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Motivation: Let $a = \cos(\alpha), b = \cos(\beta),c = \cos(\gamma)$ where $\alpha,\beta,\gamma$ are the angles in a triangle. Then by $$\alpha+\beta+\gamma = \pi$$ and using $\cos(x+y) = \cos(x)\cos(y)-\sin(x)\sin(y)$, $\sin(\operatorname{acos}(x) = \sqrt{1-x^2}$ we find that the $a,b,c$ are points on the surface:

$$a^2+b^2+c^2+2abc-1=0$$

What is known about this surface, or what is it called?

a plot of the surface with Wolfram Alpha

Edit: With the help of @GEdgar I think I found the name of this surface by looking at the plots of both surfaces. (Cayley's nodal cubic surface).

Now the question remains how to show algebraically that they are the same?

Edit: Since every three point metric space can be isometrically emdedded in $\mathbb{R}^2$, we can build the possibly to a line degenerated triangle from these three points:

Using the law of cosines to define angles, given distances, we find that the quantities:

$$a:=\frac{d(y,z)^2+d(y,x)^2-d(x,z)^2}{2d(y,z)d(x,y)}, $$ $$b:=\frac{d(y,z)^2+d(z,x)^2-d(x,y)^2}{2d(y,z)d(z,x)},$$ $$c:=\frac{d(x,z)^2+d(y,x)^2-d(y,z)^2}{2d(x,z)d(x,y)}$$ satisfy by what was given above the following equation:

$$a^2+b^2+c^2+2abc-1=0$$

hence are points on the Cayley's nodal cubic surface. For instance for the metric on natural numbers $$d(x,y) = \sqrt{\sigma(x)+\sigma(y)-2\sigma(\gcd(x,y))}$$ and for three pairwise distinct primes $p,q,r$ we get the following nice formula:

$$\pi = \operatorname{acos}(\frac{r}{\sqrt{(p+r)(q+r)}})+\operatorname{acos}(\frac{q}{\sqrt{(p+q)(q+r)}})+\operatorname{acos}(\frac{p}{\sqrt{(p+r)(p+q)}})$$

Setting

$$a = \frac{r}{\sqrt{(p+r)(q+r)}}, b= \frac{q}{\sqrt{(p+q)(q+r)}}, c=\frac{p}{\sqrt{(p+r)(p+q)}}$$

we see that those are points on the Cayley nodal cubic surface.

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  • $\begingroup$ What do you want to know about it? $\endgroup$ Oct 14 '19 at 8:35
  • $\begingroup$ @DietrichBurde: What's the name of this surface? So I can google and find out more about it. $\endgroup$
    – user276611
    Oct 14 '19 at 8:38
  • $\begingroup$ Sorry, I meant $(a+b+c)^2-2(ab+bc+ca)-2abc-1=0$. $\endgroup$
    – Toby Mak
    Oct 14 '19 at 8:39
  • $\begingroup$ @DietrichBurde: For example how to add two points if possible on this surface? (I will edit the question to show motivation for asking this). $\endgroup$
    – user276611
    Oct 14 '19 at 8:59
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    $\begingroup$ @DietrichBurde: I think this is the same as en.wikipedia.org/wiki/Cayley%27s_nodal_cubic_surface by looking at the plots of both surfaces. $\endgroup$
    – user276611
    Oct 14 '19 at 9:09

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