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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)

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 a thought Comment 7 of 7 |

I've not really followed this until now, but it seems to me that n=3 is not possible.

There must be a triangle in the middle for the 7th segment. It has to be as far away from the sides as possible. This sounds to me like an equilateral triangle lined up on a diagonal of the square.

Now assume the square is of area 49. Then the equilateral triangle is of area 7, so its sides are of length 4. This leaves nowhere near enough room for the small triangular/poly shapes around its vertices, and far too much for the polygons on the sides.

But if the triangle in the middle is not equilateral, then one of its vertices will be even nearer a side of the square, so the problem will be even worse, unless the area of the central triangle is reduced; but that is not allowed, because the middle triangle must occupy 1/7 of the area; so there seems to be a contradiction.

Edited on February 26, 2011, 4:08 am
 Posted by broll on 2011-02-26 04:07:14

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