Which convex shape requires the longest slice to break in half?

Edit: This question is discussed at length in Chords halving the area of a planar convex set (Grüne et al., 2007), along with the question of the longest possible splitting chord. However, the paper (while establishing many other bounds) leaves the relation between area and minimal halving distance open, except to note that the $\sqrt[4]{3}$ upper bound is not tight.

Notably, the circle is not optimal, and any (non-circular) convex Zindler curve serves as a counterexample. The aforementioned paper cites a 2005 preprint On geometric dilation of closed curves, graphs and point sets (Dumitrescu et al., 2005) for this fact, where perusing Section 7 we can see that the "rounded triangle" $C_\triangle$ is conjectured to have maximal cutting length $h$ for a fixed area, at $$A(C_\triangle)=\sqrt{3}\left(\log(3) - \frac{\log(3)^2}{8} - \frac12\right)h^2$$ so $h\approx 1.135547\sqrt{A}$.

The curve $C_\triangle$ is pictured below, from Dumitrescu et al.'s paper. (The central region $M$ is not a hole, but the locus of the midpoints of area-splitting chords; we have $A(Z)=\frac\pi4h^2-2A(M)$ for any Zindler curve $Z$.)

                                    

My original answer, written before locating the above papers, is given below.

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Let our convex set be $C$. Note that such a cut is at most the width $w$ of the shape, because if we place $C$ between two parallel lines, then some cut orthogonal to those lines will cut $C$ in half, and its length must be at most the distance between the lines. (This argument extends to the same question for dividing $C$ into regions of equal perimeter, or for that matter most natural measures of "size" of a convex set.)

We can then use the fact that $w^2\le \sqrt{3}A$ among bounded convex planar sets; see for instance Inequalities For Convex Sets (Scott and Awyong, 2000), which cites Yaglom and Boltyanski's book Convex Figures. This gives us a minimal cut width of at most $\sqrt[4]{3}\approx 1.316$.

Note that since the extremal convex set for this inequality is the equilateral triangle, which has a cut width of $\sqrt{2}/3^{1/4}\approx 1.075$, it does not in fact surpass the circle's $\frac{2}{\sqrt{\pi}}\approx 1.128$, but at least provides a reasonable upper bound.

As an unrelated piece of evidence, I believe I can get a cut of length around $1.1038$ on the Releaux triangle of unit area, so that particular candidate (worth checking for any question about convex sets) seems to not quite beat the circle.

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.