[ Pairing and grouping; bijections ]

What is the maximum number of pawns that can be placed on a chessboard (no more than one pawn per square), if:

1) a pawn cannot be placed on the e4 square;
2) no two pawns can stand on squares that are symmetric with respect to the e4 square?

#
Evaluation. All fields of the board except for the vertical a, the horizontal 8, and the field e4 itself can be divided into pairs symmetrical relative to e4. Such pairs amount to 24. According to the condition, no more than one pawn can be placed on the fields of each pair. Additionally, no more than one pawn can be placed on the fields of the vertical a and the horizontal 8. There are 15 such fields. Pawns cannot be placed on the field e4 according to the condition. Therefore, no more than 39 pawns can be placed in total.

An example of placing 39 pawns is shown in the figure:

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-01.jpg?height=535&width=506&top_left_y=1036&top_left_x=770)

## [ GMT - Circle or Arc  Problem 32063 Topics: Diameter, basic properties $\quad]$ [Inscribed angle, subtending the diameter ]

Through a given point on a plane, all possible lines intersecting a given circle are drawn. Find the geometric locus of the midpoints of the resulting chords.

## Solution

The sought geometric locus of the midpoints of the chords is an arc of a circle constructed on the segment connecting the given point and the center of the given circle as its diameter, lying inside the given circle (in particular, if the point lies inside the circle, then the entire circle is obtained).

![](https://cdn.mathpix.com/cropped/2024_05_06_14311972f49418f31424g-01.jpg?height=415&width=786&top_left_y=2142&top_left_x=630)

Let O be the center of the given circle, M be the given point, l be an arbitrary line intersecting the circle and passing through point M, and N be the midpoint of the chord formed by the intersection of the line with the circle (see figure). We use the known theorem: a radius is perpendicular to a chord if and only if it bisects the chord. From this theorem, it follows that angle MNO is a right angle. From the converse theorem about the measure of an angle subtending a diameter, it follows that point N lies on the circle with diameter OM. Conversely, let N be a point lying on the circle with diameter OM and inside the given circle. By the direct theorem about the measure of an angle subtending a diameter, we get that lines MN and NO are perpendicular. Applying the theorem about the radius perpendicular to a chord again, we obtain that k is the midpoint of the chord formed by the line passing through point M.

## Problem