The square base of a right pyramid has sides of integral length. The height of the pyramid is also an integer. The volume of the pyramid, which is also an integer, is equal to it's total surface area. What is the height of the pyramid?
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A square pyramid with height h and whose base has side-length a has volume 3 1 a 2 h and surface area a 2 + a a 2 + 4 h 2 .
Equating these and doing a bit of tidying, we find a 2 = h − 6 3 6 h
So we require h − 6 3 6 h , where h is a positive integer, to be a perfect square. Clearly we need h > 6 .
For h > 6 , h − 6 3 6 h is a decreasing function in h . As h → ∞ , h − 6 3 6 h → 3 6 + . So as soon as h − 6 3 6 h < 4 9 , we can stop searching, as there are no perfect squares in the interval ( 3 6 , 4 9 ) .
Solving the inequality, we find this means that h < 2 3 . We can just check the values of h − 6 3 6 h for each h from h = 7 to h = 2 2 . Several of these give integers, but the only one that gives a square is h = 8 . (The corresponding side-length is a = 1 2 ).