Difference between revisions of "1993 USAMO Problems/Problem 3"
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== Problem 3== | == Problem 3== | ||
− | Consider functions <math>f : [0, 1] \rightarrow \ | + | Consider functions <math>f : [0, 1] \rightarrow \mathbb{R}</math> which satisfy |
<table><tr> | <table><tr> | ||
<td> </td><td>(i)</td><td><math>f(x)\ge0</math> for all <math>x</math> in <math>[0, 1]</math>,</td></tr> | <td> </td><td>(i)</td><td><math>f(x)\ge0</math> for all <math>x</math> in <math>[0, 1]</math>,</td></tr> | ||
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<br/> | <br/> | ||
− | <math> | + | <math>\forall 0\le x\le 1</math>, <math> \left(\frac{1}{2}\right)^{n-1}\ge2x\ge \left(\frac{1}{2}\right)^n\ge f(x)</math>. Thus, <math>c=2</math> works. |
<br/> | <br/> | ||
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[[Category:Olympiad Algebra Problems]] | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Functional Equation Problems]] |
Latest revision as of 16:03, 13 August 2023
Problem 3
Consider functions which satisfy
(i) | for all in , | |
(ii) | , | |
(iii) | whenever , , and are all in . |
Find, with proof, the smallest constant such that
for every function satisfying (i)-(iii) and every in .
Solution
My claim:
Lemma 1) for
For , (ii)
Assume that it is true for , then
By principle of induction, lemma 1 is proven.
Lemma 2) For any , and , .
(lemma 1 and (iii) )
(because (i) )
, . Thus, works.
Let's look at a function
It clearly have property (i) and (ii). For and WLOG let ,
For , . Thus, property (iii) holds too. Thus is one of the legit function.
but approach to when is extremely close to from the right side.
See Also
1993 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.