AMC 8 · 2010 · #6

Grade 4 geometry-2d

Archetype Small-Domain Constrained Search └ Sub-archetype Answer-Choice Plug-In

line-symmetryreflection-symmetrysystematic-enumeration systematic-enumerationcasework ↑ Prerequisites: line-symmetry

📏 Short solution 💡 2 insights

Problem

Among five named figures — an equilateral triangle, a non-square rhombus, a non-square rectangle, an isosceles trapezoid, and a square — which one has the most lines of symmetry?

Pick an answer.

(A)

equilateral triangle

(B)

non-square rhombus

(C)

non-square rectangle

(D)

isosceles trapezoid

(E)

square

AMC 8 2010 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.

How to solve

Strategy Draw a Diagram

Counting lines of symmetry is a visual task, so Tool #1 (Draw a Diagram) is the natural primary tool: sketch each shape and try folding it along every candidate line — vertical, horizontal, and the two diagonals. Tool #2 (Make a Systematic List) keeps the bookkeeping clean: we list the five shapes and write down each shape's count, then read off the maximum. No algebra or coordinates are needed — this is a pure attribute-of-shapes question.

1STEP 1

Sketch the equilateral triangle: each vertex-to-opposite-midpoint fold lands it onto itself, giving 3 lines of symmetry.

equilateral triangle: 3 lines of symmetry

2STEP 2

A non-square rhombus folds onto itself along its two diagonals but not the midlines, so it has 2 lines of symmetry.

non-square rhombus: 2 lines of symmetry

3STEP 3

A non-square rectangle folds onto itself along its two midlines but not the diagonals, so it has 2 lines of symmetry.

non-square rectangle: 2 lines of symmetry

4STEP 4

An isosceles trapezoid folds onto itself only along the vertical line through the two parallel sides' midpoints — 1 line of symmetry.

isosceles trapezoid: 1 line of symmetry

5STEP 5

A square folds onto itself along both diagonals and both midlines, giving 4 lines of symmetry.

square: 4 lines of symmetry

6STEP 6

List the counts — triangle 3, rhombus 2, rectangle 2, trapezoid 1, square 4 — and the maximum is 4, the square.

max(3, 2, 2, 1, 4) = 4 → (E) square

Answer

square

A regular polygon with n sides has n lines of symmetry. The equilateral triangle is regular with n = 3, the square is regular with n = 4 — so the square should beat the triangle by exactly one, matching our counts. The non-square rhombus and non-square rectangle each keep only half of the square's symmetries (diagonals OR midlines, not both), giving 2 each. The isosceles trapezoid is the least symmetric of the four-sided figures and gets just 1. Everything fits.

💡Key takeaway

This AMC 8 problem only needs the Grade 4 "line of symmetry" idea you already know!

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