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2014 USAMO Problems/Problem 1: Difference between revisions

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==Hint==
==Hint==
Factor <math>x^2 + 1</math> as the product of two linear binomials.
Factor <math>x^2 + 1</math> as the product of two linear binomials.
==Solution==
==Solution==
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)-i(a-c))^2 \implies \boxed{16}</math>.
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)-i(a-c))^2 \implies \boxed{16}</math>.
==Solution==
The value in question is equal to
<cmath> P(i) P(-i) = \left\lvert (b-d-1) + (a-c)i \right\rvert^2= (b-d-1)^2 + (a-c)^2 \ge16 </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.

Revision as of 16:50, 23 November 2016

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 $\prod_{k=1} ^4 (x_k-i)(x_k+i) \implies P(i)P(-i) \implies ((b-d-1)-i(a-c))^2 \implies \boxed{16}$.

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