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A horse's rope (Posted on 2003-12-05) Difficulty: 2 of 5
Two ranchers shared a circular corral of 10-meter radius. One rancher wanted to tie his horse to one point on the fence and allow him to graze on just half the area of the corral. How long should the horse's rope be to make the portion available to this horse be equal to half the area of the corral?

See The Solution Submitted by Antonio    
Rating: 3.3333 (6 votes)

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Solution solution | Comment 3 of 8 |
The area to be reached by the horse is the intersection of the contents of two circles. For simplicity of calculations, let's consider the radius of the corral to be 1 unit; then we can scale up by a factor of 10 meters later. We need to find the radius of the second circle so that the area of intersection is 1/2 the area of the whole first circle (the corral).

Call the radius of the second circle r, which is to be found.

Form a triangle with vertices at the center of the corral, the center of the second circle (the hitching point), and one of the two points of intersection of the two circles. Call the angle at the center of the corral θ and that at the hitching point α. By the law of sines, since we have taken the corral's radius to be 1, r/sin θ = 1/sin α, or r=sin θ/sin α. But by the fact that the side of the triangle with length r is a chord of the corral's circle, θ = 2 arcsin(r/2). Going back to r=sin θ/sin α, we get α = arcsin(sin θ / r).

The area of intersection is the sum of the sectors of the two circles demarcated by the intersection points of the two circles minus the area where they overlap, which is two triangular areas -- the same triangle mentioned in the above paragraph, and the symetrically opposite one. We can take an altitude of one of them as sin α (which is the same as sin θ), and the base as r. As there are two of them, their combined area is r sin θ.

So the area of overlap is r²α + θ - r sin θ when angles are measured in radians

We want this to come out to pi/2.

Using Excel, allocate a cell for r, say A2, then cells for θ (B2), α (C2), the area in D2 and the area divided by pi in E2:
B2: =2*ASIN(A2/2)
C2: =ASIN(SIN(B2)/A2)
D2: =A2*A2*C2+B2-A2*SIN(C2)
E2: =D2/PI()

Then use solver to make E2 = .5 by changing cell A2.

The result is 1.158728483, which, when scaled up by a factor of 10 meters, is 11.58728483

The result is checked by using a simulation with this value:
  x = 2 * RND(1) - 1
  y = 2 * RND(1) - 1
  IF x * x + y * y < 1 THEN
    ct1 = ct1 + 1
    IF (x - 1) * (x - 1) + y * y < 1.158728 * 1.158728 THEN ct2 = ct2 + 1
    PRINT ct2 / ct1

and the ratio does indeed hover around .5.

  Posted by Charlie on 2003-12-05 11:25:56
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