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

Generalization to A Serbian Geometry Problem

by XmL, May 24, 2016, 5:06 AM

Generalization: Let $C'\in AB,B'\in AC$ such that $B,C,B',C'$ are concyclic. Suppose $\omega, \omega'$ are two circles through $B,C$ and $C',B'$ respectively, such that they are corresponding circles between similar triangles $ABC, AB'C'$ . If $\omega \cap \omega'=X,Y$ , prove that $AX,AY$ are isogonals wrt $\angle BAC$ .

Note: Another way to characterize $X,Y$ is the property $\angle BXC+\angle B'XC'=180^{\circ}, \angle BYC+\angle B'YC'=180^{\circ}$
Solution 1(Indirect):

Let $\phi$ denote the inversion transformation $\phi (A, \sqrt {AB'\cdot AC})$ , then $BC\mapsto C'B'$ under $\phi$ .

Suppose $\phi (Y)=Y'$ . Let $X'$ denote the isogonal conjugate of $Y'$ wrt $ABC$ and $\phi (X')=X''$ . We can show through angle chasing that $\angle BX'C=\angle BYC\implies X'\in \omega$ . By more angle chasing we can obtain $X''\in (BY'C)$ or $X''\in \phi (\omega')$ . Since $X''\in \phi (\omega)$ (by inversion's preservation of concylic points), therefore $X''=\phi (\omega)\cap \phi (\omega')\ne Y'$ . Hence $X'=\omega\cap \omega'\ne Y$ , so $X'=X$ and $AX,AY$ are isogonals. $\Box$

Solution 2(Direct, at last):

Lemma: Two circles $(O_1,r_1), (O_2,r_2)$ intersect at two points $A,B$ . Given a point $P$ that satisfies $\frac {PO_1}{PO_2}=\frac {r_1}{r_2}$ and doesn't lie on $O_1O_2$ . Prove that $PA,PB$ are isogonals wrt $\angle O_1PO_2$ .

Proof: Let $D$ denote the foot of the bisector of $\angle O_1PO_2$ on $O_1O_2$ . Thus the $P$ -Apollonius circle of $\triangle PO_1O_2$ (aka the circle of similitude of $(O_1),(O_2)$ passes through $P,A,D,B$ . Since $AD=BD$ , $PD$ also bisects $\angle PAB$ . Hence the lemma is proven.

Main proof: Since $ABC\cup \omega \sim AB'C'\cup \omega'$ , therefore if $\omega =(O, r), \omega' =(O',r')$ , then $\frac {AO}{AO'}=\frac {r}{r'}\implies A$ lies on the circle of similitude of $\omega, \omega'$ . By the Lemma above, this means $AX,AY$ are isogonals wrt $\angle OAO'$ . Becuase $AO,AO'$ are isogonals wrt $\angle BAC$ , therefore $AX,AY$ are isogonals wrt $\angle BAC$ . $\Box$

This post has been edited 1 time. Last edited by XmL, May 24, 2016, 5:07 AM
Reason: .

More on The Kosnita Point

by XmL, Apr 25, 2016, 10:09 PM

In this post, we dive into the realm of Kosnita point to examine its properties and relations with other special points. We will look at many interpretations of the point and how it plays a role with respect to different triangles. In particular, we will disect the significance of the intersections of line $KN$ with the circumcircle of $ABC$ .

We will make $ABC$ our main reference triangle, $O,H,N,K$ are respectively the circumcenter, orthocenter, nine-point center, and Kosnita point of $ABC$ . I will also label the propositions with trivial or non-trivial. This is done for personal note so do not be alarmed if it does not appear that way.

Section 1 Key Configuration

Proposition 1(non-trivial): Let $K_AK_BK_C$ denote the reflection triangle of the Kosnita point $K$ of $ABC$ . Reflect $K$ about the midpoints of $K_BK_C$ to get $K_1$ ; define $K_2, K_3$ similarly. Surprisingly, it is true that $AK_1\perp BC$ .

