2. Given a triangle $ABC$ and points $M, N, P$ on sides $AB, BC, CA$ respectively, such that $\overline{AM}: \overline{AB} = \overline{BN}: \overline{BC} = \overline{CP}: \overline{CA} = k$. Determine the value of $k$ for which the area of triangle $MNP$ is the smallest.
Solution. Let's denote the areas of triangles $A M P, M B N, N C P$ by $P_{1}, P_{2}, P_{3}$ respectively, see the diagram. Let $P_{4}$ be the area of triangle $M N P$. From the given ratios, according to the condition, we have $\overline{A M}: \overline{A B}=k$, that is,

$$
\begin{aligned}
\overline{A M}:(\overline{A M}+\overline{M B})=k & \Leftrightarrow(1-k) \overline{A M}=k \overline{M B} \\
& \Leftrightarrow \overline{A M}: \overline{M B}=k:(1-k)
\end{aligned}
$$

Similarly,

$$
\overline{B N}: \overline{N C}=k:(1-k) \text { and } \overline{C P}: \overline{P A}=k:(1-k)
$$

If we denote the area of triangle $A B C$ by $P$, then:

![](https://cdn.mathpix.com/cropped/2024_06_05_1d13ef53f1f77712dfd3g-4.jpg?height=106&width=586&top_left_y=1783&top_left_x=572)

where $\frac{\overline{A P}}{\overline{A C}}=\frac{\overline{A P}}{\overline{A P}+\overline{P C}}=\frac{1}{1+\frac{\overline{P C}}{\overline{A P}}}=\frac{1}{1+\frac{k}{1-k}}=1-k$. Similarly, we obtain that

$$
\frac{P_{2}}{P}=\frac{P_{3}}{P}=k(1-k)
$$

Now

$$
\frac{P_{4}}{P}=\frac{P-P_{1}-P_{2}-P_{3}}{P}=1-3 \frac{P_{1}}{P}=1-3 k(1-k)=3 k^{2}-3 k+1
$$

If $P_{4}$ takes the smallest value, then the ratio $\frac{P_{4}}{P}$ is also the smallest. The quadratic function $\frac{P_{4}}{P}=3 k^{2}-3 k+1$ is open upwards and has a minimum at the vertex of the parabola with abscissa $k=-\frac{-3}{2 \cdot 3}=\frac{1}{2}$. Then the desired value of the parameter $k$ is $k=\frac{1}{2}$.