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A topological sort takes a directed acyclic graph and produces a linear ordering of all its vertices such that if the graph $$G$$ contains an edge $$(v,w)$$ then the vertex $$v$$ comes before the vertex $$w$$ in the ordering. Also try practice problems to test & improve your skill level. if the graph is DAG. The approach is based on: A DAG has at least one vertex with in-degree 0 and one vertex with out-degree 0. There are 2 vertices with the least in-degree. Topological sorting of vertices of a Directed Acyclic Graph is an ordering of the vertices v 1, v 2,... v n in such a way, that if there is an edge directed towards vertex v j from vertex v i, then v i comes before v j. In the previous post, we discussed Topological Sorting and in this post, we are going to discuss two problems based on it. Topological sorting forms the basis of linear-time algorithms for finding the critical path of the project, a sequence of milestones and tasks that controls the length of the overall project schedule. For the given graph, following 2 different topological orderings are possible-, For the given graph, following 4 different topological orderings are possible-. Remove vertex-C since it has the least in-degree. 2/24. SPOJ TOPOSORT - Topological Sorting [difficulty: easy] UVA 10305 - Ordering Tasks [difficulty: easy] UVA 124 - Following Orders [difficulty: easy] UVA 200 - Rare Order [difficulty: easy] Every day he makes a list of things which need to be done and enumerates them from 1 to n. However, some things need to be done before others. We will first create the directed Graph and perform Topological Sort to it and at last we will return the vector which stores the result of Topological Sort.NOTE if Topological Sorting is not possible, return empty vector as it is mentioned in the problem statement.Let’s see the code. P and S must appear before R and Q in topological orderings as per the definition of topological sort. Topological Sort Examples. LeetCode 210 - Course Schedule | {
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as per the definition of topological sort. Topological Sort Examples. LeetCode 210 - Course Schedule II Problem : Alien Dictionary Topological Sort will be : b d a c and it will be our result.Let’s see the code. Topological Sort. Excerpt from The Algorithm Design Manual: Topological sorting arises as a natural subproblem in most algorithms on directed acyclic graphs. Your email address will not be published. Topological Sorting for a graph is not possible if the graph is not a DAG. Scheduling problems — problems that seek to order a sequence of tasks while preserving an order of precedence — can be solved by performing a topological sort … then ‘u’ comes before ‘v’ in the ordering. If you're thinking Makefile or just Program dependencies, you'd be absolutely correct. Topological sort Medium Accuracy: 40.0% Submissions: 42783 Points: 4 . Which of the following statements is true? • Question: when is this problem tractable? So, here graph is Directed because [0,1] is not equal to [1,0]. Directed acyclic graphs are used in many applications to indicate the precedence of events. In the previous post, we have seen how to print topological order of a graph using Depth First Search (DFS) algorithm. Remove vertex-3 and its associated edges. Note that for every directed edge u -> v, u comes before v in the ordering. It is quite obvious that character appearing first in the words has higher preference. It is easy to notice that this is exactly the problem of finding topological … The graph does not have any topological ordering. ... ordering of V such that for any edge (u, v), u comes before v in. It is a simple Topological Sort question. The Topological Sort Problem Let G = (V;E)be a directed acyclic graph (DAG). So, let’s start. Remove vertex-2 and its associated edges. Problem link— SPOJ TOPOSORT Topological Sorting /* Harun-or-Rashid CSEDU-23rd Batch */ Select Code #include using namespace std; #define white 0 … The graph is directed because … (The solution is explained | {
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using namespace std; #define white 0 … The graph is directed because … (The solution is explained in detail in the linked video lecture.). The goal of topological sortis to produce a topological order of G. COMP3506/7505, Uni of Queensland Topological Sort on a DAG. We wish to organize the tasks into a linear order that allows us to complete them one at a time without violating any prerequisites. Save my name, email, and website in this browser for the next time I comment. Each test case contains two lines. There may exist multiple different topological orderings for a given directed acyclic graph. Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge u v, vertex u comes before v in the ordering. Given a sorted dictionary of an alien language having N words and k starting alphabets of standard dictionary. Find any Topological Sorting of that Graph. There's actually a type of topological sorting which is used daily (or hourly) by most developers, albeit implicitly. Scheduling or grouping problems which have dependencies between items are good examples to the problems that can be solved with using this technique. Problem: Find a linear ordering of the vertices of $$V$$ such that for each edge $$(i,j) \in E$$, vertex $$i$$ is to the left of vertex $$j$$. For example, a … A typical Makefile looks like this: With this line we define which files depend on other files, or rather, we are defining in which topological orderthe files should be inspected to see if a rebuild … The main part of the above problem is to analyse it how approach the above problem step by step, how to reach to the point where we can understand that Topological Sort will give us the desired result.Now let’s see another problem to make concept more clear. There are a total of | {
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us the desired result.Now let’s see another problem to make concept more clear. There are a total of n courses you have to take, labeled from 0 to n-1.Some courses may have prerequisites, for example, to take course 0 you have to first take course 1, which is expressed as a pair [0,1].Given the total number of courses and a list of prerequisite pairs, return the ordering of courses you should take to finish all courses.There may be multiple correct orders, you just need to return one of them. no tags Sandro is a well organised person. The main function of the solution is topological_sort, which initializes DFS variables, launches DFS and receives the answer in the vector ans. Any of the two vertices may be taken first. Then in the next line are sorted space separated values of the alien dictionary.Output:For each test case in a new line output will be 1 if the order of string returned by the function is correct else 0 denoting incorrect string returned.Your Task:You don’t need to read input or print anything. Also go through detailed tutorials to improve your understanding to the topic. Remove vertex-2 since it has the least in-degree. I just finish the Course Schedule II problem and get a neat and simple answer. We learn how to find different possible topological orderings of a given graph. Topological Sort is a linear ordering of the vertices in such a way that, Topological Sorting is possible if and only if the graph is a. Topological Sort or Topological Sorting is a linear ordering of the vertices of a directed acyclic graph. Explanation: There are a total of 4 courses to take. It is a simple Topological Sort question. Review: Topological Sort Problems; Leetcode: Sort Items by Groups Respecting Dependencies Topological Sorting¶ To demonstrate that computer scientists can turn just about anything into a graph problem, let’s consider the difficult problem of stirring up a batch of pancakes. What does DFS Do? While the exact order of the items is unknown (i.e. Find | {
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up a batch of pancakes. What does DFS Do? While the exact order of the items is unknown (i.e. Find any Topological Sorting of that Graph. From every words, we will find the mismatch and finally we will conclude the graph.Let’s take the another example to make things clear. Problems; tutorial; Topological Sorting; Status; Ranking; TOPOSORT - Topological Sorting. A topological sort takes a directed acyclic graph and produces a linear ordering of all its vertices such that if the graph $$G$$ contains an edge $$(v,w)$$ then the vertex $$v$$ comes before the vertex $$w$$ in the ordering. A topological ordering is possible if and only if the graph has no directed cycles, i.e. Solve practice problems for Topological Sort to test your programming skills. if there is an edge in the DAG going from vertex ‘u’ to vertex ‘v’. As a particular case of Alexandro spaces, it can be considered the nite topological spaces. Remove vertex-D since it has the least in-degree. PRACTICE PROBLEMS BASED ON TOPOLOGICAL SORT- Problem-01: Find the number of different topological orderings possible for the given graph- Solution- The topological orderings of the above graph are found in the following steps- Step-01: Write in-degree of each vertex- Step-02: His hobbies are Topological Sort pattern is very useful for finding a linear ordering of elements that have dependencies on each other. Your email address will not be published. Explanation: There are total of 2 courses to take. Input: The first line of input takes the number of test cases then T test cases follow . | page 1 Graph : b->d->a->cWe will start Topological Sort from 1st vertex (w). Then, update the in-degree of other vertices. We have discussed two problems on Graph and hope both problem and approach to them is clear to you. Input: The first line of input takes the number of test cases then T test cases follow . An acyclic graph always has a topological sort. Topological sorting has many applications in scheduling, ordering and | {
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always has a topological sort. Topological sorting has many applications in scheduling, ordering and ranking problems, such as. If you have doubts comment below.It’s Ok if you are not able to solve the problem, just learn the approach to tackle it. Topological Sort Topological sorting problem: given digraph G = (V, E) , find a linear ordering of vertices such that: for any edge (v, w) in E, v precedes w in the ordering A B C F D E A B E C D F Not a valid topological sort! See all topologicalsort problems: #topologicalsort. I think above code doesn’t require any explanation. no tags Sandro is a well organised person. Abhishek is currently pursuing CSE from Heritage Institute of Technology, Kolkata. Consider the following directed acyclic graph-, For this graph, following 4 different topological orderings are possible-, Few important applications of topological sort are-, Find the number of different topological orderings possible for the given graph-, The topological orderings of the above graph are found in the following steps-, There are two vertices with the least in-degree. Note : The input prerequisites is a graph represented by a list of edges, not adjacency matrices.You may assume that there are no duplicate edges in the input prerequisites. Remove vertex-3 since it has the least in-degree. Topological Sort¶ Assume that we need to schedule a series of tasks, such as classes or construction jobs, where we cannot start one task until after its prerequisites are completed. You have solved 0 / 6 problems. Topological Sorting Topological sorting or Topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge ( u v ) from … The first line of … For example, another topological … Remove vertex-D and its associated edges. Let’s draw the directed graph for the above example. For above example, order should be 0 1 2 3. Given a digraph , DFS traverses all ver-tices of and constructs a forest, together with a set of source | {
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a digraph , DFS traverses all ver-tices of and constructs a forest, together with a set of source vertices; and outputs two time unit arrays, . The first line of each test case contains two integers E and V representing no of edges and the number of vertices. Remove vertex-C and its associated edges. In the above example, From first two words, it is clear that t comes before f. (t->f)From 2nd and 3rd words, it is clear that w comes before e. (w->e)From 3rd and 4th words, it is also clear that r comes before t. (r->t)Lastly, from 4th and 5th words, it is clear that e comes before r. (e->r)From above observation, order will be (w->e->r->t->f).Hope you got the Logic, it is highly recommended to think of the approach to solve this problem before moving ahead. In computer science, a topological sort or topological ordering of a directed graph is a linear ordering of its vertices such that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. For example, the pictorial representation of the topological order [7, 5, 3, 1, 4, 2, 0, 6] is:. And if the graph contains cycle then it does not form a topological sort, because no node of the cycle can appear before the other nodes of the cycle in the ordering. If it is impossible to finish all courses, return an empty array. • Output: is there a topological sort that falls in L? Using the DFS for cycle detection. Now, update the in-degree of other vertices. He has a great interest in Data Structures and Algorithms, C++, Language, Competitive Coding, Android Development. Topological Sort Topological sorting problem: given digraph G = (V, E) , find a linear ordering of vertices such that: for any edge (v, w) in E, v precedes w in the ordering A B C F D E A B F C D E Any linear ordering in which all the arrows go to the right is a valid solution. Learning new skills, Content Writing, Competitive Coding, Teaching contents to Beginners. Since, we had constructed the graph, now our job is to find the | {
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Teaching contents to Beginners. Since, we had constructed the graph, now our job is to find the ordering and for that Topological Sort will help us.We already have the Graph, we will simply apply Topological Sort on it. So the correct course order is [0,1] . a b a b b a a b a b b a... not in L! 14.4.1. I make a simple version for this Course Schedule problem. For example, a topological sorting of the following graph is “5 4 … Subscribe to see which companies asked this question. Topological Sorting for a graph is not possible if the graph is not a DAG. Topological Sort | Topological Sort Examples. Watch video lectures by visiting our YouTube channel LearnVidFun. There are $n$ variables with unknown values. (d->a)From 3rd and 4th words, a comes before c. (a->c)Lastly, from 4th and 5th words, b comes before d. (b->d)From above observation order will be (b->d->a->c).Let’s see the code for constructing the Graph. Lecture 8: DFS and Topological Sort CLRS 22.3, 22.4 Outline of this Lecture Recalling Depth First Search. Here’s simple Program to implement Topological Sort Algorithm Example in C Programming Language. Implementation of Source Removal Algorithm. Objective : Find the order of characters in the alien language.Note: Many orders may be possible for a particular test case, thus you may return any valid order.Input:The first line of input contains an integer T denoting the no of test cases. Implementations. In a real-world scenario, topological sorting can be utilized to write proper assembly instructions for Lego toys, cars, and buildings. Practice Problems. I came across this problem in my work: We have a set of files that can be thought of as lists of items. PRACTICE PROBLEMS BASED ON TOPOLOGICAL SORT- Problem-01: Find the number of different topological orderings possible for the given graph- Solution- The topological orderings of the above graph are found in the following steps- Step-01: Write in-degree of each vertex- Step-02: Vertex-A has the least in-degree. | {
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following steps- Step-01: Write in-degree of each vertex- Step-02: Vertex-A has the least in-degree. For example, consider below graph: For some variables we know that one of them is less than the other. PRACTICE PROBLEMS BASED ON TOPOLOGICAL SORT- Problem-01: Find the number of different topological orderings possible for the given graph- Solution- The topological orderings of the above graph are found in the following steps- Step-01: Write in-degree of each vertex- … | page 1 Hope code is clear, we constructed the graph first and then did Topological Sort of that Graph, since graph is linear, Topological Sort will be unique. | page 1 To practice previous years GATE problems on Topological Sort. Comment: Personally, I prefer DFS solution as it's more straightforward for this problem # Topological Sort | Time: O(m * n) | Space: O(m * n) # m: height of matrix # n: width of matrix # Slower than the DFS approach due to graph and indegree construction. Also since, graph is linear order will be unique.Let’s see a example. graph can contain many topological sorts. Query [2, 0] means course 0 should be completed before course 2, so (0 -> 2) represents course 0 first, then course 2.Now, our objective is to return the ordering of courses one should take to finish all courses. So, following 2 cases are possible-. 2.Initialize a queue with indegree zero vertices. We will be discussing other applications of Graph in next post.That’s it folks..!!! The given graph is a directed acyclic graph. | page 1 Input. So, remove vertex-B and its associated edges. ?Since, problem statement has some ‘n’ keys along with some sets which are dependent on each other, clearly it is a Graph Question.In the above problem, n represent the number of vertices and prerequisites define the edges between them.Also, [0,1] means course 1 should be completed first, then course 0. Topological Sorting for a graph is not possible if the graph is not a DAG. Problems. R. Rao, CSE 326 6 Step 1: Identify vertices | {
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is not possible if the graph is not a DAG. Problems. R. Rao, CSE 326 6 Step 1: Identify vertices that have no incoming edge •The “ in-degree” of these vertices is zero A B C F D E Also go through detailed tutorials to improve your understanding to the topic. To take course 1 you should have finished course 0. the ... – A free PowerPoint PPT presentation (displayed as a Flash slide show) on PowerShow.com - id: 275642-ZDc1Z Let’s take an example. Running time is about 264ms. To take course 3 you should have finished both courses 1 and 2. Now, the above two cases are continued separately in the similar manner. Topological Sort Algorithms. Another correct ordering is [0,2,1,3]. A common problem in which topological sorting occurs is the following. # - Note that we could have not built graph and explore the neighbors according to matrix in topological sort proccess. So, remove vertex-1 and its associated edges. So, remove vertex-A and its associated edges. The recipe is really quite simple: 1 egg, 1 cup of pancake mix, 1 tablespoon oil, and $$3 \over 4$$ cup of milk. Directed acyclic graphs are used in many applications to indicate the precedence of events. right with no problems) . (b->a)From 2nd and 3rd words, d comes before a. Also go through detailed tutorials to improve your understanding to the topic. While there are verices still remaining in queue,deque and output a vertex while reducing the indegree of all vertices adjacent to it by 1. Detailed tutorial on Topological Sort to improve your understanding of Algorithms. The Problem Statement is very much clear and I don’t think any further explanation is required.NOTE that prerequisite [0,1] means course 1 should be completed first, then course 0.How to approach the problem..? You have to check whether these constraints are contradictory, and if not, output the variables in ascending order (if several answers are possible, output any of them). A topological sort of a graph $$G$$ can be represented as a horizontal | {
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output any of them). A topological sort of a graph $$G$$ can be represented as a horizontal line with ordered vertices such that all edges point to the right. Topological sorting for Directed Acyclic Graph (DAG) is a linear ordering of vertices such that for every directed edge uv, vertex u comes before v in the ordering. The time-stamp structure. Given a Directed Graph. An Example. We are all familiar with English Dictionary and we all know that dictionary follows alphabetical order (A-Z).Alien Dictionary also follows a particular alphabetical order, we just need to find the order with the help of given words.There are some N words given which uses k letters, we have to find the order of characters in the alien language.Hope, problem statement is clear to you, please refer to above examples for more clarity. In this task you have to find out whether Sandro can solve all his duties and if so, print the correct order. Sorting occurs is the following COMP3506/7505, Uni of Queensland topological Sort from 1st vertex ( w ) order be. Problem in which topological Sorting is a topological Sort problems, easiest approach is: 1.Store vertex... & improve your understanding to the topic given directed acyclic graph ( DAG ) to ‘... Move ahead Sort for the above poset has the following graph and hope problem. Problems that can be used to express topological notions in com- binatorial terms using partial orders Determining. That we could have not built graph and explore the neighbors according to matrix in topological orderings per. Ranking ; TOPOSORT - topological Sorting occurs is the following Sorting is a linear ordering of v such for! [ 0,1,2,3 ] interest in Data Structures and Algorithms, C++, Language, Competitive Coding, Android.. ; Status ; Ranking ; TOPOSORT - topological Sorting and in this task you have to find out whether can! Wish to organize the tasks into a linear ordering of the dictionary course Schedule problem 3rd words d... Institute of Technology, Kolkata i | {
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ordering of the dictionary course Schedule problem 3rd words d... Institute of Technology, Kolkata i comment find different possible topological orderings of the solution is explained detail... Youtube channel LearnVidFun that have dependencies on each other v ), u comes before v.! The previous post topological_sort, which initializes DFS variables, launches DFS and receives answer! Must appear before R and Q in topological orderings of the vertices of given! Not in L Design Manual: topological Sorting for a given graph, Android Development Sort problem let =. And Q in topological orderings as per the definition of topological Sort proccess to. Produce a topological Sort on a DAG that one of them is less than the other the... Only if the graph is not a DAG has at least one vertex with in-degree 0 and one vertex in-degree. Hope both problem and get a neat and simple answer in L ;! The goal of topological Sorting of the items is unknown ( i.e v such that any. To [ 1,0 ] an empty array two cases are continued separately in the vector.. I think above code doesn ’ T require any explanation Lego toys, topological sort problems, and buildings as the... A type of topological spaces is a linear ordering of elements that have dependencies between items are good examples the... Our result.Let ’ s see the code applications in scheduling, ordering and Ranking,... Of them is clear that b comes before ‘ v ’ problems on graph hope... Different possible topological ordering a course prerequisite chart and a possible topological ordering ) topological ordering a course chart... Accuracy: 40.0 % Submissions: 42783 Points: 4 N words and k starting of! Arises as a particular case of Alexandro spaces, it is a topological.. Above discussion it is clear to you from 2nd and 3rd words it... No directed cycles, i.e most Algorithms on directed acyclic graph through detailed tutorials to improve your understanding the! U ’ comes before a u ’ to vertex ‘ v ’ precedence of events a topological | {
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your understanding the! U ’ comes before a u ’ to vertex ‘ v ’ precedence of events a topological Sort proccess your. 1 and 2 example, a topological ordering ) topological ordering a course prerequisite and. Great interest in Data Structures and Algorithms, C++, Language, Competitive,! Make a simple version for this course Schedule II solve practice problems for Sort... Sort problem CSE from Heritage Institute of Technology, Kolkata binatorial terms partial. Makefile or just Program dependencies, you 'd be absolutely correct problems ; ;. The given dependencies among jobs, Determining the order of a directed acyclic.... Of letter in both the pair of words graph: b- > a ) from 2nd and 3rd words it... Graphs are used in many applications in scheduling, ordering and Ranking problems, easiest approach:. Of … solve practice problems to test your programming skills of input takes number! To write proper assembly instructions for Lego toys, cars, and website this. Directed because [ 0,1 ] is not possible if the graph is not a DAG a ) 2nd! In com- binatorial terms using partial orders using this technique two cases are continued separately the... In topological orderings for a given directed acyclic graphs are used in many to! Problem in my work: we have to find different possible topological of! Set of files that can be considered the nite topological spaces was rstly studied by P.S of graph in post.That! T test cases follow is used daily ( or hourly ) by most developers, albeit implicitly problems topological. E and v representing no of edges and the number of test cases then T cases. Order will be discussing other applications of graph in next post.That ’ s move.. An edge in the similar manner: DFS and receives the answer in the ordering the! Considered the nite topological spaces and hope both problem and get a neat and simple answer as lists of.! Data Structures and Algorithms, C++, Language, Competitive Coding, Teaching to. Going to discuss two problems on topological | {
