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The question is to give a formula in $x$ and $y$ that gives all three sides of an equilateral triangle. The formula should not be true for points that are not part of the perimeter of the triangle. By this sub-problem, I am trying to understand deeper the page 809 and attack problem 11 on page 813 here -- there the purpose of parametrization is to find solutions to the partial differentials $\triangle u=0$, so-called boundary-problem and Dirichet's condition.

Trial 0

J.M. hinted the parametrization here, through $(1,0)$?

$$\displaystyle r=\frac{\cos\left(\frac{\pi}{n}\right)}{\cos\left(\left(\theta \mod \frac{2\pi}{n}\right) -\frac{\pi}{n}\right)} \;$$

enter image description here

I cannot understand the order of the division yet. I can understand the radius part but not why a line in the triangle, it must be due to some identity. Look I could do this with a small change (reciprocal of the radius):

n=3;
theta=(0:999)/1000;
r=cos(2*pi*(n*theta)%%1/n-pi/n)/cos(pi/n);
plot(r*cos(2*pi*theta),r*sin(2*pi*theta),asp=1,xlab="X",ylab="Y",
        main=paste("Regular ",n,"-gon",sep=""),type='l');

enter image description here

n=3;
theta=(0:999)/1000;
#r=cos(pi/n)/cos(2*pi*(n*theta)%%1/n-pi/n);
r=cos(2*pi*(n*theta)%%1/n-pi/n)/cos(pi/n);
plot(r*cos(2*pi*theta),r*sin(2*pi*theta),asp=1,xlab="X",ylab="Y",
        main=paste("Regular ",n,"-gon",sep=""),type='l');
r=cos(pi/n)/cos(2*pi*(n*theta)%%1/n-pi/n);
lines(r*cos(2*pi*theta),r*sin(2*pi*theta),asp=1,xlab="X",ylab="Y",
        main=paste("Regular ",n,"-gon",sep=""),type='l');

enter image description here

Trial 1

David Wallace instructed me here to use polar coordinates and the floor function to describe the perimeter of the triangle apparently so that

$$\begin{cases}x=r\cos(\theta)\\y=r\sin(\theta)\end{cases}$$

$$r\cos\left(\frac{2\pi}{3}\left(\frac{3\theta}{2\pi}-\left\lfloor\frac{3\theta}{2\pi}\right\rfloor\right) -\frac{\pi}{3}\right) = 1$$

...cannot understand this at all, hard to see the parametric form here.

Perhaps related

  1. Something about plotting the function here

  2. Is there an equation to describe regular polygons?

share|improve this question
    
OK, in polar co-ordinates, I think it's something like $$r\cos(\frac{2\pi}{3}(\frac{3\theta}{2\pi}-\lfloor\frac{3\theta}{2\pi}\rfloor) -\frac{\pi}{3}) = 1 $$. Still need to convert it back to rectangular co-ordinates. –  user22805 Mar 23 '12 at 9:19
2  
@David and hhh: Raskolnikov gave a derivation of the general case of a regular polygon here, which is easily adaptable to the case of an equilateral triangle. –  J. M. Mar 24 '12 at 11:36
    
What exactly isn't clear in Raskolnikov's answer? –  J. M. Mar 24 '12 at 12:57
    
That was explained in Raskolnikov's answer. Again: have you read it? –  J. M. Mar 24 '12 at 14:40

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