CEMC Fermat, 2012

We label the three points as $O(0,0)$, $P(1,4)$ and $Q(4,1)$.

There are three possible locations for the fourth vertex $R$ of the parallelogram – between $O$ and $P$ (in the second quadrant), between $P$ and $Q$ (in the first quadrant), and between $Q$ and $O$ (in the fourth quadrant).

In each of these cases, $\triangle OPQ$ will make up half of the parallelogram, and so the area of the parallelogram is twice the area of $\triangle OPQ$.

There are many ways to calculate the area of $\triangle OPQ$.

We proceed by “completing the rectangle" which includes the$x$-axis, the $y$-axis, the line $y=4$, and the line $x=4$.

We label the point $(0,4)$ as $S$, the point $(4,4)$ as $T$, and the point $(4,0)$ as $U$.

(Note that rectangle $OSTU$ is in fact a square, so we have “completed the square"!)

The area of $\triangle OPQ$ equals the area of rectangle $OSTU$ minus the combined areas of $\triangle OSP$, $\triangle PTQ$, and $\triangle QUO$.

The area of rectangle $OSTU$ is $4 \cdot 4 = 16$, since it is a square with side length $4$.

Consider $\triangle OSP$. It is right-angled at $S$, with $OS=4$ and $SP=1$.

Thus, its area is $\frac{1}{2}(OS)(SP)=\frac{1}{2}(4)(1)=2$.

Consider $\triangle PTQ$. It is right-angled at $T$, with $PT=TQ=3$.

Thus, its area is $\frac{1}{2}(PT)(TQ)=\frac{1}{2}(3)(3)=\frac{9}{2}$.

Consider $\triangle QUO$. It is right-angled at $U$, with $OU=4$ and $UQ=1$.

Thus, its area is $\frac{1}{2}(OU)(UQ)=\frac{1}{2}(4)(1)=2$.

Therefore, the area of $\triangle OPQ$ is $16 - 2 - \frac{9}{2}- 2 = \frac{15}{2}$.

Thus, the area of the parallelogram is $2\cdot \frac{15}{2} = 15$.