The major axis of an ellipse is $2a$ and the distance between the foci is $F F^{\prime}=2c$. Determine a point $M$ on the perimeter of the ellipse such that

$$
M F F^{\prime} \varangle=2 \times M F^{\prime} F \varangle
$$
Consider the problem as solved. Extend $F_{1} F$ beyond $F$ to $B$ such that $F M=F B$. If $\angle M F_{1} F = a$, then $\angle M B F = a$, because $\angle M F F_{1} = \angle F M B + \angle F B M = 2 \angle F B M = 2a$.

From triangles $M F_{1} F$ and $M F_{1} B$:

$$
M F: M F_{1} = \sin \alpha: \sin 2 \alpha
$$

and

$$
M F_{1}: F_{1} B = \sin \alpha: \sin 2 \alpha
$$

thus

$$
M F: M F_{1} = M F_{1}: 2c + M F
$$

or

$$
M F_{1}^{2} = M F^{2} + 2c M F
$$

But

$$
{\overline{M F_{1}}}^{2} = (2a - M F)^{2}
$$

so

$$
(2a - M F)^{2} = M F^{2} + 2c M F
$$

from which

$$
M F = \frac{2a^{2}}{2a + c} \quad \text{and} \quad M F_{1} = 2a \frac{a + c}{2a + c}
$$

If we substitute the values of $M F$ and $M F_{1}$ into the first proportion, we get

$$
\cos \alpha = \frac{a + c}{2a}
$$

Therefore, the construction is as follows: On the major axis $A A_{1}$, we draw a perpendicular at point $A$ and draw a circle with radius $2a$ from point $F_{1}$. This circle intersects the perpendicular at $A$ at points $C$ and $C_{1}$. The intersection points of $C F_{1}$ and $C_{1} F_{1}$ with the ellipse give the desired point.

(Pál Messer, Budapest, V. district.)

The problem was also solved by: Ádámffy E., Dömény I., Fekete M., Fuchs I., Haar A., Harsányi Z., Heimlich P., Jánosy Gy., Krampera Gy., Kürti L., Paunz A., Pichler S., Rosenberg J., Sárközy P., Schuster Gy., Schwarz Gy., Schwarz O., Tandlich E., Wáhl V., Werner M.