In right triangle $ABC$, $\angle ACB = 90^{\circ}$ and $\tan A > \sqrt{2}$. $M$ is the midpoint of $AB$, $P$ is the foot of the altitude from $C$, and $N$ is the midpoint of $CP$. Line $AB$ meets the circumcircle of $CNB$ again at $Q$. $R$ lies on line $BC$ such that $QR$ and $CP$ are parallel, $S$ lies on ray $CA$ past $A$ such that $BR = RS$, and $V$ lies on segment $SP$ such that $AV = VP$. Line $SP$ meets the circumcircle of $CPB$ again at $T$. $W$ lies on ray $VA$ past $A$ such that $2AW = ST$, and $O$ is the circumcenter of $SPM$. Prove that lines $OM$ and $BW$ are perpendicular.
Problem
Source: PMO 2021
Tags: geometry, PMO
sarjinius
13.08.2021 14:12
We claim that $\triangle WAB \sim \triangle MPS$. Since $\angle WAB = 180^{\circ} - \angle VAP = 180^{\circ} - \angle VPA = \angle MPS$, it suffices to show that $$\frac{AB}{AW} = \frac{PS}{PM} \iff \frac{2AB}{2AW} = \frac{PS}{PM} \iff \frac{2AB}{ST}= \frac{PS}{PM} \iff SP \cdot ST = 2AB \cdot PM.$$Since $BC$ is a diameter of $(CPB)$, we have $SP \cdot ST = SC^2$ and $AP \cdot AB = AC^2$. So we need to show $$SC^2 = 2AB \cdot PM \iff PM = \frac{SC^2}{2AB} \iff AM - AP = \frac{SC^2}{2AB} \iff \frac{AB}{2} - \frac{AC^2}{AB} = \frac{SC^2}{2AB}$$$$\iff \frac{AB^2-2AC^2}{2AB} = \frac{SC^2}{2AB} \iff (AB^2 - AC^2) - AC^2 = SC^2 \iff SC^2 = BC^2 - AC^2.$$Since $BCNQ$ is cyclic, $\angle PNQ = \angle PBC = \angle PCA$ and $\angle QPN = \angle APC = 90^{\circ}$, then $\triangle PNQ \sim \triangle PCA$ and $\frac{AP}{PQ} = \frac{PC}{PN} = 2$. Then $BQ = AB - AQ = AB - \frac{3}{2}AP = AB - \frac{3AC^2}{2AB} = \frac{2AB^2 - 3AC^2}{2AB}$. From $QR \parallel CP$, we have $\frac{BQ}{BP} = \frac{BR}{BC} \iff BR = \frac{BC \cdot BQ}{BP} = \frac{BC(\frac{2AB^2 - 3AC^2}{2AB})}{\frac{BC^2}{AB}} = \frac{2AB^2 - 3AC^2}{2BC}$. We calculate $SC^2$ as follows: $$SC^2 = RS^2 - RC^2 = BR^2 - RC^2 = (BR + RC)(BR - RC) = BC(2BR - BC) = BC\left(\frac{2AB^2 - 3AC^2}{BC} - BC\right)$$$$= 2AB^2 - 3AC^2 - BC^2 = (AB^2 - AC^2) + (AB^2 - BC^2) - 2AC^2 = BC^2 + AC^2 - 2AC^2 = BC^2 - AC^2$$Therefore, $\triangle WAB \sim \triangle MPS$ and $\angle WBA = \angle MSP$. Hence, $\angle PMO + \angle WBA = 90^{\circ} - \frac{\angle POM}{2} + \angle MSP = 90^{\circ} - \angle PSM + \angle MSP = 90^{\circ}$ and thus $OM \perp BW$.