1.1. A rope is stretched between the tops of two poles of height 11 and 15 meters, which are at a distance of 9 m from each other. A weight is attached to the rope and released to slide until the weight is at its lowest point. At what height will the weight be then?
1.1. The weight will descend to the lowest possible height. Let points $A, B, P, Q$ be marked as in the figure. Let $C$ be a point on $AP$ such that $BC=15 \mathrm{~m}$. The perpendicular bisector of segment $AC$ intersects segment $BC$ at point $M$, which is the equilibrium position of the weight.

Let's prove this.

![](https://cdn.mathpix.com/cropped/2024_06_04_ea7d677e29bb1e055c11g-26.jpg?height=450&width=321&top_left_y=749&top_left_x=454)

Let $M'$ be the point where the weight stops. Suppose $M'$ is lower than point $M$, and let $C'$ be a point on $AP$, different from $A$, such that $AM' = M'C'$. Then,

$AM' + M'B = C'M' + M'B = 2C'B > CB = 15 \mathrm{~m}$

(since $C'$ is lower than $C$).

Thus, $M$ is the desired position, and the desired height is easily calculated: $\quad BD=12, \quad CP=3 \Rightarrow MN=7$.