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Revision as of 21:06, 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
and
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
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
2023 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 1 |
Followed by Problem 3 |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
All AMC 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.