Proof: Let $O_A$ be the reflection of $O$ over $BC$ . Define $AO_A\cap (O)=D\ne A$ . The thing that makes Kosnita special is that we have $BCDO_A\sim K_CK_BK_1A$ . The perpendicular now follows because $\angle K_1AO_A=180-\angle OO_AA\implies (AK_1||OO_A)\perp BC$ . $\Box$

A immediate consequence of this is that triangles $ABC, K_1K_2K_3$ are perspective at $H$ . In addition, we have $K_1K_2K_3H\cong K_AK_BK_CK$ and particularly $HK_A|| KK_1$ . These facts will become important later in discussions about the perpendiculars to $HK_A$ .

Of course, if you've read my previous blogs, then these results are nothing new and were used in my proof for the infamous $IO$ problem. In my collaborated solution with Ivan Zelich, one of our Lemmas was proving $HK_A\perp AG$ , where $G$ is the intersection of $BC$ with the side of the Ceva triangle of $N$ that faces $A$ . This fact, however, is immediate from our observations above by considering the harmonic pencil $K(K_B,K_C;P_{\infty}, K_1)$ (point at infinity is on $K_BK_C$ ) and its perpendicular counterpart through $A$ .

Section 2 Perspective at $H$

For a moment we will take Kosnita out of the picture and look at somethings more intrinsic to $H$ :

Proposition 2(non-trivial): Let $X$ be on $BC$ such t hat $HX||AO$ ; define $Y,Z$ similarly. Prove that $XYZ$ and the tangential triangle of $ABC$ are orthologic to each other at $H$ , the orthocenter of $ABC$ .

Proof: It suffices to show that $HA'\perp YZ$ , where $A'$ is the intersection of the tangents to $(O)$ through $B,C$ . Perhaps unexpectedly, I will call upon poles and polar to show this.

Let $DEF$ be the orthic triangle of $ABC$ , so $H$ is the center of the incircle $\omega$ of $DEF$ . Let $E',F'$ denote the poles of $AC,AB$ wrt $\omega$ . The pole of $YZ$ is a point $P$ that satisfies $PE'||DF||A'B, PF'||DE||A'C$ . It is not hard to see that $E'F'||BC$ ; thus $PE'F', A'BC$ are homothetic through H $\implies \overline {A'HP}\perp YZ$ and we are done. $\Box$

Remark: $XYZ$ is also the Ceva triangle of the isotomic conjugate of $O$ wrt $ABC$ . Also, this might not be the easiest way to prove something about $HA'$ since there are more natural constructions that are related to that line.

Let $O_AO_BO_C$ be the reflection triangle of $O$ wrt $ABC$ . Suppose $O_AO_C\cap AB=L, O_AO_B\cap AC=J$ . Since $AJO_AL$ is clearly a parallelogram( $ABC, O_AO_BO_C$ are congruent and homothetic through $N$ ), thus $JL\cap AO_A=N$ . We now show a result that can be generalized:

Proposition 3(Trivial): $LJ||YZ$ .

Proof: Note that $AO_B||HO_C, AC||O_CO_A$ , thus $\angle O_BAJ=\angle ZO_CL, \angle AJO_B=\angle O_CLZ \implies AJO_B\sim O_CLZ$ $\implies AJ\cdot LZ=JO_B\cdot LO_C$ . Similarly, we can show that $AL\cdot JY= JO_B\cdot LO_C$ . Thus $AJ\cdot LZ=AL\cdot JY$ , which means $ZY||LJ$ . $\Box$

We now bring back the Kosnita point $K$ . Recall in our first discussion that $HK_A\perp AG$ , where $G$ is defined as follows: Let $BN\cap AC=B_N, CN\cap AB=C_N$ , $B_NC_N\cap BC=G$ . Since $A(C,B;G,N)=-1$ and $LN=NJ$ , by properties of harmonic division we have $LJ||AG$ . Because $HA'\perp (YZ||LJ)$ , we have $HA'\parallel HK_A\implies H,A',K_A$ are collinear. In other words, we have just proven the following:

Main Result 1(Non-trivial): $K_AK_BK_C$ , the reflection triangle of Kosnita point $K$ , is perspective with the tangential triangle of $ABC$ at $H$ .

