2021C AMC 12 真题（P版）

因为 \(2112 - 2021 = 91 = 7 \cdot 13\)，且 \(169 = 13^2\)，所以这个分数为 \(\dfrac{(7 \cdot 13)^2}{13^2} = 7^2 = 49\)。

所以正确答案是 C。

English

Since \(2112 - 2021 = 91 = 7 \cdot 13\) and \(169 = 13^2,\) the fraction is \(\dfrac{(7 \cdot 13)^2}{13^2} = 7^2 = 49.\)

Thus, the correct answer is C.

原卡片是 \(4 \times 6\)。把一边缩短 \(1\) 英寸后面积为 \(18\)，必须是 \(3 \times 6 = 18\)，所以被缩短的是 \(4\) 英寸的边。

若改为把另一边缩短 \(1\) 英寸，则面积为 \(4 \times 5 = 20\) 平方英寸。

所以正确答案是 E。

English

The original card is \(4 \times 6.\) Shortening a side by \(1\) inch gives area \(18,\) which requires \(3 \times 6 = 18,\) so the reduced side was the \(4\)-inch side.

Shortening the other side by \(1\) inch instead gives \(4 \times 5 = 20\) square inches.

Thus, the correct answer is E.

路线 A 用时 \(\dfrac{6}{30} = \dfrac{1}{5}\) 小时 \(= 12\) 分钟。

路线 B 中 \(4.5\) 英里按每小时 \(40\) 英里行驶，\(0.5\) 英里按每小时 \(20\) 英里行驶，用时 \(\dfrac{4.5}{40} + \dfrac{0.5}{20}\) \(= 0.1125 + 0.025\) \(= 0.1375\) 小时 \(= 8.25\) 分钟。

差值为 \(12 - 8.25 = 3.75 = 3\tfrac{3}{4}\) 分钟。

所以正确答案是 B。

English

Route A takes \(\dfrac{6}{30} = \dfrac{1}{5}\) hour \(= 12\) minutes.

Route B has \(4.5\) miles at \(40\) mph and \(0.5\) mile at \(20\) mph, taking \(\dfrac{4.5}{40} + \dfrac{0.5}{20}\) \(= 0.1125 + 0.025\) \(= 0.1375\) hour \(= 8.25\) minutes.

The difference is \(12 - 8.25 = 3.75 = 3\tfrac{3}{4}\) minutes.

Thus, the correct answer is B.

这个数是 \(202100 + A\)。任何偶数 \(A\) 都会使它为偶数，而 \(A = 5\) 时它能被 \(5\) 整除，所以 \(A\) 必须是奇数且不是 \(5\)。

\(A = 1\) 时数字和为 \(6\)（能被 \(3\) 整除）；\(A = 7\) 时数字和为 \(12\) （能被 \(3\) 整除）；并且 \(202103 = 11 \cdot 18373\)。只有 \(202109\) 通过这些检验，而它是质数。

所以正确答案是 E。

English

The number is \(202100 + A.\) Any even \(A\) makes it even, and \(A = 5\) makes it divisible by \(5,\) so \(A\) must be odd and not \(5.\)

For \(A = 1\) the digit sum is \(6\) (divisible by \(3\)); for \(A = 7\) the digit sum is \(12\) (divisible by \(3\)); and \(202103 = 11 \cdot 18373.\) Only \(202109\) survives all tests, and it is prime.

Thus, the correct answer is E.

第一根和第 \(41\) 根电线杆之间有 \(40\) 个间隔，所以每个间隔长 \(\dfrac{5280}{40} = 132\) 英尺。

Elmer 的步幅为 \(\dfrac{132}{44} = 3\) 英尺，Oscar 的跃步为 \(\dfrac{132}{12} = 11\) 英尺，差为 \(11 - 3 = 8\) 英尺。

所以正确答案是 B。

English

There are \(40\) gaps between the first and \(41\)st poles, so each gap is \(\dfrac{5280}{40} = 132\) feet.

Elmer’s stride is \(\dfrac{132}{44} = 3\) feet and Oscar’s leap is \(\dfrac{132}{12} = 11\) feet, a difference of \(11 - 3 = 8\) feet.

Thus, the correct answer is B.

