Difference between revisions of "2014 AIME I Problems/Problem 9"
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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 <cmath>(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)</cmath>we get the polynomial <cmath>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</cmath> Substituting the values obtained from Vieta's formulae, we find that this polynomial is <cmath>x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}</cmath> 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 <cmath>1007x^3 - 4029x - 2 = 0</cmath> | Now consider the polynomial with roots <math>x_1x_2, x_2x_3,</math> and <math>x_1x_3</math>. Expanding the polynomial <cmath>(x - x_1x_2)(x - x_2x_3)(x - x_1x_3)</cmath>we get the polynomial <cmath>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</cmath> Substituting the values obtained from Vieta's formulae, we find that this polynomial is <cmath>x^3 - \dfrac{8058}{2014}x - \dfrac{4}{2014}</cmath> 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 <cmath>1007x^3 - 4029x - 2 = 0</cmath> | ||
− | 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 <cmath>1007x^3 | + | 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 <cmath>1007x^3 + 4029x - 2 = (x+2)(1007x^2 - 2014x - 1)</cmath> Since the other quadratic factor clearly does not have any integer solutions and since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus, <cmath>x_1x_3 = -2</cmath> so <cmath>-x_1x_3 = \boxed{002}</cmath>and we're done. |
==Solution 3== | ==Solution 3== | ||
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We will estimate the roots of the polynomial. (This strategy normally doesn't work on AIME #9, but playing around with a function is often good strategy for getting an intuition for the problem. In this problem, estimation happens to be a valid solution path. It isn't a proof, but given the constraint that the answer is an integer, we can be certain that our answer is correct.) | We will estimate the roots of the polynomial. (This strategy normally doesn't work on AIME #9, but playing around with a function is often good strategy for getting an intuition for the problem. In this problem, estimation happens to be a valid solution path. It isn't a proof, but given the constraint that the answer is an integer, we can be certain that our answer is correct.) | ||
+ | |||
Let <math>p(x) = \sqrt{2014}x^3-4029x^2+2=0</math>. We start by estimating <math>p(-1)</math>, <math>p(0)</math>, and <math>p(1)</math> (A natural first step for function analysis.): | Let <math>p(x) = \sqrt{2014}x^3-4029x^2+2=0</math>. We start by estimating <math>p(-1)</math>, <math>p(0)</math>, and <math>p(1)</math> (A natural first step for function analysis.): | ||
− | <math>p(-1)\approx -45-4029+2 \approx -4000</math> | + | |
− | <math>p(0) = 2</math> | + | <math>p(-1)\approx -45-4029+2 \approx -4000</math> |
− | + | ||
+ | <math>p(0) = 2</math> | ||
+ | |||
+ | <math>p(1) \approx 45-4029+2 \approx -4000</math> | ||
+ | |||
We conclude by Intermediate Value Theorem (or just common sense), that there is a root on <math>(-1, 0)</math> and another root on <math>(0, 1)</math>. | We conclude by Intermediate Value Theorem (or just common sense), that there is a root on <math>(-1, 0)</math> and another root on <math>(0, 1)</math>. | ||
+ | |||
We know that <math>p(1) < 0</math> and that <math>\lim_{x\to\infty} p(x) = \infty</math>. We conclude that the third root is on <math>(1, \infty)</math>. Therefore, <math>x_1 \in (-1, 0)</math>, <math>x_2 \in (0, 1)</math>, and <math>x_3 \in (1, \infty)</math>. | We know that <math>p(1) < 0</math> and that <math>\lim_{x\to\infty} p(x) = \infty</math>. We conclude that the third root is on <math>(1, \infty)</math>. Therefore, <math>x_1 \in (-1, 0)</math>, <math>x_2 \in (0, 1)</math>, and <math>x_3 \in (1, \infty)</math>. | ||
