# Re: Math League problem

Paul Burchard (burchard@CS.Princeton.EDU)
Thu, 12 Dec 1996 17:59:25 -0500

Patrick Carney wrote:
> 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.

That's indeed the solution I was suggesting. One fun thing about doing
that way is that it provides a visual proof of the fact that the sum of
the first N odd numbers is N-squared. The number of small triangles in
each strip is 1, 3, 5, etc., while the total number of congruent small
triangles must be N-squared because of the scaling law of area (imagine
horizontal lines between the rows here):
/\ 1
/\/\ 3
/\/\/\ 5
/\/\/\/\ 7
/\/\/\/\/\ + 9 = 25 = 5*5

The league solution appears to be based on the idea that inside the big
triangle
/\ /\ /\
/ \ / \ / \
/ \ , the strip ______ = / \ - /____\ .
/ \ /______\ /______\
/________\
By the scaling law of area, the shaded strip therefore has
(4/5)^2 - (3/5)^2 = 7/25 of the total area, leaving 18/25 for the
unshaded part [where ^2 means "squared"].

PB