2022A AMC 12 真题（P版）

1 答案 D

从底部开始化简。最里面的部分是 \(3+\dfrac13=\dfrac{10}{3}\)。

下一层是 \(3+\dfrac{1}{\frac{10}{3}}=3+\dfrac{3}{10}=\dfrac{33}{10}\)。

最后，\(3+\dfrac{1}{\frac{33}{10}}=3+\dfrac{10}{33}=\dfrac{109}{33}\)。

因此，正确答案是 D。

English

Simplify from the bottom. The innermost fraction is \(3+\dfrac13=\dfrac{10}{3}.\)

The next layer is \(3+\dfrac{1}{\frac{10}{3}}=3+\dfrac{3}{10}=\dfrac{33}{10}.\)

Finally, \(3+\dfrac{1}{\frac{33}{10}}=3+\dfrac{10}{33}=\dfrac{109}{33}.\)

Thus, the correct answer is D.

2 答案 E

令第三个数为 \(t\)。那么第一个数是 \(6t\)，第二个数是 \(t+40\)。它们的和为 \[6t+(t+40)+t=8t+40=96\text{，}\] 所以 \(t=7\)。

第一个数是 \(42\)，第二个数是 \(47\)， 因此差的绝对值为 \(5\)。

因此，正确答案是 E。

English

Let the third number be \(t.\) Then the first is \(6t\) and the second is \(t+40.\) Their sum is \[6t+(t+40)+t=8t+40=96,\] so \(t=7.\)

The first number is \(42\) and the second is \(47,\) so the difference has absolute value \(5.\)

Thus, the correct answer is E.

3 答案 B

五个矩形的面积分别为 \(6,\ 8,\ 30,\ 14\)，和 \(6\)，总和为 \(64\)。所以这个正方形是 \(8\times8\)。

将 \(C\)（\(5\times6\)）放在左上方，\(D\)（\(2\times7\)）沿右边放置，\(E\) （\(2\times3\)）放在左下方，\(A\)（\(1\times6\)）沿底边放置，会留下一个中间的 \(2\times4\) 空隙，正好是矩形 \(B\)。

因此，正确答案是 B。

English

The five areas are \(6,\ 8,\ 30,\ 14,\) and \(6,\) which sum to \(64.\) So the square is \(8\times8.\)

Placing \(C\) (\(5\times6\)) across the top left, \(D\) (\(2\times7\)) up the right side, \(E\) (\(2\times3\)) in the lower left, and \(A\) (\(1\times6\)) along the bottom leaves a central \(2\times4\) gap, which is exactly rectangle \(B.\)

Thus, the correct answer is B.

4 答案 B

因为 \(180=2^2\cdot3^2\cdot5\)，且 \(18=2\cdot3^2\)，条件 \(\operatorname{lcm}(n,18)=180\) 强制 \(n\) 提供 \(2^2\) 和 \(5\)，同时 \(3\) 的幂次至多为 \(2\)。

由 \(\gcd(n,45)=\gcd(n,3^2\cdot5)\) \(=15=3\cdot5\)，可知 \(n\) 中 \(3\) 的幂次恰为 \(1\)，\(5\) 的幂次至少为 \(1\)。

因此 \(n=2^2\cdot3\cdot5=60\)，其各位数字之和为 \(6\)。

因此，正确答案是 B。

English

Since \(180=2^2\cdot3^2\cdot5\) and \(18=2\cdot3^2,\) the condition \(\operatorname{lcm}(n,18)=180\) forces \(n\) to contribute \(2^2\) and \(5,\) with its power of \(3\) at most \(2.\)

From \(\gcd(n,45)=\gcd(n,3^2\cdot5)\) \(=15=3\cdot5,\) the power of \(3\) in \(n\) is exactly \(1\) and the power of \(5\) is at least \(1.\)

Therefore \(n=2^2\cdot3\cdot5=60,\) whose digits sum to \(6.\)

Thus, the correct answer is B.

5 答案 C

对每个 \(k\ge1\)，集合 \(|x|+|y|=k\) 恰有 \(4k\) 个格点，而 \(k=0\) 时只有原点一个点。

总数为 \[ \begin{aligned} &1+\sum_{k=1}^{20}4k=1+4\cdot\frac{20\cdot21}{2} \\ &=1+840=841 \end{aligned}\text{。}\]

因此，正确答案是 C。

English

For each \(k\ge1,\) the set \(|x|+|y|=k\) contains exactly \(4k\) lattice points, and \(k=0\) gives the single origin.

