Examples of triangles, which related ellipses are perfectly packed with circles.

Question

Ellipse can be perfectly packed with $n$ circles if

\begin{align} b&=a\,\sin\frac{\pi}{2\,n} \quad \text{or equivalently, }\quad e=\cos\frac{\pi}{2\,n} , \end{align}

where $a,b$ are the major and minor semi-axis of the ellipse and $e=\sqrt{1-\frac{b^2}{a^2}}$ is its eccentricity.

Consider a triangle and any ellipse, naturally associated with it, for example, Steiner circumellipse/inellipse, Marden inellipse, Brocard inellipse, Lemoine inellipse, ellipse with the circumcenter and incenter as the foci and $r+R$ as the major axis, or any other ellipse you can come up with, which can be consistently associated with the triangle.

The question is: provide the example(s) of triangle(s) for which the associates ellipse(s) can be perfectly packed with circles.

Let's say that the max number of packed circles is 12, unless you can find some especially interesting case with more circles.

For example, the Steiner incircle for the famous $3-4-5$ right triangle can not be perfectly packed.

The example of the right triangle with the Marden inellipse, perfectly packed with six circles is given in the self-answer below.

Answer 1

We can consider of a broad family of inellipses that includes Steiner's and Mandart's.

Given $\triangle ABC$ with $A=(0,0)$, $B=(c,0)$, $C=(b\cos A, b \sin A)$, and define point $P$ by $$P = \frac{\frac1\alpha A + \frac1\beta B + \frac1\gamma C}{\frac1\alpha+\frac1\beta+\frac1\gamma} \tag{1}$$ (cumbersome reciprocals now make for cleaner expressions later). Let the cevians through $P$ meet the opposite sides in $A'$, $B'$, $C'$; that is, define $$A' := \overleftrightarrow{AP}\cap\overleftrightarrow{BC} \qquad B' := \overleftrightarrow{BP}\cap\overleftrightarrow{CA} \qquad C' := \overleftrightarrow{CP}\cap\overleftrightarrow{AB}$$ One can show that there is a unique ellipse tangent to the side-lines of the triangle at $A'$, $B'$, $C'$. In the notation of my previous answer, it has equation $$K_{20}\,x^2 + K_{11}\,x y + K_{02}\,y^2 + K_{10}\,x+K_{01}\,y+K_{00} = 0$$ with $$\begin{align} K_{20} &= \phantom{-}4\, (\alpha + \beta)^2 |\triangle ABC|^2 \\[4pt] K_{11} &= \phantom{-}4 c \left((\alpha + \beta) (a \alpha \cos B - b \beta \cos A) + (\alpha - \beta) c \gamma \,\right) |\triangle ABC| \\[4pt] K_{02} &= -4 (\alpha + \beta)^2 |\triangle ABC|^2 \\ &\phantom{=}\;+c^2 \left( a^2 \alpha^2 + b^2 \beta^2 + c^2\gamma^2 + 2 bc\beta \gamma \cos A + 2 ca\gamma \alpha \cos B + 2 ab \alpha \beta \cos C \right)\\[4pt] K_{10} &= -8 c \alpha (\alpha + \beta) |\triangle ABC|^2 \\[4pt] K_{01} &= -4 c^2 \alpha ( a \alpha \cos B - b \beta \cos A + c \gamma ) |\triangle ABC| \\[4pt] K_{00} &= \phantom{-}4 c^2 \alpha^2 |\triangle ABC|^2 \end{align}$$

From these we get (after dividing-through by a common factor of $c^4$)

$$\begin{align} p &= \left( a^2 \alpha^2 + b^2 \beta^2 + c^2\gamma^2 + 2 bc\beta \gamma \cos A + 2 ca\gamma \alpha \cos B + 2 ab \alpha \beta \cos C \right)^2 \\ q &= -64 \alpha \beta \gamma (\alpha + \beta + \gamma )|\triangle ABC|^2 \end{align} \tag{2}$$

These, in conjunction with equations $(4.x')$ in my previous answer give conditions under which the inellipse is perfectly-packable.

We recover the Steiner inellipse case (again in my previous answer) by taking $\alpha=\beta=\gamma=1$ (and dividing-through by a common factor of $4$).

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If we consider inellipses specifically of equilateral triangles, with $a=b=c=1$, then we have $$\begin{align} p &= \left( \alpha^2 + \beta^2 + \gamma^2 + \beta \gamma + \gamma \alpha + \alpha \beta \right)^2 \\ q &= -12 \alpha \beta \gamma (\alpha + \beta + \gamma ) \end{align} \tag{3}$$ For the particular case of ellipse centered on the triangle's line of symmetry (so that, say, $\alpha=\beta$) and packed with $3$ circles, equation $(4.3')$ of my previous answer reduces to $$(3 \alpha^2 - 8 \alpha\gamma - 4 \gamma^2) (12 \alpha^2 - 2 \alpha\gamma - \gamma^2) = 0$$ which we can solve to get (ignoring negative values) $$\gamma = \alpha \left(-1+\sqrt{13}\right) \qquad \gamma = \frac{\alpha}{2} \left(-2 +\sqrt{7}\right)$$ These correspond to the inellipses shown in @g.kov's answer.

