Difference between revisions of "2017 AIME II Problems/Problem 13"
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== Solution 3 == | == Solution 3 == | ||
− | We first notice that when a polygon has <math>n</math> sides where <math>n\not\equiv 0\pmod{3}</math>, there cannot exist any three vertices that form an equilateral triangle. Also, the parody of <math>n</math> and <math>n+1</math> also matters, since they influence how many isosceles triangles including equilateral triangles exist in the polygon. We can model an equation <math>2x+y=n</math>, where the lines that connect the vertices that are congruent are <math>x</math> vertices apart and the other line is <math>y</math> vertices apart. If n is even, there are <math>\frac{n-2}{2}</math> solutions for <math>(x,y)</math> which would determine the "type" of isosceles triangle. Note that we subtract two since <math>y</math> cannot be zero. | + | We first notice that when a polygon has <math>n</math> sides where <math>n\not\equiv 0\pmod{3}</math>, there cannot exist any three vertices that form an equilateral triangle. Also, the parody of <math>n</math> and <math>n+1</math> also matters, since they influence how many isosceles triangles including equilateral triangles exist in the polygon. We can model an equation <math>2x+y=n</math>, where the lines that connect the vertices that are congruent are <math>x</math> vertices apart and the other line is <math>y</math> vertices apart. If <math>n</math> is even, there are <math>\frac{n-2}{2}</math> solutions for <math>(x,y)</math> which would determine the "type" of isosceles triangle. Note that we subtract two since <math>y</math> cannot be zero. If <math>n</math> is odd, there are <math>\frac{n-1}{2}</math> solutions for <math>(x,y)</math>. |
+ | Next, we do casework on the congruence of <math>n</math> mod <math>3</math> and the parody of <math>n</math> using the information above: | ||
+ | |||
+ | Case <math>1</math>: | ||
+ | |||
+ | <math>n\not\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 <math>\frac{n-2}{2}</math> "types" of isosceles triangles in the polygon with <math>n</math> sides. Each isosceles triangle has a "unique point" which connects the two congruent sides. Therefore, for each "type" of triangle, there exists <math>n</math> of those triangles since the "unique point" can be any of the <math>n</math> vertices. There are <math>\frac{n}{2}</math> "types" of isosceles triangles in the polygon with <math>n+1</math> sides and <math>n+1</math> unique points for each "type" of triangle. Therefore, <math>\frac{n}{2}\cdot{(n+1)}-\frac{n-2}{2}\cdot{(n)}=78</math>. Solving, we get <math>n=52</math> | ||
+ | |||
+ | Case <math>2</math>: | ||
+ | |||
+ | <math>n\not\equiv 0\pmod{3}</math>, <math>n+1\not\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> possible isosceles triangles in the <math>n</math> sided polygon. There are <math>\frac{n-1}{2}\cdot{(n+1)}</math> possible isosceles triangles in the <math>n+1</math> sided polygon. The difference should be <math>78</math>, so <math>\frac{n-1}{2}\cdot{(n+1)}-\frac{n-1}{2}\cdot{(n)}=78</math>. Solving | ||
==Video Solution== | ==Video Solution== |
Revision as of 17:09, 12 May 2023
Contents
Problem
For each integer , let
be the number of
-element subsets of the vertices of the regular
-gon that are the vertices of an isosceles triangle (including equilateral triangles). Find the sum of all values of
such that
.
Solution
Considering , we have the following formulas:
Even and a multiple of 3:
Even and not a multiple of 3:
Odd and a multiple of 3:
Odd and not a multiple of 3:
To derive these formulas, we note the following:
Any isosceles triangle formed by the vertices of our regular -sided polygon
has its sides from the set of edges and diagonals of
. 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
, unless
is even when one set of diagonals (those which bisect the polygon) comes in a group of
. Three properties hold true of
:
When is odd there are
satisfactory subsets (This can be chosen with
choices for the not-base vertex, and
for the pair of equal sides as we have
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
When is even there are
satisfactory subsets (This can be chosen with
choices for the not-base vertex, and
for the pair of equal sides as we have
edges to choose from, one of them which is not satisfactory (the bisecting edge), and we must divide by 2 for over-count).
When is a multiple of three we additionally over-count equilateral triangles, of which there are
. As we count them three times, we are two times over, so we subtract
.
Considering the six possibilities and solving, we find that the only valid solutions are
, from which the answer is
.
Solution 2 (elaborates on the possible cases)
In the case that , there are
equilateral triangles. We will now count the number of non-equilateral isosceles triangles in this case.
Select a vertex of a regular
-gon. We will count the number of isosceles triangles with their vertex at
. (In other words, we are counting the number of isosceles triangles
with
among the vertices of the
-gon, and
.)
If the side spans
sides of the
-gon (where
), the side
must span
sides of the
-gon, and, thus, the side
must span
sides of the
-gon. As
has three distinct vertices, the side
must span at least one side, so
. Combining this inequality with the fact that
and
(as
cannot be equilateral), we find that there are
possible
.
As each of the vertices can be the vertex of a given triangle
, there are
non-equilateral isosceles triangles.
Adding in the equilateral triangles, we find that for
:
.
On the other hand, if , there are no equilateral triangles, and we may follow the logic of the paragraph above to find that
.
We may now rewrite the given equation, based on the remainder leaves when divided by 3.
Case 1:
The equation
for this case is
.
In this case, is of the form
or
, for some integer
.
Subcase 1:
Plugging into the equation above yields
.
Subcase 2:
Plugging into the equation above yields
, which has no integer solutions.
Case 2:
The equation
for this case is
.
In this case, is of the form
or
, for some integer
.
Subcase 1:
In this case, the equation above yields
.
Subcase 2:
In this case, the equation above yields
.
Case 3:
The equation
for this case is
.
In this case, is of the form
or
, for some integer
.
Subcase 1:
The equation above reduces to
, which has no integer solutions.
Subcase 2:
The equation above reduces to
, which does not yield a positive integer solution for
.
In summary, the possible are
, which add to
.
Solution 3
We first notice that when a polygon has sides where
, there cannot exist any three vertices that form an equilateral triangle. Also, the parody of
and
also matters, since they influence how many isosceles triangles including equilateral triangles exist in the polygon. We can model an equation
, where the lines that connect the vertices that are congruent are
vertices apart and the other line is
vertices apart. If
is even, there are
solutions for
which would determine the "type" of isosceles triangle. Note that we subtract two since
cannot be zero. If
is odd, there are
solutions for
.
Next, we do casework on the congruence of
mod
and the parody of
using the information above:
Case :
,
is even,
is odd
There are "types" of isosceles triangles in the polygon with
sides. Each isosceles triangle has a "unique point" which connects the two congruent sides. Therefore, for each "type" of triangle, there exists
of those triangles since the "unique point" can be any of the
vertices. There are
"types" of isosceles triangles in the polygon with
sides and
unique points for each "type" of triangle. Therefore,
. Solving, we get
Case :
,
is odd,
is even
There are possible isosceles triangles in the
sided polygon. There are
possible isosceles triangles in the
sided polygon. The difference should be
, so
. Solving
Video Solution
~MathProblemSolvingSkills.com
See Also
2017 AIME II (Problems • Answer Key • Resources) | ||
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 |
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