Olympiad Maths Prep

Track / Stage 6 / 149 of 400 #1149 of 2000

Problem 1149

National olympiad, first round
Geometry Difficulty 6.2 Prove it

11. Let the eccentricity of the ellipse Γ\Gamma be e,F1,F2e, F_{1}, F_{2} be its two foci, PP be any point on the ellipse (except the two vertices on the major axis), r,Rr, R be the inradius and circumradius of △PF1F2\triangle P F_{1} F_{2}, respectively. Prove: rR⩽2e(1−e)\frac{r}{R} \leqslant 2 e(1-e).

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

11. Let ∠F1PF2=α,∠PF1F2=β,∠F1F2P=γ\angle F_{1} P F_{2}=\alpha, \angle P F_{1} F_{2}=\beta, \angle F_{1} F_{2} P=\gamma, then rR=4sin⁡α2⋅sin⁡β2⋅sin⁡γ2\frac{r}{R}=4 \sin \frac{\alpha}{2} \cdot \sin \frac{\beta}{2} \cdot \sin \frac{\gamma}{2}. Also, cos⁡β+γ2=\cos \frac{\beta+\gamma}{2}= sin⁡α2\sin \frac{\alpha}{2}, then rR=2sin⁡α2(cos⁡β−γ2−sin⁡α2)\frac{r}{R}=2 \sin \frac{\alpha}{2}\left(\cos \frac{\beta-\gamma}{2}-\sin \frac{\alpha}{2}\right).

Also, e=∣F1F2∣∣PF1∣+∣PF2∣=sin⁡αsin⁡β+sin⁡γ=sin⁡α2cos⁡β−γ2e=\frac{\left|F_{1} F_{2}\right|}{\left|P F_{1}\right|+\left|P F_{2}\right|}=\frac{\sin \alpha}{\sin \beta+\sin \gamma}=\frac{\sin \frac{\alpha}{2}}{\cos \frac{\beta-\gamma}{2}}, i.e., sin⁡α2=e⋅cos⁡β−γ2\sin \frac{\alpha}{2}=e \cdot \cos \frac{\beta-\gamma}{2}, and substituting it into (*) gives rR=2e(1−e)⋅cos⁡2β−γ2⩽2e(1−e2)\frac{r}{R}=2 e(1-e) \cdot \cos ^{2} \frac{\beta-\gamma}{2} \leqslant 2 e\left(1-e^{2}\right), where equality holds if and only if β=γ\beta=\gamma.

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