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 [quote="Zag"][quote="novice"]Nice. You're constructing a right-angled triangle with hypothenuse (x+1)/2 and shorter sides of y and (x-1)/2. Pythagoras shows that y^2 = ((x+1)^2-(x-1)^2)/4 = x.[/quote] Thanks! I read esme's description 3 times and still hadn't figured out what he was trying to say. Now I get it. [quote="esme"].. intersecting the half-circle with [strike]the[/strike] a line perpendicular to the [strike]line[/strike] diameter that passes through the point that is at distances x and 1 from the [strike]circle[/strike] ends of the diameter.[/quote] This was the line that I didn't understand what he was saying. I've changed it so that I would have (in the hopes it helps anyone else who was similarly stymied). But I totally agree with novice that this is an extremely clever way to get a square root![/quote]
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esme
Posted: Sun Mar 17, 2013 11:23 pm    Post subject: 1

http://en.wikipedia.org/wiki/For_Esm%C3%A9%E2%80%94with_Love_and_Squalor

Anyways, it was a Discworld reference a long time ago, and Thok got it right.
Chuck
Posted: Sun Mar 17, 2013 8:54 pm    Post subject: 0

Maybe it was the é character. You could use Tiny URL.
Quailman
Posted: Sun Mar 17, 2013 5:33 pm    Post subject: -1

I took the whole link out. You can't link to stupid URLs with hyphens in them.
Thok
Posted: Sun Mar 17, 2013 3:37 pm    Post subject: -2

I'm pretty sure the bolded word is the reason for esme's comment.

Zag wrote:
I read esme's description 3 times and still hadn't figured out what he was trying to say.
Quailman
Posted: Sun Mar 17, 2013 1:59 pm    Post subject: -3

Oops!

Fixed it.
esme
Posted: Sun Mar 17, 2013 11:07 am    Post subject: -4

Quailman wrote:
Isn't it a Salinger reference?
Are you sure you got the right link?
Quailman
Posted: Sun Mar 17, 2013 12:48 am    Post subject: -5

Isn't it a Salinger reference?
esme
Posted: Sat Mar 16, 2013 11:58 pm    Post subject: -6

Just out of curiosity: What do you think the username "esme" means?
Zag
Posted: Sat Mar 16, 2013 3:04 pm    Post subject: -7

novice wrote:
Nice. You're constructing a right-angled triangle with hypothenuse (x+1)/2 and shorter sides of y and (x-1)/2. Pythagoras shows that y^2 = ((x+1)^2-(x-1)^2)/4 = x.

Thanks! I read esme's description 3 times and still hadn't figured out what he was trying to say. Now I get it.

esme wrote:
.. intersecting the half-circle with the a line perpendicular to the line diameter that passes through the point that is at distances x and 1 from the circle ends of the diameter.

This was the line that I didn't understand what he was saying. I've changed it so that I would have (in the hopes it helps anyone else who was similarly stymied).

But I totally agree with novice that this is an extremely clever way to get a square root!
novice
Posted: Sat Mar 16, 2013 7:23 am    Post subject: -8

esme wrote:
If you know the length 1, you can construct the square root of any given length x by drawing a line segment of length x plus 1, constructing a half-circle with the segment as diameter, intersecting the half-circle with the perpendicular to the line that passes through the point that is at distances x and 1 from the circle. The length of the perpendicular from the diameter to the circle is the square root of x.

