Home > Geometry, Olympiad, Problem Solving > Application of the isoperimetric inequality

## Application of the isoperimetric inequality

Find the shortest curve which splits an equilateral triangle with edge of length $1$ into two regions having equal area.

Hint: This is an awsome problem. Try to transform, or modify the figure such that you can use the isoperimetric inequality. The result is unusual and you cannot really expect it intuitively.
Consider the curve which splits the equilateral triangle in two regions of equal area. Since we are looking for the shortest curve, we can assume there are no autointersections and the curve touches the boundary of the equilateral triangle in two points lying on two different edges.
Then we make another 5 equilateral triangles such that the 6 congruent curves form a closed curve thus dividing the resulting regular hexagon in two regions of equal area. By the isoperimetric problem, the shortest curve which encloses a given area is the circle, and therefore our curve is an arc of a circle equal to $\frac{1}{6}$ of the length of the circle which encloses the area equal to $\frac{1}{2}\cdot 6 \cdot \frac{\sqrt{3}}{4}=\frac{3\sqrt{3}}{4}$.