Interesting Hexagon

by TelvCohl, Jun 25, 2016, 6:02 AM

In this post, I'll prove some interesting properties of the hexagon $A_1 C_2 B_1 A_2 C_1 B_2$ such that $\frac{a_1-c_2}{c_2-b_1} \cdot \frac{b_1-a_2}{a_2-c_1} \cdot \frac{c_1-b_2}{b_2-a_1}=-1 \qquad (\spadesuit )$ where $a_1,$ $a_2,$ $b_1,$ $b_2,$ $c_1,$ $c_2$ is the complex number of $A_1,$ $A_2,$ $B_1,$ $B_2,$ $C_1,$ $C_2,$ respectively. In the following discussion, I'll only prove the case when $A_1 C_2 B_1 A_2 C_1 B_2$ is a convex hexagon (other cases can be proved similarly). In this case, the condition $(\spadesuit)$ is equivalent to $\frac{A_1C_2}{C_2B_1} \cdot \frac{B_1A_2}{A_2C_1} \cdot \frac{C_1B_2}{B_2A_1} = 1$ and $\measuredangle C_1A_2B_1 + \measuredangle A_1B_2C_1 + \measuredangle B_1C_2A_1 = 360^{\circ}.$
$[asy] import graph; size(18.cm); real lsf=0.5; pen dps=linewidth(0.7)+fontsize(13); defaultpen(dps); real xmin=8.615206375429189,xmax=22.32975186944826,ymin=-9.595429169027547,ymax=2.111735923073725; pen yqqqqq=rgb(0.5019607843137255,0.,0.), uuuuuu=rgb(0.26666666666666666,0.26666666666666666,0.26666666666666666), ttttff=rgb(0.2,0.2,1.); pair A_1=(17.002961963201372,1.0063510159008429), B_1=(11.65609284806018,-6.2616562397523765), C_1=(20.02307942469264,-6.208854542397696), D_0=(-4.192264473328908,8.339457254257692), E_0=(-5.954518799948372,5.796826673092463), F_0=(-2.2280475054943683,5.7438848324763345), Q_0=(-3.5954630407492614,6.584722892746521), A_2=(14.764162301331519,-8.08604490327189), B_2=(19.682949524902266,-1.768043138082691), C_2=(10.06044609258025,-1.0967274775409979), M=(34.66597800004515,-27.343634871832254); draw(E_0--F_0,linewidth(1.1)); draw(F_0--D_0,linewidth(1.1)); draw(D_0--E_0,linewidth(1.1)); draw(E_0--Q_0,linewidth(1.1)); draw(Q_0--F_0,linewidth(1.1)); draw(Q_0--D_0,linewidth(1.1)); draw(C_2--B_1,linewidth(1.1)); draw(B_1--A_2,linewidth(1.1)); draw(A_2--C_1,linewidth(1.1)); draw(C_1--B_2,linewidth(1.1)); draw(B_2--A_1,linewidth(1.1)); draw(A_1--C_2,linewidth(1.1)); draw(B_1--C_1,linewidth(1.6)+ttttff); draw(C_1--A_1,linewidth(1.6)+ttttff); draw(A_1--B_1,linewidth(1.6)+ttttff); draw(C_2--B_2,linewidth(1.6)+yqqqqq); draw(C_2--A_2,linewidth(1.6)+yqqqqq); draw(A_2--B_2,linewidth(1.6)+yqqqqq); dot(A_1,linewidth(4.pt)+blue); label("$A_1$",(16.828610157322174,1.2995896535954954),NE*lsf,blue); dot(B_1,linewidth(4.pt)+blue); label("$B_1$",(11.128262756456147,-6.469432207262286),NE*lsf,blue); dot(C_1,linewidth(4.pt)+blue); label("$C_1$",(20.215106865901184,-6.3928146346700006),NE*lsf,blue); dot(D_0,linewidth(3.pt)+blue); dot(E_0,linewidth(3.pt)+blue); dot(F_0,linewidth(3.pt)+blue); dot(Q_0,linewidth(3.pt)+blue); dot(A_2,linewidth(4.pt)+yqqqqq); label("$A_2$",(14.6067005521459,-8.476812609180174),NE*lsf,yqqqqq); dot(B_2,linewidth(4.pt)+yqqqqq); label("$B_2$",(19.87798954649513,-1.5965545903929084),NE*lsf,yqqqqq); dot(C_2,linewidth(4.pt)+yqqqqq); label("$C_2$",(9.687852391721183,-0.8457023789885074),NE*lsf,yqqqqq); dot(M,linewidth(3.pt)+uuuuuu); label("$M$",(6.531208400919028,2.4488532424797826),NE*lsf,uuuuuu); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); currentpicture=yscale(1.0041136873597543)*currentpicture; [/asy]$
Property 1 : $\begin{align*} \frac{C_1A_1}{A_1B_1} \cdot \frac{B_1C_2}{C_2A_2} \cdot \frac{A_2B_2}{B_2C_1} = 1 & , && \measuredangle B_1A_1C_1 = \measuredangle B_1C_2A_2 + \measuredangle A_2B_2C_1 , && \measuredangle B_2A_2C_2 = \measuredangle B_2C_1A_1 + \measuredangle A_1B_1C_2 . \\ \frac{A_1B_1}{B_1C_1} \cdot \frac{C_1A_2}{A_2B_2} \cdot \frac{B_2C_2}{C_2A_1} = 1 & , && \measuredangle C_1B_1A_1 = \measuredangle C_1A_2B_2 + \measuredangle B_2C_2A_1 , && \measuredangle C_2B_2A_2 = \measuredangle C_2A_1B_1 + \measuredangle B_1C_1A_2 . \\ \frac{B_1C_1}{C_1A_1} \cdot \frac{A_1B_2}{B_2C_2} \cdot \frac{C_2A_2}{A_2B_1} = 1 & , && \measuredangle A_1C_1B_1 = \measuredangle A_1B_2C_2 + \measuredangle C_2A_2B_1 , && \measuredangle A_2C_2B_2 = \measuredangle A_2B_1C_1 + \measuredangle C_1A_1B_2 . \end{align*}$
Proof : Let $\mathcal{S}(P,\lambda , \theta)$ be the spiral similarity with center $P,$ ratio $\lambda,$ angle $\theta.$ Since $A_1$ is fixed under $\mathcal{S}(C_2,\frac{C_2A_2}{C_2B_1}, \measuredangle B_1C_2A_2)$ $\circ$ $\mathcal{S}(B_2,\frac{B_2C_1}{B_2A_2}, \measuredangle A_2B_2C_1),$ so $\mathcal{S}(C_2,\frac{C_2A_2}{C_2B_1}, \measuredangle B_1C_2A_2) \circ \mathcal{S}(B_2,\frac{B_2C_1}{B_2A_2}, \measuredangle A_2B_2C_1) = \mathcal{S}(A_1, \frac{C_2A_2}{C_2B_1} \cdot \frac{B_2C_1}{B_2A_2}, \measuredangle B_1C_2A_2 + \measuredangle A_2B_2C_1).$ Notice $B_1$ $\mapsto$ $C_1$ under $\mathcal{S}(C_2,\frac{C_2A_2}{C_2B_1}, \measuredangle B_1C_2A_2)$ $\circ$ $\mathcal{S}(B_2,\frac{B_2C_1}{B_2A_2}, \measuredangle A_2B_2C_1)$ we conclude that $\frac{B_1C_2}{C_2A_2} \cdot \frac{A_2B_2}{B_2C_1}$ $=$ $\frac{A_1B_1}{C_1A_1}$ and $\measuredangle B_1A_1C_1$ $=$ $\measuredangle B_1C_2A_2$ $+$ $\measuredangle A_2B_2C_1 .$ $\qquad \blacksquare$

