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

(Created page with "==Solution 1 (Cosine Rule) == Solution coming up soon! Don't edit now, please! ~Wilhelm Z")
 
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==Solution 1 (Cosine Rule) ==
 
==Solution 1 (Cosine Rule) ==
  
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Solution coming up soon! Don't edit now, please!
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Construct the circle that passes through <math>A</math>, <math>O</math>, and <math>C</math>, centered at <math>X</math>.
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Then connect <math>OX</math>, and notice that <math>OX</math> is the perpendicular bisector of <math>AC</math>. Let the intersection of <math>OX</math> with <math>AC</math> be <math>D</math>.
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Also notice that <math>AO</math> and <math>CO</math> are the angle bisectors of angle <math>BAC</math> and <math>BCA</math> respectively. We then deduce <math>AOC=120^\circ</math>.
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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>AMC=180^\circ-120^\circ=60^\circ</math>.
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Afterward, deduce that <math>AXC=2·AMC=120^\circ</math>.
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By the Cosine Rule, we have the equation: (where <math>r</math> is the radius of circle <math>X</math>)
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<math>2r^2(1-\cos(120^\circ))=6^2</math>
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<math>r^2=12</math>
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The area is therefore <math>12\pi \Rightarrow \boxed{\textbf{(B)}}</math>.
  
 
~Wilhelm Z
 
~Wilhelm Z

Revision as of 21:51, 23 November 2021

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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