Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

2021 Fall AMC 12B Problems/Problem 9

Revision as of 22:01, 23 November 2021 by Wilhelmz (talk | contribs) (→‎Solution 1 (Cosine Rule))

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 $X$.

Then connect $\overline{OX}$, and notice that $\overline{OX}$ is the perpendicular bisector of $\overline{AC}$. Let the intersection of $\overline{OX}$ with $\overline{AC}$ be $D$.

Also notice that $\overline{OA}$ and $\overline{OC}$ 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 $eAXC=2·AMC=120^\circ$.

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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2021_Fall_AMC_12B_Problems/Problem_9&oldid=165812"