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| − | ==Problem==
| + | #redirect[[2023 AMC 12A Problems/Problem 21]] |
| − | 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 <math>d(A, B) = 1</math>, 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>?
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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 1 ==
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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 <math>\boxed{\textbf{(A) }\frac{7}{22}}</math>. -awesomeparrot
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| − | == Solution 2 ==
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| − | 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 <math>d(Q, R)</math> and <math>d(R, S)</math> 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:
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| − | 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.
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| − | 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.
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| − | 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>
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| − | ~Ethanzhang1001
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| − | ==Solution 3==
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| − | We know that there are <math>20</math> faces. Each of those faces has <math>3</math> borders (since each is a triangle), and each edge is used as a border twice (for each face on either side). Thus, there are <math>\dfrac{20\cdot3}2=30</math> edges.
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| − | By Euler's formula, which states that <math>v-e+f=2</math> for all convex polyhedra, we know that there are <math>2-f+e=12</math> vertices.
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| − | The answer can be counted by first counting the number of possible paths that will yield <math>d(Q, R) > d(R, S)</math> and dividing it by <math>12\cdot11\cdot10</math> (or <math>\dbinom{12}3</math>, depending on the approach). In either case, one will end up dividing by <math>11</math> somewhere in the denominator. We can then hope that there will be no factor of <math>11</math> in the numerator (which would cancel the <math>11</math> in the denominator out), and answer the only option that has an <math>11</math> in the denominator: <math>\boxed{\textbf{(A) }\frac{7}{22}}</math>.
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| − | ~Technodoggo
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| − | ==See Also==
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| − | {{AMC10 box|year=2023|ab=A|num-b=24|after=Last Problem}}
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| − | {{AMC12 box|year=2023|ab=A|num-b=20|num-a=22}}
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| − | {{MAA Notice}}
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