Maths Olympiad Prep

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Combinatorics Difficulty 6.8 National olympiad Prove it

22. C2 (CAN) (a) If a 5×n5 \times n rectangle can be tiled using nn pieces like those shown in the diagram, prove that nn is even. ! (b) Show that there are more than 23k12 \cdot 3^{k-1} ways to tile a fixed 5×2k5 \times 2 k rectangle ( k3k \geq 3 ) with 2k2 k pieces. (Symmetric constructions are considered to be different.)

Solution

22. (a) Color the first, third, and fifth row red, and the remaining squares white. There in total nn pieces and 3n3 n red squares. Since each piece can cover at most three red squares, it follows that each piece colors exactly three red squares. Then it follows that the two white squares it covers must be on the same row; otherwise, the piece has to cover at least three. Hence, each white row can be partitioned into pairs of squares belonging to the same piece. Thus it follows that the number of white squares in a row, which is nn, must be even. (b) Let aka_{k} denote the number of different tilings of a 5×2k5 \times 2 k rectangle. Let bkb_{k} be the number of tilings that cannot be partitioned into two smaller tilings along a vertical line (without cutting any pieces). It is easy to see that a1=b1=2,b2=2,a2=6=23,b3=4a_{1}=b_{1}=2, b_{2}=2, a_{2}=6=2 \cdot 3, b_{3}=4, and subsequently, by induction, b3k4,b3k+12b_{3 k} \geq 4, b_{3 k+1} \geq 2, and b3k+22b_{3 k+2} \geq 2. We also have ak=bk+i=1k1biakia_{k}=b_{k}+\sum_{i=1}^{k-1} b_{i} a_{k-i}. For k3k \geq 3 we now have inductively ak>2+i=1k12aki23k1+2ak123k a_{k}>2+\sum_{i=1}^{k-1} 2 a_{k-i} \geq 2 \cdot 3^{k-1}+2 a_{k-1} \geq 2 \cdot 3^{k}

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.