3
$\begingroup$

A bisector is something that cuts some other thing into two equal pieces. More concretely, assume we are given a reasonably well-behaved (for example, compact) 3D object and we are looking for planes in space that cut this object in two equal volume bits.

For reasonable concepts of well-behavedness we can given a well-behaved object and a line in space find a bisector that contains that line. Often that bisector will be unique.

We are interested in the opposite situation: Given a well-behaved object let us call a line in space a universal bisector if any plane containing the line is a bisector of the object.

For example, the axis of a cylinder or cone is a universal bisector.

Q: Can you find universal bisectors for

  1. the regular tetrahedron?
  2. the "L" which is obtained from a cuboid by removing a full height sub-cuboid from one of its corners?

If yes:

  1. are they unique?
  2. can you give an object that does not have a universal bisector?

Note: this is not a maths question, rigorous proofs are not required. There are good intuitive arguments to answer all parts of this puzzle.

Attribution: conceived by myself

$\endgroup$
4
$\begingroup$

As long as we're not doing maths, my basic intuition about the matter is that

connecting the centre points of two opposing edges of the regular tetrahedron

should give such a line: there is definitely one bisecting plane with that line on it (cut along one edge, which is perpendicular to the other edge). Then, rotating that cutting plane about the axis of the special line seems to move symmetrical amounts of volume from one side to the other and vice versa.

As for the "L" piece,

any plane through the centre of a cuboid will bisect it by symmetry, so if we take the centre of the original cuboid, and connect it to the centre of the missing cuboid,

we should have a line that can only be included in planes that bisect the L-shape: there were equal amounts on both sides, and an equal amount was subtracted from both sides.

For 3, if I haven't gone completely astray with my earlier answers,

There are two (EDIT: duh. three.) ways to choose opposing edges of a tetrahedron, so that one's definitely not unique. The L piece's universal bisector line might be, though.

EDIT: after some prodding from OP, it now seems obvious that

there must be more such lines in the L piece: you can chop the L into 2 cuboids (in two different ways, continuing either of the cuts that took away the piece from the big cuboid to make the L), so a line connecting the centres of those cuboids

should be a universal bisector by similar logic: both parts will individually get halved by any plane containing the line. (end edit)

and for 4,

I have absolutely no idea if there might be such a line lurking in there anywhere, or even if such lines can be avoided at all, but i think I might first look at a piece that you get by

taking a unit cube, picking a corner, and then attaching another unit cube to all the three sides around that corner. (For easier visualisation, we could take a dice and glue another dice to sides marked 1, 2 and 3)

That makes an L shape, like before, but there will be an extra cube to throw off the balance, so constructing a bisector is easy, but finding a line where every plane is a bisector seems way more difficult, at least.

$\endgroup$
2
  • 2
    $\begingroup$ Good first stab, mostly spot-on intuition. (Only, there are three ways of doing the tetrahedron, not two.) Four seems to be a correct example, though I'd argue one can find examples where it is easier to be sure about it. You may want to revisit three. $\endgroup$ Mar 1 at 22:33
  • $\begingroup$ You have essentially found all the intended bits. Well done! $\endgroup$ Mar 2 at 8:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.