Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

Difference between revisions of "1990 USAMO Problems/Problem 2"

(wrote solution)
Line 36: Line 36:
 
{{USAMO box|year=1990|num-b=1|num-a=3}}
 
{{USAMO box|year=1990|num-b=1|num-a=3}}
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356624#p356624 Discussion on AoPS/MathLinks]
 
* [http://www.artofproblemsolving.com/Forum/viewtopic.php?p=356624#p356624 Discussion on AoPS/MathLinks]
 +
{{MAA Notice}}
  
  
 
[[Category:Olympiad Algebra Problems]]
 
[[Category:Olympiad Algebra Problems]]

Revision as of 19:47, 3 July 2013

Problem

A sequence of functions $\, \{f_n(x) \} \,$ is defined recursively as follows: \begin{align*} f_1(x) &= \sqrt {x^2 + 48}, \quad \text{and} \\ f_{n + 1}(x) &= \sqrt {x^2 + 6f_n(x)} \quad \text{for } n \geq 1. \end{align*} (Recall that $\sqrt {\makebox[5mm]{}}$ is understood to represent the positive square root.) For each positive integer $n$, find all real solutions of the equation $\, f_n(x) = 2x \,$.

Solution

We define $f_0(x) = 8$. Then the recursive relation holds for $n=0$, as well.

Since $f_n (x) \ge 0$ for all nonnegative integers $n$, it suffices to consider nonnegative values of $x$.

We claim that the following set of relations hold true for all natural numbers $n$ and nonnegative reals $x$: \begin{align*} f_n(x) &< 2x \text{ if }x>4 ; \\ f_n(x) &= 2x \text{ if }x=4 ; \\ f_n(x) &> 2x \text{ if }x<4 . \end{align*} To prove this claim, we induct on $n$. The statement evidently holds for our base case, $n=0$.

Now, suppose the claim holds for $n$. Then \begin{align*} f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} < \sqrt{x^2+12x} < \sqrt{4x^2} = 2x, \text{ if } x>4 ; \\ f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} = \sqrt{x^2 + 12x} = \sqrt{4x^2} = 2x, \text{ if } x=4 ; \\ f_{n+1}(x) &= \sqrt{x^2 + 6f_n(x)} > \sqrt{x^2+12x} > \sqrt{4x^2} = 2x, \text{ if } x<4 . \end{align*} The claim therefore holds by induction. It then follows that for all nonnegative integers $n$, $x=4$ is the unique solution to the equation $f_n(x) = 2x$. $\blacksquare$


Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

See also

1990 USAMO (Problems • Resources)
Preceded by
Problem 1 		Followed by
Problem 3
1 • 2 • 3 • 4 • 5
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1990_USAMO_Problems/Problem_2&oldid=53595"