$PQRS$ is a convex quadrilateral. The intersection of the two diagonals is $O$. If the areas of triangles $PQS$, $QRP$, and $SPR$ are $1, 2,$ and $3$, respectively. What is the area of triangle $PQO$?

• What is your work on the subject ? – Jean Marie Dec 9 '16 at 22:25

If you want a hint, read on.

First determine $\mathrm{Area}(\triangle QRS)$.

Now, let $x = \mathrm{Area}(\triangle SPO)$, $y = \mathrm{Area}(\triangle PQO)$, and note that $\mathrm{Area}(\triangle RSO) : \mathrm{Area}(\triangle QRO) = x : y$. (Why?)

• Very good hint. – Jean Marie Dec 9 '16 at 23:12

With the excellent hint given by @Triskele, one finds $y=2/5$ under the assumption that such a quadrilateral exists.

This existence can be established. Moreover, there is no unicity: up to a rotation, there is one degree of freedom. More precisely, taking line PR as the $x$ axis and $O$ as the origin on this axis, we can take:

$$P\left(\frac45,0\right), \ \ Q\left(a,1\right), \ \ R\left(-\frac{16}{5},0\right), \ \ S\left(-\frac32a,-\frac32\right)$$ as illustrated on figure below with the simplest choice $a=0$.