Wow! I am very late!
I have only just read this discussion! Sorry!
Yes sir, PianoExpert and Simone are right and have explained the reason with a lot of common sense and in an intuitive way.
If intuition is not enough and you also want to understand why, let's make this small model.
Let's consider two masses with weights P1 (front part of the key) and P2 (back part of the key) located at distances L1 and L2 from a center of rotation (Fulcrum); let us also imagine that between P1 and P2 there is a rigid rod connecting them, of zero weight, which passes through the Fulcrum and is free to rotate around it.
Then, physics says that if L1 x P1 = L2 x P2, the system is in equilibrium (for example, if it is stationary, it does not rotate and remains still).
It also tells us that, for example, if L1 x P1 <= L2 x P2, then the system tends to rotate (P2 goes down, P1 goes up).
And that is what happens on the piano, except that the key stops from falling with the back part, because it finds support on the bottom part of the keyboard!
To play, I must make the front part of the key go down, making the whole mechanism move all the way to the hammers; therefore, I must manage to vary P1 and make it become P'1 so that L1 x P'1 exceeds L2 x P2, and only then does the key move.
That is why the term almost in the center, which is the center in terms of "weight".
The other important thing is that, once the key starts moving, it must have low inertia, meaning that the time between touching it and it rotating at a high enough speed to produce effects is minimal.
This is achieved by containing both weights and lengths (obviously compromises must be made, because reducing weight too much would also mean keys with little resistance, reducing length would mean little movement of the back part of the key, etc.), again in light of the previous formula, which is that of the moment of inertia.