Is the mean of the squares of two numbers greater than, or less
than, the square of their means?
There are thirteen axes of rotational symmetry of a unit cube. Describe them all. What is the average length of the parts of the axes of symmetry which lie inside the cube?
weekly problem 12-2006
If we choose any two numbers, call them $a$ and $b$ $(b < a)$
and work out the arithmetic mean $ (a+b)/2 $ and the geometric mean
$ \sqrt(ab) $ then the arithmetic mean is always bigger than the
geometric mean. How can we prove it? One way is to use this
diagram. Clement Goh, age 12 years, from River Valley High,
Singapore sent a good solution.
In the diagram the measurements are:
PQ = a
PS = b
PM = PO = (a + b)/2
MQ = SO = TN = (a - b)/2
The blue rectangle measures (a - b)/2 by b ,
the same as the orange rectangle.
Rectangle PQRS, with area ab, is made up of the green plus the
The square MNOP, with area [ (a + b)/2 ] 2 , is
made up of the green plus the blue rectangles plus the yellow
area of MNOP - area PQRS = the area of the yellow square = [
(a - b)/2 ] 2 .
Hence the area of PQRS < area MNOP, that is
ab < [ (a + b)/2 ] 2
Taking square roots this shows that the geometric mean $
\sqrt(ab) $ is less than the arithmetic mean $ [(a + b)/2] $.