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In triangle $A B C$, it is known that $\angle B A C=75^{\circ}, A B=1, A C=\sqrt{6}$. A point $M$ is chosen on side $B C$, such that $\angle$ ВАМ $=30^{\circ}$. Line $A M$ intersects the circumcircle of triangle $A B C$ at point $N$, different from $A$. Find $A N$.
Denote $A N$ as $x$, express $B N$ and $C N$ through the circumradius of triangle $A B C$, and apply the Law of Cosines to triangles $A B N$ and $A C N$ (or use Ptolemy's theorem).

## Solution

## First Method.

Let $A N=x, R$ be the circumradius of the triangle. Then

$$
B N=2 R \sin 30^{\circ}=R, C N=2 R \sin 45^{\circ}=R \sqrt{2}
$$

By the Law of Cosines in triangles $A B N$ and $A C N$, we find that

$$
\begin{gathered}
B N^{2}=R^{2}=A B^{2}+A N^{2}-2 A B \cdot A N \cos 30^{\circ}=1+x^{2}-x \sqrt{3} \\
C N^{2}=2 R^{2}=6+x^{2}-2 \sqrt{6} x \cdot \frac{1}{\sqrt{2}}=6+x^{2}-2 x \sqrt{3}
\end{gathered}
$$

From the resulting system of equations, we find that $x^{2}=4$. Therefore, $A N=x=2$.

## Second Method.

Let $R$ be the radius of the circle. In the inscribed quadrilateral $A B N C$, we have:

$$
D N=2 R \sin 30^{\circ}=R, C N=2 R \sin 45^{\circ}=R \sqrt{2}, D C=2 R \sin 75^{\circ}
$$

By Ptolemy's theorem,

$$
A B \cdot C N+A C \cdot B N=A N \cdot B C \text {, or } R \sqrt{2}+R \sqrt{6}=A N \cdot 2 R \sin 75^{\circ} \text {. }
$$

Therefore,

$$
A N=\frac{\sqrt{2}+\sqrt{6}}{2 \sin 75^{\circ}}=\frac{\sqrt{2}+\sqrt{6}}{2 \cdot \frac{\sqrt{2}+\sqrt{6}}{4}}=2
$$

## Answer

2.00