, 2014

We begin by joining $Q$ to $P$. Since $Q$ and $P$ are the midpoints of $ST$ and $UV$, then $QP$ is parallel to both $SV$ and $TU$ and rectangles $SQPV$ and $QTUP$ are identical. In rectangle $SQPV$, $VQ$ is a diagonal. Similarly, since $PR$ is parallel to $VQ$ then $PR$ extended to $T$ is a diagonal of rectangle $QTUP$, as shown in Figure 1. [[IMAGE0]] In Figure 2, we label points $A, B, C, D, E$, and $F$, the midpoints of $SQ, QT, TU, UP, PV,$ and $VS$, respectively. We join $A$ to $E$, $B$ to $D$ and $F$ to $C$, with $FC$ intersecting $QP$ at the centre of the square $O$, as shown. [[IMAGE1]] Since $PR=QR$ and $R$ lies on diagonal $PT$, then both $FC$ and $BD$ pass through $R$. (That is, $R$ is the centre of $QTUP$.) The line segments $AE, QP, BD,$ and $FC$ divide square $STUV$ into 8 identical rectangles. In one of these rectangles, $QBRO$, diagonal $QR$ divides the rectangle into 2 equal areas. That is, the area of $\triangle QOR$ is half of the area of rectangle $QBRO$. Similarly, the area of $\triangle POR$ is half of the area of rectangle $PORD$. Rectangle $SQPV$ has area equal to 4 of the 8 identical rectangles. Therefore, $\triangle QPV$ has area equal to 2 of the 8 identical rectangles (since diagonal $VQ$ divides the area of $SQPV$ in half). Thus the total shaded area, which is $\triangle QOR + \triangle POR+ \triangle QPV$, is equivalent to the area of $\frac12+\frac12+2$ or 3 of the identical rectangles. Since square $STUV$ is divided into 8 of these identical rectangles, and the shaded area is equivalent to the area of 3 of these 8 rectangles, then the unshaded area occupies an area equal to that of the remaining $8-3$ or 5 rectangles. Therefore, the ratio of the shaded area to the unshaded area is $3:5$.