2021D AMC 12 真题（P版）

数字 \(1, 2, 3, 4\) 在千位、百位、十位和个位上各恰好出现一次，因此每一列的和都是 \(1 + 2 + 3 + 4 = 10\)。

所以总和为 \(10 \cdot 1111 = 11110\)。

所以正确答案是 E。

English

Each of the digits \(1, 2, 3, 4\) appears exactly once in the thousands, hundreds, tens, and units columns. So each column sums to \(1 + 2 + 3 + 4 = 10.\)

The total is therefore \(10 \cdot 1111 = 11110.\)

Thus, the correct answer is E.

外层三角形的顶点依次为 \((1, 0)\)、\((3, 5)\) 和 \((5, 0)\)，底为 \(4\)，高为 \(5\)，所以面积为 \(\tfrac12 \cdot 4 \cdot 5 = 10\)。

被挖去的三角形顶点依次为 \((1, 0)\)、\((3, 2)\) 和 \((5, 0)\)，底为 \(4\)，高为 \(2\)，所以面积为 \(\tfrac12 \cdot 4 \cdot 2 = 4\)。

阴影面积为 \(10 - 4 = 6\)。

所以正确答案是 B。

English

The outer triangle has vertices \((1, 0),\) \((3, 5),\) and \((5, 0),\) giving base \(4\) and height \(5,\) so its area is \(\tfrac12 \cdot 4 \cdot 5 = 10.\)

Removed from it is the triangle with vertices \((1, 0),\) \((3, 2),\) and \((5, 0),\) which has base \(4\) and height \(2,\) so area \(\tfrac12 \cdot 4 \cdot 2 = 4.\)

The shaded area is \(10 - 4 = 6.\)

Thus, the correct answer is B.

中午温差为 \(N\)。之后 Minneapolis 的温度下降 \(5\) 度，St. Louis 的温度上升 \(3\) 度，所以温差改变了 \(8\) 度。新的绝对温差为 \(|N - 8| = 2\)。

由此得到 \(N = 10\) 或 \(N = 6\)，乘积为 \(60\)。

所以正确答案是 C。

English

At noon the gap is \(N.\) Minneapolis then loses \(5\) degrees and St. Louis gains \(3,\) so the gap changes by \(8.\) The new absolute difference is \(|N - 8| = 2.\)

This gives \(N = 10\) or \(N = 6,\) whose product is \(60.\)

Thus, the correct answer is C.

写成二的幂，有 \(n = 8^{2022} = 2^{6066}\)。于是 \[\dfrac{n}{4} = \dfrac{2^{6066}}{2^{2}} = 2^{6064}\text{。}\]

又因为 \(2^{6064} = 4^{3032}\)，所以它对应最后一个选项。

所以正确答案是 E。

English

Write \(n = 8^{2022} = 2^{6066}.\) Then \[\dfrac{n}{4} = \dfrac{2^{6066}}{2^{2}} = 2^{6064}.\]

Since \(2^{6064} = 4^{3032},\) this matches the last option.

Thus, the correct answer is E.

对 \(1\le a\le14\) 逐一列出 \(\frac{a}{15-a}\)：其中的整数是 \(2,4,14\)；小数部分为 \(\frac{1}{2}\) 的分数是 \(\tfrac12,\tfrac32,\tfrac{13}{2}\)；而 \(\tfrac14,\tfrac{11}{4}\) 的小数部分互补，分别为 \(\frac{1}{4},\frac{3}{4}\)。

其余的小数部分是 \(\frac{1}{14},\frac{2}{13},\frac{4}{11},\frac{2}{3},\frac{7}{8},\frac{1}{7}\)，而它们各自的补数都没有出现。因此只有整数、半整数和四分之一这三类才可能凑出整数和。

整数对给出 \(4, 6, 8, 16, 18, 28\)。半整数对 \(\left(\tfrac12, \tfrac32, \tfrac{13}{2}\right)\) 给出 \(1, 2, 3, 7, 8, 13\)。四分之一型的一对 \(\tfrac14 + \tfrac{11}{4}\) 给出 \(3\)。

去重后得到 \(1, 2, 3, 4, 6, 7, 8, 13, 16, 18, 28\) 共 \(11\) 个不同整数。

所以正确答案是 C。

English

Listing \(\frac{a}{15-a}\) for \(1\le a\le14,\) the integers are \(2,4,14;\) the fractions with fractional part \(\frac{1}{2}\) are \(\tfrac12,\tfrac32,\tfrac{13}{2};\) and \(\tfrac14,\tfrac{11}{4}\) have complementary fractional parts \(\frac{1}{4},\frac{3}{4}.\)

The remaining fractional parts are \(\frac{1}{14},\frac{2}{13},\frac{4}{11},\frac{2}{3},\frac{7}{8},\frac{1}{7},\) and none of their complements occurs. Thus only the integer, half-integer, and quarter cases can give integer sums.

