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 13:26, 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