Difference between revisions of "2023 AMC 12A Problems/Problem 21"
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==Solution 2 (Cheese + Actual way)== | ==Solution 2 (Cheese + Actual way)== | ||
− | In total, there are <math>\binom{12}{3}=220</math> ways to select the points. However, if we look at the denominators of <math>B,C,D</math>, they are <math>3,8,12</math> which are not divisors of <math>220</math>. Also <math>\frac{1}{2}</math> is impossible as cases like <math>d(Q, R) = d(R, S)</math> exist. The only answer choice left is <math>\boxed{A}</math> | + | In total, there are <math>\binom{12}{3}=220</math> ways to select the points. However, if we look at the denominators of <math>B,C,D</math>, they are <math>3,8,12</math> which are not divisors of <math>220</math>. Also <math>\frac{1}{2}</math> is impossible as cases like <math>d(Q, R) = d(R, S)</math> exist. The only answer choice left is <math>\boxed{\textbf{(A) } \frac{7}{22}}</math> |
(Actual way) | (Actual way) | ||
− | Fix an arbitrary point, to select the rest <math>2</math> points, there are <math>\binom{11}{2}=55</math> ways. To make <math>d(Q, R)=d(R, S), d=1/2</math>. Which means there are in total <math>2\cdot \binom{5}{2}=20</math> ways to make the distance the same. <math>\frac{1}{2}(1-\frac{20}{55})=\frac{7}{22} | + | Fix an arbitrary point, to select the rest <math>2</math> points, there are <math>\binom{11}{2}=55</math> ways. To make <math>d(Q, R)=d(R, S), d=1/2</math>. Which means there are in total <math>2\cdot \binom{5}{2}=20</math> ways to make the distance the same. <math>\frac{1}{2}(1-\frac{20}{55})= \boxed{\textbf{(A) } \frac{7}{22}}</math> |
~bluesoul | ~bluesoul | ||
Revision as of 10:34, 10 November 2023
Contents
Problem
If and
are vertices of a polyhedron, define the distance
to be the minimum number of edges of the polyhedron one must traverse in order to connect
and
. For example, if
is an edge of the polyhedron, then
, but if
and
are edges and
is not an edge, then
. Let
,
, and
be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 20 equilateral triangles). What is the probability that
?
Solution 1
First, note that a regular icosahedron has 12 vertices. So there are ways to choose 3 distinct points.
Now, the furthest distance we can get from one point to another point in a icosahedron is 3. Which gives us a range of
With some case work, we get:
Case 1:
(ways to choose R × ways to choose Q × ways to choose S)
Case 2:
(ways to choose R × ways to choose Q × ways to choose S)
Hence,
~lptoggled
Solution 2 (Cheese + Actual way)
In total, there are ways to select the points. However, if we look at the denominators of
, they are
which are not divisors of
. Also
is impossible as cases like
exist. The only answer choice left is
(Actual way)
Fix an arbitrary point, to select the rest points, there are
ways. To make
. Which means there are in total
ways to make the distance the same.
~bluesoul
Solution 3
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 4
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 . So, there are
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 . So, there are
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 5
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
Additional note by "Fruitz": Note that one can eliminate by symmetry if you swap the ineq sign.
Solution 6 (Case Work)
WLOG, let R be at the top-most vertex of the icosahedron. There are cases where
.
Case 1: is at the bottom-most vertex
If is at the bottom-most vertex, no matter where
is,
. The probability that
is at the bottom-most vertex is
Case 2: is at the second layer
If is at the second layer,
must be at the first layer, for
to be true. The probability that
is at the second layer, and
is at the first layer is
Video Solution by epicbird08
~EpicBird08
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.