23. A cuboid has sides of lengths 22,2 and 10 . It is contained within a sphere of the smallest possible radius.
What is the side-length of the largest cube that will fit inside the same sphere?
A 10
B 11
C 12
D 13
E 14
Solution
E
If a cuboid is contained in a sphere of the smallest possible radius, the vertices of the cuboid will lie on the sphere. It follows that the diagonals of the cuboid will be diameters of the sphere. So we begin by calculating the length of the diagonals of the $22 \times 2 \times 10$ cuboid.
Consider a general cuboid with sides of lengths $a, b$ and $c$, as shown in the diagram. Let $P, Q, R$ and $S$ be the vertices shown in this diagram. Let $d$ be the length of the diagonal $R S$.

By Pythagoras' theorem, applied to the right-angled triangle $P Q R$, we have
$$
P R^{2}=a^{2}+b^{2} .
$$

Therefore, by Pythagoras' theorem, applied to the right-angled triangle, $S P R$, we have
$$
d^{2}=P R^{2}+c^{2}=a^{2}+b^{2}+c^{2} \text {. }
$$

It follows that for the cuboid given in the question,
$$
d^{2}=22^{2}+2^{2}+10^{2}=484+4+100=588 .
$$

Therefore the smallest sphere inside which the cuboid will fit has diameter $d$, where $d=\sqrt{588}$.
Let $s$ be the side length of the largest cube that will fit inside this sphere. By the equation $d^{2}=a^{2}+b^{2}+c^{2}$ for the diagonal of a cuboid, we have
$$
d^{2}=s^{2}+s^{2}+s^{2},
$$
and hence
$$
588=3 s^{2} .
$$

It follows that $s^{2}=196$, and hence $s=14$.