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2023 AMC 10A Problems/Problem 25

Revision as of 19:52, 9 November 2023 by Ethanzhang1001 (talk | contribs) (→‎Solution 2)

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, $\overline{AB}$ is an edge of the polyhedron, then d(A, B) = 1, but if $\overline{AC}$ and $\overline{CB}$ are edges and $\overline{AB}$ 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 $d(Q, R) > d(R, S)$?

$\textbf{(A) }\frac{7}{22}\qquad\textbf{(B) }\frac{1}{3}\qquad\textbf{(C) }\frac{3}{8}\qquad\textbf{(D) }\frac{5}{12}\qquad\textbf{(E) }\frac{1}{2}$


Video Solution 1 by OmegaLearn

https://youtu.be/Wc6PFNq5PAM

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 \[\frac{5}{11} \cdot \frac{6}{10} + \frac{5}{11} \cdot \frac{1}{10} = \frac{35}{110} = \frac{7}{22}\] So the answer is $\boxed{A}$. -awesomeparrot

Solution 2

We can actually see that the probability that $d(Q, R) > d(R, S)$ is the exact same as $d(Q, R) < d(R, S)$ 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 $d(Q, R) = d(R, S)$. 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 $d(Q, R) = d(R, S) = 1$. $5 \cdot 4 = 20$ 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 $d(Q, R) = d(R, S) = 2$. $5 \cdot 4 = 20$ ways to put them on the third layer.

The total number of ways to choose P and S are $11 \cdot 10 = 110$ (because there are 12 vertices), so the probability that $d(Q, R) = d(R, S)$ is $\frac{20+20}{110} = \frac{4}{11}$. Therefore, the probability that $d(Q, R) > d(R, S)$ is $\frac{1 - \frac{4}{11}}{2} = \boxed{\textbf{(A) }\frac{7}{22}}$ ~Ethanzhang1001

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