Difference between revisions of "2023 AMC 10B Problems/Problem 7"
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== Solution 3 == | == 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 <math>\overset \frown {EB}</math>, | + | 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 <math>\overset \frown {EB}</math>, <math>\angle{EAB} = 70/2 = \boxed{\textbf{(B) }35}</math>. |
~hpotter2021 | ~hpotter2021 |
Revision as of 19:36, 15 November 2023
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