Integer pairs give \(4, 6, 8, 16, 18, 28.\) Half-integer pairs \(\left(\tfrac12, \tfrac32, \tfrac{13}{2}\right)\) give \(1, 2, 3, 7, 8, 13.\) The quarter pair \(\tfrac14 + \tfrac{11}{4}\) gives \(3.\)

The distinct integers are \(1, 2, 3, 4, 6, 7, 8, 13, 16, 18, 28,\) a total of \(11.\)

Thus, the correct answer is C.

因为 \(16{,}383 = 2^{14} - 1\) \(= (2^{7} - 1)(2^{7} + 1)\) \(= 127 \cdot 129\)，而 \(129 = 3 \cdot 43\)，所以 \(16{,}383 = 3 \cdot 43 \cdot 127\)。

最大质因数是 \(127\)，其各位数字之和为 \(1 + 2 + 7 = 10\)。

所以正确答案是 C。

English

We have \(16{,}383 = 2^{14} - 1\) \(= (2^{7} - 1)(2^{7} + 1)\) \(= 127 \cdot 129.\) Then \(129 = 3 \cdot 43,\) so \(16{,}383 = 3 \cdot 43 \cdot 127.\)

The greatest prime factor is \(127,\) whose digit sum is \(1 + 2 + 7 = 10.\)

Thus, the correct answer is C.

有恒等式 \[ \small \begin{aligned} &2\bigl[x(x-y) + y(y-z) + z(z-x)\bigr] \\ &= (x-y)^2 + (y-z)^2 + (z-x)^2\text{。} \end{aligned} \] 所以原方程成立恰好当这个平方和等于 \(2\)。

由于三个差的和为 \(0\)，这要求其中两个为 \(\pm 1\)，一个为 \(0\)。

选项 D 给出 \(z - x = 0\)、\(x - y = -1\)、\(y - z = 1\)，平方和为 \(1 + 1 + 0 = 2\)。因此对所有满足该条件的整数都成立。

所以正确答案是 D。

English

The expression satisfies \[ \small \begin{aligned} &2\bigl[x(x-y) + y(y-z) + z(z-x)\bigr] \\ &= (x-y)^2 + (y-z)^2 + (z-x)^2. \end{aligned} \] So the equation holds exactly when this sum of squares equals \(2.\)

Since the three differences sum to \(0,\) this requires two of them to be \(\pm 1\) and one to be \(0.\)

Option D gives \(z - x = 0,\) \(x - y = -1,\) and \(y - z = 1,\) so the squares are \(1 + 1 + 0 = 2.\) This works for all such integers.

Thus, the correct answer is D.

设两条相等边长为 \(s\)，底边为 \(b\)，到底边的高为 \(h\)。题设条件为 \(s^2 = 2bh\)。

三角形面积既等于 \(\tfrac12 bh\)，也等于 \(\tfrac12 s^2 \sin\theta\)，其中 \(\theta\) 是顶角。因此 \(bh = s^2 \sin\theta\)。

代入 \(s^2 = 2bh\)，得到 \(bh = 2bh\sin\theta\)，所以 \(\sin\theta = \tfrac12\)。由于三角形为钝角，\(\theta = 150^\circ\)。

所以正确答案是 D。

English

Let the congruent sides have length \(s,\) the base be \(b,\) and the height to the base be \(h.\) The given condition is \(s^2 = 2bh.\)

The area equals \(\tfrac12 bh\) and also \(\tfrac12 s^2 \sin\theta,\) where \(\theta\) is the vertex angle. So \(bh = s^2 \sin\theta.\)

Substituting \(s^2 = 2bh\) gives \(bh = 2bh\sin\theta,\) so \(\sin\theta = \tfrac12.\) Since the triangle is obtuse, \(\theta = 150^\circ.\)

Thus, the correct answer is D.

对于等边三角形，\(O\) 也是外心，所以 \(OA = OC = \dfrac{6}{\sqrt3} = 2\sqrt3\)。中心角 \(\angle AOC = 120^\circ\)。

在三角形 \(AOC\) 中，边 \(AC = 6\) 所对的角为 \(120^\circ\)，因此该三角形的外接圆半径 \(R'\) 满足 \[2R' = \dfrac{6}{\sin 120^\circ} = 4\sqrt3\text{，}\] 从而 \(R' = 2\sqrt3\)。

该圆的面积为 \(\pi (2\sqrt3)^2 = 12\pi\)。

所以正确答案是 B。

English

For an equilateral triangle, \(O\) is also the circumcenter, so \(OA = OC = \dfrac{6}{\sqrt3} = 2\sqrt3.\) The central angle \(\angle AOC = 120^\circ.\)

In triangle \(AOC,\) side \(AC = 6\) is opposite the \(120^\circ\) angle, so the circumradius \(R'\) of this triangle satisfies \[2R' = \dfrac{6}{\sin 120^\circ} = 4\sqrt3,\] giving \(R' = 2\sqrt3.\)

The area of the circle is \(\pi (2\sqrt3)^2 = 12\pi.\)

Thus, the correct answer is B.

