Difference between revisions of "Kepler triangle"
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A Kepler triangle is a special right triangle with edge lengths in geometric progression. The progression can be written: <math>1:\sqrt {\varphi }:\varphi,</math> or approximately <math>1:1.272:1.618.</math> | A Kepler triangle is a special right triangle with edge lengths in geometric progression. The progression can be written: <math>1:\sqrt {\varphi }:\varphi,</math> or approximately <math>1:1.272:1.618.</math> | ||
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== | ||
| + | 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 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> | ||
| + | We need to find minimum of | ||
| + | <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> | ||
| + | <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> | ||
| + | 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> | ||
| + | <cmath>AI = \phi \cdot \sqrt{\phi}, BI = \sqrt{2} AI, IK = IM = \phi^2 \cdot \sqrt{\phi}.</cmath> | ||
| + | '''vladimir.shelomovskii@gmail.com, vvsss''' | ||
Revision as of 06:38, 5 August 2025
A Kepler triangle is a special right triangle with edge lengths in geometric progression. The progression can be written: 
or approximately 
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
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 inscribed circle of an isosceles 
touch the sides 
and 
at points 
and 
![\[\angle BAM = \alpha, \angle ABC = 2 \beta, \alpha + 2 \beta = 90^\circ, KI = MI = r.\]](http://latex.artofproblemsolving.com/f/7/e/f7ef956ab40ddce71136a6087fd25c7208558cfc.png)
We need to find minimum of
![\[\frac {AB}{r} = \cot \alpha +\cot \beta.\]](http://latex.artofproblemsolving.com/e/4/e/e4e973927aeeac6208840d5af2b4e6e05d6f52ef.png)
Let us differentiate this function with respect 
to taking into account that 
![\[\frac {-2}{\sin^2 \alpha}+\frac {1}{\sin^2 \beta} = 0 \implies \sqrt{2} \sin \beta = \sin \alpha = \cos 2 \beta = 1 - 2 \sin^2 \beta \implies \sin \alpha = 1 - \sin^2 \alpha.\]](http://latex.artofproblemsolving.com/6/c/e/6ced1366d3a41076f481ae16fc16a445dbd3fa82.png)
Therefore 
Let 
![\[AI = \phi \cdot \sqrt{\phi}, BI = \sqrt{2} AI, IK = IM = \phi^2 \cdot \sqrt{\phi}.\]](http://latex.artofproblemsolving.com/8/d/0/8d0c39b1ebdc1f315b49b7836fdb20dfa967c91a.png)
vladimir.shelomovskii@gmail.com, vvsss
