Maths Olympiad Prep

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, 2007

Combinatorics Difficulty 7.3 National Olympiad, round 2 Prove it Japan

There is a grid of 5×55 \times 5. And write the integers 1,2,,161,2,\ldots,16 in all the grids (each number can be written only once) in the upper left grid of 4×44 \times 4 (♣). About the 4 rows, write the sum of four numbers that are written in each row at the right end of each row. Similarly, about the 4 columns, write the sum of four numbers that are written in each column at the lower end of each column. And nothing is written in the lower right grid. Find the maximum integer mm which satisfies the following condition.

Conditions: About step (♣), there exists a way of writing the numbers such that you can choose two numbers a,ba, b which abm|a-b| \ge m hold from the right end column and also from the lower end row.

Solution

Define the numbers written to each grid as shown in the table 1. I can assume that A1A_1 is the minimum and A4A_4 is the maximum in A1,A2,A3,A4A_1, A_2, A_3, A_4, and that B1B_1 is the minimum and B4B_4 is the maximum in B1,B2,B3,B4B_1, B_2, B_3, B_4, by rearranging rows and columns appropriately. Now mA4A1m \leq A_4 - A_1, mB4B1m \leq B_4 - B_1, so
m(A4A1)+(B4B1)2=12(a4,4a1,1)+12(a4,4+a4,2+a4,3+a2,4+a3,4a1,1a1,2a1,3a2,1a3,1)12(161)+12(16+15+14+13+1212345)=35 \begin{aligned} m &\leq \frac{(A_4 - A_1) + (B_4 - B_1)}{2} \\ &= \frac{1}{2}(a_{4,4} - a_{1,1}) + \frac{1}{2}(a_{4,4} + a_{4,2} + a_{4,3} + a_{2,4} + a_{3,4} - a_{1,1} - a_{1,2} - a_{1,3} - a_{2,1} - a_{3,1}) \\ &\leq \frac{1}{2}(16 - 1) + \frac{1}{2}(16 + 15 + 14 + 13 + 12 - 1 - 2 - 3 - 4 - 5) \\ &= 35 \end{aligned}
And there exists a way of writing if m=35m = 35 (table 2). So the maximum mm is 3535. This is table 1.

a1,1a_{1,1}a1,2a_{1,2}a1,3a_{1,3}a1,4a_{1,4}A1A_1
a2,1a_{2,1}a2,2a_{2,2}a2,3a_{2,3}a2,4a_{2,4}A2A_2
a3,1a_{3,1}a3,2a_{3,2}a3,3a_{3,3}a3,4a_{3,4}A3A_3
a4,1a_{4,1}a4,2a_{4,2}a4,3a_{4,3}a4,4a_{4,4}A4A_4
B1B_1B2B_2B3B_3B4B_4

This is table 2.
1251119
3671228
4891435
1013151654
18293653

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