Competition problem — Geometry — Maths Olympiad Prep

Library / /350 of 520

Geometry Difficulty 6.8 National olympiad Prove it

5. (SPA 4) In the triangle $A B C$ , with $\measuredangle A=60^{\circ}$ , a parallel $I F$ to $A C$ is drawn through the incenter $I$ of the triangle, where $F$ lies on the side $A B$ . The point $P$ on the side $B C$ is such that $3 B P=B C$ . Show that $\measuredangle B F P=\measuredangle B / 2$ .

Solution

5. Let $P_{1}$ be the point on the side $B C$ such that $\angle B F P_{1}=\beta / 2$ . Then $\angle B P_{1} F=180^{\circ}-3 \beta / 2$ , and the sine law gives us $\frac{B F}{B P_{1}}=\frac{\sin (3 \beta / 2)}{\sin (\beta / 2)}=$ $3-4 \sin ^{2}(\beta / 2)=1+2 \cos \beta$ . Now we calculate $\frac{B F}{B P}$ . We have $\angle B I F=120^{\circ}-\beta / 2, \angle B F I=60^{\circ}$ and $\angle B I C=120^{\circ}, \angle B C I=\gamma / 2=60^{\circ}-\beta / 2$ . By the sine law,
$B F=B I \frac{\sin \left(120^{\circ}-\beta / 2\right)}{\sin 60^{\circ}}, \quad B P=\frac{1}{3} B C=B I \frac{\sin 120^{\circ}}{3 \sin \left(60^{\circ}-\beta / 2\right)} .$
It follows that $\frac{B F}{B P}=\frac{3 \sin \left(60^{\circ}-\beta / 2\right) \sin \left(60^{\circ}+\beta / 2\right)}{\sin ^{2} 60^{\circ}}=4 \sin \left(60^{\circ}-\beta / 2\right) \sin \left(60^{\circ}+\right.$ $\beta / 2)=2\left(\cos \beta-\cos 120^{\circ}\right)=2 \cos \beta+1=\frac{B F}{B P_{1}}$ . Therefore $P \equiv P_{1}$ .

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

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