2004B AMC 10 真题（P版）

1 答案 C

第 \(12\) 排到第 \(22\) 排共有 \(22 - 12 + 1 = 11\) 排。

每排有 \(33\) 个座位，所以总数为 \(33 \times 11 = 363\)。

所以正确答案是 C。

English

Rows \(12\) through \(22\) inclusive make up \(22 - 12 + 1 = 11\) rows.

Each row has \(33\) seats, so the total is \(33 \times 11 = 363.\)

Thus, the correct answer is C.

2 答案 B

\(70\) 到 \(79\) 给出 \(10\) 个十位为 \(7\) 的两位数。

数列 \(17, 27, \ldots, 97\) 给出 \(9\) 个个位为 \(7\) 的两位数。

因为 \(77\) 被重复计算一次，所以总数为 \(10 + 9 - 1 = 18\)。

所以正确答案是 B。

English

The numbers \(70\) through \(79\) give \(10\) with a \(7\) in the tens place.

The numbers \(17, 27, \ldots, 97\) give \(9\) with a \(7\) in the units place.

Since \(77\) is counted twice, the total is \(10 + 9 - 1 = 18.\)

Thus, the correct answer is B.

3 答案 A

从第五次训练倒推，罚中数依次为 \(48, 24, 12, 6\) 和 \(3\)，对应第五、第四、第三、第二、第一次训练。

所以正确答案是 A。

English

Working backward from the fifth practice, the counts are \(48, 24, 12, 6,\) and \(3\) at the fourth, third, second, and first practices.

Thus, the correct answer is A.

4 答案 B

因为 \(6! = 720 = 2^4 \cdot 3^2 \cdot 5\)，所以可见乘积只会用到质数 \(2\)、\(3\) 和 \(5\)。

遮住 \(4\) 时剩下的因数 \(2\) 最少，为 \(2^2\)。遮住 \(3\) 或 \(6\) 时剩下的因数 \(3\) 最少，为一个；遮住 \(5\) 时则没有因数 \(5\)。

因此 \(P\) 一定能被 \(2^2 \cdot 3 = 12\) 整除，但不一定能被更大的数整除。

所以正确答案是 B。

English

Since \(6! = 720 = 2^4 \cdot 3^2 \cdot 5,\) the visible product uses only the primes \(2,\) \(3,\) and \(5.\)

Hiding \(4\) leaves the fewest \(2\)’s, namely \(2^2.\) Hiding \(3\) or \(6\) leaves the fewest \(3\)’s, namely one. Hiding \(5\) leaves no factor of \(5.\)

Therefore \(P\) is always divisible by \(2^2 \cdot 3 = 12,\) but not necessarily by any larger number.

Thus, the correct answer is B.

5 答案 D

取 \(d = 0\) 可以去掉减法，所以只需用 \(1\)、\(2\)、\(3\) 最大化 \(c \cdot a^b\)。

取 \(c = 1\)、\(a = 3\)、\(b = 2\)，得到 \(3^2 = 9\)。另一种大幂 \(2^3 = 8\) 更小，而让 \(c \gt 1\) 会迫使幂更小，所以最大值为 \(9\)。

所以正确答案是 D。

English

Setting \(d = 0\) removes the subtraction, so we maximize \(c \cdot a^b\) using \(1,\) \(2,\) \(3.\)

Taking \(c = 1,\) \(a = 3,\) \(b = 2\) gives \(3^2 = 9.\) The alternative \(2^3 = 8\) is smaller, and any assignment with \(c \gt 1\) forces a smaller power. The maximum is \(9.\)

Thus, the correct answer is D.

6 答案 C

当 \(m \lt n\) 时，\(m! \cdot n!\) 等于 \((m!)^2 \cdot (m+1)(m+2)\cdots n\)。前面的平方因子不影响判断，因此只需检查 \((m+1)\cdots n\) 是否为完全平方数。

五个选项的剩余因子分别为 \(99\)、\(99 \cdot 100\)、\(100\)、\(100 \cdot 101\) 和 \(101\)。只有 \(100 = 10^2\) 是完全平方数。

因此 \(99! \cdot 100! = (99! \cdot 10)^2\) 是完全平方数。

所以正确答案是 C。

English

For \(m \lt n,\) \(m! \cdot n!\) equals \((m!)^2 \cdot (m+1)(m+2)\cdots n,\) which is a perfect square precisely when \((m+1)\cdots n\) is a perfect square.

