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

(→‎Solution 1 (Cosine Rule))
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~Wilhelm Z
 
~Wilhelm Z
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== Solution 2 ==
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We have <math>\angle AOC = 120^\circ</math>.
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Denote by <math>R</math> the circumradius of <math>\triangle AOC</math>.
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In <math>\triangle AOC</math>, the law of sines implies
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<cmath>
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\[
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2 R = \frac{AC}{\sin \angle AOC} = 4 \sqrt{3} .
 +
\]
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</cmath>
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Hence, the area of the circumcircle of <math>\triangle AOC</math> is
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<cmath>
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\[
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\pi R^2 = 12 \pi .
 +
\]
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</cmath>
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Therefore, the answer is <math>\boxed{\textbf{(B) }12 \pi}</math>.
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 +
~Steven Chen (www.professorchenedu.com)
  
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=10|num-b=8}}
 
{{AMC12 box|year=2021 Fall|ab=B|num-a=10|num-b=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:09, 25 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 $\pi r^2 = \boxed{\textbf{(B)}\ 12\pi}$.

~Wilhelm Z

Solution 2

We have $\angle AOC = 120^\circ$.

Denote by $R$ the circumradius of $\triangle AOC$. In $\triangle AOC$, the law of sines implies \[ 2 R = \frac{AC}{\sin \angle AOC} = 4 \sqrt{3} . \]

Hence, the area of the circumcircle of $\triangle AOC$ is \[ \pi R^2 = 12 \pi . \]

Therefore, the answer is $\boxed{\textbf{(B) }12 \pi}$.

~Steven Chen (www.professorchenedu.com)

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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All AMC 12 Problems and Solutions

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