# How to transform a triangle into a circle?

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 $$120º$$ ?

• You want to continuously deform a regular polygon (equilateral triangle in your case) out to its circumcircle? Something like this? Choose $n = 3$ for a triangle and drag the slider $0 \leq u \leq 1$ to realize the straight-line homotopy between the triangle and the circle. Commented 2 days ago
• Can you please edit the question to clarify: (i) Are you looking for a continuous mapping? (ii) Do you seek a mapping that preserves rays from the origin? (iii) Do you want a bijective (one-to-one and onto) map of the plane to itself? <> Your approach suggests answers to these are all yes, but it would be good to make any constraints explicit, since there are variously easy ways to do this if not all these conditions are required. Commented 2 days ago
• You might be interested in knowing that if we consider the plane as $\Bbb C$, there's a function that's analytic on the interior of the triangle that maps it biholomorphically onto the interior of a circle. Commented yesterday

## 1 Answer

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.