Growing Grapes in Egypt
Grapes can grow in almost any part of Egypt. When grapes vines are planted they should stand at least 2 meters away from one another whilst growing.
What is the maximum number of vines that can be planted in a square of side length 1 km.
The answer is 289290.
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1 solution
I created a configuration similar to this but the maximum i can fit is 289289. This is because the vines at corners and edges of the 1 sq.km need not occupy the 2 equilateral triangles you have mentioned. In my configuration, there are 578 rows, each 3 distance from the nearest neighbour, And half the rows have 500 vines, half the rows have 501 vines which gave me a total of 289289 vines
How do you know that this is the most optimal configuration?
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Pi Han Goh, I've got a configuration where 289290 vines can be planted, and I think that's probably the optimum. Consider 2 consecutive rows of 501 and 500 trees, all spaced 2 meters apart. Repeat these 2 rows 288 times, all the vines still spaced 2 meters apart. Then finish it off with 2 more rows of 501 vines, all the vines STILL spaced "at least" 2 meters apart. That works out to
2 8 8 ( 5 0 1 + 5 0 0 ) + 2 ( 5 0 1 ) = 2 8 9 2 9 0
The key is noting that there's enough room for a double row of 501 vines somewhere, more easily seen at one end or other, since
1 0 0 0 − 2 8 8 ( 2 3 ) = 2 . 3 3 8 7 3 . . > 2
This solution is similar to Ajinkya Shivashankar's with a total of 578 rows, except that I can stuff 1 more vine into the 1000m square plot.
The best arrangement is a triangular matrixو By applying the Pythagorean to the green triangle,
1
2
+
w
2
=
2
2
hence the smallest distance between rows
w
=
3
m
e
t
e
r
s
.
Check out my solution ...
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How do you know that the best arrangement is a "triangular matrix"?
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@Pi Han Goh – It's actually not a trivial problem to prove that a equitriangular array aligned with one side of the square offers the densest packing, but it does. This problem reminds me of another I posted a while ago Ball Packing Wars
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@Michael Mendrin – Really I am enjoying your solutions... thanks to all of you and sorry for any inconvenience ...
I think the answer is correct for highest number of trees on average per k m 2 . I got the same result as Ajinkya Shivashankar for triangular lattice in square shaped area. I am still not 100% sure that it is the maximum. I think different positions of triangular lattice within the square should be considered.
The answer will be different depending on a shape of the given area. If you take rectangle shaped area of dimension 2 m × 5 0 0 0 0 0 m , the answer will be 2 × 5 0 0 0 0 1 = 1 0 0 0 0 0 2
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I think you should of divided the length of the rectangle over 2 so the result would be 250000 and that is not the optimal solution
To get the best arrangements, each tree must occupy the same area as two equilateral triangles as shown
The best arrangement is a triangular matrixو By applying the Pythagorean to the green triangle, 1 2 + w 2 = 2 2
hence the smallest distance between rows w = 3 m e t e r s .
That is, each tree occupies 2 × 2 s i n ( 6 0 ) = 2 3 m 2 . As 1 k m 2 = 1 0 0 0 × 1 0 0 0 = 1 0 6 m 2 , the maximum number of trees that can be planted is 1 0 0 0 0 0 0 / 2 3 = 2 8 8 6 7 5 trees.