AMC 8 · 2008 · #19

Grade 8 geometry-2dcounting

Archetype Casework by Position / Vertex Type └ Sub-archetype Grid / Solid Cell Position

combinations-basicprobability-basiccoordinate-geometrysystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: combinations-basicprobability-basic

📏 Short solution 💡 2 insights 📊 Diagram

Problem

Eight points sit at one-unit intervals around the perimeter of a 2 × 2 square — the four corners plus the midpoint of each side. Two of the eight points are picked at random. What is the probability that the two chosen points are exactly one unit apart?

Pick an answer.

(A)

$\frac{1}{4}$

(B)

$\frac{2}{7}$

(C)

$\frac{4}{11}$

(D)

$\frac{1}{2}$

(E)

$\frac{4}{7}$

AMC 8 2008 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Count Without Listing One-by-One

The problem is a classic favorable-over-total probability. Tool #13 (Count Without Listing) gives the total: C(8, 2) = 28 pairs, no listing needed. Tool #15 (Visualize) handles the favorable count: by drawing the 8 points around the square, the unit-apart pairs are exactly the 8 short segments around the perimeter — easy to see and count once the picture is clear. Together they give $\frac{8}{28}$ = $\frac{2}{7}$ .

1STEP 1

Order does not matter, so the total number of pairs is the combination C(8, 2) = 28.

C(8, 2) =

\frac{8 × 7}{2}

= 28

2STEP 2

Around the perimeter the points alternate corner, midpoint, corner, …, so consecutive points make 8 unit-length gaps.

8 points around the square → 8 unit-length perimeter gaps

3STEP 3

Every non-neighbor pair is farther than one unit (√(2), 2, √(5), …), so the favorable pairs number exactly 8.

favorable pairs = 8

4STEP 4

Divide favorable by total and simplify: $\frac{8}{28}$ = $\frac{2}{7}$ , choice (B).

P =

\frac{8}{28}

\frac{2}{7}

→ (B)

Answer

\frac{2}{7}

Sanity-check with the symmetry trick: pick any single point first; among the remaining 7 points, exactly 2 are its perimeter neighbors at distance 1 (one on each side). So the probability that the second pick lands at distance 1 is

\frac{2}{7}

, matching the answer (B). This works for any starting point — corner or midpoint — because each point has exactly two unit-distance neighbors. The other choices are easy to rule out: (A)

\frac{1}{4}

\frac{7}{28}

would need 7 favorable pairs; (C)–(E) would need 10 or more, but there are only 8.

💡Key takeaway

Total pairs: C(8, 2) = 28. Unit-apart pairs: the 8 short hops around the square. Probability: $\frac{8}{28}$ = $\frac{2}{7}$ .

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