2004A AMC 12 真题（P版）

1 答案 E

因为 \(\$20\) 是 \(2000\) 美分，所以每小时的税额为 \[ 0.0145 \times 2000 = 29 \] 美分。

所以正确答案是 E。

English

Since \(\$20\) is \(2000\) cents, the tax is \[ 0.0145 \times 2000 = 29 \] cents per hour.

Thus, the correct answer is E.

2 答案 C

\(8\) 道未作答的题值 \(2.5 \times 8 = 20\) 分，所以 Charlyn 还需要从答对的题中得到至少 \(100 - 20 = 80\) 分。

每道答对的题值 \(6\) 分，而不小于 \(80\) 的最小 \(6\) 的倍数是 \(84 = 6 \times 14\)。所以她至少需要答对 \(14\) 道题。

所以正确答案是 C。

English

The \(8\) unanswered problems are worth \(2.5 \times 8 = 20\) points, so Charlyn needs at least \(100 - 20 = 80\) more points from correct answers.

Each correct answer is worth \(6\) points, and the smallest multiple of \(6\) that is at least \(80\) is \(84 = 6 \times 14.\) So she needs at least \(14\) correct answers.

Thus, the correct answer is C.

3 答案 B

写成 \(x = 100 - 2y\)。当且仅当 \(y\) 是满足 \(1 \le y \le 49\) 的正整数时，\(x\) 才是正整数。

因此共有 \(49\) 个有效有序对。

所以正确答案是 B。

English

Writing \(x = 100 - 2y,\) the value of \(x\) is a positive integer precisely when \(y\) is a positive integer with \(1 \le y \le 49.\)

This gives \(49\) valid ordered pairs.

Thus, the correct answer is B.

4 答案 E

Bertha 有 \(30 - 6 = 24\) 个外孙女，她们都没有女儿。

这些外孙女是 Bertha 的 \(\frac{24}{6} = 4\) 个女儿所生，所以恰好有 \(4\) 人有女儿。

因此没有女儿的人数是 \(30 - 4 = 26\)。

所以正确答案是 E。

English

Bertha has \(30 - 6 = 24\) granddaughters, none of whom have daughters.

These granddaughters are the children of \(\frac{24}{6} = 4\) of Bertha’s daughters, so exactly \(4\) women have daughters.

Therefore the number of women with no daughters is \(30 - 4 = 26.\)

Thus, the correct answer is E.

5 答案 B

\(y\) 截距在 \(0\) 和 \(1\) 之间，所以 \(0 \lt b \lt 1\)。

斜率为负且较平缓，在 \(-1\) 和 \(0\) 之间，所以 \(-1 \lt m \lt 0\)。

因此乘积 \(mb\) 为负，且绝对值小于 \(1\)，从而得到 \(-1 \lt mb \lt 0\)。

所以正确答案是 B。

English

The \(y\)-intercept of the line is between \(0\) and \(1,\) so \(0 \lt b \lt 1.\)

The slope is negative and shallow, between \(-1\) and \(0,\) so \(-1 \lt m \lt 0.\)

The product \(mb\) is therefore negative with absolute value less than \(1,\) giving \(-1 \lt mb \lt 0.\)

Thus, the correct answer is B.

6 答案 A

逐个因式分解这些差：

\(U - V = 2004^{2005}\)，\(V - W = 2004^{2004}\)，\(W - X = 2001 \cdot 2004^{2004}\)，\(X - Y = 2004^{2004}\)，且 \(Y - Z = 2003 \cdot 2004^{2003}\)。

因为 \(2004^{2005} = 2004 \cdot 2004^{2004}\) 大于其他各项，而其他各项都达不到 \(2004^{2005}\)，所以 \(U - V\) 最大。

所以正确答案是 A。

English

Compute each difference by factoring:

\(U - V = 2004^{2005},\) \(V - W = 2004^{2004},\) \(W - X = 2001 \cdot 2004^{2004},\) \(X - Y = 2004^{2004},\) and \(Y - Z = 2003 \cdot 2004^{2003}.\)

Since \(2004^{2005} = 2004 \cdot 2004^{2004}\) exceeds each of the others, none of which reaches \(2004^{2005},\) the difference \(U - V\) is the largest.

Thus, the correct answer is A.

