Difference between revisions of "2023 AMC 10A Problems/Problem 25"
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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> | <cmath>\frac{5}{11} \cdot \frac{6}{10} + \frac{5}{11} \cdot \frac{1}{10} = \frac{35}{110} = \frac{7}{22}</cmath> | ||
So the answer is (A). | So the answer is (A). | ||
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-awesomeparrot | -awesomeparrot | ||
Revision as of 16:00, 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 2
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
![\[\frac{5}{11} \cdot \frac{6}{10} + \frac{5}{11} \cdot \frac{1}{10} = \frac{35}{110} = \frac{7}{22}\]](http://latex.artofproblemsolving.com/9/6/c/96c3b154b3701a8bcdb566691a65802edbb3eb58.png)
So the answer is (A).
-awesomeparrot
