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 pupis 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} $