The total is \[ \begin{aligned} &1+\sum_{k=1}^{20}4k=1+4\cdot\frac{20\cdot21}{2} \\ &=1+840=841. \end{aligned} \]

Thus, the correct answer is C.

6 答案 D

已知数的和为 \(20\)，所以平均数为 \(\dfrac{20+X}{6}\)，且它必须等于数据集中的某个数。

令它等于 \(5\) 得 \(X=10\)；等于 \(7\) 得 \(X=22\)；等于 \(X\) 本身得 \(20+X=6X\)，所以 \(X=4\)。等于 \(1\) 或 \(2\) 会给出负的 \(X\)。

符合条件的正值为 \(10,22,4\)，总和为 \(36\)。

因此，正确答案是 D。

English

The known numbers sum to \(20,\) so the mean is \(\dfrac{20+X}{6},\) which must equal an element of the set.

Setting it to \(5\) gives \(X=10;\) to \(7\) gives \(X=22;\) and to \(X\) itself gives \(20+X=6X,\) so \(X=4.\) Values \(1\) and \(2\) give negative \(X.\)

The positive values are \(10,22,4,\) summing to \(36.\)

Thus, the correct answer is D.

7 答案 D

底部中间的区域与其他四个区域都共边。先给它涂色，有 \(5\) 种方法。

左上区域与它相邻，有 \(4\) 种选择。其余三个区域各与两个已经涂好的区域相邻，而这两个区域颜色不同，所以各有 \(3\) 种选择。

总数为 \(5\cdot4\cdot3\cdot3\cdot3=540\)。

因此，正确答案是 D。

English

The bottom-middle region shares a border with all four other regions. Color it first in \(5\) ways.

The top-left region borders it, giving \(4\) choices. Each of the three remaining regions borders exactly two already-colored regions, which have different colors, leaving \(3\) choices apiece.

The total is \(5\cdot4\cdot3\cdot3\cdot3=540.\)

Thus, the correct answer is D.

8 答案 A

第 \(k\) 个因子是对 \(10\) 连取 \(k\) 次立方根，也就是 \(10^{\frac{1}{3^k}}\)。

这个乘积等于 \(10\) 的如下幂：\[ \begin{aligned} &\frac13+\frac19+\frac1{27}+\cdots=\frac{\frac{1}{3}}{1-\frac{1}{3}} \\ &=\frac12 \end{aligned}\text{。}\]

所以乘积的值为 \(10^{\frac{1}{2}}=\sqrt{10}\)。

因此，正确答案是 A。

English

The \(k\)th factor is \(10\) raised to the \(k\)-fold cube root, namely \(10^{\frac{1}{3^k}}.\)

The product is \(10\) raised to \[ \begin{aligned} &\frac13+\frac19+\frac1{27}+\cdots=\frac{\frac{1}{3}}{1-\frac{1}{3}} \\ &=\frac12. \end{aligned} \]

So the value is \(10^{\frac{1}{2}}=\sqrt{10}.\)

Thus, the correct answer is A.

9 答案 A

对“你是说真话者吗？”这个问题，说真话者和说谎者都会回答“是”，而交替者中只有这次说谎的人会回答“是”。对“你是交替者吗？”这个问题，说谎者回答“是”，交替者中只有这次说真话的人回答“是”。对“你是说谎者吗？”这个问题，只有这次说谎的交替者回答“是”。

按第一次回答拆分交替者。先说谎的交替者回答模式为（谎话，真话，谎话），所以三个问题都回答“是”；先说真话的交替者回答模式为（真话，谎话，真话），所以三个问题都不回答“是”。最后一个问题的 \(9\) 个“是”正好都是先说谎的交替者，因此他们有 \(9\) 人。

第二个问题的 \(15\) 个“是”来自说谎者和这 \(9\) 个交替者，所以说谎者有 \(6\) 人。第一个问题的 \(22\) 个“是”来自说真话者、说谎者以及这 \(9\) 个交替者，所以说真话者有 \(22-6-9=7\) 人。

说真话者只会在第一个问题回答“是”，每人得到一颗糖，共 \(7\cdot1=7\) 颗。

因此，正确答案是 A。

English

To “Are you a truth-teller?” the truth-tellers and liars both answer yes, and only alternaters who lie on this question answer yes. To “Are you an alternater?” the liars answer yes, and among alternaters only those telling the truth on this question answer yes. To “Are you a liar?” only alternaters lying on this question answer yes.

Split the alternaters by first response. Those starting with a lie answer (lie, truth, lie), so they say yes to all three questions; those starting truthful answer (truth, lie, truth) and say yes to none of the three. The \(9\) yeses on the last question are exactly the lie-first alternaters, so there are \(9\) of them.

