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Difference between revisions of "Kepler triangle"

(→‎Segments of aureate triangle)
(→‎Properties of a Kepler triangle)
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<cmath>D'M \perp AB, \angle D'MB = \alpha.</cmath>
 
<cmath>D'M \perp AB, \angle D'MB = \alpha.</cmath>
  
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==Properties of a Kepler triangle==
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==Collinearity in aureate triangle==
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[[File:Triangle segments 1.png|380px|right]]
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[[File:Collinearity.png|300px|right]]
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Let <math>I,H,O,</math> and <math>F</math> be the incenter, ortocenter, circumcenter, and Feuerbach point (midpoint <math>BC),</math> respectively.
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We define <math>P = BH \cap DI.</math> Let reflections of <math>E, D, D',P</math> wrt <math>AM</math> be points <math>E_1, D_1, D'_1,P'.</math> Let <math>M'</math> be the point on incircle opposite <math>M.</math>
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<cmath>\phi = \frac {\sqrt{5} - 1}{2} = \frac {1}{\varphi}, \phi^2 + \phi = 1, AB = 1.</cmath>
 
  
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The remaining notations are shown in the figure. The properties of the triangle were found by L. Emelyanov.
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Prove:
  
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Prove :
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1. Points <math>E,M',E_1</math> and points <math>D', I, D'_1</math> are collinear.
  
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1. Points <math>E,F',E'</math> are collinear.
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2. Points <math>D,H,D_1</math> are collinear.
  
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2. <math>AH = IF.</math>
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3. <math>AM' = EH = IH</math>
  
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3. Points <math>D,H,D'</math> are collinear.
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4. <math>PI = PH = PD = P_1H=P_1I.</math>
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4. <math>AF' = EH = IH.</math>
 
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5. <math>IO = FO.</math>
 
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6. <math>P = CH \cap ID' \implies PI = PH = PD'.</math>
 
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7. <math>2 BO \cdot AF = AB^2.</math>
 
  
 
<i><b>Proof</b></i>
 
<i><b>Proof</b></i>
  
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It is known that <math>BF = BD = \phi, AF = \sqrt{\phi}, AD = \phi^2, AI = \phi \cdot \sqrt{\phi}, r=ID = IF = IF' = \phi^2 \cdot \sqrt{\phi},</math>
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1. The distance from <math>E</math> to <math>BC</math> is <math>BE \cos \alpha = 2\phi^2 \cdot \sqrt{\phi} = 2 IM = MM'.</math>
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<cmath>\sin \alpha = \phi, \cos \alpha =  \sqrt{\phi}\implies \sin 2\alpha = 2 \phi \sqrt{\phi}, \cos 2\alpha = \phi^3.</cmath>
 
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1. <math>BE = BC \sin \alpha = 2 \phi \cdot \phi = 2 \phi^2.</math>
 
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Distance from <math>E</math> to <math>BC</math> is <math>BE \cos \alpha = 2 \phi^2 \cdot \sqrt{\phi} = 2r.</math>
 
  
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2. <math>AH = \frac {AE}{\cos \alpha} = \frac{1-BE}{\sqrt{\phi}}= \frac{\phi - \phi^2}{\sqrt{\phi}}= \phi^2 \cdot \sqrt{\phi} = r.</math>
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<math>D'I</math> is the midline of trapezium <math>BMM'E.</math>
  
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3. Distance from <math>D</math> to <math>BC</math> is <math>BD \cos \alpha = \phi \cdot \sqrt{\phi} = AI = HF.</math>
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2. The distance from <math>D</math> to <math>BC</math> is <math>BD \cos \alpha = \phi \cdot \sqrt{\phi} = HM.</math>
  
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4. <math>ED = EF', \angle DHE = \angle EAF' = \alpha \implies AF' = HE.</math>
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3. <math>ED = EM', \angle DHE = \angle EAM' = \alpha \implies \triangle AEM'=\triangle HDE, AM' = HE.</math>
 
<cmath>EH = DH \cos \alpha = DH \tan \alpha = HI.</cmath>
 
<cmath>EH = DH \cos \alpha = DH \tan \alpha = HI.</cmath>
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5. <math>\angle BOF = 2 \alpha \implies FO = BF \cot 2 \alpha = \phi \cdot \frac{\phi^3}{ 2 \phi \sqrt{\phi}} = \frac {r}{2}.</math>
 
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6.<math>\angle CHD' = \angle ID'H \implies PI = PH = PD', AH = IF \implies P \in</math> midline of <math>\triangle ABC.</math>
 
  
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7. <math>R = BO = \frac{BF}{\sin 2 \alpha} = \frac{1}{2\sqrt{\phi}} \implies 2R \cdot AF = AB^2.</math>
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4.<math>\angle BHD = \angle IDH \implies PD = PH = PI, AH = IM \implies P \in</math> midline of <math>\triangle ABC.</math>
  
 
==Circles of the aureate triangle==
 
==Circles of the aureate triangle==

Revision as of 23:20, 16 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 Emelyanov, who in his works called this isosceles triangle ”aureate triangle”.

