Revisions to Six conelliptic points

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edited May 29, 2021 at 16:50

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I proveNote that they lie onthere exists a conic. Proving that it is ellipse requires checking some inequalities. I am not entirely sure that it is always so.

We starttriangle $\tilde{F}\tilde{D}\tilde{E}$ with

Lemma. Assume that on the picture side lengths $\tilde{F}\tilde{D}=\sqrt{FD}$, $$\frac{AB_1\cdot AB_2}{AC_1\cdot AC_2}\cdot \frac{CA_1\cdot CA_2}{CB_1\cdot CB_2}\cdot \frac{BC_1\cdot BC_2}{BA_1\cdot BA_2}=1.\quad\quad(\heartsuit)$$ Then$\tilde{E}\tilde{D}=\sqrt{ED}$, $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a conic$\tilde{F}\tilde{E}=\sqrt{FE}$.

Proof. Apply inverse Pascal theorem Make an affine transform $\Phi$ which maps $F,E,D$ to the hexagon $C_2C_1A_2A_1B_2B_1$$\tilde{F}$, $\tilde{E}$, $\tilde{D}$ respectively. We shouldI claim that $\Phi$-images of $J,K,M,N,O,P$ are concylic. For proving this it suffices to check the pointsthat each quadrilateral $A_0=A_1A_2\cap B_1C_2$$KJMP$,$B_0=B_1B_2\cap C_1A_2$ $POJN$, $C_0=C_1C_2\cap A_1B_2$ are collinear. By Menelaus we have $A_0B:A_0C=(C_2B:C_2A)\cdot(B_1A:B_1C)$. Applying three such relations and$NMOK$ maps to a cyclic quadrilateral: their circumcircles either all coincide (inverse) Menelausthat we see that $A_0,B_0,C_0$need), or have concurrent radical axes, but these axes are collinear if and only ifthe sides of $(\heartsuit)$ holds$\triangle \tilde{F}\tilde{D}\tilde{E}$.

On your picture six segments cancel out asThe map $NE=NM$ etc$\Phi$ divides all lengths on line $ED$ by $\sqrt{ED}$ and all lengths on line $EF$ by $\sqrt{EF}$. For segments likeThus to prove that $ND$$NMOK$ is cyclic, we denote byshould check $S$ the double area of triangle$$EN\cdot EO/ED=EM\cdot EK/EF. \quad (\diamondsuit)$$ Since $AFD$$EN=EM$, (or$(\diamondsuit)$ reads as $CFE$$EO/ED=EK/EF$, or $BED$ --- three areas are equal due to your assumptions$OD/ED=KF/EF$, or $\frac{AB}{DA}=\frac{BC}{EB}=\frac{AC}{FC}$) write$OD\cdot EF=ED\cdot KF$, or $ND=AH+ED=AH+S/FD=(AH\cdot FD+S)/ED$${\rm Area}(CEF)={\rm Area}(DEB)$. And remaining terms in This follows from $(\heartsuit)$ also cancel out$\frac{AB}{DA}=\frac{BC}{EB}=\frac{AC}{FC}$.

I prove that they lie on a conic. Proving that it is ellipse requires checking some inequalities. I am not entirely sure that it is always so.

We start with

Lemma. Assume that on the picture $$\frac{AB_1\cdot AB_2}{AC_1\cdot AC_2}\cdot \frac{CA_1\cdot CA_2}{CB_1\cdot CB_2}\cdot \frac{BC_1\cdot BC_2}{BA_1\cdot BA_2}=1.\quad\quad(\heartsuit)$$ Then $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a conic.

Proof. Apply inverse Pascal theorem to the hexagon $C_2C_1A_2A_1B_2B_1$. We should check the points $A_0=A_1A_2\cap B_1C_2$,$B_0=B_1B_2\cap C_1A_2$, $C_0=C_1C_2\cap A_1B_2$ are collinear. By Menelaus we have $A_0B:A_0C=(C_2B:C_2A)\cdot(B_1A:B_1C)$. Applying three such relations and (inverse) Menelaus we see that $A_0,B_0,C_0$ are collinear if and only if $(\heartsuit)$ holds.

