Vietnam Mathematical Olympiad

1) From the construction of the circles $(O_1)$ and $(O_2)$, we have:
$$\angle CAD = \angle DCB \ (\text{subtend the arc } CD \text{ of } (O_1)) \quad (1)$$
$$\angle ACD = \angle CBD \ (\text{subtend the arc } CD \text{ of } (O_2)) \quad (2)$$
Therefore $\triangle DAC \sim \triangle DCB$ and so
$$\frac{DA}{DC} = \frac{DB}{DB}, \text{ i.e. } CD^2 = DA \cdot DB \quad (3)$$

The law of cosines for triangle $ADB$ gives
$$AB^2 = AD^2 + DB^2 - 2AD \cdot DB \cos(\angle ADB), \text{ so} \\ AB^2 \geq 2AD \cdot DB(1 - \cos(\angle ADB)) \quad (4)$$
(3) and (4) imply that
$$AB^2 \geq 2CD^2(1 - \cos(\angle ADB)) \quad (5)$$
The law of sines for triangle $ABC$ gives $AB = 2R \sin(\angle ACB)$. So, from (5), we deduce that
$$2R^2 \sin^2(\angle ACB) \geq CD^2(1 - \cos(\angle ADB)). \quad (6)$$
From (1) and (2), we see that
$$\begin{aligned} \angle CAD + \angle ACD &= \angle DCB + \angle CBD = \angle CAD + \angle CBD = \\ &= \angle DCB + \angle DCA = \angle ACB \end{aligned} \quad (7)$$
Therefore
$$\angle ADC = \angle BDC = 180^\circ - \angle ACB \quad (8)$$
Moreover, it is easily seen that:
- $C$, $D$ lie on the same side with respect to the line $AB$
$$\Leftrightarrow \angle CAD + \angle CBD < \angle CAB + \angle CBA;$$
- $C$, $D$ lie on distinct sides with respect to the line $AB$
$$\Leftrightarrow \angle CAD + \angle CBD > \angle CAB + \angle CBA.$$
So, from (7), we deduce that:
- If $\angle ACB$ is acute then $C$, $D$ lie on the same side with respect to the line $AB$;
- If $\angle ACB$ is obtuse then $C$, $D$ lie on distinct sides with respect to the line $AB$.
Therefore, from (8),
when $\angle ACB$ is acute,
$$\angle ADB = 360^\circ - (\angle ADC + \angle CDB) = 2\angle ACB$$
when $\angle ACB$ is obtuse,
$$\angle ADB = \angle ADC + \angle CDB = 360^\circ - 2\angle ACB.$$
So, in all cases, $\cos(\angle ADB) = \cos(2\angle ACB)$.
Hence, from (6), we have $CD \le R$,
QED.

2) From the preceding results, it is easily seen that:
$$- CD \text{ is the angle bisector of } \angle ADB; \qquad (9)$$
$$- O, D \text{ lie on the same side with respect to the line AB;} \qquad (10)$$
$$- \angle ADB = \angle AOB. \qquad (11)$$
From (10), (11) we deduce that: when $C$ moves on $(O)$ (but $C$ distinct from $A, B$), the point $D$ moves on the arc $AOB$ of the fixed circle $(AOB)$. Therefore, (9) proves that the line $CD$ passes through a fixed point, namely the middle point $M$ of the arc $AB$ not containing $O$ of the circle $(AOB)$.