因为 \(ABCD\) 是正方形，\(\angle ADC = 90^\circ\)。又因为 \(E\) 与 \(A\) 在直线 \(CD\) 的两侧，射线 \(DE\) 越过了 \(DC\)，所以三角形 \(DFE\) 在 \(D\) 处的角 （其中 \(F\) 在 \(\overline{AD}\) 上）为 \(\angle FDE = 360^\circ\) \(- (\angle ADC + \angle CDE)\) \(= 360^\circ - (90^\circ + 110^\circ)\) \(= 160^\circ\)。

因为 \(DF = DE\)，三角形 \(DFE\) 是等腰三角形，其底角 \(\angle DFE = \tfrac{180^\circ - 160^\circ}{2} = 10^\circ\)。

因为 \(A\)、\(F\)、\(D\) 共线，所以 \(\angle AFE = 180^\circ - \angle DFE = 170^\circ\)。

所以正确答案是 D。

English

Because \(ABCD\) is a square, \(\angle ADC = 90^\circ.\) Since \(E\) and \(A\) lie on opposite sides of line \(CD,\) ray \(DE\) is swung past \(DC,\) so the angle of triangle \(DFE\) at \(D\) (with \(F\) on \(\overline{AD}\)) is \(\angle FDE = 360^\circ\) \(- (\angle ADC + \angle CDE)\) \(= 360^\circ - (90^\circ + 110^\circ)\) \(= 160^\circ.\)

Since \(DF = DE,\) triangle \(DFE\) is isosceles with base angles \(\angle DFE = \tfrac{180^\circ - 160^\circ}{2} = 10^\circ.\)

As \(A,\) \(F,\) \(D\) are collinear, \(\angle AFE = 180^\circ - \angle DFE = 170^\circ.\)

Thus, the correct answer is D.

老师平均值为 \(t = \dfrac{50 + 20 + 20 + 5 + 5}{5}\) \(= \dfrac{100}{5} = 20\)。

学生平均值按每个班级中的学生人数给班级人数加权：\[ \begin{aligned} s &= \small \frac{50^2 + 20^2 + 20^2 + 5^2 + 5^2}{100} \\ &= \small \frac{2500 + 400 + 400 + 25 + 25}{100} \\ &= 33.5 \end{aligned}\text{。}\]

所以 \(t - s = 20 - 33.5 = -13.5\)。

所以正确答案是 B。

English

The teacher average is \(t = \dfrac{50 + 20 + 20 + 5 + 5}{5}\) \(= \dfrac{100}{5} = 20.\)

The student average weights each class size by how many students are in it: \[ \begin{aligned} s &= \small \frac{50^2 + 20^2 + 20^2 + 5^2 + 5^2}{100} \\ &= \small \frac{2500 + 400 + 400 + 25 + 25}{100} \\ &= 33.5. \end{aligned} \]

So \(t - s = 20 - 33.5 = -13.5.\)

Thus, the correct answer is B.

\(M = \operatorname{lcm}(10, \ldots, 30)\) 含有 \(2^4\)（来自 \(16\)），\(3^3\)（来自 \(27\)），\(5^2\)（来自 \(25\)），\(7\)，以及直到 \(29\) 的每个质数。

在 \(32, \ldots, 40\) 中，唯一的新贡献是 \(32 = 2^5\)，它把 \(2\) 的幂次从 \(2^4\) 提高到 \(2^5\)，以及新的质数 \(37\)。其他数都只分解成 \(M\) 中已经有的质数和幂次。

因此 \(\dfrac{N}{M} = 2 \cdot 37 = 74\)。

所以正确答案是 D。

English

\(M = \operatorname{lcm}(10, \ldots, 30)\) contains \(2^4\) (from \(16\)), \(3^3\) (from \(27\)), \(5^2\) (from \(25\)), \(7,\) and every prime up to \(29.\)

Among \(32, \ldots, 40,\) the only new contributions are \(32 = 2^5,\) which raises the power of \(2\) from \(2^4\) to \(2^5,\) and the new prime \(37.\) Everything else factors into primes and powers already in \(M.\)

Therefore \(\dfrac{N}{M} = 2 \cdot 37 = 74.\)

Thus, the correct answer is D.

