On the extension of side $A C$ of the equilateral triangle $A B C$, a point $M$ is taken, and circles are circumscribed around triangles $A B M$ and $M B C$. Point $A$ divides the arc $M A B$ in the ratio $M A: A B=n$. In what ratio does point $C$ divide the arc $M C B$?
Express the indicated arcs in terms of $n$.

## Solution

Angle $MAB$ in the first circle subtends arc $MB$. Since $\angle MAB = 120^\circ$, then

$$
\cup MB = 240^\circ, \cup MAB = 120^\circ
$$

Therefore,

$$
\cup MA = 120^\circ \cdot \frac{n}{n+1}, \cup AB = \frac{120^\circ}{n+1}
$$

In the second circle, angle $BCM$ subtends arc $MB$. Since $\angle BCM = 60^\circ$, then $\cup BM = 120^\circ$. The angle $BMC$ subtends arc $BC$ in the second circle. Since $\angle BMC = \frac{60^\circ}{n+1}$, then

$$
\cup BC = \frac{120^\circ}{n+1}
$$

Then,

$$
\cup MC = 360^\circ - \cup MB - \cup BC = 120^\circ \cdot \frac{2n+1}{n+1}
$$

Therefore, $\cup MC : \cup CB = 2n + 1$.

![](https://cdn.mathpix.com/cropped/2024_05_06_cb0f0d605674d837f9abg-26.jpg?height=345&width=420&top_left_y=-1&top_left_x=825)

## Answer

$\cup MC : \cup CB = 2n + 1$