7 答案 B

三轮后，玩家 \(A\)、\(B\) 和 \(C\) 分别有 \(14\)、\(13\) 和 \(12\) 个筹码。之后每三轮都会使每个玩家的筹码数减少一个。

经过 \(36\) 轮后，他们分别有 \(3\)、\(2\) 和 \(1\) 个筹码；第 \(37\) 轮中，玩家 \(A\) 把自己的三个筹码全部给出并放入弃置堆后归零，游戏结束。

所以正确答案是 B。

English

After three rounds the players \(A,\) \(B,\) and \(C\) have \(14,\) \(13,\) and \(12\) tokens, respectively. Every subsequent three rounds reduces each player’s supply by one token.

After \(36\) rounds they have \(3,\) \(2,\) and \(1\) tokens. In the \(37\)th round player \(A,\) who has the most, gives away all three of their tokens and runs out, ending the game.

Thus, the correct answer is B.

8 答案 B

设 \(\triangle ADE\)，\(\triangle BDC\)，和 \(\triangle ABD\) 的面积分别为 \(x\)，\(y\)，和 \(z\)。

则 \(\triangle ABE\) 的面积为 \(\tfrac12 \cdot 4 \cdot 8 = 16 = x + z\)，而 \(\triangle ABC\) 的面积为 \(\tfrac12 \cdot 4 \cdot 6 = 12 = y + z\)。

所求差为 \[ \begin{aligned} x - y &= (x + z) - (y + z) \\ &= 16 - 12 = 4 \end{aligned}\text{。}\]

所以正确答案是 B。

English

Let \(x,\) \(y,\) and \(z\) be the areas of \(\triangle ADE,\) \(\triangle BDC,\) and \(\triangle ABD,\) respectively.

Then \(\triangle ABE\) has area \(\tfrac12 \cdot 4 \cdot 8 = 16 = x + z,\) and \(\triangle ABC\) has area \(\tfrac12 \cdot 4 \cdot 6 = 12 = y + z.\)

The requested difference is \[ \begin{aligned} x - y &= (x + z) - (y + z) \\ &= 16 - 12 = 4. \end{aligned} \]

Thus, the correct answer is B.

9 答案 C

直径乘以 \(\tfrac54\) 会使底面积乘以 \(\left(\tfrac54\right)^2 = \tfrac{25}{16}\)。

为保持体积不变，高度必须乘以 \(\tfrac{16}{25} = 0.64\)，也就是减少 \(1 - 0.64 = 0.36\)，即 \(36\%\)。

所以正确答案是 C。

English

Multiplying the diameter by \(\tfrac54\) multiplies the base area by \(\left(\tfrac54\right)^2 = \tfrac{25}{16}.\)

To keep the volume fixed, the height must be multiplied by \(\tfrac{16}{25} = 0.64.\) That is a decrease of \(1 - 0.64 = 0.36,\) or \(36\%.\)

Thus, the correct answer is C.

10 答案 C

一组连续整数的和等于项数乘以平均数，而对连续整数来说，平均数等于中位数。

所以中位数为 \[ \dfrac{7^5}{49} = \dfrac{7^5}{7^2} = 7^3 = 343\text{。}\]

所以正确答案是 C。

English

The sum of a set of consecutive integers equals the number of terms times their mean, and for consecutive integers the mean equals the median.

So the median is \[ \dfrac{7^5}{49} = \dfrac{7^5}{7^2} = 7^3 = 343. \]

Thus, the correct answer is C.

11 答案 A

如果 Paula 有 \(n\) 枚硬币，它们的总币值是 \(20n\) 美分。再加入一枚二十五美分硬币后共有 \(n + 1\) 枚硬币，总币值为 \(20n + 25\) 美分，而它也必须等于 \(21(n + 1)\) 美分。

因此 \(20n + 25 = 21(n + 1)\)，解得 \(n = 4\)。

这四枚硬币合计 \(80\) 美分，只能是三枚二十五美分硬币和一枚五美分硬币，所以十美分硬币的数量为 \(0\)。

所以正确答案是 A。

English

If Paula has \(n\) coins, their total value is \(20n\) cents. Adding a quarter gives \(n + 1\) coins worth \(20n + 25\) cents, which must also equal \(21(n + 1)\) cents.

So \(20n + 25 = 21(n + 1),\) giving \(n = 4.\)

Four coins totalling \(80\) cents must be three quarters and one nickel, so the number of dimes is \(0.\)

Thus, the correct answer is A.

