204. Given a triangle $ABC$ with area I. Let $A_{1}$, $B_{1}$, and $C_{1}$ be the midpoints of sides $BC$, $CA$, and $AB$ respectively. What is the minimum area that the common part of triangles $A_{1} B_{1} C_{1}$ and $KLM$ can have, if points $K$, $L$, and $M$ lie on segments $AB_{1}$, $CA_{1}$, and $BC_{1}$ respectively?

$$
\begin{gathered}
9 \text { th All-Union Olympiad, } \\
1975 \text { (Saratov) }
\end{gathered}
$$

| Class |  | 1st day |  | 2nd day |
| :---: | :---: | :---: | :---: | :---: |
| 8 | 205a | 206207 | 208a | 213214215 |
| 9 | 209 | 206210 | 2086 | 216215217 |
| 10 | 211 | 212205 | 208 | 214218219 |
204. Answer: 1/8. Since (notations are clear from Fig. 83)

$$
\frac{C_{1} M_{2}}{M_{2} M_{1}} \leqslant \frac{A K}{K C} \leqslant \frac{A B_{1}}{B_{1} C}=1
$$

then $C_{1} M_{2} \leqslant M_{2} M_{1}$, and therefore $S_{C_{1} M_{2} K_{1}} \leqslant S_{M_{2} M_{1} K_{1}}$. Similarly, it is proved that $S_{A L_{2} M_{1}} \leqslant S_{L_{2} L_{1} M_{1}}$ and

![](https://cdn.mathpix.com/cropped/2024_05_21_7b8a6b2625543c3dae75g-191.jpg?height=346&width=395&top_left_y=624&top_left_x=166)

Fig. 83 $S_{B_{1} K_{2} L_{1}} \leqslant S_{K_{2} K_{1} L_{1}} \cdot$

Let $S$ be the area of the common part of triangles $K L M$ and $A_{1} B_{1} C_{1}$. Adding the obtained inequalities, we get

$$
\begin{aligned}
& S_{A_{1} B_{1} C_{1}}-S \leqslant S_{K_{1} M_{2} M_{1}}+S_{M_{1} L_{2} L_{1}}+ \\
& \quad+S_{L_{1} K_{2} K_{1}}=S-S_{K_{1} M_{1} L_{1}} \leqslant S, \\
& \text { hence } 2 S \geqslant S_{A_{1} B_{1} C_{1}}=1 / 4, \text { i.e. }
\end{aligned}
$$

$S \geqslant 1 / 8$. If point $M$ coincides with $C_{\text {l }}$, point $L$ with $C$, and $K$ with $A$, then $S=1 / 8$.