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 Post subject: Triangles
PostPosted: Sat Jun 01, 2002 8:07 am 
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Knight of Daisies, Tulip Slayer
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Location: Sector ZZ9 Plural Z Alpha
A right-angled isoceles triangle has all three sides of whole number lengths.

Can you give an example? Or prove that it doesn't exist?

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PostPosted: Sat Jun 01, 2002 10:05 pm 
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Vorpal Bunny Slipper
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Can't be done.

Let's assume that the lengths of the two equal sides are given by a, and the hypotenuse is b. Since this is a right triangle, by the Pythogorean theorem, a^2 + a^2 = b^2. Thus 2a^2 = b^2. If we take the square root of both sides, we get sqrt(2) * a = b, where sqrt is the square root function. Sqrt(2) is an irrational number, so if b is an integer, a can't be, and vice versa.


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PostPosted: Sun Jun 02, 2002 3:13 am 
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Energizer Bunny
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'course, proving sqrt(2) is not rational is another sack of potatoes entirely.

it goes something like this:

all squares of rational numbers are products of even powers of prime numbers. since the square root of 2 squared is 2, if sqrt(2) is rational there must be an odd number of 2's multiplied together in either the numerator or the denominator. This is a contradiction, and working backwards reveals that the only error is assuming sqrt(2) is rational.

There is another alternative:
if you're in elliptical space, you can have an equilateral right triangle. But that's a third sack of potatoes, the size of, oh, say, Idaho.

Vorn


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 Post subject:
PostPosted: Sun Jun 02, 2002 9:53 am 
What's elliptical space?


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 Post subject:
PostPosted: Sun Jun 02, 2002 10:23 pm 
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Vorpal Bunny Slipper
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If you draw the triangle on a sphere, you can smush three right angles into a triangle.

And I could prove that sqrt(2) is irrational, but I had to do similar things several times in my number theory class last semester and I felt like it not.


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 Post subject:
PostPosted: Mon Jun 03, 2002 1:18 am 
Ah, 1/8 sphere.... yeah I tried to use that with an origami thingy but it came out... odd. :D


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 Post subject:
PostPosted: Sat Jun 08, 2002 3:03 am 
Suppose sqrt(2) is rational. Let it be expressed as p/q, where p and q are relatively prime integers. then 2 = p^2 / q^2. so p^2 = 2*(q^2). p^2 therefore has a factor of 2 in it, and since p is an integer, p must itself have a factor of 2 in it (since sqrt(2) is definitely not an integer). so p = 2*k, for some integer of k. => (2k)^2 = 2*(q^2) => 2*(k^2) = q^2. By the logic above, q also has a factor of 2. This, however, is a contradiction, since p has a factor of 2, and p and q are relatively prime. The supposition is therefore false - sqrt(2) is irrational.


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 Post subject:
PostPosted: Sat Jun 08, 2002 3:41 am 
The only thing left to prove, at least in my view, is JeremiahSmith's assumption that the right angle is indeed the angle between the two equal sides. I will proceed by contradiction.

First, let us assume that we have an isosceles right triangle ABC, in which sides a and b are equal, and angle B is right. Because opposite angles of equal sides are equal, this means that angle A is also a right angle. Now, as a result of Euclid's fifth postulate, we knwo that the sum of the angles in any triangle is 180°. This means A + B + C = 180, or A + 90 + 90 = 180, which means A = 0. The definition of a triangle states that the points A, B, and C are non-colinear. But if A = 0°, then the points must be colinear, and thus the assumption that ABC is a triangle is violated. Therefore, by contradiction, such a triangle cannot exist in Euclidean geometry. QED.

Now, to describe better what Vorn was talking about, there are fields of geometry in which people decided that Euclid was smoking crack when he came up with his fifth postulate. One such field, spherical geometry, says that the sum of the angles in a triangle can (and indeed always are) greater than 180°. In this geometry, such a triangle can, and does exist.

For a simple example, try a sphere with circumference 4. Hold it out in front of you. Now, place three points on this sphere: One at the topmost point, one at the leftmost point, and one at the point closest to you. Join these three points, and you get, in fact, an equilateral right triangle, with three sides of length 1, and three angles of 90°.

The isosceles right triangle Ogredude is talking about can, in fact, be created on any globe with two points anywhere on the equator and the third point at one of the poles. Notice that the right angle(s) in this triangle is not between the two equal sides, which is how you get around the argument that sqrt(2) is not rational.


