Cover a square with a bunch of squares

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Cover a square with a bunch of squares

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Source: Brazilian Math Olympiad, 2002

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#1 h Nov 15, 2005, 9:32 PM • 2 Y

Y by Adventure10, Mango247

A finite collection of squares has total area $4$ . Show that they can be arranged to cover a square of side $1$ .

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#2 h Jul 27, 2006, 8:21 PM • 2 Y

Y by Adventure10, Mango247

This problem i sreally old, it appeared in Russia MO for Grade 9 in 1979.
Anyways, here is the solution:
Let the sides of the squares be equal to $a_{i}$ for $i=1,2,...,N$ ( $N$ is the number of squares).
If some $a_{k}>1$ then $k$ th square will cover the unit square. Now let's assume the case when $a_{i}<1$ for all $i$
Every number $a_{i}$ must belong to some interval $(2^{-k_{i}}, 2^{-k_{i}+1})$ , ie $2^{-k_{i}}\le a_{i}< 2^{-k_{i}+1}$ , where $k_{i}$ are positive integers for all $i$ . Now let's decrease every $i$ th square to the square with side $b_{i}=\frac{1}{2^{k_{i}}}$ . Then its area would decrease by at most $4$ times, because $1\le \frac{a_{i}}{b_{i}}< 2$ . Therefore the area of all squares will be greater than $1$ .
Now let's prove we can tile the unit square fully with the new squares. Let's divide the unit square into $4$ squares of side $\frac{1}{2}$ . First place the squares with side $\frac{1}{2}$ (if they exist). Then on the non-tiled squares with side $\frac{1}{2}$ (if they exist) place the squares with side $\frac{1}{4}$ (if they exist), with dividing each non-tiled square with side $\frac{1}{2}$ into $4$ equal squares. We will continue this procedure for $k=3,4,...$ by placing squares with side $\frac{1}{2^{k}}$ on the on-tiled squares with side $\frac{1}{2^{k-1}}$ , in turn dividing them into $4$ equal squares.
Finally, because the sum of areas of squares is greater than $1$ , then on some step, we will cover the square. Then increasing the $i$ th square with side $b_{i}$ to the square with side $a_{i}$ , we will get the tiling of the unit square with the given squares.
Yay post numebr 600!

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Umut Varolgunes

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Umut Varolgunes

#3 h Sep 6, 2006, 10:55 AM • 2 Y

Y by Adventure10, Mango247

are moscow oympiads the same as russian olympiads

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#4 h Sep 6, 2006, 1:02 PM • 2 Y

Y by Adventure10, Mango247

No. Moscow olympiad is the olympiad at city level(maybe region level, I am not sure), while All-Russian olympiads are national level, one level higher than moscow olympiads.

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