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

Revision as of 18:43, 14 October 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

@@ Line 56: / Line 56: @@
 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 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>.
+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 Pythagoras, <math>3^2 + (r - \sqrt{3})^2 = r^2</math>. So <math>r = 2\sqrt{3}</math>.
 Thus the area is <math> \pi r^2 = \boxed{\textbf{(B)}\ 12\pi}</math>.

2021 Fall AMC 12B (Problems • Answer Key • Resources)
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