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| | [[Category:Geometry]] | | [[Category:Geometry]] |
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| − | Number of Chords formed by n points on a circle
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| − | The formula for finding the number of chords is <math>\dfrac{n(n+1)}{2} - n</math> or <math>\dfrac{n(n-1)}{2}</math>
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| − | Method:
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| − | Begin creating circles with an ascending number of points:
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| − | 1 point 0 chords
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| − | 2 points 1 chord
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| − | 3 points 3 chords
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| − | 4 points 6 chords
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| − | 5 points 10 chords
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| − | 6 points 15 chords
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| − | 7 points 21 chords
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| − | On the right column, the numbers (from top to bottom) are ascending in a sequence known as the triangular numbers. This occurred every time you increase the number of points on a circle by 1. However, this shows that the 1st triangular number (1 chord) was the result of 2 points, the 2nd triangular number (3 chords) was the result of 3 points, the 3rd triangular number (6 chords) was the result of 4 points, etc.
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| − | If the numbers on the left column are marking each triangular number listed as the first, second, third, etc.(1st point, 2nd point, 3rd point, etc.,), then something has been altered - the numbers on the left column have been moved backward! This means that when finding the number of chords created by "n" points on a line, you would have to subtract "n" from the "n"th triangular number! To find the "n"th triangular number, the formula "n(n+1)/2" is used.
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| − | Subtracting "n" from this formula gives a new formula:
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| − | n(n+1)/2 - n
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| − | 2(2+1)/2 - 2 = 6/2 - 2 = 3-2 = 1
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| − | 3(3+1)/2 - 3 = 12/2 - 3 = 6-3 = 3
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| − | 4(4+1)/2 - 4 = 20/2 - 4 = 10-4 = 6
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| − | 5(5+1)/2 - 5 = 30/2 - 5 = 15-5 = 10
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| − | 6(6+1)/2 - 6 = 42/2 - 6 = 21-6 = 15
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| − | 7(7+1)/2 - 7 = 56/2 - 7 = 28-7 = 21
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| − | In conclusion, if "n" points are placed on a circle, then the maximum number of chords that can connect any two points on that circle is:
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| − | n(n+1)/2 - n or n(n-1)/2 (the simplified version)
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| − | Note: In order to achieve the simplified version, use this process:
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| − | n(n+1)/2 - n = (n(n+1)-2n)/2 = (n^2-n)/2 = n(n-1)/2
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| − | This method and formula was created by AoPS member Keshav Ramesh (user kr1234)
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is a line segment joining two points on ![[asy]size(100); pair O=origin,A=dir(135),B=dir(30); D(unitcircle); D(A--B); MP("O",D(O),S); MP("A",D(A),W); MP("B",D(B),E);[/asy]](http://latex.artofproblemsolving.com/9/a/9/9a989359a781a76d4c5b53cfaac7df1a5e010d89.png)
![[asy]size(120); pair O=origin,A=dir(170),B=dir(-10); D(unitcircle); D(A--B); MP("O",D(O),N); MP("A",D(A),W); MP("B",D(B),E);[/asy]](http://latex.artofproblemsolving.com/e/2/3/e236cfbd6b8f572926d159eeb35153c720466732.png)
