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Source: Belarus TST 2024

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#1 h Jul 17, 2024, 9:36 PM • 1 Y

Y by Rounak_iitr

An acute-angled triangle $ABC$ with an altitude $AD$ and orthocenter $H$ are given. $AD$ intersects the circumcircle of $ABC$ $\omega$ at $P$ . $K$ is a point on segment $BC$ such that $KC=BD$ . The circumcircle of $KPH$ intersects $\omega$ at $Q$ and $BC$ at $N$ . A line perpendicular to $PQ$ and passing through $N$ intersects $AD$ at $T$ . Prove that the center of $\omega$ lies on line $TK$ .
U. Maksimenkau

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#2 h Jul 30, 2024, 5:25 PM • 1 Y

Y by Hamzaachak

Let $V$ be the circumcenter of triangle $KPH$ and $O$ the circumcenter of triangle $ABC$

It is a well-known lemma that $P$ is the reflection of $H$ with respect to $BC$ hence $BC$ is the side bissector of $KP$ hence $V$ lies on $BC$

Since $PQ$ is the radical axes of $(ABC)$ and $(KPH)$ we have $PQ$ is perpendicular to $OV$ thus $TN$ is parallel to $OV$

Let $KO \cap TN=T'$

Then we have $\frac{KO}{OT'}=\frac{KV}{VN}=1 \quad(KN \quad is \quad the \quad diameter)$
Hence $KO=OT'$
We know that $BD=CK \implies BK=CD$
And since $BO=CO$ and $\angle OBK=\angle OBC= \angle OCB= \angle OCD$

We get $\triangle OBK \cong \triangle OCD$ (By SAS)

Hence $OK=OD$

Thus $OK=OD=OT'$ so $O$ is the circumcenter of $(KDT')$

And since $O$ lies in $KT'$ we get $\angle T'DK=90$ hence $T=T'$

So $O$ lies on $TK$

This post has been edited 3 times. Last edited by Mrcuberoot, Jul 31, 2024, 10:03 AM

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eulerleonhardfan

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eulerleonhardfan

#3 h Aug 16, 2024, 1:29 AM

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Let $(KPH) \cap NT = S$ . Clearly, $\angle NSK = 90$ , so $PQ // SK$ and $SDKT$ is cyclic.

The perpendicular bisector of $PQ$ passes through $O$ , so the perpendicular bisector of $SK$ also passes through $O$ , therefore $SO = OK$ .

The perpendicular bisector of $BC$ passes through $O$ , so the perpendicular bisector of $DK$ also passes through $O$ , therefore $DO = OK$ .

This shows us that $O$ is the center of $(SDTK)$ , and this implies $\overline{T, O, K}$ .

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