The Shortest Waynet in the Unitsquare

by Klaus-Anton, Jun 8, 2024, 7:45 PM

The shortest waynet in the unitsquare is not twice the diagonal (not: $2\times\sqrt2=2.828\dots$ !).

But it is $\sqrt3+1=2.732\dots$ Hans Walser 2006 / 2019. Optimierung in der Geometrie (Extensive Fassung) 4.2.3.1 Minimales Wegenetz im Quadrat. Walser demonstrates this with an intuitively very well accessible physical experiment using soupsuds.

Let us try to reproduce the resulting figure with geometry.asy.

Here is the code we can take to let asy find the first Fermat-Point (Fermat Point with Asymptote).

triangle t1=triangle(A,F,D);
point F1=fermat(t1)[0];

[Our triangle t1 is here the in red drawn triangle. To find this point in the blue triangle this is in principle just the same. Look into the code, click on the figure to see it.]

$[asy] size(6cm); import geometry; point A=(0,0),B=(1,0),C=(1,1),D=(0,1) ,myE=midpoint(D--A) ,F=midpoint(B--C) ,F1p; path usquare=A--B--C--D--cycle; triangle t1=triangle(A,F,D); point F1=fermat(t1)[0]; triangle t2=triangle(B,C,myE); point F1p=fermat(t2)[0];// $F_1{^\prime}$ real gr=(sqrt(5)-1)/2; //distance("$1$",A,B,.75cm,gray); distance("$1$",A,B,.75cm,gray,Arrows(size=.25cm,FillDraw(white,gray)));//autosizes, therefore size(6cm) at the beginning draw(usquare); draw(D--F--A,red); draw(C--myE--B,blue); draw(t1,red); draw(t2,blue); draw(A--F1--F1p--C^^D--F1^^B--F1p,green+linewidth(sqrt(gr))); label("$F_1$",F1,dir(D--F1),red); label("$F_1{^\prime}$",F1p,dir(C--F1p)*dir(-1),blue); label("$A$",A,1.1*SW); label("$B$",B,.9*SE); label("$C$",C,NE); label("$D$",D,.85*NW); label("$E$",myE,1.05*W); label("$F$",F,E); dot(usquare,Fill(white)); dot(myE^^F,Fill(white)); dot(F1,red,Fill(white)); dot(F1p,blue,Fill(white)); shipout(bbox(2mm, Fill(white))); [/asy]$

Figure 1: The Shortest Waynet in the Unitsquare (green)

$\color{red}F_1$ and $\color{blue}F_1{}^\prime$ are the First Fermat Points

of the red respective the blue Isosceles $\left(1,\sqrt5/2,\sqrt5/2\right)$ Triangles

$[asy] size(7cm); usepackage("kpfonts"); import geometry; point A=(0,0),B=(1,0),C=(1,1),D=(0,1) ,myE=midpoint(D--A) ,F=midpoint(B--C) ,F1p ; path usquare=A--B--C--D--cycle; triangle t1=triangle(A,F,D); point F1=fermat(t1)[0]; triangle t2=triangle(B,C,myE); point F2=fermat(t2)[0]; point F1p=fermat(t2)[0]; real gr=(sqrt(5)-1)/2; perpfactor=.8; perpendicularmark((F1.x,0),NE,green); draw(F1--(F1.x,0),green+dotted+linewidth(gr^.5)); draw(usquare); draw(D--F--A,red); draw(C--myE--B,blue); draw(t1,red); draw(t2,blue); draw(A--F1--F2--C^^D--F1^^B--F2,green+linewidth(sqrt(gr))); distance("$1$",D,C,-.75cm,gray); label("$F_1$",F1,dir(SE)*dir(2),red); label("$F_1{^\prime}$",F2,dir(C--F2)*dir(-2),blue); label("$A$",A,1.1*SW); label("$B$",B,.9*SE); label("$C$",C,NE); label("$D$",D,.85*NW); label("$E$",myE,1.05*W); label("$F$",F,E); label(format("$A\color{red}F_1\color{black}=\sqrt3/3\approx%.3f$", arclength(A--F1)),(midpoint(A--B).x,-.25));//sqrt(3)/3=.577350269... //label(format("$F_1F_1{}^\prime\approx%.9f$", arclength(F1--F1p)),(midpoint(A--B).x,-.25));//.422649731... //label(format("$EF_1\approx%.9f$", arclength(myE--F1)),(midpoint(A--B).x,-.25));//0.288675135... //label(format("$EF_1\approx%.9f$", arclength(myE--F1)-sqrt(3)/6),(midpoint(A--B).x,-.25));//0 // sqrt(3)/6=0.2886751345948129... // 0.288675135... // arclength(myE--F1)//=F1.x label("$\sqrt3/6$",(.5*F1.x,0),gr*S,green); dot(usquare,Fill(white)); dot(myE^^F,Fill(white)); dot(F1,red,Fill(white)); dot(F2,blue,Fill(white)); dot((F1.x,0),green,Fill(white)); shipout(bbox(2mm, Fill(white))); [/asy]$

Figure 2: Further Consituting Lengths

$\begin{align*} \color{red}F_1\color{blue}F_1{}^\prime\color{black}=1-2*\sqrt3/6=1-\sqrt3/3\approx0.422\tag{$1$}\\ \end{align*}$
We can expect that also the half of this value will be of interest, therefore we keep it:

