Track / Stage 5 / 144 of 400 #744 of 1964

Problem 744

AIME late

Geometry Difficulty 5.4 Find the answer

6. $P$ is a point on the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ , $F$ is one of the foci, and $\triangle P O F$ is an isosceles triangle ( $O$ is the origin). If there are exactly 8 points $P$ that satisfy the condition, then the range of the eccentricity of the ellipse is $\qquad$
$\qquad$

Your answer

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

6. $\left(\frac{\sqrt{2}}{2}, \sqrt{3}-1\right) \cup(\sqrt{3}-1,1) \quad$ Detailed Explanation: (1) $P F=P O$ , point $P$ has 2 solutions; (2) $O P=O F$ , when $e \in$ $\left(0, \frac{\sqrt{2}}{2}\right)$ there are no points $P$ that satisfy the condition, when $e=\frac{\sqrt{2}}{2}$ there are 2 points $P$ , when $e \in\left(\frac{\sqrt{2}}{2}, 1\right)$ there are 4 points $P$ ; (3) $F P=F O$ , when $e \in\left(0, \frac{1}{2}\right]$ there are no points $P$ that satisfy the condition, when $e \in\left(\frac{1}{2}, 1\right)$ there are 2 points $P$ . Additionally, when $\triangle P O F$ is an equilateral triangle, there are 4 points $P$ , at this time $e=\sqrt{3}-1$ . Therefore, the range of the eccentricity of the ellipse is $\left(\frac{\sqrt{2}}{2}, \sqrt{3}-1\right) \cup(\sqrt{3}-1,1)$ .

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