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

Revision as of 01:17, 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

2021 Fall AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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

Revision as of 01:17, 24 November 2021

Problem

Solution 1 (Cosine Rule)