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        riddles >> medium >> Trees for Willywutang
        
(Message started by: jeremiahsmith on Sep 19^th, 2002, 7:08pm)

Title: Trees for Willywutang
Post by jeremiahsmith on Sep 19^th, 2002, 7:08pm

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.

Title: Re: PUZZLE - Trees for Willywutang
Post by Pietro K.C. on Sep 21^st, 2002, 11:01pm

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. ;D

Title: Re: PUZZLE - Trees for Willywutang
Post by jeremiahsmith on Sep 22^nd, 2002, 3:44pm

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.

Title: Re: PUZZLE - Trees for Willywutang
Post by Pietro K.C. on Sep 22^nd, 2002, 4:53pm

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!