In a large and flat grass field, there is a 30x30 feet square barn facing due North. Straight from the NorthEast corner of the barn is a fence running due East 40 feet long. There is a horse tied on the SouthEast corner of the barn with a rope that will allow him to eat grass in any direction up to 100 feet. The horse can't eat under the barn, and the rope can't pass through the barn or through the fence. The horse can walk around the barn or the fence, but is limited by the length of the rope.
How many square feet of grass can the horse reach?
(I found this problem pretty interesting, but I am not the author and have no idea who is the author).
The horse's rope is completely free (unblocked) between the direction of directly west, through south and east, up to arctan(3/4) north of east. When measured in degrees this means that the fraction of a full circle of radius 100 that this is, is (180+atn(3/4))/360, so
Part 1: pi*100^2*(180+atan(3/4))/360 = 18925.4688119154
There is also a right triangular area between the east side of the barn and the fence, with area 30*40/2, giving
Part 2: 600
On the left (west) side of the barn, there is 100-30=70 feet of rope free to pivot a quarter circle from directly west to directly north, giving
Part 3: pi*70^2 / 4 = 3848.4510006475
The remaining two pivots of the possible positions of the free end of the rope, at the northwest corner of the barn, with 40 feet of rope free, and at the eastern end of the fence, with 50 feet of rope free, allow the two swept areas to overlap north of the northeast corner of the barn, where the north side of the barn meets the fence. So the actual covered area, to avoid counting the overlap area twice, can be broken down into two circular sectors (stopping at the point where the ends of the rope, or circumferences of the circles, meet) and one triangular area formed by the final radii used (where they meet) and the line formed by the north side of the barn and the fence. These are the final three parts. First we need to figure at what heading off of north that each of the radii will stop at, by solving the triangle of the radii and barn face/fence.
That triangle has sides of 40, 50 and 70 feet, corresponding to the two remaining free ends of the rope (west and east positions) and the combined barn side and fence length.
First of all, we can get the area of this triangle via Heron's formula:
Part 4: sqrt(80*40*30*10) = 979.795897113271
Using the law of cosines and a length unit of 10-feet, the angle on the west side of this triangle is arccos((4^2+7^2-5^2)/56), while that on the right is arccos((5^2+7^2-4^2)/70). The deviations of the radii from north are the complements of these angles and so need to use arcsin rather than arccos.
The 40-foot radius at the west end is simpler in that it goes from directly north:
Part 5: pi*40^2*arcsin((4^2+7^2-5^2)/56)/360 = 636.482362787628
Since the arc on the right extends not only to the west of north, to the triangle, but also to the east of north, to pick up from where the major area (part 1) ended, its formula is the sum of these two parts:
Part 6: pi*50^2*(arcsin((5^2+7^2-4^2)/70) + atan(4/3)) / 360 = 2379.80814012789
The total comes out to 27370.0062125917 square feet.
Posted by Charlie
on 2005-04-03 04:39:49