## Task B-4.3.

Points $A_{1}, B_{1}, C_{1}$ are on the sides $\overline{B C}, \overline{C A}, \overline{A B}$ of triangle $A B C$ respectively, such that:

$$
\frac{\left|B A_{1}\right|}{|B C|}=\frac{\left|C B_{1}\right|}{|C A|}=\frac{\left|A C_{1}\right|}{|A B|}=k
$$

Determine $k$ so that the area of triangle $A_{1} B_{1} C_{1}$ is minimized.
## First Solution.

Let $\alpha, \beta$, and $\gamma$ denote the angles at vertices $A, B$, and $C$ of triangle $ABC$, respectively.

Since $P_{ABC}=\frac{|AC| \cdot |BC| \sin \gamma}{2}=\frac{|AB| \cdot |BC| \sin \beta}{2}=\frac{|AB| \cdot |AC| \sin \alpha}{2}$, we have:

$$
\begin{aligned}
P_{AC_{1}B_{1}} & =\frac{|AC_{1}| \cdot |AB_{1}| \sin \alpha}{2}=\frac{k|AB| \cdot (|AC|-k|AC|) \sin \alpha}{2}=k(1-k) P_{ABC} \\
P_{BA_{1}C_{1}} & =\frac{|BA_{1}| \cdot |BC_{1}| \sin \beta}{2}=\frac{k|BC| \cdot (|AB|-k|AB|) \sin \beta}{2}=k(1-k) P_{ABC} \\
P_{CB_{1}A_{1}} & =\frac{|CB_{1}| \cdot |CA_{1}| \sin \gamma}{2}=\frac{k|CA| \cdot (|BC|-k|BC|) \sin \gamma}{2}=k(1-k) P_{ABC}
\end{aligned}
$$

Now for the area of triangle $A_{1}B_{1}C_{1}$, we have:

$$
\begin{aligned}
P_{A_{1}B_{1}C_{1}} & =P_{ABC}-P_{AC_{1}B_{1}}-P_{BA_{1}C_{1}}-P_{CB_{1}A_{1}}= \\
& =P_{ABC}-3 k(1-k) P_{ABC}=(1-3 k(1-k)) P_{ABC}= \\
& =(3 k^{2}-3 k+1) P_{ABC}
\end{aligned}
$$

The function $f(k)=3 k^{2}-3 k+1$ achieves its minimum at $k=-\frac{-3}{2 \cdot 3}=\frac{1}{2}$.