Properties in a New Configuration Regarding Quadrilaterals

by XmL, May 30, 2016, 5:07 AM

Generalized Configuration: $ABCD$ is a quadrilateral. Define two circles $(O_1,r_1), (O_2,r_2)$ such that $A\cup B\in (O_1), C\cup D\in (O_2)$ and $\triangle ABO_1 \stackrel{+}{\sim} \triangle DCO_2$.

Note that the two circles do not necessarily have to intersect; we will discuss such cases at a later time. Also, when the notion of locus is mentioned below, interpret its "domain" as all circle pairs $(O_1), (O_2)$ that satisfy the conditions above.

Preliminary Notations: Let $AD\cap BC=E, AB\cap DC=F$. Let $M$ denote the Miquel point of $ABCD$. Define $\phi $ the spiral similarity transformation centered at $M$ that sends $AB\mapsto CD$. (I just realized I used $\phi$ twice here so please excuse that mistake)

Proposition 1: $M$ lies on the circle of similitude of $(O_1), (O_2)$.

Proof: By the definition of $O_1,O_2$, we know $O_1\mapsto O_2$ under $\phi $. Thus $\frac {MO_1}{MO_2}=\frac {r_1}{r_2}\implies M$ lies on the circle of similitude. $\Box$.

Proposition 2: Let $S_1,S_2$ denote the external and internal center of similitude of $(O_1), (O_2)$ respectively. Prove that the loci of $S_1,S_2$ (for all satisfactory $(O_1), (O_2)$ are straightlines that are perpendicular.

Proof: Because $\triangle MO_1O_2\stackrel{+}{\sim} \triangle MAD$, therefore $MO_1:MO_2:MS_1:MS_2$ is constant; generally the shape of $MO_1O_2\cup S_1\cup S_2$ stays the same up to similarity. Let $l_1$ be the perpendicular bisector of $AB$ and consider the the spiral similarity transformation $(M, \angle (MO_1, MS_1), \frac {MS_1}{MO_1})$. Suppose $l_1\mapsto s_1$ under the transformation; clearly $S_1\in s_1$. Since the transformation is independent of $(O_1), (O_2)$, therefore $s_1$ is a fixed line and the locus of $S_1$. Similarly define $s_2$ as the image of $l_1$ under $(M, \angle (MO_1, MS_2), \frac {MS_2}{MO_1})$, we know $s_2$ is the locus of $S_2$. Note that \[\angle (s_1,s_2)=\angle (s_1,l_1)+\angle (l_1,s_2)=\angle (MS_1,MO_1)+\angle (MO_1, MS_2)=\angle (MS_1,MS_2)=90^{\circ}\]thus $s_1\perp s_2$. $\Box$

Let $s_1\cap s_2=K$, then the perpendicularity implies that $K$ lies on the circle of similitude. Hence we have the following Corollary:

Corollary 1: The circle of similitude of $(O_1), (O_2)$ passes through two fixed points, namely $M$ and $K$.

Proposition 3: $F\in MK$

Proof: Let $O_S$ denote the center of the circle of similitude. Again since $O_S\cup \triangle MO_1O_2$ is invariant up to similarity, the locus of $O_S$ is the perpendicular bisector of $MK$ which satisfies \[\angle (l_1,l_s)=\angle (MO_1,MO_S)=\angle (O_2O_1, MO_2)\]where the latter follows from $MO_S$ tangent to $(MO_1O_2)$. Because \[\angle (O_2O_1, MO_2)=\angle (AD,MD)=\angle (FA,MF)\]as $M,A,D,F$ are concyclic, it follows that $l_s\perp MF\implies MF||MK\implies F\in MK$. $\Box$.

Lemma 1: Let $M_1$ be the reflection of $M$ over the perpendicular bisector of $AB$. Let $M_2$ be the reflection of $M$ over the perpendicular bisector of $DC$. Prove that $E\in M_1M_2$ and $M_1M_2$ is parallel to the midline of $ABCD$ (Here I will force the term midline to mean the Newton-Gauss line).

Proof (for now): Clearly $M_1\in (EAB), M_2\in (ECD)$ and $M_1\mapsto M_2$ under the spiral similarity transformation centered at $M$ such that $(MAB)\mapsto (MDC)$. It follows that $M_1M_2$ must pass through $E$.

