A little geometry and algebra anyone?
Bert Bates
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posted 13 years ago
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 ]
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.)
posted 13 years ago
how's this for a nonascii version? (not to scale)
updated my snazzy diagram to better meet Bert's specifications
[ November 20, 2003: Message edited by: Jessica Sant ]
updated my snazzy diagram to better meet Bert's specifications
[ November 20, 2003: Message edited by: Jessica Sant ]
 Jess
Blog:KnitClimbJava  Twitter: jsant  Ravelry: wingedsheep
Jim Yingst
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posted 13 years ago
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
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'm not back."  Bill Harding, Twister
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Bert Bates
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posted 13 years ago
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:
how did you make that leap?
[ November 19, 2003: Message edited by: Bert Bates ]
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 ]
Spot false dilemmas now, ask me how!
(If you're not on the edge, you're taking up too much room.)
Jim Yingst
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Bert Bates
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Posts: 8919
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posted 13 years ago
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?
(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?
Spot false dilemmas now, ask me how!
(If you're not on the edge, you're taking up too much room.)
Jim Yingst
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Posts: 18671
posted 13 years ago
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 ba (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 ]
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 ba (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 ]
"I'm not back."  Bill Harding, Twister
Bert Bates
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posted 13 years ago
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
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
Spot false dilemmas now, ask me how!
(If you're not on the edge, you're taking up too much room.)
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