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Difference between revisions of "2018 UMO Problems/Problem 2"

(Wrote a solution and posted the problem.)
 
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== Solution 1 ==
 
== Solution 1 ==
  
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Plugging in <math>x = 0</math>, we find that <math>abc = 1. Using AM-GM, we have that </math>a+b+c \leq 3 ^3\sqrt{abc} = \fbox{3}$
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Plugging in <math>x = 0</math>, we find that <math>abc = 1</math>. Using AM-GM, we have that <math>a+b+c \leq 3 ^3\sqrt{abc} = \fbox{3}</math>

Revision as of 16:05, 31 July 2021

Problem 2

Let $P(x)$ be a cubic polynomial $(x-a)(x-b)(x-c)$, where $a, b,$ and $c$ are positive real numbers. Let Q(x) be the polynomial with $Q(x) = (x-ab)(x-bc)(x-ac)$. If $P(x) = Q(x)$ for all $x$, then find the minimum possible value of $a + b + c$.

Solution 1

Plugging in $x = 0$, we find that $abc = 1$. Using AM-GM, we have that $a+b+c \leq 3 ^3\sqrt{abc} = \fbox{3}$

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