On your picture six segments cancel out as $NE=NM$ etc. For segments like $ND$ we denote by $S$ the double area of triangle $AFD$ (or $CFE$, or $BED$ --- three areas are equal due to your assumptions $\frac{AB}{DA}=\frac{BC}{EB}=\frac{AC}{FC}$) write $ND=AH+ED=AH+S/FD=(AH\cdot FD+S)/ED$. And remaining terms in $(\heartsuit)$ also cancel out.

Note that there exists a triangle $\tilde{F}\tilde{D}\tilde{E}$ with side lengths $\tilde{F}\tilde{D}=\sqrt{FD}$, $\tilde{E}\tilde{D}=\sqrt{ED}$, $\tilde{F}\tilde{E}=\sqrt{FE}$. Make an affine transform $\Phi$ which maps $F,E,D$ to $\tilde{F}$, $\tilde{E}$, $\tilde{D}$ respectively. I claim that $\Phi$-images of $J,K,M,N,O,P$ are concylic. For proving this it suffices to check that each quadrilateral $KJMP$, $POJN$, $NMOK$ maps to a cyclic quadrilateral: their circumcircles either all coincide (that we need), or have concurrent radical axes, but these axes are the sides of $\triangle \tilde{F}\tilde{D}\tilde{E}$.

The map $\Phi$ divides all lengths on line $ED$ by $\sqrt{ED}$ and all lengths on line $EF$ by $\sqrt{EF}$. Thus to prove that $NMOK$ is cyclic, we should check $$EN\cdot EO/ED=EM\cdot EK/EF. \quad (\diamondsuit)$$ Since $EN=EM$, $(\diamondsuit)$ reads as $EO/ED=EK/EF$, or $OD/ED=KF/EF$, or $OD\cdot EF=ED\cdot KF$, or ${\rm Area}(CEF)={\rm Area}(DEB)$. This follows from $\frac{AB}{DA}=\frac{BC}{EB}=\frac{AC}{FC}$.

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edited May 29, 2021 at 16:20

Fedor Petrov

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I prove that they lie on a conic. Proving that it is ellipse requirerequires checking some inequalities. I am not entirely sure that it is always so.

We start with

Lemma. Assume that on the picture $$\frac{AB_1\cdot AB_2}{AC_1\cdot AC_2}\cdot \frac{CA_1\cdot CA_2}{CB_1\cdot CB_2}\cdot \frac{BC_1\cdot BC_2}{BA_1\cdot BA_2}=1.\quad\quad(\heartsuit)$$ Then $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a conic.

Proof. Apply inverse Pascal theorem to the hexagon $C_2C_1A_2A_1B_2B_1$. We should check the points $A_0=A_1A_2\cap B_1C_2$,$B_0=B_1B_2\cap C_1A_2$, $C_0=C_1C_2\cap A_1B_2$ are collinear. By Menelaus we have $A_0B:A_0C=(C_2B:C_2A)\cdot(B_1A:B_1C)$. Applying three such relations and (inverse) Menelaus we see that $A_0,B_0,C_0$ are collinear if and only if $(\heartsuit)$ holds.

On your picture six segments cancel out as $NE=NM$ etc. For segments like $ND$ we denote by $S$ the double area of triangle $AFD$ (or $CFE$, or $BED$ --- three areas are equal due to your assumptions $\frac{AB}{DA}=\frac{BC}{EB}=\frac{AC}{FC}$) write $ND=AH+ED=AH+S/FD=(AH\cdot FD+S)/ED$. And remaining terms in $(\heartsuit)$ also cancel out.

I prove that they lie on a conic. Proving that is ellipse require checking some inequalities. I am not entirely sure that it is always so.

We start with

Lemma. Assume that on the picture $$\frac{AB_1\cdot AB_2}{AC_1\cdot AC_2}\cdot \frac{CA_1\cdot CA_2}{CB_1\cdot CB_2}\cdot \frac{BC_1\cdot BC_2}{BA_1\cdot BA_2}=1.\quad\quad(\heartsuit)$$ Then $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a conic.

Proof. Apply inverse Pascal theorem to the hexagon $C_2C_1A_2A_1B_2B_1$. We should check the points $A_0=A_1A_2\cap B_1C_2$,$B_0=B_1B_2\cap C_1A_2$, $C_0=C_1C_2\cap A_1B_2$ are collinear. By Menelaus we have $A_0B:A_0C=(C_2B:C_2A)\cdot(B_1A:B_1C)$. Applying three such relations and (inverse) Menelaus we see that $A_0,B_0,C_0$ are collinear if and only if $(\heartsuit)$ holds.

On your picture six segments cancel out as $NE=NM$ etc. For segments like $ND$ we denote by $S$ the double area of triangle $AFD$ (or $CFE$, or $BED$ --- three areas are equal due to your assumptions $\frac{AB}{DA}=\frac{BC}{EB}=\frac{AC}{FC}$) write $ND=AH+ED=AH+S/FD=(AH\cdot FD+S)/ED$. And remaining terms in $(\heartsuit)$ also cancel out.

I prove that they lie on a conic. Proving that it is ellipse requires checking some inequalities. I am not entirely sure that it is always so.

We start with

Lemma. Assume that on the picture $$\frac{AB_1\cdot AB_2}{AC_1\cdot AC_2}\cdot \frac{CA_1\cdot CA_2}{CB_1\cdot CB_2}\cdot \frac{BC_1\cdot BC_2}{BA_1\cdot BA_2}=1.\quad\quad(\heartsuit)$$ Then $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a conic.

Proof. Apply inverse Pascal theorem to the hexagon $C_2C_1A_2A_1B_2B_1$. We should check the points $A_0=A_1A_2\cap B_1C_2$,$B_0=B_1B_2\cap C_1A_2$, $C_0=C_1C_2\cap A_1B_2$ are collinear. By Menelaus we have $A_0B:A_0C=(C_2B:C_2A)\cdot(B_1A:B_1C)$. Applying three such relations and (inverse) Menelaus we see that $A_0,B_0,C_0$ are collinear if and only if $(\heartsuit)$ holds.

On your picture six segments cancel out as $NE=NM$ etc. For segments like $ND$ we denote by $S$ the double area of triangle $AFD$ (or $CFE$, or $BED$ --- three areas are equal due to your assumptions $\frac{AB}{DA}=\frac{BC}{EB}=\frac{AC}{FC}$) write $ND=AH+ED=AH+S/FD=(AH\cdot FD+S)/ED$. And remaining terms in $(\heartsuit)$ also cancel out.

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created May 29, 2021 at 15:58

Fedor Petrov

108.8k
9
264
459

I prove that they lie on a conic. Proving that is ellipse require checking some inequalities. I am not entirely sure that it is always so.

We start with

Lemma. Assume that on the picture $$\frac{AB_1\cdot AB_2}{AC_1\cdot AC_2}\cdot \frac{CA_1\cdot CA_2}{CB_1\cdot CB_2}\cdot \frac{BC_1\cdot BC_2}{BA_1\cdot BA_2}=1.\quad\quad(\heartsuit)$$ Then $A_1,A_2,B_1,B_2,C_1,C_2$ lie on a conic.

Proof. Apply inverse Pascal theorem to the hexagon $C_2C_1A_2A_1B_2B_1$. We should check the points $A_0=A_1A_2\cap B_1C_2$,$B_0=B_1B_2\cap C_1A_2$, $C_0=C_1C_2\cap A_1B_2$ are collinear. By Menelaus we have $A_0B:A_0C=(C_2B:C_2A)\cdot(B_1A:B_1C)$. Applying three such relations and (inverse) Menelaus we see that $A_0,B_0,C_0$ are collinear if and only if $(\heartsuit)$ holds.

On your picture six segments cancel out as $NE=NM$ etc. For segments like $ND$ we denote by $S$ the double area of triangle $AFD$ (or $CFE$, or $BED$ --- three areas are equal due to your assumptions $\frac{AB}{DA}=\frac{BC}{EB}=\frac{AC}{FC}$) write $ND=AH+ED=AH+S/FD=(AH\cdot FD+S)/ED$. And remaining terms in $(\heartsuit)$ also cancel out.

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