Long short

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This solution is from Tony Cardell, age 15, State College Area High School, PA, USA.

We investigate quadrilaterals inscribed in the unit circle. The sides of the quadrilateral $s_1$, $s_2$, $s_3$, $s_4$, subtend angles $\theta_1, \theta_2, \theta_3, \theta_4$ at the centre of the circle where $\theta_1 \leq \theta_2 \leq \theta_3 \leq \theta_4$. Now $\theta_1 \leq 90^\circ$ so the the largest value of

First investigating an inscribed quadrilateral where all sides are equal, $s = s_1 = s_1 = s_3 = s_4$ , which is a square with diagonal of length 2 units, we find that the side length of this square is given by $2s^2 = 4$, so that

$s=\sqrt{2}$.

Now imagine an equilateral triangle inscribed in the unit circle. The angle at the centre is ${120}^{\circ }$

Side 3 can approach length 2 because we can make a quadrilateral with sides three and four very close to 2 and sides one and two very close to zero. As sides one and two approach zero, sides three and four clearly approach 2, the diameter of the unit circle.

Finally, at its maximum, side 4 can be 2 when side 4 is the diameter of the unit circle. The other sides will fit into one of the halves of the circle because you can make them as small as you want. Any line longer than 2 units will extend outside the circle so side four cannot be longer than 2 units.