Competition · AMC preparation · step 4 of 4

AMC 10 · 2023A · #25

Grade 7 probability

probability-basicspatial-visualizationpolyhedron-netssymmetry-argument complementary-countingcaseworksystematic-enumeration ↑ Prerequisites: probability-basicspatial-visualization

📏 Medium solution 💡 3 insights

Problem

On a regular icosahedron (20 triangular faces, 12 vertices), define the distance d(A,B) as the minimum number of edges in a path from A to B. Three distinct vertices Q, R, S are chosen at random. Find the probability that d(Q, R) ＞ d(R, S).

Pick an answer.

(A)

$\frac{7}{22}$

(B)

$\frac{1}{3}$

(C)

$\frac{3}{8}$

(D)

$\frac{5}{12}$

(E)

$\frac{1}{2}$

AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Change Focus / Count the Complement

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.

1STEP 1

Sort the vertices by distance

By symmetry, fix any vertex R: its 11 other vertices split as 5 at distance 1, 5 at distance 2, 1 at distance 3.

From R: 5 at d=1, 5 at d=2, 1 at d=3

2STEP 2

Use the swap symmetry

Q and S are interchangeable, so P( ＞ ) = P( ＜ ); with the three cases summing to 1 this gives P( ＞ ) = $\frac{1 - P(=)}{2}$ .

P( ＞ ) = (1 - P(=))/2

3STEP 3

Count all ordered pairs

Count ordered pairs (Q, S), distinct and both different from R: 11 choices then 10, giving 110 pairs.

11 × 10 = 110

4STEP 4

Count the equal-distance pairs

Equal-distance pairs by common distance: d=1 gives 5 × 4 = 20, d=2 gives 20, d=3 gives 0 — total 40.

#{(Q,S): d(R,Q)=d(R,S)} = 20 + 20 + 0 = 40

5STEP 5

Compute the probability

Compute P(=) and finish. P(=) = $\frac{40}{110}$ = $\frac{4}{11}$ . Then P( ＞ ) = 1 - $\frac{4}{11}$ /2 = ( $\frac{7}{11}$ )/2 = $\frac{7}{22}$ , which is choice (A).

P( ＞ ) = (1 - 4/11)/2 = 7/22 → (A)

Answer

7/22

Sanity. P( ＞ ) =

\frac{7}{22}

≈ 0.318, just under

\frac{1}{3}

, which is the right ballpark: if all 11 · 10 = 110 pairs were three-way-equally split among ＞ , ＜ , =, each would be

\frac{1}{3}

, but the = event is slightly less likely than

\frac{1}{3}

(

\frac{4}{11}

≈ 0.364 is just a touch above

\frac{1}{3}

, leaving a bit less than

\frac{1}{3}

for ＞ ). The distance distribution (5, 5, 1) accounts for all 11 non-R vertices, and the count 40 properly excludes the single-antipodal class. The fraction

\frac{7}{22}

is one of the offered choices exactly.

💡Key takeaway

This AMC 10 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( ＞ ) = $\frac{1 - P(=)}{2}$ . Counting the 40 equal-distance pairs out of 110 gives $\frac{7}{22}$ .

Sort the vertices by distance
Use the swap symmetry
Count all ordered pairs
Count the equal-distance pairs
Compute the probability

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