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Uzbekistan TST 2012-PROBLEM-2
Posted: Sat Apr 14, 2012 10:31 pm
by Tahmid Hasan
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.]
Re: Uzbekistan TST 2012-PROBLEM-2
Posted: Tue May 22, 2012 3:22 am
by SANZEED
If \[f(x)=px^{2}+qx+r\]
then it is \[(p-1)(a^{2}+b^{2}+c^{2})+r\]
But what is $p,r$
Re: Uzbekistan TST 2012-PROBLEM-2
Posted: Tue May 22, 2012 9:41 am
by Nadim Ul Abrar
Re: Uzbekistan TST 2012-PROBLEM-2
Posted: Wed May 23, 2012 2:05 am
by SANZEED
Yes.My answer has come just the same.
Re: Uzbekistan TST 2012-PROBLEM-2
Posted: Wed Feb 27, 2013 12:10 pm
by zadid xcalibured
$f(x)=f(a) \frac{(x-b)(x-c)}{(a-b)(a-c)}+f(b) \frac{(x-a)(x-c)}{(b-a)(b-c)}+f(c) \frac{(x-a)(x-b)}{(c-a)(c-b)}$
Now plugging $x=a+b+c$ we get $f(a+b+c)$ .
Re: Uzbekistan TST 2012-PROBLEM-2
Posted: Wed Feb 27, 2013 12:50 pm
by Tahmid Hasan
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 $f(a+b+c)$ .
Straight-forward Lagrange's Interpolation! Hmm, impressive.
Re: Uzbekistan TST 2012-PROBLEM-2
Posted: Wed Feb 27, 2013 7:10 pm
by nayel
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
\[f(a+b+c)=\frac{(b+c)(c+a)(a+b)+abc}{a+b+c}=ab+bc+ca.\]
Re: Uzbekistan TST 2012-PROBLEM-2
Posted: Wed Feb 27, 2013 7:38 pm
by zadid xcalibured
Nayel vai's solution is more than impressive.