On Thu, 12 Dec 1996, Patrick Carney wrote:
> Just another voice on the triangle problem (and it may be what Paul had in
> mind -- if so, I apologize for the repetition).
>
> Instead of drawing just the one set of parallel lines, draw the same set of 4
> // lines in each direction -- i.e., // to AB and BC as well as to AC. That
> divides the figure into 25 congruent triangles all similar to the original.
> But the strip in question has 7 of them and does not have 18 of them. Hence
> the answer. I am also interested in what explanation the league gave. I hope
> you will eventually post it, Chuck.
>
> As ever,
> Bro. Pat Carney
>