Note to recruiters


Tuesday, April 10, 2012

Inequality of the means





The Oracle Asks

The area of the inner square is (a - b)*(a - b), which is the area of the outer square minus the area of the triangles. In other words, we have the following:
(a + b) * (a + b)  - 8. ab. (0.5) = (a - b) * (a - b) >= 0
=> (a + b) * (a + b) >= 4.a.b
=> ((a + b)/2) >= square_root(ab)
The equality discussion is also clear since the inner square has area zero precisely when a = b. The inequality is true for any finite set of non-negative numbers. Given any set of non-negative numbers a1, a2, ..., an, we always have
arithmetic_mean(a1, a2, ..., an) >= geometric_mean(a1, a2, ..., an)
This is not hard to prove algebraically but the fun starts when we try to replicate the geometric argument above. Here is the geometric problem for n = 3: given a cubical box of side length (a + b + c), can you fill it with 27 little cuboids each of length a, breadth b and width c without spilling out of the box? In concrete terms, take a cubical box of size 15 × 15 ×15 (inches for instance) and fill it with 27 little cuboids, each of size 4 × 5 × 6.
The problem is hard, ridiculously hard even. Even if you look at the solution it does not easily lend itself to memory. Here is a link to the solution worked out step by step:
Scroll down to the Javascript application or read Professor Bar-Natan’s lucid explanation of the same topic.
For larger n the problem comes in two flavors: solvable or hopeless. That is, if you can solve the problem for m and n, it is possible to bootstrap the solution for nm. Thus, there is a solution for n = 4; 6; 8 etc. But the case n = 5 is wide open. If you have some spare time, then by all means try to fit 3125 five dimensional cuboids of size (a x b x c x d x e) into a five dimensional cube of side length (a + b + c + d + e).
We end this month’s column with a question unrelated to the above inequality. This is the first puzzle for this month.
(New Dice) Take two standard six sided dice numbered 1 to 6 each. Is it possible to number them differently using positive integers so that one gets the same sums (2 through 12) with the same probabilities?
Please send in your answers with the subject “Inequality of the Means: New dice” to: