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

Latest revision as of 21:33, 10 July 2023

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

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

@@ Line 1: / Line 1: @@
+==Problem==
+Triangle <math>ABC</math> is equilateral with side length <math>6</math>. Suppose that <math>O</math> is the center of the inscribed
+circle of this triangle. What is the area of the circle passing through <math>A</math>, <math>O</math>, and <math>C</math>?
+<math>\textbf{(A)} \: 9\pi \qquad\textbf{(B)} \: 12\pi \qquad\textbf{(C)} \: 18\pi \qquad\textbf{(D)} \: 24\pi \qquad\textbf{(E)} \: 27\pi</math>
 ==Solution 1 (Cosine Rule) ==
 Construct the circle that passes through <math>A</math>, <math>O</math>, and <math>C</math>, centered at <math>X</math>.
-Then connect <math>OX</math>, and notice that <math>OX</math> is the perpendicular bisector of <math>AC</math>. Let the intersection of <math>OX</math> with <math>AC</math> be <math>D</math>.
+Also notice that <math>\overline{OA}</math> and <math>\overline{OC}</math> are the angle bisectors of angle <math>\angle BAC</math> and <math>\angle BCA</math> respectively. We then deduce <math>\angle AOC=120^\circ</math>.
-Also notice that <math>AO</math> and <math>CO</math> are the angle bisectors of angle <math>BAC</math> and <math>BCA</math> respectively. We then deduce <math>AOC=120^\circ</math>.
 Consider another point <math>M</math> on Circle <math>X</math> opposite to point <math>O</math>.
-As <math>AOCM</math> an inscribed quadrilateral of Circle <math>X</math>, <math>AMC=180^\circ-120^\circ=60^\circ</math>.
+As <math>AOCM</math> is an inscribed quadrilateral of Circle <math>X</math>, <math>\angle AMC=180^\circ-120^\circ=60^\circ</math>.
-Afterward, deduce that <math>AXC=2·AMC=120^\circ</math>.
+Afterward, deduce that <math>\angle AXC=2·\angle AMC=120^\circ</math>.
 By the Cosine Rule, we have the equation: (where <math>r</math> is the radius of circle <math>X</math>)
@@ Line 19: / Line 24: @@
 <math>r^2=12</math>
-The area is therefore <math>12\pi \Rightarrow \boxed{\textbf{(B)}}</math>.
+The area is therefore <math> \pi r^2 = \boxed{\textbf{(B)}\ 12\pi}</math>.
 ~Wilhelm Z
+== Solution 2 ==
+We have <math>\angle AOC = 120^\circ</math>.
+Denote by <math>R</math> the circumradius of <math>\triangle AOC</math>.
+In <math>\triangle AOC</math>, the law of sines implies
+<cmath>
+\[
+R = \frac{AC}{\sin \angle AOC} = 4 \sqrt{3} .
+\]
+</cmath>
+Hence, the area of the circumcircle of <math>\triangle AOC</math> is
+<cmath>
+\[
+\pi R^2 = 12 \pi .
+\]
+</cmath>
+Therefore, the answer is <math>\boxed{\textbf{(B) }12 \pi}</math>.
+~Steven Chen (www.professorchenedu.com)
+==Solution 3 ==
+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>.
+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>.
+-SharpeMind
+==Solution 5 (SIMPLE) ==
+The semiperimeter is <math>\frac{6+6+6}{2}=9</math> 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>.
+~hastapasta, bob4108
+==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==
+https://www.youtube.com/watch?v=4qgYrCYG-qw&t=1039
+~IceMatrix
+==See Also==
+{{AMC12 box|year=2021 Fall|ab=B|num-a=10|num-b=8}}
+[[Category:Introductory Geometry Problems]]
+{{MAA Notice}}

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