Question

选择正整数 $a$ 、 $b$ 、 $c$ ，使得 $a \lt b \lt c$ ，且方程组和恰有一个解。 $c$ 的最小值是多少？ $2x + y = 2003$ $\begin{aligned} &y = |x - a| + |x - b| \\ &\quad {}+ |x - c| \end{aligned}$

Positive integers $a,$ $b,$ and $c$ are chosen so that $a \lt b \lt c,$ and the system of equations $2x + y = 2003$ and $\begin{aligned} &y = |x - a| + |x - b| \\ &\quad {}+ |x - c| \end{aligned}$ has exactly one solution. What is the minimum value of $c?$

$668$

$669$

$1002$

$2003$

$2004$

答案：C

知识点：绝对值方程组

难度评级：2160

解答：

函数 $y = |x-a| + |x-b| + |x-c|$ 是分段线性的，斜率依次为 $-3, -1, 1, 3$ ，折点在 $x = a, b, c.$ 直线 $2x + y = 2003$ 的斜率为 $-2.$

斜率为 $-2$ 的直线与该图像恰好相交一次，当且仅当它经过最左边的折点 $(a,\, b + c - 2a),$ 此处图像斜率从 $-3$ 跳到 $-1.$ 代入得到 $2a + (b + c - 2a) = 2003,$ ，所以 $b + c = 2003.$

因为 $b \lt c,$ 必须有 $c \gt \dfrac{2003}{2},$ 所以 $c \ge 1002.$ 这个下界可以达到：取 $a = 1,$ $b = 1001,$ 和 $c = 1002.$ 因此最小值为 $1002.$

所以正确答案是 C。

The function $y = |x-a| + |x-b| + |x-c|$ is piecewise linear with slopes $-3, -1, 1, 3$ and corners at $x = a, b, c.$ The line $2x + y = 2003$ has slope $-2.$

A line of slope $-2$ meets this graph exactly once only if it passes through the leftmost corner $(a,\, b + c - 2a),$ where the graph's slope jumps from $-3$ to $-1.$ Substituting, $2a + (b + c - 2a) = 2003,$ so $b + c = 2003.$

Since $b \lt c,$ we need $c \gt \dfrac{2003}{2},$ so $c \ge 1002.$ This bound is attained: take $a = 1,$ $b = 1001,$ and $c = 1002.$ Therefore the minimum is $1002.$

Thus, the correct answer is C.

2003 AMC 12B 第 24 题

答案：C

难度评级：2160

解答：

其他年份的第 24 题