Orthologic Quadrilateral

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Orthologic Quadrilateral

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Source: Sharygin Lecture; Alexey Zaslavaky

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#1 h Aug 12, 2016, 7:33 PM • 5 Y

Y by mineiraojose, 62861, AmirKhusrau, Adventure10, Mango247

Two quadrilaterals $ABCD$ and $A'B'C'D'$ are called orthologic if all corresponding four triangles formed by taking three of their vertices are orthologic. Prove that there exists an affine homography which maps $ABCD$ to $A'B'C'D'$ .

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#2 h Aug 12, 2016, 7:35 PM • 1 Y

Y by Adventure10

Comment. Mr. Alexey has a solution to this, so it isn't an open synthetic problem.

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#3 h Aug 15, 2016, 9:45 PM • 8 Y

Y by mineiraojose, anantmudgal09, baopbc, baladin, AmirKhusrau, enhanced, Adventure10, Mango247

Let $P$ be the isogonal conjugate (WRT $\triangle ABC$ ) of the orthology center of $\triangle ABC$ WRT $\triangle A'B'C',$ so $\triangle A'B'C'$ and the pedal triangle $\triangle P_aP_bP_c$ of $P$ WRT $\triangle ABC$ are homothetic. Let $T$ be the point such that $A'B'C'D',$ $P_aP_bP_cT$ are homothetic and let $\triangle T_aT_bT_c$ be the pedal triangle of $T$ WRT $\triangle P_aP_bP_c.$ Let $X$ be the point such that $\triangle BXC,$ $\triangle P_bTP_c$ are inversely similar and define $Y,$ $Z,$ similarly. We divide the proof into three steps.

Claim 1. $\triangle XYZ$ and $\triangle T_aT_bT_c$ are homothetic.

Proof. Since $\frac{AY}{AZ}$ $=$ $\frac{AY}{AC}$ $\cdot$ $\frac{AC}{AB}$ $\cdot$ $\frac{AB}{AZ}$ $=$ $\frac{P_aT}{P_aP_c}$ $\cdot$ $\frac{AC}{AB}$ $\cdot$ $\frac{P_aP_b}{P_aT}$ $=$ $\frac{PC}{PB},$ so notice $\measuredangle YAZ$ $=$ $\measuredangle CPB$ we get $\triangle AYZ$ $\stackrel{+}{\sim}$ $\triangle PCB$ $\Longrightarrow$ $\measuredangle (YZ,BC)$ $=$ $\measuredangle (AY, AC)$ $+$ $\measuredangle (AC, PC)$ $=$ $\measuredangle (P_aP_c, P_aT)$ $+$ $\measuredangle (P_aP_b, P_aP)$ $=$ $\measuredangle (T_bT_c, BC),$ hence $YZ$ $\parallel$ $T_bT_c.$ Similarly, we can prove $ZX$ $\parallel$ $T_cT_a$ and $XY$ $\parallel$ $T_aT_b,$ so we conclude that $\triangle XYZ,$ $\triangle T_aT_bT_c$ are homothetic. $\qquad \blacksquare$

Claim 2. $XYZD$ and $T_aT_bT_cT$ are homothetic.

Proof. Let $D_1$ be the point such that $XYZD_1,$ $T_aT_bT_cT$ are homothetic and $X_1$ $\equiv$ $D_1X$ $\cap$ $\odot (BXC).$ Since $\measuredangle (BX_1, XX_1)$ $=$ $\measuredangle BCX$ $=$ $\measuredangle TP_cP_b,$ so notice $XX_1$ $\perp$ $P_bP_c$ we get $BX_1$ $\perp$ $TP_c.$ Similarly, we can get $CX_1$ $\perp$ $TP_b,$ so $X_1$ is the orthology center of $\triangle BD_1C$ WRT $\triangle P_bTP_c$ $\Longrightarrow$ $X_1$ is the orthology center of $\triangle BD_1C$ WRT $\triangle B'D'C'.$ By similar discussion, we can prove $\triangle CD_1A,$ $\triangle C'D'A'$ are orthologic and $\triangle AD_1B,$ $\triangle A'D'B'$ are orthologic, so we conclude that $D_1$ $\equiv$ $D.$ $\qquad \blacksquare$

Claim 3. There exists an affine homography that maps $ABCD$ to $A'B'C'D'.$

Proof. Let $\odot (AYZ)$ cuts $CA,$ $AB$ at $K,$ $L,$ respectively. Since $\measuredangle KZY$ $=$ $\measuredangle CAY$ $=$ $\measuredangle TP_aP_c$ $=$ $\measuredangle TP_cP_b$ $=$ $\measuredangle DZY,$ so $D,$ $K,$ $Z$ are collinear. Similarly, we can prove $D,$ $L,$ $Y$ are collinear. Let $\triangle D_AD_BD_C$ be the cevian triangle of $D$ WRT $\triangle ABC$ and let $\triangle T_AT_BT_C$ be the cevian triangle of $T$ WRT $\triangle P_aP_bP_c.$ Since $A'B'C'D',$ $P_aP_bP_cT$ are homothetic, so it suffices to prove $\frac{CD_A}{BD_A}$ $=$ $\frac{P_cT_A}{P_bT_A},$ $\frac{AD_B}{CD_B}$ $=$ $\frac{P_aT_B}{P_cT_B},$ $\frac{BD_C}{AD_C}$ $=$ $\frac{P_bT_C}{P_aT_C}.$ From symmetry, it suffices to prove $\frac{CD_A}{BD_A}$ $=$ $\frac{P_cT_A}{P_bT_A}.$

By Ceva's theorem we get $\frac{\sin \angle DAC}{\sin \angle DAB}$ $=$ $\frac{\sin \angle DKA}{\sin \angle DLA}$ $\cdot$ $\frac{\sin \angle DLK}{\sin \angle DKL}$ $=$ $\frac{\sin \angle PCB}{\sin \angle PBC}$ $\cdot$ $\frac{\sin \angle TP_aP_c}{\sin \angle TP_aP_b},$ so we conclude that $\frac{CD_A}{BD_A} = \frac{AC}{AB} \cdot \frac{\sin \angle DAC}{\sin \angle DAB} = \frac{AC}{AB} \cdot \frac{\sin \angle PCB}{\sin \angle PBC} \cdot \frac{\sin \angle TP_aP_c}{\sin \angle TP_aP_b} = \frac{P_aP_c}{P_aP_b} \cdot \frac{\sin \angle TP_aP_c}{\sin \angle TP_aP_b} = \frac{P_cT_A}{P_bT_A}. \qquad \blacksquare$

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