4[ Auxiliary Circle ]

Inside an angle of $60^{\circ}$, there is a point that is at distances $\sqrt{7}$ and $2 \sqrt{7}$ from the sides of the angle. Find the distance of this point from the vertex of the angle.

#
The given point, its projections on the sides of the given angle, and the vertex of the given angle lie on the same circle.

## Solution

Let $A$ and $B$ be the projections of the given point $M$ on the sides of the given angle with vertex $C$; $M A = \sqrt{7}$, $M B = 2 \sqrt{7}$. Then

$$
\angle A M B = 180^{\circ} - \angle A C B = 180^{\circ} - 60^{\circ} = 120^{\circ}.
$$

By the cosine theorem in triangle $A M B$, we find that

$$
A B = \sqrt{A M^{2} + B M^{2} - 2 A M \cdot B M \cos 120^{\circ}} = \sqrt{7 + 28 + 14} = 7
$$

The points $A, M, B$, and $C$ lie on a circle with diameter $C M$. Therefore, $A B = C M \sin \angle A C B$. From this, we find that

$$
C M = \frac{A B}{\sin 60^{\circ}} = \frac{14}{\sqrt{3}}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_6d3475182106d2e5daa6g-40.jpg?height=460&width=572&top_left_y=-1&top_left_x=743)

Answer

$\frac{14 \sqrt{3}}{3}$