Four brothers inherited a plot of land in the shape of a convex quadrilateral. By connecting the midpoints of the opposite sides of the plot, they divided the inheritance into four quadrilaterals. The first three brothers received plots of land measuring 360, 720, and $900 \mathrm{~m}^{2}$ respectively. How much land did the fourth brother receive?
Solution. Let the vertices of the quadrilateral be $A, B, C$, and $D$, and the midpoints of the sides be $E, F, G$, and $H$ as shown in Figure 1. The intersection point of the two midlines $M$ is a median in the four triangles formed by the segments connecting $M$ to the vertices, which divide the quadrilateral. These segments thus bisect the area of each of these triangles. From this, it follows that

$$
[E A F M]+[H M G C]=[D E M H]+[M F B G]=\frac{t}{2}
$$

where $t$ is the area of the quadrilateral $A B C D$. ([EAFM] denotes the area of the quadrilateral with vertices $E, A, F$, and $M$, and the same applies to the other terms.) This follows from the fact that each pair of the four triangles has one member on the left side and the other on the right side of the above sum.

![](https://cdn.mathpix.com/cropped/2024_05_02_dcad7b2fb58bed04c7aeg-1.jpg?height=290&width=312&top_left_y=603&top_left_x=901)

Figure 1

For simplicity, let's choose the unit of area to be $180 \mathrm{~m}^{2}$. Then the areas of the known quadrilaterals are 2, 4, and 5 units, respectively. According to equation (1), there are three possible cases depending on which of the three known quadrilaterals is diagonally opposite the unknown quadrilateral. These cases are shown in Figures 2/a, 2/b, and 2/c. We will show that two of these cases are not possible.

![](https://cdn.mathpix.com/cropped/2024_05_02_dcad7b2fb58bed04c7aeg-1.jpg?height=236&width=1106&top_left_y=1199&top_left_x=497)

$$
t=12, \frac{t}{8}=\frac{3}{2}>1
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_dcad7b2fb58bed04c7aeg-1.jpg?height=98&width=241&top_left_y=1496&top_left_x=931)

$t=18, \frac{t}{8}=\frac{9}{4}>2$

Figure 2/a

Figure 2/b

Figure 2/c

It is known that the midpoints of a quadrilateral's sides form a parallelogram whose area is half that of the quadrilateral. The midlines of the quadrilateral $A B C D$ are diagonals in this parallelogram $E F G H$, which divides it into four triangles of area $\frac{t}{8}$. Each of these triangles contains one of the four equal shares, so each of their areas is greater than $\frac{t}{8}$ (Figure 3). This is only true in the second arrangement, shown in Figure 2/b, where the fourth sibling's share is only 3 units, or $540 \mathrm{~m}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_dcad7b2fb58bed04c7aeg-1.jpg?height=317&width=339&top_left_y=2007&top_left_x=882)

Figure 3

We must still show that this result is possible and that there are no additional conditions that would rule it out. To do this, we provide a convex quadrilateral in which the areas of the individual sub-quadrilaterals are 2, 4, 5, and 3 units, respectively.

Start with a quadrilateral $A B C D$ whose diagonals intersect in the ratios $5: 9$ and $1: 13$ as shown in Figure 4. Such a convex quadrilateral can clearly be constructed. Then $\frac{[A B D]}{[C D B]}=\frac{1}{13}$, since the heights corresponding to the common base $B D$ of the two triangles are in the ratio $1: 13$. Similarly, we get $\frac{[A C D]}{[C A B]}=\frac{5}{9}$. If the area of the quadrilateral $A B C D$ is 14 units, then its diagonals divide it into triangles of areas 1 and 13, and 5 and 9 units, respectively, as shown in Figure 5. The segments connecting the midpoints of the sides of the quadrilateral are midlines in these triangles, so the areas of the triangles formed at the vertices of the quadrilateral $A B C D$ are one-fourth of these values. Finally, the $\frac{t}{2}=7$ unit area parallelogram $E F G H$ is divided by its diagonals into triangles of area $\frac{7}{4}$ (Figure 6). The areas of the individual parts are thus

$$
\begin{aligned}
& {[M E A F]=[E A F]+[E F M]=\frac{1}{4}+\frac{7}{4}=2} \\
& {[M F B G]=[F B G]+[F G M]=\frac{9}{4}+\frac{7}{4}=4} \\
& {[M G C H]=[G C H]+[G H M]=\frac{13}{4}+\frac{7}{4}=5, \quad \text { and finally }} \\
& {[M H D E]=[H D E]+[H E M]=\frac{5}{4}+\frac{7}{4}=3}
\end{aligned}
$$

Thus, the fourth sibling received a share of $540 \mathrm{~m}^{2}$.
![](https://cdn.mathpix.com/cropped/2024_05_02_dcad7b2fb58bed04c7aeg-2.jpg?height=828&width=532&top_left_y=926&top_left_x=791)

Figure 5

![](https://cdn.mathpix.com/cropped/2024_05_02_dcad7b2fb58bed04c7aeg-2.jpg?height=596&width=588&top_left_y=1889&top_left_x=758)

Figure 6