Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Silk Road Mathematics Competition

Let $s = (AB + BC + AC)/2$ be the semiperimeter of triangle $ABC$ . We choose two points $L$ and $N$ lying on the rays $AB$ and $CB$ satisfying $AL = CN = s$ . Let $K$ be the point symmetric to $B$ with respect to the center of the circumcircle of triangle $ABC$ . Prove that the perpendicular drawn from the point $K$ to the line $NL$ passes through the incenter of the triangle $ABC$ .

Solution

Let $BC = a$ , $AC = b$ , $AB = c$ and $I$ be the incenter of the triangle $ABC$ .

From the point $I$ let us draw parallel lines to $AB$ and $BC$ , intersecting $AK$ and $KC$ at $P$ and $Q$ , respectively. In the triangle $IPQ$ we have $\angle PIQ = \angle ABC$ , $IQ = s - c$ , $IP = s - a$ . Since $\angle NBL = \angle ABC$ , $NB = s - a$ , $LB = s - c$ , it follows that $\angle PIQ = \angle NBL$ .

On the ray $IP$ let's choose a point $L_1$ satisfying $IL_1 = IQ$ , and on the ray $IQ$ choose a point $N_1$ , such that $IN_1 = IP$ . Then $\Delta IN_1L_1 = \Delta NBL$ , $BL||IL_1$ , $BN||IN_1$ and it follows that $L_1N_1||LN$ .

$IPKQ$ is cyclic. Therefore, $\angle IKP = \angle IQP$ ,
$\angle KIP + \angle IL_1N_1 = \angle KIP + \angle IQP = \angle KIP + \angle IKP = 90^\circ$
and it follows that $IK \perp L_1N_1$ and $IK \perp LN$ .

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