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Difference between revisions of "1993 USAMO Problems"

(Created page with '== Problem 1== For each integer <math>n\ge2</math>, determine, with proof, which of the two positive real numbers <math>a</math> and <math>b</math> satisfying <center><math>a^n…')
 
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Find, with proof, the smallest constant <math>c</math> such that
 
Find, with proof, the smallest constant <math>c</math> such that
  
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<math>f(x) \le cx</math>
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<center><math>f(x) \le cx</math></center>
  
 
for every function <math>f</math> satisfying (i)-(iii) and every <math>x</math> in <math>[0, 1]</math>.
 
for every function <math>f</math> satisfying (i)-(iii) and every <math>x</math> in <math>[0, 1]</math>.
  
 
[[1993 USAMO Problems/Problem 3 | Solution]]
 
[[1993 USAMO Problems/Problem 3 | Solution]]
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== Problem 4==
 
== Problem 4==

Revision as of 17:59, 22 April 2010

Problem 1

For each integer $n\ge2$, determine, with proof, which of the two positive real numbers $a$ and $b$ satisfying

$a^n = a + 1, \quad b^{2n} = b + 3a$

is larger.

Solution

Problem 2

Let $ABCD$ be a convex quadrilateral such that diagonals $AC$ and $BD$ intersect at right angles, and let $E$ be their intersection. Prove that the reflections of $E$ across $AB$, $BC$, $CD$, $DA$ are concyclic.

Solution

Problem 3

Consider functions $f : [0, 1] \rightarrow \Re$ which satisfy

     (i)$f(x)\ge0$ for all $x$ in $[0, 1]$,
     (ii)$f(1) = 1$,
     (iii)     $f(x) + f(y) \le f(x + y)$ whenever $x$, $y$, and $x + y$ are all in $[0, 1]$.

Find, with proof, the smallest constant $c$ such that

$f(x) \le cx$

for every function $f$ satisfying (i)-(iii) and every $x$ in $[0, 1]$.

Solution

Problem 4

Let $a$, $b$ be odd positive integers. Define the sequence $(f_n)$ by putting $f_1 = a$, $f_2 = b$, and by letting fn for $n\ge3$ be the greatest odd divisor of $f_{n-1} + f_{n-2}$. Show that $f_n$ is constant for $n$ sufficiently large and determine the eventual value as a function of $a$ and $b$.

Solution

Problem 5

Let $a_0, a_1, a_2,\cdots$ be a sequence of positive real numbers satisfying $a_{i-1}a_{i+1}\le a^2_i$ for $i = 1, 2, 3,\cdots$ . (Such a sequence is said to be log concave.) Show that for each $n > 1$,

$\frac{a_0+\cdots+a_n}{n+1}\cdot\frac{a_1+\cdots+a_{n-1}}{n-1}\ge\frac{a_0+\cdots+a_{n-1}}{n}\cdot\frac{a_1+\cdots+a_{n}}{n}$.


Solution

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