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>. | ||
+ | |||
+ | |||
+ | A slightly different approach using symmetry: | ||
+ | let y = 10x -1/x | ||
+ | notice that the equation can be rewritten( after dividing across by x^3) as 2( (10x)^3 - 1/x^3 ) + (10x)^2 +1/x^2 +10 =0 | ||
+ | now it is easy to see that the equation reduces to 2(y^3+30y)+ (y^2+20) + 10 = 0 | ||
+ | OR 2y^3 +y^2 +60y+30 = 0 | ||
+ | or y^2(2y+1) + 30(2y+1) = 0 | ||
+ | which is (2y+1)(y^2+30)= 0 | ||
+ | so for real solutions we have y= -1/2 | ||
+ | and solve the quadratic in x to get final answer as \boxed{200}$. | ||
== Solution 2 (Complex Bash)== | == Solution 2 (Complex Bash)== |
Revision as of 08:57, 2 August 2022
Contents
[hide]Problem
The equation has exactly two real roots, one of which is , where , and are integers, and are relatively prime, and . Find .
Solution
We may factor the equation as:[1]
Now for real . Thus the real roots must be the roots of the equation . By the quadratic formula the roots of this are:
Thus , and so the final answer is .
^ A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of with half of the polynomial's degree (in this case, divide through by ), and then to use one of the substitutions . In this case, the substitution gives and , which reduces the polynomial to just . Then one can backwards solve for .
A slightly different approach using symmetry:
let y = 10x -1/x
notice that the equation can be rewritten( after dividing across by x^3) as 2( (10x)^3 - 1/x^3 ) + (10x)^2 +1/x^2 +10 =0
now it is easy to see that the equation reduces to 2(y^3+30y)+ (y^2+20) + 10 = 0
OR 2y^3 +y^2 +60y+30 = 0
or y^2(2y+1) + 30(2y+1) = 0
which is (2y+1)(y^2+30)= 0
so for real solutions we have y= -1/2
and solve the quadratic in x to get final answer as \boxed{200}$.
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
Notice the original expression can be written as .
Which equals to
So our solution is to find what is the root for since the determinant of (Let )
By solving the equation, we can get that for a final answer of
~bluesoul
Solution 4 (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
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.