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 Sevenths Sliced Square (Posted on 2011-02-24)
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}

 See The Solution Submitted by Jer Rating: 3.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
 n = 3 (spoiler) | Comment 5 of 7 |
No, on second thought, I don't think n = 3 is possible.  Each of the 3 slices needs to divide the square into a 3/7 piece and a 4/7 piece, and also cross the other two slices.

On a 14 X 14 square this would in one configuration involve:
a) slice 1 passes from top to bottom through (6,7)
b) slice 2 passes from left to right through (7,6)
c) slice 3 passes from left to right through (7,8)
If slice 2 and 3 cross (which they need to), then the inner triangle formed by the three slices is less than 1/7 of the area, because the three points (6,7), (7,6) and (7,8) are just too close.

Similar arguments seem to apply for other configurations, like:
a) slice 1 passes from left to top
b) slice 2 passes from left to right
c) slice 3 passes from top to bottom

I'm going to be pleasantly surprised and impressed if Jer has a configuration that works for n = 3

Edited on February 25, 2011, 4:22 pm
 Posted by Steve Herman on 2011-02-25 13:58:52

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