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/> | ||
− | Let's look at a function <math>g(x)=\left\{\begin{array}{ll}0&0\le x\le \frac{1}{2};\\1&\frac{1}{2}<x\le1;\\\end{array}\right </math> | + | Let's look at a function <math>g(x)=\left\{\begin{array}{ll}0&0\le x\le \frac{1}{2};\\1&\frac{1}{2}<x\le1;\\\end{array}\right\} </math> |
It clearly have property (i) and (ii). For <math>0\le x\le\frac{1}{2}</math> and WLOG let <math>x\le y</math>, <math>f(x)+f(y)=0+f(y)\le f(y)</math> | It clearly have property (i) and (ii). For <math>0\le x\le\frac{1}{2}</math> and WLOG let <math>x\le y</math>, <math>f(x)+f(y)=0+f(y)\le f(y)</math> | ||
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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> | ||
− | == | + | == See Also == |
{{USAMO box|year=1993|num-b=2|num-a=4}} | {{USAMO box|year=1993|num-b=2|num-a=4}} | ||
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks] | * [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356413#p356413 Discussion on AoPS/MathLinks] | ||
+ | {{MAA Notice}} | ||
+ | |||
+ | [[Category:Olympiad Algebra Problems]] | ||
+ | [[Category:Functional Equation Problems]] |
Latest revision as of 17:03, 13 August 2023
Problem 3
Consider functions which satisfy
(i) | ![]() ![]() ![]() | |
(ii) | ![]() | |
(iii) | ![]() ![]() ![]() ![]() ![]() |
Find, with proof, the smallest constant such that
![$f(x) \le cx$](http://latex.artofproblemsolving.com/b/1/e/b1e93ec7efefb0af408a87d3d13e41a9e5bd902e.png)
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.