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

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==Solution 2==
 
==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.
 
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.
 +
 +
==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}}

Revision as of 01:37, 26 September 2020

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

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.

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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