Competition · AMC preparation · step 4 of 4

AMC 10 · 2023B · #7

Grade 7 geometry-2d
rotation-isometryisosceles-triangleangle-sum-trianglesymmetry-argument identify-subproblems ↑ Prerequisites: angle-sum-triangleisosceles-triangle
📏 Medium solution 💡 3 insights 📊 Diagram
Problem
Two congruent squares ABCD and EFGH share their center O, with EFGH obtained from ABCD by a 20° clockwise rotation. Find the size of ∠ EAB.

Pick an answer.

(A)
$24^{\circ}$
(B)
$35^{\circ}$
(C)
$30^{\circ}$
(D)
$32^{\circ}$
(E)
$20^{\circ}$

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

How to solve
Strategy Draw a Diagram

The picture is everything here. Tool #1 (Draw a Diagram) says: add the center O and the segment OA (which lies along the diagonal AC) and the segment OE. That single addition exposes an isosceles triangle △ OAE with apex angle 20°, because rotation keeps OA = OE. Tool #7 (Identify Subproblems) then splits the target ∠ EAB into two clean pieces — ∠ OAE (a base angle of the isosceles triangle) minus ∠ OAB (half of the square's corner). Solve the pieces separately, subtract. No algebra is needed — just isosceles base angles, the triangle angle sum, and the diagonal-bisects-corner fact.

1STEP 1

Draw the radii to the center

Drop in the shared center O and draw OA, OE; the rotation gives OA = OE, so △ OAE is isosceles with apex ∠ AOE = 20°.

OA = OE, ∠ AOE = 20°
2STEP 2

Use the triangle angle sum

Angle sum on △ OAE with equal base angles ∠ OAE = ∠ OEA gives ∠ OAE = 80°.

2 · ∠ OAE + 20° = 180° → ∠ OAE = 160°/2 = 80°
3STEP 3

Halve the square's corner

A square's diagonal bisects its corner, so OA halves the 90° angle: ∠ OAB = 45°.

∠ OAB = 1/2 · 90° = 45°
4STEP 4

Subtract the two angles

From the figure E lies between rays AO and AB, so subtract: ∠ EAB = 80° - 45° = 35°, choice (B).

∠ EAB = 80° - 45° = 35° → (B)
Answer
35^°
Three checks. (1) Sanity of magnitude: a 20° rotation should move A by a small angle from where the diagonal sits, so ∠ EAB should be a touch under 45° — 35° fits. (2) Endpoint test: if the rotation were 0°, the formula gives ∠ OAE - ∠ OAB = 90° - 45° = 45° — exactly ∠ CAB, as expected. If the rotation were 90° (full quarter turn), E would coincide with D and ∠ EAB = 90° - 0° ·12\frac{1}{2} — also matches. (3) Eliminate distractors: (E) 20° is the bait answer (just the rotation angle); (A) 24°, (C) 30°, (D) 32° each correspond to mistakes such as forgetting the diagonal bisection or dividing 20° by 2 instead of using the triangle angle sum.
💡Key takeaway

This AMC 10 problem only needs Grade 7 angle facts you already know — draw the center, spot the isosceles triangle the rotation creates, find its base angle (80°), then subtract the square's diagonal-bisects-the-corner 45° to get 35°.

  • Draw the radii to the center
  • Use the triangle angle sum
  • Halve the square's corner
  • Subtract the two angles

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