令 \(a = \log_2 x\)，\(b = \log_3 x\)，\(c = \log_4 x\)。表面积等于体积给出 \(2(ab + bc + ca) = abc\)。两边除以 \(abc\)，得 \[ 1 = 2\left(\frac{1}{c} + \frac{1}{a} + \frac{1}{b}\right)\text{。}\]

因为 \(\dfrac{1}{a} = \log_x 2\)，等等，所以和为 \(\log_x 2 + \log_x 3 + \log_x 4 = \log_x 24\)。于是 \(1 = 2\log_x 24\)，所以 \(\log_x 24 = \tfrac{1}{2}\)，即 \(x^{\frac{1}{2}} = 24\)，从而 \(x = 576\)。

所以正确答案是 E。

English

Let \(a = \log_2 x,\) \(b = \log_3 x,\) \(c = \log_4 x.\) Surface area equals volume gives \(2(ab + bc + ca) = abc.\) Dividing by \(abc,\) \[ 1 = 2\left(\frac{1}{c} + \frac{1}{a} + \frac{1}{b}\right). \]

Since \(\dfrac{1}{a} = \log_x 2,\) etc., the sum is \(\log_x 2 + \log_x 3 + \log_x 4 = \log_x 24.\) Thus \(1 = 2\log_x 24,\) so \(\log_x 24 = \tfrac{1}{2},\) meaning \(x^{\frac{1}{2}} = 24\) and \(x = 576.\)

Thus, the correct answer is E.

因为 \(9 \equiv -1 \pmod 5\)，所以每个幂 \(9^k \equiv (-1)^k\)，因此 \(N\) 同余于其以九为底的数字的交错和。

非零数字及其从右往左的位置为：\(2\)（位置 \(0\)），\(5\)（位置 \(1\)），\(6\)（位置 \(6\)），\(7\)（位置 \(9\)），以及 \(2\)（位置 \(10\)）。交错和为 \(2 - 5 + 6 - 7 + 2 = -2\) \(\equiv 3 \pmod 5\)。

所以正确答案是 D。

English

Since \(9 \equiv -1 \pmod 5,\) each power \(9^k \equiv (-1)^k,\) so \(N\) is congruent to the alternating sum of its base-nine digits.

The nonzero digits, with their positions from the right, are \(2\) (position \(0\)), \(5\) (position \(1\)), \(6\) (position \(6\)), \(7\) (position \(9\)), and \(2\) (position \(10\)). The alternating sum is \(2 - 5 + 6 - 7 + 2 = -2\) \(\equiv 3 \pmod 5.\)

Thus, the correct answer is D.

设这条弦到共同圆心的距离为 \(d\)。它的总长度为 \(2\sqrt{361 - d^2}\)，位于较小圆内的部分长度为 \(2\sqrt{289 - d^2}\)。

因为弦的一半位于小圆内，\(2\sqrt{289 - d^2} = \tfrac{1}{2}\cdot 2\sqrt{361 - d^2}\)。平方得 \(4(289 - d^2) = 361 - d^2\)，所以 \(3d^2 = 795\)，\(d^2 = 265\)。

弦长为 \(2\sqrt{361 - 265} = 2\sqrt{96} = 8\sqrt{6}\)。

所以正确答案是 E。

English

Let the chord lie at distance \(d\) from the common center. Its total length is \(2\sqrt{361 - d^2},\) and the portion inside the smaller circle has length \(2\sqrt{289 - d^2}.\)

Since half the chord lies inside, \(2\sqrt{289 - d^2} = \tfrac{1}{2}\cdot 2\sqrt{361 - d^2}.\) Squaring gives \(4(289 - d^2) = 361 - d^2,\) so \(3d^2 = 795\) and \(d^2 = 265.\)

The chord length is \(2\sqrt{361 - 265} = 2\sqrt{96} = 8\sqrt{6}.\)

Thus, the correct answer is E.

通项为 \(\binom{1000}{k}(x\sqrt[3]{2})^{1000-k}(y\sqrt{3})^{k}\)，其系数含有 \(2^{\frac{1000-k}{3}}\) 和 \(3^{\frac{k}{2}}\)。它有理当且仅当 \((1000 - k)\) 是 \(3\) 的倍数且 \(k\) 为偶数。

因为 \(1000 \equiv 1 \pmod 3\)，所以需要 \(k \equiv 1 \pmod 3\) 且 \(k\) 为偶数，合并得 \(k \equiv 4 \pmod 6\)。有效值为 \(k = 4, 10, \ldots, 1000\)，共有 \(\dfrac{1000 - 4}{6} + 1 = 167\) 个。

所以正确答案是 C。

English

The general term is \(\binom{1000}{k}(x\sqrt[3]{2})^{1000-k}(y\sqrt{3})^{k},\) whose coefficient contains \(2^{\frac{1000-k}{3}}\) and \(3^{\frac{k}{2}}.\) This is rational exactly when \((1000 - k)\) is a multiple of \(3\) and \(k\) is even.

