Evochkiov M.A.

The distances from a certain point inside a regular hexagon to three of its consecutive vertices are 1, 1, and 2, respectively.

What is the side length of this hexagon?
Let \$A, B, C\$ be consecutive vertices of a hexagon, \$0\$ be a point inside it, and \$O = 2\$.

First method. Consider the other neighboring vertex \$F\$ of \$A\$. Then \$FABC\$ is an isosceles trapezoid (see left figure). The point \$O\$ lies on the common perpendicular bisector of its bases \$FC\$ and \$AB\$, so \$OF = OC = 2\$. Since \$FC = 2AB\$, triangles \$AOB\$ and \$FOC\$ are similar with a ratio of 2. Since point \$O\$ is the center of homothety that transforms one of these triangles into the other, it lies on the diagonal \$AC\$ (similarly, it lies on the diagonal \$BF\$). Then \$\angle OBC = 120^\circ - 30^\circ = 90^\circ\$, and \$BC = \sqrt{2^2 - 1^2} = \sqrt{3}\$.
![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-38.jpg?height=404&width=1040&top_left_y=1930&top_left_x=496)

Second method. The point \$O\$ lies on the perpendicular bisector of side \$AB\$. Construct an equilateral triangle \$ABD\$ outside the hexagon (see right figure). Clearly, then \$OD\$ is the perpendicular bisector of segment \$AB\$. Moreover, since \$DB : DC = 1 : 2 = OB : OC\$, \$OD\$ is the angle bisector of \$\angle BAO = \angle ABO = \angle BOE = \angle EOC\$, points \$A, O, C\$ lie on the same line. In triangle \$ADC\$, \$\angle D = 60^\circ\$, \$DC = 2DA\$, so angle \$\angle DAC\$ is a right angle, and \$AB = AD = AC \cot \angle D = 3 \cot 60^\circ = \sqrt{3}\$.

## Answer

\$\sqrt{3}\$.

Author: Yurran A. Yu.

Trapezoid \$ABCD\$ is inscribed in a circle. Its base \$AB\$ is three times the length of base \$CD\$. Tangents to the circumscribed circle at points \$A\$ and \$C\$ intersect at point \$K\$. Prove that angle \$KDA\$ is a right angle.

## Solution

Let \$DH\$ be the height of the trapezoid, then \$ADCH\$ is a parallelogram (see left figure). Let \$M\$ be its center. Then \$KM\$ is the perpendicular bisector of diagonal \$AC\$. By the theorem on the angle between a chord and a tangent, \(\angle KAD = \angle ABD = \angle BAC = \angle KMD\) (the last two angles are angles with mutually perpendicular sides). Therefore, points \$A, K, D\$, and \$M\$ lie on the same circle, from which \(\angle KDA = \angle KMA = 90^\circ\).
![](https://cdn.mathpix.com/cropped/2024_05_06_7adbd1f2e5e559c8684fg-39.jpg?height=444&width=1114&top_left_y=686&top_left_x=470)

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