Revision as of 17:50, 12 May 2023

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.

Case 1: $n\equiv 0\pmod 3$ 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.

Case 2: $n\equiv 1\pmod 3$ 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 $k=26\rightarrow n=157$ .

Case 3: $n\equiv 2\pmod 3$ 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 $\boxed{245}$ .

Solution 3

We first notice that when a polygon has $n$ sides where $n\not\equiv 0\pmod{3}$ , there cannot exist any three vertices that form an equilateral triangle. Also, the parody of $n$ and $n+1$ also matters, since they influence how many isosceles triangles including equilateral triangles exist in the polygon. We can model an equation $2x+y=n$ , where the lines that connect the vertices that are congruent are $x$ vertices apart and the other line is $y$ vertices apart. If $n$ is even, there are $\frac{n-2}{2}$ solutions for $(x,y)$ which would determine the "type" of isosceles triangle. Note that we subtract two since $y$ cannot be zero. If $n$ is odd, there are $\frac{n-1}{2}$ solutions for $(x,y)$ . Next, we do casework on the congruence of $n$ mod $3$ and the parody of $n$ using the information above:

Case $1$ :

$n\not\equiv 0\pmod{3}$ , $n+1\not\equiv 0\pmod{3}$

$n$ is even, $n+1$ is odd

There are $\frac{n-2}{2}$ "types" of isosceles triangles in the polygon with $n$ sides. Each isosceles triangle has a "unique point" which connects the two congruent sides. Therefore, for each "type" of triangle, there exists $n$ of those triangles since the "unique point" can be any of the $n$ vertices. There are $\frac{n}{2}$ "types" of isosceles triangles in the polygon with $n+1$ sides and $n+1$ unique points for each "type" of triangle. Therefore, $\frac{n}{2}\cdot{(n+1)}-\frac{n-2}{2}\cdot{(n)}=78$ . Solving, we get $n=52$

Case $2$ :

$n\not\equiv 0\pmod{3}$ , $n+1\not\equiv 0\pmod{3}$

$n$ is odd, $n+1$ is even

Following similar logic as above, there are $\frac{n-1}{2}\cdot{(n)}$ possible isosceles triangles in the $n$ sided polygon. There are $\frac{n-1}{2}\cdot{(n+1)}$ possible isosceles triangles in the $n+1$ sided polygon. The difference should be $78$ , so $\frac{n-1}{2}\cdot{(n+1)}-\frac{n-1}{2}\cdot{(n)}=78$ . Solving gives $n=157$ .

For both of these cases, we don't have to worry about equilateral triangles since none of these cases contain values $n$ and $n+1$ that divide $3$ .

Case $3$ :

$n\not\equiv 0\pmod{3}$ , $n+1\equiv 0\pmod{3}$

$n$ is even, $n+1$ is odd

There are $\frac{n-2}{2}\cdot{(n)}$ possible isosceles triangles in the $n$ sided polygon. The $n+1$ case is a bit more complicated, as we have to consider equilateral triangles as well. In this case, there is one solution where $x=y$ , which would give an equilateral triangle. Therefore, we subtract that case to calculate only isosceles triangles with 2 congruent sides. Only isosceles triangles with exactly 2 congruent sides have a unique point, while there exist only $\frac{n+1}{3}-1$ distinct equilateral triangles in a polygon with $n+1$ sides, the rest are equivalent and symmetrical. Therefore, there are $(\frac{n}{2}-1)\cdot{(n+1)}+\frac{n+1}{3}-1$ isosceles triangles with at least 2 congruent sides in a polygon with $n+1$ sides. Therefore, $\frac{n}{2}-1\cdot{(n+1)}+(\frac{n+1}{3}-1)-\frac{n-2}{2}\cdot{(n)}=78$ . Solving yields $5x=478$ which is impossible since $x$ has to be an integer. Therefore, this case is not valid.

