right angle wanted, incircle, circumcircle, perpendicular diagonals

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right angle wanted, incircle, circumcircle, perpendicular diagonals

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Source: 2020 Croatia MO p3

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#1 h Feb 16, 2021, 5:30 PM

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Given a triangle $ABC$ such that $AB<AC$ . On sides $AB$ and $BC$ , points $P$ and $Q$ are marked respectively such that the lines $AQ$ and $CP$ are perpendicular, and the circle inscribed in the triangle $ABC$ touches the length $PQ$ . The line $CP$ intersects the circle circumscribed around the triangle $ABC$ at the points $C$ and $T$ . If the lines $CA,PQ$ and $BT$ intersect at one point, prove that the angle $\angle CAB$ is right.

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#2 h Mar 25, 2021, 5:26 PM

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Obviously quadrilateral $APQC$ is tangential, so $AP+QC=PQ+CA$ from which $CA-AP=QC-PQ$ and also its diagonals are normal so obviously from Pythagoras $AP^2+QC^2=CA^2+PQ^2$ or $(CA+AP)(CA-AP)=(QC+PQ)(QC-PQ)$ so $CA+AP=QC+PQ$ , so obviously $AP=PQ$ and $QC=CA$ . Our quadrilateral is a deltoid. Let $CA, PQ, BT$ intersect in point $D$ . Now $\angle DQC=\alpha$ and $\angle QDC=180^\circ-\angle DQC-\angle QCD=180^\circ-\alpha-\gamma=\beta$ . Now since quadrilateral $ATBC$ is cyclic we have that $\angle ATC=\angle ATP=\beta=\angle QDC=\angle ADP$ so quadrilateral $ADTP$ is also cyclic. $\angle PDB=\angle PDT=\angle PAT=\angle BAT=\angle BCT= \frac{\gamma}{2}$ and $\angle DBA=\angle TCA=\frac{\gamma}{2}$ . Triangles $APQ$ and $DPB$ are similar and equilateral, so $DB \parallel AQ$ and $\angle CTB =90^\circ=\angle CAB$

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