Point of concurrency independent of incircle

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Point of concurrency independent of incircle

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Source: Czech-Polish-Slovak Match, 2009

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#1 h Aug 20, 2011, 2:34 AM • 2 Y

Y by Adventure10, Mango247

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.

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#2 h Aug 20, 2011, 3:12 AM • 1 Y

Y by Adventure10

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...

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#4 h Jun 11, 2016, 4:28 PM • 2 Y

Y by Adventure10, Mango247

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 $G$ .

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