to the knowledge base in this area and raises concern that using fluid boluses to maximise stroke volume during this surgery in fit patients may lead to adverse effects. As a trainee, I have worked with various anaesthetists who have used ODM and other less invasive cardiac output monitors to guide fluid management during surgery. In some situa-tions, particularly during open surgery, using the information from the ODM has led to an increase in fluid administered, while in others, particularly in laparoscopic surgery, the oppos-ite has been true. Thus, in some cases, cardiac output moni-toring has ‘directed’ increased fluid administration, and in others, it has ‘directed’ fluid restriction. From the design of your study, it appears inevitable that the goal-directed group would receive more fluid than the control group. The ODM data were only available to the investigator, who could, according to the algorithm, give boluses of colloids, until no further increase in stroke volume was recorded. At this point, they would give no further fluid. The anaesthetist without this information would continue to give maintenance and other fluid they felt necessary. The ODM algorithm did not allow the investigator to do anything except give extra fluid, and the investigator was not able to stop further fluid being given by the anaesthetist where it was not indicated in the al-gorithm. It was therefore inevitable that the goal-directed fluid therapy (GDT) group received more fluid than control.
The study design appears to lead to more fluid being given than might have been if the ODM data were visible to the an-aesthetist. A subgroup of patients may thus have had more fluid given than they would if a single person in possession of the ODM data (the normal theatre situation) had been re-sponsible for their fluid management.
It appears that the same fluid maintenance rate (10 ml kg21 h21 Hartmann’s solution) was recommended in patients undergoing laparoscopic and open surgery. It would be interesting to know whether the extra volume of colloids administered to the GDT patients within these groups differed, and also whether there was any outcome difference when comparing laparoscopic with open surgery.
Declaration of interest None declared.
J. Rivers Bath, UK E-mail: jonrivers29@hotmail.com 1 Challand C, Struthers R, Sneyd JR, et al. Randomized controlled trial of intraoperative goal-directed fluid therapy in aerobically fit and unfit patients having major colorectal surgery. Br J Anaesth 2012; 108: 53–62 doi:10.1093/bja/aes152 Letter 2 Editor—This study1 provides a surprising result demonstrat-ing a decreased time to fitness for discharge in patients with an anaerobic threshold (AT) of .11 ml O2 kg21 min21 who received goal-directed fluid therapy (GDT) compared with those who did not. This is in contrast to the large body of available data supporting the use of GDT in open surgery and the NICE technology appraisal. In view of this striking difference, we need to consider factors contributing to this result. It is interesting to note that the GDT group had almost three times the rate of admission to critical care in a relatively small total patient number; does this reflect a cause in delay for discharge or consequence of ex-cessive fluids? It is hard to see from the data presented exactly what led to the delayed readiness for discharge as tolerance of diet, bowel movement, flatus passed, renal com-plications, and postoperative complications did not show any significant difference between the two groups. It is interesting that the authors made no distinction between the use of ODM-guided GDT in the laparoscopic and open groups. The studies demonstrating the benefit of ODM GDT derive from studies focusing primarily on open surgery. Indeed, it is hard to find a trial demonstrating a proven benefit of using ODM-guided fluids in patients undergoing laparoscopic colorectal resection. It is possible to find studies using the ODM to dem-onstrate changes in cardiac index and systemic vascular re-sistance as a result of the Trendelenburg position and pneumoperitoneum associated with laparoscopic colorectal surgery.2 Could the variability of these factors during a proced-ure have their own effects on stroke volume (SV) making the following Deltex ODM SV-based algorithm difficult? Could it be the case that the extrapolation of this technique to laparo-scopic procedures without it being directly proven in this group BJA Correspondence 1036 by guest on July 20, 2013 Downloaded from Glycocalyx ~ 1 L of non flowing fluid in the glycocalyx. The glycocalyx contains water in a gel phase Capillary pressure just exceed COP down the entire length of the capillary, hence (no reabsorption rule) fluid is reabsorbed back into the capillaries and it goes to the lymphatics. There are gaps in the glycocalyx and 60% of Albumin is located in the interstitial space 40% in the plasma (i.e., if you give ex. 20% albumin, 60^ will leak out).. A jet of water passing through the small pores prevents albumin from diffusing back. So the COP gradient is between the plasma and the red. Protein free endothelial clefts. Hence, this limits (breaks) filtration to just a few mms per minute. A plasma volume leaves the circulation ~every 9 hours. When hydrostatic pressure falls (ex. hypovolemia) the COP exceeds the hydrostatic pressure and fluid is reabsorbed but this is transient as protein diffuses back from the interstitial to the sub-glycocalyx space thus abolishing the oncotic pressure gradient and stopping reabsorption. It allows ~ 500 ml of fluid to be reabsorbed before it ceases. The capillary pressure is usually low and the filtration rate of low. If the capillary pressure increases, the jet of water removes the little bit of protein in the sub-glycocalyx space until there is none hence the COP break is at its maximum and the filtration will now increase dramatically (this is the “J” point. Capillary pressures are fairly close to the J point. CLINICAL APPLICATIONS Hypervolaemia If we infuse crystalloid, it will raise capillary pressure and reduce COP . As it normally is close to the J point, it will increase filtration. Hypovolaemia Vasoconstriction drops the capillary pressure and filtration ceases. As you infuse vessels the capillary pressure will increase until the J point is exceeded. Therefore the fluid infused until you are euvolaemic will stay in the vessels. Therefore the idea that crystalloid will only stay in the vessels at a ratio of ~1:4 is untrue at least in this scenario, This is even true for 5% glucose. If you give too much fluid and pass the J point, the increase in filtration overwhelms the lymphatic system and oedema results. Leaky capillaries Albumin escape rate (in addition to increased filtration) increases from ~ 5% / hr normally to >15% in septic shock. This moves the curve to the left, I.e., for a given capillary pressure more filtration is occurring. Initially (and also after surgery or cancer), there is no oedema because the lymphatics can handle the extra filtration. It is when you “resuscitate” with lots of fluid that eventually makes you oedematous. This means that septic patients are not hypovolaemic. This means that 15% of crystalloid remains in the circulation after 3 hours in a healthy volunteer, 5% remains in a septic patient after 30 min. This means if we give our septic patient a litre of crystalloid, in ~ 1 hr, 950 ml will be in the tissues. Vasopressors, will decrease capillary pressures and as well as raising the Bp, will reduce translocation. Aim for euvolaemia and use vasopressors if you have leaky capillaries. Graphics curve shifts to left if leaky capillaries, i.e., for a given capillary pressure, there is greater filtration. |
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both the buyer and seller are presented with the Information About Brokerage Services by their respective sales agent at the time of the first substantive communication;
the seller executes a Listing Agreement or other written document with the broker that authorizes the broker to act as intermediary and specifies in conspicuous bold or underlined print the conduct that is prohibited under TRELA §1101.651(d); and
the buyer executes a Buyer Representation Agreement or other written document that authorizes the broker to act as intermediary and specifies the conduct that is prohibited under TRELA §1101.651(d) in conspicuous bold or underlined print.
What is considered an advertisement?: Under Rules 535.154 and 535.155 (effective May 15, 2018), an advertisement is defined as “any form of communication by or on behalf of a license holder designed to attract the public to use real estate brokerage services and includes, but is not limited to, all publications, brochures, radio or television broadcasts, all electronic media including email, text messages, social media, the Internet, business stationery, business cards, displays, signs and billboards." An advertisement does not include a communication from a license holder to the license holder's current client.
Are signs permitted which display the word "broker" or "agent?": Yes. Although this is not mandatory, it may still be placed on a sign.
Are the disclosure and statutory information requirements applicable to commercial transactions, new home sales, farm and ranch sales or transactions other than residential sales?: Unless an exception applies, the requirements apply to all proposed real estate transactions. The exceptions to the representation disclosure are in TRELA §1101.558(c).
Are there exceptions when the disclosure notice about agency relationships (IABS) is not required?: Yes. The Information About Brokerage Services (IABS) representations disclosure is not required when:(1) a transaction is for a residential lease less than one year and a sale is not being considered; (2) a meeting is with a party currently known to be represented by another license holder; or(3) the communication is at an open house and the communication concerns that same property.[TRELA §1101.558(c)]
In addition, the IABS is not generally required when the license holder is acting solely as a principal in the transaction.
- Can I advertise a service provider such as an inspector, moving company, or repair contractor on my website? : Yes, but if you offer, recommend, or promote the use of a service provider and expect to receive compensation from the service provider when a party uses the service, the ad must disclose that you may receive the compensation. [Rule 535.155 (effective May 15, 2018)] You may advertise an inspector’s services, however, an inspector may not pay a fee or other valuable consideration for (1) a referral, (2) inclusion on a list of inspectors or preferred providers, or a similar arrangement; or (3) inclusion on a list of inspections contingent on other financial agreements. [Rule 535.220(e)(3)] Also, acceptance of a fee from a service provider may violate the Federal Real Estate Settlement Procedures Act (RESPA), which prohibits certain referral fees and kickbacks.
- Can a broker be the designated broker for more than one business entity?: Yes.
- Can a broker pay all or a portion of a commission or fee to an unlicensed person?: In general, no. However, a license holder may rebate all or a portion of the fee or commission to the party being represented in the transaction, or, with consent of the party being represented, the license holder can also pay all or a portion to a party the license holder does not represent in the transaction. [Rule 535.147(d)]
- Can a broker use the words "team" or "group" in the broker's licensed name or assumed business name?: Yes. TREC Rules do not prohibit a broker from using the word “team” or “group” anywhere in a broker’s licensed or assumed business name.
- Can a license holder who negotiates a transaction also be employed by a lender and direct a purchaser to that lender to get a loan?: Only with appropriate disclosure and consent. If the license holder is an agent of the buyer, the license holder owes a fiduciary duty to the buyer. The license holder can work for a lender without breaching that duty, but the license holder can't direct a purchaser to any one lender. The license holder should inform the purchaser that the license holder is employed by the lender and give the names of several institutions to the purchaser. The primary duty of the agent is to represent the interests of the agent’s client, not the agent's own interests. [Rule 531.1]
- Can a listing broker share a portion of the listing broker’s commission with an attorney who represented the buyer in a real estate transaction?: No. A broker is prohibited from sharing fees with or otherwise compensating the attorney acting as a buyer’s agent unless the attorney holds an active real estate license.
- Can a name used in advertising be both an assumed business name and a team name?: No. It will be one or the other. See question regarding the difference between the types of names to figure out which one is appropriate for your situation.
- Can a sales agent be the owner of a property management company? : A: Yes. A sales agent may own the firm but the business must be conducted through the sales agent's sponsoring broker.
- Can a sales agent have a commission check written to the sales agent’s own corporation or Limited Liability Company? If so, does that entity have to be licensed as a real estate broker?: All commissions must be paid through the agent’s sponsoring broker.[TRELA §1101.651(b)] Further, a business entity that receives compensation on behalf of a license holder must be licensed as a broker. [TRELA §1101.355(c)]
- Can a sales agent have an escrow account?: No. Rule 535.146(b)(2) prohibits a sales agent from having an escrow account. The sales agent must turn all money received over to the sales agent's sponsoring broker.
- Can a sales agent own his or her own real estate business entity?: Yes, a sales agent can own his or her own business entity. The entity must, if it engages in real estate brokerage, hold a separate license. The entity must have a designated broker through whom all transactions must be handled and who is responsible for the entity's (and any sponsored sales agent’s) actions. The specific details of the supervision that the sales agent’s sponsoring broker exercises over the sales agent’s actions should take into consideration the sales agent’s experience and ability, acknowledging the fact that the broker is responsible for the sales agent’s actions, and should be described in a written agreement between the sales agent and the sales agent’s sponsoring broker. The sales agent’s agreement should also address how compensation is handled with the broker. For additional information, read Rule 535.2 regarding broker responsibility. See also Rules 535.154 and 535.155 regarding advertising.
- Can a sales agent receive or pay a commission to a party in a real estate transaction?: No. A sales agent may not accept compensation for a real estate transaction from anyone other than the broker the sales agent was associated with at the time the commission was earned and may not pay a commission to a person except through the sales agent’s sponsoring broker. [TRELA §1101.651(b) and (c)]
- Can a sales agent work in a different office than the sponsoring broker’s office?: A sales agent may work from an office location different from the main office of the sales agent's sponsoring broker, but the sponsoring broker is still responsible for the sales agent's actions. [TRELA §1101.803, Rule 535.2(a)] Despite this flexibility, a sales agent may not lawfully engage in brokerage activity unless the sales agent is associated with, and acting for, the sponsoring broker at all times. [TRELA §1101.351(c)] Thus, a sales agent may not work for a broker who is not the sales agent’s sponsoring broker or work for another broker or out of another broker's office.
- Can an associated broker refer to himself or herself as a "broker" in an advertisement?: No. This could “tend to mislead” the public that the associated broker is in charge of the brokerage operation. The public needs to know the broker with whom they will have a legal agreement (remember listing agreements, etc. are taken in the name of the broker, not the sales agent or the associated broker). The associated broker could use “associated broker”, “broker associate” or “associate broker”, since that distinguishes his or her role. [TRELA §1101.652(b)(23)]
- Can an associated broker use his or her own assumed business name in advertising?: No.A name used in advertising by an associated broker that is not the associated broker’s licensed name must be registered as a team name by the broker he or she is associated with and meet all the requirements for a team name. [TRELA §1101.652(b)(23) and Rule 535.154(a)(5)]
- Can an attorney get a broker license without first being licensed as a sales agent?: No. A licensed attorney will need to meet all the standard requirements, including education, examination and experience, for issuance of first a real estate sales agent license and then a broker license. However, many of the college and law school courses completed by the attorney could count toward the education requirements. Transcripts would need to be evaluated to determine whether the attorney may receive credit for any applicable courses.