Property 2 : There exists a point $M$ such that $\triangle MB_1C_1$ $\stackrel{+}{\sim}$ $\triangle MC_2B_2,$ $\triangle MC_1A_1$ $\stackrel{+}{\sim}$ $\triangle MA_2C_2,$ $\triangle MA_1B_1$ $\stackrel{+}{\sim}$ $\triangle MB_2A_2.$

Proof : Let $M$ be the Miquel point of the complete quadrilateral with the sides $B_1C_1,$ $B_1C_2,$ $B_2C_1,$ $B_2C_2$ and $A_1^*$ be the point such that $\triangle MC_1A_1^*$ $\stackrel{+}{\sim}$ $\triangle MA_2C_2,$ $\triangle MA_1^*B_1$ $\stackrel{+}{\sim}$ $\triangle MB_2A_2.$ Since $\measuredangle B_1A_1^*C_1$ $=$ $\measuredangle B_1A_1^*M$ $+$ $\measuredangle MA_1^*C_1$ $=$ $\measuredangle A_2B_2M$ $+$ $\measuredangle MC_2A_2$ $=$ $\measuredangle A_2B_2C_1$ $+$ $\measuredangle C_1B_2M$ $+$ $\measuredangle MC_2B_1$ $+$ $\measuredangle B_1C_2A_2$ $=$ $\measuredangle B_1C_2A_2$ $+$ $\measuredangle A_2B_2C_1$ (similar for others), so from Property 1 we conclude that $A_1^*$ $\equiv$ $A_1.$ $\qquad \blacksquare$

