Difference between revisions of "Mock AIME 1 2007-2008 Problems/Problem 13"
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== Problem == | == Problem == | ||
| − | Let <math>F(x)</math> be a polynomial such that <math>F(6) = 15</math> and | + | Let <math>F(x)</math> be a [[polynomial]] such that <math>F(6) = 15</math> and |
<cmath>\frac{F(3x)}{F(x+3)} = 9-\frac{48x+54}{x^2+5x+6}</cmath> | <cmath>\frac{F(3x)}{F(x+3)} = 9-\frac{48x+54}{x^2+5x+6}</cmath> | ||
for <math>x \in \mathbb{R}</math> such that both sides are defined. Find <math>F(12)</math>. | for <math>x \in \mathbb{R}</math> such that both sides are defined. Find <math>F(12)</math>. | ||
== Solution == | == Solution == | ||
| − | Combining | + | Combining [[denominator]]s and simplifying, |
<cmath>\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)}</cmath> | <cmath>\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)}</cmath> | ||
| − | 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> | ||
| − | ( | + | |
| + | 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>. | ||
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| + | 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>. | ||
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== See also == | == See also == | ||
Latest revision as of 21:10, 22 July 2008
Problem
Let 
be a polynomial such that 
and
![\[\frac{F(3x)}{F(x+3)} = 9-\frac{48x+54}{x^2+5x+6}\]](http://latex.artofproblemsolving.com/f/d/0/fd0069334a3b7ec601f2dad037ad38f1d982847c.png)
for 
such that both sides are defined. Find 
.
Solution
Combining denominators and simplifying,
![\[\frac{F(3x)}{F(x+3)} = \frac{9(x^2+5x+6)-48x-54}{x^2+5x+6} = \frac{9x^2 - 3x}{x^2 + 5x + 6}= \frac{3x(3x-1)}{(x+3)(x+2)}\]](http://latex.artofproblemsolving.com/6/a/7/6a74eac97c248bfddd0a2076642ef35a39c3cb6e.png)
It becomes obvious that 
, for some constant 
, matches the definition of the polynomial. To prove that must have this form, note that
![\[(x+3)(x+2)F(3x) = 3x(3x-1)F(x+3)\]](http://latex.artofproblemsolving.com/a/4/2/a42343839d84b14f115b4d70320ceea03008bbc5.png)
Since 
and 
divides the right side of the equation, and divides the left side of the equation. Thus 
divides 
, so 
divides .
It is easy to see that is a quadratic, thus 
as desired.
By the given, 
. Thus, 
.
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 | ||
