2. (15 points) Two medians of a triangle, equal to 18 and 24, are perpendicular to each other. Find the length of the third median of this triangle.
Solution. Let $\mathrm{AP}=18$ and $\mathrm{BH}=24$. We will find the length of the median $\mathrm{CM}$. By the property of medians in $\triangle A B C$ we have:

$A O: O P=B O: O H=2: 1 \Rightarrow A O=$

$=\frac{2}{3} A P=\frac{2}{3} \cdot 18=12 ; B O=\frac{2}{3} B H=\frac{2}{3} \cdot 24=16$.

Then, by the Pythagorean theorem in the right-angled

$\triangle A O B: A B=\sqrt{A O^{2}+B O^{2}}==\sqrt{12^{2}+16^{2}}=20$. Since $O M$ is the median of this triangle, drawn from the vertex of the right angle,

$O M=\frac{1}{2} A B=\frac{1}{2} \cdot 20=10$. Considering that $C O: O M=2: 1$,

![](https://cdn.mathpix.com/cropped/2024_05_06_aadf511cc872419fdfecg-02.jpg?height=480&width=365&top_left_y=357&top_left_x=1508)

we get: $C M=3 O M=3 \cdot 10=30$.

Answer: 30.

| Criterion | Points |
| :--- | :---: |
| The correct answer is obtained with justification. | 15 |
| The solution contains a computational error, possibly leading to an incorrect answer, but the solution has a correct sequence of all steps. | 10 |
| In the solution of the problem, one or two formulas are correctly written, which could be the beginning of a possible solution. | 5 |
| The solution does not meet any of the criteria described above. | 0 |
| Maximum score | 15 |