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| − | 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, <math>\overline{AB}</math> is an edge of the polyhedron, then d(A, B) = 1, but if <math>\overline{AC}</math> and <math>\overline{CB}</math> are edges and <math>\overline{AB}</math> 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 <math>d(Q, R) > d(R, S)</math>?
| + | #redirect[[2023 AMC 12A Problems/Problem 21]] |
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| − | <math>\textbf{(A) }\frac{7}{22}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{3}{8}\qquad\textbf{(D) }\frac{5}{12}\qquad\textbf{(E) }\frac{1}{2}</math>
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| − | == Video Solution 1 by OmegaLearn ==
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| − | https://youtu.be/Wc6PFNq5PAM
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| − | == Solution 2 ==
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| − | 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
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| − | <cmath>\frac{5}{11} \cdot \frac{6}{10} + \frac{5}{11} \cdot \frac{1}{10} = \frac{35}{110} = \frac{7}{22}</cmath>
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| − | So the answer is (A).
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