The second question’s \(15\) yeses are the liars plus these \(9,\) so there are \(6\) liars. The first question’s \(22\) yeses are truth-tellers plus liars plus the \(9,\) so the truth-tellers number \(22-6-9=7.\)

Truth-tellers answer yes only to the first question, receiving one candy each, for \(7\cdot1=7\) pieces.

Thus, the correct answer is A.

10 答案 E

任何大于或等于 \(8\) 的数都不能作为较小元素（它的两倍超过 \(14\)），所以 \(8\) 到 \(14\) 都是较大元素，\(1\) 到 \(7\) 都是较小元素。

将每个较小数 \(s\) 配到一个较大数 \(g\ge2s\)。从限制最强的开始：\(s=7\) 强制 \(g=14\)（\(1\) 种）；然后 \(s=6\) 可配剩下的 \(\{12,13\}\)（\(2\) 种）；\(s=5\) 有 \(3\) 种；\(s=4\) 有 \(4\) 种；\(s=3\) 有 \(3\) 种；\(s=2\) 有 \(2\) 种；\(s=1\) 有 \(1\) 种。

匹配数为 \(1\cdot2\cdot3\cdot4\cdot3\cdot2\cdot1=144\)。

因此，正确答案是 E。

English

Any number \(8\) or larger cannot be a smaller element (its double exceeds \(14\)), so \(8\)–\(14\) are all larger elements and \(1\)–\(7\) are all smaller elements.

Match each smaller \(s\) to a larger \(g\ge2s.\) Processing from the most restrictive: \(s=7\) forces \(g=14\) (\(1\) way); then \(s=6\) has \(\{12,13\}\) left (\(2\)); \(s=5\) has \(3;\) \(s=4\) has \(4;\) \(s=3\) has \(3;\) \(s=2\) has \(2;\) \(s=1\) has \(1.\)

The number of matchings is \(1\cdot2\cdot3\cdot4\cdot3\cdot2\cdot1=144.\)

Thus, the correct answer is E.

11 答案 E

右侧的距离为 \(|\log_6 10-1|=\log_6\dfrac53\)，所以它的两倍为 \(2\log_6\dfrac53=\log_6\dfrac{25}{9}\)。

因此 \(\left|\log_6\dfrac{x}{9}\right|=\log_6\dfrac{25}{9}\)，得 \(\dfrac{x}{9}=\dfrac{25}{9}\) 或 \(\dfrac{x}{9}=\dfrac{9}{25}\)，所以 \(x=25\) 或 \(x=\dfrac{81}{25}\)。

它们的乘积为 \(25\cdot\dfrac{81}{25}=81\)。

因此，正确答案是 E。

English

The right-hand distance is \(|\log_6 10-1|=\log_6\dfrac53,\) so twice it is \(2\log_6\dfrac53=\log_6\dfrac{25}{9}.\)

Thus \(\left|\log_6\dfrac{x}{9}\right|=\log_6\dfrac{25}{9},\) giving \(\dfrac{x}{9}=\dfrac{25}{9}\) or \(\dfrac{x}{9}=\dfrac{9}{25},\) so \(x=25\) or \(x=\dfrac{81}{25}.\)

Their product is \(25\cdot\dfrac{81}{25}=81.\)

Thus, the correct answer is E.

12 答案 B

取棱长为 \(1\)。因为 \(M\) 是 \(\overline{AB}\) 的中点，线段 \(CM\) 和 \(DM\) 是等边三角形面的高，长度都为 \(\dfrac{\sqrt3}{2}\)。另外 \(CD=1\)。

在 \(\triangle CMD\) 中由余弦定理，\[ \begin{aligned} \cos(\angle CMD) &=\frac{\frac34+\frac34-1}{2\cdot\frac34} \\ &=\frac{\frac{1}{2}}{\frac{3}{2}}=\frac13 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Take edge length \(1.\) Since \(M\) is the midpoint of \(\overline{AB},\) segments \(CM\) and \(DM\) are altitudes of the equilateral faces, each of length \(\dfrac{\sqrt3}{2}.\) Also \(CD=1.\)

By the Law of Cosines in \(\triangle CMD,\) \[ \begin{aligned} \cos(\angle CMD) &=\frac{\frac34+\frac34-1}{2\cdot\frac34} \\ &=\frac{\frac{1}{2}}{\frac{3}{2}}=\frac13. \end{aligned} \]

Thus, the correct answer is B.