- Can an inactive license holder negotiate the purchase, sale, or lease of real property between third parties?: No. An active real estate license is required to negotiate a real estate transaction between third parties. Conductingreal estate brokerage activity with an inactive license is considered a violation subject to sanctions. [TRELA §1101.351(c)]
- Can an individual broker use an assumed business name in advertising?: Yes, as long as the broker has the legal authority to use that name in the State of Texas and it is registered with TREC before it is used in advertisements. [Rule 535.154(d)]
- Can an individual sales agent use an assumed business name in advertising?: No. Any name used by an individual sales agent, other than the name on the license or a registered alternate name, is considered a team name under TREC rules and must meet the team name requirements. [Rule 535.154(a)(5)]
- Can an unlicensed person own a real estate company and receive all or a portion of a commission paid to a licensed broker?: Yes, within certain limitations. The unlicensed person may share in the income earned by a real estate brokerage if the person engages in no acts for which a license is required. [Rule 535.147(b)]
- Can the intermediary delegate to another license holder the authority to appoint license holders associated with the broker intermediary?: The intermediary may delegate to another license holder the authority to appoint license holders. If the intermediary authorizes another license holder to appoint associated license holders to work with the respective principals, that license holder cannot designate himself/herself as one of the appointed license holders. This is an improper combination of the different functions of intermediary and appointed license holder. It is important to remember that there will always be a single intermediary broker even if another license holder has been authorized to make the specific appointments. The intermediary is prohibited from acting so as to favor one principal over the other, and may not reveal confidential information obtained from one principal without the written instructions of that principal, unless disclosure is required by TRELA, court order, or the information materially relates to the condition of the property. The intermediary and any associated license holders appointed by the intermediary broker are prohibited from disclosing, without the written authorization from the seller, that the seller will accept a price less than the asking price or that the buyer will pay a price greater than the price submitted in a written offer. [TRELA § 1101.558]
- Can the team name used by an associated broker be the same as the assumed name used by the associated broker when he or she is not associating with another broker?: Yes, if the assumed business name of the associated broker meets the requirements for a team name since a team name must be used when the associated broker is associating with another broker. Keep in mind that the associated broker registers the assumed business name that is used when they are not associating with another broker and the broker that they are associating with registers the team name being used when associating with that broker. [Rule 535.154(a)(5)]
Note on Exclusive Name Use: TREC does not, and is not, required to vet names submitted for registration as an assumed name or team name for exclusivity. A brokerage should decide whether it makes good sense from a liability exposure standpoint to allow an associated broker (or a sales agent who owns a business entity) to use the same name for a team name under the brokerage that they have registered as an assumed name under their own licensed business entity.
- Do partnerships need to be licensed if real estate brokerage activity is performed under the partnership name?: Yes. All business entities engaged in real estate brokerage activity, including partnerships, need to be licensed. [TRELA §1101.002] Further to receive or maintain a license, a business entity must designate an individual holding an active Texas real estate broker license, in good standing, who is an officer, manager, or general partner of the entity to act for it. [TRELA §1101.355 and Rules 535.50(5) and 535.53]
- Does TREC consider a license holder’s URL or domain name or email address to be advertising?: TREC does not consider URLs or email addresses to be advertisements in and of themselves. However, an advertisement that contains a URL or email address of a sales agent that includes a title that implies responsibility for a brokerage violates TREC Rule 535.155(d)(4). Further, TREC will consider all advertisements in their entirety and if an email address or URL] makes the advertisement misleading or deceptive, it violates the law. [Rule 535.155(b)(4) and TRELA §1101.652(b)(23)]
- Does TREC consider a logo that contains the broker’s full name sufficient for the broker name requirement? : Yes, as long as the size of the broker’s name itself (not the whole logo) is at least ½ the size of the largest contact information. [Rule 535.155(b)(3)]
- Does TREC consider a sign on a building to be an advertisement? : Yes. If a sales agent’s name or team name is on a building sign, the broker’s name must also be present (in at least half the size). A broker’s name alone is okay. [Rule 535.155(b)(1)]
- Does TREC consider promotional items and sponsorships to be advertising? : Yes. If a sales agent’s name or team name is on them, the broker’s name must also be present (in at least half the size). A broker’s name alone is okay. [Rule 535.155(b)(1)]
- Does a broker have to supervise the daily activities of a sales agent? : A broker is not required to directly supervise sponsored sales agents; this responsibility may be delegated to another person with the required level of experience and expertise to provide proper supervision under the law. [Rule 535.2(e)] However, the broker remains responsible for the authorized acts of the broker’s sales agents [TRELA §1101.803, Rule 535.2(a) ] and has many affirmative duties regarding written policies, record keeping and advising, training, approving advertising for and responding to sales agents. Brokers should be familiar with and follow the duties set out in Rule 535.2 . A broker should also be aware that a complaint filed against a sponsored sales agent is also a complaint against the broker for the purpose of determining the broker’s involvement in the alleged violation and whether the broker properly supervised the sales agent. [Rule 535.141(c) and (d)] A business entity can only provide the proper supervision by the personal involvement of the broker or by the broker’s delegation to a qualified supervisor.
- Does the designated broker own 10% or more of a licensed business entity if the designated broker owns 10% or more of another business entity that owns 100% of the licensed business entity?: Yes because the designated broker own 10% or more of the licensed business entity through the broker's ownership of the other entity.
- How can I start a real estate brokerage referral business or Limited Function Referral Office (LFRO)?: The real estate brokerage referral business, commonly called a Limited Function Referral Office (LFRO), must have a business entity brokerage license because the business is assisting in procuring property or clients to effect the sale, exchange, or lease of real estate [TRELA §§1101.001(A)(viii) and (ix)]. Only an active licensed sales agent sponsored by a licensed business entity may make a referral on behalf of the brokerage, and any referral fee must be paid to the sponsoring broker. A key distinction between a limited function referral office and another brokerage is the agreement between the agent and the broker to limit the agent’s actions to generating referrals. We do not provide advice on how to run or set up a referral-based brokerage business (or LFRO).
- How do I change my business physical address?: Log on to your My License Services account.
From the Start Menu page, click on the to change your business physical address. If you prefer, you can also submit the Change of Main Address form by email.
Log on to your My License Services account.
From the Start Menu page, click on the dropdown menu under the "Change your license information and manage relationships" category.
Click "Change/Verify Contact Information" and click "Select".
Read the information on the "Change/Verify Contact Information" introduction page and click “Next”.
Update contact information and “Place of Business” (business physical) address on the Contact Information page and click "Next".
Verify contact information on the Application Summary page and click “Submit”.
On the Statement of Applicant page, select "Yes" to certify the address information is accurate and correct, and click "Next" to process the request.
How do I request inactive status for my sales agent license?: Request inactive status by following the steps below:
Log on to your My License Services account.
From the Start Menu page, click on the dropdown menu under the "Change your license information and manage relationships" category.
Click on the "Manage my Sponsorship (Sales)" and click on the "Select" tab.
Click on the "Next" tab after reading the information on the "Manage my Sponsorship (Sales)" introduction page.
Under Actions for the sponsoring broker, click "Terminate" and then click "Next".
Verify the sponsoring broker information on the Summary page and click "Submit".
On the Attestation page, read the certification, select "Yes", and click "Submit" to process the request.
If you prefer, you can also submit the Application for Inactive Broker or Sales Agent Status form by email.
- How long does a license holder have to keep financial and real estate transactions on file?: When a broker maintains a trust account, documentary records of each deposit or withdrawal for that account must be retained for four years. [Rules 535.146((c)(6) and (e)] TREC requires a broker to maintain for at least four years from the date of a closing or termination of a contract eight specific types of records in a format that can be readily made available to the Commission. [Rule 535.2(h)]
- I am a 25% stockholder in a corporation that holds title to a number of investment properties. I typically negotiate the resale of the properties for the corporation. Must I disclose my licensed status?: Yes. If you participate in a transaction and hold more than a 10% interest in the corporation or other business entity, you are subject to the disclosure requirements. Additionally, you may not use your expertise to the disadvantage of others with whom you deal. [Rule 535.144(b)]
- I am a broker licensed in another state and would like to apply in Texas. Can I apply online?: No. To apply as an out of state broker, you must submit the paper Application for Broker Licenseby an Individual along with requested documents and the applicable fee.
- I am renewing my license for the first time. Am I required to complete the Legal Update I and Legal Update II courses?: Yes. You are under the Sales Apprentice Education (SAE) requirements. Before you can renew your license, you are required to have completed a total of 270 hours of qualifying course hours, and the Legal Update I and II courses.
To determine how many hours have posted to your license record, you may visit the license holder info search feature on our website.
- I am renewing my license. How do I know how many classes I need to take for renewal?: License holders must meet Continuing Education (CE) requirements during each two- year license period. CE must include 4 hours of Legal Update I, 4 hours of Legal Update II, 3 hours of Contract-Related coursework, and 7 hours of elective CE courses for a total of 18 hours. To determine how many hours have posted to your license record, you can visit our license holder search.
A broker or designated broker of a business entity who sponsor one or more sales agents or a delegated supervisor of one or more license holders must complete the six-hour Broker Responsibility Course as part of the 18-hour CE requirement.
- I have completed several courses for my law degree. Can any of these courses be used towards the required qualifying courses to obtain a sales agent license?: No. All sales agent applicants must complete the required education.
- I have moved my sponsorship to a new broker, can my prior sponsoring broker pay me a commission directly for a transaction or referral?: Yes, Commission Rule 535.3 permits you to receive any compensation through your current sponsoring broker or the broker who sponsored you at the time you earned the right to the compensation.
- I only represent buyers. Can I advertise that I will rebate a part of my compensation to the buyer?: Yes, as long as the advertisement complies with Rule 535.155 (effective May 15, 2018) regarding any restrictions that might apply. However, a rebate to a buyer from a license holder may be subject to restrictions by the buyer’s lender. You should contact your broker or private attorney to find out how you should notify and obtain the consent of the buyer’s lender to address any impact the rebate may have on the determination regarding the buyer’s creditworthiness.
- I want to renew my sales agent or broker license active but am unable to complete my CE hours by the license expiration date. Can I get an extension to complete my hours?: Yes. If TREC records do not show that you have completed your CE at the time you submit your renewal application, you must pay a $200 CE deferral fee, or renew in inactive status. Paying the deferral fee allows you to continue to be active in real estate activities and provides an additional 60 days from your expiration date to complete your CE.
- If I have a licensed limited partnership, can the general partner be an LLC if a manager of the LLC is a designated broker?: No. The designated broker acting as a general partner must be an individual, not another business entity. [See TRELA §1101.355 and Rules 535.50(5) and §535.53(a)]
- If I represent the seller, can I advertise that I will rebate part of my compensation to the buyer? : Yes, but the ad must disclose that payment of the rebate is subject to the consent of the seller and if the rebate is contingent upon certain restrictions, such as the use of a particular service provider, the ad must contain a disclosure that payment of the rebate is subject to restrictions. [Rule 535.155 (effective May 15, 2018)] A sales agent must also have their sponsoring broker's authorization to offer a rebate.
- If a broker does not have a written agreement to represent the buyer, what recourse does the broker have if another broker “steals” a client? : A buyer can choose the broker with whom the buyer wants to work. TREC does not determine what constitutes "procuring cause" or who is entitled to a commission or other compensation. Like a listing agreement, the buyer representation agreement must be in writing and signed by the buyer to be binding.
- If a real estate broker has an escrow account, can the broker keep any interest that is earned on the money on deposit?: Not unless the person depositing the money has signed an agreement authorizing the broker to keep the interest. Otherwise, the interest must be treated in the same manner as the deposited money. The broker is responsible for accounting for the interest and disbursing it to the person whose money is held by the broker. [Rule 535.146(c)(3)] Accounting is more simple if the broker puts all escrow money into a non-interest bearing account
- If a sales agent’s last name is contained in the broker’s licensed or assumed business name, how can the sales agent let the consumer know the agent is not the broker?: To avoid an advertisement that implies the sales agent is responsible for the operation of the brokerage in this situation, the sales agent should make sure that the ad clearly indicates that the sales agent is not the broker. One simple way to accomplish this is to put “sales agent” next to the agent’s name. Using “REALTOR” or “agent” is insufficient to distinguish the license status of sales agent. [TRELA §1101.652(b)(23) and Rule 535.155(d)(7)]
- Is a broker required to act as an escrow agent or hold the money of others?: No, not unless the broker agrees to do so.
- Is a broker responsible for the actions of a sales agent who owns his or her own real estate business entity? : Yes, if the real estate business entity holds a real estate license and the sales agent is sponsored by that entity. In such a situation, the designated broker for the entity is still responsible for the sales agent's actions, even when the sales agent owns the licensed business entity. [TRELA §1101.803, Rule 535.2(l), ] A sales agent may not engage in real estate brokerage activity unless the sales agent is associated with, and acting for, a sponsoring broker. [TRELA §1101.351(c)]. When the sponsoring broker is a licensed business entity, it must have a designated broker to be active.
- Is a broker responsible for the actions of a sales agent who transacts business from a separate location? : Yes. The sponsoring broker is still responsible for the sales agent's actions, even when the sales agent does not work out of the broker’s main office. [TRELA §1101.803, Rule 535.2(a)] A sales agent may not lawfully engage in brokerage activity unless the sales agent is associated with, and acting for, a sponsoring broker at all times. [TRELA §1101.351(c)]
- Is a license holder acting as a principal required to provide a party with written information relating to agency?: No, but a license holder is required to provide a written notice to the other party that the license holder is licensed as a real estate broker or sales agent before entering into a contract, including a lease. Additionally, the license holder may not use the license holder’s expertise to the disadvantage of the other party. [See Rule 535.144]
- Is a license holder required to provide a party with written information relating to agency?: Yes. At the time of first substantive communication with a party relating to a proposed transaction regarding specific real property, a license holder must provide a written notice regarding agency, commonly referred to as “Information About Brokerage Services” or IABS Form. [TRELA §1101.558(b-1)]. TREC publishes the Information About Brokerage Services Form, TREC No. IABS 1-0, that license holders must use to comply with the statute.
- Is a license holder required to put their license number on a sign or other advertising?: No.
- Is a link to the IABS form required on my website?: Yes. Rule 531.20(b) states that each broker and sales agent must provide a link on its homepage to the IABS Form labeled "Texas Real Estate Commission Information About Brokerage Services". The link must be in at least a 10 point font and in a readily noticeable place on the homepage of the business website of the broker and sales agent. The link can also be "TREC Information About Brokerage Services," in at least 12 point font.
- Is a real estate license necessary in order to be an apartment locator? : Yes. TREC requires licensure if the person seeks or has an expectation of compensation for offering to locate a unit in an apartment complex to a prospective tenant. [TRELA §1101.002(6)] In addition, a person may not engage in business as a residential rental locator (apartment locator) unless the person is licensed as a real estate broker or sales agent. [TRELA §1101.351(a)(2), Rule 535.4(k)]
- Is it illegal or unethical for an agent to present two offers to the seller at the same time? : The listing agent represents the seller and has a duty to present all offers in a timely manner to the seller. There is no prohibition against a license holder presenting more than one offer at a time to a seller. A seller may receive, review and negotiate several offers simultaneously.
