Straightedge and graph paper geometry

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Aldarion
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Straightedge and graph paper geometry

Remember those compass-and-straightedge constructions? Now, suppose I take your compass away, but instead give you a sheet of ordinary graph paper: essenstialy, a grid of unit-size squares. You're free to assume the sheet is infinitely big, of course.
Now, what will you be able to do with a sheet of graph paper and a straightedge?

Can you find me a segment of any rational length?
Can you find me a sum of two provided segments of any length?
Can you find the product of two lengths? Can they be arbitrary, or only whole/rational/something else?
Can you find me a square root of a segment? Of any segment? What about a cube root? Arbitrary root?
Can you find me a segment with a length which is irrational? Transcendental?

Can you construct an equilateral triangle?
*If you can - do so.
*If you can't - prove it, and construct a good approximation with difference between sides of at most... let's make it 5%.
Same with a pentagon?
Hexagon?

Generally, which points can you construct with such a method?
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Cleverbeans
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Re: Straightedge and graph paper geometry

Aldarion wrote:Can you find me a segment with a length which is irrational? Transcendental?

Draw a diagonal through a unit square for an irrational.
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Aldarion
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Re: Straightedge and graph paper geometry

Cleverbeans wrote:
Aldarion wrote:Can you find me a segment with a length which is irrational? Transcendental?

Draw a diagonal through a unit square for an irrational.

Correct!
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The Pathological Case
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Re: Straightedge and graph paper geometry

Am I granted the ability to lay the straightedge down perfectly parallel with one of the grid directions? This is crucial to being able to translate line segments of arbitrary length. How about parallel to any existing line segment?
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antonfire
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Re: Straightedge and graph paper geometry

What operations are allowed?

If all you can do is draw a line between two points and take the intersection of two lines, then:
Spoiler:
You can only construct rational points (points with rational coordinates). Given two rational points, the line through them has a presentation ax+by=c where a, b, and c are rational (a rational line), and given two such lines, the intersection between them is a rational point. You can construct any rational point, since given a rational point, any line through it and some lattice point is rational, and contains another lattice point, so any two such lines are constructable and intersect at that point.

So the lengths you can do are rational numbers times numbers which are representable as a sum of two perfect squares.

In particular, you can't always add two of these, because [imath]1+\sqrt 2[/imath] is not such a number.

You can always multiply two such numbers, and there's probably a nice construction for it, maybe someone can work out what that is.

You can't always find a square root, because you can't construct [imath]\sqrt[4] 2[/imath], same for other roots.

Irrational yes, Transcendental, no.

Equilateral triangle, no, because you can't do [imath]\sqrt 3[/imath]. (To construct approximations, just use rational approximations of sqrt(3)/2.)

While you can construct [imath]\sqrt 5[/imath], you can't make a pentagon, because if the two coordinates of one side of a pentagon are rational points, the opposite one is not. (Similar reasoning for all regular polygons other than a square.)

If you could do a hexagon, you could do an equilateral triangle.

Moral of the story, do the last exercise in a long chain first.
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achan1058
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Re: Straightedge and graph paper geometry

To get a line segment of a/b, draw a right triangle of length b and height a. Now, look at the vertical segment that is 1 unit away from the right angle.

Aldarion
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Re: Straightedge and graph paper geometry

The Pathological Case wrote:Am I granted the ability to lay the straightedge down perfectly parallel with one of the grid directions? This is crucial to being able to translate line segments of arbitrary length. How about parallel to any existing line segment?

Nope.
antonfire wrote:If all you can do is draw a line between two points and take the intersection of two lines

Yep.

antonfire wrote:then:

A nice analysis, antonfire!
I hereby formally applaud you.
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Cosmologicon
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Re: Straightedge and graph paper geometry

antonfire wrote:So the lengths you can do are rational numbers times numbers which are representable as a sum of two perfect squares....

Equilateral triangle, no, because you can't do [imath]\sqrt 3[/imath]. (To construct approximations, just use rational approximations of sqrt(3)/2.)

What am I missing here? An equilateral triangle only has 3 points, and they're all the same distance from each other.

I know you can't construct an equilateral triangle with rational coordinates, but I don't see how it follows from the fact that you can't construct a length of sqrt(3).

antonfire
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Re: Straightedge and graph paper geometry

The distance from a vertex to the midpoint of the opposite edge is the side length times [imath]\sqrt 3 / 2[/imath]. Though, yeah, the argument that generalizes to all polygons is probably better.
Jerry Bona wrote:The Axiom of Choice is obviously true; the Well Ordering Principle is obviously false; and who can tell about Zorn's Lemma?

Buttons
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Re: Straightedge and graph paper geometry

One nice fact is that any polygon you create must have rational area. (Hint: Pick's Theorem.) I'm not quite sure how, but I suspect it's possible to turn that into another proof that you can't make an n-gon (for n > 4).

t0rajir0u
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Re: Straightedge and graph paper geometry

Yep, that works fine. The area of a regular polygon with [imath]n > 4[/imath] is an irrational multiple of the square of the side length, which has to be rational here.

Luppoewagan
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Re: Straightedge and graph paper geometry

I constructed an equilateral triangle just now using a pencil, a straightedge, and a piece of paper.

