4.4. Two circles $k_{1}$ and $k_{2}$ with centers at $O_{1}$ and $O_{2}$ with radii 1 and $\sqrt{2}$ intersect in two points $A$ and $B$, and $O_{1} O_{2}=2$. Let $A C$ be a chord on $k_{2}$. Find the length of $A C$, if the midpoint of $A C$ lies on $k_{1}$.

## 5th BALKANIAN MATHEMATICAL OLYMPIAD

NICOSIA, Cyprus, May 1988.
4.4. Let $M$ be the midpoint of $A C$ and $A M=x, \angle A M O_{1}=\varphi$. As the angle $\angle A M O_{2}$ is equal to $\pi / 2$, we have $M O_{2}=\sqrt{2-x^{2}}$. From the cosine theorem for triangle $\triangle O_{1} O_{2} M$ we get:

$$
\sin \varphi=-\cos \left(\frac{\pi}{2}+\varphi\right)=\frac{1+x^{2}}{2 \sqrt{2-x^{2}}}
$$

As the triangle $A M O_{1}$ is isosceles, we also have $\cos \varphi=\frac{x}{2}$, i.e. $\sin \varphi=\sqrt{1-x^{2} / 4}$ (note that $0<\varphi<\pi / 2$ ). Introduce a new variable $t=x^{2}, 0<t<2$. The unique solution of

$$
\frac{1+t}{2 \sqrt{2-t}}=\sqrt{1-\frac{t}{4}}
$$

is $t=7 / 8$. Therefore $x=\sqrt{t}=\sqrt{7 / 8}$ and

$$
A C=2 x=\sqrt{\frac{7}{2}}
$$