Difference between revisions of "2023 AMC 10B Problems/Problem 7"
Countmath1 (talk | contribs) (→Solution 3) |
Megahertz13 (talk | contribs) (→Video Solution 1 by OmegaLearn) |
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+ | https://www.youtube.com/watch?v=KsAxW53-P0A&t=4s | ||
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+ | ~megahertz13 | ||
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+ | ==Video Solution 2 by OmegaLearn== | ||
https://youtu.be/LI1Xq2onHHg | https://youtu.be/LI1Xq2onHHg | ||
Revision as of 21:00, 18 November 2023
Contents
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 .
~yourmomisalosinggame (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)
Video Solution by MegaMath
https://www.youtube.com/watch?v=KsAxW53-P0A&t=4s
~megahertz13
Video Solution 2 by OmegaLearn
Video Solution 2 by SpreadTheMathLove
https://www.youtube.com/watch?v=cT-0V4a3FYY
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.