2017 AMC 10 A

Complete problem set with solutions and individual problem pages

Problem 22 Hard

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? (2017 AMC 10A Problem, Question#22)

  • A.

    \dfrac{4 \sqrt{3} \pi }{27}- \dfrac{1}{3}

  • B.

    \dfrac{ \sqrt{3}}{2}- \dfrac{ \pi }{8}

  • C.

    \dfrac{1}{2}

  • D.

    \sqrt{3}- \dfrac{2 \sqrt{3} \pi }{9}

  • E.

    \dfrac{4}{3}- \dfrac{4 \sqrt{3} \pi }{27}

Answer:E

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 \dfrac{(r \sqrt{3})^{2} \sqrt{3}}{4}= \dfrac{3r^{2} \sqrt{3}}{4} and the area of the sector we are subtracting from it is \dfrac{1}{3} \pi r^{2}- \dfrac{1}{2}r \cdot r \cdot \dfrac{ \sqrt{3}}{2}= \dfrac{ \pi r^{2}}{3}- \dfrac{r^{2} \sqrt{3}}{4}. The area outside of the circle \dfrac{3r^{2} \sqrt{3}}{4}-( \dfrac{ \pi r^{2}}{3}- \dfrac{r^{2} \sqrt{3}}{4})=r^{2} \sqrt{3}- \dfrac{ \pi r^{2}}{3}. Therefore, the answer is

\dfrac{r^{2}\sqrt{3}-\dfrac{\pi r^{2}}{3}}{\dfrac{3r^{2}\sqrt{3}}{4}}=(\rm E) \dfrac{4}{3}- \dfrac{4 \sqrt{3} \pi }{27}.

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