Difference between revisions of "2017 AIME II Problems/Problem 13"
Yojan sushi (talk | contribs) (→Solution) |
Yojan sushi (talk | contribs) (→Solution 2 (elaborates on the possible cases)) |
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| − | + | Case 1: <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>. | 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>. | ||
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Plugging into the equation above yields <math>7k=75</math>, which has no integer solutions. | Plugging into the equation above yields <math>7k=75</math>, which has no integer solutions. | ||
| − | + | Case 2: <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>. | 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>. | ||
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In this case, the equation above yields <math>k=26\rightarrow n=157</math>. | In this case, the equation above yields <math>k=26\rightarrow n=157</math>. | ||
| − | + | Case 3: <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>. | 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>. | ||
Revision as of 15:55, 14 February 2018
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 
.
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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
