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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)
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==Solution==
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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 01:25, 14 February 2021

Problem

Semicircle $\Gamma$ has diameter $\overline{AB}$ of length $14$. Circle $\Omega$ lies tangent to $\overline{AB}$ at a point $P$ and intersects $\Gamma$ at points $Q$ and $R$. If $QR=3\sqrt3$ and $\angle QPR=60^\circ$, then the area of $\triangle PQR$ equals $\tfrac{a\sqrt{b}}{c}$, where $a$ and $c$ are relatively prime positive integers, and $b$ is a positive integer not divisible by the square of any prime. What is $a+b+c$?

$\textbf{(A) } 110\qquad\textbf{(B) } 114\qquad\textbf{(C) } 118\qquad\textbf{(D) } 122\qquad\textbf{(E) } 126\qquad$

Diagram

File:2021 AMC 12A Problem 24.png

~MRENTHUSIASM (by Geometry Expressions)

Solution

Let $O=\Gamma$ be the center of the semicircle, $X=\Omega$ be the center of the circle, and $M$ be the midpoint of $\overline{QR}.$ By the Perpendicular Chord Theorem Converse, we have $\overline{XM}\perp\overline{QR}$ and $\overline{OM}\perp\overline{QR}.$ Together, points $O, X,$ and $M$ must be collinear.

Applying the Extended Law of Sines on $\triangle PQR,$ we have \[XP=\frac{QR}{2\cdot\sin \angle QPR}=\frac{3\sqrt3}{2\cdot\frac{\sqrt3}{2}}=3,\] in which the radius of $\circ \Omega$ is $3.$

~MRENTHUSIASM

Video Solution by Punxsutawney Phil

https://youtube.com/watch?v=cEHF5iWMe9c

Video Solution by OmegaLearn (Similar Triangles, Law of Sines, Law of Cosines )

https://youtu.be/j965v6ahUZk

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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