13 答案 A

给线段上的每个点加上半径为 \(1\) 的圆盘，会扫出所有到该线段距离不超过 \(1\) 的点。从 \(3\) 到 \(4i\) 的线段长度为 \(\sqrt{3^2+4^2}=5\)。

这个“体育场”形状由一个 \(5\times2\) 的矩形和两个半径为 \(1\) 的半圆组成，面积为 \[5\cdot2+\pi(1)^2=10+\pi\approx13.14\text{。}\]

最接近的整数是 \(13\)。

因此，正确答案是 A。

English

Adding a disk of radius \(1\) to every point of the segment sweeps out all points within distance \(1\) of it. The segment from \(3\) to \(4i\) has length \(\sqrt{3^2+4^2}=5.\)

This “stadium” is a \(5\times2\) rectangle plus two half-disks of radius \(1,\) with area \[5\cdot2+\pi(1)^2=10+\pi\approx13.14.\]

The closest integer is \(13.\)

Thus, the correct answer is A.

14 答案 C

令 \(u=\log 2\)。则 \(\log 5=1-u\)，\(\log 20=1+u\)，\(\log 8=3u\)，且 \(\log 0.25=-2u\)。

设 \(a=1-u,\ b=1+u\)，则 \(a+b=2\)，\(ab=1-u^2\)，所以 \[ \begin{gathered} a^3+b^3 \\ =(a+b)\big((a+b)^2-3ab\big) \\ =2\big(4-3(1-u^2)\big) \\ =2+6u^2 \end{gathered}\text{。}\]

最后一项是 \((3u)(-2u)=-6u^2\)，因此总和为 \(2+6u^2-6u^2=2\)。

因此，正确答案是 C。

English

Let \(u=\log 2.\) Then \(\log 5=1-u,\) \(\log 20=1+u,\) \(\log 8=3u,\) and \(\log 0.25=-2u.\)

With \(a=1-u,\ b=1+u,\) we have \(a+b=2\) and \(ab=1-u^2,\) so \[ \begin{gathered} a^3+b^3 \\ =(a+b)\big((a+b)^2-3ab\big) \\ =2\big(4-3(1-u^2)\big) \\ =2+6u^2. \end{gathered} \]

The last term is \((3u)(-2u)=-6u^2,\) so the total is \(2+6u^2-6u^2=2.\)

Thus, the correct answer is C.

15 答案 D

设根为 \(r,s,t\)。由韦达定理，\(r+s+t=\dfrac{39}{10}\)，\(rs+rt+st=\dfrac{29}{10}\)，且 \(rst=\dfrac{6}{10}=\dfrac35\)。

新体积为 \[ \begin{gathered} (r+2)(s+2)(t+2) \\ =rst+2(rs+rt+st) \\ \quad {}+4(r+s+t)+8 \\ =\frac35+\frac{58}{10} \\ \quad {}+\frac{156}{10}+8 \\ =30 \end{gathered}\text{。}\]

因此，正确答案是 D。

English

Let the roots be \(r,s,t.\) By Vieta’s formulas, \(r+s+t=\dfrac{39}{10},\) \(rs+rt+st=\dfrac{29}{10},\) and \(rst=\dfrac{6}{10}=\dfrac35.\)

The new volume is \[ \begin{gathered} (r+2)(s+2)(t+2) \\ =rst+2(rs+rt+st) \\ \quad {}+4(r+s+t)+8 \\ =\frac35+\frac{58}{10} \\ \quad {}+\frac{156}{10}+8 \\ =30. \end{gathered} \]

Thus, the correct answer is D.

16 答案 D

若 \(t_n=y^2\)，则 \(\frac{n(n+1)}{2}=y^2\)，即 \((2n+1)^2-8y^2=1\)。正的佩尔方程解可依次通过将 \((2n+1)+y\sqrt8\) 乘以 \(3+\sqrt8\) 得到。

从 \((2n+1,y)=(3,1)\) 开始，依次得到 \((17,6)\)，\((99,35)\)，然后是 \((577,204)\)。因此第四个值对应 \(n=288\)，并且等于 \(204^2=41616\)。

它的数位和为 \(4+1+6+1+6=18\)。

因此，正确答案是 D。

English

If \(t_n=y^2,\) then \(\frac{n(n+1)}{2}=y^2,\) or \((2n+1)^2-8y^2=1.\) The positive Pell solutions occur successively by multiplying \((2n+1)+y\sqrt8\) by \(3+\sqrt8.\)

Starting from \((2n+1,y)=(3,1),\) this gives \((17,6),\) \((99,35),\) and then \((577,204).\) Thus the fourth value has \(n=288\) and equals \(204^2=41616.\)

The sum of its digits is \(4+1+6+1+6=18.\)

Thus, the correct answer is D.