三个点位于单位圆上，对应角分别为 \(40^\circ\)，\(60^\circ\)，\(t^\circ\)。

若第三点到其余两点的距离相等，它必在这两点连线的垂直平分线上，故 \(t = 50\) 或 \(t = 230\)。

若它到 \(40^\circ\) 所对点的距离等于两个已知点间的弦长（角度差为 \(20^\circ\)），则 \(t = 20\)（\(t = 60\) 时三角形退化）。同理，若它到 \(60^\circ\) 所对点的距离相等，则 \(t = 80\)（\(t = 40\) 时退化）。

有效值为 \(50, 230, 20, 80\)，总和为 \(380\)。

所以正确答案是 E。

English

The three points lie on the unit circle at angles \(40^\circ,\) \(60^\circ,\) and \(t^\circ.\) A chord’s length depends only on the angular separation of its endpoints.

If the third point is equidistant from the other two, it lies on the perpendicular bisector: \(t = 50\) or \(t = 230.\)

If its distance to \(40^\circ\) equals the fixed chord (separation \(20^\circ\)), then \(t = 20\) (since \(t = 60\) is degenerate). If its distance to \(60^\circ\) matches, then \(t = 80\) (since \(t = 40\) is degenerate).

The valid values are \(50, 230, 20, 80,\) summing to \(380.\)

Thus, the correct answer is E.

当乘积中至多含一个因子 \(2\) 时，它不能被 \(4\) 整除。每个骰子掷出奇数的概率为 \(\tfrac12\)；恰好贡献一个因子 \(2\)（即掷出 \(2\) 或 \(6\)）的概率为 \(\tfrac13\)；贡献两个因子（即掷出 \(4\)）的概率为 \(\tfrac16\)。

六个骰子全为奇数：\(\left(\tfrac12\right)^6 = \tfrac{1}{64}\)。恰有一个骰子为 \(2\) 或 \(6\)，其余为奇数：\(6 \cdot \tfrac13 \cdot \left(\tfrac12\right)^5 = \tfrac{1}{16} = \tfrac{4}{64}\)。

不能被四整除的概率为 \(\tfrac{1}{64} + \tfrac{4}{64} = \tfrac{5}{64}\)，因此所求概率为 \(1 - \tfrac{5}{64} = \tfrac{59}{64}\)。

所以正确答案是 C。

English

The product fails to be divisible by \(4\) when it has at most one factor of \(2.\) Each die is odd with probability \(\tfrac12,\) contributes exactly one factor of \(2\) (a \(2\) or \(6\)) with probability \(\tfrac13,\) and two factors (a \(4\)) with probability \(\tfrac16.\)

All six odd: \(\left(\tfrac12\right)^6 = \tfrac{1}{64}.\) Exactly one die a \(2\) or \(6\) and the rest odd: \(6 \cdot \tfrac13 \cdot \left(\tfrac12\right)^5 = \tfrac{1}{16} = \tfrac{4}{64}.\)

The complement is \(\tfrac{1}{64} + \tfrac{4}{64} = \tfrac{5}{64},\) so the answer is \(1 - \tfrac{5}{64} = \tfrac{59}{64}.\)

Thus, the correct answer is C.

因为 \(768 = 2^8 \cdot 3\)，它的因数和是 \((2^9 - 1)(1 + 3)\) \(= 511 \cdot 4 = 2044\)，所以 \(f(768) = \dfrac{2044}{768} = \dfrac{511}{192}\)。

因为 \(384 = 2^7 \cdot 3\)，它的因数和是 \((2^8 - 1)(1 + 3)\) \(= 255 \cdot 4 = 1020\)，所以 \(f(384) = \dfrac{1020}{384} = \dfrac{510}{192}\)。

差为 \(\dfrac{511 - 510}{192} = \dfrac{1}{192}\)。

所以正确答案是 B。

English

Since \(768 = 2^8 \cdot 3,\) its divisor sum is \((2^9 - 1)(1 + 3)\) \(= 511 \cdot 4 = 2044,\) so \(f(768) = \dfrac{2044}{768} = \dfrac{511}{192}.\)

Since \(384 = 2^7 \cdot 3,\) its divisor sum is \((2^8 - 1)(1 + 3)\) \(= 255 \cdot 4 = 1020,\) so \(f(384) = \dfrac{1020}{384} = \dfrac{510}{192}.\)

The difference is \(\dfrac{511 - 510}{192} = \dfrac{1}{192}.\)

Thus, the correct answer is B.

