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Here's two puzzles for the weekend. (If you've seen them before, don't
post a solution right away so others get a chance to try them.)
(a) Place 10 coins on a flat surface so that there are 5 straight rows with
4 coins in each row.
(b) Place 13 coins on a flat surface so that there are 9 straight rows with
4 coins in each row.
You're not allowed to stack coins.
Enjoy!

> (a) Place 10 coins on a flat surface so that there are 5 straight rows with
> 4 coins in each row.
I have discovered a truly wondrous configuration, but the pico buffer is
too small to contain it. ;-)
Seriously though, I believe I've found an answer to this one via trial and
error. If you want to figure this out on your own, avert your eyes!
(that's easier said than done, heh..)
Due to the limitations of ASCII, this may not look quite perfect, but here
goes:
o
o o
o
o o o o
o o
If I have more free time I may give the other arrangement a shot...

Here's my solutions to this coin-arrangement problem, in case anyone's
still puzzling over it.
For the 10-coin 5-row problem, a 5-pointed star arrangement is all you need
(or you can use Jason's construction). Excuse the crude ASCII art---it
doesn't line up quite perfectly here like it would on paper.
o
o o o o
o o
o
o o
The 13-coin 9-row problem is a bit trickier. Here are the steps to create
it:
1. Make a line segment AC, and choose a point B on it.
2. Make a line segment CE (not colinear with AC), and choose a point D on
it.
3. All the rest of the points are now determined by intersecting line
segments as shown below. (Place remaining points in the order
F,G,H,I,J,K,L,M). Note that this is hard to draw using ASCII; you can
confirm it, however, by drawing it on paper.
_C
_-//|
_-- //|
_-- _// |
_-- / / |
_-- / / |
_-I _/ / |
_-- | / / |
_-B_ | / _/ |
_-- |\--_ | _/ / |
_-- \\_ -_| / / |
_-- \ \ -G_ / |
_-- | \J/| L-_ |
-- \ / \|/ --_ |
A--------------------F---H_------D
| / |\ |
\/ | \_ |
M\ | \ |
| | K
\ | |
\ | |
| | |
\ ||
\||
E
The 9 lines are AC, AD, BE, BK, BD, CE, CM, CF, IE.
Note that to figure this out I started with the 3x3 diamond at the lower
part of the diagram (8 rows of 3 coins each) and then "skewed" the diamond
so it was possible to add the "top" row of 4 coins so that each of the 8
lines in the diamond had an extra coin from this new row. The above
construction method generates a properly skewed 3x3 diamond as well as the
top row AC.
Mike