Section 3 $KO$ and $P_K$

We now go deeper into how Kosnita relates to various triangles. With just Proposition 1, we can actually prove:

Main Result 2(Non-trivial):The orthocenter of $K_AK_BK_C$ , denoted by $H_K$ , lies on $OK$ . In fact, $K$ is the midpoint of $OH_K$ .

Proof: By properties of isogonal conjugates, $N$ is the circumcenter of $K_AK_BK_C$ . Let $S$ denote the midpoint of $K_BK_C$ . Reflect $H$ over $S$ to get $H'$ . Thus $OH'=2SN=K_AH_K$ , $H'K=HK_1=KK_A$ , and the corresponding parallels give us $KH'O\cong KK_AH_K\implies O,K,H_K$ are collinear and $KO=KH_K$ . $\Box$

A corollary of this is, if $H'$ is the reflection of $K_A$ over $K$ , then $OH'||AN$ . The construction of $H'$ is analogous to the perspective approach in proving the Liang-Zelich Theorem, so it is safe to say that $H'$ is a very special point and we can show the following:

Proposition 4(Trivial): $A,K,O,H'$ are concyclic and the isogonal of $AH'$ wrt $\angle BAC$ is parallel to $KO$ .

Proof: We showed that $OH'||AN$ , so $\angle (H'K,OH')=\angle (HA, AN)=\angle (AK,AO)$ because $AK,AN$ are isogonal wrt $\angle HAO$ . Thus the cyclic follows. Suppose $\ell, AH'$ are isogonal wrt $\angle BAC$ . Since $AH,AO$ are isogonals of the same angle, we have $\angle (KH',KO)=\angle (AH',AO)=\angle (AH, \ell)$ $\implies \ell || KO$ because $KH'||AH$ . $\Box$

If we construct the analogs of $AH'$ for $B,C$ , then they all concur on $(O)$ at $P_K$ , specifically at the isogonal conjugate of the point at infinity on $KO$ (this then implies that the Simson's line of $P_K$ wrt $ABC$ is perpendicular to $KO$ .) Furthermore, $P_K, K,N$ are collinear by Sondat's Theorem. To see this, relabel $H'$ as $K_A'$ and define $K_B', K_C'$ similarly. We showed that $K_A',K_B'K_C'$ and $ABC$ are perspective at $P_K$ (which is actually immediate by Liang-Zelich). Since $N,K$ are the clearly the orthologic centers for the two triangles, Sondat's gives $P_K\in KN$ . What a found! The following result summarizes this discussion:

Main result 3(non-trivial): $P_K, N,K$ are collinear, where $P_K$ is the isogonal conjugate of the point at infinity on $PK$ wrt $ABC$ .

To prepare the construction for investigating the point $P_K$ , we need the following:

Lemma(Trivial): $O$ is the circumcenter of $ABC$ . Let $P$ be an arbitrary point. Define $D,E,F$ on the circle $\odot (O,OP)$ such that $PD\perp BC, PE\perp AC, PF\perp AB$ . Prove that the orthologic center of $ABC$ wrt $DEF$ lies on $(ABC)$ and is the isogonal conjugate of the point at infinity on $PO$ wrt $ABC$ .

Proof: Let $\ell$ denote the line through $A$ and parallel to $PO$ , it suffices to show that the isogonal of $\ell$ wrt $\angle BAC$ is perpendicular to $EF$ . Select a point $Q\ne A\in \ell$ and let $Y,Z$ be its feet of projection on $AC,AB$ . By property of isogonals, it amounts to showing $ZY||EF$ , which is equivalent to $QYZ, PEF$ are homothetic. If we let $PO\cap \odot (O,OP)=R\ne P$ , then with $PO||\ell$ we can easily show that $QZAY\sim PFRE$ and the homothety follows. $\Box$

We now introduce the crux construction:

Proposition 5(non-trivial): Reflect $A$ over $K_BK_C$ to get $D$ ; similarly construct $E,F$ . From our first discussion, it is clear that $D\in KK_A$ . We will now show that $DEFK$ is a cyclic quadrilateral with center $H_K$ .

