Difference between revisions of "2014 AIME I Problems/Problem 9"
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+ | From Vieta's formulae, we know that <math>x_1x_2x_3 = \dfrac{-2}{\sqrt{2014}}, x_1 + x_2 + x_3 = \dfrac{4029}{\sqrt{2014}},</math> and <math>x_1x_2 + x_2x_3 + x_1x_3 = 0.</math> Thus, we know that <math>x_2(x_1 + x_3) = -x_1x_3.</math> | ||
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+ | Now consider the polynomial with roots <math>x_1x_2, x_2x_3,</math> and <math>x_1x_3</math>. Expanding the polynomial <math>(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)</math>, we get the polynomial <math>x^3 - (x_1x_2 + x_2x_3 + x_1x_3)x^2 + (x_1x_2x_3)(x_1 + x_2 + x_3)x - (x_1x_2x_3)^2.</math> Substituting the values obtained from Vieta's formulae, we find that this polynomial is <math>x^3 - \dfrac{8028}{2014}x - \dfrac{4}{2014}</math>. We know <math>x_1x_3</math> is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to <math>1007x^3 - 4029x - 2 = 0</math>. | ||
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+ | Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the <math>x^3</math> term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that <math>x = -2</math> is a solution. Factoring it out, we get that <math>1007x^3 - 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)</math>. Since the other quadratic factor clearly does not have any integer solutions and this is an AIME problem, we know that this must be the answer they are looking for. Thus, <math>x_1x_3 = -2</math>, so <math>-x_1x_3 = \boxed{002}</math>. | ||
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=8|num-a=10}} | {{AIME box|year=2014|n=I|num-b=8|num-a=10}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:36, 15 March 2014
Contents
Problem 9
Let be the three real roots of the equation . Find .
Solution
Substituting for , we get . Noting that factors as a difference of squares to , we can factor the left side as . This means that is a root, and the other two roots are the roots of . Note that the constant term of the quadratic is negative, so one of the two roots is positive and the other is negative. In addition, by Vieta's Formulas, the roots sum to , so the positive root must be greater than in order to produce this sum when added to a negative value. Since is clearly true, and . Multiplying these values together, we find that .
Solution 2
From Vieta's formulae, we know that and Thus, we know that
Now consider the polynomial with roots and . Expanding the polynomial , we get the polynomial Substituting the values obtained from Vieta's formulae, we find that this polynomial is . We know is a root of this polynomial, so we set it equal to 0 and simplify the resulting expression to .
Given the problem conditions, we know there must be at least 1 integer solution, and that it can't be very large (because the term quickly gets much larger/smaller than the other 2). Trying out some numbers, we quickly find that is a solution. Factoring it out, we get that . Since the other quadratic factor clearly does not have any integer solutions and this is an AIME problem, we know that this must be the answer they are looking for. Thus, , so .
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 8 |
Followed by Problem 10 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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