每个角都可写成 \(kc = \dfrac{2\pi k}{11}\)。按 \(2\pi\) 取模后，\(\sin 12c = \sin c\)，且 \(\sin 15c = \sin 4c\)。

因此分子为 \(\sin 3c \cdot \sin 6c \cdot \sin 9c\) \(\cdot \sin c \cdot \sin 4c\)。约去公共因子 \(\sin c\)、\(\sin 3c\)、\(\sin 4c\)，剩下 \(\dfrac{\sin 6c \cdot \sin 9c}{\sin 2c \cdot \sin 5c}\)。

又有 \(\sin 9c = \sin\left(2\pi - 2c\right) = -\sin 2c\)，并且 \(\sin 6c = \sin\left(2\pi - 5c\right) = -\sin 5c\)，所以比值为 \(\dfrac{(-\sin 5c)(-\sin 2c)}{\sin 2c \cdot \sin 5c} = 1\)。

所以正确答案是 E。

English

Write each angle as \(kc = \dfrac{2\pi k}{11}.\) Reducing modulo \(2\pi,\) \(\sin 12c = \sin c\) and \(\sin 15c = \sin 4c.\)

So the numerator is \(\sin 3c \cdot \sin 6c \cdot \sin 9c\) \(\cdot \sin c \cdot \sin 4c.\) Cancelling the common factors \(\sin c,\) \(\sin 3c,\) \(\sin 4c\) leaves \(\dfrac{\sin 6c \cdot \sin 9c}{\sin 2c \cdot \sin 5c}.\)

Now \(\sin 9c = \sin\left(2\pi - 2c\right) = -\sin 2c\) and \(\sin 6c = \sin\left(2\pi - 5c\right) = -\sin 5c,\) so the ratio equals \(\dfrac{(-\sin 5c)(-\sin 2c)}{\sin 2c \cdot \sin 5c} = 1.\)

Thus, the correct answer is E.

令 \(D(z) = P(z)Q(z) - R(z)\)。因为 \(P Q\) 的次数为 \(5\)，而 \(R\) 的次数为 \(6\)，所以 \(D\) 的次数为 \(6\)。它的常数项为 \(1 \cdot 2 - 3 = -1 \neq 0\)。

由于 \(R\) 除这些限制外没有其他限制，\(D\) 可以成为任意次数为 \(6\)、常数项为 \(-1\) 的实系数多项式，例如 \(-(z - 1)^6\)。

这样的多项式只有一个不同的根，所以最小值为 \(N = 1\)。

所以正确答案是 B。

English

Let \(D(z) = P(z)Q(z) - R(z).\) Since \(P Q\) has degree \(5\) and \(R\) has degree \(6,\) the degree of \(D\) is \(6.\) Its constant term is \(1 \cdot 2 - 3 = -1 \neq 0.\)

Because \(R\) is otherwise unconstrained, \(D\) can be made equal to any real degree-\(6\) polynomial with constant term \(-1,\) for instance \(-(z - 1)^6.\)

Such a polynomial has a single distinct root, so the minimum is \(N = 1.\)

Thus, the correct answer is B.

因为三个正方形分别旋转 \(0^\circ\)，\(30^\circ\)，\(60^\circ\)，图形有 \(12\) 重对称。它的 \(24\) 个顶点每隔 \(15^\circ\) 交替出现：外侧顶点到中心的距离为 \(3\sqrt2\)，内侧交点到中心的距离为 \(2\sqrt3\)。

连接中心与全部 \(24\) 个顶点，把多边形分成 \(24\) 个三角形，每个三角形的两边为 \(3\sqrt2\) 和 \(2\sqrt3\)，夹角为 \(15^\circ\)。总面积是 \[ \begin{aligned} &24 \cdot \tfrac12 (3\sqrt2)(2\sqrt3)\sin 15^\circ \\ &= 72\sqrt6 \cdot \dfrac{\sqrt6 - \sqrt2}{4} \end{aligned}\text{。}\]

这化简为 \(18\sqrt6(\sqrt6 - \sqrt2) = 108 - 36\sqrt3\)，所以 \(a + b + c = 108 + 36 + 3 = 147\)。

所以正确答案是 E。

English

Because the three squares are rotated by \(0^\circ,\) \(30^\circ,\) and \(60^\circ,\) the figure has \(12\)-fold symmetry. Its \(24\) vertices alternate every \(15^\circ\): outer vertices are the square corners at distance \(3\sqrt2\) from the center, and inner vertices are edge crossings at distance \(2\sqrt3.\)

Connecting the center to all \(24\) vertices splits the polygon into \(24\) triangles, each with sides \(3\sqrt2\) and \(2\sqrt3\) and included angle \(15^\circ.\) The total area is \[ \begin{aligned} &24 \cdot \tfrac12 (3\sqrt2)(2\sqrt3)\sin 15^\circ \\ &= 72\sqrt6 \cdot \dfrac{\sqrt6 - \sqrt2}{4}. \end{aligned} \]

This simplifies to \(18\sqrt6(\sqrt6 - \sqrt2) = 108 - 36\sqrt3,\) so \(a + b + c = 108 + 36 + 3 = 147.\)

Thus, the correct answer is E.

