2003B AMC 12 真题（P版）

1 答案 C

从分子提出 \(2\)，从分母提出 \(3\) 得到 \[\frac{2(1 - 2 + 3 - 4 + 5 - 6 + 7)}{3(1 - 2 + 3 - 4 + 5 - 6 + 7)}\text{。}\]

相同的括号和可以约去，剩下 \(\dfrac{2}{3}\)。

因此，正确答案是 C。

English

Factor \(2\) from the numerator and \(3\) from the denominator: \[\frac{2(1 - 2 + 3 - 4 + 5 - 6 + 7)}{3(1 - 2 + 3 - 4 + 5 - 6 + 7)}.\]

The equal parenthesized sums cancel, leaving \(\dfrac{2}{3}.\)

Thus, the correct answer is C.

2 答案 D

共 \(14\) 天，两粒药丸的每日费用为 \[ \frac{546}{14} = 39\text{。} \]

设一粒绿色药丸价格为 \(g\)。粉色药丸价格为 \(g - 1\)，所以 \[ g + (g - 1) = 39\text{，} \] 解得 \(g = 20\)。

因此，正确答案是 D。

English

Over \(14\) days the daily cost of the two pills is \[ \frac{546}{14} = 39. \]

Let \(g\) be the cost of a green pill. The pink pill costs \(g - 1,\) so \[ g + (g - 1) = 39, \] giving \(g = 20.\)

Thus, the correct answer is D.

3 答案 A

五个区域的面积分别为 \(4\)、\(6\)、\(15\)、\(20\) 和 \(21\) 平方英尺。

为了使费用最小，应把越贵的花种在越小的区域。最小可能费用为 \[ \begin{aligned} &3(4) + 2.5(6) + 2(15) \\ &\quad {}+ 1.5(20) + 1(21) = 108 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

The five regions have areas \(4,\) \(6,\) \(15,\) \(20,\) and \(21\) square feet.

To minimize the cost, plant the most expensive flowers in the smallest regions. The least possible cost is \[ \begin{aligned} &3(4) + 2.5(6) + 2(15) \\ &\quad {}+ 1.5(20) + 1(21) = 108. \end{aligned} \]

Thus, the correct answer is A.

4 答案 C

因为重叠，每次新增的草带宽度为 \(28 - 4 = 24\) 英寸 \(= 2\) 英尺。所以 Moe 每走一英尺就割 \(2\) 平方英尺，也就是每小时割 \(2 \cdot 5000 = 10000\) 平方英尺。

草坪面积为 \(90 \cdot 150 = 13500\) 平方英尺，所以所需时间为 \[ \frac{13500}{10000} = 1.35 \] 小时。

因此，正确答案是 C。

English

Because of the overlap, each pass adds a strip \(28 - 4 = 24\) inches \(= 2\) feet wide. So each foot Moe walks mows \(2\) square feet, that is, \(2 \cdot 5000 = 10000\) square feet per hour.

The lawn has area \(90 \cdot 150 = 13500\) square feet, so the time is \[ \frac{13500}{10000} = 1.35 \] hours.

Thus, the correct answer is C.

5 答案 D

边长比为 \(4 : 3\) 的长方形，其高度、水平长度和对角线之比为 \(3 : 4 : 5\)。对角线为 \(27\)，所以水平长度为 \[ \frac{4}{5}(27) = 21.6\text{，} \] 最接近 \(21.5\)。

因此，正确答案是 D。

English

A rectangle with side ratio \(4 : 3\) has height, length, and diagonal in ratio \(3 : 4 : 5.\) With diagonal \(27,\) the horizontal length is \[ \frac{4}{5}(27) = 21.6, \] which is closest to \(21.5.\)

Thus, the correct answer is D.

6 答案 B

设第一项为 \(a\)，公比为 \(r\)，则 \(ar = 2\)，且 \(ar^3 = 6\)，因此 \(r^2 = 3\)，\(r = \pm\sqrt{3}\)。

第一项为 \[ a = \frac{2}{r} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}\text{。} \] 选项中出现的是 \(-\dfrac{2\sqrt{3}}{3}\)。

因此，正确答案是 B。

English

Let the first term be \(a\) and the common ratio \(r.\) Then \(ar = 2\) and \(ar^3 = 6,\) so \(r^2 = 3\) and \(r = \pm\sqrt{3}.\)

The first term is \[ a = \frac{2}{r} = \pm\frac{2}{\sqrt{3}} = \pm\frac{2\sqrt{3}}{3}. \] The choice \(-\dfrac{2\sqrt{3}}{3}\) appears among the options.

Thus, the correct answer is B.

7 答案 D

设镍币、角币、二十五美分硬币数分别为 \(n\)、\(d\)、\(q\)，则 \(n + d + q = 100\)，且把总价值方程除以 \(5\) 得 \(n + 2d + 5q = 167\)。

两式相减得 \(d + 4q = 67\)，所以 \(d = 67 - 4q\)。

当 \(q = 0\) 时 \(d\) 最大，得到 \(d = 67\)（此时 \(n = 33\)）。当 \(q = 16\) 时，角币数最小，得到 \(d = 3\)（此时 \(n = 81\)）。差为 \(67 - 3 = 64\)。

因此，正确答案是 D。

English

Let \(n,\) \(d,\) \(q\) be the numbers of nickels, dimes, quarters. Then \(n + d + q = 100\) and \(n + 2d + 5q = 167\) (dividing the value equation by \(5\)).

Subtracting gives \(d + 4q = 67,\) so \(d = 67 - 4q.\)

The largest \(d\) is at \(q = 0,\) giving \(d = 67\) (with \(n = 33\)). The smallest occurs at \(q = 16,\) giving \(d = 3\) (with \(n = 81\)). The difference is \(67 - 3 = 64.\)

Thus, the correct answer is D.

8 答案 E

令 \(y = \clubsuit(x)\)。因为 \(x \le 99\)，所以 \(y \le 18\)。因此 \(\clubsuit(y) = 3\) 要求 \(y = 3\) 或 \(y = 12\)。

数字和为 \(3\) 的两位数是 \(12, 21, 30\)（\(3\) 个），数字和为 \(12\) 的两位数是 \(39, 48, 57, 66, 75, 84, 93\)（\(7\) 个），共 \(10\) 个。

因此，正确答案是 E。

English

Let \(y = \clubsuit(x).\) Since \(x \le 99,\) we have \(y \le 18.\) Then \(\clubsuit(y) = 3\) requires \(y = 3\) or \(y = 12.\)

The two-digit numbers with digit sum \(3\) are \(12, 21, 30\) (\(3\) values), and those with digit sum \(12\) are \(39, 48, 57, 66, 75, 84, 93\) (\(7\) values), for \(10\) in all.

Thus, the correct answer is E.

9 答案 D

\(f\) 的斜率为 \[ \frac{f(6) - f(2)}{6 - 2} = \frac{12}{4} = 3\text{。}\]

因此 \[ f(12) - f(2) = 3(12 - 2) = 30\text{。}\]

因此，正确答案是 D。

English

The slope of \(f\) is \[ \frac{f(6) - f(2)}{6 - 2} = \frac{12}{4} = 3. \]

Therefore \[ f(12) - f(2) = 3(12 - 2) = 30. \]

Thus, the correct answer is D.

10 答案 B

假设一个三角形接在边 \(AB\) 上。第二个三角形可以接在与 \(AB\) 相隔一步或两步的边上。

接在 \(BC\) 或 \(CD\) 上得到两种图形；接在 \(AE\) 或 \(DE\) 上得到的图形分别是前两种关于五边形对称轴的镜像。

因此只有 \(2\) 个互不全等的图形。

因此，正确答案是 B。

English

Assume one triangle is attached to side \(AB.\) The second triangle can be attached to a side that is one step away or two steps away from \(AB.\)

Attaching it to \(BC\) or \(CD\) gives two figures; attaching it to \(AE\) or \(DE\) gives figures that are mirror images of these across the pentagon’s axis of symmetry.

So there are only \(2\) non-congruent figures.

Thus, the correct answer is B.

11 答案 C

实际经过 \(60\) 分钟时，手表只前进了 \(57\) 分 \(36\) 秒 \(= 57.6\) 分钟。因此当手表显示中午后 \(t\) 分钟时，实际经过时间为 \(\dfrac{60}{57.6}t = \dfrac{25}{24}t\) 分钟。

手表显示晚上 \(10{:}00\) 整时，记录经过了 \(600\) 分钟，所以实际经过时间为 \[ \frac{25}{24}(600) = 625 \] 分钟 \(= 10\) 小时 \(25\) 分钟。因此实际时间是晚上 \(10{:}25\)。

因此，正确答案是 C。

English

In \(60\) real minutes the watch advances only \(57\) minutes \(36\) seconds \(= 57.6\) minutes. So when the watch shows \(t\) minutes past noon, the real elapsed time is \(\dfrac{60}{57.6}t = \dfrac{25}{24}t\) minutes.