Since \(1000 \equiv 1 \pmod 3,\) we need \(k \equiv 1 \pmod 3\) and \(k\) even, which combine to \(k \equiv 4 \pmod 6.\) The valid values \(k = 4, 10, \ldots, 1000\) number \(\dfrac{1000 - 4}{6} + 1 = 167.\)

Thus, the correct answer is C.

角平分线沿两条直线的单位方向向量之和：\(\dfrac{(1,1)}{\sqrt2} + \dfrac{(1,3)}{\sqrt{10}}\)。它的斜率为 \[ k = \frac{\tfrac{1}{\sqrt2} + \tfrac{3}{\sqrt{10}}}{\tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt{10}}} = \frac{\sqrt5 + 3}{\sqrt5 + 1}\text{。}\]

分子分母同乘 \(\sqrt5 - 1\)，得 \(\dfrac{(\sqrt5 + 3)(\sqrt5 - 1)}{4}\) \(= \dfrac{2 + 2\sqrt5}{4}\) \(= \dfrac{1 + \sqrt5}{2}\)。

所以正确答案是 A。

English

The bisector points along the sum of the unit vectors of the two lines: \(\dfrac{(1,1)}{\sqrt2} + \dfrac{(1,3)}{\sqrt{10}}.\) Its slope is \[ k = \frac{\tfrac{1}{\sqrt2} + \tfrac{3}{\sqrt{10}}}{\tfrac{1}{\sqrt2} + \tfrac{1}{\sqrt{10}}} = \frac{\sqrt5 + 3}{\sqrt5 + 1}. \]

Multiplying numerator and denominator by \(\sqrt5 - 1\) gives \(\dfrac{(\sqrt5 + 3)(\sqrt5 - 1)}{4}\) \(= \dfrac{2 + 2\sqrt5}{4}\) \(= \dfrac{1 + \sqrt5}{2}.\)

Thus, the correct answer is A.

设公共边长为 \(s\)。三个锐角顶点是三个等腰三角形的尖端，这些三角形的两边为 \(s\)，顶角为 \(30^\circ\)；每个面积为 \(\tfrac12 s^2 \sin 30^\circ = \tfrac{s^2}{4}\)。

三个凹角顶点形成一个内侧等边三角形，边长为 \(2s\sin 15^\circ\)，面积为 \(\sqrt3\,s^2\sin^2 15^\circ\)。利用 \(\sin^2 15^\circ = \tfrac{2 - \sqrt3}{4}\)，总面积为 \[ \frac{3s^2}{4} + \sqrt3\,s^2\cdot\frac{2 - \sqrt3}{4} = \frac{s^2\sqrt3}{2}\text{。}\]

令 \(\tfrac{s^2\sqrt3}{2} = 6\sqrt3\)，得 \(s^2 = 12\)，所以 \(s = 2\sqrt3\)，周长为 \(6s = 12\sqrt3\)。

所以正确答案是 E。

English

Let the common side length be \(s.\) The three acute vertices are the tips of isosceles triangles with two sides \(s\) and apex \(30^\circ;\) each has area \(\tfrac12 s^2 \sin 30^\circ = \tfrac{s^2}{4}.\)

The three reflex vertices form an inner equilateral triangle with side \(2s\sin 15^\circ,\) whose area is \(\sqrt3\,s^2\sin^2 15^\circ.\) Using \(\sin^2 15^\circ = \tfrac{2 - \sqrt3}{4},\) the total area is \[ \frac{3s^2}{4} + \sqrt3\,s^2\cdot\frac{2 - \sqrt3}{4} = \frac{s^2\sqrt3}{2}. \]

Setting \(\tfrac{s^2\sqrt3}{2} = 6\sqrt3\) gives \(s^2 = 12,\) so \(s = 2\sqrt3\) and the perimeter is \(6s = 12\sqrt3.\)

Thus, the correct answer is E.

