Inscribing three balls in a cone, what is the radius ?

Consider a side and top view,

In the side view, we see $\triangle ABC$ with sides $AC=200$ , $BC=100$ , and $AB=100\sqrt{5}$ the sphere with radius $r$ center $D$ is tangent to $BC$ and $AB$ at $E$ and $F$ respectively. Draw $DG$ parallel to $AB$ then draw $GH$ parallel to $DF$ then draw $DI$ parallel to $BC$ .

It's simple to show $\triangle ABC \sim \triangle GDI \sim \triangle AGH$ . Then using proportions $GH=r$ , $AH=2r$ , $AG=r\sqrt{5}$ . Also $IC=r$ so subtracting $AG$ and $IC$ from $AC$ gives $GI=200-r-r\sqrt{5}$ . By similar triangles again $DI=100-\frac{1+\sqrt{5}}{2}r$

Now we turn our attention to the top view. $DI$ is the same as before but this now part of $30-60-90$ triangle $DIJ$ . Since $DJ=r$ , $DI=\frac{2}{\sqrt{3}}r$

So now we have an equation to solve for $r$

$100-\frac{1+\sqrt{5}}{2}r=\frac{2}{\sqrt{3}}r$

$r=\frac{600}{3+4\sqrt{3}+3\sqrt{5}} \approx \boxed{36.065479}$