For the five choices this leftover factor is \(99,\) \(99 \cdot 100,\) \(100,\) \(100 \cdot 101,\) and \(101.\) Only \(100 = 10^2\) is a perfect square.

Therefore \(99! \cdot 100! = (99! \cdot 10)^2\) is the perfect square.

Thus, the correct answer is C.

7 答案 A

Isabella 得到 \(\dfrac{10}{7}d\) 加元，花掉 \(60\) 加元后剩 \(d\) 加元，所以 \[ \dfrac{10}{7}d - 60 = d\text{。}\]

即 \(\dfrac{3}{7}d = 60\)，所以 \(d = 140\)。数字和为 \(1 + 4 + 0 = 5\)。

所以正确答案是 A。

English

Isabella received \(\dfrac{10}{7}d\) Canadian dollars and spent \(60,\) leaving \(d.\) So \[ \dfrac{10}{7}d - 60 = d. \]

Then \(\dfrac{3}{7}d = 60,\) so \(d = 140.\) The sum of its digits is \(1 + 4 + 0 = 5.\)

Thus, the correct answer is A.

8 答案 A

两个给定方向互相垂直，因此机场位于一个直角三角形的直角顶点，两条直角边为 \(8\) 和 \(10\)。

两个市中心之间的距离为 \(\sqrt{8^2 + 10^2} = \sqrt{164} \approx 12.8\)，最接近 \(13\)。

所以正确答案是 A。

English

The two given directions are perpendicular, so the airport sits at the right angle of a right triangle with legs \(8\) and \(10.\)

The distance between the downtowns is \(\sqrt{8^2 + 10^2} = \sqrt{164} \approx 12.8,\) which is closest to \(13.\)

Thus, the correct answer is A.

9 答案 B

正方形面积为 \(10^2 = 100\)，圆面积为 \(\pi(10)^2 = 100\pi\)。

因为圆心在正方形顶点，正好有四分之一圆位于正方形内部，面积为 \(25\pi\)。

并集面积为 \(100 + 100\pi - 25\pi = 100 + 75\pi\)。

所以正确答案是 B。

English

The square has area \(10^2 = 100\) and the circle has area \(\pi(10)^2 = 100\pi.\)

Since the circle is centered at a vertex of the square, exactly one quarter of the circle, area \(25\pi,\) lies inside the square.

The union has area \(100 + 100\pi - 25\pi = 100 + 75\pi.\)

Thus, the correct answer is B.

10 答案 D

各排罐头数为 \(1, 3, 5, \ldots\)，前 \(n\) 个奇数的和为 \(n^2\)。

令 \(n^2 = 100\)，得 \(n = 10\)。

所以正确答案是 D。

English

The rows hold \(1, 3, 5, \ldots\) cans, and the sum of the first \(n\) odd numbers is \(n^2.\)

Setting \(n^2 = 100\) gives \(n = 10.\)

Thus, the correct answer is D.

11 答案 C

共有 \(8 \cdot 8 = 64\) 个有序结果。条件 \(mn \gt m + n\) 等价于 \((m-1)(n-1) \gt 1\)。

不满足条件的情况只有 \(m = 1\)、\(n = 1\)，或 \(m = n = 2\)。这些情况共有 \(8 + 8 - 1 + 1 = 16\) 个。

因此概率为 \(\dfrac{64 - 16}{64} = \dfrac{48}{64} = \dfrac{3}{4}\)。

所以正确答案是 C。

English

There are \(8 \cdot 8 = 64\) ordered pairs. The inequality \(mn \gt m + n\) is equivalent to \((m-1)(n-1) \gt 1.\)

This fails only when \(m = 1,\) \(n = 1,\) or \(m = n = 2,\) which account for \(8 + 8 - 1 + 1 = 16\) pairs.

The probability is \(\dfrac{64 - 16}{64} = \dfrac{48}{64} = \dfrac{3}{4}.\)

Thus, the correct answer is C.

12 答案 A

圆环面积是两个圆面积之差，即 \(\pi b^2 - \pi c^2\)。

因为 \(\overline{XZ}\) 在 \(Z\) 点与小圆相切，所以它垂直于半径 \(\overline{OZ}\)。在直角三角形 \(OZX\) 中，\(OX = b\)、\(OZ = c\)、\(XZ = a\)，所以 \(b^2 - c^2 = a^2\)。

因此圆环面积为 \(\pi a^2\)。

所以正确答案是 A。

English

The annulus is the difference of the two circular areas, \(\pi b^2 - \pi c^2.\)

Because \(\overline{XZ}\) is tangent to the small circle at \(Z,\) it is perpendicular to the radius \(\overline{OZ}.\) In right triangle \(OZX\) with \(OX = b,\) \(OZ = c,\) and \(XZ = a,\) we get \(b^2 - c^2 = a^2.\)

Therefore the area of the annulus is \(\pi a^2.\)

Thus, the correct answer is A.

