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Difference between revisions of "2011 AIME I Problems/Problem 10"

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== Solution 1 ==
 
== Solution 1 ==
  
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Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle.
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Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater).
  
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Break up the problem into two cases: an even number of sides <math>2n</math>, or an odd number of sides <math>2n-1</math>. After looking at a few diagrams, it becomes apparent that there are exactly <math>n</math> points on one side of a diameter.
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Break up the problem into two cases: an even number of sides <math>2n</math>, or an odd number of sides <math>2n-1</math>. For polygons with <math>2n</math> sides, the circumdiameter has endpoints on <math>2</math> vertices. There are <math>n-1</math> points on one side of a diameter, plus <math>1</math> of the endpoints of the diameter for a total of <math>n</math> points. For polygons with <math>2n - 1</math> points, the circumdiameter has <math>1</math> endpoint on a vertex and <math>1</math> endpoint on the midpoint of the opposite side. There are also <math>n - 1</math> points on one side of the diameter, plus the vertex for a total of <math>n</math> points on one side of the diameter.
  
 
Case 1: <math>2n</math>-sided polygon. There are clearly <math> \binom{2n}{3}</math> different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously <math>2n</math> choices for this point. From there, the other two points must be within the <math>n-1</math> points remaining on the same side of the diameter. So our desired probability is
 
Case 1: <math>2n</math>-sided polygon. There are clearly <math> \binom{2n}{3}</math> different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously <math>2n</math> choices for this point. From there, the other two points must be within the <math>n-1</math> points remaining on the same side of the diameter. So our desired probability is
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Adding, <math>127+376=\boxed{503}</math>
 
Adding, <math>127+376=\boxed{503}</math>
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−
 
  
 
== Solution 2 ==
 
== Solution 2 ==
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We use casework on the locations of the vertices, if we choose the locations of vertices <math>v_a, v_b, v_c</math> on the n-gon (where the vertices of the n-gon are <math>v_0, v_1, v_2, ... v_{n-1},</math> in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that <math>a<b<c</math>.
 
We use casework on the locations of the vertices, if we choose the locations of vertices <math>v_a, v_b, v_c</math> on the n-gon (where the vertices of the n-gon are <math>v_0, v_1, v_2, ... v_{n-1},</math> in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that <math>a<b<c</math>.
  
−
By symmetry, we can assume W/O LOG that the location of vertex A is vertex <math>v_0</math>.
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By symmetry, we can assume WLOG that the location of vertex A is vertex <math>v_0</math>.
  
 
Now, vertex B can be any of <math>v_1, v_2, ... v_{n-2}</math>.  We start in on casework.
 
Now, vertex B can be any of <math>v_1, v_2, ... v_{n-2}</math>.  We start in on casework.
  
−
Case 1: vertex B is at one of the locations <math>v_{n-2}, v_{n-3}, ... v_{\lfloor n/2 \rfloor +1}</math>. (The floor function is necessary for the cases in which n is odd.)
+
Case 1: vertex B is at one of the locations <math>v_{n-2}, v_{n-3}, ... v_{\lfloor n/2 \rfloor +1}</math>. (The ceiling function is necessary for the cases in which n is odd.)
  
 
Now, since the clockwise arc from A to B measures more than 180 degrees; for every location of vertex C we can choose in the above restrictions, angle C will be an obtuse angle.
 
Now, since the clockwise arc from A to B measures more than 180 degrees; for every location of vertex C we can choose in the above restrictions, angle C will be an obtuse angle.
  
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There are <math>\lceil n/2 \rceil - 2</math> choices for vertex B now (again, the floor function is necessary to satisfy both odd and even cases of n).  If vertex B is placed at <math>v_m</math>, there are <math>n - m - 1</math> possible places for vertex C.
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There are <math>\lceil n/2 \rceil - 2</math> choices for vertex B now (again, the ceiling function is necessary to satisfy both odd and even cases of n).  If vertex B is placed at <math>v_m</math>, there are <math>n - m - 1</math> possible places for vertex C.
  
 
Summing over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is <math>\frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil) - 1}{2}</math>.
 
Summing over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is <math>\frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil) - 1}{2}</math>.
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We find that <math>n = 127</math> and <math>n = 376</math> both work, so the final answer is <math>127 + 376 = \boxed{503}</math>.
 
We find that <math>n = 127</math> and <math>n = 376</math> both work, so the final answer is <math>127 + 376 = \boxed{503}</math>.
 +
 +
==Video Solution==
 +
https://youtu.be/FyTEjRjW_pQ
 +
 +
~IceMatrix
  
 
== See also ==
 
== See also ==
 
{{AIME box|year=2011|n=I|num-b=9|num-a=11}}
 
{{AIME box|year=2011|n=I|num-b=9|num-a=11}}
 +
{{MAA Notice}}

Latest revision as of 08:13, 9 August 2024

Problem

The probability that a set of three distinct vertices chosen at random from among the vertices of a regular n-gon determine an obtuse triangle is $\frac{93}{125}$ . Find the sum of all possible values of $n$.

