1
$\begingroup$

Suppose we have a pyramid with a triangular base and there is another pyramid with a triangular base that is inside the first one I mentioned. So you can imagine a smaller sphere being in a big sphere and how it works. Now try picturing it with pyramids with triangular bases.

Prove that the surface area of the outside pyramid is greater then the surface area of the inside pyramid and prove that the perimeter of the outside pyramid is greater then the perimeter of the inside pyramid.

This problem seems so obvious visually that I am unsure how to show it. A couple of ideas I had was to unfold each pyramid or flatten it onto a plane then try to get some pattern I can find from the pyramid to the one inside it. Another idea was to consider the formulas for the surface area of a pyramid with a triangular base and see if that would show anything.

Any ideas?

$\endgroup$
  • $\begingroup$ "pyramid with triangular base" = "tetrahedron". :) If you're assuming that the tetrahedra are regular (that is, all the faces are equilateral triangles), then the problem is easy ... but perhaps too easy. $\endgroup$ – Blue Sep 8 '14 at 2:43
0
$\begingroup$

First the perimeter:

Let inner tetrahedron have edge lengths (a,b,c,d) and outer one ( A,B,C,D).

Place the smaller tetrahedron comfortably inside the bigger one so that one vertex of the outer coincides with an inner tetrahedron vertex and other vertices do not touch any of outer tetrahedron faces. By comfortably I mean avoid touching of outer solid by rotation of inner solid with as much uniform clearance as possible.

In such an enclosing arrangement it would be possible to find corresponding sides:

$ (a<A, b<B, c<C , d<D) $ and so, $ (a+b+c+d) < (A+B+C+D)$.

Next the surface area:

Since each face area =$\sqrt{ s(s-a)(s-b)(s-c)}$ where $2s= (a+b+c)$,

Smaller corresponding tetrahedron face areas are less in area than corresponding bigger tetrahedron face areas, so their sum is also less.

$\endgroup$
  • $\begingroup$ You make it sound easy but I am trying to prove that it is less, not assume that it is which then the results follow. $\endgroup$ – Nick Freeman Sep 9 '14 at 0:13

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.