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

(Solution)
(Solution)
Line 4: Line 4:
 
==Solution==
 
==Solution==
 
The value in question is equal to
 
The value in question is equal to
<cmath> P(i) P(-i) = \left\lvert (b-d-1) + (a-c)i \right\rvert= (b-d-1)^2 + (a-c)^2 \ge16 </cmath>
+
<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.
 
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 17:04, 30 April 2014

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.

Solution

The value in question is equal to \[P(i) P(-i) = \left\lvert (b-d-1) + (a-c)i \right\rvert^2= (b-d-1)^2 + (a-c)^2 \ge16\] where $i = \sqrt{-1}$. Equality holds if $x_1 = x_2 = x_3 = x_4 = 1$, so this bound is sharp.