Remark : Notice if we are given five points, then the sixth point such that the hexagon formed by these six points satisfy $(\spadesuit)$ is unique, so the converse of Property 2 is also true. i.e. If there is a point $M$ and six points $A_1,$ $A_2,$ $B_1,$ $B_2,$ $C_1,$ $C_2$ such that $\triangle MB_1C_1$ $\stackrel{+}{\sim}$ $\triangle MC_2B_2,$ $\triangle MC_1A_1$ $\stackrel{+}{\sim}$ $\triangle MA_2C_2,$ $\triangle MA_1B_1$ $\stackrel{+}{\sim}$ $\triangle MB_2A_2,$ then the hexagon $A_1 C_2 B_1 A_2 C_1 B_2$ satisfy $(\spadesuit).$

For a given point $P$ and a given line $\ell,$ we denote $\Phi_{(\varsigma, \ell)}^P$ as the inversion with center $P,$ power $\varsigma$ followed by reflection in $\ell.$ From Property 2 we get $MA_1$ $\cdot$ $MA_2$ $=$ $MB_1$ $\cdot$ $MB_2$ $=$ $MC_1$ $\cdot$ $MC_2$ $=$ $\eta$ and $\angle A_1MA_2,$ $\angle B_1MB_2,$ $\angle C_1MC_2$ share the same bisector $\tau$ $\Longrightarrow$ $\Phi_{(\eta, \tau)}^M$ swaps $(A_1,A_2),$ $(B_1,B_2),$ $(C_1,C_2),$ so we get the following important corollary.

Corollary 3 : A hexagon $A_1 C_2 B_1 A_2 C_1 B_2$ satisfy $(\spadesuit)$ $\Longleftrightarrow$ there exists a transformation $\Phi_{(\eta, \tau)}^M$ swaps $(A_1,A_2),$ $(B_1,B_2),$ $(C_1,C_2).$

Remark : From Corollary 3 (or Property 1) $\Longrightarrow$ the hexagon $A_1C_1B_1A_2C_2B_2,$ $A_1C_1B_2A_2C_2B_1,$ $A_1C_2B_2A_2C_1B_1$ also satisfy $(\spadesuit).$

Property 4 : $\odot (A_1B_2C_2),$ $\odot (A_2B_1C_2),$ $\odot (A_2B_2C_1),$ $\odot (A_1B_1C_1)$ are concurrent at $D_1.$ $\odot (A_2B_1C_1),$ $\odot (A_1B_2C_1),$ $\odot (A_1B_1C_2),$ $\odot (A_2B_2C_2)$ are concurrent at $D_2.$ Furthermore, if $A,$ $B,$ $C,$ $D$ is the midpoint of $A_1A_2,$ $B_1B_2,$ $C_1C_2,$ $D_1D_2,$ respectively, then $A,$ $B,$ $C,$ $D$ lie on a circle.

Proof : Let $D_1$ be the second intersection of $\odot (A_2B_1C_2),$ $\odot (A_2B_2C_1).$ Since $\measuredangle B_1D_1C_1$ $=$ $\measuredangle B_1D_1A_2$ $+$ $\measuredangle A_2D_1C_1$ $=$ $\measuredangle B_1C_2A_2$ $+$ $\measuredangle A_2B_2C_1$ $=$ $\measuredangle B_1A_1C_1$ (Property 1), so $D_1$ lies on $\odot (A_1B_1C_1).$ Similarly, we can prove $D_1$ lies on $\odot (A_1B_2C_2),$ so $\odot (A_1B_2C_2),$ $\odot (A_2B_1C_2),$ $\odot (A_2B_2C_1),$ $\odot (A_1B_1C_1)$ are concurrent at $D_1.$ Analogously, we can prove $\odot (A_2B_1C_1),$ $\odot (A_1B_2C_1),$ $\odot (A_1B_1C_2),$ $\odot (A_2B_2C_2)$ are concurrent at $D_2.$

Let $A_2^*,$ $B_2^*,$ $C_2^*$ be the reflection of $A_2,$ $B_2,$ $C_2$ in the midpoint of $B_1C_1,$ $C_1A_1,$ $A_1B_1,$ respectively. If $J$ is the midpoint of $B_1C_1,$ then $JB$ $\stackrel{\parallel}{=}$ $\tfrac{1}{2} A_1B_2^*,$ $JC$ $\stackrel{\parallel}{=}$ $\tfrac{1}{2} A_1C_2^*,$ so $BC$ $\stackrel{\parallel}{=}$ $\tfrac{1}{2} B_2^*C_2^*.$ Similarly, we can prove $CA$ $\stackrel{\parallel}{=}$ $\tfrac{1}{2} C_2^*A_2^*$ and $AB$ $\stackrel{\parallel}{=}$ $\tfrac{1}{2} A_2^*B_2^*.$ On the other hand, it's easy to see the hexagon $A_1C_2^*B_1A_2^*C_1B_2^*$ satisfy $(\spadesuit),$ so from Property 1 we get $\measuredangle CAB$ $=$ $\measuredangle C_2^*A_2^*B_2^*$ $=$ $\measuredangle C_2^*B_1A_1$ $+$ $\measuredangle A_1C_1B_2^*$ $=$ $\measuredangle C_2A_1B_1$ $+$ $\measuredangle C_1A_1B_2.$ From Corollary 3 we know the hexagon $D_1B_1C_2D_2B_2C_1$ satisfy $(\spadesuit),$ so by similar discussion we can prove $\measuredangle CDB$ $=$ $\measuredangle C_2D_2B_1$ $+$ $\measuredangle C_1D_2B_2,$ hence notice $D_2$ lies on $\odot (A_1B_2C_1),$ $\odot (A_1B_1C_2)$ we conclude that $A,$ $B,$ $C,$ $D$ are concyclic. $\qquad \blacksquare$

