2023 AMC 12A Problems/Problem 21

The following problem is from both the 2023 AMC 10A #25 and 2023 AMC 12A #21, so both problems redirect to this page.

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, if $\overline{AB}$ is an edge of the polyhedron, then $d(A, B) = 1$ , but if $\overline{AC}$ and $\overline{CB}$ are edges and $\overline{AB}$ 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 $d(Q, R) > d(R, S)$ ?

$\textbf{(A) } \frac{7}{22} \qquad \textbf{(B) } \frac{1}{3} \qquad \textbf{(C) } \frac{3}{8} \qquad \textbf{(D) } \frac{5}{12} \qquad \textbf{(E) } \frac{1}{2}$

Solution 1

To find the total amount of vertices we first find the amount of edges, and that is $\frac{20 \times 3}{2}$ . Next, to find the amount of vertices we can use Euler's characteristic, $V - E + F = 2$ , and therefore the amount of vertices is $12$

So there are $P(12,3)$ = 1320$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$ (Error compiling LaTeX. Unknown error_msg)1 \leq d(Q, R), d(R, S) \leq 3$With some case work, we get:

Case 1:$ (Error compiling LaTeX. Unknown error_msg)d(Q, R) = 3; d(R, S) = 1, 2$Since we have only one way to choose Q, that is, the opposite point from R, we have one option for Q and any of the other points could work for S.

Then, we get$ (Error compiling LaTeX. Unknown error_msg)12 \times 1 \times 10 = 120$(ways to choose R × ways to choose Q × ways to choose S)

Case 2:$ (Error compiling LaTeX. Unknown error_msg)d(Q, R) = 2; d(R, S) = 1$We can visualize the icosahedron as 4 rows, first row with 1 vertex, second row with 5 vertices, third row with 5 vertices and fourth row with 1 vertex. We set R as the one vertex on the first row, and we have 12 options for R. Then, Q can be any of the 5 points on the third row and finally S can be one of the 5 points on the second row.

Therefore, we have$ (Error compiling LaTeX. Unknown error_msg)12 \times 5 \times 5 = 300$(ways to choose R × ways to choose Q × ways to choose S)

Hence,$ (Error compiling LaTeX. Unknown error_msg)P(d(Q, R)>d(R, S)) = \frac{120+300}{1320} = \boxed{\textbf{(A) } \frac{7}{22}}$

~lptoggled, edited by ESAOPS

Solution 2 (Cheese + Actual way)

In total, there are $\binom{12}{3}=220$ ways to select the points. However, if we look at the denominators of $B,C,D$ , they are $3,8,12$ which are not divisors of $220$ . Also $\frac{1}{2}$ is impossible as cases like $d(Q, R) = d(R, S)$ exist. The only answer choice left is $\boxed{\textbf{(A) } \frac{7}{22}}$

(Actual way)

Fix an arbitrary point, to select the rest $2$ points, there are $\binom{11}{2}=55$ ways. To make $d(Q, R)=d(R, S), d=1/2$ . Which means there are in total $2\cdot \binom{5}{2}=20$ ways to make the distance the same. $\frac{1}{2}(1-\frac{20}{55})= \boxed{\textbf{(A) } \frac{7}{22}}$ ~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 $\frac{5}{11} \cdot \frac{6}{10} + \frac{5}{11} \cdot \frac{1}{10} = \frac{35}{110} = \frac{7}{22}$ So the answer is $\boxed{\textbf{(A) }\frac{7}{22}}$ . -awesomeparrot

Solution 4

We can actually see that the probability that $d(Q, R) > d(R, S)$ is the exact same as $d(Q, R) < d(R, S)$ because $d(Q, R)$ and $d(R, S)$ have no difference. (In other words, we can just swap Q and S, meaning that can be called the same probability-wise.) Therefore, we want to find the probability that $d(Q, R) = d(R, S)$ .

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 $d(Q, R) = d(R, S) = 1$ . So, there are $5 \cdot 4 = 20$ 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 $d(Q, R) = d(R, S) = 2$ . So, there are $5 \cdot 4 = 20$ ways to put them on the third layer.

The total number of ways to choose P and S are $11 \cdot 10 = 110$ (because there are 12 vertices), so the probability that $d(Q, R) = d(R, S)$ is $\frac{20+20}{110} = \frac{4}{11}$ .

Therefore, the probability that $d(Q, R) > d(R, S)$ is $\frac{1 - \frac{4}{11}}{2} = \boxed{\textbf{(A) }\frac{7}{22}}$

~Ethanzhang1001

Solution 5

We know that there are $20$ faces. Each of those faces has $3$ borders (since each is a triangle), and each edge is used as a border twice (for each face on either side). Thus, there are $\dfrac{20\cdot3}2=30$ edges.

By Euler's formula, which states that $v-e+f=2$ for all convex polyhedra, we know that there are $2-f+e=12$ vertices.

The answer can be counted by first counting the number of possible paths that will yield $d(Q, R) > d(R, S)$ and dividing it by $12\cdot11\cdot10$ (or $\dbinom{12}3$ , depending on the approach). In either case, one will end up dividing by $11$ somewhere in the denominator. We can then hope that there will be no factor of $11$ in the numerator (which would cancel the $11$ in the denominator out), and answer the only option that has an $11$ in the denominator: $\boxed{\textbf{(A) }\frac{7}{22}}$ .

~Technodoggo

Additional note by "Fruitz": Note that one can eliminate $1/2$ 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 $2$ cases where $d(Q, R) > d(R, S)$ .

Case 1: $Q$ is at the bottom-most vertex

If $Q$ is at the bottom-most vertex, no matter where $S$ is, $d(Q, R) > d(R, S)$ . The probability that $Q$ is at the bottom-most vertex is $\frac{1}{11}$

Case 2: $Q$ is at the second layer

If $Q$ is at the second layer, $S$ must be at the first layer, for $d(Q, R) > d(R, S)$ to be true. The probability that $Q$ is at the second layer, and $S$ is at the first layer is $\frac{5}{11} \cdot \frac{5}{10} = \frac{5}{22}$

$\frac{1}{11} + \frac{5}{22} = \boxed{\textbf{(A) }\frac{7}{22}}$

~isabelchen

Video Solution by epicbird08

https://youtu.be/s_6q0C0z6Ug

~EpicBird08

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