Uzbekistan TST 2012-PROBLEM-2

For discussing Olympiad Level Algebra (and Inequality) problems
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Tahmid Hasan
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Uzbekistan TST 2012-PROBLEM-2

Unread post by Tahmid Hasan » Sat Apr 14, 2012 10:31 pm

Let $a,b,c$ be real numbers and $f(x)$ a quadratic polynomial.$f(a)=bc,f(b)=ac,f(c)=ab$.Find $f(a+b+c)$.[This problem was very fun to solve but may be too easy for expert FE/polynomial solvers despite being a TST problem.]
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SANZEED
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Re: Uzbekistan TST 2012-PROBLEM-2

Unread post by SANZEED » Tue May 22, 2012 3:22 am

If \[f(x)=px^{2}+qx+r\]
then it is \[(p-1)(a^{2}+b^{2}+c^{2})+r\]
But what is $p,r$ :?: :?: :?
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Nadim Ul Abrar
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Re: Uzbekistan TST 2012-PROBLEM-2

Unread post by Nadim Ul Abrar » Tue May 22, 2012 9:41 am

Is a,b,c are distinct ??
let
\[ f(x)=px^2+qx+r \]

Now $ f(a)+f(b+f(c)+p(2ab+2bc+2ca)-2r=p(a^2+b^2+c^2+2ab+2bc+2ca)+q(a+b+c)+r$

so $f(a+b+c)=(2p+1)(ab+bc+ca)-2r $

We know that
$pa^2+qa+r=bc$ .. .. $(i)$
$pb^2+qb+r=ca$ .. .. $(ii)$
$pc^2+qc+r=ab$ .. .. $(iii)$

$(i)-(ii),(ii)-(iii)$ imply
$(a+b)p+q+c=0$ .. .. $(iv)$
$(b+c)p+q+a=0$ .. .. $(v)$

$(iv)-(v)$ imply
$p=1$
So $q=-(a+b+c)$
And $r=ab+bc+ca$ .

So that $f(a+b+c)=ab+bc+ca$
$\frac{1}{0}$

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SANZEED
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Re: Uzbekistan TST 2012-PROBLEM-2

Unread post by SANZEED » Wed May 23, 2012 2:05 am

Yes.My answer has come just the same. :roll:
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zadid xcalibured
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Re: Uzbekistan TST 2012-PROBLEM-2

Unread post by zadid xcalibured » Wed Feb 27, 2013 12:10 pm

$f(x)=f(a) \frac{(x-b)(x-c)}{(a-b)(a-c)}+f(b) \frac{(x-a)(x-c)}{(b-a)(b-c)}+f(c) \frac{(x-a)(x-b)}{(c-a)(c-b)}$
Now plugging $x=a+b+c$ we get $f(a+b+c)$ . :mrgreen:

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Tahmid Hasan
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Re: Uzbekistan TST 2012-PROBLEM-2

Unread post by Tahmid Hasan » Wed Feb 27, 2013 12:50 pm

zadid xcalibured wrote:$f(x)=f(a) \frac{(x-b)(x-c)}{(a-b)(a-c)}+f(b) \frac{(x-a)(x-c)}{(b-a)(b-c)}+f(c) \frac{(x-a)(x-b)}{(c-a)(c-b)}$
Now plugging $x=a+b+c$ we get $f(a+b+c)$ . :mrgreen:
Straight-forward Lagrange's Interpolation! Hmm, impressive. ;)
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nayel
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Re: Uzbekistan TST 2012-PROBLEM-2

Unread post by nayel » Wed Feb 27, 2013 7:10 pm

Let $g(x)=xf(x)-abc$. $g$ is a cubic polynomial with $g(a)=g(b)=g(c)=0$, hence
\[g(x)=k(x-a)(x-b)(x-c)\]
for some constant $k$, and so $f(x)=[k(x-a)(x-b)(x-c)+abc]/x$, which is a polynomial in $x$ only if $k=1$. Then
\[f(a+b+c)=\frac{(b+c)(c+a)(a+b)+abc}{a+b+c}=ab+bc+ca.\]
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zadid xcalibured
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Re: Uzbekistan TST 2012-PROBLEM-2

Unread post by zadid xcalibured » Wed Feb 27, 2013 7:38 pm

Nayel vai's solution is more than impressive. :mrgreen:

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