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 $\angle EAB$?

[asy] size(170); defaultpen(linewidth(0.6)); real r = 25; draw(dir(135)--dir(45)--dir(315)--dir(225)--cycle); draw(dir(135-r)--dir(45-r)--dir(315-r)--dir(225-r)--cycle); label("$A$",dir(135),NW); label("$B$",dir(45),NE); label("$C$",dir(315),SE); label("$D$",dir(225),SW); label("$E$",dir(135-r),N); label("$F$",dir(45-r),E); label("$G$",dir(315-r),S); label("$H$",dir(225-r),W); [/asy]

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

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 $\boxed{\text{(B)}   35}$.

~Stead (a.k.a. Aaron)

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

Video Solution by MegaMath

https://www.youtube.com/watch?v=KsAxW53-P0A&t=4s

~megahertz13

Video Solution 2 by OmegaLearn

https://youtu.be/LI1Xq2onHHg

Video Solution 3 by SpreadTheMathLove

https://www.youtube.com/watch?v=cT-0V4a3FYY

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/EuLkw8HFdk4?si=Te_9kmP_bmBoKrTn&t=1393

~Math-X

Video Solution

https://youtu.be/R9uCV2KsXc8

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

Video Solution by Interstigation

https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L

See also

2023 AMC 10B (ProblemsAnswer KeyResources)
Preceded by
Problem 6
Followed by
Problem 8
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
All AMC 10 Problems and Solutions

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