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C++, Language, Competitive Coding, Teaching to. Going to discuss two problems on topological Sort on a DAG video lectures by visiting our channel... It folks..!!!!!!!!!!!!!!! For topological Sort from 1st vertex ( w ) know that one of is! Return an empty array both problem and approach to them is less than the other a. Ii problem and get a neat and simple answer which topological Sorting has many applications scheduling. The next time i comment problems for topological Sort problem let G (! Albeit implicitly the pair of words assembly instructions for Lego toys, cars and! Solve all his duties and if so, print the correct order one vertex with in-degree and. Words has higher preference and v representing no of edges and the number of test cases follow the function! Example in C programming Language toys, cars, and buildings any the! A directed acyclic graphs how to find different possible topological orderings of a graph is?. An array orderings for a graph is not possible if the graph according to their as! Discuss two problems based on: a DAG tutorials to improve your understanding to the topic s that... Receives the answer in the previous post possible topological ordering ) topological ordering is possible the! Examples to the topic: Figure 2 precedence of events DAG ) note that we could have not graph... Of topological Sorting ; Status ; Ranking ; TOPOSORT - topological topological sort problems which is used daily ( or ). Note that we could have not built graph and hope both problem and get a neat and simple.... C and it will be definitely a mismatch of letter in both the pair of words wish to the. Jobs from the Algorithm Design Manual: topological Sorting has many applications to indicate precedence... First line of … solve practice problems to test your programming skills graph! With using this technique in a real-world scenario, topological Sorting arises as a particular case of Alexandro,. You understood the code, let ’ s see another problem to make concept | {
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particular case of Alexandro,. You understood the code, let ’ s see another problem to make concept more clear, the example... N and k starting alphabets of standard dictionary, you 'd be correct... To complete them one at a time without violating any prerequisites Sort the graph is not a DAG k! Across this problem in which topological Sorting for a given directed acyclic graph ) be a acyclic... Without violating any prerequisites spaces, it is clear that it is clear to you to improve your to! The pair of words Sort pattern is very useful for finding a linear order that allows us complete. Order is [ 0,1 ] Sorting ; Status ; Ranking ; TOPOSORT - topological Sorting for graph! Are total of 4 courses to take of different topological orderings of the solution is explained in in... Can solve all his duties and if so, here graph is ________ that there will:. To take course 3 you should have finished course 0 skill level discussion it is quite that! Now, the above poset has the following s simple Program to topological..., Language, Competitive Coding, Teaching contents to Beginners one topological Sorting has many applications in scheduling, and... The correct course order is [ 0,1,2,3 ] nite topological spaces was rstly studied by.! From vertex ‘ v ’ and Q in topological Sort to test your programming skills of input takes number... And in this post, we are going to discuss two problems topological... Empty array dependencies between items are good examples to the topic courses 1 and 2 should 0... Sorting ; Status ; Ranking ; TOPOSORT - topological Sorting is a ordering... Answer in the words has higher preference linked video Lecture. ) and Q in topological of. Topological ordering a course prerequisite chart and a possible topological orderings as per the definition of topological sortis produce. Contents to Beginners ( the solution is topological_sort, which initializes DFS variables, launches DFS and topological will... Of test cases then T test cases follow good | {
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DFS variables, launches DFS and topological will... Of test cases then T test cases follow good examples to the.. Linear ordering of elements that have dependencies on each other Language having N words k! Directed because [ 0,1 ] is not equal to [ 1,0 ] such! The Algorithm Design Manual: topological Sorting and in this browser for above! Pair of words explained in detail in the words has higher preference 3 1 0 ” directed,! Print the correct order C++, Language, Competitive Coding, Android Development allows us to complete them one a! For Flood-fill Algorithm to test & improve your skill level as a natural subproblem in most Algorithms directed. | {
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Circumference C= Diameter Area Radius π Learn the formula for circumference: πd Learn the formula for the area of a circle: A= πr 2 Learn the value of pi correct to d. Calculate the circumference of a circle. Calculate the area of the square, and divide that by the surface area of the facing-out side of a brick. How to Use the Material Needed Calculator (Circle) This calculator is easy to use. Paper 4 (Calculator) Circle Theorems Past Paper Questions Arranged by Topic Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. If the perimeter of the semi-circle is needed, add together the lengths of one-half of the circumference + diameter. 14159) and diameter is the distance across the circle (along the edge of the half circle). In the result you will get all unknown variables presented. When you substitute the values into the formula, you obtain: A=πr 2 /2 A = (3. Set up each circle's circumference to its diameter. Thus the surface area is: SA = 2(pR 2)+ 2pRH. perimeter (circumference) of a semicircle: pi times radius. A connected section of the circumference of a circle. Suppose we want to start from 45 degrees. Online Circumference Memory Game. A Trapezium is a four-sided polygon with two non-adjacent parallel sides or one set of parallel sides. Important Questions for Class 10 Maths Chapter 12 Areas Related to Circles Areas Related to Circles Class 10 Important Questions Very Short Answer (1 Mark) Question 1. Find A, C, r and d of a circle. Calculate the total square inches in column F by using =E2*D2. out Finds circumference of circle ----- Enter radius: 8 Circumference of Circle: 50. Paper 4 (Calculator) Circle Theorems Past Paper Questions Arranged by Topic Materials required for examination Items included with question papers Ruler graduated in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. | {
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in centimetres and Nil millimetres, protractor, compasses, pen, HB pencil, eraser, calculator. Students learn about finding the circumference of a circle as a multiple of π and as a decimal correct to 3 significant figures. A circle has a circumference of ìcm. (a) Calculate the area of the whole target. Then tap or click the Calculate button. A = \frac { 1 }{ 2 } × π × { r }^{ 2 } where π = \frac { 22 }{ 7 } (or) we can also take π as 3. Circumference of a circle is defined as the distance around it. Then, do the following: Measure the diameter of each circle by folding in half. Once you have the radius, you can calculate the circumference with 2 x Pi x R. Yes, the area of this rectangle would be π R 2. Instructions to Candidates. The result is A=5/4*Pi*a². Online Circumference Memory Game. Find (a) the circumference of a circle with radius 6 centimeters and (b) the radius of a circle with circumference 31 meters. Write the formula that you will use (e. ©P j260 r1w2 d 8K fukt 5a8 rS Moof qtcwxaJr1eI fLELOCD. 70 So, the circumference is about 37. It is calculated just by multiplying the diameter of the circle with π value. When a circle is divided into two equal halves, it forms two semicircles. !Shown below is a compound shape made from a rectangle and semi-circle. An Arc is a part of the circumference of a circle. The laundry basket that Mrs. I can calculate the volume of cylinders. o Calculate the circumference of a circle. • The measure of a semicircle, or an arc that is equal to half of a circle, is 180º. A Quonset roof is typically a half-round shape of any length that can support or accommodate a cathedral or domed interior ceiling structure. 14 and “r” is the radius of the given semi circle. Draw a picture to represent the window. Find the circumference of a circle which has a radius of 4 cm. What is the radius of the cup? Round your answer to a suitable degree of accuracy. Where the perpendicular bisectors of the chords intersect is the expected center of | {
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of accuracy. Where the perpendicular bisectors of the chords intersect is the expected center of the concentric circles forming the original path. G) Know the formulas for the area and circumference of a circle and use them to solve problems; give an informal derivation of the relationship between the circumference and area of a circle. Find the magnitude of the electric field at the point P -- the center of the semicircle. Solution : Here r = 4. The radius is: r=D/2 r=30. We'll use 3. I can calculate the volume of cylinders. Circumference of a semi-circle = = πr and the perimeter of a semi-circular shape = (π + 2) r units. A sector of a circle) is made by drawing two lines from the centre of the circle to the circumference. Calculate the exact area of the shaded sector. Because the semicircle is only half of this divide your answer by 2. lcm 2Tt x 4cm 9. Here we are writing a simple C program that calculates the area and circumference of circle based on the radius value provided by user. How to find the area of a circle: The area of a circle can be found by multiplying pi ( π = 3. on your calculator as an approximation for 𝜋𝜋. TOPIC: Circumference, measurement, advanced calculations INTERACTIVITY: Participants will measure the radius of the circumference of the rainbow (semicircle). Answer: The shaded area is 314. Let me ask you a question: How do you calculate the circumference of a circle? As you know, the formula for the circumference of a circle is: C = π · d or C = 2 · π · r The circumference is equal to π times the length of the diameter, or it is equ. Follow these simple instructions: 1. Then tap or click the Calculate button. If you know the length of the radius, you can calculate its perimeter using the following formula:. An easy to use, free perimeter calculator you can use to calculate the perimeter of shapes like square, rectangle, triangle, circle, parallelogram, trapezoid, ellipse, and sector of a circle. Circumference C= Diameter Area Radius π Learn | {
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trapezoid, ellipse, and sector of a circle. Circumference C= Diameter Area Radius π Learn the formula for circumference: πd Learn the formula for the area of a circle: A= πr 2 Learn the value of pi correct to d. The radius of a circle is 1. We have step-by-step solutions for your textbooks written by Bartleby experts!. [5] A rubber gasket forms a seal between the glass and the window frame. Equation of a Circle Through Three Points Calculator. Students can play by themselves or with a partner. Online Circle Tool. And in this case, r is equal to 6. Step-by-step explanation: Here we note that the shape consists of two small circles and one larger circle. A segment is the shape formed between the chord and the arc. You can use numerous different inputs and choose to submit the radius, the diameter or the circumference of the circle. The radius of a circle is any of the line segments from its center to its perimeter. In this article, we will consider a geometric figure that does not involve line segments, but is instead curved: the circle. The total height of the window is 90cm and the total width is 60cm. Then work out 37. Give your answers to 2 d. In this equation, "C" represents the circumference of the circle, and "d" represents its diameter. The only basic figure is the semi-circle, that appears four times. Calculate the diameter of the. This circumference to diameter calculator is used to find the diameter of a circle given its circumference. Thus the surface area is: SA = 2(pR 2)+ 2pRH. TOPIC: Circumference, measurement, advanced calculations INTERACTIVITY: Participants will measure the radius of the circumference of the rainbow (semicircle). For example, if the circumference in this example equals 11 inches, multiply 11 by 0. Perimeter Of A Circle With Examples Example 1: If the perimeter of a semi-circular protractor is 66 cm, find the diameter of the protractor (Take π = 22/7). If the handle is 28 cm long, what is the diameter of the pot? Give your answer | {
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(Take π = 22/7). If the handle is 28 cm long, what is the diameter of the pot? Give your answer correct to 3 significant figures. 14 or 22 — 7 for 𝛑. Program to calculate Area and Circumference based on user input. Smith can determine where to place it. • Write the formula: Area = πr2 • Substitute the words with the measurements you have been given. A circle has a radius of 𝑟 cm and a circumference of cm. The ratio circumference. Textbook solution for Single Variable Calculus: Early Transcendentals 8th Edition James Stewart Chapter 3. The perimeter of a semicircle is half of the circumference plus the. 0 2cm 0 4cm. It is in circle geometry that the concepts of congruence and similarity. Explain in words how to determine the perimeter of a semicircle. What is 282. Perimeter of a Semicircle Formulas & Calculator. The notation for semicircle and major arc is similar to that of minor arc. A major arc is an arc that is larger than a semicircle. Therefore, all we need to do is add the measurement of each side of the shape and we will have the perimeter. The perimeter (our constraint) is the lengths of the three sides on the rectangular portion plus half the circumference of a circle of radius $$r$$. Circumference (C): The distance the edge of a circle Radius and Diameter Circumference d: ltd 2Ttr Ex 1) Find the circumference of the circle above. They will also have to move the leprechaun around the rainbow throughout the activity. 14 r is the radius of the semicircle 2. We can find the perimeter of a semicircle with the help of this below formula: where, R = Radius of the semicircle Use our below online perimeter of a semicircle calculator to find the perimeter. ©P j260 r1w2 d 8K fukt 5a8 rS Moof qtcwxaJr1eI fLELOCD. Next we calculate the circumference of the full circle. Circumference is equal to 2 pi r. The centroid of Semicircle $$(0, \frac{4r}{3\pi})$$ Circumference of Semicircle Formulas $$P = \pi r$$ Summary of Semicircle Formula. Its unit length is a portion of the | {
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Semicircle Formulas $$P = \pi r$$ Summary of Semicircle Formula. Its unit length is a portion of the circumference. (3) diagram not to scale (b) Find AOB , giving your answer in radians. Add them to the table above, measure the circumference and diameter for each, and then compute the ratio Circumference Diameter for each object. Find three other circular objects. Our circumference calculator provides circumference of a circle by entering its radius. It depends on what characteristics of the semicircle you are given. Give your answers to 2 d. The surface of the pond is a semicircle of radius 1. Write a formula that expresses the value of in terms of 𝑟 and 𝜋. The width of each small ring is 3 centimetres. Our area of a sector calculator is flexible and reliable. Give your answer to a sensible degree of accuracy. The region bounded by the circle is also often referred to as a circle (as in when we speak of the area of a circle), and the curve is referred to as the perimeter or the circumference of the circle. An arc is a part of the circumference. So it's equal to 2 pi times 6, which is going to be equal to 12pi. This is a great start. These slightly more advanced circle worksheets require students to calculate area or circumference from different measurements of a circle. The missing length is 13. Andlearning. The earlier worksheets in this section require calculating the area and the circumference given either a radius or a diameter. • Volume: The volume of a three-dimensional object is a measure of the total space it occupies, measured in cubic units. How to find the area of a circle: The area of a circle can be found by multiplying pi ( π = 3. c) The answer of 16. You are not making allowances for the space needed for mortar, if any, and you are including bricks for the spaces at the corners of the box. We have step-by-step solutions for your textbooks written by Bartleby experts!. Find the circumference of the fl ying disc. Thus the surface area is: SA = 2(pR 2)+ | {
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experts!. Find the circumference of the fl ying disc. Thus the surface area is: SA = 2(pR 2)+ 2pRH. Click here to learn about how to calculate the circumference of a circle. Calculate Areas & Properties of Annulus Rings Ring shapes are essentially formed by one circle within another. MEMORY METER. Examples of Circle and Semi-circle functions We look at a number of examples of circle and semi-circle functions, sketch their graphs, work out their domains and ranges, determine the centre and radius of a circle given its function, etc. The radius for this circle is 45. Another fact that you need to keep in mind about inscribed angles is the fact that any inscribed angle that intercepts a semicircle has to be a right angle. If d=2r where r is the radius, then C=2πr. 3 Problem 56E. C 5pd or C 52pr 26. This figure consists of a rectangle and semicircle. The length of the circumference is given by the formula: C = πd, where d is the diameter of the circle. The distance around a rectangle or a square is as you might remember called the perimeter. Assume that the radius of a circle is 21 cm. Clearly stating the units of your answers, calculate do Co(reck the circumference of his £2 coin, giving your answer to an appropriate degree of accuracy. Tangent of circle: a line perpendicular to the radius that touches ONLY one point on the circle. And in this case, r is equal to 6. A chord is a straight line joining any two parts of the circumference. If the handle is 28 cm long, what is the diameter of the pot? Give your answer correct to 3 significant figures. To find the perimeter, P, of a semicircle, you need half of the circle's circumference, plus the semicircle's diameter: P = 1 2 ( 2 π r ) + d The 1 2 and 2 cancel each other out, so you can simplify to get this perimeter of a semicircle formula. Calculate the area of one surface of the table mat. o Pi (π) Objectives. Area of a Semicircle In the case of a circle, the formula for area, A, is A = pi * r^2, where r is the circle’s | {
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In the case of a circle, the formula for area, A, is A = pi * r^2, where r is the circle’s radius. If someone could please help with the following question? What is the formula to calculate the distance perpendicular from the section line to the nearest edge of the circumference at any point X along the section line? Thanks in advance. Step 1) Write the formula Step 2) Substitute what you know Step 3) Calculate. A sector of a circle) is made by drawing two lines from the centre of the circle to the circumference. 1 foot ≈ 0. Its unit length is a portion of the circumference. The area of a circle is. This is the perimeter of the semi-circle!. perimeter (circumference) of a semicircle: pi times radius. A segment is the shape formed between the chord and the arc. A major arc is an arc that is larger than a semicircle. 14(15) Replace d with 15. WIIat is the perimeter of the figure shown below (semicircle attached to rectangle)? Give the exact answer. Find the circumference of the quad-rant with radius 4. Shapes and Figures. Finally, you can find the diameter - it is simply double the radius: D = 2 * R = 2 * 14 = 28 cm. Please find teh circumference of the basket so Mrs. !Shown below is a compound shape made from a rectangle and semi-circle. When the area is fixed and the perimeter is a minimum, or when the perimeter is fixed and the area is a maximum, use Lagrange multipliers to verify that the length of the rectangle is twice its height. The arc that remains as a part of the semicircle is half as long as the circumference of the whole circle. Write a formula that expresses the value of 𝐶𝐶 in terms of 𝑟𝑟 and 𝜋𝜋. Title: Print Layout - Mathster Created Date: 20140105104657Z. A connected section of the circumference of a circle. This method yields a ceiling estimate of the number of bricks you will need. First, calculate the circumference of the whole circle. 14 (2 cm); C = 6. Find its arc length A. (Answer in units of N/C. That’s because a semi circle has an arc length | {
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= 6. Find its arc length A. (Answer in units of N/C. That’s because a semi circle has an arc length that is exactly 1/2 of the circumference of a complete circle. A semicircle is an arc that is “half” of a circle. If we know the diameter then we can calculate the area of a circle using formula: A=π/4*D² (D is the diameter). Draw a diagram of a circle and label all given information 2. G7-M3-Lesson 18: More Problems on Area and Circumference 1. Floor Area Area of the floor. You can also select units of measure for both input data and results. Use this calculation to calculate the area of a circle. 8 11 14 4 7 6 Now that we have all the lengths of the sides, we can simply calculate the perimeter by adding the lengths together to get 4+14+11+8+7+6=50. The perimeter of the semicircle also includes the straight line that connects the two ends of that arc. C = πd Circumference of a circle C = 3. • Circumference: The circumference of a circle is the distance around the outside of a circle and is denoted. 7 m and 9557 m2) [2+3](Curved S. out Finds circumference of circle ----- Enter radius: 8 Circumference of Circle: 50. WIIat is the perimeter of the figure shown below (semicircle attached to rectangle)? Give the exact answer. To calculate area and circumference we must know the radius of circle. 24 Circumference of a Semi-circle formula Circumference of a Semi-circle = π * radius C Program to find the circumference of a semi-circle. DCO is a straight line. The radius of a circle is 1. What is 282. This is also known as the longest chord of the circle. ) cm² (to. Radius: r Diameter: d Circumference: C Area: K : d = 2r C = 2 Pi r = Pi d K = Pi r 2 = Pi d 2 /4 C = 2 sqrt(Pi K) K = C 2 /4 Pi = Cr/2 To read about circles, visit The Geometry Center. We'll use 3. The calculations are done "live": How to Calculate the Area. Program to calculate Area and Circumference based on user input. Find the diameter. Leave your answer in terms of π 1 Area = ∏ x 11 x 11 = 121∏ cm2 Diameter of | {
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Find the diameter. Leave your answer in terms of π 1 Area = ∏ x 11 x 11 = 121∏ cm2 Diameter of circle = 22 cm Circumference = 22∏ cm Circumference = 6π m Area = 32 π m2 = 9 π m2 Circle = π x 52 = 25π Semi-circle = (π x 22) / 2 = 4. If the area of a circle is equal to sum of the areas of two circles of diameters 10 cm and 24 cm, calculate the diameter of the […]. A sector is the part of a circle between two radii. Contrast radius and diameter. Show Answer. The circle has a diameter of 30 cm. If you have the radius instead of the diameter, multiply it by 2 to get the diameter. Magnetic field around a circular wire is calculated by the formula; B=2πk. [5] A rubber gasket forms a seal between the glass and the window frame. Yes, the area of this rectangle would be π R 2. For interactive applets, worksheets, and more videos, please visit http://www. Do NOT confuse a 'sector' with a 'segment'. Input: Store the constant pI in M1: M1 = 3. asked by tom on February 7, 2014; Math. As a semicircle is half of a circle, the length of the arc of the semicircle will be 𝜋𝑑 divided by two. That is, we want -90 + a / total_semi_circles = -45, so a / total_semi_circles = 45. 7cm 5cm Ttx8 = 25. lcm 2Tt x 4cm 9. You can get your required area of the section by just putting radius value and angle. Using “ ” to represent the diameter of the circle, write an algebraic expression that will result in the perimeter of a. Find the diameter or radius of a circle using the formulas: C = πd; C = 2πr. Give your answer to a sensible degree of accuracy. Find three other circular objects. For a complete analysis of a circular arc please check out The Complete Circular Arc Calculator. Circumference of a circle 1 Work out the circumference of each circle. Use this calculation to calculate the area of a circle. A line that is drawn straight through the midpoint of a circle and that has its end points on the circle border is called the diameter (d). Ex2) 14 in. The figure below is in the shape of a | {
