Difference between revisions of "2010 AMC 10B Problems/Problem 19"

Revision as of 22:50, 8 March 2021

Problem

A circle with center $O$ has area $156\pi$ . Triangle $ABC$ is equilateral, $\overline{BC}$ is a chord on the circle, $OA = 4\sqrt{3}$ , and point $O$ is outside $\triangle ABC$ . What is the side length of $\triangle ABC$ ?

$\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18$

Solution 1

The formula for the area of a circle is $\pi r^2$ so the radius of this circle is $\sqrt{156}.$

Because $OA=4\sqrt{3} < \sqrt{156}, A$ must be in the interior of circle $O.$

$[asy] unitsize(3mm); defaultpen(linewidth(.8pt)+fontsize(11pt)); dotfactor=3; real r=sqrt(156); pair A=(0,sqrt(48)), B=(-3,sqrt(147)), C=(3,sqrt(147)); pair O=(0,0); pair X=(0,7sqrt(3)); path outer=Circle(O,r); draw(outer); draw(A--B--C--cycle); draw(O--X); draw(O--B); pair[] ps={A,B,C,O,X}; dot(ps); label("$A$",A,SE); label("$B$",B,NW); label("$C$",C,NE); label("$O$",O,S); label("$X$",X,N); label("$s$",A--C,SE); label("$\frac{s}{2}$",B--X,N); label("$\frac{s\sqrt{3}}{2}$",A--X,NE); label("$\sqrt{156}$",O--B,SW); label("$4\sqrt{3}$",A--O,E); [/asy]$

Let $s$ be the unknown value, the sidelength of the triangle, and let $X$ be the point on $BC$ where $OX \perp BC.$ Since $\triangle ABC$ is equilateral, $BX=\frac{s}{2}$ and $AX=\frac{s\sqrt{3}}{2}.$ We are given $AO=4\sqrt{3}.$ Use the Pythagorean Theorem and solve for $s.$

$\begin{align*} (\sqrt{156})^2 &= \left(\frac{s}{2}\right)^2 + \left( \frac{s\sqrt{3}}{2} + 4\sqrt{3} \right)^2\\ 156 &= \frac14s^2 + \frac34s^2 + 12s + 48\\ 0 &= s^2 + 12s - 108\\ 0 &= (s-6)(s+18)\\ s &= \boxed{\textbf{(B)}\ 6} \end{align*}$

Solution 2

We can use the same diagram as Solution 1 and label the side length of $\triangle ABC$ as $s$ . Using congruent triangles, namely the two triangles $\triangle BOA$ and $\triangle COA$ , we get that $\angle BAO = \angle CAO \implies \angle BAO = \frac{360-60}{2} = 150$ . From this, we can use the Law of Cosines, to get $s^2 + (4 \sqrt{3})^2 - 2 \times s \times 4 \sqrt{3} \times - \frac{\sqrt{3}}{2} = (2 \sqrt{39})^2$ Simplifying, we get $s^2 + 12s + 48 = 156 \implies s^2 + 12s - 108 = 0$ We can factor this to get $(x-6)(x+18)$ Obviously, we want the positive solution to get $\boxed{\textbf{(B)}\ 6}$ ~hansenhe

Video Solution

https://youtu.be/FQO-0E2zUVI?t=906

~IceMatrix

@@ Line 50: / Line 50: @@
 ==Solution 2==
-Using the diagram in solution 1, we can instead do the law of cosines. We know that angle OAB is 150 degrees, and the measurements of each side (excluding side A), so we just plug the values in to the law of cosines. Doing so gives us 6, which is answer choice B.
+We can use the same diagram as Solution 1 and label the side length of <math>\triangle ABC</math> as <math>s</math>. Using congruent triangles, namely the two triangles <math>\triangle BOA</math> and <math>\triangle COA</math>, we get that <math>\angle BAO = \angle CAO \implies \angle BAO = \frac{360-60}{2} = 150</math>. From this, we can use the Law of Cosines, to get <cmath>s^2 + (4 \sqrt{3})^2 - 2 \times s \times 4 \sqrt{3} \times - \frac{\sqrt{3}}{2} = (2 \sqrt{39})^2</cmath> Simplifying, we get <cmath>s^2 + 12s + 48 = 156 \implies s^2 + 12s - 108 = 0</cmath> We can factor this to get <cmath>(x-6)(x+18)</cmath> Obviously, we want the positive solution to get <math>\boxed{\textbf{(B)}\ 6}</math>
+~hansenhe
 ==Video Solution==

2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by Problem 18	Followed by Problem 20
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