What? Let's guess Mr. Zeus has a cubic alter?

If the original has volume a^3, the new one will need volume b^3 - a^3 = 2a^3 for some b. A little algebra suggests b^3 = 3a^3, which probably could have been guessed.

Results 1 to 6 of 6

- Feb 11th 2011, 07:53 PM #1

- Joined
- Feb 2011
- Posts
- 3

## Zeus' altar.

Make Zeus' altar double as big using equal parts of it enclosing a hollow altar of the original. What I have so far is-

Where a equals the divisor of a cube, so that if a = 2, the cube is divided into 8 parts, more straightforward that a is the number of squares along a side:

6a^2 + 12a + 8

A phrase in it which you will probably use in it. I have a lightning bolt of a headache and can't finish it.

- Feb 11th 2011, 10:24 PM #2

- Joined
- Aug 2007
- From
- USA
- Posts
- 3,111
- Thanks
- 2

- Feb 11th 2011, 11:01 PM #3

- Joined
- Feb 2011
- Posts
- 3

- Feb 12th 2011, 01:10 AM #4

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5

Your wording is almost as obscure as that of Apollo. If you mean divide the original cubical altar into equal sub-cubes such that they can be reassembles into an altar who's enclosed volume is double that of the original we would have

where is the side of the new alter and the side of the original. But this is imposible since would have to be an rational multiple of and so imply that is rational

CB

- Feb 12th 2011, 07:19 AM #5

- Joined
- Feb 2011
- Posts
- 3

- Feb 12th 2011, 11:05 AM #6

- Joined
- Nov 2005
- From
- someplace
- Posts
- 14,972
- Thanks
- 5