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

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label("O", (4, 3.4));
 
label("O", (4, 3.4));
 
</asy>
 
</asy>
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Let the radius of the circle be <math>r</math>, and let its center be <math>O</math>. Since <math>\overline{AB}</math> and <math>\overline{AC}</math> are tangent to circle <math>O</math>, 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 <math>r</math>, 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 - 2 \cdot \frac 12 \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 circle 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>
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Let the radius of the circle be <math>r</math>, and let its center be <math>O</math>. Since <math>\overline{AB}</math> and <math>\overline{AC}</math> are tangent to circle <math>O</math>, 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 <math>r</math>, 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 \cdot \cdot \frac{{r}2 r\sqrt{3}\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 circle 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>
  
 
==Video Solution==
 
==Video Solution==

Revision as of 02:05, 24 July 2021

Problem

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

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

Solution

[asy] real sqrt3 = 1.73205080757; draw(Circle((4, 4), 4)); draw((4-2*sqrt3,6)--(4,4)--(4+2*sqrt3,6)--(4-2*sqrt3,6)--(4,12)--(4+2*sqrt3,6)); label("A", (4, 12.4)); label("B", (-.3, 6.3)); label("C", (8.3, 6.3)); label("O", (4, 3.4)); [/asy] 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 $\frac 13 \pi r^2 - \frac 12 \cdot \cdot \frac{{r}2 r\sqrt{3}\frac 12 r \cdot r \cdot \frac{\sqrt{3}}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4$ (Error compiling LaTeX. Unknown error_msg). The area outside of the circle 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 \[\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}\]

Video Solution

https://www.youtube.com/watch?v=GnJDNtjd57k&feature=youtu.be

https://youtu.be/ADDAOhNAsjQ -Video Solution by Richard Rusczyk

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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