1. [4] The distance from a certain point inside a regular hexagon to three of its consecutive vertices is 1, 1, and 2, respectively. What is the side length of this hexagon?

M. Evdokimov
Answer: $\sqrt{3}$.

Let $A, B, C$ be consecutive vertices of a hexagon, $O$ be a point inside it, and let $O A=O B=1, O C=2$.

Solution 1. Consider the other neighboring vertex $F$ of $A$. Then $F A B C$ is an isosceles trapezoid (see figure). The point $O$ lies on the common perpendicular bisector of its bases $F C$ and $A B$, so $O F=O C=2$. But $F C=2 A B$ (in a regular hexagon, the main diagonal is twice the side length), from which triangles $A O B$ and $F O C$ are similar with a ratio of 2. Since $A B$ and $F C$ are parallel and $\angle B A O=\angle O C F$, the point $O$ lies on the diagonal $A C$ (and, similarly, the point $O$ lies on the diagonal $B F$). Then $\angle O B C=120^{\circ}-30^{\circ}=90^{\circ}$, and $B C=\sqrt{2^{2}-1^{2}}=\sqrt{3}$ by the Pythagorean theorem for the right triangle $O B C$.

Solution 2. The point $O$ lies on the perpendicular bisector of the side $A B$. Construct an equilateral triangle $A B D$ outside the hexagon (see figure on the right). Clearly, then $O D$ is the perpendicular bisector of the segment $A B$. Moreover, since $D B: D C=1: 2=O B: O C$, $O D$ is the bisector of the external angle $O$ of triangle $B O C$. Therefore, it is perpendicular to the bisector $O E$ of angle $B O C$. Due to the equality $\angle B A O=\angle A B O=\angle B O E=\angle E O C$, the points $A, O, C$ lie on the same line. In triangle $A D C, \angle D=60^{\circ}, D C=2 D A$, from which angle $D A C$ is a right angle, and $A B=A D=A C \operatorname{ctg} \angle D=3 \operatorname{ctg} 60^{\circ}=\sqrt{3}$.

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