Difference between revisions of "Kepler triangle"

Latest revision as of 08:42, 14 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

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

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

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 $\cos 2 \angle ABC = 2 - \sqrt{5}.$

Properties of a Kepler triangle

Let $I,H,O,$ and $F$ be the incenter, ortocenter, circumcenter, and Feuerbach point (midpoint $BC),$ respectively. $\phi = \frac {\sqrt{5} - 1}{2} = \frac {1}{\varphi}, \phi^2 + \phi = 1, AB = 1.$

The remaining notations are shown in the figure. The properties of the triangle were found by L. Emelyanov.

Prove :

1. Points $E,F',E'$ are collinear.

2. $AH = IF.$

3. Points $D,H,D'$ are collinear.

4. $AF' = EH = IH.$

5. $IO = FO.$

6. $P = CH \cap ID' \implies PI = PH = PD'.$

7. $2 BO \cdot AF = AB^2.$

Proof

It is known that $BF = BD = \phi, AF = \sqrt{\phi}, AD = \phi^2, AI = \phi \cdot \sqrt{\phi}, r=ID = IF = IF' = \phi^2 \cdot \sqrt{\phi},$ $\sin \alpha = \phi, \cos \alpha = \sqrt{\phi}\implies \sin 2\alpha = 2 \phi \sqrt{\phi}, \cos 2\alpha = \phi^3.$ 1. $BE = BC \sin \alpha = 2 \phi \cdot \phi = 2 \phi^2.$ Distance from $E$ to $BC$ is $BE \cos \alpha = 2 \phi^2 \cdot \sqrt{\phi} = 2r.$

2. $AH = \frac {AE}{\cos \alpha} = \frac{1-BE}{\sqrt{\phi}}= \frac{\phi - \phi^2}{\sqrt{\phi}}= \phi^2 \cdot \sqrt{\phi} = r.$

3. Distance from $D$ to $BC$ is $BD \cos \alpha = \phi \cdot \sqrt{\phi} = AI = HF.$

4. $ED = EF', \angle DHE = \angle EAF' = \alpha \implies AF' = HE.$ $EH = DH \cos \alpha = DH \tan \alpha = HI.$ 5. $\angle BOF = 2 \alpha \implies FO = BF \cot 2 \alpha = \phi \cdot \frac{\phi^3}{ 2 \phi \sqrt{\phi}} = \frac {r}{2}.$

6. $\angle CHD' = \angle ID'H \implies PI = PH = PD', AH = IF \implies P \in$ midline of $\triangle ABC.$

7. $R = BO = \frac{BF}{\sin 2 \alpha} = \frac{1}{2\sqrt{\phi}} \implies 2R \cdot AF = AB^2.$

