Competition problem — Geometry — Maths Olympiad Prep

Library / /438 of 520

Geometry Difficulty 5.8 AIME, harder Prove it

Example 2 As shown in Figure $6, \triangle A B C$ has internal angle bisectors $B E$ and $C F$ intersecting at point $I, I Q \perp E F$ intersects $B C$ at point $P$ , and $I P=2 I Q$ . Prove: $\angle B A C=60^{\circ}$ .

Solution

Prove as shown in Figure 6, construct $AX \perp EF$ intersecting $BC$ at point $Y$ .
By Property 4, $A, D', I, D$ form a harmonic range.
Thus, $\frac{IQ}{AX}=\frac{D'I}{D'A}=\frac{DI}{DA}=\frac{PI}{YA}$ .
Since $IP = 2IQ$ , then $AX = XY$ , meaning $EF$ is the median of $AY$ .
By the Law of Sines,
$\begin{array}{l} \frac{CF}{\sin \angle FYC}=\frac{FY}{\sin \angle 1} \\ =\frac{FA}{\sin \angle 2}=\frac{CF}{\sin \angle FAC}. \end{array}$

Thus, $A, F, Y, C$ are concyclic.
Similarly, $A, E, Y, B$ are concyclic.
Therefore, $\angle BYF = \angle BAC = \angle CYE = \angle EYF$ .
Hence, $\angle BAC = 60^{\circ}$ .

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.