4. Triangle $ABC$ has an area of $1, A_{1}$ is the midpoint of $BC, B_{1}$ is the midpoint of $AC$ and $C_{1}$ is the midpoint of $AB$. Let $M$ be a point on segment $C_{1} B, L$ be a point on segment $A_{1} C$ and $K$ be a point on segment $B_{1} A$. What is the minimum common area of triangles $A_{1} B_{1} C_{1}$ and $K L M$?
Solution. We denote the points $M_{1}$ and $M_{2}, L_{1}$ and $L_{2}, K_{1}$ and $K_{2}$ as shown in the diagram. Note that $\frac{\overline{C_{1} M_{1}}}{\overline{M_{1} L_{2}}} \leq \frac{\overline{A K}}{\overline{K C}} \leq \frac{\overline{A B_{1}}}{\overline{B_{1} C}}=1$.

From this, we get that $P_{\Delta C_{1} M_{1} M_{2}} \leq P_{\Delta M_{1} L_{2} M_{2}}$. Similarly, we obtain

$$
P_{\Delta A_{1} L_{1} L_{2}} \leq P_{\Delta L_{1} L_{2} K_{2}} \text { and } P_{\Delta B_{1} K_{1} K_{2}} \leq P_{\Delta K_{1} K_{2} M_{2}}
$$

Let $P$ denote the common area of triangles $A_{1} B_{1} C_{1}$ and $K L M$. Then

![](https://cdn.mathpix.com/cropped/2024_06_05_c02eb5ea46120e551070g-8.jpg?height=372&width=507&top_left_y=1811&top_left_x=972)

$$
\begin{aligned}
P_{\Delta A_{1} B_{1} C_{1}}-P & =P_{\Delta C_{1} M_{1} M_{2}}+P_{\Delta A_{1} L_{1} L_{2}}+P_{\Delta B_{1} K_{1} K_{2}} \\
& \leq P_{\Delta M_{1} L_{2} M_{2}}+P_{\Delta L_{1} L_{2} K_{2}}+P_{\Delta K_{1} K_{2} M_{2}} \\
& =P-P_{\Delta M_{2} L_{2} K_{2}} \leq P
\end{aligned}
$$

Therefore, $P \geq \frac{1}{2} P_{\triangle A_{1} B_{1} C_{1}}=\frac{1}{8}$. Note that the value $\frac{1}{8}$ is achieved, for example, if we choose $M \equiv C_{1}, L \equiv C$ and $K \equiv B_{1}$.