13 答案 B

把每个厚度都用百分之一毫米表示。四种可能的厚度为 \(135\)、\(155\)、\(175\)、\(195\)，它们都同余于 \(15\pmod{20}\)。若一摞有 \(k\) 枚硬币且高度为整数毫米，则以百分之一毫米表示的总厚度能被 \(20\) 整除，所以 \(15k\) 能被 \(20\) 整除。因此 \(k\) 必须是 \(4\) 的倍数。

\(4\) 枚硬币的高度至多为 \(4(1.95) = 7.8\) 毫米，\(12\) 枚硬币的高度至少为 \(12(1.35) = 16.2\) 毫米，所以只有 \(8\) 枚硬币可能恰好高 \(14\) 毫米。

事实上，\(8\) 枚二十五美分硬币的高度为 \(8(1.75) = 14\) 毫米。

所以正确答案是 B。

English

Measure every thickness in hundredths of a millimeter. The four possible thicknesses are \(135,\) \(155,\) \(175,\) \(195,\) all congruent to \(15\pmod{20}.\) If the stack has \(k\) coins and an integer height, its total in hundredths is divisible by \(20,\) so \(15k\) is divisible by \(20.\) Therefore \(k\) must be a multiple of \(4.\)

A stack of \(4\) coins is at most \(4(1.95) = 7.8\) mm, and a stack of \(12\) coins is at least \(12(1.35) = 16.2\) mm, so only \(8\) coins can total \(14\) mm.

Indeed, \(8\) quarters give \(8(1.75) = 14\) mm.

Thus, the correct answer is B.

14 答案 C

设蓝色弹珠数为 \(B\)。加入红色弹珠后，总数为 \(3B\)；加入黄色弹珠后，总数为 \(5B\)，蓝色仍为 \(B\)。

将蓝色弹珠数量加倍后，有 \(2B\) 个蓝色弹珠，总数变为 \(6B\)，所以蓝色所占比例为 \(\dfrac{2B}{6B} = \dfrac{1}{3}\)。

所以正确答案是 C。

English

Let there be \(B\) blue marbles. After adding red marbles the total is \(3B;\) after adding yellow marbles the total is \(5B,\) still with \(B\) blue.

Doubling the blue marbles gives \(2B\) blue out of \(6B\) total, which is \(\dfrac{2B}{6B} = \dfrac{1}{3}.\)

Thus, the correct answer is C.

15 答案 A

交换后价值增加，所以 Patty 的五分硬币比十分硬币多。每枚硬币交换后价值变化 \(5\) 美分，所以五分硬币比十分硬币多 \(\frac{70}{5} = 14\) 枚。

两种硬币的数量满足 \(n + d = 20\) 和 \(n - d = 14\)，解得她有 \(17\) 枚五分硬币和 \(3\) 枚十分硬币。

她的硬币总值为 \(17 \cdot 5 + 3 \cdot 10 = 115\) 美分，即 \(\$1.15\)。

所以正确答案是 A。

English

Swapping increases the value, so Patty has more nickels than dimes. Each swapped coin changes the total by \(5\) cents, so she has \(\frac{70}{5} = 14\) more nickels than dimes.

With \(n + d = 20\) and \(n - d = 14,\) she has \(17\) nickels and \(3\) dimes.

Her coins are worth \(17 \cdot 5 + 3 \cdot 10 = 115\) cents, or \(\$1.15.\)

Thus, the correct answer is A.

16 答案 D

三个单位圆的圆心形成边长为 \(2\) 的等边三角形；它的中心就是大圆的圆心。

对于边长为 \(2\) 的等边三角形，中心到顶点的距离为 \(\dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}\)。

再加上单位圆半径，大圆半径为 \(1 + \dfrac{2\sqrt3}{3} = \dfrac{3 + 2\sqrt3}{3}\)。

所以正确答案是 D。

English

The centers of the three unit circles form an equilateral triangle with side \(2.\) Its center is the center of the large circle.