条件是对称的，所以不妨设 \(a\le b\le c\)。于是 \(a\le7\)，且对每个固定的 \(a\)，有 \(a\le b\le\lfloor\frac{23-a}{2}\rfloor\) 以及 \(c=23-a-b\)。最大公因数之和可以写成\[ \begin{aligned} &\gcd(a,b)+\gcd(b,23-a) \\ &\quad {}+\gcd(a,23-b) \end{aligned}\text{。}\]

逐一检查这些不长的范围，\(a=1,2,\ldots,7\) 时可能的最大公因数之和分别是 \(\{3,4,13\}\)、\(\{4,6,10\}\)、\(\{3,5,6,8,9,12\}\)、\(\{4,6\}\)、\(\{3,7,8,11\}\)、\(\{4,6,8\}\) 和 \(\{9,10\}\)。取值 \(9\) 只在 \((a,b,c)=(3,5,15)\) 和 \((7,7,9)\) 处出现。

三元组 \((7, 7, 9)\) 的 \(\gcd\) 之和为 \(7 + 1 + 1 = 9\)，并且 \(a^2 + b^2 + c^2 = 49 + 49 + 81\) \(= 179\)。三元组 \((3, 5, 15)\) 的 \(\gcd\) 之和为 \(1 + 5 + 3 = 9\)，并且 \(a^2 + b^2 + c^2 = 9 + 25\) \(+ 225 = 259\)。

不同值之和为 \(179 + 259 = 438\)。

所以正确答案是 B。

English

The condition is symmetric, so assume \(a\le b\le c.\) Then \(a\le7,\) and for each fixed \(a\) we have \(a\le b\le\lfloor\frac{23-a}{2}\rfloor\) and \(c=23-a-b.\) The gcd sum can be evaluated as \[ \begin{aligned} &\gcd(a,b)+\gcd(b,23-a) \\ &\quad {}+\gcd(a,23-b). \end{aligned} \]

Checking these short ranges, the possible gcd sums for \(a=1,2,\ldots,7\) are, respectively, \(\{3,4,13\},\) \(\{4,6,10\},\) \(\{3,5,6,8,9,12\},\) \(\{4,6\},\) \(\{3,7,8,11\},\) \(\{4,6,8\},\) and \(\{9,10\}.\) The value \(9\) occurs only at \((a,b,c)=(3,5,15)\) and \((7,7,9).\)

The triple \((7, 7, 9)\) has \(\gcd\) sum \(7 + 1 + 1 = 9\) and \(a^2 + b^2 + c^2 = 49 + 49 + 81\) \(= 179.\) The triple \((3, 5, 15)\) has \(\gcd\) sum \(1 + 5 + 3 = 9\) and \(a^2 + b^2 + c^2 = 9 + 25\) \(+ 225 = 259.\)

The sum of the distinct values is \(179 + 259 = 438.\)

Thus, the correct answer is B.

保持在距离 \(1\) 以内意味着虫子始终在原点或其 \(6\) 个相邻顶点上。从原点出发，所有 \(6\) 个方向都可行；从相邻顶点出发，只有 \(3\) 个方向可行：回到原点，或到两个相邻的邻点之一。

令 \(a_k\) 和 \(b_k\) 分别表示有效的 \(k\) 步路径中，结束在原点和结束在某个相邻点的路径数。则 \(a_{k+1} = b_k\)、\(b_{k+1} = 6a_k + 2b_k\)，初值为 \(a_0 = 1\)、\(b_0 = 0\)。

迭代得 \(b_1 = 6\)，\((a_2, b_2) = (6, 12)\)、\((a_3, b_3) = (12, 60)\)、\((a_4, b_4) = (60, 192)\)、\((a_5, b_5) = (192, 744)\)。有效路径总数为 \(192 + 744 = 936\)。

所求概率为 \(\dfrac{936}{6^5} = \dfrac{936}{7776} = \dfrac{13}{108}\)。

所以正确答案是 A。

English

Staying within distance \(1\) means the bug is always at the origin or one of its \(6\) neighbors. From the origin, all \(6\) moves are allowed. From a neighbor, only \(3\) moves keep it in range: back to the origin, or to either of the two adjacent neighbors.