Corollary 5 : Following 12 hexagons satisfy $(\spadesuit).$ $\begin{align*} D_1B_1C_1D_2B_2C_2, && D_1B_1C_2D_2B_2C_1, && D_1B_2C_1D_2B_1C_2, && D_1B_2C_2D_2B_1C_1, \\ D_1C_1A_1D_2C_2A_2, && D_1C_1A_2D_2C_2A_1, && D_1C_2A_1D_2C_1A_2, && D_1C_2A_2D_2C_1A_1, \\ D_1A_1B_1D_2A_2B_2, && D_1A_1B_2D_2A_2B_1, && D_1A_2B_1D_2A_1B_2, && D_1A_2B_2D_2A_1B_1. \end{align*}$
Proof : This is the direct consequence of Corollary 3. $\qquad \blacksquare$

By Property 1 we know there exists a point $P_{111}$ such that $\triangle P_{111}B_1C_1$ $\stackrel{+}{\sim}$ $\triangle A_1C_2B_2,$ $\triangle P_{111}C_1A_1$ $\stackrel{+}{\sim}$ $\triangle B_1A_2C_2,$ $\triangle P_{111}A_1B_1$ $\stackrel{+}{\sim}$ $\triangle C_1B_2A_2.$

Property 6 : Let $Q_{111}$ be the isogonal conjugate of $P_{111}$ WRT $\triangle A_1B_1C_1.$ Let $A_3,$ $B_3,$ $C_3$ be the reflection of $A_1,$ $B_1,$ $C_1$ in $B_2C_2,$ $C_2A_2,$ $A_2B_2,$ respectively. Then $\triangle A_1B_1C_1$ $\cup$ $Q_{111}$ $\stackrel{-}{\sim}$ $\triangle A_3B_3C_3$ $\cup$ $Q_{111}.$

Proof : Let $B_1^*$ $\equiv$ $B_1Q_{111}$ $\cap$ $A_2B_2,$ $C_1^*$ $\equiv$ $C_1Q_{111}$ $\cap$ $C_2A_2.$ From $\measuredangle C_1B_1B_1^*$ $=$ $\measuredangle P_{111}B_1A_1$ $=$ $\measuredangle C_1A_2B_1^*$ $\Longrightarrow$ $B_1^*$ lies on $\odot (A_2B_1C_1).$ Analogously, we can prove $C_1^*$ lies on $\odot (A_2B_1C_1).$ Since $\measuredangle B_3C_1^*Q_{111}$ $=$ $\measuredangle A_2C_1^*B_1$ $+$ $\measuredangle A_2C_1^*C_1$ $=$ $\measuredangle A_2B_1^*B_1$ $+$ $\measuredangle A_2B_1^*C_1$ $=$ $\measuredangle C_3B_1^*Q_{111},$ so combining $\tfrac{C_1^*B_3}{C_1^*Q_{111}}$ $=$ $\tfrac{C_1^*B_1}{C_1^*Q_{111}}$ $=$ $\tfrac{B_1^*C_1}{B_1^*Q_{111}}$ $=$ $\tfrac{B_1^*C_3}{B_1^*Q_{111}}$ we get $\triangle Q_{111}C_1^*B_3$ $\stackrel{+}{\sim}$ $\triangle Q_{111}B_1^*C_3$ $\Longrightarrow$ $\triangle Q_{111}B_1C_1$ $\stackrel{-}{\sim}$ $\triangle Q_{111}C_1^*B_1^*$ $\stackrel{+}{\sim}$ $\triangle Q_{111}B_3C_3.$ Similarly, we can prove $\triangle Q_{111}C_1A_1$ $\stackrel{-}{\sim}$ $\triangle Q_{111}C_3A_3$ and $\triangle Q_{111}A_1B_1$ $\stackrel{-}{\sim}$ $\triangle Q_{111}A_3B_3,$ so we conclude that $\triangle A_1B_1C_1$ $\cup$ $Q_{111}$ $\stackrel{-}{\sim}$ $\triangle A_3B_3C_3$ $\cup$ $Q_{111}.$ $\qquad \blacksquare$

