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Mathematical Olympiad in Ukraine, Third Round (January 23, ), 2010

Combinatorics Difficulty 6.4 National Olympiad Prove it Ukraine

A rectangle with the lengths of sides $2010$ and $11$ is partitioned into unit squares. The external slice of squares is colored with yellow, the next slice (all squares that share a vertex with external slice) are colored with blue. A slice of squares that touches the previous slice is colored with yellow and so on. Find the number of yellow and blue squares.

Solution

Do allocate external slices of squares of each color step by step:
$2010 \times 11$ - yellow; $2008 \times 9$ - blue; $2006 \times 7$ - yellow; $2004 \times 5$ - blue; $2002 \times 3$ - yellow;
$2000 \times 1$ - blue.

Now calculate the number of squares in each slice as difference between numbers of squares of external and internal rectangles. Bottom-up:

Blue: $2000 \cdot 1 = 2000$ .
Yellow: $2002 \cdot 3 - 2000 \cdot 1 = 4006$ .
Blue: $2004 \cdot 5 - 2002 \cdot 3 = 4014$ .
Yellow: $2006 \cdot 7 - 2004 \cdot 5 = 4022$ .
Blue: $2008 \cdot 9 - 2006 \cdot 7 = 4030$ .
Yellow: $2010 \cdot 11 - 2008 \cdot 9 = 4038$ .

And simultaneously: yellow $4038 + 4022 + 4006 = 12066$ , blue $4030 + 4014 + 2000 = 10044$ .

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