## [ Product of lengths of chord segments and lengths of secant segments ]

The radii of two concentric circles are in the ratio $1: 2$. A chord of the larger circle is divided into three equal parts by the smaller circle. Find the ratio of this chord to the diameter of the larger circle.
Apply the theorem of the products of segments of intersecting chords.

## Solution

Let the chord $A D$ of the larger circle with center $O$ intersect the smaller circle at points $B$ and $C$ ($B$ between $A$ and $C$). Consider the diameter $P Q$ of the larger circle passing through point $C$ ($C$ between $O$ and $P$). Let $O C=x$, $A B=B C=C D=a$. Then

$$
C D \cdot A C=P C \cdot C Q \text {, or } a \cdot 2 a=x \cdot 3 x \text {, }
$$

from which $\frac{a}{x}=\sqrt{\frac{3}{2}}$. Therefore,

$$
\frac{A D}{P Q}=\frac{3 a}{4 x}=\frac{3}{4} \cdot \frac{a}{x}=\frac{3}{4} \sqrt{\frac{2}{3}}=\frac{3 \sqrt{6}}{8}
$$

Let the chord $A D$ of the larger circle with center $O$ intersect the smaller circle at points $B$ and $C$ ($B$ between $A$ and $C$). Consider the diameter $P Q$ of the larger circle passing through point $C$ ($C$ between $O$ and $P$). Let $O C=x$, $A B=B C=C D=a$. Then

$$
C D \cdot A C=P C \cdot C Q \text {, or } a \cdot 2 a=x \cdot 3 x,
$$

from which $\frac{a}{x}=\sqrt{\frac{3}{2}}$. Therefore,

$$
\frac{A D}{P Q}=\frac{3 a}{4 x}=\frac{3}{4} \cdot \frac{a}{x}=\frac{3}{4} \sqrt{\frac{2}{3}}=\frac{3 \sqrt{6}}{8}
$$

Let the chord $A D$ of the larger circle with center $O$ intersect the smaller circle at points $B$ and $C$ ($B$ between $A$ and $C$). Consider the diameter $P Q$ of the larger circle passing through point $C$ ($C$ between $O$ and $P$). Let $O C=x$, $A B=B C=C D=a$. Then

$$
C D \cdot A C=P C \cdot C Q \text {, or } a \cdot 2 a=x \cdot 3 x,
$$[^0]

$$
\frac{A D}{P Q}=\frac{3 a}{4 x}=\frac{3}{4} \cdot \frac{a}{x}=\frac{3}{4} \sqrt{\frac{2}{3}}=\frac{3 \sqrt{6}}{8}
$$

## Answer

$\frac{3 \sqrt{6}}{8}$.