Property 7 : Let $A_4,$ $B_4,$ $C_4$ be the reflection of $A_2,$ $B_2,$ $C_2$ in $B_1C_1,$ $C_1A_1,$ $A_1B_1,$ respectively. Then $\left | [\triangle A_1B_1C_1] - [\triangle A_3B_3C_3] \right |$ $=$ $\left | [\triangle A_2B_2C_2] - [\triangle A_4B_4C_4] \right |.$

Proof : Let $X_1,$ $Y_1,$ $Z_1,$ $X_2,$ $Y_2,$ $Z_2$ be the projection of $A_1,$ $B_1,$ $C_1,$ $A_2,$ $B_2,$ $C_2$ on $B_2C_2,$ $C_2A_2,$ $A_2B_2,$ $B_1C_1,$ $C_1A_1,$ $A_1B_1,$ respectively. Clearly, $B_1,$ $C_2,$ $Y_1,$ $Z_2$ are concyclic, so $\measuredangle (Y_1Z_2, A_1B_1)$ $=$ $\measuredangle A_2C_2B_1.$ Similarly, we can prove $\measuredangle (C_1A_1, Y_2Z_1)$ $=$ $\measuredangle C_1B_2A_2,$ so from Property 1 we get $\measuredangle (Y_1Z_2, Y_2Z_1)$ $=$ $\measuredangle (Y_1Z_2, A_1B_1)$ $+$ $\measuredangle B_1A_1C_1$ $+$ $\measuredangle (C_1A_1, Y_2Z_1)$ $=$ $\measuredangle A_2C_2B_1$ $+$ $\measuredangle B_1A_1C_1$ $+$ $\measuredangle C_1B_2A_2$ $=$ $0^{\circ}$ $\Longrightarrow$ $Y_1Z_2$ $\parallel$ $Y_2Z_1.$ Similarly, we can prove $Z_1X_2$ $\parallel$ $Z_2X_1,$ $X_1Y_2$ $\parallel$ $X_2Y_1.$ From the generalization at here (post #3) we conclude that $\left | [ \triangle A_1B_1C_1 ] - [ \triangle A_3B_3C_3 ] \right | = 4 [ \triangle X_1Y_1Z_1 ] = 4 [ \triangle X_2Y_2Z_2 ] = \left | [ \triangle A_2B_2C_2 ] - [ \triangle A_4B_4C_4 ] \right |. \qquad \blacksquare$
If a hexagon $V_1V_6V_3V_2V_5V_4$ satisfy $(\spadesuit),$ then we construct the reflection $V_1^*,$ $V_3^*,$ $V_5^*$ of $V_1,$ $V_3,$ $V_5$ in $V_4V_6,$ $V_6V_2,$ $V_2V_4,$ respectively and call the fixed point of the inversely similar triangle $\triangle V_1V_3V_5,$ $\triangle V_1^*V_3^*V_5^*$ the T-center of $\triangle V_1V_3V_5$ WRT $\triangle V_2V_4V_6.$ By this definition $\Longrightarrow$ $Q_{111}$ is the T-center of $\triangle A_1B_1C_1$ WRT $\triangle A_2B_2C_2.$ Similarly, we denote $Q_{211},$ $Q_{121},$ $Q_{112},$ $Q_{122},$ $Q_{212},$ $Q_{221},$ $Q_{222}$ as the T-center of $\triangle A_2B_1C_1,$ $\triangle A_1B_2C_1,$ $\triangle A_1B_1C_2,$ $\triangle A_1B_2C_2,$ $\triangle A_2B_1C_2,$ $\triangle A_2B_2C_1,$ $\triangle A_2B_2C_2$ WRT $\triangle A_1B_2C_2,$ $\triangle A_2B_1C_2,$ $\triangle A_2B_2C_1,$ $\triangle A_2B_1C_1,$ $\triangle A_1B_2C_1,$ $\triangle A_1B_1C_2,$ $\triangle A_1B_1C_1,$ respectively.

