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

(Solution 3)
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==Solution 3 ==
 
==Solution 3 ==
  
As before, construct the circle that passes through <math>A</math>, <math>O</math>, and <math>C</math>, centered at <math>X</math>. Let <math>Y</math> be the intersection of <math>\overline{OX}</math> and <math>\overline{AB}</math>.
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As in the previous solution, construct the circle that passes through <math>A</math>, <math>O</math>, and <math>C</math>, centered at <math>X</math>. Let <math>Y</math> be the intersection of <math>\overline{OX}</math> and <math>\overline{AB}</math>.
  
 
Note that since <math>\overline{OA}</math> is the angle bisector of <math>\angle BAC</math> that <math>\angle OAC=30^\circ</math>. Also by symmetry, <math>\overline{OX}</math> <math>\perp</math> <math>\overline{AB}</math> and <math>AY = 3</math>. Thus <math>\tan(30^\circ) = \frac{OY}{3}</math> so <math>OY = \sqrt{3}</math>.
 
Note that since <math>\overline{OA}</math> is the angle bisector of <math>\angle BAC</math> that <math>\angle OAC=30^\circ</math>. Also by symmetry, <math>\overline{OX}</math> <math>\perp</math> <math>\overline{AB}</math> and <math>AY = 3</math>. Thus <math>\tan(30^\circ) = \frac{OY}{3}</math> so <math>OY = \sqrt{3}</math>.
  
Let <math>r</math> be the radius of circle <math>X</math>, and note that <math>AX = OX = r</math>. So <math>\triangle AYX</math> is a right triangle with legs  
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Let <math>r</math> be the radius of circle <math>X</math>, and note that <math>AX = OX = r</math>. So <math>\triangle AYX</math> is a right triangle with legs of length <math>3</math> and <math>r - \sqrt{3}</math> and hypotenuse <math>r</math>. By Pythagorus, <math>3^2 + (r - \sqrt{3})^2 = r^2</math>. So <math>r = 2\sqrt{3}</math>.
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Thus the area is <math> \pi r^2 = \boxed{\textbf{(B)}\ 12\pi}</math>.
  
  

Revision as of 19:41, 9 January 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 Pythagorus, $3^2 + (r - \sqrt{3})^2 = r^2$. So $r = 2\sqrt{3}$.

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


-SharpeMind

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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