Kepler triangle

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

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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

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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

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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

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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

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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

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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 $K = I_AI_C \cap MG.$ Then $D_0, K,$ and $F$ lies on circle $\gamma$ with diameter $MI_C.$ Point $K \in \Omega.$

Proof

1. $BD' = \phi^2 = BD_0, \angle BI_CD_0 = \beta \implies r_C = \frac{BD'}{\tan \beta } = \sqrt{\phi} = AM \implies$ distance from $I_C(I_B)$ to $BC$ is equal $AM \implies I_CA ||BC \implies A,I_B,I_C$ are collinear. $\angle AI_CD' = \alpha \implies AI_C = \frac{r_C}{\cos \alpha} = 1 = AB \implies A$ is the center of the circle $BCI_BI_C.$

2. $\angle BI_AM = \beta \implies r_A = \frac{BM}{\tan \beta} = \frac{1}{\sqrt {\phi}}.$

$D'I_C \perp AB, D'M \perp AB \implies$ points $M, D',$ and $I_C$ are collinear.

$\angle BMI_C = \alpha \implies MI_C = \frac{r_C}{\sin \alpha} = \frac{1}{\sqrt {\phi}} = MI_A \implies M$ is the center of the circle $\Theta = \odot I_AI_BI_C.$

3. $GL \perp AM, BC \perp AM \implies GL||BC, L \in MI_A \implies \angle MGL = \beta \implies$ $ML = GL \tan \beta = \phi \sqrt {\phi} \implies OM + ML = R.$

4. $MI_AGI_C$ is rhomb with $\angle MI_CD_0 = 2 \beta, \angle MNI_A = 90^\circ MN = r_A \sin 2 \beta = 1.$ $BD' = ED' \implies \angle BMD' = \angle EMD', FI_C = D_0I_C \implies$ $\triangle MFI_C = \triangle MD_0I_C \implies MF = MD_0 = MN.$

5. $\angle MD_0I_C = 90^\circ = \angle MKI_C = \angle MFI_C.$ In triangle $\triangle KMO$ $\angle KMO = 90^\circ + \beta, KM = MI_A \sin \beta = \sqrt {\frac {\phi}{2}}, OM = \frac { \phi^2 \cdot \sqrt{\phi}}{2}.$

$OK^2 = KM^2 + MO^2 - 2 KM \cdot MO \cdot \cos (90^\circ + \beta) =$ $= \frac {\phi}{2} + \frac{\phi^5}{4} - 2 \frac{\phi^2 \sqrt{\phi}}{2} \frac{\sqrt{\phi}}{\sqrt{2}} \cdot (-\frac{\phi}{\sqrt{2}}) = \frac{1}{4 \phi} = R^2.$ vladimir.shelomovskii@gmail.com, vvsss

Rhombs in the aureate triangle

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Let $F_0$ be the foot from $F$ to $BC, L' = I_AI_C \cap GL,$

$G_1$ is symmetrical to $G$ with respect $AM.$

Prove that $I_CMI_AG, GBML', ID'BF_0$ are similar rhombs, $MI_A = \sqrt {\varphi}, BM = \phi, BD' = \phi^2.$

Proof

$MI_C = MI_A = GI_A = 2R = AL = GI_C = \sqrt {\varphi}$ $\implies MI_AGI_C$ is the rhomb.

$GI_C \perp BC, MI_C \perp AB \implies$ $\angle MI_CG = \angle ABC = 2 \beta.$

$B$ is the orthocenter of $\triangle I_CGM,$

$L'$ is the orthocenter of $\triangle I_AGM \implies$

$BM = BG = GL' = ML' \implies MBGL'$ is the rhomb with side $BM = \phi.$

$\angle FMB = 2 \alpha \implies MF_0 = MF \cos 2 \alpha = \phi^3 \implies$

$BF_0 = BM - MF_0 = \phi - \phi^3 = \phi^2 = BD' = DI, BF || D'I \implies ID'BF_0$ is the rhomb.

