Hong Kong Preliminary Selection Contest

Let $r$ be the radius of the circle. Denote by $x$ and $y$ the distances from the two chords to the centre of the circle (see the figure). Since the perpendicular from the centre to a chord bisects this chord, by Pythagoras' Theorem we see that ${\left(\frac{24}{2}\right)}^2+x^2=r^2$, i.e. $12^2+x^2=r^2$. Similarly, we have $16^2+y^2=r^2$. Combining the two equations gives $12^2+x^2=16^2+y^2$, which, upon simplification, becomes $(x-y)(x+y)=x^2-y^2=112$. There are two possibilities.

If the two chords lie on the same half of the circle, then $x-y=14$, which gives $x+y=8$ and $y=-3$. This is clearly impossible.

If the two chords lie on opposite halves of the circle, then $x+y=14$, which gives $x-y=8$. Solving the equations, we get $x=11$ and $y=3$. Hence $r=\sqrt{265}$. It is now clear that the desired chord is at a distance of $\frac{11-3}{2}=4$ from the centre. Using Pythagoras' Theorem again, its length is $2\times \sqrt{{\sqrt{265}}^2-4^2}=2\sqrt{249}$.