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

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==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)^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>.
 
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>.
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 +
A more detailed version goes as follows:
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 +
Observe that <math>x^2+1=x^2-(-1)=x^2-i^2=(x-i)(x+i).</math> Now, notice that:
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<cmath>\prod_{k=1} ^4 (x_k-i)(x_k+i)=\prod_{k=1}^4(x_k-i)\cdot\prod_{k=1}^4(x_k+i).</cmath>
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The definition of <math>P(x)</math> is <math>(x-x_1)(x-x_2)(x-x_3)(x-x_4)</math> where the leading coeffecient is <math>1</math> since <math>P</math> is a monic polynomial as given in the problem. Then, substituting in <math>i,</math> we have:
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<cmath>P(i)=(i-x_1)\cdots(i-x_4)=(-1)^4(x_1-i)\cdots(x_4-i).</cmath>
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This is exactly our first product. Our second product can be found as follows:
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<cmath>P(-i)=(-i-x_1)\cdots(-i-x_4)=(-1)^4(x_1+i)\cdots(x_4+i).</cmath>
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Hence, what we want to find is <math>P(i)P(-i).</math> But substituting in <math>i</math> into our expressoin <math>x^4+ax^3+bx^2+cx+d</math> gets us <math>P(i)=(1-b+d)+(c-a)i,</math> and <math>P(-i)=(1-b+d)+(a-c)i.</math> Hence:
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<cmath>P(i)P(-i)=((1-b+d)+(c-a)i)((1-b+d)+(a-c)i)=(1-b+d)^2-(a-c)^2=(-1)^2(b-d-1)^2-(a-c)^2 \ge 4^2 -0^2=16</cmath>
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where equality holds at <math>a=c</math> and <math>b-d=5 \Rightarrow b=d+5.</math> Hence, the minimum is <math>\boxed{16}.</math>
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To finish this off we find a construction for this minimum. We know that <math>a=c</math> and <math>b=d+5.</math> Hence
 +
<cmath>P(x)=x^4+cx^3+(d+5)x^2+cx+d.</cmath>
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We set <math>d=1</math> to get as much symmetry as possible within our polynomial. This leads to <math>x^4+cx^3+6x^2+cx+1.</math> However, note that <math>\binom{4}{2}=6</math> and <math>\binom{4}{0}=\binom{4}{4}=1,</math> so with some wishful thinking this leads us to think about the binomial theorem. We can try <math>(x\pm 1)^4</math> and realize that those solutions do work.
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Hence, <math>16</math> is obtained when all of <math>x_i</math> are equal to <math>\pm 1.</math>
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~mathboy282

Revision as of 21:57, 20 November 2024

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 $1$.

A more detailed version goes as follows:

Observe that $x^2+1=x^2-(-1)=x^2-i^2=(x-i)(x+i).$ Now, notice that: \[\prod_{k=1} ^4 (x_k-i)(x_k+i)=\prod_{k=1}^4(x_k-i)\cdot\prod_{k=1}^4(x_k+i).\] The definition of $P(x)$ is $(x-x_1)(x-x_2)(x-x_3)(x-x_4)$ where the leading coeffecient is $1$ since $P$ is a monic polynomial as given in the problem. Then, substituting in $i,$ we have: \[P(i)=(i-x_1)\cdots(i-x_4)=(-1)^4(x_1-i)\cdots(x_4-i).\] This is exactly our first product. Our second product can be found as follows: \[P(-i)=(-i-x_1)\cdots(-i-x_4)=(-1)^4(x_1+i)\cdots(x_4+i).\] Hence, what we want to find is $P(i)P(-i).$ But substituting in $i$ into our expressoin $x^4+ax^3+bx^2+cx+d$ gets us $P(i)=(1-b+d)+(c-a)i,$ and $P(-i)=(1-b+d)+(a-c)i.$ Hence: \[P(i)P(-i)=((1-b+d)+(c-a)i)((1-b+d)+(a-c)i)=(1-b+d)^2-(a-c)^2=(-1)^2(b-d-1)^2-(a-c)^2 \ge 4^2 -0^2=16\] where equality holds at $a=c$ and $b-d=5 \Rightarrow b=d+5.$ Hence, the minimum is $\boxed{16}.$

To finish this off we find a construction for this minimum. We know that $a=c$ and $b=d+5.$ Hence \[P(x)=x^4+cx^3+(d+5)x^2+cx+d.\] We set $d=1$ to get as much symmetry as possible within our polynomial. This leads to $x^4+cx^3+6x^2+cx+1.$ However, note that $\binom{4}{2}=6$ and $\binom{4}{0}=\binom{4}{4}=1,$ so with some wishful thinking this leads us to think about the binomial theorem. We can try $(x\pm 1)^4$ and realize that those solutions do work.

Hence, $16$ is obtained when all of $x_i$ are equal to $\pm 1.$

~mathboy282

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