Difference between revisions of "2017 AIME II Problems/Problem 13"

m (Solution)
(Solution)
Line 6: Line 6:
  
 
Even and a multiple of 3: <math>\frac{n(n-4)}{2} + \frac{n}{3}</math>
 
Even and a multiple of 3: <math>\frac{n(n-4)}{2} + \frac{n}{3}</math>
 +
 
Even and not a multiple of 3: <math>\frac{n(n-2)}{2}</math>
 
Even and not a multiple of 3: <math>\frac{n(n-2)}{2}</math>
 +
 
Odd and a multiple of 3: <math>\frac{n(n-3)}{2} + \frac{n}{3}</math>
 
Odd and a multiple of 3: <math>\frac{n(n-3)}{2} + \frac{n}{3}</math>
 +
 
Odd and not a multiple of 3: <math>\frac{n(n-1)}{2}</math>
 
Odd and not a multiple of 3: <math>\frac{n(n-1)}{2}</math>
 +
 +
To derive these formulas, we note the following:
 +
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 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 a multiple of three we additionally over-count equilateral triangles, of which there are <math>\frac{n}{3}</math>. As we count them three times, we are two times over, so we subtract <math>\frac{2n}{3}</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>.
 
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>.

Revision as of 18:49, 23 March 2017

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).

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 $36 + 52 + 157 = \boxed{245}$.

See Also

2017 AIME II (ProblemsAnswer KeyResources)
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. AMC logo.png

Invalid username
Login to AoPS