Sevenths sliced square
The square in the diagram has been divided into 7 pieces by 4 slices, and note that the pieces are not equal size in this example.
As illustrated, define a slice of a square to be a line segment with ends on
- two different sides whether opposite or adjacent (in red),
- one corner and an opposing side (in blue), or
- two opposite corners of the square (in green).
Sequential slices may or may not cross previous ones, but a set of slices will subdivide the square into polygonal regions.
Let denote the number of slices. Then for what values of is it possible to slice a square into exactly 7 pieces of equal areas?
Pick the set of all possible values of
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1 solution
n=5 and n=6 are easy enough that they are part of every choice. It is pretty easy to see that n=7 is impossible. You'd get 8 pieces minimum.
n=3 is impossible.
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 is a contradiction.
n=4 is possible. Here is a concrete example
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 ∗ 2 3 / 1 0 ∗ x + 1 / 2 ∗ a ( 1 / 5 x + 2 3 / 1 0 )
1 4 = 1 / 2 ∗ 4 7 / 1 0 ∗ x + 1 / 2 ∗ b ( 7 − ( 1 / 5 + 2 3 / 1 0 ) )
2 1 = 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)