Kozhevnikov P.A.

On each of two parallel lines $a$ and $b$, 50 points were marked.

What is the maximum possible number of acute-angled triangles with vertices at these points?
Obviously, the maximum is achieved when the points on both lines are sufficiently close, so that each segment with marked ends on one of the lines is seen from any point on the other line at an acute angle. Let's paint the points on one of the lines blue, and the projections of the points of the second line onto this line - red. Now the problem can be reformulated as follows.

On a line, there are 50 blue and 50 red points. Find the maximum possible number of triples of points, in which the middle one is of one color, and the extreme ones are of another. Let $A_{1}, \ldots, A_{50}, B_{1}, \ldots, B_{50}$ be red and blue points,

ordered from left to right. Consider a pair of adjacent points $A_{\mathrm{i}}$ and $B_{j}$. If $A_{i}$ lies to the left of $B_{j}$, then these points form a good triple with points $B_{1}, \ldots, B_{j-1}$ and $A_{i+1}, \ldots, A_{50}$, that is, $49+(j-i)$ good triples. If $i>j$, we can swap $A_{i}$ and $B_{j}$ to increase the number of such triples, while the other good triples will remain good. Therefore, the optimal arrangement will be such that each point $A_{i}$ lies to the right of $B_{i-1}$, but to the left of $B_{i+1}$ (the order of points $A_{i}$ and $B_{i}$ can be arbitrary). In particular, the optimal arrangement will be achieved by alternating the colors of the points.

In this case, the number of good triples with a red point in the middle is

$1 \cdot 49+2 \cdot 48+\ldots+49 \cdot 1=50(1+2+\ldots+49)+1^{2}+2^{2}+\ldots+49^{2}=1 / 2502 \cdot 49-1 / 649 \cdot 50 \cdot 99=1 / 649 \cdot 50 \cdot 51=20825$.

There are as many good triples with a blue point in the middle.

## Answer

41650.