\left[\begin{array}{l}\text { [Properties of medians. Center of gravity of a triangle.] ] } \\ {[\text { Cosine theorem }}\end{array}\right]

Let $O$ be the center of an equilateral triangle $ABC$, with a side length of 10. Point $K$ divides the median $BM$ of triangle $BOC$ in the ratio 3:1, counting from point $B$. Which is greater: $BO$ or $BK$?
Apply the Law of Cosines to triangle ВОМ.

## Solution

Let $D$ be the midpoint of $A C$. Then

$$
\begin{gathered}
B O=\frac{2}{3} B D=\frac{2}{3} D C \operatorname{tg} 60^{\circ}=\frac{10 \sqrt{3}}{3} \\
O M=\frac{1}{2} O C=\frac{1}{2} B O=\frac{5 \sqrt{3}}{3}
\end{gathered}
$$

$$
\begin{gathered}
B M^{2}=B O^{2}+O M^{2}-2 B O \cdot O M \cos 120^{\circ}= \\
=\frac{100}{3}+\frac{25}{3}+\frac{50}{3}=\frac{175}{3} \\
B K=\frac{3}{4} B M=\frac{3}{4} \sqrt{\frac{175}{3}}=\frac{5 \sqrt{21}}{4}
\end{gathered}
$$

Since $\frac{10 \sqrt{3}}{3}>\frac{5 \sqrt{21}}{4}$, then $B O>B K$.

![](https://cdn.mathpix.com/cropped/2024_05_06_96d74c2697cd905bd948g-04.jpg?height=414&width=518&top_left_y=502&top_left_x=770)

Answer

$B O>B K$.