Chinese Dumbass Notation

by v_Enhance, Aug 7, 2012, 3:05 AM

\[ \begin{tabular}{rccccccccccccc}
   &    &    &    &    &    &  0\\\noalign{\smallskip\smallskip}
   &    &    &    &    &  0 &	&  0\\\noalign{\smallskip\smallskip}
   &    &    &    &  1 &	& -2 &	&  1\\\noalign{\smallskip\smallskip}
   &    &    & -2 &	&  2 &	&  2 &	& -2\\\noalign{\smallskip\smallskip}
   &    &  1 &	&  2 &	& -6 &	&  2 &	&  1\\\noalign{\smallskip\smallskip}
   &  0 &	& -2 &	&  2 &	&  2 &	& -2 &	&  0\\\noalign{\smallskip\smallskip}
 0 &	&  0 &	&  1 &	& -2 &	&  1 &	&  0 &	&  0\\\noalign{\smallskip\smallskip}
\end{tabular} \]

(I heard this was actually given as a problem at MOP '03.)

Comment

9 Comments

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darn this thing is pretty sharp

by dinoboy, Aug 7, 2012, 3:58 AM

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I'm surprised it's not widely recognized...

by v_Enhance, Aug 7, 2012, 4:38 AM

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If this was a MOP '03 problem, then I probably would've cried during the test xD

If I'm not mistaken, this is equivalent to $(a-b)^2(b-c)^2(c-a)^2$ (first time I tried standard Schur/AM-GM but realized the signs were switched).
This post has been edited 1 time. Last edited by GlassBead, Aug 7, 2012, 1:20 PM

by GlassBead, Aug 7, 2012, 1:20 PM

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lol it looks my horrible intuition for inequalities was correct : you need to use some cruel sum of squares to do this problem since it was to sharp for schur/muirhead/etc. I searched for 5 minutes but then gave up, oops this problem is trivial :blush:

by dinoboy, Aug 7, 2012, 5:47 PM

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Yeah I tried for a while too and then was like this has got to be troll. So I plugged in a=b and did derivatives. Then then used Wolfram to verify. xD Man I'm so bad at inequalities.
This post has been edited 1 time. Last edited by GlassBead, Aug 7, 2012, 5:53 PM

by GlassBead, Aug 7, 2012, 5:53 PM

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Wait this problem is definitely not trivial ^^

The tip-off is that $a=b$ leads to equality; you can see this by noting that the sum of the entries in each row is zero. Hence, if this problem is going to be true, $(a-b)^2$ had better divide the polynomial, and then cyclically for the others, and since this is degree six we see that this must be the factorization (possibly times a constant).

by v_Enhance, Aug 7, 2012, 10:16 PM

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there's a problem in mildorf that basically requires dumbassing and this identity

by Mewto55555, Aug 8, 2012, 7:31 PM

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This was a problem in Mildorf; by an odd coincidence, I had expanded this the same morning that I saw it while surfing through my resources.

by v_Enhance, Aug 8, 2012, 8:17 PM

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@bzprules haha are there people who actually cry during tests?

On a more serious note, it seems like not in the set of stuff of what people do when faced with an ineq is to plug in a=b (especially in a nice, refined symmetric one like this). BUt hey, if nothing else works, then you might just try that and hope for the best. v_Enhance, you probably had a similar experience with USAMO 2014 #3 hahaha :-) lolololol

by Konigsberg, Dec 24, 2014, 7:16 PM

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