Consider a semicircle with diameter $AB$. From a point $C$ on the circumference, we draw a perpendicular to $AB$, whose foot is $M$; from $C$ we also draw the perpendicular $CP$ to the tangent at $B$, and finally, we draw a tangent at $C$, which intersects $BP$ at $D$. Determine the point $C$ such that the triangles $CPD$ and $CAM$ have equal areas.
Let's consider the problem as solved and mark the center of the semicircle as $O$, the intersection of $BC$ and $OD$ as $H$, then:

$$
A M C \Delta = C P D \Delta
$$

and

$$
B M C \Delta = C P B \Delta
$$

thus

$$
A B C \Delta = B C D \Delta
$$

which means

$$
A C \cdot B C = D H \cdot B C
$$

from which

$$
A C = D H
$$

Extend $AC$ and place the segment $CL = AC$ on it, and denote the intersection of $LD$ and $AB$ as $S$; then

$$
A S : A B = A L : A C
$$

which means

$$
\begin{gathered}
A S = 2 A B \\
D O S \Delta \sim C A B \Delta
\end{gathered}
$$

and

$$
D O B \Delta \sim C A M \Delta
$$

thus

$$
A B : S O = A M : O B
$$

But

$$
A B = \frac{A S}{2} = \frac{S O + O A}{2} = \frac{S O + \frac{S O}{3}}{2} = \frac{2}{3} S O
$$

so

$$
A M = \frac{2}{3} O B = \frac{1}{3} A B
$$

(Kornél Riesz, Budapest.)

The problem was also solved by: Aczél F., Bartók I., Bayer B., Blau A., Bogdán G., Dessauer A., Deutsch I., Goldstein A., Haar A., Hirschfeld Gy., Jánosy J., Kalmár S., Kertész F., Klein A., König D., Lázár L., Ligeti P., Papp F., Pilczer P., Pintér M., Póka Gy., Raab R., Sasvári J., Schlesinger A., Selényi P., Sümegi Gy., Szmodics H., Wohlstein S.