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
15.05.2022 11:37
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 $(EH{N}_{A}C)$.