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

(→‎Solution: the beginning of a solution to 2011 AIME I question 10 using casework)
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<math>\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}</math>.
 
<math>\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}</math>.
 +
 +
So, <math> \frac{(n- \lfloor n/2 \rfloor - 1)(n - \lfloor n/2 \rfloor)}{(n-2)(n-1)}  = \frac{31}{125}</math>.
  
 
Now, we have that <math>(n-2)(n-1)</math> is divisible by <math>125 = 5^3</math>.  It is now much easier to perform trial-and-error on possible values of n, because we see that <math>n \equiv 1,2 mod 125</math>.
 
Now, we have that <math>(n-2)(n-1)</math> is divisible by <math>125 = 5^3</math>.  It is now much easier to perform trial-and-error on possible values of n, because we see that <math>n \equiv 1,2 mod 125</math>.

Revision as of 14:48, 31 March 2011

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

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.

See also

2011 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 9 		Followed by
Problem 11
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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