Proof: It suffices to prove that $DH_K=KH_K$ . Reflect $H_K$ over $S$ , the midpoint of $K_BK_C$ to get $T$ . Since $N$ is the circumcenter of $K_AK_BK_C$ (isogonal conjugates), we know $K_AT$ is a diameter of $(K_AK_BK_C)$ $\implies N$ is the midpoint of $K_AT$ . Note that $N$ is also the midpoint of $AO_A$ , so $DH_K=AT=K_AO_A=KO=KH_K$ and we are done. $\Box$

Now consider the Lemma with $P=K$ . Observe that since $O,H_K$ are reflexive about $K$ , the corresponding triangle $DEF$ in the Lemma is congruent and homothetic to $DEF$ in the current diagram. Because $P_K$ is the isogonal conjugate of the point ot infinity on $KO$ wrt $ABC$ , therefore $P_K$ is the orthologic center of $ABC$ wrt $DEF$ . In particular, $EF\perp AK_A'$ $\implies EF\perp DH\implies H$ is the orthocenter of $DEF$ .

It is not hard to see that $DEFK\sim ABCP_K$ . Hence we have

Main result 4(Very Non-trivial): $KHH_K\sim P_KHO$

From this we can obtain some pretty extraordinary facts:

1. $\frac {OH}{R}=\frac {HH_K}{KH_K}=\frac {2KN}{KO}\implies \frac {ON\cdot OK}{KN}=R$ , where $R$ is the circumradius of $ABC$ .
2. $\angle HKO=180-\angle OP_KH$
3. We can avoid Sondats and use this to deduce $P_K,K,N$ are collinear
4. $OHP_KK$ is similar to well known configuration, in which $(HKP_K), (OKP_K)$ are tangent to $HO$ . It is surprising for a bunch of special points to exhibit this beautiful property!

To end this section, we reveal the identity of the second intersection between $NK$ and $(ABC)$ :

Main Result 5(Non-trivial):Let $KN\cap (ABC)=E_r\ne P_K$ . We claim that $E_r$ is the Euler reflection point of $ABC$ , i.e. $\overline {E_rH_A}, \overline {OH}$ are reflexive over $BC$ .

Proof: Since $AO$ is parallel to the reflection of $\overline {OH_A}$ over $BC$ , thus it suffices to show that $\angle AOH=\angle OH_AE_r=\angle OE_rH_A$ .

We have shown before that $\angle P_KAH_A=\angle AOK$ , thus $\angle P_KE_rH_A=\angle H_AAP_K=\angle AOK$ . By the tangent result (fact 4) above, we have $\angle OE_rP_K=\angle OP_KN=\angle KON$ ; hence our claim follows. $\Box$

This post has been edited 6 times. Last edited by XmL, May 19, 2016, 4:10 AM
Reason: Thank You Telv!

Kosnita

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

Number Theory Problems

by XmL, Oct 8, 2014, 10:52 PM

Problem: Find all polynomials $f(x)$ with integer coefficients such that $f(n)$ and $f(2^{n})$ are co-prime for all natural numbers $n$

Solution

Not DONE

This post has been edited 2 times. Last edited by XmL, Oct 10, 2014, 1:42 PM

A Friend's Geometric Inequality(or is it?)

by XmL, Aug 20, 2014, 4:37 AM

If $a,b,c$ are the sides of a triangle, prove the inequality:
$\frac {a^2+b^2+c^2}{16}\Sigma \frac {a^2}{M_b^2M_c^2}\ge 1$ .

