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

Still an immature proof writer :)

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  • Good blog! I hope to see more posts here!

    by WizardMath, Feb 3, 2018, 6:57 PM

  • Thanks Harry Potter :D

    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..:D

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  • Check your private massage :)

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