Difference between revisions of "2021 AMC 12A Problems/Problem 24"

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<math>\textbf{(A) } 110\qquad\textbf{(B) } 114\qquad\textbf{(C) } 118\qquad\textbf{(D) } 122\qquad\textbf{(E) } 126\qquad</math>
 
<math>\textbf{(A) } 110\qquad\textbf{(B) } 114\qquad\textbf{(C) } 118\qquad\textbf{(D) } 122\qquad\textbf{(E) } 126\qquad</math>
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==Diagram==
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[[File:2021 AMC 12A Problem 24.png|center]]
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~MRENTHUSIASM (by Geometry Expressions)
  
 
==Video Solution by Punxsutawney Phil==
 
==Video Solution by Punxsutawney Phil==

Revision as of 00:52, 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

~MRENTHUSIASM (by Geometry Expressions)

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 (ProblemsAnswer KeyResources)
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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