- Is the Information About Brokerage Services Form promulgated by TREC mandatory?: For all practical purposes, yes. A license holder may reproduce the IABS Form for the purpose of prefilling the Broker Contact Information section. If a license holder prefills this information, the license holder must ensure that the text of the IABS Form is copied verbatim and that spacing, borders and placement of text on the page appear identical to that in the promulgated IABS Form. [Rule 531.20(e)]
- Is the use of a “net” listing agreement a breach of the broker’s fiduciary duty?: It could be. The use of net listing agreement places the broker’s interest above the principal’s interest with regards to obtaining the best possible price. A broker may not enter into a net listing agreement unless the principal requires a net listing and the principal is clearly familiar with the current market values of real property. [Rule 535.16(b)]
- May a broker act as a dual agent?: No. Texas law does not permit dual agency. A license holder may not represent both principals as a dual agent under the revisions to TRELA. Under the current law, a broker must agree to act as an intermediary in accordance with the statute if the broker agrees to represent more than one party in a transaction. [TRELA §1101.561(b)] To the extent a dual agency relationship is created by accident or otherwise, a license holder must resolve the matter by immediate compliance with the notice and consent requirements under TRELA §§1101.558-561 and act as either an intermediary or represent only one of the principals in a transaction while working with the other principal only as a customer.
- May a license holder who is a rental locator advertise that they will pay a prospective tenant a portion of their fee received from an apartment complex if the tenant uses the locator’s services?: Yes, as long as the ad complies with Rule 535.155 (effective May 15, 2018), which requires the consent of the party the license holder represents in a transaction. When a rental locator represents an apartment complex, the locator needs the consent of the apartment complex. When the rental locator represents a tenant and not an apartment complex, as demonstrated by a written representation agreement or other evidence of representation, the locator is not required to obtain the consent of the apartment complex because the complex is not his client. Regardless of representation, however, it is misleading advertising to advertise a rebate for an apartment complex that the locator knows has a “no rebate” policy.
- Must a person be licensed to locate apartment units for prospective tenants and be paid by the owner of the apartments? : Yes, unless the person is an employee of the owner of the apartments or otherwise exempt, residential rental locators are required to be licensed as either a real estate broker or sales agent.
- My business entity is a series LLC. What documents are required to be submitted with the Franchise Tax Account Status page?: If the business entity is a series LLC, you must include a copy of the Certificate of Filing from the Secretary of State's Office with the Franchise Tax Account Status page. The Certificate of Filing must show that the business entity is a registered series of the business entity listed on the Franchise Tax Account Status page.
- Online Services is not accepting my temporary password. What do I do?: Copy the temporary password from the email message you received when you registered. Paste it into the “Online Services – Login and Registration” web page when you log in. You will be prompted to create a new password of your own. Once you create your password, you must save it to have future access to your online account.
- What are the agency disclosure requirements for a real estate license holder?: A license holder must disclose the fact that he or she represents a party upon the first contact with another party or a license holder representing another party. This disclosure may be oral or in writing.
- What can unlicensed office personnel or an unlicensed assistant do?: An unlicensed person may not engage in any activity for which a license is required. [TRELA §1101.002(1) and Rules 535.4 and 535.5]. For a more detailed discussion of what an unlicensed person can and cannot do, see the article on our website titled “Use of Unlicensed Assistants in Real Estate Transactions.”
- What if a license holder does not comply with the requirements for forming an intermediary relationship?: Failure of the intermediary broker or the sponsored sales agents to comply with the Intermediary Provisions of TRELA §§ 1101.558-561 may subject them to disciplinary sanctions by the TREC, including but not limited to, revocation, suspension, reprimand and/or an administrative penalty.
- What is an associated broker?: Associated broker means a broker who associates with and is paid through another broker under a relationship that is intended to be a continuous relationship, including but not limited to, an employment or ongoing independent contractor relationship. [Rule 535.154(a)(3)]
- What is an intermediary?: An intermediary is a broker who negotiates the transaction between the parties when the broker or a sales agent sponsored by the broker has obtained consent from the parties to represent both the buyer and the seller. The broker intermediary may, with the written consent of the parties, appoint separate individual license holder associated with the broker to work with and advise the party to whom they have been appointed. [TRELA §§ 1101.558-1101.561 and §1101.651(d)]
- What is proof of legal authority to use an assumed business name in Texas?: Generally, in Texas, filing an assumed business name is required to put the public on notice that you are doing business under a name other than your legal name. For most business entities, the assumed business name is filed with the Secretary of State. For a general partnership or individual broker, the assumed business name is filed with the county clerk in the county or counties where you do business. See
Evidence of registration of the assumed business name with the Secretary of State or in the county or counties where the broker does business is adequate proof of authority to do business under that name. [Rule 535.154(d)]
- What is the difference between an assumed business name and a team name?: Assumed Business Name (also known as a dba) is another name for the broker’s business that can be used by all sponsored sales agents and brokers associated with this broker. Another name that indicates a broker’s line of business, e.g. property management or commercial, is also an assumed business name of the broker, even though it may not be available for use by all sponsored agents and associated brokers. [Rule 535.154(a)(4)]
Team Name is a name used only by a team or group of one or more sponsored sales agents or brokers associated with this broker. Team names must end in “team” or “group” and cannot contain the words “brokerage”, “company”, “associates” or other similar terms. [Rule 535.154(a)(5)]
- What is the difference if a broker appoint sales agents to represent the buyer and the seller or if the broker makes no appointments to the principals in a real estate transaction?: If the broker appoints an associated license holder to represent the seller and another associated license holder to represent the buyer, the individual agents may offer advice and opinions regarding the real estate transaction to the party each has been appointed to represent. If the broker does not appoint associated license holders to represent the buyer and seller respectively, then the broker and/or agent may not offer advice and opinions relevant to the real estate transaction to either party and must not favor one principal in the transaction over the other principal. Appointments provide the agents the opportunity to provide a higher level of service to their clients.
- What is the required information that must be provided in advertisements such as signs, email and business cards?: All advertisements must comply with TRELA §1101.652(b)(23) and Rules 535.154 and 535.155.
Rule 535.155 requires each advertisement to include the following items in a readily noticeable location in the advertisement:
The name of the license holder or team placing the advertisement; and
The broker’s name in at least half the size of the largest contact information for any sales agent, associated broker, or team name contained in the advertisement.
Notes:
A broker’s name includes a broker’s assumed business name that has been registered with TREC.
Requirements for an assumed business name and team name are set out in Rule 535.154
There is no requirement that a phone number or email address included in an advertisement belong to the broker.
What qualifies as proof of ownership of the business entity?: A resolution, minutes or other official record of the business entity. We also accept copies of tax records which indicate ownership.
When can an agent say that they “sold” a property in an advertisement?: Any agent who worked with the seller or the buyer in a transaction that resulted in the sale of a property may correctly state in an advertisement that they “sold” the property. If the license holder did not participate in that specific transaction, he cannot state or imply that his actions resulted in the sale of that property. An example of a misleading advertisement of this nature would be if a license holder sent out “Just Sold” postcards with her contact information and a picture of a recently sold property that she did not help to sell. She didn’t state that she sold it but an average person reading the card could surely and reasonably imply an erroneous claim of involvement. Another potential example of a misleading advertisement is a license holder who included a list of “Recently Sold Homes” in his advertisement that included many properties where he had no role in the transaction, but he failed to make it clear in the ad which – if any – of those transactions he was involved in. Under Texas law, a license holder may not “create a misleading impression” in their advertisement. A broker must review all ads to ensure this result is avoided.
When does a license holder dealing in property in the license holder's own name have to disclose the fact that they have a real estate license?: It is best to disclose it as early as possible but it must be disclosed in the contract, lease, or in another written document given to the other principal before the agreement is signed. [Rule 535.144(b)] The disclosure is required even if the license holder is on inactive status.
When should a license holder provide the Disclosure of Relationship with Residential Service Company form?: A license holder should fully complete and provide the Disclosure of Relationship with Residential Service Company (RSC-3) when the license holder will receive a fee from a residential service company because a party to the transaction purchases a contract from that company.
Many transactions do not require use of this form. For example,
If there is no residential service company contract as part of the transaction, you do not need to provide this form.
If you are not being paid a fee by a residential service company, you do not need to provide this form.
If a residential service company contract is part of the transaction but you do not receive a fee from that specific company, you do not need to provide this form.
Generally, the only reason you might fill out this form if you do not receive a fee from a residential service company is because the other agent or broker in your transaction is providing their own disclosure. In this situation, you would fill out the appropriate portion of the form and check the box that says you "will receive no compensation from a residential service company." Your signature in this situation is merely disclosure and is not an endorsement, approval, or otherwise binding.
- Who registers an assumed business name, team, or alternate name?: Assumed Business Name: Broker [Rule 535.154(d)(1)]
Team Name: Broker [Rule 535.154(c)(3)]
Alternate Name: Individual License Holder [Rule 535.154(b)(1)]
Remember -- all these types of names must be registered with TREC before use in advertising.
More FAQs
Information About Brokerage Services
Form Name:
Information About Brokerage Services (Form ID: IABS 1-1)
Effective Date:
04/01/2025
Read more about the form and how to provide it.
Related Terms:
Contracts, Forms & Applications, Notices, Real Estate Sales Agent, Real Estate Broker Individual, Business Entity Broker
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188594 | http://ncvm4.books.nba.co.za/chapter/unit-4-complete-contingency-tables-to-solve-probability-problems/ | Unit 4: Complete contingency tables to solve probability problems – National Curriculum (Vocational) Mathematics Level 4
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Contents
Introduction
I. Complex Numbers: Working with complex numbers
Unit 1: Revise the basic operations with complex numbers in standard form
Unit 2: Revise the polar form of complex numbers
Unit 3: Revise the multiplication and division of complex numbers in polar form
Unit 4: Raise complex numbers to exponents using De Moivre’s Theorem
II. Complex Numbers: Solve problems with complex numbers
Unit 1: Solve complex number problems
III. Functions and algebra: Work with algebraic expressions using the remainder and the factor theorems
Unit 1: Use the remainder and factor theorem to factorise third degree polynomials
IV. Functions and algebra: Use a variety of techniques to sketch and interpret information for graphs of the inverse of a function
Unit 1: Determine and sketch the inverse of a linear function
Unit 2: Determine and sketch the inverse of a quadratic function
Unit 3: Determine and sketch the inverse of the exponential function
V. Functions and algebra: Use mathematical models to investigate linear programming problems
Unit 1: Solve linear programming problems by optimising a function
VI. Functions and algebra: Investigate and use instantaneous rate of change of a variable when interpreting models both in mathematical and real life situations
Unit 1: Use first principles to find the derivative
Unit 2: Work with rules for differentiation
Unit 3: Find the equation of a tangent to a curve
Unit 4: Derivatives as rates of change
Unit 5: Sketch cubic functions
VII. Functions and algebra: Analyse and represent mathematical and contextual situations using integrals and find areas under curves by using integration rules
Unit 1: Introduction to integration
Unit 2: Rules for integration
Unit 3: Determine the area under a curve
VIII. Space, shape and measurement: Use the Cartesian co-ordinate system to derive and apply equations
Unit 1: Determine the equation of a circle with any centre
Unit 2: Find the equation of a tangent to a circle
IX. Space, shape and measurement: Explore, interpret and justify geometric relationships
Unit 1: Finding angles of straight lines and triangles
Unit 2: Circle geometry theorems
Unit 3: Properties of cyclic quadrilaterals
Unit 4: Tangent to circle theorems
X. Space, shape and measurement: Solve problems by constructing and interpreting trigonometric models
Unit 1: Work with compound angles
Unit 2: Solve trigonometric equations
Unit 3: Solve 2-D and 3-D trigonometry problems using sine and cosine rules
Unit 4: An introduction to radians
XI. Data handling: Represent, analyse and interpret data using various techniques
Unit 1: Use various techniques for data collection, representation and interpretation
XII. Data handling: Use variance and regression analysis to interpolate and extrapolate bivariate data
Unit 1: Calculate variance and standard deviation
Unit 2: Represent data using a scatter plot
Unit 3: Linear regression analysis
XIII. Data handling: Use experiments, simulation and probability distribution to set and explore probability models
Unit 1: Understand probability and make predictions
Unit 2: Draw Venn diagrams to solve probability problems
Unit 3: Draw tree diagrams to solve probability problems
Unit 4: Complete contingency tables to solve probability problems
XIV. Financial mathematics: Use mathematics to plan and control financial instruments
Unit 1: Work with simple and compound growth formulae
Unit 2: Interpret tax tables
Unit 3: Work with simple and compound depreciation
Appendix
National Curriculum (Vocational) Mathematics Level 4
Data handling: Use experiments, simulation and probability distribution to set and explore probability models
Unit 4: Complete contingency tables to solve probability problems
Natashia Bearam-Edmunds
Unit outcomes
By the end of this unit you will be able to:
Draw and complete contingency tables.
Use contingency tables to solve probability problems.
What you should know
Before you start this unit, make sure you can:
Calculate the probability of equally likely outcomes. You can revise finding probabilities in unit 1 of this subject outcome.
Introduction
A two-way contingency table is also called a two-way table or a cross tabulation table. Contingency tables are used to examine relationships between categorical data. ‘Contingency’ means ‘possibilities’. Contingency tables help you work out all the possible outcomes of combined events.
Constructing contingency tables
Contingency tables are especially helpful for figuring out whether events are dependent or independent.
In a two-way contingency table one event is written down the side of the table and the other event is written along the top of the table. The results are written in the cells of the table. The combined result of the activities is found by working two ways: across and then down.
We count the number of outcomes for two events and their complements, when working with two-way contingency tables, making four events in total. A two-way contingency table always shows the counts for the four possible combinations of events, as well as the totals for each event and its complement.
Example 4.1
A coin is tossed and a die is rolled simultaneously. Draw a table to show all possible pairs of outcomes. From your table calculate the probability of getting a head and a two and state if these events are independent.
Solution
When you toss a coin there are two possible outcomes heads (H) or tails (T). When you roll a die there are six possible outcomes, {1;2;3;4;5;6}{1;2;3;4;5;6}.
We will draw up the contingency table with the event rolling a die at the top of the table and the event toss a coin along the side.
Roll a die
1 1 2 2 3 3 4 4 5 5 6 6
Toss a coin
H H H;1 H;1 H; 2 H; 2 H; 3 H; 3 H; 4 H; 4 H; 5 H; 5 H; 6 H; 6
T T T;1 T;1 T; 2 T; 2 T; 3 T; 3 T; 4 T; 4 T; 5 T; 5 T; 6 T; 6
In the cells of the table we list the possible combination of outcomes. There are 12 12 outcomes in total.
To find the probability of getting a head and a two, we find the row with H H and the column with 2 2 and read across from the H H and down from 2 2, the cell where they meet shows the outcome H; 2 H; 2 . This outcome occurs once.