17 答案 A

在 \((0,\pi)\) 上，\(\sin x\ne0\)，所以两边除以 \(\sin x\)：\[ \begin{gathered} a(1+2\cos x)=4\cos^2 x-1 \\ =(2\cos x-1)(2\cos x+1) \end{gathered}\text{。}\]

当 \(\cos x=-\tfrac12\)（即 \(x=\tfrac{2\pi}{3}\)）时，两边都为零，因此这对每个 \(a\) 都是一个解。否则可以约去 \(1+2\cos x\) 得 \(a=2\cos x-1\)，即 \(\cos x=\dfrac{a+1}{2}\)。

它给出 \((0,\pi)\) 中第二个解的条件正是 \(-1\lt\dfrac{a+1}{2}\lt1\)，即 \(a\in(-3,1)\)，且当 \(a=-2\) 时这个解与 \(x=\tfrac{2\pi}{3}\) 相同，不是不同的解。

因此有多于一个解时，\(a\in(-3,-2)\cup(-2,1)\)，所以 \(p+q+r=-3-2+1=-4\)。

因此，正确答案是 A。

English

Since \(\sin x\ne0\) on \((0,\pi),\) divide by \(\sin x:\) \[ \begin{gathered} a(1+2\cos x)=4\cos^2 x-1 \\ =(2\cos x-1)(2\cos x+1). \end{gathered} \]

When \(\cos x=-\tfrac12\) (that is, \(x=\tfrac{2\pi}{3}\)) both sides vanish, so this is a solution for every \(a.\) Otherwise we may cancel \(1+2\cos x\) to get \(a=2\cos x-1,\) i.e. \(\cos x=\dfrac{a+1}{2}.\)

This yields a second solution in \((0,\pi)\) exactly when \(-1\lt\dfrac{a+1}{2}\lt1,\) that is \(a\in(-3,1),\) and it is distinct from \(x=\tfrac{2\pi}{3}\) unless \(a=-2.\)

So more than one solution occurs for \(a\in(-3,-2)\cup(-2,1),\) giving \(p+q+r=-3-2+1=-4.\)

Thus, the correct answer is A.

18 答案 A

角度为 \(\theta\) 的点旋转 \(k^\circ\) 后角度变为 \(\theta+k\)，再关于 \(y\) 轴反射会把角度 \(\phi\) 变为 \(180-\phi\)。所以 \(T_k\) 把 \(\theta\) 变为 \((180-k)-\theta\)。

从角度 \(0\) 的点 \((1,0)\) 出发，依次应用 \(T_1,T_2,\ldots\) 得到的角度为 \(179,-1,178,-2,177,\ldots\)。经过偶数 \(2m\) 步后角度为 \(-m\)，经过奇数 \(2m+1\) 步后角度为 \(179-m\)。

点要回到原位，角度必须是 \(360^\circ\) 的倍数。偶数情形要求 \(m=360\)，即 \(n=720\)。奇数情形要求 \(179-m=0\)，即 \(m=179\) 且 \(n=359\)，此时总反射固定 \((1,0)\)。

满足条件的最小 \(n\) 是 \(359\)。

因此，正确答案是 A。

English

Rotating a point at angle \(\theta\) by \(k^\circ\) gives \(\theta+k,\) and reflecting across the \(y\)-axis sends angle \(\phi\) to \(180-\phi.\) So \(T_k\) sends \(\theta\) to \((180-k)-\theta.\)

Starting from \((1,0)\) at angle \(0,\) applying \(T_1,T_2,\ldots\) gives angles \(179,-1,178,-2,177,\ldots.\) After an even number \(2m\) of steps the angle is \(-m,\) and after an odd number \(2m+1\) it is \(179-m.\)

For the point to return, the angle must be a multiple of \(360^\circ.\) The even case needs \(m=360,\) i.e. \(n=720.\) The odd case needs \(179-m=0,\) i.e. \(m=179\) and \(n=359,\) where the net reflection fixes \((1,0).\)

The least such \(n\) is \(359.\)

Thus, the correct answer is A.

