2023 AMC 10B Problem 7

Let $O$ be the center of the squares. Turning a shape around $O$ keeps each corner the same distance from $O$, so we can use a circle.

1. Draw the circle. All eight corners $A, B, C, D, E, F, G, H$ are the same distance from $O$, so they all lie on one circle with center $O$.
2. Arc $AB$. $A$ and $B$ are neighboring corners of a square, and the four corners split the circle into four equal arcs. So arc $AB$ is $360^\circ \div 4 = 90^\circ$.
3. Where $E$ is. $E$ is where $A$ lands after turning $20^\circ$ clockwise, so $E$ is on arc $AB$, $20^\circ$ past $A$. That leaves
$$\text{arc } EB = 90^\circ - 20^\circ = 70^\circ.$$
4. Angle at $A$. $\angle EAB$ has its corner $A$ on the circle, and its two sides cut off the arc $EB$. An angle like this (called an inscribed angle) is half of the arc it cuts off:
$$\angle EAB = \frac{70^\circ}{2} = 35^\circ.$$

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

When a shape is turned around its center, every point stays the same distance from the center, so a circle through the corners is the natural thing to draw. On that circle, the turning angle becomes an arc, and the rule "an angle with its corner on the circle is half its arc" changes arcs into the angle you want.

Skip the circle and use a triangle instead.

1. Triangle $AOE$ has $OA = OE$ (both are the distance from the center to a corner), and the angle at $O$ is the turn, $20^\circ$.
2. The two equal angles at $A$ and $E$ share the remaining $180^\circ - 20^\circ = 160^\circ$, so each is $80^\circ$. That means $\angle OAE = 80^\circ$.
3. The line $AO$ runs along the diagonal of square $ABCD$, and a diagonal cuts the $90^\circ$ corner in half, so $\angle OAB = 45^\circ$.
4. Seen from $A$, the side $AB$ lies between the lines $AO$ and $AE$, so $\angle EAB = \angle OAE - \angle OAB = 80^\circ - 45^\circ = 35^\circ$.

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