Further correct solutions were also received from Silvain Vinassac, Al Zimmermann, Burkart Venzke, Philippe Fondanaiche, William Webb.

The easiest solution was presented by Nathan Pauli.  It has the advantage of showing that the fact that the triangle is a right triangle is irrelevant since that fact is never used in the solution!

Consider the diagram on the right. The fence, f,  is in blue; the units for the lengths of the sides are in hundreds of feet. The area of the entire field is 6 square units, so the fence is to subdivide the field into parts of 3 square units each.  Look at the point C which is b units from the corner labeled A.  The fence goes to a point labeled B which is c units from A.  The angle CAB is labeled q; this gives tan(q) = 3/4, sin(q) = 3/5, and cos(q) = 4/5.

Now, the area of triangle ABC, having base AB and altitude drawn from C, is cbsin(q) / 2, so cbsin(q) / 2 = 3 and so c = 10/b.  By Pythagoras, f ² = x² + h², while b + x = c cos(q) gives x = 8/b - b and h = 6/b.  Thus

A few of points to note:

First, to be totally at ease with this solution, you also need to verify that this minimum is less than what you would get working from the other sides, but it's fairly clear that the shortest fence will have to join these two sides of the field. Working out the algebra verifies this.

Second, since c = 10/b it's also the case that c = Ö10 so the smaller triangle will turn out to be isosceles; in particular, the fence will not form a right angle with any of the sides of the field. Given the solution, the diagram above doesn't accurately depict the location of the fence!

Finally, if the fence is not required to be straight, one can do better than the solution above. A shorter fence can be achieved if you build it in the shape of an arc of circle. This brings up the interesting question: What is the absolutely shortest possible fence that bisects the area of the field?

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Page last updated 14 May 1999.