XLII OM - I - Problem 5

Given a segment $ AD $. Find points $ B $ and $ C $ on it, such that the product of the lengths of the segments $ AB $, $ AC $, $ AD $, $ BC $, $ BD $, $ CD $ is maximized.
We consider such positions of points $ B $ and $ C $, for which

this means, we assume that points $ A $, $ B $, $ C $, $ D $ lie on a line in this exact order (we do not lose generality, as the product being studied in the problem does not change value when the roles of $ B $ and $ C $ are swapped); the weak inequalities in (1) indicate that we allow for boundary cases where some of the points $ A $, $ B $, $ C $, $ D $ coincide.
Let us take the length of segment $ AD $ as the unit: $ |AD| = 1 $. Furthermore, let us denote:

then $ |AB| + |CD| = 1 - x $, from which

The product of the lengths of the segments in question is

this inequality becomes an equality if and only if $ y = 0 $ - that is, when $ |AB| = |CD| $.
The maximum value of the polynomial

on the interval $ \langle 0;\ 1\rangle $ can be found by differentiation:

we see that $ f $ achieves its maximum at $ x = \frac{1}{5} \sqrt{5} $.
[Instead of differentiating, one can apply a more elementary (and thus more elegant) method: using the inequality between the geometric mean and the arithmetic mean of five numbers. These five numbers will be: $ 4x^2 $ and the number $ 1-x^2 $ repeated four times. We obtain the inequality

or

with equality achieved when the averaged numbers are equal; that is, when
$ 4x^2 = 1 - x^2 $, or for $ x = \frac{1}{5} \sqrt{5} $.]
In accordance with the estimate (2) and the remark following it, the expression $ I $ achieves its maximum for $ x = \frac{1}{5} \sqrt{5} $, $ y = 0 $. This corresponds to placing segment $ BC $ of length $ x = \frac{1}{5} \sqrt{5} $ symmetrically relative to the midpoint of segment $ AD $ - while maintaining the order (1), or by swapping the roles of points $ B $ and $ C $.