Some More (Basic) Properties of the configuration of orthic axis

by AlastorMoody, Jan 28, 2019, 10:11 PM

Here are some more basic properties!

Define:
$\text{(i) } X_A \longrightarrow \text{ as the A-Ex Point}$
$\text{(ii) } Q_A \longrightarrow \text{ as the A-Queue Point}$
Credits to math_pi_rate for giving such amazing names
$\text{(iii) } H_A \longrightarrow \text{ as the feet of altitude from A}$
$\text{(iv) } M_A \longrightarrow \text{ midpoint of BC}$

Pre Eliminaries

$\longrightarrow $ $X_A$ is defined as $H_BH_C \cap BC$
$\longrightarrow $ $Q_A$ is defined as $\odot (AH_BH_C) \cap \odot (ABC)$
Note: This implies that $A,H_B,H,H_C,Q_A$ lie on the same circle
Also Define:
$K_A$ as the Radical Center of $\odot (ABC) , \odot (AH_BH_C) , \odot (BH_CH_A)$
Similarly, define $K_B$ and $K_C$

Lemma 1: $X_A,X_B,X_C$ lie on the same line (This line is known as the orthic axis)

Proof: Note, $AH_A \cap BH_B \cap CH_C =H$, hence, $H$ is the center of perspectivity of $\Delta ABC$ and $\Delta H_AH_BH_C$, applying Desargues' Theorem on perspective triangles $\Delta ABC$ and $\Delta H_AH_BH_C$ implies $X_A,X_B,X_C$ lie on the same line $\qquad \blacksquare$
Note(1): Infact the above proof follows from the fact that the Orthic Axis is defined as the Axis of Perspectivity of the orthic triangle and reference triangle
Note(2): Another way of defining the Orthic Axis, Let the reflections of $H$ over $BC,CA,AB$ be $H_A' , H_B' ,  H_C'$, Denote, $X,Y,Z$ as the midpoints of $H_AH_A' , H_BH_B' , H_CH_C'$, then the polar of $H$ WRT $\odot (XYZ)$ is the Orthic Axis WRT $\Delta ABC$

Lemma 2: $AH$ is the diameter of $\odot (AH_BH_C)$

Proof: Easy to see that $\measuredangle AH_BH=\measuredangle AH_CH=\measuredangle AQ_AH=90^{\circ}$ finishes the lemma $\qquad \blacksquare$
Note: If $M_{AH}$ is the midpoint of $AH$, then, $(M_{AH}) \equiv \odot (AH_BH_C)$

Lemma 3: Let $X'$ be any point on $AH_A$, and Let $BX' \cap AC =X_2$ and $CX' \cap AB=X_1$, then, $\angle X_1H_AA =\angle X_2H_AA$

Proof: Let $AH_A \cap X_1X_2 =Y$ and $X_1X_2 \cap BC=Z$, then, $-1=(C,B;H_A,Z) \overset{A}{=} (X_2,X_1;Y,Z) \overset{D}{=} (X_2,X_1H_A \cap X_2B; X',B) $ $\implies$ $\angle X_1H_AA$ $=$ $\angle X_2H_AA$ $\qquad \blacksquare$

Lemma 4: Let $\ell || H_BH_C$ intersect $\odot (H_AH_BH_C) \text{ (Nine-Point Circle)}$ at $\mathcal{H}_1, \mathcal{H}_2$, then $\Delta \mathcal{H}_1 \mathcal{H}_2M_{AH}$ is isosceles triangle

Proof: It's Well-Known that $M_{AH} \in \odot (H_AH_BH_C)$, then using (Lemma 3) and the fact that $H_B \mathcal{H}_2 \mathcal{H}_1 H_C$ is cyclic trapezoid $\implies$ $\Delta \mathcal{H}_1 \mathcal{H}_2M_{AH}$ is isosceles $\qquad \blacksquare$

Generalization(Lemma 3): Let $\ell || H_BH_C$ intersect $\odot (H_AH_BH_C) \text{ (Nine-Point Circle)}$ at $\mathcal{H}_1, \mathcal{H}_2$, then, $AH_A$ is the angle bisector of $\angle \mathcal{H}_1 H_A \mathcal{H}_2$
Proof: Follows from (Lemma 4) $\qquad \blacksquare$
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Main (Basic) Properties

Property 1: $Q_A$ is the foot of altitude from $H$ in $\Delta AHX_A$ OR (Restatement) $Q_A$ lies on $HM_A$

Proof: Let $A^*$ be the antipode of $A$ WRT $\odot (ABC)$, hence, it easily follows that $Q_A-H-M_A-A^*$ and then, Notice that $M_A$ is the center of $\odot (BH_CH_BC)$,and finish off with Brokard's Theorem $\qquad \blacksquare$
Consequence(1): From (Property 1), we can show that, $Q_A$ lies on $AX_A$
Consequence(2): From (Property 1), $Q_AHH_AX_A$ is a cyclic quadrilateral

