Revision as of 15:54, 14 February 2018

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

Solution 2 (elaborates on the possible cases)

In the case that $n\equiv 0\pmod 3$ , there are $\frac{n}{3}$ equilateral triangles. We will now count the number of non-equilateral isosceles triangles in this case.

Select a vertex $P$ of a regular $n$ -gon. We will count the number of isosceles triangles with their vertex at $P$ . (In other words, we are counting the number of isosceles triangles $\triangle APB$ with $A, B, P$ among the vertices of the $n$ -gon, and $AP=BP$ .)

If the side $AP$ spans $k$ sides of the $n$ -gon (where $k<\frac{n}{2}$ ), the side $BP$ must span $k$ sides of the $n$ -gon, and, thus, the side $AB$ must span $n-2k$ sides of the $n$ -gon. As $\triangle ABP$ has three distinct vertices, the side $AB$ must span at least one side, so $n-2k \ge 1$ . Combining this inequality with the fact that $1\le k<\frac{n}{2}$ and $k\not = \frac{n}{3}$ (as $\triangle ABP$ cannot be equilateral), we find that there are $\lceil\frac{n}{2}\rceil-2$ possible $k$ .

As each of the $n$ vertices can be the vertex of a given triangle $\triangle ABP$ , there are $\left(\lceil \frac{n}{2} \rceil -2 \right)\cdot n$ non-equilateral isosceles triangles.

Adding in the $\frac{n}{3}$ equilateral triangles, we find that for $n\equiv 0\pmod 3$ : $f(n) = \frac{n}{3}+\left(\lceil \frac{n}{2} \rceil -2\right)\cdot n$ .

On the other hand, if $n\equiv 1, 2\pmod 3$ , there are no equilateral triangles, and we may follow the logic of the paragraph above to find that $f(n)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n$ .

We may now rewrite the given equation, based on the remainder $n$ leaves when divided by 3.

[b]Case 1:[/b] $n\equiv 0\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\left(\lceil \frac{n+1}{2} \rceil -1\right)\cdot (n+1)=\frac{n}{3}+\left(\lceil \frac{n}{2} \rceil -2\right)\cdot n+78$ .

In this case, $n$ is of the form $6k$ or $6k+3$ , for some integer $k$ .

Subcase 1: $n=6k$ Plugging into the equation above yields $k=6\rightarrow n=36$ .

Subcase 2: $n=6k+3$ Plugging into the equation above yields $7k=75$ , which has no integer solutions.

[b]Case 2:[/b] $n\equiv 1\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\left(\lceil \frac{n+1}{2} \rceil -1\right)\cdot (n+1)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n+78$ .

In this case, $n$ is of the form $6k+1$ or $6k+4$ , for some integer $k$ .

Subcase 1: $n=6k+1$ In this case, the equation above yields $k=8\rightarrow n=52$ .

Subcase 2: $n=6k+4$ In this case, the equation above yields $k=26\rightarrow n=157$ .

[b]Case 3:[/b] $n\equiv 2\pmod 3$ The equation $f(n+1)=f(n)+78$ for this case is $\frac{n+1}{3}+\left(\lceil \frac{n+1}{2} \rceil -2\right)\cdot (n+1)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n+78$ .

In this case, $n$ is of the form $6k+2$ or $6k+5$ , for some integer $k$ .

Subcase 1: $n=6k+2$ The equation above reduces to $5k=77$ , which has no integer solutions.

Subcase 2: $n=6k+5$ The equation above reduces to $k=-80$ , which does not yield a positive integer solution for $n$ .

In summary, the possible $n$ are $36, 52, 157$ , which add to $\boxed{245}$ .

