Let $\omega$ denote the excircle tangent to side $BC$ of triangle $ABC$. A line $\ell$ parallel to $BC$ meets sides $AB$ and $AC$ at points $D$ and $E$, respectively. Let $\omega'$ denote the incircle of triangle $ADE$. The tangent from $D$ to $\omega$ (different from line $AB$) and the tangent from $E$ to $\omega$ (different from line $AC$) meet at point $P$. The tangent from $B$ to $\omega'$ (different from line $AB$) and the tangent from $C$ to $\omega'$ (different from line $AC$) meet at point $Q$. Prove that, independent of the choice of $\ell$, there is a fixed point that line $PQ$ always passes through.
Problem
Source: Czech-Polish-Slovak Match, 2009
Tags: geometry
oneplusone
20.08.2011 06:12
Note that $\omega$ is tangent to the extensions of $AD,AE,EP,DP$. So $AD+DP=AE+EP$. Let $\omega'$ tangents $DE$ at $Y$, $\omega$ tangents $BC$ at $X$. Then $DY-YE=EP-PD$. This means that the excircle $s$ of $\triangle DPE$ opposite $P$ is tangent to $DE$ at $Y$. Note that $\omega,s$ are homothetic about $P$, so $X,P,Y$ are collinear. Similarly $X,Q,Y$ are collinear. Thus $PQ$ passes through the fixed point $X$. This is quite similar to IMO 08 Q6 actually...
Number1
11.06.2016 19:28
I don't know if the following is actually solution, so someone please check. First we se that map $\pi: D\mapsto E$ is projective from line $AB$ to line $AC$. A map $\pi _E$ which takes $E$ to it's tangent $EP$ on $\omega $ and map $\pi _D$ which takes $D$ to tangent $DP$ are both projective. Thus composition $\pi ' = \pi _E\circ \pi \circ \pi _D^{-1}$ is also projective and takes line $DP$ to line $EP$. So $P$ must be on some fixed conic (or line) $\mathcal{C}_1$. It's obviously that $BQ|| EP$ and $CQ|| DP$, so map $CQ \mapsto BQ$ is also projective and thus $Q$ must also be on some fixed conic (or line) $\mathcal{C}_2$. Now let $\omega $ touch $BC$ at $F$ and let $\omega' $ touch $BC$ at $G$. Also let $B'C'$ be parallel tangent on $\omega$ to $BC$. If $E =C$ we see $P=F$ (and $Q = G$) and if $E = C'$ we see $Q = F$. So $F$ is on both $\mathcal{C}_1$ and $\mathcal{C}_2$ so map $\rho: P\mapsto Q$ is also projective from $\mathcal{C}_1$ to $\mathcal{C}_2$. Now if $E= A$ then $Q=P =A$ and so $P,Q$ and $G$ are collinear; and if $P=F$ then $Q=G$ so $P,Q$ and $G$ are again collinear, thus $QP$ always passes through point $G$.