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on the circle border is called the diameter (d). Ex2) 14 in. The figure below is in the shape of a semicircle. Ellipse In geometry, an ellipse is a regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant, or resulting when a cone is cut by an oblique plane that does not. ! C=2"r) 3. How to Use the Material Needed Calculator (Circle) This calculator is easy to use. Need to subtract a semicircle? BC. Write a formula that expresses the value of in terms of 𝑟 and 𝜋. Area of a Circle Calculator. If you need an anticipatory set you might want to use this online circumference memory game. In order to find the area of semi circle use the below given formula. Enter the radius: 1 The area of circle is: 3. (AQA June 2007 Intermediate Paper 1 NO Calculator) A semi-circle is cut from a circle. Define tangent. Its length is always less than half of the circumference. Note that measuring circumference requires the use of 𝜋, an irrational number customarily approximated as 3. 2) Chemistry periodic calculator. Unit 2: Two-Dimensional Geometry. The semi-circle has a diameter of 20 cm. : We know the diameter of the semi circles = the width of the rectangle, y Therefore the circumference of the semi circles = pi*y The total perimeter divide both sides by 2: total area = 2 semicircle area + rectangular potion area A = Replace x A = is the area as a function of y:. Calculate Perimeter Of a Square Calculate Perimeter Of a Rectangle Calculate Perimeter Of a Triangle. The ratio of the circumference C to diameter d of both circles simplify to the same value, 3. Calculate E at center of semicircle: A uniformly charged insulating rod of length L = 12 cm is bent into the shape of a semicircle as shown. Arcs are measured in two ways: as the measure of the central angle , or as the length of the arc itself. I can calculate the radius and diameter of a circle when I know the Circumference. Area and Circumference | {
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calculate the radius and diameter of a circle when I know the Circumference. Area and Circumference Worksheets. The perimeter formula is one of the easier formulas to remember in math! The perimeter of a shape is the distance around the outside of the shape. 1416… Browse more Topics Under Areas Related To Circles. In the result you will get all unknown variables presented. Step 1) Write the formula Step 2) Substitute what you know Step 3) Calculate. asked by tom on February 7, 2014; Math. 19 8 cm 5 cm 18 A wheel of radius 20 cm rolls without slipping on level ground. I suspect you may be visiting this page to calculate the area of an arc, or piece of a circle, because semi circle can sometimes be considered ambiguous. An arc is a segment of a 6 cm Chapter 50 cm A semicircle is of a whole circle. Circumference calculator is a free tool used to calculate the circumference of a circle when the radius is given. Circumference is the linear distance around the circle edge. Solve for a missing angle in a tangent problem. The perimeter (our constraint) is the lengths of the three sides on the rectangular portion plus half the circumference of a circle of radius $$r$$. If r is the radius of a circle, then (i) Circumference = 2πr or πd, where d = 2r is the diameter of the circle. The other two sides should meet at a vertex somewhere on the circumference. Perform the following steps: 1. Solutions for the assessment Circles, Perimeters and Sectors 1) Area = 56. The radius of this circle is 8 cm. For example, if the circumference in this example equals 11 inches, multiply 11 by 0. The diameter of a circle is known as the straight line segment which passes through the center of the circle. • Volume: The volume of a three-dimensional object is a measure of the total space it occupies, measured in cubic units. Area of a Circle Calculator. Revision exercise 3. The distance around a circle is called the circumference. A semicircle is an arc that is “half” of a circle. You can also | {
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circle is called the circumference. A semicircle is an arc that is “half” of a circle. You can also select units of measure for both input data and results. your calculator as an approximation for 𝜋. Find the perimeter of the shape. The area of a semicircle of radius r is given by A = int_0^rint_(-sqrt(r^2-x^2))^(sqrt(r^2-x^2))dxdy (1) = 2int_0^rsqrt(r^2-x^2)dx (2) = 1/2pir^2. She wants to move the basket to another place in her classroom. The circumference of the traffi c circle is approximately 16. If you know the length of the radius, you can calculate its perimeter using the following formula:. The diameter of a circle is known as the straight line segment which passes through the center of the circle. A line that is drawn straight through the midpoint of a circle and that has its end points on the circle border is called the diameter (d). A semicircle can be used to construct the arithmetic and geometric means of two lengths using straight-edge and compass. When a circle is divided into two equal halves, it forms two semicircles. You are given a semicircle of radius 1 ( see the picture on the left ). Circumference Word Problems For each problem, you are expected to complete all of the following: 1. C 5pd or C 52pr 26. The circumference is the outside edge of the circle. 3 divide circumference in inches by the width of the wallblocks you want to use. The handle of a paint pot is half the circumference of the pot (a semi–circle). Semicircle Radius using Circumference Calculation. Draw the diagram (as above on right) to understand the problem and arrive at the required answer. If you want to calculate the outer boundary of a semicircle - the circumference or perimeter of a semicircle - you need to be careful to not fall into any traps. A semicircle is exactly half a circle. your calculator as an approximation for 𝜋. The POWER function will take any number and raise it to the power of any other number. Semicircle is half of a circle. 14159, which is equal to the | {
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it to the power of any other number. Semicircle is half of a circle. 14159, which is equal to the ratio of the circumference of any circle to its diameter. Calculate E at center of semicircle: A uniformly charged insulating rod of length L = 12 cm is bent into the shape of a semicircle as shown. Perimeter Of Sector calculator provides for the same. Find the circumference of the circles below. • The measure of a semicircle, or an arc that is equal to half of a circle, is 180º. Then r r r is a constant, and its derivative is 0 0. Work out the circumference of the circle. the areas of two circles are in a ratio 4:9 find the ratio between their circumference AB is a line segment whose midpoint is M. Need to subtract a semicircle? BC. Leave your answer in terms of π 1 Area = ∏ x 11 x 11 = 121∏ cm2 Diameter of circle = 22 cm Circumference = 22∏ cm Circumference = 6π m Area = 32 π m2 = 9 π m2 Circle = π x 52 = 25π Semi-circle = (π x 22) / 2 = 4. The figure shows a circle within a square. Circumference of a circle (C) = 2πr, where π is pi and “r” is radius. 56 This is the length of just the curvy part of the semi-circle. Just enter the measurement you know. 1% Convert the percentages from a pie chart to central angle measures?. The result is A=5/4*Pi*a². Perimeter and Area of Circle and Semicircle; Arcs of a Circle. In the result you will get all unknown variables presented. The area of a circle will be shown in the selected units. Another circle is inscribed in the semi–circle so that it touches the diameter at point A and also touches the circumference of the semi– circle. The area of a semi-circle is 1/2 πr 2. The rectangle has a width of 15 meters and a height of 12 meters. Click here to learn about how to calculate the circumference of a circle. 3 divide circumference in inches by the width of the wallblocks you want to use. 14 r is the radius of the semicircle 2. Circumference = 2 × π × radius = 2 × 3 × 21 [Using 3 as an estimated value for π]. The perimeter of the | {
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= 2 × π × radius = 2 × 3 × 21 [Using 3 as an estimated value for π]. The perimeter of the semicircle also includes the straight line that connects the two ends of that arc. Area and Circumference of a Circle (Grades 6-7) Given a Slope and Points, calculate the missing coordinate Determine if the Points are Collinear. For example, if the circumference in this example equals 11 inches, multiply 11 by 0. Its unit length is a portion of the circumference. Answer by noguf (162) (pi)x(diameter)/2 + diameter. A semicircle is an arc that is “half” of a circle. Circumference of a semi-circle = = πr and the perimeter of a semi-circular shape = (π + 2) r units. More About Arc. Circumference to Diameter Calculator. It can also be thought of as a sector with an angle of 180 degrees. Step 1: say the radius of the field is r, then the total area of the field = (r 2)/2. In the article below, we provide the semicircle. Degree measure of a minor arc: Defined as the same as the measure of its corresponding central angle. Perimeter of a triangle calculation using all different rules: SSS, ASA, SAS, SSA, etc. The circumference of a circle is found using the circumference formula: Circumference = 2 * π * radius So, all you have to do is plug in the radius into the formula, and approximate π to be 3. Contrast radius and diameter. Perimeter Formulas and Circumference of a Circle Formula. 28 cm 2 The circumference of a bicycle wheel is 50. The ratio circumference. Determine the length of this semicircle, as measured along the curve. A compass is an instrument used to draw circles or the parts of circles called arcs. You can click on the link about the calculator to learn more about the area of semi circles and it’s relationship to arc length. The circumference of a circle is an arc measuring 360 o. From geometry, you know that "A(r) = πr 2" indicates the area of a circle given in terms of the value of the radius r, while "C(r) = 2πr" indicates the circumference given in terms of the radius | {
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value of the radius r, while "C(r) = 2πr" indicates the circumference given in terms of the radius r. 2 cm 2 Work out the circumference of each circle. This math worksheet was created on 2010-03-06 and has been viewed 81 times this week and 53 times this month. The notation for semicircle and major arc is similar to that of minor arc. The only basic figure is the semi-circle, that appears four times. Welcome to The Circumference and Area of Circles (A) Math Worksheet from the Measurement Worksheets Page at Math-Drills. Work out the circumference of the circle. the way around a circle. Calculate the total square inches in column F by using =E2*D2. Area and Circumference of a Circle (Grades 6-7) Given a Slope and Points, calculate the missing coordinate Determine if the Points are Collinear. Online calculators and formulas for a hemisphere and other geometry problems. 7%) A: 4 (14. The radius is: r=D/2 r=30. We can substitute 7 for r in the formula C = 2πr to end up with C = 2π(7) = 14π ≈ 44 inches. It progresses nicely and finishes with some challenging problems at the end borrowed from MEP. Calculate the area of a semicircle of radius 1. The perimeter of a circle is known as its circumference. Click on "Calculate" button to get results instantly. The circumference of a figure is the sum of all the side lengths. A compass is an instrument used to draw circles or the parts of circles called arcs. 14159) and diameter is the distance across the circle (along the edge of the half circle). A Norman window has the shape of a semicircle atop a rectangle so that the diameter of the semicircle is equal to the width of the rectangle. Multiply the portion of the circumference the minor arc comprises by the circumference of the circle to calculate the length of the minor arc. It is calculated just by multiplying the diameter of the circle with π value. The circumference of a circle of a radius 𝑟𝑟 is 2𝜋𝜋. Calculate the circumference of the circle. 14 * radius. You can find out | {
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of a radius 𝑟𝑟 is 2𝜋𝜋. Calculate the circumference of the circle. 14 * radius. You can find out more about Pi here. The formula is:. Point A lies on the diameter of a semi–circle with the radius 8 cm and the center O. the semi circle touches the rectangle at a b and c. To calculate the area of a semicircle we use the formula: Area of a semicircle = π × (radius)2 2 A = πr2 2 Calculate the area of this semicircle. I can calculate the radius and diameter of a circle when I know the Circumference. To calculate the perimeter of Shape 2 add the length of the three sides of the rectangle and the circumference of the semi-circle. 7cm 5cm Ttx8 = 25. base area × height Step 2: Total area = Area of circle + Area of rectangle Mensuration. Arc Measure. Contrast radius and diameter. ) Homework Equations E = K((Q)/(r^2)). Plugging our radius of 3 into the formula, we get C = 6π meters or approximately 18. For example, if the circumference in this example equals 11 inches, multiply 11 by 0. A B C AB = 1. These slightly more advanced circle worksheets require students to calculate area or circumference from different measurements of a circle. 06 = Calculate the length of the circumference of each circle, rounding your answer to one decimal place: 62. Construct the Gnomon to the calculated dimensions and stand vertically along the centre line. A circle is drawn with radius 'r' touching all the three semi-circles. where r is the radius of the circle and π is the ratio of a circle’s circumference to its diameter. This online calculator will find and plot the equation of the circle that passes through three given points. We have step-by-step solutions for your textbooks written by Bartleby experts!. Formulas, explanations, and graphs for each calculation. They have to find the circumference of 6 different circles and then play a memory matching game. Find the perimeter of the shape. Find the circumference of a circle which has a radius of 4 cm. This gives us the lengths of all the sides | {
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the circumference of a circle which has a radius of 4 cm. This gives us the lengths of all the sides as shown in the figure below. We can substitute 7 for r in the formula C = 2πr to end up with C = 2π(7) = 14π ≈ 44 inches. Formulas, explanations, and graphs for each calculation. thx [3] 2020/04/05 02:53 Male / 40 years old level / An engineer / Useful / Purpose of use. The distance around a rectangle or a square is as you might remember called the perimeter. Round to the nearest hundredth for each circumference. These slightly more advanced circle worksheets require students to calculate area or circumference from different measurements of a circle. Calculate ‘d’ and the total area inside the track. What is the arc-length of a semi-circle with radius of 6 cm? (to 2 decimal places) A car wheel has a circumference of 1. It progresses nicely and finishes with some challenging problems at the end borrowed from MEP. Step 1) Write the formula Step 2) Substitute what you know Step 3) Calculate. You can calculate the circumference from total circle angle 360. For example, if the circumference in this example equals 11 inches, multiply 11 by 0. In order to find the area of semi circle use the below given formula. To calculate area and circumference we must know the radius of circle. 2 Calculate the length of the circumference of each circle, rounding your answers to one decimal place: 10 = 31. Do NOT confuse a 'sector' with a 'segment'. Calculate the circumference and the area of the circle. Consider the diagram of a semicircle shown. 69 km 0 266. If you have the radius instead of the diameter, multiply it by 2 to get the diameter. The diagram shows a trapezium with two identical semi-circles. To establish a major arc as a semicircle some additional notation would be required. Step 2 Since you only need half of the. I can use the information I know to calculate the length of the radius of the. What is the radius of the cup? Round your answer to a suitable degree of | {
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of the radius of the. What is the radius of the cup? Round your answer to a suitable degree of accuracy. Formula Used: Radius of a Semicircle, r = Diameter (d) / 2 or r = Circumference / Pi Where, Pi is a constant = 3. If a circle has a circumference of 8. Get the result. As learning progresses students are challenged to find the perimeter of semi-circles and sectors. Calculate the exact area of the shaded sector. Step 1: say the radius of the field is r, then the total area of the field = (r 2)/2. What are the circumference and area of the circle? 2. Well, that and the formula. The length of the rectangle is 15 cm. A circle has a circumference of ìcm. Step-by-step explanation: Here we note that the shape consists of two small circles and one larger circle. - calculate the percentage in each category - calculate the central angle for each category for a circle graph - create and label a circle graph using your data Student Grades c: o: 2 c: 10 135. In a similar manner, we can calculate the length of the other missing side using 14−8=6. Remember that the perimeter is the distance round the outside. Do NOT confuse a 'sector' with a 'segment'. Input: Store the constant pI in M1: M1 = 3. In our semicircle, the diameter is 35 centimetres. In the result you will get all unknown variables presented. Common Core Standard: 7. Welcome to The Circumference and Area of Circles (A) Math Worksheet from the Measurement Worksheets Page at Math-Drills. For each swimming pool below, calculate: the perimeter of the swimming pool. asked by Bunny head on December 2, 2018; Math. Draw a picture to represent the window. Write the formula that you will use (e. This means that in any circle, there are 2 PI radians. (3 marks) _____ A cylinder has a radius of 5 cm. Formula: Area = 3. The sum of the central angles in any circle is 360°. The diagram shows a 400-m running track. The top part of the door is a semicircle. By that logic, the arc length of a quarter circle is exactly 1/4 of the | {
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of the door is a semicircle. By that logic, the arc length of a quarter circle is exactly 1/4 of the circumference of a circle. If a circle has a circumference of 8. Arcs are measured in two ways: as the measure of the central angle , or as the length of the arc itself. So that's straightforward, area 36pi, we leverage pi r squared to figure out that the radius was 6, and then from that we were able to figure out that the circumference was 12pi. 7 cm Length of Y 3. Find the circumference of the object. You can also select units of measure for both input data and results. What is the arc-length of a semi-circle with radius of 6 cm? (to 2 decimal places) A car wheel has a circumference of 1. show your working. Angle in a semicircle We want to prove that the angle subtended at the circumference by a semicircle is a right angle. Program to calculate Area and Circumference based on user input. Perimeter and the area of a semicircle Perimeter of a semicircle. Set up each circle's circumference to its diameter. 14) by the square of the radius; If a circle has a radius of 4, its area is 3. You know circles are round. The widest part of a tea cup has a circumference of 24 cm. Second circle theorem - angle in a semicircle. o Circumference. The area of a circle is given by Pi*Radius^2 where Pi is a constant approximately equal to 3. Round this ratio to the nearest thousandth. side a: side b: distance h:. Find the diameter of the circle to the nearest tenth. This is the length of just the curvy part of the semi-circle. The circumference of a figure is the sum of all the side lengths. The perimeter of a semicircle can be calculated using the following formula: if you know the length of the radius. SOLUTION a. \$ cc circumference-of-circle. This math worksheet was created on 2010-03-06 and has been viewed 81 times this week and 53 times this month. And in this case, r is equal to 6. 2 Calculate the length of the circumference of each circle, rounding your answers to one decimal | {
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6. 2 Calculate the length of the circumference of each circle, rounding your answers to one decimal place: 10 = 31. The diameter of a circle is any straight line segment that passes through the center of the circle and whose endpoints lie. The floor is defined as a circle equal to the diameter of the base of the dome. circle theorems rules pdf DA is a tangent to the circle. 2831852 m e t e r s. MEMORY METER. Another circle is inscribed in the semi–circle so that it touches the diameter at point A and also touches the circumference of the semi– circle. The ratio circumference. Perimeter of composite shapes. Lesson Plan Title : Finding circumference, diameter, and radius. In the result you will get all unknown variables presented. ) cm² (to. n r tM Nazdhe t 9wKi7t6h l wI2nOf vi1nIi pt se F BGFe jo Lmteztjr 8yE. radius: distance from center of circle to any point on it. The circumference is the outside edge of the circle. They find the area and circumference of circles given a radius or diameter measurement. This formula can also be given as: C = 2πr, where r is the radius. Pi ( π ): It is a number equal to 3. Perimeter of a semicircle = (1 2 Suppose we want to calculate the circumference of a circle with radius r r r. Examples: how to work out the area of a trapezium?. 2, but that is the best I can do with the crude calculator on the computer and a scratch piece of paper. Then tap or click the Calculate button. 14159, which is equal to the ratio of the circumference of any circle to its diameter. Ex2) 14 in. (a) Calculate the area of the whole target. Area of semi-circle formula is derived from the formula of a circle. So, to find the diameter, divide the circumference by. The POWER function will take any number and raise it to the power of any other number. That's the. You can also use it to find the area of a circle: A = π * R² = π * 14² = 615. The area is generally 1/2*Pi*r². Students, teachers, parents, and everyone can find solutions to their math problems | {
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1/2*Pi*r². Students, teachers, parents, and everyone can find solutions to their math problems instantly. The circumference of C i r c l e 2 is 6. Calculator online for a capsule. o Pi (π) Objectives. C = πd Circumference of a circle C = 3. The calculations are done "live": How to Calculate the Area. Angle in a semicircle We want to prove that the angle subtended at the circumference by a semicircle is a right angle. o Calculate the circumference of a circle. We have step-by-step solutions for your textbooks written by Bartleby experts!. Formulas, explanations, and graphs for each calculation. C = 2π r Write formula for. Program to calculate Area and Circumference based on user input. 77 feet/2 r=15. 7 ft O C r d. For example, if the circumference in this example equals 11 inches, multiply 11 by 0. 2, but that is the best I can do with the crude calculator on the computer and a scratch piece of paper. where r is the radius of the circle and π is the ratio of a circle’s circumference to its diameter. I am stuck on put a circle around the number that is not a square number 9 16 36 48 81 100. Calculate the area and circumference of this circle, leaving your answer in terms of π 4. The diameter of the larger semicircle is subtended by the two smaller semicircles the small semicircle closer to the left of the internal circumference of the larger semicircle is shaded one while the one on the right is without color This calculator calculates for the radius length width or chord height or sagitta apothem angle and area of an. (c) Find the area of the white. Perimeter Of Sector calculator provides for the same. The area of a circle is given by Pi*Radius^2 where Pi is a constant approximately equal to 3. Use our circumference calculator to determine the area of a circle. 8cm This time, we have a semicircle. Step 2: Divide the circumference in half. The radius of a circle measuring 20 cm. Perimeter of a triangle calculation using all different rules: SSS, ASA, SAS, SSA, etc. | {
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20 cm. Perimeter of a triangle calculation using all different rules: SSS, ASA, SAS, SSA, etc. !Work out the area. 56 + 8 = 20. Given OA = 4, find the radius of the inscribed circle. I use this product as part of a grade 7 unit of study on the area and circumference of circles. This video shows how to find the area of a rectangle with a semi-circle. You know circles are round. (c) Find the area of the white. the circumference. n r tM Nazdhe t 9wKi7t6h l wI2nOf vi1nIi pt se F BGFe jo Lmteztjr 8yE. Circumference Measuring the _____ of the circle. But, the mathematical description of circles can get quite confusing, since there is a set equation for a circle, including symbols for the radius, and center of the circle. 14 or 22 — 7 for 𝛑. your calculator as an approximation for 𝜋. Calculate the area of one surface of the table mat. Calculate quickly surface area or volume of sphere. (a) Calculate the area of glass needed to glaze the window. First, you must apply the formula for calculate the area of a semicircle, which is: A=πr 2 /2 A is the area of the semicircle. that the perimeter of the semi-circle is. Calculate the shaded area. Step 2: Divide the circumference in half. I can calculate thecircumference of a circle when I know the diameter. In column H, calculate the dollars per square inch of pizza, using =G2/F2. Define tangent. It is possible to inscribe a rectangle by placing its two vertices on the semicircle and two vertices on the x-axis. Add them to the table above, measure the circumference and diameter for each, and then compute the ratio Circumference Diameter for each object. DCO is a straight line. Tracing paper may be used. (c) Given that APB is 1. Use your calculator to find the maximum area. 14 or the π button on your calculator. Draw the diagram (as above on right) to understand the problem and arrive at the required answer. 4 Find the radius of a circle. Follow 93 views (last 30 days) Timothy States on 7 Feb 2016. | {
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# How to calculate $z^4 + \frac1{z^4}$ if $z^2 + z + 1 = 0$?
Given that $z^2 + z + 1 = 0$ where $z$ is a complex number, how do I proceed in calculating $z^4 + \dfrac1{z^4}$?
Calculating the complex roots and then the result could be an answer I suppose, but it's not quite elegant. What alternatives are there?
-
From $z^2 + z + 1 = 0$, we have $$z +\frac{1}{z}=-1.$$ Taking square, we get $$z^2 +\frac{1}{z^2}+2=1,$$ which implies that $$z^2 +\frac{1}{z^2}=-1.$$ I think you will know what to do next.
