Track / Stage 6 / 125 of 400 #1125 of 2000

Problem 1125

National olympiad, first round

Geometry Difficulty 6.1 Prove it

\left.\begin{array}{l}{\left[\begin{array}{l}\text { Auxiliary circle } \\ \text { [angles subtending equal arcs and equal chords] }\end{array}\right]}\end{array}\right]

The center $I$ of the inscribed circle of an acute triangle $ABC$ lies on the bisector of the acute angle between the altitudes $AA_1$ and $CC_1$ . Prove that $IA_1 = IC_1 = IL$ , where $L$ is the foot of the bisector of angle $B$ of triangle $ABC$ .

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

Let $H$ be the orthocenter of triangle $ABC$ . Suppose point $I$ lies inside angle $A_1HC$ . Denote $\angle ABC = \alpha$ . Then,

$\angle A_1HC = \alpha, \angle A_1HI = \frac{\alpha}{2} = \angle A_1BI$

This means that segment $A_1I$ is seen from points $H$ and $B$ at the same angle. Therefore, points $B, H, I, A_1$ lie on the same circle, and $BH$ is the diameter of this circle because $\angle BA_1H = 90^\circ$ . Since segment $BH$ is seen from point $C_1$ at a right angle, point $C_1$ also lies on this circle. Since $BI$ is the bisector of the inscribed angle $A_1BC_1$ , point $I$ is the midpoint of the arc $A_1IC_1$ , so $IA_1 = IC_1$ . Inscribed angles $\angle BIA_1$ and $\angle BHA_1$ subtend the same arc, so

$\angle BIA_1 = \angle BHA_1 = \angle ACB,$

which means quadrilateral $ILCA_1$ is cyclic. Since $CI$ is the bisector of angle $LCA_1$ , $IL = IA_1$ . This completes the proof.

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