
2. Find the greatest possible area of a triangle ABC with medians satisfying $t_{a} \leqslant 2$, $t_{b} \leqslant 3, t_{c} \leqslant 4$.

(Pavel Novotný)


Solution. Let $T$ be the centroid of $A B C$ and $K, L, M$ be the midpoints of $B C$, $C A, A B$. Medians cut $A B C$ into six smaller triangles, each with the same area: for example in the triangle $A M T$ we have $A M=\frac{1}{2} c$, its altitude through $T$ is $\frac{1}{3} v_{c}$ long, that is $S_{A M T}=\frac{1}{2} \cdot \frac{1}{2} c \cdot \frac{1}{3} v_{c}=\frac{1}{6} \cdot \frac{1}{2} c \cdot v_{c}=\frac{1}{6} S_{A B C}$. Analogously for the other triangles.

![](https://cdn.mathpix.com/cropped/2024_04_17_010c1b06d84ea080b2e5g-13.jpg?height=375&width=503&top_left_y=1389&top_left_x=451)

Fig. 1

Thus we will seek for the greatest possible area of one of the triangles, say $A T L$ (Fig. 1), and then we multiply the result by six. There is

$$
A T=\frac{2}{3} t_{a} \leqslant \frac{4}{3}, \quad T L=\frac{1}{3} t_{b} \leqslant 1
$$

Therefore we can constrain the area:

$$
S_{A T L}=\frac{1}{2} A T \cdot T L \cdot \sin \angle A T L \leqslant \frac{1}{2} \cdot \frac{4}{3}=\frac{2}{3} .
$$

Thus the area of the $A B C$ can be at most $6 \cdot \frac{2}{3}=4$, where the equality holds iff $t_{a}=2, t_{b}=3$ and $\angle A T L=90^{\circ}$.

There is a triangle satisfying the conditions with the area 6 indeed: first we construct right triangle $A T L$ with the legs $A T=\frac{4}{3}$ and $T L=1$. Then $C$ is the symmetric image of $A$ with respect to $L$ and $B$ is the image of $L$ under the homothety with center $T$ and coefficient -2 (Fig. 2). It is easy to count the length of $A B$. For

![](https://cdn.mathpix.com/cropped/2024_04_17_010c1b06d84ea080b2e5g-14.jpg?height=325&width=486&top_left_y=661&top_left_x=512)

Fig. 2

example the theorem of Pythagoras in $A B T$ gives

$$
A B=\sqrt{A T^{2}+T B^{2}}=\sqrt{\frac{16}{9}+4}=\sqrt{\frac{52}{9}}=\frac{2}{3} \sqrt{13} .
$$

Since $M$ is on the Thales' circle with the diameter $A B$, there is $M T=\frac{1}{2} A B=\frac{1}{3} \sqrt{13}$. Then $t_{c}=3 \cdot M T=\sqrt{13}<4$.

Conclusion. The greatest possible area of the triangle $A B C$ is 4 .
