Difference between revisions of "2005 AIME I Problems/Problem 8"
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== Solution == | == Solution == | ||
| − | Let <math>y = 2^{111x}</math>. Then our equation reads <math>\frac{1}{4}y^3 + 4y = 2y^2 + 1</math> or <math>y^3 - 8y^2 + 16y - 4 = 0</math>. Thus, if this equation has roots <math>r_1, r_2</math> and <math>r_3</math>, by [[Vieta's formulas]] we have <math>r_1\cdot r_2\cdot r_3 = 4</math>. Let the corresponding values of <math>x</math> be <math>x_1, x_2</math> and <math>x_3</math>. Then the previous statement says that <math>2^{111\cdot(x_1 + x_2 + x_3)} = 4</math> so taking a [[logarithm]] of that gives <math>111(x_1 + x_2 + x_3) = 2</math> and <math>x_1 + x_2 + x_3 = \frac{2}{111}</math>. Thus the answer is 111 + 2 = 113 | + | Let <math>y = 2^{111x}</math>. Then our equation reads <math>\frac{1}{4}y^3 + 4y = 2y^2 + 1</math> or <math>y^3 - 8y^2 + 16y - 4 = 0</math>. Thus, if this equation has roots <math>r_1, r_2</math> and <math>r_3</math>, by [[Vieta's formulas]] we have <math>r_1\cdot r_2\cdot r_3 = 4</math>. Let the corresponding values of <math>x</math> be <math>x_1, x_2</math> and <math>x_3</math>. Then the previous statement says that <math>2^{111\cdot(x_1 + x_2 + x_3)} = 4</math> so taking a [[logarithm]] of that gives <math>111(x_1 + x_2 + x_3) = 2</math> and <math>x_1 + x_2 + x_3 = \frac{2}{111}</math>. Thus the answer is <math>111 + 2 = \boxed{113}</math>. |
== See also == | == See also == | ||
Latest revision as of 16:46, 17 September 2024
Problem
The equation 
has three real roots. Given that their sum is 
where 
and 
are relatively prime positive integers, find 
Solution
Let 
. Then our equation reads 
or 
. Thus, if this equation has roots 
and 
, by Vieta's formulas we have 
. Let the corresponding values of 
be 
and 
. Then the previous statement says that 
so taking a logarithm of that gives 
and 
. Thus the answer is 
.
See also
| 2005 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
