Difference between revisions of "2000 AIME II Problems/Problem 13"
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{{note|1}} A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of <math>x</math> with half of the polynomial's degree (in this case, divide through by <math>x^3</math>), and then to use one of the substitutions <math>t = x \pm \frac{1}{x}</math>. In this case, the substitution <math>t = x\sqrt{10} - \frac{1}{x\sqrt{10}}</math> gives <math>t^2 + 2 = 10x^2 + \frac 1{10x^2}</math> and <math>2\sqrt{10}(t^3 + 3t) = 200x^3 - \frac{2}{10x^3}</math>, which reduces the polynomial to just <math>(t^2 + 3)\left(2\sqrt{10}t + 1\right) = 0</math>. Then one can backwards solve for <math>x</math>. | {{note|1}} A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of <math>x</math> with half of the polynomial's degree (in this case, divide through by <math>x^3</math>), and then to use one of the substitutions <math>t = x \pm \frac{1}{x}</math>. In this case, the substitution <math>t = x\sqrt{10} - \frac{1}{x\sqrt{10}}</math> gives <math>t^2 + 2 = 10x^2 + \frac 1{10x^2}</math> and <math>2\sqrt{10}(t^3 + 3t) = 200x^3 - \frac{2}{10x^3}</math>, which reduces the polynomial to just <math>(t^2 + 3)\left(2\sqrt{10}t + 1\right) = 0</math>. Then one can backwards solve for <math>x</math>. | ||
+ | Note: After dividing the equation with <math>x^3</math>, divide again with <math>10</math> before substituting it with <math>t</math> to get it right. | ||
+ | |||
+ | |||
+ | A slightly different approach using symmetry: | ||
+ | <br /> | ||
+ | Let <math>y = 10x - 1/x</math>. | ||
+ | Notice that the equation can be rewritten (after dividing across by <math>x^3</math>) as | ||
+ | <cmath> 2( (10x)^3 - 1/x^3 ) + (10x)^2 + (1/x)^2 + 10 = 0 </cmath> | ||
+ | Now it is easy to see that the equation reduces to | ||
+ | <cmath> | ||
+ | \begin{align*} | ||
+ | 2(y^3+30y)+ (y^2+20) + 10 = 0 \\ | ||
+ | 2y^3 + y^2 + 60y + 30 = 0 \\ | ||
+ | y^2(2y+1) + 30(2y+1) = 0 \\ | ||
+ | (2y+1)(y^2+30)= 0 \\ | ||
+ | \end{align*} | ||
+ | </cmath> | ||
+ | so for real solutions we have <math>y = -1/2</math>. Solve the quadratic in <math>x</math> to get the final answer as <math>\boxed{200}</math>. | ||
== Solution 2 (Complex Bash)== | == Solution 2 (Complex Bash)== | ||
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-Grizzy | -Grizzy | ||
+ | |||
+ | ==Solution 3 (Geometric Series)== | ||
+ | Observe that the given equation may be rearranged as | ||
+ | <math>2000x^6-2+(100x^5+10x^3+x)=0</math>. | ||
+ | The expression in parentheses is a geometric series with common factor <math>10x^2</math>. Using the geometric sum formula, we rewrite as | ||
+ | <math>2000x^6-2+\frac{1000x^7-x}{10x^2-1}=0, 10x^2-1\neq0</math>. | ||
+ | Factoring a bit, we get | ||
+ | <math>2(1000x^6-1)+(1000x^6-1)\frac{x}{10x^2-1}=0, 10x^2-1\neq0 \implies</math> | ||
+ | <math>(1000x^6-1)(2+\frac{x}{10x^2-1})=0, 10x^2-1\neq0</math>. | ||
+ | Note that setting <math>1000x^6-1=0</math> gives <math>10x^2-1=0</math>, which is clearly extraneous. | ||
+ | Hence, we set <math>2+\frac{x}{10x^2-1}=0</math> and use the quadratic formula to get the desired root | ||
+ | <math>x=\frac{-1+\sqrt{161}}{40} \implies -1+161+40=\boxed{200}</math> | ||
+ | |||
+ | ~keeper1098 | ||
+ | |||
+ | ==Solution 4== | ||
+ | If we look at the polynomial's terms, we can see that the number of zeros in a term more or less correlates to the power of <math>x^2</math>. Thus, we let <math>y=10x^2</math>. The equation then becomes <math>2y^3+xy^2+xy+x-2=0</math>, or <math>x(y^2+y+1)=2(1-y^3)</math>. | ||
+ | |||
+ | By the difference of cubes formula, <math>2(1-y^3)=2(1-y)(1+y+y^2)</math>, so we have two cases: either <math>y^2+y+1=0</math>, or <math>x=2(1-y)</math>. We start with the second formula as it is simpler. | ||
