2023 AMC 10A Problem 25

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

AB

is an edge of the polyhedron, then

d(A, B) = 1,

but if

AC

and

CB

are edges and

AB

is not an edge, then

d(A, B) = 2.

Let

Q,

R,

and

S

be randomly chosen distinct vertices of a regular icosahedron (a regular polyhedron made up of

20

equilateral triangles). What is the probability that

d(Q, R) \gt d(R, S)?

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Solution

Fix

R.

Of the other

11

vertices,

5

sit at distance

1,

5

at distance

2,

and

1

(the opposite vertex) at distance

3.

Pick ordered distinct

Q, S

from these

11:

that’s

11 \cdot 10 = 110

pairs. The ones with

d(R,Q) = d(R,S)

number

5\cdot4 + 5\cdot4 + 1\cdot0 = 40,

P(\text{equal}) = \frac{40}{110} = \frac{4}{11}.

By the symmetry between

Q

and

S,

the

\gt

and

\lt

cases split the rest evenly, so

P(d(Q,R) \gt d(R,S))

= \frac{1 - \frac4{11}}{2}

= \frac{7}{22}.

Thus, A is the correct answer.