P(H; 2)=1 12 P(H; 2)=1 12
For independent events:
P(A and B)=P(A)⋅P(B)P(A and B)=P(A)⋅P(B)
From the table we see that:
P(H)=6 12=1 2 P(H)=6 12=1 2
P(2)=2 12=1 6 P(2)=2 12=1 6
P(H)⋅P(2)=1 2⋅1 6=1 12 P(H)⋅P(2)=1 2⋅1 6=1 12
We have shown that P(H; 2)=1 12 P(H; 2)=1 12.
Since, P(H; 2)=P(H)⋅P(2)P(H; 2)=P(H)⋅P(2)the events are independent.
Example 4.2
Example taken from from Siyavula Maths Grade 12
The table below shows the results of testing two different treatments on 240 240 fruit trees which have a disease causing the trees to die. Treatment A involves the careful removal of infected branches and treatment B involves removing infected branches as well as spraying the trees with antibiotics.
Tree dies within four yearsTree lives for more than four yearsTOTAL
Treatment A70 70 50 50
Treatment B
TOTAL90 90 150 150
Fill in the missing values on the table.
What is the probability a tree received treatment B?
What is the probability that a tree will live beyond four years?
What is the probability that a tree is given treatment B and lives beyond four years?
Of the trees that were given treatment B, what is the probability that a tree lives beyond four years?
Are a tree given treatment B and living beyond four years independent events? Justify your answer with a calculation.
Solutions
Since each column has to add up to its total, we can work out the number of trees which fall into each category for treatments A and B. Then, we can add each row to get the totals on the right-hand side of the table.
Tree dies within four yearsTree lives for more than four yearsTOTAL
Treatment A70 70 50 50 120 120
Treatment B20 20 100 100 120 120
TOTAL90 90 150 150 240 240
2. The probability that treatment B is given to a tree is the number of trees that received treatment B divided by the total number of trees.
P(B)=120 240=1 2 P(B)=120 240=1 2
3. To find the probability that a tree lives beyond four years, we use the total of the column ‘tree lives for more than four years’ and divide that by the total number of trees.
P(tree lives more than 4 years)=150 240=5 8 P(tree lives more than 4 years)=150 240=5 8
4. To determine the probability that a tree receives treatment B and lives beyond four years, we must find the cell within the table that shows the combination of these events (row 3 3, column 3 3) and divide that by the total number of trees.
P(B and tree lives more than 4 years)=100 240=5 12 P(B and tree lives more than 4 years)=100 240=5 12
5. Here, we are restricted to only the trees that received treatment B, living beyond four years. This means we no longer need to include the trees given treatment A, so the denominator needs to be adjusted accordingly.
P(lives beyond 4 years having received B)=100 120=5 6 P(lives beyond 4 years having received B)=100 120=5 6
6. .
P(tree lives more than 4 years)×P(B)=5 8×1 2=5 16 P(tree lives more than 4 years)×P(B)=5 8×1 2=5 16
P(B and tree lives more than 4 years)=5 12 P(B and tree lives more than 4 years)=5 12
So we see that P(B and tree lives more than 4 years)≠P(B)⋅P(lives more than 4 years)P(B and tree lives more than 4 years)≠P(B)⋅P(lives more than 4 years). Therefore, the treatment of a tree with treatment B and living beyond four years are dependent events.
Exercise 4.1
1. Use the contingency table below to answer the following questions:
Brown eyesNot brown eyesTOTAL
Black hair50 50 30 30 80 80
Red hair70 70 80 80 150 150
TOTAL120 120 110 110 230 230
1. What is the probability that someone with black hair has brown eyes?
2. What is the probability that someone has black hair?
3. What is the probability that someone has brown eyes?
4. Are having black hair and having brown eyes dependent or independent events?
5. What is the probability of having brown eyes or red hair?
You are given the following information:
Events A and B are independent.
P(not A)=0.3 P(not A)=0.3
P(B)=0.4 P(B)=0.4
Complete the contingency table below.
ANot ATOTAL
B
Not B
TOTAL50 50
3. A new treatment for influenza (the flu) was tested on a number of patients to determine if it was better than a placebo (a pill with no therapeutic value). The table below shows the results three days after treatment:
FluNo fluTOTAL
Placebo228 228 60 60
Treatment
TOTAL240 240 312 312
1. Complete the table.
2. Calculate the probability of a patient receiving the treatment.
3. Calculate the probability of a patient having no flu after three days.
4. Calculate the probability of a patient receiving the treatment and having no flu after three days.
5. Using a calculation, determine whether a patient receiving the treatment and having no flu after three days are dependent or independent events.
6. Calculate the probability that a patient receiving treatment will have no flu after three days.
7. Calculate the probability that a patient receiving a placebo will have no flu after three days.
8. Comparing you answers in f. and g., would you recommend the use of the new treatment for patients suffering from influenza?
The full solutions are at the end of the unit.
Summary
In this unit you have learnt the following:
How to construct a two-way contingency table.
How to complete a two-way contingency table.
How to calculate probabilities from a contingency table.
Unit 4: Assessment
Suggested time to complete: 25 minutes
Researchers conducted a study to test how effective a certain inoculation is at preventing malaria. Part of their data is shown below:
MalariaNo malariaTOTAL
MaleA B 216 216
FemaleC D 648 648
TOTAL108 108 756 756 864 864
1. Calculate the probability that a randomly selected study participant will be female.
2. Calculate the probability that a randomly selected study participant will have malaria.
3. If being female and having malaria are independent events, calculate the value C.
4. Using the value of C, fill in the missing values on the table.
A rare kidney disease affects only one in 1 000 1 000 people and the test for this disease has a 99%99% accuracy rate.
Draw a two-way contingency table showing the results if 100 000 100 000 of the general population are tested.
Calculate the probability that a person who tests positive for this rare kidney disease is sick with the disease, correct to two decimal places.
The Clueless Club consists of 500 500 members. In order to be part of this club you have to be a lawyer, a teacher or an engineer. Given below is an incomplete contingency table that shows the distribution of 500 500 members, in terms of their profession and the type of hot drinks they drink.
TeaCoffeeHot chocolateTOTAL
Lawyer52 52 41 41 80 80 173 173
Teacher48 48 69 69 A 137 137
Engineer100 100 B 10 10 190 190
TOTAL200 200 190 190 110 110 500 500
1. Determine the values of A and B.
2. If a member is selected at random, what is the probability of selecting a teacher who drinks coffee?
3. If a member is selected at random, what is the probability of selecting an engineer or a person who drinks hot chocolate?
The full solutions are at the end of the unit.
Unit 4: Solutions
Exercise 4.1
.
.
P(with black hair has brown eyes)=50 80=5 8 P(with black hair has brown eyes)=50 80=5 8
2. .
P(black hair)=80 230=8 23 P(black hair)=80 230=8 23
3. .
P(brown eyes)=120 230=12 23 P(brown eyes)=120 230=12 23
4. .
P(black hair)⋅P(brown eyes)=8 23⋅12 23=96 529 P(black hair and brown eyes)=50 230=5 23 P(black hair and brown eyes)≠P(black hair)⋅P(brown eyes)P(black hair)⋅P(brown eyes)=8 23⋅12 23=96 529 P(black hair and brown eyes)=50 230=5 23 P(black hair and brown eyes)≠P(black hair)⋅P(brown eyes)
The events are dependent.
5. The probability of having brown eyes or red hair is the union of the two events.
P(brown eyes or red hair)=P(brown eyes)+P(red hair)−P(brown eyes and red hair)=120 230+150 230−70 230=200 230=20 23 P(brown eyes or red hair)=P(brown eyes)+P(red hair)−P(brown eyes and red hair)=120 230+150 230−70 230=200 230=20 23
.
P(not A)=0.3∴n(not A)=0.3×50=15 P(not A)=0.3∴n(not A)=0.3×50=15
P(B)=0.4∴n(B)=0.4×50=20 P(B)=0.4∴n(B)=0.4×50=20
.
P(A and B)=P(A)⋅P(B)=35 50×0.4=7 25∴n(A and B)=7 25×50=14 P(A and B)=P(A)⋅P(B)=35 50×0.4=7 25∴n(A and B)=7 25×50=14
ANot ATOTAL
B14 14 6 6 20 20
Not B21 21 9 9 30 30
TOTAL35 35 15 15 50 50
3. .
1. .
FluNo fluTOTAL
Placebo228 228 60 60 288 288
Treatment12 12 252 252 264 264
TOTAL240 240 312 312 552 552
2. .
P(receiving the treatment)=264 552=11 23 P(receiving the treatment)=264 552=11 23
3. .
P(no flu)=312 552=13 23 P(no flu)=312 552=13 23
4. .
P(receiving the treatment and having no flu)=252 552=21 46 P(receiving the treatment and having no flu)=252 552=21 46
5. .
P(receiving the treatment)⋅P(no flu)=11 23⋅13 23=143 529 P(receiving the treatment and having no flu)=21 46 P(receiving the treatment)⋅P(no flu)=11 23⋅13 23=143 529 P(receiving the treatment and having no flu)=21 46
P(receiving the treatment and having no flu)≠P(receiving the treatment)⋅P(no flu)P(receiving the treatment and having no flu)≠P(receiving the treatment)⋅P(no flu)
Therefore, a patient receiving the treatment and having no flu after three days are dependent events.
6. .
P(having no flu if received the treatment)=252 264=21 22 P(having no flu if received the treatment)=252 264=21 22
7. .
P(having no flu if received the placebo)=60 288=5 24 P(having no flu if received the placebo)=60 288=5 24
8. Yes, I would recommend the use of the new treatment for patients with the flu as there is a 95.5%95.5% chance of not getting the flu if they have been treated compared to the 20.8%20.8% of getting the flu if they were given the placebo.
Back to Exercise 4.1
Unit 4: Assessment
.
.
P(female)=648 864=3 4 P(female)=648 864=3 4
2. .
P(malaria)=108 864=1 8 P(malaria)=108 864=1 8
3. .
P(female and malaria)=P(female)⋅P(malaria)since these are independent events=3 4⋅1 8∴c=3 32×864=81 P(female and malaria)=P(female)⋅P(malaria)since these are independent events=3 4⋅1 8∴c=3 32×864=81
4. .
MalariaNo malariaTOTAL
Male27 27 189 189 216 216
Female81 81 567 567 648 648
TOTAL108 108 756 756 864 864
.
.
Number of people with disease=1 1 000×100 000=100 Number of people with disease=1 1 000×100 000=100
.
Test for this disease has 99%99% accuracy rate:
Number of people with disease and test is positive=0.99×100=99 Number of people with disease and test is positive=0.99×100=99
Number of people with no disease and test is negative=0.99×99 9 01=98 901 Number of people with no disease and test is negative=0.99×99 9 01=98 901
DiseaseNo diseaseTOTAL
Test is positive99 99 999 999 1 098 1 098
Test is negative1 1 98 901 98 901 98 902 98 902
TOTAL100 100 99 900 99 900 100 000 100 000
2. .
P(have the disease if test is positive)=99 1 098=11 122=0.09 P(have the disease if test is positive)=99 1 098=11 122=0.09
.
.
A=137−48−69=20 A=137−48−69=20
B=190−41−69=80 B=190−41−69=80
2. .
P(teacher and coffee)=69 500 P(teacher and coffee)=69 500
3. .
P(engineer or hot chocolate)=P(engineer)+P(hot chocolate)−P(engineer and hot chocolate)=190 500+110 500−10 500=290 500=29 50 P(engineer or hot chocolate)=P(engineer)+P(hot chocolate)−P(engineer and hot chocolate)=190 500+110 500−10 500=290 500=29 50
Back to Unit 4: Assessment
Previous: Unit 3: Draw tree diagrams to solve probability problems
Next: Financial mathematics: Use mathematics to plan and control financial instruments
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National Curriculum (Vocational) Mathematics Level 4 by Natashia Bearam-Edmunds is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
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Critical Point and Triple Point
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$\begingroup$
Triple Point vs. Critical Point
Why is specific volume unchanged at Critical Point but can have a range of values of Triple Point?
thermodynamics
phase-transition
phase-diagram
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The triple point is the temperature in which the three phases ( solid, liquid, gas) can coexist but each phase will have a different density. The critical point is the boundary between liquid and gas where the density of the two phases are equal and therefore, the volumes will be equal ( since V = mass/density). So 1 gram of liquid will occupy the same volume as 1 gram of gas at the critical point.
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Because these two points are very different in kind:
The triple point is where three different phases meet. As such, you find all three different phases in the vicinity. And, when you do the experiment, you will have some of each phase. Each phase has its own density, so the specific volume of the three phases put together depends on their relative amounts.
The critical point is where a phase change line ends. It's the point where the two phases on either side of the line become equal. As such, you only have a single phase at the critical point, and thus only a single possible specific volume.
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Neutron Stars
Neutron stars are formed when a massive star runs out of fuel and collapses. The very central region of the star – the core – collapses, crushing together every proton and electron into a neutron. If the core of the collapsing star is between about 1 and 3 solar masses, these newly-created neutrons can stop the collapse, leaving behind a neutron star. (Stars with higher masses will continue to collapse into stellar-mass black holes.)
This collapse leaves behind the most dense object known – an object with the mass of a sun squished down to the size of a city. These stellar remnants measure about 20 kilometers (12.5 miles) across. One sugar cube of neutron star material would weigh about 1 trillion kilograms (or 1 billion tons) on Earth – about as much as a mountain.
Since neutron stars began their existence as stars, they are found scattered throughout the galaxy in the same places where we find stars. And like stars, they can be found by themselves or in binary systems with a companion.
Many neutron stars are likely undetectable because they simply do not emit enough radiation. However, under certain conditions, they can be easily observed. A handful of neutron stars have been found sitting at the centers of supernova remnants quietly emitting X-rays. More often, though, neutron stars are found spinning wildly with extreme magnetic fields as pulsars or magnetars. In binary systems, some neutron stars can be found accreting materials from their companions, emitting electromagnetic radiation powered by the gravitational energy of the accreting material. Below we introduce two general classes of non-quiet neutron star – pulsars and magnetars.
Pulsars
Most neutron stars are observed as pulsars. Pulsars are rotating neutron stars observed to have pulses of radiation at very regular intervals that typically range from milliseconds to seconds. Pulsars have very strong magnetic fields which funnel jets of particles out along the two magnetic poles. These accelerated particles produce very powerful beams of light. Often, the magnetic field is not aligned with the spin axis, so those beams of particles and light are swept around as the star rotates. When the beam crosses our line-of-sight, we see a pulse – in other words, we see pulsars turn on and off as the beam sweeps over Earth.
