AMC 10 · 2023 · #7
Grade 7 geometry-2dProblem
Square ABCD is rotated 20∘ clockwise about its center to obtain square EFGH, as shown below. What is the degree measure of ∠EAB?
Pick an answer.
AMC 10 2023 problem © Mathematical Association of America (MAA AMC). Reproduced for educational use.
Try it yourself first — the explanation is most useful after you’ve attempted it.
Toolkit + CCSS Solution
Understand
Restated: Two congruent squares $ABCD$ and $EFGH$ share their center $O$, with $EFGH$ obtained from $ABCD$ by a $20^\circ$ clockwise rotation. Find the size of $\angle EAB$.
Givens: Square $ABCD$ and its rotated image $EFGH$ share the center $O$; Rotation angle is $20^\circ$ clockwise; Diagram fixes the labels: $E$ is the image of $A$, $F$ of $B$, etc.; Answer choices: (A) $24^\circ$, (B) $35^\circ$, (C) $30^\circ$, (D) $32^\circ$, (E) $20^\circ$
Unknowns: The measure of $\angle EAB$ at vertex $A$
Understand
Restated: Two congruent squares $ABCD$ and $EFGH$ share their center $O$, with $EFGH$ obtained from $ABCD$ by a $20^\circ$ clockwise rotation. Find the size of $\angle EAB$.
Givens: Square $ABCD$ and its rotated image $EFGH$ share the center $O$; Rotation angle is $20^\circ$ clockwise; Diagram fixes the labels: $E$ is the image of $A$, $F$ of $B$, etc.; Answer choices: (A) $24^\circ$, (B) $35^\circ$, (C) $30^\circ$, (D) $32^\circ$, (E) $20^\circ$
Plan
Primary tool: #1 Draw a Diagram
Secondary: #7 Identify Subproblems
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 $\triangle OAE$ with apex angle $20^\circ$, because rotation keeps $OA = OE$. Tool #7 (Identify Subproblems) then splits the target $\angle EAB$ into two clean pieces — $\angle OAE$ (a base angle of the isosceles triangle) minus $\angle OAB$ (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.
Execute — Answer: B
4.G.A.1 Step 1 - Mark the shared center $O$ and draw the segments $OA$, $OE$, and $OB$.
- Note that $A, O, C$ are collinear (the diagonal of the original square) and that the angle $\angle AOE$ is the rotation angle, $20^\circ$.
- The picture now shows the isosceles triangle $\triangle OAE$ with $OA = OE$, and the right angle $\angle DAB = 90^\circ$ of the square cut in half by diagonal $AC$.
💡 Sketching the rotation center and the two equal radii surfaces the hidden isosceles triangle — Grade 4 segment and angle drawing.
7.G.B.5 Step 2 - Use the triangle angle sum on $\triangle OAE$.
- Because $OA = OE$, the two base angles are equal: $\angle OAE = \angle OEA$.
- Together with $\angle AOE = 20^\circ$ they must add to $180^\circ$.
💡 Triangle angle facts plus equal base angles of an isosceles triangle — Grade 7 angle reasoning.
4.G.A.2 Step 3 The diagonal of a square bisects each $90^\circ$ corner, so $OA$ splits $\angle DAB$ in half: $\angle OAB$ is exactly half of $90^\circ$.
💡 A square's diagonal cutting the corner in half is a Grade 4 shape-property fact.
4.MD.C.7 Step 4 - From the picture, $E$ sits between $A$ and $B$ as seen from $A$, so $\angle EAB = \angle OAE - \angle OAB$.
- Subtract.
💡 Adding and subtracting angles meeting at a point — Grade 4 angle additivity.
4.G.A.1 Mark the shared center $O$ and draw the segments $OA$, $OE$, and $OB$. Note that 7.G.B.5 Use the triangle angle sum on $\triangle OAE$. Because $OA = OE$, the two base a 4.G.A.2 The diagonal of a square bisects each $90^\circ$ corner, so $OA$ splits $\angle 4.MD.C.7 From the picture, $E$ sits between $A$ and $B$ as seen from $A$, so $\angle EAB Review
Reasonableness: Three checks. (1) Sanity of magnitude: a $20^\circ$ rotation should move $A$ by a small angle from where the diagonal sits, so $\angle EAB$ should be a touch under $45^\circ$ — $35^\circ$ fits. (2) Endpoint test: if the rotation were $0^\circ$, the formula gives $\angle OAE - \angle OAB = 90^\circ - 45^\circ = 45^\circ$ — exactly $\angle CAB$, as expected. If the rotation were $90^\circ$ (full quarter turn), $E$ would coincide with $D$ and $\angle EAB = 90^\circ - 0^\circ \cdot\tfrac{1}{2}$ — also matches. (3) Eliminate distractors: (E) $20^\circ$ is the bait answer (just the rotation angle); (A) $24^\circ$, (C) $30^\circ$, (D) $32^\circ$ each correspond to mistakes such as forgetting the diagonal bisection or dividing $20^\circ$ by $2$ instead of using the triangle angle sum.
Alternative: Tool #13 (Convert to Algebra): place the squares on a coordinate plane with $O$ at the origin and side length $\sqrt{2}$, so $A = (-1, 1)$ and $B = (1, 1)$. Rotate $A$ by $-20^\circ$ to get $E = (-\cos 20^\circ - \sin 20^\circ, \cos 20^\circ - \sin 20^\circ)$. Then $\angle EAB$ comes from the dot product of $\vec{AE}$ and $\vec{AB}$. The trigonometry collapses to the same $35^\circ$, but the picture-and-subtraction path is one short paragraph while the coordinate path drags in trig identities.
CCSS standards used (min grade 7)
4.G.A.1Draw points, lines, line segments, rays, angles, and identify in figures (Adding the segments $OA$ and $OE$ from the shared center $O$ to expose the isosceles triangle hidden inside the diagram.)4.G.A.2Classify two-dimensional figures based on presence of parallel or perpendicular lines (Using the square property that its diagonal bisects each $90^\circ$ vertex angle, giving $\angle OAB = 45^\circ$.)4.MD.C.7Recognize angle measure as additive and solve addition and subtraction problems (Computing $\angle EAB = \angle OAE - \angle OAB$ — the angles at vertex $A$ behave additively.)7.G.B.5Use facts about supplementary, complementary, vertical, and adjacent angles (Combining the isosceles triangle's equal base angles with the triangle angle sum to get $\angle OAE = 80^\circ$.)
⭐ 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 ($80^\circ$), then subtract the square's diagonal-bisects-the-corner $45^\circ$ to get $35^\circ$.
⭐ 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 ($80^\circ$), then subtract the square's diagonal-bisects-the-corner $45^\circ$ to get $35^\circ$.