I have some lecture notes where the following is claimed:

Suppose that three points $0, p_1, p_2$ constitute a triangle $T$ in $\mathbb{R}^2$. Suppose the angle between $p_1$ and $p_2$ is $\theta$ and $0$ and $p_2$ is $\varphi$ and that the center of the circumball around $T$ has center $(r, \tau)$ written in polar coordinates. Then we may express the points $p_1,p_2$ in cartesian coordinates as $$ p_1 = -2r \big(\sin(\varphi) \cos(\tau+\varphi), \sin(\varphi) \sin(\tau+\varphi) \big) $$ and $$ p_2 = -2r \big(\sin(\theta + \varphi) \cos(\theta+\tau+\varphi), \sin(\theta+\varphi) \sin(\theta+\tau+\varphi) \big) $$

I have attached a drawing of the setup here (for me it suffices to look at the case of acute triangles):

enter image description here

To begin with we just assume $r=1$ and scale everything up after (corresponding to multiplication with $2r$). My attempts so far have been trying to using Thales theorem to create some right angles in order to have expression with sine and cosine appear. However I can't for the life of me make sense of what "$\varphi + \tau$" actually represent and how to connect it to the first and second coordinates of the points.

Update: I think the formula in the notes might be wrong. Consider the case of $\theta=\varphi=\pi/3$, $\tau=0$ and $r=1$. Then $p_1$ lies in the first quadrant and should have positive coordinates, yet the formula yields $p_1=(-\sqrt{2}/3,-4/9)$.

As a separate attempt to come up with a parametrization of $p_1$ myself, I have drawn the following:

enter image description here

Appealing to the law of sines, we would then have

$$ p_1 = 2r\sin(\varphi)(\cos(\pi/2-\alpha+\tau),\sin(\pi/2-\alpha+\tau)). $$

However I need $\alpha$ to be an expression involving only $\theta, \phi, r, \tau$. If one loads up Geogebra to play around it actually looks like $\alpha = \varphi$, at least in the case of an acute triangle.

enter image description here

enter image description here

However I cannot see why this is true but this would certainly solve the problem if it was. I have tried to look at so many angle identities around triangles and parallel lines as I could find.

Can anyone help me solve this?

  • $\begingroup$ "Suppose the angle between $p_1$ and $p_2$ is $\theta$ and $0$ and $p_2$ is $\varphi$" As an aside, I hope this isn't the literal wording in the notes, as the interpretation is unclear. You can't have an angle between points; only edges or lines. Sometimes people refer to the point vector from origin by just the point, and that's fine. But between $0$ and $p_2$ doesn't make sense in that context. $\endgroup$ Commented Aug 10, 2023 at 11:47
  • 1
    $\begingroup$ No for the second one it is worded a bit more precise in the notes as the angle between $-p_2$ and $p_1-p_2$ $\endgroup$
    – Jacobiman
    Commented Aug 10, 2023 at 12:15
  • $\begingroup$ What do you mean by Thales theorem? $\endgroup$
    – DK2412
    Commented Aug 10, 2023 at 18:13
  • $\begingroup$ @DK2412 That three points on a circle with two of the points being exactly distance radius apart creates a right-angle triangle. en.wikipedia.org/wiki/Thales%27s_theorem $\endgroup$
    – Jacobiman
    Commented Aug 10, 2023 at 20:26

1 Answer 1


You are almost there.

it actually looks like $\alpha = \varphi$

$\alpha=\varphi$ does hold.

(Proof : In your drawing, $\triangle{ADB}$ is an isosceles triangle with $DA=DB$ and $\angle{ADB}=2\angle{ACB}$ (see here). Let $E$ be a point on $AB$ such that $AB\perp DE$. Then, since $DA=DB$ and $\angle{DAE}=\angle{DBE}$, we have $\triangle{DAE}\equiv\triangle{DBE}$. So, it follows from $\angle{EDA}=\angle{EDB}$ that $\alpha=\angle{EDA}=\frac{1}{2}\angle{ADB}=\angle{ACB}=\varphi$.$\ \square$)

So, as you wrote, we have


$$p_2\bigg(2r\sin(\pi-\theta-\varphi)\cos\bigg(\frac{\pi}{2}-\varphi+\tau-\theta\bigg),2r\sin(\pi-\theta-\varphi)\sin\bigg(\frac{\pi}{2}-\varphi+\tau-\theta\bigg)\bigg)$$ i.e. $$\color{red}{p_1\bigg(2r\sin(\varphi)\sin(\varphi-\tau),2r\sin(\varphi)\cos(\varphi-\tau)\bigg)}$$ $$\color{red}{p_2\bigg(2r\sin(\theta+\varphi)\sin(\theta+\varphi-\tau),2r\sin(\theta+\varphi)\cos(\theta+\varphi-\tau)\bigg)}$$

which is not the same as the formula in the notes.


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