HMMT-Feb 2014 guts, Problem 9

Compute the side length of the largest cube contained in the region \[ \left\{(x, y, z): x^{2}+y^{2}+z^{2} \leq 25 \text { and } x \geq 0\right\} \] of three-dimensional space.

Solution: The given region is a hemisphere, so the largest cube that can fit inside it has one face centered at the origin and the four vertices of the opposite face on the spherical surface. Let the side length of this cube be $s$. Then, the radius of the circle is the hypotenuse of a triangle with side lengths $s$ and $\frac{\sqrt{2}}{2} s$. So, by the Pythagorean Theorem, the radius equals $\frac{\sqrt{6}}{2} s$. Since the radius of the hemisphere is 5 , the side length of the cube is $\frac{5 \sqrt{6}}{3}$. $\fbox{\sqrt{\frac{5 \sqrt{6}}{3}}}$.

Answer: \sqrt{\frac{5 \sqrt{6}}{3}}