# 2023 AMC 10B Problems/Problem 7

## Problem

Square $ABCD$ is rotated $20^{\circ}$ clockwise about its center to obtain square $EFGH$, as shown below.

What is the degree measure of $\angle EAB$?

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

## Solution 1

First, let's call the center of both squares $I$. Then, $\angle{AIE} = 20$, and since $\overline{EI} = \overline{AI}$, $\angle{AEI} = \angle{EAI} = 80$. Then, we know that $AI$ bisects angle $\angle{DAB}$, so $\angle{BAI} = \angle{DAI} = 45$. Subtracting $45$ from $80$, we get $\boxed{\text{(B)} 35}$

~jonathanzhou18

## Solution 2

First, label the point between $A$ and $H$ point $O$ and the point between $A$ and $H$ point $P$. We know that $\angle{AOP} = 20$ and that $\angle{A} = 90$. Subtracting $20$ and $90$ from $180$, we get that $\angle{APO}$ is $70$. Subtracting $70$ from $180$, we get that $\angle{OPB} = 110$. From this, we derive that $\angle{APE} = 110$. Since triangle $APE$ is an isosceles triangle, we get that $\angle{EAP} = (180 - 110)/2 = 35$. Therefore, $\angle{EAB} = 35$. The answer is $\boxed{\text{(B)} 35}$.

## Solution 3

Call the center of both squares point $O$, and draw circle $O$ such that it circumscribes the squares. $\angle{EOF} = 90$ and $\angle{BOF} = 20$, so $\angle{EOB} = 70$. Since $\angle{EAB}$ is inscribed in arc $\overset \frown {EB}$, $\angle{EAB} = 70/2 = \boxed{\textbf{(B) }35}$.

~hpotter2021

## Solution 4

Draw $EA$: we want to find $\angle EAB$. Call $P$ the point at which $AB$ and $EH$ intersect. Reflecting $\triangle APE$ over $EA$, we have a parallelogram. Since $\angle EPB = 70^{\circ}$, angle subtraction tells us that two of the angles of the parallelogram are $110^{\circ}$. The other two are equal to $2\angle EAB$ (by properties of reflection).

Since angles on the transversal of a parallelogram sum to $180^{\circ}$, we have $2\angle EAB + 110 = 180$, yielding $\angle EAB = \boxed{\textbf{(B) }35}$

-Benedict T (countmath1)

## Solution 5 (Educated Guess)

We call the point where $AB$ and $EH$ intersect I. We can make an educated guess that triangle AEI is isosceles so $AI=EI$, $\angle AIE = 110^{\circ}$ , $\angle AIH = 20^{\circ}$ , and $\angle EIB = 70^{\circ}$ . So, we get $\angle EAI$ is $(180^{\circ} - 110^{\circ})/2 = \boxed{\textbf{(B) }35}$.

~aleyang

~megahertz13

~Math-X

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)