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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:
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| | | X
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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. |
