# Geometry

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• Dec 2nd 2009, 08:24 AM
aman_cc
Quote:

Originally Posted by Drexel28
Question:

Spoiler:
What do you mean "lying in $\mathbb{Q}^2$? Do you mean that the verticies of the triangle are rational points or that literally every point of the triangle is rational? Because I have a problem witht the second one...

Guess, you can't have that. Points in a line sengment are not countable - so every point can't be rational. Correct?

Also, what do you mean by tangent of the angle?
• Dec 2nd 2009, 08:35 AM
Unbeatable0
Quote:

Originally Posted by aman_cc
Guess, you can't have that. Points in a line sengment are not countable - so every point can't be rational. Correct?

Also, what do you mean by tangent of the angle?

Sure you're right, I meant the vertices are in http://www.mathhelpforum.com/math-he...fe4dca53-1.gif, my mistake for no paying enough attention to wording.
If a is an interior angle of the triangle, tan(a) is what I meant by the tangent of the angle, and the proposition claims that it is rational.
In the original case of equilateral triangle, it turns out no equilateral triangle
exist with vertices in http://www.mathhelpforum.com/math-he...fe4dca53-1.gif, as tan(60 deg) = sqrt(3)
• Dec 2nd 2009, 01:01 PM
Drexel28
Quote:

Originally Posted by aman_cc
Guess, you can't have that. Points in a line sengment are not countable - so every point can't be rational. Correct?

Also, what do you mean by tangent of the angle?

That is precisely why! When speaking of triangles, we must first talk about the edges of the triangle. These edges are connected and may be interpreted as being homeomorphic to some interval in $\mathbb{R}$ and since any interval in $\mathbb{R}$ is uncountable we can't speak of lines in a countable spac such as $\mathbb{Q}^2$

P.S. I didn't check that my language above was precise, but you get the idea.
• Dec 3rd 2009, 07:05 AM
Unbeatable0
I almost forgot about the first problem you stated.
The solution:

Spoiler:

Let O be the center of the regular polygon (the center of the circumcenter), and let R be the distance from the O to the vertices.
The angle between two lines beginning in O and ending in two adjacent vertices is thus $\frac{2\pi}{n}$, and from the sinus law we get that the area of a triangle formed by such two lines is given by $\frac{1}{2}R^2\sin{\frac{2\pi}{n}} = R^2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}$
But there are n such triangles in the regular polygon. Using our last result this gives us $n R^2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}$
Furthermore, if we drop an altitude to the base in one of those isosceles triangles, we come up with $R = \frac{\frac{1}{2} a}{\sin{\frac{\pi}{n}}}$ , and after subtitution in the expression for the area we get $n (\frac{\frac{1}{2} a}{\sin{\frac{\pi}{n}}})^2\sin{\frac{\pi}{n}}\cos{ \frac{\pi}{n}} = \frac{1}{4} a^2 n \cot{\frac{\pi}{n}}$ .
Q.E.D
• Dec 3rd 2009, 11:57 AM
Drexel28
Quote:

Originally Posted by Unbeatable0
I almost forgot about the first problem you stated.
The solution:

Spoiler:

Let O be the center of the regular polygon (the center of the circumcenter), and let R be the distance from the O to the vertices.
The angle between two lines beginning in O and ending in two adjacent vertices is thus $\frac{2\pi}{n}$, and from the sinus law we get that the area of a triangle formed by such two lines is given by $\frac{1}{2}R^2\sin{\frac{2\pi}{n}} = R^2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}$
But there are n such triangles in the regular polygon. Using our last result this gives us $n R^2\sin{\frac{\pi}{n}}\cos{\frac{\pi}{n}}$
Furthermore, if we drop an altitude to the base in one of those isosceles triangles, we come up with $R = \frac{\frac{1}{2} a}{\sin{\frac{\pi}{n}}}$ , and after subtitution in the expression for the area we get $n (\frac{\frac{1}{2} a}{\sin{\frac{\pi}{n}}})^2\sin{\frac{\pi}{n}}\cos{ \frac{\pi}{n}} = \frac{1}{4} a^2 n \cot{\frac{\pi}{n}}$ .
Q.E.D

Note now that with this and Pick's theorem we can say any $n$ such that $\cot\left(\frac{\pi}{n}\right)\notin\mathbb{Q}$ cannot have all vertices on lattice points. I just came to that conclusion! Sweet! Suck that pentagon!
• Dec 4th 2009, 01:32 AM
Unbeatable0
Quote:

Originally Posted by Drexel28
Note now that with this and Pick's theorem we can say any $n$ such that $\cot\left(\frac{\pi}{n}\right)\notin\mathbb{Q}$ cannot have all vertices on lattice points. I just came to that conclusion! Sweet! Suck that pentagon!

Actually, from the proposition I posted before, it follows that for any regular n-sided polygon, if $\cot{\frac{\pi}{n}}$ is irrational, then it can't have three vertices in http://www.mathhelpforum.com/math-he...fe4dca53-1.gif, two of which are adjacent.
Nobody is up to prove the claim in post #14?
• Dec 4th 2009, 06:53 AM
Drexel28
Quote:

Originally Posted by Unbeatable0
Actually, from the proposition I posted before, it follows that for any regular n-sided polygon, if $\cot{\frac{\pi}{n}}$ is irrational, then it can't have three vertices in http://www.mathhelpforum.com/math-he...fe4dca53-1.gif, two of which are adjacent.
Nobody is up to prove the claim in post #14?

I would like to give it a try, but I am swamped right now. Give me, or someone else, a day or so and then post up the solution otherwise!
• Dec 6th 2009, 04:41 AM
Unbeatable0
Quote:

Originally Posted by Unbeatable0
Prove that in any triangle in $\mathbb{R}^2$ whose vertices are in $\mathbb{Q}^2$, the tangents
of the angles are rational.

I allow myself to post a solution (not mine), after seeing that the problem has not caught too much interest. The proof is in the attached image.
• Dec 7th 2009, 09:02 PM
Drexel28
Quote:

Originally Posted by Unbeatable0
I allow myself to post a solution (not mine), after seeing that the problem has not caught too much interest. The proof is in the attached image.

That was close to my solution! Which I actually logged onto to post! I didn't have a picture though..so this one is decidedly better. Sorry about not doing it. I had Putnam, finals, etc. Thanks for the solution!
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