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 ]-> ]-> ]->