Polyhedron Math Problem: Determining Regularity and Type

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Homework Statement



P is a polyhedron defined by (+/-) x (+/-) z <= 1
(+/-) x (+/-) y <= 1
(+/-) y (+/-) z <= 1

These are 12 inequalities with every possible sign choice taken.

Is P a regular polyhedron? If so, which type?



Homework Equations



If we change one inequality to an equation, we get a face of the polyhedron. In order to see if the face is a regular polygon, work out its vertices and/or edge-lengths and/or angles


The Attempt at a Solution



I got an octahedron but I see a counterexample of (0.5,0.5,0.5), which seems to satisfy our equations, but not the formula for an octahedron.
 
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Your 12 inequalities give 12 faces. This is a dodecahedron, not an octahedron.
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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