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Difficult math question
By farmersrice, history, 7 years ago, In English

Describe the curve of minimal length that divides an equilateral triangle into two sections of equal area.

I don't know the solution and would appreciate any help, thanks!

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7 years ago, # |
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If my intuition is correct, it’s a circular arc of length slicing off a sector around one of the triangle’s corners.

The circle is the shape with the smallest circumference for a fixed area, but apparently proving this is not trivial.

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7 years ago, # |
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Hello, yesterday I came across this same question on Quora. Someone in the comments gave a link to this pdf. The answer is on page 7.

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7 years ago, # |
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Without loss of generality, assume the area of the triangle to be equal to 1.

I will be using Isoperimetric inequality. Specifically in , the isoperimetric inequality states, for the length L of a closed curve and the area A of the planar region that it encloses, that , and that equality holds if and only if the curve is a circle. There are three possible cases:

Case 1: The curve is closed and inside the triangle.
Here . We have, , and the shortest curve is in this case of length .

Case 2: The curve has both its endpoints on the same side of the triangle.
Denote to be the equilateral triangle and suppose that , the endpoints of the curve lie on . Take the reflection of the triangle about the line and get a triangle . The two symmetric arcs which have endpoints in enclose an area of 1 (two halves). We have , and the shortest curve is in this case of length .

Case 3: The curve has its endpoints on two different sides of the triangle. Denote to be the equilateral triangle and suppose that , the endpoints of the curve lie on two different sides. Take six symmetric triangles in order to get a regular hexagon. Then . We have , and the shortest curve is in this case of length .

So the curve of minimal length that divides an equilateral triangle into two sections of equal area has the length equal to where is area of the equilateral triangle.

You can read more about it here.

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