## Maximum Radius of Sphere Inside a Square Based Pyramid

For a square based pyramid with apex above the centre of the base, a sphere of maximum radius will meet the base at the centre, and each slanted face tangentially.
For the diagram below, if

is the angle in the triangle ABC subtended by the radius of the circle from the midpoint of a side of the base, then
$tan \alpha = \frac{r}{x/2} = \frac{2r}{x}$
. The triangle ACD has a side AC in common, both have right angles opposite this side and radius of the sphere as another side, so are congruent. Angle DAC is the
$\alpha$
and angle DAB is
$2 \alpha$
.
From the diagram,
$tan 2 \alpha = \frac{h}{2/x} = \frac{2h}{x}$
.
Now use the identity
$tan 2 \alpha = \frac{2 tan \alpha}{1- tan^2 \alpha}$
to obtain
$\frac{2h}{x} = \frac{4r/x}{1-(2r/x)^2} = \frac{4rx}{x^2 - 4r^2}$

We can rearrange this as a quadratic in
$r$
, obtaining
$8hr^2+4rx^2-2hx^2=0$
.
Solving this gives
$r=\frac{-4x^2 \pm \sqrt{16x^4 +64h^2x^2}}{16h} = \frac{-x^2 \pm x \sqrt{x^2+4h^2}}{4h}$

Only the
$+$
option gives a valid value of
$r$
so
$r= \frac{-x^2 + x \sqrt{x^2+4h^2}}{4h}$ 