# 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 non-ascii 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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Posts: 18671

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*

HS Thomas

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Bert Bates

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Posts: 8905

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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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Sheriff

Posts: 18671

Bert Bates

author

Sheriff

Sheriff

Posts: 8905

5

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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Sheriff

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 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 ]

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 ]

"I'm not back." - Bill Harding, *Twister*

Bert Bates

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Posts: 8905

5

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.)