Question

设实数 $a, b, c$ 和 $d$ 满足方程组存在互质正整数 $m$ 和 $n$ ，使得求 $m + n$ 。 $\begin{aligned} a + b &= -3, \\ ab + bc + ca &= -4, \end{aligned}$ $\begin{aligned} abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{aligned}$ $a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.$

Let $a, b, c,$ and $d$ be real numbers that satisfy the system of equations $\begin{aligned} a + b &= -3, \\ ab + bc + ca &= -4, \end{aligned}$ $\begin{aligned} abc + bcd + cda + dab &= 14, \\ abcd &= 30. \end{aligned}$ There exist relatively prime positive integers $m$ and $n$ such that $a^2 + b^2 + c^2 + d^2 = \frac{m}{n}.$ Find $m + n.$

答案

答案：145

知识点：方程组对称性（代数）因式分解

难度评级：2650

解答：

因为 $a + b = -3$ ，第二个方程为 $ab + c(a + b) = ab - 3c = -4$ ，所以 $ab = 3c - 4$ 。把第三个方程分组为 $ab(c + d) + cd(a + b) = 14$ ，得到 $(3c - 4)(c + d) - 3cd = 14$ ，化简为 $3c^2 - 4c - 4d = 14$ ，所以 $d = \frac{3c^2 - 4c - 14}{4}$ 。第四个方程变为 $(3c - 4)\,cd = 30$ 。

代入 $d$ 得 $c(3c - 4)(3c^2 - 4c - 14) = 120$ ，即二次因式判别式为负，所以 $c = -2$ 或 $c = \frac{10}{3}$ 。若 $c = \frac{10}{3}$ ，则 $ab = 6$ 且 $a + b = -3$ ，因为 $9 - 24 \lt 0$ 。这不可能对应实数 $a, b$ 。因此 $c = -2$ ，给出 $ab = -10$ 和 $d = \frac{12 + 8 - 14}{4} = \frac{3}{2}$ 。 $\begin{aligned} &9c^4 - 24c^3 - 26c^2 \\ &\quad {}+ 56c - 120 \\ &= (c + 2)(3c - 10) \\ &\quad {}\cdot (3c^2 - 4c + 6) \\ &= 0. \end{aligned}$

于是 $a^2 + b^2$ $= (a + b)^2 - 2ab$ $= 9 + 20 = 29$ ，且 $c^2 + d^2 = 4 + \frac{9}{4} = \frac{25}{4}$ ，所以 $a^2 + b^2 + c^2 + d^2 = \frac{141}{4}$ ，从而 $m + n = 141 + 4 = 145$ 。

Since $a + b = -3,$ the second equation reads $ab + c(a + b) = ab - 3c = -4,$ so $ab = 3c - 4.$ Grouping the third equation as $ab(c + d) + cd(a + b) = 14$ gives $(3c - 4)(c + d) - 3cd = 14,$ which simplifies to $3c^2 - 4c - 4d = 14,$ so $d = \frac{3c^2 - 4c - 14}{4}.$ The fourth equation becomes $(3c - 4)\,cd = 30.$

Substituting for $d$ yields $c(3c - 4)(3c^2 - 4c - 14) = 120,$ i.e. $\begin{aligned} &9c^4 - 24c^3 - 26c^2 \\ &\quad {}+ 56c - 120 \\ &= (c + 2)(3c - 10) \\ &\quad {}\cdot (3c^2 - 4c + 6) \\ &= 0. \end{aligned}$ The quadratic factor has negative discriminant, so $c = -2$ or $c = \frac{10}{3}.$ If $c = \frac{10}{3},$ then $ab = 6$ with $a + b = -3,$ impossible for real $a, b$ since $9 - 24 \lt 0.$ So $c = -2,$ giving $ab = -10$ and $d = \frac{12 + 8 - 14}{4} = \frac{3}{2}.$

Then $a^2 + b^2$ $= (a + b)^2 - 2ab$ $= 9 + 20 = 29$ and $c^2 + d^2 = 4 + \frac{9}{4} = \frac{25}{4},$ so $a^2 + b^2 + c^2 + d^2 = \frac{141}{4}$ and $m + n = 141 + 4 = 145.$

2021 AIME II 第 7 题

答案：145

难度评级：2650

解答：

其他年份的第 7 题