In the triangle $ABC$, what fraction of the area of the triangle is occupied by the points $K$ that lie inside the triangle and for which the central similarity with center $K$ and ratio $-1/2$ maps $A$ and $B$ to points inside the triangle and $C$ to a point outside the triangle?
Let the points dividing the sides of triangle $ABC$ into thirds be denoted as $A_{B}, A_{C}, B_{A}, B_{C}, C_{A}$, and $C_{B}$ as shown in the first figure, and let the centroid of the triangle be denoted by $S$. Since $S$ divides the medians of triangle $ABC$ in the ratio $2:1$, the triangles $A B_{A} C_{A}$, $B A_{B} C_{B}$, and $C A_{A} B_{C}$ can be obtained by a $\frac{2}{3}$ reduction of triangle $ABC$ from the corresponding vertices. Therefore, the midpoint of each of the segments $B_{A} C_{A}$, $A_{B} C_{B}$, and $A_{C} B_{C}$ is $S$.

## 2nd Figure

A similarity transformation with center $K$ and ratio $\frac{1}{2}$ will map $A$ to an interior point of triangle $ABC$ if and only if, on the line $AK$, measuring $\frac{1}{2}AK$ from $K$ in the direction opposite to $A$, the point $A_{K}$ thus obtained is an interior point of triangle $ABC$ (2nd figure). This is precisely the case if the point $A_{K}$, which is the $\frac{2}{3}$ reduction of $A$, is an interior point of the reduced image of triangle $ABC$, that is, of triangle $A B_{A} C_{A}$. Therefore, the points satisfying the condition for $A$ are the interior points of triangle $A B_{A} C_{A}$. Similarly, we can show that the points satisfying the condition for $B$ are the interior points of triangle $B A_{B} C_{B}$, while the points satisfying the condition for $C$ are those that lie outside triangle $C A_{C} B_{C}$, including its boundary points. Thus, the points satisfying the condition for $C$ are those that lie outside triangle $C A_{C} B_{C}$.

Therefore, the points satisfying all three conditions are the interior points of triangle $A_{B} B_{A} S$. Due to the parallelism of their corresponding sides, triangles $A_{B} B_{A} S$ and $ABC$ are similar, with the similarity ratio $\frac{A_{B} B_{A}}{AB} = \frac{1}{3}$. Hence, the ratio of their areas is $\frac{1}{9}$. Therefore, the points $K$ form $\frac{1}{9}$ of the area of triangle $ABC$.