Tokarev S.i.

On the first horizontal line of the chessboard, there are 8 black queens, and on the last - 8 white queens. In what minimum number of moves can the white queens exchange places with the black ones? White and black take turns, moving one queen per move.
The rook on the squares b1 and b8, which moved first, could not move to the opposite rank with that move, meaning these two rooks together made at least three moves. Similarly, pairs of rooks standing on the files c, ..., g each made at least three moves. The first of the four corner rooks to move did not land on the same square with that move, so they made at least five moves in total. Thus, a total of at least $6 \cdot 3 + 5 = 23$ moves are required.

An example of 23 moves is shown in the figure (the black rook moves first):
![](https://cdn.mathpix.com/cropped/2024_05_06_c9b930b00ea53112180dg-31.jpg?height=846&width=420&top_left_y=1857&top_left_x=384)
![](https://cdn.mathpix.com/cropped/2024_05_06_c9b930b00ea53112180dg-31.jpg?height=842&width=420&top_left_y=1858&top_left_x=819)
![](https://cdn.mathpix.com/cropped/2024_05_06_c9b930b00ea53112180dg-31.jpg?height=846&width=426&top_left_y=1858&top_left_x=1253)

Author: Shapovalov A.V.

On the coordinate plane, four graphs of functions of the form $y=x^{2}+a x+b$ are drawn, where $a, b$ are numerical coefficients. It is known that there are exactly four intersection points, and in each point, exactly two graphs intersect. Prove that the sum of the largest and smallest of the x-coordinates of the intersection points is equal to the sum of the other two x-coordinates.

## Solution

The x-coordinates of the intersection points are the roots of equations of the form $x^{2}+a x+b=x^{2}+c x+d$, which are equivalent to equations $a x+b=c x+d$. By replacing each parabola of the form $y=x^{2}+a x+b$ with the line $y=a x+b$, we not only preserve the x-coordinates of the intersection points but also do not change their number (the y-coordinates of points with the same x-coordinates will decrease by the same amount, $x^{2}$, and different points will not merge). Thus, we have reduced the problem to an analogous problem about four lines. Four pairwise intersections (out of six possible) can only occur in the case of two pairs of parallel lines, so the intersection points are the vertices of a parallelogram. The smallest and largest x-coordinates are at opposite vertices, and the half-sum of the x-coordinates of each pair of opposite vertices is equal to the x-coordinate of the center of the parallelogram. The statement is now obvious.