For an equilateral triangle of side length \(2,\) the distance from its center to a vertex is \(\dfrac{2}{\sqrt3} = \dfrac{2\sqrt3}{3}.\)

Adding the unit radius, the large radius is \(1 + \dfrac{2\sqrt3}{3} = \dfrac{3 + 2\sqrt3}{3}.\)

Thus, the correct answer is D.

17 答案 B

设 Jack 的年龄为 \(10x + y\)，Bill 的年龄为 \(10y + x\)。条件给出 \(10x + y + 5 = 2(10y + x + 5)\)，即 \(8x = 19y + 5\)。

由于 \(x\) 和 \(y\) 是数字，唯一解为 \(y = 1\)、\(x = 3\)。

所以 Jack 为 \(31\) 岁，Bill 为 \(13\) 岁，差为 \(18\)。

所以正确答案是 B。

English

Let Jack’s age be \(10x + y\) and Bill’s be \(10y + x.\) In five years \(10x + y + 5 = 2(10y + x + 5),\) which simplifies to \(8x = 19y + 5.\)

Since \(x\) and \(y\) are digits, the only solution is \(y = 1,\) \(x = 3.\)

So Jack is \(31\) and Bill is \(13,\) a difference of \(18.\)

Thus, the correct answer is B.

18 答案 E

\(\triangle ACE\) 的面积是 \(\tfrac12(12)(16) = 96\)。

三个角上的三角形 \(\triangle ABF\)、\(\triangle BCD\) 和 \(\triangle DEF\) 的一组底和高，分别是 \(\triangle ACE\) 中对应底和高的 \(\tfrac34\) 和 \(\tfrac14\)。所以每个小三角形的面积都是 \(\triangle ACE\) 面积的 \(\tfrac14 \cdot \tfrac34 = \tfrac{3}{16}\)。

因此 \[ \begin{aligned} \dfrac{[BDF]}{[ACE]} &= 1 - 3 \cdot \dfrac{3}{16} \\ &= 1 - \dfrac{9}{16} = \dfrac{7}{16} \end{aligned}\text{。}\]

所以正确答案是 E。

English

The area of \(\triangle ACE\) is \(\tfrac12(12)(16) = 96.\)

Each corner triangle \(\triangle ABF,\) \(\triangle BCD,\) and \(\triangle DEF\) has a base and an altitude that are \(\tfrac34\) and \(\tfrac14\) of a corresponding base and altitude of \(\triangle ACE.\) So each has area \(\tfrac14 \cdot \tfrac34 = \tfrac{3}{16}\) of \(\triangle ACE.\)

Hence \[ \begin{aligned} \dfrac{[BDF]}{[ACE]} &= 1 - 3 \cdot \dfrac{3}{16} \\ &= 1 - \dfrac{9}{16} = \dfrac{7}{16}. \end{aligned} \]

Thus, the correct answer is E.

19 答案 C

递推式 \(a_{k+1} = a_{k-2} + a_{k-1} - a_k\) 给出 \(a_{k+1} - a_{k-1} = -(a_k - a_{k-2})\)。数列开头为 \(2001\)、\(2002\)、\(2003\)、\(2000\)、\(2005\)、\(1998\)、\(\ldots\)。

因此偶数位置项形成等差数列 \(2002, 2000, 1998, \ldots\)，公差为 \(-2\)。第 \(2004\) 项是这个等差数列的第 \(1002\) 项，即 \(2002 + 1001(-2) = 0\)。

所以正确答案是 C。

English

The recurrence \(a_{k+1} = a_{k-2} + a_{k-1} - a_k\) gives \(a_{k+1} - a_{k-1} = -(a_k - a_{k-2}).\) The sequence begins \(2001,\) \(2002,\) \(2003,\) \(2000,\) \(2005,\) \(1998,\) \(\ldots\)

So the even-position terms form the arithmetic sequence \(2002, 2000, 1998, \ldots\) with common difference \(-2.\) The \(2004\)th term is its \(1002\)nd term, \(2002 + 1001(-2) = 0.\)

Thus, the correct answer is C.

