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2021 Fall AMC 12B Problems/Problem 9

Revision as of 21:51, 23 November 2021 by Wilhelmz (talk | contribs)

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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