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

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~hastapasta, bob4108
 
~hastapasta, bob4108
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==Video Solution by TheBeautyofMath==
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https://youtu.be/kuZXQYHycdk?t=1039
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~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{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}}
 
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{{MAA Notice}}

Revision as of 23:41, 29 December 2022

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)

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

-SharpeMind

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

Video Solution by TheBeautyofMath

https://youtu.be/kuZXQYHycdk?t=1039

~IceMatrix

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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