− | We will estimate <math>x_3</math>. For <math>x > 1</math>, the constant term of <math>p(x)</math> is negligible. We simplify and get <math>p( | + | |
+ | We will estimate <math>x_3</math>. For <math>x > 1</math>, the constant term of <math>p(x)</math> is negligible. We simplify and get <math>p(x_3) \approx \sqrt{2014}x_{3}^3 - 4029x_{3}^2 = 0</math>. Solving for <math>x_3</math> (We can divide by <math>x_{3}^2</math> because we know <math>x_3 \neq 0</math>), we get <math>x_3 \approx \frac{4029}{\sqrt{2014}} \approx 2\sqrt{2014} \approx 90</math>. We can intuitively bound <math>x_3</math> between <math>88</math> and <math>92</math>. | ||
+ | |||
We will now estimate <math>x_1</math> and <math>x_2</math>. <math>x_1</math> and <math>x_2</math> are close to <math>0</math>. As a result, the <math>\sqrt{2014}x^3</math> term is negligible. We simplify and get <math>p(x) \approx -4029x^2 + 2 = 0</math>. Solving for <math>x</math>, we get <math>x \approx \pm \sqrt{\frac{2}{4029}} \approx \pm \sqrt{\frac{1}{2014.5}} \approx \pm \frac{1}{45}</math>. We can intuitively bound <math>x_1</math> between <math>-\frac{1}{43}</math> and <math>-\frac{1}{47}</math>. Similarly, we can intuitively bound <math>x_2</math> between <math>\frac{1}{47}</math> and <math>\frac{1}{43}</math>. | We will now estimate <math>x_1</math> and <math>x_2</math>. <math>x_1</math> and <math>x_2</math> are close to <math>0</math>. As a result, the <math>\sqrt{2014}x^3</math> term is negligible. We simplify and get <math>p(x) \approx -4029x^2 + 2 = 0</math>. Solving for <math>x</math>, we get <math>x \approx \pm \sqrt{\frac{2}{4029}} \approx \pm \sqrt{\frac{1}{2014.5}} \approx \pm \frac{1}{45}</math>. We can intuitively bound <math>x_1</math> between <math>-\frac{1}{43}</math> and <math>-\frac{1}{47}</math>. Similarly, we can intuitively bound <math>x_2</math> between <math>\frac{1}{47}</math> and <math>\frac{1}{43}</math>. | ||
+ | |||
We calculate that the minimum possible value of <math>x_2(x_1 + x_3)</math> is <math>\frac{88-\frac{1}{43}}{47}</math> and the maximum possible value is <math>\frac{92-\frac{1}{47}}{43}</math>. The only integer that falls is this range is <math>\boxed{002}</math>. | We calculate that the minimum possible value of <math>x_2(x_1 + x_3)</math> is <math>\frac{88-\frac{1}{43}}{47}</math> and the maximum possible value is <math>\frac{92-\frac{1}{47}}{43}</math>. The only integer that falls is this range is <math>\boxed{002}</math>. | ||
+ | |||
~numerophile | ~numerophile | ||
+ | |||
+ | == Solution 8 == | ||
+ | Let <math>a=x\sqrt{2014}</math>. We have <math>\sqrt{2014}x^3-4029x^2+2=ax^2-2a^2-x^2+2=x^2(a-1)-2(a^2-1)=x^2(a-1)-2(a-1)(a+1)=(a-1)(x^2-2a-2)=0</math>, so <math>a=x\sqrt{2014}=1</math> or <math>x^2-2a-2=x^2-2\sqrt{2014}x-2=0</math>. As such, the solutions are at <math>x=\frac{1}{\sqrt{2014}}</math> and <math>x=\frac{2\sqrt{2014}\pm\sqrt{4\cdot2014+8}}{2}</math>. Note that <math>\frac{2\sqrt{2014}-\sqrt{4\cdot2014+8}}{2}<0<\frac{1}{\sqrt{2014}}<\sqrt{2014}<\frac{2\sqrt{2014}+\sqrt{4\cdot2014+8}}{2}</math>, so <math>x_2(x_1+x_3)=\frac{x_1+x_3}{\sqrt{2014}}=\frac{2\sqrt{2014}}{\sqrt{2014}}=\boxed{2}</math>. (<math>x_1+x_3=2\sqrt{2014}</math> by vietas.) | ||
+ | |||
+ | -zhoujef000 | ||
+ | |||
+ | == Solution 9 (cheese) == | ||
+ | Rewrite the polynomial as <math>nx^3 - (2n^2+1)x^2 + 2 = 0</math>. The answer probably holds for general <math>n</math>, so letting <math>n=0</math>, we have <math>x^2+2=0</math>. We'll assume that the missing root doesn't impact the final answer, so by Vietas, we have <math>\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}} |
Latest revision as of 13:03, 17 August 2024
Contents
Problem 9
Let be the three real roots of the equation
. Find
.
