Find the curve of minimum length that divides an equilateral triangle into two figures of equal area.

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Consider a curve dividing an equilateral triangle $ABC$ into two figures of area $S / 2$. There are two cases: either the curve separates one of the vertices of the triangle (for definiteness, vertex $A$) from the opposite side, or the curve is closed. In the second case, according to problem 22.BIs14, the length of the curve does not exceed $\sqrt{2 \pi S}$. Now consider the first case. The images of the curve under successive reflections about the lines $AC, AB_{1}, AC_{2}, AB_{2}$, and $AC_{1}$ (see figure) form a closed curve bounding a figure of area $3S$. Therefore, the desired curve is an arc of a circle with radius $\sqrt{3 S / \pi}$ centered at point $A$. Its length is $\sqrt{\pi S / 3} < \sqrt{2 \pi S}$.

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