Let $ABC$ be an acute scalene triangle and let $\omega$ be its Euler circle. The tangent $t_A$ of $\omega$ at the foot of the height $A$ of the triangle ABC, intersects the circle of diameter $AB$ at the point $K_A$ for the second time. The line determined by the feet of the heights $A$ and $C$ of the triangle $ABC$ intersects the lines $AK_A$ and $BK_A$ at the points $L_A$ and $M_A$, respectively, and the lines $t_A$ and $CM_A$ intersect at the point $N_A$. Points $K_B, L_B, M_B, N_B$ and $K_C, L_C, M_C, N_C$ are defined similarly for $(B, C, A)$ and $(C, A, B)$ respectively. Show that the lines $L_AN_A, L_BN_B,$ and $L_CN_C$ are concurrent.
Problem
Source: Romania TST 2022
Tags: geometry, Euler Circle, romania, Romanian TST
VicKmath7
15.05.2022 11:37
Note that $ABK_AH_A$ is cyclic, where $AH_A$ is altitude. So $\angle BAK_A= \angle P_AH_BH_A =\gamma - \beta$, where $P_A$ is the midpoint of $BC$. Hence $\angle K_AAH_A = 90-\gamma=\angle H_AAC$, so if we let $AK_A$ intersect $BC$ at $X_A$, then $X_AAC$ is isosceles and $X_AH_A=H_AC$. Now, note that $\angle K_AH_AH_C=\beta - \gamma + \alpha = 180 - 2\beta$ and $\angle H_AK_AM_A= \angle BAH_A=90-\beta$, so $H_A$ is midpoint of $M_AL_A$ since $\angle L_AK_AM_A=90$. So we obtain $L_AX_AM_AC$ is a parallelogram, so $\angle BCM_A=\angle L_AX_AC=\gamma$. Therefore, $\angle H_AM_AC=\beta$, since $\angle M_AH_AC=\alpha$. Hence, $\angle K_AM_AN_A=90$, so $H_A$ is midpoint of $K_AN_A$, so $L_AK_AM_AN_A$ is a square and $L_AN_A||BK_A$.
Also, note that $\angle L_AH_CH=90-\gamma = \angle L_AAH$, so $L_A$ lies on $(AH)$ and $\angle AL_AH=90= \angle BK_AA$, so $L_AH||BK_A$. So $L_AN_A$ passes through the orthocenter $H$ and we are done.
Cyf0
17.05.2022 22:18
Cute problem! We will prove that all lines pass through the orthocenter of the triangle. Let $AD$, $BE$, and $CF$ be the altitudes of the triangle and $H$ be the orthocenter. Claim 1 $A, E, H, F$ and $ {L}_{A}$ are concyclic Proof: Obviously, $A, E, D, B$ and ${K}_{A}$ and $A, E, H$ and $ F$ are concyclic, respectively. Now, since ${K}_{A}D$ is tangent to $(DEF)$ we have that $\angle {K}_{A}DE=180^\circ-\angle EFD=\angle{L}_{A}FE$. On the other hand, $\angle EA{K}_{A}=180^\circ-\angle {K}_{A}DE$. Therefore $\angle EA{L}_{A}+\angle EF{L}_{A}=180^\circ$, thus proving the claim. Claim 2 $E, F, B, {M}_{A} $ and $C$ are concyclic Proof:We have that $E$ is the intersection of circles $(A{K}_{A}B)$ and $(A{L}_{A}F)$. Therefore, it is the $Miquel$ $Point$ of complete quadrilateral ${L}_{A}FB{K}_{A}A{M}_{A}$. Thus, $E, F, B, {M}_{A} $ and $C$ are concyclic. Claim 3 $A{K}_{A}$ is parallel to $C{M}_{A}$ Proof: This follows from Reim's theorem applied to $(AEB{K}_{A})$ and $(CEB{M}_{A})$. Claim 4 $E,H,D,{N}_{A}$ and $C$ are concyclic Proof: We already know that $(EHDC)$ is inscribed, therefore it is sufficient to prove that $(ED{N}_{A}C)$ is inscribed. But we have that $A{K}_{A}$ is antiparallel to $DE$ w.r.t. ${K}_{A}{N}_{A}$ and $AC$. Since $A{K}_{A}$ is parallel to $C{N}_{A}$, we have that $DE$ is antiparallel to ${N}_{A}C$. Thus the claim is proved. Claim 5 ${L}_{A}, H$ and ${N}_{A}$ are collinear Proof: This just follows from the converse of Reim's Theorem applied to $(A{L}_{A}HE)$ and $(EH{N}_{A}C)$.