Two squares of side 6. One has a corner on the centre O of the other. Drag the tilt, then step through the argument. Every formula below is evaluated at the angle you choose.
Why the side length matters: the farthest point of the fixed square from O is a corner, at distance 3√2 ≈ 4.24. That is less than 6, so the tilted square always covers the whole wedge between its two sides at O. With a tilted square of side 2, the overlap would be only 4.