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2023 AMC 10A Problem 25

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2023 AMC 10A Problem 25AMC 10 Step by Step
Name:                             Date: September 13, 2026
Basic Probability · difficulty 4 of 5 · about 8 min · position P21-25

If AA and BB are vertices of a polyhedron, define the distance d(A,B)d(A, B) to be the minimum number of edges of the polyhedron one must traverse in order to connect AA and BB . For example, AB‾\overline{AB} is an edge of the polyhedron, then d(A,B)=1d(A, B) = 1 , but if AC‾\overline{AC} and CB‾\overline{CB} are edges and AB‾\overline{AB} is not an edge, then d(A,B)=2d(A, B) = 2 . Let QQ , RR , and SS be randomly chosen distinct vertices of a regular icosahedron (regular polyhedron made up of 2020 equilateral triangles). What is the probability that d(Q,R)>d(R,S)d(Q, R) > d(R, S) ?

  1. A)

    722\frac{7}{22}

  2. B)

    13\frac{1}{3}

  3. C)

    38\frac{3}{8}

  4. D)

    512\frac{5}{12}

  5. E)

    12\frac{1}{2}

Answer

A

Key insight

By symmetry greater and less are equally likely, so find the tie probability: from R, 5 vertices lie at distance 1, 5 at distance 2, 1 at distance 3.

Solution

Fix RR. The roles of QQ and SS are interchangeable, so d(Q,R)>d(R,S)d(Q,R) > d(R,S) and d(Q,R)<d(R,S)d(Q,R) < d(R,S) are equally likely. Therefore
P(d(Q,R)>d(R,S))=1−P(tie)2.P\big(d(Q,R) > d(R,S)\big) = \frac{1 - P(\text{tie})}{2}.

Distances from a vertex of the icosahedron: it has 1212 vertices, and each vertex has 55 neighbors (distance 11). The vertex directly opposite is at distance 33, and the remaining 12−1−5−1=512 - 1 - 5 - 1 = 5 vertices are at distance 22. So from RR the other 1111 vertices split as 5,5,15, 5, 1 at distances 1,2,31, 2, 3.

QQ and SS are two distinct vertices chosen from these 1111, giving 11⋅10=11011 \cdot 10 = 110 ordered pairs. Ties occur when both lie in the same layer: 5⋅4=205 \cdot 4 = 20 pairs at distance 11, 5⋅4=205 \cdot 4 = 20 at distance 22, and 1⋅0=01 \cdot 0 = 0 at distance 33. So
P(tie)=40110=411,P(d(Q,R)>d(R,S))=1−4112=722.P(\text{tie}) = \frac{40}{110} = \frac{4}{11}, \qquad P\big(d(Q,R) > d(R,S)\big) = \frac{1 - \frac{4}{11}}{2} = \frac{7}{22}.

The answer is (A) 722\boxed{\textbf{(A)}\ \frac{7}{22}}.

Why this works

When two random quantities are identically distributed and exchangeable, "greater than" and "less than" have the same probability, so only the tie probability needs computing. The geometry enters only through the layer sizes 5,5,15, 5, 1, which follow from the icosahedron's vertex count and degree plus its antipodal symmetry. Look for this "half of one minus the tie" structure whenever a comparison of two symmetric random variables is asked.

Alternative approach

Direct count of ordered pairs (Q,S)(Q, S) with d(Q,R)>d(R,S)d(Q,R) > d(R,S): QQ at distance 22, SS at distance 11: 5⋅5=255 \cdot 5 = 25; QQ at distance 33, SS at distance 11 or 22: 1⋅10=101 \cdot 10 = 10. Total 3535 out of 110110, which is 722\frac{7}{22}.

The trap

Assuming the answer is 1/2 by ignoring ties, or forgetting that Q and S must be distinct when counting equal-distance pairs.

Common mistakes

  • Assuming the answer is 1/2 by ignoring ties, or forgetting that Q and S must be distinct when counting equal-distance pairs.
  • Getting the distance layers wrong, e.g. thinking the farthest vertex is at distance 22 or that there are 66 vertices at distance 22.

Techniques

Count the complement and subtract from the total · Exploit symmetry to reduce work or pair up objects