Geometric Pattern in Triangle Construction with Squares

Question

I've been exploring some constructions in geometry recently and have been looking at the following algorithm:

1. Start with an arbitrary triangle (red)
2. Construct 3 squares from each of the sides of the triangle where the squares do not intersect
3. Connect the centers of each square to form a new triangle (green)
4. Repeat ad infinitum

It appears that this process will converge to an equilateral triangle after infinite repetitions. I'm curious as to why that is. I started brute forcing the calculation algebraically with arbitrary coordinates for the vertices of the original triangle, but it gets messy quickly. I'm wondering if there's a more elegant explanation.

Answer 1

Let the side lengths of the original triangle be $BC=a,AC=b,AB=c$, and let its area be $K$. Let the center of the square from side $BC$ be $S_{a}$, etc. and let $S_{b}S_{c}=x, S_{a}S_{c}=y,S_{a}S_{b}=z$.

Note that $\angle S_{b}AS_{c} = \frac{\pi}{4}+\angle A + \frac{\pi}{4} = \angle A + \frac{\pi}{2}$. Thus, by the Law of Cosines:

$$\begin{align}x^{2} &= AS_{b}^{2}+AS_{c}^{2} - 2AS_{b}AS_{c}\cos(\angle S_{b}AS_{c}) \\ &= \frac{b^{2}}{2} + \frac{c^{2}}{2} - bc\cos\left(\angle A + \frac{\pi}{2}\right)\\ & = \frac{b^{2}}{2} + \frac{c^{2}}{2} + bc\sin(\angle A)\\ &= \frac{b^{2}}{2} + \frac{c^{2}}{2} + 2K\end{align}$$

We find $y^{2},z^{2}$ similarly. You might already see that the differences between the sidelengths are being "smoothed out".

To formalize this, let's measure the "equilateralness" of a triangle as $\text{eq}(\triangle ABC) = (a^{2}-b^{2})^{2} + (b^{2}-c^{2})^{2} + (c^{2}-a^{2})^{2}$. You can check that this is zero iff $\triangle ABC$ is equilateral, and positive otherwise. We have:

$$\begin{align}\text{eq}(\triangle S_{a}S_{b}S_{c}) &= (x^{2}-y^{2})^{2} + (x^{2}-z^{2})^{2} + (y^{2}-z^{2})^{2}\\ &= \left(\frac{a^{2}-b^{2}}{2}\right)^{2} + \left(\frac{a^{2}-c^{2}}{2}\right)^{2} + \left(\frac{b^{2}-c^{2}}{2}\right)^{2}\\ &= \frac{1}{4}\text{eq}(\triangle ABC)\end{align}$$

Thus, the equilateralness converges geometrically to 0, as desired. Note that this could also signify the triangle shrinking to a point, but we can rule this out because the perimeter is increasing.

Answer 2

Geometry with complex numbers (+ linear algebra) provides an interesting proof that I am going to develope.

I will use the same notations for the points of the plane and for their associated complex numbers in a fixed reference frame.

Notations :

$$\omega:= e^{2 i \pi/3} \ \ \text{and} \ \ \alpha = \frac12(1+i)\tag{0}$$

(please note that $1,\omega,\omega^2$ are the 3 third roots of unity, and as such form an equilateral triangle)

Let $A_1,B_1,C_1$ be the vertices of the initial triangle.

Let $A_2,B_2,C_2$ be the centers of the squares built on sides $B_1C_1, C_1A_1, A_1B_1$ resp.

As triangle $B_1A_2C_1$ is a right isosceles triangle, one can
write :

$$B_1-A_2=i(C_1-A_2) \ \iff \ (i-1)A_1=B_1-iC_1 \ \iff \ A_2=\alpha(B_1-iC_1)$$

or $$A_2=\alpha B_1+\bar{\alpha}C_1\tag{1}$$

where $\alpha=\frac{1}{1-i}=\frac12(1+i)$ has been introduced in (0).

