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

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Properties in a New Configuration Regarding Quadrilaterals

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

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