2023 AMC 10A Problem 25

Attempt Problem 25 of the 2023 AMC 10A 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 10A solutions, or check the answer key.

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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 B.B. For example, if ABAB is an edge of the polyhedron, then d(A,B)=1,d(A, B) = 1, but if ACAC and CBCB are edges and ABAB 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 (a 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 theorysymmetrybasic probability
Difficulty rating: 2600
Solution:

Fix R.R. Of the other 1111 vertices, 55 sit at distance 1,1, 55 at distance 2,2, and 11 (the opposite vertex) at distance 3.3. Pick ordered distinct Q,SQ, S from these 11:11: that's 1110=11011 \cdot 10 = 110 pairs. The ones with d(R,Q)=d(R,S)d(R,Q) = d(R,S) number 54+54+10=40,5\cdot4 + 5\cdot4 + 1\cdot0 = 40, so P(equal)=40110=411.P(\text{equal}) = \frac{40}{110} = \frac{4}{11}. By the symmetry between QQ and S,S, the >\gt and <\lt cases split the rest evenly, so P(d(Q,R)>d(R,S))P(d(Q,R) \gt d(R,S)) =14112= \frac{1 - \frac4{11}}{2} =722.= \frac{7}{22}. Thus, A is the correct answer.

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