Difference between revisions of "Complete Quadrilateral"

Latest revision as of 03:42, 21 November 2024

Complete quadrilateral

Let four lines made four triangles of a complete quadrilateral. In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$ One can see some of the properties of this configuration and their proof using the following links.

Radical axis

Let four lines made four triangles of a complete quadrilateral. In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$

Let points $H,$ and $H_A$ be the orthocenters of $\triangle ABC$ and $\triangle ADE,$ respectively.

Let circles $\omega, \theta,$ and $\Omega$ be the circles with diameters $CD, BE,$ and $AF,$ respectively. Prove that Steiner line $HH_A$ is the radical axis of $\omega, \theta,$ and $\Omega.$

Proof

Let points $G, K, L, N, P,$ and $Q$ be the foots of perpendiculars $AH_A, CH, DH_A, BH, AH,$ and $EH_A,$ respectively.

Denote $Po(X)_{\omega}$ power of point $X$ with respect the circle $\omega.$ $\angle AGF = 90^\circ \implies G \in \Omega \implies Po(H_A)_{\Omega} = AH_A \cdot GH_A.$ $\angle APF = 90^\circ \implies P \in \Omega \implies Po(H)_{\Omega} = AH \cdot PH_A.$ $\angle CLD = 90^\circ \implies L \in \omega \implies Po(H_A)_{\omega} = DH_A \cdot LH_A = AH_A \cdot GH_A = Po(H_A)_{\Omega}.$ $\angle EQB = 90^\circ \implies Q \in \theta \implies Po(H_A)_{\theta} = EH_A \cdot QH_A = AH_A \cdot GH_A = Po(H_A)_{\Omega}.$

$\angle BNE = 90^\circ \implies N \in \theta \implies Po(H)_{\theta} = BH \cdot NH = AH \cdot PH = Po(H)_{\Omega}.$ $\angle CKD = 90^\circ \implies K \in \theta \implies Po(H)_{\omega} = CH \cdot KH = AH \cdot PH = Po(H)_{\Omega}.$ Therefore power of point $H (H_A)$ with respect these three circles is the same. These points lies on the common radical axis of $\omega, \theta,$ and $\Omega \implies$ Steiner line $HH_A$ is the radical axis as desired.

vladimir.shelomovskii@gmail.com, vvsss

Areas in complete quadrilateral

Let complete quadrilateral $ABCDEF$ be given $(E = AC \cap BD, F = AB \cap CD)$ . Let $X, Y,$ and $Z$ be the midpoints of $BC, AD,$ and $EF,$ respectively.

Prove that $\frac {[ADFE]}{[ADC]} = \frac {ZY}{XY},$ where $[t]$ is the area of $t.$

Proof

Let $X_1, Y_1, D_1, C_1, Z_1,$ and $F_1$ be the projections of $X, Y, A, B, Z,$ and $E,$ respectively onto $CD.$

Let $\omega, \theta,$ and $\Omega$ be the circles with diameters $AD, BC,$ and $EF,$ respectively. $\angle EF_1F = 90^\circ \implies F_1 \in \Omega.$ Similarly $D_1 \in \omega, C_1 \in \theta.$

Let common radical axes of $\omega, \Omega,$ and $\theta$ cross $CD$ at point $G.$ $EF_1 || AD_1 \implies \frac {EC}{AC} = \frac {F_1C}{D_1C}.$ The power of the point $G$ with respect $\omega, \theta,$ and $\Omega$ is the same, therefore $GF \cdot GF_1 = GC \cdot GC_1 = GD \cdot GD_1 \implies$ $(\vec G - \vec F) \cdot (\vec G - \vec F_1) = (\vec G - \vec C) \cdot (\vec G - \vec C_1) = (\vec G - \vec D) \cdot (\vec G – \vec D_1) \implies$ $\vec F \cdot \vec F_1 – \vec G \cdot (\vec F + \vec F_1) = \vec F \cdot \vec F_1 – \vec G \cdot 2\vec Z_1 = \vec C \cdot \vec C_1 – \vec G \cdot (\vec C + \vec C_1) =$ $= \vec C \cdot \vec C_1 - \vec G \cdot 2\vec X_1 = \vec D \cdot \vec D_1 – \vec G \cdot (\vec D + \vec D_1) = \vec D \cdot \vec D_1 - \vec G \cdot 2\vec Y_1 \implies$

