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Difference between revisions of "2017 AMC 10A Problems/Problem 22"

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<math> \mathrm{(A) \ }\dfrac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \mathrm{(B) \ } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \mathrm{(C) \ } \frac{1}{2} \qquad \mathrm{(D) \ }\sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \mathrm{(E) \ } \frac{4}{3}-\dfrac{4\sqrt{3}\pi}{27}</math>
 
<math> \mathrm{(A) \ }\dfrac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \mathrm{(B) \ } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \mathrm{(C) \ } \frac{1}{2} \qquad \mathrm{(D) \ }\sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \mathrm{(E) \ } \frac{4}{3}-\dfrac{4\sqrt{3}\pi}{27}</math>
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==Solution==
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Let the radius of the circle be r, and let its center be O. Since <math>\overline{AB}</math> and <math>\overline{AC}</math> are tangent to circle O, then <math>\angle OBA = \angle OCA = 90^{\circ}</math>, so <math>\angle BOC = 120^{\circ}</math>. Therefore, since <math>\overline{OB}</math> and <math>\overline{OC}</math> are equal to r, then (pick your favorite method) <math>\overline{BC} = r\sqrt{3}</math>. The area of the equilateral triangle is <math>\frac{(r\sqrt{3})^2 \sqrt{3}}4 = \frac{3r^2 \sqrt{3}}4</math>, and the area of the sector we are subtracting from it is <math>\frac 13 \pi r^2 - \frac 12 r \cdot r \cdot \frac{\sqrt{3}}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4</math>. The area outside of the triangle is <math> \frac{3r^2 \sqrt{3}}4-\left(\frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4\right) = r^2 \sqrt{3} - \frac{\pi r^2}3</math>. Therefore, the answer is <cmath>\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}</cmath>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=A|num-b=21|num-a=23}}
 
{{AMC10 box|year=2017|ab=A|num-b=21|num-a=23}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 18:56, 8 February 2017

Problem

Sides $\overline{AB}$ and $\overline{AC}$ of equilateral triangle $ABC$ are tangent to a circle as points $B$ and $C$ respectively. What fraction of the area of $\triangle ABC$ lies outside the circle?

$\mathrm{(A) \ }\dfrac{4\sqrt{3}\pi}{27}-\frac{1}{3}\qquad \mathrm{(B) \ } \frac{\sqrt{3}}{2}-\frac{\pi}{8}\qquad \mathrm{(C) \ } \frac{1}{2} \qquad \mathrm{(D) \ }\sqrt{3}-\frac{2\sqrt{3}\pi}{9}\qquad \mathrm{(E) \ } \frac{4}{3}-\dfrac{4\sqrt{3}\pi}{27}$

Solution

Let the radius of the circle be r, and let its center be O. Since $\overline{AB}$ and $\overline{AC}$ are tangent to circle O, then $\angle OBA = \angle OCA = 90^{\circ}$, so $\angle BOC = 120^{\circ}$. Therefore, since $\overline{OB}$ and $\overline{OC}$ are equal to r, then (pick your favorite method) $\overline{BC} = r\sqrt{3}$. The area of the equilateral triangle is $\frac{(r\sqrt{3})^2 \sqrt{3}}4 = \frac{3r^2 \sqrt{3}}4$, and the area of the sector we are subtracting from it is $\frac 13 \pi r^2 - \frac 12 r \cdot r \cdot \frac{\sqrt{3}}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4$. The area outside of the triangle is $\frac{3r^2 \sqrt{3}}4-\left(\frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4\right) = r^2 \sqrt{3} - \frac{\pi r^2}3$. Therefore, the answer is \[\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}\]

See Also

2017 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 21 		Followed by
Problem 23
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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