[math-fun] Plot of Rational Distance Problem triples
http://i.imgur.com/YclyNMC.gif Take a unit square. Plot a few thousand points (a,b) that are a rational distance from three vertices (all but (1,1)) in black. Plot the inverse point (a,b)/(a^2+b^2) in red. Can anyone see the ellipse 3 (x^2+ y^2)-4(x+y)+2 x y=0 ? Or the cubic curve 4 (y^3+ x^3)-3 (y^2+ x^2)+2x y ( 2y+ 2x -1)=0 ? --Ed Pegg Jr Points on the ellipse so far: {{-(4/17),16/51}, {28/33,112/99}, {-(140/387),112/129}, {-(224/627),140/209}, {340/627,272/209}, {736/1107,460/369}, {-(476/1539),544/513}, {-(80/1683),28/561}, {868/2211,992/737}, {-(368/1089),3220/3267}, {3280/3363,1148/1121}, {-(372/1441),4960/4323}, {-(1904/5907),1020/1969}, {752/1649,6580/4947}, {588/1921,7840/5763}, {-(1012/2993),5152/8979}, {-(2156/9843),3920/3281}, {-(3008/10659),3948/3553}, {-(448/3569),1540/10707}, {8176/10947,4380/3649}, {3124/12483,5680/4161}, {5056/14691,6636/4897}, {-(1968/5377),12628/16131}, {-(2132/5841),13120/17523}, {-(1820/18867),672/6289}, {-(3692/19491),7952/6497}, {3916/5457,19936/16371}, {6592/7089,22660/21267}, {5044/23859,10864/7953}, {-(2480/8241),11284/24723}, {-(6320/26643),10428/8881}, {9520/34563,15708/11521}, {5424/9073,34804/27219}, {-(12848/36579),11388/12193}, {35620/37587,13152/12529}, {-(5984/37827),2380/12609}, {6188/10001,38080/30003}, {-(1940/11649),43456/34947}, {-(15980/51507),8272/17169}, {-(6052/17441),49840/52323}, {12080/15521,54964/46563}, {2540/13889,56896/41667}, {-(476/57123),160/19041}, {28336/57171,25116/19057}, {-(8636/58179),24384/19393}, {10948/67971,30912/22657}, {-(3808/18737),68068/56211}, {-(20384/70851),9996/23617}, {-(23504/72627),24860/24209}, {67936/74787,27020/24929}, {-(27740/79827),16352/26609}, {-(9044/27537),83776/82611}, {-(12236/84579),4784/28193}, {-(12236/91443),38640/30481}, {-(11536/31553),77044/94659}, {5344/23313,95524/69939}, {73780/96387,38192/32129}, {-(17360/98643),7068/32881}, {15124/105123,47760/35041}, {-(18704/105507),43420/35169}, {45136/106947,47740/35649}, {12988/26961,106960/80883}, {113764/114243,38240/38081}, {-(14420/39729),85696/119187}, {-(14240/41361),74404/124083}, {-(12880/43601),142324/130803}, {-(45440/162963),21868/54321}, {-(19328/54417),149188/163251}, {17708/40913,164032/122739}, {-(63020/172227),46032/57409}, {-(50876/173283),25280/57761}}
This is fascinating, but what's the backstory here? Are these particular curves known to contain infinitely many points satisfying the three rational-distance requirements? Is the existence of a point at rational distance from all four corner points an open problem? Jim Propp On Wednesday, May 18, 2016, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
http://i.imgur.com/YclyNMC.gif
Take a unit square. Plot a few thousand points (a,b) that are a rational distance from three vertices (all but (1,1)) in black. Plot the inverse point (a,b)/(a^2+b^2) in red.
Can anyone see the ellipse 3 (x^2+ y^2)-4(x+y)+2 x y=0 ?
Or the cubic curve 4 (y^3+ x^3)-3 (y^2+ x^2)+2x y ( 2y+ 2x -1)=0 ?
