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      /  環球城市數學競賽2008秋季賽高級卷國中組第一題
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註冊日: 2002-07-30
發表數: 1094


 環球城市數學競賽2008秋季賽高級卷國中組第一題

1. 在一塊100×100的棋盤上放入100枚皇后,使得它們之間都互相不攻擊。若將此棋盤切割成四塊50×50的小棋盤,試證每塊小棋盤上至少有一枚皇后。(註:西洋棋皇后可攻擊東、西、南、北、東南、西南、東北、西北方向的棋子。)(四分)


_________________
孫文先 敬上

 2009-01-12 10:27個人資料傳送 Email 給 孫文先
colfulus
Not too shy to talk



註冊日: 2008-08-29
發表數: 27


 Re: 環球城市數學競賽2008秋季賽高級卷國中組第一題

First assume the opposite: it is possible to put all 100 queens on the chessboard so that in one 50x50 subboard there are no queens.

The chessboard can roughly be drawn as follows, where no queens can go in the X'ed part.

XXXXX-----
XXXXX-----
XXXXX--A--
XXXXX-----
XXXXX-----
----------
----------
--B----C--
----------
----------


Clearly, in each row there must be exactly one queen. If there are two or more, those queens would attack each other; if there are none, by the pigeonhole principle some other row will have two or more queens. In particular, in the top 50 rows there must be 50 queens, and since they can't go in the X'ed subboard they must all go in the A subboard.

By the same reasoning with columns, subboard B must also have 50 queens so each column will have a queen in it.

Therefore, there are no queens left for subboard C; it must also be empty.


      50
     +---+
     |   |
XXXXX/////
XXXXX/////-+
XXXXX//A// | 49
XXXXX///// |
XXXXX/////-+
/////XXXXX
/////XXXXX
//B//XXCXX
/////XXXXX
/////XXXXX


Now consider the diagonals that run from A to B. There are only 99 of them that go through A and B. The queens can't be on any other diagonals, because they are limited to the subboards A and B. By the pigeonhole principle at least one of the diagonals must have two or more queens on it; those queens attack each other.
Contradiction.
Therefore it is impossible to put all 100 queens on the board this way.

 2009-01-27 21:21個人資料


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