Problem — Shortest chord through an interior point

Text

Let $P$ be a point inside a circle with centre $O$ and radius $r$. Prove that, among all the chords passing through $P$, the shortest is the one perpendicular to the diameter through $P$. Then compute the length of this shortest chord in the case $r = 10$ and $\overline{OP} = 6$.

Solution

A generic chord through $P$ lies at a distance $h = \overline{OM}$ from the centre, where $M$ is the foot of the perpendicular from $O$ to the chord. In the right triangle $OMP$ (right-angled at $M$) the hypotenuse is $\overline{OP}$, so $h = \overline{OM} \le \overline{OP}$, with equality only when $M \equiv P$, that is when the chord is perpendicular to $OP$. The length of a chord at distance $h$ from the centre is $2\sqrt{r^2 - h^2}$, a decreasing function of $h$. The shortest chord therefore corresponds to the greatest value of $h$, namely $h = \overline{OP}$: it is the chord perpendicular to the diameter through $P$. With $r = 10$ and $\overline{OP} = 6$: $L_{\min} = 2\sqrt{r^2 - \overline{OP}^{\,2}} = 2\sqrt{100 - 36} = 2\sqrt{64} = 16.$ $\boxed{L_{\min} = 16}$