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Re: greetings and a question!
*To*: mathgroup at smc.vnet.net
*Subject*: [mg83134] Re: [mg83116] greetings and a question!
*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>
*Date*: Mon, 12 Nov 2007 05:17:13 -0500 (EST)
*References*: <200711110802.DAA04679@smc.vnet.net> <5A0A3CCA-C056-4322-94FC-2E7FEFFC173A@mimuw.edu.pl>
On 11 Nov 2007, at 20:01, Andrzej Kozlowski wrote:
>
> On 11 Nov 2007, at 17:02, dimitris wrote:
>
>> 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
>>
>>
>
>
> I believe Reduce calls on CylindricalDecomposition:
>
> CylindricalDecomposition[
> x^2 + y^2 + (a + b)*x - (a - b)*y + a^2 + b^2 -
> a - b + 1 == 0, {x, y, a, b}]
> x == -1 && y == 0 && a == 1 && b == 1
>
> However, I do not think I want to try to explain here what
> CylindricalDecomposition (an algorithm in real algebraic Geometery
> due to G. Collins, usually known as Cylindrical Algebraic
> Decomposition) does. There are lots of books where you can find it
> explained in detail. A comparatively simple explanation can be
> found in Chapter 8 of B. Mishra, "Algorithmic Algebra", Springer
> 1993. If you are only interested in the general idea without
> proofs take a look at Mishra's article in the Handbook of
> Discrete and Computational Geometry, edited by Goodman and
> O'Rourke, CRC Press, 1997.
>
> Andrzej Kozlowski
In fact, although Mathematica certianly does this by calling on
CylindricalDecomposition, one cna easily do it by hand by elementary
means. To do this just observe that your expression can be written as:
(x + (a + b)/2)^2 + (y - (a - b)/2)^2 + (a - 1)^2/2 + (b - 1)^2/2 ==0.
On the left hand side you have a sum of non-negative expressions so
each of them has to be 0. This tells you at once that a=1, b=1, x=-1
and y=0.
Andrzej Kozlowski
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