A Short Result on Extension of Orthic Sides & Reflection of Vertex

by AlastorMoody, Feb 6, 2019, 10:49 AM

We'll discuss some remarkable properties of a well-known and common configuration
Thanks to TDP's solution to this post which inspired/boosted me to research more properties

Let's Start!
Define: $H_A,H_B,H_C$ as the feet of perpendiculars from $A,B,C$ to $BC,CA,AB$, respectively,

Lemma 1: Let $\omega$ be a circle at $A$ with radius $\sqrt{AH \cdot AH_A}$ which intersects $\odot (ABC)$ at $X,Y$, respectively, such, $CX>BX$, then $X,Y \in H_BH_C$
Proof: Perform Inversion $\Phi$ aroud $\omega$ with radius $\sqrt{AH \cdot AH_A}$, then, $\Phi : B \longleftrightarrow H_C , C \longleftrightarrow H_B $ and ofcourse, $X,Y$ remain invariant under $\Phi$ $\implies$ $X,Y \in H_BH_C$ $\qquad \blacksquare$

Now if we let $AH \cap \omega =D,E$ ,such, $EH_A >DH_A$, then we have the following interesting/but trivial results!
$\text{(a) } -1=(E,D;H,H_A)$ $~~~~~~~~~$ $\text{(b) } AX^2=AH \cdot AH_A =AY^2 $ $~~~~~~~~~$ $\text{(c) } \Delta AXH \sim \Delta AH_AX$ $~~~~~~~~~~$
$\text{(d) }$ If $AH_A \cap \odot (ABC)=H_A'$, then, $\angle XH_A'A=\angle YH_A'A=\angle XCA=\angle YCA$ ........ and many more results (such as midpoints of arcs and bisectors.....but not required here!)

Lemma 2: Let $X',Y'$ be the reflections of $X,Y$ over $AB$ and $AC$, then, $X',Y' \in \omega$

Basic Properties
The configuration holds many cyclic quadrilaterals, and we shall not state the obvious ones over here!

Property 1: Quadrilateral $AXY'H_B$ is cyclic
Proof: Just Notice, $\Phi : H_A \longleftrightarrow H ; H_B \longleftrightarrow C$ and ofcourse, $X,Y,X',Y'$ are invariant under $\Phi$, now note, $\angle XCA=\angle YCA=\angle Y'CA$ $\implies$ $Y'$ lies on $XC$ $\implies$ $AXY'H_B$ is cyclic $\qquad \blacksquare$

Some Trivial Consequences:
$\text{(a) } X',Y'$ lie on $H_AH_C,H_AH_B$ respectively
$\text{(b) } X',Y'$ lie on $BY,CX$ respectively
$\text{(c) } AYX'H_C$ and $XX'YY'$ are cyclic quadrilaterals
$\text{(d) } AHY'C$ and $AHX'B$ are cyclic quadrilaterals

Property 2: $HY' ||H_A'X$ and $HX' || H_A'Y$
Proof:Followed from one of the cyclic quadrilateral mentioned above and then, just some angle chasing

Property 3: Quadrilaterals $BXY'H_A$ and $CH_AX'Y$ are cyclic
Proof: $\angle DBC=\angle ACD+B$ and $\angle CY'H_A=180^{\circ} -(A+C-\angle ACD)=\angle DBC$ $\implies$ $BXY'H_A$ is a cyclic quadrilateral, the other one follows similarly! $\qquad \blacksquare$

Some more Trivial Consequences:
$\longrightarrow $ The following all are cyclic quadrilaterals:
$\text{(i) } XY'HH_C$ $~~~~~$ $\text{(iv) } X'HH_BY$
$\text{(ii) } H_BHX'Y$ $~~~~~$ $\text{(v) } Y'HH_CX$
$\text{(iii) } H_AX'YC$ $~~~~~$ $\text{(vi) } H_AY'XB$
____________________________________________________________________________________________________________________________________________________________________________________________________________
And Here's the problem from which I got motivated/interested to do this post
Sharygin Finals 2017 Grade 9 Problem 5 wrote:
Let $BH_B, CH_C$ be altitudes of an acute-angled triangle $ABC$. The line $H_BH_C$ meets the circumcircle of $ABC$ at points $X$ and $Y$. Points $P,Q$ are the reflections of $X,Y$ about $AB,AC$ respectively. Prove that $PQ \parallel BC$.

Proposed by Pavel Kozhevnikov
Let's solve it using our lemmas and properties developed above!
Proof: $\angle PQX=\frac{1}{2} \angle XAP$ and $\angle BCX=\angle BAX=\frac{1}{2} \angle XAP$ $\implies$ $PQ||BC$ $\qquad \blacksquare$ (The other case can be solved similarly)

Q1)
CentroAmerican 2000 P5 wrote:
Let $ ABC$ be an acute-angled triangle. $ C_{1}$ and $ C_{2}$ are two circles of diameters $ AB$ and $ AC$, respectively. $ C_{2}$ and $ AB$ intersect again at $ F$, and $ C_{1}$ and $ AC$ intersect again at $ E$. Also, $ BE$ meets $ C_{2}$ at $ P$ and $ CF$ meets $ C_{1}$ at $ Q$. Prove that $ AP=AQ$.
Solution: If $D$ is the feet of perpendicular from $A$ to $BC$, then, $C_1,C_2$ pass through $D$, Using (Lemma 1), Invert around the circle $\omega$ at $A$ with radius $\sqrt{AH \cdot AD}$ and then notice that $P,Q$ remain invariant, hence they lie on $\omega$ $\qquad \blacksquare$

