Given two points, the distance between which is 1 unit. Using only a compass, construct two points such that the distance between them is $\sqrt{2}$ units.
Solution. Let the two given points be denoted by $A$ and $B$. First, draw the circles $k_{1}$ and $k_{2}$ with centers at $A$ and $B$ respectively, and radius $1 = AB$. Let the intersection points of these two circles be $C_{1}$ and $C_{2}$. Then, construct the circle with center $C_{1}$ and radius $C_{1}A = 1$, and let the intersection point of this circle and $k_{1}$, different from $B$, be $D$. Next, construct the circle with center $D$ and radius $DA = 1$. Let the intersection point of this circle and $k_{1}$, different from $C_{1}$, be $E$.

Then,

$$
1 = AB = AC_{1} = AC_{2} = AD = AE = BC_{1} = BC_{2} = DC_{1} = DE
$$

Thus, the triangles $ABC_{1}$, $ABC_{2}$, $AC_{1}D$, and $ADE$ are all equilateral. Therefore, $\angle EAD + \angle DAC_{1} + \angle C_{1}AB = 180^{\circ}$, which means the points $E$, $A$, and $B$ are collinear and $EB = 2$. Moreover, $C_{1}C_{2}$ is perpendicular to $AB$, and its length is twice the height of an equilateral triangle with side length 1, i.e., $C_{1}C_{2} = \sqrt{3}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_32de401a37410ce726f6g-1.jpg?height=602&width=591&top_left_y=569&top_left_x=756)

Next, draw the circles with centers at $B$ and $E$ and radius $\sqrt{3}$. Let the intersection points of these circles be $F_{1}$ and $F_{2}$. Then, $EBF_{i} (i=1,2)$ are isosceles triangles with base 2 and legs of length $\sqrt{3}$. The height corresponding to the base, by the Pythagorean theorem, is

$$
\sqrt{\sqrt{3}^{2} - \left(\frac{2}{2}\right)^{2}} = \sqrt{2}
$$

The height connects the midpoint of the base to the opposite vertex, which is exactly the point $A$ with $F_{i}$.

Thus, the distance between the points $A$ and $F_{2}$, constructed only with a compass, is $\sqrt{2}$, which is one solution to the problem.

Note. The problem can be solved in many ways, and the construction presented here is, in our opinion, the simplest.