Difference between revisions of "2023 AMC 10B Problems/Problem 7"

Revision as of 09:08, 16 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)

Video Solution 1 by OmegaLearn

https://youtu.be/LI1Xq2onHHg

Video Solution 2 by SpreadTheMathLove

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

@@ Line 24: / Line 24: @@
 ~hpotter2021
+== Solution 4 ==
+Draw <math>EA</math>: we want to find <math>\angle EAB</math>. Call <math>P</math> the point at which <math>AB</math> and <math>EH</math> intersect. Reflecting <math>\triangle APE</math> over <math>EA</math>, we have a parallelogram. Since <math>\angle EPB = 70^{\circ}</math>, angle subtraction tells us that two of the angles of the parallelogram are <math>110^{\circ}</math>. The other two are equal to <math>2\angle EAB</math> (by properties of reflection).
+Since angles on the transversal of a parallelogram sum to <math>180^{\circ}</math>, we have <math>2\angle EAB + 110 = 180</math>, yielding <math>\angle EAB = \boxed{\textbf{(B) }35}</math>
+-Benedict T (countmath1)
 ==Video Solution 1 by OmegaLearn==

2023 AMC 10B (Problems • Answer Key • Resources)
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

Page

Toolbox

Search