-Draw a line of arbitrary length from the end of the straightedge.
-Rotate the straightedge about the corner you started at whilst holding your pencil at the length of the line you made about 70 degrees, just eyeballing it.
-Do that again from the other end of your first line.
-Connect the arcs' intersection to the ends of the first segment.
- ???
-Profit!

Even though I bet using the straightedge as a compass was cheating.

jestingrabbit
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Re: Straightedge and graph paper geometry

J Spade wrote:Even though I bet using the straightedge as a compass was cheating.

Yeah, it was. Using stuff like this you can trisect an angle with straightedge and compass.
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Excalibur0998
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Re: Straightedge and graph paper geometry

You can create your infinite sheet of graph paper using a straightedge and compass.
Therefore anything you can't do with a straightedge and compass you also can't do with a sheet of graph paper.

Contrapositive isn't true though.

HenryGifford
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Re: Straightedge and graph paper geometry

You can create your infinite sheet of graph paper using a straightedge and compass.
Therefore anything you can't do with a straightedge and compass you also can't do with a sheet of graph paper.

Contrapositive isn't true though.

If a conditional is true, the contrapositive is always true, if I'm not mistaken.

Contrapositive of your conditional would be "Anything you can do with a sheet of graph paper you can do with a straightedge and compass."
Which would be true, if your conditional is true.
Did you mean the converse?

Excalibur0998
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Re: Straightedge and graph paper geometry

Yeah, I meant the converse, my bad.

PM 2Ring
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Re: Straightedge and graph paper geometry

jestingrabbit wrote:
J Spade wrote:Even though I bet using the straightedge as a compass was cheating.

Yeah, it was. Using stuff like this you can trisect an angle with straightedge and compass.

Indeed. "Eyeballing" the straightedge is equivalent to having a marked straightedge. Also, the compass used in traditional straightedge & compass geometry is a collapsible compass: when you lift it off the page, it collapses. If you want to use the compass to transport a given length to another part of the page, you have to "walk" it there.
Last edited by PM 2Ring on Mon Feb 23, 2009 4:33 am UTC, edited 1 time in total.

Yakk
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Re: Straightedge and graph paper geometry

What if you can re-apply the (unmarked) grid such that one point is at the new origin, and the other is along the x axis of your grid?

This is stronger than the original construction: you can express the point (1+1/sqrt(2), 1+1/sqrt(2)) easily.
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Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Cycle
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Re: Straightedge and graph paper geometry

Doing things that way, I'm pretty sure you can construct all numbers which are constructable with ruler and compass. This basically says any number which is a constructable length is constructable on the x-axis, right? Suppose we have a constructable length, a. Then the remaining leg a right triangle with hypotenuse (1+a) and leg a, has length (1+2a)1/2. The remaining leg of the triangle with hypotenuse (1+2a)1/2 and leg 1, has length (2a)1/2. We can divide by 21/2, since it is a constructable number. This proves that the constructable numbers are square root closed. But that's the exact description of Euclidean constructable numbers.

quintopia
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Re: Straightedge and graph paper geometry

J Spade wrote:I constructed an equilateral triangle just now using a pencil, a straightedge, and a piece of paper.

-Draw a line of arbitrary length from the end of the straightedge.
-Rotate the straightedge about the corner you started at whilst holding your pencil at the length of the line you made about 70 degrees, just eyeballing it.
-Do that again from the other end of your first line.
-Connect the arcs' intersection to the ends of the first segment.
- ???
-Profit!

Even though I bet using the straightedge as a compass was cheating.

Well, if we can cheat, here's my pentagon construction:

1) Crease a 2-unit wide, 15-unit long strip of the graph paper, and further score it with the straightedge.
2) Carefully tear out the straightedge
3) Tie a straight flat knot in the strip.
4) Trace the knot onto the sheet.

You now have a pentagon of side length 2.

Token
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Re: Straightedge and graph paper geometry

quintopia wrote:Well, if we can cheat, here's my pentagon construction:

1) Crease a 2-unit wide, 15-unit long strip of the graph paper, and further score it with the straightedge.
2) Carefully tear out the straightedge
3) Tie a straight flat knot in the strip.
4) Trace the knot onto the sheet.

You now have a pentagon of side length 2.

If your original strip is width 2, wouldn't that mean the pentagon had side length greater than 2? I'm too lazy to work out what it would actually be.
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quintopia
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Re: Straightedge and graph paper geometry

I suppose I should say "arbitrarily close to 2" depending on the thickness of the paper and how well you can fold.

jaap
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Re: Straightedge and graph paper geometry

quintopia wrote:I suppose I should say "arbitrarily close to 2" depending on the thickness of the paper and how well you can fold.

If the paper strip has width 2, the pentagon formed from knotting it has side length 2/cos(pi/10). The distance between a diagonal and the parallel edge is 2. The side length is slightly more.

quintopia
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Re: Straightedge and graph paper geometry

but the sides where the ends of the strip emerge have to have width 2. I guess it means it's not a regular pentagon?

Token
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Re: Straightedge and graph paper geometry

The sides of the pentagon won't be perpendicular to the edges of the strip.
All posts are works in progress. If I posted something within the last hour, chances are I'm still editing it.

quintopia
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Re: Straightedge and graph paper geometry

oh. right you are.