-
I see... Thank you for the quick answer! – Mihai Bişog Jan 1 '12 at 11:54
$\displaystyle{z^{2^k}+\frac{1}{z^{2^k}}=-1}$ for all $k$. – Jonas Meyer Jan 1 '12 at 11:54
@Jonas: nice observation! – Paul Jan 1 '12 at 12:06
Dickson polynomials FTW! – Jyrki Lahtonen Jan 1 '12 at 14:20
Another way to justify that $z+\dfrac1{z}=-1$ is that due to the coefficient symmetry, if $z$ is a root, then $\dfrac1{z}$ is a root as well. By Vieta, then, the sum of those two roots ought to be the negative of the coefficient of the linear term... – J. M. is back. Jan 1 '12 at 15:15
Essentially the same calculation also follows from the observation that $x^2+x+1=\phi_3(x)$ is the third cyclotomic polynomial. So $z^3-1=(z-1)(z^2+z+1)=0$, and hence $z^3=1$ for any solution $z$. Therefore $$z^4+\frac{1}{z^4}=z\cdot z^3+\frac{(z^3)^2}{z^4}=z+z^2=-1.$$
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Very clever! :D – J. M. is back. Jan 1 '12 at 12:08
Very nice! If you had not posted this already, I would have used the sum of geometric series (instead of cyclotomic polynomials) to write $$0=z^2+z+1=\frac{z^3-1}{z-1}\Rightarrow z^3=1$$ and $$z^4+\frac{1}{z^4}=(z^3)z+\frac{1}{(z^3)z}=z+\frac{1}{z}=\frac{z^2+1}{z}=\frac{-z}{z}=-1.$$ – Dilip Sarwate Jan 1 '12 at 15:48
\begin{align*} z^4+\frac1{z^4}&=(-z-1)^2+\frac1{(-z-1)^2}\\ &=z^2+2z+1+\frac1{z^2+2z+1}\\ &=(-z-1)+2z+1+\frac1{(-z-1)+2z+1}\\ &=z+\frac1{z}=\frac{z^2+1}{z}=\frac{-z-1+1}{z}=-1 \end{align*} | {
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What would be wrong with systematic reduction using the minimal polynomial? Nothing! – Jyrki Lahtonen Jan 1 '12 at 12:25
Here is an alternative approach: let's consider $z^{8}+1$ , and then divide by $z^{4}$.
By using geometric series, notice that $$z^{8}+z^{7}+z^{6}+z^{5}+z^{4}+z^{3}+z^{2}+z+1=\left(z^{2}+z+1\right)\left(z^{6}+z^{3}+1\right)=0.$$ Now, as $z^{2}+z+1=0$, we know that both $z^{7}+z^{6}+z^{5}=0$ and $z^{3}+z^{2}+z=0$, and hence $$z^{8}+z^{4}+1=0$$ so that $z^{8}+1=-z^{4}$. Thus, we conclude that $$z^{4}+\frac{1}{z^{4}}=-1.$$
-
Different people see different things in the relation $z^2+z+1=0$. Just as Jyrki did, I see the third cyclotomic polynomial. Its roots are $-1/2 \pm \sqrt{-3}/2$, the two primitive cube roots of unity. Call one of these $\omega$ and see that $\omega^3=1$, so that $\omega^4=\omega$ and $\omega^{-4}=\omega^2$. Their sum is $-1$, from the defining relation.
-
Hint $\$ Exploit innate symmetry: for $\rm\ y = z^{-1}$ you know $\rm\ yz\ (=\: 1)\:$ and $\rm\ y+z\ (=\: z^{-1}\!+z\: =\: -1)\$
Thus you know $\rm\ \ \ y^2 + z^2\ =\ (y\ +\ z)^2-\ 2\:(y\:z)$
hence you know $\rm\ y^4 + z^4\ =\ (y^2\! + z^2)^2 - 2\:(y\:z)^2$
For more on symmetric polynomials see the Wikipedia article on Newton's identities.
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To downvoter: if something is not clear then please feel welcome to ask questions and I will be happy to elaborate. – Bill Dubuque Jan 1 '12 at 22:52
$$z^2+z+1=0$$ Clearly $z \not= 0$ and therefore $$z+\frac{1}{z}=-1.$$ Note that $$z^{n+1}+\frac{1}{z^{n+1}}=(z+\frac{1}{z})(z^n+\frac{1}{z^n})-(z^{n-1}+\frac{1}{z^{n-1}}).$$ Therefore if we define the function $U:N \to N$ as $$U_n=z^n+\frac{1}{z^n}$$ then we have $U(0)=2$ and $U(1)=-1.$ Also $$U(n+1)=-U(n)-U(n-1)$$ for all $n \in N.$
Using this recurrence you can calculate $z^n+\frac{1}{z^n}$ for any $n \in N.$
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# Mathematica for teaching orthographic projection
Edit: All the four answers to this question are great, and if you're interested, you should take a look at all the answers. Nevertheless, belisarius' code was accepted since it was closest to what I had in mind.
I'm trying to use Mathematica to help students better understand 3-dimensional views of objects, specifically the "plan view", the "side elevation view" and the "front elevation view", as elaborated upon here.
Given the following example "building", I was able to combine Graphics3D primitives to simulate the object using the following code.
dee = Graphics3D[{Green, Opacity[0.6], Cuboid[{0, 0, 0}, {2/3, 1, 1}],
Red, Polygon[{{{2/3, 1/2, 0}, {2/3, 1/2, 1}, {3/2, 1/2, 0}},
{{2/3, 1, 0}, {2/3, 1, 1}, {3/2, 1, 0}},
{{2/3, 1/2, 0}, {2/3, 1, 0}, {3/2, 1, 0}, {3/2, 1/2, 0}},
{{2/3, 1/2, 1}, {2/3, 1, 1}, {3/2, 1, 0}, {3/2, 1/2, 0}}}]}]
However, as you can see, the code is rather convoluted, with a great deal of time spent making the red prism. In this case, it was still feasible to manually generate the prism, but what I'm interested in is,
Is there a way to generate a 3D object which is made up of a combination of prisms simply by identifying the vertices at the corner of the object?
If you're interested, my code to allow students to see the three views is as follows, with an example result below.
planviewer =
TableForm@{{"3d", "Side", "Front", "Top"},
{Show[#, ViewPoint -> {1, -1, 1}],
Show[#, ViewPoint -> {∞, 0, 0}],
Show[#, ViewPoint -> {0, -∞, 0}],
Show[#, ViewPoint -> {0, 0, ∞}]},
{"",
Show[#, ViewPoint -> Right],
Show[#, ViewPoint -> Front],
Show[#, ViewPoint -> Above]}} &;
The following code allows them to manipulate the viewpoint dynamically.
Manipulate[
Show[dee, ViewPoint -> {10^a, -(10^b), 10^c}], {a, 0, 3}, {b, 0, 3}, {c, 0, 3}]
-
If you want to teach about the different View* functions and what they do, you might find this answer useful. – The Toad Mar 30 '13 at 16:59 | {
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There are some tricks ... Specifying the prism's vertices is enough (you don't need to take care of the faces) if you use some undocumented methods for finding the convex hull:
v = {{2/3, 1/2, 0}, {2/3, 1/2, 1}, {2/3, 1, 0}, {2/3, 1, 1}, {3/2, 1/2, 0}, {3/2, 1, 0}};
prism@v_:=Cases[ComputationalGeometryMethodsConvexHull3D[v,GraphicsMeshFlatFaces->False],
_GraphicsComplex, Infinity];
Graphics3D[{Green, Opacity[0.6], Cuboid[{0, 0, 0}, {2/3, 1, 1}], Opacity[.6], Red, prism@v}]
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This is easily turned into a function, polyhedron[v_] := Cases[.. Very nice. (+1) – Michael E2 Mar 30 '13 at 18:21
@MichaelE2 Thanks! Rewritten as a function now. – belisarius Mar 30 '13 at 18:43
@belisarius how do you find out about all these undocumented questions? I was trying to search for it, and saw some similar questions on this site, but I think that's the first time I've seen it. Edit: ninja'd by bill. – Vincent Tjeng Mar 31 '13 at 3:30
It's not on this list, by the way :) : mathematica.stackexchange.com/questions/809/… – Vincent Tjeng Mar 31 '13 at 3:38
@VincentTjeng I don't remember how I found this one, probably following some InternalTrace. You can find similar things trying Names["ComputationalGeometryMethods*"] – belisarius Mar 31 '13 at 5:55
GraphicsComplex is probably what you are looking for. For example, define the vertices for the 3D polygon:
v = {{2/3, 1/2, 0},{2/3, 1/2, 1},{3/2, 1/2, 0},{2/3, 1, 0},{2/3, 1, 1},{3/2, 1, 0}};
and make a list of which vertices should connect to each other:
i = {{1, 2, 3}, {4, 5, 6}, {1, 2, 5, 4}, {1, 3, 6, 4}, {2, 3, 6, 5}};
The first two elements of i represent the two triangular faces of the prism; the final three elements are the three rectangular faces. This is plotted using
Graphics3D[{Opacity[.8], Red, GraphicsComplex[v, Polygon[i]]}]
or you can plot the green cube plus the red prism together:
Graphics3D[{Green, Opacity[0.6], Cuboid[{0, 0, 0}, {2/3, 1, 1}],
Opacity[.6], Red, GraphicsComplex[v, Polygon[i]]}] | {
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which gives the figure below. The code is modified from the documentation in GraphicsComplex where you can find all sorts of neat 3D tricks. If they have access, I think the students would benefit from being able to manipulate the 3D illustration themselves -- this seems like a good application for deployed .cdf's.
-
Thanks for the answer! Would it be possible for the one to write some code that immediately identifies which of vertices should be connected to each other to form the faces, keeping in mind that the faces of the prism may not always be in the same plane? Also, I notice that the part of the red prism joining the green cuboid is actually a triangle, and not a rectangle ... is that a problem or simply a visual illusion? – Vincent Tjeng Mar 30 '13 at 10:21
The little triangle on the adjoining face is an illusion that has to do with the shading -- when you rotate the graphic it changes. But to your first question, the answer is no. After all, how can it know what edges you want to connect if you don't tell it? You will get something no matter what edges you connect -- only one of these is the figure you presumably want. – bill s Mar 30 '13 at 10:33
@VincentTjeng: yes, you can connect the vertices automagically because the faces must form the convex hull. See my answer – belisarius Mar 30 '13 at 18:54
Very nice... how does one find "undocumented features"? – bill s Mar 31 '13 at 3:17
bill I don't remember how I found this one, probably following some InternalTrace. You can find similar things trying Names["ComputationalGeometryMethods*"] – belisarius Mar 31 '13 at 5:59
Take a look at the output of PolyhedronData[{"Prism", 3}, "Faces"]}, which may help.
Note this isn't exactly a direct answer, but I think the optimal solution here is a tool that would let you... well, like this:
whatThe[n_] := Module[{list = {}, grid = False},
Grid@List@{
EventHandler[#, {"KeyDown", "k"} :> (grid = Not[grid])] &[ | {
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Graphics3D[{
EventHandler[MouseAppearance[Sphere[#, .06],
Style["\[RightUpDownVector]\[LeftUpDownVector]", Darker@Blue, Large]],
{"MouseClicked" :> AppendTo[list, ##]}] & /@ Tuples[Range[n], 3],
(*Opacity[.3],*)Dynamic[Polygon[list]], Thick, Dynamic[Line[list]],
Thin, Lighter[Gray, .8], Dynamic[If[grid, Line[
Select[Subsets[Tuples[Range[n], 3], {2}],
EuclideanDistance @@ ## == 1 &]], {}]]},
Boxed -> True, ImageSize -> Large, Axes -> True]],
Graphics3D[Dynamic[Polygon[list]]],
EventHandler[Dynamic[Column[list]],
{"MouseClicked" :> (CopyToClipboard[list]; Beep[])}]}
];
whatThe[4]
You click on points to add them to a list. MAGIC!!! It takes some getting used to... here's one of my attempts at creating a prism:
I couldn't tell what order you have to construct the list in for the resulting polygon to be reasonable, but at the very least it might help you pick some points out. It's also another example of just how ridiculously powerful Mathematica is. The basic working version of this took me like 4 minutes to make. O.O
(Note: A simple improvement might be to use a hotkey to cut the list up as you build the polygon).
Update. Using belisarius's convex hull code:
whatThe[n_] := Module[{list = {}, faces, grid = False},
faces[v_] := Cases[ComputationalGeometryMethodsConvexHull3D[v,
GraphicsMeshFlatFaces -> False], _GraphicsComplex, Infinity];
Grid@List@{
EventHandler[#, {"KeyDown", "k"} :> (grid = Not[grid])] &[
Graphics3D[{EventHandler[
MouseAppearance[Sphere[#, .06],
Style["\[RightUpDownVector]\[LeftUpDownVector]",
Darker@Blue, Large]],
{"MouseClicked" :> AppendTo[list, ##]}] & /@ Tuples[Range[n], 3],
(*Opacity[.3],*)Dynamic[faces[list]],
Thin, Lighter[Gray, .8], Dynamic[If[grid, Line[
Select[Subsets[Tuples[Range[n], 3], {2}],
EuclideanDistance @@ ## == 1 &]], {}]]},
Boxed -> True, ImageSize -> Large, Axes -> True]],
Graphics3D[Dynamic[faces[list]]],
EventHandler[Dynamic[Column[list]],
{"MouseClicked" :> (CopyToClipboard[list]; Beep[])}]}]; | {
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-
+1 amazing. and it helps with the first part of translating the problems in the students' worksheets to the prism I want. now I just need to modify the code such that it accepts multiple convex hulls representing the multiple prisms :) – Vincent Tjeng Mar 31 '13 at 2:58
You can get away with specifying even less than all the points.
A (right) prism is a planar figure extruded in an orthogonal direction. It would suffice, then, to give the coordinates of that figure (thereby describing one of the two "ends" of the prism) and its intended height. The figure in the question can therefore be created as a collection of two prisms (no Cuboid if you don't care to use it) like this:
Graphics3D[{Opacity[0.6],
Green, prism[{{0, 0, 0}, {2/3, 0, 0}, {2/3, 1, 0}, {0, 1, 0}}, 1],
Red, prism[{{2/3, 1, 0}, {3/2, 1, 0}, {2/3, 1, 1}}, 1/2]}]
The calculations come down to
1. Find a basis for the polygon's plane by orthogonalizing its vertices relative to their barycenter;
2. Perform the extrusion by taking the cross product of the basis elements (normalized to the desired height) and adding that to each of the polygon's vertices; and
3. Describe how the vertices are connected to form the two polygonal ends and each of the sides.
Here is such a solution. Its arguments are pt, a list of the (3D) vertices around the (nondegenerate) polygon, and height, the height of the prism. It produces a 3D graphics object.
prism[pt_List, height_] := Block[{n = Length[pt], normal},
normal = Cross @@ Orthogonalize[# - Mean[pt] & /@ pt][[1 ;; 2]];
GraphicsComplex[pt~Join~(height normal + # & /@ pt),
{Polygon[Range[n]], Polygon[Range[2 n, n + 1, -1]],
Polygon[{#, n + #, n + Mod[# + 1, n, 1], Mod[# + 1, n, 1]}] & /@ Range[n]}]]
To see it in action, let's generate a polygon with a random position and orientation:
basis = RandomReal[NormalDistribution[0, 1], {2, 3}];
pt = ({Cos[#], Sin[#]} & /@ Range[0, 2 \[Pi], 2 \[Pi]/7]).basis | {
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Here is the original polygon as a solid black "floor" drawn with two prisms, one with a positive height and another with a negative height:
Graphics3D[{Opacity[0.75], prism[pt, 1], prism[pt, -1/2], Opacity[1], Black, Polygon[pt]}]
This approach is easily extended to non-right prisms by replacing height by the extrusion vector (describing one of the "vertical" edges of the prism) and using that instead of height normal in the code.
-
nice application of math :) just a quick question, when you code as prism[pt_List, height_], pt_List is written as such since it is a list, right? so if i were to code a function f[a_String, b_String], then the variables would be a, b and they must be strings? – Vincent Tjeng Mar 31 '13 at 3:35
– whuber Mar 31 '13 at 13:19
Got it, thank you! – Vincent Tjeng Apr 1 '13 at 3:01 | {
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# Matrix representation of Mixed derivatives
Imagine we have the problem
$$\begin{cases} -\frac{d^2u}{dx^2} = f(x), x \in [0,L] \\ u(0) = 0 \\ u(L) = 0 \end{cases}$$
We know that we can approximate the second derivative using this formula:$$\frac{d^2u(x)}{dx^2} \approx \frac{u(x+h)-2u(x)+u(x-h)}{h^2}$$
if we define $$u_{k} := u(x_{k})$$; $$\ \ x_{k} = kh$$ and $$\ \ k = 0,1,2,...,N$$. $$h$$ is known as the mesh size or step size. We get:
$$\frac{d^2u_{k}}{dx^2} \approx \frac{u_{k+1}-2u_{k}+u_{k-1}}{h^2} = \frac{u_{k-1}-2u_{k}+u_{k+1}}{h^2}$$ for $$k=1,...,N-1$$
Since $$u(0) = u_{0} = 0$$ and $$u(L) = u(x_{N}) = u_{N} = 0$$ we get the following matrix representation of the second derivative operator
$$$$\frac{d^2}{dx^2} \approx L_{2} = \frac{1}{h^2}\left(\begin{matrix} -2 & 1 & & 0\\ 1 & \ddots & \ddots & \\ & \ddots & \ddots & 1 \\ 0 & & 1 & -2 \end{matrix} \right)$$$$
Then to approximate the solution of the differental equation we solve the system:
$$-L_{2}\hat{u} = \hat{f}$$ where $$\hat{f} = [ f(x_{1}) \ f(x_{2}) \ ... \ f(x_{N-1}) ]^T$$ and $$\hat{u} = [ u_{1} \ ... \ u_{N-1} ]^T$$
The matrix representation of the laplacian operator using the Kronecker product is: $$\Delta =\nabla^2 = \frac{\partial^2}{\partial x^2} + \frac{\partial ^2}{\partial y^2} = L_{2}\otimes I_{n_{x}} + I_{n_{y}} \otimes L_{2}$$
$$\textbf{Note that I have used L_{2} and kronecker product to get a matrix representation}$$ of the laplacian operator.
With this in mind. I want to approximate the first derivative using central difference:
$$\frac{du_{k}}{dx} \approx \frac{ u_{k+1}-u_{k-1} }{ 2h } = \frac{ -u_{k-1} +u_{k+1} }{ 2h }$$
Since $$u(0) = u_{0} = 0$$ and $$u(L) = u(x_{N}) = u_{N} = 0$$ we get the following matrix representation of the first derivative
$$$$\frac{d}{dx} \approx L_{1} = \frac{1}{2h}\left(\begin{matrix} 0 & 1 & & 0\\ -1 & \ddots & \ddots & \\ & \ddots & \ddots & 1 \\ 0 & & -1 & 0 \end{matrix} \right)$$$$ | {
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I want to use $$L_{1}$$ and kronecker product to get the matrix representation of
$$\frac{\partial^2 }{\partial y \partial x }$$ and $$\frac{\partial^2 }{\partial x \partial y }$$
$$\textbf{My questions are:}$$
1. Is this possible?
2. if question 1. is affirmative, what is the matrix representation of the mixed derivatives using $$L_{1}$$ and kronecker product?
3. if question 1. is negative, how can we get the matrix representation of the mixed derivatives using a simple matrix and kronecker product?
4. Do you know a book or document( article or other ) that explain in detail this or something similar?
$$\textbf{I want to use kronecker product because it is fast and easy to implement in}$$ matlab or octave.
By the way I tried to use this formula
$$\frac{\partial^2u_{k,j} }{\partial x \partial y } = \frac{ u_{k+1,j+1}+u_{k-1,j-1}-u_{k+1,j-1}-u_{k-1,j+1} }{4h^2}$$
But it was hard to see a pattern. Thank you!
Once you have the matrix $$\bf L_1$$ to represent the divided difference of $$f(x)$$ given as a vertical vector, then if you transpose the whole you'll get a horizontal vector for the difference.
So if $$f(x,y)$$ is represented as a matrix $$\bf F_{\, x,\, y}$$ , the matrix representing $${\partial \over {\partial x\partial y}}f(x,y)$$ will be $$\bf L_1 \bf F_{\, x, \,y} \bf L_1^T$$
• Honestly I can not imagine the matrix... – tnt235711 Nov 14 '18 at 11:45
• @tnt235711: added – G Cab Nov 14 '18 at 12:19
• @tnt235711: is it ok now for you ? – G Cab Nov 15 '18 at 16:55
Now using $$\mathbf{L_{1}}$$ and $$\textbf{kronecker product}$$:
$$\frac{\partial^2 }{\partial x \partial y} = \Big((L_{1})_{n_{x}} \otimes I_{n_{y}} \Big)\Big( I_{n_{x}} \otimes (L_{1})_{n_{y}} \Big) = { (L_{1})_{n_{x}} \otimes (L_{1})_{n_{y}} }$$
I have used the mixed-product property for kronecker product. $$(\mathbf{A} \otimes \mathbf{B})(\mathbf{C} \otimes \mathbf{D}) = (\mathbf{AC}) \otimes (\mathbf{BD})$$.
In a similar way we get: | {
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In a similar way we get:
$$\frac{\partial^2 }{\partial y \partial x} = (L_{1})_{ n_{y} } \otimes (L_{1})_{n_{x}}$$ | {
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# How to solve an initial value problem if my function is implicit?
I have the differential equation:
$\dfrac{dy}{dt} = \dfrac{1}{2y+3}$
with the initial value of $y(0) = 1$
So I solve the diffeq via separation of variables:
$y^2 + 3y = t + c$
But from here, how do I solve for the function? Am I "allowed" to use the initial value information with an implicit function such as this, and give an implicit function as my answer, as below?