Interestingly, all six equations $(4.x')$ factor when $\alpha=\beta$.

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Ceva Ratios. If we define $$\delta := \frac{|A'C|}{|BA'|} \qquad \epsilon := \frac{|B'A|}{|CB'|} \qquad \phi := \frac{|C'B|}{|AC'|}$$

Ceva's Theorem states that cevians $\overleftrightarrow{AA'}$, $\overleftrightarrow{BB'}$, $\overleftrightarrow{CC'}$ concur if and only if $\delta\epsilon\phi = 1$. The point of concurrence can be written in the form $(1)$ with $$\alpha:\beta:\gamma \;=\; 1 : \phi : \frac1\epsilon \qquad \left( = \delta^0 : \phi^1 : \epsilon^{-1} \right) \tag{4}$$

(The relation $\delta\epsilon\phi=1$ means that $(4)$ can be written in a variety of ways, none of which is satisfyingly symmetric. However, I think of $(4)$ as being written "relative to" vertex $A$. The parameter $\alpha$ is naturally assigned its corresponding Ceva ratio, $\delta$, raised to the $0$-th power; but $\beta$ steals $\gamma$'s Ceva ratio, $\phi$, which is raised to the $1$-th power (because it's "looking forward"); likewise, $\gamma$ gets $\beta$'s Ceva ratio, $\epsilon$, which is raised to the $(-1)$-th power (because it's "looking backward").)

Some ellipses, such as Mandart's, may be easier to describe using Ceva ratios $\delta:\epsilon:\phi$ than reciprocal-barycentric coordinates $\alpha:\beta:\gamma$.

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Actual Examples! Here's a family of the first six perfectly-packed ellipses in the equilateral triangle whose major axes align with a side of the triangle. Since the concurrence point $P$ lies on an axis of symmetry, so we have $$\alpha:\beta:\gamma = 1 : \kappa: \kappa$$ for values of $\kappa$ we give below.

First, a family portrait ...

... and now the individual cases:

$$\begin{array}{ccc} n = 1 & \qquad & n = 2 \\ \kappa = 1 & & \kappa = \frac16\left(1 + \sqrt{7}\right)= 0.607\ldots \end{array}$$

$$\begin{array}{ccc} n = 3 & \qquad & n = 4 \\ \kappa = \frac1{12}\left(1 + \sqrt{13}\right) = 0.383\ldots & & \kappa = 0.275\ldots \\ & & 1 + 4 \kappa - 20 \kappa^2 - 48 \kappa^3 + 72 \kappa^4 = 0 \end{array}$$

$$\begin{array}{ccc} n = 5 & \qquad & n = 6 \\ \kappa = 0.213\ldots & & \kappa = 0.173\ldots \\ 1 + 4 \kappa - 32 \kappa^2 - 72 \kappa^3 + 144 \kappa^4 = 0 & & 1 + 4 \kappa - 44 \kappa^2 - 96 \kappa^3 + 144 \kappa^4 = 0 \end{array}$$

Answer 2

An example of the right triangle, which Mandart inellipse can be perfectly packed with six circles:

\begin{align} |BC|&=6\,\sqrt3-9-\sqrt{108\,\sqrt3-187}\approx 1.14 ,\\ |AC|&=6\,\sqrt3-9+\sqrt{108\,\sqrt3-187}\approx 1.64 ,\\ |AB|&=2 , \end{align}

major and minor semi-axes of the Mandart inellipse \begin{align} s_a&=\tfrac23\,(5-\sqrt3)\,\sqrt{7\,\sqrt3-12} \approx 0.768275 ,\\ s_b&=\tfrac13\,(5-\sqrt3)\,\sqrt{26\,\sqrt3-45} \approx0.198844 ,\\ \frac{s_b}{s_a}&= \frac{\sqrt2}4\,(\sqrt3-1) =\sin\frac\pi{12} ,\\ |F_1F_2|&=\frac23\,\sqrt{86\,\sqrt3-144} \approx 1.48419 . \end{align}

Answer 3

Here's a way to proceed with the investigation:

1. Determine the minimal polynomial of $e = \cos\frac{\pi}{2n}$ for a desired chain-length $n$.
2. Determine a formula for the eccentricity $e$ of the desired type of ellipse in terms of a triangle's side-lengths $a$, $b$, $c$.
3. Use, for instance, the method of resultants or Groebner bases to eliminate $e$ from the equations in (1) and (2), leaving a polynomial relation among the side-lengths. (I used Mathematica's Resultant[] function.)

