2. Determine the maximum possible area of triangle $ABC$, whose medians satisfy the inequalities $t_{a} \leqq 2, t_{b} \leqq 3, t_{c} \leqq 4$.

(Pavel Novotný)
Solution. Let $T$ be the centroid of triangle $ABC$ and $K, L, M$ the midpoints of sides $BC, CA, AB$. The medians divide triangle $ABC$ into six smaller triangles of equal area: For example, triangle $AMT$ has side $|AM|=\frac{1}{2} c$ and its height to side $AM$ has a length of $\frac{1}{3} v_{c}$, so $S_{AMT}=\frac{1}{2} \cdot \frac{1}{2} c \cdot \frac{1}{3} v_{c}=\frac{1}{6} \cdot \frac{1}{2} c \cdot v_{c}=\frac{1}{6} S_{ABC}$, and we get the same result analogously for the remaining five triangles.

The given problem is thus equivalent to determining the maximum possible area of one of the six smaller triangles - the result just needs to be multiplied by six.

![](https://cdn.mathpix.com/cropped/2024_04_17_3b5139769eeeeb546b50g-3.jpg?height=522&width=709&top_left_y=1672&top_left_x=682)

Fig. 1

Consider, for example, triangle $ATL$ (Fig. 1). For its two sides, we have

$$
|AT|=\frac{2}{3} t_{a} \leqq \frac{4}{3}, \quad|TL|=\frac{1}{3} t_{b} \leqq 1
$$

Therefore, for its area, we get

$$
S_{ATL}=\frac{1}{2}|AT| \cdot|TL| \cdot \sin |\angle ATL| \leqq \frac{1}{2} \cdot \frac{4}{3}=\frac{2}{3} .
$$

This proves that the area of triangle $ABC$, whose medians satisfy the given inequalities, cannot be greater than $6 \cdot \frac{2}{3}=4$. Moreover, the equality $S_{ATL}=\frac{2}{3}\left(\text{i.e., } S_{ABC}=4\right)$ occurs precisely when $t_{a}=2, t_{b}=3$ and $|\angle ATL|=90^{\circ} .{ }^{1}$

We can construct a triangle $ABC$ with these properties: First, we draw a right triangle $ATL$ in which we know the lengths of both legs $|AT|=\frac{4}{3},|TL|=1$, and then we construct point $C$ as the image of point $A$ under central symmetry with center $L$ and point $B$ as the image of point $L$ under a homothety with center $T$ and coefficient -2 (Fig. 2). It remains to verify that in such a triangle $ABC$, $t_{c} \leqq 4$.

![](https://cdn.mathpix.com/cropped/2024_04_17_3b5139769eeeeb546b50g-4.jpg?height=443&width=671&top_left_y=612&top_left_x=698)

Fig. 2

The length $t_{c}$ can be calculated in various ways. For example, in the right triangle $ABT$, the hypotenuse $AB$ has a length, according to the Pythagorean theorem,

$$
|AB|=\sqrt{|AT|^{2}+|TB|^{2}}=\sqrt{\frac{16}{9}+4}=\sqrt{\frac{52}{9}}=\frac{2}{3} \sqrt{13},
$$

so the radius of the Thales circle over the diameter $AB$ is

$$
|MT|=\frac{1}{2}|AB|=\frac{1}{3} \sqrt{13} .
$$

Thus, $t_{c}=3|MT|=\sqrt{13}<4 .^{2}$

Answer. The maximum possible area of triangle $ABC$ is 4.