Vietnam Mathematical Olympiad

1) From the construction of the circles $(O_1)$ and $(O_2)$, we have: $$\angle CAD=\angle DCB(\text{subtend the arc }CD\text{ of }(O_1))(1)$$ $$\angle ACD=\angle CBD(\text{subtend the arc }CD\text{ of }(O_2))(2)$$ Therefore $△DAC\sim △DCB$ and so $$\frac{DA}{DC}=\frac{DB}{DB},\text{ i.e. }C{D}^2=DA\cdot DB(3)$$ The law of cosines for triangle $ADB$ gives $$\begin{gathered} A{B}^2=A{D}^2+D{B}^2-2AD\cdot DB\cos (\angle ADB),\text{ so} \\ A{B}^2\ge 2AD\cdot DB(1-\cos (\angle ADB))(4) \end{gathered}$$ (3) and (4) imply that $$A{B}^2\ge 2C{D}^2(1-\cos (\angle ADB))(5)$$ The law of sines for triangle $ABC$ gives $AB=2R\sin (\angle ACB)$. So, from (5), we deduce that $$2R^2{\sin }^2(\angle ACB)\ge C{D}^2(1-\cos (\angle ADB)).(6)$$ From (1) and (2), we see that $$\begin{array}{rl}\angle CAD+\angle ACD& =\angle DCB+\angle CBD=\angle CAD+\angle CBD=\\ & =\angle DCB+\angle DCA=\angle ACB\end{array}(7)$$ Therefore $$\angle ADC=\angle BDC=180^{\circ }-\angle ACB(8)$$ Moreover, it is easily seen that: - $C$, $D$ lie on the same side with respect to the line $AB$ $$\iff \angle CAD+\angle CBD<\angle CAB+\angle CBA;$$ - $C$, $D$ lie on distinct sides with respect to the line $AB$ $$\iff \angle CAD+\angle CBD>\angle CAB+\angle CBA.$$ 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: $$-CD\text{ is the angle bisector of }\angle ADB;(9)$$ $$-O,D\text{ lie on the same side with respect to the line AB;}(10)$$ $$-\angle ADB=\angle AOB.(11)$$ 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 $(AOB)$.