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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 9"

(→‎Solution 1 (Cosine Rule))
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Consider another point <math>M</math> on Circle <math>X</math> opposite to point <math>O</math>.  
 
Consider another point <math>M</math> on Circle <math>X</math> opposite to point <math>O</math>.  
  
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As <math>AOCM</math> an inscribed quadrilateral of Circle <math>X</math>, <math>\angle AMC=180^\circ-120^\circ=60^\circ</math>.
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As <math>AOCM</math> is an inscribed quadrilateral of Circle <math>X</math>, <math>\angle AMC=180^\circ-120^\circ=60^\circ</math>.
  
 
Afterward, deduce that <math>\angle AXC=2·\angle AMC=120^\circ</math>.
 
Afterward, deduce that <math>\angle AXC=2·\angle AMC=120^\circ</math>.

Revision as of 00:31, 24 November 2021

Problem

Triangle $ABC$ is equilateral with side length $6$. Suppose that $O$ is the center of the inscribed circle of this triangle. What is the area of the circle passing through $A$, $O$, and $C$?

$\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi$

Solution 1 (Cosine Rule)

Construct the circle that passes through $A$, $O$, and $C$, centered at $X$.

Also notice that $\overline{OA}$ and $\overline{OC}$ are the angle bisectors of angle $\angle BAC$ and $\angle BCA$ respectively. We then deduce $\angle AOC=120^\circ$.

Consider another point $M$ on Circle $X$ opposite to point $O$.

As $AOCM$ is an inscribed quadrilateral of Circle $X$, $\angle AMC=180^\circ-120^\circ=60^\circ$.

Afterward, deduce that $\angle AXC=2·\angle AMC=120^\circ$.

By the Cosine Rule, we have the equation: (where $r$ is the radius of circle $X$)

$2r^2(1-\cos(120^\circ))=6^2$

$r^2=12$

The area is therefore $12\pi \Rightarrow \boxed{\textbf{(B)}}$.

~Wilhelm Z

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