Solution 1

Inscribe the regular polygon inside a circle. A triangle inside this circle will be obtuse if and only if its three vertices lie on one side of a diameter of the circle. (This is because if an inscribed angle on a circle is obtuse, the arc it spans must be 180 degrees or greater).

Break up the problem into two cases: an even number of sides $2n$, or an odd number of sides $2n-1$. For polygons with $2n$ sides, the circumdiameter has endpoints on $2$ vertices. There are $n-1$ points on one side of a diameter, plus $1$ of the endpoints of the diameter for a total of $n$ points. For polygons with $2n - 1$ points, the circumdiameter has $1$ endpoint on a vertex and $1$ endpoint on the midpoint of the opposite side. There are also $n - 1$ points on one side of the diameter, plus the vertex for a total of $n$ points on one side of the diameter.

Case 1: $2n$-sided polygon. There are clearly $\binom{2n}{3}$ different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously $2n$ choices for this point. From there, the other two points must be within the $n-1$ points remaining on the same side of the diameter. So our desired probability is $\frac{2n\binom{n-1}{2}}{\binom{2n}{3}}$ $=\frac{n(n-1)(n-2)}{\frac{2n(2n-1)(2n-2)}{6}}$ $=\frac{6n(n-1)(n-2)}{2n(2n-1)(2n-2)}$ $=\frac{3(n-2)}{2(2n-1)}$

so $\frac{93}{125}=\frac{3(n-2)}{2(2n-1)}$

$186(2n-1)=375(n-2)$.

$372n-186=375n-750$

$3n=564$

$n=188$ and so the polygon has $376$ sides.

Case 2: $2n-1$-sided polygon. Similarly, $\binom{2n-1}{3}$ total triangles. Again choose the leftmost point, with $2n-1$ choices. For the other two points, there are again $\binom{n-1}{2}$ possibilities.

The probability is $\frac{(2n-1)\binom{n-1}{2}}{\binom{2n-1}{3}}$

$=\frac{3(2n-1)(n-1)(n-2)}{(2n-1)(2n-2)(2n-3)}$

$=\frac{3(n-2)}{2(2n-3)}$

so $\frac{93}{125}=\frac{3(n-2)}{2(2n-3)}$

$186(2n-3)=375(n-2)$

$375n-750=372n-558$

$3n=192$

$n=64$ and our polygon has $127$ sides.

Adding, $127+376=\boxed{503}$

Solution 2

We use casework on the locations of the vertices, if we choose the locations of vertices $v_a, v_b, v_c$ on the n-gon (where the vertices of the n-gon are $v_0, v_1, v_2, ... v_{n-1},$ in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that $a<b<c$.

By symmetry, we can assume WLOG that the location of vertex A is vertex $v_0$.

Now, vertex B can be any of $v_1, v_2, ... v_{n-2}$. We start in on casework.

Case 1: vertex B is at one of the locations $v_{n-2}, v_{n-3}, ... v_{\lfloor n/2 \rfloor +1}$. (The ceiling function is necessary for the cases in which n is odd.)

Now, since the clockwise arc from A to B measures more than 180 degrees; for every location of vertex C we can choose in the above restrictions, angle C will be an obtuse angle.

There are $\lceil n/2 \rceil - 2$ choices for vertex B now (again, the ceiling function is necessary to satisfy both odd and even cases of n). If vertex B is placed at $v_m$, there are $n - m - 1$ possible places for vertex C.

Summing over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is $\frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil) - 1}{2}$.


Case 2: vertex B is at one of the locations not covered in the first case.

Note that this will result in the same number of obtuse triangles as case 1, but multiplied by 2. This is because fixing vertex B in $v_0$, then counting up the cases for vertices C, and again for vertices C and A, respectively, is combinatorially equivalent to fixing vertex A at $v_0$, then counting cases for vertex B, as every triangle obtained in this way can be rotated in the n-gon to place vertex A at $v_0$, and will not be congruent to any obtuse triangle obtained in case 1, as there will be a different side opposite the obtuse angle in this case.


Therefore, there are $\frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{2}$ total obtuse triangles obtainable.

The total number of triangles obtainable is $1+2+3+...+(n-2) = \frac{(n-2)(n-1)}{2}$.

The ratio of obtuse triangles obtainable to all triangles obtainable is therefore

$\frac{\frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{2}}{\frac{(n-2)(n-1)}{2}} = \frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{(n-2)(n-1)} = \frac{93}{125}$.

So, $\frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{(n-2)(n-1)} = \frac{31}{125}$.

Now, we have that $(n-2)(n-1)$ is divisible by $125 = 5^3$. It is now much easier to perform trial-and-error on possible values of n, because we see that $n \equiv 1,2 \pmod{125}$.

We find that $n = 127$ and $n = 376$ both work, so the final answer is $127 + 376 = \boxed{503}$.

Video Solution

https://youtu.be/FyTEjRjW_pQ

~IceMatrix

See also

2011 AIME I (Problems • Answer Key • Resources)
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Problem 9 		Followed by
Problem 11
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All AIME Problems and Solutions

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