Minimum boundary when dividing eqilateral triangle in 4 equal sized parts

1. Sep 29, 2007

jonas.hall

So I have an equilateral triangle an I want to divide it in 4 parts, all having the same area. This can be done in a multitude of ways of course. But assuming it's a garden and the division is about putting up a fence, which division uses the least fencing?

Now I have two alternatives so far.

The first is to create a cirle in the middle and add three short segments from the circle to the midpoints of the sides. There should be a uniqe such solution and I just haven't bothered calculating it yet.

The second is, in my opinion more interesting. Cut of the corners with a curve symmetrical around the bisectors. Now if this curve was a straight line at right angles to the bisector it would be uniquely determined. Also if it was a circle centered in the vertex. After calculating some special cases one might be satisfied and pick the best but I was thinking one might set up a differential equation to find the best possible curve. But how would I set this up? Perhaps this should be posted in Differential equations instead?

2. Oct 1, 2007

mathman

Here's a simple alternative - I don't know if it is minimal, but it might be. Bisect each side and connect the three midpoints to form a triangle. You will have four equilateral triangles, each having 1/4 of the area.

3. Oct 2, 2007

jonas.hall

I tried that and the fencing needed is then exactly 0.5 of the total perimeter. Using a circle to cut of the corners lowers this to approximately 0.475. I have som argument that points towards thie circle being best, but no proofs or counterexamples.

Your example is actually the special case with a perpendicular line being the curve.

The circle bulges inward in the middle and comes out at the edges but because of the slope of the triangle the endpoints then come closer to each other, making the distance shorter.

I've been thinking about coshyp curves and parabolas but I wonder if the curve has to be orthogonal to any ray from the vertex, forcing it to be a circle. This might be so because this condition would minimize the distance for a given area locally, or infinitesimally. But I don't know if this holds "globally" i.e. when inegrated.