A letter F is composed of 6 unit squares and two rectangles of unit width as in the figure:

Find the lengths of the two rectangles such that the center of gravity is at the center of the middle square.

(In reply to re: maybe -- numerical confirmation and question by Charlie)

Solving for the two quadratics is not hard.  Starting with:
8 + 7*y/2 + y^2 / 2 = 2*(6 + x + y)
2 + 3*x/2 + x^2 / 2 = 6 + x + y

First do a little simplification and rearranging:
y^2 + 3y - 8 = 4x
(x^2)/2 + x/2 - 4 = y

Now substitute the second into the first:
((x^2)/2 + x/2 - 4)^2 + 3*((x^2)/2 + x/2 - 4) - 8 = 4x

Rearranging and simplifying yields a quartic:
x^4 + 2x^3 - 9x^2 - 26x - 16 = 0

This quartic factors into:
(x^2 - x - 8)*(x + 2)*(x + 1) = 0

The only positive root is a root from the quadratic factor: x = (1+sqrt(33))/2.  Substituting into the expression for y also gives y = (1+sqrt(33))/2.

Posted by Brian Smith on 2007-05-18 20:08:58