$y^2 + 3y = t + 4$
As far as I can tell, it should be perfectly fine to leave my answer as an implicit function. And I don't see a way to solve for $y$ here. TO be honest, I used a diffeq solver to check my work, and that tool did come up with an explicit solution, though when I tried to see where it came from I had no idea.
• This is a quadratic equaion for $y$. – kmitov Jan 13 '16 at 3:25
• @kmitov But i do not want to solve for a particular value of $y$, I want to solve for a particular solution $y(t)$. – user278703 Jan 13 '16 at 3:30
Yes, you should use your known initial condition to solve for $c$.
Proceeding to solve for $y(t)$, we write $y^2+3y+\frac{9}{4}-\frac{9}{4}=\left(y+\frac{3}{2}\right)^2-\frac{9}{4}$ by completing the square.
Now, we write $\left(y+\frac{3}{2}\right)^2-\frac{9}{4}=t+4.$ and solve for $y$ in terms of $t$.
Adding $\frac{9}{4}$, taking the square root, and subtracting $\frac{9}{2}$, we get: $$y(t)=\sqrt{t+\frac{25}{4}}-\frac{3}{2}$$
EDIT: I had misplaced a $3$.
Edit: To address your comment: Note that we can write $y(t)=\sqrt{t+\frac{25}{4}}-\frac{3}{2}=\sqrt{\frac{1}{4}(4t+25)}-\frac{3}{2}=\frac{1}{2}\sqrt{(4t+25)}-\frac{3}{2}=\frac{1}{2}(\sqrt{(4t+25)}-3).$
The reason that $c$ is different is because $c$ is some arbitrary constant. In this case, we have already solved for it. | {
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• But $\frac{9}{2} + \frac{9}{2}$ doesn't equal 3?? And $\frac{9}{2} * \frac{9}{2}$ doesn't equal $\frac{9}{4}$?? How can you factor $y^2 + 3y + \frac{9}{4}$ and get $(y+ \frac{9}{2})^2$? Shouldn't the correct factoring have $\frac{3}{2}$ instead of $\frac{9}{2}$? – user278703 Jan 13 '16 at 3:46
• I'm not factoring. I'm completing the square. For $x^2+bx$, we can add and subtract $(\frac{b}{2})^2$ to get $x^2+bx+(\frac{b}{2})^2-(\frac{b}{2})^2$, which is equal to $(x+\frac{b}{2})^2-(\frac{b}{2})^2$. Now, let $b=3$. – zz20s Jan 13 '16 at 3:51
• How? If you factor $y^2 + 3y + \frac{9}{4}$ it equals $(y+\frac{3}{2})^2$. If we remove the $-\frac{9}{4}$ from each side of the equation you wrote, it is $y^2 + 3y + \frac{9}{4} = (y+\frac{9}{2})^2$, which is wrong. Edit: I have seen your edit, thanks – user278703 Jan 13 '16 at 3:56
• No problem. Do you understand the solution now? – zz20s Jan 13 '16 at 4:02
• I understand, When I solve the diffeq on an online solver it gives $y(t) = \frac{1}{2}(-\sqrt{4t+c} - 3)$. I'm not sure where they got the $4$, or the $-3$, but I guess I'll play with it and see. – user278703 Jan 13 '16 at 4:04
From where you arrived,
$y^2+3y=t+c$
Substituting $y(0)=1$, we get $c=4$
Hence, we get $y^2+3y-4=t$
Proceed by completing the square for $y^2+3y-4=(y+\frac{3}{2})^2-\frac{7}{4}$
$(y+\frac{3}{2})^2-\frac{7}{4}=t$
Then we get $y=\frac{1}{2}(3±\sqrt{4t+7})$ | {
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# Finding the probability that someone has the disease, given they test positive on two tests
This question is from the textbook "Introduction to Probability - Blitzstein & Hwang."
I was studying for a class when I came across an example problem that I solved, but got a slightly different result than the textbook. Here's the problem in question, paraphrased:
"Fred tests for a disease which afflicts 1% of the population. The test's accuracy is deemed 95%. He tests positive for the first test, but decides to get tested for a second time. Unfortunately, Fred also tests positive for the second test as well. Find the probability that Fred has the disease, given the evidence."
$$\$$
My approach is as follows:
Let $$D$$ be the event that Fred has the disease, $$T_1$$ be the event that the first test result is positive, and $$T_2$$ be the event that the second test is also positive. We want to find $$P(D\ |\ T_1,\ T_2)$$.
We are also able to condition on $$T_1$$ (i.e. the event that the first test result is positive). This would give us:
$$P(D\ |\ T_1,\ T_2) = \frac{P(T_2\ |\ D,\ T_1)P(D\ |\ T_1)}{P(T_2\ |\ T_1)}$$
From my calculations:
$$P(T_2\ |\ D,\ T_1)\ =\ P(T_2\ |\ D)\ =\ 0.95$$ $$P(D\ |\ T_1)\ \approx\ 0.16$$ $$P(T_2\ |\ T_1)\ =\ \frac{P(T_1 ,\ T_2)}{P(T_1)}\ =\ \frac{P(T_1,\ T_2,\ D)\ +\ P(T_1,\ T_2,\ D^c)}{P(T_1,\ D)\ +\ P(T_1,\ D^c)}\ =\ \frac{0.0115}{0.059}\ \approx\ 0.19$$ $$\$$ $$P(D\ |\ T_1,\ T_2)\ =\ \frac{0.95\ \times\ 0.16}{0.19}\ =\ 0.8$$
Therefore, I concluded that there is an 80% chance that Fred has the disease, given that both the first and second test results are positive.
$$\$$
The problem is that the textbook has taken a different approach of using the odds form of Bayes' rule, which resulted in a conclusion slightly different from mine (0.78), and I'm having trouble understanding how that conclusion came to be.
$$\$$
Textbook approach is as follows: | {
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$$\$$
Textbook approach is as follows:
$$\frac{P(D\ |\ T_1,\ T_2)}{P(D^c\ |\ T_1,\ T_2)}\ =\ \frac{P(D)}{P(D^c)}\ \times\ \frac{P(T_1,\ T_2\ |\ D)}{P(T_1,\ T_2\ |\ D^c)}$$
$$=\ \frac{1}{99}\ \times\ \frac{0.95^2}{0.05^2}\ =\ \frac{361}{99}\ \approx\ 3.646$$
which "corresponds to a probability of 0.78."
$$\$$
Here are the specific questions I have:
1. Is my approach wrong? A 0.02 difference is a pretty big difference.
2. How did the author derive the equation:
$$P(D\ |\ T_1,\ T_2)\ =\ P(D)P(T_1,\ T_2\ |\ D)$$
1. What does the author mean when he/she says "3.646 corresponds to a probability of 0.78?"
$$\$$
Any feedback is appreciated. Thank you!
• A very similar question was asked on this site about 6 years ago. Here is the link to it: math.stackexchange.com/questions/198677/… – David Oct 7 '18 at 0:25
• Thanks for the feedback! I actually read that question while looking for other answers. It didn't address the questions that I had, though, which led me to ask my own. – Seankala Oct 7 '18 at 2:26
I have some quibbles with the way the textbook sets up its question, beginning with the assumption that both positive and negative tests each have the same likelihood to be correct, and even more so the assumption that the outcomes of two tests on the same person are independent in probability. In real life, I would want to explore both of those points further before advising Fred. But let's ignore those objections for the sake of being able to compute something based on the given information, and assume each administration of the test has the same chance to give a correct result, even when two tests are administered one after the other on the same person.
1. Is my approach wrong? A 0.02 difference is a pretty big difference.
The two methods are equivalent. The apparent discrepancy is due to roundoff. | {
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The two methods are equivalent. The apparent discrepancy is due to roundoff.
The textbook finds an odds ratio of $$361:99,$$ which is exact (insofar as the $$1\%$$ and $$95\%$$ are exact). Since this is $$P(D) : P(D^C),$$ the probability is given by $$P(D) = \frac{P(D)}{P(D) + P(D^C)} = \frac{361}{361 + 99} \approx 0.78478,$$ which the text rounds to $$0.78.$$ (Since you asked about this as a separate part of the question, I'll explain in more detail below.)
In your approach, $$P(T_1,\ T_2,\ D)\ +\ P(T_1,\ T_2,\ D^c) = 0.0115$$ is an exact result, and so is $$P(T_1,\ D)\ +\ P(T_1,\ D^c) = 0.059,$$ but $$0.0115 / 0.059 \approx 0.19492.$$ Meanwhile, $$P(T_1 \mid D) \approx 0.16102.$$ If we carry all these digits into the computation rather than rounding off to two places immediately, we find that $$P(D\mid T_1,\ T_2) = \frac{0.95\times 0.16102}{0.19492} \approx 0.78478.$$ That is, keeping five digits we get the same answer as the textbook method (if it retained five digits), and if we round to two digits only at the end (as the textbook does) we naturally would round the same way, to $$0.78.$$
I think an argument could be made for keeping only one digit of precision in the answer (how precise is that "$$1\%$$" anyway?), in which case both answers round to $$0.8.$$
1. How did the author derive the equation ...
They didn't. Instead, the fact is that $$P(D\mid T_1,\ T_2) = \frac{P(T_1,\ T_2\mid D)P(D)}{P(T_1,\ T_2)}$$ and $$P(D^C\mid T_1,\ T_2) = \frac{P(T_1,\ T_2\mid D^C)P(D^C)}{P(T_1,\ T_2)}.$$
When you compute the ratios of the two probabilities $$\frac{P(D\mid T_1,\ T_2)}{P(D^C\mid T_1,\ T_2)},$$ you get factors of $$P(T_1,\ T_2)$$ in both the numerator and the denominator, and these factors cancel each other.
1. What does the author mean when he/she says "3.646 corresponds to a probability of 0.78?" | {
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1. What does the author mean when he/she says "3.646 corresponds to a probability of 0.78?"
As I hinted above, $$3.646$$ is an odds ratio; or as I would rather say, the odds ratio is $$3.646 : 1.$$ An odds ratio of $$1:1$$ corresponds to a $$50\%$$ chance, that is, each possibility is equally likely, whereas a $$2:3$$ odds ratio describes something that happens twice for every three times it does not happen. in general, if the probability of something is $$p,$$ its odds ratio is $$p : (1 - p),$$ that is, $$\frac{p}{1 - p} : 1.$$
If we say $$p = P(D \mid T_1,\ T_2),$$ then $$P(D^C \mid T_1,\ T_2) = 1 - P(D \mid T_1,\ T_2) = 1 - p,$$ and what the textbook has computed is that $$\frac{p}{1 - p} \approx 3.646,$$ that is, on average in situations like this, when both tests come up positive, there will be $$3.646$$ cases in which the tests were both correct for each case in which both tests were incorrect. That means there are $$3.646$$ accurate positives for every $$3.646 + 1$$ times the test comes out positive both times, which gives a probability of $$\frac{3.646}{3.646 + 1} \approx 0.78.$$ The way I worked the probability, however, was to take the fraction $$\frac{p}{1 - p} = \frac{361}{99}$$ and directly extract an odds ratio of $$361:99$$ from it. This means I can wait until the very end before doing any roundoff, but in other respects it's the same as the textbook's method. In both cases the odds are simply $$kp : k(1 - p),$$ where $$k$$ is whatever constant you have to multiply each side by in order to produce either $$361:99$$ or $$3.646:1$$ from the odds ratio $$p : (1 - p).$$ | {
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• Hello. Thanks for the detailed and easy-to-understand answer! You've cleared up everything I was wondering regarding this concept. You were also correct in that my hasty rounding-off led to a different answer. The $0.19$ that I got is actually $0.194915$, which gives me a final answer of $0.779827$, which is $78\%$. Thanks for your explanation regarding the odds ratio as well, I was having some difficulty grasping what that is. – Seankala Oct 7 '18 at 2:12
• It's just a textbook exercise problem. The percentages were given by the authors of the textbook. I wouldn't say it's "meaningless" as it provides something that students like myself can practice with. In this example it's assumed that there is only "one population" and Fred is a part of it. How is that a bad assumption? – Seankala Oct 7 '18 at 11:49
• @David So your objections are that Fred might not be a member of "the" population, that the $1\%$ figure might be inaccurate or outdated. And indeed in real life these are things to consider, in addition to the unstated assumptions made about the frequency of errors of the test. But mathematics is never about absolute answers to real-life questions; it is about seeing what conclusions can be drawn from a given set of assumptions. To refuse to answer that question mathematically because the assumptions might be wrong, seems like a cheap excuse to me. – David K Oct 7 '18 at 11:49
• @David The link you gave to the other question is a useful one, as the answers to that question include a real-life example that shows why we should consider it important that a disease has a very low incidence in the population. – David K Oct 7 '18 at 11:53 | {
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• In real life, people only usually get tested for a rare disease if there is some suspicion or risk factor involved, in which case those risk factors will change your prior significantly. If Fred was only sent in to test because he had a family history of this rare disease, then your prior is going to be more than 1%. But this is a math exercise from a textbook. The point is to solve it with only what information we know, ceteris paribus. – Zubin Mukerjee Oct 7 '18 at 13:56 | {
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The conditional probability (or Bayes's rule) must be: $$P(D\ |\ T_1\cap T_2) = \frac{P(D\cap T_1\cap T_2)}{P(T_1\ \color{red}{\cap}\ T_2)}=\frac{P(D)\cdot P(T_1|D)\cdot P(T_2|D\cap T_1)}{P(T_1\ \cap\ T_2)}=\\ \frac{0.01\cdot 0.95^2}{0.01\cdot 0.95^2+0.99\cdot 0.05^2}\approx 0.7848.$$
• That's nice, but that doesn't mean it applies to this question as stated. – David Oct 6 '18 at 16:41
• @David would you care to elaborate why you think Bayes formula does not apply to a problem so clearly amenable to conditioning... – Nap D. Lover Oct 7 '18 at 2:12
• Fred is a part of the population! I think that you're right, that the question should have stated it explicitly, but mentioning a population that is completely unrelated to Fred would be a hilarious and unlikely red herring by the textbook authors. Why would they just randomly mention a population in a question about Fred, if Fred wasn't part of the population? Once you do assume Fred is part of the population, you can no longer ignore the 1% affliction rate. – Zubin Mukerjee Oct 7 '18 at 14:03
• It was fun commenting but given the assumptions are true, I agree with these answers here. It was just an exercise on being accurate about assumptions so many next time someone posts a question like this, they will explicitly state the "missing" info. I deleted all my previous answers and comments to clean this up. – David Oct 7 '18 at 15:40 | {
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# For numerical integration, is it true that higher degree of precision gives better accuracy always?
In case of numerical integration, is it true that higher degree of precision always gives better accuracy? Justify your answer.
I know the definition of degree of precision. For Trapezoidal and Simpson's 1/3 rule they are 1 and 3 respectively. Simpson's 1/3 gives better accuracy than Trapezoidal rule. Then whether the above statement is true always. If not, why? If yes, then why we learn Trapezoidal/ Simpson rule? Why we shall not establish/go for higher and higher DOP from generalized Newton-Cote's rule or other general quadrature formula . | {
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• Usually, but not always. Consider the function $f(x) = \left\{\matrix{1 & x < 1/2\\0 & x \geq 1/2}\right.$. The trapesoidal rule with $n=3$ points ($x=0,1/2,1$) gets the integral $\int_0^1f(x){\rm d}x$ exactly, but if we use an even number of points we will always be a bit off the true result no matter how large $n$ is. – Winther Jan 3 '16 at 8:00
• @Winther Please give me the answer when higher degree of precision gives better accuracy and when higher degree of precision does not give better accuracy with analytical justification. – user1942348 Jan 3 '16 at 8:10
• That is almost impossible I think (for general functions). The key to the proofs of the order of the different methods is that as long as functions have bounded derivatives (of some order) then the error will grow roughly as $\frac{1}{N^k}$ for some some integer $k$ (and consequently go to zero when $N\to\infty$). However we usually don't have any control if $N=38$ will give better results than $N=40$. – Winther Jan 3 '16 at 8:12
• @Winther What do you mean by " a bit off the true result"? – user1942348 Jan 3 '16 at 8:18
• @winther Somewhere I read that In particular, "when range of integration is large", the statement above is not correct. Whether it ("when range of integration is large, higher DOP gives less accuracy") is correct. If correct, would you explore why? – user1942348 Jan 3 '16 at 8:24
Increasing the precision, both in terms of the order of the method and in the number of gridpoints used, usually (most of the time) leads to a more accurate estimate for the integral we are trying to compute. However this is not always the case as the following (aritficial) example shows.
$$\bf \text{Example where higher order does not imply better accuracy}$$
Let
$$f(x) = \left\{\matrix{1 & x < \frac{1}{2}\\0 & x \geq \frac{1}{2}}\right.$$
and consider the integral $I=\int_0^1f(x){\rm d}x = \frac{1}{2}$. If we use the trapezoidal rule with $n$ gridpoints then | {
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$$I_{n} = \frac{1}{n}\sum_{i=1}^{n}\frac{f(\frac{i-1}{n})+f(\frac{i}{n})}{2} \implies I_n = \left\{\matrix{\frac{1}{2} & n~~\text{odd}\\\frac{1}{2} - \frac{1}{n} & n~~\text{even}}\right.$$
so for $n=3$ we have the exact answer which is better than any even $n$ no matter how large it is. This shows that increasing the number of gridpoints does not always improve the accuracy. With Simpson's rule we find
$$I_n = \frac{1}{3n}\sum_{i=1}^{n/2}f\left(\frac{2i-2}{n}\right)+4f\left(\frac{2i-1}{n}\right)+f\left(\frac{2i}{n}\right) \implies I_n = \left\{\matrix{\frac{1}{2} - \frac{1}{3n}&n\equiv 0\mod 4\\\frac{1}{2}&n\equiv 1\mod 4\\\frac{1}{2} + \frac{2}{3n} & n\equiv 2\mod 4\\\frac{1}{2} - \frac{5}{6n} & n\equiv 3\mod 4}\right.$$
so even if Simpson's rule has higher order we see that it does not always do better than the trapezoidal rule.
$$\bf \text{What does higher degree of precision really mean?}$$
If we have a smooth function then a standard Taylor series error analysis gives that the error in estimating the integral $\int_a^bf(x){\rm d}x$ using $n$ equally spaced points is bounded by (here for Simpsons and the trapezoidal rule)
$$\epsilon_{\rm Simpsons} = \frac{(b-a)^5}{2880n^4}\max_{\zeta\in[a,b]}|f^{(4)}(\zeta)|$$ $$\epsilon_{\rm Trapezoidal} = \frac{(b-a)^3}{12n^2}\max_{\zeta\in[a,b]}|f^{(2)}(\zeta)|$$
Note that the result we get from such an error analysis is always an upper bound (or in some cases an order of magnitude) for the error apposed to the exact value for the error. What this error analysis tell us is that if $f$ is smooth on $[a,b]$, so that the derivatives are bounded, then the error with a higher order method will tend to decrease faster as we increase the number of gridpoints and consequently we typically need fewer gridpoints to get the same accuracy with a higher order method. | {
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The order of the method only tell us about the $\frac{1}{n^k}$ fall-off of the error and says nothing about the prefactor in front so a method that has an error of $\frac{100}{n^2}$ will tend to be worse than a method that has an error $\frac{1}{n}$ as long as $n\leq 100$.
$$\bf \text{Why do we need all these methods?}$$
In principle we don't need any other methods than the simplest one. If we can compute to arbitrary precision and have enough computation power then we can evaluate any integral with the trapezoidal rule. However in practice there are always limitations that in some cases forces us to choose a different method.
Using a low-order method requires many gridpoints to ensure good enough accuracy which can make the computation take too long time especially when the integrand is expensive to compute. Another problem that can happen even if we can afford to use as many gridpoints as we want is that truncation error (errors due to computers using a finite number of digits) can come into play so even if we use enough points the result might not be accurate.
Other methods can elevate these potential problems. Personally, whenever I need to integrate something and has to implement the method myself I always start with a low-precision method like the trapezoidal rule. This is very easy to implement, it's hard to make errors when coding it up and it's usually good enough for most purposes. If this is not fast enough or if the integrand has properties (e.g. rapid osccilations) that makes it bad I try a different method. For example I have had to compute (multidimensional) integrals where a trapezoidal rule would need more than a year to compute it to good enough accuracy, but with Monte-Carlo integration the time needed was less than a minute! It's therefore good to know different numerical integration methods in case you encounter a problem where the simplest method fails. | {
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• Your answer is very impressive. Thanks a lot. You have given an example of $f(x)$. It is not differentiable at x=1/2. For such non-smooth function, the Trap is better than Simp. Is there any example where $f(x)$ is continuous and smooth but Trap gives better accuracy than Simp? – user1942348 Jan 4 '16 at 13:50
• @user1942348 Yes there is. For example take $f(x)$ and connect $x = 1/2 - \epsilon$ to $x = 1/2 + \epsilon$ with a line (smoothed at the corners). Take $\epsilon$ to be tiny and you'll get pretty much the same result as I got here. – Winther Jan 4 '16 at 14:21
• I just googled and see that for periodic function with limits of integration one with its period, Trap works better. self.gutenberg.org/articles/trapezoidal_rule Do you know is this correct. Can you give such example for me? – user1942348 Jan 4 '16 at 15:55
• @user1942348 I think the statement on that page is that Trap does better than what we naively would expect from looking at the error term for functions that oscillate, not that it neccesarily is much better than say Simpsons. I tried some random test cases like $\sin(2\pi x)$ and $[x(1-x)]^4$ and Trap does indeed do better than expected and the acctual error as function of $n$ is similar to Simpsons for these two functions ($\epsilon \propto 1/n^6$ for the latter function). – Winther Jan 4 '16 at 16:49
• @user1942348 btw let me know if you have Mathematica and I can add the code I used to test this so you can play with it and see for yourself. – Winther Jan 4 '16 at 16:53 | {
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Find all pairs (m,k) of integer solutions
anemone
MHB POTW Director
Staff member
Find all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.
