Maths Olympiad Prep

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Geometry Difficulty 4.3 AIME Prove it United States

Problem:
Show that for each n17n \geq 17 one can cut a square into nn smaller squares.

Solution

Solution:
One can cut a square into nn smaller squares for all n6n \geq 6. Here is how.
For n=4n=4
Figure 1

For n=6n=6
Figure 2

For n=8n=8
Figure 3

Suppose we know how to cut a square into nn smaller squares. If we take one of the square "pieces" and cut it onto four smaller squares (as in the picture above for n=4n=4). Now we have cut our original square into n+3n+3 pieces. Since we know how to cut a square into 4 pieces, we can cut it into 7,10,,3k+1,7,10, \ldots, 3k+1, \ldots pieces. Similarly, starting with 6 pieces we can get 6,9,,3k,6,9, \ldots, 3k, \ldots pieces, and from 8 pieces 8,11,,3k+2,8,11, \ldots, 3k+2, \ldots pieces. Hence we can get any number of pieces bigger or equal to 6.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.