Difference between revisions of "2021 AMC 12A Problems/Problem 24"
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[[File:2021 AMC 12A Problem 24.png|center]] | [[File:2021 AMC 12A Problem 24.png|center]] | ||
~MRENTHUSIASM (by Geometry Expressions) | ~MRENTHUSIASM (by Geometry Expressions) | ||
+ | |||
+ | ==Solution== | ||
+ | Let <math>O=\Gamma</math> be the center of the semicircle, <math>X=\Omega</math> be the center of the circle, and <math>M</math> be the midpoint of <math>\overline{QR}.</math> By the Perpendicular Chord Theorem Converse, we have <math>\overline{XM}\perp\overline{QR}</math> and <math>\overline{OM}\perp\overline{QR}.</math> Together, points <math>O, X,</math> and <math>M</math> must be collinear. | ||
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+ | Applying the Extended Law of Sines on <math>\triangle PQR,</math> we have <cmath>XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3,</cmath> in which the radius of <math>\circ \Omega</math> is <math>3.</math> | ||
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+ | ~MRENTHUSIASM | ||
==Video Solution by Punxsutawney Phil== | ==Video Solution by Punxsutawney Phil== |
Revision as of 02:25, 14 February 2021
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
~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.