There is a circle and two points on it: $A$ and $B$. Construct the point $P$ on the circle such that

$$
\frac{P A}{P B}=\frac{2}{3}
$$
I. Solution: The geometric locus of points $P$ in the plane for which $\frac{P A}{P B}=\frac{2}{3}$ is an Apollonian circle, whose diameter $X Y$ lies on the line $A B$, where $\frac{A X}{X B}=\frac{2}{3}$ and $\frac{A Y}{Y B}=-\frac{2}{3}$. By constructing the points $X$ and $Y$ (Figure 1), the Apollonian circle drawn over $X Y$ as a diameter intersects the given circle at the sought points $P_{1}$ and $P_{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a133b9de3db06dcd0e4ag-1.jpg?height=629&width=941&top_left_y=371&top_left_x=576)

Figure 1

Since the point $A$ is necessarily between $X$ and $Y$, and the point $X$ is between $A$ and $B$, there are always 2, and only 2 solutions.

Bartha László (Balassagyarmat, Balassa g. II. o. t.)

II. Solution: It is known that the angle bisector of a triangle divides the opposite side in the ratio of the adjacent sides. Therefore, if the angle bisector from the vertex $P$ of $\triangle APB$ passes through the point $X$ on the side $AB$ such that $\frac{A X}{X B}=\frac{2}{3}$, it also bisects the arc $\overparen{A B}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_a133b9de3db06dcd0e4ag-1.jpg?height=694&width=637&top_left_y=1466&top_left_x=730)

Figure 2

Accordingly, the construction process is as follows: connect the midpoint of the arc $\overparen{A B}$ to the second division point of the segment $AB$ divided into five parts, starting from $A$, which is point $X$. The line thus obtained intersects the circle at the sought point $P$. Since there are two arcs $\overparen{A B}$ and thus two midpoints: $F_{1}$ and $F_{2}$ (Figure 2), there are two solutions: $P_{1}$ and $P_{2}$.

Simonfai László (Bp. II., Rákóczi g. II. o. t.)

Note: Both solutions can naturally be generalized to the case where the given ratio $\frac{2}{3}$ is replaced by any arbitrary $\frac{m}{n}$.