For instance, we have these minimal polynomials for the first few $e = \cos\frac{\pi}{2n}$ (computed using Mathematica's MinimalPolynomial[] function):

$$\begin{align} n = 1: &\qquad \phantom{16e^4+12e^2-}\;\,e &= 0 \tag{1.1}\\ n = 2: &\qquad \phantom{16e^4+1}2e^2-1 &= 0 \tag{1.2}\\ n = 3: &\qquad \phantom{16e^4+1}4e^2-3 &= 0 \tag{1.3}\\ n = 4: &\qquad \phantom{1}8 e^4 - \phantom{1}8 e^2 + 1 &= 0 \tag{1.4}\\ n = 5: &\qquad 16 e^4- 20 e^2+ 5 &= 0 \tag{1.5}\\ n = 6: &\qquad 16 e^4 - 16 e^2 + 1 &= 0 \tag{1.6} \end{align}$$

And for, say, the Steiner (circum)ellipse and inellipse, which are similar, the eccentricity satisfies $$(1-e^2)s - 3 e^4t = 0 \tag{2}$$ with $$\begin{align} s &:= a^4 + b^4 + c^4 - a^2 b^2 - b^2 c^2 - c^2 a^2 \\[4pt] t &:= \frac1{16}(a+b+c)(-a+b+c)(a-b+c)(a+b-c) \end{align} \tag{3}$$ (Note that $\sqrt{t}$ is the area of the triangle, by Heron's formula.)

So, a triangle whose Steiner circumellipse/inellipse is perfectly-packed by $n$ circles has sides $a$, $b$, $c$ satisfying ...

$$\begin{align} n = 1: &\; \phantom{1}1s &= 0 \tag{4.1}\\[6pt] n = 2: &\; \phantom{1}2s - \phantom{2}3 t &= 0 \tag{4.2} \\[6pt] n = 3: &\; \phantom{1}4s - 27t &= 0 \tag{4.3} \\[6pt] n = 4: &\; \phantom{1}8s^2 - 120 s t + \phantom{22}9 t^2 &= 0 \tag{4.4} \\[6pt] n = 5: &\; 16 s^2 - 420 s t + 225 t^2 &= 0 \tag{4.5} \\[6pt] n = 6: &\; 16 s^2- 624 s t + \phantom{22}9 t^2 &= 0 \tag{4.6} \end{align}$$

Finding sample triangles is left as an exercise to the reader. $\square$

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More generally, in a coordinate-based context, the general form of the ellipse (or any conic) is

$$K_{20}\,x^2 + K_{11}\,x y + K_{02}\,y^2 + K_{10}\,x + K_{01}\,y + K_{00} = 0$$

One can show that the eccentricity of the represented conic satisfies $$4 p(1 - e^2) + q(2-e^2)^2 = 0 \tag{2'}$$ where $$p := ( K_{20} + K_{02} )^2 \qquad\qquad q := K_{11}^2 - 4 K_{20} K_{02} \;(<0\;\text{for ellipse})\tag{3'}$$

(The $p$ value is squared to match the dimension of $q$.) The represented ellipse is perfectly-packable with $n$ circles when the coefficients satisfy

$$\begin{align} n = 1: &\quad \phantom{12}p + \phantom{12}q &= 0 \tag{4.1'}\\[2pt] n = 2: &\quad \phantom{1}8 p + \phantom{1}9q &= 0 \tag{4.2'}\\[2pt] n = 3: &\quad 16 p + 25 q &= 0 \tag{4.3'}\\[2pt] n = 4: &\quad 128 p^2 + \phantom{1}416 p q + \phantom{1}289 q^2 &= 0 \tag{4.4'}\\[2pt] n = 5: &\quad 256 p^2 + 1072 p q + \phantom{1}841 q^2 &= 0 \tag{4.5'}\\[2pt] n = 6: &\quad 256 p^2 + 1344 p q + 1089 q^2 &= 0 \tag{4.6'} \end{align}$$

For instance, given $\triangle ABC$ with $A=(0,0)$, $B=(c,0)$, $C=(b\cos A,b\sin A)$, the Steiner inellipse is tangent to midpoints $D=\frac12(B+C)$, $E=\frac12(C+A)$, $F=\frac12(A+B)$. We find that the equation of the ellipse is

$$\begin{align} 0 &= 16 x^2 |\triangle ABC|^2 + 8 x y (a - b) (a + b)|\triangle ABC| + y^2 (a^4 - 2 a^2 b^2 + b^4 + 3 c^4) \\[2pt] &- 16 x |\triangle ABC|^2 - 4 y c (a^2 - b^2 + c^2) |\triangle ABC| + 4 c^2 |\triangle ABC|^2 \end{align}$$ so that

$$\begin{align} p &= \phantom{-3}\left(a^2+b^2+c^2\right)^2 \\ q &= -3 (a+b+c)(-a + b + c) (a + b - c) (a - b + c) = -48 |\triangle ABC|^2 \end{align}$$

(where I have divided-out a common factor of $4c^4$). The reader can verify that equations $(4.x)$ and $(4.x')$ are equivalent via the substitutions $s \to (p+q)/4$ and $t\to -q/48$.