Let \(a_k\) and \(b_k\) count valid \(k\)-step paths ending at the origin and at a neighbor. Then \(a_{k+1} = b_k\) and \(b_{k+1} = 6a_k + 2b_k,\) starting from \(a_0 = 1,\) \(b_0 = 0.\)

Iterating gives \(b_1 = 6,\) then \((a_2, b_2) = (6, 12),\) \((a_3, b_3) = (12, 60),\) \((a_4, b_4) = (60, 192),\) \((a_5, b_5) = (192, 744).\) The total number of valid paths is \(192 + 744 = 936.\)

The probability is \(\dfrac{936}{6^5} = \dfrac{936}{7776} = \dfrac{13}{108}.\)

Thus, the correct answer is A.

极限满足 \(L = 2L - 2L^2\)，得 \(L = \tfrac12\)。令 \(v_k = 1 - 2u_k\)。于是 \[ \begin{aligned} v_{k+1} &= 1 - 2u_{k+1} \\ &= 1 - 4u_k + 4u_k^2 \\ &= (1 - 2u_k)^2 = v_k^2 \end{aligned}\text{。}\]

由 \(v_0 = 1 - 2 \cdot \tfrac14 = \tfrac12\)，可得 \(v_k = \left(\tfrac12\right)^{2^k}\)，所以 \(|u_k - L| = \dfrac{|v_k|}{2} = 2^{-2^k - 1}\)。

需要 \(2^k + 1 \ge 1000\)，也就是 \(2^k \ge 999\)。满足条件的最小 \(k\) 是 \(10\)，因为 \(2^{10} = 1024\)。

所以正确答案是 A。

English

The limit satisfies \(L = 2L - 2L^2,\) giving \(L = \tfrac12.\) Let \(v_k = 1 - 2u_k.\) Then \[ \begin{aligned} v_{k+1} &= 1 - 2u_{k+1} \\ &= 1 - 4u_k + 4u_k^2 \\ &= (1 - 2u_k)^2 = v_k^2. \end{aligned} \]

Since \(v_0 = 1 - 2 \cdot \tfrac14 = \tfrac12,\) we get \(v_k = \left(\tfrac12\right)^{2^k},\) so \(|u_k - L| = \dfrac{|v_k|}{2} = 2^{-2^k - 1}.\)

We need \(2^k + 1 \ge 1000,\) i.e. \(2^k \ge 999.\) The least such \(k\) is \(10,\) since \(2^{10} = 1024.\)

Thus, the correct answer is A.

对同圆内接且没有共顶点的两个凸多边形，较小多边形的每条边会与较大多边形的边界相交两次，因此交点数为 \(2\min(m, n)\)。

对所有多边形对求和：\((5,6), (5,7), (5,8)\) 各给出 \(10\) 个交点；\((6,7), (6,8)\) 各给出 \(12\) 个；\((7,8)\) 给出 \(14\) 个。

总数为 \(3 \cdot 10 + 2 \cdot 12 + 14 = 68\)。

所以正确答案是 E。

English

For two convex polygons inscribed in the same circle with no shared vertices, each side of the smaller polygon crosses the larger polygon’s boundary exactly twice, so they meet at \(2\min(m, n)\) points.

Summing over all pairs: \((5,6), (5,7), (5,8)\) give \(10\) each; \((6,7), (6,8)\) give \(12\) each; \((7,8)\) gives \(14.\)

The total is \(3 \cdot 10 + 2 \cdot 12 + 14 = 68.\)

Thus, the correct answer is E.

由 Burnside 引理，答案是 \(24\) 个旋转作用于 \(8\) 个小立方体位置时，每个旋转固定的“恰有 \(4\) 个蓝色”的构造数的平均值。

恒等旋转固定 \(\binom{8}{4} = 70\) 种。\(6\) 个面四分之一转各固定 \(2\) 种，共 \((12)\) 种；\(3\) 个面半转各固定 \(6\) 种，共 \((18)\) 种；\(8\) 个顶点旋转各固定 \(4\) 种，共 \((32)\) 种；\(6\) 个棱半转各固定 \(6\) 种，共 \((36)\) 种。

固定构造数之和为 \(70 + 12 + 18 + 32 + 36 = 168\)，平均数为 \(\dfrac{168}{24} = 7\)。

所以正确答案是 A。

English

By Burnside’s lemma, the count is the average number of \(4\)-blue colorings fixed by each of the \(24\) rotations acting on the \(8\) cubies.

The identity fixes \(\binom{8}{4} = 70.\) The \(6\) face quarter-turns fix \(2\) each \((12).\) The \(3\) face half-turns fix \(6\) each \((18).\) The \(8\) vertex rotations fix \(4\) each \((32).\) The \(6\) edge half-turns fix \(6\) each \((36).\)

The total is \(70 + 12 + 18 + 32 + 36 = 168,\) and \(\dfrac{168}{24} = 7.\)

Thus, the correct answer is A.

