For a circle with a radius of $7 \mathrm{~m}$. from a point located $11 \mathrm{~m}$. away from the circle's center, we need to draw a secant such that the segment outside the circle is as long as the segment inside the circle. How long is the entire secant and what angle does it form with the line that connects the point outside the circle to the circle's center?
Let the center of the circle be $O$, a point $A$ at a distance of $11 \mathrm{~m}$ from it, and the intersection points of the secant with the circle be $B$ and $C$. Then

$$
A B \cdot A C=2 A B^{2}=4 \times 18=72
$$

that is,

$$
A B=6 \mathrm{~m}
$$

and the entire secant

$$
A C=12 \mathrm{~m}
$$

The angle formed by $A C$ and $A O$ can be calculated from the following equation:

$$
\begin{gathered}
\cos \alpha=\frac{A B+\frac{B C}{2}}{A O}=\frac{9}{11} \\
\log \cos \alpha=9.91285-10 \\
\alpha=35^{\circ} 6^{\prime}
\end{gathered}
$$

neglecting the seconds.

(Schulhof Gábor, real gym. VIII. class, Pécs.)

The problem was also solved by: Bolemann Béla, Budapest; ifj. Breznyik János, Selmeczbánya; Engel Richárd, Győr; Grossmann Gusztáv, Budapest; Hertzka Róbert, Budapest; Hirschler Ármin, Győr; ifj. Imre János, Nyíregyháza; Jankovich György, Losoncz; Jorga Gergely, Gilad; Kiss Jenő, Budapest; Meitner Elemér, Budapest; Pilczer Ignácz, Kaposvár; Prónai Győző, Banská Bystrica; Suták Sándor, Nyíregyháza; Szabó Gusztáv, Győr; Unger Jenő, Győr; Visnya Aladár, Pécs.