Applying the median length formula should be the obvious first step:

LHS= $\frac {a^2+b^2+c^2}{16}\Sigma \frac {a^2}{(\frac {a^2+b^2}{2}-\frac {c^2}{4})(\frac {a^2+c^2}{2}-\frac {b^2}{4})}$

$=(a^2+b^2+c^2)\Sigma \frac {a^2}{(2a^2+2b^2-c^2)(2a^2+2c^2-b^2)}$

To make things easier we apply substitution $x=a^2,y=b^2,z=c^2$ :

$=(x+y+z)\Sigma \frac {x}{(2x+2y-z)(2x+2z-y)}$

Since the expression is homogenous, we can let $x+y+z=1$ , so the left-hand side becomes:

$=\Sigma \frac {x}{(2-3z)(2-3y)}=\frac {-3\Sigma x^2+2\Sigma x}{(2-3x)(2-3y)(2-3z)}=\frac {-3\Sigma x^2+2}{-4+18\Sigma xy-27xyz}$

Recall that $(2-3x)(2-3y)(2-3z)>0$ since the terms each represents a length, so after some expansion and rearranging the inequality is equivalent to:

$9xyz+2\ge 6\Sigma xy+\Sigma x^2=4\Sigma xy+1$

$\Leftrightarrow 9xyz+(x+y+z)^3\ge 4(x+y+z)\Sigma xy$

$\Leftrightarrow x^3+y^3+z^3+3xyz\ge \Sigma x^2y$

which is true by schur's. $_\Box$ .

This post has been edited 2 times. Last edited by XmL, Aug 20, 2014, 7:01 AM

A New Proof to the IO Nine-point center Problem

by XmL, Aug 17, 2014, 12:17 AM

In this Blog I will present a new synthetic proof to a problem that IDMasterz and I had worked on a while ago. The old proof can be found here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=550000

Problem: The nine-point center of the incentral triangle of a triangle lies on the line connecting the incenter and the circumcenter of the triangle.

Before we begin, I will get the preliminary notations out of the way: $\triangle ABC$ will be the main triangle. $H,O,N,K$ are respectively the orthocenter, cirumcenter, nine-point center, kosnita's point(Isogonal conjugate of $N$ ) of $\triangle ABC$ .

Lemma 1: Let $\triangle A_KB_KC_K$ be the pedal triangle of $K$ , the Kosnita's point $K$ of $ABC$ . Prove that the Euler line of $\triangle AC_KB_K$ is perpendicular to $BC$ .

Proof: Since $AK$ is the diameter of circumcircle of the cyclic quadrilateral $AC_KKB_K$ , therefore the midpoint of $AK$ ,denoted by $O_{Ka}$ , is the circumcenter of $\triangle AC_KB_K$ . It's well known that the orthocenter of the triangle, denoted by $H_{Ka}$ is the reflection of $K$ over the midpoint of $C_KB_K$ , hence if we reflection $O_{Ka}$ over $C_KB_K$ to obtain $O_{Ka}'$ , then $KO_{Ka}'H_{Ka}O_{Ka}$ is a parallelogram $\Rightarrow KO_{Ka}'\parallel O_{Ka}H_{Ka}$ $\Rightarrow$ now it suffices to prove $O_{Ka}'\in KA_K$ because $KA_K\perp BC$ .

Let $AN\cap (O)=A_N$ , $O'$ is the reflection of $O$ over $BC$ . It's a property of the nine-point center that $O'\in AN$ . Since $AN,AK$ are isogonals wrt $\angle BAC$ , and $\angle C_KO_{Ka}'B_K=\angle BO'C$ we can see that $BCA_NO'\sim B_KC_KKO_{Ka}'$ $\Rightarrow \angle C_KKO_{Ka}'=\angle CA_NO'=180-\angle B$ $=\angle C_KKA_K \Rightarrow O_{Ka}'\in KA_K$ which is what we need to prove. $\Box$

To discover more properties regarding the current diagram, we similarly define $O_{Kb},O_{Kc},H_{Kb},H_{Kc}$ . Note that since $\triangle O_{Ka}O_{Kb}O_{Kc}$ is the result of the homothety $(K,\frac {1}{2})$ applied to $\triangle ABC$ . Hence the midpoint of $HK$ , denoted by $H_K$ is the orthocenter of $\triangle O_{Ka}O_{Kb}O_{Kc}$ . Moreover, recall that $H_{Ka}$ is simply the reflection of $K$ over the midpoint of $C_KB_K$ and likewise for $H_{Kb},H_{Kc}$ , hence by congruence $\triangle A_KB_KC_K$ , $\triangle H_{Ka}H_{Kb}H_{Kc}$ are homothetically congruent. By Lemma 1 we know that $H_K\in O_{Ki}H_{Ki}$ with $i=a,b,c$ , therefore $H_K,K$ are two corresponding points wrt the two congruent triangles, hence $KA_KH_KH_{Ka}$ is a parallelogram.

