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 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 (regular polyhedron made up of 20 equilateral triangles). What is the probability that d(Q,R)>d(R,S)?
Answer Choices:
A. 227​
B. 31​
C. 83​
D. 125​
E. 21​
Solution:
One way to envision a regular icosahedron is with pentagonal pyramids on top and bottom, joined by a pentagonal antiprism. Let vertex A be at the top, as illustrated.
There are five vertices on the base of the pyramid with apex A, so they are all at distance 1 from A. There are five vertices on the base of the pyramid with apex B, which are all at distance 1 from the previous five vertices, so they are at distance 2 from A. Finally, only vertex B is at distance 3 from A. Thus from any vertex, there are 5 vertices at distance 1,5 at distance 2 , and 1 at distance 3 . Randomly select Q,R, and S from the 12 vertices. Let P[E] denote the probability of event E, and let P[E∣F] denote the probability of event E given event F. By symmetry
P[d(Q,R)>d(R,S)]=P[d(Q,R)<d(R,S)]
so it suffices to compute P[d(Q,R)=d(R,S)] and divide its complement by 2 to find P[d(Q,R)> d(R,S)]. Note that P[d(Q,R)=1]=115​ and
P[d(R,S)=1∣d(Q,R)=1]=104​=52​,
because vertex Q is not available. Therefore
P[d(Q,R)=d(R,S)=1]=115​⋅52​=112​.
Because the number of vertices that are at distance 2 from Q is also 5,P[d(Q,R)=d(R,S)=2] is also 112​.\
Note that d(Q,R)=d(R,S)=3 is impossible, because Q and S are distinct. Therefore
P[d(Q,R)>d(R,S)]=21​(1−112​−112​)=(A)227​​.
OR
Use the same setting and notation as in the first solution. If d(Q,R)=1, then d(Q,R)≯d(R,S). If d(Q,R)=3, then d(Q,R)>d(R,S), and this happens with probability 111​. If d(Q,R)=2, then d(Q,R)>d(R,S) exactly when d(R,S)=1, which happens for 5 vertices of the remaining 10 from which to choose. Because P[d(Q,R)=2]=115​,
P[d(R,S)=1∣d(Q,R)=2]=105​=21​,
so
P[d(Q,R)>d(R,S) and d(Q,R)=2]=115​⋅21​=225​.
The two ways found are mutually exclusive, so P[d(Q,R)>d(R,S)]=111​+225​=(A)227​​.
The problems on this page are the property of the MAA's American Mathematics Competitions
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