2023 AMC 12A Problem 21

Attempt Problem 21 of the 2023 AMC 12A below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2023 AMC 12A solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

21.

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 B.B. For example, if AB‾\overline{AB} is an edge of the polyhedron, then d(A,B)=1,d(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)=2.d(A,B)=2. Let Q,Q, R,R, 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)\gt d(R,S)?
722\dfrac{7}{22}
13\dfrac{1}{3}
38\dfrac{3}{8}
512\dfrac{5}{12}
12\dfrac{1}{2}
Answer: A
Concepts:graph theorybasic probabilitysymmetry
Small hint:
Fix R.R. The other 1111 vertices split into 55 at distance 1,1, 55 at distance 2,2, and 11 at distance 33
Big hint:
By symmetry P(d(Q,R)>d(R,S))P(d(Q,R)\gt d(R,S)) =1−P(equal)2=\dfrac{1-P(\text{equal})}{2}
Solution:
Fix R.R. Among the other 1111 vertices of the icosahedron, 55 are at distance 1,1, 55 are at distance 2,2, and 11 (the antipode) is at distance 3.3.
Choosing ordered distinct Q,S,Q,S, the probability that d(Q,R)=d(R,S)d(Q,R)=d(R,S) is 5⋅4+5⋅411⋅10=40110=411. \dfrac{5\cdot 4+5\cdot 4}{11\cdot 10}=\dfrac{40}{110}=\dfrac{4}{11}.
By the symmetry between QQ and S,S, P(d(Q,R)>d(R,S))=1−4112=722. \begin{gathered} P(d(Q,R)\gt d(R,S))\\ {}=\dfrac{1-\tfrac{4}{11}}{2}\\ {}=\dfrac{7}{22}. \end{gathered}
Thus, the correct answer is A.
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