12 答案 B

直线 \(AC\) 经过 \((0, 9)\)，斜率为 \(\tfrac{8 - 9}{2 - 0} = -\tfrac12\)，所以方程是 \(y = -\tfrac12 x + 9\)。令 \(y = x\)，得到 \(A' = (6, 6)\)。

直线 \(BC\) 经过 \((0, 12)\)，斜率为 \(-2\)，所以方程是 \(y = -2x + 12\)。令 \(y = x\)，得到 \(B' = (4, 4)\)。

因此 \[ \begin{aligned} A'B' &= \sqrt{(6 - 4)^2 + (6 - 4)^2} \\ &= 2\sqrt{2} \end{aligned}\text{。}\]

所以正确答案是 B。

English

Line \(AC\) passes through \((0, 9)\) with slope \(\tfrac{8 - 9}{2 - 0} = -\tfrac12,\) so its equation is \(y = -\tfrac12 x + 9.\) Setting \(y = x\) gives \(A' = (6, 6).\)

Line \(BC\) passes through \((0, 12)\) with slope \(-2,\) so \(y = -2x + 12.\) Setting \(y = x\) gives \(B' = (4, 4).\)

Then \[ \begin{aligned} A'B' &= \sqrt{(6 - 4)^2 + (6 - 4)^2} \\ &= 2\sqrt{2}. \end{aligned} \]

Thus, the correct answer is B.

13 答案 B

共有 \(\binom{9}{2} = 36\) 对点，每一对点确定一条直线。

但是有三条水平线、三条竖直线和两条对角线各经过 \(S\) 中的 \(3\) 个共线点，因此每条这样的直线被多算 \(2\) 次。

这样的直线共有 \(8\) 条，所以不同直线的数量为 \(36 - 2 \cdot 8 = 20\)。

所以正确答案是 B。

English

There are \(\binom{9}{2} = 36\) pairs of points, and each pair determines a line.

However, there are three horizontal, three vertical, and two diagonal lines that each pass through three collinear points of \(S.\) Each such line is counted \(3\) times, an overcount of \(2\) per line.

With \(8\) such lines, the number of distinct lines is \(36 - 2 \cdot 8 = 20.\)

Thus, the correct answer is B.

14 答案 A

等差数列为 \(9\)，\(9 + d\)，\(9 + 2d\)。加上指定数后，等比数列为 \(9\)，\(11 + d\)，\(29 + 2d\)。

等比条件给出 \((11 + d)^2 = 9(29 + 2d)\)，化简为 \(d^2 + 4d - 140 = 0\)，所以 \(d = 10\) 或 \(d = -14\)。

对应的第三项 \(29 + 2d\) 分别为 \(49\) 和 \(1\)，所以最小可能值为 \(1\)。

所以正确答案是 A。

English

The arithmetic progression is \(9,\) \(9 + d,\) \(9 + 2d.\) After the additions, the geometric progression is \(9,\) \(11 + d,\) \(29 + 2d.\)

The geometric condition gives \((11 + d)^2 = 9(29 + 2d),\) which simplifies to \(d^2 + 4d - 140 = 0,\) so \(d = 10\) or \(d = -14.\)

The corresponding third terms \(29 + 2d\) are \(49\) and \(1,\) so the smallest possible value is \(1.\)

Thus, the correct answer is A.

15 答案 C

从相对两端出发时，她们第一次相遇时合计跑了半个跑道长度。

第一次和第二次相遇之间，她们合计跑了一个完整跑道长度。因为 Brenda 第一次相遇前跑了 \(100\) 米，所以这段时间她跑了 \(2 \cdot 100 = 200\) 米。

同一时间段内 Sally 跑了 \(150\) 米，因此跑道长度为 \(200 + 150 = 350\) 米。

所以正确答案是 C。

English

Starting at opposite ends, when they first meet they have together run half the track. Between the first and second meetings, they together run a full track length.

Since Brenda runs at a constant speed and covered \(100\) meters before the first meeting, she covers \(2 \cdot 100 = 200\) meters between the two meetings.

Adding Sally’s \(150\) meters over that same interval gives a track length of \(200 + 150 = 350\) meters.

Thus, the correct answer is C.

