Many thanks to those of you who sent in solutions to thie problem including Mai of The Chinese High School Singapore and Andrei of School 205 Bucharest. Thanks also to the pupils of Madras College who answered the first part of this problem as part of their solution to tilting triangles (November 2002).

$P$ is the centre of the square $ABCD$

## First

If the sides $PQ$ and $PS$ of the square $PQRS$ conincide with $PB$ and $PC$ then the overlapped area is $PBC$.

By symmetry (or congruence) $PBC$ has an area which is one quarter of the area of $ABCD$.

## Second

Rotating the square $PQRS$ about $P$ gives a diagram equivalent to the one opposite.

We know that $PBC$ is a quarter of $ABCD$ so if we can show that the area of $PXCY =$ area of $PBC$ then the overlap will always be a quarter of the square.

To prove this we need to show that triangle $PBX$ is congruent to triangle $PYC$.

$PB = PC$ ( half the diagonal of the square)

angle $PBX =$ angle $PCY$ (diagonals of square bisect the angles).

angle $CPY =$ angle $BPX =$ angle of rotation

Therefore triangles $BPX$ and $CPY$ are congruent (ASA)

Area $PXCY =$ Area $XPC +$ Area $CPY =$ Area $XPC +$ Area $BPX =$ Area $BPC$

Therefore $PXCY$ is a quarter of the square.

## Lastly

Let the length of the side of the large square is $x$ and the the length of the diagonal is $d$.

Using Pythagoras Theorem

$ d^2 = x^2 + x^2 = 2x^2 $

From this we know that the limit of the length of the side of the small square is:

$ \frac {d}{2} = \frac {x\sqrt 2}{2} $