Olympiad Maths Prep

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Geometry Difficulty 5.9 AIME, harder Prove it Ukraine

One corner cell is removed from a 2011×20112011 \times 2011 square. Is it possible to cut the obtained figure along the lines of the grid into less than 121121 squares?

Solution

We will write axa \to x, if an a×aa \times a square without one corner cell can be cut along the lines of the grid into xx squares.

Evidently, 232 \to 3. Also note that 787 \to 8 and 999 \to 9 (fig. 24 and 25). We will show that if axa \to x and byb \to y, then abx+yab \to x+y. Indeed, a square with the side length abab without a corner cell can be cut into two parts: a square with the side length bb without a corner cell, and a square with the side length abab, with a b×bb \times b square removed from its corner. The first part can be cut into yy squares, and the second part can be cut into xx squares, since it is a square with the side length aa without a corner, magnified by a factor of bb. So,
ax2ax+3.() a \to x \Rightarrow 2a \to x+3. \quad (*)
We will also show that
ax2a12x+2.() a \to x \Rightarrow 2a-1 \to 2x+2. \quad (**)
Indeed, an "incomplete" square can be partitioned into 2 squares with the side length a1a-1 and 2 "incomplete" squares with the side length aa (fig. 26).
So: 787 \to 8, 999 \to 9, 631763 \to 17, 12620126 \to 20, 25223252 \to 23, 50348503 \to 48, 1006511006 \to 51, 20111042011 \to 104, i.e., we have shown that our original figure can be cut into 104<121104 < 121 squares.

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