--Ed Pegg Jr
Points on the ellipse so far:
{{-(4/17),16/51}, {28/33,112/99}, {-(140/387),112/129}, {-(224/627),140/209}, {340/627,272/209}, {736/1107,460/369}, {-(476/1539),544/513}, {-(80/1683),28/561}, {868/2211,992/737}, {-(368/1089),3220/3267}, {3280/3363,1148/1121}, {-(372/1441),4960/4323}, {-(1904/5907),1020/1969}, {752/1649,6580/4947}, {588/1921,7840/5763}, {-(1012/2993),5152/8979}, {-(2156/9843),3920/3281}, {-(3008/10659),3948/3553}, {-(448/3569),1540/10707}, {8176/10947,4380/3649}, {3124/12483,5680/4161}, {5056/14691,6636/4897}, {-(1968/5377),12628/16131}, {-(2132/5841),13120/17523}, {-(1820/18867),672/6289}, {-(3692/19491),7952/6497}, {3916/5457,19936/16371}, {6592/7089,22660/21267}, {5044/23859,10864/7953}, {-(2480/8241),11284/24723}, {-(6320/26643),10428/8881}, {9520/34563,15708/11521}, {5424/9073,34804/27219}, {-(12848/36579),11388/12193}, {35620/37587,13152/12529}, {-(5984/37827),2380/12609}, {6188/10001,38080/30003}, {-(1940/11649),43456/34947}, {-(15980/51507),8272/17169}, {-(6052/17441),49840/52323}, {12080/15521,54964/46563}, {2540/13889,56896/41667}, {-(476/57123),160/19041}, {28336/57171,25116/19057}, {-(8636/58179),24384/19393}, {10948/67971,30912/22657}, {-(3808/18737),68068/56211}, {-(20384/70851),9996/23617}, {-(23504/72627),24860/24209}, {67936/74787,27020/24929}, {-(27740/79827),16352/26609}, {-(9044/27537),83776/82611}, {-(12236/84579),4784/28193}, {-(12236/91443),38640/30481}, {-(11536/31553),77044/94659}, {5344/23313,95524/69939}, {73780/96387,38192/32129}, {-(17360/98643),7068/32881}, {15124/105123,47760/35041}, {-(18704/105507),43420/35169}, {45136/106947,47740/35649}, {12988/26961,106960/80883}, {113764/114243,38240/38081}, {-(14420/39729),85696/119187}, {-(14240/41361),74404/124083}, {-(12880/43601),142324/130803}, {-(45440/162963),21868/54321}, {-(19328/54417),149188/163251}, {17708/40913,164032/122739}, {-(63020/172227),46032/57409}, {-(50876/173283),25280/57761}} _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
This is indeed very interesting! But a) I don't understand why, with the problem having the symmetry of the square, the curves seem to have only (x,y) <—> (y,x) symmetry and b) that doesn't seem to be the right equation of the cubic curve that I see, both because it lacks (x,y) <—> (y,x) symmetry, and its shape doesn't seem right. —Dan
On May 18, 2016, at 12:35 PM, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
http://i.imgur.com/YclyNMC.gif
Take a unit square. Plot a few thousand points (a,b) that are a rational distance from three vertices (all but (1,1)) in black. Plot the inverse point (a,b)/(a^2+b^2) in red.
Can anyone see the ellipse 3 (x^2+ y^2)-4(x+y)+2 x y=0 ?
Or the cubic curve 4 (y^3+ x^3)-3 (y^2+ x^2)+2x y ( 2y+ 2x -1)=0 ?
--Ed Pegg Jr
Points on the ellipse so far:
. . . . . . . . .
Dan, You probably already figured this out, but the point (1,1) is treated differently from the other three vertices of the square; this breaks the symmetry (and explains why only bilateral symmetry is observed). Jim On Wednesday, May 18, 2016, Dan Asimov <dasimov@earthlink.net> wrote:
This is indeed very interesting!
But
a) I don't understand why, with the problem having the symmetry of the square, the curves seem to have only (x,y) <—> (y,x) symmetry
and
b) that doesn't seem to be the right equation of the cubic curve that I see, both because it lacks (x,y) <—> (y,x) symmetry, and its shape doesn't seem right.
—Dan
On May 18, 2016, at 12:35 PM, Ed Pegg Jr <ed@mathpuzzle.com <javascript:;>> wrote:
http://i.imgur.com/YclyNMC.gif
Take a unit square. Plot a few thousand points (a,b) that are a rational distance from three vertices (all but (1,1)) in black. Plot the inverse point (a,b)/(a^2+b^2) in red.
