In triangle $A B C$, given: $\angle A C B=60^{\circ}, \angle A B C=45^{\circ}$. On the extension of $A C$ beyond vertex $C$, a point $K$ is taken such that $A C=C K$. On the extension of $B C$ beyond vertex $C$, a point $M$ is taken such that the triangle with vertices $C, M$, and $K$ is similar to the original triangle. Find $B C: M K$, given that $C M: M K<1$.
Prove that $\angle K C M=75^{\circ}$.

## Solution

Since $C M<M K$ and the angles of triangle $C M K$ are $60^{\circ}, 45^{\circ}$, and $75^{\circ}$, $C M$ is the smallest side of the triangle. Therefore,

$$
\angle M K C=45^{\circ}, \angle K M C=75^{\circ} .
$$

Using the Law of Sines in triangles $A B C$ and $M K C$, we find that

$$
A C=\frac{B C \sin 45^{\circ}}{\sin 75^{\circ}}, M K=\frac{K C \sin 60^{\circ}}{\sin 75^{\circ}}=\frac{A C \sin 60^{\circ}}{\sin 75^{\circ}}=\frac{B C \sin 45^{\circ} \sin 60^{\circ}}{\sin ^{2} 75^{\circ}}
$$

Therefore,

$$
\frac{B C}{M K}=\frac{\sin ^{2} 75^{\circ}}{\sin 45^{\circ} \sin 60^{\circ}}=\frac{1-\cos 150^{\circ}}{2 \sin 45^{\circ} \sin 60^{\circ}}=\frac{2\left(1+\frac{\sqrt{3}}{2}\right)}{\sqrt{2} \cdot \sqrt{3}}=\frac{2+\sqrt{3}}{\sqrt{6}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_df2bbd76ac4dd2f0f20fg-56.jpg?height=452&width=728&top_left_y=966&top_left_x=665)

## Answer

$\frac{2+\sqrt{3}}{\sqrt{6}}$.