In case you haven’t heard, Rip It Up has just come out in paperback. Just thought I would mention it. No pressure.

On Friday I posted this puzzle….

A man buys 10 trees. Can he plant the 10 trees in 5 rows with 4 trees in each row?

If you have not tried to solve it, have a go now. For everyone else the answer if after the break.

Yes he can. Just imagine a 5 pointed star, and then plant one tree at each point, and one tree where the sides intersect.

Any other solutions?

I have produced an ebook containing 101 of the previous Friday Puzzles! It is called

**PUZZLED** and is available for the

**Kindle **(UK

here and USA

here) and on the

**iBookstore** (UK

here in the USA

here). You can try 101 of the puzzles for free

here.

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Clever puzzle. More like this one, please!

Another solution: http://imgur.com/a/hBR0G

Very good. Didn’t get it.

Too bad for me I’d seen this one before.

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weird, all of my spaces were parsed.

i am not following your soln. i think perhaps you have clever rewording so if (say) five trees in one line you call that several different lines of four trees that just happen to be colinerar?

is this?

let’s try that again; maybe i can outfox the autoformatting:

X

0

0

0

X

0 0

0 0 X

0 0 0

X 0 0 0

0 0 0 X

0 0 0 0 0 0 X

0 0 0 0 0 0 0 0

X 0 0 0 X 0 0 0 X 0 0 0 X

since i will never win, make two perpendicular lines of four trees each that share a tree in the corner. label the column trees 1, 2, 3, and 4. label the row of trees 4, 5, 6, and 7. draw the following straight lines: 1-5, 3-7, and 2-6. the three points where these lines intersect are trees 8, 9, and 10.

Nice solution!

(Looks like a stealth bomber)

I’m really impressed with this – and I understood the second diagram as well !

Nice one!

nice!

Very nice i like it…

http://i.imgur.com/obJ4Zu9.png

(paint + imgur.com is a nice way to quickly show pictures by the way)

Also found the star a while after. But generally I think the other one is easier to expand on (I made it to 17 lines of 4 using 22 trees).

Nope! Need more then 1 to make a row

I can now look at my roughly scribbled solution and realise that if I’d made it more regularl it would be a nice symmetric star, rather than the contorted mess it is.

Found it. But took me too long…

star

The more of these you do the more you realise how you have to think outside the box

I note Richard uses squared note paper. I never buy notebooks without squared paper. So much better than lined/ruled.

Shouldn’t rows be parallel. I’d say the star is made of five lines, not five rows.

Duncan….I’d say the same thing.

That would leave an alternative answer: “no”

Had a look at http://www.rdzl.nl/bomen_planten_raadsel/uitleg.html in Dutch but nice images… explaining it all…:-)

The problem was :”A man buys 10 trees. Can he plant the 10 trees in 5 rows with 4 trees in each row?” But the star is not correponding to word “row”.You were supposed simply to ask if it is possible to plant 10 trees in 5 lines!!! Because “row”as per law of math is always parallel and can not be crossed with another one.

There are actually several distinct solutions; the star is only one. All of them can be constructed as follows:

- Draw a nice long straight line.

- Draw a second straight line that intersects the first.

- Draw three more straight lines making sure each line intersects all the lines you’ve already drawn and avoiding any of the previous points of intersection. That is, no three lines should intersect at the same point.

With the first four lines, there’s only one topologically distinct configuration, but by varying the position of the fifth line, several different distinct configurations can be created.

Amazing! – I truly thought the star was the only solution. But you and ctj have completely blown that idea out of the water.

Blimey!

Someone who claims on Friday to have an alternative solution and demonstrates on Monday that he actually doea have one!

A first for this website surely?

Well done Ken.

Ken, just worked out a formula for the no. of intersecion points, P, based on the no. of lines, n, for the general case in your model. It’s P = 0.5(n^2 – n). E.g. for n = 5, P = 10.

However it should be possible on Earth with only 8 Trees too, …

Place each one at north and south pole, and 4 on the equator.. on 0, 180 90W and 90E

That make 6 trees and 3 rows…

Place two additional trees somewhere diametrically on the planet in an antipodal relationship but not on the equator or the above meridians, and you have 8 trees and 6 lines..

Nobody told that the trees have to be placed on land on a plane or very near to each other…

This dates from my childhood. I’m 86.

Awesome. I was stumped. Who won the caption contest again? I don’t think I got an email about it.

Awesome. I was stumped. Who won the caption contest again? I don’t think I got an email about it. Did anyone else get it?

re-read the Friday Puzzle (Friday’s version).

And well done for being ‘stumped’ :-)

My ideas were either the pentagram or one of the following

0 0 0 0 0 0 0 0 0 0 Trees

——— Row 1

——— Row 2

———- Row 3

———- Row 4

——— Row 5

0 0 0 0 0 0 0 0 0 0 Trees

x x x x Row 1

x x x x Row 2

x x x x Row 3

x x x x Row 4

x x x x Row5

So the 5 rows are hypothetical. Got these Ideas on the train as i tried to count the trees passing by (don’t ask why i would do such things). The problem was that i kept on counting trees twice or missed one. I realized, that this is a solution for the riddle.

For something went wrong here once again:

My ideas were either the pentagram or one of the following:

All ten trees are standing in one row, but you only count those marked with an x

0000000000 Trees

xxxx000000 row 1

0xxxx00000 row 2

00xxxx0000 row 3

000xxxx000 row 4

0000xxxx00 row 5

or

0000000000 Trees

xxxx000000 row 1

0000xxxx00 row 2

x0x0x0x000 row 3

0x0x0x0x00 row 4

x00x00x00x row 5

VrrChurr: Haven’t you got too many trees there? Or am I being a cheese-eater?