2023 AMC 10B Problems/Problem 7: Difference between revisions

Revision as of 01:05, 27 November 2023

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

~yourmomisalosinggame (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 $AI=EI$ , $\angle AIE$ is 110 degrees, $\angle AIH$ is 20 degrees, and $\angle EIB$ is 70 degrees. So, we get $\angle EAI$ is $(180°-110°)/2$ = \boxed{\textbf{(B) }35}$.

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

@@ Line 32: / Line 32: @@
 -Benedict T (countmath1)
+== Solution 5 (Educated Guess) ==
+We call the point where <math>AB</math> and <math>EH</math> intersect I. We can make an educated guess that triangle AEI is isosceles so <math>AI=EI</math>, <math>\angle AIE</math> is 110 degrees, <math>\angle AIH</math> is 20 degrees, and <math>\angle EIB</math> is 70 degrees. So, we get <math>\angle EAI</math> is <math>(180°-110°)/2</math> = \boxed{\textbf{(B) }35}$.
 ==Video Solution by MegaMath==

2023 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 6	Followed by Problem 8
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