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A little geometry and algebra anyone?

 
Bert Bates
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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 ]
 
Jessica Sant
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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 ]
 
Jim Yingst
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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
 
HS Thomas
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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
 
Bert Bates
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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 ]
 
Jim Yingst
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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.
 
Bert Bates
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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?
 
Jim Yingst
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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 ]
 
Jessica Sant
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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.
 
Bert Bates
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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
 
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