Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.3 AIME, harder Find the answer

For example, let $F_{1}, F_{2}$ be the two foci of the hyperbola $\frac{x^{2}}{4}-\frac{y^{2}}{45}=1$ , and $P$ be a point on the hyperbola. It is known that $\left|P F_{2}\right|$ , $\left|P F_{1}\right|,\left|F_{1} F_{2}\right|$ form an arithmetic sequence (or $2\left|P F_{1}\right|=\left|P F_{2}\right|+\left|F_{1} F_{2}\right|$ ), and the common difference is greater than 0. Try to find $\angle F_{1} P F_{2}$ .

Your answer

A number or a short expression. Spacing and $ signs are ignored.

Solution

From the given, we know that $a^{2}=4, b^{2}=45$ , then $c=7$ .
Also, $2\left|P F_{1}\right|=\left|P F_{2}\right|+2 c$ , so $2\left|P F_{1}\right|-\left|P F_{2}\right|=14$ .
And $\left|P F_{1}\right|-\left|P F_{2}\right|=2 a=4$ , thus we find $\left|P F_{1}\right|=10,\left|P F_{2}\right|=6$ . Therefore, by property 3(II), we know $60=\left|P F_{1}\right| \cdot\left|P F_{2}\right|=\frac{b^{2}}{\sin ^{2} \theta}=\frac{2 b^{2}}{1-\cos 2 \theta}$ , which gives $\cos \theta=-\frac{1}{2}$ .
Thus, $\theta=120^{\circ}$ , i.e., $\angle F_{1} P F_{2}=120^{\circ}$ .

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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.