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Problems of Ukrainian Authors

Geometry Difficulty 5.2 AIME, harder Prove it Ukraine

On the sides $AB$ , $BC$ , $CA$ of triangle $ABC$ such points $C_1$ , $A_1$ , $B_1$ are chosen that the lines $AA_1$ , $BB_1$ and $CC_1$ meet at point $K$ . The perpendiculars are dropped from the point $K$ onto the sides of triangle. Lines $l_1$ , $l_2$ , $l_3$ are drawn through the feet of these perpendiculars, parallel to lines, symmetrical to $AA_1$ , $BB_1$ and $CC_1$ with respect to angular bisectors of $∠A$ , $∠B$ and $∠C$ respectively. Prove that the lines $l_1$ , $l_2$ , $l_3$ meet at a point.

Fig.28

Solution

The points $A$ , $M$ , $K$ , $N$ are cyclic, therefore

$\begin{aligned} \angle KAN &= \angle NMK = \\ &= 90^\circ - \angle AMN = \angle MAP, \text{ and so } AP \perp MN. \text{ Due to } l_1 \parallel AP = u_1 \text{ we have } l_1 \perp MN, \\ \text{therefore the lines } l_1, l_2, l_3 \text{ are altitudes of } \triangle MNK \text{ (Fig.28).} \end{aligned}$

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.