2023 AMC 10B Problem 7

Let $O$ be the common center. Rotating about $O$ keeps every vertex at the same distance from $O$, so all eight vertices lie on one circle centered at $O$. Draw that circle.

On this circle, $A$ and $B$ are adjacent vertices of a square, so arc $AB$ measures $90^\circ$. The vertex $E$ is the image of $A$ under a $20^\circ$ clockwise rotation, so it sits on arc $AB$ with arc $AE = 20^\circ$. Hence arc $EB = 90^\circ - 20^\circ = 70^\circ$.

$\angle EAB$ is an inscribed angle intercepting arc $EB$, so
$$\angle EAB = \frac{1}{2}\cdot 70^\circ = 35^\circ.$$

The answer is $\boxed{\textbf{(B)}\ 35^{\circ}}$.

A rotation about a point preserves distances to that point, so the circumcircle is the natural auxiliary figure whenever a shape is rotated about its center. On the circle, rotation angles become arcs and the inscribed angle theorem turns arcs into the angles you are asked for.

Triangle $AOE$ is isosceles with $OA = OE$ and apex angle $\angle AOE = 20^\circ$, so $\angle OAE = 80^\circ$. Since $\angle OAB = 45^\circ$ (half a right angle, along the diagonal), $\angle EAB = 80^\circ - 45^\circ = 35^\circ$.

• Answering the rotation angle 20 degrees itself, or subtracting it from 45 degrees to get 25.
• Placing $E$ on the wrong side of $A$ (a counterclockwise rotation) and computing with arc $EB = 110^\circ$.