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Sevenths Sliced Square (Posted on 2011-02-24) Difficulty: 3 of 5
Define a slice of a square to be a line segment with ends on

two different sides,
one corner and an opposing side, or
two opposite corners of the square.

Sequential slices may or may not cross previous ones, but a set of slices will subdivide the square into polygonal regions.

Find (or prove impossible) a way to slice a square into 7 pieces of equal area with n distinct slices for each n={3,4,5,6,7}

  Submitted by Jer    
Rating: 3.0000 (1 votes)
Solution: (Hide)
n=3 may be impossible. It would require a central triangle but this doesn't seem to leave room for the shapes off of its vertices.

n=4 is quite difficult. Make a square with corners (0,0) (7,0) (0,7) and (7,7). The square has area 49 so each region must then have an area of 7.
Make the first cut the line x=5. To the right of this line the area is 14. The second cut will make two equal regions as long as it goes through (6,3.5). Make this cut have a total area of 21 below it. A bit of calculation shows this line should be y = x/5 + 23/10.
The third cut is the hardest one. It goes from (a,0) to (b,7) and leaves an area of 21 to its left, 7 below the second cut and 14 above the second cut. Call the intersection of the second and third cuts (x,x/5+23/10). This yields a set of 3 equations with 3 unknowns:

7=1/2*23/10*x + 1/2*a(1/5x+23/10)
14=1/2*47/10*x + 1/2*b(7-(1/5+23/10))
21=1/2*7*(a+b)

Solving for a gives the immense quadratic:
140a^2 + 1602a - 4946 = 0
With solution
a = (-801+sqrt(1334041))/140 = 2.5862
The third equation of the system simplifies to b=6-a so
b = (1641-sqrt(1334041))/140 = 3.4138

The final cut has one degree of freedom to split the upper left corner. I chose to have it hit the left edge in the same spot as the second cut (0,23/10) and so its other end is at (140/47,7)

n=5 is an X and 3 parallel lines. Picture the following inside a square:

 | | |\ /
 | | | X
 | | |/ \
Using the same square as before the 3 vertical cuts are the lines with equations x=1, x=2 and x=3 and the diagonal ones go from (3,0) to (7,7) and from (3,7) to (7,0).

n=6 is just 6 evenly spaced parallel slices.

n=7 is not possible as you get a minimum of 8 pieces.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some Thoughtsa thoughtbroll2011-02-26 04:07:14
re: n = 4 (spoiler)Brian Smith2011-02-25 20:24:47
Some Thoughtsn = 3 (spoiler)Steve Herman2011-02-25 13:58:52
n = 3Steve Herman2011-02-25 10:43:55
Some Thoughtsn = 4 (spoiler)Steve Herman2011-02-25 10:34:41
Some Thoughtsn = 5 (spoiler)Steve Herman2011-02-25 01:03:41
Some Thoughts7 easy pieces (n = 6,7)Steve Herman2011-02-25 00:52:54
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