A rhombus with a side length of one unit has an acute angle of $60^{\circ}$. How many circular arcs are there such that the vertices of the rhombus are equidistant from them? What is the radius of the circles?
Solution. We know that the set of points at an equal distance $d$ from a circle $k$ with center $O$ and radius $R$ consists of two circles concentric with $k$, with radii $R+d$ and $R-d$. The first circle is outside the circular region, while the second is inside, possibly just a single point.

Consider a rhombus $ABCD$ with unit side length and one of its acute angles being $60^\circ$.

1. Case. Choose three vertices of the rhombus that form an equilateral triangle. Let these be, for example, $A, B$, and $D$. Draw the circumcircle of the equilateral triangle $ABD$, with center $O$ and radius (since the median and the altitude coincide) the altitude $\frac{2}{3}$, i.e.,

$$
r_{1}=O D=\frac{2}{3} \cdot \frac{\sqrt{3}}{2}=\frac{\sqrt{3}}{3}
$$

Draw the circle with center $O$ and radius $OC$, where

$$
r_{2}=O C=\frac{1}{3} \cdot \frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}=\frac{4 \sqrt{3}}{6}=\frac{2 \sqrt{3}}{3}
$$

The sought circle with center $O$—that is, the circle whose points are equidistant from the two concentric circles—is:

$$
r=\frac{r_{1}+r_{2}}{2}=\frac{\frac{\sqrt{3}}{3}+\frac{2 \sqrt{3}}{3}}{2}=\frac{\sqrt{3}}{2}
$$

![](https://cdn.mathpix.com/cropped/2024_05_02_6027dd34a414923c5821g-1.jpg?height=574&width=553&top_left_y=957&top_left_x=772)

We get two such circles depending on whether we first choose the equilateral triangle $ABD$ or $BCD$. The vertices of the rhombus will be equidistant from the drawn circle.

2. Case. Now consider the triangle $ACD$ and draw its circumcircle. The vertex $B$ of the rhombus is the center of the circumcircle, since

$$
B A=B D=B C=1
$$

and thus $r_{1}=1, r_{2}=0$; the radius of the sought circle is $r=\frac{1}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_6027dd34a414923c5821g-1.jpg?height=512&width=545&top_left_y=1907&top_left_x=779)

The vertices of the rhombus will be equidistant from the drawn circle at a distance of $\frac{1}{2}$. We get two circles depending on whether we start from triangle $ACD$ or $ABC$.
3. Case. Two concentric circles contain two pairs of vertices of the rhombus. This can only be the two pairs of opposite vertices of the rhombus, with the common center of the two circles being the intersection of the diagonals, $O$. The radius of the circle passing through $B$ and $D$ is $r_{1}=\frac{1}{2}$. The radius of the circle with center $O$ passing through $A$ and $C$ is $r_{2}=\frac{\sqrt{3}}{2}$. The sought geometric locus is a circle concentric with the previous circles with radius

$$
r=\frac{\frac{1}{2}+\frac{\sqrt{3}}{2}}{2}=\frac{\sqrt{3}+1}{4}
$$

In this case, we get only one solution.

![](https://cdn.mathpix.com/cropped/2024_05_02_6027dd34a414923c5821g-2.jpg?height=513&width=526&top_left_y=562&top_left_x=797)

Therefore, there are a total of 5 circles such that the vertices of the rhombus are equidistant from them.