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

Revision as of 03:08, 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 $12\pi \Rightarrow \boxed{\textbf{(B)}}$ .

~Wilhelm Z

2021 Fall AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 8	Followed by Problem 10
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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

@@ Line 24: / Line 24: @@
 <math>r^2=12</math>
-The area is therefore <math>\boxed{\textbf{(B)</math>12\pi<math>}}</math>.
+The area is therefore <math>12\pi \Rightarrow \boxed{\textbf{(B)}}</math>.
 ~Wilhelm Z

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

Revision as of 03:08, 25 November 2021

Problem

Solution 1 (Cosine Rule)