4. Variant 1.

Given an equilateral triangle $A B C$ with an area of 210. Inside it, points for which vertex $A$ is neither the closest nor the farthest are painted red. What is the area of the painted part of the triangle?
Answer: 41.

Solution.
![](https://cdn.mathpix.com/cropped/2024_05_06_0d3dbdd740b551c86dd8g-3.jpg?height=328&width=766&top_left_y=1760&top_left_x=629)

Consider a point $D$ for which vertex $B$ is the nearest, and vertex $C$ is the farthest. Let $M, N, K$ be the midpoints of sides $AB, BC$, and $AC$ respectively. $O$ is the point of intersection of the medians $AN, BK, CM$, which are also the perpendicular bisectors of the sides of the triangle. Since point $D$ is closer to $B$ than to $A$, $D$ lies on the sides or inside triangle $CMB$. Since point $D$ is closer to $A$ than to $C$, $D$ lies on the sides or inside triangle $ABK$. Therefore, point $D$ lies in the intersection of these triangles, i.e., in triangle $BOM$. Similarly, points for which vertex $B$ is the farthest and vertex $C$ is the nearest lie in triangle $COK$. Since $CO: OM = 2: 1$, the area of triangle $BOM$ is one-third of the area of triangle $CMB$, which is half the area of triangle $ABC$. Therefore, the area of triangle $BOM$ is $\frac{1}{6}$ of the area of triangle $ABC$. Applying similar reasoning to triangle $COK$, we finally get that the sum of the areas of triangles $BOM$ and $COK$ is $\frac{1}{6} + \frac{1}{6} = \frac{1}{3}$ of the area of triangle $ABC$, i.e., it is equal to 41.