20 答案 D

设 \(F\) 在 \(\overline{AC}\) 上，且 \(DF \parallel BE\)。令 \(ET = x\)，则 \(BT = 4x\)。

由 \(\triangle ATE \sim \triangle ADF\)，得 \(\dfrac{DF}{x} = \dfrac{AD}{AT} = \dfrac{4}{3}\)，所以 \(DF = \dfrac{4x}{3}\)。

由 \(\triangle BEC \sim \triangle DFC\)，得 \(\dfrac{CD}{BC} = \dfrac{DF}{BE} = \dfrac{\frac{4x}{3}}{5x} = \dfrac{4}{15}\)。

因此 \[ \begin{aligned} \dfrac{CD}{BD} &= \dfrac{\frac{CD}{BC}}{1 - \frac{CD}{BC}} \\ &= \dfrac{\frac{4}{15}}{\frac{11}{15}} = \dfrac{4}{11} \end{aligned}\text{。}\]

所以正确答案是 D。

English

Let \(F\) be on \(\overline{AC}\) with \(DF \parallel BE,\) and write \(ET = x,\) \(BT = 4x.\)

From \(\triangle ATE \sim \triangle ADF,\) \(\dfrac{DF}{x} = \dfrac{AD}{AT} = \dfrac{4}{3},\) so \(DF = \dfrac{4x}{3}.\)

From \(\triangle BEC \sim \triangle DFC,\) \(\dfrac{CD}{BC} = \dfrac{DF}{BE} = \dfrac{\frac{4x}{3}}{5x} = \dfrac{4}{15}.\)

Therefore \[ \begin{aligned} \dfrac{CD}{BD} &= \dfrac{\frac{CD}{BC}}{1 - \frac{CD}{BC}} \\ &= \dfrac{\frac{4}{15}}{\frac{11}{15}} = \dfrac{4}{11}. \end{aligned} \]

Thus, the correct answer is D.

21 答案 A

第一个数列为 \(1 + 3k\)，最大项为 \(6010\)。第二个数列为 \(9 + 7j\)，其最后一项更大，所以公共项上界由 \(6010\) 限制。

公共项形如 \(16 + 21m\)，因为第一个公共项为 \(16\)，且 \(\mathrm{lcm}(3, 7) = 21\)。由 \(16 + 21m \le 6010\) 得 \(0 \le m \le 285\)，共有 \(286\) 个公共项。

因此不同数的个数为 \(2004 + 2004 - 286 = 3722\)。

所以正确答案是 A。

English

The first sequence is \(1 + 3k\) with largest term \(6010,\) and the second is \(9 + 7j\) with a much larger last term, so the binding limit is \(6010.\)

A common value has the form \(16 + 21m\) (the first shared term is \(16,\) spaced by \(\mathrm{lcm}(3, 7) = 21\)). Requiring \(16 + 21m \le 6010\) gives \(0 \le m \le 285,\) that is \(286\) common numbers.

The number of distinct values is \(2004 + 2004 - 286 = 3722.\)

Thus, the correct answer is A.

22 答案 D

因为 \(5^2 + 12^2 = 13^2\)，这是直角三角形。将顶点放在 \((0, 0)\)、\((5, 0)\)、\((0, 12)\)，则外心为斜边中点 \(\left(\tfrac52, 6\right)\)。

内切圆半径满足 \((12 - r) + (5 - r) = 13\)，所以 \(r = 2\)，内心为 \((2, 2)\)。

两圆心之间的距离为 \[ \begin{gathered} \sqrt{\left(\tfrac52 - 2\right)^2 + (6 - 2)^2} \\ = \sqrt{\tfrac14 + 16} \\ = \dfrac{\sqrt{65}}{2} \end{gathered}\text{。}\]

所以正确答案是 D。

English

Since \(5^2 + 12^2 = 13^2,\) the triangle is right. Place it at \((0, 0),\) \((5, 0),\) \((0, 12).\) The circumcenter is the midpoint of the hypotenuse, \(\left(\tfrac52, 6\right).\)

The inradius satisfies \((12 - r) + (5 - r) = 13,\) so \(r = 2\) and the incenter is \((2, 2).\)

The distance is \[ \begin{gathered} \sqrt{\left(\tfrac52 - 2\right)^2 + (6 - 2)^2} \\ = \sqrt{\tfrac14 + 16} \\ = \dfrac{\sqrt{65}}{2}. \end{gathered} \]

Thus, the correct answer is D.

23 答案 B

固定立方体方向，共有 \(2^6 = 64\) 种涂色。

可行情况包括：六面同色，有 \(2\) 种；恰好五面同色，有 \(\binom{6}{5} \cdot 2 = 12\) 种；以及四个面同色而剩下的一对相对面为另一种颜色，有 \(3\) 组相对面、\(2\) 种颜色，共 \(6\) 种。

总共有 \(2 + 12 + 6 = 20\) 种，所以概率为 \(\dfrac{20}{64} = \dfrac{5}{16}\)。

所以正确答案是 B。

English

Fixing the orientation, there are \(2^6 = 64\) colorings.

