Difference between revisions of "2000 AIME II Problems/Problem 13"
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Thus <math>r=\frac{-1+\sqrt{161}}{40}</math>, and so the final answer is <math>-1+161+40 = \boxed{200}</math>. | Thus <math>r=\frac{-1+\sqrt{161}}{40}</math>, and so the final answer is <math>-1+161+40 = \boxed{200}</math>. | ||
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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>. | ||
+ | == Solution 2 (Complex Bash)== | ||
+ | It would be really nice if the coefficients were symmetrical. What if we make the substitution, <math>x = -\frac{i}{\sqrt{10}}y</math>. The the polynomial becomes <math>\\$ | ||
+ | </math>-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2\\$ | ||
+ | It's symmetric! Dividing by <math>y^3</math> and rearranging, we get <math>\\$ | ||
+ | </math>-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})\\$ | ||
+ | Now, if we let <math>z = y + \frac{1}{y}</math>, we can get the equations <math>\\$ | ||
+ | </math>z = y + \frac{1}{y}\\$ | ||
+ | <math>z^2 - 2 = y^2 + \frac{1}{y^2}\\$ | ||
+ | </math>z^3 - 3z = y^3 + \frac{1}{y^3}\\$ | ||
+ | (These come from squaring <math>z</math> and subtracting <math>2</math>, then multiplying that result by <math>z</math> and subtracting <math>z</math>) <math>\\$ | ||
+ | Plugging this into our polynomial, expanding, and rearranging, we get </math>\\$ | ||
+ | <math>-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})\\$ | ||
+ | Now, we see that the two </math>i<math> terms must cancel in order to get this polynomial equal to </math>0<math>, so what squared equals 3? Plugging in </math>z = \sqrt{3}<math> into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying </math>z = -\sqrt{3}<math>, we see that it also works! Great, we use long division on the polynomial by </math>(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)<math> and we get </math>\\$ | ||
+ | <math>2z -(\frac{i}{\sqrt{10}}) = 0</math>, we know that the other two solutions for z wouldn't result in real solutions for <math>x</math> since we have to solve a quadratic with a negative discriminant, then multiply by <math>-(\frac{i}{\sqrt{10}})</math>. We get that <math>z = (\frac{i}{-2\sqrt{10}})</math>. Solving for <math>y</math> (using <math>y + \frac{1}{y} = z</math>) we get that <math>y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}</math>, and multiplying this by <math>-(\frac{i}{\sqrt{10}})</math> (because <math>x = -(\frac{i}{\sqrt{10}})y</math>) we get that <math>x = \frac{-1 \pm \sqrt{161}}{40}</math> for a final answer of <math>-1 + 161 + 40 = \boxed{200}</math> | ||
== See also == | == See also == | ||
{{AIME box|year=2000|n=II|num-b=12|num-a=14}} | {{AIME box|year=2000|n=II|num-b=12|num-a=14}} |
Revision as of 21:14, 5 October 2019
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 .
Solution 2 (Complex Bash)
It would be really nice if the coefficients were symmetrical. What if we make the substitution, . The the polynomial becomes $\$ (Error compiling LaTeX. ! Missing $ inserted.)-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2\$ It's symmetric! Dividing by and rearranging, we get $\$ (Error compiling LaTeX. ! Missing $ inserted.)-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})\$ Now, if we let , we can get the equations $\$ (Error compiling LaTeX. ! Missing $ inserted.)z = y + \frac{1}{y}\$ $z^2 - 2 = y^2 + \frac{1}{y^2}\$ (Error compiling LaTeX. ! Missing $ inserted.)z^3 - 3z = y^3 + \frac{1}{y^3}\$ (These come from squaring and subtracting , then multiplying that result by and subtracting ) $\$ Plugging this into our polynomial, expanding, and rearranging, we get$ (Error compiling LaTeX. ! Missing $ inserted.)\$ $-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})\$ Now, we see that the two$ (Error compiling LaTeX. ! Missing $ inserted.)i0z = \sqrt{3}z = -\sqrt{3}(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)\$ , 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
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 |
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