$|\vec G| = \frac {\vec F \cdot\vec F_1 - \vec C \cdot \vec C_1}{2|\vec Z_1 - \vec X_1|} = \frac {\vec D \cdot \vec D_1 - \vec C \cdot \vec C_1}{2|\vec Y_1 – \vec X_1|} \implies$ $\frac {|\vec Z_1 - \vec X_1|}{|\vec Y_1 - \vec X_1|} = \frac {\vec F \cdot \vec F_1 - \vec C \cdot \vec C_1} {\vec D \cdot \vec D_1 - \vec C \cdot\vec C_1} = \frac {Z_1X_1}{Y_1X_1} = \frac {ZX}{YX} .$ Let $\vec C = \vec 0 \implies \vec C \cdot\vec C_1 = 0, \vec F \cdot \vec F_1 = FC \cdot F_1C, \vec D \cdot \vec D_1 = DC \cdot D_1C.$ $\frac {ZX}{YX} = \frac {FC \cdot F_1C} {DC \cdot D_1C} = \frac {FC \cdot EC} {DC \cdot AC} = \frac {[CEF]}{[CAD]}.$

Therefore $\frac {[ADEF]}{[CAD]} = \frac {[CEF]-[CAD]}{[CAD]} =\frac {ZX - YX}{YX} = \frac {ZY}{YX}.$

vladimir.shelomovskii@gmail.com, vvsss

Newton–Gauss line

Let four lines made four triangles of a complete quadrilateral.

In the diagram these are $\triangle ABC, \triangle ADE, \triangle CEF, \triangle BDF.$

Let points $K, L,$ and $N$ be the midpoints of $BE, CD,$ and $AF,$ respectively.

Let points $H$ and $H_A$ be the orthocenters of $\triangle ABC$ and $\triangle ADE,$ respectively.

Prove that Steiner line $HH_A$ is perpendicular to Gauss line $KLN.$

Proof

Points $K, L,$ and $N$ are the centers of circles with diameters $BE, CD,$ and $AF,$ respectively.

Steiner line $HH_A$ is the radical axis of these circles.

Therefore $HH_A \perp KL$ as desired.

vladimir.shelomovskii@gmail.com, vvsss

Shatunov-Tokarev line

Let the complete quadrilateral ABCDEF be labeled as in the diagram. Quadrilateral $BCED$ is not cyclic.

Let points $H, H_A, H_B, H_C$ be the orthocenters and points $O, O_A, O_B, O_C$ be the circumcenters of $\triangle ABC, \triangle ADE, \triangle BDF,$ and $\triangle CEF,$ respectively.

Let bisector $BD$ cross bisector $CE$ at point $Q.$ Let bisector $BC$ cross bisector $DE$ at point $P.$

Prove that

a) points $P$ and $Q$ lie on circumcircle of $\triangle OO_AO_C,$

b) line $PQ$ is symmetric to Steiner line with respect centroid of $BDEC.$

I suppose that this line was found independently by two young mathematicians Leonid Shatunov and Alexander Tokarev in 2022. I would be grateful for information on whether this line was previously known.

Proof

a) Points $P$ and $O$ lies on bisector of $BC,$ points $P$ and $O_A$ lies on bisector of $DE \implies$

$\angle OPO_A + \angle OO_CO_A = 180^\circ \implies P \in$ circle $OO_AO_C.$

Similarly $P \in$ circle $OO_BO_C$ as desired.

Miquel's point

b) Let $M, M', P'$ and $G$ be midpoints of $CE, BD, HH_A,$ and $MM',$ respectively.

It is clear that $G$ is centroid of $BDEC.$

$M'P \perp BD, EH_A \perp AD, CH \perp AB \implies MP' \perp AB$ (midline of trapezium $ECHH_A) \implies EH_A||MP'||CH||M'P.$ $MP \perp CE, DH_A \perp AE, BH \perp AC \implies$ $M'P' \perp AB$ (midline of trapezium $HBDH_A) \implies$ $DH_A||M'P'||BH||MP \implies M'P'MP$ is parallelogram.

Similarly one can prove that point $Q',$ the midpoint of $H_BH_C,$ is symmetric to $Q$ with respect $G.$

Therefore line $P'Q'$ coincide with Steiner line and line $PQ$ is symmetric to Steiner line with respect $G$ and is parallel to this line.

vladimir.shelomovskii@gmail.com, vvsss

Complete quadrilateral theorem

Let points $A, B, C, D,$ no three of which are collinear, be given. $P = AD \cap BC, Q = AC \cap BD, E = AB \cap PQ, F = CD \cap PQ.$ Prove that $\frac{QE \cdot PF}{PE \cdot QF} = 1.$

Proof

We make the projective transformation of the vertices of $ABCD$ into vertices of the square. (Projecting non-convex quadrilateral into rectangle) Then image of the point $P$ is the point at infinity, image of $Q$ is the center of the square, images of $AD, PQ,$ and $BC$ are parallel, so for images $QF = QE$ and $\frac {PE}{PF} = 1 \implies \frac{QE \cdot PF}{PE \cdot QF} = 1.$

The double ratio $\frac{QE \cdot PF}{PE \cdot QF}$ is the projective invariant of a quadruple of collinear points $(P,Q;E,F)$ so the equality also holds for the preimages.

