There are three points on a straight line: $A, B, C$; let's find the geometric place of points from which the distances $A B$ and $B C$ are seen at equal angles.
I. Solution. Let $P$ be a point that meets the requirements of the problem. Since the line $PB$ bisects the angle $APC$, we can say that:

$$
AP: CP = AB: BC
$$

Thus, the problem can also be stated as: Find the geometric locus of the vertices of triangles whose common base is $AC$, and the ratio of the other two sides is $AB: BC$.

Let $A$ be the origin of the coordinate system, and: $AB = p, BC = q$; the coordinates of point $P$ are: $AD = x, PD = y$. From the right triangle $ADP$:

$$
\overline{AP}^2 = x^2 + y^2
$$

From the triangle $CDP$:

$$
\overline{CP}^2 = (p+q-x)^2 + y^2
$$

Divide (1) by (2) and note that $\frac{AP}{CP} = \frac{p}{q}$, then:

$$
\frac{p^2}{q^2} = \frac{x^2 + y^2}{(p+q-x)^2 + y^2}
$$

From this equation, it follows that:

$$
\begin{gathered}
x^2 - \frac{2p^2 x}{p-q} + y^2 = -\frac{p^2(p+q)}{p+q} \\
\left(x - \frac{p^2}{p-q}\right)^2 + (y-0)^2 = \left(\frac{pq}{p-q}\right)^2
\end{gathered}
$$

This equation shows that the sought geometric locus is a circle whose center lies on the line $AC$, and whose radius is $+\frac{pq}{p-q}$. Let's find the abscissas of the intersection points of the circle and the line $AC$. If $y=0$, then $x_1 = \frac{p(p+q)}{p-q}$ and $x_2 = p$; we see that the circle always passes through the point $B$.

If $p=q$, then $\frac{p^2}{p-q} = \infty$, meaning the geometric locus is a line, specifically the perpendicular line raised at the midpoint of the line $AC$.

(Friedmann Bernát.)

II. Solution. Draw the triangle $APC$ over the distances $AB$ and $BC$, such that $AP: CP = AB: BC$. Raise a perpendicular to $PB$ at point $P$, which intersects the line $AC$ at point $F$. The sought geometric locus is the circle determined by the points $F$, $P$, and $B$.

Proof. $PB$ is the angle bisector because the ratio of $AB$ and $BC$ is equal to the ratio of $AP$ and $CP$. The geometric locus of points whose distances from two points ($A$ and $C$) have a constant ratio is the circle passing through points $B$, $F$, and $P$. This circle is called the Apollonian circle. The points $A$, $B$, $C$, and $F$ are harmonic points.

(Roth Miksa, 6th grade, Pécs.)

The problem was also solved by: Beck F., Dénes A., Devecis M., Goldstein Zs., Goldziher K., and Kármán T., Kornis Ö., Misángyi V., Prakatur T., Spitzer Ö., Szabó I., Szabó K., Weisz Á.