All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Ratio of Volumes (Posted on 2003-04-25)
What is the ratio of volumes of a regular tetrahedron and the smallest cube that can encompass it?

 See The Solution Submitted by Bryan Rating: 2.0000 (3 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 solution | Comment 1 of 5
The largest tetrahedron that will fit within a given cube has each of its 6 edges as a diagonal of one of the cube's 6 faces. Suppose one edge of the tetrahedron is a diagonal of the top face of the cube. Another edge of that tetrahedron would be a diagonal of the bottom face, but at right angles to the one on top (but of course skew as they are on parallel planes). The other 4 edges of the tetrahedron are then the diagonals of the side, front and back faces of the cube that connect up each of the two ends of the top diagonal to each of the two ends of the bottom diagonal.

Half the vertices of the cube are also vertices of the tetrahedron. The other half (numbering 4) are the right-angled vertices of right tetrahedra. Each of these smaller tetrahedra has 3 isosceles right triangular faces (each of which is half a face of the cube), and one equilateral triangular face where it adjoins the central regular tetrahedron.

Let the edges of the cube have length 1, so the cube's volume is 1.

Consider one right triangular face of a small tetrahedron to be the base. It's two equal legs have the same length as the edge of the cube, which we have taken as 1. The area of the base is therefore 1/2. The altitude of the small tetrahedron is one of the edges of the cube, and therefore 1. As it is a pyramid, the volume is equal to the area of the base times the height divided by 3, which therefore comes out to 1/6.

As there are 4 of these small tetrahedra, their total volume is 4/6, or 2/3. Together with the regular tetrahedron whose volume we seek, that makes up the whole of the volume of the cube. So our regular tetrahedron has volume 1 - 2/3 = 1/3. That is the tetrahedron has 1/3 the volume of the cube.

I don't know of a mathematical proof that this tetrahedron is the largest that will fit in the cube -- the same as saying the cube is the smallest that will fit around the tetrahedron-- but it is hard to imagine a larger tetrahedron fitting.
 Posted by Charlie on 2003-04-25 09:56:25
Please log in:
 Login: Password: Remember me: Sign up! | Forgot password

 Search: Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (4)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2019 by Animus Pactum Consulting. All rights reserved. Privacy Information