19 答案 D

设 \(\text{pos}(k)\) 为卡片 \(k\) 的位置。正好在 \(\text{pos}(k+1)\lt\text{pos}(k)\) 时需要开始新一轮扫描，因此扫描次数等于序列 \(\text{pos}(1),\text{pos}(2),\ldots,\text{pos}(13)\) 的下降数加一。

要构造至多有一个下降的排列，只需选出排在可能的下降之前的那些元素，并把两块都按递增顺序写出。这样的子集共有 \(2^{13}\) 个。其中 \(14\) 个初始段 \(\varnothing,\{1\},\ldots,\{1,\ldots,13\}\) 不产生下降；其余每个子集都恰好给出一个含一个下降的排列。因此总数为 \(2^{13}-14=8178\)。

因此，正确答案是 D。

English

Let \(\text{pos}(k)\) be the position of card \(k.\) A fresh pass is needed exactly when \(\text{pos}(k+1)\lt\text{pos}(k),\) so the number of passes is one more than the number of descents in the sequence \(\text{pos}(1),\text{pos}(2),\ldots,\text{pos}(13).\)

To build a permutation with at most one descent, choose the entries before the possible descent and write both chosen blocks in increasing order. There are \(2^{13}\) subsets. The \(14\) initial segments \(\varnothing,\{1\},\ldots,\{1,\ldots,13\}\) produce no descent; every other subset produces a unique permutation with one descent. Hence the count is \(2^{13}-14=8178.\)

Thus, the correct answer is D.

20 答案 B

将梯形放成关于 \(y\) 轴对称：\(A=(-p,0)\)，\(D=(p,0)\)，\(B=(-q,h)\)，\(C=(q,h)\)，并设 \(P=(x,y)\)。

则 \(PA^2-PD^2=4px\) \(=1-16=-15\) 而 \(PB^2-PC^2=4qx\) \(=4-9=-5\)。两式相除得 \(\dfrac{p}{q}=3\)。

因为 \(AD=2p\) 且 \(BC=2q\)，所以 \(\dfrac{BC}{AD}=\dfrac{q}{p}=\dfrac13\)。

因此，正确答案是 B。

English

Place the trapezoid symmetric about the \(y\)-axis: \(A=(-p,0),\) \(D=(p,0),\) \(B=(-q,h),\) \(C=(q,h),\) with \(P=(x,y).\)

Then \(PA^2-PD^2=4px\) \(=1-16=-15\) and \(PB^2-PC^2=4qx\) \(=4-9=-5.\) Dividing gives \(\dfrac{p}{q}=3.\)

Since \(AD=2p\) and \(BC=2q,\) we get \(\dfrac{BC}{AD}=\dfrac{q}{p}=\dfrac13.\)

Thus, the correct answer is B.

21 答案 E

如果 \(\zeta\) 是某个因子的根，则必须有 \(P(\zeta)=\zeta^{2022}+\zeta^{1011}+1=0\)。

\(x^6+x^3+1\) 的根是本原 \(9\) 次单位根，因此 \(\zeta^9=1\)。将指数对 \(9\) 取模，由 \(2022\equiv6\) 和 \(1011\equiv3\) 可得 \(\zeta^6+\zeta^3+1\)。这里 \(\omega=\zeta^3\) 是本原三次单位根，所以该式等于 \(\omega^2+\omega+1=0\)。

其他四个选项不成立：代入它们的根会得到非零值（例如本原三次单位根会给出 \(P=3\)）。

因此，正确答案是 E。

English

If \(\zeta\) is a root of a divisor, then \(P(\zeta)=\zeta^{2022}+\zeta^{1011}+1=0\) is required.

The roots of \(x^6+x^3+1\) are the primitive \(9\)th roots of unity, so \(\zeta^9=1.\) Reducing exponents modulo \(9,\) \(2022\equiv6\) and \(1011\equiv3,\) giving \(\zeta^6+\zeta^3+1.\) Here \(\omega=\zeta^3\) is a primitive cube root of unity, so this equals \(\omega^2+\omega+1=0.\)

The other four options fail: substituting their roots yields nonzero values (for instance, the primitive cube roots of unity give \(P=3\)).

Thus, the correct answer is E.

22 答案 A

若根不是实数，则 \(z_1=\sqrt{10}\,e^{i\theta}\) 且 \(z_2=\overline{z_1}\)，因为 \(z_1z_2=10\)。于是 \(\dfrac{1}{z_1}=\dfrac{1}{\sqrt{10}}e^{-i\theta}\)，\(\dfrac{1}{z_2}=\dfrac{1}{\sqrt{10}}e^{i\theta}\)。

这个梯形的两条竖边长度分别为 \(2\sqrt{10}\sin\theta\) 和 \(\frac{2\sin\theta}{\sqrt{10}}\)，它们之间的水平距离为 \((\sqrt{10}-\frac{1}{\sqrt{10}})\cos\theta\)。因此它的面积为 \[ \frac{99}{10}\sin\theta\cos\theta =\frac{99}{20}\sin2\theta\text{，}\] 在 \(\theta=45^\circ\) 时取得最大值。

