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Difference between revisions of "2019 USAMO Problems/Problem 6"

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==Solution==
 
==Solution==
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If <math>P(x)=c</math> for a constant <math>c,</math> then <math>\dfrac{c(x+y+z)}{xyz}=3c</math>. We have <math>2c=3c.</math> Therefore <math>c=0.</math>
  
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Now consider the case of non-constant polynomials.
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First we have <cmath>xP(x)+yP(y)+zP(z)=xyz(P(x-y)+P(y-z)+P(z-x))</cmath> for all nonzero real numbers <math>x,y,z</math> satisfying <math>2xyz=x+y+z</math>. Both sides of the equality are polynomials (of <math>x,y,z</math>). They have the same values on the 2-dimensional surface <math>2xyz=x+y+z</math>, except for some 1-dimensional curves in it. By continuity, the equality holds for all points on the surface, including those with <math>z=0.</math> Let <math>z=0, </math> we have <math>y=-x</math> and <math>x(P(x)-P(-x))=0.</math> Therefore <math>P</math> is an even function. 
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(Here is a sketch of an elementary proof. Let <math>z=\dfrac{x+y}{2xy-1}.</math> We have
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<cmath>xP(x)+yP(y)+\dfrac{x+y}{2xy-1}P(\dfrac{x+y}{2xy-1})=xy\dfrac{x+y}{2xy-1}(P(x-y)+P(y-\dfrac{x+y}{2xy-1})+P(\dfrac{x+y}{2xy-1}-x)).</cmath>
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This is an equality of rational expressions. By multiplying <math>(2xy-1)^N</math> on both sides for a sufficiently large <math>N</math>, they become polynomials, say <math>A(x,y)=B(x,y)</math> for all real <math>x, y</math> with <math>x\ne 0, y\ne 0, x+y\ne 0</math> and <math>2xy-1\ne 0.</math> For a fixed <math>x,</math> we have two polynomials (of <math>y</math>) having same values for infinitely many <math>y</math>. They must be identical. Let <math>y=0,</math> we have <math>x^{N+1}(P(x)-P(-x))=0.</math> )
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Notice that if <math>P(x)</math> is a solution,  then is <math>cP(x)</math> for any constant <math>c.</math> For simplicity, we assume the leading coefficient of <math>P</math> is <math>1</math>: <cmath>P(x)=x^n+a_{n-2}x^{n-2}+\cdots +a_2x^2+a_0,</cmath> where <math>n</math> is a positive even number.
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Let <math>y=\dfrac{1}{x}</math>, <math>z=x+\dfrac{1}{x}.</math> we have <cmath>xP(x)+\dfrac{1}{x}P\left (\dfrac{1}{x}\right )+\left ( x+\dfrac{1}{x}\right ) P\left ( x+\dfrac{1}{x}\right ) =\left (x+\dfrac{1}{x}\right )\left ( P\left (x-\dfrac{1}{x}\right )+P(-x)+P\left (\dfrac{1}{x}\right )\right ).</cmath>
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Simplify using <math>P(x)=P(-x),</math>
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<cmath>\left (x+\dfrac{1}{x}\right ) \left (P\left (x+\dfrac{1}{x}\right )-P\left (x-\dfrac{1}{x}\right )\right )=\dfrac{1}{x}P(x)+xP\left (\dfrac{1}{x}\right ).</cmath>
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Expand and combine like terms, both sides are of the form <cmath>c_{n-1}x^{n-1}+c_{n-3}x^{n-3}+\cdots+c_1x+c_{-1}x^{-1}+\cdots+c_{-n+1}x^{-n+1}.</cmath>
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They have the same values for infinitely many <math>x.</math> They must be identical. We just compare their leading terms. On the left hand side it is <math>2nx^{n-1}</math>. There are two cases for the right hand sides: If <math>n>2</math>, it is <math>x^{n-1}</math>; If <math>n=2</math>, it is <math>(1+a_0)x.</math> It does not work for <math>n>2.</math> When <math>n=2,</math> we have <math>4=1+a_0.</math> therefore <math>a_0=3.</math>
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The solution: <math>P(x)=c(x^2+3)</math> for any constant <math>c.</math>
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-JZ
 
==See also==
 
==See also==
 
{{USAMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}}
 
{{USAMO newbox|year=2019|num-b=5|aftertext=|after=Last Problem}}
  
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:10, 20 February 2020

Problem

Find all polynomials $P$ with real coefficients such that \[\frac{P(x)}{yz}+\frac{P(y)}{zx}+\frac{P(z)}{xy}=P(x-y)+P(y-z)+P(z-x)\]holds for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$.

