How to transform a triangle into a circle?

Question

I am trying to come up with a function that allows me to "transform a triangle into a circle".

For example, given a circumference with radius $1$ and center $(0,0)$ , imagine a triangle inscribed in the circumference such that these are its vertexes and one of the $3$ middle points of the $3$ sides:

$$P_{1}(x,y) = (0,1)$$ $$P_{2}(x,y) = (\frac{\sqrt3}{2},-\frac{1}{2})$$ $$P_{3}(x,y) = (-\frac{\sqrt3}{2},-\frac{1}{2})$$ $$P_{4}(x,y) = (0,-\frac{1}{2})$$

After applying the transform, all of these points should end up in the circumference with radius $1$.

What I think I must do is:

• Transform points to polar coordinates $r,\sigma$ $$P_{1}(r,\sigma) = (1,90º)$$ $$P_{2}(r,\sigma) = (1,210º)$$ $$P_{3}(r,\sigma) = (1,330º)$$ $$P_{4}(r,\sigma) = (0.5,270º)$$
• Multiply the r by a function $F(\sigma)$, such that:

$F(90º) = 1$ such that $P_{1}(r,\sigma)·1 = (1,90º)$

$F(210º) = 1$ such that $P_{2}(r,\sigma)·1 = (1,210º)$

$F(330º) = 1$ such that $P_{3}(r,\sigma)·1 = (1,330º)$

$F(270º) = 2$ such that $P_{4}(r,\sigma)·2 = (1,270º)$

• Transform the points back to cartesian coordinates $x,y$

So I thought, given a trajectory starting at the origin, and with a direction defined by the angle $\sigma$, I could compute the distance $D(\sigma)$ from the origin to where such trajectory intersects the triangle (positive direction only), and invert it ($F(\sigma) = D(\sigma)^{-1}$).

Is my assumption correct? I can't figure out how to find such a function $D(\sigma)$ that expresses the distance from the origin to the triangle for a given angle $\sigma$. Will it be a sinusoidal oscillating from $1$ to $0.5$ with a period of $120º$ ?

Answer 1

Let the Circle have the triangle inside it.

We can make the Equations for the 3 lines , like $y=mx+c$ , using the known Points.
We have to identify the angle each side take up.
Eg let $\theta_1 \le \theta \le \theta_2$ be the angle range for the side shown here.

At angle $\theta$ , we can easily get the Point $P_1$ on the line.
The Point $P_2$ on the Circle is $(\cos \theta,\sin \theta)$
We have $P_1 \leftrightarrow P_2$ , hence we have the transformation in terms of Parameter $\theta$.
When we want to , we can eliminate $\theta$.
We will get a continuous , though piece-wise , mapping.
When we handle negative angle ranges suitably , the 3 pieces with be

• $\theta_1 \le \theta \le \theta_2$
• $\theta_2 \le \theta \le \theta_3$
• $\theta_3 \le \theta \le \theta_1+2\pi$

Do let me know whether this is what you want. You can easily continue from here. Do let me know whether you want a little more calculations here.