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

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==Solution 5 (SIMPLE) ==
==Solution 5 (SIMPLE) ==


The semiperimeter is <math>\frac{6+6+6}{2}=9</math> units.
The semiperimeter is <imath>\frac{6+6+6}{2}=9</imath> units.
The area of the triangle is <math>9\sqrt{3}</math> units squared.  By the formula that says that the area of the triangle is its semiperimeter times its inradius, the inradius <math>r=\sqrt{3}</math>. As <math>\angle{AOC}=120^\circ</math>, we can form an altitude from point <math>O</math> to side <math>AC</math> at point <math>M</math>, forming two 30-60-90 triangles. As <math>CM=MA=3</math>, we can solve for <math>OC=2\sqrt{3}</math>. Now, the area of the circle is just <math>\pi*(2*\sqrt{3})^2 = 12\pi</math>. Select <math>\boxed{B}</math>.
The area of the triangle is <imath>9\sqrt{3}</imath> units squared.  By the formula that says that the area of the triangle is its semiperimeter times its inradius, the inradius <imath>r=\sqrt{3}</imath>. As <imath>\angle{AOC}=120^\circ</imath>, we can form an altitude from point <imath>O</imath> to side <imath>AC</imath> at point <imath>M</imath>, forming two 30-60-90 triangles. As <imath>CM=MA=3</imath>, we can solve for <imath>OC=2\sqrt{3}</imath>. Now, call the center of the circle we are looking for <imath>X</imath>. From here we can now utilize the fact that since the inscribed angle <imath>\angle{AOC}=120^\circ</imath> that the arclength that the angle describes is <imath>240^\circ</imath>. This leaves us <imath>120^\circ</imath> (<imath>360-240</imath>) left to be formed by the central angle, which by the Central Angle Theorem must be <imath>120^\circ</imath> as well. Note that now we have a parallelogram with the side <imath>OC</imath> (which is <imath>2\sqrt{3}</imath>) being opposite and therefor equivalent to the radius <imath>XA</imath>. Now, the area of the circle is just <imath>\pi*(2*\sqrt{3})^2 = 12\pi</imath>. Select <imath>\boxed{B}</imath>.


~hastapasta, bob4108
~hastapasta, bob4108, TheHuskyKing
==Note For Solution 5==
 
One must be careful as one can easily accidentally get this problem CORRECT by prematurely saying the radius is <imath>2\sqrt{3}</imath> BEFORE proving that <imath>OC</imath> is a part of a parallelogram. Otherwise, your statement is simply saying that the Circumcircle's area is <imath>12\pi</imath>. Meaning that although your answer was correct the reasoning was not.
 
~TheHuskyKing
 
==Solution 6 (Ptolemy) ==
 
Call the diameter of the circle <math>d</math>. If we extend points <math>A</math> and <math>C</math> to meet at a point on the circle and call it <math>E</math>, then <math>\bigtriangleup OAE=\bigtriangleup OCE</math> . Note that both triangles are right, since their hypotenuse is the diameter of the circle. Therefore, <math>CE=AE=\sqrt{d^2-12}</math>. We know this since <math>OC=OA=OB</math> and <math>OC</math> is the hypotenuse of a <math>30-60-90</math> right triangle, with the longer leg being <math>\frac{6}{2}=3</math> so <math>OC=2\sqrt{3}</math>. Applying Ptolemy's Theorem on cyclic quadrilateral <math>OCEA</math>, we get <math>2({\sqrt{d^2-12}})\cdot{2\sqrt{3}}=6d</math>. Squaring and solving we get <math>d^2=48 \Longrightarrow (2r)^2=48</math> so <math>r^2=12</math>. Therefore, the area of the circle is <math>\boxed{12\pi}</math>
 
~[https://artofproblemsolving.com/wiki/index.php/User:Magnetoninja Magnetoninja]
 
 
==Video Solution (Just 3 min!)==
https://youtu.be/kkm1d09bVpM
 
<i>~Education, the Study of Everything</i>


==Video Solution by TheBeautyofMath==
==Video Solution by TheBeautyofMath==

Latest revision as of 18:47, 11 November 2025

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, call the center of the circle we are looking for $X$. From here we can now utilize the fact that since the inscribed angle $\angle{AOC}=120^\circ$ that the arclength that the angle describes is $240^\circ$. This leaves us $120^\circ$ ($360-240$) left to be formed by the central angle, which by the Central Angle Theorem must be $120^\circ$ as well. Note that now we have a parallelogram with the side $OC$ (which is $2\sqrt{3}$) being opposite and therefor equivalent to the radius $XA$. Now, the area of the circle is just $\pi*(2*\sqrt{3})^2 = 12\pi$. Select $\boxed{B}$.

~hastapasta, bob4108, TheHuskyKing

Note For Solution 5

One must be careful as one can easily accidentally get this problem CORRECT by prematurely saying the radius is $2\sqrt{3}$ BEFORE proving that $OC$ is a part of a parallelogram. Otherwise, your statement is simply saying that the Circumcircle's area is $12\pi$. Meaning that although your answer was correct the reasoning was not.

~TheHuskyKing

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 $\boxed{12\pi}$

~Magnetoninja


Video Solution (Just 3 min!)

https://youtu.be/kkm1d09bVpM

~Education, the Study of Everything

Video Solution by TheBeautyofMath

https://www.youtube.com/watch?v=4qgYrCYG-qw&t=1039

~IceMatrix

See Also

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

These problems are copyrighted © by the Mathematical Association of America.

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