In a separate answer, I'll consider a broader family of inellipses.

Answer 4

This is example of the ellipse, perfectly packed with three circles, inscribed into the equilateral triangle $ABC$.

Let the center of the ellipse be $M=0$ and its semi-axes defined as \begin{align} s_a=|DF_1|=|DF_2|&=1 ,\\ s_b=|MD|=|ME|&=\sin\frac\pi{2\cdot3}=\frac12 , \end{align} locations of the top and bottom points are \begin{align} D&=(0,-\tfrac12) ,\quad E=(0,\tfrac12) ,\\ F_1&=(-\tfrac{\sqrt3}2,0) ,\quad F_2=(\tfrac{\sqrt3}2,0) . \end{align}

\begin{align} \text{Then the equation of the ellipse is } \quad x^2+2\,y^2&=0 \end{align} and for the upper arc we have \begin{align} y(x)&=\tfrac12\,\sqrt{1-x^2} ,\\ y'(x)&=-\tfrac12\,\frac{x}{\sqrt{1-x^2}} , \end{align}

so ve can find the point $K$, tangent to the circumscribed equilateral $\triangle ABC$: \begin{align} -\tfrac12\,\frac{x}{\sqrt{1-x^2}} &=\tan\tfrac\pi6=\tfrac{\sqrt3}3 ,\\ x&=-\tfrac{2}{13}\sqrt{39} ,\\ y(x)&=\tfrac{1}{26}\sqrt{13} . \end{align} So, the tangential points are \begin{align} K&=(-\tfrac{2}{13}\sqrt{39},\tfrac{1}{26}\sqrt{13} ) ,\quad L=(\tfrac{2}{13}\sqrt{39},\tfrac{1}{26}\sqrt{13} ) , \end{align} and the location of the vertices of $\triangle ABC$ can be easily found as \begin{align} A&=(-\tfrac16\,\sqrt3\,(\sqrt{13}+1), -\tfrac12) ,\quad B=(\tfrac16\,\sqrt3\,(\sqrt{13}+1), -\tfrac12) ,\\ C&=(0,\tfrac12\,\sqrt{13}) , \end{align} the side length of the triangle is thus \begin{align} |AB|=|BC|=|CA|=a&=\tfrac13\,\sqrt3\,(\sqrt{13}+1) , \end{align}

and the tangential points $D,K,L$ divide the side segments as follows: \begin{align} |AK|=|BL|&=\sqrt{\tfrac2{39}(7+\sqrt{13})} ,\quad |CK|=|CL|=\tfrac4{13}\,\sqrt{39} . \end{align}

This ellipse is neither Steiner nor Mandart inellipse, it is essentially a [![generalized Steiner inellipse]][Linfield1920], for which the foci are the roots of the derivative of the rational function \begin{align} f(z)&=(z-A)^u (z-B)^v (z-C)^w , \end{align}

where $A,B,C$ are the coordinates of the vertices of the triangle, and the tangent points $L,K,D$ divide the segments $BC,CA$ and $AB$ as $v:w,w:u$ and $u:v$, respectively. In this case \begin{align} u=v&=\tfrac1{12}\,(\sqrt{13}-1) ,\\ w&=\tfrac16\,(7-\sqrt{13}) . \end{align}

Semi-axes of such ellipse, expressed in terms of the side length $a$ of the equilateral triangle, are

\begin{align} s_a&=\frac{a\sqrt3}{12}\,(\sqrt{13}-1) ,\quad s_b=\frac{a\sqrt3}{24}\,(\sqrt{13}-1) . \end{align}

Similarly, for another orientation,

we have

\begin{align} s_a&=\frac{a\sqrt3}{3}\,(\sqrt{7}-2) ,\quad s_b=\frac{a\sqrt3}{6}\,(\sqrt{7}-2) \end{align}

and \begin{align} u&=\tfrac{11}3-\tfrac43\,\sqrt7 ,\quad v=\tfrac23\,\sqrt7-\tfrac43 ,\quad w=v . \end{align}

Reference

[Linfield1920]: Ben-Zion Linfield. “On the relation of the roots and poles of a rational function to the roots of its derivative”. In: Bulletin of the American Mathematical Society 27.1 (1920), pp. 17-21