对 \(P\)，使用韦达定理，\(\sum_{i\lt j} z_iz_j = 3\)，且 \(\prod z_j = 1\)，二者都是实数，所以它们的共轭仍为 \(3\) 和 \(1\)。

\(Q\) 的根为 \(4i\overline{z_j}\)。于是 \(B\) 是两两乘积之和：\(B = (4i)^2 \sum_{i\lt j}\overline{z_i}\,\overline{z_j}\) \(= -16 \cdot 3 = -48\)。且 \(D = (4i)^4 \prod \overline{z_j} = 256 \cdot 1 = 256\)。

所以 \(B + D = -48 + 256 = 208\)。

所以正确答案是 D。

English

By Vieta on \(P,\) \(\sum_{i\lt j} z_iz_j = 3\) and \(\prod z_j = 1,\) both real, so their conjugates are also \(3\) and \(1.\)

The roots of \(Q\) are \(4i\overline{z_j}.\) Then \(B\) is the sum of products of pairs: \(B = (4i)^2 \sum_{i\lt j}\overline{z_i}\,\overline{z_j}\) \(= -16 \cdot 3 = -48.\) And \(D = (4i)^4 \prod \overline{z_j} = 256 \cdot 1 = 256.\)

So \(B + D = -48 + 256 = 208.\)

Thus, the correct answer is D.

技术员不断加电缆，直到图变为连通。为了最大化电缆数，要让网络尽可能久地保持不连通：留下一台 A 品牌电脑孤立，并把其余 \(19\) 台 A 品牌电脑与全部 \(10\) 台 B 品牌电脑完全相连。

这样在仍不连通时使用了 \(19 \cdot 10 = 190\) 根电缆。下一根电缆连接最后一台 A 品牌电脑，使所有人连通，总数为 \(190 + 1 = 191\)。

所以正确答案是 B。

English

The technician keeps adding cables until the graph becomes connected. To maximize the count, keep the network disconnected for as long as possible: leave a single brand A computer isolated and fully connect the remaining \(19\) brand A computers to all \(10\) brand B computers.

That uses \(19 \cdot 10 = 190\) cables while still disconnected. The next cable connects the last brand A computer, joining everyone, for a total of \(190 + 1 = 191.\)

Thus, the correct answer is B.

两个二次方程都没有两个不同实根，当且仅当两个判别式都非正：\(b^2 \le 4c\) 且 \(c^2 \le 4b\)。

把 \(c\ge \frac{b^2}{4}\) 与 \(c\le2\sqrt b\) 结合，得到 \(b^{\frac{3}{2}}\le8\)，所以 \(b\le4\)。逐一检查：\(b = 1\) 时 \(c \in \{1,2\}\)；\(b = 2\) 时 \(c \in \{1,2\}\)；\(b = 3\) 时 \(c = 3\)；\(b = 4\) 时 \(c = 4\)。

这些为 \((1,1)\)、\((1,2)\)、\((2,1)\)、\((2,2)\)、\((3,3)\)、\((4,4)\)，共 \(6\) 个有序对。

所以正确答案是 B。

English

Neither quadratic has two distinct real roots exactly when both discriminants are nonpositive: \(b^2 \le 4c\) and \(c^2 \le 4b.\)

Combining \(c\ge \frac{b^2}{4}\) with \(c\le2\sqrt b\) gives \(b^{\frac{3}{2}}\le8,\) so \(b\le4.\) Checking: \(b = 1\) gives \(c \in \{1,2\};\) \(b = 2\) gives \(c \in \{1,2\};\) \(b = 3\) gives \(c = 3;\) and \(b = 4\) gives \(c = 4.\)

That is \((1,1),\) \((1,2),\) \((2,1),\) \((2,2),\) \((3,3),\) \((4,4)\) — \(6\) ordered pairs.

Thus, the correct answer is B.

两个概率都除以 \(5^{20}\)，所以 \(\dfrac{p}{q}\) 是排列计数之比。

对于 \(q\)，所有箱子都有 \(4\) 个球，计数为 \(\dfrac{20!}{(4!)^5}\)。对于 \(p\)，选择哪个箱子有 \(3\) 个、哪个箱子有 \(5\) 个有 \(5\cdot 4 = 20\) 种，再乘以 \(\dfrac{20!}{3!\,5!\,(4!)^3}\)。因此 \[ \begin{aligned} \frac{p}{q} &= 20 \cdot \frac{(4!)^5}{3!\,5!\,(4!)^3} \\ &= 20 \cdot \frac{(4!)^2}{3!\,5!} \\ &= 20 \cdot \frac{576}{720} \\ &= 16 \end{aligned}\text{。}\]

所以正确答案是 E。

English

Both probabilities divide by \(5^{20},\) so \(\dfrac{p}{q}\) is a ratio of arrangement counts.