For similar reasons, by circular permutation, we have

$$\begin{cases}B_2=\alpha C_1+\bar{\alpha}A_1\\C_2=\alpha A_1+\bar{\alpha}B_1\end{cases}\tag{2}$$

We can now gather (1) and (2) under a matrix-vector form that we can at once make general in order to understand the whole process :

$$\underbrace{\pmatrix{A_{n+1}\\B_{n+1}\\C_{n+1}}}_{V_{n+1}}=\underbrace{\pmatrix{0&\alpha&\bar{\alpha}\\ \bar{\alpha}&0&\alpha\\ \alpha&\bar{\alpha}&0}}_M\underbrace{\pmatrix{A_{n}\\B_{n}\\C_{n}}}_{V_{n}}\tag{3}$$ Matrix $M$ has two properties :

• (a) $M$ is hermitian ($M^H=M$ where symbol $H$ means transposition + conjugation);

• (b) $M$ is circulant.

As a consequence of (a), $M$ has real eigenvalues :

$$\begin{cases} \lambda_1&=& \ \ \ 1\\ \lambda_2 &=& - \frac12 (1+\sqrt{3}) &\approx &-1.366 \\ \lambda_3&=& \ \ \ \frac12 (-1+\sqrt{3}) &\approx & \ \ \ 0.366\end{cases}\tag{4}$$

Please note that $\lambda_2$, that we will write $\lambda_{max}$, is the dominant eigenvalue.

As a consequence of (b), $M$ is diagonalized by Discrete Fourier matrix $F$ (see here) in this way :

$$M=\underbrace{\frac{1}{\sqrt{3}}\pmatrix{1&1&1\\1&\omega&\omega^2\\1&\omega^2&\omega}}_{F}\underbrace{\pmatrix{\lambda_1&&\\ &\lambda_2&\\ &&\lambda_3}}_{D}\underbrace{\frac{1}{\sqrt{3}}\pmatrix{1&1&1\\1&\omega^2&\omega\\1&\omega& \omega^2}}_{F^H}\tag{5}$$

Remark : the columns of $F$ are, of course, unit-norm eigenvectors associated to the $\lambda_k$s in their resp. order.

(5) gives, for large values of $n$, the (rather well known) approximation:

$$M^n=F D^n F^H \approx F \pmatrix{1&&\\ &(\lambda_{max})^n&\\ &&0} F^H \approx (\lambda_{max})^n UU^H, \tag{6}$$

where $U$ is the second column of matrix $F$ ($U$ is a unit-norm eigenvector associated with $\lambda_{max}$), namely :

$$U=\frac{1}{\sqrt{3}}\pmatrix{1 \\ \omega \\ \omega^2}\tag{7},$$

$U^H$ being the second line of matrix $F^H$.

Therefore :

$$V_{n+1}=M^n V_1 \approx (\lambda_{max})^n \frac13 \pmatrix{1 \\ \omega \\ \omega^2}\pmatrix{1&\omega^2&\omega}\pmatrix{A_1 \\ B_1 \\ C_1}$$

$$V_{n+1}\approx (\lambda_{max})^n \frac13 (A_1+\omega^2 B_1+\omega C_1) \pmatrix{1 \\ \omega \\ \omega^2}$$

Otherwise said, $V_{n+1}:=\pmatrix{A_{n+1}\\B_{n+1}\\C_{n+1}}$ is arbitrarily close to a (complex) multiple of $\pmatrix{1 \\ \omega \\ \omega^2}$ which is an equilateral triangle.

MATLAB program :

 a=(1+i)/2;b=conj(a);
 M=[0,a,b;
    b,0,a;
    a,b,0];
 A=1,B=exp(i*pi*rand);C=B*exp(i*pi*rand);
 V=[A;B;C];
 for k=1:7
    plot([V;V(1)]);hold on;axis equal;
    K=num2str(k);
    text(real(V),imag(V),{['A_',K],['B_',K],['C_',K]})
    V=M*V;
 end;

Answer 3

Here's a slightly different approach starting from Jean-Marie's excellent proof, but avoiding the "it is well known that..." step. It's about 80% copied from that answer, so Jean-Marie deserves all the credit. I'm marking it 'community wiki' because it really is a community effort.

I'll use the notation of that answer up through step 3, where it's shown that

$$\underbrace{\pmatrix{A_{n+1}\\B_{n+1}\\C_{n+1}}}_{V_{n+1}}=\underbrace{\pmatrix{0&\alpha&\bar{\alpha}\\ \bar{\alpha}&0&\alpha\\ \alpha&\bar{\alpha}&0}}_M\underbrace{\pmatrix{A_{n}\\B_{n}\\C_{n}}}_{V_{n}}\tag{3}$$

From this one can immediately conclude that $$V_{n+1}=M^n V_1 \tag{4}$$.

