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 ]

Spot false dilemmas now, ask me how!
(If you're not on the edge, you're taking up too much room.)

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

Bert Bates
author
Sheriff

Joined: Oct 14, 2002
Posts: 8898

5

posted

0

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
Wanderer
Sheriff

Joined: Jan 30, 2000
Posts: 18671

posted

0

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
author
Sheriff

Joined: Oct 14, 2002
Posts: 8898

5

posted

0

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
Wanderer
Sheriff

Joined: Jan 30, 2000
Posts: 18671

posted

0

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.

Bert Bates
author
Sheriff

Joined: Oct 14, 2002
Posts: 8898

5

posted

0

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