@@ Line 24: / Line 24: @@
 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>.
+==Solution 2 (elaborates on the possible cases)==
+In the case that <math>n\equiv 0\pmod 3</math>, there are <math>\frac{n}{3}</math> equilateral triangles. We will now count the number of non-equilateral isosceles triangles in this case.
+Select a vertex <math>P</math> of a regular <math>n</math>-gon. We will count the number of isosceles triangles with their vertex at <math>P</math>. (In other words, we are counting the number of isosceles triangles <math>\triangle APB</math> with <math>A, B, P</math> among the vertices of the <math>n</math>-gon, and <math>AP=BP</math>.)
+If the side <math>AP</math> spans <math>k</math> sides of the <math>n</math>-gon (where <math>k<\frac{n}{2}</math>), the side <math>BP</math> must span <math>k</math> sides of the <math>n</math>-gon, and, thus, the side <math>AB</math> must span <math>n-2k</math> sides of the <math>n</math>-gon. As <math>\triangle ABP</math> has three distinct vertices, the side <math>AB</math> must span at least one side, so <math>n-2k \ge 1</math>. Combining this inequality with the fact that <math>1\le k<\frac{n}{2}</math> and <math>k\not = \frac{n}{3}</math> (as <math>\triangle ABP</math> cannot be equilateral), we find that there are <math>\lceil\frac{n}{2}\rceil-2</math> possible <math>k</math>.
+As each of the <math>n</math> vertices can be the vertex of a given triangle <math>\triangle ABP</math>, there are <math>\left(\lceil \frac{n}{2} \rceil -2 \right)\cdot n</math> non-equilateral isosceles triangles.
+Adding in the <math>\frac{n}{3}</math> equilateral triangles, we find that for <math>n\equiv 0\pmod 3</math>: <math>f(n) = \frac{n}{3}+\left(\lceil \frac{n}{2} \rceil -2\right)\cdot n</math>.
+On the other hand, if <math>n\equiv 1, 2\pmod 3</math>, there are no equilateral triangles, and we may follow the logic of the paragraph above to find that <math>f(n)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n</math>.
+We may now rewrite the given equation, based on the remainder <math>n</math> leaves when divided by 3.
+[b]Case 1:[/b] <math>n\equiv 0\pmod 3</math>
+The equation <math>f(n+1)=f(n)+78</math> for this case is <math>\left(\lceil \frac{n+1}{2} \rceil -1\right)\cdot (n+1)=\frac{n}{3}+\left(\lceil \frac{n}{2} \rceil -2\right)\cdot n+78</math>.
+In this case, <math>n</math> is of the form <math>6k</math> or <math>6k+3</math>, for some integer <math>k</math>.
+Subcase 1: <math>n=6k</math>
+Plugging into the equation above yields <math>k=6\rightarrow n=36</math>.
+Subcase 2: <math>n=6k+3</math>
+Plugging into the equation above yields <math>7k=75</math>, which has no integer solutions.
+[b]Case 2:[/b] <math>n\equiv 1\pmod 3</math>
+The equation <math>f(n+1)=f(n)+78</math> for this case is <math>\left(\lceil \frac{n+1}{2} \rceil -1\right)\cdot (n+1)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n+78</math>.
+In this case, <math>n</math> is of the form <math>6k+1</math> or <math>6k+4</math>, for some integer <math>k</math>.
+Subcase 1: <math>n=6k+1</math>
+In this case, the equation above yields <math>k=8\rightarrow n=52</math>.
+Subcase 2: <math>n=6k+4</math>
+In this case, the equation above yields <math>k=26\rightarrow n=157</math>.
+[b]Case 3:[/b] <math>n\equiv 2\pmod 3</math>
+The equation <math>f(n+1)=f(n)+78</math> for this case is <math>\frac{n+1}{3}+\left(\lceil \frac{n+1}{2} \rceil -2\right)\cdot (n+1)=\left(\lceil \frac{n}{2} \rceil -1\right)\cdot n+78</math>.
+In this case, <math>n</math> is of the form <math>6k+2</math> or <math>6k+5</math>, for some integer <math>k</math>.
+Subcase 1: <math>n=6k+2</math>
+The equation above reduces to <math>5k=77</math>, which has no integer solutions.
+Subcase 2: <math>n=6k+5</math>
+The equation above reduces to <math>k=-80</math>, which does not yield a positive integer solution for <math>n</math>.
+----------------------
+In summary, the possible <math>n</math> are <math>36, 52, 157</math>, which add to <math>\boxed{245}</math>.
 =See Also=
 {{AIME box|year=2017|n=II|num-b=12|num-a=14}}
 {{MAA Notice}}

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

Revision as of 15:54, 14 February 2018

Contents

Problem

Solution

Solution 2 (elaborates on the possible cases)

See Also