Difference between revisions of "2019 USAMO Problems/Problem 6"
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==Solution== | ==Solution== | ||
+ | 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> | ||
+ | Now consider the case of non-constant polynomials. | ||
+ | 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. | ||
+ | |||
+ | (Here is a sketch of an elementary proof. Let <math>z=\dfrac{x+y}{2xy-1}.</math> We have | ||
+ | <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> | ||
+ | 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> ) | ||
+ | |||
+ | 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. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | Simplify using <math>P(x)=P(-x),</math> | ||
+ | <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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | 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> | ||
+ | |||
+ | The solution: <math>P(x)=c(x^2+3)</math> for any constant <math>c.</math> | ||
+ | |||
+ | -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 16:10, 20 February 2020
Problem
Find all polynomials with real coefficients such that holds for all nonzero real numbers satisfying .
Solution
If for a constant then . We have Therefore
Now consider the case of non-constant polynomials. First we have for all nonzero real numbers satisfying . Both sides of the equality are polynomials (of ). They have the same values on the 2-dimensional surface , except for some 1-dimensional curves in it. By continuity, the equality holds for all points on the surface, including those with Let we have and Therefore is an even function.
(Here is a sketch of an elementary proof. Let We have This is an equality of rational expressions. By multiplying on both sides for a sufficiently large , they become polynomials, say for all real with and For a fixed we have two polynomials (of ) having same values for infinitely many . They must be identical. Let we have )
Notice that if is a solution, then is for any constant For simplicity, we assume the leading coefficient of is : where is a positive even number.
Let , we have
Simplify using
Expand and combine like terms, both sides are of the form
They have the same values for infinitely many They must be identical. We just compare their leading terms. On the left hand side it is . There are two cases for the right hand sides: If , it is ; If , it is It does not work for When we have therefore
The solution: for any constant
-JZ
See also
2019 USAMO (Problems • Resources) | ||
Preceded by Problem 5 |
Last Problem | |
1 • 2 • 3 • 4 • 5 • 6 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.