Let $A$ be the vertex of a given angle $\varphi$. On the ray $AX$ of this angle, we take points $A_{1}, A_{2}, A_{3}, \ldots A_{n}$ such that $A A_{1}=a, A_{1} A_{2}=2 a, A_{2} A_{3}=3 a, \ldots A_{n-1} A_{n}=n a$. Then, perpendiculars to $AX$ are drawn at points $A_{1}, A_{2}, A_{3}, \ldots A_{n}$, intersecting the other ray of the angle $\varphi$ at points $B_{1}, B_{2}, B_{3}, \ldots B_{n}$. What is the area of

(1) the trapezoid $A_{n-1} A_{n} B_{n} B_{n-1}$ and

(2) the triangle $A A_{n} B_{n}$?
We will need the length of $A A_{n}$, so we calculate it first:

$$
A A_{n}=a+2 a+\ldots+n a=\frac{n}{2}[a+n a]=\frac{1}{2} n \cdot(n+1) \cdot a
$$

$1^{\circ}$. Denoting the area of the trapezoid $A_{n-1} A_{n} B_{n} B_{n-1}$ by $T_{n}$:

$$
\begin{gathered}
2 T_{n}=\left(A_{n} B_{n}+A_{n-1} B_{n-1}\right) \cdot A_{n-1} A_{n}= \\
=\left(A A_{n} \cdot \operatorname{tg} \varphi+A_{n} A_{n-1} \operatorname{tg} \varphi\right) \cdot A_{n-1} A_{n}= \\
\left.=\left[\frac{1}{2} \cdot n(n+1) a \operatorname{tg} \varphi\right)+\frac{1}{2} \cdot n(n-1) n a \operatorname{tg} \varphi\right] \cdot n a
\end{gathered}
$$

thus

$$
T_{n}=\frac{1}{2} a^{2} n^{3} \operatorname{tg} \varphi
$$

$2^{\circ}$. Denoting the area of the triangle $A A_{n} B_{n}$ by $t_{n}$:

$$
t_{n}=\frac{1}{2} \cdot A A_{n} \cdot A_{n} B_{n}=\frac{1}{2} \cdot \frac{1}{2} n \cdot(n+1) a \cdot \frac{1}{2} a \cdot n \cdot(n+1) \operatorname{tg} \varphi
$$

thus

$$
t_{n}=\frac{1}{8} a^{2} n^{2}(n+1)^{2} \operatorname{tg} \varphi
$$

(József Kiss, Pápa.)

The problem was also solved by: L. Bánó, J. Blum, E. Bauer, J. Csada, N. Ehrenfeld, K. Epstein, M. Fekete, H. Fodor, R. Földes, P. Füstös, A. Haar, P. Heimlich, V. Horti, E. Kirchknopf, Gy. Kovács, M. Lusztig, I. Martini, P. Merse, A. Paunz, M. Rosenthal, S. Ruvald, P. Sárközy, Gy. Schuster, L. Spitzer, J. Székely, O. Szilas, E. Tandlich, B. Tóth, J. Tóth, V. Wáhl.