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Difference between revisions of "2021 Fall AMC 12B Problems/Problem 9"
(→‎Solution 1 (Cosine Rule))
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==Problem==
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Triangle <math>ABC</math> is equilateral with side length <math>6</math>. Suppose that <math>O</math> is the center of the inscribed
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circle of this triangle. What is the area of the circle passing through <math>A</math>, <math>O</math>, and <math>C</math>?
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<math>\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi</math>
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==Solution 1 (Cosine Rule) ==
 
==Solution 1 (Cosine Rule) ==
  

Revision as of 21:53, 23 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 $X$.

Then connect $OX$, and notice that $OX$ is the perpendicular bisector of $AC$. Let the intersection of $OX$ with $AC$ be $D$.

Also notice that $AO$ and $CO$ are the angle bisectors of angle $BAC$ and $BCA$ respectively. We then deduce $AOC=120^\circ$.

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

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

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

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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