+ | |||
+ | Solving with the quadratic formula after re-substitution, we see that <math>x=\frac{-1\pm\sqrt{161}}{40}</math>, so the answer is <math>-1+161+40=\boxed{200}</math>. | ||
+ | |||
+ | For the sake of completeness, if we check the other equation, we come to the conclusion that <math>y=\frac{-1\pm i\sqrt{3}}{2}</math>, so no real solution exists for <math>x</math>. Thus our solution is correct. | ||
+ | |||
+ | ~eevee9406 | ||
==Video solution== | ==Video solution== |
Latest revision as of 17:36, 26 April 2024
Contents
Problem
The equation has exactly two real roots, one of which is
, where
,
and
are integers,
and
are relatively prime, and
. Find
.
Solution
We may factor the equation as:[1]
Now for real
. Thus the real roots must be the roots of the equation
. By the quadratic formula the roots of this are:
Thus , and so the final answer is
.
^ A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of with half of the polynomial's degree (in this case, divide through by
), and then to use one of the substitutions
. In this case, the substitution
gives
and
, which reduces the polynomial to just
. Then one can backwards solve for
.
Note: After dividing the equation with
, divide again with
before substituting it with
to get it right.
A slightly different approach using symmetry:
Let .
Notice that the equation can be rewritten (after dividing across by
) as
Now it is easy to see that the equation reduces to
so for real solutions we have
. Solve the quadratic in
to get the final answer as
.
Solution 2 (Complex Bash)
It would be really nice if the coefficients were symmetrical. What if we make the substitution, . The the polynomial becomes
It's symmetric! Dividing by and rearranging, we get
Now, if we let , we can get the equations
and
(These come from squaring and subtracting
, then multiplying that result by
and subtracting
)
Plugging this into our polynomial, expanding, and rearranging, we get
Now, we see that the two terms must cancel in order to get this polynomial equal to
, so what squared equals 3? Plugging in
into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying
, we see that it also works! Great, we use long division on the polynomial by
and we get
.
We know that the other two solutions for z wouldn't result in real solutions for since we have to solve a quadratic with a negative discriminant, then multiply by
. We get that
. Solving for
(using
) we get that
, and multiplying this by
(because
) we get that
for a final answer of
-Grizzy
Solution 3 (Geometric Series)
Observe that the given equation may be rearranged as
.
The expression in parentheses is a geometric series with common factor
. Using the geometric sum formula, we rewrite as
.
Factoring a bit, we get
.
Note that setting
gives
, which is clearly extraneous.
Hence, we set
and use the quadratic formula to get the desired root
~keeper1098
Solution 4
If we look at the polynomial's terms, we can see that the number of zeros in a term more or less correlates to the power of . Thus, we let
. The equation then becomes
, or
.
By the difference of cubes formula, , so we have two cases: either
, or
. We start with the second formula as it is simpler.
Solving with the quadratic formula after re-substitution, we see that , so the answer is
.
For the sake of completeness, if we check the other equation, we come to the conclusion that , so no real solution exists for
. Thus our solution is correct.
~eevee9406
Video solution
https://www.youtube.com/watch?v=mAXDdKX52TM
See also
2000 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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.