Equilibrium of a tower of 2-d blocks In the image below, 


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*All the blocks are frictionless & identical with side of unit length, height $h$, weight $w$ & center of gravity at their geometric centers.

*The 2 lowest blocks are on solid ground.

*The distance from the corner of each block to the midpoint of the bottom side of the box above it is given (namely $a1,a2,b1,b2$).

*forces $F_{ac}$ and $F_{bc}$ are the resultant reaction forces exerted by blocks $A$ & $B$ on $C$.
I am interested in the behavior of these blocks immediately after setting them in this configuration and releasing them, or more specifically: For what relation between $a1,a2,b1$ & $b2$: 


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*do $A,B$ and $C$ move?- case 1

*do only $A$ and $C$ move?- case 2

*do only $B$ and $C$ move?- case 3

*is the configuration stable (does not change at all once set under this condition and then left.)?- case 4 


For those interested, here's my approach & what (I think) I know already:
In an attempt to find the limiting conditions (the borderline conditions between equilibrium and non equilibrium), I assumed that initially $C$ will tend to be in equilibrium.(I have no rigorous justification for this assumption, just a hunch that "this isn't where the trouble is").Under this condition, $F_{ac}$ & $F_{bc}$ can be calculated, and the moments due to their "equal and opposites" ($F_{ca}$ & $F_{cb}$) about $P1$ and $P2$ can be obtained as: $$ M_{ca}(x,y)=w(.5+a1-a2-x)/(x/y+1)$$ $$ M_{cb}(y,x)=w(.5+b1-b2-y)/(y/x+1) $$ where $x$ & $y$ are the perpendicular distances of the respective forces from the center of $C$. 
With some ad-hoc and shaky logic here's what I arrived at:


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*when $$w\cdot a_2 <M_{ca}(.5,.5)\quad \& \quad w\cdot b_2 <M_{cb}(.5,.5)$$ case-1 occurs with $A$ & $B$ touching $C$ only through it's vertices.

*when $$w\cdot a_2 =M_{ca}(x1,.5)\quad \& \quad w\cdot b_2 <M_{cb}(.5,x1)$$ where $a1\leqslant x1< .5$ case-1 occurs with $C$ rotating with $A$ maintaining a surface of contact with $A$ but only a point contact with $B$, but if $$\boldsymbol{w\cdot a_2 =M_{ca}(x1,.5)\quad \& \quad w\cdot b_2 \geqslant M_{cb}(.5,x1)}$$ the configuration is stable (case-4).

*when $$w\cdot a_2 >M_{ca}(a1,.5)\quad \& \quad w\cdot b_2 <M_{cb}(.5,a1)$$ case-3 occurs, but if $$\boldsymbol{w\cdot a_2 >M_{ca}(a1,.5)\quad \& \quad w\cdot b_2 <M_{cb}(.5,a1)}$$ the configuration is stable (case-4).

*The above conditions with $a$ & $b$ exchanged along with their corresponding variables.
But I have no way to check this or provide a satisfactory argument for these conditions, especially the parts written in bold (I arrived at it by putting various combinations of arguments for $M_{ca}$ & $M_{cb}$ & thinking about what would happen in each case). Is this set of conditions right?. What would be a good approach with a logical progression of steps to solve it?
 A: Revised Answer

Blocks A, B and C are identical, so cases 2 & 3 are the same. In case 1 blocks A and B turn about P1 and P2 at the same time, while C is supported equally at P5 and P3. In case 2 (illustrated in the diagram above) block C turns about P4 while also touching A at P5. I suppose case 4 means that no blocks move, so this is the same as saying that neither of cases 1 or 2 applies.
In case 1, if the 2 blocks at the base are not further apart than $1+2h$, where $h$ is the height of each block - ie if $(a1+b1)-(a2+b2) < 2h$ - then block C will not fall through to the ground. Likewise in case 2 blocks A and C may become jammed when the end face of C is flat against the top face of A. These complications add further restrictions if "unstable" means that block C reaches the ground. For simplicity, I assume that "unstable" means that the blocks move from the initial position to some different configuaration.
Your approach is correct but you need to complete it. Eliminate $w$ from your inequalities by substituting for $M$, assign $x$ and $y$ their extreme values as in the 1st paragraph, and rearrange to get inequalities relating only $a1$, $a2$, $b1$ and $b2$.  
I think your concern is that there are not enough equations to enable you to find limiting values of each of the 4 variables independently. This is inevitable because there are not enough restrictions in the problem. Each of the 4 variables can be adjusted independently, whereas there are only 2 inequalities for each mode (case) of toppling, arising from the balancing of moments at P1 and P2. 
For case 1 the ratios $r_a=\frac{a1}{a2}$ and $r_b=\frac{b1}{b2}$ must each be less than the same constant integer.
For case 2 the ratios $r_a$ and $r_b$ must each be less than a critical value which depends on $b1$ alone, a different critical value for each.
The final question is how to represent the stability conditions. This can be done either on a 3D plot of $(r_a,r_b,b1)$, or on 2D plots of $(r_a,r_b)$ for selected values of $b1$, which can vary from $0$ to $\frac12$, or 2D plots of $(r_a,b1)$ and $(r_b,b1)$. The boundaries are provided by the above inequalities. The stable region lies between the boundaries and the origin.
A: Not an answer
This is very interesting problem, but I think finding the configurations where the "back" contact point lift is not sufficient for instability. For example the configuration below is stable:

This is because the middle block on the right is "pinched" by the other blocks and friction can hold it up. Is there a reason you are looking at frictionless only? I think this is too much of a simplification without friction.
You also have to consider that small angle rotations change the problem considerably. Again, in the situation above if the top block is not pushing down on the middle block on the right then it is going to fall. But, if the top block has a small angle and establishes contact you can end up in a stable scenario.