16 答案 B

该表达式有定义当且仅当 \(\log_{2003}(\log_{2002}(\log_{2001} x)) \gt 0\)，也就是 \(\log_{2002}(\log_{2001} x) \gt 1\)。

这当且仅当 \(\log_{2001} x \gt 2002\)，等价于 \(x \gt 2001^{2002}\)。

因此 \(c = 2001^{2002}\)。

所以正确答案是 B。

English

The expression is defined if and only if \(\log_{2003}(\log_{2002}(\log_{2001} x)) \gt 0,\) that is, \(\log_{2002}(\log_{2001} x) \gt 1.\)

This holds if and only if \(\log_{2001} x \gt 2002,\) which is equivalent to \(x \gt 2001^{2002}.\)

Therefore \(c = 2001^{2002}.\)

Thus, the correct answer is B.

17 答案 D

在 \(f(2n) = n \cdot f(n)\) 中令 \(n = 2^{k}\) 得到 \(f(2^{k+1}) = 2^{k} \cdot f(2^{k})\)。

从 \(f(2^1) = f(2) = 1 \cdot f(1) = 2^0\)，展开，指数累加为 \[ \begin{aligned} f(2^n) &= 2^{0 + 1 + 2 + \cdots + (n-1)} \\ &= 2^{\frac{n(n-1)}{2}} \end{aligned}\text{。}\]

因此 \(f(2^{100}) = 2^{\frac{100 \cdot 99}{2}} = 2^{4950}\)。

所以正确答案是 D。

English

Applying \(f(2n) = n \cdot f(n)\) with \(n = 2^{k},\) we get \(f(2^{k+1}) = 2^{k} \cdot f(2^{k}).\)

Unwinding from \(f(2^1) = f(2) = 1 \cdot f(1) = 2^0,\) the exponents accumulate: \[ \begin{aligned} f(2^n) &= 2^{0 + 1 + 2 + \cdots + (n-1)} \\ &= 2^{\frac{n(n-1)}{2}}. \end{aligned} \]

Therefore \(f(2^{100}) = 2^{\frac{100 \cdot 99}{2}} = 2^{4950}.\)

Thus, the correct answer is D.

18 答案 D

设 \(F\) 为 \(CE\) 与半圆的切点。由于 \(CB\) 和 \(CF\) 都是从 \(C\) 作出的切线，所以 \(CF = CB = 2\)。同理，令 \(x = AE\)，从 \(E\) 作出的切线给出 \(EF = EA = x\)。

因此 \(CE = CF + FE = 2 + x\)。在直角三角形 \(CDE\) 中，\(CD = 2\) 且 \(DE = 2 - x\)，所以 \[ (2 - x)^2 + 2^2 = (2 + x)^2\text{。}\]

展开得 \(8x = 4\)，所以 \(x = \tfrac12\)，并且 \(CE = 2 + \tfrac12 = \tfrac52\)。

所以正确答案是 D。

English

Let \(F\) be the point where \(CE\) touches the semicircle. Since \(CB\) and \(CF\) are both tangents from \(C,\) we have \(CF = CB = 2.\) Similarly, with \(x = AE,\) the tangents from \(E\) give \(EF = EA = x.\)

Thus \(CE = CF + FE = 2 + x.\) In right triangle \(CDE,\) where \(CD = 2\) and \(DE = 2 - x,\) \[ (2 - x)^2 + 2^2 = (2 + x)^2. \]

Expanding gives \(8x = 4,\) so \(x = \tfrac12\) and \(CE = 2 + \tfrac12 = \tfrac52.\)

Thus, the correct answer is D.

19 答案 D

圆 \(A\) 半径为 \(1\)，经过圆 \(D\) 的圆心并与圆 \(D\) 内切，所以圆 \(D\) 的半径为 \(2\)。

将圆 \(D\) 的圆心放在原点，圆 \(A\) 的圆心在 \((-1, 0)\)。由对称性，\(B\) 的圆心为 \((x, y)\)，半径为 \(r\) 而 \(C\) 是它关于水平轴的镜像，所以两圆在该轴上相切，且 \(y = r\)。

与圆 \(D\) 内切给出 \(x^2 + y^2 = (2 - r)^2\)，与圆 \(A\) 外切给出 \((x + 1)^2 + y^2 = (1 + r)^2\)。

两式相减并使用 \(y = r\) 得到 \(x = \tfrac23\) 且 \(r = \tfrac89\)。因此圆 \(B\) 的半径是 \(\tfrac89\)。

