2017 AMC 10A Problems/Problem 22

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 1

[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 $\frac 13 \pi r^2 - \frac 12 \cdot r\sqrt{3} \cdot \frac{r}2 = \frac{\pi r^2}3 -\frac{r^2 \sqrt{3}}4$. 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 \[\frac{r^2 \sqrt{3} - \frac{\pi r^2}3}{\frac{3r^2 \sqrt{3}}4} = \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}\]

Note

The sector angle is $120$ because $\angle B$ and $\angle C$ are both 90 degrees meaning $\angle B + \angle C = 180^\circ$, so $ABCO$ is cyclic. Thus, the angle is $180-60=120^\circ$

~mathboy282

Solution 2

[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)); draw((4, 6)--(4, 4)); label("A", (4, 12.4)); label("B", (-.3, 6.3)); label("C", (8.3, 6.3)); label("O", (4, 3.4)); label("D", (4, 6.6)); [/asy] (same diagram as Solution 1)

Without the Loss of Generality, let the side length of the triangle be $1$.

Then, the area of the triangle is $\frac{\sqrt{3}}{4}$. We are looking for the area of the portion inside the triangle but outside the circle divided by the area of the triangle. Since $m\angle ABO = 90^{\circ}=m\angle ACO = 90^{\circ}$, and $m\angle ABC = m\angle ACB = 60^{\circ}$, we know $m\angle OBC=m\angle OCB=30^{\circ}$, and $m\angle BOC = 120^{\circ}$. Drop an angle bisector of $O$ onto $BC$, call the point of intersection $D$. By SAS congruence, $\triangle BDO \cong \triangle CDO$, by CPCTC (Congruent Parts of Congruent Triangles are Congruent) $BD \cong DC$ and they both measure $\frac{1}{2}$. By 30-60-90 triangle, $OC = BO = \frac{\sqrt{3}}{3}$. The area of the sector bounded by arc BC is one-third the area of circle O, whose area is $\left(\frac{\sqrt{3}}{3}\right)^{2}\pi=\frac{1}{3}\pi$. Therefore, the area of the sector bounded by arc BC is $\frac{1}{3}\cdot\frac{1}{3}\pi=\frac{\pi}{9}$.

We are nearly there. By 30-60-90 triangle, we know $DO = \frac{\sqrt{3}}{6}$, so the area of $\triangle BOC$ is $\frac{1\cdot\frac{\sqrt{3}}{6}}{2}=\frac{\sqrt{3}}{12}$. The area of the region inside both the triangle and circle is the area of the sector bounded by arc BC minus the area of $\triangle BOC$: $\frac{\pi}{9}-\frac{\sqrt{3}}{12}$. The area of the region outside of the circle but inside the triangle is $\frac{\sqrt{3}}{4}-\left(\frac{\pi}{9}-\frac{\sqrt{3}}{12}\right)=\frac{\sqrt{3}}{4}-\frac{\pi}{9}+\frac{\sqrt{3}}{12}=\frac{\sqrt{3}}{3}-\frac{\pi}{9}$ and the ratio is $\frac{\frac{\sqrt{3}}{3}-\frac{\pi}{9}}{\frac{\sqrt{3}}{4}}=\left(\frac{\sqrt{3}}{3}-\frac{\pi}{9}\right)\cdot\frac{4}{\sqrt{3}}=\frac{4\sqrt{3}}{3\sqrt{3}}-\frac{4\pi}{9\sqrt{3}}=\frac{4}{3}-\frac{4\sqrt{3}\pi}{27} \Longrightarrow \boxed{\textbf{(E) } \frac 43 - \frac{4\sqrt 3 \pi}{27}}$.

~JH. L

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 (ProblemsAnswer KeyResources)
Preceded by
Problem 21
Followed by
Problem 23
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All AMC 10 Problems and Solutions

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