Definition of doubled Kepler triangle

Definition.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 $\triangle ABC (AB = AC, I$ be the incenter) touch the side $BC$ at point $M, \angle BAM = \alpha,$ $\angle ABC = 2 \beta, \alpha + 2 \beta = 90^\circ, MI = r($ inradius).

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 $d \alpha + 2 d \beta = 0:$ \[\frac {-2}{\sin^2 \alpha}+\frac {1}{\sin^2 \beta} = 0 \implies\] \[\sqrt{2} \sin \beta = \sin \alpha = \cos 2 \beta = 1 - 2 \sin^2 \beta \implies \sin \alpha = 1 - \sin^2 \alpha.\]

Therefore $\sin \alpha = \cos 2 \beta = \frac {\sqrt{5} - 1}{2} = \phi = \varphi.$

Let $AB = 1 \implies BM = \phi, AM = \sqrt{\phi}, IM = \phi^2 \cdot \sqrt{\phi}.$

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

Construction of a Kepler triangle

Triangle construction.png

Let $M$ be the midpoint of the base $BC,$ $DM \perp BC, DM = BC.$ Point $E \in BD, DE = BC.$

The point $F$ is the intersection of a circle with diameter $BM$ and a circle centered at point $B$ and radius $BE$, which is located in the half-plane $BC$ where there is no point $D$.

The bisector of the obtuse angle between lines $BF$ and $BC$ intersects bisector $BC$ at the vertex $A$ of the Kepler triangle.

The construction is based on the fact that \[\cos 2 \angle ABC = 2 - \sqrt{5}.\]

Segments of aureate triangle

Segments.png

We call the doubled Kepler triangle the aureate triangle according to Lev Emelyanov. Let $I,I_C,H,O,$ and $M$ be the incenter, C-excenter, ortocenter, circumcenter, and midpoint $BC$ of aureate $\triangle ABC (AB = AC).$

Let $E,D,D'$ be the foots from $C, I,I_C$ to $AB.$

Find segments, prove $IO = OM, MD' \perp AB.$

Proof

\[\sin \alpha = \cos 2 \beta = \frac {\sqrt{5} - 1}{2} = \phi = \frac {1}{\varphi} \implies\] \[\cos \alpha = \sin 2 \beta = \sqrt{\phi} = \tan \alpha = \cot 2 \beta,\] \[\cos \beta = \frac{1}{\sqrt{2 \phi}}, \sin \beta = \frac {\phi}{\sqrt{2}}, \tan \beta = \phi \cdot \sqrt{\phi},\] \[\sin 2 \alpha = 2 \phi \cdot \sqrt{\phi}, \cos 2 \alpha = \phi^3, \tan 2\alpha = 2 \varphi \sqrt{\varphi}.\] \[AB = AC = 1 \implies BD = BM = AD' = \phi, AD = BD' =1 - \phi = \phi^2,\] \[AE = AC \cos 2 \alpha = \phi^3 = DD' = 1 - 2 \phi^2 = \phi^3,\] \[DE = AD - AE = \phi^4,BE = AB - AE = 2 \phi^2.\]

\[AH = \frac{AE}{\cos \alpha} = \phi^2 \cdot \sqrt{\phi} = IM = DI, HM = AM - AH = \phi \cdot \sqrt{\phi}.\] \[AI = AM - IM = \sqrt{\phi} - \phi^2 \cdot \sqrt{\phi} = \phi \cdot \sqrt{\phi}.\] \[CE = \frac{BC \cdot AM}{AB} = 2 \phi \cdot \sqrt{\phi} = 2 AI, EH = AH \sin \alpha = \phi^3 \cdot \sqrt{\phi}.\] \[CH = CE - EH = \sqrt{\phi} = AM, HI = AI – AH = \phi^3 \cdot \sqrt{\phi} = EH.\] \[BO = \frac{BM}{\sin 2 \alpha} = \frac{1}{2 \sqrt{\phi}} = AO, 2 BO \cdot AF = AB^2.\] \[OM = BO \cos 2 \alpha = \frac { \phi^2 \cdot \sqrt{\phi}}{2} \implies IM = 2 MO.\] \[BI = \frac {BM}{\cos \beta} = \phi \cdot \sqrt{2 \phi} = AI \cdot \sqrt{\phi}.\] \[D'M^2 = BD'^2 + BM^2 - 2 BM \cdot BD' \cos 2 \beta = \phi^3 \implies D'M = \phi \cdot \sqrt{\phi} \implies\] \[D'M \perp AB, \angle D'MB = \alpha.\]

Collinearity in aureate triangle

Collinearity.png

We define $P = BH \cap DI.$ Let reflections of $E, D, D',P$ wrt $AM$ be points $E_1, D_1, D'_1,P'.$ Let $M'$ be the point on incircle opposite $M.$

