66th Belarusian Mathematical Olympiad

Let $A(a;1/a)$, $B(b;1/b)$, $C(c;1/c)$. It is easy to see that $y=-\frac{x}{ab}+\frac{1}{a}+\frac{1}{b}$ is the equation of the line $AB$ and $y=abx$ is the equation of the line $OD$ because $OD$ is perpendicular to $AB$. Then we easily calculate the coordinates of $C$: $C=(\frac{1}{\sqrt{ab}},\sqrt{ab})$.

Let $(x-t{)}^2+(y-l{)}^2=r^2$ be the equation of the circle $S$. Since $A,B,C$ lie on the hyperbola, we have $$(x-t{)}^2+{\left(\frac{1}{x}-l\right)}^2=r^2⟹x^4-2t{x}^3+(t^2+l^2-r^2)x^2-2lx+1=0.(1)$$ Equation (1) has even number of real roots. Since $S$ passes through $A,B,C$, equation (1) has four real roots (not necessarily different). Let $\alpha$ be the fourth root of (1), then $$1=\alpha abc=\alpha \frac{ab}{\sqrt{ab}}=\alpha \sqrt{ab}⟹\alpha =\frac{1}{\sqrt{ab}}.$$ By Vieta's theorem, $$2t=a+b+2\frac{1}{\sqrt{ab}},2l=\frac{1}{a}+\frac{1}{b}+2\sqrt{ab},t^2+l^2-r^2=ab+\frac{1}{ab}+2\frac{a+b}{\sqrt{ab}}.$$ Hence $$r^2={\left(\frac{a+b}{2}+\frac{1}{\sqrt{ab}}\right)}^2+{\left(\frac{a+b}{2ab}+\sqrt{ab}\right)}^2-ab-\frac{1}{ab}-2\frac{a+b}{\sqrt{ab}}=$$ $$={\left(\frac{a+b}{2}\right)}^2\left(1+\frac{1}{a^2{b}^2}\right).$$

Let $a+b=n$, $\sqrt{ab}=m$, then $${\left(x-\frac{n}{2}-\frac{1}{m}\right)}^2+{\left(y-\frac{n}{2m^2}-m\right)}^2={\left(\frac{n}{2}\right)}^2\left(1+\frac{1}{m^4}\right).$$ Find the coordinates of the intersection points of $S$ and the line $OD$. Since the equation of $OD$ is $y=abx=m^{2}x$, we have $${\left(x-\frac{n}{2}-\frac{1}{m}\right)}^2+{\left(m^{2}x-\frac{n}{2m^2}-m\right)}^2=r^2.(2)$$ This equation has two roots, one of them is the abscissa of $C$, i.e., $\frac{1}{m}$ and the other is the abscissa of $F$ (it is easy to verify that $x=\frac{1}{m}$ satisfies (2)). Let $F=(f;g)$. Then, by Vieta's theorem, $$\frac{1}{m}+f=\frac{2\left(\frac{n}{2}+\frac{1}{m}+\frac{n}{2}+m^3\right)}{1+m^4}=\frac{2n}{m^4+1}+\frac{2}{m}⟹f=\frac{2n}{m^4+1}+\frac{1}{m}.$$ Then $g=abf=m^{2}f=m+\frac{2n{m}^2}{m^4+1}$. Therefore $$\begin{gathered} C{F}^2={\left(\frac{1}{m}-\frac{1}{m}-\frac{2n}{m^4+1}\right)}^2+{\left(m-m-\frac{2n{m}^2}{m^4+1}\right)}^2= \\ ={\left(\frac{2n}{m^4+1}\right)}^2(m^4+1)=\frac{4n^2}{m^4+1}. \end{gathered}$$ Let $D=(d;e)$. Since $D$ is the intersection point of $OD$ and $AB$, we have $$abd=-\frac{d}{ab}+\frac{1}{a}+\frac{1}{b}⟹\frac{a+b}{ab}=d\left(ab+\frac{1}{ab}\right)⟹$$ $$d=\frac{a+b}{1+a^2{b}^2}=\frac{n}{m^4+1},$$ $$e=abd=m^{2}d=\frac{m^{2}n}{m^4+1}.$$ Therefore $$O{D}^2=\frac{n^2}{(m^4+1{)}^2}+\frac{m^4{n}^2}{(m^4+1{)}^2}=\frac{n^2}{m^4+1}.$$ Then $$\frac{O{D}^2}{C{F}^2}=\frac{n^2}{m^4+1}\cdot \frac{m^4+1}{4n^2}=\frac{1}{4}⟹\frac{OD}{CF}=\frac{1}{2}.$$