For \(q,\) all bins have \(4:\) \(\dfrac{20!}{(4!)^5}.\) For \(p,\) choose which bin has \(3\) and which has \(5\) in \(5\cdot 4 = 20\) ways, times \(\dfrac{20!}{3!\,5!\,(4!)^3}.\) Therefore \[ \begin{aligned} \frac{p}{q} &= 20 \cdot \frac{(4!)^5}{3!\,5!\,(4!)^3} \\ &= 20 \cdot \frac{(4!)^2}{3!\,5!} \\ &= 20 \cdot \frac{576}{720} \\ &= 16. \end{aligned} \]

Thus, the correct answer is E.

正弦值相等要求对某个整数 \(k\) 有 \(x^2 = x + 360k\) 或 \(x^2 = 180 - x + 360k\)。

方程族 \(x^2 = x + 360k\) 第一次超过 \(1\) 出现在 \(k = 1\)，给出 \(x \approx 19.5\)。方程族 \(x^2 = 180 - x + 360k\) 在 \(k = 0\) 时给出 \(x^2 + x - 180 = 0\)，所以 \(x = \dfrac{-1 + \sqrt{721}}{2} \approx 12.93\)，这更小。

向上取整后，\(x = 13\)。

所以正确答案是 B。

English

Equal sines require \(x^2 = x + 360k\) or \(x^2 = 180 - x + 360k\) for some integer \(k.\)

The family \(x^2 = x + 360k\) first exceeds \(1\) at \(k = 1,\) giving \(x \approx 19.5.\) The family \(x^2 = 180 - x + 360k\) with \(k = 0\) gives \(x^2 + x - 180 = 0,\) so \(x = \dfrac{-1 + \sqrt{721}}{2} \approx 12.93,\) which is smaller.

Rounded up, \(x = 13.\)

Thus, the correct answer is B.

\(8\) 和 \(12\) 都是不动点。对 \(n\le50\)，第一个值 \(f_1(n)=2d(n)\) 至多为 \(20\)。检查不超过 \(20\) 的偶数可知，一个轨道恰好在第一个值为 \(12,18\) 或 \(20\) 时到达 \(12\)。所以需要 \(d(n)=6,9\) 或 \(10\)。

不超过 \(50\) 且有 \(6\) 个因数的数是 \(12,18,20,28,32,44,45,50\)；有 \(9\) 个因数的唯一一个数是 \(36\)，有 \(10\) 个因数的唯一一个数是 \(48\)。这 \(10\) 个值都到达不动点 \(12\)。

因此，正确答案是 D。

English

Both \(8\) and \(12\) are fixed. For \(n\le50,\) the first value \(f_1(n)=2d(n)\) is at most \(20.\) Checking the even values through \(20\) shows that an orbit reaches \(12\) exactly when its first value is \(12,18,\) or \(20.\) Thus we need \(d(n)=6,9,\) or \(10.\)

The numbers at most \(50\) with \(6\) divisors are \(12,18,20,28,32,44,45,50;\) the only one with \(9\) divisors is \(36,\) and the only one with \(10\) divisors is \(48.\) These \(10\) values all reach the fixed point \(12.\)

Thus, the correct answer is D.

令 \(A = (0,0)\)，\(X = (3,0)\)，\(Y = (4,0)\)，\(C = (6,0)\)。直角条件给出 \(D = (3, t)\)，且 \(B = (4, s)\)，它们在 \(AC\) 的两侧。

平行关系 \(\overline{AD}\parallel\overline{BC}\) 强制 \(t = -\tfrac{3}{2}s\)，而 \(AB = CD\) 给出 \(16 + s^2 = 9 + t^2\)，所以 \(t^2 - s^2 = 7\)。代入得 \(s^2 = \tfrac{28}{5}\)。

鞋带公式给出面积 \(= 3\,|t - s|\) \(= 3\cdot\tfrac{5}{2}s\) \(= \tfrac{15}{2}s\) \(= \tfrac{15}{2}\sqrt{\tfrac{28}{5}}\) \(= 3\sqrt{35}\)。

所以正确答案是 C。

English

Put \(A = (0,0),\) \(X = (3,0),\) \(Y = (4,0),\) \(C = (6,0).\) The right angles give \(D = (3, t)\) and \(B = (4, s)\) on opposite sides of \(AC.\)

Parallelism \(\overline{AD}\parallel\overline{BC}\) forces \(t = -\tfrac{3}{2}s,\) and \(AB = CD\) gives \(16 + s^2 = 9 + t^2,\) so \(t^2 - s^2 = 7.\) Substituting yields \(s^2 = \tfrac{28}{5}.\)

The shoelace formula gives area \(= 3\,|t - s|\) \(= 3\cdot\tfrac{5}{2}s\) \(= \tfrac{15}{2}s\) \(= \tfrac{15}{2}\sqrt{\tfrac{28}{5}}\) \(= 3\sqrt{35}.\)

Thus, the correct answer is C.