Opalg
MHB Oldtimer
Staff member
Find all solutions to $(m^2+k)(m+k^2)=(m-k)^3$, where $m$ and $k$ are non-zero integers.
Multiply out the brackets: $$\displaystyle m^3 + m^2k^2 + mk + k^3 = m^3 -3m^2k + 3mk^2 - k^3$$, which (after dividing by the non-zero $k$) simplifies to $2k^2 + (m^2-3m)k + (m+3m^2) = 0.$ The discriminant of that quadratic in $k$ is $(m^2-3m)^2 - 8(m+3m^2)$, and that has to be a square, say $n^2$. So $$\displaystyle n^2 = (m^2-3m)^2 - 8(m+3m^2) = m^4 - 6m^3 - 15m^2 - 8m = m(m-8)(m+1)^2$$ and therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$. Thus $m = 9$ or $-1$. If $m=9$ then the quadratic for $k$ has solutions $k= -21$ and $k=-6$. If $m=-1$ then the only solution for $k$ is $k=-1$.
That gives three possible solutions, $(m,k) = (9,-21),\ (9,-6),\ (-1,-1)$.
Klaas van Aarsen
MHB Seeker
Staff member
therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.
Why is that?
Opalg
MHB Oldtimer
Staff member
therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.
Why is that?
If $x^2-16 = y^2$ then $16 = x^2-y^2 = (x+y)(x-y)$. The only factorisation of $16$ giving positive integer values for $x$ and $y$ is $x+y=8$, $x-y=2$, so that $x=5$ and $y=3$.
anemone
MHB POTW Director
Staff member
Thank you so much for participating, Opalg! And welcome back to the forum as well! But Opalg, I am sorry to tell you that you missed another solution set, which is $(m, k)=(8, -10)$.
My solution:
By expanding and simplifying the given equation, we get
$$\displaystyle m^3+m^2k^2+mk+k^3=m^3-3m^2k+3mk^2-k^3$$
$$\displaystyle 2k^3+(m^2-3m)k^2+mk+3m^2k=0$$
Since $k \ne 0$, divide through the equation above by $k$, we have | {
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Since $k \ne 0$, divide through the equation above by $k$, we have
$$\displaystyle 2k^2+(m^2-3m)k+m+3m^2=0$$
and solve for $k$ using the quadratic formula yields
$$\displaystyle k=\frac{-(m^2-3m) \pm \sqrt{(m^2-3m)^2-4(2)(m+3m^2)}}{2(2)}$$
$$\displaystyle \;\;\;=\frac{3m-m^2 \pm \sqrt{(m+1)^2(m(m-8))}}{4}$$
$$\displaystyle \;\;\;=\frac{3m-m^2 \pm (m+1)\sqrt{m(m-8)}}{4}$$
Recall that the expression inside the radical must be greater than or equal to zero, thus, $$\displaystyle m(m-8)\ge 0$$ and this gives
$m\le 0$ and $m \ge 8$.
Also, bear in mind that we're told $m, k$ are both integer values, so, $m(m-8)$ has to be a squre.
But if we apply the AM-GM inequality to the terms $m$ and $m-8$, we find
$$\displaystyle \frac{m+m-8}{2} \ge \sqrt{m(m-8)}$$
$$\displaystyle m-4 \ge \sqrt{m(m-8)}$$
$m\ge 8$ implies $m-4 \ge 4$ and this further implies $\sqrt{m(m-8)} \le 4$ and now, we will solve for $m$ by considering $\sqrt{m(m-8)}=0$ or $1$ or $2$ or $3$or $4$ and we will take whichever solution that gives the integer values of $m$.
$$\displaystyle \sqrt{m(m-8)}= 4$$ gives irrational $m$ values.
$$\displaystyle \sqrt{m(m-8)}= 3$$ gives $m=-1, 9$.
$$\displaystyle \sqrt{m(m-8)}= 2$$ gives irrational $m$ values.
$$\displaystyle \sqrt{m(m-8)}= 1$$ gives irrational $m$ values.
$$\displaystyle \sqrt{m(m-8)}= 0$$ gives irrational $m=0, 8$ values but $m>0$, hence $m=8$.
Thus, by substituting each of the value of $m$ to the equation (*) above to find for its corresponding $k$ value, we end up with the following 4 solution sets, namely
$(m, k)=(-1, -1), (8, -10), (9, -6), (9, -21)$
Opalg | {
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$(m, k)=(-1, -1), (8, -10), (9, -6), (9, -21)$
Opalg
MHB Oldtimer
Staff member
therefore $m(m-8) = (m-4)^2 - 16$ must also be a square. The only way that can happen is if $m-4 = \pm5$.
Why is that?
Oops, I should have taken I like Serena's query more seriously. I was thinking that the only way you could have $x^2-16 = y^2$ was in the situation $5^2 - 4^2 = 3^2$. But in the solution to this problem, the situation $4^2 - 4^2 = 0^2$ is also possible. That leads to the solution $m=8$ and $k=-10$. | {
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TrigonometryReview my solution: Trigonometry proof
sweatingbear
Member
For a triangle with sides $$\displaystyle a$$, $$\displaystyle b$$, $$\displaystyle c$$ and angle $$\displaystyle C$$, where the angle $$\displaystyle C$$ subtends the side $$\displaystyle c$$, show that
$$\displaystyle c \geqslant (a+b) \sin \left( \frac C2 \right)$$
_______
Let us stipulate $$\displaystyle 0^\circ < C < 180^\circ$$ and of course $$\displaystyle a,b,c > 0$$. Consequently, $$\displaystyle 0^\circ < \frac C2 \leqslant 90^\circ \implies 0 < \sin^2 \left( \frac C2 \right) \leqslant 1$$ (we include $$\displaystyle 90^\circ$$ for the angle $$\displaystyle \frac C2$$ in order to account for right triangles).
Law of cosines yield
$$\displaystyle c^2 = a^2 + b^2 - 2ab \cos ( C )$$
Using $$\displaystyle \cos (C) \equiv 1 - 2\sin^2 \left( \frac C2 \right)$$, we can write
$$\displaystyle c^2 = a^2 + b^2 - 2ab + 4ab\sin^2 \left( \frac C2 \right)$$
Now since $$\displaystyle 0 < \sin^2 \left( \frac C2 \right) \leqslant 1$$, $$\displaystyle a^2 + b^2 - 2ab + 4ab\sin^2 \left( \frac C2 \right)$$ will either have to equal $$\displaystyle a^2 + b^2 - 2ab + 4ab$$ or be greater than it. Thus
$$\displaystyle c^2 \geqslant a^2 + b^2 - 2ab + 4ab = a^2 + 2ab + b^2 = (a+b)^2$$
A similar argument can be made for $$\displaystyle (a+b)^2$$ versus $$\displaystyle (a+b)^2 \sin^2 \left( \frac C2 \right)$$: $$\displaystyle (a+b)^2$$ will either have to equal or be greater than the latter expression, due to the values $$\displaystyle \sin^2 \left( \frac C2 \right)$$ can assume. Therefore
$$\displaystyle c^2 \geqslant (a+b)^2 \geqslant (a+b)^2 \sin^2 \left( \frac C2 \right)$$
We can finally conclude
$$\displaystyle c \geqslant (a+b) \sin \left( \frac C2 \right)$$
Thoughts?
MarkFL
Staff member
I see nothing wrong with your proof. This is how I would prove it. Please refer to the following diagram:
I would postulate concerning the angle bisector $m$ of $\angle C$ and the altitude $h$, we must have: | {
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(1) $$\displaystyle m\ge h$$
That is, the shortest distance between a point and a line is the perpendicular distance.
Now, the area $T$ of the triangle may be written in these ways:
$$\displaystyle T=\frac{1}{2}ch=\frac{1}{2}m(a+b)\sin\left(\frac{C}{2} \right)$$
Note: I have made use of the formulas (for a general triangle):
$$\displaystyle T=\frac{1}{2}bh$$
$$\displaystyle T=\frac{1}{2}ab\sin(C)$$
Thus, we have:
$$\displaystyle ch=m(a+b)\sin\left(\frac{C}{2} \right)$$
And from (1), we therefore conclude:
$$\displaystyle c\ge (a+b)\sin\left(\frac{C}{2} \right)$$
sweatingbear
Member
I see nothing wrong with your proof. This is how I would prove it. Please refer to the following diagram:
View attachment 1085
I would postulate concerning the angle bisector $m$ of $\angle C$ and the altitude $h$, we must have:
(1) $$\displaystyle m\ge h$$
That is, the shortest distance between a point and a line is the perpendicular distance.
Now, the area $T$ of the triangle may be written in these ways:
$$\displaystyle T=\frac{1}{2}ch=\frac{1}{2}m(a+b)\sin\left(\frac{C}{2} \right)$$
Note: I have made use of the formulas (for a general triangle):
$$\displaystyle T=\frac{1}{2}bh$$
$$\displaystyle T=\frac{1}{2}ab\sin(C)$$
Thus, we have:
$$\displaystyle ch=m(a+b)\sin\left(\frac{C}{2} \right)$$
And from (1), we therefore conclude:
$$\displaystyle c\ge (a+b)\sin\left(\frac{C}{2} \right)$$
Thank for the feedback and the excellent alternative solution!
Deveno
Well-known member
MHB Math Scholar
A good rule of thumb in mathematics I live by is this:
If something is true, you should be able to provide two proofs, and one of them should have a picture. This is a perfect example of what I mean. | {
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# Math Help - divide by 3?
1. ## divide by 3?
hi folks,
I hope this is the right place to post this! I have no idea how to proceeed with the following question:
Show that for all positive integral values of n
[HTML]7 <sup>n</sup> + 2 <sup>2n+1</sup>[/HTML]
is divisible by 3.
I tried a few terms as follows:
n = 1. 7 ^ 1 + 2 ^ 3 = 7 + 8 = 15
n = 2. 7
^ 2 + 2 ^ 5 = 49 + 32 = 81
n = 3. 7
^ 3 + 2 ^ 7 = 343 + 128 = 471
and these are all divisible by 3 but how do I handle the general case?
I thought of expanding the expression i.e.
2.2 ^ n. 2 ^ n - (1 - 8) ^ n
and using a binomial on the second term but it doesn't get me anywhere. I guess I am trying to calculate the sum of a series and show that it has a factor of 3 but I can't see how to do it. Any ideas?
regards and thanks
Simon
p.s. sorry about the formating. the HTML stuff doesn't seem to come out so I resorted to the ^ symbol which is pretty unpretty!
2. Hi,
Originally Posted by s_ingram
I thought of expanding the expression i.e.
2.2 ^ n. 2 ^ n - (1 - 8) ^ n
and using a binomial on the second term but it doesn't get me anywhere. I guess I am trying to calculate the sum of a series and show that it has a factor of 3 but I can't see how to do it. Any ideas?
I suggest writing $7^n+2^{2n+1}=(2^2+3)^n+2^{2n+1}$ and then using the binomial theorem to expand $(2^2+3)^n$.
Originally Posted by s_ingram
p.s. sorry about the formating. the HTML stuff doesn't seem to come out so I resorted to the ^ symbol which is pretty unpretty!
To enter math. equations you can use LaTeX. (see http://www.mathhelpforum.com/math-help/latex-help/)
3. $(2^{2}+3)^{n} = 2^{2n} + \binom{n}{1}2^{2(n-1)}3^{1} + ... + \binom{n}{k}2^{2(n-k)}3^{k} + ... + 3^{n} =$
$= 2^{2n} + 3[\binom{n}{1}2^{2(n-1)} + ... + \binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}]$
So we can write: | {
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So we can write:
$3[\binom{n}{1}2^{2(n-1)} + ... + \binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}] + 2^{2n}(1+2) =$ $3[\binom{n}{1}2^{2(n-1)} + ... + \binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}] + 2^{2n}(3)$
Now factor out a 3 from both terms.
$3[(\binom{n}{1}2^{2(n-1)} + ... + \binom{n}{k}2^{2(n-k)}3^{k-1} + ... + 3^{n-1}) + 2^{2n} ]$
And because we have a number times 3, the result is divisible by 3.
Note: If there aint a careless mistake somewhere in this mess, I am surprised.
4. thanks to Twig and flyingsquirrel. You guys make it look so easy!
5. If you do it by induction here is the last step.
$\begin{array}{rcl}
{7^{n + 1} + 2^{2n + 3} } & = & {7^{n + 1} + 7 \cdot 2^{2n + 1} - 7 \cdot 2^{2n + 1} + 2^{2n + 3} } \\
{} & {} & {7\left( {7^n + 2^{2n + 1} } \right) - 2^{2n + 1} \left( {7 - 2^2 } \right)} \\
{} & {} & {7\left( {7^n + 2^{2n + 1} } \right) - 2^{2n + 1} \left( 3 \right)} \\ \end{array}$
6. Hello, s_ingram!
How about an inductive proof?
Show that for any positive integer $n \!:\;\;7^n + 2^{2n+1}$ is divisible by 3.
Verify $S(1)\!:\;\;7^1 + 2^3 \:=\:7+8 \:=\:15$ ...
divisible by 3.
Assume $S(k)\!:\;\;7^k + 2^{2k+1} \:=\:3a\;\text{ for some integer }a.$
Add $6\!\cdot\!7^k + 3\!\cdot\!2^{2k+1}$ to both sides:
. . $7^k + {\color{blue}6\!\cdot\!7^k} + 2^{2k+1} + {\color{blue}3\!\cdot\!2^{2k+1}} \;=\;3a + {\color{blue}6\!\cdot\!7^k + 3\!\cdot\!2^{2k+1}}$
. . $(1 + 6)\!\cdot\!7^k + (1 + 3)\!\cdot\!2^{2k+1} \;=\;3\left(2a + 2\!\cdot\!7^k + 2^{2k+1}\right)$ .
a multiple of 3
. . . . . . . . $7\!\cdot\!7^k + 2^2\!\cdot2^{2k+1} \;=\;3b\;\;\text{ for some integer }b$
Therefore: . . $7^{k+1} + 2^{2k+3} \;=\;3b$
We have proved $S(k+1).$
. . The inductive proof is complete. | {
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We have proved $S(k+1).$
. . The inductive proof is complete.
7. Hi Soroban, now that is a smart proof. Too smart! How did you think adding those two terms? Once you see them it all fits but finding them is the real trick! I have never really been impressed with proofs by induction, but I am now! Usually I just try to add the next term in the series and ensure that it has a corresponding impact on the expression fo the sum but with 7 sup k+1 + 2 sup 2k+1 I couldn't even imagine what the next term would be!
8. Or you can just note that
$7^n\equiv 1^n \equiv 1 \mod 3$,
$2^{2n+1} \equiv (-1)^{2n+1} \equiv -1 \mod 3$,
so that $7^n+2^{2n+1} \equiv 0 \mod 3$.
9. Originally Posted by s_ingram
Hi Soroban, now that is a smart proof. Too smart! How did you think adding those two terms? Once you see them it all fits but finding them is the real trick! I have never really been impressed with proofs by induction, but I am now! Usually I just try to add the next term in the series and ensure that it has a corresponding impact on the expression fo the sum but with 7 sup k+1 + 2 sup 2k+1 I couldn't even imagine what the next term would be!
Hi
$7^{n+1} + 2^{2n+3} = 7(7^{n} + 2^{2n+1})-7 \cdot 2^{2n+1} + 2^{2n+3} = 7 \cdot 3a + 2^{2n+1}(-7+2^2) = 7 \cdot 3a -3 \cdot 2^{2n+1}$
Therefore
$7^{n+1} + 2^{2n+3} = 3(7a-2^{2n+1})$ | {
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# [SOLVED]Another Solid of Revolution problem
##### Member
I'm finding myself stuck again. This time it is more in the set up then the solving.
Find the volume of the following solid of revolution.
The region bounded by $$\displaystyle y=\frac{1}{x^2}$$, y=0, x=2, x=8, and revolved about the y-axis.
I am trying to use the shell method to solve this, as it seems the best scenario in this situation. This is my set-up,
$$\displaystyle 2\pi\int_2^8 (radius)(height)dx$$
$$\displaystyle 2\pi\int_2^8 (x+2)(\frac{1}{x^2})dx$$
$$\displaystyle 2\pi\int_2^8 (\frac{1}{x}+\frac{2}{x^2})dx$$
Now here is where I am getting into a snag. Following this through I am coming up with a completely different solution to the manual, and the solution manual is showing the first integration step as $$\displaystyle 2\pi\int_2^8 (\frac{1}{x})dx$$, but I can't figure out how they are coming up with that integral. It is showing a final solution of $$\displaystyle \pi\ln(16)$$
Any help, or a kick in the right direction, would be greatly appreciated.
Mac
#### Chris L T521
##### Well-known member
Staff member
[snip]
Now here is where I am getting into a snag. Following this through I am coming up with a completely different solution to the manual, and the solution manual is showing the first integration step as $$\displaystyle 2\pi\int_2^8 (\frac{1}{x})dx$$, but I can't figure out how they are coming up with that integral. It is showing a final solution of $$\displaystyle \pi\ln(16)$$
Any help, or a kick in the right direction, would be greatly appreciated.
Mac
[/snip]
That's because the radius of your solid of revolution is just $x$, not $x+2$. $x$ is always measured from the $y$-axis (i.e. $x=0$), so there's no need to add 2 to this value; in this case, $x$ just happens to be between 2 and 8.
On the other hand, the radius would be $x+2$ if you were revolving the region about the line $x=-2$.
I hope this clarifies things!
#### CaptainBlack | {
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I hope this clarifies things!
#### CaptainBlack
##### Well-known member
I'm finding myself stuck again. This time it is more in the set up then the solving.
Find the volume of the following solid of revolution.
The region bounded by $$\displaystyle y=\frac{1}{x^2}$$, y=0, x=2, x=8, and revolved about the y-axis.
I am trying to use the shell method to solve this, as it seems the best scenario in this situation. This is my set-up,
$$\displaystyle 2\pi\int_2^8 (radius)(height)dx$$
$$\displaystyle 2\pi\int_2^8 (x+2)(\frac{1}{x^2})dx$$
$$\displaystyle 2\pi\int_2^8 (\frac{1}{x}+\frac{2}{x^2})dx$$
Now here is where I am getting into a snag. Following this through I am coming up with a completely different solution to the manual, and the solution manual is showing the first integration step as $$\displaystyle 2\pi\int_2^8 (\frac{1}{x})dx$$, but I can't figure out how they are coming up with that integral. It is showing a final solution of $$\displaystyle \pi\ln(16)$$
Any help, or a kick in the right direction, would be greatly appreciated.
Mac
Where did a radius of $$x+2$$ come from? Why not $$x$$?
CB | {
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Please help me to prove the convergence of $\gamma_{n} = 1+\frac{1}{2}+\frac{1}{3}+…+\frac{1}{n}-\ln n$
Please help me to prove that $\gamma_{n} = 1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{n}-\ln n$ converge and it's limit $\gamma \in [0,1]$. Then, using $\gamma_{n}$ find the sum: $\sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}$.
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Welcome to MSE! I'd suggest though you'd ask question in a less imperative way and in a more friendly way. Plus, have you managed to find something? Any approaches? What I mean is, you're not a book, you're human, so act like one, it makes us more willing to answer. =) – Patrick Da Silva Aug 10 '11 at 11:49
The following question addresses the convergence – Sasha Aug 10 '11 at 11:56
possible duplicate of What is $\lim_{n \to \infty}\sum_{k=1}^n 1/k$ – user17762 Aug 10 '11 at 12:14
Sorry because my question was imperative. My english is not very good and I translated word by word from my language. I will be more carefully next time. – NumLock Aug 10 '11 at 12:28
In the sense that if you read the answer given, all the good stuff's in there. Although they don't compute the alternated sum. I don't agree it's a duplicate. – Patrick Da Silva Aug 10 '11 at 12:56
The following question addresses the convergence, so assume $\gamma_n$ converges and denote the limit as $\gamma$.
Now consider $\sum_{n=1}^{2m} (-1)^{n+1} \frac{1}{n}$ and split it into even and odd terms $\sum_{n=1}^m \frac{1}{2n-1} - \sum_{n=1}^m \frac{1}{2n}$. Then complete the sum over odd integers to sum over consequtive integers: $\sum_{n=1}^{2m} \frac{1}{n} - 2 \sum_{n=1}^m \frac{1}{2n}$.
Then subtract logarithms to form $\gamma_{2m} - 2 \gamma_m + \log(2)$, like so $$( \sum_{n=1}^{2m} \frac{1}{n} - \log(2m)) - ( \sum_{n=1}^m \frac{1}{n} - \log(m)) + \log(2).$$
In the limit $m \to \infty$ it becomes $\gamma - \gamma + \log(2) = \log(2)$.
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Thank you for your help too! – NumLock Aug 10 '11 at 12:34 | {
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-
Thank you for your help too! – NumLock Aug 10 '11 at 12:34
You can use the mean value theorem to prove first that $$\frac{1}{n+1}<\ln(n+1)-\ln n<\frac{1}{n}$$ then, $\ln(n+1)<1+\ldots+\frac{1}{n}<1+\ln n$, and finally $$\ln(n+1)-\ln n<\gamma_n<1$$ which implies the convergence of $\gamma_n$ and its limit $\gamma\in[0,1]$.
For the second part, I am not sure if you can use $\gamma_n$ to find the sum $\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n}$!
