Sigh
by djmathman, Oct 28, 2013, 11:04 PM
I haven't been doing many math problems lately darn darn darn
and the ones I do try take forever darn I have to learn to get faster at these if I have any hope of trying to do more of them throughout junior year
like this one took two hours and I'm not even sure if it's correct gg
EDIT: Oops yea it's sort-of wrong; easily patchable since the only real thing I do is switch the meanings of
and 
toward the end of the solution
and the ones I do try take forever darn I have to learn to get faster at these if I have any hope of trying to do more of them throughout junior year
like this one took two hours and I'm not even sure if it's correct gg
EDIT: Oops yea it's sort-of wrong; easily patchable since the only real thing I do is switch the meanings of
and 
toward the end of the solutionIran 2009 Round 2.1 wrote:
Let 
be a quadratic polynomial for which
![\[ |p(x)| \leq 1 \qquad \forall x \in \{-1,0,1\}.\]](//latex.artofproblemsolving.com/b/d/6/bd65826d2e04eac0b928389890acc3cd1bbc4e8d.png)
Prove that
![\[ \ |p(x)|\leq\frac{5}{4} \qquad \forall x \in [-1,1].\]](//latex.artofproblemsolving.com/d/4/d/d4d1db6787fb7aa0b24243fb2494c6ab8f112a01.png)
be a quadratic polynomial for which![\[ |p(x)| \leq 1 \qquad \forall x \in \{-1,0,1\}.\]](http://latex.artofproblemsolving.com/b/d/6/bd65826d2e04eac0b928389890acc3cd1bbc4e8d.png)
Prove that
![\[ \ |p(x)|\leq\frac{5}{4} \qquad \forall x \in [-1,1].\]](http://latex.artofproblemsolving.com/d/4/d/d4d1db6787fb7aa0b24243fb2494c6ab8f112a01.png)
,
.
-
,
and 
respectively. Otherwise, let 
and 
.
and 
,![\[x_{\text{vertex}}=\dfrac{-b}{2a}=\dfrac{-(m+n)/2}{2(m-n)/2}=\dfrac{-m-n}{2(m-n)}.\]](http://latex.artofproblemsolving.com/5/7/e/57efbe3491f7439d19894ed49e3ad8bcbd2f5edc.png)
and 
,
and 
.
-![\[a\left(\dfrac{-b}{2a}\right)^2+b\left(\dfrac{-b}{2a}\right)+c=c-\dfrac{b^2}{4a}=c-\dfrac{(m+n)^2/4}{4(m-n)/2}=c-\dfrac{(m+n)^2}{8(m-n)}.\]](http://latex.artofproblemsolving.com/b/4/e/b4e5cf5b32ba9cac7d048af5527e857e6128dd95.png)
,
,
that makes 
.
.
are essentially independent, so we shall maximize our expression as though they were independent. Our expression becomes 
.
is negative, but this is absurd, as 
,
.
and 
,
gives the value of the expression to be![\[2+\dfrac{2^2}{8(2\cdot 2-2)}=2+\dfrac{4}{8\cdot 2}=\dfrac94.\]](http://latex.artofproblemsolving.com/a/1/6/a1621dab5637d800ff55c24b991046600a534c0c.png)
,
(as then 
)
has to be 
,
along the interval as desired. 
This post has been edited 3 times. Last edited by djmathman, Oct 31, 2013, 3:08 AM
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