All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Getting Sheepish (Posted on 2003-11-12)
Within a farmer's pastures, there is a one-acre 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?

 No Solution Yet Submitted by DJ Rating: 4.1818 (11 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Mathematical / Graphical Solution | Comment 6 of 10 |
Let's call one leg of the triangle B (for Base), the other leg of the triangle L, and the hypotenuse H.

And let's call the area of the Dog's semi-circle D, the area of one sheep's semi-circle S1, the area of the other sheep's semi-circle S2, and the area of the triangular field T.

Since we know the field is a right triangle, we know that B^2 + L^2 = H^2

The dog's semi-circle intersects with all three vertices of the triangle (I'll prove this later on, I don't want to bog down the solution so soon). This is very important because otherwise the math won't work out nicely.

So now we can define the area of freedom for the sheep as:
A = T + S1 + S2 - D
A = T + pi/2*(B/2)^2 + pi/2*(L/2)^2 - pi/2*(H/2)^2
A = T + pi/8*(B^2 + L^2 - H^2)
A = T + pi/8*0
A = T = 1 acre

To prove that the Dog's semi-circle intersects with all three vertices, I can approach this one of two ways. First, there is some geometric proof that if you have a circle and draw it's diameter, then pick any point on the circumference (other than the end points of the diameter, of course) you will always get a right triangle with the diameter being the hypotenuse. Since the dog's post would be located at the center of the circle, and the circle represents the boundary of the dog's reach, we see that the vertex of the right angle will always be on this boundary.

The other approach is more graphical. On a cartesian plane, place the right angle vertex at (0,0), another vertex at (B,0) and the last vertex at (0,L) to represent the triangle. Clearly the posts of the sheep will be at (B/2,0) and (0,L/2). The coordinates of the midpoint of the hypotenuse is found by averaging the coordinates of the endpoints of the hypotenuse. So the X coordinate is the average of B and 0, which is B/2, and the Y coordinate is the average of 0 and L, which is L/2. This means the dog's post is exactly horizontal and exactly vertical to the other two posts. Well, draw a triangle with vertices at (0,0), (B/2,0), and (B/2,L/2). This is a right triangle, and the hypotenuse is sqrt[(B/2)^2 + (L/2)^2] = sqrt[(B^2 + L^2)/4] = sqrt(H^2/4) = H/2 which is exactly the length of the dog's tether. So again, the right angle vertex is just exactly on boundary of the dog's reach.
Edited on November 12, 2003, 5:34 pm
 Posted by nikki on 2003-11-12 17:33:15

 Search: Search body:
Forums (0)