Some formulas for $\pi$?

Let $$0 < x < y < z$$. Then we have:

$$\pi = \arccos\left(-\frac{\frac{x^{2}}{y^{2}} + \frac{x^{2}}{z^{2}} - \frac{y^{2}}{z^{2}} - 1}{2 \, \sqrt{-\frac{x^{2}}{y^{2}} + 1} \sqrt{-\frac{x^{2}}{z^{2}} + 1}}\right) + \arccos\left(-\frac{\frac{x^{2}}{y^{2}} - \frac{x^{2}}{z^{2}} + \frac{y^{2}}{z^{2}} - 1}{2 \, \sqrt{-\frac{x^{2}}{y^{2}} + 1} \sqrt{-\frac{y^{2}}{z^{2}} + 1}}\right) + \arccos\left(\frac{\frac{x^{2}}{y^{2}} - \frac{x^{2}}{z^{2}} - \frac{y^{2}}{z^{2}} + 1}{2 \, \sqrt{-\frac{x^{2}}{z^{2}} + 1} \sqrt{-\frac{y^{2}}{z^{2}} + 1}}\right)$$

I have a proof for this, but I would like to know if there are other proofs for this.

Here are some formulas I derived using the technique above, which I find amusing:

$$\pi = \arccos\left(\frac{1}{12} \, \sqrt{5} \sqrt{3}\right) + \arccos\left(\frac{5}{32} \, \sqrt{5} \sqrt{2}\right) + \arccos\left(\frac{13}{48} \, \sqrt{3} \sqrt{2}\right)$$

$$\pi = \arccos\left(\frac{1}{24} \, \sqrt{7} \sqrt{5}\right) + \arccos\left(\frac{13}{108} \, \sqrt{7} \sqrt{3}\right) + \arccos\left(\frac{25}{144} \, \sqrt{5} \sqrt{3}\right)$$

$$\pi = \arccos\left(\frac{75}{128}\right) + \arccos\left(\frac{41}{160} \, \sqrt{7}\right) + \arccos\left(\frac{3}{40} \, \sqrt{7}\right)$$

$$\pi = \arccos\left(\frac{41}{500} \, \sqrt{11} \sqrt{5}\right) + \arccos\left(\frac{1}{20} \, \sqrt{11}\right) + \arccos\left(\frac{61}{200} \, \sqrt{5}\right)$$

$$\pi = \arccos\left(\frac{1}{84} \, \sqrt{13} \sqrt{11}\right) + \arccos\left(\frac{61}{864} \, \sqrt{13} \sqrt{6}\right) + \arccos\left(\frac{85}{1008} \, \sqrt{11} \sqrt{6}\right)$$

$$\pi = \arccos\left(\frac{1}{112} \, \sqrt{15} \sqrt{13}\right) + \arccos\left(\frac{85}{1372} \, \sqrt{15} \sqrt{7}\right) + \arccos\left(\frac{113}{1568} \, \sqrt{13} \sqrt{7}\right)$$

$$\pi = \arccos\left(\frac{1}{144} \, \sqrt{17} \sqrt{15}\right) + \arccos\left(\frac{113}{1024} \, \sqrt{17} \sqrt{2}\right) + \arccos\left(\frac{145}{1152} \, \sqrt{15} \sqrt{2}\right)$$

$$\pi = \arccos\left(\frac{1}{180} \, \sqrt{19} \sqrt{17}\right) + \arccos\left(\frac{145}{972} \, \sqrt{19}\right) + \arccos\left(\frac{181}{1080} \, \sqrt{17}\right)$$

$$\pi = \arccos\left(\frac{1}{220} \, \sqrt{21} \sqrt{19}\right) + \arccos\left(\frac{181}{4000} \, \sqrt{21} \sqrt{10}\right) + \arccos\left(\frac{221}{4400} \, \sqrt{19} \sqrt{10}\right)$$

Edit: The proof is based on noticing that every $$3$$ point metric space can be embedded in $$\mathbb{R}^2$$ as a triangle and then using trigonometry. (See Whats the name of this surface $a^2+b^2+c^2+2abc-1=0$? )

