At most how many points $P$ are there in the plane of the convex quadrilateral $A B C D$ such that the areas of the triangles $P A B, P B C, P C D$ and $P D A$ are equal?
We will prove that in the plane of the quadrilateral, there is at most one point $P$ that meets the conditions of the problem. In the 1989/2 issue of our journal, starting from page 50, the solutions to the problems of the József Kürschák Mathematical Competition were published. Here, in the first solution of the 1st problem, we read that if the point $P$ is inside the convex quadrilateral, then on the one hand it is the midpoint of one of the diagonals, and on the other hand, there can be at most one such point inside the quadrilateral. Let us now assume that there is a point $P$ that meets the conditions outside the quadrilateral. The various possible positions of the point can be seen in the 1st, 2nd, and 3rd figures.

![](https://cdn.mathpix.com/cropped/2024_05_02_78b92cdf8213c1464237g-1.jpg?height=352&width=382&top_left_y=431&top_left_x=861)

1st figure

![](https://cdn.mathpix.com/cropped/2024_05_02_78b92cdf8213c1464237g-1.jpg?height=342&width=288&top_left_y=924&top_left_x=905)

2nd figure

![](https://cdn.mathpix.com/cropped/2024_05_02_78b92cdf8213c1464237g-1.jpg?height=358&width=371&top_left_y=1407&top_left_x=861)

3rd figure

It is easy to show that we do not need to deal with the cases according to the 2nd and 3rd figures, because in these cases one of the triangles $PAB, PBC, PCD, PDA$ contains another, and thus has a larger area. In the 1st figure, the triangles $PBA$ and $PBC$ have equal areas, and they share the side $PB$, so the heights corresponding to this side are also equal. Therefore, $PB$ is parallel to $AC$. For similar reasons, $AP$ and $BD$ are also parallel. Thus, the quadrilateral $APBE$ is a parallelogram, in which $P$ is the reflection of $E$ over the midpoint of $AB$. (Here, $E$ is the intersection of the diagonals of the quadrilateral.) Let the area of the triangle $PAB$ be denoted by $t_{PAB}$. According to the previous observation, $t_{PAB} = t_{EAB}$, and this area is exactly half of the area of the quadrilateral. Indeed, by our assumption, $P$ is such that $t_{PAB} = t_{PBC} = t_{PCD} = t_{PDA}$, and the sum of the areas of the three triangles outside $PAB$ is exactly $t_{PAD}$ more than the area of the quadrilateral.

Since $t_{EAB}$ is half the area of the quadrilateral, neither of the diagonals $AC$ and $BD$ bisects the area of the quadrilateral. Therefore, according to the solution of the mentioned Kürschák competition problem, there is no suitable point inside the quadrilateral, and there is also only one external point.

Thus, the answer to the question of the problem is that there is at most one suitable point $P$.

The question may arise whether there exists a convex quadrilateral $ABCD$ for which a point $P$ can be found outside the quadrilateral that meets the requirements of the problem. The attentive reader can find a construction for such a quadrilateral on page 51 of the 1989/2 issue of KöMaL, in the 3rd remark after the first solution.