vladimir.shelomovskii@gmail.com, vvsss

@@ Line 5: / Line 5: @@
 *[[Simson line]]
 *[[Steiner line]]
+*[[Gauss line]]
 ==Radical axis==
@@ Line 28: / Line 29: @@
 <cmath>\angle CKD = 90^\circ \implies K \in \theta \implies Po(H)_{\omega} = CH \cdot KH = AH \cdot PH = Po(H)_{\Omega}.</cmath>
 Therefore power of point <math>H (H_A)</math> with respect these three circles is the same. These points lies on the common radical axis of <math>\omega, \theta,</math> and <math>\Omega \implies</math> Steiner line <math>HH_A</math> is the radical axis as desired.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Areas in complete quadrilateral==
+[[File:Complete areas.png|400px|right]]
+Let complete quadrilateral <math>ABCDEF</math> be given <math>(E = AC \cap BD, F = AB \cap CD)</math>. Let <math>X, Y,</math> and <math>Z</math> be the midpoints of <math>BC, AD,</math> and <math>EF,</math> respectively.
+Prove that <math>\frac {[ADFE]}{[ADC]} = \frac {ZY}{XY},</math> where <math>[t]</math> is the area of <math>t.</math>
+<i><b>Proof</b></i>
+[[File:Complete areas 1.png|400px|right]]
+Let <math>X_1, Y_1, D_1, C_1, Z_1,</math> and <math>F_1</math> be the projections of <math>X, Y, A, B, Z,</math> and <math>E,</math> respectively onto <math>CD.</math>
+Let <math>\omega, \theta,</math> and <math>\Omega</math> be the circles with diameters <math>AD, BC,</math> and <math>EF,</math> respectively.
+<cmath>\angle EF_1F = 90^\circ \implies F_1 \in \Omega.</cmath>
+Similarly <math>D_1 \in \omega, C_1 \in \theta.</math>
+Let common radical axes of <math>\omega, \Omega,</math> and <math>\theta</math> cross <math>CD</math> at point <math>G.</math>
+<cmath>EF_1 || AD_1 \implies \frac {EC}{AC} = \frac {F_1C}{D_1C}.</cmath>
+The power of the point <math>G</math> with respect <math>\omega, \theta,</math> and <math>\Omega</math> is the same, therefore
+<cmath>GF \cdot GF_1 = GC \cdot GC_1 = GD \cdot GD_1 \implies </cmath>
+<cmath>(\vec G - \vec F) \cdot (\vec G - \vec F_1) = (\vec G - \vec C) \cdot (\vec G - \vec C_1) = (\vec G - \vec D) \cdot (\vec G – \vec D_1) \implies </cmath>
+<cmath>\vec F \cdot \vec F_1 – \vec G \cdot (\vec F + \vec F_1) = \vec F \cdot \vec F_1 – \vec G \cdot 2\vec Z_1 = \vec C \cdot \vec C_1 – \vec G \cdot (\vec C + \vec C_1) =</cmath>
+<cmath>= \vec C \cdot \vec C_1 - \vec G \cdot 2\vec X_1 = \vec D \cdot \vec D_1 – \vec G \cdot (\vec D + \vec D_1) = \vec D \cdot \vec D_1 - \vec G \cdot 2\vec Y_1 \implies</cmath>
+<cmath>|\vec G| = \frac {\vec F \cdot\vec  F_1 - \vec C \cdot \vec C_1}{2|\vec Z_1 - \vec X_1|} = \frac {\vec D \cdot \vec D_1 - \vec C \cdot \vec C_1}{2|\vec Y_1 – \vec X_1|} \implies</cmath>
+<cmath>\frac {|\vec Z_1 - \vec X_1|}{|\vec Y_1 - \vec X_1|} =  \frac {\vec F \cdot \vec F_1 - \vec C \cdot \vec C_1} {\vec D \cdot \vec D_1 - \vec C \cdot\vec C_1} = \frac {Z_1X_1}{Y_1X_1} = \frac {ZX}{YX} .</cmath>
+Let <math>\vec C = \vec 0 \implies \vec C \cdot\vec C_1 = 0, \vec F \cdot \vec F_1 = FC \cdot F_1C, \vec D \cdot \vec D_1 = DC \cdot D_1C.</math>
+<cmath>\frac {ZX}{YX} = \frac {FC \cdot F_1C} {DC \cdot D_1C} = \frac {FC \cdot EC} {DC \cdot AC} = \frac {[CEF]}{[CAD]}.</cmath>
+Therefore  <cmath>\frac {[ADEF]}{[CAD]} =  \frac {[CEF]-[CAD]}{[CAD]} =\frac {ZX - YX}{YX} = \frac {ZY}{YX}.</cmath>
 '''vladimir.shelomovskii@gmail.com, vvsss'''
@@ Line 45: / Line 78: @@
 <i><b>Proof</b></i>
-Points <math>K, L,</math> and <math>N</math> are the centers of circles with diameters  <math>CD, BE,</math> and <math>AF,</math> respectively.
+Points <math>K, L,</math> and <math>N</math> are the centers of circles with diameters  <math>BE, CD,</math> and <math>AF,</math> respectively.
 Steiner line <math>HH_A</math> is the radical axis of these circles.
 Therefore <math>HH_A \perp KL</math> as desired.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Shatunov-Tokarev line==