Carl 在第三个 \(O\) 时获胜，所以棋盘上有三个 \(O\) 形成 \(8\) 条线之一，另有三个 \(X\) 在其余六个格子中。\(X\) 不能成一行（否则 Azar 已经先赢了）。

若 \(O\) 的线是某一行或某一列（\(6\) 种选择），剩下六个格子包含两条完整线，所以有效 \(X\) 放法数为 \(\binom{6}{3} - 2 = 18\)。若 \(O\) 的线是对角线（\(2\) 种选择），剩下六个格子不包含完整线，有 \(\binom{6}{3} = 20\) 种。

总数为 \(6\cdot 18 + 2\cdot 20 = 108 + 40 = 148\)。

所以正确答案是 D。

English

Carl wins on his third \(O,\) so the board has three \(O\)s forming one of the \(8\) lines and three \(X\)s in the other six cells. The \(X\)s must not form a line (else Azar would have won first).

If the \(O\) line is a row or column (\(6\) choices), the remaining six cells contain two full lines, so valid \(X\) placements number \(\binom{6}{3} - 2 = 18.\) If the \(O\) line is a diagonal (\(2\) choices), the remaining six cells contain no full line, giving \(\binom{6}{3} = 20.\)

The total is \(6\cdot 18 + 2\cdot 20 = 108 + 40 = 148.\)

Thus, the correct answer is D.

设 \(p\) 的根为 \(r\) 和 \(s\)。那么 \(p(p(x)) = 0\) 分解成 \(p(x) = r\) 与 \(p(x) = s\)，两个判别式分别为 \((r - s)^2 + 4r\) 和 \((r - s)^2 + 4s\)。恰有三个实根意味着一个判别式为 \(0\)，另一个为正。

取 \((r - s)^2 + 4s = 0\)，并令 \(u = r - s\)。则 \(s = -\tfrac{u^2}{4}\)，另一个判别式为 \(4u\)，所以 \(u>0\)。此外，\(r+s=-\tfrac{u^2}{2}+u\)，它在 \(u=1\) 时达到最大值，此时 \(r=\tfrac34\) 且 \(s=-\tfrac14\)。

所以 \(\tilde{p}(x) = \left(x - \tfrac34\right)\left(x + \tfrac14\right)\)，并且 \(\tilde{p}(1) = \tfrac14\cdot\tfrac54 = \tfrac{5}{16}\)。

因此，正确答案是 A。

English

Let \(p\) have roots \(r\) and \(s.\) Then \(p(p(x)) = 0\) splits into \(p(x) = r\) and \(p(x) = s,\) with discriminants \((r - s)^2 + 4r\) and \((r - s)^2 + 4s.\) Exactly three real roots means one discriminant is \(0\) and the other positive.

Take \((r - s)^2 + 4s = 0\) and set \(u = r - s.\) Then \(s = -\tfrac{u^2}{4}\) and the other discriminant is \(4u,\) so \(u>0.\) Also \(r+s=-\tfrac{u^2}{2}+u,\) maximized at \(u=1,\) giving \(r=\tfrac34\) and \(s=-\tfrac14.\)

So \(\tilde{p}(x) = \left(x - \tfrac34\right)\left(x + \tfrac14\right),\) and \(\tilde{p}(1) = \tfrac14\cdot\tfrac54 = \tfrac{5}{16}.\)

Thus, the correct answer is A.