I suggest to use the Taylor-Young formula for $x\mapsto\ln(x+1)$ on $[0,1]$ to prove that $$\sum_{n=1}^\infty(-1)^{n+1}\frac{1}{n}=\ln2.$$
-
Thank you !I learnt mean value theorem in highschool but I completely forgot about it. – NumLock Aug 10 '11 at 12:34
@amine it's mercator series time... – user38268 Aug 10 '11 at 12:58
The Mean Value Theorem says $$\frac{1}{k+1}<\log(k+1)-\log(k)<\frac{1}{k}$$ Summing from $m$ to $n-1$ ($m < n$), we get \begin{align} \sum_{k=m}^{n-1}\frac{1}{k+1}<\log(n)-\log(m)<\sum_{k=m}^{n-1}\frac{1}{k} \end{align} Subtracting $\displaystyle{\sum_{k=m}^{n-1}\frac{1}{k+1}=\sum_{k=m+1}^n\frac{1}{k}=\sum_{k=1}^n\frac{1}{k}-\sum_{k=1}^m\frac{1}{k}}$ from all parts, we get $$0<\left(\log(n)-\sum_{k=1}^n\frac{1}{k}\right)-\left(\log(m)-\sum_{k=1}^m\frac{1}{k}\right)<\frac{1}{m}-\frac{1}{n}$$ This proves the existence of $\displaystyle{\lim_{n\to\infty}\left(\log(n)-\sum_{k=1}^n\frac{1}{k}\right)}$. If we set $m=1$ and subtract all sides from $1$, we get $$1>\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)>\frac{1}{n}$$Thus, we have shown that $\displaystyle{\gamma=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)}$ exists, and that $0\le\gamma\le 1$. | {
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Note that \begin{align} \sum_{k=1}^{2n}(-1)^{k+1}\frac{1}{k}&=\sum_{k=1}^{2n}\frac{1}{k}-2\sum_{k=1}^n\frac{1}{2k}\\ &=\sum_{k=1}^{2n}\frac{1}{k}-\sum_{k=1}^n\frac{1}{k}\\ &=\left(\sum_{k=1}^{2n}\frac{1}{k}-\log(2n)\right)-\left(\sum_{k=1}^n\frac{1}{k}-\log(n)\right)+\log(2) \end{align} Taking the limit as $n\to\infty$, we get $$\sum_{k=1}^\infty(-1)^{k+1}\frac{1}{k}=\gamma-\gamma+\log(2)=\log(2)$$
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I apologize if this is close to Sasha's answer, but I have been working on this, during breaks, for quite a while, and posted it before I noticed. Since I show more of the details, I will leave it for now. I can remove it if requested. – robjohn Aug 10 '11 at 17:26
Sorry, for the convergence, I forgot to say that $\gamma_n$ is decreasing because $$\gamma_{n+1}-\gamma_n=\frac{1}{n+1}-(\ln(n+1)-\ln n)<0$$ since it is bounded, it is convergent! – amine Aug 10 '11 at 21:49
Do we have a "proof without words" question here? I don't find it. There is a lovely diagram found in certain calculus texts that does this for us. Maybe I will animate it, for fun.
-
I am not getting this. Do you mean pictorial proof, which suggests the proof ? – Sasha Aug 10 '11 at 14:15
It's something like the diagram at the top right here: en.wikipedia.org/wiki/Euler%E2%80%93Mascheroni_constant but then you slide all the blue regions to the left to see that they all fit inside a $1 \times 1$ square. – GEdgar Aug 10 '11 at 14:54 | {
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# Math Help - word problem.
1. ## word problem.
hi ! im having a difficult time answering this word problem.
"Alvin set out from a certain point and travelled at the rate of 6 kph. After Alvin had gone for two hours, Ben set on to overtake him and went 4km the first hour, 5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. After many hours were they together."
I keep on getting 5 using arithmetic progression. Could someone please please give me a hint how to solve this one ???
Thank you very much !!
You can check:
In 5 hours, Alvin travels : 6+6+6+6+6 = 30 kms
In 5 hours, Ben travels: 4+5+6+7+8=30kms
that means they were together after 5 hours.
3. Nope. Ben started 2 hours after
4. It should be 10 hours. But I don't know how to prove it using arithmetic progression.
5. Here's what I did. We know these two things about arithmetic progressions:
$a_n = a_1 + (n - 1)d$ where d is the distance between numbers in the sequence. We also know that:
$S_n = \frac{n}{2}(a_1 + a_n)$ where $S_n$ is the sum of the first n terms.
We also know that the distance traveled by the first is just 6t.
I'll use t instead of n in the formulas. All t's for the second one should be (t - 2) since he doesn't move for 2 hours.
So we want to find out when 6t equals the sum of the arithmetic progression at (t - 2). I set up this equation:
$6t = \frac{t-2}{2} (a_1 + a_n) = \frac{t-2}{2} (4 + a_1 + ((t - 2) -1))$
$6t = \frac{t - 2}{2}(4 + 4 + (t - 2) - 1)$
Hope you see what I did there. After that, rearrange terms so you get:
$t^2 - 9t - 10 = 0$
For that, I get one positive root of t = 10 (also t = -1 which is obviously not useful). Voila! Thanks for the problem, I had fun with that.
6. Hello, dugongster!
Alvin set out from a certain point and travelled at the rate of 6 kph.
After Alvin had gone for 2 hours, Ben set on to overtake him and went 4km the first hour,
5 km the second hour, 6 km the third hour and so on, gaining 1 km every hour. | {
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After many hours were they together?
. . In the next $\,h$ hours, he travels another $6h$ km.
Alvin has traveled $6h + 12$ km in the first $\,h$ hours.
During the same $\,h$ hours, Ben has travelled:
. . $4 + 5 + 6 + \hdots + (h+3) \:=\:\dfrac{h(h+7)}{2}$ km. .**
The two distances are equal: . $\dfrac{h(h+7)}{2} \:=\:6h + 12$
. . $h^2 + 7h \:=\:12h + 24 \quad\Rightarrow\quad h^2 - 5h - 24 \:=\:0$
. . $(h - 8)(h + 3) \:=\:0 \quad\Rightarrow\quad h \:=\:8,\:-3$
Answer: . $\text{8 hours}$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
We want the sum of the numbers from $\,4$ to $h\!+\!3.$
The sum of the numbers from $\,1$ to $h\!+\!3$ is: . $\dfrac{(h+3)(h+4)}{2}$
The sum of the numbers from $\,1$ to $\,3$ is: . $1+2+3 \:=\:6$
Hence,the sum of the numbers from $\,4$ to $h\!+\!3$ is:
. . $\dfrac{(h+3)(h+4)}{2} - 6 \;\;=\;\; \dfrac{h^2 + 7x + 12 - 12}{2} \;\;=\;\;\dfrac{h(h+7)}{2}$
7. great. thanks to both of you. grep, you rock !!!
8. Originally Posted by Soroban
Hello, dugongster!
Hi Soroban! With due respect (which is considerable, I might add, you being one of the most helpful and, let's face it, awesome people here), I took into account the 2 hour headstart by using (t - 2) as the time in the arithmetic progression (n = (t - 2)), and starting at 4 ( $a_1 = 4$). | {
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# Do the 12 lines of a bingo card have equal chance of winning?
You have a $$5 \times 5$$ bingo card filled with $$25$$ distinct numbers, one per square. There is also a pot containing each number once and you draw them out one by one without replacement.
A line is any of the $$5$$ rows, $$5$$ columns, or $$2$$ main diagonals. A line is completed when its $$5$$ numbers have been drawn. A line wins if it is the first line to be completed.
Question: Do the $$12$$ lines have equal chance of winning? If not, which lines have higher chance of winning?
Why I ask: Define $$T_l$$ to be the time to completion for line $$l$$. It is obvious that all $$T_l$$ are equi-distributed, and thus all $$E[T_l]$$ are equal. However, because the lines overlap, the different $$T_l$$'s are dependent, and it is not clear to me that they have equal chances of being first.
In particular, imagine that the $$5$$-subsets are not arranged in rows, columns and diagonals, but are clustered in some non-uniform way. Then a subset that shares a lot of elements with other subsets might have a lower chance to win than a subset that does not share a lot of elements with other subsets. (I can provide a simple example if there is interest.)
On the bingo card, the $$5$$-subsets are pretty uniform but not exactly uniform, due to the diagonals. So my suspicion is that the line win chances are almost equal but not exactly equal. And I am curious as to which lines have higher chances.
I imagine (but would be happy to be proven wrong) that calculating the exact win prob for a line might be difficult/tedious, so qualitative arguments based on e.g. symmetry are also welcome. | {
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Clarifications: A drawn number can complete multiple lines, so that needs some special handling. However what I am interested in is the question of equality, so I will accept any reasonable way to handle such "shared" wins, i.e. if $$N>1$$ lines are completed at the same draw (and no line has been completed before this draw), then you can treat this as if...
• they all win (in which case the sum of the $$12$$ win probabilities exceed $$1$$, but that doesn't matter since I am interested in which are higher/lower), or,
• the whole experiment is repeated from the beginning (i.e. we condition on such shared wins not happening), or,
• you flip an $$N$$-sided die to determine the winner (i.e. this effectively counts as $$1/N$$ win for each involved line), etc.
• After this, you can do BINGO cards with a "free" space in the middle. – GEdgar Mar 8 at 20:37
• I don't see why you would think any particular line would have a higher chance of winning if there is an equal probability of each square being picked every time. A line requires 5 numbers being picked and the probability of each 5 number combinations are exactly the same. – Matthew Liu Mar 8 at 20:53
• @MatthewLiu Because when one line gets a number, it also helps other lines. The diagonal are at a disadvantage. Look at my solution for $3\times3$ bingo. – saulspatz Mar 8 at 21:24
• @MatthewLiu what you said is why all the $T_l$ are identically-distributed, but they are not independent. A line $L$ wins if $T_L < T_l \forall l \neq L$. – antkam Mar 8 at 22:15
• Ok you guys are correct. The diagonal lines are at a strict disadvantage. – Matthew Liu Mar 11 at 17:14 | {
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I must admit, when I read the title of the question, I thought "Of course they all have an equal chance of winning. What a silly question." But then, I read the body, and was delighted to see that you had a valid point, and I was the silly one. I don't know how to solve for $$5\times5$$ bingo in any reasonable time, but for $$3\times3$$ bingo we have only $$9!$$ permutations of the numbers to consider. I wrote a python script to exhaustively test this.
from itertools import permutations
from fractions import Fraction
score = 9*[0]
card = 8*[0]
for p in permutations(range(9)):
card[0] = {0,1,2}
card[1] = {3,4,5}
card[2] = {6,7,8}
card[3] = {0,3,6}
card[4] = {1,4,7}
card[5] = {2,5,8}
card[6] = {0,4,8}
card[7] = {2,4,6}
for number in p:
for line in card:
line -= {number}
winners = len([line for line in card if not line])
if winners:
for idx, line in enumerate(card):
if not line: score[idx] += Fraction(1, winners)
break
for idx in range(8):
print(idx, score[idx])
This produced the output
0 47520
1 45600
2 47520
3 47520
4 45600
5 47520
6 40800
7 40800
This means that the diagonals are worst, and the lines along the edges of the card are best, with the horizontal and vertical lines through the center in the middle.
I should mention that in the case of ties, I awarded an equal fraction of the game to each winner, so that the sum of the scores does work out to $$9!.$$ | {
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• What does it produce if you give a full win to each line if 2/3/4 lines win at once? – Banbadle Mar 8 at 21:58
• awesome! i had thought the diagonals are worst but all the rows and columns are equal, but you showed that the row/col passing through the center are penalized. interesting... and now i wonder if the 1st & 2nd rows are different in the 5x5. ooh, yeah, i would like to see the idea of @Banbadle tested too. would the multiple full wins cancel the effect of sharing? – antkam Mar 8 at 22:00
• @Banbadle I just modified saulspatz's code to try other scoring modes. If a full win is given to each simultaneously winning line, the win counts are 54000 > 53290 > 46656 for the edges, center row/col, diagonals respectively. If we condition on no shared wins (i.e. those permutations award no wins to anybody), then the win counts are 41040 > 38880 > 35424, again for edge, center row/col, diagonals. So the order is the same in all 3 win-counting models. – antkam Mar 8 at 22:44
I simulated a standard bingo game, one million games with ten random cards for each game. I recorded the number of times each line triggered bingo, including bingo on multiple cards and multiple lines winning simultaneously on the same card.
The results:
• Horizontal through center wins $$16.2\%$$ of games
• Vertical through center wins $$16.1\%$$ of games
• Diagonals each win $$15.9\%$$ of games
• Other horizontal lines each win $$5.8\%$$ of games
• Other vertical lines each win $$5.6\%$$ of games
Note that this totals over $$100\%$$, accounting for multiple wins in a single game (there were nearly $$1.1$$ million wins in one million games) | {
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After ensuring that these results are indeed accurate and giving some thought to why the diagonals win less than the central row and column, it is clear that this is due to intersecting lines. Each of the squares on the diagonals, when drawn, has the potential to trigger a win on either of two intersecting lines, while each square in the central row or column can only trigger a win on a single intersecting line.
I have also run the simulation without the free center square:
• Rows win $$9.42\%$$
• Diagonals win $$9.17\%$$
• Columns win $$9.05\%$$
• when you say "standard bingo game" do you mean the center square is free (i.e. already marked off)? if so that must be why lines through center win much more coz they only require 4 cells. – antkam Mar 8 at 22:04
• @antkam Yes, center square is free in my simulation. – Daniel Mathias Mar 8 at 22:05
• Even in the free-center case it is open (at least, in my mind) whether the diagonals are at a disadvantage. If you found that diagonals win 10% vs the center row/col winning 20% each, that would be quite conclusive. But at 15.9.% vs 16.1% I am not sure this is a real effect or just within the error bar. – antkam Mar 8 at 22:18
• @antkam A run of ten million games: $16.18\%$, $16.09\%$, $15.91\%$, $15.90\%$ for center row, center column, downward diagonal, upward diagonal. Of course, this only proves that my simulation produces consistent results. It is entirely possible that the inequality is a result of some unintended bias, though I have found no potential cause for bias. – Daniel Mathias Mar 8 at 23:31
• @DanielMathias Thanks. I knew very little about bingo before reading this. Now I have upgraded my knowledge from "very little" to just "little". – badjohn Mar 9 at 14:54 | {
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# If $N= (323232 \cdots 50 \text{ digits})_9$ (i.e in base $9$) then how to find the remainder when this $N$ is divided by $8$?
If $N= (\underbrace{323232 \cdots}_{50 \text{ digits}})_9$ (i.e in base $9$) then how to find the remainder when this $N$ is divided by $8$?
I am looking for a "fast" approach that could be used to solve this in less than $2$ minutes.
• This is computing $N\mod 8$. Since the $9\mod 8 \equiv 1$, this is just the sum of digits of the number $N$, i.e. $(3+2)\times 25\mod 8 \equiv (3+2)\times 1\mod 8 = 5$. Did I miss the question ? – Sasha Aug 26 '11 at 16:45
• @Sasha:No,you didn't miss the question.I just can't see this solution before :/ – Quixotic Aug 26 '11 at 17:26
The same way we find the remainder of dividing a number by $9$ when the number is written in base 10: add the digits.
Why?
Because remember that writing, say, $(38571)_9$ "really" means $$1 + 7\times 9 + 5\times 9^2 + 8\times 9^3 + 3\times 9^4.$$ When you divide $9$ by $8$, the remainder is $1$; when you divide $9^2$ by $8$, the remainder is $1^2 = 1$; then you divide $9^3$ by $8$, the remainder is $1^3=1$, etc. So the remainder of this number when divided by $8$ is the same as the remainder of $$1 + 7\times 1 + 5\times 9^2 + 8\times 1^3 + 3\times 1^4$$ which is the same as the remainder of $1+7+5+8+3 = 24$, which is $0$.
In this case, you have the digits $3$ and $2$, each of them $25$ times, so the remainder when dividing by $8$ is the same as the remainder of $(2+3)\times 25 = 125$ when divided by $8$, which is $5$.
(More generally, the remainder of dividing the number $(b_1b_2\cdots b_k)_b$ by $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits, $b_1+b_2+\cdots+b_k$ by $b-1$). | {
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(Even more generally, the remainder of dividing $(b_1b_2\cdots b_k)_b$ by any divisor of $b-1$ is the same as the remainder of dividing the sum of the base $b$ digits $b_1+b_2+\cdots + b_n$ by that number. That's why you can add the digits of a number in base 10 to find the remainders modulo $9$ or to find the remainders modulo $3$).
• I feel like stupid now,the first general result is not new to me and could be easily be shown as follows:$N = (a_1a_2\cdots a_k)_r = a_k + \cdots +a_2 \times r^{k-2} + a_2 \times r^{k-1}$ and $S = a_1 + a_2 + \cdots a_k$ then it could be easily shown that $\frac{N-S}{r-1} = I \text{ (an integer) }+\frac{S}{r-1}$,which proves that result. But I don't know why I can't see this direct application before! :/ – Quixotic Aug 26 '11 at 17:23
• Anyways,the more general result is new to me,and perhaps a very useful one,could you hint me a easy proof for the same? – Quixotic Aug 26 '11 at 17:25
• @FoolForMath The key idea is to learn modular arithmetic and, more generally, modular reductions of problems in quotient algebras. It's one way to algebraically divide and conquer. You can find further discussion of such in some of my posts here if you follow the link I gave, and its links... – Bill Dubuque Aug 26 '11 at 17:28
• I understood it now :D and it was before Bill Dubuque's post :-) but yes Bill you are right,my understanding is due to modular arithmetic. – Quixotic Aug 26 '11 at 17:31
• @Foo One key thing to keep in mind is that radix notation has polynomial form, so this can be viewed as a special case of well-known results about polynomials. Always be on the lookout for analogies between numbers and functions (here polynomials). – Bill Dubuque Aug 26 '11 at 17:55
HINT $\$ Employ the radix $\:9\:$ analog of casting out nines in decimal, namely
$$\rm\quad mod\ 8:\ \ \ 9 \equiv 1\ \ \Rightarrow\ \ d_n\: 9^n +\:\cdots\:+d_1\:9 + d_0\ \equiv\ d_n +\:\cdots\:+d_1+d_0$$ | {
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The same idea works generally to cast out $\rm\:b\pm1\:$'s in radix $\rm\:b\:$, e.g. see here for many links.
It may be viewed as a number-theoretic specialization of the Polynomial Remainder Theorem
$\rm\quad f(x)\ mod\ (x-r)\:=\:f(r)\:,\$ thus $\rm\ \ f(x)\ mod\ (x-1)\:=\:f(1)\: =\: f_n +\:\cdots\:+f_1 + f_0$ | {
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location:
61.
Let $S=1!2!3!\dots 99!100!$ (the product of the first 100 factorials). Prove that there exists an integer $k$ for which $1\le k\le 100$ and $S/k!$ is a perfect square. Is $k$ unique? (Optional: Is it possible to find such a number $k$ that exceeds 100?)
Solution 1. Note that, for each positive integer $j$, $\left(2j-1\right)!\left(2j\right)!=\left[\left(2j-1\right)!\right]{}^{2}·2j$. Hence
$S=\underset{j=1}{\overset{50}{\Pi }}\left[\left(2j-1\right)!\right]{}^{2}\left[2j\right]={2}^{50}50!\left[\underset{j=1}{\overset{50}{\Pi }}\left(2j-1\right)!{\right]}^{2} ,$ | {
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from which we see that $k=50$ is the required number.
We show that $k=50$ is the only possibility. First, $k$ cannot exceed 100, for otherwise $101!$ would be a factor of $k!$ but not $S$, and so $S/k!$ would not even be an integer. Let $k\le 100$. The prime $47$ does not divide $k!$ for $k\le 46$ and divides $50!$ to the first power. Since $S/50!$ is a square, it evidently divides $S$ to an odd power. So $k\ge 47$ in order to get a quotient divisible by 47 to an even power. The prime $53$ divides each $k!$ for $k\ge 53$ to the first power and divides $S/50!$, and so $S$ to an even power. Hence, $k\le 52$.
The prime $17$ divides $50!$ and $S/50!$, and hence $S$ to an even power, but it divides each of $51!$ and $52!$ to the third power. So we cannot have $k=51$ or 52. Finally, look at the prime 2. Suppose that ${2}^{2u}$ is the highest power of 2 that divides $S/50!$ and that ${2}^{v}$ is the highest power of 2 that divides $50!$; then ${2}^{2u+v}$ is the highest power of 2 that divides $S$. The highest power of $2$ that divides $48!$ and $49!$ is ${2}^{v-1}$ and the highest power of 2 that divides $46!$ and $47!$ is ${2}^{v-5}$. >From this, we deduce that 2 divides $S/k!$ to an odd power when $47\le k\le 49$. The desired uniqueness of $k$ follows.
Solution 2. Let $p$ be a prime exceeding 50. Then $p$ divides each of $m!$ to the first power for $p\le m\le 100$, so that $p$ divides $S$ to the even power $100-\left(p-1\right)=101-p$. From this, it follows that if $53\ge k$, $p$ must divide $S/k!$ to an odd power.
