Difference between revisions of "Kimberling’s point X(24)"
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<cmath>k = \frac {DE}{KL} = 4\cos \alpha \cdot \cos \beta \cdot \cos \gamma \implies</cmath> | <cmath>k = \frac {DE}{KL} = 4\cos \alpha \cdot \cos \beta \cdot \cos \gamma \implies</cmath> | ||
<math>\frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C \implies P</math> is the point <math>X(24).</math> | <math>\frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C \implies P</math> is the point <math>X(24).</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
+ | ==Theorem 3== | ||
+ | [[File:X24 as Exeter.png|500px|right]] | ||
+ | Let <math>\triangle ABC</math> be the reference triangle (other than a right triangle). Let the altitudes through the vertices <math>A, B, C</math> meet the circumcircle <math>\Omega</math> of triangle <math>ABC</math> at <math>A_0, B_0,</math> and <math>C_0,</math> respectively. Let <math>A'B'C'</math> be the triangle formed by the tangents at <math>A, B,</math> and <math>C</math> to <math>\Omega.</math> (Let <math>A'</math> be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through <math>A_0A', B_0B',</math> and <math>C_0C'</math> are concurrent, the point of concurrence <math>X_24</math> lies on Euler line of triangle <math>ABC, X_{24} = O + \frac {2}{J^2 + 1} (H – O), J = \frac {|OH|}{R}.</math> | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | At first one can prove that lines <math>A_0A', B_0B',</math> and <math>C_0C'</math> are concurrent. This follows from the fact that lines <math>AA_0, BB_0,</math> and <math>CC_0</math> are concurrent at point <math>H</math> and <i><b>Mapping theorem</b></i> (see Exeter point). | ||
+ | Let <math>A_1, B_1,</math> and <math>C_1</math> be the midpoints of <math>BC, AC,</math> and <math>AB,</math> respectively. | ||
+ | Let <math>A_2, B_2,</math> and <math>C_2</math> be the midpoints of <math>AH, BH,</math> and <math>CH,</math> respectively. | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 14:31, 25 November 2022
Kimberling's point X(24)
Kimberling defined point X(24) as perspector of and Orthic Triangle of the Orthic Triangle of
.
Theorem 1
Denote obtuse or acute
Let
be the base triangle,
be Orthic triangle of
be Orthic Triangle of
. Let
and
be the circumcenter and orthocenter of
Then and
are homothetic, the point
center of this homothety lies on Euler line
of
The ratio of the homothety is
Proof
WLOG, we use case
Let be reflection
in
In accordance with Claim,
and
are collinear.
Similarly, and
were
is reflection
in
are collinear.
Denote
and
are concurrent at point
In accordance with Claim, points
and
are isogonal conjugate with respect
Claim
Let be an acute triangle, and let
and
denote its altitudes. Lines
and
meet at
Prove that
Proof
Let be the circle
centered at
is midpoint
Let meet
at
Let
be the circle centered at
with radius
Let be the circle with diameter
Well known that is the polar of point
so
Let be inversion with respect
Denote
Theorem 2
Let be the base triangle,
be orthic triangle of
be Kosnita triangle of
Then
and
are homothetic, the point
center of this homothety lies on Euler line of
the ratio of the homothety is
We recall that vertex of Kosnita triangle are:
is the circumcenter of
is the circumcenter of
is the circumcenter of
where
is circumcenter of
Proof
Let be orthocenter of
be the center of Nine-point circle of
is the Euler line of
Well known that
is antiparallel
with respect
is the bisector of
therefore
is antiparallel
with respect
Similarly,
and
are homothetic.
Let be the center of homothety.
is
-excenter of
is
-excenter of
Denote circumradius
is the point
vladimir.shelomovskii@gmail.com, vvsss
Theorem 3
Let be the reference triangle (other than a right triangle). Let the altitudes through the vertices
meet the circumcircle
of triangle
at
and
respectively. Let
be the triangle formed by the tangents at
and
to
(Let
be the vertex opposite to the side formed by the tangent at the vertex A). Prove that the lines through
and
are concurrent, the point of concurrence
lies on Euler line of triangle
Proof
At first one can prove that lines and
are concurrent. This follows from the fact that lines
and
are concurrent at point
and Mapping theorem (see Exeter point).
Let
and
be the midpoints of
and
respectively.
Let
and
be the midpoints of
and
respectively.
vladimir.shelomovskii@gmail.com, vvsss