此时 \(c=z_1+z_2\) \(=2\sqrt{10}\cos45^\circ\) \(=2\sqrt5\approx4.47\)，最接近 \(4.5\)。

因此，正确答案是 A。

English

If the roots are non-real, then \(z_1=\sqrt{10}\,e^{i\theta}\) and \(z_2=\overline{z_1},\) since \(z_1z_2=10.\) Then \(\dfrac{1}{z_1}=\dfrac{1}{\sqrt{10}}e^{-i\theta}\) and \(\dfrac{1}{z_2}=\dfrac{1}{\sqrt{10}}e^{i\theta}.\)

The two vertical sides of this trapezoid have lengths \(2\sqrt{10}\sin\theta\) and \(\frac{2\sin\theta}{\sqrt{10}},\) and their horizontal separation is \((\sqrt{10}-\frac{1}{\sqrt{10}})\cos\theta.\) Hence its area is \[ \frac{99}{10}\sin\theta\cos\theta =\frac{99}{20}\sin2\theta, \] which is maximized at \(\theta=45^\circ.\)

Then \(c=z_1+z_2\) \(=2\sqrt{10}\cos45^\circ\) \(=2\sqrt5\approx4.47,\) closest to \(4.5.\)

Thus, the correct answer is A.

23 答案 D

\(L_n\) 总是 \(k_n\) 的倍数，所以 \(k_n\lt L_n\) 恰好在某个质数 \(p\) 同时整除 \(L_n\) 和分子 \(N=\sum_{k=1}^n \tfrac{L_n}{k}\) 时成立（即有质数被约掉）。

对具有最大幂 \(p^a\le n\) 的质数 \(p\)，只有满足 \(v_p(k)=a\) 的项会使 \(\frac{L_n}{k}\) 不含 \(p\)，其余项都能被 \(p\) 整除。因此 \(p\) 被约掉，当且仅当 \(\sum_{v_p(k)=a}\tfrac{L_n}{k}\equiv0\pmod p\)。

利用 \(H_n=H_{n-1}+\frac{1}{n}\) 递推应用这个判定，得到：当 \(1\le n\le5\) 时，\(k_n=L_n\)；当 \(6\le n\le8\) 时，\(k_n=\frac{L_n}{3}\)；当 \(9\le n\le17\) 时，再次有 \(k_n=L_n\)。最后，对 \(n=18,19,20,21,22\)，比值 \(\frac{L_n}{k_n}\) 分别为 \(3,3,15,45,45\)，因此恰好在 \(n=6,7,8,18,19,20,21,22\) 时发生约分，共有 \(8\) 个值。

因此，正确答案是 D。

English

\(L_n\) is always divisible by \(k_n,\) so \(k_n\lt L_n\) exactly when some prime \(p\) divides both \(L_n\) and the numerator \(N=\sum_{k=1}^n \tfrac{L_n}{k}\) (i.e. a prime cancels).

For a prime \(p\) with maximal power \(p^a\le n,\) only the terms with \(v_p(k)=a\) keep \(p\) out of \(\frac{L_n}{k};\) all others are divisible by \(p.\) So \(p\) cancels iff \(\sum_{v_p(k)=a}\tfrac{L_n}{k}\equiv0\pmod p.\)

Applying this test recursively with \(H_n=H_{n-1}+\frac{1}{n}\) gives \(k_n=L_n\) for \(1\le n\le5,\) \(k_n=\frac{L_n}{3}\) for \(6\le n\le8,\) and \(k_n=L_n\) again for \(9\le n\le17.\) Finally, the ratios \(\frac{L_n}{k_n}\) for \(n=18,19,20,21,22\) are \(3,3,15,45,45,\) respectively. Thus cancellation occurs precisely for \(n=6,7,8,18,19,20,21,22,\) which is \(8\) values.

Thus, the correct answer is D.

