Difference between revisions of "2011 AIME I Problems/Problem 10"
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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 <math>\frac{93}{125}</math> . Find the sum of all possible values of <math>n</math>. | 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 <math>\frac{93}{125}</math> . Find the sum of all possible values of <math>n</math>. | ||
− | == Solution == | + | == Solution 1 == |
− | |||
− | |||
− | The probability of this | + | 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 <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 | ||
+ | <math>\frac{2n\binom{n-1}{2}}{\binom{2n}{3}}</math> | ||
+ | <math>=\frac{n(n-1)(n-2)}{\frac{2n(2n-1)(2n-2)}{6}}</math> | ||
+ | <math>=\frac{6n(n-1)(n-2)}{2n(2n-1)(2n-2)}</math> | ||
+ | <math>=\frac{3(n-2)}{2(2n-1)}</math> | ||
+ | |||
+ | so <math>\frac{93}{125}=\frac{3(n-2)}{2(2n-1)}</math> | ||
+ | |||
+ | <math>186(2n-1)=375(n-2)</math>. | ||
+ | |||
+ | <math>372n-186=375n-750</math> | ||
+ | |||
+ | <math>3n=564</math> | ||
+ | |||
+ | <math>n=188</math> and so the polygon has <math>376</math> sides. | ||
+ | |||
+ | Case 2: <math>2n-1</math>-sided polygon. Similarly, <math>\binom{2n-1}{3}</math> total triangles. Again choose the leftmost point, with <math>2n-1</math> choices. For the other two points, there are again <math>\binom{n-1}{2}</math> possibilities. | ||
+ | |||
+ | The probability is <math>\frac{(2n-1)\binom{n-1}{2}}{\binom{2n-1}{3}}</math> | ||
+ | |||
+ | <math>=\frac{3(2n-1)(n-1)(n-2)}{(2n-1)(2n-2)(2n-3)}</math> | ||
+ | |||
+ | <math>=\frac{3(n-2)}{2(2n-3)}</math> | ||
+ | |||
+ | so <math>\frac{93}{125}=\frac{3(n-2)}{2(2n-3)}</math> | ||
+ | |||
+ | <math>186(2n-3)=375(n-2)</math> | ||
+ | |||
+ | <math>375n-750=372n-558</math> | ||
+ | |||
+ | <math>3n=192</math> | ||
+ | |||
+ | <math>n=64</math> and our polygon has <math>127</math> sides. | ||
+ | |||
+ | Adding, <math>127+376=\boxed{503}</math> | ||
+ | |||
+ | == Solution 2 == | ||
+ | |||
+ | |||
+ | |||
+ | |||
+ | 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 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. | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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>. | ||
+ | |||
+ | |||
+ | 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 <math>v_0</math>, then counting up the cases for vertices C, and again for vertices C and A, respectively, is combinatorially equivalent to fixing vertex A at <math>v_0</math>, 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 <math>v_0</math>, 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 <math>\frac{3(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{2}</math> total obtuse triangles obtainable. | ||
+ | |||
+ | The total number of triangles obtainable is <math>1+2+3+...+(n-2) = \frac{(n-2)(n-1)}{2}</math>. | ||
+ | |||
+ | The ratio of obtuse triangles obtainable to all triangles obtainable is therefore | ||
+ | |||
+ | <math>\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}</math>. | ||
+ | |||
+ | So, <math> \frac{(n- \lceil n/2 \rceil - 2)(n - \lceil n/2 \rceil - 1)}{(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 \pmod{125}</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 == | ||
+ | {{AIME box|year=2011|n=I|num-b=9|num-a=11}} | ||
+ | {{MAA Notice}} |
Latest revision as of 09:13, 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 . Find the sum of all possible values of
.
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 , or an odd number of sides
. For polygons with
sides, the circumdiameter has endpoints on
vertices. There are
points on one side of a diameter, plus
of the endpoints of the diameter for a total of
points. For polygons with
points, the circumdiameter has
endpoint on a vertex and
endpoint on the midpoint of the opposite side. There are also
points on one side of the diameter, plus the vertex for a total of
points on one side of the diameter.
Case 1: -sided polygon. There are clearly
different triangles total. To find triangles that meet the criteria, choose the left-most point. There are obviously
choices for this point. From there, the other two points must be within the
points remaining on the same side of the diameter. So our desired probability is
so
.
and so the polygon has
sides.
Case 2: -sided polygon. Similarly,
total triangles. Again choose the leftmost point, with
choices. For the other two points, there are again
possibilities.
The probability is
so
and our polygon has
sides.
Adding,
Solution 2
We use casework on the locations of the vertices, if we choose the locations of vertices on the n-gon (where the vertices of the n-gon are
in clockwise order) to be the vertices of triangle ABC, in order, with the restriction that
.
By symmetry, we can assume WLOG that the location of vertex A is vertex .
Now, vertex B can be any of . We start in on casework.
Case 1: vertex B is at one of the locations . (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 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
, there are
possible places for vertex C.
Summing over all these possibilities, we obtain that the number of obtuse triangles obtainable from this case is .
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 , then counting up the cases for vertices C, and again for vertices C and A, respectively, is combinatorially equivalent to fixing vertex A at
, 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
, 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 total obtuse triangles obtainable.
The total number of triangles obtainable is .
The ratio of obtuse triangles obtainable to all triangles obtainable is therefore
.
So, .
Now, we have that is divisible by
. It is now much easier to perform trial-and-error on possible values of n, because we see that
.
We find that and
both work, so the final answer is
.
Video Solution
~IceMatrix
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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