$\begin{align*} \dfrac{1-\sqrt3/3}{2}\approx0.211\tag{$2$} \\ \end{align*}$
In our waynet of the unitsquare we have four segments of the same size. And a fifth segment, which is some what smaller.
So now we are able to reproduce:
$\begin{align*} \left(4*\sqrt3/3+\left(1-\sqrt3/3\right)\right)-\left(\sqrt3+1\right)=0\tag{$3a$}\\ \end{align*}$
Or just the same in an equivalent form:

$\begin{align*} \frac{4\times\sqrt3}{3}+1-\frac{\sqrt3}{3}\approx2.732\tag{$3b$}\\ \frac{\sqrt3}{3}+1\approx2.732\tag{$3c$}\\ \end{align*}$
You see that $AB\color{blue}F_1{}^\prime\color{red}F_1$ form an isosceles trapezoid. The length of the baseline of this trapezoid is 1. The baseangle is $60^\circ$ , the angle from the leg to the topline is $120^\circ$ , the length of the topline is $\color{red}F_1\color{blue}F_1{}^\prime\color{black}\approx0.422$ , the height of the trapezoid is $.5$ .

And clearly, we have symmetry. You find this trapezoid twice. The other one is $CD\color{red}F_1\color{blue}F_1{}^\prime\color{black}\kern-1pt.$

Note 1:
You do not find the keyword Fermat in geometry_en.pdf, but in geometry.asy.

Note 2:
The red or blue isosceles triangle with the sidelengths $\left(1,\sqrt5/2,\sqrt5/2\right)$ also is known as "Knauth’sches Dreieck (nach Knauth)" (Alfred Hoehn 2012. Pythagoras.pdf, Figur 9. and Klaus-Anton's blog The Knauth-Triangle, and Klaus-Anton's blog Half-Circle-Tangent in Square, 3-4-5-Triangle, Knauthian Fig).

For your memorization and for you to be able to compare here again the complete Knauthian Figure:

Figure 3: Knauthian Figure

code, click me

size(7cm);

import geometry;
point A=(0,0),B=(1,0),C=(1,1),D=(0,1)
,myE=midpoint(D--A)
,F=midpoint(B--C)
,G=midpoint(C--D)
,H=midpoint(A--B)  
;

path usquare=A--B--C--D--cycle;

triangle t1=triangle(A,F,D)
, t2=triangle(B,C,myE)
, t3=triangle(A,G,B)
, t4=triangle(C,H,D)
  ;

real gr=(sqrt(5)-1)/2;

filldraw((path)t1,blue+opacity(gr^2),black+linewidth(gr));
filldraw((path)t2,red+opacity(gr^2),black+linewidth(gr));
filldraw((path)t3,yellow+opacity(gr^2),black+linewidth(gr));
filldraw((path)t4,cyan+opacity(gr^2),black+linewidth(gr));

draw(usquare,linewidth(gr));

shipout(bbox(2mm, Fill(white)));

/*
online compile: http://asymptote.ualberta.ca/
download as svg
load in gimp
export from gimp to png
*/

Note 3:
Look here. This is the from Walser mentioned Coxeter way to find the first Fermat Point. As the example we do it with the red triangle of our Figure 1.
$[asy] import geometry; size(8cm); point A=(0,0), B=(1,.5), C=(0,1); triangle t=triangle(A,B,C); point F1=fermat(t)[0]; point mpAB=midpoint(A--B); point mpBC=midpoint(B--C); point mpAC=midpoint(A--C); real degAB=degrees(dir(A--B)); real degAC=degrees(dir(A--C)); real degBC=degrees(dir(B--C)); real a=arclength(B--C); real b=arclength(A--C); real c=arclength(A--B); point Ap=shift(B)*(a*dir(degBC-60)); point Bp=b*dir(degAC+60); point Cp=c*dir(degAB-60); label("$A^\prime$",Ap,dir(mpBC--Ap)); label("$B^\prime$",Bp,dir(mpAC--Bp)); label("$C^\prime$",Cp,dir(mpAB--Cp)); draw(Ap--F1^^Bp--F1^^Cp--F1,red+dashed); draw(A--Cp--B^^A--Bp--C^^C--Ap--B,dotted); draw(A--F1^^B--F1^^C--F1); draw(t,red); label(t, alignFactor=1.25,red); dot("$F_1$",F1, dir(C--F1),red,Fill(white)); dot(Ap^^Bp^^Cp,black,Fill(white)); dot(t.A^^t.B^^t.C,red,Fill(white)); shipout(bbox(2mm, FillDraw(white,white))); [/asy]$

Figure 4: The First Fermat Point in the Isosceles $\left(1,\sqrt5/2,\sqrt5/2\right)$ Triangle

The Coxeter Way of Construction

[Sidenote: Gentle Reader! - Please do not be too angry with me. Here i label this red triangle with $ABC$ but in Figure 1 it is labelled with other letters. As i here have begun a new figure without the other points there, i think - you will can agree with me. This is okay so. Trust your eyes! It is just the same triangle.]

You draw over the three sides of the triangle ABC equilateral triangles. This produces you the three points $A^\prime,\ B^\prime$ and $C^\prime$ . Now connect $A^\prime$ with $A$ , $B^\prime$ with $B$ , and $C^\prime$ with $C$ . The intersectionpoint of these three segemnts leads you to $F_1$ . So you have your "Mercedes Star" in the triangle.

My opinion is that this way of construction is for the intuition quite easier to reproduce than the way with the circles in the above quoted link Fermat Point with Asymptote.

Note 4:
The green path remembers a little bit asy home gallery eetomumu.svg coming from eetomumu.asy. Compare: cern.ch eemumu.png. But i think here especially the angles do not have to significate much, they seem me to have only a simple schematical meaning, and not more. Okay so.

Literature: