2021 AMC 12A Problems/Problem 24
Contents
Problem
Semicircle has diameter of length . Circle lies tangent to at a point and intersects at points and . If and , then the area of equals , where and are relatively prime positive integers, and is a positive integer not divisible by the square of any prime. What is ?
Diagram
~MRENTHUSIASM (by Geometry Expressions)
Solution
Let be the center of the semicircle, be the center of the circle, and be the midpoint of By the Perpendicular Chord Theorem Converse, we have and Together, points and must be collinear.
Applying the Extended Law of Sines on we have in which the radius of is
By the SAS Congruence, we have both of which are -- triangles. By the side-length ratios, and By the Pythagorean Theorem in we get and By the Pythagorean Theorem on we obtain
As shown above, we construct an altitude of Since and we know that We construct on such that Clearly, is a rectangle. Since by alternate interior angles, we have by the AA Similarity, with ratio of similitude Therefore, we get that and
The area of is and the answer is
~MRENTHUSIASM
Video Solution by Punxsutawney Phil
https://youtube.com/watch?v=cEHF5iWMe9c
Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines )
See also
2021 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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 |
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