# Zeus' altar.

• Feb 11th 2011, 06:53 PM
Zeus
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, 09:24 PM
TKHunny
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.
• Feb 11th 2011, 10:01 PM
Zeus
What's the answer? How many times do you have to divide a cube to form another one around the former? It's somewhere between 7 and 8.
• Feb 12th 2011, 12:10 AM
CaptainBlack
Quote:

Originally Posted by Zeus
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.

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

\$\displaystyle b^3=2a^3\$

where \$\displaystyle \$\$b\$ is the side of the new alter and \$\displaystyle \$\$a\$ the side of the original. But this is imposible since \$\displaystyle \$\$b\$ would have to be an rational multiple of \$\displaystyle \$\$a\$ and so imply that \$\displaystyle \root 3 \of 2\$ is rational

CB
• Feb 12th 2011, 06:19 AM
Zeus
I'm sorry. It would work if rectangle sided shapes were used. Then a decimal divisor could be used. Can you get that?
• Feb 12th 2011, 10:05 AM
CaptainBlack
Quote:

Originally Posted by Zeus
I'm sorry. It would work if rectangle sided shapes were used. Then a decimal divisor could be used. Can you get that?

No if the altar is a cube the volume of the big altar built from congruent bricks produced by a dissection of the original alter must always be a rational multiple of the volume of the original altar, but it is not.

CB