Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 10B Problems/Problem 7"
Line 17: Line 17:
  
 
~yourmomisalosinggame (a.k.a. Aaron)
 
~yourmomisalosinggame (a.k.a. Aaron)
 +
 +
== Solution 3 ==
 +
 +
Call the center of both squares point <math>O</math>, and draw circle <math>O</math> such that it circumscribes the squares. <math>\angle{EOF} = 90</math> and <math>\angle{BOF} = 20</math>, so <math>\angle{EOB} = 70</math>. Since <math>\angle{EAB}</math> is inscribed in arc EB, <math>\angle{EAB} = 70/2 = \boxed{\text{(B)}  35}</math>.

Revision as of 19:26, 15 November 2023

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

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 EB, $\angle{EAB} = 70/2 = \boxed{\text{(B)} 35}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_10B_Problems/Problem_7&oldid=203518"