The metric space I am considering is a Hilbert space:

$$k(x,y) = \frac{\min(x,y)^2}{\max(x,y)^2}$$

with metric:

$$d(x,y) = \sqrt{2(1-k(x,y))}$$

For three points $$x,y,z$$ in a metric space, we can define (using the law of cosines) the following quantity:

$$s(x,y,z) = \frac{d(x,y)^2+d(y,z)^2-d(x,z)^2}{2d(x,y)d(y,z)}$$

Then we can embedd $$X \times X \times X$$ to the Cayleys surface ( https://mathcurve.com/surfaces.gb/cayley/cayley.shtml ) through the mapping:

$$f(x,y,z) = (s(x,y,z), s(z,x,y),s(y,z,x))$$

We then have:

$$\pi = \arccos(s(x,y,z))+\arccos(s(z,x,y))+\arccos(s(y,z,x))$$

which proves the claim.

• Maybe add a few details about your proof so that people don't reinvent the wheel when answering this question? Jun 25, 2021 at 18:55
• If you have proof, you should provide it; otherwise, how will anyone else know what constitutes an "other" proof? ... Also, your formula is unnecessarily complicated. The first $\arccos$ argument, for instance, reduces to $$\frac{\sqrt{(x^2-y^2)(z^2-x^2)}}{2xz}$$ (Your parameter order would seem to need adjustment to avoid negatives under the square roots.)
– Blue
Jun 25, 2021 at 18:56
• @Blue: Thanks, The complicated part is a point of view... But I will otuline the proof. Jun 25, 2021 at 18:57
• @NoName: Ok, I will outline the proof. Jun 25, 2021 at 18:57
• @Blue: What do you mean with parameter order. If there is a bug in the formula, please feel free to edit it. Jun 25, 2021 at 19:10

Defining $$\sigma_{uv} := \sqrt{u^2-v^2}$$, OP asserts that $$\pi = \alpha+\beta+\gamma \tag{1}$$ for angles in $$[0,\pi]$$ satisfying $$\cos\alpha = \frac{y^2 + z^2}{2yz}\frac{\sigma_{yx}}{\sigma_{zx}}\qquad\cos\beta=\frac{\sigma_{yx}\sigma_{zy}}{2yz} \qquad \cos\gamma=\frac{x^2+y^2}{2y^2}\frac{\sigma_{zy}}{\sigma_{zx}} \tag{2}$$ Note that, since the cosines are non-negative, we may actually take $$\alpha$$, $$\beta$$, $$\gamma$$ to be first-quadrant angles. Calculating $$\sqrt{1-\cos^2}$$ to get the corresponding sines (which are non-negative for first-quadrant angles), we (and by "we", I mean Mathematica) obtain $$\sin\alpha=\frac{\tau\sigma_{zy}}{2yz\sigma_{zx}} \qquad \sin\beta= \frac{\tau}{2yz}\qquad \sin\gamma=\frac{\tau\sigma_{yx}}{2y^2\sigma_{zx}} \tag{3}$$ where $$\tau:= \sqrt{4y^2z^2 - \sigma_{yx}^2\sigma_{zy}^2} \tag{4}$$ Now, let's take the cosine of the right-hand side of $$(1)$$: \begin{align}\cos(\alpha+\beta+\gamma)&= \cos\alpha\cos\beta\cos\gamma \\ &\quad - \cos\alpha\sin\beta\sin\gamma - \sin\alpha\cos\beta\sin\gamma - \sin\alpha\sin\beta\cos\gamma \tag{5}\\[4pt] &= \frac{(y^2+z^2)(x^2+y^2)}{8y^4z^2}\frac{\sigma_{yx}^2\sigma_{zy}^2}{\sigma_{zx}^2} \\ &\quad-\frac{\tau^2}{8y^4z^2\sigma_{zx}^2}\left((y^2+z^2)\sigma_{yx}^2 +\sigma_{yx}^2\sigma_{zy}^2 +(x^2+y^2)\sigma_{zy}^2\right) \tag{6} \\[4pt] &= \cdots \tag{7} \\[4pt] &= -1 \tag{8} \end{align} Thus, $$\alpha+\beta+\gamma$$ must be an odd multiple of $$\pi$$; since no one angle exceeds $$\pi/2$$, we conclude that the sum is specifically $$\pi$$. $$\square$$