The watch reads \(10{:}00\) PM after \(600\) recorded minutes, so the real elapsed time is \[ \frac{25}{24}(600) = 625 \] minutes \(= 10\) hours \(25\) minutes past noon. The actual time is \(10{:}25\) PM.

Thus, the correct answer is C.

12 答案 D

当 \(n\) 为偶数时，这五个因数是连续奇数。任意五个连续奇数中至少有一个能被 \(3\) 整除，且有一个能被 \(5\) 整除，所以乘积总能被 \(15\) 整除。

没有更大的整数总是可行：当 \(n = 20\) 和 \(n = 10\) 时，乘积分别为 \(21 \cdot 23 \cdot 25 \cdot 27 \cdot 29\) 和 \(11 \cdot 13 \cdot 15 \cdot 17 \cdot 19\)，它们的最大公因数是 \(15\)。

因此，正确答案是 D。

English

For even \(n,\) the five factors are consecutive odd numbers. Among any five consecutive odd numbers, at least one is divisible by \(3\) and exactly one by \(5,\) so the product is always divisible by \(15.\)

No larger divisor always works: the products for \(n = 10\) and \(n = 20\) are \(11 \cdot 13 \cdot 15 \cdot 17 \cdot 19\) and \(21 \cdot 23 \cdot 25 \cdot 27 \cdot 29,\) whose greatest common divisor is \(15.\)

Thus, the correct answer is D.

13 答案 B

设共同半径为 \(r\)，圆锥高为 \(h\)。融化后的冰淇淋装满圆锥，所以 \[ \frac{3}{4}\cdot\frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h\text{。}\]

化简得 \(\pi r^3 = \dfrac{1}{3}\pi r^2 h\)，所以 \(h = 3r\)，比为 \(3 : 1\)。

因此，正确答案是 B。

English

Let \(r\) be the common radius and \(h\) the cone’s height. The melted ice cream fills the cone, so \[ \frac{3}{4}\cdot\frac{4}{3}\pi r^3 = \frac{1}{3}\pi r^2 h. \]

This simplifies to \(\pi r^3 = \dfrac{1}{3}\pi r^2 h,\) so \(h = 3r,\) a ratio of \(3 : 1.\)

Thus, the correct answer is B.

14 答案 D

因为 \(FG = 5 - 1 - 2 = 2\)，且 \(\overline{FG} \parallel \overline{AB}\)，所以三角形 \(FEG\) 与 \(AEB\) 相似，相似比为 \(\dfrac{FG}{AB} = \dfrac{2}{5}\)。

设 \(E\) 到直线 \(CD\) 的距离为 \(k\)，则 \(E\) 到 \(AB\) 的距离为 \(k + 3\)，且 \[ \frac{k}{k + 3} = \frac{2}{5}\text{，} \] 解得 \(k = 2\)。

\(\triangle AEB\) 的高为 \(k + 3 = 5\)，所以面积为 \[ \frac{1}{2}(5)(5) = \frac{25}{2}\text{。} \]

因此，正确答案是 D。

English

Since \(FG = 5 - 1 - 2 = 2\) and \(\overline{FG} \parallel \overline{AB},\) triangles \(FEG\) and \(AEB\) are similar with ratio \(\dfrac{FG}{AB} = \dfrac{2}{5}.\)

Let the distance from \(E\) to line \(CD\) be \(k.\) Then the distance from \(E\) to \(AB\) is \(k + 3,\) and \[ \frac{k}{k + 3} = \frac{2}{5}, \] giving \(k = 2.\)

The height of \(\triangle AEB\) is \(k + 3 = 5,\) so its area is \[ \frac{1}{2}(5)(5) = \frac{25}{2}. \]

Thus, the correct answer is D.

