Within a farmer's pastures, there is a oneacre tract of land shaped like a right triangle. At the midpoint of the hypotenuse is a post, to which a dog is tethered with just enough rope to reach the endpoints of the hypotenuse. There are also posts at the midpoints of the legs of the triangle; to each of which is tethered a sheep with just enough rope to reach the endpoints of its respective leg of the triangle.
How much space do the sheep have to graze in (collectively) without having to worry about the dog reaching them?
The dog's circle will indeed intersect the right angled vertex as well as the other two vertices and thus the endpoints of the sheep's two lines will lie on the circle.
By the pythagorean theorem, the sum of the squares on the two sides will equal the square on the hypotenuse. As the areas of any two similar shapes are proportional to the square of a corresponding side, this applies to semicircles erected on the outside of each side of the triangle.
The inner semicircle of the dog's circle is of course equal to the outer semicircle, and further, encompasses the triangle plus a two small areas, one cut out of each of the sheep's outer semicircle.
Let t be the area of the triangle; let s be the area of the outer semicircle on the hypotenuse, which is equal to the inner semicircle on the hypotenuse as well as the sum of the two outer semicircles on the legs; let o be the area of overlap between the dog's (hypotenuse's) inner semicircle and the combined pair of outer semicircles on the legs (the sheep's semicircles). As the sheep's inner semicircles are completely covered by the dog's circle, only the outer semicircles count. Then:
s = t + o (considering s as the area of the dog's semicircle)
so
so = t
But s, being the total of the sheep's outer semicircles, minus the area of those semicircles' overlap with the dog's semicircle, is the area the sheep have free of the dog, so that is t, which in this case is 1 acre.

Posted by Charlie
on 20031112 15:12:18 