AMC 10 · 2023 · #21
Grade 7 probabilityPick an answer.
Directly counting triples with d(Q,R) ＞ d(R,S) would require casework on three distance values for each of Q and S. Tool #16 (Change Focus) is the smart move: by symmetry, P( ＞ ) = P( ＜ ), so 1 = P( ＞ ) + P( ＜ ) + P(=) = 2 P( ＞ ) + P(=). We just need P(=). Tool #10 (Create a Physical Representation) — build or visualize a model of the icosahedron — gives the vertex-distance distribution (5, 5, 1) from any fixed vertex. Then Tool #2 (Systematic List) counts ordered pairs (Q, S) with d(R, Q) = d(R, S) by summing over distance classes.
Count the distance distribution
Count the distance distribution from one vertex.
Grade 6 "represent 3D figures via nets/surface" — picture or build the icosahedron and count neighbors at each ring.
6.G.A.4Create A Physical RepresentationReduce by symmetry
The two directions are symmetric.
Grade 7 probability — by symmetry the ＞ and ＜ events are equally likely, so we only need P(=).
A symmetry that swaps the two outcomes forces them to be equally likely.
▸ Why?
The symmetry moves the solid onto itself without stretching, so the picture is unchanged afterwards.
▸ Why?
Matched outcomes carry equal weight, so the two shares must be the same.
Count all the cases
Count the ordered pairs of the other two.
Grade 7 counting principle — multiply choices.
7.SP.C.8Make A Systematic ListCount the ties
Count the ties.
Grade 7 "sample space via organized list" — sum over the three possible common distances.
7.SP.C.8Make A Systematic ListTake the probability
Dividing gives seven twenty-seconds.
Plug P(=) = 4/11 into the symmetry formula from step 2 — done.
7.SP.C.7Change Focus Count The ComplementThis AMC 12 problem only needs Grade 7 probability you already know — picture the icosahedron, see that from any vertex the other 11 split 5 + 5 + 1, then use P( ＞ ) = P( ＜ ) to write P( ＞ ) = (1 - P(=))/2. Counting the 40 equal-distance pairs out of 110 gives 7/22.