Property 2: Let the reflections of $H$ over $BC,CA,AB$ be $H_A' , H_B' ,  H_C'$, Denote, $X,Y,Z$ as the midpoints of $H_AH_A' , H_BH_B' , H_CH_C'$, then the polar of $H$ WRT $\odot (XYZ)$ is the Orthic Axis WRT $\Delta ABC$

Proof:(enhanced) Define: $H_B'H_C' \cap BC =T_1$ and let a parallel line ($\ell$) to $H_BH_C$ through $H$ intersect $BC$ at $T_2$, Let $\ell \cap \odot (H_AH_BH_C) =\mathcal{P}_1 , \mathcal{P}_2$, Applying (Lemma 4) $\implies$ $-1$ $=$ $(\mathcal{P}_1 ,\mathcal{P}_2 ; T_2 , H) $ $\implies$ $T_2 $ lies on the polar of $H$ WRT $\odot (H_AH_BH_C)$, also it is trivial to see that, $X_AT_1=X_AT_2$ and the fact that $H-N_9-O$ $\implies$ the polar of $H$ WRT $\odot (ABC)$ and $\odot (H_AH_BH_C)$ are parallel, now take homothety($\mathcal{X}$) centered at $H$, such, $\mathcal{X} : \Delta H_AH_BH_C \mapsto \Delta XYZ$, then, it's clear that the polar of $H$ WRT $\odot (XYZ)$ will be the Orthic Axis WRT $\Delta ABC$ $\qquad \blacksquare$

Property 3: The Euler line WRT $\Delta ABC$ is orthogonal to Orthic Axis WRT $\Delta ABC$

Proof: Notice that, if, $(X_9) \equiv \odot (XYZ)$, then $X_9 \in \text{ Euler line WRT } \Delta ABC $, then Apply (Property 2), $\implies$ the orthogonality $\qquad \blacksquare$

Property 4: $X_A,K_A,Q_A,A,K_C$ lie on the same line

Proof: This follows from the definition of $K_A,K_C$ combined with (Consequence(1) from Property(1)) $\qquad \blacksquare$

Property 5: Let $A-\text{symmedian} \cap \odot (ABC)=F$, then, $Q_A$ lies on $H_AF$

Proof: Let $A^*$ be the antipode of $A$ WRT $\odot (ABC)$, Hence, from (Property 1), $Q_A-H-M_A-A^*$, and then, it is easy to see, $\Delta AH_AF \sim \Delta AM_A$ and using the fact that $\angle Q_AM_AA=\angle Q_AH_AA$ $\implies$ $Q_A-H_A-F$ $\qquad \blacksquare$

Property 6: Finding Some Cyclic Quadrilaterals!
(1) Points $K_A,Q_A,H,Q_B$ lie on a circle
(2) Points $X_A, Q_A, H , H_A$ lie on a circle
(3) Points $X_B,Q_C,B,H_A$ lie on a circle

Note: Adding the $A_{HM}$ ($A-$ Humpty Point), $\implies$ $A_{HM} \in \odot (AH_BH_C) $ and $A_{HM} \in HX_A$
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And....of course, there are many more properties that I found, But I haven't listed them over here, because many of them are pretty trivial and kinda useless(not so interesting), So I will limit these properties so as not make this post boring
This post has been edited 47 times. Last edited by AlastorMoody, Feb 5, 2019, 2:12 PM

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Great work! Keep doing such posts, man!

by PhysicsMonster_01, Jan 31, 2019, 1:52 PM

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@above Thanks......stay tuned !

by AlastorMoody, Feb 2, 2019, 5:46 PM

I'll talk about all possible non-sense :D

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    by bookstuffthanks, Jul 31, 2024, 12:05 PM

  • hello fellow moody!!

    by crazyeyemoody907, Oct 31, 2023, 1:55 AM

  • @below I wish I started earlier / didn't have to do JEE and leave oly way before I could study conics and projective stuff which I really wanted to study :( . Huh, life really sucks when u are forced due to peer pressure to read sh_t u dont want to read

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  • Lots of good stuffs here.

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  • But even if he went to jee he could continue with this.

    Doing JEE(and completely leaving oly) seems like a insult to the oly math he knows

    by HoRI_DA_GRe8, Feb 11, 2022, 2:11 PM

  • Ohhh did he go for JEE? Good for him, bad for us :sadge:. Hmmm so that is the reason why he is inactive
    Btw @below finally everyone falls to the monopoly of JEE :) Coz IIT's are the best in India.

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  • Kukuku first shout of 2022,why did this guy left this and went for trashy JEE

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  • nice blog :)

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