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Difference between revisions of "Kepler triangle"
(→‎Sides and angles of doubled Kepler triangle)
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When an isosceles triangle is formed from two Kepler triangles, reflected across their long sides, it has the maximum possible inradius among all isosceles triangles having legs of a given size. Most of the properties described below were discovered by the famous Russian mathematician Lev Emelianov.
 
When an isosceles triangle is formed from two Kepler triangles, reflected across their long sides, it has the maximum possible inradius among all isosceles triangles having legs of a given size. Most of the properties described below were discovered by the famous Russian mathematician Lev Emelianov.
 
==Sides and angles of doubled Kepler triangle==
 
==Sides and angles of doubled Kepler triangle==
 +
[[File:Triangle segments.png|300px|right]]
 
Let’s define the doubled Kepler triangle as triangle which has the maximum possible inradius among all isosceles triangles having legs of a given size.  
 
Let’s define the doubled Kepler triangle as triangle which has the maximum possible inradius among all isosceles triangles having legs of a given size.  
  
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Let the inscribed circle of an isosceles <math>\triangle ABC (AB = AC)</math> touch the sides <math>AB</math> and <math>BC</math> at points <math>K</math> and <math>M,</math> <cmath>\angle BAM = \alpha, \angle ABC = 2 \beta, \alpha + 2 \beta = 90^\circ, KI = MI = r.</cmath>
+
Let the incircle of an isosceles <math>\triangle ABC (AB = AC)</math> touch the sides <math>AB</math> and <math>BC</math> at points <math>K</math> and <math>M, KI = MI = r,</math>
 +
<cmath>\angle BAM = \alpha, \angle ABC = 2 \beta, \alpha + 2 \beta = 90^\circ.</cmath>
 
We need to find minimum of
 
We need to find minimum of
 
<cmath>\frac {AB}{r} = \cot \alpha +\cot \beta.</cmath>
 
<cmath>\frac {AB}{r} = \cot \alpha +\cot \beta.</cmath>
−
Let us differentiate this function with respect <math>\beta</math> to taking into account that <math>0<\alpha,2 \beta < 90^\circ, \frac {d \alpha}{d \beta} = -2:</math>
+
Let us differentiate this function with respect <math>\beta</math> to taking into account that
−
<cmath>\frac {-2}{\sin^2 \alpha}+\frac {1}{\sin^2 \beta} = 0 \implies \sqrt{2} \sin \beta = \sin \alpha = \cos 2 \beta  = 1 - 2 \sin^2 \beta \implies \sin \alpha = 1 - \sin^2 \alpha.</cmath>
+
<cmath>0<\alpha,2 \beta < 90^\circ, \frac {d \alpha}{d \beta} = -2: \frac {-2}{\sin^2 \alpha}+\frac {1}{\sin^2 \beta} = 0 \implies</cmath>
 +
<cmath>\sqrt{2} \sin \beta = \sin \alpha = \cos 2 \beta  = 1 - 2 \sin^2 \beta \implies \sin \alpha = 1 - \sin^2 \alpha.</cmath>
 
Therefore <math>\sin \alpha = \cos 2 \beta = \frac {\sqrt{5} - 1}{2} = \phi = \frac{1}{\varphi}.</math>
 
Therefore <math>\sin \alpha = \cos 2 \beta = \frac {\sqrt{5} - 1}{2} = \phi = \frac{1}{\varphi}.</math>
 +
 
Let <math>AB = 1 \implies BM = BK = \phi, AM = \sqrt{\phi}, AK = \phi^2,</math>
 
Let <math>AB = 1 \implies BM = BK = \phi, AM = \sqrt{\phi}, AK = \phi^2,</math>
 
<cmath>AI = \phi \cdot \sqrt{\phi}, BI = \sqrt{2} AI, IK = IM = \phi^2 \cdot \sqrt{\phi}.</cmath>
 
<cmath>AI = \phi \cdot \sqrt{\phi}, BI = \sqrt{2} AI, IK = IM = \phi^2 \cdot \sqrt{\phi}.</cmath>
 
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
'''vladimir.shelomovskii@gmail.com, vvsss'''

Revision as of 06:46, 5 August 2025

A Kepler triangle is a special right triangle with edge lengths in geometric progression. The progression can be written: $1:\sqrt {\varphi }:\varphi,$ or approximately $1:1.272:1.618.$ When an isosceles triangle is formed from two Kepler triangles, reflected across their long sides, it has the maximum possible inradius among all isosceles triangles having legs of a given size. Most of the properties described below were discovered by the famous Russian mathematician Lev Emelianov.

Sides and angles of doubled Kepler triangle

Triangle segments.png

Let’s define the doubled Kepler triangle as triangle which has the maximum possible inradius among all isosceles triangles having legs of a given size.

Let the incircle of an isosceles $\triangle ABC (AB = AC)$ touch the sides $AB$ and $BC$ at points $K$ and $M, KI = MI = r,$ \[\angle BAM = \alpha, \angle ABC = 2 \beta, \alpha + 2 \beta = 90^\circ.\] We need to find minimum of \[\frac {AB}{r} = \cot \alpha +\cot \beta.\] Let us differentiate this function with respect $\beta$ to taking into account that \[0<\alpha,2 \beta < 90^\circ, \frac {d \alpha}{d \beta} = -2: \frac {-2}{\sin^2 \alpha}+\frac {1}{\sin^2 \beta} = 0 \implies\] \[\sqrt{2} \sin \beta = \sin \alpha = \cos 2 \beta = 1 - 2 \sin^2 \beta \implies \sin \alpha = 1 - \sin^2 \alpha.\] Therefore $\sin \alpha = \cos 2 \beta = \frac {\sqrt{5} - 1}{2} = \phi = \frac{1}{\varphi}.$

Let $AB = 1 \implies BM = BK = \phi, AM = \sqrt{\phi}, AK = \phi^2,$ \[AI = \phi \cdot \sqrt{\phi}, BI = \sqrt{2} AI, IK = IM = \phi^2 \cdot \sqrt{\phi}.\] vladimir.shelomovskii@gmail.com, vvsss

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