由 Euler 公式，\(P(x) = 1 + e^{ix} - e^{2ix} + e^{3ix}\)。其虚部为 \[ \begin{gathered} \sin x - \sin 2x + \sin 3x \\ = (\sin x + \sin 3x) - \sin 2x \\ = \sin 2x(2\cos x - 1) \end{gathered}\text{。}\]

它在 \(\sin 2x = 0\) 时为零，此时 \(x = 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}\)；或者在 \(\cos x = \tfrac12\) 时为零，此时 \(x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}\)。

逐一检查实部 \(1 + \cos x - \cos 2x + \cos 3x\)，在这些值处结果为 \(\pm 2\) 或 \(1\)，从不为 \(0\)。所以不存在 \(x\) 使 \(P(x) = 0\)。

所以正确答案是 A。

English

Group by Euler’s formula: \(P(x) = 1 + e^{ix} - e^{2ix} + e^{3ix}.\) The imaginary part is \[ \begin{gathered} \sin x - \sin 2x + \sin 3x \\ = (\sin x + \sin 3x) - \sin 2x \\ = \sin 2x(2\cos x - 1). \end{gathered} \]

This vanishes when \(\sin 2x = 0\) (so \(x = 0, \tfrac{\pi}{2}, \pi, \tfrac{3\pi}{2}\)) or \(\cos x = \tfrac12\) (so \(x = \tfrac{\pi}{3}, \tfrac{5\pi}{3}\)).

Checking the real part \(1 + \cos x - \cos 2x + \cos 3x\) at each of these values gives \(\pm 2\) or \(1,\) never \(0.\) So no \(x\) makes \(P(x) = 0.\)

Thus, the correct answer is A.

令 \(C = (0, 0)\)，\(B = (6, 0)\)，\(A = (0, 8)\)，于是直角在 \(C\)。

圆 \(O\) 与直线 \(BC\)（\(x\)-轴）在 \(B\) 点相切，所以 \(O = (6, k)\)。令 \(OA = OB\)，得 \(36 + (k - 8)^2 = k^2\)，所以 \(k = \tfrac{25}{4}\)，即 \(O = \left(6, \tfrac{25}{4}\right)\)。

圆 \(P\) 与直线 \(AC\)（\(y\)-轴）在 \(A\) 点相切，所以 \(P = (h, 8)\)。由 \(PB = PA\) 得 \((h - 6)^2 + 64 = h^2\)，所以 \(h = \tfrac{25}{3}\)，且 \(P = \left(\tfrac{25}{3}, 8\right)\)。

因此 \(OP = \sqrt{\left(\tfrac73\right)^2 + \left(\tfrac74\right)^2}\) \(= 7\sqrt{\tfrac{25}{144}}\) \(= \dfrac{35}{12}\)。

所以正确答案是 C。

English

Place \(C = (0, 0),\) \(B = (6, 0),\) and \(A = (0, 8),\) so the right angle is at \(C.\)

Circle \(O\) is tangent to line \(BC\) (the \(x\)-axis) at \(B,\) so \(O = (6, k).\) Setting \(OA = OB\) gives \(36 + (k - 8)^2 = k^2,\) so \(k = \tfrac{25}{4}\) and \(O = \left(6, \tfrac{25}{4}\right).\)

Circle \(P\) is tangent to line \(AC\) (the \(y\)-axis) at \(A,\) so \(P = (h, 8).\) Setting \(PB = PA\) gives \((h - 6)^2 + 64 = h^2,\) so \(h = \tfrac{25}{3}\) and \(P = \left(\tfrac{25}{3}, 8\right).\)

Then \(OP = \sqrt{\left(\tfrac73\right)^2 + \left(\tfrac74\right)^2}\) \(= 7\sqrt{\tfrac{25}{144}}\) \(= \dfrac{35}{12}.\)

Thus, the correct answer is C.

对 \(29\) 个相邻对中的每一对 \((i, i+1)\)，设指标变量在两数都被选中时为 \(1\)。这个事件的概率为 \(\dfrac{5}{30} \cdot \dfrac{4}{29} = \dfrac{2}{87}\)。

由期望的线性性，相邻整数对的期望数为 \(29 \cdot \dfrac{2}{87} = \dfrac{2}{3}\)。

所以正确答案是 A。

English

For each of the \(29\) adjacent pairs \((i, i+1),\) let an indicator be \(1\) if both are in the subset. The probability of this is \(\dfrac{5}{30} \cdot \dfrac{4}{29} = \dfrac{2}{87}.\)

By linearity of expectation, the expected number of consecutive pairs is \(29 \cdot \dfrac{2}{87} = \dfrac{2}{3}.\)

Thus, the correct answer is A.

