Force distribution on corner supported plane This question has been annoying me for a while. If you have a completely ridged rectangular plate of width and height x and y that is supported on each corner (A,B,C,D) and has force (F) directly in its center then I think the force on each corner support will be F/4.

What I want to know is how to prove this? Obviously there are 4 unknowns so we require 4 equations. 
$$\sum F_z = 0$$
$$\therefore F_A+F_B+F_C+F_D=F$$
Also
$$\sum M = 0$$
Now taking the moments about point A 
\begin{equation*}
\begin{vmatrix}
        i & j & k \\
        x & 0 & 0 \\
        0 & 0 & F_B \\
        \end{vmatrix}
+
\begin{vmatrix}
        i & j & k \\
        x & y & 0 \\
        0 & 0 & F_C \\
        \end{vmatrix}
+\begin{vmatrix}
        i & j & k \\
        0 & y & 0 \\
        0 & 0 & F_D \\
        \end{vmatrix}+
\begin{vmatrix}
        i & j & k \\
        x/2 & y/2 & 0 \\
        0 & 0 & -F \\
        \end{vmatrix}=0
\end{equation*}
As the sum of moments equals 0, let i and j = 0
$$\therefore F_C+F_D=F/2$$ and $$F_B+F_C=F/2$$
Then taking the moments about another point to get 4th equation. But no matter what location I use the equations will not solve. I have been using a matrix to find $$F_A, F_B, F_C, F_D$$  See below. But when getting the det of the first matrix the answer is always equal to 0.
\begin{equation*}
\begin{vmatrix}
        1 & 1 & 1 & 1 \\
        0 & 0 & 1 & 1 \\
        0 & 1 & 1 & 0 \\
        ? & ? & ? & ? \\
        \end{vmatrix}
*
\begin{vmatrix}
        F_A \\
        F_B \\
        F_C \\
        F_D \\
        \end{vmatrix}
=\begin{vmatrix}
        1 \\
        0.5 \\
        0.5 \\
        ? \\
        \end{vmatrix}*F
\end{equation*}
Can someone please tell me what I am doing wrong? Thank you in advance.
 A: Disproof by counterexample
Let's see if we can construct a counterexample to the fully-symmetric solution, where the forces at $A$ and $C$ are equal to each other and the forces at $B$ and $D$ are equal to each other but there is a different force on each diagonal, $A \neq B$.  Does such an arrangement exist where the rectangle is stable?
It does.  We can build it by changing our rectangle resting points to a rectangular table with springs for legs. Let's attach to each corner a spring with Hooke constant $k$, which exerts a restoring force $\vec F_H = -k\Delta\vec x$ when compressed by a distance $\Delta x$.  However due to some quality control issues at the spring factory we have two long springs, which we attach at the corners $A,C$, and two short springs which we attach at $B,D$.  Now we lower our table gently towards the level ground, keeping the table level.  What do the springs do as the distance between the table and the ground changes?
Clearly the long springs at $A,C$ touch the ground and begin to compress while the short springs at $B,D$ still dangle uselessly in the air.  This gives us an equilibrium configuration where the supporting force at $B,D$ is zero and the supporting forces at $A,C$ are not.  It's an unstable equilibrium, since any deviation of the weight of the table $F$ from the axis of rotation $AC$ will cause an unbalanced torque, but it's an equilibrium.
Now push down on the table some more, so that the legs at $B$ and $D$ also touch the ground.  You have forces at all four corners, but they are manifestly unequal:  the compression $\Delta x$ is greater for the long springs at $A,C$ than for the short springs at $B,D$.  This configuration is stable against small excursions of $F$ from the center of the rectangle --- and it gets stabler as the magnitude of $F$ gets larger.
I was planning to relate this to the wobbly table theorem, a consequence of the Intermediate Value Theorem in calculus, but I got sleepy while writing.
Disproof of uniqueness for forces supporting all four-legged tables
In fact it's relatively straightforward to argue that any four-legged support has this indeterminacy.  Here's an outline of the argument.


*

*It's possible to support a plane at exactly three points.  The engineering solution is use two of those points to define an axis of rotation and the third to define an orientation about that axis.  If you're more of a classical-mechanics person you can think of one point as defining the location of your object and the other two as specifying the two Euler angles.

*Any rigid quadrilateral $ABCD$ supported at four points, if the weight force $F$ of the object is not on either of the diagonals $AB$ or $CD$, could therefore be supported just as well at only three of the points.  Therefore it is always possible to set the supporting force at one corner to zero.

*If the supporting force $F$ does lie along one of the diagonals (without loss of generality: $AC$) we see from the construction above that the torques due to $B$ and $D$ must be equal, but we have no constraint on their magnitude.
Or in linear-algebra language: since the forces supporting a triangle are uniquely determined, the forces supporting a quadrilateral are always underdetermined.
A: I think I have a solution.  Considering the corner forces $A$, $B$, $C$ and $D$ you have a system of 3 equations and 4 unknowns
$$\begin{align} 
 A + B + C + D & = F \\
 \frac{y}{2} \left(C+D-A-B\right) &= 0 \\
 \frac{x}{2} \left(A+D-B-C\right) & = 0 
\end{align}$$
$$\begin{vmatrix}
   1 & 1 & 1 & 1 \\
  -\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
  \frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} A \\ B \\ C \\ D \end{vmatrix} = \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} $$
What if consider the forces as deviation from $\frac{F}{4}$ such that
$$\begin{align}
  A &  = \frac{F}{4} + U \\
  B &  = \frac{F}{4} + V \\
  C &  = \frac{F}{4} + W \\
  D &  = \frac{F}{4} + G \end{align} $$
$$ \begin{vmatrix}
   1 & 1 & 1 & 1 \\
  -\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
  \frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} \frac{F}{4} \\ \frac{F}{4} \\ \frac{F}{4} \\ \frac{F}{4} \end{vmatrix} +
\begin{vmatrix}
   1 & 1 & 1 & 1 \\
  -\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
  \frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} U \\ V \\ W \\ G \end{vmatrix} = \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} $$
$$ \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} +
\begin{vmatrix}
   1 & 1 & 1 & 1 \\
  -\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
  \frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} U \\ V \\ W \\ G \end{vmatrix} = \begin{vmatrix} F \\ 0 \\ 0 \end{vmatrix} $$
$$ \begin{vmatrix}
   1 & 1 & 1 & 1 \\
  -\frac{y}{2} & -\frac{y}{2} & \frac{y}{2} & \frac{y}{2} \\
  \frac{x}{2} & -\frac{x}{2} & -\frac{x}{2} & \frac{x}{2}
\end{vmatrix}
\begin{vmatrix} U \\ V \\ W \\ G \end{vmatrix} = \begin{vmatrix} 0 \\ 0 \\ 0 \end{vmatrix} $$
Now a possible solution is $U=0$, $V=0$, $W=0$ and $G=0$
I think to beyond this method, you need to assume a non-rigid frame, with some simple stiffness matrix, and in the end find the limit where $k \rightarrow \infty$.
