Let's mark points $C_{1}$ and $B_{1}$ on the sides $AB$ and $AC$ of triangle $ABC$, respectively, such that the ratio $BC_{1} / CB_{1}$ is equal to a given number, and the segment $B_{1}C_{1}$ is as short as possible.
Let the given ratio be $k$, and let $B_{1}^{\prime}, C_{1}^{\prime}$ be a pair of points such that $B C_{1}^{\prime} / C B_{1}^{\prime}=k$. We get a clear overview of the segments $B_{1}^{\prime} C_{1}^{\prime}$ if we draw a segment $C D$ that is equal, parallel, and in the same direction as each such segment. Then $B_{1}^{\prime} C_{1}^{\prime} D C$ forms a parallelogram, or if $C_{1}^{\prime}=A$, then $D$ is the reflection of $B_{1}^{\prime}$ over the midpoint of $A C$; thus $C_{1}^{\prime} D$ is parallel, in the same direction, and equal to $B_{1}^{\prime} C$. Let us assume for now that $k$ is positive and choose the directions on the sides to be positive in the directions of $B A$ and $C A$. Therefore,

$$
\angle D C_{1}^{\prime} B = \angle C A B \quad \text{and} \quad \frac{B C_{1}^{\prime}}{D C_{1}^{\prime}} = \frac{B C_{1}^{\prime}}{C B_{1}^{\prime}} = k,
$$

both independent of the choice of points $B_{1}^{\prime}, C_{1}^{\prime}$.

![](https://cdn.mathpix.com/cropped/2024_05_02_c7aed32f2a7706901222g-1.jpg?height=602&width=1220&top_left_y=542&top_left_x=450)

The resulting triangles $D C_{1}^{\prime} B$ are therefore similar, and thus the angle $\angle C_{1}^{\prime} B D$ is constant, and the points $D$ lie on a line $d$ passing through $B$. Conversely, let $d$ have an arbitrary point $D$ (different from $B$ and the intersection point $D_{0}$ of $d$ with $A C$), draw a line through $D$ parallel to $A C$, let its intersection with $A B$ be $C_{1}^{\prime}$, and the line through $C_{1}^{\prime}$ parallel to $D C$ intersect $A C$ at $B_{1}^{\prime}$; then $B C_{1}^{\prime} / C B_{1}^{\prime} = k$. If $D = D_{0}$, it is easy to see that $C_{1}^{\prime} = A$, and $B_{1}^{\prime}$ is the reflection of $D_{0}$ over the midpoint of $A C$.

$B_{1}^{\prime} C_{1}^{\prime}$ will be the shortest when $C D$, i.e., when $D$ is at the orthogonal projection $D_{1}$ of $C$ onto $d$. From this, the desired points $C_{1}, B_{1}$ follow according to the above.

In the case of a negative $k$, only the angle $\angle D C_{1} B$ will be equal to $180^{\circ} - \angle C A D$. $D$ and $d$, and from them $D_{1}, C_{1}, B_{1}$, can always be constructed uniquely because $D$ cannot coincide with $B$, and because $A B$ and $A C$ are not parallel; however, if $D_{1}$ coincides with $B$, then the desired position of $C_{1}$ is $B$, and that of $B_{1}$ is $C$, and in this case the ratio has no meaning.

József Berács (Győr, Czuczor G. Bencés g. II. o. t.) László Hárs (Budapest, Kölesey F. g. I. o. t.)

Note. The problem can also be solved by calculation. According to the condition - denoting the (signed) length of the segment $C B_{1}$ by $t$, we have $-B C_{1} = k t$. In the triangle $A B_{1} C_{1}$, $A B_{1} = b - t, A C_{1} = c - k t$, so $B_{1} C_{1}^{2} - t$ can be determined using the cosine theorem, yielding a quadratic polynomial in $t$, whose extremal value can be determined in the usual way by completing the square.

János Pintz (Budapest, Fazekas M. Gyak. g. I. o. t.) László Kóczy (Budapest, XI., Bocskai úti ált. isk. 8. o. t.)