[ Combinatorics (miscellaneous). ] [ [Geometry on grid paper] 

Given a table of size $m \times n$ ( $m, n>1$ ). In it, the centers of all cells are marked. What is the maximum number of marked centers that can be chosen so that no three of them form the vertices of a right-angled triangle

#
It is possible to reduce the table by crossing out a row in which more than two selected points are located.

## Solution

Select the centers of all cells in the bottom row and the right column, except for the bottom right corner cell. In total, $m+n-2$ points are selected, and every triplet of marked points forms an obtuse triangle.

We will prove that it is impossible to select more than $m+n-2$ cell centers. For each marked center, there are no other marked centers in its row or column. Mark this row. If all rows are marked, then no more than $m \leq m+n-2$ centers are selected. The situation is similar when all columns are marked.

If not all rows and not all columns are marked, then the total number of marked rows (and thus marked centers) is no more than

$(m-1)+(n-1)=m+n-2 \ldots$

## Answer

$m+n-2$.

| on 35227 Topics: | ![](https://cdn.mathpix.com/cropped/2024_05_06_36c4956010bd91c4b4eeg-01.jpg?height=110&width=721&top_left_y=1413&top_left_x=459) |
| :---: | :---: |

It is known that in the tetrahedron $A B C D$, the circles inscribed in the faces $A B C$ and $B C D$ touch.

Prove that the circles inscribed in the faces $A B D$ and $A C D$ also touch.

## Hint

Both conditions mentioned in the problem are equivalent to the equality $A B + C D = A C + B D$.

## Solution

Let the circles inscribed in the faces $A B C$ and $B C D$ touch at point $O$. Thus, point $O$ is the point of tangency of the inscribed circles of triangles $A B C$ and $B C D$ with side $B C$. By the known formula for the distance from the vertex of a triangle to the point of tangency with the inscribed circle,

$1 / 2(A B + B C - A C) = B O = 1 / 2(B D + B C - C D)$, which means $A B - B D = A C - C D$. Let $P$ and $Q$ be the points of tangency of side $A D$ with the inscribed circles of triangles $A B D$ and $A C D$, respectively. Then $A P = 1 / 2(A B + A D - B D) = 1 / 2(A C + A D - C D) = A Q$. Therefore, points $P$ and $Q$ coincide, which means that the inscribed circles of triangles $A B D$ and $A C D$ touch.

## Problem