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).

I used the same basic approach of breaking up the area into chunks as the other two folks, and I got the same answer as Charlie, but I solved it just a little differently and thought it might help some people out.

Ok, so I wanted to break this up into 5 sections, but two of those sections overlap. So I’ll actually have to analyze 6 sections.

The first section (A) is bounded by where the horse can reach with all 100 ft of rope taut. So he can move West 100 ft from his anchor point, but cannot move NE without wrapping the rope on the SW corner of the barn. From there he can walk in a circle CCW until his rope touches the end of the fence. At that point the angle his rope makes with the "horizontal" (E-W line) is atan(3/4) = 0.6435011 radians. So that means Section A is a wedge of a circle (of radius 100 ft) of angle pi+0.6435011. So AreaA = pi*100^2*( pi+0.6435011)/(2pi).
AreaA = 5000(pi+0.6435011)

Section B is the triangle bounded by the barn, the fence, and the border of Section A (which is the diagonal connecting the anchor and the fence post). In other words, Section B is a triangle of base=40 and height =30. So AreaB = 40*30/2.
AreaB = 600

Section C is bounded by where the horse can reach around the West side of the barn with 70 ft (100-30) of rope taut. So he can move from having the rope perfectly horizontal to perfectly vertical before he would start to wrap around the NW corner of the barn, which is a quarter circle. So AreaC = pi*70^2*1/4.
AreaC = 1225pi

Section D is bounded by where the horse can reach around the North side of the barn with 40 ft (70-30) of rope taut. So he can move from having the rope perfectly vertical to perfectly horizontal where he runs into the fence, which is also a quarter circle. So AreaD = pi*40^2*1/4.
AreaD = 400pi

Section E is bounded by where the horse can reach around the end of the fence with 50 ft of rope (the barn and fence create a 3-4-5 triangle, so we know 50 ft of rope is connecting the anchor to the fence, and 100ft-50ft = 50ft). So Section E is a wedge of a circle (of radius 50 ft) of angle pi-0.6435011. So AreaE = pi*50^2(pi-0.6435011)/(2pi).
AreaE = 1250(pi-0.6435011)

So, so far the area the horse can cover is

A = 5000(pi+0.6435011) + 600 + 1225pi + 400pi + 1250(pi-0.6435011) – overlap.
A = 7875pi + 3750(0.6435011) + 600 – overlap.

Now comes the harder part. Sections D and E overlap a bit, the area of which I will call Section F. The way I will find out this area is by doing the following: First, I will define a coordinate plane, with center at the NW corner of the building (so the coordinates of the centers of the two arcs are 0,0 and 70,0). Second, I will find the coordinates of the intersection of the two arcs that bound each section. Third, I can use the y coordinate of the intersection point to find the area of a section of a circle bounded by a chord of (2y), and I will do this for each circle. The area of these two sections, divided by 2, will be the area of Section F.

The equation for the arc for Section C is x^2 + y^2 = 40^2, and the equation for the arc for Section D is (70-x)^2 + y^2 = 50^2.

So y^2 = 40^2 – x^2 = 50^2 – (70-x)^2
40^2 – x^2 = 50^2 – (70-x)^2 = 50^2 – (70^2 – 140x + x^2)
70^2 – 50^2 + 40^2 = 140x
4000 = 140x
x = 200/7

y^2 = 40^2 – x^2 = 1600 – 40000/49 = 38400/49
y = sqrt(38400/49) = sqrt(6400*6/49) = 80/7*sqrt(6)

So a chord length of 2y in the 40 ft circle would bound an area of
40^2[2arcsin(2y/2/40) – sin(2arcsin(2y/2/40))]/2
800[2arcsin(y/40) – sin(2arcsin(y/40))]
But we only want half this area, so:
AreaF1 = 400[2arcsin(y/40) – sin(2arcsin(y/40))]

And a chord length of 2y in the 50 ft circle would bound an area of
50^2[2arcsin(2y/2/50) – sin(2arcsin(2y/2/50))]/2
1250[2arcsin(y/50) – sin(2arcsin(y/50))]
Again, we only want half this area, so:
AreaF2 = 625[2arcsin(y/50) – sin(2arcsin(y/50))]

So the total area is
A = 7875pi + 3750(0.6435011) + 600 - 400[2arcsin(y/40) – sin(2arcsin(y/40))] - 625[2arcsin(y/50) – sin(2arcsin(y/50))]
With y = 80/7*sqrt(6)

A = 27370.00621

Posted by nikki on 2005-04-04 16:38:36