Question

设 $P(x) = x^2 - 3x - 7$ ，且 $Q(x)$ 、 $R(x)$ 是两个二次多项式，它们的 $x^2$ 系数也都为 $1$ 。David 计算三个和 $P + Q$ 、 $P + R$ 、 $Q + R$ ，惊讶地发现这三个和中任意两个都有一个公共根，并且这三个公共根互不相同。若 $Q(0) = 2$ ，则 $R(0) = \frac{m}{n}$ ，其中 $m$ 与 $n$ 是互质正整数。求 $m + n$ 。

Let $P(x) = x^2 - 3x - 7,$ and let $Q(x)$ and $R(x)$ be two quadratic polynomials also with the coefficient of $x^2$ equal to $1.$ David computes each of the three sums $P + Q,$ $P + R,$ and $Q + R$ and is surprised to find that each pair of these sums has a common root, and these three common roots are distinct. If $Q(0) = 2,$ then $R(0) = \frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m + n.$

答案

答案：71

知识点：多项式韦达定理方程组

难度评级：2920

解答：

令 $a$ 为 $P+Q$ 与 $P+R$ 的公共根， $b$ 为 $P+Q$ 与 $Q+R$ 的公共根， $c$ 为 $P+R$ 与 $Q+R$ 的公共根。每个和都是首项系数为 $2$ 的二次式，所以 $P+Q = 2(x-a)(x-b)$ 、 $P+R = 2(x-a)(x-c)$ ，且 $Q+R = 2(x-b)(x-c)$ 。写 $Q = x^2 + qx + 2$ ， $R = x^2 + rx + s$ 。由韦达定理，根和分别为 $a + b = \frac{3-q}{2}$ 、 $a + c = \frac{3-r}{2}$ 、 $b + c = -\frac{q+r}{2}$ ，所以 $\begin{aligned} 2a &= (a+b) + (a+c) - (b+c) \\ &= \frac{3-q}{2} + \frac{3-r}{2} \\ &\quad {}+ \frac{q+r}{2} = 3, \end{aligned}$ 得到 $a = \frac{3}{2}$ 。

$P + Q$ 的常数项为 $-7 + 2 = -5$ ，所以 $2ab = -5$ ，从而 $b = -\frac{5}{3}$ 。另外 $2ac = P(0) + R(0) = s - 7$ ，且 $2bc = Q(0) + R(0) = s + 2$ 。相减得 $2c(b - a) = 9$ ，所以 $c = \frac{9}{2\left(-\frac{5}{3} - \frac{3}{2}\right)} = -\frac{27}{19}$ 。

因此 $s = 2bc - 2$ $= 2 \cdot \left(-\frac{5}{3}\right)\left(-\frac{27}{19}\right) - 2$ $= \frac{90}{19} - \frac{38}{19} = \frac{52}{19}$ ，所以 $R(0) = \frac{52}{19}$ ， $m + n = 52 + 19 = 71$ 。

Let $a$ be the common root of $P+Q$ and $P+R,$ let $b$ be that of $P+Q$ and $Q+R,$ and let $c$ be that of $P+R$ and $Q+R.$ Each sum is quadratic with leading coefficient $2,$ so $P+Q = 2(x-a)(x-b),$ $P+R = 2(x-a)(x-c),$ and $Q+R = 2(x-b)(x-c).$ Write $Q = x^2 + qx + 2$ and $R = x^2 + rx + s.$ By Vieta's formulas, the root sums are $a + b = \frac{3-q}{2},$ $a + c = \frac{3-r}{2},$ and $b + c = -\frac{q+r}{2},$ so $\begin{aligned} 2a &= (a+b) + (a+c) - (b+c) \\ &= \frac{3-q}{2} + \frac{3-r}{2} \\ &\quad {}+ \frac{q+r}{2} = 3, \end{aligned}$ giving $a = \frac{3}{2}.$

The constant term of $P + Q$ is $-7 + 2 = -5,$ so $2ab = -5$ and $b = -\frac{5}{3}.$ Also $2ac = P(0) + R(0) = s - 7$ and $2bc = Q(0) + R(0) = s + 2.$ Subtracting, $2c(b - a) = 9,$ so $c = \frac{9}{2\left(-\frac{5}{3} - \frac{3}{2}\right)} = -\frac{27}{19}.$

Then $s = 2bc - 2$ $= 2 \cdot \left(-\frac{5}{3}\right)\left(-\frac{27}{19}\right) - 2$ $= \frac{90}{19} - \frac{38}{19} = \frac{52}{19},$ so $R(0) = \frac{52}{19}$ and $m + n = 52 + 19 = 71.$

2020 AIME II 第 11 题

答案：71

难度评级：2920

解答：

其他年份的第 11 题