Difference between revisions of "Spieker center"
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We use notation of previous proof. <math>DG</math> is the segment contains the Spieker center, <math>G \in AB, \angle EDG = \angle FDG, H = DG \cap AC.</math> WLOG, <math>AC > AB.</math> | We use notation of previous proof. <math>DG</math> is the segment contains the Spieker center, <math>G \in AB, \angle EDG = \angle FDG, H = DG \cap AC.</math> WLOG, <math>AC > AB.</math> | ||
<cmath>DF||AC \implies \angle AHG = \angle FDG.</cmath> | <cmath>DF||AC \implies \angle AHG = \angle FDG.</cmath> | ||
− | Similarly, <math> \angle AGH = \angle EDG = \angle AHG | + | Similarly, <math> \angle AGH = \angle EDG = \angle AHG.</math> |
+ | |||
+ | So <math>AH = AG \implies CH = AB + AH \implies DH</math> is cleaver. | ||
+ | |||
Therefore, the three cleavers meet at the Spieker center. | Therefore, the three cleavers meet at the Spieker center. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Radical center of excircles== | ||
+ | [[File:Radical center.png|370px|right]] | ||
+ | Prove that the Spieker center of triangle is the radical center of the three excircles. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>\triangle ABC</math> be given,<math>M_A, M_B, M_C</math> be the midpoints of <math>BC, AC, BC,</math> respectively. | ||
+ | |||
+ | Let <math>\omega_A, \omega_B, \omega_C</math> be A-excircle, B-excircle, C-excircle centered at <math>X,Y,Z,</math> respectively. | ||
+ | |||
+ | Let <math>I</math> be the incenter of <math>\triangle ABC.</math> | ||
+ | Let <math>R_A</math> be the radical axis of <math>\omega_B</math> and <math>\omega_C, R_B</math> be the radical axis of <math>\omega_A</math> and <math>\omega_C, R_C</math> be the radical axis of <math>\omega_B</math> and <math>\omega_A,</math> respectively. | ||
+ | |||
+ | It is known that the distances from <math>B</math> to the tangent points of <math>\omega_C</math> is equal to the distances from <math>C</math> to the tangent points of <math>\omega_B, BM_A = CM_A</math> therefore <math>M_A</math> lies on the radical axis <math>R_A</math> of <math>\omega_B</math> and <math>\omega_C.</math> Similarly, <math>M_B \in R_B, M_C \in R_C.</math> | ||
+ | |||
+ | <math>AI \perp ZY, R_A \perp XY \implies AI || RA \implies R_A</math> is cleaver. Similarly, <math>R_B</math> and <math>R_C</math> are cleavers. | ||
+ | |||
+ | Therefore the radical center of the three excircles coinside with the intersection of the three cleavers of triangle. | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Nagel line== | ||
+ | [[File:Nagel line.png|400px|right]] | ||
+ | Let points <math>I, G, S</math> be the incenter, the centroid and the Spieker center of triangle <math>\triangle ABC,</math> respectively. Prove that points <math>I, G, S</math> are collinear, <math>IG = 2 GS,</math> and the barycentric coordinates of S are <math>{ b+c : c+a : a+b.}</math> | ||
+ | |||
+ | The Nagel line is the line on which points <math>I, G, S,</math> and Nagel point <math>N</math> lie. | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | Let <math>D, E, F</math> be the midpoints of <math>BC, AC, BC,</math> respectively. | ||
+ | Bisector <math>AI</math> is parallel to cleaver <math>DS, BI || ES.</math> | ||
+ | <cmath>AB ||ED, \frac {AB}{DE} = 2 \implies \triangle ABI \sim \triangle DES \implies \frac {AI}{DS} = 2.</cmath> | ||
+ | Centroid <math>G</math> divide the median <math>AD</math> such that <math>\frac {AG}{DG} = 2 \implies </math> | ||
+ | |||
+ | <math>\triangle AGI \sim \triangle DGS \implies \frac {GI}{SG} = 2,</math> and points <math>I, G, S</math> are collinear. | ||
+ | |||
+ | The barycentric coordinates of <math>I</math> are <math>{a : b : c}.</math> The barycentric coordinates of <math>G</math> are <math>{1 : 1 : 1}.</math> | ||
+ | |||
+ | <cmath>\vec {GI} = 2 \vec {SG} \implies \vec {G} - \vec {I} = 2(\vec {S} - \vec {G}) \implies</cmath> | ||
