We know about a convex quadrilateral that if we divide it with diagonals, the area of three of the resulting four triangles is either $24 \mathrm{~cm}^{2}$ or $25 \mathrm{~cm}^{2}$. What is the maximum possible area of the quadrilateral?
Since the quadrilateral $ABCD$ is convex, it contains the intersection point $O$ of its diagonals. Thus, the area $T$ of the quadrilateral is given by the sum of the areas $T_{1}, T_{2}, T_{3}, T_{4}$ of the triangles $BOC$, $COD$, $DOA$, and $AOB$.

![](https://cdn.mathpix.com/cropped/2024_05_02_da0f7823f521c9fd90a2g-1.jpg?height=610&width=715&top_left_y=256&top_left_x=694)

Express the areas using the diagonal $DB = DO + OB$ and the corresponding heights.

$$
T_{1}=\frac{OB \cdot m_{1}}{2}, \quad T_{2}=\frac{OD \cdot m_{1}}{2}, \quad T_{3}=\frac{OD \cdot m_{2}}{2}, \quad T_{4}=\frac{OB \cdot m_{2}}{2}
$$

Notice that $T_{1} T_{3}=T_{2} T_{4}$, or

$$
T_{4}=\frac{T_{1} T_{3}}{T_{2}}
$$

Let $T_{4}$ be the unknown area, then we know the values of the other three areas are either $24 \, \text{cm}^2$ or $25 \, \text{cm}^2$. We seek the maximum value of the sum $T=T_{1}+T_{2}+T_{3}+T_{4}$. We might think this is the largest when each of $T_{1}, T_{2}, T_{3}$ is $25 \, \text{cm}^2$. In that case, by (1),

$$
T_{4}=\frac{T_{1} T_{3}}{T_{2}}=\frac{25 \cdot 25}{25}=25 \, \text{cm}^2, \quad \text{and thus}
$$

$T=100 \, \text{cm}^2$ would result.

Actually, this is not the maximum value. This becomes clear when we examine when the fraction in (1) is maximized. Clearly, this happens when the numerator is as large as possible and the denominator is as small as possible, i.e., when $T_{1}=T_{3}=25 \, \text{cm}^2$ and $T_{2}=24 \, \text{cm}^2$.

Then $T_{4}=\frac{25 \cdot 25}{24}=26 \frac{1}{24} \, \text{cm}^2$ and $T=25+24+25+26 \frac{1}{24}=100 \frac{1}{24} > 100 \, \text{cm}^2$.

We will show that this is the maximum possible value of $T$. Since $T_{1} \leq 25 \, \text{cm}^2$ and $T_{3} \leq 25 \, \text{cm}^2$, it follows that $T_{1}+T_{3} \leq 50 \, \text{cm}^2$, and $T_{1} \cdot T_{3} \leq W=625 \, \text{cm}^4$. Therefore,

$$
T \leq 50 \, \text{cm}^2 \left(T_{2}+\frac{W}{T_{2}}\right)
$$

Here, $T_{2}$ has two possible values. We have already calculated the value of the right-hand side for both, and saw that the value corresponding to $T_{2}=24 \, \text{cm}^2$ is larger than the value corresponding to $T_{2}=25 \, \text{cm}^2$.

F. G.

Remarks. 1. It is easy to show that the quadrilateral with the maximum area is a trapezoid. If $T_{1}=T_{3}=25 \, \text{cm}^2$, then $T_{ACD}=T_{BCD}$. Since the base $DC$ of $\triangle ACD$ and $\triangle BCD$ is the same and their areas are equal, their heights must also be equal. This means that $AB \parallel DC$, or that the quadrilateral is a trapezoid.

2. Many solvers converted the fraction $\frac{1}{24}$ to a decimal and rounded the value 0.0416 to 0.041 or 0.04. In such cases, strictly speaking, they did not provide the correct answer to the question "What is the maximum possible value...?"