Circles of the aureate triangle

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Prove that:

1. Circle $\nu$ centered at $D$ with radius $FD$ is tangent to $BC$ and $\omega_B.$

2. Circle $\mu$ with diameter $D'L$ is tangent to lines $D_0G'$ and $NL.$ This circle contain points $I, L, L', B, D'.$

3. Let $R = BI \cap FD, R'$ be the point of $NM$ such that $RR' || BC.$ Then $R'$ lies on circles $\mu, \nu,$ and circle with diameter $LD_{01},$ lines $LG', D'D_{01}, FH, DG_1, F_0D'_1.$

Proof

1. $FF_0 = MF \sin 2 \alpha = 2 \phi \sqrt{\phi},$ $DF_0 = HM = \phi \sqrt{\phi}, DF \perp BC \implies \nu$ is tangent to $BC.$ $D_{01} = \omega_B \cap BC \implies I_BD_{01} = \sqrt{\phi}$ $F_0D_{01} = F_0M + MD_{01} = \phi^3 + 1 = 2 \phi = 2 \sqrt{DF_0 \cdot I_BD_{01}}$ so $\nu$ is tangent to $\omega_B.$

2. $D'M = LM = NG = GD_0 = \phi \sqrt{\phi},$ $I_CM = I_AM = I_AG = GI_C = r_A=\sqrt{\varphi} = \frac{1}{\sqrt{\phi}},$ $I_CI_A = 2 r_A \cos \beta = \sqrt{2} \varphi \implies D'L = \frac{I_AI_C \cdot ML}{r_A} = \sqrt{2} \phi,$ $D'L || D_0N || I_AI_C \perp GM || D'D_0 ||NL \implies D'D_0$ and $LN$ are tangent line to $\mu.$

Let point $Q$ be the midpoint of $D'L.$ Then $Q$ lies on $GM.$

$\angle MD_{01}L = \arctan{\frac {ML}{MD_{01}}} = \beta = \arcsin{\frac {ND_0}{2 MD_{01}}} \implies$ points $N,L,$ and $D_{01}$ are collinear.

Points $B, I,$ and $I_B$ are collinear (lie on bisector of $\angle ABC), \frac{DD'}{BD'} = \frac{DG'}{I_BG'} \implies D'G' || BI.$ $\angle D'D_0M = \beta = \angle HD_0M = \angle (D'G')(BC) \implies$ points $D_0, D', H,$ and $G'$ are collinear. $D'I = LL', D'I || LL', D'I \perp IL \implies$ $D'ILL'$ is the rectangle, so points $I$ and $L'$ belong $\mu.$ $BM = \phi, \angle BMQ = \beta, QM = ML \sin \beta = \phi \sqrt{\frac{\phi}{2}}\implies QB = \frac{\phi}{\sqrt{2}} \implies B \in \mu.$

3. $RF_0 = BF_0 \tan \beta = \phi^3 \sqrt {\phi}, \angle R'MC = \angle ML'L = 2 \beta \implies$ $R'M = \frac{RF_0}{\sin 2 \beta} = \phi^3 = F_0M \implies \angle F_0R'M = \angle R'F_0M = \beta = \angle F_0R'R.$ We use $\triangle FD'_1M$ where $CD'_1 = BD' = \phi^2, CF_0 = F_0M + MC = \phi^3 + \phi = 3\phi - 1,$

$\angle ACB = 2 \beta$ and get $\angle D'_1F_0C = \beta,$ so points $D'_1,F_0,$ and $R'$ are collinear.

$F_0R' || BR \perp FH, F_0R' = \frac{F_0R}{\sin \beta} = \phi^2 \sqrt{2 \phi}.$ $\angle FF_0R' = 90^\circ - \beta, FF_0 = 2 \phi \sqrt{\phi} \implies F_0R' = FF_0 \cos FF_0R' \implies R' \in FH.$ $I_CF = I_CD_0 = I_AN = I_AN_1, \angle FI_CB = 3 \beta = \angle BI_AN_1 \implies FN_1 || I_AI_C.$ $FH \perp HD', DD' \perp I_AI_C \implies FH || I_AI_C \implies$ points $F, H, R',$ and $N_1$ are collinear.

$D'R = BD' \cdot \tan \beta = \phi^3 \sqrt{\phi}, D'M = BD' \cdot \tan 2 \beta = \phi \sqrt{\phi},$ $MD_{01} = 1, RR' = \phi^2, RR' || MD_{01} \implies R' \in D'D_{01}.$ $GG_1 = D_0D_{01} = 2, GB = MB = DB = \phi, \angle DGG_1 = 2 \beta \implies$ $G_1D \perp AB, ID \perp AB \implies \angle IDR = 2 \beta.$ $DR = DF_0 - RF_0 = BF_0 (\tan 2 \beta - \tan \beta) = \phi^2 \sqrt{\phi}, \cot \angle R'DR = \frac{DR}{RR'} = \sqrt{\phi} \implies$ $\angle R'DR = 2 \beta \implies$ points $D, I, R',$ and $G_1$ are collinear. $DR' = \frac {DR}{\cos 2 \beta} = \frac{ \phi^2 \sqrt{\phi}}{\phi} = DF_0 \implies R' \in \nu.$ $I_BG_1 = I_BD = I_BM = \frac{1}{\sqrt{\phi}}, DG_1^2 = (DF_0+ML)^2+(G_1L+F_0M)^2 = 4\phi, DG_1 = 2\sqrt{\phi} \implies$ $\angle G'DR' = 2 \beta = \angle R'DF_0 \implies \angle F_0R'G = 2 \angle DR'G' = 180^\circ - 2 \beta,$ $F_0R' = \phi^2 \sqrt{2 \phi}, F_0L = \phi^2 \sqrt{2}, F_0R' \perp D'L \implies$ $\angle F_0R'L = 2 \beta \implies$ points $G', R',$ and $L$ are collinear.

