AMC 8 · 2005 · #3

Grade 5 geometry-2d

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

reflection-symmetryline-symmetrycoordinate-geometrysystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: line-symmetrycoordinate-geometry

📏 Medium solution 💡 3 insights 📊 Diagram

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Problem

A 4 × 4 grid sits inside square ABCD, and five small squares are already shaded black. Find the smallest number of additional small squares that must be shaded so the diagonal BD becomes a line of symmetry of the black pattern.

Pick an answer.

(A)

(B)

(C)

(D)

(E)

AMC 8 2005 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Find an Invariant

A line of symmetry is exactly the statement that the black pattern is invariant under reflection across BD — tool #11. The reflection across BD swaps each square with its mirror partner, so the unknowns aren't numbers but pairings. Tool #1 (Draw a Diagram) lets us mark every black square right on the grid, and tool #2 (Make a Systematic List) walks through each black square one at a time to record its mirror image. Any mirror image that isn't already black is a square we must add — counting those new squares gives the minimum.

1STEP 1

Put D at the origin so BD is the line y = x; then reflecting any square swaps its column and row: (c, r) → (r, c).

Reflection across BD: (c, r) ⟷ (r, c)

2STEP 2

List the five already-black squares by their (column, row) coordinates by reading them off the figure.

Black squares = {(0,3), (1,0), (2,3), (3,1), (3,3)}

3STEP 3

Swap each black square's coordinates to get its mirror image; (3,3) lies on the diagonal, so it mirrors to itself.

(0,3) → (3,0), (1,0) → (0,1), (2,3) → (3,2), (3,1) → (1,3), (3,3) → (3,3)

4STEP 4

Check each mirror image against the black list: (3,3) is already there, but (3,0), (0,1), (3,2), (1,3) are missing.

(3,0), (0,1), (3,2), (1,3) ∉ Black squares; (3,3) ∈ Black squares

5STEP 5

Shade exactly those four missing partners; each new square's mirror is already black, so nothing more is needed — answer (D).

Minimum squares to add = 4 → (D)

Answer

Sanity-check by pairing up all 9 black squares after the fix: {(0,3),(3,0)}, {(1,0),(0,1)}, {(2,3),(3,2)}, {(3,1),(1,3)}, and the self-mirror (3,3). Every pair is a mirror image across y=x, so the diagonal BD is indeed a line of symmetry. Could we do it with fewer? No — each of the four original off-diagonal black squares had a missing partner, and a single new square can fix at most one missing partner, so we need at least 4. That rules out (A), (B), (C), and (E).

💡Key takeaway

A line of symmetry just means every black square has a black mirror twin. Pair up the black squares, find the lonely ones, and color in their twins — four were missing, so the answer is (D).

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