Problem 12.2. Let $k$ be a circle with center $O$ and radius $\sqrt{5}$ and let $M$ and $N$ be points on a diameter of $k$ such that $M O=N O$. The chords $A B$ and $A C$, passing through $M$ and $N$, respectively, are such that

$$
\frac{1}{M B^{2}}+\frac{1}{N C^{2}}=\frac{3}{M N^{2}}
$$

Find the length of $M O$.

Oleg Mushkarov
12.2. Let $M$ and $N$ lie on the diameter $P Q$ $(M \in P O, N \in Q O)$ of $k$. Set $x=M O=N O$, $0 \leq x \leq \sqrt{5}$. Then

$M A . M B=M P . M Q=(\sqrt{5}-x)(\sqrt{5}+x)=5-x^{2}$.

Analogously $N$ A. NC $=5-x^{2}$. It follows that $\frac{1}{M B^{2}}+\frac{1}{N C^{2}}=\frac{M A^{2}+N A^{2}}{\left(5-x^{2}\right)^{2}}$.

![](https://cdn.mathpix.com/cropped/2024_06_03_f8ccd71b4db367e7c2a9g-177.jpg?height=436&width=468&top_left_y=1705&top_left_x=1179)

Using the median formula we get

$$
5=A O^{2}=\frac{1}{4}\left[2\left(M A^{2}+N A^{2}\right)-4 x^{2}\right]
$$

i.e. $M A^{2}+N A^{2}=2\left(5+x^{2}\right)$. Therefore

$$
\frac{1}{M B^{2}}+\frac{1}{N C^{2}}=\frac{2\left(5+x^{2}\right)}{\left(5-x^{2}\right)^{2}}=\frac{3}{4 x^{2}}
$$

Hence $x^{4}+14 x^{2}-15=0$, i.e. $x=1$.