Est. 1998 — 27 years of woodworking knowledge
Tuesday puzzle -- volunteer needed
Alex Y
Can someone please step in to this role pending Dan's return. I am running out of source material (as seen by a recently skipped Friday and Tuesday) to be able to keep two days going. So someone (or someones) please post on Tuesdays, or raise a plea to Dan.
Alex
Tuesday puzzle
Bill Earl
You have done an outstanding job Alex. I don't know where you come up with all the material. Here's something for Tuesday:
How many trees do you need to plant to create 10 rows of 3 trees each?
Re: Tuesday puzzle
Gary Smyth
I don't need double digits.
Re: Tuesday puzzle
Alex Y
Why should some 3rd century Greek limit my thinking on this?
My best answer for 10 rows gives me 12 rows, and I can't see any way to eliminate a tree and still get ten rows.
Can the rows overlap? E.g., if I plant four trees in a row in my field, does that count as two rows of 3, i.e., 1,2,3 and 2,3,4? Or do I also get credit for 1,3,4 and 1,2,4, even in that Greek's world?
Re: Tuesday puzzle
John Veerkamp
10 is the best I can do
Re: Tuesday puzzle
Bill Earl
All rows are exactly 3 trees in length. They may intersect, but cannot be co-linear.
Before or after thinning?
Alex Y
The "not colinear" condition shot down my solution for twelve rows with only six trees.
Tuesday Answer
Bill Earl
Gary got this one. The answer is 9 trees arranged as follows:
Re: Tuesday Answer
Alex Y
Very good, and Euclidean to boot! Never did come up with the idea of squeezing the "waist" in like that.
My solution for six trees in twelve rows of three (some colinear) required spherical geometry. Plant one tree each at the north pole, south pole, and along the equator at 0, 90, 180, and 270.
On a sphere, trees are in a row if they are on the same great circle.
I think I was making it too complicated! 