In the triangle $ABC$ the relationship $AB+AC = 2BC$ holds. Let $I$ and $M$ be the incenter and intersection point of the medians of triangle $ABC$ respectively, $AL$ its angle bisector, and point $P$ the orthocenter of triangle $BIC$. Prove that the points $L, M, P$ lie on a straight line. (Matvii Kurskyi)
Problem
Source: 2022 Yasinsky Geometry Olympiad X-XI advanced p4 , Ukraine
Tags: geometry, collinear
iammocnhan
09.11.2022 05:39
Let $P'$ be the symmetric point of $P$ about $BC$, $F$ be the second intersection of $AI$ and $(BIC)$, $E$ be the midpoint of $BC$, $G$ be the point on $BC$ that $IG \perp BC$ It's simple result that $P' \in (BIC)$ and $F$ is $A$-excenter of $\Delta ABC$ Therefore $\dfrac{FL}{FA}=\dfrac{IL}{IA}=\dfrac{BC}{AB+AC}=\dfrac{1}{2}$ Hence, $L$ is the midpoint of $AF$. Also $\widehat{IP'F}=90^o \Rightarrow LG \parallel FP'$ Using Thales theorem we have $\dfrac{IG}{GP'}=\dfrac{IL}{LF}=\dfrac{1}{3}$ $\Rightarrow GP'=3IG \Rightarrow GP=3IG \Rightarrow PI=2IG$ Therefore $\dfrac{PI}{IG}=\dfrac{AI}{IL}$, so $AP \parallel LG$ Now it's sufficient to prove $\dfrac{EL}{AP}=\dfrac{EM}{MA}=\dfrac{1}{2}$ Using Thales theorem for $AP \parallel FP'$ we have $\dfrac{AP}{FP'}=\dfrac{AI}{IF}=\dfrac{1}{2}$ Let $H$ be the midpoint of $FP'$. Since $BCP'F$ is isosceles trapezoid, $EH$ is the perdipencular bisector of $FP'$ Therefore $\widehat{EHP'}=90^o$ or $EHP'G$ is a rectangle. Hence $EG=HP'$ Also, $\dfrac{CL}{CB}=\dfrac{AC}{AC+AB}=\dfrac{AC}{2BC} \Rightarrow CL=\dfrac{AC}{2}$ $CG=\dfrac{BC+CA-AB}{2}, CE=\dfrac{BC}{2}$ $\Rightarrow CG+CE=2CL$, therefore $L$ is midpoint of $EG$ $\Rightarrow EL=\dfrac{1}{2}EG=\dfrac{1}{2}HP'=\dfrac{1}{4}FP'=\dfrac{1}{2}AP$ (Q.E.D)