Let $ABC$ be a triangle with circumcentre $O$ and incentre $I$ such that $O$ is different from $I$. Let $AK, BL, CM$ be the altitudes of $ABC$, let $U, V , W$ be the mid-points of $AK, BL, CM$ respectively. Let $D, E, F$ be the points at which the in-circle of $ABC$ respectively touches the sides $BC, CA, AB$. Prove that the lines $UD, VE, WF$ and $OI$ are concurrent.
Problem
Source: Indian Postal Coaching 2009 set 1 p3
Tags: geometry, Concyclic, Circumcenter, altitudes, incenter
Prod55
17.08.2020 13:39
I suppoce that $b>c$.Let $R\equiv DU\cap OI,ON\parallel DU,N\in DI,T\equiv ON \cap BC$ and $X$ the midpoint of $BC$. It suffice to show that $\dfrac{RI}{IO}=ct$.Set $R=OA=OB=OC,r=ID$ $\dfrac{RI}{IO}=\dfrac{ID}{IN}=\dfrac{r}{NI}$.So we must show that $DN=ct$ $\dfrac{DN}{OX}=\dfrac{DT}{XT}\Leftrightarrow ND=\dfrac{DT}{OX}{XT}$.Also $\dfrac{OX}{XT}=\dfrac{KU}{KD}=\dfrac{AK}{2KD}\Leftrightarrow XT=\dfrac{OX2KD}{AK}$. So $ND=(DX+XT)\dfrac{AK}{2KD}=(DX+\dfrac{OX2KD}{AK})\dfrac{AK}{2KD}$ $DX=\dfrac{a}{2}-\dfrac{b+c-a}{2}=\dfrac{b-c}{2}$ $KD=\dfrac{a+c-b}{2}-c\cos\angle B=..=\dfrac{(b-c)(b+c-a)}{2a}$ $AK=\dfrac{bc}{2R},OX=R\cos \angle A$ So $ND=(\dfrac{b-c}{2}+\dfrac{R\cos \angle A\cdot 2\dfrac{(b+c-a)(b-c)}{a}}{\dfrac{bc}{2R}})\cdot \dfrac{bc}{2R}\dfrac{1}{\dfrac{2(b-c)(b+c-a)}{2a}}=..=$ $=..=\dfrac{4R^2\cos\angle A(b+c-a)+abc}{4R(b+c-a)}=R\cos A+\dfrac{abc}{4R(b+c-a)}$ Now we have to show that $R\cos\angle A+\dfrac{abc}{4R(b+c-a)}=R\cos \angle B+\dfrac{abc}{4R(a+c-b)}\Leftrightarrow $ $\Leftrightarrow R(\cos\angle A-\cos\angle B)=\dfrac{abc}{4R}\cdot\dfrac{2(b-a)}{(a+c-b)(b+c-a)}$ Since $\dfrac{abc}{4R}=(ABC)$ we have that $\dfrac{4R^2}{abc}=\dfrac{abc}{4(ABC)^2}$ So we must show that $\dfrac{abc}{4(ABC)^2}(\dfrac{b^2+c^2-a^2}{2bc}-\dfrac{a^2+c^2-b^2}{2ac})=\dfrac{2(b-a)}{(a+c-b)(b+c-a)}\Leftrightarrow$ $\Leftrightarrow \dfrac{abc}{4\dfrac{a+b+c}{2}\dfrac{a+b-c}{2}\dfrac{b+c-a}{2}\dfrac{a+c-b}{2}}\cdot \dfrac{(b-a)(b+a+c)(b+a-c)}{2abc}=\dfrac{2(b-a)}{(a+c-b)(b+c-a)}\Leftrightarrow 2=2$ The proof is complete.
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