As Jean-Marie observes, $M$ is hermitian. In particular, it has real eigenvalues :

$\lambda_{max} = - \frac12 (1+\sqrt{3}) \approx -1.366 $ (dominant) , the other eigenvalues being $1$ and $\approx 0.366$.

A this point I diverge from Jean-Marie, and calling the eigenvalues $\lambda_1, \lambda_2, \lambda_3$ in decreasing order of size, with corresponding unit eigenvectors $u_1, u_2, u_3$, we have (because $M$ is Hermitian) that $u_i \cdot u_j = 0$ for $i \ne j$. If we make a matrix $U$ with columns $u_1, u_2,$ and $u_3$, then this says that $$ U^{*} U = I $$ the $3 \times 3$ identity, i.e., $U^{1} = U^{*}$.

Because the $u_i$ are eigenvectors, we have $$ MU = U\Lambda $$ where $\Lambda$ is diagonal with nonzero entries $\lambda_1, \lambda_2, \lambda_3$. That means that $$ M = U \Lambda U^{-1}. $$ That means that \begin{align} M^2 &= (U \Lambda U^{-1})(U \Lambda U^{-1})\\ &= U \Lambda (U^{-1}U) \Lambda U^{-1})\\ &= U \Lambda^2 U^{-1}\\ &= U \Lambda^2 U^{*}\\ \end{align} and similarly for higher powers. If you write this out in terms of the individual eigenvectors $u_i$, it says that $$ M^n = u_1\lambda_1^n u_1^{*} + u_2 \lambda_2^n u_2^{*} + u_3 \lambda_1^n u_3^{*}. $$ Dividing through the $\lambda_1^n$, this gives us \begin{align} \frac{1}{\lambda_1^n} M^n &= u_1 u_1^{*} + u_2 \frac{\lambda_2^n}{\lambda_1^n} u_2^{*} + u_3 \frac{\lambda_1^n }{\lambda_1^n}u_3^{*}\\ &= u_1 u_1^{*} + u_2 \left(\frac{\lambda_2}{\lambda_1}\right)^n u_2^{*} + u_3 \left(\frac{\lambda_1 }{\lambda_3}\right)^n u_3^{*}\\ \end{align} Now because $|\lambda_2| < |\lambda_1|$, and similarly for $\lambda_3$, the fractions in the second and third terms tend to zero as $n$ gets large, so we have $$ \lim_{n \to \infty} \frac{1}{\lambda_1^n} M^n = u_1 u_1^{*}. $$

where

$$u_1=\frac{1}{\sqrt{3}}\pmatrix{1 \\ \omega \\ \omega^2}\tag{6}$$

as Jean-Marie notes.

Now \begin{align} \frac{1}{\lambda_1^n}V_{n}&=\frac{1}{\lambda_1^n}M^n V_1 \end{align} so \begin{align} \lim_{n \to \infty} \frac{1}{\lambda_1^n}V_{n} &=\frac{1}{\lambda_1^n}M^n V_1\\ &=\frac{1}{\lambda_1^n}u_1 u_1^{*} V_1 \end{align} Thus if we scale down by a factor of $\lambda_1$ after each iteration of the "erect squares and connect centers" operation, the resulting sequence of triangles will have a limiting triangle. At this point, Jean-Marie's approximation steps can be replaced with equalities; because

V_1 \approx (\lambda_{max})^n \frac13 \pmatrix{1 \ \omega \ \omega^2}\pmatrix{1&\bar{\omega}&\bar{\omega}^2}\pmatrix{A_1 \ B_1 \ C_1}$$

$$ \lim_{n \to \infty} \frac{1}{\lambda_1}^{n+1}) V_{n+1} = \frac13 (A_1+\omega^2 B_1+\omega C_1) \pmatrix{1 \\ \omega \\ \omega^2}$$

(NB: I've probably lost a complex conjugate somewhere above, and maybe shifted an $n$ for an $n+1$; my apologies for that. I lack the time to go back and double-check every word, of which there are too many.)

I've edited the matlab program slightly to show the convergence of the triangles when you include the division.

