# Express corners in a triangle from two angles and angle and polar coordinates of circumcenter

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):

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:

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.

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?

• "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. Commented Aug 10, 2023 at 11:47
• No for the second one it is worded a bit more precise in the notes as the angle between $-p_2$ and $p_1-p_2$ Commented Aug 10, 2023 at 12:15
• What do you mean by Thales theorem? Commented Aug 10, 2023 at 18:13
• @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 Commented Aug 10, 2023 at 20:26

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

$$p_1\bigg(2r\sin(\varphi)\cos\bigg(\frac{\pi}{2}-\varphi+\tau\bigg),2r\sin(\varphi)\sin\bigg(\frac{\pi}{2}-\varphi+\tau\bigg)\bigg)$$

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