, 2019

Begin by constructing $AV$ perpendicular to $TX$ and $UB$ perpendicular to $YW$. The four segments $TX, UB, AV$, and $YW$ divide $PQRS$ into 9 identical squares. Label the intersections of the perpendicular pairs of these four segments as points $C, D, E$, and $F$. [[IMAGE0]] $A$ lies on $PS$ such that $PA = AY = YS$. $B$ lies on $SR$ such that $SX = XB = BR$. $C$ is the intersection of segments $UB$ and $YW$. $D$ is the intersection of segments $TX$ and $YW$. $E$ is the intersection of segments $TX$ and $AV$. $F$ is the intersection of $UB$ and $AV$. Segment $UY$ is a diagonal of square $PUCY$ and so $UY$ passes through $E$, the centre of square $PUCY$. Segment $UE$ is a diagonal of square $TUFE$. Segment $TW$ is a diagonal of square $TQWD$ and $TF$ is a diagonal of square $TUFE$. The diagonals in any square divide the square into 4 identical triangles. For example, the diagonals $TF$ and $UE$ divide the square $TUFE$ into 4 identical triangles, 3 of which are shaded. Similarly, we can show that diagonals $FW$ and $VC$ divide square $FVWC$ into 4 identical triangles, 3 of which are shaded. We may construct the missing diagonals in each of the 9 squares. [[IMAGE1]] These diagonals divide square $PQRS$ into $9\times4=36$ identical triangles. Since 10 of these triangles are shaded, then $\frac{10}{36}=\frac{5}{18}$ of square $PQRS$ is shaded.