所以正确答案是 D。

English

Circle \(A\) has radius \(1\) and passes through the center of \(D\) while being internally tangent to \(D,\) so \(D\) has radius \(2.\)

Place the center of \(D\) at the origin, with \(A\) centered at \((-1, 0).\) By symmetry, \(B\) has center \((x, y)\) and radius \(r,\) with \(C\) its mirror image across the horizontal axis, so the two congruent circles touch on that axis and \(y = r.\)

Internal tangency to \(D\) gives \(x^2 + y^2 = (2 - r)^2,\) and external tangency to \(A\) gives \((x + 1)^2 + y^2 = (1 + r)^2.\)

Subtracting and using \(y = r\) yields \(x = \tfrac23\) and \(r = \tfrac89.\) The radius of circle \(B\) is \(\tfrac89.\)

Thus, the correct answer is D.

20 答案 E

将选择表示为单位正方形中的点 \((a, b)\)。\(a\) 和 \(b\) 小于 \(\tfrac12\) 时四舍五入为 \(0\)，否则四舍五入为 \(1\)；而 \(c = a + b\) 的分界点为 \(\tfrac12\) 和 \(\tfrac32\)。

这个等式恰好在两个区域内不成立。若 \(a, b \lt \tfrac12\)，则当 \(a + b \ge \tfrac12\) 时不成立，这是一个面积为 \(\tfrac18\) 的直角三角形。若 \(a, b \ge \tfrac12\)，则当 \(a + b \lt \tfrac32\) 时不成立，这是另一个面积为 \(\tfrac18\) 的直角三角形。当 \(a, b\) 中恰好有一个不小于 \(\tfrac12\) 时，等式总是成立。

因此不成立的概率为 \(\tfrac18 + \tfrac18 = \tfrac14\)，所以所求概率为 \(1 - \tfrac14 = \tfrac34\)。

所以正确答案是 E。

English

Represent the choices as a point \((a, b)\) in the unit square. Each of \(a\) and \(b\) rounds to \(0\) if below \(\tfrac12\) and to \(1\) otherwise, while \(c = a + b\) rounds based on \(\tfrac12\) and \(\tfrac32.\)

The equation fails in exactly two regions. If \(a, b \lt \tfrac12,\) it fails when \(a + b \ge \tfrac12;\) this is a right triangle of area \(\tfrac18.\) If \(a, b \ge \tfrac12,\) it fails when \(a + b \lt \tfrac32;\) this is another right triangle of area \(\tfrac18.\) When exactly one of \(a, b\) is at least \(\tfrac12,\) the equation always holds.

Thus the failure probability is \(\tfrac18 + \tfrac18 = \tfrac14,\) so the requested probability is \(1 - \tfrac14 = \tfrac34.\)

Thus, the correct answer is E.

21 答案 D

该级数首项为 \(1\)，公比为 \(\cos^2 \theta\)，所以和为 \(\dfrac{1}{1 - \cos^2 \theta} = \dfrac{1}{\sin^2 \theta} = 5\)。

因此 \(\sin^2 \theta = \tfrac15\)，并且 \[ \begin{aligned} \cos 2\theta &= 1 - 2\sin^2 \theta \\ &= 1 - \tfrac25 = \tfrac35 \end{aligned}\text{。}\]

所以正确答案是 D。

English

The series is geometric with first term \(1\) and ratio \(\cos^2 \theta,\) so its sum is \(\dfrac{1}{1 - \cos^2 \theta} = \dfrac{1}{\sin^2 \theta} = 5.\)

Thus \(\sin^2 \theta = \tfrac15,\) and \[ \begin{aligned} \cos 2\theta &= 1 - 2\sin^2 \theta \\ &= 1 - \tfrac25 = \tfrac35. \end{aligned} \]

Thus, the correct answer is D.

22 答案 B

设三个单位球的球心为 \(A\)，\(B\)，\(C\)，它们在离平面高度 \(1\) 处形成边长为 \(2\) 的等边三角形；设 \(E\) 为大球球心，位于 \(\triangle ABC\) 的重心 \(D\) 正上方。

从顶点到重心的距离为 \(AD = \tfrac{2\sqrt3}{3}\)，且 \(AE = 1 + 2 = 3\)，所以 \[ \begin{aligned} DE &= \sqrt{3^2 - \left(\tfrac{2\sqrt3}{3}\right)^2} \\ &= \sqrt{9 - \tfrac{4}{3}} \\ &= \dfrac{\sqrt{69}}{3} \end{aligned}\text{。}\]

