
Problem 3. A square with side length 5 is divided into unit squares by parallel to its sides lines. Let $A$ be the set of the vertexes of the unit squares which are not on the sides of the given square. How many points from $A$ can be chosen at most in a way that no three of them are vertexes of isosceles rectangle triangle?


Solution. We shall prove that the maximal number is 6 . Let us enumerate the points in the way, shown on the table 1 .

It is easy to be seen that no 3 of the points $1,2,3,8,12$ and 16 are vertexes of a isosceles rectangle triangle. Assume that there exists a set of 7 points with the desired property. Note that if 4 points form a square, then at most 2 of them can be among the already chosen ones. The points $1,4,16$ and $13 ; 2,8,15$ and $9 ; 3,12,14$ and 5 form squares.

| 13 | 14 | 15 | 16 |
| ---: | ---: | ---: | ---: |
| 9 | 10 | 11 | 12 |
| 5 | 6 | 7 | 8 |
| 1 | 2 | 3 | 4 |

Table 1:
Consequently, at most 6 of the chosen points lie on the contour. It follows from here that at least one of the points $6,7,10$ and 11 is from the chosen ones. Due to the symmetry we may assume that this is the point 7 . Since the points 7,16 and $9 ; 1,7$ and 14 form isosceles rectangle triangles, then at most two of the points $1,9,14$ and 16 are from the chosen ones. The points $5,7,13$ and 15 form a square and therefore at most one of the points 5,13 and 15 is from the chosen ones. It follows from here that at least 3 points are chosen from $2,3,4,6,8,10,11$ and 12. By the pigeonhole principle we deduce that at least two points are chosen in one of the sets $3,6,11,8$ and $2,4,10,12$. It is easy to see that if the two points are in the first set, then we have two possibilities -3 and 11 or 6 and 8 (in both
cases it is not possible to chose more points on the square which encounters 7). Analogously, if the two points are in the second set, then the possibilities are two again -2 and 12 or 4 and 10 (in both cases it is not possible to chose more points on the square which encounters 7 ). The contradiction shows that the maximal number of points which can be chosen is equal to 6 .

## Grade 10
