really hard center of gravity and equilibrium problem, physics?

Question

Four bricks of length L, identical and uniform, are stacked on top of one another , in such a way that part of each extends beyond the one beneath.
The distance from the edge of the table to the edge of the bottom block is a1, from the edge of that block to the edge of the next block is a2, and so forth until a4 which is the distance from the edge of the top block to the edge of the one directly below it.
h is the distance from the farthes t edge to the table, namely a1+a2+a3+a4
L is smaller than h



Find, in terms of L, the maximum values of the following, such that the stack is in equilibrium.

a1=_______*L
a2=_______*L
a3=_______*L
a4=_______*L
h=________*L

Answer 1

The condition for equilibrium for the lowest brick is (done by balancing moments around the edge of the table):

a1^2 + (a1+a2)^2 + (a1+a2+a3)^2 + (a1+a2+a3+a4)^2 =
(L-a1)^2 + (L-a1-a2)^2 + (L-a1-a2-a3)^2 + (L-a1-a2-a3-a4)^2

for the next (balance moments around edge of first brick):

a2^2 + (a2+a3)^2 + (a2+a3+a4)^2 =
(L-a2)^2 + (L-a2-a3)^2 + (L-a2-a3-a4)^2

etc.

a3^2 + (a3+a4)^2 = (L-a2)^2 + (L-a3-a4)^2

a4^2 = (L-a4)^2

4 eq in 4 unknowns. I solved numerically for L=1 to get the following answers

a1 = 0.125*L
a2 = 0.167*L
a3 = 0.25*L
a4 = 0.50*L

h = 1.042*L

More details here:
http://img264.imageshack.us/img264/3955/bricksequilibrium1fc5.png

Also the analytical solution to the simultaneous equations:
http://img404.imageshack.us/img404/7645/brickequilibrium2jr0.png