Property 8 : $\begin{align*} Q_{111} \text{ is the T-center of } \triangle B_1C_1D_2, && \triangle A_1C_1D_2, && \triangle A_1B_1D_2 \text{ WRT } \triangle B_2C_2D_1, && \triangle A_2C_2D_1, && \triangle A_2B_2D_1, \text{ respectively}. \\ Q_{211} \text{ is the T-center of } \triangle B_1C_1D_1, && \triangle A_2C_1D_1, && \triangle A_2B_1D_1 \text{ WRT } \triangle B_2C_2D_2, && \triangle A_1C_2D_2, && \triangle A_1B_2D_2, \text{ respectively}. \\ Q_{121} \text{ is the T-center of } \triangle B_2C_1D_1, && \triangle A_1C_1D_1, && \triangle A_1B_2D_1 \text{ WRT } \triangle B_1C_2D_2, && \triangle A_2C_2D_2, && \triangle A_2B_1D_2, \text{ respectively}. \\ Q_{112} \text{ is the T-center of } \triangle B_1C_2D_1, && \triangle A_1C_2D_1, && \triangle A_1B_1D_1 \text{ WRT } \triangle B_2C_1D_2, && \triangle A_2C_1D_2, && \triangle A_2B_2D_2, \text{ respectively}. \\ Q_{122} \text{ is the T-center of } \triangle B_2C_2D_2, && \triangle A_1C_2D_2, && \triangle A_1B_2D_2 \text{ WRT } \triangle B_1C_1D_1, && \triangle A_2C_1D_1, && \triangle A_2B_1D_1, \text{ respectively}. \\ Q_{212} \text{ is the T-center of } \triangle B_1C_2D_2, && \triangle A_2C_2D_2, && \triangle A_2B_1D_2 \text{ WRT } \triangle B_2C_1D_1, && \triangle A_1C_1D_1, && \triangle A_1B_2D_1, \text{ respectively}. \\ Q_{221} \text{ is the T-center of } \triangle B_2C_1D_2, && \triangle A_2C_1D_2, && \triangle A_2B_2D_2 \text{ WRT } \triangle B_1C_2D_1, && \triangle A_1C_2D_1, && \triangle A_1B_1D_1, \text{ respectively}. \\ Q_{222} \text{ is the T-center of } \triangle B_2C_2D_1, && \triangle A_2C_2D_1, && \triangle A_2B_2D_1 \text{ WRT } \triangle B_1C_1D_2, && \triangle A_1C_1D_2, && \triangle A_1B_1D_2, \text{ respectively}. \end{align*}$
Proof : Since $\measuredangle Q_{111}B_1C_1$ $=$ $\measuredangle B_2A_2C_1$ $=$ $\measuredangle B_2D_1C_1$ and $\measuredangle Q_{111}C_1B_1$ $=$ $\measuredangle C_2A_2B_1$ $=$ $\measuredangle C_2D_1B_1,$ so we conclude that $Q_{111}$ is the T-center of $\triangle D_2B_1C_1$ WRT $\triangle D_1B_2C_2.$ $\qquad \blacksquare$

Property 9 : $\begin{align*} A_1Q_{111} \parallel A_2Q_{222}, && B_1Q_{111} \parallel B_2Q_{222}, && C_1Q_{111} \parallel C_2Q_{222}, && D_2Q_{111} \parallel D_1Q_{222},\\ A_1Q_{112} \parallel A_2Q_{221}, && B_1Q_{112} \parallel B_2Q_{221}, && C_1Q_{112} \parallel C_2Q_{221}, && D_1Q_{112} \parallel D_2Q_{221},\\ A_1Q_{121} \parallel A_2Q_{212}, && B_1Q_{121} \parallel B_2Q_{212}, && C_1Q_{121} \parallel C_2Q_{212}, && D_1Q_{121} \parallel D_2Q_{212}, \\ A_1Q_{122} \parallel A_2Q_{211}, && B_1Q_{122} \parallel B_2Q_{211}, && C_1Q_{122} \parallel C_2Q_{211}, && D_2Q_{122} \parallel D_1Q_{211}. \end{align*}$
Proof : Since $\measuredangle (A_1B_1, A_1Q_{111})$ $=$ $\measuredangle B_1C_2A_2$ $=$ $\measuredangle (A_1B_1, Y_1Z_2),$ so $A_1Q_{111}$ $\parallel$ $Y_1Z_2.$ Similarly, we can prove $A_2Q_{222}$ $\parallel$ $Y_2Z_1,$ so we conclude that $A_1Q_{111}$ $\parallel$ $A_2Q_{222}.$ $\qquad \blacksquare$

Corollary 10 : The midpoint of $Q_{111}Q_{222},$ $Q_{211}Q_{122},$ $Q_{121}Q_{212},$ $Q_{112}Q_{221}$ lie on $\odot (ABCD).$

Proof : If $T$ is the midpoint of $Q_{111}Q_{222},$ then by Property 9 we get $BT$ $\parallel$ $B_1Q_{111}$ $\parallel$ $B_2Q_{222},$ $CT$ $\parallel$ $C_1Q_{111}$ $\parallel$ $C_2Q_{222},$ so $\measuredangle CTB$ $=$ $\measuredangle C_1Q_{111}B_1$ $=$ $\measuredangle (C_1Q_{111}, C_1A_1)$ $+$ $\measuredangle C_1A_1B_1$ $+$ $\measuredangle (A_1B_1, B_1Q_{111})$ $=$ $\measuredangle C_2B_2A_1$ $+$ $\measuredangle C_1A_1B_1$ $+$ $\measuredangle A_1C_2B_2$ $=$ $\measuredangle C_2A_1B_1$ $+$ $\measuredangle C_1A_1B_2$ $=$ $\measuredangle CAB,$ hence we conclude that $T$ lies on $\odot (ABCD).$ $\qquad \blacksquare$

