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        riddles >> medium >> What does this expression equal?
        
(Message started by: pcbouhid on Nov 25^th, 2005, 10:20am)

Title: What does this expression equal?
Post by pcbouhid on Nov 25^th, 2005, 10:20am

If a + b + c = 0, what does the following expression equal?

[(b-c)/a + (c-a)/b + (a-b)/c] * [a/(b-c) + b/(c-a) + c/(a-b)].

Title: Re: What does this expression equal?
Post by Michael_Dagg on Nov 25^th, 2005, 10:31am

Looks like it = 9

Title: Re: What does this expression equal?
Post by pcbouhid on Nov 25^th, 2005, 10:58am

I read in a comment of William Wu something like this: "I don´t want just an answer; Í want to learn how to think with you".

And I agree totally with him!

Title: Re: What does this expression equal?
Post by Michael_Dagg on Nov 25^th, 2005, 11:13am

Well, then think about how I arrived at 9 by simple inspection.

Title: Re: What does this expression equal?
Post by JohanC on Nov 25^th, 2005, 11:55am

Is there another way than just [hide]substitute c with -a-b and then do a lengthy but straightforward simplification[/hide] ?

Title: Re: What does this expression equal?
Post by Michael_Dagg on Nov 25^th, 2005, 12:35pm

Investigate manipulating the condition a + b + c = 0 to form fractional expressions instead of using substitution.

Title: Re: What does this expression equal?
Post by Eigenray on Nov 25^th, 2005, 8:39pm

Substituting c=-(a+b),
[(b-c)/a + (c-a)/b + (a-b)/c] = f(a,b)/[ab(a+b)],
for some polynomial f of total degree 3. But we don't need to compute it to know that it's divisible by (a-b), (a+2b), and (2a+b), because it becomes 0 upon setting a=b, b=c, or c=a, respectively. It follows that
[(b-c)/a + (c-a)/b + (a-b)/c] = C(a-b)(a+2b)(2a+b)/[ab(a+b)].
Similarly,
[a/(b-c) + b/(c-a) + c/(a-b)] = g(a,b)/[(a+2b)(2a+b)(a-b)],
for some g. But if a=0, this becomes b/c-c/b = 0, for b=-c; thus g(a,b) is divisible by a. Similarly, it follows g(a,b) is divisible by a*b*(a+b). So
[a/(b-c) + b/(c-a) + c/(a-b)] = C'ab(a+b)/[(a+2b)(2a+b)(a-b)].
It follows
[(b-c)/a + (c-a)/b + (a-b)/c] * [a/(b-c) + b/(c-a) + c/(a-b)] = CC'
is constant. To evaluate the constant, one may take a=1, b=e^{2pi i/3}, c=b²; then (b-c)/a = (c-a)/b = (a-b)/c, so the result is 9.

Title: Re: What does this expression equal?
Post by Eigenray on Nov 25^th, 2005, 9:03pm

Maybe this is more constructive: let
x=a/(b-c), y=b/(c-a), z=c/(a-b).
Then
x+y = (ac-a²+b²-bc)/[(b-c)(c-a)] = [c(a-b)-(a+b)(a-b)]/[(b-c)(c-a)] = 2c(a-b)/[(b-c)(c-a)],
(x+y)/z = 2(a-b)²/[(b-c)(c-a)].
Taking cyclic permutations, it follows
[x+y+z][1/x+1/y+1/z] = 3 + 2[(a-b)³+(b-c)³+(c-a)³]/[(a-b)(b-c)(c-a)].
The numerator is a polynomial in a,b,c, of degree 3, and vanishes when a=b, b=c, or c=a. So the above is constant. Or to see this directly, use
u³ + v³ + (-u-v)³ = 3uv(-u-v).

Of course one can also do
1/y+1/z = (c-a)/b + (a-b)/c = [c²-ac+ab-b²]/[bc] = [(b-c)(a - (b+c))]/[bc] = 2a(b-c)/[bc],
x(1/y + 1/z) = 2a²/[bc],
so
[x+y+z][1/x+1/y+1/z] = 3 + 2[a³+b³+c³]/[abc] = 9.
(Another way to see the last equality is to note that (a³+b³+c³) is a symmetric polynomial, so is a polynomial in the elementary symmetric polynomials. Since it's degree 3, we have
a³+b³+c³ = A(a+b+c)³ + B(a+b+c)(ab+bc+ca) + C*abc = C*abc
for some C.)

Title: Re: What does this expression equal?
Post by Aryabhatta on Nov 26^th, 2005, 1:30am

Isn't (b-c)/a = (c-a)/b ?