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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)==
−
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>\\$
+
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>-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>-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2</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>\\$
+
It's symmetric! Dividing by <math>y^3</math> and rearranging, we get  
−
</math>z = y + \frac{1}{y}\\$
+
 
−
<math>z^2 - 2 = y^2 + \frac{1}{y^2}\\$
+
<math>-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})</math>
−
</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>\\$
+
Now, if we let <math>z = y + \frac{1}{y}</math>, we can get the equations  
−
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}})\\$
+
<math>z = y + \frac{1}{y}</math>
−
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>
+
<math>z^2 - 2 = y^2 + \frac{1}{y^2}</math>
 +
 
 +
and
 +
 
 +
<math>z^3 - 3z = y^3 + \frac{1}{y^3}</math>
 +
 
 +
(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>)  
 +
Plugging this into our polynomial, expanding, and rearranging, we get  
 +
 
 +
<math>-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})</math>
 +
 
 +
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>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>
 +
 
 +
-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==
 +
 
 +
https://www.youtube.com/watch?v=mAXDdKX52TM
 +
 
 
== 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}}

Latest revision as of 16:36, 26 April 2024

Problem

The equation $2000x^6+100x^5+10x^3+x-2=0$ has exactly two real roots, one of which is $\frac{m+\sqrt{n}}r$, where $m$, $n$ and $r$ are integers, $m$ and $r$ are relatively prime, and $r>0$. Find $m+n+r$.

Solution

We may factor the equation as:[1]

\begin{align*} 2000x^6+100x^5+10x^3+x-2&=0\\ 2(1000x^6-1) + x(100x^4+10x^2+1)&=0\\ 2[(10x^2)^3-1]+x[(10x^2)^2+(10x^2)+1]&=0\\ 2(10x^2-1)[(10x^2)^2+(10x^2)+1]+x[(10x^2)^2+(10x^2)+1]&=0\\ (20x^2+x-2)(100x^4+10x^2+1)&=0\\ \end{align*}

Now $100x^4+10x^2+1\ge 1>0$ for real $x$. Thus the real roots must be the roots of the equation $20x^2+x-2=0$. By the quadratic formula the roots of this are:

\[x=\frac{-1\pm\sqrt{1^2-4(-2)(20)}}{40} = \frac{-1\pm\sqrt{1+160}}{40} = \frac{-1\pm\sqrt{161}}{40}.\]

Thus $r=\frac{-1+\sqrt{161}}{40}$, and so the final answer is $-1+161+40 = \boxed{200}$.


^ A well-known technique for dealing with symmetric (or in this case, nearly symmetric) polynomials is to divide through by a power of $x$ with half of the polynomial's degree (in this case, divide through by $x^3$), and then to use one of the substitutions $t = x \pm \frac{1}{x}$. In this case, the substitution $t = x\sqrt{10} - \frac{1}{x\sqrt{10}}$ gives $t^2 + 2 = 10x^2 + \frac 1{10x^2}$ and $2\sqrt{10}(t^3 + 3t) = 200x^3 - \frac{2}{10x^3}$, which reduces the polynomial to just $(t^2 + 3)\left(2\sqrt{10}t + 1\right) = 0$. Then one can backwards solve for $x$. Note: After dividing the equation with $x^3$, divide again with $10$ before substituting it with $t$ to get it right.


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 \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*} so for real solutions we have $y = -1/2$. Solve the quadratic in $x$ to get the 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, $x = -\frac{i}{\sqrt{10}}y$. The the polynomial becomes

$-2y^6 - (\frac{i}{\sqrt{10}})y^5 + (\frac{i}{\sqrt{10}})y^3 - (\frac{i}{\sqrt{10}})y - 2$

It's symmetric! Dividing by $y^3$ and rearranging, we get

$-2(y^3 + \frac{1}{y^3}) - (\frac{i}{\sqrt{10}})(y^2 + \frac{1}{y^2}) + (\frac{i}{\sqrt{10}})$

Now, if we let $z = y + \frac{1}{y}$, we can get the equations

$z = y + \frac{1}{y}$

$z^2 - 2 = y^2 + \frac{1}{y^2}$

and

$z^3 - 3z = y^3 + \frac{1}{y^3}$

(These come from squaring $z$ and subtracting $2$, then multiplying that result by $z$ and subtracting $z$) Plugging this into our polynomial, expanding, and rearranging, we get

$-2z^3 - (\frac{i}{\sqrt{10}})z^2 + 6z + (\frac{3i}{\sqrt{10}})$

Now, we see that the two $i$ terms must cancel in order to get this polynomial equal to $0$, so what squared equals 3? Plugging in $z = \sqrt{3}$ into the polynomial, we see that it works! Is there something else that equals 3 when squared? Trying $z = -\sqrt{3}$, we see that it also works! Great, we use long division on the polynomial by

$(z - \sqrt{3})(z + \sqrt{3}) = (z^2 - 3)$ and we get

$2z -(\frac{i}{\sqrt{10}}) = 0$.

We know that the other two solutions for z wouldn't result in real solutions for $x$ since we have to solve a quadratic with a negative discriminant, then multiply by $-(\frac{i}{\sqrt{10}})$. We get that $z = (\frac{i}{-2\sqrt{10}})$. Solving for $y$ (using $y + \frac{1}{y} = z$) we get that $y = \frac{-i \pm \sqrt{161}i}{4\sqrt{10}}$, and multiplying this by $-(\frac{i}{\sqrt{10}})$ (because $x = -(\frac{i}{\sqrt{10}})y$) we get that $x = \frac{-1 \pm \sqrt{161}}{40}$ for a final answer of $-1 + 161 + 40 = \boxed{200}$

-Grizzy

Solution 3 (Geometric Series)

Observe that the given equation may be rearranged as $2000x^6-2+(100x^5+10x^3+x)=0$. The expression in parentheses is a geometric series with common factor $10x^2$. Using the geometric sum formula, we rewrite as $2000x^6-2+\frac{1000x^7-x}{10x^2-1}=0, 10x^2-1\neq0$. Factoring a bit, we get $2(1000x^6-1)+(1000x^6-1)\frac{x}{10x^2-1}=0, 10x^2-1\neq0 \implies$ $(1000x^6-1)(2+\frac{x}{10x^2-1})=0, 10x^2-1\neq0$. Note that setting $1000x^6-1=0$ gives $10x^2-1=0$, which is clearly extraneous. Hence, we set $2+\frac{x}{10x^2-1}=0$ and use the quadratic formula to get the desired root $x=\frac{-1+\sqrt{161}}{40} \implies -1+161+40=\boxed{200}$

~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 $x^2$. Thus, we let $y=10x^2$. The equation then becomes $2y^3+xy^2+xy+x-2=0$, or $x(y^2+y+1)=2(1-y^3)$.

By the difference of cubes formula, $2(1-y^3)=2(1-y)(1+y+y^2)$, so we have two cases: either $y^2+y+1=0$, or $x=2(1-y)$. We start with the second formula as it is simpler.

Solving with the quadratic formula after re-substitution, we see that $x=\frac{-1\pm\sqrt{161}}{40}$, so the answer is $-1+161+40=\boxed{200}$.

For the sake of completeness, if we check the other equation, we come to the conclusion that $y=\frac{-1\pm i\sqrt{3}}{2}$, so no real solution exists for $x$. Thus our solution is correct.

~eevee9406

Video solution

https://www.youtube.com/watch?v=mAXDdKX52TM

See also

2000 AIME II (Problems • Answer Key • Resources)
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