Q2)
Epsilon 1.1, LOG by Titu wrote:
In $\Delta ABC$, Let the perpendicular from $B$ to $AC$ intersect circle with diameter $AC$ at points $P$ and $Q$, and Let the perpendicular from $C$ to $AB$ intersect circle with diameter $AB$ at points $R$ and $S$, then Points $P,Q,R,S$ are concyclic
Solution: From the above Q1) and (Lemma 1), $P,R$ lie on circle $\omega$ centered at $A$ with radius $\sqrt{AH \cdot AD}$, now notice that $AB \perp RS$ and center of circle with diameter lies on $AB$, hence, $AS=AR$ and similarly, $AP=AQ$ $\implies$ $S,Q$ also lie on $\omega$ $\implies$ Points $P,Q,R,S$ are concyclic $\qquad \blacksquare$

Q3)
Japan Junior MO Finals 2017 P5 wrote:
Let $ABC$ be an acute-angled triangle with orthocenter $H$. Let $D,E$ and $F$ be the feet of the altitudes from $A,B$ and $C$, respectively. The circumcircle of $ACD$ and $BE$ meet at $P$, the circumcircle of $ABD$ and $CF$ meet at $Q$, the circumcircle of $ABH$ and $DF$ meet at $S$, and the circumcircle of $ACH$ and $DE$ meet at $T$. Prove that $P,Q,S,T$ are concyclic.
Solution: From Q1) and (Lemma 1), we already have, $P,Q$ lie on a circle $\omega$ with radius $\sqrt{AH \cdot AD}$ centered at $A$, Perform Inversion $\Psi$ around $\omega$, then trivial to see, $\Psi : {B,D,C} \mapsto {F,H,E}$, and $P,Q$ remain invariant, $\Psi (S)=S^*=\Psi (\odot (ABH)) \cap \Psi (DF)=DF \cap \odot (ABH)=S$ and similarly, trivial to see, $S,R$ also remain invariant hence, lie on $\omega$ $\implies$ that $P,Q,S,T$ are concyclic $\qquad \blacksquare$

_____________________________________________________________________________________________________________________________________________________________________________________________________________
Let me know any other problem that involves some similar configuration! (below)
_____________________________________________________________________________________________________________________________________________________________________________________________________________

Define: $A'$ as the reflection of $A$ over $BC$
It's quite obvious that $A'$ lies on the $AH_A$

Some Properties

Lemma 1: Let $\odot (BHC) \cap AB,AC=D,E$ respectively, then, $H_B,H_C$ are midpoints of $AD,AE$
Proof: Notice, that the Simson Line of $\odot (BHC)$ WRT $E,D$ passes through $H_C,H_B$, hence, using the Simson Line Bisection Lemma, the Simson Line bisects $AE$ and $AD$ $\qquad \blacksquare$

Result #1: As a result, $AC=DC$ and $AB=BE$
Construct a circle $\omega_1$ and $\omega_2$ with center $B,C$ and radius $AB,AC$, then,
Result #2: $\omega_1$ and $\omega_2$ pass through $A'$
Result #3: $A'$ is the radical center WRT $\odot (ABC), \omega_1, \omega_2$

I'll write down some (trivial) properties....(I'm not at all in a mood to write their proofs down) HINT
Let $A'D,A'E$ intersect $\odot (ADE)$ at $X_1,Y_1$ and Let $A'D,A'E$ intersect $\odot (ABA'), \odot (ACA')$ at $X_2,Y_2$
Property 1: $X_1E||DY_1||BC$
Property 2: $AD=AY_1$, $X_1D=Y_1E$ and $AX_1=AE$
Property 3: $A$ is the mid-point of the arc $X_1DE, X_1Y_1E$
Property 4: Lines $AD,AY$ are Isogonal WRT $\angle X_1AE$
Property 5: $ED||H_BH_C||X_2Y_2$
Property 6: $X_2,Y_2$ lie on $HC.HB$
Property 7: $HA'$ is the angle bisector of $\angle DA'E$

The following are some problems that can be solved using these results and properties:
Q1) Indian RMO 2013 Region 4 P5In a triangle $ABC$, let $H$ denote its orthocentre. Let $P$ be the reflection of $A$ with respect to $BC$. The circumcircle of triangle $ABP$ intersects the line $BH$ again at $Q$, and the circumcircle of triangle $ACP$ intersects the line $CH$ again at $R$. Prove that $H$ is the incentre of triangle $PQR$.

Q2) Japan MO Finals 2018 P2 Given a scalene triangle $\triangle ABC$, $D,E$ lie on segments $AB,AC$ respectively such that $CA=CD, BA=BE$. Let $\omega$ be the circumcircle of $\triangle ADE$. $P$ is the reflection of $A$ across $BC$, and $PD,PE$ meets $\omega$ again at $X,Y$ respectively. Prove that $BX$ and $CY$ intersect on $\omega$.
_____________________________________________________________________________________________________________________________________________________________________________________________________________
Let me know any other problem that involves some similar configuration! (below)
_____________________________________________________________________________________________________________________________________________________________________________________________________________
This post has been edited 31 times. Last edited by AlastorMoody, Feb 12, 2019, 9:38 AM

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  • what a goat, u used to be friends with my brother :)

    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

    by kamatadu, Jan 3, 2023, 1:25 PM

  • Lots of good stuffs here.

    by amar_04, Dec 30, 2022, 2:31 PM

  • 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.

    by BVKRB-, Feb 1, 2022, 12:57 PM

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

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  • When are you going to br alive again ,we miss you

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  • nice blog :)

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  • Hello everyone, nice blog :)

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  • pro blogo

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