Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 6.1 National olympiad Prove it Belarus

Points $K$ and $L$ are marked on the side $AB$ of the triangle $ABC$ so that $\angle ACK = \angle KCL = \angle LCB$ . Point $M$ is marked on the side $BC$ so that $\angle MKC = \angle BKM$ .
Find the value of $\angle MLC$ , if $ML$ is a bisector of $\angle KMB$ .
(S. Mazanik)

Solution

Answer: $30^\circ$ .

Since $L$ lies on the bisector of $\angle KCB$ , $L$ is equidistant from the lines $CP$ and $CB$ . Similarly, since $L$ lies on the bisector of $\angle KMB$ , $L$ is equidistant from the rays $MK$ and $MB$ . Therefore, $L$ is an equidistant point for the rays $KM$ and $KP$ , so $L$ lies on the bisector of $\angle PKM$ . Thus, $\angle PKL = \angle BKM = \angle MKC$ . Therefore, each of these angles is equal to $60^\circ$ . Let $\angle ACB = 3x$ . Since $\angle LKC = 120^\circ$ , we have $\angle KAC = \angle LKC - \angle ACK = 120^\circ - x$ . Therefore
$\angle ABC = 180^\circ - \angle BAC - \angle BCA = 180^\circ - (120^\circ - x) - 3x = 60^\circ - 2x.$
So,
$\begin{aligned} \angle LMB &= 0.5\angle KMB = 0.5(180^\circ - \angle LKM - \angle ABC) = \\ &= 0.5(180^\circ - 60^\circ - (60^\circ - 2x)) = 30^\circ + x. \end{aligned}$
Since $\angle LMB = \angle LCM + \angle MLC$ , we obtain $\angle MLC = 30^\circ + x - x = 30^\circ$ .

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