With these properties in mind, it's time to prove this big result:

Lemma 2: Let $N_K$ denote the nine-point center of $A_KB_KC_K$ , the pedal triangle of the $K$ , prove that $KN_K$ is parallel to the Euler line of $\triangle ABC$

Proof: Let $M$ denote the midpoint of $C_KB_K$ . It's a well known property of isogonal conjugates that the circumcenter of $A_KB_KC_K$ ,denoted by $O_K$ , is the midpoint of $KN$ .

We apply homothety $(A_K,2)$ and it maps $N_K,K\rightarrow O_K',K'$ . Hence $O_K'$ is the reflection of $O_K$ over $C_KB_K$ . Recall our newly discovered property that $KA_KH_KH_{Ka}$ is a parallelogram, which implies $MK\parallel H_KA_K, 2MK=H_KA_K$ , since $K'K=KA_K$ , therefore $M$ is the midpoint of $N_KK'$ $\Rightarrow O_K'K'O_KH_K$ is a parallelogram. Recall that $H_K$ is the midpoint of $KH$ , hence $H_KO_K$ is the $K-$ midline of $KHN$ and altogether we have $KN_K\parallel K'O_K'\parallel H_KO_K\parallel NK$ , the Euler line of $\triangle ABC$ and we are done. $\Box$

Remark: In addition to the parallelism, we've also proven the metric property that $4KN_K=HN$ or $8KN_K=HO$ .

Main Proof: Now let us consider the orthic triangle $DEF$ of $\triangle ABC$ , let $A_1B_1C_1$ be the cevian triangle of $H$ wrt $DEF$ (So $A_1=AD\cap EF$ and etc.) It's well known that $B_1C_1\perp AN$ and since $B_KC_K\perp AN$ , therefore $B_1C_1\parallel B_KC_K$ and consequently $\triangle A_1B_1C_1, \triangle A_KB_KC_K$ are homothetic with respective corresponding points $H,K$ . Let $N^*$ denote the nine-point center of $\triangle A_1B_1C_1$ . By Lemma 2, $HN^*\parallel KN_K\parallel HN$ $\Rightarrow N^*$ lies on the Euler line.

Note that $H,N$ are the incenter and circumcenter of $\triangle DEF$ , therefore we've proven the original problem. $\Box$ .

Nine-point Center is awesome

by XmL, Aug 15, 2014, 9:43 PM

As we all know, a lot of properties concerning the nine-point center are hard to prove synthetically. An example of this can be found here: http://www.artofproblemsolving.com/Forum/viewtopic.php?f=46&t=550000&p=3549742#p3549742. Anyway I've stumble upon a few more while working on a problem that Arab gave me(Correlary 3) and here they are:

We start with a pretty awesome Generalization:

Generalization 1: $A'B'C',DEF$ are the medial, altitude triangle of $ABC$ , it's well known that the nine-point circle is their circumcircle. Let $P\in l$ with $l$ the Euler line of $ABC$ , $XYZ$ is the circumcevian triangle of $P$ wrt $A'B'C'$ . Prove that $XYZ$ is the circumcevian triangle of some point on $l$ wrt $DEF$ .

Proof: Let $XYZ$ be the circumcevian triangle. It's well known that $FB'\cap EC'=W\in$ the Euler line of $ABC$ , hence by Pascal's theorem $FZ\cap EY\in PW=l\Rightarrow DX,EY,FZ$ concur on the Euler line. $\Box$ .

