Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.0 AIME, harder Find the answer

8. Given $F_{1}$ and $F_{2}$ are the foci of an ellipse, $P$ is a point on the ellipse, $\angle F_{1} P F_{2}=90^{\circ}$ , and $\left|P F_{2}\right|<\left|P F_{1}\right|$ , the eccentricity of the ellipse is $\frac{\sqrt{6}}{3}$ . Then $\angle P F_{1} F_{2}: \angle P F_{2} F_{1}$ $=(\quad)$ .

Pick one

Solution

8. A.

Let $\left|P F_{2}\right|=x$ , then $\left|P F_{1}\right|=2 a-x$ .
Given $\angle F_{1} P F_{2}=90^{\circ}$ , we know $\sin \angle P F_{1} F_{2}=\frac{x}{2 c}, \cos \angle P F_{1} F_{2}=\frac{2 a-x}{2 c}$ . Therefore, $\left(\frac{x}{2 c}\right)^{2}+\left(\frac{2 a-x}{2 c}\right)^{2}=1$ . Given $e=\frac{\sqrt{6}}{3}$ , we know $c=\frac{\sqrt{6}}{3} a$ , solving for $x$ yields $x=\left(1 \pm \frac{\sqrt{3}}{3}\right) a$ . Since $\left|P F_{2}\right|<\left|P F_{1}\right|$ , we have $\left|P F_{2}\right|=\left(1-\frac{\sqrt{3}}{3}\right) a$ .
Thus, $\sin \angle P F_{1} F_{2}=\frac{\sqrt{6}-\sqrt{2}}{4}$ . Hence, $\angle P F_{1} F_{2}=15^{\circ}, \angle P F_{2} F_{1}=75^{\circ}$ .

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.