24 答案 E

将五个数字排序为 \(d_{(1)}\le d_{(2)}\le\cdots\le d_{(5)}\)。条件“至少有 \(j\) 个数字小于 \(j\)”等价于对 \(j=1,2,3,4\)，有 \(d_{(j)}\le j-1\)，即 \(d_{(1)}=0,\ d_{(2)}\le1\)，\(\ d_{(3)}\le2,\ d_{(4)}\le3\)（而 \(d_{(5)}\le4\) 自动成立）。

这些字符串恰好是长度为 \(5\) 的停车函数。为计数，将 \(6\) 个停车位排成圆形，让 \(5\) 辆有标号的汽车任意选择首选车位。每辆车向前行驶到第一个空位。在 \(6^5\) 个偏好字符串中，同时旋转所有偏好会使唯一的空位依次经过全部 \(6\) 个位置。因此，恰有 \(\frac{6^5}{6}=6^4=1296\) 个字符串使指定车位为空。将这个车位选作额外的第六个车位，恰好得到上述排序不等式。

因此，正确答案是 E。

English

Sort the five digits as \(d_{(1)}\le d_{(2)}\le\cdots\le d_{(5)}.\) The requirement “at least \(j\) digits less than \(j\)” is equivalent to \(d_{(j)}\le j-1\) for \(j=1,2,3,4,\) i.e. \(d_{(1)}=0,\ d_{(2)}\le1,\) \(\ d_{(3)}\le2,\ d_{(4)}\le3\) (with \(d_{(5)}\le4\) automatic).

These strings are exactly the parking functions of length \(5.\) To count them, arrange \(6\) parking spaces in a circle and let \(5\) labeled cars choose arbitrary preferred spaces. Each car moves forward to the first open space. Among the \(6^5\) preference strings, rotating all preferences cycles the unique empty space through all \(6\) positions. Therefore exactly \(\frac{6^5}{6}=6^4=1296\) strings leave a specified space empty. Choosing that space as the extra sixth space gives precisely the sorted inequalities above.

Thus, the correct answer is E.

25 答案 E

圆心为 \((r,r)\)、半径为 \(r\) 的圆与两条坐标轴都相切。从 \((0,a)\) 到 \((b,0)\) 的线段满足 \(a^2+b^2=c^2\)；当 \(r\) 等于以 \(a,b\) 为直角边的直角三角形的内切圆半径 \(\tfrac{a+b-c}{2}\) 或半周长 \(\tfrac{a+b+c}{2}\) 时，该线段与圆相切。

在内切圆半径情形中，令 \(x=a-2r\) 且 \(y=b-2r\)。则 \(xy=2r^2\)，每个正因数 \(x\) 确定一条有方向的线段，其中 \(a=x+2r\)，\(b=\frac{2r^2}{x}+2r\)，\(c=x+\frac{2r^2}{x}+2r\)。因此恰有 \(d(2r^2)\) 条这样的线段。

当 \(r=1,2,3,4,5\) 时，数量分别为 \(2,4,6,6,6\)，且半周长情形不可能，因为最小的整数直角三角形半周长为 \(6\)。当 \(r=6\) 时，\(d(72)=12\)，而 \(3\)-\(4\)-\(5\) 三角形再贡献两条有方向的线段。因此 \(6\) 是最小可能半径，并且恰好给出 \(14\) 条线段。

两个半周长情形的线段满足 \(c=5\)。在内切圆半径族中，\(c=x+\frac{72}{x}+12\) 在 \(x=1\) 或 \(72\) 时最大，得到 \(c=85\)。因此 \(c_1=5\)，\(c_{14}=85\)，并且 \(\frac{c_{14}}{c_1}=17\)。

因此，正确答案是 E。

English

The circle centered \((r,r)\) with radius \(r\) is tangent to both axes. A segment from \((0,a)\) to \((b,0)\) with \(a^2+b^2=c^2\) is tangent to it when \(r\) equals either the inradius \(\tfrac{a+b-c}{2}\) or the semiperimeter \(\tfrac{a+b+c}{2}\) of the right triangle with legs \(a,b.\)

In the inradius case, put \(x=a-2r\) and \(y=b-2r.\) Then \(xy=2r^2,\) and every positive divisor \(x\) determines one oriented segment, with \(a=x+2r,\) \(b=\frac{2r^2}{x}+2r,\) and \(c=x+\frac{2r^2}{x}+2r.\) Thus there are exactly \(d(2r^2)\) such segments.

For \(r=1,2,3,4,5,\) these counts are \(2,4,6,6,6,\) and no semiperimeter case is possible because the smallest integer right triangle has semiperimeter \(6.\) At \(r=6,\) \(d(72)=12,\) and the \(3\)-\(4\)-\(5\) triangle contributes two more oriented segments. Hence \(6\) is the least possible radius and gives exactly \(14\) segments.

The two semiperimeter segments have \(c=5.\) In the inradius family, \(c=x+\frac{72}{x}+12\) is largest at \(x=1\) or \(72,\) giving \(c=85.\) Therefore \(c_1=5,\) \(c_{14}=85,\) and \(\frac{c_{14}}{c_1}=17.\)

Thus, the correct answer is E.