$\angle LR'D_{01} = \alpha + 2 \beta = 90^\circ \implies R'$ lies on the circle with diameter $L_{01}.$ $\angle D'R'L = \angle D'R'F_0 + \angle LR'F0 = 90^\circ \implies$ $R'$ lies on the circle with diameter $D'L.$

vladimir.shelomovskii@gmail.com, vvsss

Collinearity

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1. Collinear groups of point are $N,U,X,X',D,$

$M,R,X',D', I_C$

$T',X,R, R',$

$D_0,J,T',I,$

$D',R',V_1,D_{01}$ ,...

2. $CI_C$ is tangent to circle $\odot DIL.$

3. $J$ is the common point of $\omega_C, \odot FG'R', \odot DIL.$

4. $CI_C$ is tangent to $\odot DIL$ at point $I.$

5. $YY'$ is the diameter of $\odot DIL.$

One can prove these using method shown before.

Bevan point’s problems

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The Bevan point, named after Benjamin Bevan, is a center of the Bevan circle, that is the circle through the centers of the three excircles of a triangle.

Let Bevan point $Q$ lies on side $BC$ of the triangle $ABC.$ Find the ratio $\frac{BC}{AB+AC}$ if $BQ : QC$ is:

1. $1 : 1.$

2. $6+\sqrt{15} : 6 - \sqrt{15}.$

Solution

Denote $I,I_A, I_B, I_C$ the incenter and A-,B-,C-excenters of $\triangle ABC, A',B',C'$ the midpoints of sides of $\triangle I_AI_BI_C, A'' -$ the midpoint of $II_A, Q$ the Bevan point, $O$ the circumcenter.

Circumcircle $\Omega$ of $\triangle ABC$ is the Euler circle of $\triangle I_AI_BI_C,$ so $A',B',C',A''$ lies at $\Omega, A'A''$ is the diameter of $\Omega, AA'' \perp I_BI_C.$

1. $QA' \perp I_BI_C \implies A = A' \implies AB = AC.$

Let $D$ be the foot from $Q$ to $AB \implies D \in \odot BB'QD.$

$AB' = CB' \implies AD = QC$ ( Sparrow's Lemma 1).

$\triangle BDQ \sim \triangle BQA \implies \frac {BD}{BQ} = \frac {BQ}{AB} = \frac {BQ}{BD + BQ} \implies \triangle ABC$ is the aureate triangle.

2. $2Rr = 2 \frac{abc}{4S} \cdot \frac{2S}{a+b+c} = \frac{abc}{a+b+c},$ where $S$ is the area of $\triangle ABC.$

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$OI^2 = R^2 - 2Rr = R^2 - \frac{abc}{a+b+c}.$

$OM^2 = OB^2 - BM^2 = R^2 - a^2/4,$ $OM^2 = OQ^2 - MQ^2 = OI^2 - MQ^2 = R^2 - \frac{abc}{a+b+c} - MQ^2,$ $MQ^2 = 5a^2/48 \implies 7(a+b+c) = 48bc,$ $bc \sin \alpha = (a+b+c)r \implies \sin \alpha = \frac{4 \sqrt{3}}{7} \implies \cos\alpha = \frac{1}{7}.$ $a^2 = b^2+c^2 - 2bc/7 = (b+c)^2 - 16bc/7 = (b+c)^2 - a(a+b+c)/3 \implies$ $b+c-a = a/3, 3(b+c) = 4a \implies \frac{BC}{AB+AC} =0.75.$

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Kepler triangle

Contents

Definition of doubled Kepler triangle

Sides and angles of doubled Kepler triangle

Construction of a Kepler triangle

Segments of aureate triangle

Collinearity in aureate triangle

Excircles of the aureate triangle

Rhombs in the aureate triangle

Circles of the aureate triangle

Collinearity

Bevan point’s problems