15 答案 D

设 \(O\) 为八边形中心。连接 \(O\) 与各顶点，把八边形分成 \(8\) 个全等三角形，所以 \(\triangle AOB\) 的面积为 \(\dfrac{1}{8}\)。

因为 \(O\) 是 \(\overline{AE}\) 的中点，三角形 \(AOB\) 和 \(BOE\) 面积相等，所以 \(\triangle ABE\) 的面积为 \(\dfrac{1}{4}\)。

长方形 \(ABEF\) 被对角线 \(\overline{BE}\) 分成两个全等三角形，所以 \(\triangle ABE\) 是它的一半。因此 \(ABEF\) 的面积为 \(\dfrac{1}{2}\)。

因此，正确答案是 D。

English

Let \(O\) be the center of the octagon. Joining \(O\) to the vertices splits the octagon into \(8\) congruent triangles, so \(\triangle AOB\) has area \(\dfrac{1}{8}.\)

Since \(O\) is the midpoint of \(\overline{AE},\) triangles \(AOB\) and \(BOE\) have equal areas, so \(\triangle ABE\) has area \(\dfrac{1}{4}.\)

The rectangle \(ABEF\) is split by diagonal \(\overline{BE}\) into two congruent triangles, so \(\triangle ABE\) is half of it. Hence \(ABEF\) has area \(\dfrac{1}{2}.\)

Thus, the correct answer is D.

16 答案 E

大半圆面积为 \(\dfrac{1}{2}\pi(2)^2 = 2\pi\)。

在计入重叠之前，三个小半圆的总面积为 \(\dfrac{3\pi}{2}\)。每相邻两个小半圆的圆心相距 \(1\)，所以它们的公共部分由两个 \(60^\circ\) 扇形和一个等边三角形围成。因此两块重叠部分的面积各为 \(\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}\)。

阴影面积为 \[ \begin{aligned} &2\pi - \left[\frac{3\pi}{2} - 2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)\right] \\ &= \frac{7}{6}\pi - \frac{\sqrt{3}}{2} \end{aligned}\text{。}\]

因此，正确答案是 E。

English

The large semicircle has area \(\dfrac{1}{2}\pi(2)^2 = 2\pi.\)

The three small semicircles have total area \(\dfrac{3\pi}{2}\) before their overlaps are accounted for. The centers of each adjacent pair are \(1\) unit apart, so their intersection is bounded by two \(60^\circ\) sectors and an equilateral triangle. Each of the two overlaps therefore has area \(\dfrac{\pi}{3} - \dfrac{\sqrt{3}}{4}.\)

The shaded area is \[ \begin{aligned} &2\pi - \left[\frac{3\pi}{2} - 2\left(\frac{\pi}{3} - \frac{\sqrt{3}}{4}\right)\right] \\ &= \frac{7}{6}\pi - \frac{\sqrt{3}}{2}. \end{aligned} \]

Thus, the correct answer is E.

17 答案 D

令 \(X = \log x\)，\(Y = \log y\)，则 \[X + 3Y = 1\] 且 \[2X + Y = 1\text{。}\]

解得 \(X = \dfrac{2}{5}\)，\(Y = \dfrac{1}{5}\)，所以 \[ \log(xy) = X + Y = \frac{3}{5}\text{。} \]

因此，正确答案是 D。

English

Let \(X = \log x\) and \(Y = \log y.\) Then \[X + 3Y = 1\] and \[2X + Y = 1.\]

Solving gives \(X = \dfrac{2}{5}\) and \(Y = \dfrac{1}{5},\) so \[ \log(xy) = X + Y = \frac{3}{5}. \]

Thus, the correct answer is D.

18 答案 B

要使 \(x\) 最小，\(x\) 和 \(y\) 都不应含有 \(7\) 与 \(11\) 以外的素因数。写成 \(x = 7^c 11^d\)，则 \(7x^5 = 7^{5c+1} 11^{5d}\)。再写成 \(y = 7^m 11^n\)，需要 \(7^{5c+1}11^{5d} = 7^{13m}11^{13n+1}\)。

比较指数：\(5c + 1 \equiv 0 \pmod{13}\) 给出最小的 \(c = 5\)，而 \(5d \equiv 1 \pmod{13}\) 给出最小的 \(d = 8\)。所以 \(a = 7\)，\(b = 11\)，\[ \begin{aligned} &a + b + c + d \\ &= 7 + 11 + 5 + 8 = 31 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

For the minimum \(x,\) neither \(x\) nor \(y\) has prime factors other than \(7\) and \(11.\) Write \(x = 7^c 11^d,\) so \(7x^5 = 7^{5c+1} 11^{5d}.\) Writing \(y = 7^m 11^n,\) we need \(7^{5c+1}11^{5d} = 7^{13m}11^{13n+1}.\)

Matching exponents: \(5c + 1 \equiv 0 \pmod{13}\) gives the least \(c = 5,\) and \(5d \equiv 1 \pmod{13}\) gives the least \(d = 8.\) So \(a = 7,\) \(b = 11,\) and \[ \begin{aligned} &a + b + c + d \\ &= 7 + 11 + 5 + 8 = 31. \end{aligned} \]

Thus, the correct answer is B.