One way to think of a pulsar is like a lighthouse. At night, a lighthouse emits a beam of light that sweeps across the sky. Even though the light is constantly shining, you only see the beam when it is pointing directly in your direction. The video below is an animation of a neutron star showing the magnetic field rotating with the star. Partway through, the point-of-view changes so that we can see the beams of light sweeping across our line of sight – this is how a pulsar pulses.
This animation takes us into a spinning pulsar, with its strong magnetic field rotating along with it. Clouds of charged particles move along the field lines and their gamma-rays are beamed like a lighthouse beacon by the magnetic fields. As our line of sight moves into the beam, we see the pulsations once every rotation of the neutron star.Credit: NASA/Goddard/ CI Lab
Magnetars
Another type of neutron star is called a magnetar. In a typical neutron star, the magnetic field is trillions of times that of the Earth's magnetic field; however, in a magnetar, the magnetic field is another 1000 times stronger.
In all neutron stars, the crust of the star is locked together with the magnetic field so that any change in one affects the other. The crust is under an immense amount of strain, and a small movement of the crust can be explosive. But since the crust and magnetic field are tied, that explosion ripples through the magnetic field. In a magnetar, with its huge magnetic field, movements in the crust cause the neutron star to release a vast amount of energy in the form of electromagnetic radiation. A magnetar called SGR 1806-20 had a burst where in one-tenth of a second it released more energy than the sun has emitted in the last 100,000 years!
Text updated: March 2017
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188597 | https://www.combinatorics.org/ojs/index.php/eljc/article/download/v32i2p33/pdf/ | A Max-Min Problem on Spectral Radius and Connectedness of Graphs
Zhenzhen Lou a Changxiang He a,b
Submitted: Jan 29, 2024; Accepted: Apr 16, 2025; Published: May 23, 2025
© The authors. Released under the CC BY-ND license (International 4.0).
Abstract
In the past decades, many scholars have been concerned with the question of which edge-extremal problems have spectral analogues. Recently, Wang, Kang, and Xue established an interesting result on F -free graphs [ J. Combin. Theory Ser. B 159 (2023) 20–41]. In this paper, we investigate this problem in the context of critical graphs. Let P be a property defined on a family G of graphs. A graph
G ∈ G is said to be P -critical if it satisfies P but G − e does not satisfy P for any edge e ∈ E(G). Specifically, a graph is minimally k-(edge)-connected if it is
k-connected (respectively, k-edge-connected) and the deletion of any edge results in a graph that is not k-connected (respectively, k-edge-connected). A natural max-min problem is to determine the maximum spectral radius of minimally k-(edge)-connected graphs with n vertice. In 2019, Chen and Guo [ Discrete Math. 342 (2019) 2092–2099] resolved the case k = 2. In 2021, Fan, Goryainov, and Lin [ Discrete Appl. Math. 305 (2021) 154–163] determined the extremal spectral radius for minimally 3-connected graphs. In this paper, we establish structural properties of minimally
k-(edge)-connected graphs. Furthermore, we solve the max-min problem for the case
k > 3, demonstrating that any minimally k-(edge)-connected graph attaining the maximum spectral radius simultaneously achieves the maximum number of edges.
Mathematics Subject Classifications: 05C50, 05C75
1 Introduction
Perhaps the most basic property a graph may blue posses is that of being connected. At a more refined level, there are various functions that may be said to measure the connectedness of a connected graph . A graph is said to be connected if for every pair of vertices there is a path joining them. Otherwise the graph is disconnected. The
connectivity (or vertex-connectivity ) κ(G) of a graph G is the minimum number of vertices
a
College of Science, University of Shanghai for Science and Technology, Shanghai, 200093, China (xjdxlzz@163.com ,cxhe@usst.edu.cn ).
b
Corresponding author.
the electronic journal of combinatorics 32(2) (2025), #P2.33
whose removal results in a disconnected graph or in a trivial graph. The edge-connectivity
κ′(G) is defined analogously, only instead of vertices we remove edges. A graph is k-connected if its connectivity is at least k and k-edge-connected if its edge-connectivity is at least k. It is almost as simple to check that the minimal degree δ(G), the edge-connectivity and vertex-connectivity satisfy the following inequality:
δ(G) > κ′(G) > κ(G).
A number of extremal problems related to graph connectivity have been studied in recent years. One of the most important tasks for characterization of k-connected graphs is to give a certain operation such that they can be produced from simple k-connected graphs by repeatedly applying this operation . This goal has been accomplished by Tutte for 3-connected graphs, by Dirac and Plummer for 2-connected graphs, and by Slater for 4-connected graphs. A graph is said to be minimally k-(edge)-connected if it is k-(edge)-connected but omitting any of edges the resulting graph is no longer k-(edge)-connected. Clearly, a
k-(edge)-connected graph whose every edge is incident with one vertex of degree k is minimally k-(edge)-connected. Especially, a k-regular and k-(edge)-connected graph is minimally k-(edge)-connected. One of the central problems in this area is to determine the number of vertices of degree
k in a minimally k-edge-connected graph. In 1972, Lick showed that every minimally
k-edge-connected finite graph has at least two vertices of degree k (see also Lemma 13 in ), which is clearly best possible. But for simple graphs, this was improved in as follows: every minimally k-edge-connected finite simple graph has at least k + 1 vertices of degree k. It was proved in that for every k / ∈ { 1, 3} there exists a ck > 0 such that every minimally k-edge-connected finite simple graph G has at least ck|G| vertices of degree k. The value of the constant ck was improved in and , and a rather good estimate for ck was given by Cai . In 1995, Mader further improved the value ck
and gave the best possible linear bound for k ≡ 3( mod 4). Another interesting problem is to determine the maximum number of edges in a min-imally k-(edge)-connected graph. Mader proved that e(G) 6 kn − (k+1 2
) for every minimally k-connected graph G of order n, and if n > 3k − 2 then e(G) 6 k(n − k), where the equality is uniquely attained by the complete bipartite graph Kk,n −k provided that k > 2 and n > 3k − 1. Cai proved that e(G) 6 b(n+k)2
8
c for every minimally
k-connected graph G of order n < 3k − 2. Mader also proved that every minimally
k-edge-connected graph on n vertices has at most k(n−k) edges provided n > 3k −2. The complete bipartite graph Kk,n −k shows that this bound is tight. Dalmazzo proved that every minimally k-edge-connected multidigraph on n vertices has at most 2 k(n − 1) edges. In 2005, Berg and Jord´ an showed that if multiple edges are not allowed then Dalmazzo’s bound can be improved to 2 k(n − k) for n sufficiently large. In this paper, we first obtain an extremal result for every subgraph of a minimally k-(edge)-connected graph.
the electronic journal of combinatorics 32(2) (2025), #P2.33 2
Theorem 1. Let G be a minimally k-(edge)-connected graph and let H be a subgraph of
G. Then e(H) 6 k(|H| − 1) . Moreover, if |H| > 12 k(k + 5) , then e(H) 6 k(|H| − k),where the equality holds if and only if H ∼= Kk, |H|− k.
Let A(G) be the adjacency matrix of a graph G. The largest eigenvalue of A(G) is called the spectral radius of G, and denoted by ρ(G). In classical theory of graph spectra, many scholars are interested in an extremal problem, that is, what is the maximal spectral radius of a family G of graphs, where graphs in G have a common property P . A graph is said to be P -saturated , if it has the property P but adding an edge between an arbitrary pair of non-adjacent vertices results in a graph which does not have the property. It is known that A(G) is a non-negative matrix, and adding an edge in G always increases the spectral radius provided that G is connected. Therefore, most of the spectral extremal problems have saturated extremal graphs (see for example, [8, 9, 23, 15, 28, 22, 30, 31, 32]). Particularly, we have the following problem.
Problem 2. What is the maximal spectral radius among all n-vertex saturated graphs with fixed vertex-connectivity or edge-connectivity? Ye, Fan and Wang showed that among all graphs of order n with vertex (edge)-connectivity k, K(n − 1, k ) has the maximal spectral radius, where K(n − 1, k ) is obtained from the complete graph Kn−1 by adding a new vertex of degree k. Clearly, K(n−1, k ) has the same vertex-connectivity, edge-connectivity and minimum degree. Ning, Lu and Wang proved that for all graphs of order n with minimum degree δ and edge connectivity
κ′ < δ , the maximal spectral radius is attained by joining κ′ edges between two disjoint complete graphs Kδ+1 and Kn−δ−1, and they also determined the unique extremal graph with minimum degree δ and edge-connectivity κ′ ∈ { 0, 1, 2, 3}. Very recently, Fan, Gu and Lin determined the unique spectral extremal graph over all n-vertex graphs with minimum degree δ and edge connectivity κ′ ∈ { 4, . . . , δ − 1}.A graph G is said to be P -critical , if it admits a property P but G − e does not have it for any edge e ∈ E(G). Clearly, every minimally k-(edge)-connected graph is a connectivity-critical graph. Comparing with Problem 2, the following problem also attracts interest of scholars.
Problem 3. What is the maximal spectral radius among all n-vertex critical graphs with fixed vertex-connectivity or edge-connectivity? Obviously, every minimally 1-(edge)-connected graph is a tree. Furthermore, it is known that the maximal spectral radius of a tree is attained uniquely by a star (see ). In 2019, Chen and Guo showed that K2,n −2 attains the maximal spectral radius among all minimally 2-connected graphs and minimally 2-edge-connected graphs, respectively. Subsequently, Fan, Goryainov and Lin proved that K3,n −3 attains the largest spectral radius over all minimally 3-connected graphs. Now let k > 3 be a fixed integer and α = 124 k(k+1) . Let X = ( x1, x 2, . . . , x n)T be a non-negative eigenvector with respect to ρ(G). We may assume that xu∗ = max 16i6n xi
for some u∗ ∈ V (G). In this paper, we prove the following result, which implies that every
the electronic journal of combinatorics 32(2) (2025), #P2.33 3
minimally k-(edge)-connected graph with large spectral radius contains a certain number of vertices of high degrees.
Theorem 4. Let G be an n-vertex minimally k-(edge)-connected graph, where n > 18 kα2 .If ρ2(G) > k(n − k), then G contains a k-vertex subset L such that xv > (1 − 12k )xu∗ and
dG(v) > (1 − 23k )n for each vertex v ∈ L.
The main result of the paper is the following Max-Min theorem, which implies that every minimally k-(edge)-connected graph with maximal spectral radius also has maximal number of edges.
Theorem 5. For n > 18 kα2 , the maximal spectral radius of an n-vertex minimally k-(edge)-connected graph is attained uniquely by the complete bipartite graph Kk,n −k.
Finally, we present the following problem.
Problem 6. Consider a given property P . Does an edge-extremal problem on P -critical graphs possess a spectral analogue? The rest of the paper is organized as follows. In Section 2, we give some structural properties of a minimally k-(edge)-connected graph as well as the proof of Theorem 1. In Section 3, we use Theorem 1 to show Theorems 4 and 5.
2 Structural properties
Let G be a graph with vertex set V (G) and edge set E(G). We write |G| for the number of vertices and e(G) the number of edges in G. For a vertex v ∈ V (G), let NG(v) be the neighborhood of v. For S ⊆ V (G), we denote NS (v) = N (v)∩S and dS (v) = |NS (v)|. The subgraph of G induced by S and V (G) \ S are denoted by G[S] and G − S, respectively. Let eG(S) be the number of edges within S, and let eG(S, V (G) \ S) be the number of edges between S and V (G) \ S. All the subscripts defined here will be omitted if it is clear from the context. We start with the following lemma.
Lemma 7. Every k-(edge)-connected subgraph of a minimally k-(edge)-connected graph is minimally k-(edge)-connected. Proof. We first prove that for every subgraph of a minimally k-edge-connected graph, if it is k-edge-connected then it is minimally k-edge-connected. Let G be a minimally
k-edge-connected graph. Suppose to the contrary that H is a k-edge-connected subgraph of G but it is not minimally k-edge-connected. Then there exists an edge, say u1u2, of H
such that H − u1u2 is also k-edge-connected. Notice that G is a minimally k-edge-connected graph. Hence, G − u1u2 is ( k − 1)-edge-connected. Thus, there exists a partition V (G) = V1 ∪ V2 such that u1 ∈ V1, u2 ∈ V2
and e(V1, V 2) = k. Now, let Vi(H) = V (H) ∩ Vi for i ∈ { 1, 2}. Clearly,
e(V1(H), V 2(H)) 6 e(V1, V 2
) = k,
the electronic journal of combinatorics 32(2) (2025), #P2.33 4
and thus e(V1(H), V 2(H)) 6 k − 1 in H − u1u2, which contradicts the fact that H − u1u2
is k-edge-connected. Therefore, the result follows. The vertex-connected case of the lemma is an exercise of Chapter one in . Hence, we omit its proof here. Next, we give the maximal number of edges in every subgraph of a minimally k-edge-connected graph. Before proceeding, we need two more lemmas due to Mader .
Lemma 8 () . Let G be a graph of order n > k. If G does not contain any (k + 1) -edge-connected subgraph, then
e(G) 6 k(n − k) + (k
2
).
Furthermore, this bound is best possible.
Lemma 9 () . Let G be a minimally k-edge-connected graph of order n > 3k. Then
e(G) 6 k(n − k),
with equality if and only if G ∼= Kk,n −k.
Theorem 10. Let G be a minimally k-edge-connected graph and let H be a subgraph of
G. Then e(H) 6 k(|H| − 1) . Moreover, if |H| > 12 k(k + 5) , then e(H) 6 k(|H| − k),where the equality holds if and only if H ∼= Kk, |H|− k.Proof. Firstly, we will show that e(H) 6 k(|H| − 1). If |H| < k , then e(H) 6 12 |H|(|H| −
1) 6 k(|H| − 1), as desired. Now assume that |H| > k. It suffices to show e(H) 6
k(|H| − k+1 2 ). By Lemma 7, every k-edge-connected subgraph of G is minimally k-edge-connected, and thus has edge-connectivity k. Hence, G contains no ( k+1)-edge-connected subgraphs. By Lemma 8, we have e(H) 6 k(|H| − k) + (k
2
) = k(|H| − k+1 2 ), as required. In the following, we prove that e(H) 6 k(|H| − k) for |H| > 12 k(k + 5). The proof should be distinguished into two cases.
Case 1: H contains no k-edge-connected subgraphs. By Lemma 8, we know that e(H) 6 (k − 1)( |H| − k
2
). Note that |H| > 12 k(k + 5). It is easy to see that (k − 1)( |H| − k
2
) < k (|H| − k), and the result follows.