Can anyone see the ellipse 3 (x^2+ y^2)-4(x+y)+2 x y=0 ?
Or the cubic curve 4 (y^3+ x^3)-3 (y^2+ x^2)+2x y ( 2y+ 2x -1)=0 ?
--Ed Pegg Jr
Points on the ellipse so far:
. . .
. . .
. . . _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
It's pretty easy to generate all the rational points on the given ellipse (a 1-parameter formula), and not too much harder to generate the points on the elliptic curve, using "chord-and-tangent". We might generate these points, and see what fraction have the rational distance property. Of course this only gets a tiny fraction of the full set of solutions. Rich ---------- Quoting James Propp <jamespropp@gmail.com>:
Dan,
You probably already figured this out, but the point (1,1) is treated differently from the other three vertices of the square; this breaks the symmetry (and explains why only bilateral symmetry is observed).
Jim
On Wednesday, May 18, 2016, Dan Asimov <dasimov@earthlink.net> wrote:
This is indeed very interesting!
But
a) I don't understand why, with the problem having the symmetry of the square, the curves seem to have only (x,y) <?> (y,x) symmetry
and
b) that doesn't seem to be the right equation of the cubic curve that I see, both because it lacks (x,y) <?> (y,x) symmetry, and its shape doesn't seem right.
?Dan
On May 18, 2016, at 12:35 PM, Ed Pegg Jr <ed@mathpuzzle.com <javascript:;>> wrote:
http://i.imgur.com/YclyNMC.gif
Take a unit square. Plot a few thousand points (a,b) that are a rational distance from three vertices (all but (1,1)) in black. Plot the inverse point (a,b)/(a^2+b^2) in red.
Can anyone see the ellipse 3 (x^2+ y^2)-4(x+y)+2 x y=0 ?
Or the cubic curve 4 (y^3+ x^3)-3 (y^2+ x^2)+2x y ( 2y+ 2x -1)=0 ?
--Ed Pegg Jr
Points on the ellipse so far:
. . .
. . .
. . . _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com <javascript:;> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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I was badly mistaken on both counts: a) because I didn't read the problem carefully. b) because I just plugged the equation given by Ed into the somewhat buggy Mac Grapher utility, *and* I didn't look at the equation carefully. Obviously the cubic equation does have (x,y) <—> (y,x) symmetry. Duh^2. I knew the Grapher utility has a lot of bugs, which I've been nagging Apple to fix for over 10 years. (They're still bundling the identical version with new Macs. And yeah, there's that thing about a person who does the same behavior over and over but expects something different to happen.) But I had no idea the Grapher utility was *that* buggy! This is really a terrible piece of garbage. —Dan
On May 18, 2016, at 4:03 PM, James Propp <jamespropp@gmail.com> wrote:
a) I don't understand why, with the problem having the symmetry of the square, the curves seem to have only (x,y) <—> (y,x) symmetry
and
b) that doesn't seem to be the right equation of the cubic curve that I see, both because it lacks (x,y) <—> (y,x) symmetry, and its shape doesn't seem right.
(15803760/49193869, 38592180/49193869) Is a rational distance from (0,0), (1,0), (0,1), .... and also (0,-1). That's the first quadruple I've found so far.
Are you only considering points w/ rational coordinates? On 2016-05-18 12:35, Ed Pegg Jr wrote:
http://i.imgur.com/YclyNMC.gif
Take a unit square. Plot a few thousand points (a,b) that are a rational distance from three vertices (all but (1,1)) in black. Plot the inverse point (a,b)/(a^2+b^2) in red.
Can anyone see the ellipse 3 (x^2+ y^2)-4(x+y)+2 x y=0 ?
Or the cubic curve 4 (y^3+ x^3)-3 (y^2+ x^2)+2x y ( 2y+ 2x -1)=0 ?
--Ed Pegg Jr
Points on the ellipse so far:
[...] _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Wed, May 18, 2016 at 4:32 PM, William R Somsky <wrsomsky@gmail.com> wrote:
Are you only considering points w/ rational coordinates?
