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

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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>
 
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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Because <math>OA=4\sqrt{3}=\sqrt{48} < \sqrt{156}, A</math> must be in the interior of circle <math>O.</math>
  
 
<center><asy>
 
<center><asy>
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</asy></center>
 
</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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Let <math>s</math> be the side length of the triangle, the unknown value, 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>
  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
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==Solution 2==
 
==Solution 2==
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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.
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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^\circ-60^\circ}{2} = 150^\circ</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> Lengths must be non-negative, so the answer is <math>\boxed{\textbf{(B)}\ 6}</math>
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~bryan gao
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 +
==Video Solution==
 +
https://youtu.be/FQO-0E2zUVI?t=906
 +
 
 +
~IceMatrix
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2010|ab=B|num-b=18|num-a=20}}
 
{{AMC10 box|year=2010|ab=B|num-b=18|num-a=20}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:34, 4 May 2024

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 $OA=4\sqrt{3}=\sqrt{48} < \sqrt{156}, A$ 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 side length of the triangle, the unknown value, 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 $\angle BAO = \angle CAO \implies \angle BAO = \frac{360^\circ-60^\circ}{2} = 150^\circ$. 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 \[(x-6)(x+18)\] Lengths must be non-negative, so the answer is $\boxed{\textbf{(B)}\ 6}$ ~bryan gao

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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