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

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== Problem ==
  
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A circle with center <math>O</math> has area <math>156\pi</math>. Triangle <math>ABC</math> is equilateral, <math>\overline{BC}</math> is a chord on the circle, <math>OA = 4\sqrt{3}</math>, and point <math>O</math> is outside <math>\triangle ABC</math>. What is the side length of <math>\triangle ABC</math>?
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<math>\textbf{(A)}\ 2\sqrt{3} \qquad \textbf{(B)}\ 6 \qquad \textbf{(C)}\ 4\sqrt{3} \qquad \textbf{(D)}\ 12 \qquad \textbf{(E)}\ 18</math>
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==Solution 1==
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The formula for the area of a circle is <math>\pi r^2</math> so the radius of this circle is <math>\sqrt{156}.</math>
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Because <math>OA=4\sqrt{3} < \sqrt{156}, A</math> must be in the interior of circle <math>O.</math>
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<center><asy>
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unitsize(3mm);
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defaultpen(linewidth(.8pt)+fontsize(11pt));
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dotfactor=3;
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real r=sqrt(156);
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pair A=(0,sqrt(48)), B=(-3,sqrt(147)), C=(3,sqrt(147));
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pair O=(0,0);
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pair X=(0,7sqrt(3));
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path outer=Circle(O,r);
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draw(outer);
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draw(A--B--C--cycle);
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draw(O--X); draw(O--B);
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pair[] ps={A,B,C,O,X};
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dot(ps);
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label("$A$",A,SE);
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label("$B$",B,NW);
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label("$C$",C,NE);
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label("$O$",O,S);
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label("$X$",X,N);
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label("$s$",A--C,SE);
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label("$\frac{s}{2}$",B--X,N);
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label("$\frac{s\sqrt{3}}{2}$",A--X,NE);
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label("$\sqrt{156}$",O--B,SW);
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label("$4\sqrt{3}$",A--O,E);
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</asy></center>
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Let <math>s</math> be the unknown value, the sidelength of the triangle, and let <math>X</math> be the point on <math>BC</math> where <math>OX \perp BC.</math> Since <math>\triangle ABC</math> is equilateral, <math>BX=\frac{s}{2}</math> and <math>AX=\frac{s\sqrt{3}}{2}.</math> We are given <math>AO=4\sqrt{3}.</math> Use the [[Pythagorean Theorem]] and solve for <math>s.</math>
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<cmath>\begin{align*}
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(\sqrt{156})^2 &= \left(\frac{s}{2}\right)^2 + \left( \frac{s\sqrt{3}}{2} + 4\sqrt{3} \right)^2\\
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156 &= \frac14s^2 + \frac34s^2 + 12s + 48\\
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0 &= s^2 + 12s - 108\\
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0 &= (s-6)(s+18)\\
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s &= \boxed{\textbf{(B)}\ 6}
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\end{align*} </cmath>
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==Solution 2==
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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>
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~hansenhe
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==Video Solution==
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https://youtu.be/FQO-0E2zUVI?t=906
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 +
~IceMatrix
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==See Also==
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{{AMC10 box|year=2010|ab=B|num-b=18|num-a=20}}
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{{MAA Notice}}

Revision as of 23:51, 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

See Also

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

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