Difference between revisions of "1987 USAMO Problems/Problem 3"
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==Solution== | ==Solution== | ||
− | {{ | + | 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> | ||
+ | |||
+ | 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 | ||
+ | <math>s(x)-t(x)=x(s_1(x)-t_1(x))\ne 0</math> for <math>0<x<1.</math> | ||
+ | |||
+ | 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 | ||
+ | <math>s(x)-t(x)=(1-x)(s_1(x)-t_1(x))\ne 0</math> for <math>0<x<1.</math> | ||
+ | |||
+ | The proof is done by an induction. | ||
+ | |||
+ | J.Z. | ||
==See Also== | ==See Also== |
Latest revision as of 15:49, 18 May 2018
Problem
is the smallest set of polynomials such that:
- 1. belongs to .
- 2. If belongs to , then and both belong to .
Show that if and are distinct elements of , then for any .
Solution
Let be an arbitrary polynomial in Then when Define for some and for some
If and we have for all with Therefore for any
For any , Let and for If for then for
Similarly, for any , Let and for If for then for
The proof is done by an induction.
J.Z.
See Also
1987 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.