In the triangle $A B C$, from vertex $B$ we draw a secant that is antiparallel to $B C$, which intersects side $b$ at point $B_{1}$. From $B_{1}$, we draw another such secant, which is now parallel to $B C$ and intersects side $c$ at point $C_{1}$. This procedure can be repeated indefinitely.

What is the value of $\overline{B C}+\overline{B B_{1}}+B_{1} C_{1}+\ldots$ in the limit, if $b>c . ?$[^0]

[^0]:    ${ }^{1}$ Editor's note: A secant antiparallel to $B C$ is a line $B B_{1}$ such that $A C \cdot A B_{1}=A B^{2}$.
According to the condition:

$$
b \cdot A B_{1}=c^{2}
$$

or

$$
A B_{1}: c=c: b
$$

Furthermore, since the angle $BAC$ is common in the triangles $BAB_{1}$ and $BAC$, it follows that the triangles $BAB_{1}$ and $BAC$ are similar.

Therefore:

$$
B B_{1}: c=a: b
$$

from which

$$
B B_{1}=a \cdot \frac{c}{b}
$$

And thus we have:

$$
B_{1} C_{1}=B B_{1} \cdot \frac{c}{b}=a \cdot\left(\frac{c}{b}\right)^{2}
$$

We see, therefore, that

$$
B C+B B_{1}+B_{1} C_{1}+\ldots \text { to infinity }
$$

is the sum of an infinite geometric progression whose first term is $a$, and the common ratio is $\frac{c}{b}$; and since $b>c$, or $\frac{c}{b}<1$, the sought sum is:

$$
\frac{a}{1-\frac{c}{b}}=\frac{a b}{b-c}
$$

(Bernát Friedmann, 7th grade, S.-A.-Ujhely). The problem was also solved by: Béla Grünhut, Pécs; Ervin Hofbauer, Budapest; Aladár Visnya, Pécs.