2023 AMC 10B Problems/Problem 7

Sqrt $ABCD$ is rotated $20^{\circ}$ clockwise about its center to obtain square $EFGH$ , as shown below(Please help me add diagram and then remove this). <img src="https://wiki-images.artofproblemsolving.com//9/9d/IMG_1031.jpeg" alt="IMG 1031.jpeg"/> 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 $\boxed{\text{(B)} 35}$ .

~yourmomisalosinggame (a.k.a. Aaron)

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2023 AMC 10B Problems/Problem 7

Solution 1

Solution 2