SOLUTION: How far from the vertex of a right circular cone must a plane be passed in order to divide the volume into two equal parts? three equal parts?

Question 1029812: How far from the vertex of a right circular cone must a plane be passed in order to divide the volume into two equal parts? three equal parts?
Answer by KMST(5328) (Show Source): You can put this solution on YOUR website!
It is not clear to me how are you going to angle that plane you are using to slice the cone into two parts. Parallel to the base of the cone?
Also, are you intending to use two parallel planes to slice that cone into three equal parts?
A plane is parallel to the base of the right circular cone of height ,
at a distance from the vertex,
it would delimit a smaller cone similar to the original cone.
The ratio of their heights is .
The ratio of the volume of the original, larger cone, , ,
to the volume of the smaller cone,
is .

If we want the plane to divide the volume into two equal parts,
we want <---> <---> .
For that we need
--> --> .
For a more elegant expression, .
For an approximate solution, .
To divide the volume into two equal parts, a plane parallel to the base of a right circular cone should pass at a distance from the vertex of about times the height of the cone.

If you use two planes parallel to the base of a cone,
at distances and from the vertex,
with
to slice that cone into three parts of equal volume,
The volume, , of the small cone with its base from the vertex will be
.
Since , , then
--> .
The approximate value is (to four decimal places).
So, to lop off the top times the height of the cone.
The cone you would get slicing only at a distance ,
would have a volume twice as large as the cone above from the vertex:
<--> , with
<--> <--> .
Since we had <--->. ,
.
For an approximate value, .
So, to slice a cone into three equal volume parts using two slicing planes parallel to the base,
cut at distances from the vertex of times the cone's height and times the cone's height.
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