11. Three Triangles (8th grade, 3 points). Inside triangle $ABC$, a random point $M$ is chosen. What is the probability that the area of one of the triangles $ABM$, $BCM$, and $CAM$ will be greater than the sum of the areas of the other two?
Answer: $0.75$.

Solution. Draw the midlines $A_{1} B_{1}, B_{1} C_{1}$, and $C_{1} A_{1}$ in the triangle. For the area of triangle $A C M$ to be greater than half the area of triangle $A B C$, point $M$ must lie inside triangle $A_{1} B C_{1}$. Similarly, two other cases are considered. The desired probability is

$$
\frac{S_{A B_{1} C_{1}}+S_{B_{C} C A_{1}}+S_{A_{1} B C_{1}}}{S_{A B C}}=\frac{3}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_142b35c881fc974fcd25g-07.jpg?height=485&width=465&top_left_y=1885&top_left_x=1452)

Fig. 5