Wikipedia:Reference desk/Archives/Mathematics/2024 January 31

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January 31

Where would this cone be cut?

If a cone with height H and width W is stood up vertically, how would I determine where along the height to cut it so that the two pieces produced as a result are of equal volume (originally said “area” by accident) to each other? Primal Groudon (talk) 19:45, 31 January 2024 (UTC)

That's gonna depend on whether or not you want to include the circular base as part of the surface area. The non-base portion of a cone of radius $r$ and height $h$ has area $\pi r{\sqrt {r^{2}+h^{2}}}=\pi r^{2}{\sqrt {1+(h/r)^{2}}}$. The base naturally has area $\pi r^{2}$. The nice thing is that if you cut along the height, the leftover conic piece with radius $r'$ and height $h'$ still has the same slope $h'/r'=h/r$. So if you consider a cone with no bottom:

${\frac {\pi r'^{2}{\sqrt {1+(h'/r')^{2}}}}{\pi r^{2}{\sqrt {1+(h/r)^{2}}}}}={\frac {1}{2}}\Rightarrow {\frac {r'^{2}}{r^{2}}}={\frac {1}{2}}\Rightarrow r'={\sqrt {\frac {1}{2}}}*r$

In other words, you cut a conic piece with height ${\sqrt {1/2}}$ that of the original in order to get two pieces of equal area, assuming the cone's base does not matter. If it does matter though, then cutting off the top yields a conic piece with no base:

${\frac {\pi r'^{2}{\sqrt {1+(h'/r')^{2}}}}{\pi r^{2}+\pi r^{2}{\sqrt {1+(h/r)^{2}}}}}={\frac {1}{2}}\Rightarrow {\frac {r'^{2}}{r^{2}}}*{\frac {\sqrt {1+(h/r)^{2}}}{1+{\sqrt {1+(h/r)^{2}}}}}={\frac {1}{2}}\Rightarrow {\frac {r'^{2}}{r^{2}}}={\frac {1+{\sqrt {1+(h/r)^{2}}}}{2{\sqrt {1+(h/r)^{2}}}}}\Rightarrow r'=r*{\sqrt {\frac {1+{\sqrt {1+(h/r)^{2}}}}{2{\sqrt {1+(h/r)^{2}}}}}}$

This time the expression is a lot messier, and it does depend on slope, although again you cut a conic piece with height a constant (up to the slope of the cone) factor times that of the original in order to get two pieces of equal area (which makes sense, since the height you cut at should scale perfectly with the cone itself.) GalacticShoe (talk) 20:32, 31 January 2024 (UTC)

For halving the lateral surface area, this cut at a fraction of ${\sqrt {\tfrac {1}{2}}}$ from the apex works not only for cones with circular bases, but for any solid formed by connecting a fixed apex by straight line segments to a piecewise smooth Jordan curve (a non-self-intersecting continuous loop) in a base plane. This includes pyramids, for which that loop is a polygon. The surface obtained by extending the line segments to whole lines is a conical surface.  --Lambiam 21:12, 31 January 2024 (UTC)

Sorry, I meant volume, but accidentally said area. Primal Groudon (talk) 21:23, 31 January 2024 (UTC)

Some additional clarification, I'm assuming that the cut is meant to be parallel to the base, otherwise there are many possible cuts that divide it into two equal volumes. If so then one of the two pieces is similar to the original cone. The volume of two similar cones is proportional to the cube of one of their dimensions, say height. So you'd want the height of the smaller cone to be ${\frac {1}{\sqrt[{3}]{2}}}$ the height of the original cone. Note that this would work for any cone, not just circular ones. --RDBury (talk) 22:17, 31 January 2024 (UTC)

For halving the volume cut it at the height times the cube root of a half from the apex. That gives a cone similar to the original and the volume ratio is as the cube of the heights. NadVolum (talk) 22:24, 31 January 2024 (UTC)