19 答案 E

集合 \(S\) 中有 \(4 \cdot 4! = 96\) 个排列，因为第一项有 \(4\) 种选择，剩下四项可按 \(4!\) 种方式任意排列。

若第二项为 \(2\)，第一项必须是 \(3, 4\) 或 \(5\)（不能是 \(1\) 也不能是 \(2\)），有 \(3\) 种选择，剩下三项有 \(3!\) 种排列，共 \(3 \cdot 3! = 18\) 个。

概率为 \(\dfrac{18}{96} = \dfrac{3}{16}\)，所以 \(a + b = 3 + 16 = 19\)。

因此，正确答案是 E。

English

The set \(S\) contains \(4 \cdot 4! = 96\) permutations, since the first term has \(4\) choices and the remaining four terms can be arranged in \(4!\) ways.

For the second term to be \(2,\) the first term must be \(3, 4,\) or \(5\) (not \(1,\) not \(2\)), giving \(3\) choices, and the remaining three terms can be arranged in \(3!\) ways: \(3 \cdot 3! = 18.\)

The probability is \(\dfrac{18}{96} = \dfrac{3}{16},\) so \(a + b = 3 + 16 = 19.\)

Thus, the correct answer is E.

20 答案 B

图像经过 \((-1, 0)\)，\((1, 0)\)，和 \((0, 2)\)。所以 \(f(0) = d = 2\)。

将两式相加：\[ \begin{aligned} &f(1) + f(-1) \\ &= (a + b + c + d) \\ &\quad {}+ (-a + b - c + d) \\ &= 2b + 2d = 0\text{，} \end{aligned} \] 因此 \(b = -d = -2\)。

因此，正确答案是 B。

English

The graph passes through \((-1, 0),\) \((1, 0),\) and \((0, 2).\) So \(f(0) = d = 2.\)

Adding \[ \begin{aligned} &f(1) + f(-1) \\ &= (a + b + c + d) \\ &\quad {}+ (-a + b - c + d) \\ &= 2b + 2d = 0, \end{aligned} \] so \(b = -d = -2.\)

Thus, the correct answer is B.

21 答案 D

令 \(\beta = \pi - \alpha\) 为 \(\triangle ABC\) 在 \(B\) 处的内角。由余弦定理，\[ \begin{aligned} &AC^2 = 8^2 + 5^2 \\ &\quad {}- 2(8)(5)\cos\beta \\ &= 89 - 80\cos\beta \end{aligned}\text{。}\]

于是 \(AC \lt 7\) 表示 \(89 - 80\cos\beta \lt 49\)，即 \(\cos\beta \gt \dfrac{1}{2}\)，也就是 \(\beta \lt \dfrac{\pi}{3}\)。

因为 \(\alpha\) 在 \((0, \pi)\) 上均匀分布，所以 \(\beta\) 也均匀分布。概率为 \[ \frac{\frac{\pi}{3}}{\pi} = \frac{1}{3}\text{。}\]

因此，正确答案是 D。

English

Let \(\beta = \pi - \alpha\) be the interior angle of \(\triangle ABC\) at \(B.\) By the Law of Cosines, \[ \begin{aligned} &AC^2 = 8^2 + 5^2 \\ &\quad {}- 2(8)(5)\cos\beta \\ &= 89 - 80\cos\beta. \end{aligned} \]

Then \(AC \lt 7\) means \(89 - 80\cos\beta \lt 49,\) i.e. \(\cos\beta \gt \dfrac{1}{2},\) i.e. \(\beta \lt \dfrac{\pi}{3}.\)

As \(\alpha\) is uniform on \((0, \pi),\) so is \(\beta.\) The probability is \[ \frac{\frac{\pi}{3}}{\pi} = \frac{1}{3}. \]

Thus, the correct answer is D.