Solution 1
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 since the AIME has only positive integer answers, we know that this must be the answer they are looking for. Thus,
so
and we're done.
Solution 3
Observing the equation, we notice that the coefficient for the middle term is equal to
.
Also notice that the coefficient for the term is
. Therefore, if the original expression was to be factored into a linear binomial and a quadratic trinomial, the
term of the binomial would have a coefficient of
. Similarly, the
term of the trinomial would also have a coefficient of
. The factored form of the expression would look something like the following:
where
are all positive integers (because the
term of the original expression is negative, and the constant term is positive), and
Multiplying this expression out gives Equating this with the original expression gives
The only positive integer solutions of this expression is
or
. If
then setting
yields
and therefore
which clearly isn't equal to
as the constant term. Therefore,
and the factored form of the expression is:
Therefore, one of the three roots of the original expression is
Using the quadratic formula yields the other two roots as
and
Arranging the roots in ascending order (in the order
),
Therefore,
Solution 4
By Vieta's, we are seeking to find . Substitute
and
. Substituting this back into the original equation, we have
, so
. Hence,
, and
. But since
because it is our desired answer, the only possible value for
is
BEST PROOOFFFF
Stormersyle & mathleticguyyy
Solution 5
Let The original equation simplifies to
Here we clearly see that
is a root. Dividing
from the sum we find that
From simple bounding we see that
is the middle root. Therefore
Solution 6
occurs multiple times, so let k =
.
The equation becomes . Since we want to relate k and x, we should solve for one of them. We can't solve for x, since that would require the cubic formula, so we solve for k, and express it in terms of a quadratic, and apply the quadratic formula.
We get the roots are:
, and
.
In the first case, .
In the second case, . The solutions are
. The sum of these 2 solutions is
, and
is the middle solution, and thus,
Solution 7 (Estimation and Intuitive Function Analysis)
We will estimate the roots of the polynomial. (This strategy normally doesn't work on AIME #9, but playing around with a function is often good strategy for getting an intuition for the problem. In this problem, estimation happens to be a valid solution path. It isn't a proof, but given the constraint that the answer is an integer, we can be certain that our answer is correct.)
Let . We start by estimating
,
, and
(A natural first step for function analysis.):
We conclude by Intermediate Value Theorem (or just common sense), that there is a root on and another root on
.
We know that and that
. We conclude that the third root is on
. Therefore,
,
, and
.
We will estimate . For
, the constant term of
is negligible. We simplify and get
. Solving for
(We can divide by
because we know
), we get
. We can intuitively bound
between
and
.
We will now estimate and
.
and
are close to
. As a result, the
term is negligible. We simplify and get
. Solving for
, we get
. We can intuitively bound
between
and
. Similarly, we can intuitively bound
between
and
.
We calculate that the minimum possible value of is
and the maximum possible value is
. The only integer that falls is this range is
.
~numerophile
Solution 8
Let . We have
, so
or
. As such, the solutions are at
and
. Note that
, so
. (
by vietas.)
-zhoujef000
Solution 9 (cheese)
Rewrite the polynomial as . The answer probably holds for general
, so letting
, we have
. We'll assume that the missing root doesn't impact the final answer, so by Vietas, we have
.
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.