Problem

For each integer $n\geq3$ , let $f(n)$ be the number of $3$ -element subsets of the vertices of the regular $n$ -gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of $n$ such that $f(n+1)=f(n)+78$ .

Solution

Considering $n \pmod{6}$ , we have the following formulas:

Even and a multiple of 3: $\frac{n(n-4)}{2} + \frac{n}{3}$

Even and not a multiple of 3: $\frac{n(n-2)}{2}$

Odd and a multiple of 3: $\frac{n(n-3)}{2} + \frac{n}{3}$

Odd and not a multiple of 3: $\frac{n(n-1)}{2}$

To derive these formulas, we note the following: Any isosceles triangle formed by the vertices of our regular $n$ -sided polygon $P$ has its sides from the set of edges and diagonals of $P$ . 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 $n$ , unless $n$ is even when one set of diagonals (those which bisect the polygon) comes in a group of $\frac{n}{2}$ . Three properties hold true of $f(n)$ :

When $n$ is odd there are $\frac{n(n-1)}{2}$ satisfactory subsets (This can be chosen with $n$ choices for the not-base vertex, and $\frac{n-1}{2}$ for the pair of equal sides as we have $n-1$ edges to choose from, and we must divide by 2 for over-count).*

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 $\dbinom{n}{2}=\dfrac{n(n-1)}{2}.$

When $n$ is even there are $\frac{n(n-2)}{2}$ satisfactory subsets (This can be chosen with $n$ choices for the not-base vertex, and $\frac{n-2}{2}$ for the pair of equal sides as we have $n-1$ edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count).

When $n$ is a multiple of three we additionally over-count equilateral triangles, of which there are $\frac{n}{3}$ . As we count them three times, we are two times over, so we subtract $\frac{2n}{3}$ .

Considering the six possibilities $n \equiv 0,1,2,3,4,5 \pmod{6}$ and solving, we find that the only valid solutions are $n = 36, 52, 157$ , from which the answer is $36 + 52 + 157 = \boxed{245}$ .

2017 AIME II (Problems • Answer Key • Resources)
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2017 AIME II Problems/Problem 13

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