Case 2: H contains k-edge-connected subgraphs. Let H0 be a maximal k-edge-connected subgraph of H. Then H0 is a vertex-induced subgraph with |H0| > k + 1. If
H = H0, then by Lemma 7, H is minimally k-edge-connected. Since |H| > 12 k(k+5) > 3k,by Lemma 9 we have e(H) 6 k(|H| − k), with equality if and only if H ∼= Kk, |H|− k.Now we may assume that H0 is a proper induced subgraph of H. Then κ′(H) 6 k − 1, and thus we can find a partition V (H) = V0 ∪ V1 such that e(H) 6 e(V0) + e(V1) + ( k − 1) .
One can observe that H0 is a subgraph of H[V0] or H[V1] (otherwise, write Ui = V (H0)∩Vi
for i ∈ { 0, 1}, then e(U0, U 1) > k as H0 is k-edge-connected, consequently, e(V0, V 1) > k,a contradiction). For i ∈ { 0, 1}, if κ′(H[Vi]) 6 k − 1 and |Vi| > 2, then we can find a partition Vi = V ′
i
∪ V ′′
i
such that e(Vi) 6 e(V ′
i
) + e(V ′′
i
) + ( k − 1). Similarly, every
k-edge-connected subgraph of H[Vi] can only be a subgraph of H[V ′
i
] or H[V ′′
i
].
the electronic journal of combinatorics 32(2) (2025), #P2.33 5
By a series of above iterative operations (say s steps), we can obtain a partition
V (H) = ∪si=0 Vi satisfying that
e(H) 6 s∑
i=0
e(Vi) + ( k − 1) s (1) and every H[Vi] is either k-edge-connected or a single vertex. Recall that G contains no (k + 1)-edge-connected subgraphs. If H[Vi] is k-edge-connected, then |Vi| > k + 1 and
e(H[Vi]) 6 k(|Vi| − k+1 2 ) by Lemma 8. Let S1 = {i | | Vi| = 1 } and S2 = {0, . . . , s } \ S1.Then s = |S1| + |S2| − 1 and |H| = ∑
i∈S2
|Vi| + |S1|. In view of (1), we have
e(H) 6 ∑
i∈S2
k(|Vi| − k+1 2 ) + ( k − 1)( |S1| + |S2| − 1) = k|H| − 12 (k2 − k + 2) |S2| − | S1| − (k − 1) .
(2) If |S2| > 2, then 12 (k2 − k + 2) |S2| + ( k − 1) > k 2, and so e(H) < k (|H| − k), as desired. Now assume that |S2| = 1 (say S2 = {0} and H[V0] = H0). Then S1 6 = ∅ as H0
is a proper induced subgraph of H. By Lemma 7, H0 is minimally k-edge-connected. If
|H0| > 3k, then by Lemma 9, we have e(H0) 6 k(|V0| − k). Combining (1), we obtain
e(H) 6 k(|V0| − k) + ( k − 1) |S1| = k(|H| − k) − | S1|. The result follows. If |H0| < 3k,then |S1| = |H| − | H0| > 12 k(k − 1), and by (2) we have e(H) 6 k|H| − 12 (k2 − k + 2) −|S1| − (k − 1) < k (|H| − k). This completes the proof. Now we give a vertex-connected version of Theorem 10, which will be proved by a different approach.
Lemma 11 () . Let G be a minimally k-connected graph and let S be the set of vertices of degree k in G. Then G − S is empty or a forest.
Recall that e(G) 6 k(n − k) for n > 3k − 2 and every n-vertex minimally k-connected graph G. We also want to know the maximal number of edges in every subgraph of a minimally k-connected graph.
Theorem 12. Let G be a minimally k-connected graph and let H be a subgraph of G.Then e(H) 6 k(|H| − 1) . Moreover, if |H| > 5k − 4, then e(H) 6 k(|H| − k), where the equality holds if and only if H ∼= Kk, |H|− k.Proof. Firstly, we show e(H) 6 k(|H| − 1). We partition V (H) into two parts: V (H) =
V1 ∪ V2, where V1 is the set of vertices of degree k in G. If |V2| = 0, then e(H) 6 k|H|
2
6
k(|H| − 1), as desired. If |V2| > 1, from Lemma 11 we know that G[V2] is a forest, and so
e(V2) 6 |V2| − 1. Thus, we can get an upper bound of e(H) as below:
e(H) = e(V1) + e(V1, V 2) + e(V2) 6 k|V1| + ( |V2| − 1) , (3) where the equality holds if and only if G[V2] is a tree and NG(v) ⊆ V2 for each v ∈ V1. It is clear that k|V1| + |V2| − 1 6 k(|V1| + |V2| − 1), and hence e(H) 6 k(|H| − 1).
the electronic journal of combinatorics 32(2) (2025), #P2.33 6
Next, we shall distinguish three cases to show e(H) 6 k(|H| − k) for |H| > 5k − 4. If
k = 1, then G is a tree. Clearly, the result holds. In the following, we may assume k > 2.
Case 1: |V2| > k + 1. From (3) we have
e(H) 6 k|V1| + |V2| − 1 < k (|V1| + |V2| − k) = k(|H| − k).
The result follows.
Case 2: |V2| = k. Then |V1| > 4( k − 1). If e(V2) = 0, then by (3), we have
e(H) 6 k|V1| = k(|V1| + |V2| − k) = k(|H| − k), with equality if and only if H ∼= Kk, |H|− k.Now, assume that e(V2) > 1, and let V ′
1
= {v ∈ V1 | NG(v) = V2}. Then K|V ′
1|,|V2|
⊆
G[V ′
1
∪ V2]. We will see that |V ′
1
| 6 k − 1. Otherwise, if |V ′
1
| > k, then G[V ′
1
∪ V2]is k-connected. By Lemma 7, G[V ′
1
∪ V2] is minimally k-connected, which implies that
G[V ′
1
∪ V2] ∼= K|V ′
1|,|V2|
and so e(V2) = 0, a contradiction. Hence, |V ′
1
| 6 k − 1. On the other hand, let V ′′
1
= V1 \ V ′
1
, then
e(V ′′
1
) + e(V ′′
1
, V 2
) 6 (|V2| − 1)|V ′′
1
| + 12 |V ′′
1
| = (k − 12
)|V ′′
1
|.
Since |V ′
1
| 6 k − 1 and |V ′
1
| + |V ′′
1
| = |V1|, we further obtain
e(V1
) + e(V1, V 2
) 6 k|V ′
1
| + e(V ′′
1
) + e(V ′′
1
, V 2
) 6 (k − 12
)|V1| + 12
(k − 1).
Recall that |V1| > 4( k − 1) and e(V2) 6 k − 1. Thus we also have
e(H) 6 (k − 12
)|V1| + 32
(k − 1) < k |V1| = k(|H| − k).
Case 3: |V2| 6 k − 1. Then |V1| > 4k − 3. Let |V1| = x and |V2| = y. Then
e(H) = e(V1, V 2) + e(V1) + e(V2)
6 |V1|| V2| + 12 |V1|(k − | V2|) + ( |V2| − 1) = 12 xy + 12 kx + ( y − 1) ,
Notice that k(|H| − k) = k(x + y − k). Let
f (x, y ) = 12 xy + 12 kx + ( y − 1) − k(x + y − k).
It suffices to show f (x, y ) < 0 for x > 4k−3 and y 6 k−1. Note that ∂f (x,y )
∂x
= 12 (y−k) < 0and ∂f (x,y )
∂y
= 12 x + 1 − k > 0. Hence, f (x, y ) is decreasing with respect to x and increasing with respect to y. Therefore, f (x, y ) |max = f (4 k − 3, k − 1) = −12 , as desired. Observe that 12 k(k + 5) > 5k − 4 for every positive integer k. Combining Theorems 10 and 12, we immediately obtain Theorem 1.
the electronic journal of combinatorics 32(2) (2025), #P2.33 7
3 Spectral extremal results
Let G be a minimally k-(edge)-connected graph of order n. By Perron-Frobenius theorem, there exists a positive unit eigenvector with respect to ρ(G), which is called the Perron vector of G. Let X = ( x1, x 2, . . . , x n)T be the Perron vector with coordinate xu∗ =max {xi | i ∈ V (G)}. In this section, we first show Theorem 4, that is, if ρ2(G) > k(n−k), then G contains a k-vertex subset L such that xv > (1 − 12k )xu∗ and d(v) > (1 − 23k )n for each vertex v ∈ L. Before proceeding, we define three subsets of V (G).
Lα = {v ∈ V (G) | xv > αx u∗ }, where 0 < α 6 124 k(k+1) ;
Lβ = {v ∈ V (G) | xv > βx u∗ }, where 53 α 6 β 6 16k2 ;
Lγ = {v ∈ V (G) | xv > γx u∗ }, where 12k 6 γ 6 1.
Clearly, Lγ ⊆ Lβ ⊆ Lα. In the following, assume that k > 3 and n > 18 kα2 . We shall prove some lemmas on these three subsets.
Lemma 13. |Lα| < √4kn .Proof. For every v ∈ Lα, we have ρx v = ∑
u∈N(v)
xu, and thus
ρx v = ∑
u∈N(v)∩Lα
xu + ∑
u∈N(v)\Lα
xu 6
(
dLα (v) + α · dV (G)\Lα (v)
)
xu∗ . (4) Since ρx v > √k(n − k)αx u∗ for v ∈ Lα, from (4) we have
√k(n − k)α 6 dLα (v) + α · dV (G)\Lα (v). (5) Summing (5) over all v ∈ Lα, we have
|Lα|√k(n − k)α 6 2e(Lα) + α · e(Lα, V (G) \ Lα). (6) By Theorem 1, we have e(Lα) 6 k|Lα| and e(Lα, V (G)\Lα) 6 e(G) 6 k(n−k). Combining (6), we get that
|Lα|√k(n − k) 6 2kα |Lα| + k(n − k). (7) Since n > 18 kα2 , we have n − k > 16 kα2 , and hence 2kα < 12
√k(n − k). Combining (7), we obtain that |Lα| < 2√k(n − k), and thus |Lα| < √4kn , as desired. For a vertex v ∈ V (G), let N [v] = N (v) ∪ { v} and N 2(v) denote the set of vertices at distance two from v.
Lemma 14. |Lβ | < 12 kα .
the electronic journal of combinatorics 32(2) (2025), #P2.33 8
Proof. We proceed the proof by contradiction. Suppose that |Lβ | > 12 kα . Recall that
Lβ ⊆ Lα and α 6 124 k(k+1) . Then |Lα| > 12 kα > max {5k − 4, 12 k(k + 5) }. We first prove that d(v) > α
12
n + k for each vertex v ∈ Lβ .By Theorem 1, we get that e(G) 6 kn , e(N [v]) 6 k(|N [v]| − 1) = kd (v) and e(N (v) ∪
Lα) 6 k(d(v) + |Lα| − k). Since v ∈ Lβ , we can easily see that v ∈ Lα. Let S =
N (v) ∪ (Lα \ { v}). Then e(S) = e(N (v) ∪ Lα) − d(v) 6 (k − 1) d(v) + k|Lα| − k2, where
|Lα| < √4kn < α
2
n by Lemma 13 and the assumption that n > 18 kα2 .It is easy to see that
d(v)xv + ∑
u∈N(v)
dN (v)(u)xu 6
(
d(v) + 2 e(N (v))
)
xu∗ =
(
e(N [v]) + e(N (v))
)
xu∗ .
Note that S = N (v) ∪ (Lα \ { v}). Then e(N 2(v) ∩ Lα, N (v)) 6 e(S) − e(N (v)) and
∑
u∈N2(v)
dN (v)(u)xu = ∑
u∈N2(v)∩Lα
dN (v)(u)xu + ∑
u∈N2(v)\Lα
dN (v)(u)xu
6
(
e(S) − e(N (v)) + α · e(G)
)
xu∗ .
Combining the above two inequalities, we obtain
ρ2xv = d(v)xv + ∑
u∈N(v)
dN (v)(u)xu + ∑
u∈N2(v)
dN (v)(u)xu
6
(
e(N [v]) + e(S) + α · e(G)
)
xu∗ .
6
(
(2 k − 1) d(v) + 3α
2
kn − k2
)
xu∗ .
Notice that 53 α 6 β < 1 and ρ2xv > k(n − k)βx u∗ > (βkn − k2)xu∗ for each vertex v ∈ Lβ .In view of the above inequality, we have ( β − 32 α)kn < (2 k − 1) d(v), which yields that
d(v) > k
2k−1
(β − 32 α)n > α
12
n + k for each vertex v ∈ Lβ .By Theorem 1, we also have e(Lβ ) 6 k|Lβ |. Observe that ∑
u∈V(G)\Lβ
d(u) > e(Lβ , V (G)\
Lβ ) = ∑
v∈Lβ
d(v) − 2e(Lβ ). Therefore, 2e(G) = ∑
v∈Lβ
d(v) + ∑
u∈V(G)\Lβ
d(u) > 2 ∑
v∈Lβ
d(v) − 2e(Lβ ) > |Lβ |α
6
n.
Combining e(G) 6 kn , we obtain |Lβ | < 12 kα . This completes the proof.
Lemma 15. d(v) > (γ − 16k
)n for each v ∈ Lγ .Proof. Suppose to the contrary that there exists a vertex v0 ∈ Lγ with d(v0) 6 (γ − 16k )n.We may assume that xv0 = γ0xu∗ . By the definition of Lγ , we know that 12k 6 γ 6 γ0 6 1, and thus d(v0) 6 (γ0 − 16k )n.
the electronic journal of combinatorics 32(2) (2025), #P2.33 9
Set R = N (v0) ∪ N 2(v0). Then xv 6 βx u∗ for each v ∈ R \ Lβ . Therefore,
ρ2xv0 = d(v0)xv0 + ∑
v∈R
dN (v0)(v)xv
= d(v0)xv0 + ∑
v∈R\Lβ
dN (v0)(v)xv + ∑
v∈R∩Lβ
dN (v0)(v)xv
6
(
γ0d(v0) + β ∑
v∈R\Lβ
dN (v0)(v) + ∑
v∈R∩Lβ
dN (v0)(v)
)
xu∗ .
(8) Since N (v0) ⊆ R, we can see that
∑
v∈R\Lβ
dN (v0)(v) 6 ∑
v∈R
dR(v) = 2 e(R) 6 2e(G) 6 2kn. (9) Observe that R ∩ Lβ ⊆ Lβ \ { v0}. We also have
∑
v∈R∩Lβ
dN (v0)(v) 6 ∑
v∈Lβ{ v0}
dN (v0)∩Lβ (v) + ∑
v∈Lβ{ v0}
dN (v0)\Lβ (v)
6 2e(Lβ
) + e(Lβ , N (v0) \ Lβ
) − | N (v0) \ Lβ |.