+ | <cmath>\vec {G} = \frac {3 \vec {G} - \vec {I} }{2} = {1 - \frac{a}{a+ b+c} : 1 - \frac {b}{a+b+c} : 1 - \frac{c}{a+b+c}} = { b+c : c+a : a+b.}</cmath> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Shatunov triangle== | ||
+ | [[File:Shatunov triangle A.png|400px|right]] | ||
+ | Let <math>\triangle ABC</math> be given. Let <math>\omega, \omega_A, \omega_B, \omega_C</math> be incircle, A-excircle, B-excircle, C-excircle centered at points <math>I,X,Y,Z,</math> respectively. | ||
+ | |||
+ | Let <math>r_A = EF, r_B = DF, r_C = DE</math> be the radical axes of the inscribed circle and one of the excircles of <math>\triangle ABC.</math> | ||
+ | |||
+ | The triangle <math>\triangle DEF</math> whose sides are <math>r_A, r_B, r_C</math> we name the Shatunov triangle. Accordingly, the vertices of the Shatunov triangle are the radical centers of a pair of excircles and an inscribed circle. | ||
+ | |||
+ | Prove: | ||
+ | |||
+ | a) the heights of the Shatunov triangle lie on the bisectors of the medial triangle. The orthocenter of the Shatunov triangle is the Steiner point of <math>\triangle ABC.</math> | ||
+ | |||
+ | b) The Shatunov triangle is homothetic to the anticomplementary triangle of <math>\triangle ABC</math> with respect to the centroid <math>ABC</math> with coefficient <math>\frac {1}{2}.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | a) Let <math>M_A, M_B, M_C</math> be the midpoints of <math>BC, AC, AB,</math> respectively. | ||
+ | |||
+ | The distances from <math>M_A</math> to the tangent points of <math>\omega</math> and <math>\omega_A</math> are the same, so <math>M_A \in r_A.</math> Similarly <math>M_B \in r_B, M_C \in r_C.</math> | ||
+ | |||
+ | Let <math>T</math> and <math>T_Z</math> be the points of tangency of <math>BC</math> and <math>\omega_B</math> and <math>\omega_C,</math> respectively. | ||
+ | |||
+ | It is clear that <math>TC = BT_Z, BM_A = CM_A \implies M_A</math> lies on the radical axis <math>R_A.</math> | ||
+ | |||
+ | Similarly, <math>M_B</math> lies on the radical axis <math>R_B, M_C</math> lies on the radical axis <math>R_C.</math> | ||
+ | |||
+ | <math>D = r_B \cap r_C \implies</math> D is the radical center of <math>\omega, \omega_B, \omega_C \implies D \in R_A \implies R_A = DM_A.</math> | ||
+ | |||
+ | <math>AI \perp ZY, R_A \perp ZY, r_A \perp AI \implies DM_A \perp EF.</math> Similarly <math>EM_B \perp DF, FM_C \perp DE.</math> | ||
+ | |||
+ | Therefore <math>DM_A, EM_B, FM_C</math> are the heights of <math>\triangle DEF \implies S</math> is the orthocenter of <math>\triangle DEF.</math> | ||
+ | |||
+ | <math>\triangle M_AM_BM_C</math> is the medial triangle of <math>\triangle ABC, DSM_A</math> | ||
+ | is the bisector of <math>\angle M_BM_AM_C \implies S</math> is the Steiner point of <math>\triangle ABC.</math> | ||
+ | |||
+ | b) <math>ZY || EF, XY || ED, ZX || FD \implies \triangle DEF \sim \triangle XYZ.</math> | ||
+ | |||
+ | <math>I</math> is the orthocenter of <math>\triangle XYZ, S</math> is the orthocenter of <math>\triangle DEF.</math> | ||
+ | |||
+ | Points <math>I, G,</math> and <math>S,</math> where <math>G</math> is the centroid <math>ABC,</math> are collinear, sides of is the <math>\triangle DEF</math> are parallel to the respective sides of <math>\triangle XYZ \implies \triangle DEF</math> is homothetic to the <math>\triangle XYZ</math> with respect to <math>G.</math> | ||
+ | |||
+ | <math>IG = 2CS \implies</math> the coefficient of homothety is <math>\frac {1}{2}.</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Latest revision as of 06:48, 8 August 2023
The Spieker center is defined as the center of mass of the perimeter of the triangle. The Spieker center of a is the center of gravity of a homogeneous wire frame in the shape of
The Spieker center is a triangle center and it is listed as the point
Contents
Incenter of medial triangle
Prove that the Spieker center of triangle is the incenter of the medial triangle
of a
Proof
Let's hang up the in the middle of side
Side
is balanced.
