Inconvenient points

First note that inconvenient points are not limited to the ground.

Any point $Q$ on the sphere with $AB$ as diameter (apart from the points $A,B$ themselves) subtends a right angle - that is, $\angle AQB=90^\circ$ . This is the boundary of the set inconvenient points.

We want the area of the intersection of this sphere with the ground.

The sphere has radius $R=\frac12 \sqrt{(4.4-3.1)^2+8.4^2}=4.25$

and its centre is at a height $h=\frac{4.4+3.1}{2}=3.75$

The radius of the circular section made by the ground is then $r=\sqrt{R^2-h^2}=2$

and hence its area is $\boxed{4\pi}$ .

This region of inconvenient points in 3-D space is a sphere where the line $AB$ is an axis through the center of the sphere. Therefore the radius of the sphere $R = \dfrac {\sqrt{8.4^2+(4.4-3.1)^2}}2 = 4.25 \ \rm m$ . The center $O$ of the sphere is horizontally midway between the two poles and $h = \dfrac {4.4+3.1}2 = 3.75 \ \rm m$ above the ground. The portion of the sphere underground would have a top radius of $r = \sqrt{R^2-h^2} = \sqrt{4.25^2-3.75^2} = 2 \ \rm m$ .

Therefore the area of the inconvenient region is $\pi r^2 = 4 \pi \implies k = \boxed 4$ .

Let $A'$ be $(0, 0, 0)$ , $A$ be $(0, 0, 4.4)$ , $B'$ be $(8.4, 0, 0)$ , $B$ be $(8.4, 0, 3.1)$ , and $P$ be $(x, y, 0)$ .

Now $\cos \angle APB = \frac{\overrightarrow{PA} \cdot \overrightarrow{PB}}{|\overrightarrow{PA}| |\overrightarrow{PA}|} \leq \cos 90° = 0$ , so $\overrightarrow{PA} \cdot \overrightarrow{PB} \leq 0$ , or $(-x, -y, 4.4) \cdot (8.4 - x, -y, 3.1) \leq 0$ , which rearranges to $(x - 4.2)^2 + y^2 \leq 4$ , a circle with a radius of $2$ .

Therefore, the area of the region is $4\pi$ , and $k = \boxed{4}$ .