Let $ABC$ be a triangle. The midpoints of the sides $BC$, $AC$, and $AB$ are denoted by $D$, $E$, and $F$, respectively.

The two medians $AD$ and $BE$ are to be perpendicular to each other and have lengths $\overline{AD}=18$ and $\overline{BE}=13.5$.

Calculate the length of the third median $CF$ of this triangle.

K. Czakler, GRG 21, Vienna
The three medians $A D, B E$, and $C F$ intersect at the centroid $S$. It is known that the centroid divides the medians in the ratio $2: 1$. Therefore, we get

$$
\overline{A S}=\frac{2}{3} \cdot \overline{A D}=12, \quad \overline{B S}=\frac{2}{3} \cdot \overline{B E}=9 .
$$

According to the problem, $A S B$ is a right triangle, and by the Pythagorean theorem, we have

$$
\overline{A B}=\sqrt{\overline{A S}^{2}+\overline{S B}^{2}}=\sqrt{12^{2}+9^{2}}=15 .
$$

![](https://cdn.mathpix.com/cropped/2024_04_17_9412ce3c68f39656e41ag-08.jpg?height=460&width=597&top_left_y=364&top_left_x=752)

Figure 3: Problem 4, Solution 1.

By the Thales' theorem, $S$ lies on the semicircle with diameter $A B$ and center $F$. Thus,

$$
\overline{S F}=\overline{F A}=\frac{1}{2} \cdot \overline{A B}=\frac{15}{2} .
$$

Since $\overline{F S}: \overline{S C}=1: 2$, we get

$$
\overline{F C}=3 \cdot \overline{F S}=\frac{45}{2}
$$

(C. Heuberger)

![](https://cdn.mathpix.com/cropped/2024_04_17_9412ce3c68f39656e41ag-08.jpg?height=466&width=583&top_left_y=1569&top_left_x=751)

Figure 4: Problem 4, Solutions 1a and 1b.