Difference between revisions of "Kimberling’s point X(24)"
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Denote <math>a = BC, \alpha = \angle A, \beta = \angle B, \gamma = \angle C, R</math> circumradius <math>\triangle ABC.</math> | Denote <math>a = BC, \alpha = \angle A, \beta = \angle B, \gamma = \angle C, R</math> circumradius <math>\triangle ABC.</math> | ||
<math>\angle EHF = 180^\circ - \alpha, EF = BC |\cos \alpha| = 2R \sin \alpha |\cos \alpha|.</math> | <math>\angle EHF = 180^\circ - \alpha, EF = BC |\cos \alpha| = 2R \sin \alpha |\cos \alpha|.</math> | ||
− | <math>LM = \frac {R}{2} (\tan \beta + \tan \gamma) = \frac {R \sin (\beta + \gamma)}{2 \cos \beta \cdot \cos \gamma} \implies k = \frac {DE}{KL} = 4\cos \alpha \cdot \cos \beta \cdot \cos \gamma.</ | + | <math>LM = \frac {R}{2} (\tan \beta + \tan \gamma) = \frac {R \sin (\beta + \gamma)}{2 \cos \beta \cdot \cos \gamma} \implies</math> |
+ | <cmath>k = \frac {DE}{KL} = 4\cos \alpha \cdot \cos \beta \cdot \cos \gamma.</cmath> | ||
<math>\frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C \implies P</math> is point <math>X(24).</math> | <math>\frac {\vec {PH}}{\vec {OP}}= 4 \cos A \cos B \cos C \implies P</math> is point <math>X(24).</math> | ||
'''vladimir.shelomovskii@gmail.com, vvsss''' | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 14:26, 12 October 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 point
vladimir.shelomovskii@gmail.com, vvsss