In an acute triangle $ABC$ with $AC<BC$, lines $m_a$ and $m_b$ are the perpendicular bisectors of sides $BC$ and $AC$, respectively. Further, let $M_c$ be the midpoint of side $AB$. The Median $CM_c$ intersects $m_a$ in point $S_a$ and $m_b$ in point $S_b$; the lines $AS_b$ und $BS_a$ intersect in point $K$. Prove: $\angle ACM_c = \angle KCB$.
Problem
Source: German Bundeswettbewerb 2022, Round 2, Problem 3
Tags: geometry, perpendicular bisector
Zena_B
08.01.2023 19:10
By Menelaus we have: $\frac {KS_a}{ BS_a} \cdot \frac {BM_c}{ AM_c} \cdot \frac {AS_b}{KS_b}=1$ $\implies$ $\frac {KS_a}{KS_b} = \frac {BS_a}{AS_b} = \frac{CS_a}{CS_b} $ $\frac{CS_a}{CS_b}=\frac{S(KCS_a)}{S(KCS_b)}=\frac{KC\cdot sin(CKS_a)\cdot KS_a}{KC\cdot sin(CKS_b)\cdot KS_b}=\frac {KS_a}{KS_b}$ $\implies$ $ sin(CKS_a)= sin(CKS_b)$ Hence $\angle CKS_a + \angle CKS_b = 180$ $\angle S_aKS_b = 180 - (\angle KS_aS_b + \angle KS_bS_a) = 180 - ( 2 \angle ACS_b +2\angle BCS_b) =180 - 2\angle ACB$ $180=\angle CKS_a + \angle CKS_b = \angle S_aKS_b + 2\angle CKS_a = 180 - 2\angle ACB + 2\angle CKS_a $ $\implies \angle ACB = \angle CKS_a = \angle KCB +\angle KBC= \angle KCB+\angle S_aCB = \angle KCB +\angle ACB - \angle ACM $ Thus, we have $ \angle KCB = \angle ACM$
PROF65
08.01.2023 23:45
Simple angle chase leads to $ \angle KS_aO= 90^{\circ} +\angle BCM_c, \angle S_aOS_b=180^{\circ}_\angle A, \angle OS_bK=90^{\circ}+\angle M_c CA$ then $ \angle S_AKB=2\angle C $ thus $ K\in (OBA) $ let $CB\cap (AOB)=\{BD\}$ hence $\angle S_bOK =\angle DAK=\angle DAS_b=\angle A _ \angle S_bAC=\angle A - \angle ACS_b=\angle M_cCB$ consider the point $K'$ the intersection of $SK$ and the $C-$ symmedian , $I$ the midpoint of $AC$ then $\angle K'OI =\angle KOS_b= \angle M_cCB =\angle K'CA $ therfore $K'$ is on the circle of diameter $AO$ and the symmedian so it s also on $(ABO)$ which means $K=K' $ since a line intersects a circle at most two points.