A coloring works if all six faces match (\(2\) ways), exactly five match (\(\binom{6}{5} \cdot 2 = 12\) ways), or four faces share a color with the remaining pair being opposite faces of the other color (\(3\) opposite pairs, \(2\) colors, giving \(6\) ways).

The total is \(2 + 12 + 6 = 20,\) so the probability is \(\dfrac{20}{64} = \dfrac{5}{16}.\)

Thus, the correct answer is B.

24 答案 B

设 \(\overline{AD}\) 交 \(\overline{BC}\) 于 \(E\)。因为 \(\angle ABC\) 与 \(\angle ADC\) 截同弧，所以它们相等；又有 \(\angle EAB = \angle CAD\)，因此 \(\triangle ABE \sim \triangle ADC\)。

因此 \(\dfrac{AD}{CD} = \dfrac{AB}{BE}\)。

由角平分线定理，\(\dfrac{BE}{EC} = \dfrac{AB}{AC}\)，所以 \(BE = \dfrac{AB \cdot BC}{AB + AC} = \dfrac{7 \cdot 9}{15}\)。

因此 \[ \begin{aligned} \dfrac{AD}{CD} &= \dfrac{AB}{BE} = \dfrac{AB + AC}{BC} \\ &= \dfrac{15}{9} = \dfrac{5}{3} \end{aligned}\text{。}\]

所以正确答案是 B。

English

Let \(\overline{AD}\) meet \(\overline{BC}\) at \(E.\) Since \(\angle ABC\) and \(\angle ADC\) subtend the same arc, they are equal, and \(\angle EAB = \angle CAD,\) so \(\triangle ABE \sim \triangle ADC.\)

Hence \(\dfrac{AD}{CD} = \dfrac{AB}{BE}.\)

By the Angle Bisector Theorem, \(\dfrac{BE}{EC} = \dfrac{AB}{AC},\) so \(BE = \dfrac{AB \cdot BC}{AB + AC} = \dfrac{7 \cdot 9}{15}.\)

Therefore \[ \begin{aligned} \dfrac{AD}{CD} &= \dfrac{AB}{BE} = \dfrac{AB + AC}{BC} \\ &= \dfrac{15}{9} = \dfrac{5}{3}. \end{aligned} \]

Thus, the correct answer is B.

25 答案 B

设两个大圆圆心为 \(A\) 和 \(B\)，小圆圆心为 \(C\)，两个大圆的一个交点为 \(D\)。

则 \(\triangle ACD\) 是直角三角形，且 \(AC = 1\)、\(AD = 2\)，所以 \(CD = \sqrt3\)、\(\angle CAD = 60^\circ\)，其面积为 \(\dfrac{\sqrt3}{2}\)。

阴影区域的四分之一等于半径为 \(2\) 的大圆中一个 \(60^\circ\) 扇形面积 \(\dfrac{2\pi}{3}\)，减去 \(\triangle ACD\) 的面积 \(\dfrac{\sqrt3}{2}\)，再减去小圆四分之一的面积 \(\dfrac{\pi}{4}\)，得到 \(\dfrac{2\pi}{3} - \dfrac{\sqrt3}{2} - \dfrac{\pi}{4} = \dfrac{5\pi}{12} - \dfrac{\sqrt3}{2}\)。

乘以 \(4\)，阴影总面积为 \(\dfrac{5\pi}{3} - 2\sqrt3\)。

所以正确答案是 B。

English

Let the large circles have centers \(A\) and \(B,\) let \(C\) be the center of the small circle, and let \(D\) be a point where the two large circles meet.

Then \(\triangle ACD\) is right with \(AC = 1\) and \(AD = 2,\) so \(CD = \sqrt3,\) \(\angle CAD = 60^\circ,\) and its area is \(\dfrac{\sqrt3}{2}.\)

One quarter of the shaded region equals the \(60^\circ\) sector of the radius-\(2\) circle (area \(\dfrac{2\pi}{3}\)) minus \(\triangle ACD\) (area \(\dfrac{\sqrt3}{2}\)) minus a quarter of the small circle (area \(\dfrac{\pi}{4}\)), giving \(\dfrac{2\pi}{3} - \dfrac{\sqrt3}{2} - \dfrac{\pi}{4} = \dfrac{5\pi}{12} - \dfrac{\sqrt3}{2}.\)

Multiplying by \(4,\) the shaded area is \(\dfrac{5\pi}{3} - 2\sqrt3.\)

Thus, the correct answer is B.