2023 AMC 10A Problems/Problem 25
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[hide]Problem
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
Solution 3
We know that there are faces. Each of those faces has
borders (since each is a triangle), and each edge is used as a border twice (for each face on either side). Thus, there are
edges.
By Euler's formula, which states that for all convex polyhedra, we know that there are
vertices.
The answer can be counted by first counting the number of possible paths that will yield and dividing it by
(or
, depending on the approach). In either case, one will end up dividing by
somewhere in the denominator. We can then hope that there will be no factor of
in the numerator (which would cancel the
in the denominator out), and answer the only option that has an
in the denominator:
.
~Technodoggo
See Also
2023 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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 20 |
Followed by Problem 22 |
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 |
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