Difference between revisions of "1987 USAMO Problems/Problem 3"

(Created page with "==Problem== <math>X</math> is the smallest set of polynomials <math>p(x)</math> such that: : 1. <math>p(x) = x</math> belongs to <math>X</math>. : 2. If <math>r(x)</math> be...")
 
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==Solution==
 
==Solution==
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Let <math>s(x)</math> be an arbitrary polynomial in <math>X.</math> Then <math>0<s(x)<1</math> when <math>0<x<1.</math> Define <math>X_1=\{s(x)\in X:s(x)=x\cdot s_1(x)</math> for some <math>s_1(x)\in X\},</math> and <math>X_2=\{t(x)\in X: t(x)=x+(1-x)t_1(x)</math> for some <math>t_1(x)\in X\}.</math>
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If <math>s(x) \in X_1</math> and <math>t(x)\in X_2,</math> we have <math>s(x) <x<t(x)</math> for all <math>x</math> with <math>0<x<1.</math> Therefore <math>s(x)\ne t(x)</math> for any <math>0<x<1.</math>
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For any <math>s(x), t(x) \in X_1</math>, Let <math>s(x)=x\cdot s_1(x)</math> and <math>t(x)=x\cdot t_1(x)</math> for <math>s_1(x), t_1(x) \in X.</math> If <math>s_1(x) \ne t_1(x)</math> for <math>0<x<1,</math> then
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<math>s(x)-t(x)=x(s_1(x)-t_1(x))\ne 0</math> for <math>0<x<1.</math>
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Similarly, for any <math>s(x), t(x) \in X_2</math>, Let <math>s(x)=x+(1-x) s_1(x)</math> and <math>t(x)=x+(1-x) t_1(x)</math> for <math>s_1(x), t_1(x) \in X.</math> If <math>s_1(x) \ne t_1(x)</math> for <math>0<x<1,</math> then
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<math>s(x)-t(x)=(1-x)(s_1(x)-t_1(x))\ne 0</math> for <math>0<x<1.</math>
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The proof is done by an induction.
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J.Z.
  
 
==See Also==
 
==See Also==

Latest revision as of 15:49, 18 May 2018

Problem

$X$ is the smallest set of polynomials $p(x)$ such that:

1. $p(x) = x$ belongs to $X$.
2. If $r(x)$ belongs to $X$, then $x\cdot r(x)$ and $(x + (1 - x) \cdot r(x) )$ both belong to $X$.

Show that if $r(x)$ and $s(x)$ are distinct elements of $X$, then $r(x) \neq s(x)$ for any $0 < x < 1$.

Solution

Let $s(x)$ be an arbitrary polynomial in $X.$ Then $0<s(x)<1$ when $0<x<1.$ Define $X_1=\{s(x)\in X:s(x)=x\cdot s_1(x)$ for some $s_1(x)\in X\},$ and $X_2=\{t(x)\in X: t(x)=x+(1-x)t_1(x)$ for some $t_1(x)\in X\}.$

If $s(x) \in X_1$ and $t(x)\in X_2,$ we have $s(x) <x<t(x)$ for all $x$ with $0<x<1.$ Therefore $s(x)\ne t(x)$ for any $0<x<1.$

For any $s(x), t(x) \in X_1$, Let $s(x)=x\cdot s_1(x)$ and $t(x)=x\cdot t_1(x)$ for $s_1(x), t_1(x) \in X.$ If $s_1(x) \ne t_1(x)$ for $0<x<1,$ then $s(x)-t(x)=x(s_1(x)-t_1(x))\ne 0$ for $0<x<1.$

Similarly, for any $s(x), t(x) \in X_2$, Let $s(x)=x+(1-x) s_1(x)$ and $t(x)=x+(1-x) t_1(x)$ for $s_1(x), t_1(x) \in X.$ If $s_1(x) \ne t_1(x)$ for $0<x<1,$ then $s(x)-t(x)=(1-x)(s_1(x)-t_1(x))\ne 0$ for $0<x<1.$

The proof is done by an induction.

J.Z.

See Also

1987 USAMO (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions

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