6
$\begingroup$

When I try to sleep at night, I find myself envisioning topological objects and wondering about the behavior of material in the abstract. Here's my latest quandary​:

Let an equilateral triangle be constrained so that one vertex lies on a surface, at some point. Is there a surface such that the other two points of the triangle cannot also be placed upon the surface?

More intuitively, is there any finite object upon which a pizza saver/tripod table cannot be set to rest?

More precisely, in 3-dimensional Euclidean space, is there a surface with a point at which one of an equilateral triangle's vertices can be placed, while the other two vertices cannot be be simultaneously also placed onto some other points of the surface?

Note that the the plane of the triangle may be pierced by the surface (alternatively, the length of the pizza saver's legs may be arbitrarily long).

Pizza Saver

Fig. 1 A trivial case. Credit

.

Notes:

  • The triangle's sides can be made arbitrarily small.

.

After more careful thought:

Perhaps what I'm really wondering is What limitations should be set upon this scenario so that the problem is non-trivial?

$\endgroup$
  • 2
    $\begingroup$ What about a long pole, with one vertex of the triangle constrained to be at the end of the pole? $\endgroup$ – angryavian May 31 '17 at 4:54
  • 3
    $\begingroup$ Well, a sphere whose diameter is smaller than the triangle's sides... $\endgroup$ – Chris Culter May 31 '17 at 4:54
  • $\begingroup$ The triangle's sides can be made arbitrarily small, I will update to make this explicit. $\endgroup$ – electronpusher May 31 '17 at 4:55
  • $\begingroup$ As this is a pizza saver, is gravity a consideration? If the pizza saver is placed on a plane of sufficient incline it may slide or tip over. $\endgroup$ – Doug M May 31 '17 at 5:06
  • 1
    $\begingroup$ By changing the triangle, you are changing the rules set out above: you are not trying to place a tripod, but one of an infinite collection of tripods. You can validly ask if for every surface there is a subset of three of its points that form an equilateral triangle; the answer will depend on exactly how you define "surface", but for a reasonable choice for that definition the answer will be "yes". $\endgroup$ – Marc van Leeuwen May 31 '17 at 5:08
0
$\begingroup$

Yes, there are such surfaces. For example, take the sphere that you pictured, and shrink it down so that the other two vertices are much farther apart than the edges of the sphere. Then, if one vertex has been constrained to lie on the sphere, then the other two can't, because the sphere just isn't big enough.

However, there are further versions of this which ARE true, in a sense. Look up the Wobbly Table Theorem.

$\endgroup$
  • $\begingroup$ Yes, the Wobbly Table situation was one of my inspirations. If the side length of the equilateral triangle may vary, then it may be made small enough to lie with all three points upon the sphere. $\endgroup$ – electronpusher May 31 '17 at 5:05
0
$\begingroup$

Imagine 2 points on the surface, $d$ distance apart. Imagine a circle in the plane perpendicular to the segment joining them, centered at the midpoint of the segment and having radius $\frac{\sqrt{3}}{2}d$. If this circle intersects the surface, the point of intersection is the third point, otherwise we can make $d$ arbitrarily smaller until it intersects.

$\endgroup$
  • $\begingroup$ How do we know we can do this? What if, for example, there is a slit between the two points you chose at the start? $\endgroup$ – Duncan Ramage Jun 2 '17 at 18:42
  • $\begingroup$ We can make the triangle smaller until there is no such slit in the neighbourhood. We can always select 2 points such that this works, otherwise it can't be a "surface" in the usual sense. This is not a formal proof, just an intuition. $\endgroup$ – Meet Taraviya Jun 5 '17 at 6:25

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.