(10) Furthermore, e(Lβ , N (v0) \ Lβ
) 6 e(Lβ ∪ N (v0)) − e(Lβ
). Notice that e(Lβ ) 6 k|Lβ | and
e(Lβ ∪ N (v0)) 6 k(|Lβ | + d(v0)). Combining (10), we obtain
∑
v∈R∩Lβ
dN (v0)(v) 6 e(Lβ ∪ N (v0)) − | N (v0) \ Lβ | + e(Lβ )
6 (k − 1) d(v0) + ( k + 1) |Lβ | + e(Lβ )
6 (k − 1) d(v0) + (2 k + 1) |Lβ |.
(11) Substituting (9) and (11) into (8), we get that
ρ2xv0 6
(
γ0d(v0) + 2 kβn + ( k − 1) d(v0) + (2 k + 1) |Lβ |
)
xu∗
=
(
(γ0 + k − 1) d(v0) + 2 kβn + (2 k + 1) |Lβ |
)
xu∗ .
(12) Since n > 18 kα2 and α < 124 k2 , we have 12 kα 6 23 αn < n
(6 k)2
. Moreover, by Lemma 14, we have |Lβ | < 12 kα . Thus, we can check that (2 k + 1) |Lβ | < n
6k
− k2 6 n
6k
− k2γ0. Recall that
ρ2xv0 > k(n − k)γ0xu∗ and d(v0) 6 (γ0 − 16k )n. Combining (12), we obtain that
k(n − k)γ0 < (γ0 + k − 1)( γ0 − 16k
)n + 2 kβn + n
6k
− k2γ0,
which gives kγ 0 < (γ0 + k − 1)( γ0 − 16k ) + 2 kβ + 16k . Recall that β 6 16k2 . It follows that
(γ0 − 1)( γ0 − 16k
) > k−16k − 2kβ > k−36k > 0. (13) Now let f (γ) = ( γ − 1)( γ − 16k ), where 12k 6 γ 6 1. Obviously, f (γ)|max = f (1) = 0, which contradicts (13). The proof is completed.
the electronic journal of combinatorics 32(2) (2025), #P2.33 10
Recall that Lγ = {u ∈ V (G) | xu > γx u∗ }, where 12k 6 γ 6 1. Let γ0 := 12k . Clearly,
L1−γ0 ⊆ Lγ0 . We will see that every vertex u ∈ Lγ0 has a larger value xu.
Lemma 16. Lγ0 = L1−γ0 .Proof. Suppose to the contrary that there exists a vertex u0 ∈ Lγ0 \ L1−γ0 . Assume that
xu0 = γx u∗ . Then γ0 6 γ < 1 − γ0. Set R = N [u∗] ∪ N 2(u∗). Then we have
ρ2xu∗ = ∑
u∈R
dN (u∗)(u)xu = ∑
u∈R\Lβ
dN (u∗)(u)xu + ∑
u∈R∩Lβ
dN (u∗)(u)xu. (14) Recall that e(G) 6 kn and xu 6 βx u∗ for each u ∈ R \ Lβ . Then
∑
u∈R\Lβ
dN (u∗)(u)xu 6 ∑
u∈R
dR(u)βx u∗ 6 2e(G)βx u∗ 6 2βknx u∗ . (15) On the other hand, since u0 ∈ Lγ0 and Lγ0 ⊆ Lβ , we have u0 ∈ Lβ , and thus
∑
u∈R∩Lβ
dN (u∗)(u)xu 6 ∑
u∈Lβ
dN (u∗)(u)xu∗ + dN (u∗)(u0)( xu0 − xu∗ ), (16) where xu0 − xu∗ = ( γ − 1) xu∗ and
∑
u∈Lβ
dN (u∗)(u) = ∑
u∈Lβ
dN (u∗)\Lβ (u) + ∑
u∈Lβ
dN (u∗)∩Lβ (u)
6 e(Lβ , N (u∗) \ Lβ ) + 2 e(Lβ )
6 e(G) + e(Lβ ).
Recall that e(G) 6 k(n − k) and e(Lβ ) 6 k|Lβ | < 12
α
k2. Consequently, ∑
u∈Lβ
dN (u∗)(u) 6
k(n − k) + 12
α
k2. Combining (14)-(16), we obtain
ρ2xu∗ 6
(
2βkn + k(n − k) + 12
α
k2 + ( γ − 1) dN (u∗)(u0)
)
xu∗ . (17) By Lemma 15, we have d(u∗) > (1 − 16k )n and d(u0) > (γ− 16k )n. Thus, |V (G)\N (u∗)| 6 n
6k
and dN (u∗)(u0) > (γ − 13k )n. Notice that ρ2 > k(n − k). It follows from (17) that
(γ − 1)( γ − 13k
)n > −(2βkn + 12
α
k2).
Recall that α 6 124 k(k+1) , β > 53 α and n > 18 kα2 . Then 12
α
k2 6 23 αkn. Now choose β = 53 α.Then we have 2 βkn + 12
α
k2 6 4αkn , and hence ( γ − 1)( γ − 13k ) > −4αk > − 16( k+1) . Let
f (γ) = ( γ −1)( γ − 13k ), where γ0 6 γ 6 1−γ0 and γ0 = 12k . Obviously, f (γ)|max = f (γ0) =
−2k−112 k2 < − 16( k+1) for k > 3, a contradiction. With the above lemmas in hand, we now provide the proof of Theorem 4.
the electronic journal of combinatorics 32(2) (2025), #P2.33 11
Proof. Choose L = Lγ0 in Theorem 4. Given an arbitrary vertex v ∈ L. By Lemma 16, we have v ∈ L1−γ0 , and thus xv > (1 − γ0)xu∗ = (1 − 12k )xu∗ . Furthermore, by Lemma 15 we have d(v) > (1 − γ0 − 16k )n = (1 − 23k )n.In the following, it remains to show that |L| = k. Firstly, suppose that |L| > k + 1. Taking v1, v 2, . . . , v k+1 from L, we have
∣∣ k+1 ⋂
i=1
N (vi)∣∣ > k+1 ∑
i=1
∣∣N (vi)∣∣ − k∣∣ k+1 ⋃
i=1
N (vi)∣∣ > (k + 1 )( 1 − 23k
)n − kn = k−23k n > k + 1 .
Thus, G contains a copy of Kk+1 ,k +1 , which is clearly a ( k +1)-(edge)-connected subgraph. However, by Lemma 7, every k-(edge)-connected subgraph of G is minimally k-(edge)-connected, which implies that G contains no any ( k + 1)-(edge)-connected subgraph. We get a contradiction. Therefore, |L| 6 k.Finally, suppose that |L| 6 k − 1. Since L = Lγ0 , we have xv < γ 0xu∗ = 12k xu∗ for every v ∈ V (G) \ L. Setting R = N [u∗] ∪ N 2(u∗), we have
ρ2xu∗ = ∑
u∈R
dN (u∗)(u)xu 6
( ∑
u∈R∩L
dN (u∗)(u) + 12k
∑
u∈R\L
dN (u∗)(u)
)
xu∗ , (18) Let E0 be the set of edges incident to vertices of L. Then, every edge in E0 can not be counted twice in ∑
u∈R\L
dN (u∗)(u). Moreover, it is easy to see that u∗ ∈ L and every edge incident to u∗ can not be counted in ∑
u∈R\L
dN (u∗)(u). Consequently, ∑
u∈R\L
dN (u∗)(u) 6
2e(G) − d(u∗) − | E0|. Note that e(G) 6 kn and
d(u∗) + |E0| = d(u∗) + ∑
v∈L
d(v) − e(L) > (|L| + 1 )( 1 − 23k
)n − 12 k2 > |L|n.
It follows that ∑
u∈R\L
dN (u∗)(u) 6 (2 k − | L|)n. Observe that ∑
u∈R∩L
dN (u∗)(u) 6 |L|n.Combining (18) and |L| 6 k − 1, we obtain
ρ2 6 |L|n + 12k
(2k − | L|)n 6 (k − 1) n + (k+1) 2k n = kn − k−12k n,
which contradicts ρ2 > k(n − k). Therefore, |L| = k. This completes the proof. At the end of this section, we give the proof of Theorem 5.
Proof. Let G∗ be a graph that has the maximal spectral radius among all minimally k-(edge)-connected graphs of order n, where n > 18 kα2 and α = 124 k(k+1) . Since Kk,n −k is also minimally k-(edge)-connected, we have ρ2(G∗) > ρ2(Kk,n −k) = k(n − k). Furthermore, by Theorem 4, G∗ contains a k-vertex subset L such that xv > (1 − 12k )xu∗ and d(v) > (1 − 23k )n
for each vertex v ∈ L, where L = L 12k
.
Denote by V the common neighbourhood of vertices in L, and let U = V (G∗)(L∪V ). Since |L| = k and every vertex in L has at most 23k n non-neighbors, we can see that
|L ∪ V | > n − k · 23k n = n
3
12 k(k + 5) .
the electronic journal of combinatorics 32(2) (2025), #P2.33 12
The key point is to show that U = ∅. Suppose to the contrary that |U | = t 6 = 0. By Theorem 1, we have e(G∗) 6 k(n − k) = k(|V | + |U |). Now, define G0 = G∗ and U0 = U . Moreover, let E0 be the subset of E(G0) in which every edge is incident to at least one vertex from U0. Then |E0| 6 e(G∗)−e(L, V ) 6 k|U0|,as e(L, V ) = |L|| V | = k|V |. It follows that ∑
u∈U0
dG0 (u) 6 2|E0| 6 2k|U0|, which implies that there exists a vertex u0 ∈ U0 such that dG0 (u0) 6 2k.Then, let G1 = G0 −{ u0}, U1 = U0 { u0} and E1 be the subset of E(G1) in which every edge is incident to some vertices from U1. Similarly as above, we have e(G1) 6 k(|V |+|U1|)and |E1| 6 e(G1) − e(L, V ) 6 k|U1|. Thus, we can find a vertex u1 ∈ U1 such that
dG1 (u1) 6 2k. Consequently, we can obtain a vertex ordering u0, u 1, . . . , u t−1 such that
Gi = Gi−1 − { ui−1}, Ui = Ui−1 \ { ui−1} and dGi (ui) 6 2k for each i ∈ { 1, . . . , t − 1}. For simplicity, we denote dL(ui) = di and dGi−L(ui) = d′
i
. Then di 6 k − 1 by the definition of U , and di + d′
i
= dGi (ui) 6 2k for i ∈ { 0, . . . , t − 1}.We shall construct a new graph G from G∗ in the following way. For each vertex ui
(0 6 i 6 t − 1), we delete all d′
i
edges from ui to V (Gi − L), and then add all possible
k − di edges from ui to L. Denote N L(ui) = L \ NL(ui). Then, we can see that
ρ(G) − ρ(G∗) > ∑
uv ∈E(G)
xuxv − ∑
uv ∈E(G∗)
xuxv = t−1∑
i=0
xui
( ∑
v∈NL(ui)
xv − ∑
v∈NGi−L(ui)
xv
)
. (19) Recall that xv > (1 − 12k
)xu∗ for each v ∈ L. Moreover, since we choose L = L 12k
, it is obvious that xv < 12k xu∗ for each v / ∈ L. In view of (19), we obtain
ρ(G) − ρ(G∗) > t−1∑
i=0
xui xu∗
(
(k − di)(1 − 12k ) − d′
i
· 12k
)
.
Recall that di + d′
i
6 2k and di 6 k − 1 for each i ∈ { 0, . . . , t − 1}. Thus, (k − di)(1 − 12k ) − d′
i
· 12k > (k − di)(1 − 12k ) − (2 k − di) 12k > 1 − k+2 2k > 0.
It follows that ρ(G) > ρ (G∗).
Observe that NG(ui) = L for each ui ∈ U . We will further see that G ∼= Kk,n −k.Indeed, otherwise, G Kk,n −k, then either eG(L) 6 = 0 or eG(V ) 6 = 0. However, G∗[L ∪ V ]contains a spanning subgraph K|L|,|V |, where |L| = k and |L ∪ V | > n
3
. Hence, G∗[L ∪ V ]is clearly k-(edge)-connected. By Lemma 7, G∗[L ∪ V ] is minimally k-(edge)-connected, which implies that G∗[L ∪ V ] ∼= K|L|,|V |. Since G[L ∪ V ] = G∗[L ∪ V ], we have G[L ∪ V ] ∼=
K|L|,|V |, and thus eG(L) = eG(V ) = 0, a contradiction. Hence, G ∼= Kk,n −k. But now, the inequality ρ(G∗) < ρ (G) contradicts the assumption that G∗ has maximal spectral radius. Therefore, U = ∅ and G∗ ∼= Kk,n −k. This completes the proof.
