Difference between revisions of "2023 AMC 10A Problems/Problem 25"
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== Solution 2 == | == Solution 2 == | ||
We can actually see that the probability that <math>d(Q, R) > d(R, S)</math> is the exact same as <math>d(Q, R) < d(R, S)</math> because QR and RS have no difference. (In other words, we can just swap Q and S, meaning that can be called the same.) Therefore, we want to find the probability that <math>d(Q, R) = d(R, S)</math>. | We can actually see that the probability that <math>d(Q, R) > d(R, S)</math> is the exact same as <math>d(Q, R) < d(R, S)</math> because QR and RS have no difference. (In other words, we can just swap Q and S, meaning that can be called the same.) Therefore, we want to find the probability that <math>d(Q, R) = d(R, S)</math>. | ||
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WLOG, we can rotate the icosahedron so that R is the top of the icosahedron. Then we can divide this into 2 cases: | WLOG, we can rotate the icosahedron so that R is the top of the icosahedron. Then we can divide this into 2 cases: | ||
1. They are on the second layer | 1. They are on the second layer | ||
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There are 5 ways to put one point, and 4 ways to put the other point such that <math>d(Q, R) = d(R, S) = 1</math>. <math>5 \cdot 4 = 20</math> ways to put them on the second layer. | There are 5 ways to put one point, and 4 ways to put the other point such that <math>d(Q, R) = d(R, S) = 1</math>. <math>5 \cdot 4 = 20</math> ways to put them on the second layer. | ||
2. They are on the third layer | 2. They are on the third layer | ||
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There are 5 ways to put one point, and 4 ways to put the other point such that <math>d(Q, R) = d(R, S) = 2</math>. <math>5 \cdot 4 = 20</math> ways to put them on the third layer. | There are 5 ways to put one point, and 4 ways to put the other point such that <math>d(Q, R) = d(R, S) = 2</math>. <math>5 \cdot 4 = 20</math> ways to put them on the third layer. | ||
The total number of ways to choose P and S are <math>11 \cdot 10 = 110</math> (because there are 12 vertices), so the probability that <math>d(Q, R) = d(R, S)</math> is <math>\frac{20+20}{110} = \frac{4}{11}</math>. | The total number of ways to choose P and S are <math>11 \cdot 10 = 110</math> (because there are 12 vertices), so the probability that <math>d(Q, R) = d(R, S)</math> is <math>\frac{20+20}{110} = \frac{4}{11}</math>. | ||
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Therefore, the probability that <math>d(Q, R) > d(R, S)</math> is <math>\frac{1 - \frac{4}{11}}{2} = \boxed{\textbf{(A) }\frac{7}{22}}</math> | Therefore, the probability that <math>d(Q, R) > d(R, S)</math> is <math>\frac{1 - \frac{4}{11}}{2} = \boxed{\textbf{(A) }\frac{7}{22}}</math> | ||
~Ethanzhang1001 | ~Ethanzhang1001 |
Revision as of 19:53, 9 November 2023
If A and B are vertices of a polyhedron, define the distance d(A, B) to be the minimum number of edges of the polyhedron one must traverse in order to connect A and B. For example, is an edge of the polyhedron, then d(A, B) = 1, but if and are edges and is not an edge, then d(A, B) = 2. Let Q, R, and S be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that ?
Video Solution 1 by OmegaLearn
Solution 1
We can imagine the icosahedron as having 3 layers. 1 vertex at the top, 5 vertices below connected to the top vertex, 5 vertices below that which are 2 edges away from the top vertex, and one vertex at the bottom that is 3 edges away. WLOG because the icosahedron is symmetric around all vertices, we can say that R is the vertex at the top. So now, we just need to find the probability that S is on a layer closer to the top than Q. We can do casework on the layer S is on to get So the answer is . -awesomeparrot
Solution 2
We can actually see that the probability that is the exact same as because QR and RS have no difference. (In other words, we can just swap Q and S, meaning that can be called the same.) Therefore, we want to find the probability that .
WLOG, we can rotate the icosahedron so that R is the top of the icosahedron. Then we can divide this into 2 cases:
1. They are on the second layer
There are 5 ways to put one point, and 4 ways to put the other point such that . ways to put them on the second layer.
2. They are on the third layer
There are 5 ways to put one point, and 4 ways to put the other point such that . ways to put them on the third layer.
The total number of ways to choose P and S are (because there are 12 vertices), so the probability that is .
Therefore, the probability that is
~Ethanzhang1001