Re: [math-fun] Triangles question
Is the "distance from the corners" question equivalent to this? Take two pairs of Pythagorean triples, (a1, b1, c1) and (a2, b2, c2). Let the point in question be at the origin, with square corners at (a1, b1), (-a2, b1), (-a2, -b2), and (a1, -b2). Then the problem is to find triples such that (a1, b2, c3) and (a2, b1, c4) are Pythagorean triples too. I looked at that problem briefly for triples up to 20,000 on a side. The closest I came was a square of side 4063 from triples (3, 4, 5) and (4060, 4059, 5741). In that case, the distance to the quadrant 2 corner was 4060.002 and the distance to the quadrant 4 corner was 4059.001. This is a neat problem visually: draw all the arcs of integer radius from each of the four corners and look for points where four arcs intersect. After minutes of laborious effort, I have decided that the problem has no solution. :-) Kerry -------------- Original message from "Daniel Asimov" <dasimov@earthlink.net>: --------------
I heard this question in the immediately equivalent version:
< Is there a point in the plane at rational distances from each corner of a unit square?
What if we inscribe a triangle in a square so that the base of the triangle coincides with one edge of the square, and the apex lies on the opposite edge of the square. Can this figure be made to have all integer segments? ----- Original Message ----- From: "Kerry Mitchell" <lkmitch@att.net> To: <ham>; "math-fun" <math-fun@mailman.xmission.com> Sent: Friday, April 29, 2005 6:59 PM Subject: Re: [math-fun] Triangles question
Is the "distance from the corners" question equivalent to this? Take two pairs of Pythagorean triples, (a1, b1, c1) and (a2, b2, c2). Let the point in question be at the origin, with square corners at (a1, b1), (-a2, b1), (-a2, -b2), and (a1, -b2). Then the problem is to find triples such that (a1, b2, c3) and (a2, b1, c4) are Pythagorean triples too.
I looked at that problem briefly for triples up to 20,000 on a side. The closest I came was a square of side 4063 from triples (3, 4, 5) and (4060, 4059, 5741). In that case, the distance to the quadrant 2 corner was 4060.002 and the distance to the quadrant 4 corner was 4059.001.
This is a neat problem visually: draw all the arcs of integer radius from each of the four corners and look for points where four arcs intersect. After minutes of laborious effort, I have decided that the problem has no solution. :-)
Kerry
-------------- Original message from "Daniel Asimov" <dasimov@earthlink.net>: --------------
I heard this question in the immediately equivalent version:
< Is there a point in the plane at rational distances from each corner of a unit square?
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To prove that this is impossible rears itself as a problem from time to time. There are references and proofs in the papers cited. R. On Fri, 29 Apr 2005, David Wilson wrote:
What if we inscribe a triangle in a square so that the base of the triangle coincides with one edge of the square, and the apex lies on the opposite edge of the square. Can this figure be made to have all integer segments?
participants (3)
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David Wilson -
lkmitch＠att.net -
Richard Guy