Case $4$ :

$n\not\equiv 0\pmod{3}$ , $n+1\equiv 0\pmod{3}$

$n$ is odd, $n+1$ is even

There are $\frac{n-1}{2}\cdot{(n)}$ isosceles triangles in the $n$ sided polygon. There are $(\frac{n-1}{2}-1)\cdot{(n+1)}$ isosceles triangles with 2 congruent sides in an $n+1$ sided polygon. There are $\frac{n+1}{3}-1$ equilateral triangles. Therefore, $(\frac{n-1}{2}-1)\cdot{(n+1)}+(\frac{n+1}{3}-1)-\frac{n-1}{2}\cdot{(n)}=75$ . Looking at the left hand side, it is clear that $n$ has to be negative, which is not valid. Therefore, this case is not valid.

Case $5$ :

$n\equiv 0\pmod{3}$ , $n+1\not\equiv 0\pmod{3}$

$n$ is even, $n+1$ is odd

There are a total of $(\frac{n-2}{2}-1)\cdot{(n)}+(\frac{n}{3}-1)$ isosceles triangles in a polygon with $n$ sides. There are a total of $\frac{n}{2}\cdot{(n+1)}$ isosceles triangles in a polygon with $n+1$ sides. Therefore, $(\frac{n-2}{2}-1)\cdot{(n)}+(\frac{n}{3}-1)-\frac{n}{2}\cdot({(n+1)})=75$ For this case as well, it is clear that $n$ would have to be extremely negative, so it will not suffice. Therefore, this case is also invalid.

Case $6$ :

Video Solution

https://youtu.be/fFgakiw66WY

~MathProblemSolvingSkills.com

@@ Line 101: / Line 101: @@
 Case <math>3</math>:
-<math>n\not\equiv 0\pmod{3}</math>, <math>n+1\equiv\pmod{3}</math>
+<math>n\not\equiv 0\pmod{3}</math>, <math>n+1\equiv 0\pmod{3}</math>
 <math>n</math> is even, <math>n+1</math> is odd
@@ Line 109: / Line 109: @@
 Case <math>4</math>:
-<math>n\not\equiv 0\pmod{3}</math>, <math>n+1\equiv\pmod{3}</math>
+<math>n\not\equiv 0\pmod{3}</math>, <math>n+1\equiv 0\pmod{3}</math>
 <math>n</math> is odd, <math>n+1</math> is even
+There are  <math>\frac{n-1}{2}\cdot{(n)}</math> isosceles triangles in the <math>n</math> sided polygon. There are <math>(\frac{n-1}{2}-1)\cdot{(n+1)}</math> isosceles triangles with 2 congruent sides in an <math>n+1</math> sided polygon. There are <math>\frac{n+1}{3}-1</math> equilateral triangles. Therefore,  <math>(\frac{n-1}{2}-1)\cdot{(n+1)}+(\frac{n+1}{3}-1)-\frac{n-1}{2}\cdot{(n)}=75</math>. Looking at the left hand side, it is clear that <math>n</math> has to be negative, which is not valid. Therefore, this case is not valid.
+Case <math>5</math>:
+<math>n\equiv 0\pmod{3}</math>, <math>n+1\not\equiv 0\pmod{3}</math>
+<math>n</math> is even, <math>n+1</math> is odd
+There are a total of <math>(\frac{n-2}{2}-1)\cdot{(n)}+(\frac{n}{3}-1)</math> isosceles triangles in a polygon with <math>n</math> sides. There are a total of <math>\frac{n}{2}\cdot{(n+1)}</math> isosceles triangles in a polygon with <math>n+1</math> sides. Therefore, <math>(\frac{n-2}{2}-1)\cdot{(n)}+(\frac{n}{3}-1)-\frac{n}{2}\cdot({(n+1)})=75</math> For this case as well, it is clear that <math>n</math> would have to be extremely negative, so it will not suffice. Therefore, this case is also invalid.
+Case <math>6</math>:
 ==Video Solution==

2017 AIME II (Problems • Answer Key • Resources)
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Difference between revisions of "2017 AIME II Problems/Problem 13"

Revision as of 17:50, 12 May 2023

Contents

Problem

Solution

Solution 2 (elaborates on the possible cases)

Solution 3

Video Solution

See Also