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Problem 1812

National Olympiad, first round

Combinatorics Difficulty 6.8 Prove it Slovenia — National Math Olympiad · Slovenia

Let $ABCD$ be a square with the side of $20$ units. $Vid$ divides this square into $400$ unit squares. Eva then picks $4$ of the vertices of these unit squares. These vertices lie inside the square $ABCD$ and define a rectangle with the sides parallel to the sides of the square $ABCD$ . There are exactly $24$ unit squares which have at least one point in common with the sides of this rectangle. Find all possible values for the area of a rectangle with these properties.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Let $a$ and $b$ be the lengths of the sides of the rectangle. We may assume that $a \ge b$ . The unit squares that have at least one point in common with the rectangle are marked in the figure.

The number of the unit squares is equal to the area of this part. We can find this area by subtracting the area of the white rectangle from the area of the rectangle with the sides of length $a+2$ and $b+2$ . We consider two cases.

If $b \le 2$ then there is no white rectangle and the number of the unit squares is equal to the area of the enlarged rectangle, i.e. $(a+2)(b+2)$ . From $b \le 2$ and $(a+2)(b+2) = 24$ we get $b=1, a=6$ and $b=2, a=4$ .

If $b \ge 3$ then the number of the unit squares which have at least one point in common with the sides of the rectangle is equal to $(a+2)(b+2)-(a-2)(b-2) = 4a+4b$ . Hence, $a+b = \frac{24}{4} = 6$ . The only possibility is $a=b=3$ .

All possible areas of the rectangle are $1 \cdot 6 = 6$ , $2 \cdot 4 = 8$ and $3 \cdot 3 = 9$ .

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