Two vertical poles stand 8.4m apart. AA' is 4.4m high, BB' is 3.1m high (A' and B' lying on the ground). A point P on the ground is defined to be a “convenient point”, if the viewing angle of points A and B from P is an acute one. If you move away from the poles, you can certainly find convenient points. There is a region of points on the ground, however, where all points are inconvenient.
Find the area of this region of inconvenient points.
The locus of points P meeting the criterion is the set of points within a sphere that has segment AB as a diameter.
So, as to use integers, I'll convert the measurements into decimeters. The two poles are 84 dm apart and differ in height by 13 dm. Points A and B are therefore 85 dm apart, which is therefore the diameter of the containing sphere, making its radius 42.5 dm.
The center of the sphere is 37.5 dm above the ground (the average of the heights of A and B). The radius of the circle that's the intersection of the surface of the sphere with the ground is then sqrt(42.5^2  37.5^2) = 20 dm.
The circular area sought is therefore pi * 400 dm ~= 1256.63706143592 dm^2 or about 12.5663706143592 square meters, being a circle of 2 meters radius, centered midway between the two poles.

Posted by Charlie
on 20210823 11:57:46 