点 \(A\)，\(D\)，\(E\) 共线于角平分线上，点 \(A\)，\(B\)，\(F\) 共线于直线 \(AB\)。对经过 \(B\)，\(E\)，\(D\) 的圆，点 \(A\) 的幂给出 \(AB \cdot AF = AD \cdot AE\)。

因为 \(\angle BAE = \angle DAC\) 且 \(\angle AEB = \angle ACB\)（同弦 \(AB\) 所对的角），三角形 \(ABE\) 与 \(ADC\) 相似，所以 \(AD \cdot AE = AB \cdot AC\)。因此 \(AF = AC = 20\)。

取 \(A = (0, 0)\)、\(B = (11, 0)\)。由 \(CA = 20\)、\(CB = 24\)，可得 \(C = \left(-\tfrac52, \tfrac{\sqrt{1575}}{2}\right)\)。点 \(F\) 在射线 \(AB\) 上，且 \(AF = 20\)，所以 \(F = (20, 0)\)。

因此 \(CF^2 = \left(20 + \tfrac52\right)^2 + \tfrac{1575}{4}\) \(= \tfrac{2025}{4} + \tfrac{1575}{4}\) \(= 900\)，所以 \(CF = 30\)。

所以正确答案是 C。

English

Points \(A,\) \(D,\) \(E\) are collinear on the bisector, and \(A,\) \(B,\) \(F\) are collinear on line \(AB.\) The power of \(A\) with respect to the circle through \(B,\) \(E,\) \(D\) gives \(AB \cdot AF = AD \cdot AE.\)

Since \(\angle BAE = \angle DAC\) and \(\angle AEB = \angle ACB\) (subtending \(AB\)), triangles \(ABE\) and \(ADC\) are similar, so \(AD \cdot AE = AB \cdot AC.\) Therefore \(AF = AC = 20.\)

Place \(A = (0, 0),\) \(B = (11, 0).\) From \(CA = 20,\) \(CB = 24,\) point \(C = \left(-\tfrac52, \tfrac{\sqrt{1575}}{2}\right).\) Point \(F\) lies on ray \(AB\) with \(AF = 20,\) so \(F = (20, 0).\)

Then \(CF^2 = \left(20 + \tfrac52\right)^2 + \tfrac{1575}{4}\) \(= \tfrac{2025}{4} + \tfrac{1575}{4}\) \(= 900,\) so \(CF = 30.\)

Thus, the correct answer is C.

从 \(n\) 变为 \(n + 1\) 时，每个余数 \(n \bmod m\) 都增加 \(1\)，除非 \(n + 1\) 是 \(m\) 的倍数，此时它从 \(m - 1\) 降为 \(0\)。所以 \[ \begin{aligned} &R(n+1) - R(n) \\ &= 9 - \sum_{\substack{2 \le m \le 10 \\ m \mid n+1}} m \end{aligned}\text{。}\]

从 \(2\) 到 \(10\) 的互异整数之和等于 \(9\) 的方式只有 \(\{9\},\{2,7\},\{3,6\},\{4,5\}\)，或 \(\{2,3,4\}\)。因数的闭包性质排除了除 \(\{2,7\}\) 外的所有情形：例如，\(9\) 的倍数也有因数 \(3\)，而 \(6\) 的倍数也有因数 \(2\) 和 \(3\)。因此 \(n+1\) 必须是 \(14\) 的倍数，并且在 \(\{2,\ldots,10\}\) 中没有其他因数。

检查所需范围内的倍数 \(14,28,42,56,70,84,98\)，只留下 \(14\) 和 \(98\)。所以 \(n=13\) 或 \(n=97\)，共有 \(2\) 个值。

因此，正确答案是 C。

English

Going from \(n\) to \(n + 1,\) each remainder \(n \bmod m\) increases by \(1\) unless \(n + 1\) is divisible by \(m,\) in which case it drops from \(m - 1\) to \(0.\) So \[ \begin{aligned} &R(n+1) - R(n) \\ &= 9 - \sum_{\substack{2 \le m \le 10 \\ m \mid n+1}} m. \end{aligned} \]

Distinct integers from \(2\) through \(10\) can sum to \(9\) only as \(\{9\},\{2,7\},\{3,6\},\{4,5\},\) or \(\{2,3,4\}.\) Divisibility closure eliminates every case except \(\{2,7\}:\) for example, a multiple of \(9\) also has divisor \(3,\) and a multiple of \(6\) also has divisors \(2\) and \(3.\) Thus \(n+1\) must be a multiple of \(14\) with no other divisor in \(\{2,\ldots,10\}.\)

Testing the multiples \(14,28,42,56,70,84,98\) in the required range leaves only \(14\) and \(98.\) Hence \(n=13\) or \(n=97,\) giving \(2\) values.

Thus, the correct answer is C.