Perpendicular to Euler Line

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Perpendicular to Euler Line

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#1 h Aug 4, 2019, 10:26 PM • 6 Y

Y by Pluto1708, amar_04, Adventure10, Mango247, Mango247, Mango247

Let $O$ , $H$ be the circumcenter, orthocenter of $\triangle ABC$ . Let $A'$ be the reflexion of $A$ with respect to $BC$ . Let $P = A'B \cap \odot O$ , $Q= A'C \cap \odot O$ , $T = PQ \cap BC$ . Prove that $AT \perp OH$ .

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#2 h Aug 5, 2019, 4:09 AM • 3 Y

Y by Flash_Sloth, Adventure10, Mango247

Recall that $(BHC)$ is the reflection of $(ABC)$ over $BC$ , so if we define $P' = (BHC) \cap AB$ and $Q' = (BHC) \cap AC$ then $T \in P'Q'$ .

Let $E, F$ be the feet of the altitudes from $B, C$ . Observe that $\angle BQ'A = \angle CP'A = \angle A$ so $P', Q'$ are reflections of $A$ over $F, E$ . Now let $M, N$ be the midpoints of $AC, AB$ , and let $T' = MN \cap EF$ . By homothety at $A$ , it suffices to show that $AT' \perp OH$ .

Define $X = EF \cap BC$ and let the tangent to $(AMN)$ at $A$ meet $MN$ at $Y$ . It is easy to see that $XY$ is the radical axis of $(ABC)$ and the nine-point circle, i.e. $XY \perp OH$ . (see also here) But $EF \parallel AY$ and $\text{d}(A, MN) = \text{d}(X, MN)$ , so $AYXT'$ is a parallelogram. So $XY \parallel AT' \perp OH$ , as desired.

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#3 h Aug 5, 2019, 4:28 AM • 2 Y

Y by Adventure10, Mango247

rocketscience wrote:

Recall that $(BHC)$ is the reflection of $(ABC)$ over $BC$ , so if we define $P' = (BHC) \cap AB$ and $Q' = (BHC) \cap AC$ then $T \in P'Q'$ .

Let $E, F$ be the feet of the altitudes from $B, C$ . Observe that $\angle BQ'A = \angle CP'A = \angle A$ so $P', Q'$ are reflections of $A$ over $F, E$ . Now let $M, N$ be the midpoints of $AC, AB$ , and let $T' = MN \cap EF$ . By homothety at $A$ , it suffices to show that $AT' \perp OH$ .

Define $X = EF \cap BC$ and let the tangent to $(AMN)$ at $A$ meet $MN$ at $Y$ . It is easy to see that $XY$ is the radical axis of $(ABC)$ and the nine-point circle, i.e. $XY \perp OH$ . (see also here) But $EF \parallel AY$ and $\text{d}(A, MN) = \text{d}(X, MN)$ , so $AYXT'$ is a parallelogram. So $XY \parallel AT' \perp OH$ , as desired.

Thank you for the nice solution and resource! I didn't manage to find a geo solution myself, but it's pretty easy to bash by complex number.

This post has been edited 1 time. Last edited by Flash_Sloth, Aug 5, 2019, 4:32 AM

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#4 h Aug 5, 2019, 6:27 AM • 3 Y

Y by Flash_Sloth, Adventure10, Mango247

Another solution:
As shown above it is enough to prove that $AT' \perp OH$ where $T'=MN \cap EF$ . Note that $MNFE$ is cyclic with centre on the Euler line (nine point centre, in fact) and $EN \cap MF$ also lies on the Euler line by Pappus theorem. So, by Brokard's theorem, we are done.

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#5 h Aug 5, 2019, 6:34 AM • 2 Y

Y by Flash_Sloth, Adventure10

Let $DEF$ be orthic triangle of $ABC,$ $M, N$ be the midpoint of $AB, AC$
Let $T=EF \cap MN,$ then $AT \perp OH$ (radical axis of circle with diameter $AH$ and $AO$
Use homothety center $A$ ratio $\frac{1}{2}$
$P', Q' = DM, DN \cap (AMN),$ we need to prove $P'Q'$ passes through $T,$ which is trivial since $(DMN), (AMN)$ are reflection wrt $MN \Rightarrow P'Q', EF$ are reflection wrt $MN$

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#6 h Nov 7, 2019, 2:33 PM • 1 Y

Y by Adventure10

Easy yet nice.
Let $P',Q'$ be reflections of $P,Q$ over $BC$ . Let $AT \cap \odot (ABC) =$ { $A,R$ }
Then $P' \in AB$ and $Q' \in AC$ and $T = P'Q' \cap BC$ .

Lemma 1: $H$ is circumcentre of $\triangle AP'Q'$ .
Proof: We see by easy angle chasing that $AB = Q'B$ and $BH \perp AQ'$ . So $BH$ is $\perp$ bisector of $AQ'$ $\implies HA=HQ'$ . Similarly $HA=HP'$ . So $HA=HP'=HQ'$ as desired.

Lemma 2: $ARP'Q'$ is cyclic.
Proof: $TR.TA = TP.TQ = TP'.TQ'$ .
So $ARP'Q'$ is cyclic.

Now both the lemmas together imply that $H$ is circumcentre of $ARP'Q'$ .
So $HA=HR$ and also $OA=OR$ .
Thus $OH$ is $\perp$ bisector of $AR$ $\implies OH \perp AR \implies OH \perp AT$ . Q.E.D.

This post has been edited 4 times. Last edited by Euler365, Nov 9, 2019, 9:51 AM

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#10 h Jun 21, 2021, 12:11 AM

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Reflect literally everything over $BC$ , i.e. reflect $H$ and $O$ over $BC$ , call those points $H'$ and $O'$ .

Claim. $H'$ is the centre of $(A'QP)$ .
Proof. $\measuredangle BPH'=\measuredangle BAH'=\measuredangle H'A'B\implies H'P=H'A$ , do the same with the point $Q$ . $\square$

Let $(A'BC)$ and $(A'PQ)$ intersect at $L$ . By radical axis, $L$ lies on $A'T$ . Also, as $LH'=HA=OO'$ and $H'O=HO'=AO=LO'$ , hence we conclude that $H'O'OL$ is an isosceles trapezoid.
Let $R$ be the reflection of $H'$ over the midpoint of $BC$ , then it's well-known that $A'R$ is the diameter of $(A'BC)$ . Also, $H'O'RO$ is parallelogram, hence $OL\parallel H'O' \parallel OR\implies L,O$ and $R$ are collinear. Hence, $A'T\perp LR\parallel H'O'\implies AT\perp OH$ . $\blacksquare$
Easier way.

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