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Ci ∩ Cj
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KY
Nov 22, 2009, 9:58:56 AM11/22/09
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Four planar curves are defined as follows:
C1:x^3 + x^5 - 3*x*y^2 - 10*x^3*y^2 + 5*x*y^4 = x/(x^2+y^2),
C2:3*x^2*y + 5*x^4*y - y^3 - 10*x^2*y^3 + y^5 + y/(x^2 + y^2) = 3
C1:x^3 + x^5 - 3*x*y^2 - 10*x^3*y^2 + 5*x*y^4 = x/(x^2+y^2),
C2:3*x^2*y + 5*x^4*y - y^3 - 10*x^2*y^3 + y^5 + y/(x^2 + y^2) = 3
C3:-1 + x^4 + x^6 + 3*y - 6*x^2*y^2 - 15*x^4*y^2 + y^4 + 15*x^2*y^4 - y^6 =0,
C4:-3*x + 4*x^3*y + 6*x^5*y - 4*x*y^3 - 20*x^3*y^3 + 6*x*y^5 = 0
Prove that C1 ∩ C2 = C3 ∩ C4.
KY
Nov 22, 2009, 10:01:06 AM11/22/09
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intersection of C1 and C2=
intersection of C3 and C4
intersection of C3 and C4
Boner J. Skidmark
Nov 22, 2009, 12:41:28 PM11/22/09
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Blithering mathforum fuckwit.
"KY" <wkfk...@yahoo.co.jp> wrote in message
news:272515270.20197.12589...@gallium.mathforum.org...
Blithering mathforum fuckwit.
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