Collinearity with orthocenter

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Collinearity with orthocenter

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#1 h Mar 30, 2025, 4:42 PM

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Given scalene triangle $ABC$ with circumcenter $(O)$ . Let $H$ be a point on $(BOC)$ such that $\angle AOH = 90^{\circ}$ . Denote $N$ the point on $(O)$ satisfying $AN \parallel BC$ . If $L$ is the projection of $H$ onto $BC$ , show that $LN$ passes through the orthocenter of $\triangle ABC$ .

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soryn

#2 h Mar 31, 2025, 12:03 AM

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Any ideas?...

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#3 h Mar 31, 2025, 1:41 AM • 2 Y

Y by Retemoeg, soryn

Since I believe that the orthocenter being $H$ is more clean, we redefine points. Let $T$ be a point on $(BOC)$ that $AO \perp AT$ , and $H$ be the orthocenter of $ABC$ . $H^*$ is the second intersection of $(BOC)$ and $AO$ , and $L'$ be the projection of $H^*$ to $BC$ . $D$ is the intersection of $AH$ and $BC$ .

Since $AO \perp OT$ , we have that $H^*$ is the antipode of $T$ wrt $(BOC)$ . This implies that $BL = L'C$ , so it suffices to prove that $AN : AH = DL' : DA$ . From $\sqrt{bc}$ inversion, we can easily see that $AH \cdot AH^* = AB \cdot AC = 2 AD \cdot AO$ . This implies that
$\dfrac{AD}{L'D} = \dfrac{AD}{AH^* \cdot \sin \angle OAD} = \dfrac{AH}{2 AO \cdot \sin \angle OAD} = \dfrac{AH}{AN}$ Therefore, $N, L, H$ are collinear.

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#4 h Mar 31, 2025, 5:04 PM • 1 Y

Y by soryn

@above nice solution!
I actually reformulated this from a relatively well known incircle result. So the introduction of points $H*, L'$ is basically working backwards to the original problem lol

This is my solution: Let $R$ be the second intersection of $AO$ with $BOC$ and $G$ the intersection of $AO$ with $BC$ . Denote $A'$ the antipode of $A$ in $(O)$ and $D$ the intersection of $AH$ and $BC$ . Let $E, F$ be orthogonal projections from $A', R$ onto $BC$ and $Z$ the intersection of $TA$ with $EN$ .
We readily notice that $T, R$ are diametrically opposite in $(BOC)$ , thus $BL = CF$ . Same follows with $A, A'$ in $(O)$ so $BD = CE$ . Thus we have symmetry between $A, D, L$ and $N, E, F$ w.r.t perpidencular bisector of $BC$ .
Now:
$F(AG,OR) = (AG,OR) = C(AG,OR) = -1$ Implying that $FG$ bisects $\angle AFA'$ . In other words, $\triangle ZFA'$ is isoceles at $F$ , hence $EZ = EA'$ . Now, since $BHCA'$ is a parallelogram, it follows from symmetry that $HD = EA'$ . All this gives us $HD = EZ$ . With the above symmetrical relations, one can conclude that $L, H, N$ are collinear.

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