Let's construct a circle that passes through one vertex of a given square and is tangent to the perpendicular bisectors of the two square sides associated with that vertex.
I. Solution: Let the sought circle $k$ pass through the vertex $B$ of the square $ABCD$ with center $K$ (see Figure 1).

![](https://cdn.mathpix.com/cropped/2024_05_02_64072c0dc596b023b870g-1.jpg?height=594&width=637&top_left_y=237&top_left_x=736)

Figure 1

Thus, due to the required tangencies (the non-crossing of $e$ and $f$) and the properties of the square, the center $O$ of $k$ can only be on the ray $KB$ with endpoint $K$, because this is the angle bisector of the sector containing $B$ among the four sectors determined by $e$ and $f$. Therefore, the diameter of $k$ through $B$ will lie on $KB$, and $k$ will also be tangent to the line $g$ perpendicular to $KB$ at $B$. In other words, $k$ is a tangent circle of the triangle $KEF$ whose center lies on the internal angle bisector at $K$. There are two such circles, with centers $O_1$ and $O_2$, which can be found by intersecting the angle bisectors of the angles at $E$ or $F$ with $KB$, or by constructing the perpendiculars to $e$ at the points of tangency $E_1$ and $E_2$ of $k_1$ and $k_2$ with $e$.

Miklós Budai (Budapest, II., Rákóczi F. g. I. o. t.)

II. Solution: The center of the circumscribed circle $c$ of the square is $K$, so the line $KB$ is a common central line of $c$ and the sought circle $k$. Moreover, $c$ and the sought circle $k$ are tangent at $B$, so $B$ is their similarity point, and it is an internal similarity point if $k$ and $c$ are externally tangent, and an external similarity point if the tangency is internal. Since the lines connecting the points of tangency of the parallel tangents of two circles pass through the corresponding similarity point, by the converse of this theorem, we can use the intersection points of $c$ and $f$, the points $T_1$ and $T_2$ (Figure 1 and Figure 2), as the points of tangency of the tangents parallel to $e$ of $c$.

![](https://cdn.mathpix.com/cropped/2024_05_02_64072c0dc596b023b870g-1.jpg?height=583&width=764&top_left_y=1579&top_left_x=648)

Figure 2

These lines intersect $B T_1$ at $E_1$ and $B T_2$ at $E_2$.

Ferenc Mezey (Budapest, II., Rákóczi F. g. II. o. t.)

III. Solution: Calculate the radii $r$ and $R$ of the sought circles, and then construct the segments $r$ and $R$ to determine $O_1$ and $O_2$ from $B$ or $E_1$ and $E_2$ from $K$. According to Figure 1,

$$
KB = r(\sqrt{2} + 1) = R(\sqrt{2} - 1)
$$

These can also be written as:

$$
r : KB = 1 : (\sqrt{2} + 1), \quad \text{or} \quad R : KB = 1 : (\sqrt{2} - 1)
$$

The construction is shown in Figure 3.

![](https://cdn.mathpix.com/cropped/2024_05_02_64072c0dc596b023b870g-2.jpg?height=916&width=591&top_left_y=125&top_left_x=775)

(The I., II., III., and IV. marks indicate the chronological order of the steps.)

Béla Bollobás (Budapest V., Apáczai Csere J. g. I. o. t.)

Remark. We can also arrive at the construction using the circle centered at $E$ by calculation: the radii of the desired tangent circles of the isosceles right triangle $KEF$ are, based on known formulas, and $KE = KF = \sqrt{2} KB$ and $EF = 2 KB$,

$$
\begin{gathered}
r = \frac{KE + KF - EF}{2} = (\sqrt{2} - 1) KB = KE - KB = KE - EB = KE - EE_1 = KE_1 \\
R = \frac{KE + KF + EF}{2} = (\sqrt{2} + 1) KB = KE + EB = KE + EE_2 = KE_2
\end{gathered}
$$