22 答案 C

设 \(O\) 为两条对角线的交点。则 \(\triangle AOB\) 在 \(O\) 处为直角，且 \(OA = 8\)，\(OB = 15\)。四边形 \(OPNQ\) 在 \(O\)，\(P\)，\(Q\) 处均为直角，所以它是长方形，且 \(PQ = ON\)。

\(ON\) 的最小值是在 \(\triangle AOB\) 中从 \(O\) 到 \(\overline{AB}\) 的高。因为 \(AB = \sqrt{8^2 + 15^2} = 17\)，由两种面积表达式相等，得 \[ \begin{aligned} ON &= \frac{OA \cdot OB}{AB} \\ &= \frac{8 \cdot 15}{17} \\ &= \frac{120}{17} \approx 7.06 \end{aligned}\text{。}\]

这最接近 \(7\)。

因此，正确答案是 C。

English

Let \(O\) be the intersection of the diagonals. Then \(\triangle AOB\) is right-angled at \(O\) with legs \(OA = 8\) and \(OB = 15.\) Quadrilateral \(OPNQ\) has right angles at \(O,\) \(P,\) and \(Q,\) so it is a rectangle and \(PQ = ON.\)

The minimum of \(ON\) is the altitude from \(O\) to \(\overline{AB}\) in \(\triangle AOB.\) Since \(AB = \sqrt{8^2 + 15^2} = 17,\) equating the two area expressions gives \[ \begin{aligned} ON &= \frac{OA \cdot OB}{AB} \\ &= \frac{8 \cdot 15}{17} \\ &= \frac{120}{17} \approx 7.06. \end{aligned} \]

This is closest to \(7.\)

Thus, the correct answer is C.

23 答案 A

截距出现在 \(\frac{1}{x} = k\pi\) 时，也就是 \(x = \dfrac{1}{k\pi}\)，其中 \(k\) 为非零整数。

条件 \(0.0001 \lt \dfrac{1}{k\pi} \lt 0.001\) 化为 \[ \frac{1000}{\pi} \lt k \lt \frac{10000}{\pi}\text{。}\]

这样的整数个数为 \[ \begin{aligned} &\left\lfloor \frac{10000}{\pi} \right\rfloor - \left\lfloor \frac{1000}{\pi} \right\rfloor \\ &= 3183 - 318 = 2865\text{，} \end{aligned} \]最接近 \(2900\)。

因此，正确答案是 A。

English

The intercepts occur where \(\frac{1}{x} = k\pi,\) that is \(x = \dfrac{1}{k\pi}\) for a nonzero integer \(k.\)

The condition \(0.0001 \lt \dfrac{1}{k\pi} \lt 0.001\) becomes \[ \frac{1000}{\pi} \lt k \lt \frac{10000}{\pi}. \]

The number of such integers is \[ \begin{aligned} &\left\lfloor \frac{10000}{\pi} \right\rfloor - \left\lfloor \frac{1000}{\pi} \right\rfloor \\ &= 3183 - 318 = 2865, \end{aligned} \] closest to \(2900.\)

Thus, the correct answer is A.

24 答案 C

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 \[ 2a + (b + c - 2a) = 2003\text{，}\]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。

English

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.

25 答案 D

一条弦的长度小于半径，当且仅当它所对的弧小于 \(60^\circ\)，因为 \(60^\circ\) 弧对应的弦长正好等于半径。

三条两两弦都短于半径，恰好等价于三个点都位于某个 \(60^\circ\) 弧内。

在任何成功的情形中，恰好有一个点是这样一段包含弧的逆时针端点（概率为零的边界情形除外）。这个端点有 \(3\) 种选法；另外两个点各自独立地以概率 \(\dfrac{1}{6}\) 落在其后的 \(60^\circ\) 弧内。因此所求概率为 \[ 3\left(\frac{1}{6}\right)^2 = \frac{1}{12}\text{。}\]

因此，正确答案是 D。

English

A chord has length less than the radius exactly when the arc it subtends is less than \(60^\circ,\) since a chord of a \(60^\circ\) arc equals the radius.

All three pairwise chords are shorter than the radius precisely when the three points all lie within some arc of \(60^\circ.\)

For any successful configuration, exactly one of the three points is the counterclockwise endpoint of such a containing arc (apart from probability-zero boundary cases). Choose that endpoint in \(3\) ways; each of the other two points independently has probability \(\dfrac{1}{6}\) of lying in the next \(60^\circ.\) Hence the probability is \[ 3\left(\frac{1}{6}\right)^2 = \frac{1}{12}. \]

Thus, the correct answer is D.