2021 AMC 12A Problems/Problem 24

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 $\odot \Omega$ is $3.$

By the SAS Congruence, we have $\triangle QXM\cong\triangle RXM,$ both of which are $30^\circ$ - $60^\circ$ - $90^\circ$ triangles. By the side-length ratios, $RM=\frac{3\sqrt3}{2}, RX=3,$ and $MX=\frac{3}{2}.$ By the Pythagorean Theorem in $\triangle ORM,$ we get $OM=\frac{13}{2}$ and $OX=OM-XM=5.$ By the Pythagorean Theorem on $\triangle OXP,$ we obtain $OP=4.$

[...]

As shown above, we construct an altitude $\overline{PC}$ of $\triangle PQR.$ Since $\overline{PC}\perp\overline{RQ}$ and $\overline{OM}\perp\overline{RQ},$ we know that $\overline{PC}\parallel\overline{OM}.$ We construct $D$ on $\overline{PC}$ such that $\overline{XD}\perp\overline{PC}.$ Clearly, $MXDC$ is a rectangle. Since $\angle XPD=\angle OXP$ by alternate interior angles, we have $\triangle XPD\sim\triangle OXP$ by the AA Similarity, with ratio of similitude $\frac{XP}{OX}=\frac 35.$ Therefore, we get that $PD=\frac 95$ and $PC=PD+DC=PD+MX=\frac 95 + \frac 32 = \frac{33}{10}.$

The area of $\triangle PQR$ is $\frac12(RQ)(PC)=\frac12(3\sqrt3)(\frac{33}{10})=\frac{99\sqrt3}{20},$ and the answer is $99+3+20=\boxed{\textbf{(A) } 110}.$

~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

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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