[ Extreme properties of a triangle (miscellaneous).]

The area of triangle $ABC$ is 1. Let $A_1, B_1, C_1$ be the midpoints of sides $BC, CA, AB$ respectively. Points $K, L, M$ are taken on segments $AB_1, CA_1, BC_1$ respectively. What is the minimum area of the common part of triangles $KLM$ and $A_1B_1C_1$?
Let the intersection point of lines $K M$ and $B C$ be denoted as $T$, and the intersection points of the sides of triangles $A_{1} B_{1} C_{1}$ and $K L M$ as shown in the figure. Then, $T L: R Z = K L: K Z = L C: Z B_{1}$. Since $T L \geq B A_{1} = A_{1} C \geq L C$, it follows that $R Z \geq Z B_{1}$, i.e., $S_{\mathrm{RZQ}} \geq S_{\mathrm{ZB}_{1} \mathrm{Q}}$. Similarly, $S_{\mathrm{QYP}} \geq S_{\mathrm{YA}_{1} \mathrm{P}}$ and $S_{\mathrm{PXR}} \geq S_{\mathrm{XC}_{1} \mathrm{R}}$. Adding all these inequalities and the inequality $S_{\mathrm{PQR}} \geq 0$, we get that the area of the hexagon $P X R Z Q Y$ is not less than the area of the remaining part of triangle $A_{1} B_{1} C_{1}$, i.e., its area is not less than $S_{\mathrm{A}_{1} \mathrm{~B}_{1} \mathrm{C}_{1}} / 2 = 1 / 8$. Equality is achieved, for example, if point $K$ coincides with $B_{1}$, and point $M$ with $B$.

![](https://cdn.mathpix.com/cropped/2024_05_06_992517ed05f335573ca2g-26.jpg?height=635&width=920&top_left_y=1532&top_left_x=569)

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