Solution

If $P(x)=c$ for a constant $c,$ then $\dfrac{c(x+y+z)}{xyz}=3c$. We have $2c=3c.$ Therefore $c=0.$

Now consider the case of non-constant polynomials. First we have \[xP(x)+yP(y)+zP(z)=xyz(P(x-y)+P(y-z)+P(z-x))\] for all nonzero real numbers $x,y,z$ satisfying $2xyz=x+y+z$. Both sides of the equality are polynomials (of $x,y,z$). They have the same values on the 2-dimensional surface $2xyz=x+y+z$, except for some 1-dimensional curves in it. By continuity, the equality holds for all points on the surface, including those with $z=0.$ Let $z=0,$ we have $y=-x$ and $x(P(x)-P(-x))=0.$ Therefore $P$ is an even function.

(Here is a sketch of an elementary proof. Let $z=\dfrac{x+y}{2xy-1}.$ We have \[xP(x)+yP(y)+\dfrac{x+y}{2xy-1}P(\dfrac{x+y}{2xy-1})=xy\dfrac{x+y}{2xy-1}(P(x-y)+P(y-\dfrac{x+y}{2xy-1})+P(\dfrac{x+y}{2xy-1}-x)).\] This is an equality of rational expressions. By multiplying $(2xy-1)^N$ on both sides for a sufficiently large $N$, they become polynomials, say $A(x,y)=B(x,y)$ for all real $x, y$ with $x\ne 0, y\ne 0, x+y\ne 0$ and $2xy-1\ne 0.$ For a fixed $x,$ we have two polynomials (of $y$) having same values for infinitely many $y$. They must be identical. Let $y=0,$ we have $x^{N+1}(P(x)-P(-x))=0.$ )

Notice that if $P(x)$ is a solution, then is $cP(x)$ for any constant $c.$ For simplicity, we assume the leading coefficient of $P$ is $1$: \[P(x)=x^n+a_{n-2}x^{n-2}+\cdots +a_2x^2+a_0,\] where $n$ is a positive even number.

Let $y=\dfrac{1}{x}$, $z=x+\dfrac{1}{x}.$ we have \[xP(x)+\dfrac{1}{x}P\left (\dfrac{1}{x}\right )+\left ( x+\dfrac{1}{x}\right ) P\left ( x+\dfrac{1}{x}\right ) =\left (x+\dfrac{1}{x}\right )\left ( P\left (x-\dfrac{1}{x}\right )+P(-x)+P\left (\dfrac{1}{x}\right )\right ).\]

Simplify using $P(x)=P(-x),$ \[\left (x+\dfrac{1}{x}\right ) \left (P\left (x+\dfrac{1}{x}\right )-P\left (x-\dfrac{1}{x}\right )\right )=\dfrac{1}{x}P(x)+xP\left (\dfrac{1}{x}\right ).\]

Expand and combine like terms, both sides are of the form \[c_{n-1}x^{n-1}+c_{n-3}x^{n-3}+\cdots+c_1x+c_{-1}x^{-1}+\cdots+c_{-n+1}x^{-n+1}.\]

They have the same values for infinitely many $x.$ They must be identical. We just compare their leading terms. On the left hand side it is $2nx^{n-1}$. There are two cases for the right hand sides: If $n>2$, it is $x^{n-1}$; If $n=2$, it is $(1+a_0)x.$ It does not work for $n>2.$ When $n=2,$ we have $4=1+a_0.$ therefore $a_0=3.$

The solution: $P(x)=c(x^2+3)$ for any constant $c.$

-JZ

See also

2019 USAMO (Problems • Resources)
Preceded by
Problem 5 		Last Problem
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