## Task B-1.5.

A circle with a radius of $3 \text{~cm}$ is inscribed in a parallelogram such that it touches three of its sides. The measure of the acute angle of the parallelogram is $60^{\circ}$, and one side of the parallelogram is $2 \sqrt{3} \text{~cm}$ longer than the other side. Determine the distance from the center of the circle to the farthest vertex of the parallelogram.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_5ec97620b8cc23e4b297g-04.jpg?height=377&width=745&top_left_y=577&top_left_x=196)

The sketch shows at least one right angle, radius, and/or height, and the sought length $d$.

From the figure, vertex $C$ lies on side $\overline{B C}$ which the given circle does not touch, so it is clear that vertex $C$ is the farthest from the center. First, let's determine the lengths of the sides of the parallelogram.

From triangle $A N D$, we have $|A D|=\frac{|D N|}{\sin 60^{\circ}}=\frac{6}{\frac{\sqrt{3}}{2}}=4 \sqrt{3} \mathrm{~cm}$, and then

$|C D|=|A B|=|A D|+2 \sqrt{3}=6 \sqrt{3} \mathrm{~cm}$.

$D S$ is the angle bisector of $\angle F D E$ with a measure of $120^{\circ}$, so $\angle S D E=60^{\circ}$.

From triangle $D E S$, we have

$|D E|=\frac{3}{\tan 60^{\circ}}=\sqrt{3} \mathrm{~cm}$, so $|E C|=6 \sqrt{3}-\sqrt{3}=5 \sqrt{3} \mathrm{~cm}$.

Finally, applying the Pythagorean theorem, we get the sought distance

$d=\sqrt{|E S|^{2}+|E C|^{2}}=\sqrt{9+75}=\sqrt{84} \mathrm{~cm}$ or $2 \sqrt{21} \mathrm{~cm}$.