Let's replace side with point
(the center of mass of
the midpoint
Denote
the linear density of a homogeneous wire frame.
The mass of point is equal to
the shoulder of the gravity force is
The moment of this force is
Similarly the moment gravity force acting on AB is
Therefore, equilibrium condition is and the center of gravity of a homogeneous wire frame
lies on each bisector of
This point is the incenter of the medial triangle
vladimir.shelomovskii@gmail.com, vvsss
Intersection of three cleavers
Prove that the Spieker center is located at the intersection of the three cleavers of triangle. A cleaver of a triangle is a line segment that bisects the perimeter of the triangle and has one endpoint at the midpoint of one of the three sides.
Proof
We use notation of previous proof. is the segment contains the Spieker center,
WLOG,
Similarly,
So is cleaver.
Therefore, the three cleavers meet at the Spieker center.
vladimir.shelomovskii@gmail.com, vvsss
Radical center of excircles
Prove that the Spieker center of triangle is the radical center of the three excircles.
Proof
Let be given,
be the midpoints of
respectively.
Let be A-excircle, B-excircle, C-excircle centered at
respectively.
Let be the incenter of
Let
be the radical axis of
and
be the radical axis of
and
be the radical axis of
and
respectively.
It is known that the distances from to the tangent points of
is equal to the distances from
to the tangent points of
therefore
lies on the radical axis
of
and
Similarly,
is cleaver. Similarly,
and
are cleavers.
Therefore the radical center of the three excircles coinside with the intersection of the three cleavers of triangle.
vladimir.shelomovskii@gmail.com, vvsss
Nagel line
Let points be the incenter, the centroid and the Spieker center of triangle
respectively. Prove that points
are collinear,
and the barycentric coordinates of S are
The Nagel line is the line on which points and Nagel point
lie.
Proof
Let be the midpoints of
respectively.
Bisector
is parallel to cleaver
Centroid
divide the median
such that
and points
are collinear.
The barycentric coordinates of are
The barycentric coordinates of
are
vladimir.shelomovskii@gmail.com, vvsss
Shatunov triangle
Let be given. Let
be incircle, A-excircle, B-excircle, C-excircle centered at points
respectively.
Let be the radical axes of the inscribed circle and one of the excircles of
The triangle whose sides are
we name the Shatunov triangle. Accordingly, the vertices of the Shatunov triangle are the radical centers of a pair of excircles and an inscribed circle.
Prove:
a) the heights of the Shatunov triangle lie on the bisectors of the medial triangle. The orthocenter of the Shatunov triangle is the Steiner point of
b) The Shatunov triangle is homothetic to the anticomplementary triangle of with respect to the centroid
with coefficient
Proof
a) Let be the midpoints of
respectively.
The distances from to the tangent points of
and
are the same, so
Similarly
Let and
be the points of tangency of
and
and
respectively.
It is clear that lies on the radical axis
Similarly, lies on the radical axis
lies on the radical axis
D is the radical center of
Similarly
Therefore are the heights of
is the orthocenter of
is the medial triangle of
is the bisector of
is the Steiner point of
b)
is the orthocenter of
is the orthocenter of
Points and
where
is the centroid
are collinear, sides of is the
are parallel to the respective sides of
is homothetic to the
with respect to
the coefficient of homothety is
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