Difference between revisions of "Complete Quadrilateral"
(→Shatunov line) |
(→Areas in complete quadrilateral) |
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*[[Simson line]] | *[[Simson line]] | ||
*[[Steiner line]] | *[[Steiner line]] | ||
+ | *[[Gauss line]] | ||
==Radical axis== | ==Radical axis== | ||
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<cmath>\angle CKD = 90^\circ \implies K \in \theta \implies Po(H)_{\omega} = CH \cdot KH = AH \cdot PH = Po(H)_{\Omega}.</cmath> | <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. | 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''' | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
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b) line <math>PQ</math> is symmetric to Steiner line with respect centroid of <math>BDEC.</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 | + | 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> | <i><b>Proof</b></i> |
Latest revision as of 16:16, 5 May 2023
Contents
Complete quadrilateral
Let four lines made four triangles of a complete quadrilateral. In the diagram these are
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
Let points and
be the orthocenters of
and
respectively.
Let circles and
be the circles with diameters
and
respectively.
Prove that Steiner line
is the radical axis of
and
Proof
Let points and
be the foots of perpendiculars
and
respectively.
Denote power of point
with respect the circle
Therefore power of point
with respect these three circles is the same. These points lies on the common radical axis of
and
Steiner line
is the radical axis as desired.
vladimir.shelomovskii@gmail.com, vvsss
Areas in complete quadrilateral
Let complete quadrilateral be given
. Let
and
be the midpoints of
and
respectively.
Prove that where
is the area of
Proof
Let and
be the projections of
and
respectively onto
Let and
be the circles with diameters
and
respectively.
Similarly
Let common radical axes of and
cross
at point
The power of the point
with respect
and
is the same, therefore
Let
Therefore
vladimir.shelomovskii@gmail.com, vvsss
Newton–Gauss line
Let four lines made four triangles of a complete quadrilateral.
In the diagram these are
Let points and
be the midpoints of
and
respectively.
Let points and
be the orthocenters of
and
respectively.
Prove that Steiner line is perpendicular to Gauss line
Proof
Points and
are the centers of circles with diameters
and
respectively.
Steiner line is the radical axis of these circles.
Therefore as desired.
vladimir.shelomovskii@gmail.com, vvsss
Shatunov-Tokarev line
Let the complete quadrilateral ABCDEF be labeled as in the diagram. Quadrilateral is not cyclic.
Let points be the orthocenters and points
be the circumcenters of
and
respectively.
Let bisector cross bisector
at point
Let bisector
cross bisector
at point
Prove that
a) points and
lie on circumcircle of
b) line is symmetric to Steiner line with respect centroid of
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 and
lies on bisector of
points
and
lies on bisector of
circle
Similarly circle
as desired.
b) Let and
be midpoints of
and
respectively.
It is clear that is centroid of
(midline of trapezium
(midline of trapezium
is parallelogram.
Similarly one can prove that point the midpoint of
is symmetric to
with respect
Therefore line coincide with Steiner line and line
is symmetric to Steiner line with respect
and is parallel to this line.
vladimir.shelomovskii@gmail.com, vvsss