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RE: greetings and a question!

• To: mathgroup at smc.vnet.net
• Subject: [mg83136] RE: [mg83116] greetings and a question!
• From: "Tony Harker" <a.harker at ucl.ac.uk>
• Date: Mon, 12 Nov 2007 05:18:15 -0500 (EST)

Rearrange the expression as

  (((x + 1) + (a + b - 2)/2)^2 + (y - (a - b)/2)^2) + ((a - 1)^2)/
  2 + ((b - 1)^2/)2,

 which, if a, b, x and y are real, can only be zero if each term is zero.
The result follows.

  As to how Mathematica does it -- over to the experts!

   Tony

A.H. Harker
112 Cumnor Hill
Oxford
OX2 9HY
UK 

]-> -----Original Message-----
]-> From: dimitris [mailto:dimmechan at yahoo.com] 
]-> Sent: 11 November 2007 08:03
]-> To: mathgroup at smc.vnet.net
]-> Subject: [mg83116] greetings and a question!
]-> 
]-> Hello to all of you!
]-> 
]-> Unfortunately, family, working and research issues prevent 
]-> me from participating to the forum as frequent as I used to.
]-> 
]-> Anyway...
]-> 
]-> It is my first post since a long time so everybody be patient!
]-> 
]-> A student of mine came across the following equation in a 
]-> mathematical contest:
]-> 
]-> In[1]:=
]-> eq = x^2 + y^2 + (a + b)*x - (a - b)*y + a^2 + b^2 - a - b + 1==0;
]-> 
]-> (all variables are assumed real)
]-> 
]-> Of course for Mathematica the solution is rather trivial.
]-> 
]-> In[1]:=
]-> $Version
]-> 
]-> Out[1]=
]-> "5.2 for Microsoft Windows (June 20, 2005)"
]-> 
]-> In[2]:=
]-> eq = x^2 + y^2 + (a + b)*x - (a - b)*y + a^2 + b^2 - a - b + 1==0;
]-> 
]-> In[3]:=
]-> Reduce[eq, {a, b, x, y}, Reals]
]-> ToRules[%]
]-> eq /. %
]-> 
]-> Out[3]=
]-> a == 1 && b == 1 && x == -1 && y == 0
]-> 
]-> Out[4]=
]-> {a -> 1, b -> 1, x -> -1, y -> 0}
]-> 
]-> Out[5]=
]-> True
]-> 
]-> Can somebody explain concisely the mathematica concept 
]-> behind this solution? In fact I would be much obliged if 
]-> somebody pointed me out how to obtain the result by hand. 
]-> Also, by curiosity, how Mathematica reaches the result?
]-> 
]-> Dimitris
]-> 
]-> 
]->