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Difference between revisions of "Kepler triangle"

(→‎Properties of a Kepler triangle)
(→‎Circles of the aureate triangle)
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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>
 
4.<math>\angle BHD = \angle IDH \implies PD = PH = PI, AH = IM \implies P \in</math> midline of <math>\triangle ABC.</math>
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==Excircles of the aureate triangle==
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[[File:Aureate circles3.png|300px|right]]
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Let <math>\omega_A, \omega_B,\omega_C</math> centered at points <math>I_A, I_B, I_C,</math> respectively be the excircles of aureate <math>\triangle ABC.</math> Prove that:
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1. Let <math>D_0</math> and <math>F</math> be the point of tangency of <math>\omega_C</math> with  <math>BC</math> and <math>ME,</math> respectively. Then radius <math>D'I_C = r_C = \sqrt{\phi}.</math>
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<math>AI_C || BC, A</math> is the center of the circle <math>BCI_BI_C.</math>
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2. Let <math>G</math> be the point of tangency of  <math>AB</math> and <math>\omega_A. </math> Then radii <math>MI_A = GI_A = r_A = \frac{1}{\sqrt{\phi}}, M</math> is the center of the circle <math>\Theta = \odot I_AI_BI_C.</math>
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3. Let <math>L</math> be the foot from <math>G</math> to <math>AM.</math> Then <math>GL</math> is tangent to <math>\Omega = \odot ABC.</math>
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4. Let <math>N</math> be the foot from <math>M</math> to <math>GI_A.</math> Then <math>MN = MF = MD_0 = 1.</math>
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5. Let <math>K = I_A \cap MG.</math> Then <math>D_0, K,</math> and <math>F</math> lies on circle with diameter <math>MI_C.</math>
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<i><b>Proof</b></i>
  
 
==Circles of the aureate triangle==
 
==Circles of the aureate triangle==

Revision as of 13:28, 18 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 $\triangle ABC.$

Excircles of the aureate triangle

Aureate circles3.png

Let $\omega_A, \omega_B,\omega_C$ centered at points $I_A, I_B, I_C,$ respectively be the excircles of aureate $\triangle ABC.$ Prove that:

1. Let $D_0$ and $F$ be the point of tangency of $\omega_C$ with $BC$ and $ME,$ respectively. Then radius $D'I_C = r_C = \sqrt{\phi}.$ $AI_C || BC, A$ is the center of the circle $BCI_BI_C.$

2. Let $G$ be the point of tangency of $AB$ and $\omega_A.$ Then radii $MI_A = GI_A = r_A = \frac{1}{\sqrt{\phi}}, M$ is the center of the circle $\Theta = \odot I_AI_BI_C.$

3. Let $L$ be the foot from $G$ to $AM.$ Then $GL$ is tangent to $\Omega = \odot ABC.$

4. Let $N$ be the foot from $M$ to $GI_A.$ Then $MN = MF = MD_0 = 1.$

5. Let $K = I_A \cap MG.$ Then $D_0, K,$ and $F$ lies on circle with diameter $MI_C.$

Proof

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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