Acknowledgments
The authors would like to express their sincere gratitude to the anonymous referees for their careful reading and invaluable suggestions. The authors were supported by National Natural Science Foundation of China (Nos. 12271362 and 12171154).
the electronic journal of combinatorics 32(2) (2025), #P2.33 13
References
A.R. Berg, T. Jord´ an, Minimally k-edge-connected directed graphs of maximal size,
Graphs Combin. 21(1) (2005) 39–50. B. Bollob´ a, Extremal graph theory, Dover publications, New York, 1978. B. Bollob´ as, D.L. Goldsmith, D.R. Woodall, Indestructive deletions of edges from graphs, J. Combin. Theory Ser. B 30(3) (1981) 263–275. M.C. Cai, Minimally k-connected graphs of low order and maximal size, Discrete Math. 41 (1982) 229–234. M.C. Cai, A remark on the number of vertices of degree k in a minimally k-edge-connected graph, Discrete Math. 104(2) (1992) 221–226. M.C. Cai, The number of vertices of degree k in a minimally k-edge-connected graph,
J. Combin. Theory Ser. B 58(2) (1993) 225–239. X.D. Chen, L.T. Guo, On minimally 2-(edge)-connected graphs with extremal spec-tral radius, Discrete Math. 342(6) (2019) 2092–2099. S. Cioab˘ a, D.N. Desai, M. Tait, A spectral Erd˝ os–S´ os theorem, SIAM J. Discrete Math. 37(3) (2023) 2228–2239. S. Cioab˘ a, L.H. Feng, M. Tait, X.D. Zhang, The maximum spectral radius of graphs without friendship subgraphs, Electron. J. Combin. 27(4) (2020), #P4.22. D. Cvetkovi´ c, P. Rowlinson, S. Simi´ c, An introduction to the theory of graph spectra, Cambrige University Press, Cambridge, 2010. M. Dalmazzo, Nombre d ′arcs dans les graphes k-arc-fortement connexes minimaux,
C. R. Acad. Sci. Paris S´ er. A-B 285 (1977), no. 5, A341–A344. G.A. Dirac, Minimally 2-connected graphs, J. Reine Angew. Math. 228 (1967) 204– 216. D.D. Fan, S. Goryainov, H.Q. Lin, On the (signless Laplacian) spectral radius of minimally k-(edge)-connected graphs for small k, Discrete Appl. Math. 305 (2021) 154–163. D.D. Fan, X.F Gu, H.Q. Lin, Spectral radius and edge-disjoint spanning trees, J. Graph Theory 104 (2023) 697–711. Y.T. Li, Y.J. Peng, The spectral radius of graphs with no intersecting odd cycles,
Discrete Math. 345 (8) (2022), Paper No. 112907, 16 pp. D.R. Lick, Minimally n-line-connected graphs, J. Reine Angew. Math. 252 (1972) 178–182. W. Mader, Minimale n-fach kantenzusammenh¨ angende Graphen, Math. Ann. 191 (1971) 21–28. W. Mader, ¨Uber minimal n-fach zusammenh¨ angende, unendliche Graphen und ein Extremal problem, Archiv. Math. (Basel) 23 (1972) 553–560. W. Mader, Kantendisjunkte Wege in Graphen, Monatsh. Math. 78 (1974) 395–404.
the electronic journal of combinatorics 32(2) (2025), #P2.33 14
W. Mader, A reduction method for edge-connectivity in graphs, Ann. Discrete Math.
3 (1978) 145–164. W. Mader, On vertices of degree n in minimally n-edge-connected graph, Combin. Probab. Comput. 4 (1995) 81–95. Z.Y. Ni, J. Wang, L.Y. Kang, Spectral extremal graphs for disjoint cliques, Electron. J. Combin. 30(1) (2023), #P1.20. V. Nikiforov, Spectral saturation: inverting the spectral Tur´ an theorem, Electron J. Combin. 16(1) (2009), #R33. W.J. Ning, M. Lu, K. Wang, Maximizing the spectral radius of graphs with fixed minimum degree and edge connectivity, Linear Algebra Appl. 540 (2018) 138–148. M. Plummer, On minimal blocks, Trans. Amer. Math. Soc. 134 (1968) 85–94. P.J. Slater, A classification of 4-connected graphs, J. Combin. Theory Ser. B 17 (1974) 281–298. W.T. Tutte, A theory of 3-connected graphs, Indag. Math. 23 (1961) 441–455. J. Wang, L.Y. Kang, Y.S. Xue, On a conjecture of spectral extremal problems, J. Combin. Theory Ser. B 159 (2023) 20–41. M.L. Ye, Y.Z. Fan, H.F. Wang, Maximizing signless Laplacian or adjacency spectral radius of graphs subject to fixed connectivity, Linear Algebra Appl. 433 (2010) 1180– 1186. M.Q. Zhai, R.F. Liu, J. Xue, A unique characterization of spectral extrema for friendship graphs, Electron. J. Combin. 29 (2022), #P3.32. M.Q. Zhai, H.Q. Lin, Spectral extrema of graphs: forbidden hexagon, Discrete Math.
343 (10) (2020) 112028, 6 pp. M.Q. Zhai, H.Q. Lin, Spectral extrema of Ks,t -minor free graphs–on a conjecture of M. Tait, J. Combin. Theory Ser. B 157 (2022) 184–215. |
188598 | https://www.geeksforgeeks.org/machine-learning/dirichlet-distribution/ | Dirichlet Distribution
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23 Jul, 2025
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The Dirichlet distribution is a multivariate extension of the Beta distribution and is extensively applied in Bayesian statistics and machine learning. It is used to model categorical data, proportions, and probabilities and acts as a conjugate prior for multinomial distributions. Some of its important properties are its use in Bayesian inference and parameter estimation simplicity.
The Dirichlet distribution is a multivariate continuous family of probability distributions parameterized by a vector α (alpha), with each αᵢ > 0. It is a probability simplex, i.e., the components of a Dirichlet-distributed random variable add up to 1.
Mathematically
if X = (X₁, X₂,., Xₖ) is Dirichlet distributed with parameter vector α = (α₁, α₂,., αₖ), then its probability density function (PDF) is:
f(X1,X2,...,Xk;α1,α2,...,αk)=B(α)1∏i=1kXiαi−1
where X₁, X₂, ..., Xₖ are non-negative and satisfy:
∑i=1kXi=1
The normalization constant B(α) (also called the multivariate Beta function) ensures that the probability density integrates to 1 and is defined as:
B(α)=Γ(∑i=1kαi)∏i=1kΓ(αi)
where Γ(αᵢ) is the Gamma function, which generalizes the factorial function.
Properties of Dirichlet Distribution
1. Expectation (Mean)
The mean of each component Xᵢ in a Dirichlet-distributed random variable is given by:
E[Xi]=∑j=1kαjαi
This shows that each expected proportion depends on its corresponding parameter αᵢ relative to the sum of all α parameters.
2. Variance and Covariance
The variance of each component is:
Var(Xi)=(∑j=1kαj)2(∑j=1kαj+1)αi(∑j=1kαj−αi)
The covariance between two components Xᵢ and Xⱼ (for i ≠ j) is:
Cov(Xi,Xj)=−(∑j=1kαj)2(∑j=1kαj+1)αiαj
This negative covariance indicates that an increase in one component leads to a decrease in others due to the constraint that their sum is 1.
3. Relationship with Beta Distribution
The Dirichlet distribution is a generalization of the Beta distribution. Specifically, for k = 2, the Dirichlet distribution reduces to a Beta distribution:
Dirichlet(α1,α2)∼Beta(α1,α2)
Practical Example
We are a market analyst estimating the proportions of customer preferences for three brands. We believe that the prior probabilities for the market share are approximately:
Brand A: 50%
Brand B: 30%
Brand C: 20%
These prior beliefs can be modeled using a Dirichlet distribution with the parameter vector:
α = ( 5, 3, 2)
Here:
𝛼1 = 5 corresponds to Brand A
𝛼2 = 3 corresponds to Brand B
𝛼3 = 2 corresponds to Brand C
The values of 𝛼 reflect the concentration of beliefs:
Larger values → more confident priors
Smaller values → more uncertain priors
1. Dirichlet Distribution PDF
The probability density function (PDF) for the three-brand scenario is given by:
f(X1,X2,X3;α1,α2,α3)=B(α)1X1α1−1X2α2−1X3α3−1
Where: X1+X2+X3=1
The normalization constant B(α) is the multivariate Beta function:
B(5,3,2)=Γ(5+3+2)Γ(5)Γ(3)Γ(2)
Using the Gamma function values:
B(5,3,2)=4032024×2×1=0.000297
2. Sampling from the Distribution
If we sample from this Dirichlet distribution, we get probability vectors representing possible market share proportions.
For example, some sampled vectors could be:
(0.52,0.28,0.20)→ Brand A dominates
(0.48, 0.32, 0.20) → Brand B has increased slightly
(0.50,0.30,0.20)→ Even split according to prior belief
These vectors indicate the probabilistic nature of the Dirichlet distribution, where the proportions vary but still sum to 1.
Practical Implementation in Python
The scipy.stats library provides functions to work with the Dirichlet distribution.
Sampling from a Dirichlet Distribution
C++
````
import numpy as np
from scipy.stats import dirichlet
Define Dirichlet parameters
alpha = [2, 3, 5] # Example parameters
Generate 5 samples
samples = dirichlet.rvs(alpha, size=5)
print(samples)
````
import numpy as np import numpy as np
from scipy.stats import dirichlet from scipy stats import dirichlet
```
Define Dirichlet parameters # Define Dirichlet parameters
```
alpha = [2, 3, 5] # Example parameters alpha = 2 3 5 # Example parameters
```
Generate 5 samples # Generate 5 samples
```
samples = dirichlet.rvs(alpha, size=5) samples = dirichlet rvs alpha size = 5
print(samples) print samples
Output
Plotting a 2D Dirichlet Distribution
For visualization, we can plot the probability simplex:
Python
````
import matplotlib.pyplot as plt
Generate random samples
samples = dirichlet.rvs([2, 3, 5], size=500)
Plot in 2D simplex
plt.scatter(samples[:, 0], samples[:, 1], alpha=0.5)
plt.xlabel("X1")
plt.ylabel("X2")
plt.title("Dirichlet Distribution Samples")
plt.show()
````
import matplotlib.pyplot as plt import matplotlib pyplot as plt
```
Generate random samples
Generate random samples
```
samples = dirichlet.rvs([2, 3, 5], size=500) samples = dirichlet rvs 2 3 5 size = 500
```
Plot in 2D simplex
Plot in 2D simplex
```
plt.scatter(samples[:, 0], samples[:, 1], alpha=0.5) plt scatter samples 0 samples 1 alpha =0.5
plt.xlabel("X1") plt xlabel "X1"
plt.ylabel("X2") plt ylabel "X2"
plt.title("Dirichlet Distribution Samples") plt title "Dirichlet Distribution Samples"
plt.show() plt show
This provides insight into how probability distributions are generated from the Dirichlet prior.
Output:
Parameter Estimation for Dirichlet Distribution
Having a sample of observed probability vectors X₁, X₂,., Xₙ, we would like to estimate the Dirichlet distribution parameters α.
Method of Moments Estimation
The method of moments estimates α based on sample means:
α^i=EXi
This approach is simple but less accurate than maximum likelihood estimation.
Maximum Likelihood Estimation (MLE)
The MLE approach finds α by maximizing the likelihood function:
L(α)=∏n=1Nf(X(n);α)
However, this requires numerical optimization techniques, such as Newton’s method or the fixed-point iteration method, because there is no closed-form solution.
Applications of Dirichlet Distribution
Bayesian Inference: Used as a conjugate prior for the multinomial distribution, ensuring that both prior and posterior distributions remain Dirichlet in Bayesian modeling.
Topic Modeling (LDA): Applied in Latent Dirichlet Allocation (LDA) to model topic distributions in documents and word distributions within topics, making it essential in natural language processing.
Mixture Models & Clustering: Utilized in Dirichlet Process Mixture Models (DPMMs), extending Gaussian Mixture Models (GMMs) to handle an unknown number of clusters.
Probability Estimation: Helps model uncertainty over categorical probabilities in applications like genetics, polls, and marketing
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Vandita Gupta
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188599 | https://betaprojects.com/solutions/project-euler/project-euler-problem-092-solution | Project Euler Problem 92 Solution | {Beta Projects}
Project Euler Problem 92 Solution
Square Digit Chains
by {BetaProjects} | Project Euler & HackerRank
Difficulty: Easy
Project Euler Problem 92 Statement
A number chain is created by continuously adding the square of the digits in a number to form a new number until it has been seen before.
For example, 44→32→13→10→1→1 85→89→145→42→20→4→16→37→58→89\begin{align} &44 \to 32 \to 13 \to 10 \to \mathbf 1 \to \mathbf 1\ &85 \to \mathbf{89} \to 145 \to 42 \to 20 \to 4 \to 16 \to 37 \to 58 \to \mathbf{89} \end{align}44→32→13→10→1→1 85→89→145→42→20→4→16→37→58→89
Therefore any chain that arrives at 1 1 1 or 89 89 89 will become stuck in an endless loop. What is most amazing is that EVERY starting number will eventually arrive at 1 1 1 or 89 89 89.
How many starting numbers below ten million will arrive at 89 89 89?
Solution
Project Euler Problem 92 overview:
Determine how many starting numbers below 1 0 K 10^K 1 0 K (original problem uses K=7 K=7 K=7) will eventually arrive at 89 through the iterative process of replacing a number with the sum of the squares of its digits. Given the potentially large range of K K K (up to 200), the final count needs to be presented modulo 1 0 9+7 10^9+7 1 0 9+7
Preprocessing with Digit Square Sums:
Maximum Possible Sum Calculation:
For any number with K K K digits, the maximum possible sum of the squares of its digits is 81×K 81 \times K 81×K (since 9 2=81 9^2 = 81 9 2=81). The script initializes a range C[] from 0 to 81 K+1 81K + 1 81 K+1 to cover all possible sums.
Square List Initialization:
The list S contains all possible single-digit squares: {0,1,4,9,16,25,36,49,64,81}{0, 1, 4, 9, 16, 25, 36, 49, 64, 81}{0,1,4,9,16,25,36,49,64,81}. This set is crucial for efficiently calculating possible sums when adding new digits.
The algorithm works in two main phases. In the first phase, it uses dynamic programming to count how many K−d i g i t K-digit K−d i g i t numbers produce each possible digit square sum. The array p p p starts by marking which single-digit square values are possible (0,1,4,9,16,25,36,49,64,81 0, 1, 4, 9, 16, 25, 36, 49, 64, 81 0,1,4,9,16,25,36,49,64,81). Then, for each additional digit position, it updates p p p to count how many ways you can form each sum by adding another digit. The line
p = [sum(p[j-s] for s in S if j>=s) % M for j in C] essentially says: "the number of ways to get sum j j j with k k k digits equals the sum of ways to get (j j j minus each possible digit square) with k−1 k-1 k−1 digits." This builds up the complete distribution of digit square sums for K−d i g i t K-digit K−d i g i t numbers.
In the second phase, the solution determines which digit square sums eventually lead to 89 89 89. The recursive lambda function d d d checks whether a given number converges to 89 89 89 by repeatedly computing the sum of its digits' squares. Once the algorithm knows both how many K−d i g i t K-digit K−d i g i t numbers map to each sum and which sums lead to 89 89 89, it simply adds up the counts for all sums that converge to 89 89 89. The modulo operation with M=1 0 9+7 M=10^9+7 M=1 0 9+7 keeps the numbers manageable.
HackerRank version
Solves all test cases for Hackerrank Project Euler Problem 92
HackerRank Project Euler Problem 92: Changes the upper limit from 1 0 7 10^7 1 0 7 to 1 0 K 10^K 1 0 K, where 1≤K≤200 1\le K\le 200 1≤K≤200.
Python Source Code
```php
K = int(input())
M, C, S = 109+7, range(81K+1), [0,1,4,9,16,25,36,49,64,81]
d = lambda n: n>1 and (n==89 or d(sum(S[int(c)] for c in str(n))))
p = [i in S for i in C]
for _ in range(K-1):
p = [sum(p[j-s] for s in S if j>=s) % M for j in C]
print(sum(v for i,v in enumerate(p) if i and d(i)) % M)
```
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Next » solution Project Euler Problem 93: Arithmetic Expressions
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