MATLAB program :

 a=(1+i)/2;b=conj(a);
 M=[0,a,b;
    b,0,a;
    a,b,0];
 A=1;B=exp(i*pi*rand);C=B*exp(i*pi*rand);
 V=[A;B;C];
 Lmax = -(1./2.) * (1.0+sqrt(3.0));
 for k=1:7
    plot([V;V(1)]);hold on;axis equal;
    K=num2str(k);
    text(real(V),imag(V),{['A_',K],['B_',K],['C_',K]})
    V=(1/Lmax) * M*V;
 end;

Sample output:

Answer 4

Follows a MATHEMATICA script which generates a sequence of triangles approaching an equilateral structure.

$$ \Delta_k = \Phi\left(\Delta_{k-1}\right) $$

Here $\Phi$ is defined as the module gTri[tri, f, sign] so we have equivalently tri = gTri[tri, f, sign]. The parameters are: sign which calculates the triangle orientation, and f = 0.5 for the square, as the semi-length of the side which is the square basis.

gTri[tri_, f_, sign_] := Module[{u, v, w, du, dv, dw, nu, nv, nw, c1, c2, c3},
  u = tri[[1]] - tri[[2]];
  v = tri[[2]] - tri[[3]];
  w = tri[[3]] - tri[[1]];
  du = Norm[u];
  dv = Norm[v];
  dw = Norm[w];
  nu = {u[[2]], -u[[1]]}/du;
  nv = {v[[2]], -v[[1]]}/dv;
  nw = {w[[2]], -w[[1]]}/dw;
  c1 = (tri[[1]] + tri[[2]])/2;
  c2 = (tri[[2]] + tri[[3]])/2;
  c3 = (tri[[3]] + tri[[1]])/2;
  Return[{c1 - nu du f sign, c2 - nv dv f sign, c3 - nw dw f sign}]
  ]

SeedRandom[10];
tri0 = RandomReal[{-1, 1}, {3, 2}];
u = tri0[[1]] - tri0[[2]];
v = tri0[[2]] - tri0[[3]];
sign = Sign[u[[1]] v[[2]] - u[[2]] v[[1]]];
grtris = {ListLinePlot[Join[tri0, {tri0[[1]]}], PlotStyle -> Red]};
tri = tri0;
ERR = {Abs[Norm[tri[[1]] - tri[[2]]] - (Norm[tri[[1]] - tri[[2]]] + 
       Norm[tri[[2]] - tri[[3]]] + Norm[tri[[3]] - tri[[1]]])/3]/Norm[tri[[1]] - tri[[2]]]};

error = 1000;
f = 0.285;
f = 0.5;
While[error > 10^-10,
 tri = gTri[tri, f, sign];
 error = Abs[Norm[tri[[1]] - tri[[2]]] - (Norm[tri[[1]] - tri[[2]]] + 
         Norm[tri[[2]] - tri[[3]]] + Norm[tri[[3]] - tri[[1]]])/3]/Norm[tri[[1]] - tri[[2]]];
 AppendTo[grtris, ListLinePlot[Join[tri, {tri[[1]]}]]];
 AppendTo[ERR, error]
 ]


Show[grtris, PlotRange -> All, AspectRatio -> 1, PlotRange -> {{-1, 1}, {-1, 1}}, Axes -> False]
ListLinePlot[ERR, PlotRange -> All]

Those plots show a typical case with f = 0.5

Now changing to f = 0.287 we obtain

which suggests the existence of a value for f that leads to the equilateral triangle instantaneously.

NOTE

Applying symbolically this iteration to a generic triangle given by

$$ \Delta_0=((x_1,y_1),(x_2,y_2),(x_3,y_3)) $$

we obtain

$$ \Delta_1 = \cases{ ((x_1 + x_2)/2 - f (y_1 - y_2),(y_1+y_2)/2-f(-x_1+x_2))+\\ ((x_2 + x_3)/2 - f (y_2 - y_3), (y_2 + y_3)/2-f (-x_2 + x_3)) +\\ ((x_1 + x_3)/2 - f (-y_1 + y_3), (y_1 + y_3)/2-f (x_1 - x_3)) } $$

and after imposing same length for the sides of $\Delta_1$ we got

$$ f = \frac{1}{2\sqrt{3}} = 0.2886751345948129... $$

and thus we obtain the equilateral property instantaneously. Is this a known fact? The following plot is for SeedRandom[12]; and f = 0.2886751345948129; just in one shot. In red the initial triangle and in blue the iterated result.