Nice. You're constructing a right-angled triangle with hypothenuse (x+1)/2 and shorter sides of y and (x-1)/2. Pythagoras shows that y^2 = ((x+1)^2-(x-1)^2)/4 = x.
esme
Posted: Fri Mar 15, 2013 11:48 pm    Post subject: -9

If you know the length 1, you can construct the square root of any given length x by drawing a line segment of length x plus 1, constructing a half-circle with the segment as diameter, intersecting the half-circle with the perpendicular to the line that passes through the point that is at distances x and 1 from the circle. The length of the perpendicular from the diameter to the circle is the square root of x.
Zag
Posted: Fri Mar 15, 2013 9:53 pm    Post subject: -10

novice wrote:
Finding the square root of an integer (given a unit square):
https://dl.dropbox.com/u/15215428/gl/construction/squareroots.jpg

The observation I made which prompted this thread was an easier way to find the square root of an integer. I realized that you can find the square root of any number which is the difference of two squares.

if x = h 2 - a 2 , then construct a right angle, measure out a from the angle. From that point set your compass to h and find the intersection of the other side of the angle. i.e. you're constructing a right triangle with one side length a and the hypotenuse length h. The other side will be sqrt(x).

Observe the difference between successive squares, and you'll realize that all odd numbers are the difference of two squares. So you can use the above method to get the square root of any odd number. You can use the multiplication technique times some multiple of sqrt(2) to get the square root of any even number.
esme
Posted: Wed Mar 13, 2013 8:42 pm    Post subject: -11

With ruler and compass, you can construct exactly the numbers built with basic arithmetic and square roots (e.g. sqrt( 10 - sqrt(3)) / 2 ) The positive ones among these can be the side length of a constructible square.
novice
Posted: Wed Mar 13, 2013 8:24 pm    Post subject: -12

To solve this we need to construct multiplication, division, and taking the square root.

https://dl.dropbox.com/u/15215428/gl/construction/multiplication.jpg

Division is basically the same, except another line gives you your answer:
https://dl.dropbox.com/u/15215428/gl/construction/division.jpg

Finding the square root of an integer (given a unit square):
https://dl.dropbox.com/u/15215428/gl/construction/squareroots.jpg

So, to construct a square with area p/q where p and q are any integers, given a unit square:

1. Construct q by repeating the unit square q times.
2. Construct sqrt(q) using the method above.
3. Construct sqrt(q) / q using the division method above. This equals 1/sqrt(q).
4. Construct sqrt(p) using the method above.
5. Construct sqrt(p) * (1/sqrt(q)) = sqrt(p/q) using the multiplication method above.
6. Construct a square with sides sqrt(p/q).
Zag
Posted: Tue Mar 12, 2013 10:19 pm    Post subject: -13

No takers, so I'll give a small hint.

Obviously, your goal is to construct a line segment that is [sqrt(A) / sqrt(B)]. Then you just need to make a square with sides equal to that.

If you can construct a line segment that is sqrt(B), then you only need to divide it into B equal parts to have [sqrt(B) / B], which equals [1 / sqrt(B)]

Code:
.

1           sqrt(B)          sqrt(B)
----------  *  ----------   =   ----------
sqrt(B)        sqrt(B)            B

Trojan Horse
Posted: Mon Mar 11, 2013 11:43 pm    Post subject: -14

Hello, it's your friendly neighborhood abstract algebra specialist. I can confirm that Zag is correct; it is possible to construct a square whose area is any positive rational number.

Don't worry, I won't spoil it for the rest of you guys.
Zag
Posted: Mon Mar 11, 2013 10:23 pm    Post subject: -15

He definitely comes into play.
Coyote
Posted: Mon Mar 11, 2013 9:22 pm    Post subject: -16

Interesting question, and I'm also interested in hearing what line of thought led you to think it is in fact possible to construct any such square.

Just to get the ball rolling here, can you give a construction for a 1/7 area square?

 Okay, now that I've thought about it for a bit, I'm thinking you were thinking of Pythagoras in your line of thought. Am I right about that?
pun unintended
Zag
Posted: Mon Mar 11, 2013 7:17 pm    Post subject: -17

I put this problem on the board at work: Constructing a square that is 1/5 the area of a given square

It got me to thinking what fractions (or multiples) of area it is possible to construct. After some thought, I'm pretty sure that it is possible to construct any rational number multiple. Or, more precisely:

Given a unit square, show that it is possible to construct (using only compass and straight edge) a square with area A/B, where A and B are any integer.