## Uzbekistan TST 2012-PROBLEM-2

For discussing Olympiad Level Algebra (and Inequality) problems
Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Uzbekistan TST 2012-PROBLEM-2

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.]
বড় ভালবাসি তোমায়,মা

SANZEED
Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm

### Re: Uzbekistan TST 2012-PROBLEM-2

If $f(x)=px^{2}+qx+r$
then it is $(p-1)(a^{2}+b^{2}+c^{2})+r$
But what is $p,r$   $\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

Posts: 244
Joined: Sat May 07, 2011 12:36 pm
Location: B.A.R.D , kotbari , Comilla

### Re: Uzbekistan TST 2012-PROBLEM-2

Is a,b,c are distinct ??
$\frac{1}{0}$

SANZEED
Posts: 550
Joined: Wed Dec 28, 2011 6:45 pm

### Re: Uzbekistan TST 2012-PROBLEM-2

Yes.My answer has come just the same. $\color{blue}{\textit{To}} \color{red}{\textit{ problems }} \color{blue}{\textit{I am encountering with-}} \color{green}{\textit{AVADA KEDAVRA!}}$

Posts: 217
Joined: Thu Oct 27, 2011 11:04 am
Location: mymensingh

### Re: Uzbekistan TST 2012-PROBLEM-2

$f(x)=f(a) \frac{(x-b)(x-c)}{(a-b)(a-c)}+f(b) \frac{(x-a)(x-c)}{(b-a)(b-c)}+f(c) \frac{(x-a)(x-b)}{(c-a)(c-b)}$
Now plugging $x=a+b+c$ we get $f(a+b+c)$ . Tahmid Hasan
Posts: 665
Joined: Thu Dec 09, 2010 5:34 pm

### Re: Uzbekistan TST 2012-PROBLEM-2

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. বড় ভালবাসি তোমায়,মা

nayel
Posts: 268
Joined: Tue Dec 07, 2010 7:38 pm
Location: Dhaka, Bangladesh or Cambridge, UK

### Re: Uzbekistan TST 2012-PROBLEM-2

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
$f(a+b+c)=\frac{(b+c)(c+a)(a+b)+abc}{a+b+c}=ab+bc+ca.$
"Everything should be made as simple as possible, but not simpler." - Albert Einstein

Nayel vai's solution is more than impressive. 