Prove:

1. Points $E,M',E_1$ and points $D', I, D'_1$ are collinear.

2. Points $D,H,D_1$ are collinear.

3. $AM' = EH = IH$

4. $PI = PH = PD = P_1H=P_1I.$

Proof

1. The distance from $E$ to $BC$ is $BE \cos \alpha = 2\phi^2 \cdot \sqrt{\phi} = 2 IM = MM'.$

$D'I$ is the midline of trapezium $BMM'E.$

2. The distance from $D$ to $BC$ is $BD \cos \alpha = \phi \cdot \sqrt{\phi} = HM.$

3. $ED = EM', \angle DHE = \angle EAM' = \alpha \implies \triangle AEM'=\triangle HDE, AM' = HE.$ \[EH = DH \cos \alpha = DH \tan \alpha = HI.\]

4.$\angle BHD = \angle IDH \implies PD = PH = PI, AH = IM \implies P \in$ midline of $\triangle ABC.$

Circles of the aureate triangle

Aureate circles.png

Prove that:

1. Exradius $D'I_C = D''I_C = r_C = \sqrt{\phi}, AI_C || BC, A$ is the center of the circle $BCI_BI_C.$

2. Exradius $FI_A = N'I_A = r_A = \frac{1}{\sqrt{\phi}}, F$ is the circumcenter of the circle $\Theta = \odot I_AI_BI_C.$

3. $FI_AN'I_C, FI_AN''I_B$ are rhombs. $N'NN''$ is tangent to $\Omega = \odot ABC.$

4. A circle $\theta$ with center at point $F$ and radius $|AB|$ touches $IN'.$

5. $K = FN' \cap I_AI_C$ lies on $\Omega = \odot ABC, KO \perp AC.$

Proof

1. $BD' = AD = \phi^2 = BD'', FD'' = BD'' + FB = AD + BD = AB.$

Let line $I_CD'$ cross $BC$ at point $F' \implies BF' = \frac {BD'}{\sin \alpha} = \phi = BF \implies F = F'.$ \[r_C = I_CD'' = FD'' \cdot \tan \alpha = \sqrt{\phi} = AF \implies I_C A I_B ||BC.\] \[I_CF = \frac {FD''}{\cos \alpha} = \frac{1}{\sqrt{\phi}}.\] $AI_C = FD'' = AB \implies A$ is the center of the circle $BCI_BI_C.$

2. Denote $\angle ABC = 2\beta \implies \angle D''I_CD' = 2\beta \implies \angle FI_CB = \beta \implies$ $\angle FI_AI_C = 180^\circ - \beta - (90^\circ + \angle BFI_C) = \beta \implies FI_A = FI_C = FI_B \implies$ $F$ is the center of $\odot I_AI_BI_C.$

3. Let $N'$ be point of tangency line $AB$ and A-excircle. \[BD'' = BD', BN' = BF, \angle D''BN' = \angle D'BF \implies\] \[N'D'' = FD', \angle BD''N = \angle BD'F = 90^\circ \implies N'I_C = FI_C \implies\] $FI_AN'I_C$ is the rhomb.

$FN = D''N' = FD' = \phi \cdot \sqrt{\phi}, OF + FN = \frac{r}{2} + \phi \cdot \sqrt{\phi} = \frac{1}{2 \sqrt{\phi}}= R \implies$ $N \in \odot ABC,$ so $NN'$ is tangent to circumcircle of $\triangle ABC.$

4. $\angle G'FI_A = \alpha \implies FI_A \cos \alpha = \frac{1}{\sqrt{\phi}} \cdot \sqrt{\phi} = 1 = |AB|,$ so circle $\theta$ touches $IN'.$ $\angle GFG' = \angle G'FG'' = 90^\circ + \alpha.$

5. $\alpha + 2 \beta = 90^\circ \implies \phi = \sin \alpha = \cos 2 \beta = 1 - 2 \sin^2 \beta \implies \sin \beta = \frac {\phi}{\sqrt{2}}.$ \[K = FN' \cap I_AI_C \implies \angle BKF = 90^\circ \implies\] $\angle BFK = 90^\circ - \angle BI_CF - \angle BFI_C = 90^\circ - \beta - \alpha = \beta.$

$OK^2 = KF^2 + FO^2 - 2 KF \cdot FO \cdot \cos (90^\circ + \beta) = \frac{\phi^5}{4} + \frac{\phi^5}{4} - 2 \frac{\phi^2 \sqrt{\phi}}{2} \frac{\sqrt{\phi}}{\sqrt{2}} \cdot (-\frac{\phi}{\sqrt{2}}) =$ \[= \frac{\phi}{2} + \frac{\phi^5}{4} + \frac{\phi^4 }{2} = \frac{1}{4 \phi} = R^2.\] We know all sides of $\triangle FKO$ and can find $\angle KOF = 2 \beta \implies KO \perp AC.$

vladimir.shelomovskii@gmail.com, vvsss

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