With this Generalization in mind, we examine the special case where $P$ becomes the nine-point center of $A'B'C'$ . If we let $N$ denote the nine-point center of $ABC$ , then $AN\parallel A'X$ . It's easy to prove by Ceva's theorem that $DX,EY,FZ$ concur implies $AX,BY,CZ$ concur.

Correlary 1: The reflection of the $X$ -altitude over $XN$ concur, where $X=A,B,C$ .

Proof: Let $H_A$ be the foot of $A$ -altitude on $BC$ , we will show that $\angle NAX=\angle NAH_A$ and our assertion follows. Let $O_A$ denote the reflection of the circumcenter of $ABC$ over $BC$ , hence $N$ is the midpoint of $AO_A$ and that $AX,O_AA'$ are reflesive over the perpendicular bisector of $AO_A$ , therefore $\angle XAN=\angle A'O_AA=\angle NAH_A$ and we are done. $\Box$ .

What a beautiful property regarding the nine-point center!

Correlary 2: Let $N_A,N_B,N_C$ denote the circumcenter of $NBC,NAC,NAB$ , let $k_A$ denote the reflection of $ON_A$ over $N_BN_C$ , similarly define $k_B,k_C$ . Prove that $k_A,k_B,k_C$ concur.

Proof: Since $O_A\in ON_A$ , we reflect $O_A,A'$ over $N_BN_C$ to get $O_A',A''$ , hence $O_A'A''=k_A$ . Since $N_CN_B$ perpendicularly bisects $AN$ , hence $AN=NO_A=AO_A'$ . Moreover it's easy to prove that $A''X=AO_A'$ , hence $AXA''O_A'$ is parallelogram $\Rightarrow AX\parallel O_A'A''$ If we similarly define $O_B',O_C',B'',C''$ , then $ABC,O_A'O_B'O_C'$ are homothetic and $A''O_A',B''O_B',C''O_C'$ concur $\iff AX,BY,CZ$ concur and we are done. $\Box$ .

Note that $N_BN_AN_C$ is homothetic to the pedal triangle of the Konita's point of $ABC$ and $O$ corresponds to the Konita's points, therefore we now have a cool property regarding Konita's point!

Also if we let $U'$ denote the point at the concurrence, then the circumcenter of $N_AN_BN_C$ lies on $NU'$ . To prove this, just note that $U'$ lies on the Neuberg Cubic and the assertion is simply a property of the cubic.

Correlary 3: Reflect $N_A$ over $BC$ to obatin $N_A'$ , similarly define $N_B',N_C'$ , then $AN_A',BN_B',CN_C'$ concur.

Proof: Construct $A^*B^*C^*$ such that its medial triangle is $ABC$ , hence now we just have to prove $A^*N_A, B^*N_B,C^*N_C$ concur. By Desargue's theorem this is equivalent to $N_BN_C\cap B^*C^*$ and the cyclic contructions are collinear. Apply an inversion at $N$ with an arbitrary radius, Let $A,B,C\rightarrow A_1,B_1,C_1$ . Hence the pole of $N_BN_C$ wrt the inversion circle is the reflection of $N$ over $A_1$ and the pole of $B^*C^*$ is the interestion of the polar of $A$ with $NA_2$ where $A_2,B_2,C_2$ are the poles of $BC,AC,AB$ . Together the pole of $N_BN_C\cap B^*C^*$ is the reflection of $NA_2$ over $B_2C_2$ . Since it's easy to prove that $A_2B_2C_2$ is homothetic to $N_AN_BN_C$ and $O,N$ are obviously two corresponding points, therefore we have the poles of $N_BN_C\cap B^*C^*$ and the cyclic contructions concur $\Rightarrow$ the intersection points are collinear and we are done. $\Box$ .

This post has been edited 2 times. Last edited by XmL, Aug 16, 2014, 2:22 AM

Fontene Theorems Elementary Proofs

by XmL, Aug 12, 2014, 10:26 PM

In this blog we will prove all three of the Fontene theorems in an elementary fashion with a different point of view. Either you want to know what they are beforehand or try proving them yourself they are here: http://mathworld.wolfram.com/FonteneTheorems.html

Let's begin by defining an important point:

Let $l$ be an arbitrary line that passes through the circumcenter of $\triangle ABC$ , define $l\cap (O)=Y,Z$ . The Simson lines of $Y,Z$ wrt $ABC$ meet at $P$ .

