Consider a triangle ABC where

$$\mathrm { AB } = 10 , \quad \angle \mathrm { B } = 30 ^ { \circ }$$

and the radius of its inscribed circle O is 1.

(1) Set $a = \mathrm { BC }$ and $b = \mathrm { CA }$. Finding the area $S$ of the triangle ABC by two different methods, we have

$$S = \frac { \mathbf { A } } { \mathbf{B} } \mathbf { B } ,$$

and

$$S = \frac { \mathbf { C } } { \mathbf{C} } \mathbf { D } ( a + b + \mathbf { E C } ) .$$

Hence we obtain

$$b = \mathbf { G } a - \mathbf { H I } .$$

Since the relationship between $a$ and $b$ can also be expressed by the equation

$$b ^ { 2 } = a ^ { 2 } - \mathbf { J K } \sqrt { \mathbf { L } } a + \mathbf { M N O } ,$$

we have

$$a = \frac { \mathbf { P Q } - \mathbf { R } } { 3 } \sqrt { \mathbf { S } } , \quad b = \frac { \mathbf { T U } } { \mathbf{P} } \cdot \frac { \mathbf { V } } { 3 } \sqrt { \mathbf { W } } .$$

(2) Let D denote the point of intersection of the segment BC and the straight line which passes through the two points A and O. We denote the area of the triangle OBC by $S ^ { \prime }$. Since

$$S : S ^ { \prime } = \mathbf { X } : 1 ,$$

it follows that

$$\mathrm { AO } : \mathrm { OD } = \mathbf { Y } : 1 .$$