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
Reason: edited title

geometry

Consequences of Tangent Circles Configurations

by XmL, Dec 5, 2015, 6:30 AM

Proposition1(Sawayama): Circle $\omega$ is internally tangent to circle $\Omega$ at $P$ . $AC,BD$ are two chords of $\Omega$ that are tangent to $\omega$ at $Y,X$ respectively. Prove that the incenters of $ABC,BDC$ lie on $XY$ .

Lemma1: If $\frac {BX}{BP}$ is constant for $B\in \Omega$ , and $BX$ is tangent to $\omega$ at $X$ .

Let $PB\cap \omega=B'$ . $BX^2=BP'\cdot BP\implies \frac {BX}{BP}^2=\frac {BP'}{BP}=1-\frac {R(\omega)}{R(\Omega)}$ where $R(X)$ is the radius of circle $X$ , and this proves our lemma.

Correlary1: $PX$ bisects $\angle BPD$ by the converse of angle bisector theorem: $\frac {BX}{BP}=\frac {DX}{DP}$ .

Proof of our proposition: Let $I_1$ denote the incenter of $ABC$ , $BI_1\cap \Omega=E\ne B$ . By Correlary1 $E,Y,P$ are collinear since $E$ is the midpoint of arc $AC$ . By similarity, $\frac {EA}{EP}=\frac {AY}{AP}$ , which is the constant described in Lemma1, therefore $EA^2=EY\cdot EP$ , the power of $E$ wrt $\omega$ . Since $EI_1=EA$ , thus $\frac {EI_1}{EP}=\frac {DX}{DP}$ and $\triangle EPI_1\sim\triangle DPX$ $\implies BXP=\angle BI_1P$ . Since $\triangle EYI_1\sim \triangle EI_1P$ , therefore $\angle I_1YP=\angle BI_1P=\angle BXP=\angle XYP$ . The last angle equality implies $Y,I_1,X$ are collinear and we are done because the incenter of $BDC$ lies on $XY$ by symmetry.

Some properties I want to highlight and are trivial from the proof above:
Let $I_2$ denote the incenter of $BCD$ .
1. $\odot (E,EA)$ and $\omega$ are orthogonal
2. $XBPI_1, YCPI_2$ are cyclic quadrilaterals.
3. $\triangle PBI_1 \sim \triangle PXY \sim \triangle PI_2C$
4. $PI_1,PD$ are isogonal wrt $\angle XPY$ .
5. $B,C,I_2,I_1$ are concyclic.

Property6: $K$ is the midpoint of arc $BC$ that doesn't contain $A$ . Prove that $I_1,I_2,K,P$ are concyclic.

Proof: By property3 $PI_1,PI_2$ are isogonal wrt $\angle BPC$ . Since $PK$ externally bisects $\angle BPC$ , it also externally bisects $\angle I_1PI_2$ . As $KI_1=KI_2=KB$ by properties of incenters, therefore $PKI_2I_1$ is cyclic.

Correlary2: $PK,BC,XY$ are concurrent by radical axis theorem.

Property7: $M$ is the midpoint of arc $BC$ that does not contain $A$ . Prove that $PM\cap CD,X,Y$ are collinear. This is immediate by pascal's theorem.

Proposition 2: $\omega _1, \omega _2$ are internally tangent to $\Omega$ at $P_1,P_2$ . $AC,BD$ are chords of $\Omega$ that are also internal common tangents of $\omega _1, \omega _2$ like so in the diagram. Prove that the external tangent of these two circles closer to $DC$ than $AB$ is parallel to $DC$ .

Proof:

Let $M$ be the midpoint of arc $DC$ that doesn't contain $A$ . Define $H_1,Z_1=MP_1\cap \omega _1, DC$ and $H_2,Z_2$ similarly. Thus $Z_1Z_2$ is antiparallel to $P_1P_2$ wrt $\angle P_1MP_2$ . It suffices to show that $H_1H_2||Z_1Z_2$ , which implies $H_1H_2\perp (OM||O_1H_1||O_2H_2)\implies H_1H_2$ is the external tangent. We will show that $\frac {Z_1H_1}{Z_1M}=\frac {Z_2H_2}{Z_2M}$ .

$AC$ is tangent to $\omega _1, \omega _2$ at $Y_1,X_2$ , $DB$ is tangent to the same circles at $X_1,Y_2$ . By property7 we know $X_1,Y_1,Z_1$ are collinear. Thus $Z_1H_1\cdot Z_1P_1=Z_1Y_1\cdot Z_1X_1$ . In addition, $Z_1M\cdot Z_1P_1=CZ_1\cdot DZ_1$ . Dividing these two gives us $\frac {Z_1H_1}{Z_1M}=\frac {Z_1Y_1}{CZ_1}\cdot \frac {Z_1X_1}{DZ_1}$ . Since $X_1Y_1||X_2Y_2$ , the analogous expression for $\frac {Z_2H_2}{Z_2M}$ equals $\frac {Z_1H_1}{Z_1M}$ and we are done.

This post has been edited 2 times. Last edited by XmL, Dec 5, 2015, 6:36 AM
Reason: .

geometry

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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
Check your private massage
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

Consequences of Tangent Circles Configurations

by XmL, Dec 5, 2015, 6:30 AM