Difference between revisions of "2020 AMC 12B Problems/Problem 12"
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<math>\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100</math> | <math>\textbf{(A)}\ 96 \qquad\textbf{(B)}\ 98 \qquad\textbf{(C)}\ 44\sqrt5 \qquad\textbf{(D)}\ 70\sqrt2 \qquad\textbf{(E)}\ 100</math> | ||
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+ | ==Diagram== | ||
+ | <b>DIAGRAM IN PROGRESS. NO EDIT PLEASE.</b> | ||
+ | |||
+ | ~MRENTHUSIASM (by Geometry Expressions) | ||
==Solution 1 (Pythagorean Theorem) == | ==Solution 1 (Pythagorean Theorem) == |
Revision as of 05:53, 22 April 2021
Contents
Problem
Let be a diameter in a circle of radius Let be a chord in the circle that intersects at a point such that and What is
Diagram
DIAGRAM IN PROGRESS. NO EDIT PLEASE.
~MRENTHUSIASM (by Geometry Expressions)
Solution 1 (Pythagorean Theorem)
Let be the center of the circle, and be the midpoint of . Let and . This implies that . Since , we now want to find . Since is a right angle, by Pythagorean theorem . Thus, our answer is .
~JHawk0224
Solution 2 (Power of a Point)
Let be the center of the circle, and be the midpoint of . Draw triangle , and median . Because , is isosceles, so is also an altitude of . , and because angle is degrees and triangle is right, . Because triangle is right, . Thus, .
We are looking for + which is also .
Because , .
By Power of a Point, , so .
Finally, .
Solution 3 (Law of Cosines)
Let be the center of the circle. Notice how , where is the radius of the circle. By applying the law of cosines on triangle , . Similarly, by applying the law of cosines on triangle , . By subtracting these two equations, we get . We can rearrange it to get . Because both and are both positive, we can safely divide both sides by to obtain . Because , . Through power of a point, we can find out that , so .
~Math_Wiz_3.14
Solution 4 (Reflections)
Let be the center of the circle. By reflecting across the line to produce , we have that . Since , . Since , by the Pythagorean Theorem, our desired solution is just . Looking next to circle arcs, we know that , so . Since , and , . Thus, . Since , by the Pythagorean Theorem, the desired .
~sofas103
Video Solutions
https://www.youtube.com/watch?v=h-hhRa93lK4
See Also
2020 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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 12 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.