AIME 2005 I · 第 6 题

Solution 1

The left-hand side of that equation is nearly equal to $(x - 1)^4$. Thus, we add 1 to each side in order to complete the fourth power and get $(x - 1)^4 = 2006$.

Let $r = \sqrt[4]{2006}$ be the positive real fourth root of 2006. Then the roots of the above equation are $x = 1 + i^n r$ for $n = 0, 1, 2, 3$. The two non-real members of this set are $1 + ir$ and $1 - ir$. Their product is $P = 1 + r^2 = 1 + \sqrt{2006}$. $44^2 = 1936 < 2006 < 2025 = 45^2$ so $\lfloor P \rfloor = 1 + 44 = \boxed{045}$.

Solution 2

Starting like before, $(x-1)^4= 2006$ This time we apply differences of squares. $(x-1)^4-2006=0$ so $((x-1)^2+\sqrt{2006})((x-1)^2 -\sqrt{2006})=0$ If you think of each part of the product as a quadratic, then $((x-1)^2+\sqrt{2006})$ is bound to hold the two non-real roots since the other definitely crosses the x-axis twice since it is just $x^2$ translated down and right. Therefore $P$ is the product of the roots of $((x-1)^2+\sqrt{2006})$ or $P=1+\sqrt{2006}$ so

$\lfloor P \rfloor = 1 + 44 = \boxed{045}$.

Solution 3

If we don't see the fourth power, we can always factor the LHS to try to create a quadratic substitution. Checking, we find that $x=0$ and $x=2$ are both roots. Synthetic division gives $(x^2-2x)(x^2-2x+2)=2005$. We now have our quadratic substitution of $y=x^2-2x+1=(x-1)^2$, giving us $(y-1)(y+1)=2005$. From here we proceed as in Solution 1 to get $\boxed{045}$.

Solution 4

Realizing that if we add 1 to both sides we get $x^4-4x^3+6x^2-4x+1=2006$ which can be factored as $(x-1)^4=2006$. Then we can substitute $(x-1)$ with $y$ which leaves us with $y^4=2006$. Now subtracting 2006 from both sides we get some difference of squares $y^4-2006=0 \rightarrow (y-\sqrt[4]{2006})(y+\sqrt[4]{2006})(y^2+\sqrt{2006})=0$. The question asks for the product of the complex roots so we only care about the last factor which is equal to zero. From there we can solve $y^2+\sqrt{2006}=0$, we can substitute $(x-1)$ for $y$ giving us $(x-1)^2+\sqrt{2006}=0$, expanding this we get $x^2-2x+1+\sqrt{2006}=0$. We know that the product of a quadratics roots is $\frac{c}{a}$ which leaves us with $\frac{1+\sqrt{2006}}{1}=1+\sqrt{2006}\approx\boxed{045}$.

Solution 5

As in solution 1, we find that $(x-1)^4 = 2006$. Now $x-1=\pm \sqrt[4]{2006}$ so $x_1 = 1+\sqrt[4]{2006}$ and $x_2 = 1-\sqrt[4]{2006}$ are the real roots of the equation. Multiplying, we get $x_1 x_2 = 1 - \sqrt{2006}$. Now transforming the original function and using Vieta's formula, $x^4-4x^3+6x^2-4x-2005=0$ so $x_1 x_2 x_3 x_4 = \frac{-2005}{1} = -2005$. We find that the product of the nonreal roots is $x_3 x_4 = \frac{-2005}{1-\sqrt{2006}} \approx 45.8$ and we get $\boxed{045}$.

Note: $\frac{2005}{\sqrt[4]{2006}-1}=\frac{2005(1+\sqrt[4]{2006})}{2005} = 1+\sqrt[4]{2006}.$

Solution 6 (De Moivre's Theorem)

As all the other solutions, we find that $(x-1)^4 = 2006$. Thus $x=\sqrt[4]{2006}+1$. Thus $x= \sqrt[4]{2006}(\cos(\frac{2\pi(k)}{4}+i\sin(\frac{2\pi(k)}{4}))+1$ when $k=0,1,2,3$. The complex values of $x$ are the ones where $i\sin(\frac{2\pi(k)}{4})$ does not equal 0. These complex roots are $1+\sqrt[4]{2006}(i)$ and $1-\sqrt[4]{2006}(i)$. The product of these two nonreal roots is ($1+\sqrt[4]{2006}(i)$)($1-\sqrt[4]{2006}(i)$) which is equal to $1+\sqrt {2006}$. The floor of that value is $\boxed{045}$.

Solution 7 (The Floor Divisor Method)

Continue Solution 1 until we obtain the equation $(x-1)^4 = 2006$. Then, we see that $x = 1 \pm \sqrt[4]{2006}$. While one could see the possible roots are $1 \pm \sqrt[4]{2006}$, and $1 \pm \sqrt[4]{2006}i$, and obtain 45, one could have also seen the product of all roots is $-2005$ through Vieta's. Then, the product of the two remaining real roots is $1 - \sqrt{2006}$, and thus the wanted value is $\left\lfloor \frac{-2005}{1 - \sqrt{2006}} \right\rfloor$ or $\left\lfloor \frac{2005}{\sqrt{2006} - 1} \right\rfloor$, which through the divisor method is $\left\lfloor \frac{2005}{\lfloor \sqrt{2006} - 1 \rfloor} \right\rfloor +1$, or $\left\lfloor \frac{2005}{43} \right\rfloor +1$, whose floor is $\boxed{045}$.

~Pinotation

The formal Floor Divisor Method states that for all $a$ and $b$, where $A = \lfloor a \rfloor, \quad B = \lfloor b \rfloor$, the floor of $\frac{a}{b}$ is

$$\lfloor a/b \rfloor = \left\lfloor \frac{A}{B} - \frac{\mathbf{1}_{(a \bmod b) < (A \bmod B)}}{1} \right\rfloor$$ More informally, without the bolded 1 (conditional 1),

For all $A = \lfloor a \rfloor, \quad B = \lfloor b \rfloor$, \begin{array}{c} \lfloor a/b \rfloor = \begin{cases} \lfloor A/B \rfloor & \text{if } a \ge \lfloor A/B \rfloor \cdot b \\[2mm] \lfloor A/B \rfloor - 1 & \text{if } a < \lfloor A/B \rfloor \cdot b \end{cases} \end{array}

As a problem to the reader, try to figure out why I simplified this formula above into $\left\lfloor \frac{a}{b} \right\rfloor +1$ for this problem, and when should and should not simplify to this form.