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

Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 13"

m (<blank edit> solution incomplete, rigor necessary. Could someone provide?)
 
Line 9: Line 9:
 
It becomes obvious that <math>F(x) = ax(x-1)</math>, for some [[constant]] <math>a</math>, matches the definition of the polynomial. To prove that <math>F(x)</math> must have this form, note that  
 
It becomes obvious that <math>F(x) = ax(x-1)</math>, for some [[constant]] <math>a</math>, matches the definition of the polynomial. To prove that <math>F(x)</math> must have this form, note that  
 
<cmath>(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)</cmath>
 
<cmath>(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)</cmath>
−
(rigor needed)
+
 
 +
Since <math>3x</math> and <math>3x-1</math> divides the right side of the equation, <math>3x</math> and <math>3x-1</math> divides the left side of the equation. Thus <math>3x(3x-1)</math> divides <math>F(3x)</math>, so <math>x(x-1)</math> divides <math>F(x)</math>.
 +
 
 +
It is easy to see that <math>F(x)</math> is a [[quadratic]], thus <math>F(x)=ax(x-1)</math> as desired.
  
 
By the given, <math>F(6) = a(6)(5) = 15 \Longrightarrow a = \frac 12</math>. Thus, <math>F(12) = \frac{1}{2}(12)(11) = \boxed{066}</math>.
 
By the given, <math>F(6) = a(6)(5) = 15 \Longrightarrow a = \frac 12</math>. Thus, <math>F(12) = \frac{1}{2}(12)(11) = \boxed{066}</math>.
−
 
−
{{incomplete}}
 
  
 
== See also ==
 
== See also ==

Latest revision as of 21:10, 22 July 2008

Problem

Let $F(x)$ be a polynomial such that $F(6) = 15$ and \[\frac{F(3x)}{F(x+3)} = 9-\frac{48x+54}{x^2+5x+6}\] for $x \in \mathbb{R}$ such that both sides are defined. Find $F(12)$.

Solution

Combining denominators and simplifying, \[\frac{F(3x)}{F(x+3)} = \frac{9(x^2+5x+6)-48x-54}{x^2+5x+6} = \frac{9x^2 - 3x}{x^2 + 5x + 6}= \frac{3x(3x-1)}{(x+3)(x+2)}\] It becomes obvious that $F(x) = ax(x-1)$, for some constant $a$, matches the definition of the polynomial. To prove that $F(x)$ must have this form, note that \[(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)\]

Since $3x$ and $3x-1$ divides the right side of the equation, $3x$ and $3x-1$ divides the left side of the equation. Thus $3x(3x-1)$ divides $F(3x)$, so $x(x-1)$ divides $F(x)$.

It is easy to see that $F(x)$ is a quadratic, thus $F(x)=ax(x-1)$ as desired.

By the given, $F(6) = a(6)(5) = 15 \Longrightarrow a = \frac 12$. Thus, $F(12) = \frac{1}{2}(12)(11) = \boxed{066}$.

See also

Mock AIME 1 2007-2008 (Problems, Source)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=Mock_AIME_1_2007-2008_Problems/Problem_13&oldid=27135"