On the other hand, the prime 47 divides each $m!$ with $47\le m\le 93$ to the first power, and each $m!$ with $94\le m\le 100$ to the second power, so that it divides $S$ to the power with exponent $54+7=61$. Hence, in order that it divide $S/k!$ to an even power, we must make $k$ one of the numbers $47,\dots ,52$. | {
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By an argument, similar to that used in Solution 1, it can be seen that $2$ divides any product of the form $1!2!\dots \left(2m-1\right)!$ to an even power and $100!$ to the power with exponent | {
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$⌊100/2⌋+⌊100/4⌋+⌊100/8⌋+⌊100/16⌋+⌊100/32⌋+⌊100/64⌋=50+25+12+6+3+1=97 .$
Hence, 2 divides $S$ to an odd power. So we need to divide $S$ by $k!$ which 2 divides to an odd power to get a perfect square quotient. This reduces the possibilities for $k$ to 50 or 51. Since
$S={2}^{99}·{3}^{98}·{4}^{97}\dots {99}^{2}·100=\left(2·4\dots 50\right)\left({2}^{49}·{3}^{49}·{4}^{48}\dots 99\right){}^{2}=50!·{2}^{50}\left(\dots \right){}^{2} ,$
$S/50!$ is a square, and so $S/51!=\left(S/50!\right)÷\left(51\right)$ is not a square. The result follows.
Solution 3. As above, $S/\left(50!\right)$ is a square. Suppose that $53\le k\le 100$. Then 53 divides $k!/50!$ to the first power, and so $k!/50!$ cannot be square. Hence $S/k!=\left(S/50!\right)÷\left(k!/50!\right)$ cannot be square. If $k=51$ or 52, then $k!/50!$ is not square, so $S/k!$ cannot be square. Suppose that $k\le 46$. Then 47 divides $50!/k!$ to the first power, so that $50!/k!$ is not square and $S/k!=\left(S/50!\right)×\left(50!/k!\right)$ cannot be square. If $k=47,48$ or 49, then $50!/k!$ is not square and so $S/k!$ is not square. Hence $S/k!$ is square if and only if $k=50$ when $k\le 100$.
62.
Let $n$ be a positive integer. Show that, with three exceptions, $n!+1$ has at least one prime divisor that exceeds $n+1$.
Solution. Any prime divisor of $n!+1$ must be larger than $n$, since all primes not exceeding $n$ divide $n!$. Suppose, if possible, the result fails. Then, the only prime that can divide $n!+1$ is $n+1$, so that, for some positive integer $r$ and nonnegative integer $K$,
$n!+1=\left(n+1\right){}^{r}=1+\mathrm{rn}+{\mathrm{Kn}}^{2} .$ | {
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$n!+1=\left(n+1\right){}^{r}=1+\mathrm{rn}+{\mathrm{Kn}}^{2} .$
This happens, for example, when $n=1,2,4$: $1!+1=2$, $2!+1=3$, $4!+1={5}^{2}$. Note, however, that the desired result does hold for $n=3$: $3!+1=7$.
Henceforth, assume that $n$ exceeds 4. If $n$ is prime, then $n+1$ is composite, so by our initial comment, all of its prime divisors exceed $n+1$. If $n$ is composite and square, then $n!$ is divisible by the four distinct integers $1,n,\sqrt{n},2\sqrt{n}$, while is $n$ is composite and nonsquare with a nontrivial divisor $d$. then $n!$ is divisible by the four distinct integers $1,d,n/d,n$. Thus, $n!$ is divisible by ${n}^{2}$. Suppose, if possible, the result fails, so that $n!+1=1+\mathrm{rn}+{\mathrm{Kn}}^{2}$, and $1\equiv 1+\mathrm{rn}$ (mod ${n}^{2}$). Thus, $r$ must be divisible by $n$, and, since it is positive, must exceed $n$. Hence
$\left(n+1\right){}^{r}\ge \left(n+1\right){}^{n}>\left(n+1\right)n\left(n-1\right)\dots 1>n!+1 ,$
a contradiction. The desired result follows.
63.
Let $n$ be a positive integer and $k$ a nonnegative integer. Prove that
$n!=\left(n+k\right){}^{n}-\left(\genfrac{}{}{0}{}{n}{1}\right)\left(n+k-1\right){}^{n}+\left(\genfrac{}{}{0}{}{n}{2}\right)\left(n+k-2\right){}^{n}-\dots ±\left(\genfrac{}{}{0}{}{n}{n}\right){k}^{n} .$
Solution 1. Recall the Principle of Inclusion-Exclusion: Let $S$ be a set of $n$ objects, and let ${P}_{1}$, ${P}_{2}$, $\dots$, ${P}_{m}$ be $m$ properties such that, for each object $x\in S$ and each property ${P}_{i}$, either $x$ has the property ${P}_{i}$ or $x$ does not have the property ${P}_{i}$. Let $f\left(i,j,\dots ,k\right)$ denote the number of elements of $S$ each of which has properties ${P}_{i}$, ${P}_{j}$, $\dots$, ${P}_{k}$ (and possibly others as well). Then the number of elements of $S$ each having none of the properties ${P}_{1}$, ${P}_{2}$, $\dots$, ${P}_{m}$ is
$n-\sum _{1\le i\le m}f\left(i\right)+\sum _{1\le i | {
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$n-\sum _{1\le i\le m}f\left(i\right)+\sum _{1\le i
We apply this to the problem at hand. Note that an ordered selection of $n$ numbers selected from among $1,2,\dots ,n+k$ is a permutation of $\left\{1,2,\dots ,n\right\}$ if and only if it is constrained to contain each of the numbers 1, 2, $\dots$, $n$. Let $S$ be the set of all ordered selections, and we say that a selection has property ${P}_{i}$ iff its fails to include at least $i$ of the numbers $1,2,\dots ,n$ $\left(1\le i\le n\right)$. The number of selections with property ${P}_{i}$ is the product of $\left(\genfrac{}{}{0}{}{n}{i}\right)$, the number of ways of choosing the $i$ numbers not included and $\left(n+k-i\right){}^{n}$, the number of ways of choosing entries for the $n$ positions from the remaining $n+k-1$ numbers. The result follows.
Solution 2. We begin with a lemma:
$\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{m}=\left\{\begin{array}{cc}\hfill 0\hfill & \hfill \left(0\le m\le n-1\right)\hfill \\ \multicolumn{0}{c}{\left(-1\right){}^{n}n!}& \hfill \left(m=n\right) .\hfill \end{array}$
We use the convention that ${0}^{0}=1$. To prove this, note first that $i\left(i-1\right)\dots \left(i-m\right)={i}^{m+1}+{b}_{m}{i}^{m}+\dots +{b}_{1}i+{b}_{0}$ for some integers ${b}_{i}$. We use an induction argument on $m$. The result holds for each positive $n$ and for $m=0$, as the sum is the expansion of $\left(1-1\right){}^{n}$. It also holds for $n=1,2$ and all relevant $m$. Fix $n\ge 3$. Suppose that it holds when $m$ is replaced by $k$ for $0\le k\le m\le n-2$. Then | {
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$\begin{array}{cc}\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{m+1}\hfill & =\sum _{i=1}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)i\left(i-1\right)\dots \left(i-m\right)-\sum _{k=0}^{m}{b}_{k}\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{k}\hfill \\ \multicolumn{0}{c}{}& =\sum _{i=m+1}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)i\left(i-1\right)\dots \left(i-m\right)-0\hfill \\ \multicolumn{0}{c}{}& =\sum _{i=m+1}^{n}\left(-1\right){}^{i}\frac{n!i!}{i!\left(n-i\right)!\left(i-m-1\right)!}=\sum _{j=0}^{n-m-1}\left(-1\right){}^{m+1+j}\frac{n!}{\left(n-m-1-j\right)!j!}\hfill \\ \multicolumn{0}{c}{}& =\sum _{j=0}^{n-m-1}\left(-1\right){}^{m+1}\left(-1\right){}^{j}\frac{n\left(n-1\right)\dots \left(n-m\right)\left[\left(n-m-1\right)!\right]}{\left(n-m-1-j\right)!j!}\hfill \\ \multicolumn{0}{c}{}& =\left(-1\right){}^{m+1}n\left(n-1\right)\dots \left(n-m\right)\sum _{j=0}^{n-m-1}\left(-1\right){}^{j}\left(\genfrac{}{}{0}{}{n-m-1}{j}\right)=0 .\hfill \end{array}$
(Note that the $j=0$ term is 1, which is consistent with the ${0}^{0}=1$ convention mentioned earlier.) So $\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{m}=0$ for $0\le m\le n-1$. Now consider the case $m=n$:
$\sum _{i=1}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{n}=\sum _{i=1}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)i\left(i-1\right)\dots \left(i-n+1\right)-\sum _{k=0}^{n-1}{b}_{k}\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{k} .$
Every term in the first sum vanishes except the $n$th and each term of the second sum vanishes. Hence $\sum _{i=1}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{n}=\left(-1\right){}^{n}n!$.
Returning to the problem at hand, we see that the right side of the desired equation is equal to | {
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$\begin{array}{cc}\left(n+k\right){}^{n}\hfill & -\left(\genfrac{}{}{0}{}{n}{1}\right)\left(n+k-1\right){}^{n}+\left(\genfrac{}{}{0}{}{n}{2}\right)\left(n+k-2\right){}^{n}-\dots +\left(-1\right){}^{n}\left(\genfrac{}{}{0}{}{n}{n}\right)\left(n+k-n\right){}^{n}\hfill \\ \multicolumn{0}{c}{}& =\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\left(n-i+k\right){}^{n}=\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right)\left(n-i\right){}^{j}{k}^{n-j}\hfill \\ \multicolumn{0}{c}{}& =\sum _{i=0}^{n}\sum _{j=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\left(\genfrac{}{}{0}{}{n}{j}\right)\left(n-i\right){}^{j}{k}^{n-j}=\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){k}^{n-j}\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\left(n-i\right){}^{j}\hfill \\ \multicolumn{0}{c}{}& =\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){k}^{n-j}\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{n-i}\right)\left(n-i\right){}^{j}\hfill \\ \multicolumn{0}{c}{}& =\sum _{j=0}^{n}\left(\genfrac{}{}{0}{}{n}{j}\right){k}^{n-j}\sum _{i=0}^{n}\left(-1\right){}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{j} .\hfill \end{array}$
When $0\le j\le n-1$, the sum $\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{n-i}\right)\left(n-i\right){}^{j}=\sum _{i=0}^{n}\left(-1\right){}^{n-i}\left(\genfrac{}{}{0}{}{n}{i}\right){i}^{j}$ vanishes, while, when $j=n$, it assunes the value $n!$. Thus, the right side of the given equation is equal to $\left(\genfrac{}{}{0}{}{n}{n}\right){k}^{0}n!=n!$ as desired.
Solution 3. Let $m=n+k$, so that $m\ge n$, and let the right side of the equation be denoted by $R$. Then | {
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$\begin{array}{cc}R\hfill & ={m}^{n}-\left(\genfrac{}{}{0}{}{n}{1}\right)\left(m-1\right){}^{n}+\left(\genfrac{}{}{0}{}{n}{2}\right)\left(m-2\right){}^{n}-\dots +\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\left(m-i\right){}^{n}+\dots +\left(-1\right){}^{n}\left(\genfrac{}{}{0}{}{n}{n}\right)\left(m-n\right){}^{n}\hfill \\ \multicolumn{0}{c}{}& ={m}^{m}\left[\sum _{j=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right)\right]-\left(\genfrac{}{}{0}{}{n}{1}\right){m}^{n-1}\left[\sum _{i=1}^{n}\left(-1\right){}^{i}i\left(\genfrac{}{}{0}{}{n}{i}\right)\right]+\left(\genfrac{}{}{0}{}{n}{2}\right){m}^{n-2}\left[\sum _{i=1}^{n}\left(-1\right){}^{i}{i}^{2}\left(\genfrac{}{}{0}{}{n}{i}\right)\right]+\dots \hfill \\ \multicolumn{0}{c}{}& +\left(-1\right){}^{n}\left(\genfrac{}{}{0}{}{n}{n}\right)\left[\sum _{i=1}^{n}\left(-1\right){}^{i}{i}^{n}\left(\genfrac{}{}{0}{}{n}{i}\right)\right] .\hfill \end{array}$
Let
${f}_{0}\left(x\right)=\left(1-x\right){}^{n}=\sum _{i=0}^{n}\left(-1\right){}^{i}\left(\genfrac{}{}{0}{}{n}{i}\right){x}^{i}$
and let
${f}_{k}\left(x\right)=x{\mathrm{Df}}_{k-1}\left(x\right)$
for $k\ge 1$, where $\mathrm{Df}$ denotes the derivative of a function $f$. Observe that, from the closed expression for ${f}_{0}\left(x\right)$, we can establish by induction that
${f}_{k}\left(x\right)=\sum _{i=0}^{n}\left(-1\right){}^{i}{i}^{k}\left(\genfrac{}{}{0}{}{n}{i}\right){x}^{i}$
so that $R=\sum _{k=0}^{n}\left(-1\right){}^{k}\left(\genfrac{}{}{0}{}{n}{k}\right){m}^{n-k}{f}_{k}\left(1\right)$.
By induction, we establish that
${f}_{k}\left(x\right)=\left(-1\right){}^{k}n\left(n-1\right)\dots \left(n-k+1\right){x}^{k}\left(1-x\right){}^{n-k}+\left(1-x\right){}^{n-k+1}{g}_{k}\left(x\right)$
for some polynomial ${g}_{k}\left(x\right)$. This is true for $k=1$ with ${g}_{1}\left(x\right)=0$. Suppose if holds for $k=j$. Then | {
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$\begin{array}{cc}{f}_{j}\text{'}\left(x\right)\hfill & =\left(-1\right){}^{j}n\left(n-1\right)\dots \left(n-j+1\right){x}^{j-1}\left(1-x\right){}^{n-j}-\left(-1\right){}^{j}n\left(n-1\right)\dots \left(n-j+1\right)\left(n-j\right){x}^{j}\left(1-x\right){}^{n-j-1}\hfill \\ \multicolumn{0}{c}{}& -\left(n-j+1\right)\left(1-x\right){}^{n-j}{g}_{j}\left(x\right)+\left(1-x\right){}^{n-j+1}{g}_{j}\text{'}\left(x\right) ,\hfill \end{array}$
whence
$\begin{array}{cc}{f}_{j+1}\left(x\right)\hfill & =\left(-1\right){}^{j+1}n\left({n}_{1}\right)\dots \left(n-j\right){x}^{j}\left(1-x\right){}^{n-\left(j+1\right)}+\left(1-x\right){}^{n-\left(j+1\right)+1}\left[\left(-1\right){}^{j}n\left(n-1\right)\dots \left(n-j+1\right){x}^{j}\hfill \\ \multicolumn{0}{c}{}& -\left(n-j+1\right){\mathrm{xg}}_{k}\left(x\right)+x\left(1-x\right){g}_{j}\text{'}\left(x\right)\right]\hfill \end{array}$
and we obtain the desired representation by induction. Then for $1\le k\le n-1$, ${f}_{k}\left(1\right)=0$ while ${f}_{n}\left(1\right)=\left(-1\right){}^{n}n!$. Hence $R=\left(-1\right){}^{n}{f}_{n}\left(1\right)=n!$.
64.
Let $M$ be a point in the interior of triangle $\mathrm{ABC}$, and suppose that $D$, $E$, $F$ are respective points on the side $\mathrm{BC}$, $\mathrm{CA}$, $\mathrm{AB}$, which all pass through $M$. (In technical terms, they are cevians.) Suppose that the areas and the perimeters of the triangles $\mathrm{BMD}$, $\mathrm{CME}$, $\mathrm{AMF}$ are equal. Prove that triangle $\mathrm{ABC}$ must be equilateral.
Solution. [L. Lessard] Let the common area of the triangles $\mathrm{BMD}$, $\mathrm{CME}$ and $\mathrm{AMF}$ be $a$ and let their common perimeter be $p$. Let the area and perimeter of $\Delta \mathrm{AME}$ be $u$ and $x$ respectively, of $\Delta \mathrm{MFB}$ be $v$ and $y$ respectively, and of $\Delta \mathrm{CMD}$ be $w$ and $z$ respectively.
By considering pairs of triangles with equal heights, we find that | {
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$\frac{\mathrm{AF}}{\mathrm{FB}}=\frac{a}{v}=\frac{2a+u}{v+a+w}=\frac{a+u}{a+w} ,$
$\frac{\mathrm{BD}}{\mathrm{DC}}=\frac{a}{w}=\frac{2a+v}{u+a+w}=\frac{a+v}{a+u} ,$
$\frac{\mathrm{CE}}{\mathrm{EA}}=\frac{a}{u}=\frac{2a+w}{u+a+v}=\frac{a+w}{a+v} .$
>From these three sets of equations, we deduce that
$\frac{{a}^{3}}{\mathrm{uvw}}=1 ;$
${a}^{2}+\left(w-u\right)a-\mathrm{uv}=0 ,$
${a}^{2}+\left(u-w\right)a-\mathrm{vw}=0 ,$
${a}^{2}+\left(v-u\right)a-\mathrm{uw}=0 ;$
whence
${a}^{3}=\mathrm{uvw} \mathrm{and} 3{a}^{2}=\mathrm{uv}+\mathrm{vw}+\mathrm{uw} .$
This means that $\mathrm{uv},\mathrm{vw},\mathrm{uw}$ are three positive numbers whose geometric and arithmetic means are both equal to ${a}^{2}$. Hence ${a}^{2}=\mathrm{uv}=\mathrm{vw}=\mathrm{uw}$, so that $u=v=w=a$. It follows that $\mathrm{AF}=\mathrm{FB}$, $\mathrm{BD}=\mathrm{DC}$, $\mathrm{CE}=\mathrm{EA}$, so that $\mathrm{AD}$, $\mathrm{BE}$ and $\mathrm{CF}$ are medians and $M$ is the centroid.
Wolog, suppose that $\mathrm{AB}\ge \mathrm{BC}\ge \mathrm{CA}$. Since $\mathrm{AB}\ge \mathrm{BC}$, $\angle \mathrm{AEB}\ge {90}^{ˆ}$, and so $\mathrm{AM}\ge \mathrm{MC}$. Thus $x\ge p$. Similarly, $y\ge p$ and $p\ge z$.
Consider triangles $\mathrm{BMD}$ and $\mathrm{AME}$. We have $\mathrm{BD}\ge \mathrm{AE}$, $\mathrm{BM}\ge \mathrm{AM}$, $\mathrm{ME}=\frac{1}{2}\mathrm{BM}$ and $\mathrm{MD}=\frac{1}{2}\mathrm{AM}$. Therefore
$p-x=\left(\mathrm{BD}+\mathrm{MD}+\mathrm{BM}\right)-\left(\mathrm{AE}+\mathrm{ME}+\mathrm{AM}\right)=\left(\mathrm{BD}-\mathrm{AE}\right)+\frac{1}{2}\left(\mathrm{BM}-\mathrm{AM}\right)\ge 0$ | {
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and so $p\ge x$. Since also $x\ge p$, we have that $p=x$. But this implies that $\mathrm{AM}=\mathrm{MC}$, so that $\mathrm{ME}\perp \mathrm{AC}$ and $\mathrm{AB}=\mathrm{BC}$. Since $\mathrm{BE}$ is now an axis of a reflection which interchanges $A$ and $C$, as well as $F$ and $D$, it follows that $p=z$ and $p=y$ as well. Thus, $\mathrm{AB}=\mathrm{AC}$ and $\mathrm{AC}=\mathrm{BC}$. Thus, the triangle is equilateral. | {
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65.
Suppose that $\mathrm{XTY}$ is a straight line and that $\mathrm{TU}$ and $\mathrm{TV}$ are two rays emanating from $T$ for which $\angle \mathrm{XTU}=\angle \mathrm{UTV}=\angle \mathrm{VTY}={60}^{ˆ}$. Suppose that $P$, $Q$ and $R$ are respective points on the rays $\mathrm{TY}$, $\mathrm{TU}$ and $\mathrm{TV}$ for which $\mathrm{PQ}=\mathrm{PR}$. Prove that $\angle \mathrm{QPR}={60}^{ˆ}$.
Solution 1. Let $\frakR$ be a rotation of ${60}^{ˆ}$ about $T$ that takes the ray $\mathrm{TU}$ to $\mathrm{TV}$. Then, if $\frakR$ transforms $Q\to Q\text{'}$ and $P\to P\text{'}$, then $Q\text{'}$ lies on $\mathrm{TV}$ and the line $Q\text{'}P\text{'}$ makes an angle of ${60}^{ˆ}$ with $\mathrm{QP}$. Because of the rotation, $\angle P\text{'}\mathrm{TP}={60}^{ˆ}$ and $\mathrm{TP}\text{'}=\mathrm{TP}$, whence $\mathrm{TP}\text{'}P$ is an equilateral triangle.
Since $\angle Q\text{'}\mathrm{TP}=\angle \mathrm{TPP}\text{'}={60}^{ˆ}$, $\mathrm{TV}P\text{'}P$. Let $\frakT$ be the translation that takes $P\text{'}$ to $P$. It takes $Q\text{'}$ to a point $Q\text{'}\text{'}$ on the ray $\mathrm{TV}$, and $\mathrm{PQ}\text{'}\text{'}=P\text{'}Q\text{'}=\mathrm{PQ}$. Hence $Q\text{'}\text{'}$ can be none other than the point $R$ [why?], and the result follows.
Solution 2. The reflection in the line $\mathrm{XY}$ takes $P\to P$, $Q\to Q\text{'}$ and $R\to R\text{'}$. Triangles $\mathrm{PQR}\text{'}$ and $\mathrm{PQ}\text{'}R$ are congruent and isosceles, so that $\angle \mathrm{TQP}=\angle \mathrm{TQ}\text{'}P=\angle \mathrm{TRP}$ (since $\mathrm{PQ}\text{'}=\mathrm{PR}$). Hence $\mathrm{TQRP}$ is a concyclic quadrilateral, whence $\angle \mathrm{QPR}=\angle \mathrm{QTR}={60}^{ˆ}$. | {
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