Cone and sphere.

This can be done in two-dimensions, the sphere becomes a circle, the cone becomes a triangle and the radii of the circles become line segments.

To solve with a graph, let the circle be $x^{2}+y^{2}=1$ and choose the line with positive slope, it contains $(-1,-1)$ and $(0,h-1)$ where $h$ is the height of the cone so its equation is $y=hx+h-1$ . The point of intersection $A$ and $B$ need to satisfy $(\frac{x_{A}}{x_{B}})=\sqrt{3}$

Substituting the line into the circle and solving for $x$ gives

$x^{2}+(hx+h-1)^{2}=1$

$(h^{2}+1)x^{2}+(2h^{2}-2h)x+(h^{2}-2h)=0$

$x=\frac{h-h^{2}\pm \sqrt{2h}}{h^{2}+1}$ Of the solutions, the one with the " $+$ " is the smaller so dividing to get the required ratio gives

$\large \frac{h-h^{2}-\sqrt{2h}}{h-h^2+\sqrt{2h}}=\sqrt{3}$

This can lead to a cubic whose only real solution is $h=2+\sqrt{3}$

so $a+b+c=2+3+1=\boxed{6}$