因为 \(D\) 位于平面上方 \(1\) 个单位，而大球顶部在 \(E\) 上方 \(2\) 个单位，所以总高度为 \[ 1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}\text{。}\]

所以正确答案是 B。

English

Let the centers of the three unit spheres be \(A,\) \(B,\) \(C,\) forming an equilateral triangle of side \(2\) at height \(1\) above the plane, and let \(E\) be the center of the large sphere directly above the centroid \(D\) of \(\triangle ABC.\)

The distance from a vertex to the centroid is \(AD = \tfrac{2\sqrt3}{3},\) and \(AE = 1 + 2 = 3,\) so \[ \begin{aligned} DE &= \sqrt{3^2 - \left(\tfrac{2\sqrt3}{3}\right)^2} \\ &= \sqrt{9 - \tfrac{4}{3}} \\ &= \dfrac{\sqrt{69}}{3}. \end{aligned} \]

Since \(D\) is \(1\) unit above the plane and the top of the large sphere is \(2\) units above \(E,\) the total height is \[ 1 + \dfrac{\sqrt{69}}{3} + 2 = 3 + \dfrac{\sqrt{69}}{3}. \]

Thus, the correct answer is B.

23 答案 E

因为 \(z_1 = a_1 + b_1 i = 0\) 是根，所以 \(c_0 = P(0) = 0\)。

非实零点成共轭对出现，所以 \(\sum b_k = 0\) 而题设于是强制 \(\sum a_k = 0\)。系数 \(c_{2003}\) 等于 \(-c_{2004}\) 乘以根之和 \(\sum a_k + i \sum b_k = 0\)，所以 \(c_{2003} = 0\)。

因为次数为偶数，\(z_2, \ldots, z_{2004}\) 中至少有一个实根，使得某个 \(b_k = 0\)，所以 \(b_2 b_3 \cdots b_{2004} = 0\)。因此 (A) 到 (D) 都必须为 \(0\)。

另一方面，\(\sum_{k=1}^{2004} c_k = P(1)\)。一个有效多项式例如 \(P(x) = x(x - 2)(x - 3) \cdots\) \(\cdot (x - 2003)\) \(\cdot \left(x + \sum_{k=2}^{2003} k\right)\) 满足 \(P(1) \ne 0\)，所以只有 \(\sum c_k\) 可能非零。

所以正确答案是 E。

English

Since \(z_1 = a_1 + b_1 i = 0\) is a root, \(c_0 = P(0) = 0.\)

The nonreal zeros occur in conjugate pairs, so \(\sum b_k = 0,\) and the hypothesis then forces \(\sum a_k = 0.\) The coefficient \(c_{2003}\) equals \(-c_{2004}\) times the sum of the roots \(\sum a_k + i \sum b_k = 0,\) so \(c_{2003} = 0.\)

Because the degree is even, at least one of \(z_2, \ldots, z_{2004}\) is real, making one \(b_k = 0,\) so \(b_2 b_3 \cdots b_{2004} = 0.\) Thus (A) through (D) all must be \(0.\)

On the other hand, \(\sum_{k=1}^{2004} c_k = P(1),\) and a valid polynomial such as \(P(x) = x(x - 2)(x - 3) \cdots\) \(\cdot (x - 2003)\) \(\cdot \left(x + \sum_{k=2}^{2003} k\right)\) has \(P(1) \ne 0.\) So only \(\sum c_k\) can be nonzero.

Thus, the correct answer is E.

24 答案 C

一个半径为 \(1\) 的圆盘覆盖线段 \(\overline{AB}\)，当且仅当它的圆心到 \(A\) 和 \(B\) 的距离都不超过 \(1\)。这个区域 \(R\) 是以 \(A\) 和 \(B\) 为圆心的两个单位圆的重叠透镜。

每个单位圆都经过另一个圆的圆心，所以透镜由两段 \(120^\circ\) 圆弧围成。两个 \(120^\circ\) 扇形的面积各为 \(\tfrac{\pi}{3}\)，重叠中扣掉的两个等边三角形总面积为 \(\tfrac{\sqrt3}{2}\)，所以 \(R\) 的面积为 \(\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}\)。