Property 11 : The points in the following 4-points sets are concyclic. $\begin{align*} (B_1, C_1, Q_{111}, Q_{211}), && (B_1, C_2, Q_{112}, Q_{212}), && (B_2, C_1, Q_{121}, Q_{221}), && (B_2, C_2, Q_{122}, Q_{222}), \\ (C_1, A_1, Q_{111}, Q_{121}), && (C_1, A_2, Q_{211}, Q_{221}), && (C_2, A_1, Q_{112}, Q_{122}), && (C_2, A_2, Q_{212}, Q_{222}), \\ (A_1, B_1, Q_{111}, Q_{112}), && (A_1, B_2, Q_{121}, Q_{122}), && (A_2, B_1, Q_{211}, Q_{212}), && (A_2, B_2, Q_{221}, Q_{222}), \\ (A_1, D_1, Q_{112}, Q_{121}), && (A_1, D_2, Q_{111}, Q_{122}), && (A_2, D_1, Q_{211}, Q_{222}), && (A_2, D_2, Q_{212}, Q_{221}), \\ (B_1, D_1, Q_{112}, Q_{211}), && (B_1, D_2, Q_{111}, Q_{212}), && (B_2, D_1, Q_{121}, Q_{222}), && (B_2, D_2, Q_{122}, Q_{221}), \\ (C_1, D_1, Q_{121}, Q_{211}), && (C_1, D_2, Q_{111}, Q_{221}), && (C_2, D_1, Q_{112}, Q_{222}), && (C_2, D_2, Q_{122}, Q_{212}). \end{align*}$
Proof : Since $\measuredangle B_1Q_{111}C_1$ $=$ $\measuredangle Q_{111}B_1C_1$ $+$ $\measuredangle B_1C_1Q_{111}$ $=$ $\measuredangle B_2A_2C_1$ $+$ $\measuredangle B_1A_2C_2$ $=$ $\measuredangle B_1A_2C_1$ $-$ $\measuredangle C_2A_2B_2$ $=$ $\measuredangle B_1C_1Q_{211}$ $+$ $\measuredangle Q_{211}B_1C_1$ $=$ $\measuredangle B_1Q_{211}C_1,$ so we conclude that $B_1,$ $C_1,$ $Q_{111},$ $Q_{211}$ are concyclic. $\qquad \blacksquare$

Property 12 : $\begin{align*} \text{There exists a transformation } \Phi_{(\eta_{111}, \tau_{111})}^{M_{111}} \text{ swaps } (A_1, Q_{211}), && (B_1, Q_{121}), && (C_1, Q_{112}), && (D_1, Q_{111}). \\ \text{There exists a transformation } \Phi_{(\eta_{211}, \tau_{211})}^{M_{211}} \text{ swaps } (A_2, Q_{111}), && (B_1, Q_{221}), && (C_1, Q_{212}), && (D_2, Q_{211}). \\ \text{There exists a transformation } \Phi_{(\eta_{121}, \tau_{121})}^{M_{121}} \text{ swaps } (A_1, Q_{221}), && (B_2, Q_{111}), && (C_1, Q_{122}), && (D_2, Q_{121}). \\ \text{There exists a transformation } \Phi_{(\eta_{112}, \tau_{112})}^{M_{112}} \text{ swaps } (A_1, Q_{212}), && (B_1, Q_{122}), && (C_2, Q_{111}), && (D_2, Q_{112}). \\ \text{There exists a transformation } \Phi_{(\eta_{122}, \tau_{122})}^{M_{122}} \text{ swaps } (A_1, Q_{222}), && (B_2, Q_{112}), && (C_2, Q_{121}), && (D_1, Q_{122}). \\ \text{There exists a transformation } \Phi_{(\eta_{212}, \tau_{212})}^{M_{212}} \text{ swaps } (A_2, Q_{112}), && (B_1, Q_{222}), && (C_2, Q_{211}), && (D_1, Q_{212}). \\ \text{There exists a transformation } \Phi_{(\eta_{221}, \tau_{221})}^{M_{221}} \text{ swaps } (A_2, Q_{121}), && (B_2, Q_{211}), && (C_1, Q_{222}), && (D_1, Q_{221}). \\ \text{There exists a transformation } \Phi_{(\eta_{222}, \tau_{222})}^{M_{222}} \text{ swaps } (A_2, Q_{122}), && (B_2, Q_{212}), && (C_2, Q_{221}), && (D_2, Q_{222}). \end{align*}$
Proof : Since $\begin{align*} \measuredangle Q_{211}B_1C_1 = \measuredangle B_2A_1C_1, && \measuredangle Q_{121}C_1A_1 = \measuredangle C_2B_1A_1, && \measuredangle Q_{112}A_1B_1 = \measuredangle A_2C_1B_1, \\ \measuredangle Q_{211}C_1B_1 = \measuredangle C_2A_1B_1, && \measuredangle Q_{121}A_1C_1 = \measuredangle A_2B_1C_1, && \measuredangle Q_{112}B_1A_1 = \measuredangle B_2C_1A_1, \end{align*}$ so we get the hexagon $A_1Q_{121}C_1Q_{211}B_1Q_{112}$ satisfy $(\spadesuit)$ $\Longrightarrow$ there exists a transformation $\Phi_{(\eta_{111}, \tau_{111})}^{M_{111}}$ swaps $(A_1, Q_{211}),$ $(B_1, Q_{121}),$ $(C_1, Q_{112})$ (Property 3). Notice $\odot (A_1Q_{121}Q_{112}),$ $\odot (B_1Q_{112}Q_{211}),$ $\odot (C_1Q_{211}Q_{121})$ are concurrent at $D_1$ and $\odot (Q_{211}B_1C_1),$ $\odot (Q_{121}C_1A_1),$ $\odot (Q_{112}A_1B_1)$ are concurrent at $Q_{111}$ (Property 11), so we conclude that $\Phi_{(\eta_{111}, \tau_{111})}^{M_{111}}$ also swaps $D_1$ and $Q_{111}.$ $\qquad \blacksquare$