For the second assertion, let $H_1 H_2$ be the orthocenter of $\triangle EAB, \triangle EDC$ respectively. It is not hard to show that $\triangle EH_1H_2\stackrel{+}{\sim} \triangle MM_1M_2$, thus $EH_1\perp MM_1\implies H_1H_2\perp M_1M_2$. Since $H_1H_2$ is also perpendicular to the midline of $ABCD$, therefore $M_1M_2$ is parallel to the midline. $\Box$.

Proposition 4: $EK$ is parallel to the midline of $ABCD$.

Proof: Suppose we reflect $M$ across the perpendicular bisector of $AB,DC$ to obtain $M_1,M_2$. It is not hard to see that $\triangle MKM_1\sim \triangle MO_SO_1, \triangle MKM_2\sim \triangle MO_SO_2$, thus $MKM_1M_2\sim MO_SO_1O_2\implies M,M_1,M_2$ are collinear. By Lemma 1, $E\in M_1M_2$, thus $\overline {EKM_1M_2}$ is parallel to the midline of $ABCD$. $\Box$

Lemma 2: $ABCD$ is a quadrilateral. Define $(O_1), (O_2)$ as in the introduction. Let $L$ be a point outside $ABCD$ such that $\triangle LAD\sim \triangle LCB$(note that they are inversely similar). Prove that the radical axis of $(O_1), (O_2)$ passes through $L$.

Proof: Let $LB\cap (O_1)=E\ne B, LC\cap (O_2)=F\ne C$. By the construction of $(O_1),(O_2)$ we have $\angle LEA=\angle LFD$. Because $\angle DLA=\angle CLB$, therefore $\angle ELA=\angle FLD$. Hence $\triangle LEA\sim \triangle LFD\implies \triangle LAD\sim \triangle LFE\implies FEBC$ is cyclic. It follows that $L$ is the radical axis of the two circles. $\Box$

Main result 1: The radical axis of $(O_1), (O_2)$ passes through a fixed point. In particular, the fixed point is the intersection of $ME$ with the line through $F$ parallel to the midline of $ABCD$.

Proof: Let $L\in ME$ such that $LF$ is parallel to the midline. already gave a characterization of the fixed point, so it suffices to prove that $\triangle LAD\sim \triangle LCB$. We first show that $A,C$ are isogonal conjugates wrt $\triangle LFM$.

By properties of Miquel point it's clear that $MA,MC$ are isogonals wrt $\angle EMF$ and consequently $\angle LMF$. Define $M^*$ on $(MADF)$ such that $MM^*||AD$. By Lemma1 we know $M^*\in LF$. Since $AM=DM^*$, it follows that $FA,FD$ are isogonals wrt $\angle LFM$. Hence $A,C$ are isogonal conjugates of $\triangle LFM$. Similarly we can show that $B,D$ are isogonal conjugates.

It is a property of isogonal conjugates that \[\frac {LA}{LC}=\frac {\sin \angle MAF}{\sin \angle MCF}, \frac {LD}{LB}=\frac {\sin \angle FDM}{\sin \angle MBF}\]thus \[\frac {LA}{LC}=\frac {LD}{LB}\]. From the isogonals we can obtain $\angle ALD=\angle BLC$; it follows that $\triangle LAD\sim \triangle LCB$. $\Box$

Let $(I)$ be the circle that passes through $E$ and is coaxal with $(O_1), (O_2)$. We will develop some more properties before proving the following main results:

Main result 2: $(I)$ passes through another fixed point on $EF$
Main result 3: $\angle IMO_1=\angle EFA$.

It seems the most natural to investigate $(I)$ using inversion. Let $\phi$ denote the inversion through $E$ with radius $\sqrt {EM\cdot EL}$. Besides $M\mapsto L$, other existing points $X$ get mapped to $X'$.

It is clear that $A'B'\cap D'C'=L$. Also, the Miquel point of $A'B'C'D'$ is $\phi (F)=F'$. Now let the inversion transform $(O_1), (O_2)$ into $(O_1'), (O_2')$. We can check with angle properties of inversion that these two circles still satisfy the special conditions stated in the introduction. This prompts us to construct $S$, the analog of $L$ wrt $ABCD'$ in $A'B'C'D'$, i.e. $S\in EF$ such that $LS,LF'$ are isogonals wrt $\angle A'LC'$.

