Precalculus in a Nutshell, Geometry, Appendix B, Section 5, Question 14. A sphere is circumscribed about a cube. Find the ratio of the volume of the cube to the volume of the sphere.
So I drew this diagram:

Next, I want to relate s and r.
I apply Pythagorean theory to my diagram.
$$\left(\frac12 s\right)^2+\left(\frac12s\right)^2=r^2$$
Or more simply:
$$\frac12 s^2=r^2$$
This gives me the identity
$$s=r\times \sqrt2$$
So far so good, I think.
$$V_{\text{cube}}=s^3$$
$$V_{\text{sphere}}=\frac{4\pi r^3}{3}$$
Substituting for $s$,
$$V_{\text{cube}}=r^3\times2\sqrt2$$
Ratio of $V_{\text{cube}}$ to $V_{\text{sphere}}$ is:
$$r^3\times2\sqrt2/4\pi r^3/3$$
Simply:
$$\frac{6\sqrt2r^3}{4\pi r^3}=\frac{6\sqrt2}{4\pi}=\color{red}{\boxed{\displaystyle\color{black}{\frac{3\sqrt2}{2\pi}}}}$$
That's the solution I keep getting.
The answer key stays the solution is actually $\displaystyle\frac{6\sqrt2}{3\pi}$
Can someone please tell me where I'm going wrong?
 A: Let $O$ be the centre of the cube, and let $C$ be one of the corners of the cube. Drop a perpendicular from $O$ to the centre $M$ of one of the faces that contains the vertex $C$. 
Then $\triangle OMC$ is right-angled at $M$.
We have, in your notation, $OM=\frac{s}{2}$. Distance $OC$ is half a face diagonal of the cube. A face diagonal has length $\sqrt{2}{s}$. Half a face diagonal has therefore length $\frac{\sqrt{2} s}{2}$. 
By the Pythagorean Theorem, we have
$$(OC)^2 =\left(\frac{s}{2}\right)^2+ \left(\frac{\sqrt{2}s}{2}\right)^2.$$
It follows that $OC$, the radius of the circle, is $\frac{\sqrt{3}}{2}s$. The rest of the calculation is along lines you are familiar with.
Remark: The above argument was used because it seemed closest in spirit to your approach. To me, the following seems simpler. 
Grab a cube, or something close enough. As a distant second best, draw a picture of a cube in the conventional way. Let $P$ and $Q$ be opposite each other along the long diagonal of the cube. Let $S$ be a vertex on a face that contains $P$, with $S$ diagonally across from $P$.  Then $\triangle PSQ$ is right-angled at $S$. We have $(PQ)^2=(PS)^2+(SQ)^2=2s^2+s^2=3s^2$. The long diagonal of a cuve of side $s$ has length $\sqrt{3} s$.
