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Topic: Trees for Willywutang (Read 4074 times) |
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Jeremiah Smith
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Trees for Willywutang
« on: Sep 19th, 2002, 7:08pm » |
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So, Willywutang has (somehow) managed to get himself a nice big mansion. The mansion has a nice huge yard in front. However, the yard is completely flat and boring, so Willy decides it'd look nice with a few trees in front. So, he has a landscaper come in to put in some trees. Being the puzzlemeister that he is, Willy decides to give the landscaper a riddle: Plant 9 trees in the yard, so that there are 10 rows of three trees each. Help the poor landscaper decide how to place the trees.
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| « Last Edit: Oct 23rd, 2003, 7:42pm by Icarus » |
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Pietro K.C.
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Re: PUZZLE - Trees for Willywutang
« Reply #1 on: Sep 21st, 2002, 11:01pm » |
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We assume the yard is so big that in fact if goes all the way around the Earth. So the landscaper just has to place the 10 trees in a straight line, equally spaced. Let them be T1, ... T10. Then (T1,T2,T3) is a row, (T2,T3,T4) is a row, ..., (T10,T1,T2) is a row, and so there are 10 three-tree rows in all. 
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"I always wondered about the meaning of life. So I looked it up in the dictionary under 'L' and there it was --- the meaning of life. It was not what I expected." (Dogbert)
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Jeremiah Smith
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Re: PUZZLE - Trees for Willywutang
« Reply #2 on: Sep 22nd, 2002, 3:44pm » |
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Oh. Hm.
I should say that you can't count partial rows, as it were. If you had five in a row, you couldn't count three separate three-tree rows. For a row of three to count, there can only be three collinear trees.
Plus, Willy's yard isn't the size of the world.
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Pietro K.C.
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Re: PUZZLE - Trees for Willywutang
« Reply #3 on: Sep 22nd, 2002, 4:53pm » |
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Well, that DOES make it a lot more complicated... I've given it some thought, how about this:
-place 3 collinearly;
-make that the side of an equilateral triangle of 6 trees, like the following:
.
. .
. . .
this gives us 3 rows;
-add a 7th tree in the middle of the triangle, making 6 rows (the sides plus the three medians/angle bisectors/heights)
Now we have 2 more trees to make four more rows, which should be feasible... I just don't see how right now. If you have a quadrilateral with tree vertices, adding a tree at the diagonals' intersection gives you an additional 2 rows, but you said we can't count partial rows.
Wait! I got it! Go like this:
. . .
...
. . .
Ha-ha!  Notice that I no longer mean the second dot on the second line to mean the center of the equilateral triangle. I mean it just where it is. Numbering the dots in reading order, we have the 3-tree rows:
123, 456, 789 (across)
148, 159 (diagonals starting at 1)
247, 258, 269 (at 2)
357, 368 (at 3)
Yay!
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"I always wondered about the meaning of life. So I looked it up in the dictionary under 'L' and there it was --- the meaning of life. It was not what I expected." (Dogbert)
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