I tried using conservation of areas instead of volumes to solve this but no such luck, the answer doesn't come out right. The equations I was using were:
>$(h-2)b_{1} = (h-8)(b_{2} - b_{0})$
>$b_{0} + b_{1} = 4b_{2}$
>$\frac{h}{b_{0}} = \frac{h-2}{b_{1}}$
>$\frac{h}{b_{0}} = \frac{h-8}{b_{2}}$
By any means I didn't think shit would go sideways. Doesn't the radial symmetry of the cone allow for this to be done with area instead of volume? Like if you just looked at it from the side in 2D?