Difference between revisions of "Tucker circles"
(Created page with "The Tucker circles are a generalization of the cosine circle and first Lemoine circle.") |
|||
Line 1: | Line 1: | ||
The Tucker circles are a generalization of the cosine circle and first Lemoine circle. | The Tucker circles are a generalization of the cosine circle and first Lemoine circle. | ||
+ | ==Tucker circle== | ||
+ | [[File:Tucker circle.png|450px|right]] | ||
+ | [[File:Tucker circle A.png|450px|right]] | ||
+ | Let triangle <math>ABC</math> be given. <math>O</math> is it’s circumcenter, <math>L</math> is it’s Lemoine point. | ||
+ | |||
+ | Let homothety centered at <math>L</math> with factor <math>k</math> maps <math>\triangle ABC</math> into <math>\triangle DEF</math>. | ||
+ | |||
+ | Denote the crosspoints of sidelines these triangles as | ||
+ | <cmath>A' = BC \cap DF, B' = AC \cap DE, C' = AB \cap EF,</cmath> | ||
+ | <cmath> A'' = BC \cap DE, B'' = AC \cap EF, C'' = AB \cap DF.</cmath> | ||
+ | |||
+ | Prove that points <math>A', B', C', A'', B'',</math> and <math>C''</math> lies on the circle centered at <math>LO</math> (Tucker circle). | ||
+ | |||
+ | <i><b>Proof</b></i> | ||
+ | |||
+ | <math>AB' || C''D, AC'' || B'D \implies AB'DC''</math> is the parallelogram. | ||
+ | |||
+ | Denote <math>K = AD \cap B'C'', AK = KD, B'K = KC'' \implies</math> | ||
+ | <math>B'C''</math> is antiparallel to <math>BC.</math> | ||
+ | |||
+ | Similarly, <math>A''C'</math> is antiparallel to <math>AC, A'B''</math> is antiparallel to <math>AB.</math> | ||
+ | |||
+ | <math>M = BE \cap A''C'</math> is midpoint <math>BE, N = A'B'' \cap CF</math> is the midpoint <math>CF.</math> | ||
+ | |||
+ | <math>\triangle AB'C'' \sim \triangle ABC.</math> | ||
+ | |||
+ | <math>\frac {LD}{AL} = k, AL = LD \implies \frac {AK}{AL} = \frac{2}{k+1}, \frac {KL}{AL} = \frac{1-k}{1+k}.</math> | ||
+ | |||
+ | Similarly, <math>\frac {BM}{BL} = \frac {CN}{CL} = \frac {AK}{AL} =\frac{2}{k+1}, \frac {KL}{AL} = \frac{ML}{BM}= \frac {NL}{CL} = \frac{1-k}{1+k}.</math> | ||
+ | |||
+ | Let <math>B'''C'''</math> be the symmedian <math>BC</math> through <math>L.</math> | ||
+ | <cmath>B'''C''' || B'C'' \implies B'C'' = B'''C''' \cdot \frac {AK}{AL} =\frac{2B'''C'''}{k+1}.</cmath> | ||
+ | |||
+ | It is known that three symmedians through <math>L</math> are equal, so <math>A''C' = C''B' = B''A' = \frac{2B'''C'''}{k+1}.</math> | ||
+ | |||
+ | <math>\triangle KMN</math> is homothetic to <math>\triangle ABC</math> with center <math>L</math> and factor <math>\frac{1-k}{1+k}.</math> | ||
+ | |||
+ | So segments <math>A''C' = C''B' = B''A'</math> are tangents to <math>\odot KMN</math> and points of contact are the midpoints of these segments. | ||
+ | |||
+ | Denote <math>Q</math> the circumcenter of <math>\triangle KMN, Q \in LO.</math> | ||
+ | |||
+ | Therefore <math>A'Q = B'Q = C'Q = A''Q = B''Q = C''Q. \blacksquare</math> | ||
+ | |||
+ | '''vladimir.shelomovskii@gmail.com, vvsss''' |
Revision as of 06:32, 4 August 2024
The Tucker circles are a generalization of the cosine circle and first Lemoine circle.
Tucker circle
Let triangle be given.
is it’s circumcenter,
is it’s Lemoine point.
Let homothety centered at with factor
maps
into
.
Denote the crosspoints of sidelines these triangles as
Prove that points and
lies on the circle centered at
(Tucker circle).
Proof
is the parallelogram.
Denote
is antiparallel to
Similarly, is antiparallel to
is antiparallel to
is midpoint
is the midpoint
Similarly,
Let be the symmedian
through
It is known that three symmedians through are equal, so
is homothetic to
with center
and factor
So segments are tangents to
and points of contact are the midpoints of these segments.
Denote the circumcenter of
Therefore
vladimir.shelomovskii@gmail.com, vvsss