A Cute Problem about Square OPHQ

by TelvCohl, Aug 15, 2016, 3:02 AM

Problem : Given a $\triangle ABC$ with circumcenter $O$ and orthocenter $H.$ Let $P,$ $Q$ be the points such that $OPHQ$ is a square and let $P^*,$ $Q^*$ be the isogonal conjugate of $P,$ $Q$ WRT $\triangle ABC,$ respectively. Prove that (1) $PQ$ $\parallel$ $P^*Q^*.$ (2) $PQ^*,$ $P^*Q$ pass through the Euler reflection point $T$ of $\triangle ABC.$ (3) $PP^*,$ $QQ^*$ pass through the Kosnita point $K_o$ of $\triangle ABC.$

Proof : This problem is a nice application of the Properties of Hagge circle. Let the circle $\odot (Q)$ with center $Q$ passing through $O,$ $H$ cuts $AH,$ $BH,$ $CH$ at $A_P^*,$ $B_P^*,$ $C_P^*,$ respectively. Clearly, $\odot (Q)$ is the Hagge circle of $P^*$ WRT $\triangle ABC,$ so combining $\measuredangle OQA_P^*$ $=$ $2 \measuredangle (OH, \perp BC)$ $=$ $\measuredangle AOT$ we get $\triangle ABC$ $\cup$ $O$ $\cup$ $T$ $\cup$ $P^*$ $\stackrel{-}{\sim}$ $\triangle A_P^*B_P^*C_P^*$ $\cup$ $Q$ $\cup$ $O$ $\cup$ $P^*,$ hence $\measuredangle OP^*T$ $=$ $\measuredangle OP^*Q$ $\Longrightarrow$ $T$ lies on $P^*Q.$ Similarly, we can prove $T$ lies on $PQ^*,$ so $T$ is the intersection of $PQ^*,$ $PQ^*.$ On the other hand, it's well-known that the intersection of $PQ,$ $P^*Q^*$ is the isogonal conjugate of $T$ WRT $\triangle ABC,$ so $PQ$ $\parallel$ $P^*Q^*.$

Let $K$ $\equiv$ $PP^*$ $\cap$ $QQ^*$ and let $M,$ $N$ be the midpoint of $P^*Q^*,$ $PQ,$ respectively. Let $OT$ cuts the perpendicular bisector of $P^*Q^*$ at $U.$ Clearly, $K,$ $M,$ $N,$ $T$ are collinear and $\triangle OPQ,$ $\triangle UQ^*P^*$ are homothetic with center $T,$ so from $O(K,T;M,N)$ $=$ $-1$ we know $OK$ passes through the reflection $V$ of $U$ in $M.$ From $OQ$ $\parallel$ $UP^*$ $\Longrightarrow$ $\measuredangle UP^*O$ $=$ $\measuredangle QOP^*$ $=$ $\measuredangle P^*TO$ $=$ $\measuredangle P^*TU,$ so we get $UO$ $\cdot$ $UT$ $=$ ${UP^*}^2$ $=$ $UM$ $\cdot$ $UV$ $\Longrightarrow$ $M,$ $O,$ $T,$ $V$ are concyclic, hence $\measuredangle OTK$ $=$ $\measuredangle OTM$ $=$ $\measuredangle OVM$ $=$ $\measuredangle KON$ $\Longrightarrow$ $OK$ is the tangent of $\odot (NOT).$ $\qquad (\dagger)$

Let the circle $\odot (N)$ with diameter $OH$ cuts $AH,$ $BH,$ $CH$ at $A^*,$ $B^*,$ $C^*,$ respectively. Clearly, $\odot (N)$ is the Hagge circle of $K_o$ WRT $\triangle ABC,$ so combining $\measuredangle ONA^*$ $=$ $2\measuredangle (OH, \perp BC)$ $=$ $\measuredangle AOT$ we get $\triangle ABC$ $\cup$ $O$ $\cup$ $T$ $\cup$ $K_o$ $\stackrel{-}{\sim}$ $\triangle A^*B^*C^*$ $\cup$ $N$ $\cup$ $O$ $\cup$ $K_o$ , hence $\measuredangle OK_oN$ $=$ $\measuredangle OK_oT$ $\Longrightarrow$ $K_o,$ $M,$ $N,$ $T$ are collinear. Furthermore, from $\measuredangle OTK_o$ $=$ $\measuredangle K_oON$ we know $OK_o$ is tangent to $\odot (NOT)$ at $O,$ so combining $(\dagger)$ we conclude that $K$ coincide with $K_o.$ i.e. the Kosnita point $K_o$ of $\triangle ABC$ lies on $PP^*$ and $QQ^*.$ $\qquad \blacksquare$

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A Cute Problem about Square OPHQ

by TelvCohl, Aug 15, 2016, 3:02 AM

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