Difference between revisions of "1993 USAMO Problems/Problem 3"
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<math>\frac{1}{2}c>1</math> | <math>\frac{1}{2}c>1</math> | ||
− | <math>c>2</math> but approach to <math>2</math> when <math>x</math> is | + | <math>c>2</math> but approach to <math>2</math> when <math>x</math> is extremely close to <math>\frac{1}{2}</math> from the right side. |
<P align="right"><math>\mathbb{Q.E.D}</math></P> | <P align="right"><math>\mathbb{Q.E.D}</math></P> |
Revision as of 00:25, 14 April 2012
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 $g(x)=\left\{\begin{array}{ll}0&0\le x\le \frac{1}{2};\\1&\frac{1}{2}<x\le1;\\\end{array}\right$ (Error compiling LaTeX. Unknown error_msg)
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.
Resources
1993 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |