6. Two parallel chords of a circle have lengths 24 and 32 respectively, and the distance between them is 14 . What is the length of another parallel chord midway between the two chords?
(1 mark)
某圓中兩條互相平行的弦的長度分別為 24 和 32，而它們之間的距離為 14 。
那麼，剛好位於該兩條弦中間且與其平行的另一條弦的長度是多少？
(1 分)
6. $2 \sqrt{249}$
6. 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. Since the perpendicular from the centre to a chord bisects the 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 also have $16^{2}+y^{2}=r^{2}$. Combining the two equations gives $12^{2}+x^{2}=16^{2}+y^{2}$, which upon rearranging 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 we have $x-y=14$, which gives $x+y=8$ and $y=-3$. This is clearly impossible.
- If the two chords lie on opposite halves of the circle, then we have $x+y=14$, 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, we find that its length is $2 \times \sqrt{\sqrt{265}^{2}-4^{2}}=2 \sqrt{249}$.