[hide=This should not have been posted]Upvote if you got IMO 2011 vibes  :rotfl: [/hide]

Let $ABCD$ be a square. For a point $P$ inside it, a [i]windmill[/i] centred at $P$ consists of 2 perpendicular lines $l_1$ and $l_2$ passing through $P$ , such that 
$\bullet ~$ $l_1$ intersects the sides AB and CD at W and Y, respectively, and
$\bullet ~$ $l_2$ intersects the side BC and DA at X and Z, respectively.

A windmill is called [i]round[/i] if the quadrilateral WXYZ is [b]cyclic.[/b] Determine all points P inside the square such that every windmill centred at P is round.
1. **Understanding the Problem:**
   We need to determine all points \( P \) inside a square \( ABCD \) such that every windmill centered at \( P \) is round. A windmill consists of two perpendicular lines \( l_1 \) and \( l_2 \) passing through \( P \), intersecting the sides of the square at points \( W, X, Y, \) and \( Z \). The quadrilateral \( WXYZ \) must be cyclic for the windmill to be round.

2. **Claim:**
   \( P \) produces a round windmill if and only if it lies on the diagonals of the square.

3. **Proof of the "if" part:**
   - Assume \( P \) lies on one of the diagonals of the square \( ABCD \). Without loss of generality, let \( P \) lie on diagonal \( AC \).
   - Since \( P \) is on the diagonal, the angles \( \angle PDA \) and \( \angle PBA \) are equal because \( \triangle PDA \) and \( \triangle PBA \) are congruent by the properties of the square and the symmetry about the diagonal.
   - Similarly, \( \angle PDB \) and \( \angle PBD \) are equal.
   - Therefore, the quadrilateral \( WXYZ \) is cyclic because the opposite angles of \( WXYZ \) sum to \( 180^\circ \).

4. **Proof of the "only if" part:**
   - Assume \( P \) does not lie on either diagonal of the square.
   - Consider the angles \( \angle PDA \) and \( \angle PBA \). If \( P \) does not lie on the diagonal, these angles are not equal.
   - For \( WXYZ \) to be cyclic, the opposite angles must sum to \( 180^\circ \). However, if \( \angle PDA \neq \angle PBA \), this condition is violated.
   - Therefore, \( P \) must lie on the diagonal for \( WXYZ \) to be cyclic in every windmill configuration.

5. **Conclusion:**
   - The point \( P \) must lie on the diagonals of the square \( ABCD \) for every windmill centered at \( P \) to be round.

\(\blacksquare\)

The final answer is \( \boxed{ P } \) must lie on the diagonals of the square \( ABCD \).