Difference between revisions of "2014 USAMO Problems/Problem 1"

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==Problem==
 
==Problem==
 
Let <math>a,b,c,d</math> be real numbers such that <math>b-d \ge 5</math> and all zeros <math>x_1, x_2, x_3,</math> and <math>x_4</math> of the polynomial <math>P(x)=x^4+ax^3+bx^2+cx+d</math> are real. Find the smallest value the product <math>(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)</math> can take.
 
Let <math>a,b,c,d</math> be real numbers such that <math>b-d \ge 5</math> and all zeros <math>x_1, x_2, x_3,</math> and <math>x_4</math> of the polynomial <math>P(x)=x^4+ax^3+bx^2+cx+d</math> are real. Find the smallest value the product <math>(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)</math> can take.
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==Hint==
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Factor <math>x^2 + 1</math> as the product of two linear binomials.
  
 
==Solution==
 
==Solution==
The value in question is equal to
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Using the hint we turn the equation into <math>\prod_{k=1} ^4 (x_k-i)(x_k+i) \implies P(i)P(-i) \implies (b-d-1)^2 + (a-c)^2 \implies \boxed{16}</math>. This minimum is achieved when all the <math>x_i</math> are equal to <math>1</math>.
<cmath> P(i) P(-i) = \left[ (b-d-1) + (a-c)i \right][ (b-d-1) - (a-c)i \right] = (b-d-1)^2 + (a-c)^2 \ge (5-1)^2 + 0^2 = 16 </cmath>
 
where <math>i = \sqrt{-1}</math>. Equality holds if <math>x_1 = x_2 = x_3 = x_4 = 1</math>, so this bound is sharp.
 

Latest revision as of 02:14, 21 June 2019

Problem

Let $a,b,c,d$ be real numbers such that $b-d \ge 5$ and all zeros $x_1, x_2, x_3,$ and $x_4$ of the polynomial $P(x)=x^4+ax^3+bx^2+cx+d$ are real. Find the smallest value the product $(x_1^2+1)(x_2^2+1)(x_3^2+1)(x_4^2+1)$ can take.

Hint

Factor $x^2 + 1$ as the product of two linear binomials.

Solution

Using the hint we turn the equation into $\prod_{k=1} ^4 (x_k-i)(x_k+i) \implies P(i)P(-i) \implies (b-d-1)^2 + (a-c)^2 \implies \boxed{16}$. This minimum is achieved when all the $x_i$ are equal to $1$.

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