Problem:
For each integer n≥3, let f(n) be the number of 3-element subsets of the vertices of a 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:
Let O be the center of the regular polygon. Note that O is the circumcenter of any isosceles triangle whose vertices are vertices of the polygon. Moreover, if P is the vertex incident to the two congruent sides of the isosceles triangle, then the other two vertices are symmetric with respect to the line PO. For every vertex P of the polygon, there are exactly ⌊2n−1⌋ vertices of the polygon on one side of ray PO, and by symmetry, for every one of these vertices Q, there is a vertex Q′ in the polygon on the other side of ray PO such that △PQQ′ is isosceles. In this way every non-equilateral isosceles triangle is counted once, and for n≡0(mod3), every equilateral triangle is counted 3 times. Thus
f(n)={n⌊2n−1⌋−32nn⌊2n−1⌋ if n≡0(mod3) otherwise.
Therefore
f(n+1)−f(n)=⎩⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎧13k3k5k+17k+39k+6−(k+2) if n=6k if n=6k+1 if n=6k+2 if n=6k+3 if n=6k+4 if n=6k+5
The equations 5k+1=78,7k+3=78, and −(k+2)=78 have no positive integer solutions. If 13k=78, then k=6 and n=6k=36. If 3k=78, then k=26 and n=6k+1=157. If 9k+6=78, then k=8 and n=6k+4=52. The sum of all values of n with the required property is 36+157+52=245.
The problems on this page are the property of the MAA's American Mathematics Competitions