TITLE: Which convex shape requires the longest slice to break in half?
QUESTION [5 upvotes]: Suppose you are given a convex unit area region of the plane, and you wish to minimize the length of a line segment that cuts the shape into two regions of equal area. Which shape results maximizes this length, i.e., is the hardest to slice into two equal regions. Is it the circle?
For generalizations, does anything change if we allow the slice to be any curve? What if we want to chop it into $n$ equal regions for $n>2$? What if we are minimizing the area of planes that slice higher dimensional regions in half? I'm curious in any generalizations along these lines.

REPLY [2 votes]: I'll only consider the particular case of an acute angled triangle with area $1$ and with angles $\alpha \le \beta \le \gamma$.
Let's first find the arc of circle inside an angle of angle $\phi$ and of area $\frac{1}{2}$.
If the radius of the arc is $r$ then we have
$$\frac{1}{2} r^2 \phi = \frac{1}{2}$$
so
$r= \frac{1}{\sqrt{\phi}}$,
and therefore the length of the arc is
$$r \phi = \sqrt{\phi}$$
Let us see when the arc corresponding to the angle $\alpha$ lies inside the triangle. This is true if and only if the distance from the vertex to the opposite side $h$ is $\ge \frac{1}{\sqrt{\alpha}}$.
Now, we calculate the area of the triangle
$$1 = \frac{1}{2} h^2 ( \cot \beta + \cot \gamma)$$
and therefore
$$h = \sqrt{\frac{2}{\cot \beta + \cot \gamma}}$$
and so the condition is
$$\frac{\cot \beta + \cot \gamma}{2}\le \alpha$$
Now, since the function $\cot $ is decreasing, we get
$$\frac{\cot \beta + \cot \gamma}{2}\le \cot \alpha$$
Therefore, if
$$\cot \alpha \le \alpha$$
(equivalently, $\alpha \ge \approx 49^{\circ}.\ldots$)
then we have the inequality.
Therefore, if the angle $\alpha$ is not too small, the shortest dividing line is of length $\sqrt{\alpha}$.
In the particular case of an equilateral triangle, the shortest line is $\sqrt{\frac{\pi}{3}} = 1.0233\ldots$
For the equilateral triangle, the radius halving the area equals
$$\frac{\sqrt{\pi}}{2 \sqrt[4]{3}}\cdot l= 0.6733\ldots \cdot l$$
where $l$ is the side of the triangle.