AMC 8 · 2019 · #6

Grade 7 geometry-2dprobability

Archetype Casework by Position / Vertex Type └ Sub-archetype Vertex or Element Role

line-symmetryprobability-basicsystematic-enumeration caseworksystematic-enumeration ↑ Prerequisites: line-symmetryprobability-basic

📏 Medium solution 💡 2 insights 📊 Diagram

Problem

A square contains 81 evenly spaced grid points arranged as a 9 × 9 array (including the edge points). Point P sits at the very center. Another point Q is chosen at random from the remaining 80 grid points. What is the probability that the segment PQ lies along a line of symmetry of the square?

Pick an answer.

(A)

$\frac{1}{5}$

(B)

$\frac{1}{4}$

(C)

$\frac{2}{5}$

(D)

$\frac{9}{20}$

(E)

$\frac{1}{2}$

AMC 8 2019 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

The problem already gives us a 9 × 9 grid with P at the center, so Tool #1 (Draw a Diagram) is the natural entry point — sketch the four lines of symmetry of the square right on top of the grid and the favorable Q points become visible as the dots those lines hit. Tool #7 (Identify Subproblems) then breaks the count into four independent pieces (one per symmetry axis), and Tool #2 (Make a Systematic List) tallies the grid points along each axis without double-counting P. The final probability is just (favorable points) ÷ 80.

1STEP 1

A square has exactly four lines of symmetry, all through P — horizontal, vertical, two diagonals — so PQ is an axis only when Q sits on one.

Symmetry axes through P: horizontal, vertical, diagonal_↗, diagonal_↖

2STEP 2

Count each axis separately; the four axes meet only at P (excluded), so the four subcounts are disjoint and simply add.

favorable = (horiz) + (vert) + (diag_↗) + (diag_↖)

3STEP 3

Each axis is a line of 9 grid points, so each gives 9 - 1 = 8 after dropping P; four disjoint axes give 32 favorable points for Q.

8 + 8 + 8 + 8 = 32 favorable points for Q

4STEP 4

Divide favorable by the 80 equally likely points: $\frac{32}{80}$ , which reduces (dividing by 16) to $\frac{2}{5}$ — choice (C).

P(symmetry) =

\frac{32}{80}

\frac{(32 ÷ 16)}{(80 ÷ 16)}

\frac{2}{5}

→ (C)

Answer

\frac{2}{5}

Sanity check the number 32 against the grid: each symmetry axis is a chord of the square that obviously hits 9 collinear grid points (one per row or per column or per diagonal step), and the four axes share only the center P. So 4 × 9 = 36 total hits, but P is counted four times and we want to exclude it entirely, giving 36 - 4 = 32 — the same number. The probability

\frac{2}{5}

= 0.4 also passes the gut check: 4 special directions out of a roughly evenly distributed cloud of 80 points should land somewhere noticeably below

\frac{1}{2}

but not as small as

\frac{1}{5}

, which matches (C) nicely.

💡Key takeaway

This AMC 8 problem only needs Grade 7 probability — favorable outcomes over total outcomes — that you already know!

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