smc

To find the total amount of vertices we first find the amount of edges, and that is $\frac{20\times 3}{2}$. Next, to find the amount of vertices we can use Euler's characteristic, $V-E+F=2$, and therefore the amount of vertices is $12$ So there are $P(12,3)=1320$ ways to choose 3 distinct points. Now, the furthest distance we can get from one point to another point in an icosahedron is 3. Which gives us a range of $1\le d(Q,R),d(R,S)\le 3$ With some case work, we get two cases: Case 1: $d(Q,R)=3;d(R,S)=1,2$ Since we have only one way to choose Q, that is, the opposite point from R, we have one option for Q and any of the other points could work for S. Then, we get $12\times 1\times 10=120$ (ways to choose R × ways to choose Q × ways to choose S) Case 2: $d(Q,R)=2;d(R,S)=1$ We can visualize the icosahedron as 4 rows, first row with 1 vertex, second row with 5 vertices, third row with 5 vertices and fourth row with 1 vertex. We set R as the one vertex on the first row, and we have 12 options for R. Then, Q can be any of the 5 points on the third row and finally S can be one of the 5 points on the second row. Therefore, we have $12\times 5\times 5=300$ (ways to choose R × ways to choose Q × ways to choose S) Hence, $P(d(Q,R)>d(R,S))=\frac{120+300}{1320}=\boxed{\frac{7}{22}}$