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 Post subject:
PostPosted: Sat Jun 08, 2002 10:52 am 
sqrt(2) being rational or irrational becomes irrelevant at that point anyway, because the Pythagorean Theorem itself doesn't hold. As Pi mentioned, it's in fact possible to make an equilateral right triangle. The Pythagorean Theorem would need to be reformulated in order to hold on a sphere. I, for one, have no idea where to start, but it seems clear that such a theorem would need to involve the size of the triangle with respect to the size of the sphere.


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 Post subject:
PostPosted: Sun Jun 09, 2002 4:40 pm 
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Vorpal Bunny Slipper
Vorpal Bunny Slipper

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Wow. You guys overanalyzed this to death.


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 Post subject:
PostPosted: Sat Aug 17, 2002 1:49 am 
jeremiahsmith wrote:
Can't be done.

Let's assume that the lengths of the two equal sides are given by a, and the hypotenuse is b. Since this is a right triangle, by the Pythogorean theorem, a^2 + a^2 = b^2. Thus 2a^2 = b^2. If we take the square root of both sides, we get sqrt(2) * a = b, where sqrt is the square root function. Sqrt(2) is an irrational number, so if b is an integer, a can't be, and vice versa.


Should I point out the flaw here? My maths isn't up to scratch, but as I recall irrational numbers can exist, they simply can't be expressed with perfect accuracy. You've proven that a right-angled triangle cannot be an iscosceles triangle, which isn't true. sqrt(2) is a ratio, here. If you multiply a rational number by an irrational number, the result can be either. How on earth does a number being irrational preclude its use as a ratio? The ratio is (sqrt)2 : 1.

The most famous example is the irrational number Pi. You've proven that a circle cannot exist. Pi is an irrational number, defined as the ratio between the diameter of a circle and its perimeter.

2Pi r = P
but Pi is irrational . . . follow?
Pi * D = P
The ratio would be
Pi : D = 1

But both the diameter and the perimeter of a circle exist. It looks like you're confusing the properties of irrational and *imaginary* numbers, to me.

Here's a working proof, however:

given that a^2 + b^2 = c^2, and a = b = c, we have
2a^2 = a^2
this statement is only true at a = 0, at which time the shape described is not a triangle. It's a point.

This of course assumes we're dealing with plane geometry . . . I wouldn't have a clue how to deal with anything else ;)


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 Post subject:
PostPosted: Sat Aug 17, 2002 7:57 am 
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Knight of Daisies, Tulip Slayer
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Location: Sector ZZ9 Plural Z Alpha
Nobody said you couldn't have an isoceles right triangle. We said you couldn't have one with INTEGER LENGTH SIDES

_________________
Fandemonium 2010 -- No Boundaries.
http://www.fandemonium.org
Friday - Sunday, August 6th - 8th, 2010
Nampa Civic Center - Nampa, Idaho (Only 20 minutes from the airport!)
(Idaho: It ain't just potatoes anymore.)


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 Post subject:
PostPosted: Sat Aug 17, 2002 9:33 am 
eeeeep!

/me rereads the question

ooooops . . .

:oops:

I'm gonna, erm, leave this thread now. Yeah. That's what I'll do.

For some reason I read that as prove that a right-angled equilateral triangle cannot exist . . . I must be insane ;)


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 Post subject:
PostPosted: Mon Mar 10, 2003 12:27 am 
*sighs softly as she goes into your internal brains and sets a new init string for you...ATZ*


Sorry, can't help but feel stupid when reading this one guys. Think I'll still with mindtrap.


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 Post subject:
PostPosted: Mon Mar 10, 2003 2:18 am 
Yeah, although mindtrap seems decidedly below the level of nightstar users. :)


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 Post subject:
PostPosted: Mon Mar 10, 2003 2:29 am 
We need some sort of hard version of MindTrap.


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 Post subject:
PostPosted: Mon Mar 10, 2003 2:34 am 
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Vorpal Bunny Slipper
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Joined: Sun May 12, 2002 2:54 am
Posts: 2707
Shut your pie holes. All of you. INFIDELS! You will burn in eternity for your transgressions!

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 Post subject:
PostPosted: Wed Mar 12, 2003 7:39 pm 
A version of mindtrap targetted toward mathematicians and engineers has indeed been suggested and seconded. :twisted:


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 Post subject:
PostPosted: Sat Mar 15, 2003 3:32 pm 
You mean to mathmeticians and engineers with way to much time on their hands.


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