wu :: forums
wu :: forums - Line Segments in a Square

Welcome, Guest. Please Login or Register.
Jan 23rd, 2022, 4:25am

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   hard
(Moderators: ThudnBlunder, Eigenray, Grimbal, Icarus, towr, SMQ, william wu)
   Line Segments in a Square
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: Line Segments in a Square  (Read 1018 times)
cool_joh
Newbie
*





   
WWW

Gender: male
Posts: 50
Line Segments in a Square  
« on: Jul 5th, 2008, 7:41pm »
Quote Quote Modify Modify

A finite set of line segments (each including its endpoints) of total length 18 is placed on a square of side 1 (the square includes the boundary and the interior). Each segment is parallel to one of the sides and does not extend beyond the boundary of the square. Show that the lines divide the square into regions one of which has area at least 1/100.
IP Logged

MATH PRO
towr
wu::riddles Moderator
Uberpuzzler
*****



Some people are average, some are just mean.

   


Gender: male
Posts: 13730
Re: Line Segments in a Square  
« Reply #1 on: Jul 6th, 2008, 7:11am »
Quote Quote Modify Modify

I don't see any reason why the line-segments would have to divide the square at all.
For example, you can have 36 parallel pieces 0.5 units long spaced 0.02 units apart, somewhere in the middle of the square.
IP Logged

Wikipedia, Google, Mathworld, Integer sequence DB
cool_joh
Newbie
*





   
WWW

Gender: male
Posts: 50
Re: Line Segments in a Square  
« Reply #2 on: Jul 6th, 2008, 7:14am »
Quote Quote Modify Modify

In that case, I think the region is the whole square, whose area is 1 > 1/100.  Grin
IP Logged

MATH PRO
NightBreeze
Newbie
*





   
Email

Posts: 26
Re: Line Segments in a Square  
« Reply #3 on: Jul 6th, 2008, 8:59am »
Quote Quote Modify Modify

Indeed. The problem is to proof that for every possible division, there is a part of the square whose area is > 1/100
 
Edit: the inequality can't be strict. Since 9 horizontal and 9 vertical lines, equally spaced apart will divide the square into 100 parts of equal surface.
« Last Edit: Jul 6th, 2008, 11:07am by NightBreeze » IP Logged
Grimbal
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 7517
Re: Line Segments in a Square  
« Reply #4 on: Jul 6th, 2008, 12:20pm »
Quote Quote Modify Modify

Suppose we manage to create a division where all areas are <1/100 unit.
 
The total perimeter of the regions equals the perimeter of the square plus twice the lines dividing the square (twice because they contribute on both sides). So the total perimeter is
  P = 4 + 218 = 40
 
Since all lines are horizontal and vertical, the minimum perimeter for one region of area ai is achieved with a square, where the perimeter is 4sqrt(ai). So
  pi >= 4sqrt(ai)
 
Since ai<1/100 for all regions, we have
  pi >= 4sqrt(ai) = 4ai/sqrt(ai) > 4ai/sqrt(1/100) = 40ai.
  pi > 40ai.
 
Since the total perimeter P=sum(pi) and the total area A = sum(ai), we have
  P > 40A
But A = 1 and P = 40, so we have a contradiction.
« Last Edit: Jul 6th, 2008, 12:22pm by Grimbal » IP Logged
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright 2000-2004 Yet another Bulletin Board