From Main result 1 we know $S$ lies on the radical axis of $(O_1'), (O_2')$. However, $(I)$ is mapped to this radical axis under $\phi$; therefore $\phi (S)=S'=(I)\cap EF\ne E$. Since the construction of $S'$ is independent of $(O_1), (O_2)$, therefore it is the fixed point on $(I)$. We have just proven Main result 2.

Lemma 3: $ABCD$ is a quadrilateral, Let $AD\cap  DC=E, AB\cap DC=F$ and $M$ the Miquel point of $ABCD$. If $O$ is the circumcenter of $EDC$ and $h$ is the line through $E$ perpendicular to $AB$. Prove that $\angle (EO, h)=\angle (ME,MF)$.

Proof: Let $MF\cap (EDC)=P\ne M$. Note that \[\angle (PE,PM)=\angle (DE,DM)=\angle (DA,DM)=\angle (FA,FM)\], thus $PE||AB$. Let $PO\cap (EDC)=P'\ne P$, thus $\overline {EP'}\equiv h$. Hence \[\angle (EO,h)=\angle (P'E,P'O)=\angle (P'E,P'P)=\angle (ME,MP)=\angle (ME,MF)\]. $\Box$

Proposition 5: Let $D'C'\cap AB=G\ne L, A'B'\cap DC=H\ne L$. Prove that $G,H\in (LFM)$.

Proof: $G\in (LFM)\iff \angle (GL,GF)=\angle (ML,MF)$. Define $O,h$ as in Lemma 3, then $h\perp AB, EO\perp D'C'$. Thus by Lemma 3, $\angle (GL,GF)=\angle (EO, h)=\angle (ML,MF)$ and we are done(Note that we can show $H\in (LFM)$ in an analogous fashion. $\Box$

Let $(EMS')\cap AD, BC=U,V$ respectively.

Proposition 6: $\angle UMA=\angle VMB=\angle EFA$

Proof: Suppose $\phi (U)=U'$. Because $E,U,S',M$ are concyclic, therefore $U'\in LS$. Thus $\angle UMA=\angle A'LU'$ (property of inversion). Since $LS,LF'$ are isogonals wrt $\angle A'LC'$ and $G\in (LMFF')$ by Proposition 5, we have \[\angle UMA=\angle A'LU'=\angle F'LG=\angle F'FG=\angle EFA\]We can similarly show that $\angle MVB=\angle EFA$. $\Box$.

It follows from Proposition 6 that $\triangle MUA\sim \triangle MVB$, thus there exists a spiral similarity transfomation about $M$ that sends $AB\mapsto UV$. Explicitly, the transformaiton is $(M, \angle (MA,MU), \frac {MU}{MA})$. Note that\[\angle (AB,UV)=\angle (MA,MU)=\angle (AB,EF)\], thus $EF||UV$. Since $U,V\in (MES')$, it follows that $ES'UV$ is an isosceles trapezoid.

Now suppose $O_1\mapsto I'$ under the spiral similarity defined above; it is clear that $I'\in O_1O_2$. In addition, $I$ lies on the perpendicular bisector of $UV$, and thus $ES'$. Therefore $I\equiv I'$ and we have proven that $IMO_1=\angle EFA$, which is Main result 3.

Ending Remarks: The motivation to study this configuration came from the content of the previous post. I suspected that it would contain many fixed points and was lucky enough to find them by experimenting on my drawing software. This post examines the significance of the point $L$, whose definition makes me want to name it "anti-Miquel point". The connection of $L$ with the vertices of the quadrilateral as isogonal conjugates is really surprising. The circle $(I)$ is of great interest in the second part of the post. I seldem use inversion, but I realized the equivalence between the fixed point of $(I)$ and $L$ and it made the use of inversion natural (it also supports the post's organization.) Main result 3 is truly beautiful and I wonder if any proof exists without developing $L$.

Acknowledgement: I want to thank User Telv Cohl for reviewing the content of this post.
This post has been edited 7 times. Last edited by XmL, Jun 7, 2016, 11:16 PM
Reason: Telv

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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

    by vanu1996, Jun 4, 2014, 1:27 PM

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    by zabihpourmahdi, Jan 11, 2013, 5:34 PM

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