The Largest Sphere Within A Pyramid

First of all, $|AS|\cdot |DS| \leq \frac{1}{2}$ by the Arithmetic-Geometric Mean Inequality. At the same time, $[ASD]=\frac{1}{2}\sin \angle ADS \cdot (|AS|\cdot |DS| ) .$ So $[ASD]\leq \frac{1}{4}.$ Because $BD$ divides the area of $ABCD$ in half, the volume of $ABDS$ is exactly half of the volume of the pyramid, so it equals $\frac{1}{12}.$ On the other hand, consider $ABDS$ as a pyramid with base $ASD.$ Its volume is $\frac{1}{3}\cdot [ASD]\cdot |AB|\sin \alpha,$ where $\alpha$ is the angle between $AB$ and the plane $ASD.$ Since $|AB|=1,$ all these inequalities must be equalities! Thus, $AB \perp ASD,$ $|AS|=|DS|,$ and $\angle ASD = 90^\circ.$

By the Pythagorean theorem, $|AD|=1,$ so $ABD$ is an isosceles right triangle. Since $|BC|=|CD|$ and $[BCD]=[BAD],$ $ABCD$ is a square, with side $1$ . Because $AB \perp ASD,$ the planes $ABCD$ and $ASD$ are perpendicular. If $E$ is the midpoint of $AD$ and $F$ is the midpoint of $BC,$ then $\angle FES = 90^\circ,$ $|EF|=1,$ $|ES|=\frac{1}{2}.$

Note that the given conditions define the pyramid uniquely, we just need to find the largest radius of a ball that fits in it. Consider the infinite cylinder with the perpendicular cross-section $FES$ . Clearly, any ball that fits inside $ABCDS$ will fit inside this cylinder. The largest possible radius of a ball inside this prism cannot be larger than the in-radius of the triangle $FES.$

Lemma. The in-radius of a right triangle $UVW,$ with the right angle $V,$ equals $\frac{|UV|\cdot |VW|}{|UV|+|VW|+|UW|}.$

Proof. If the in-radius is $r$ , then $\frac{1}{2}|UV|\cdot |VW| =[UVW]=r\cdot \frac{|UV|+|VW|+|UW| }{2}$ and the formula easily follows.

From the above lemma, the in-radius of $ASD$ equals $\frac{1}{2}/(1+\sqrt{2})=\frac{1}{2+2\sqrt{2}}.$ Similarly, the in-radius of the triangle $FES$ is $\left( \frac{1}{2}\right) /\left( \frac{\sqrt{5}}{2} +\frac{1}{2}+1\right)=\frac{1}{3+\sqrt{5}}$ One can easily check that $2+2\sqrt{2}<3+\sqrt{5}$ , so the in-radius of $ASD$ is greater than the in-radius of $FES.$ Clearly, the largest radius of a ball inside $ABCDS$ is no larger than $\frac{1}{3+\sqrt{5}}.$ On the other hand, place a ball of this radius in the in-center of $FES.$ Because the distance from this in-center to $EF$ is $\frac{1}{3+\sqrt{5}},$ the distance from it to the planes $ABS$ and $DSC$ equals $\frac{1}{2\sqrt{2}}-\frac{1}{\sqrt{2}} \cdot \frac{1}{3+\sqrt{5}}=\frac{\sqrt{5}+1}{2\sqrt{2}(3+\sqrt{5})}$
One can easily check that $\sqrt{5}+1>2\sqrt{2},$ so this distance is greater than the in-radius of $FES,$ thus the ball fits! So the largest possible radius is $r=\frac{1}{3+\sqrt{5}}=\frac{3-\sqrt{5}}{4}.$ The surface area is $4\pi r^2=\frac{14-6\sqrt{5}}{4}\pi= \frac{7-\sqrt{45}}{2}\pi .$ So $(a,b,c)=(7,45,2),$ thus $a+b+c=54.$