By the way, one can build representation of acos(x) + acos(y) + acos(z) through acos(w) + something. Below I send this formula with verification of its correction everywhere, including branch cuts. All components for deriving such type summation formulas for ArcTrigs one can find on our Functions Site http://functions.wolfram.com/ There you can find also exact formulas for a ArcTrig[x]+ b OtherArcTrig[y] where ArcTrig ia ArcSin,..., ArcSinh,..., Log . Oleg Marichev In[13]:= Union[ Chop[Flatten[Table[With[{x = (u + Random[])*Exp[Pi*I*(k/4)], y = (v + Random[])*Exp[Pi*I*(j/4)], z = (w + Random[])*Exp[Pi*I*(s/4)]}, {ArcCos[x] + ArcCos[y] + ArcCos[z] - (2*Pi - 2*Pi*Sqrt[((I*Sqrt[1 - x] + Sqrt[1 + x])*(I*Sqrt[1 - y] + Sqrt[1 + y])*(I*Sqrt[1 - z] + Sqrt[1 + z]))/ (1 - (I*(Sqrt[1 - x] - I*Sqrt[1 + x])*(Sqrt[1 - y] - I*Sqrt[1 + y])*( Sqrt[1 - z] - I*Sqrt[1 + z]))/(2* Sqrt[2]))]* Sqrt[-((I*(-1 + ((Sqrt[1 - x] - I*Sqrt[1 + x])*(I*Sqrt[1 - y] + Sqrt[1 + y])*(Sqrt[1 - z] - I*Sqrt[1 + z]))/(2*Sqrt[2])))/ ((Sqrt[1 - x] - I*Sqrt[1 + x])*(Sqrt[1 - y] - I*Sqrt[1 + y])* (Sqrt[1 - z] - I*Sqrt[1 + z])))] - (4*Sqrt[2]* Sqrt[((-I)*x + Sqrt[1 - x^2])*(y + I*Sqrt[1 - y^2])* ((-I)*z + Sqrt[1 - z^2])]* Sqrt[(-1 - (I*(Sqrt[1 - x] - I*Sqrt[1 + x])* (Sqrt[1 - y] - I*Sqrt[1 + y])*(Sqrt[1 - z] - I*Sqrt[1 + z]))/(2* Sqrt[2]))/(1 - (I*(Sqrt[1 - x] - I*Sqrt[1 + x])*(Sqrt[1 - y] - I*Sqrt[1 + y])*(Sqrt[1 - z] - I*Sqrt[1 + z]))/(2*Sqrt[2]))]* Sqrt[(1 - (I*(Sqrt[1 - x] - I*Sqrt[1 + x])*(Sqrt[1 - y] - I*Sqrt[1 + y])* (Sqrt[1 - z] - I*Sqrt[1 + z]))/(2* Sqrt[2]))/ (-1 - (I*(Sqrt[1 - x] - I*Sqrt[1 + x])*(Sqrt[1 - y] - I*Sqrt[1 + y])* (Sqrt[1 - z] - I*Sqrt[1 + z]))/(2* Sqrt[2]))]* ArcCos[(I* Sqrt[2]*(1 - (1/8)*(Sqrt[1 - x] - I*Sqrt[1 + x])^2* (Sqrt[1 - y] - I*Sqrt[1 + y])^2*(Sqrt[1 - z] - I*Sqrt[1 + z])^2))/ ((Sqrt[1 - x] - I*Sqrt[1 + x])*(Sqrt[1 - y] - I*Sqrt[1 + y])* (Sqrt[1 - z] - I*Sqrt[1 + z]))])/((Sqrt[ 1 - x] - I*Sqrt[1 + x])* (Sqrt[1 - y] - I*Sqrt[1 + y])*(Sqrt[1 - z] - I*Sqrt[1 + z])) - 4*Pi*Floor[(-(1/(2*Pi)))*(-Pi + Arg[I*Sqrt[1 - x] + Sqrt[1 + x]] + Arg[I*Sqrt[1 - y] + Sqrt[1 + y]] + Arg[I*Sqrt[1 - z] + Sqrt[ 1 + z]])])}], {u, 0, 1}, {v, 0, 1}, {w, 0, 1}, {k, 0, 7}, {j, 0, 7}, {s, 0, 7}]]]] Out[13]= {0} On 2/2/2013 7:48 PM, Bill Gosper wrote:

DanA>

Actually RWG's samosa, and the locus of (*) below in [-1,1]^3, should be identical, both corresponding to all triples of cosines of 3 angles that add up to 2pi.

But it seems that peculiarities of the way acos() is interpreted might mess things up. This is the case in the Mac Grapher utility, where only one face of the pillow is displayed for the equation

(**) acos(x) + acos(y) + acos(z) = 2pi.

PUZZLE: Determine whether any planar geometric circle lies on the locus of (*).

--Dan

I'll bet not. The surface contains four equilateral triangles.Resting it on

a "face" and considering horizontal plane sections, we see this triangle deforming continuously toward infinitesimal circles at the rounded bottom

and pointy top. I think other sections would lack symmetry. But we shouldn't forgetVillarceau.

You say "puzzle". Have you proof?

On 2013-02-01, at 2:25 AM, Dan Asimov wrote:

(since (x,y,z) must lie in [-1,1]^3) so that leaves only the fact that (*) 2xyz + 1 - (x^2 + y^2 + z^2) >= 0 whose locus fits RWG's samosa description to a T.

Like I said, "This can't be new." In fact, both definitions lead to the same (intractable?) area integral,

Integrate[ Sqrt[Sin[t]^2*Sin[u]^2 + (Sin[t]^2 + Sin[u]^2)*Sin[t + u]^2], {t, 0, Pi}, {u, 0, Pi}]==

Integrate[ Sqrt[1 + ((-2 + x^2 + y^2)*(-y^2 - 2*x*y*Sqrt[(-1 + x^2)*(-1 + y^2)] +

x^2*(-1 + 2*y^2)))/((-1 + x^2)*(-1 + y^2))], {x, -1, 1}, {y, -1, 1}]

In[135]:= 4*N[List@@%]

Out[135]= {31.4159,31.4159}

so5 pi /2 remains unproven. Will this be another "Coxeter's Challenge Integral"? In[136]:= 2*Pi^2/15 == Integrate[ArcSec[t]*(2 + 1/Sqrt[t + 3])/Sqrt[t + 1]/(t + 2), {t, 1, 6}]

In[137]:= N[%]

Out[137]= True

Except Coxeter had a geometric proof. We have only a numerical tease.

Are there lots of theseconjectural integral identities? --rwg