因为 \(AB = 18\) 是最大边，四条边为 \(18, 18 - d, 18 - 2d, 18 - 3d\)。取 \(A = (0,0)\)，\(B = (18,0)\)，\(D = \left(\tfrac{m}{2}, \tfrac{m\sqrt3}{2}\right)\)，其中 \(m=DA\)，令 \(n=CD\) 且 \(\ell=BC\)。则 \(C=(\tfrac m2+n,\tfrac{m\sqrt3}{2})\)，所以 \[ \begin{aligned} \ell^2&=m^2+(18-n)^2 \\ &\quad {}-m(18-n) \end{aligned}\text{。}\]

令 \(u_j=18-jd\)，其中 \(j=1,2,3\)。将 \((u_1,u_2,u_3)\) 的排列 \(123,132,213,231,312,321\) 依次代入 \((m,n,\ell)\)，得到非零候选值 \(d=18,2,9,5,6,6\)，因为 \(u_3\) 为正要求 \(d<6\)，所以只留下 \(d=2\) 和 \(d=5\)。情形 \(d=0\) 也给出一个有效的菱形。

因此边长集合为 \(\{18,16,14,12\}\)，\(\{18,13,8,3\}\)，和 \(\{18,18,18,18\}\)。

非 \(AB\) 边长的可能值为 \(\{3, 8, 12, 13, 14, 16, 18\}\)，其和为 \(84\)。

因此，正确答案是 E。

English

Since \(AB = 18\) is the largest, the four sides are \(18, 18 - d, 18 - 2d, 18 - 3d.\) Placing \(A = (0,0),\) \(B = (18,0),\) and \(D = \left(\tfrac{m}{2}, \tfrac{m\sqrt3}{2}\right)\) with \(m=DA,\) let \(n=CD\) and \(\ell=BC.\) Then \(C=(\tfrac m2+n,\tfrac{m\sqrt3}{2}),\) so \[ \begin{aligned} \ell^2&=m^2+(18-n)^2 \\ &\quad {}-m(18-n). \end{aligned} \]

Write \(u_j=18-jd\) for \(j=1,2,3.\) Substituting the permutations \(123,132,213,231,312,321\) of \((u_1,u_2,u_3)\) for \((m,n,\ell)\) gives the nonzero candidates \(d=18,2,9,5,6,6,\) respectively. Positivity of \(u_3\) requires \(d<6,\) leaving only \(d=2\) and \(d=5.\) The case \(d=0\) also gives a valid rhombus.

Thus the side sets are \(\{18,16,14,12\},\) \(\{18,13,8,3\},\) and \(\{18,18,18,18\}.\)

The possible values of a non-\(AB\) side length are \(\{3, 8, 12, 13, 14, 16, 18\},\) whose sum is \(84.\)

Thus, the correct answer is E.

把 \(1,2,\ldots,m\) 看作模 \(m\) 的全部剩余类。若不要求各分量互不相同，则前三个分量可以任取，第四个分量随之唯一确定，于是共有 \(m^3\) 个有序四元组。

按哪些坐标相等来划分并做容斥，即可得到其余各项。恰有一对坐标相等的选法有 \(6\) 种，每种留下 \(m^2\) 组解，共贡献 \(-6m^2\)。\(3\) 种“两对相等”的划分贡献 \(+3m\)。\(4\) 种“三个相等加一个单独”的划分的容斥系数为 \(2\)，贡献 \(+8m\)。最后，四个全相等的划分系数为 \(-6\)，且只有一组解。（这里 \(m\) 为奇数保证了 \(2\) 和 \(4\) 模 \(m\) 可逆。）因此\[ \begin{aligned} D(m)&=m^3-6m^2 \\ &\quad {}+11m-6 \\ &=(m-1)(m-2) \\ &\quad {}\cdot(m-3) \end{aligned}\text{。}\] 于是 \(c_1=11\)。

所以正确答案是 E。

English

Regard \(1,2,\ldots,m\) as all residues modulo \(m.\) Without the distinctness condition, the first three entries are arbitrary and the fourth is determined, giving \(m^3\) ordered quadruples.

Inclusion-exclusion over equal-coordinate partitions gives the remaining terms. There are \(6\) choices of one equal pair, each leaving \(m^2\) solutions, for \(-6m^2.\) The \(3\) two-pair partitions contribute \(+3m.\) The \(4\) triple-and-single partitions have inclusion-exclusion coefficient \(2\) and contribute \(+8m.\) Finally, the all-equal partition has coefficient \(-6\) and one solution. (Here odd \(m\) makes \(2\) and \(4\) invertible modulo \(m.\)) Therefore \[ \begin{aligned} D(m)&=m^3-6m^2 \\ &\quad {}+11m-6 \\ &=(m-1)(m-2) \\ &\quad {}\cdot(m-3). \end{aligned} \] Hence \(c_1=11.\)

Thus, the correct answer is E.