HMMT November

Geometry Difficulty 4.8 AIME Find the answer

Points $D, E, F$ lie on circle $O$ such that the line tangent to $O$ at $D$ intersects ray $\overrightarrow{E F}$ at $P$ . Given that $P D=4, P F=2$ , and $\angle F P D=60^{\circ}$ , determine the area of circle $O$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

By the power of a point on $P$ , we get that $16=P D^{2}=(P F)(P E)=2(P E) \Rightarrow P E=8$ . However, since $P E=2 P D$ and $\angle F P D=60^{\circ}$ , we notice that $P D E$ is a $30-60-90$ triangle, so $D E=4 \sqrt{3}$ and we have $E D \perp D P$ . It follows that $D E$ is a diameter of the circle, since tangents the tangent at $D$ must be perpendicular to the radius containing $D$ . Hence, the area of the circle is $\left(\frac{1}{2} D E\right)^{2} \pi=12 \pi$

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