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Print59th Ukrainian National Mathematical Olympiad
Ukraine counting and probability
Problem
Determine if there exists a connected figure that consists of squares and satisfies the following conditions: it consists of squares and is not a rectangle; any rectangle, where can be split into figures that coincide with ? A figure that consists of squares is called connected, if one can get from any cell to any cell by moving to adjacent cells.

Fig. 16
Fig. 17

Fig. 16
Fig. 17
Solution
Yes, such figure exists.
Consider an L-shaped tromino that consists of three cells. , rectangles can be easily split into such figures, as well as rectangle (Fig. 16). Now let every square be a rectangle. Then the tromino became a new figure that will be denoted as (Fig. 17), then .
An rectangle is called nice, if it can be split into figures . Clearly, if a rectangle can be split into nice rectangles, it is nice itself. Since , and rectangles are nice, then so are , and rectangles.
We will continue the proof with induction. Consider rectangle.
, we can split into six .
, firstly, from we are cutting off, then the remainder can be split into two .
, can split into ten .
, firstly, from we are cutting off, then the remainder can be split into three .
Suppose all rectangles of the type for are nice. Consider rectangle. Split it into and .
Fig. 17 Both are nice by the induction hypothesis and by an easy split into other nice rectangles. This completes the proof.
Consider an L-shaped tromino that consists of three cells. , rectangles can be easily split into such figures, as well as rectangle (Fig. 16). Now let every square be a rectangle. Then the tromino became a new figure that will be denoted as (Fig. 17), then .
An rectangle is called nice, if it can be split into figures . Clearly, if a rectangle can be split into nice rectangles, it is nice itself. Since , and rectangles are nice, then so are , and rectangles.
We will continue the proof with induction. Consider rectangle.
, we can split into six .
, firstly, from we are cutting off, then the remainder can be split into two .
, can split into ten .
, firstly, from we are cutting off, then the remainder can be split into three .
Suppose all rectangles of the type for are nice. Consider rectangle. Split it into and .
Fig. 17 Both are nice by the induction hypothesis and by an easy split into other nice rectangles. This completes the proof.
Final answer
Yes
Techniques
Induction / smoothing