[math-fun] What's your proof?
Prove the following: If a polygon in R^2 has an odd number of equal-length sides, it has a nonrational vertex (not in Q^2). I just wanted to know if there was a proof very different from my own.
At 01:49 AM 6/11/03, David Wilson wrote:
Prove the following:
If a polygon in R^2 has an odd number of equal-length sides, it has a nonrational vertex (not in Q^2).
I just wanted to know if there was a proof very different from my own.
If there is an equilateral polygon with vertices in Q^2, it can be scaled so that its vertices are in Z^2. If such a polygon has sides of even length, they can be halved (as any Pythagorean triangle with an even hypotenuse has even legs). Thus the sides can be taken odd, which means the parity of the sum of the coordinates changes between each consecutive pair of vertices. (Odd hypotenuse implies one odd leg, one even leg.) Which is impossible in a circuit with an odd number of edges. -- Fred W. Helenius <fredh@ix.netcom.com>
You wrote:
If there is an equilateral polygon with vertices in Q^2, it can be scaled so that its vertices are in Z^2.
If such a polygon has sides of even length, they can be halved (as any Pythagorean triangle with an even hypotenuse has even legs).
What if the sides are irrational?
At 02:29 AM 6/11/03, David Wilson wrote:
You wrote:
If there is an equilateral polygon with vertices in Q^2, it can be scaled so that its vertices are in Z^2.
If such a polygon has sides of even length, they can be halved (as any Pythagorean triangle with an even hypotenuse has even legs).
What if the sides are irrational?
Oops, I skipped a step. If the sides are irrational, we can scale and rotate to make them integers. Imagine Z^2 to be the Gaussian integers, and suppose one vertex is at 0 and the next is a + b i. Multiply all the vertices by a - b i; they remain Gaussian integers, and the side lengths are now a^2 + b^2 instead of sqrt(a^2 + b^2). -- Fred W. Helenius <fredh@ix.netcom.com>
With the additional step, Fred's proof coincides with mine. So, there is no such thing as an equilateral oddgon with rational vertices. Here's a quick application: If a regular polygon has rational vertices, it's a power-of-2-gon.
A polygon can be scaled to have vertices at lattice points if and only if all its angles have rational tangents.
At 09:26 AM 6/11/03, John McCarthy wrote:
A polygon can be scaled to have vertices at lattice points if and only if all its angles have rational tangents.
I don't think so. That would mean there was a lattice-point regular octagon, but there isn't, as JHC has just confirmed. For another example, consider a parallelogram with angles that are alternately arctan(2) and arctan(-2), and with sides alternately 1 and pi. The tangents are rational, but lattice-point distances can't be in the ratio 1:pi. -- Fred W. Helenius <fredh@ix.netcom.com>
On Wed, 11 Jun 2003, Fred W. Helenius wrote:
At 09:26 AM 6/11/03, John McCarthy wrote:
A polygon can be scaled to have vertices at lattice points if and only if all its angles have rational tangents.
I don't think so. That would mean there was a lattice-point regular octagon, but there isn't, as JHC has just confirmed. For another example, consider a parallelogram with angles that are alternately arctan(2) and arctan(-2), and with sides alternately 1 and pi. The tangents are rational, but lattice-point distances can't be in the ratio 1:pi.
But the angle of a regular octagon DOESN'T have a rational tangent. The "rational tangents" condition is exactly right. John conway
I obviously have to modify my assertion. That the angles of a polygon have rational tangents is necessary for it to be similar to a polygon with vertices at lattice points, because any angle of such a polygon is the difference of two angles of rational tangent and the formula tan(x - y) = (tan x - tan y)/(1 + tan x tan y) preserves rationality. I think the converse is true for completely triangulated polygons. Thus if you put in the diagonals of the 1 by e square you get an angle with an irrational tangent, and the regular octagon is shown to be non-lattice by the same device. Anything that can be built from the (0,0)-(1,0) line segment by gluing triangles with rational tangents onto edges will have rational vertices, again because of the addition formula for the tangent. Any finite figure with rational vertices can be expanded to a figure with integer vertices. The remaining gap is on whether every completely triangulated polygon can be built up in this way.
On Wed, 11 Jun 2003, David Wilson wrote:
With the additional step, Fred's proof coincides with mine.
So, there is no such thing as an equilateral oddgon with rational vertices.
Here's a quick application:
If a regular polygon has rational vertices, it's a power-of-2-gon.
Yes, since in fact it's a 4-gon, and 4 is indeed a power of 2. John Conway
Yes, I knew that power-of-2-greater-than-4-gons could not have rational vertices. I was just observing that the equilateral oddgon theorem gets us most of the way there. That with the regular octogon theorem gets us home. ----- Original Message ----- From: John Conway To: math-fun Sent: Wednesday, June 11, 2003 9:45 AM Subject: Re: [math-fun] What's your proof? On Wed, 11 Jun 2003, David Wilson wrote:
With the additional step, Fred's proof coincides with mine.
So, there is no such thing as an equilateral oddgon with rational vertices.
Here's a quick application:
If a regular polygon has rational vertices, it's a power-of-2-gon.
Yes, since in fact it's a 4-gon, and 4 is indeed a power of 2. John Conway _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Thu, 12 Jun 2003, David Wilson wrote:
Yes, I knew that power-of-2-greater-than-4-gons could not have rational vertices. I was just observing that the equilateral oddgon theorem gets us most of the way there. That with the regular octogon theorem gets us home.
I'm not sure which theorem you're talking about. It's a rather trivial remark that a polygon is embeddable (after a scale-change) with rational vertices if and only if every triple of vertices defines a triangle whose angles have rational tangents. Here's the proof. If a triangle ABC has such angles, put A at (0,0) and B at (1,0). Then C is that the intersection of two lines through these points with rational slope, so has rational vertices. For a quadrilateral ABCD this will be true of both C and D, etc. John Conway
participants (4)
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David Wilson -
Fred W. Helenius -
John Conway -
John McCarthy