IMC 2013 Q13

The diagram shows the top-right-hand portion of the square. The shaded trapezium is labelled \(QXYZ\) and \(W\) is the point at which \(ZY\) produced meets \(PQ\).
As \(QXYZ\) is an isosceles trapezium, \(\angle QZY = \angle ZQX = 45^\circ\).
Also, as \(YX\) is parallel to \(ZQ\), \(\angle XYW = \angle WXY = 45^\circ\). So \(WYX\) and \(WZQ\) are both isosceles right-angled triangles. As \(\angle ZWQ = 90^\circ\) and \(Z\) is at the centre of square \(PQRS\), we deduce that \(W\) is the midpoint of \(PQ\). Hence \(WX = XQ = \tfrac{1}{4}PQ\). So the ratio of the side-lengths of similar triangles \(WYX\) and \(WZQ\) is 1 : 2 and hence the ratio of their areas is 1 : 4.
Therefore the area of trapezium \(QXYZ\) = \(\tfrac{3}{4}\) × area of triangle \(ZWQ\) = \(\tfrac{3}{32}\) × area \(PQRS\) since triangle \(ZWQ\) is one-eighth of \(PQRS\). So the fraction of the square which is shaded is 4 × \(\tfrac{3}{32}\) = \(\tfrac{3}{8}\).