A little geometry and algebra anyone?

Once again stretching (or maybe breaking) the limits of ascii art:

If 1,2,3,4,5 are squares (yes even the ones that look like parallelograms

), and join at the corners (that maybe you can imagine :roll: ), prove that the area of square 3 is equal to the area of triangle T.
p.s. I extended the bases of 2 and 5 to indicate they are on the same line as the corner of 3.
[ November 19, 2003: Message edited by: Bert Bates ]
[ November 19, 2003: Message edited by: Bert Bates ]

how's this for a non-ascii version? (not to scale)

updated my snazzy diagram to better meet Bert's specifications

[ November 20, 2003: Message edited by: Jessica Sant ]

Hmmm, I think I'll go insane if I try show diagrams of each step, so I'll just outline the process.
If we let
a = sides of square 2
b = sides of square 5
c = sides of square 3
Then it's easy to see that the two triangles under 3 are identical (though in different orientation). So we have

Area of 3 = c^2 = a^2 + b^2
Now let
d = sides of square 1
e = sides of square 4
It can be shown that
d^2 = 4a^2 + b^2
e^2 = a^2 + 4b^2
Now to get area of T we need
T = de sin θ / 2
where θ is the bottom angle of T. If we draw a vertical line through that bottom corner we can divide θ into two parts α and β. which are actually the same as the angles made by the bottommost edges of 1 and 4 with the horizontal. So we can say
sin θ = sin(α + &beta

= sin α cos β + cos α sin β
T = (de/2) sin θ = (de/2)[(b/d)(2b/e) + (2a/d)(a/e)] = (1/2)[2b^2 + 2a^2] = a^2 + b^2

I'd print out the diagram. Take a scissors to triangle T and cut it straight down the middle and fit the pieces on to square 3.

regards

Jessica -
cool pix - but... :
- lower corner of 1 meets upper left corner of 2
- right corner of 3 meets upper left of 5
- lower right of 4 meets upper right of 5
Jim - OK, I'll see if I can figure out what you just said... the good news is, I don't think your answer gave it away for many people

OK Jim, first question:

It can be shown that
d^2 = 4a^2 + b^2
e^2 = a^2 + 4b^2

how did you make that leap?
[ November 19, 2003: Message edited by: Bert Bates ]

That would be telling.

I thought you were happy my answer didn't give too much away? This way there are still puzzles within puzzles.

As master and commander of this particular puzzle, on this forum, I hereby declare Jim's answer null and void

(Actually, I'm sure he got it right, in about 1/20th the time it took me :roll: )
However, no "then at step 2 a miracle occurs" solutions will be accepted.
Jess, any chance you'll update your fine picture?

Well, do you want it "given away", or not?

Just draw in two horizontal and two vertical lines from the four corners of square 3, toward the center. You thus form a spiral of four triangles, each with side lengths a & b (plus hypotenuse c) and a square in the middle with side length b-a (assuming b > a, otherwise it's a - b). From this it's easy to find two right triangles which demonstrate that
d^2 = (2a)^2 + b^2
e^2 = a^2 + (2b)^2
Those same two right triangles can be used in a subsequent step to figure out what the sine and cosine of α and β are.
[ November 20, 2003: Message edited by: Jim Yingst ]

Originally posted by Bert Bates:
Jess, any chance you'll update your fine picture?

I did, I did!! -- Hit reload -- I just replaced the file so it will show up in my 1st post up there

Lemme know if it needs to be adjusted anymore.

Jessica, thanks for your amazing picture!
Jim, my hat's off to you! Took me about two pages of formulas and equations...
It just encourages me however, to find a puzzle that will slow you down