New Results for Orthocenters

by XmL, Dec 22, 2015, 7:30 PM

Given $ABC$ with isogonal conjugates $P,Q$ and orthocenter $H$ .

Property1: Reflect $P$ over $BC$ to get $P_A$ , Prove that the lines through $H,Q$ perpendicular to $AP,HP_A$ resp. are concurrent with $BC$ .

Proof: Denote the two lines defined in the statement $l,j$ resp. Reflect $H$ over $BC$ to get $H_A$ which lies on the circumcircle of $ABC$ , with center $O$ . Let $AQ\cap (O)=D\ne A$ . It is not hard to show that $l,H_AD$ are symmetric over $BC$ . If $j\cap (BQC)=E\ne Q$ , then it suffices to show $QE,BC,DH_A$ concur, which is equivalent to $H_A,D,Q,E$ are concyclic by the radical axis theorem

Let $F$ be the isogonal conjugate of $P$ wrt $BH_AC$ . By angle relations of conjugates, $F\in (QBC)$ . Define $E'=H_AF\cap (QBC)\ne F$ , thus $\angle (E'Q,E'H_A)=\angle (CQ,CF)=\angle (CA,CH_A)=\angle (DA,DH_A)$ because $\angle (CH_A,CF)=\angle (CP,CB)=\angle (CA,CQ)$ . Hence $H_AEQD$ is cyclic, and it suffices to show $E\equiv E'\iff QE'\perp HP_A$ .

Define $k$ the reflection of $\overline {OH_A}$ over $AH_A$ . Since $(H_AH,H_AO), (H_AE',H_AP)$ are isogonals wrt $\angle BH_AC$ , therefore $\angle (H_AO,H_AE')=\angle (H_AP,H_A)=\angle (H_AH,k)$ . Hence $\angle (H_AE,k)=\angle (H_AO,H_AH)$ $\implies \angle (E'Q,E'H_A)+ \angle (E'H_A,k)=90^{\circ}$ , i.e $k\perp QE'$ . Because $k\parallel HP_A$ , it has been shown that $QE'\perp HP_A \Box$ .

Property2: If $X=HP_A\cap AP$ , then $X$ lies on the radical axis of $(H,HP), (Q,QP_A)$ . The second circle is also known as the circumcircle of the reflection triangle of $P$ wrt $ABC$ .

Proof: It suffices to show that the intersections of $AP$ with $(H,HP)$ and $HP_A$ with $(Q,QP_A)$ are concyclic $\iff$ the perpendicular bisectors of the two intersection segments is concurrent with the perpendicular bisector of $PP_A$ , i.e $BC$ . This is equivalent to Property1, thus Property2 is proven. $\Box$

Property3: Define $X=HP_A\cap AP$ , $Y=HP_B\cap BP$ , $Z=HP_C\cap CP$ . Then $X,Y,Z$ are collinear and $HQ\perp \overline {XYZ}$ .

This is immediate from property2, since $X,Y,Z$ lie on the radical axis of $(H,HP), (Q,QP_A)$ . $\Box$

Theorem1: $R_{PQ}^2-HP^2-HQ^2=~pot~(H,(O))$ , where $R_{PQ}$ denote the circumradius of the reflection triangle of $P$ wrt $ABC$ .

Proof: Let the radical axis of $(H,HP), (Q,QP_A)$ intersect $QH$ at $K$ . Then $KH^2-HP^2=KQ^2-QP_A^2=KQ^2-R_{PQ}^2$ $\implies R_{PQ}^2-HP^2-HQ^2=KQ^2-KH^2-HQ^2=QH[KQ+KH-HQ]=2QH\cdot HK$ . If $AH\cap BC=D$ , then it suffices to show that $QH\cdot HK=\frac {~pot~(H,(O))}{2}=AH\cdot HD$ .

Let the lines through $H,Q$ perpendicular to $AP, HP_A$ resp. meet at $L$ ; from property1, $L\in BC$ . Define $M=AP\cap HL, N=HP_A\cap QL$ , $X=AP\cap HP_A$ , then $XK\perp QH$ by property3. Hence the multiple cyclic quadrilaterals give $QH\cdot HK=HN\cdot HX=HM\cdot HL=AH\cdot HD$ . $\Box$

This post has been edited 1 time. Last edited by XmL, Jul 28, 2017, 7:08 AM
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Another proof to (1) :

Let $M$ $\equiv$ $AP$ $\cap$ $BC,$ $P_a$ $\equiv$ $AP$ $\cap$ $\odot (ABC),$ $Q_a$ $\equiv$ $AQ$ $\cap$ $\odot (ABC)$ . From $\triangle ACQ_a$ $\sim$ $\triangle CMP_a$ and $\triangle CQQ_a$ $\sim$ $\triangle PCP_a$ we get $\frac{AQ}{QQ_a}=\frac{PM}{MP_a} .$ Let $D$ $\equiv$ $AH$ $\cap$ $\odot (ABC)$ and let $N$ $\in$ $BC$ be the orthocenter of $\triangle AHM$ . Let $R$ be the reflection of $P_a$ in $BC$ . Clearly, $HN,$ $HR$ is the Steiner line of $Q_a,$ $P_a$ WRT $\triangle ABC$ , respectively, so we get $HR$ $\perp$ $AQ_a$ and $D,$ $N,$ $Q_a$ are collinear, hence notice $MR$ $\perp$ $NQ_a$ (from symmetry) we get $\triangle ANQ_a \sim \triangle HMR.$ From the discussion above, if $S$ is the point on $HR$ s.t. $MS$ $\parallel$ $HP_A$ , then $\triangle ANQ_a$ $\cup$ $Q$ $\sim$ $\triangle HMR$ $\cup$ $S$ $\Longrightarrow$ $HP_A$ ( $\parallel$ $MS$ ) $\perp$ $NQ$ .

by TelvCohl, Dec 23, 2015, 8:36 AM

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Good blog! I hope to see more posts here!
by WizardMath, Feb 3, 2018, 6:57 PM
Thanks Harry Potter
by XmL, Jul 30, 2014, 10:47 PM
Hey dude. Nice blog. I like your work.
by gamjawon, Jul 30, 2014, 7:43 AM
Thanks...although most of the posts here won't be about geometry...rightnow I'm trying to improve my skills in other areas such as number theory and combo.
by XmL, Jun 5, 2014, 1:11 AM
Hi Geometer,,nice blog..
by vanu1996, Jun 4, 2014, 1:27 PM
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by zabihpourmahdi, Jan 11, 2013, 5:34 PM

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Geometry Treasures

New Results for Orthocenters

by XmL, Dec 22, 2015, 7:30 PM

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