+[[File:Shatunov line 3.png|500px|right]]
+Let the complete quadrilateral ABCDEF be labeled as in the diagram. Quadrilateral <math>BCED</math> is not cyclic.
+Let points <math>H, H_A, H_B, H_C</math> be the orthocenters and points <math>O, O_A, O_B, O_C</math> be the circumcenters of <math>\triangle ABC, \triangle ADE, \triangle BDF,</math> and <math>\triangle CEF,</math> respectively.
+Let bisector <math>BD</math> cross bisector <math>CE</math> at point <math>Q.</math> Let bisector <math>BC</math> cross bisector <math>DE</math> at point <math>P.</math>
+Prove that
+a) points <math>P</math> and <math>Q</math> lie on circumcircle of <math>\triangle OO_AO_C,</math>
+b) line <math>PQ</math> is symmetric to Steiner line with respect centroid of <math>BDEC.</math>
+I suppose that this line was found independently by two young mathematicians Leonid Shatunov and Alexander Tokarev in 2022. I would be grateful for information on whether this line was previously known.
+<i><b>Proof</b></i>
+[[File:Shatunov line 2.png|500px|right]]
+a) Points <math>P</math> and <math>O</math> lies on bisector of <math>BC,</math> points <math>P</math> and <math>O_A</math> lies on bisector of <math>DE \implies</math>
+<math>\angle OPO_A + \angle OO_CO_A = 180^\circ \implies P \in </math> circle <math>OO_AO_C.</math>
+Similarly <math>P \in </math> circle <math>OO_BO_C</math> as desired.
+*[[Miquel's point]]
+b) Let <math>M, M', P'</math> and <math>G</math> be midpoints of <math>CE, BD, HH_A,</math> and <math>MM',</math> respectively.
+It is clear that <math>G</math> is centroid of <math>BDEC.</math>
+<math>M'P \perp BD, EH_A \perp AD, CH \perp AB \implies MP' \perp AB</math> (midline of trapezium <math>ECHH_A) \implies EH_A||MP'||CH||M'P.</math>
+<cmath>MP \perp CE, DH_A \perp AE, BH \perp AC \implies</cmath>
+<math>M'P' \perp AB</math> (midline of trapezium <math>HBDH_A) \implies</math>
+<math>DH_A||M'P'||BH||MP \implies M'P'MP</math> is parallelogram.
+Similarly one can prove that point <math>Q',</math> the midpoint of <math>H_BH_C,</math> is symmetric to <math>Q</math> with respect <math>G.</math>
+Therefore line <math>P'Q'</math> coincide with Steiner line and line <math>PQ</math> is symmetric to Steiner line with respect <math>G</math> and is parallel to this line.
+'''vladimir.shelomovskii@gmail.com, vvsss'''
+==Complete quadrilateral theorem==
+[[File:Complete quadrilateral map.png|350px|right]]
+Let points <math>A, B, C, D,</math> no three of which are collinear, be given.
+<cmath>P = AD \cap BC, Q = AC \cap BD, E = AB \cap PQ, F = CD \cap PQ.</cmath>
+Prove that <math>\frac{QE \cdot PF}{PE \cdot QF} = 1.</math>
+<i><b>Proof</b></i>
+We make the projective transformation of the vertices of <math>ABCD</math> into vertices of the square. ([[Projective geometry (simplest cases) |Projecting non-convex quadrilateral into rectangle]])
+Then image of the point <math>P</math> is the point at infinity, image of <math>Q</math> is the center of the square, images of <math>AD, PQ,</math> and <math>BC</math> are parallel, so for images <math>QF = QE</math> and <cmath>\frac {PE}{PF} = 1  \implies \frac{QE \cdot PF}{PE \cdot QF} = 1.</cmath>
+The double ratio <math>\frac{QE \cdot PF}{PE \cdot QF}</math> is the projective invariant of a quadruple of collinear points <math>(P,Q;E,F)</math> so the equality also holds for the preimages.
 '''vladimir.shelomovskii@gmail.com, vvsss'''

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Difference between revisions of "Complete Quadrilateral"

Latest revision as of 03:42, 21 November 2024

Contents

Complete quadrilateral

Radical axis

Areas in complete quadrilateral

Newton–Gauss line

Shatunov-Tokarev line

Complete quadrilateral theorem