Difference between revisions of "2023 AMC 10B Problems/Problem 7"
(→Solution 2) |
m (Change formatting of question + figure to match the "2023 AMC 10B Problems" page) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
− | Square <math>ABCD</math> is rotated <math>20^{\circ}</math> clockwise about its center to obtain square <math>EFGH</math>, as shown below. | + | Square <math>ABCD</math> is rotated <math>20^{\circ}</math> clockwise about its center to obtain square <math>EFGH</math>, as shown below. What is the degree measure of <math>\angle EAB</math>? |
− | |||
− | + | <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> | ||
<math>\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}</math> | <math>\text{(A)}\ 24^{\circ} \qquad \text{(B)}\ 35^{\circ} \qquad \text{(C)}\ 30^{\circ} \qquad \text{(D)}\ 32^{\circ} \qquad \text{(E)}\ 20^{\circ}</math> |
Latest revision as of 11:22, 20 October 2024
Contents
[hide]- 1 Problem
- 2 Solution 1
- 3 Solution 2
- 4 Solution 3
- 5 Solution 4
- 6 Solution 5 (Educated Guess)
- 7 Video Solution by MegaMath
- 8 Video Solution 2 by OmegaLearn
- 9 Video Solution 3 by SpreadTheMathLove
- 10 Video Solution by Math-X (First understand the problem!!!)
- 11 Video Solution
- 12 Video Solution by Interstigation
- 13 See also
Problem
Square is rotated clockwise about its center to obtain square , as shown below. What is the degree measure of ?
Solution 1
First, let's call the center of both squares . Then, , and since , . Then, we know that bisects angle , so . Subtracting from , we get
~jonathanzhou18
Solution 2
First, label the point between and point and the point between and point . We know that and that . Subtracting and from , we get that is . Subtracting from , we get that . From this, we derive that . Since triangle is an isosceles triangle, we get that . Therefore, . The answer is .
~Stead (a.k.a. Aaron)
Solution 3
Call the center of both squares point , and draw circle such that it circumscribes the squares. and , so . Since is inscribed in arc , .
~hpotter2021
Solution 4
Draw : we want to find . Call the point at which and intersect. Reflecting over , we have a parallelogram. Since , angle subtraction tells us that two of the angles of the parallelogram are . The other two are equal to (by properties of reflection).
Since angles on the transversal of a parallelogram sum to , we have , yielding
-Benedict T (countmath1)
Solution 5 (Educated Guess)
We call the point where and intersect I. We can make an educated guess that triangle AEI is isosceles so , , , and . So, we get is .
~aleyang
Video Solution by MegaMath
https://www.youtube.com/watch?v=KsAxW53-P0A&t=4s
~megahertz13
Video Solution 2 by OmegaLearn
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
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by Interstigation
https://youtu.be/gDnmvcOzxjg?si=cYB6uChy7Ue0UT4L
See also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.