# Hardest question IMO I had ever seen!

Let $\alpha$ and $\beta$ and $\gamma$ be 3 real numbers.

Prove that there exist only one polynomial $P(x)$ of the second degree such that $$\begin{cases}P(1)=\alpha \\ P(2)=\beta \\ P(3)=\gamma\end{cases}$$

I don't even know how to start?? Perhaps this is a new type of question to me and so if I know how to solve this one I will be able to solve similar questions!

Also this is second part of the question maybe that would help: Determine $P(x)$ in every case: $$\alpha=\beta=\gamma=2500\text{ and } \alpha=3;\beta=6;\gamma=9$$

Thanks!!

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$p(x)=ax^2+bx+c$, so we get a system of equations $a+b+c=\alpha$, $4a+2b+c=\beta$ and $9a+3b+c=\gamma$. Solving this we get $a=\gamma-\alpha$ and so on... If I understood the problem correctly. –  Poppy Dec 21 '13 at 17:43
Perhaps your last equation should have right side equal to $\;\gamma\;$ , @Vospost ? –  DonAntonio Dec 21 '13 at 17:45
Yes my friend DonAntonio –  VosPost Dec 21 '13 at 17:48
I've seen IMO questions that are quite a bit harder than this one. But I can only speak for myself. –  Marc van Leeuwen Dec 21 '13 at 17:49

The statement is false. For $\alpha=1$, $\beta=2$, $\gamma=3$ there is only one polynomial of degree at most $2$ that satifies the evaluations, namely $P(x)=x$; it is not of degree$~2$. There are many similar counterexamples.

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See my recent edit –  VosPost Dec 21 '13 at 17:59
@VosPost It still says "of the second degree". And the added examples actually give polynomials of degree $0$ and $1$, respectively. –  Marc van Leeuwen Dec 21 '13 at 18:14
All the time I thought it was given either "of second degree at most" or else "of second degree" being defined in the loose sense of the word, say like taking a straight line as a general quadratic with quadratic coefficient equal to zero. –  DonAntonio Dec 21 '13 at 18:38
+100000000000000000000000000000000000000000000000000 Mr Marc van Leeuwen Thank you so much!!!! –  VosPost Dec 21 '13 at 18:56
IMO means In My Opinion :p –  VosPost Dec 21 '13 at 20:19

Hints:

Solve the system

\begin{align*}a+b+c&=\alpha\\ 4a+2b+c&=\beta\\ 9a+3b+c&=\gamma\end{align*}

and show the polynomial $\;p(x)=ax^2+bx+c\;$ fulfills the conditions. And about uniqueness: suppose there's another polynomial $\;g(x)\;$ that fulfills the same conditions. What can you say about the roots of $\;h(x):=f(x)-g(x)\;$ ?

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" there's another polynomial g(x) that fulfills the same conditions. What can you say about the roots of h(x):=f(x)−g(x) ?" Euuuuuuuuuuuuuuh.... No idea.... –  VosPost Dec 21 '13 at 17:49
And how can I show that the polynomial p(x)=ax2+bx+c fulfills the conditions. –  VosPost Dec 21 '13 at 17:50
Why should there be any other discussion about uniqueness? Isn't a second degree polynomial uniquely specified by its coefficients? –  hhsaffar Dec 21 '13 at 17:51
And how do you intend to show that the given conditions uniquely specify the coefficients of quadratic polynomial, @hhsaffar ? –  DonAntonio Dec 21 '13 at 18:11

assume the quadratic equation $$ax^2+bx+c= P(x)$$ when you substitute the values of $x$ into the function. you get $3$ equations in $a,b$ and $c$ which has unique solutions.

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This simply restates what needs to be proved, without actually proving it. –  Greg Martin Dec 21 '13 at 18:44
rather than giving a readymade solution i prefer such hints so that the person who asks the question can put some effort himself to get it. and also i dont this is a restating of the question. –  Suraj M S Dec 21 '13 at 19:23

Assume that there are 2 polynomials $p(x)$ and $g(x)$
Then $q(x)=p(x)-g(x)$ is of degree at most $2$ and the equation $q(x)=0$ has $3$ solutions
the numbers $1,2,3$ which is a contradiction.
So, $g(x)=0$ which means $p(x)=g(x)$

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Way better than my friend DonAntonio ! –  VosPost Dec 21 '13 at 18:03
Thanks friend!! –  VosPost Dec 21 '13 at 18:04
you are welcome! –  Konstantinos Gaitanas Dec 21 '13 at 18:04
Very nice indeed, @VosPost...isn't this exactly what is written in my answer? Of course, you may prefer this one as you didn't understand my hint... –  DonAntonio Dec 21 '13 at 18:12
@Don Antonio actually i believe that you did not need to prove the existence of such a polynomial. i think that we suppose that there exists one polynomial taking these values $α,β,γ$ because the statement is not true in general for every triple of real numbers.(take $p(x)=x^2+1$ and $α=-1,β=-2,γ=0$ for example)But basically it is the same answer yes. –  Konstantinos Gaitanas Dec 21 '13 at 18:23