Difference between revisions of "1998 USAMO Problems/Problem 4"

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== Solution ==
 
== Solution ==
{{solution}}
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Answer: <math>98</math>.
  
Answer: 98.
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There are <math>4\cdot97</math> adjacent pairs of squares in the border and each pair has one black and one white square. Each move can fix at most <math>4</math> pairs, so we need at least <math>97</math> moves. However, we start with two corners one color and two another, so at least one rectangle must include a corner square. But such a rectangle can only fix two pairs, so at least <math>98</math> moves are needed.
  
There are 4·97 adjacent pairs of squares in the border and each pair has one black and one white square. Each move can fix at most 4 pairs, so we need at least 97 moves. However, we start with two corners one color and two another, so at least one rectangle must include a corner square. But such a rectangle can only fix two pairs, so at least 98 moves are needed.
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It is easy to see that 98 moves suffice: take 49 <math>1\times98</math> rectangles (alternate rows), and 49 <math>98\times1</math> rectangles (alternate columns).
 
 
It is easy to see that 98 suffice: take 49 1x98 rectangles (alternate rows), and 49 98x1 rectangles (alternate columns).
 
  
 
credit: https://mks.mff.cuni.cz/kalva/usa/usoln/usol984.html
 
credit: https://mks.mff.cuni.cz/kalva/usa/usoln/usol984.html
  
 
editor: Brian Joseph
 
editor: Brian Joseph
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second editor: integralarefun
  
 
==See Also==
 
==See Also==

Latest revision as of 17:56, 10 May 2023

Problem

A computer screen shows a $98 \times 98$ chessboard, colored in the usual way. One can select with a mouse any rectangle with sides on the lines of the chessboard and click the mouse button: as a result, the colors in the selected rectangle switch (black becomes white, white becomes black). Find, with proof, the minimum number of mouse clicks needed to make the chessboard all one color.

Solution

Answer: $98$.

There are $4\cdot97$ adjacent pairs of squares in the border and each pair has one black and one white square. Each move can fix at most $4$ pairs, so we need at least $97$ moves. However, we start with two corners one color and two another, so at least one rectangle must include a corner square. But such a rectangle can only fix two pairs, so at least $98$ moves are needed.

It is easy to see that 98 moves suffice: take 49 $1\times98$ rectangles (alternate rows), and 49 $98\times1$ rectangles (alternate columns).

credit: https://mks.mff.cuni.cz/kalva/usa/usoln/usol984.html

editor: Brian Joseph

second editor: integralarefun

See Also

1998 USAMO (ProblemsResources)
Preceded by
Problem 3
Followed by
Problem 5
1 2 3 4 5 6
All USAMO Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png