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Math Help - Zeus' altar.

  1. #1
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    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.
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  2. #2
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    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.
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  3. #3
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    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.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by Zeus View Post
    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

    b^3=2a^3

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

    CB
    Last edited by CaptainBlack; February 12th 2011 at 10:55 AM. Reason: spelling
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  5. #5
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    I'm sorry. It would work if rectangle sided shapes were used. Then a decimal divisor could be used. Can you get that?
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Zeus View Post
    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
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