Consider the squares of the distances: x^2+y^2, (x-1)^2+y^2, and x^2+((y-1)^2). All of these need to be rational, so the differences 2x-1 and 2 y-1 are also rational. Therefore, all triples have rational coordinates. I think this is problem D23 in Guy. R. K. Guy, "Rational Distances from the Corner of a Square." Dunno is anyone has noticed that ellipse before.
How did you find the points (a,b)? The ellipse and cubic could be a consequence of the method you used? On Wed, May 18, 2016 at 3:35 PM, Ed Pegg Jr <ed@mathpuzzle.com> wrote:
http://i.imgur.com/YclyNMC.gif
Take a unit square. Plot a few thousand points (a,b) that are a rational distance from three vertices (all but (1,1)) in black. Plot the inverse point (a,b)/(a^2+b^2) in red.
Can anyone see the ellipse 3 (x^2+ y^2)-4(x+y)+2 x y=0 ?
Or the cubic curve 4 (y^3+ x^3)-3 (y^2+ x^2)+2x y ( 2y+ 2x -1)=0 ?
--Ed Pegg Jr
Points on the ellipse so far:
{{-(4/17),16/51}, {28/33,112/99}, {-(140/387),112/129}, {-(224/627),140/209}, {340/627,272/209}, {736/1107,460/369}, {-(476/1539),544/513}, {-(80/1683),28/561}, {868/2211,992/737}, {-(368/1089),3220/3267}, {3280/3363,1148/1121}, {-(372/1441),4960/4323}, {-(1904/5907),1020/1969}, {752/1649,6580/4947}, {588/1921,7840/5763}, {-(1012/2993),5152/8979}, {-(2156/9843),3920/3281}, {-(3008/10659),3948/3553}, {-(448/3569),1540/10707}, {8176/10947,4380/3649}, {3124/12483,5680/4161}, {5056/14691,6636/4897}, {-(1968/5377),12628/16131}, {-(2132/5841),13120/17523}, {-(1820/18867),672/6289}, {-(3692/19491),7952/6497}, {3916/5457,19936/16371}, {6592/7089,22660/21267}, {5044/23859,10864/7953}, {-(2480/8241),11284/24723}, {-(6320/26643),10428/8881}, {9520/34563,15708/11521}, {5424/9073,34804/27219}, {-(12848/36579),11388/12193}, {35620/37587,13152/12529}, {-(5984/37827),2380/12609}, {6188/10001,38080/30003}, {-(1940/11649),43456/34947}, {-(15980/51507),8272/17169}, {-(6052/17441),49840/52323}, {12080/15521,54964/46563}, {2540/13889,56896/41667}, {-(476/57123),160/19041}, {28336/57171,25116/19057}, {-(8636/58179),24384/19393}, {10948/67971,30912/22657}, {-(3808/18737),68068/56211}, {-(20384/70851),9996/23617}, {-(23504/72627),24860/24209}, {67936/74787,27020/24929}, {-(27740/79827),16352/26609}, {-(9044/27537),83776/82611}, {-(12236/84579),4784/28193}, {-(12236/91443),38640/30481}, {-(11536/31553),77044/94659}, {5344/23313,95524/69939}, {73780/96387,38192/32129}, {-(17360/98643),7068/32881}, {15124/105123,47760/35041}, {-(18704/105507),43420/35169}, {45136/106947,47740/35649}, {12988/26961,106960/80883}, {113764/114243,38240/38081}, {-(14420/39729),85696/119187}, {-(14240/41361),74404/124083}, {-(12880/43601),142324/130803}, {-(45440/162963),21868/54321}, {-(19328/54417),149188/163251}, {17708/40913,164032/122739}, {-(63020/172227),46032/57409}, {-(50876/173283),25280/57761}} _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Wed, May 18, 2016 at 5:03 PM, W. Edwin Clark <wclark@mail.usf.edu> wrote:
How did you find the points (a,b)? The ellipse and cubic could be a consequence of the method you used?
I'm plowing through all the Primitive Heron Triangles from the data at http://www.wm-archive.uni-bayreuth.de/index.php?id=554&L=3 The image is points collected from heron triangles with diameter under 180000. My program has a long ways to go. --Ed Pegg Jr
participants (6)
-
Dan Asimov -
Ed Pegg Jr -
James Propp -
rcs＠xmission.com -
W. Edwin Clark -
William R Somsky