No longer certain if that is the minimal required, however you’ll be able to do it in 6 cuts.
Lower the primary cake into quarters – two cuts.
Lower the second one cake into quarters – two cuts.
Lower the 3rd cake into quarters – two cuts.
Trade resolution (nonetheless 6 cuts):
The full quantity of the truffles is $(10^2 + 8^2 + 6^2)*pi*h = 200 pi h$. So $4$ equivalent items may have quantity of $50 pi h$ (you’ll be able to believe simply floor space if you happen to like, since $h$ is similar for all). We will be able to get two of the ones simply by slicing the $20$ diameter cake in part ($1$ minimize). Subsequent we would want to upload $14 pi h$ to the $12$ diameter cake to get any other $50 pi h$ team. $14$ out of $64$ is $frac 7{32}$, which will require $5$ cuts around the center of the $16$ diameter cake (to chop it into $32$nds) – yielding $6$ general cuts additionally.
EDIT:
Here is the 2 minimize resolution:
1) Lower the $20$ cm cake in part. That offers you two items of $50*pi*h$ quantity. The opposite two truffles are a complete of $100*pi*h$ in quantity.
2) Position the $12$ cm cake on most sensible of and tangent to the $16$ cm cake. Get started slicing the $16$ cm cake alongside the threshold of the $12$ cm cake. Whilst you succeed in what will be the finish of the diameter for the $12$ cm cake, move perpendicularly out to the threshold of the $16$ cm cake. Position the smaller piece of the $16$ cm cake with the $12$ cm cake and you’ll be able to have two items which are reflect pictures of one another. Each and every might be part of $100*pi*h$.
(Inspiration taken from hkboy’s resolution – see visible there, however opposite his minimize to head from the tangent indicate.)