Property 13 : $\begin{align*} A_1B_1C_1D_1 \stackrel{-}{\sim} Q_{122}Q_{212}Q_{221}Q_{222} , && A_2B_1C_1D_2 \stackrel{-}{\sim} Q_{222}Q_{112}Q_{121}Q_{122} , && A_1B_2C_1D_2 \stackrel{-}{\sim} Q_{112}Q_{222}Q_{211}Q_{212} , && A_1B_1C_2D_2 \stackrel{-}{\sim} Q_{121}Q_{211}Q_{222}Q_{221}, \\ A_2B_2C_2D_2 \stackrel{-}{\sim} Q_{211}Q_{121}Q_{112}Q_{111} , && A_1B_2C_2D_1 \stackrel{-}{\sim} Q_{111}Q_{221}Q_{212}Q_{211} , && A_2B_1C_2D_1 \stackrel{-}{\sim} Q_{221}Q_{111}Q_{122}Q_{121} , && A_2B_2C_1D_1 \stackrel{-}{\sim} Q_{212}Q_{122}Q_{111}Q_{112}. \end{align*}$
Proof : Since the hexagon $A_2Q_{212}C_2Q_{122}B_2Q_{221}$ satisfy $(\spadesuit),$ so from Property 1 we get $\measuredangle Q_{122}Q_{212}Q_{221}$ $=$ $\measuredangle Q_{122}C_2B_2$ $+$ $\measuredangle B_2A_2Q_{221}$ $=$ $\measuredangle C_1A_2B_2$ $+$ $\measuredangle B_2C_2A_1$ $=$ $\measuredangle C_1B_1A_1.$ Similarly, we can prove $\measuredangle Q_{212}Q_{221}Q_{122}$ $=$ $\measuredangle A_1C_1B_1,$ so $\triangle A_1B_1C_1$ $\stackrel{-}{\sim}$ $\triangle Q_{122}Q_{212}Q_{221}.$ Analogously, we can prove $\triangle B_1C_1D_1$ $\stackrel{-}{\sim}$ $\triangle Q_{212}Q_{221}Q_{222},$ so we conclude that $A_1B_1C_1D_1$ $\stackrel{-}{\sim}$ $Q_{122}Q_{212}Q_{221}Q_{222}.$ $\qquad \blacksquare$

Corollary 14 : The points in the following 4-points sets are concyclic. $\begin{align*} (Q_{122}, Q_{212}, Q_{221}, Q_{222}) , && (Q_{222}, Q_{112}, Q_{121}, Q_{122}) , && (Q_{112}, Q_{222}, Q_{211}, Q_{212}) , && (Q_{121}, Q_{211}, Q_{222}, Q_{221}), \\ (Q_{211}, Q_{121}, Q_{112}, Q_{111}) , && (Q_{111}, Q_{221}, Q_{212}, Q_{211}) , && (Q_{221}, Q_{111}, Q_{122}, Q_{121}) , && (Q_{212}, Q_{122}, Q_{111}, Q_{112}). \end{align*}$

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Telv Cohl's Geometry Blog

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