Given a point $P$, lines $e, f$, and acute angles $\varepsilon, \varphi$. The circles to be constructed are those that pass through $P$ and intersect $e$ at angle $\varepsilon$ and $f$ at angle $\varphi$. (By the angle between a circle and a line, we mean the angle between the circle's tangent at the intersection point and the line.)
I. solution. First, let's assume that $e$ and $f$ intersect at point $M$. Let $k$ be any circle with radius $r$, center $O$, and $P$ any point on its circumference. The distance from $O$ to a line passing through $P$ and forming an angle $\varepsilon$ with the tangent to $k$ at $P$ is independent of the choice of $P$ and the line, since it is the leg adjacent to $\varepsilon$ in a right triangle with hypotenuse $r$. Similarly, let $r_{2}$ be the distance from $O$ to a line intersecting $k$ at an angle $\varphi$ (1st figure).

![](https://cdn.mathpix.com/cropped/2024_05_02_13904485add084cb2235g-1.jpg?height=748&width=1179&top_left_y=393&top_left_x=454)

A circle with radius $r$ intersects $e$ at an angle $\varepsilon$ and $f$ at an angle $\varphi$ if and only if its center is at a distance $r_{1}$ from $e$ and $r_{2}$ from $f$. There are four such points: the vertices of the parallelogram determined by the lines $e_{1}, e_{2}$, which are at a distance $r_{1}$ from $e$, and the lines $f_{1}, f_{2}$, which are at a distance $r_{2}$ from $f$. Let these points be $O_{1}, O_{2}, O_{3}, O_{4}$. Thus, drawing circles with radius $r$ around these four points, the circles intersect $e$ at an angle $\varepsilon$ and $f$ at an angle $\varphi$; denote these circles by $k_{1}, k_{2}, k_{3}, k_{4}$ (2nd figure).

If we map the obtained circles to new circles using a similarity transformation centered at $M$, the new circles also intersect the lines at the appropriate angles. We can achieve that the circles pass through $P$ using such a similarity transformation. We get as many suitable circles as there are intersection points on the line $MP$, and due to symmetry with respect to $M$, these circles come in pairs, so the number of solutions is equal to the number of intersection points of the circles $k_{1}$ and $k_{2}$ on the line $MP$, which can be 4, 3, or 2.

If $e$ and $f$ are parallel, and a circle intersects them at the given angles, then the ratio of the distances from the center of the circle to $e$ and $f$ is equal to $r_{1}: r_{2}$. These distances are, if the distance between $e$ and $f$ is $d$,

$$
\frac{r_{1} d}{r_{1}+r_{2}}, \quad \frac{r_{2} d}{r_{1}+r_{2}} \quad \text { or } \quad \frac{r_{1} d}{r_{1}-r_{2}}, \quad \frac{r_{2} d}{r_{1}-r_{2}}
$$

depending on whether the center is between the two lines or not. (Here we assumed that $\varepsilon<\varphi$; if $\varepsilon=\varphi$, the center can only be between the two lines.) Based on this, we can construct circles that intersect the lines at the given angles, and by translating them parallel to $e$ and $f$, we can generally achieve in 2 ways that they also pass through $P$.

Beck József

II. solution. Let $k$ be any circle with center $P$. Inverting with respect to $k$, the lines $e, f$ map to circles $k_{1}, k_{2}$ passing through $P$, and the sought circle maps to a line that intersects $k_{1}$ at an angle $\varepsilon$ and $k_{2}$ at an angle $\varphi$. The lines intersecting $k_{1}$ at an angle $\varepsilon$ all touch the circle $k_{1}^{\prime}$ concentric with $k_{1}$ (which can be constructed using a line intersecting $k_{1}$ at an angle $\varepsilon$ and otherwise arbitrarily chosen). Similarly, we obtain the circle $k_{2}^{\prime}$ from $k_{2}$: the inverse of the sought circle $x^{\prime}$ is thus a common tangent to the circles $k_{1}^{\prime}$ and $k_{2}^{\prime}$, which, when inverted with respect to $k$, gives the sought circle $x$ (3rd figure, on the next page).

![](https://cdn.mathpix.com/cropped/2024_05_02_13904485add084cb2235g-2.jpg?height=830&width=1163&top_left_y=130&top_left_x=470)

3rd figure

Since the number of common tangents to two circles is 4, 3, 2, 1, or 0, the number of sought circles is also 4, 3, 2, 1, or 0.

Horváth Miklós (Veszprém, Lovassy L. Gym., I. o. t.)


[^0]:    ${ }^{1}$ See: Surányi János-Tusnády Gábor: On inversion, K. M. L. 37 (1968) 97-101.