Baltic Way 2023 Shortlist

Let $K$ be the orthogonal projection of $B$ onto the angle bisector of $\angle BAC$. Now by the Iran lemma and its analogue for the excircle (both easy angle chases), $K$ lies on lines $DE$ and $PQ$. Now let $L$ be the orthogonal projection of $K$ onto line $AB$. Our main claim is that $L$ lies on both $XY$ and $WY$ and thus $X$, $Y$, $W$ are colinear. This also suffices since then a symmetric claim with respect to $C$ gives $X$, $Y$, $Z$ colinear. We use directed angles.

Claim: $L$ lies on $XY$. Proof: Clearly $KLFY$ is cyclic. Also letting $I$ be the incenter of $△ABC$, we see that $BFIK$ is cyclic. Now $$\angle XYD=\angle XFE=\angle CDE$$ which is quite easy to chase to be $$\angle BIK=\angle BFK=\angle LFK=\angle LYK=\angle LYD.$$

Claim: $L$ lies on $WY$. Proof: First note that $\angle QKE=90^{\circ }$ as $$\angle KEQ+\angle EQK=\angle DEC+\angle CQP=90^{\circ }$$ and since $FR$ and $EQ$ are symmetric with respect to $AK$, we also have $\angle FKR=90^{\circ }$. Now as $KLRW$ is cyclic, we have $$\angle WLR=\angle WKR=\angle QKR$$ and by the perpendicularities, we have that this is just $$\angle EKF=\angle YKF=\angle YLF.$$

So $L$ lies on both $XY$ and $WY$, and thus $X$, $Y$, $W$ are collinear. By symmetry, $X$, $Y$, $Z$ are also collinear, so $X$, $Y$, $Z$, $W$ are collinear.

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Alternative solution.

First we show that $XY$ and $BC$ are parallel. Indeed, since $EFXY$ is cyclic and $BC$ is tangent to the incircle at $D$, we have $$\angle DXY=\angle FED=\angle FDB=\angle XDB.$$ Similarly, $ZW$ and $BC$ are parallel. Moreover, the points $X$, $Y$, $Z$ and $W$ all lie on the same side of $BC$ because, for example, $X$ lies in the interior of segment $DF$ (since $\angle EDF<90^{\circ }$), and similarly for the other points. Therefore it suffices to show that the distance from $BC$ to $XY$ and $ZW$ is the same.

We will use the following Lemma: Lemma. Let $△ABC$ be a non-rectangular triangle with altitudes $BU$ and $CV$. Then $d(A,UV)=d(A,BC)\cdot |\cos \angle A|$, where $d(P,\ell )$ denotes the distance from point $P$ to line $\ell$. Proof. Let $\varphi$ be the homothety with center $A$ and factor $\cos \angle A$, followed by reflection across the internal angle bisector of $\angle A$. Then $\varphi$ maps $B$ to $U$ and $C$ to $V$. Hence $△ABC$ and $△AUV$ are similar with factor $|\cos \angle A|$, from which the Lemma follows.

After applying the Lemma to $△DEF$ and $△PQR$ and noting that $$\cos \angle RPQ=\cos (180^{\circ }-\angle EDF)=-\cos \angle EDF,$$ it remains to show that $d(D,EF)=d(P,QR)$. However, since $EF$ and $RQ$ are parallel and $|BD|=|PC|$, this is equivalent to showing that $d(B,EF)=d(C,QR)$. But this can be done as follows: $$\begin{array}{rl}d(B,EF)& =|BF|\cdot \sin \angle BFE=|BD|\cdot \sin \angle EFA\\ & =|PC|\cdot \sin \angle AEF=|CQ|\cdot \sin \angle AQR\\ & =d(C,QR).\end{array}$$

Therefore, the lines $XY$ and $ZW$ are parallel and equidistant from $BC$, so the points $X$, $Y$, $Z$, $W$ are collinear.