, 2015

Solution 1

Since $PQRS$ is a square and $TX$ and $UY$ are perpendicular to $QR$, then $TX$ and $UY$ are parallel to $PQ$ and $SR$. Similarly, $VY$ and $WX$ are parallel to $PS$ and $QR$. Therefore, if we extend $WX$ and $VY$ to meet $PQ$ and extend $TX$ and $UY$ to meet $PS$, then square $PQRS$ is divided into 9 rectangles. Since $QT=TU=UR=1$ and $RV=VW=WS=1$, then in fact $PQRS$ is divided into 9 squares, each of which is 1 by 1. Of these 9 squares, 6 are shaded and 3 are unshaded. [[IMAGE0]] Therefore, the ratio of the shaded area to the unshaded area is $6:3$, which equals $2:1$. Solution 2 Consider quadrilateral $YURV$. $YURV$ has three right angles: at $U$ and $V$ because $UY$ and $VY$ are perpendicular to $QR$ and $RS$, respectively, and at $R$ because $PQRS$ is a square. Since $YURV$ has three right angles, then it has four right angles and so is a rectangle. Since $RV=UR=1$, then $YURV$ is actually a square and has side length 1, and so has area $1^2$, or 1. Similarly, $XTRW$ is a square of side length 2, and so has area $2^2$, or 4. [[IMAGE1]] Since square $PQRS$ is $3 \times 3$, then its area is $3^2$, or 9. The area of the unshaded region is equal to the difference between the areas of square $XTRW$ and square $YURV$, or $4-1=3$. Since square $PQRS$ has area 9 and the area of the unshaded region is 3, then the area of the shaded region is $9-3=6$. Finally, the ratio of the shaded area to the unshaded area is $6:3$, which equals $2:1$.