A chord intersecting a diameter

Text

In a circle of radius $4$, a chord $AB$ of length $7$ meets a diameter $CD$ at a point $K$, dividing it into two parts, one triple the other. Determine $\overline{AK}$, $\overline{BK}$, $\overline{CK}$, $\overline{DK}$. What is the minimum length of the chord $AB$ for the problem to have solutions?

Solution

The diameter is $CD=2\cdot4=8$. Split into two parts with one triple the other: $x+3x=8\Rightarrow x=2$, so $\{CK,DK\}=\{2,6\}$. By the intersecting chords theorem (power of the point $K$): $AK\cdot BK=CK\cdot DK=12.$ With $AK+BK=AB=7$, the two lengths are roots of $t^2-7t+12=0$, i.e. $t=3$ and $t=4$. Hence $AK=4,\quad BK=3,\quad CK=2,\quad DK=6.$ Minimum chord. $K$ is at distance $\overline{OK}=4-2=2$ from the centre. The shortest chord through $K$ is the one perpendicular to $OK$: $AB_{\min}=2\sqrt{r^2-\overline{OK}^2}=2\sqrt{16-4}=4\sqrt3\approx6{,}93.$ $\boxed{AK=4,\ BK=3,\ CK=2,\ DK=6;\qquad AB_{\min}=4\sqrt3\approx6{,}93.}$