Circles of the aureate triangle

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

@@ Line 1: / Line 1: @@
 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, who in his works called this isosceles triangle ”aureate triangle”.
+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==
+[[File:Definition.png|250px|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 the incircle of an isosceles <math>\triangle ABC (AB = AC, I</math> be the incenter) touch the side <math>BC</math> at point <math>M, \angle BAM = \alpha,</math>
+<math>\angle ABC = 2 \beta, \alpha + 2 \beta = 90^\circ, MI = r(</math> inradius).
+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>d \alpha + 2 d \beta = 0:</math>
+<cmath>\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 = \varphi.</math>
+Let <math>AB = 1 \implies BM = \phi, AM = \sqrt{\phi}, IM = \phi^2 \cdot \sqrt{\phi}.</math>
 ==Sides and angles of doubled Kepler triangle==
 [[File:Triangle segments.png|300px|right]]
@@ Line 17: / Line 34: @@
 <cmath>AI = \phi \cdot \sqrt{\phi}, BI = \sqrt{2} AI, IK = IM = \phi^2 \cdot \sqrt{\phi}.</cmath>
 '''vladimir.shelomovskii@gmail.com, vvsss'''
 ==Construction of a Kepler triangle==
 [[File:Triangle construction.png|300px|right]]
@@ Line 33: / Line 51: @@
 <cmath>\phi = \frac {\sqrt{5} - 1}{2} = \frac {1}{\varphi}, \phi^2 + \phi = 1, AB = 1.</cmath>
-The remaining notations are shown in the figure.
+The remaining notations are shown in the figure. The properties of the triangle were found by L. Emelyanov.
-Prove:
+Prove :
 . Points <math>E,F',E'</math> are collinear.
@@ Line 69: / Line 87: @@
 . <math>R = BO = \frac{BF}{\sin 2 \alpha} = \frac{1}{2\sqrt{\phi}} \implies 2R \cdot AF = AB^2.</math>
 ==Circles of the aureate triangle==
-[[File:Aureate excircles.png|300px|right]]
+[[File:Aureate circles.png|300px|right]]
 Prove that:
 . Exradius <math>D'I_C = D''I_C = r_C = \sqrt{\phi}, AI_C || BC, A</math> is the center of the circle <math>BCI_BI_C.</math>
-. Exradius <math>FI_A = N'I_A = r_A = \frac{1}{\sqrt{\phi}}, F</math> is the circumcenter of the circle <math>\odot I_AI_BI_C.</math>
+. Exradius <math>FI_A = N'I_A = r_A = \frac{1}{\sqrt{\phi}}, F</math> is the circumcenter of the circle <math>\Theta = \odot I_AI_BI_C.</math>
-. <math>FI_AN'I_C, FI_AN''I_B</math> are rhombs. <math>N'NN''</math> is tangent to <math>\odot ABC.</math>
+. <math>FI_AN'I_C, FI_AN''I_B</math> are rhombs. <math>N'NN''</math> is tangent to <math>\Omega = \odot ABC.</math>
+. A circle <math>\theta</math> with center at point <math>F</math> and radius <math>|AB|</math> touches <math>IN'.</math>
+. <math>K = FN' \cap I_AI_C</math> lies on <math>\Omega = \odot ABC, KO \perp AC.</math>
 <i><b>Proof</b></i>
@@ Line 99: / Line 122: @@
 <math>FN = D''N' = FD' = \phi \cdot \sqrt{\phi}, OF + FN = \frac{r}{2} + \phi \cdot \sqrt{\phi} = \frac{1}{2 \sqrt{\phi}}= R \implies</math>
 <math>N \in \odot ABC,</math> so <math>NN'</math> is tangent to circumcircle of <math>\triangle ABC.</math>
+. <math>\angle G'FI_A = \alpha \implies FI_A \cos \alpha =  \frac{1}{\sqrt{\phi}} \cdot \sqrt{\phi} = 1 = |AB|,</math> so circle <math>\theta</math> touches <math>IN'.</math>
+<math>\angle GFG' = \angle G'FG'' = 90^\circ + \alpha.</math>
+. <math>\alpha + 2 \beta = 90^\circ \implies \phi = \sin \alpha = \cos 2 \beta = 1 - 2 \sin^2 \beta \implies \sin \beta = \frac {\phi}{\sqrt{2}}.</math>
+<cmath>K = FN' \cap I_AI_C \implies \angle BKF = 90^\circ \implies</cmath>
+<math>\angle BFK = 90^\circ - \angle BI_CF - \angle BFI_C =  90^\circ - \beta - \alpha = \beta.</math>
+<math>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}}) =</math>
+<cmath>= \frac{\phi}{2} +  \frac{\phi^5}{4} + \frac{\phi^4 }{2} =  \frac{1}{4 \phi} = R^2.</cmath>
+We know all sides of <math>\triangle FKO</math> and can find <math>\angle KOF = 2 \beta \implies KO \perp AC.</math>
 '''vladimir.shelomovskii@gmail.com, vvsss'''

Page

Toolbox

Search