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Question

A wooden right circular cone of altitude h and radius r is to be sawed into 3 parts of equal volume. How far from the vertex must the cuts be made? The cuts must be parallel to the base.

Answer 1

Let be V1,V2,V3 the volume of the 3 parts,then V=V1+V2+V3 is the volume of right circular cone.Then V1/V=1/3;V1+ V2/V=2/3.
Let be h1,h2,h the altitude of the 3 circular cone and h1/h=k1, h2/h=k2 then we have V1/V=k1^3=1/3=>k1=(1/3)^1/3=>h1=k1h
V1+V2/V=k2^3=2/3=>k2=(2/3)^1/3 =>h2=k2h
the cuts must be made of h1 and h2 far from the vertex. OK

Answer 2

Go off and look up the volume formula for a cone.

Then you can work out the overall (uncut) volume.

The top part will be a smaller cone, which you need to be 1/3rd of the overall volume - solve this bit.

The middle part, if you were to glue it to the top part would make a mid-size cone which needs to be 2/3 the overall volume. Solve this, then you can take away the height of the small top cone to find the height of the middle part.

Then you can work out the height of the lowest part.

Answer 3

Volume of right cone is (pi/3)*h*r^2
Volume of truncated cone is Volume of large cone minus volume of removed cone.

Volume of total cone = (pi/3)*h*r^2
Part to be removed = (pi/3)*xh*((xr)^2) = (2/3)*(pi/3)*h*r^2
(x is the part to be removed, as a fraction of the whole cone)

(pi/3), h and r^2 cancel out, leaving

x^3 = 2/3

therefore, the height of the first cut, measured from the base, is
1 - CubeRoot(2/3) = 0.1264...