Difference between revisions of "2017 AIME II Problems/Problem 13"
(→Solution) |
(→Solution) |
||
| Line 16: | Line 16: | ||
Any isosceles triangle formed by the vertices of our regular <math>n</math>-sided polygon <math>P</math> has its sides from the set of edges and diagonals of <math>P</math>. Notably, as two sides of an isosceles triangle must be equal, it is important to use the property that same-lengthed edges and diagonals come in groups of <math>n</math>, unless <math>n</math> is even when one set of diagonals (those which bisect the polygon) comes in a group of <math>\frac{n}{2}</math>. Three properties hold true of <math>f(n)</math>: | Any isosceles triangle formed by the vertices of our regular <math>n</math>-sided polygon <math>P</math> has its sides from the set of edges and diagonals of <math>P</math>. Notably, as two sides of an isosceles triangle must be equal, it is important to use the property that same-lengthed edges and diagonals come in groups of <math>n</math>, unless <math>n</math> is even when one set of diagonals (those which bisect the polygon) comes in a group of <math>\frac{n}{2}</math>. Three properties hold true of <math>f(n)</math>: | ||
| − | When <math>n</math> is odd there are <math>\frac{n(n-1)}{2}</math> satisfactory subsets (This can be chosen with <math>n</math> choices for the not-base vertex, and <math>\frac{n-1}{2}</math> for the pair of equal sides as we have <math>n-1</math> edges to choose from, and we must divide by 2 for over-count). | + | When <math>n</math> is odd there are <math>\frac{n(n-1)}{2}</math> satisfactory subsets (This can be chosen with <math>n</math> choices for the not-base vertex, and <math>\frac{n-1}{2}</math> for the pair of equal sides as we have <math>n-1</math> edges to choose from, and we must divide by 2 for over-count).* |
When <math>n</math> is even there are <math>\frac{n(n-2)}{2}</math> satisfactory subsets (This can be chosen with <math>n</math> choices for the not-base vertex, and <math>\frac{n-2}{2}</math> for the pair of equal sides as we have <math>n-1</math> edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count). | When <math>n</math> is even there are <math>\frac{n(n-2)}{2}</math> satisfactory subsets (This can be chosen with <math>n</math> choices for the not-base vertex, and <math>\frac{n-2}{2}</math> for the pair of equal sides as we have <math>n-1</math> edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count). | ||
| Line 23: | Line 23: | ||
Considering the six possibilities <math>n \equiv 0,1,2,3,4,5 \pmod{6}</math> and solving, we find that the only valid solutions are <math>n = 36, 52, 157</math>, from which the answer is <math>36 + 52 + 157 = \boxed{245}</math>. | Considering the six possibilities <math>n \equiv 0,1,2,3,4,5 \pmod{6}</math> and solving, we find that the only valid solutions are <math>n = 36, 52, 157</math>, from which the answer is <math>36 + 52 + 157 = \boxed{245}</math>. | ||
| + | |||
| + | *Another explanation: For any diagonal or side of the polygon chosen as the base of the isosceles triangle, there is exactly 1 isosceles triangle that can be formed. So, the total number of satisfactory subsets is <math>\dbinom{n}{2}=\dfrac{n(n-1)}{2}.</math> | ||
=See Also= | =See Also= | ||
{{AIME box|year=2017|n=II|num-b=12|num-a=14}} | {{AIME box|year=2017|n=II|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 22:32, 29 March 2017
Problem
For each integer 
, let 
be the number of 
-element subsets of the vertices of the regular 
-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of such that 
.
Solution
Considering 
, we have the following formulas:
Even and a multiple of 3: 
Even and not a multiple of 3: 
Odd and a multiple of 3: 
Odd and not a multiple of 3: 
To derive these formulas, we note the following:
Any isosceles triangle formed by the vertices of our regular -sided polygon 
has its sides from the set of edges and diagonals of . Notably, as two sides of an isosceles triangle must be equal, it is important to use the property that same-lengthed edges and diagonals come in groups of , unless is even when one set of diagonals (those which bisect the polygon) comes in a group of 
. Three properties hold true of :
When is odd there are satisfactory subsets (This can be chosen with choices for the not-base vertex, and 
for the pair of equal sides as we have 
edges to choose from, and we must divide by 2 for over-count).*
When is even there are satisfactory subsets (This can be chosen with choices for the not-base vertex, and 
for the pair of equal sides as we have edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count).
When is a multiple of three we additionally over-count equilateral triangles, of which there are 
. As we count them three times, we are two times over, so we subtract 
.
Considering the six possibilities 
and solving, we find that the only valid solutions are 
, from which the answer is 
.
- Another explanation: For any diagonal or side of the polygon chosen as the base of the isosceles triangle, there is exactly 1 isosceles triangle that can be formed. So, the total number of satisfactory subsets is
See Also
| 2017 AIME II (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