Result: If $X\in l$ and $X\ne Y,Z$ . Then the pedal circle of $X$ wrt $ABC$ passes through $P$ .

Before reading on

Diagram

Proof: Let the pedal triangle of $P$ be $DEF$ , The $Y,Z$ Simson lines $\cap BC=Y_A,Z_A$ , $Y_B,Y_C,Z_B,Z_C$ are similarly defined.

Since $ZY$ is the diameter, we know that the Simson lines are orthogonal. From this we can discover a lot of similar triangles in the diagram(that can be proven by simple angle chasings). In particular, $PZ_BY_B\sim BZY, PZ_CY_C\sim CZY$ . Note that since $ZZ_B\parallel XE\parallel YY_B$ , therefore $\frac {Z_BE}{EY_B}=\frac {ZX}{YX}$ , hence $E,X$ are corresponding points in the two similar triangles $\Rightarrow \angle EPY_B=\angle XBY$ , symmetrically we have $\angle YPZ_C=\angle XCZ$ . Hence $\angle YPE=90+\angle XBY+\angle XCZ=$ $90+(90-\angle ZBX)+(90-\angle YCX)=180-\angle ABX-\angle XCA$ $=180-\angle FDE$ $\Rightarrow P,E,D,F$ are concyclic which means $P\in (DEF)$ and we are done. $\Box$ .

Correlary-2nd Fontene theorem:
If a point moves on a fixed line through the circumcenter, then its pedal circle passes through a fixed point on the nine-point circle.

Proof: Clearly $P$ defined above is the fixed point, moreover since the nine-point circle is the pedal circle of the circumcenter therefore $P$ lies on it. $\Box$ .

1st Fontene theorem(Rephrased): The $A-$ midline, $NM,DP$ concur.

Proof: Let $M,N$ be the midpoints of $AC,AB$ , $NM\cap FE=W$ . Since $\frac {Z_BM}{Y_BM}=\frac {ZO}{YP}=1$ , therefore $M$ is the midpoint of $Z_BY_B$ , likewise $N$ is the midpoint of $Y_CZ_C$ . Now construct $P'=(M,MP)\cap (N,NP)$ , which is the miquel point of complete quad. $ANWEMF$ and the reflection of $P$ over $NM$ . Hence we have $\angle FP'E=\angle NP'M=\angle NPM=180-\angle A=\angle FXE$ $\Rightarrow P'\in (AEFX)$

Next we will prove that $P'\in l$ : It's easy to show that $P'Y_B\parallel CZ,P'Z_C\parallel BY, CYY_B\sim ZBZ_C$ . Let $P'Y_B\cap CY,ZY=P_Y,U_Y, P'Z_C\cap U_Z,P_Y$ . Since $U_Y,U_Z$ are two corresponding points of the similar triangles mentions, therefore $\frac {YP_Y}{ZY}=\frac {YU_Y}{CY}=\frac {BU_Z}{ZB}=\frac {YP_Z}{ZY}$ and hence $P_Y\equiv P_Z\equiv P'$ .

Note that $AZY\sim PZ_AY_A$ and that $X,D$ are two corresponding points, therefore $\angle PDC=\angle AXP'=\angle MEP'=\angle MWP'=\angle PWM$ , since $WM\parallel DC$ , therefore $P,W,D$ are collinear.(Let $PD\cap NM=W'\Rightarrow \angle PW'M=\angle PWM\Rightarrow W\equiv W'$ ), and we are done. $\Box$ .

Note: $l$ is the dilation of $P$ 's Simson line wrt the medial triangle by a factor of $2$ .

*To be continued*

This post has been edited 7 times. Last edited by XmL, Aug 12, 2014, 11:54 PM

Submit

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

6 shouts

Geometry Treasures