2021 Fall AMC 12B Problems/Problem 9


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 $X$)



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)

Solution 3

As in the previous solution, construct the circle that passes through $A$, $O$, and $C$, centered at $X$. Let $Y$ be the intersection of $\overline{OX}$ and $\overline{AB}$.

Note that since $\overline{OA}$ is the angle bisector of $\angle BAC$ that $\angle OAC=30^\circ$. Also by symmetry, $\overline{OX}$ $\perp$ $\overline{AB}$ and $AY = 3$. Thus $\tan(30^\circ) = \frac{OY}{3}$ so $OY = \sqrt{3}$.

Let $r$ be the radius of circle $X$, and note that $AX = OX = r$. So $\triangle AYX$ is a right triangle with legs of length $3$ and $r - \sqrt{3}$ and hypotenuse $r$. By Pythagoras, $3^2 + (r - \sqrt{3})^2 = r^2$. So $r = 2\sqrt{3}$.

Thus the area is $\pi r^2 = \boxed{\textbf{(B)}\ 12\pi}$.


Solution 5 (SIMPLE)

The semiperimeter is $\frac{6+6+6}{2}=9$ units. The area of the triangle is $9\sqrt{3}$ units squared. By the formula that says that the area of the triangle is its semiperimeter times its inradius, the inradius $r=\sqrt{3}$. As $\angle{AOC}=120^\circ$, we can form an altitude from point $O$ to side $AC$ at point $M$, forming two 30-60-90 triangles. As $CM=MA=3$, we can solve for $OC=2\sqrt{3}$. Now, the area of the circle is just $\pi*(2*\sqrt{3})^2 = 12\pi$. Select $\boxed{B}$.

~hastapasta, bob4108

Solution 6 (Ptolemy)

Call the diameter of the circle $d$. If we extend points $A$ and $C$ to meet at a point on the circle and call it $E$, then $\bigtriangleup OAE=\bigtriangleup OCE$ . Note that both triangles are right, since their hypotenuse is the diameter of the circle. Therefore, $CE=AE=\sqrt{d^2-12}$. We know this since $OC=OA=OB$ and $OC$ is the hypotenuse of a $30-60-90$ right triangle, with the longer leg being $\frac{6}{2}=3$ so $OC=2\sqrt{3}$. Applying Ptolemy's Theorem on cyclic quadrilateral $OCEA$, we get $2({\sqrt{d^2-12}})\cdot{2\sqrt{3}}=6d$. Squaring and solving we get $d^2=48 \Longrightarrow (2r)^2=48$ so $r^2=12$. Therefore, the area of the circle is $\boxed{12\pi}$


Video Solution (Just 3 min!)


~Education, the Study of Everything

Video Solution by TheBeautyofMath



See Also

2021 Fall AMC 12B (ProblemsAnswer KeyResources)
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

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