集合 \(S\) 由所有到 \(R\) 距离不超过 \(1\) 的点组成。除 \(R\) 本身外，还增加两个 \(60^\circ\) 半径 \(1\) 的扇形（每个面积 \(\tfrac{\pi}{6}\)）和两个 \(120^\circ\) 外半径 \(2\)、内半径 \(1\) 的环形区域（每个面积 \(\pi\)）。

因此 \(S\) 的面积为 \[ \begin{aligned} &\left(\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}\right) + 2 \cdot \tfrac{\pi}{6} \\ &\quad {}+ 2\pi = 3\pi - \tfrac{\sqrt3}{2} \end{aligned}\text{。}\]

所以正确答案是 C。

English

A radius-\(1\) disk covers segment \(\overline{AB}\) exactly when its center is within \(1\) of both \(A\) and \(B.\) That region \(R\) is the lens where the two unit circles centered at \(A\) and \(B\) overlap.

Each unit circle passes through the other’s center, so the lens is bounded by two \(120^\circ\) arcs. Two \(120^\circ\) sectors of area \(\tfrac{\pi}{3}\) overlap in two equilateral triangles of total area \(\tfrac{\sqrt3}{2},\) giving \(R\) area \(\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}.\)

The set \(S\) consists of all points within \(1\) of \(R.\) Beyond \(R\) itself, this adds two \(60^\circ\) sectors of radius \(1\) (each area \(\tfrac{\pi}{6}\)) and two \(120^\circ\) annuli of outer radius \(2\) and inner radius \(1\) (each area \(\pi\)).

Therefore the area of \(S\) is \[ \begin{aligned} &\left(\tfrac{2\pi}{3} - \tfrac{\sqrt3}{2}\right) + 2 \cdot \tfrac{\pi}{6} \\ &\quad {}+ 2\pi = 3\pi - \tfrac{\sqrt3}{2}. \end{aligned} \]

Thus, the correct answer is C.

25 答案 E

因为 \(n^3 \cdot a_n = 133.\overline{133}_n\) \(= a_n + n^2 + 3n + 3\)，得到 \[ \begin{aligned} a_n &= \dfrac{n^2 + 3n + 3}{n^3 - 1} \\ &= \dfrac{(n+1)^3 - 1}{n(n^3 - 1)} \end{aligned}\text{。}\]

写成 \(n^3 - 1 = (n - 1)(n^2 + n + 1)\)，以及 \((n+1)^3 - 1\) \(= n\big((n+1)^2 + (n+1) + 1\big)\)，乘积 \(a_4 a_5 \cdots a_{99}\) 裂项相消为 \[ \begin{gathered} \dfrac{3!}{99!} \cdot \dfrac{100^3 - 1}{4^3 - 1} \\ {}= \dfrac{3!}{99!} \cdot \dfrac{99(100^2 + 100 + 1)}{63} \end{gathered}\text{。}\]

它化简为 \(\dfrac{(2)(10101)}{(21)(98!)} = \dfrac{962}{98!}\)。若 \(n \le 97\)，把这个分数改写成以 \(n!\) 为分母会要求 \(98\) 整除 \(962\)，但并不成立。因此最小可能的 \(n\) 是 \(98\)，且 \(m = 962\)。

所以正确答案是 E。

English

Since \(n^3 \cdot a_n = 133.\overline{133}_n\) \(= a_n + n^2 + 3n + 3,\) we get \[ \begin{aligned} a_n &= \dfrac{n^2 + 3n + 3}{n^3 - 1} \\ &= \dfrac{(n+1)^3 - 1}{n(n^3 - 1)}. \end{aligned} \]

Writing \(n^3 - 1 = (n - 1)(n^2 + n + 1)\) and \((n+1)^3 - 1\) \(= n\big((n+1)^2 + (n+1) + 1\big),\) the product \(a_4 a_5 \cdots a_{99}\) telescopes to \[ \begin{gathered} \dfrac{3!}{99!} \cdot \dfrac{100^3 - 1}{4^3 - 1} \\ {}= \dfrac{3!}{99!} \cdot \dfrac{99(100^2 + 100 + 1)}{63}. \end{gathered} \]

This simplifies to \(\dfrac{(2)(10101)}{(21)(98!)} = \dfrac{962}{98!}.\) If \(n \le 97,\) then rewriting this fraction with denominator \(n!\) would require \(98\) to divide \(962,\) which it does not. Hence the smallest possible \(n\) is \(98,\) and \(m = 962.\)

Thus, the correct answer is E.