2011 AIME I Problems/Problem 10

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.

Break up the problem into two cases: an even number of sides $2n$ , or an odd number of sides $2n-1$ . After looking at a few diagrams, it becomes apparent that there are exactly $n$ points on one side of a 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, $127+376=\boxed{503}$

Incomplete Solution

NOTE: This is not complete; but it can probably become a viable solution. If you have a different one, please put it under an alternate solution until it's been verified.

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 W/O LOG 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 floor 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 $\lfloor n/2 \rfloor - 1$ 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 $v_m$ , 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 $\frac{(n- \lfloor n/2 \rfloor - 1)(n - \lfloor n/2 \rfloor)}{2}$ .

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 $\frac{3(n- \lfloor n/2 \rfloor - 1)(n - \lfloor n/2 \rfloor)}{2}$ 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

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

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

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 $n \equiv 1,2 mod 125$ .

However, there is no solution to this that is less than 1000. I must have made an error somewhere; if someone could find and fix it, I would be much obliged.

2011 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 9	Followed by Problem 11
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2011 AIME I Problems/Problem 10

Contents

Problem

Solution 1

Incomplete Solution

See also