Indiatimes is running an edited version of
this article, which is the next installment of the puzzle column. This article is in collaboration with Prof. Krishnan Shankar of the University of Oklahoma.
The main problem in this article, the geometric version of the Inequality of the Means, has been chosen for its simplicity and elegance. It is one of those mathematical problems that is easy to state but ridiculously hard to prove. There are two puzzles this month. Please be sure to use the two different Subject lines mentioned in the article to help us distinguish which puzzle it is you are replying to.
This article is in collaboration
with Prof. Krishnan Shankar, Professor of Mathematics at the University of
Oklahoma.
This article’s material came
forth from the fertile mind of the extraordinary John Conway. References to the
mathematics of the problem are listed at the end. It was a pleasure to discuss
this problem with Prof. Dror Bar-Natan. Prof. Bar-Natan’s exposition and
Javascript applet make the subject come alive on his website
(http://www.math.toronto.edu/~drorbn/), which is certainly worth a visit.
We start with a well-known
inequality from high school algebra: let a and b be any two
non-negative numbers. Then, their arithmetic mean is at least as large as their
geometric mean, i.e.,
square_root(a *b ) <= ((a + b)/2)
Equality occurs precisely when a
= b. This is not hard to prove algebraically, but here is a nice geometric
proof. Consider the following figure where a square of side length a + b
encloses 8 right angled triangles of orthogonal sides a and b each.