2023A AMC 12 真题（P版）

1 答案 E

两人之间的距离以每小时 \(18+12=30\) 英里的速度缩短，所以她们在 \(\dfrac{45}{30}=1.5\) 小时后相遇。

在这段时间里，Alicia 从城市 \(A\) 出发骑了 \(18\cdot 1.5=27\) 英里。

所以正确答案是 E。

English

The gap between them closes at \(18+12=30\) miles per hour, so they meet after \(\dfrac{45}{30}=1.5\) hours.

In that time Alicia has ridden \(18\cdot 1.5=27\) miles from City \(A.\)

Thus, the correct answer is E.

2 答案 A

设披萨重 \(P\) 磅。则 \(3\tfrac12\) 杯重 \(\tfrac72\cdot\tfrac14=\tfrac78\) 而 \(\tfrac12\) 杯重 \(\tfrac12\cdot\tfrac14=\tfrac18\)。

方程为 \[ \tfrac13 P+\tfrac78=\tfrac34 P+\tfrac18\text{。}\]

相减得 \(\tfrac78-\tfrac18=\left(\tfrac34-\tfrac13\right)P\)，所以 \(\tfrac34=\tfrac{5}{12}P\)，从而 \(P=\dfrac{9}{5}=1\tfrac45\)。

所以正确答案是 A。

English

Let the pizza weigh \(P\) pounds. Then \(3\tfrac12\) cups weigh \(\tfrac72\cdot\tfrac14=\tfrac78\) and \(\tfrac12\) cup weighs \(\tfrac12\cdot\tfrac14=\tfrac18.\)

The equation is \[ \tfrac13 P+\tfrac78=\tfrac34 P+\tfrac18. \]

Subtracting gives \(\tfrac78-\tfrac18=\left(\tfrac34-\tfrac13\right)P,\) so \(\tfrac34=\tfrac{5}{12}P\) and \(P=\dfrac{9}{5}=1\tfrac45.\)

Thus, the correct answer is A.

3 答案 A

一个完全平方数能被 \(5\) 整除，只有它的平方根也能被五整除，因此这些平方数形如 \((5k)^2=25k^2\)。

因为 \(44^2=1936\lt 2023\)，而且这个数 \(\lt 2025=45^2\)，所以平方根可以是 \(5,10,\ldots,40\)，也就是 \(k=1\) 到 \(8\)。

所以正确答案是 A。

English

A perfect square is divisible by \(5\) only if its root is, so the squares are \((5k)^2=25k^2.\)

Since \(44^2=1936\lt 2023\) \(\lt 2025=45^2,\) the root can be \(5,10,\ldots,40,\) which is \(k=1\) through \(8.\)

Thus, the correct answer is A.

4 答案 E

全部分解为质因数：\[ \begin{gathered} 8^5\cdot 5^{10}\cdot 15^5\\ {}=2^{15}\cdot 5^{10}\cdot 3^5\cdot 5^5\\ {}=3^5\cdot 2^{15}\cdot 5^{15} \end{gathered}\text{。}\]

这等于 \(3^5\cdot 10^{15}=243\cdot 10^{15}\) 也就是 \(243\) 后接 \(15\) 个零，总共有 \(18\) 位数字。

所以正确答案是 E。

English

Writing everything in primes, \[ \begin{gathered} 8^5\cdot 5^{10}\cdot 15^5\\ {}=2^{15}\cdot 5^{10}\cdot 3^5\cdot 5^5\\ {}=3^5\cdot 2^{15}\cdot 5^{15}. \end{gathered} \]

This equals \(3^5\cdot 10^{15}=243\cdot 10^{15},\) which is \(243\) followed by \(15\) zeros, for a total of \(18\) digits.

Thus, the correct answer is E.

5 答案 B

累加和递增，所以它恰好到达 \(3\) 当且仅当出现以下互斥开头之一：第一次掷出 \(3\)；前两次为 \(1,2\)；前两次为 \(2,1\)；或前三次为 \(1,1,1\)。

它们的概率之和为 \[ \begin{gathered} \dfrac16+\dfrac1{36}+\dfrac1{36}+\dfrac1{216}\\ {}=\dfrac{36+6+6+1}{216}\\ {}=\dfrac{49}{216} \end{gathered}\text{。}\]

所以正确答案是 B。

English

The running total is increasing, so it hits \(3\) exactly when one of these disjoint openings occurs: a first roll of \(3;\) rolls \(1,2;\) rolls \(2,1;\) or rolls \(1,1,1.\)

Their probabilities are \[ \begin{gathered} \dfrac16+\dfrac1{36}+\dfrac1{36}+\dfrac1{216}\\ {}=\dfrac{36+6+6+1}{216}\\ {}=\dfrac{49}{216}. \end{gathered} \]

Thus, the correct answer is B.

6 答案 D

设两个 \(x\) 坐标为 \(x_1\) 和 \(x_2\)。由中点可得 \(x_1+x_2=12\)，由两个 \(y\) 值的平均数可得 \(\log_2 x_1+\log_2 x_2=4\)，所以 \(x_1x_2=2^4=16\)。

因此 \[ \begin{gathered} |x_1-x_2|\\ {}=\sqrt{(x_1+x_2)^2-4x_1x_2}\\ {}=\sqrt{144-64}\\ {}=\sqrt{80}\\ {}=4\sqrt5 \end{gathered}\text{。}\]

所以正确答案是 D。

English

Let the \(x\)-coordinates be \(x_1\) and \(x_2.\) The midpoint gives \(x_1+x_2=12,\) and the average of the \(y\)-values gives \(\log_2 x_1+\log_2 x_2=4,\) so \(x_1x_2=2^4=16.\)

Then \[ \begin{gathered} |x_1-x_2|\\ {}=\sqrt{(x_1+x_2)^2-4x_1x_2}\\ {}=\sqrt{144-64}\\ {}=\sqrt{80}\\ {}=4\sqrt5. \end{gathered} \]

Thus, the correct answer is D.

7 答案 E

年份贡献数字 \(2,0,2,3\)，所以 \(2\) 出现两次，而 \(0\) 和 \(3\) 各出现一次。为了让每个数字最终出现偶数次，月份和日期的四个数字必须提供奇数个 \(0\)、奇数个 \(3\)，以及偶数个其他每种数字。

只有四个数字可用，因此月份日期串必须恰好包含一个 \(0\)、一个 \(3\)，以及由某个数字组成的一对。检查合法月份和日期，得到九个日期：\(01\text{-}13\)、\(01\text{-}31\)、\(02\text{-}23\)、\(03\text{-}11\)、\(03\text{-}22\)、\(10\text{-}13\)、\(10\text{-}31\)、\(11\text{-}03\) 和 \(11\text{-}30\)。

所以正确答案是 E。

English

The year contributes the digits \(2,0,2,3,\) so \(2\) appears twice while \(0\) and \(3\) each appear once. For every digit to end up with an even count, the four digits of the month and day must supply an odd number of \(0\)’s, an odd number of \(3\)’s, and an even number of every other digit.

With only four digits available, the month-day string must use exactly one \(0,\) one \(3,\) and a repeated pair of some digit. Checking valid months and days leaves nine dates: \(01\text{-}13,\) \(01\text{-}31,\) \(02\text{-}23,\) \(03\text{-}11,\) \(03\text{-}22,\) \(10\text{-}13,\) \(10\text{-}31,\) \(11\text{-}03,\) and \(11\text{-}30.\)

Thus, the correct answer is E.

8 答案 D

设当前 \(n\) 次小测的平均分为 \(m\)，所以总分为 \(S=mn\)。加入一次 \(11\) 分后，有 \[ \dfrac{mn+11}{n+1}=m+1\text{，} \] 化简得 \(m+n=10\)。

加入三次 \(11\) 分后，有 \[ \dfrac{mn+33}{n+3}=m+2\text{，} \] 化简得 \(3m+2n=27\)。

解 \(m+n=10\) 和 \(3m+2n=27\) 得到 \(m=7\)。

所以正确答案是 D。

English

Let the current mean be \(m\) over \(n\) quizzes, so the total is \(S=mn.\) Adding one \(11\) gives \[ \dfrac{mn+11}{n+1}=m+1, \] which simplifies to \(m+n=10.\)

Adding three \(11\)’s gives \[ \dfrac{mn+33}{n+3}=m+2, \] which simplifies to \(3m+2n=27.\)

Solving \(m+n=10\) and \(3m+2n=27\) gives \(m=7.\)

Thus, the correct answer is D.

9 答案 C

外正方形边长为 \(\sqrt3\)，内正方形边长为 \(\sqrt2\)。每个三角形是直角三角形，其两条直角边 \(p\) 和 \(q\) 沿着外正方形的一条边，所以 \(p+q=\sqrt3\)；其斜边是内正方形的一条边，所以 \(p^2+q^2=2\)。

由 \((p+q)^2=3\) 可得 \(2pq=1\)，所以 \(p\) 和 \(q\) 是 \(t^2-\sqrt3\,t+\tfrac12=0\) 的两个根，即 \(\dfrac{\sqrt3\pm 1}{2}\)。

较短边与较长边之比为 \[ \begin{gathered} \dfrac{\sqrt3-1}{\sqrt3+1}\\ {}=\dfrac{(\sqrt3-1)^2}{2}\\ {}=2-\sqrt3 \end{gathered}\text{。}\]

所以正确答案是 C。

English

The outer square has side \(\sqrt3\) and the inner square has side \(\sqrt2.\) Each triangle is right, with legs \(p\) and \(q\) along an outer side, so \(p+q=\sqrt3,\) and with hypotenuse an inner side, so \(p^2+q^2=2.\)

Then \((p+q)^2=3\) gives \(2pq=1,\) so \(p\) and \(q\) are the roots of \(t^2-\sqrt3\,t+\tfrac12=0,\) namely \(\dfrac{\sqrt3\pm 1}{2}.\)

The ratio of shorter to longer leg is \[ \begin{gathered} \dfrac{\sqrt3-1}{\sqrt3+1}\\ {}=\dfrac{(\sqrt3-1)^2}{2}\\ {}=2-\sqrt3. \end{gathered} \]

Thus, the correct answer is C.

10 答案 D

由 \((y-x)^2=4y^2\)，可得 \(y-x=\pm 2y\)。若 \(y-x=2y\)，则 \(x=-y\lt 0\)，不可能；因此 \(y-x=-2y\)，也就是 \(x=3y\)。

代入 \(y^3=x^2=9y^2\) 得到 \(y=9\)，所以 \(x=27\)，\(x+y=36\)。

所以正确答案是 D。

English

From \((y-x)^2=4y^2\) we get \(y-x=\pm 2y.\) The choice \(y-x=2y\) gives \(x=-y\lt 0,\) impossible, so \(y-x=-2y,\) meaning \(x=3y.\)

Substituting into \(y^3=x^2=9y^2\) gives \(y=9,\) hence \(x=27\) and \(x+y=36.\)

Thus, the correct answer is D.

11 答案 C

两直线夹角的正切为 \[ \left|\dfrac{2-\tfrac13}{1+2\cdot\tfrac13}\right| =\left|\dfrac{\tfrac53}{\tfrac53}\right|=1\text{。}\]

正切为 \(1\) 的锐角是 \(45^\circ\)。

所以正确答案是 C。

English

The tangent of the angle between the lines is \[ \left|\dfrac{2-\tfrac13}{1+2\cdot\tfrac13}\right| =\left|\dfrac{\tfrac53}{\tfrac53}\right|=1. \]

The acute angle with tangent \(1\) is \(45^\circ.\)

Thus, the correct answer is C.

12 答案 D

配成 \((2k)^3-(2k-1)^3\)，其中 \(k=1,\ldots,9\)。展开得 \[ \begin{gathered} (2k)^3-(2k-1)^3\\ {}=12k^2-6k+1 \end{gathered}\text{。}\]

对 \(k=1\) 到 \(9\)，求和，并用 \(\sum k^2=285\)、\(\sum k=45\) 得到 \[ \begin{gathered} 12\cdot 285-6\cdot 45+9\\ {}=3420-270+9\\ {}=3159 \end{gathered}\text{。}\]

所以正确答案是 D。

English

Group into pairs \((2k)^3-(2k-1)^3\) for \(k=1,\ldots,9.\) Expanding, \[ \begin{gathered} (2k)^3-(2k-1)^3\\ {}=12k^2-6k+1. \end{gathered} \]

Summing for \(k=1\) to \(9,\) with \(\sum k^2=285\) and \(\sum k=45,\) gives \[ \begin{gathered} 12\cdot 285-6\cdot 45+9\\ {}=3420-270+9\\ {}=3159. \end{gathered} \]

Thus, the correct answer is D.

13 答案 B

设有 \(L\) 名左手选手和 \(2L\) 名右手选手，则共有 \(3L\) 人，总场数为 \(\binom{3L}{2}\)。

若右手选手赢 \(R\) 场，则左手选手赢 \(1.4R\) 场，所以总场数为 \(2.4R=\tfrac{12}{5}R\)。要使它是整数场数，总场数必须是 \(12\) 的倍数。

左手选手至多能赢下所有至少有一名左手选手参加的比赛，即 \(\binom{L}{2}+2L^2=\tfrac{5L^2-L}{2}\) 场。另一方面，他们必须赢 \(\tfrac{7}{12}\binom{3L}{2}=\tfrac{7L(3L-1)}8\) 场。比较这两个量可得 \(21L-7\le 20L-4\)，所以 \(L\le 3\)。

依次试 \(L=1,2,3\)，总场数为 \(3,15,36\)；其中只有 \(36\) 是 \(12\) 的倍数，且可实现：\(3\) 名左手选手可以赢下全部 \(18\) 场混合对局以及左手内部的 \(3\) 场，共 \(21=1.4\cdot 15\) 场。

所以正确答案是 B。

English

Let there be \(L\) left-handed and \(2L\) right-handed players, for \(3L\) players and \(\binom{3L}{2}\) games total.

If right-handers win \(R\) games, left-handers win \(1.4R,\) so the total is \(2.4R=\tfrac{12}{5}R.\) For this to be an integer count, the total number of games must be a multiple of \(12.\)

The left-handers can win at most every game involving at least one left-hander, namely \(\binom{L}{2}+2L^2=\tfrac{5L^2-L}{2}.\) On the other hand, they must win \(\tfrac{7}{12}\binom{3L}{2}=\tfrac{7L(3L-1)}8\) games. Comparing these quantities gives \(21L-7\le 20L-4,\) so \(L\le 3.\)

Testing \(L=1,2,3\) gives totals \(3,15,36;\) only \(36\) is a multiple of \(12.\) It is achievable: the \(3\) left-handers can win all \(18\) mixed games and their \(3\) internal games, giving \(21=1.4\cdot 15\) wins.

Thus, the correct answer is B.

14 答案 E

取模长得 \(|z|^5=|z|\)，所以 \(|z|=0\) 或 \(|z|=1\)。\(z=0\) 满足方程，给出一个解。

若 \(|z|=1\)，把原方程乘以 \(z\) 得到 \(z^6=z\overline{z}=|z|^2=1\)。这有 \(6\) 个不同的根，并且这些根的模长都等于 \(1\)。

总共有 \(1+6=7\) 个解。

所以正确答案是 E。

English

Taking magnitudes gives \(|z|^5=|z|,\) so \(|z|=0\) or \(|z|=1.\) The value \(z=0\) works, giving one solution.

If \(|z|=1,\) multiply the equation by \(z\) to get \(z^6=z\overline{z}=|z|^2=1.\) This has \(6\) distinct roots, all of modulus \(1.\)

Altogether there are \(1+6=7\) solutions.

Thus, the correct answer is E.

15 答案 A

之字形的每一段都与场地的水平边成角 \(\theta\)。因此，长度为 \(s\) 的一段在水平方向前进 \(s\cos\theta\) 米。即使最后一段在横跨场地完整宽度之前结束，这一点仍然成立。

将整条 \(120\) 米路径的水平投影相加，得到 \(120\cos\theta=100\)。

所以 \(\cos\theta=\dfrac56\)，从而 \(\theta=\arccos\dfrac56\)。

因此，正确答案是 A。

English

Every segment of the zigzag makes angle \(\theta\) with a horizontal side of the field. Therefore a segment of length \(s\) advances \(s\cos\theta\) meters horizontally. This remains true for the last segment even if it ends before crossing the full width of the field.

Adding the horizontal projections over the entire \(120\)-meter path gives \(120\cos\theta=100.\)

Therefore \(\cos\theta=\dfrac56,\) so \(\theta=\arccos\dfrac56.\)

Thus, the correct answer is A.

16 答案 B

令 \(z=x+yi\)。则 \(1+z+z^2\) \(=(1+x+x^2-y^2)\) \(+y(1+2x)i\)，约束条件为 \[ \begin{gathered} (1+x+x^2-y^2)^2\\ {}+y^2(1+2x)^2=16 \end{gathered}\text{。}\]

在这条闭合有界曲线上 \(y\) 取最大值的点处，隐函数求导（或拉格朗日乘数法）给出 \((1+2x)\bigl(P+2y^2\bigr)=0\)，其中 \(P=1+x+x^2-y^2\)。但 \(P+2y^2=x^2+x+1+y^2\gt 0\)，所以 \(x=-\tfrac12\)。

此时 \(1+2x=0\)，约束化为 \(\left(\tfrac34-y^2\right)^2=16\)。取 \(\tfrac34-y^2=-4\)，得 \(y^2=\tfrac{19}{4}\)，所以最大值为 \(y=\dfrac{\sqrt{19}}{2}\)。

这里 \(m=19\)、\(n=2\)，所以 \(m+n=21\)。

所以正确答案是 B。

English

Write \(z=x+yi.\) Then \(1+z+z^2\) \(=(1+x+x^2-y^2)\) \(+y(1+2x)i,\) and the constraint is \[ \begin{gathered} (1+x+x^2-y^2)^2\\ {}+y^2(1+2x)^2=16. \end{gathered} \]

At a point where \(y\) is maximal on this closed, bounded curve, implicit differentiation (or Lagrange multipliers) gives \((1+2x)\bigl(P+2y^2\bigr)=0,\) where \(P=1+x+x^2-y^2.\) But \(P+2y^2=x^2+x+1+y^2\gt 0,\) so \(x=-\tfrac12.\)

Then \(1+2x=0,\) so the constraint reduces to \(\left(\tfrac34-y^2\right)^2=16.\) Taking \(\tfrac34-y^2=-4\) gives \(y^2=\tfrac{19}{4},\) so the maximum is \(y=\dfrac{\sqrt{19}}{2}.\)

Here \(m=19\) and \(n=2,\) so \(m+n=21.\)

Thus, the correct answer is B.

17 答案 E

设 \(a_n\) 为 Flora 曾经恰好落在 \(n\) 上的概率，其中 \(a_0=1\)。按第一次跳跃分类，\[ a_n=\sum_{k=1}^{n}\dfrac{1}{2^k}\,a_{n-k}\text{。}\]

我们用归纳法证明对每个 \(n\ge 1\) 都有 \(a_n=\tfrac12\)。\(n=1\) 的情形是显然的。若结论直到 \(n-1\) 都成立，则 \[ \begin{aligned} a_n&=\frac{1}{2^n}a_0 +\sum_{k=1}^{n-1}\frac{1}{2^k}\cdot\frac12\\ &=\frac{1}{2^n} +\frac12\left(1-\frac{1}{2^{n-1}}\right)\\ &=\frac12 \end{aligned}\text{。}\]

因此落在 \(10\) 上的概率是 \(\dfrac12\)。

所以正确答案是 E。

English

Let \(a_n\) be the probability that Flora ever lands exactly on \(n,\) with \(a_0=1.\) Conditioning on the first jump, \[ a_n=\sum_{k=1}^{n}\dfrac{1}{2^k}\,a_{n-k}. \]

We prove by induction that \(a_n=\tfrac12\) for every \(n\ge 1.\) The case \(n=1\) is immediate. If the claim holds through \(n-1,\) then \[ \begin{aligned} a_n&=\frac{1}{2^n}a_0 +\sum_{k=1}^{n-1}\frac{1}{2^k}\cdot\frac12\\ &=\frac{1}{2^n} +\frac12\left(1-\frac{1}{2^{n-1}}\right)\\ &=\frac12. \end{aligned} \]

Hence the probability of landing on \(10\) is \(\dfrac12.\)

Thus, the correct answer is E.

18 答案 D

把圆心放在 \(O_1=\left(-\tfrac14,0\right)\) 和 \(O_2=\left(\tfrac14,0\right)\)。由对称性，\(C_3\) 以原点为圆心；与 \(C_1\) 内切给出其半径为 \(1-\tfrac14=\tfrac34\)。

设 \(C_4\) 的半径为 \(r\)，圆心为对称轴上的 \((0,k)\)。与 \(C_3\) 外切可得 \(k=\tfrac34+r\)，与 \(C_1\) 内切可得 \(\sqrt{\tfrac{1}{16}+k^2}=1-r\)。

代入得 \(\tfrac{1}{16}+\left(\tfrac34+r\right)^2=(1-r)^2\)，化简为 \(\tfrac72 r=\tfrac38\)，所以 \(r=\dfrac{3}{28}\)。

所以正确答案是 D。

English

Put the centers at \(O_1=\left(-\tfrac14,0\right)\) and \(O_2=\left(\tfrac14,0\right).\) By symmetry \(C_3\) is centered at the origin, and internal tangency to \(C_1\) gives radius \(1-\tfrac14=\tfrac34.\)

Let \(C_4\) have radius \(r,\) centered at \((0,k)\) on the axis of symmetry. External tangency to \(C_3\) gives \(k=\tfrac34+r,\) and internal tangency to \(C_1\) gives \(\sqrt{\tfrac{1}{16}+k^2}=1-r.\)

Substituting, \(\tfrac{1}{16}+\left(\tfrac34+r\right)^2=(1-r)^2,\) which simplifies to \(\tfrac72 r=\tfrac38,\) so \(r=\dfrac{3}{28}.\)

Thus, the correct answer is D.

19 答案 C

设 \(a=\log_{2023}7\)，\(b=\log_{2023}289\)。因为 \(2023=7\cdot 289\)，所以 \(a+b=1\)。令 \(t=\log_{2023}x\)，每个对数都变成倒数，方程化为 \[ (1+t)=(a+t)(b+t)\text{。}\]

展开并使用 \(a+b=1\)，一次项相消，留下 \(t^2+(ab-1)=0\)。两个根满足 \(t_1+t_2=0\)。

对应的解乘积为 \(x_1x_2=2023^{t_1}\cdot 2023^{t_2}\) \(=2023^{\,t_1+t_2}\) \(=2023^0=1\)。

所以正确答案是 C。

English

Let \(a=\log_{2023}7\) and \(b=\log_{2023}289.\) Since \(2023=7\cdot 289,\) we have \(a+b=1.\) Writing \(t=\log_{2023}x,\) each logarithm becomes a reciprocal, and the equation turns into \[ (1+t)=(a+t)(b+t). \]

Expanding and using \(a+b=1,\) the linear terms cancel, leaving \(t^2+(ab-1)=0.\) Its two roots satisfy \(t_1+t_2=0.\)

The corresponding solutions multiply to \(x_1x_2=2023^{t_1}\cdot 2023^{t_2}\) \(=2023^{\,t_1+t_2}\) \(=2023^0=1.\)

Thus, the correct answer is C.

20 答案 C

设 \(S_n\) 为第 \(n\) 行的和。每个内部项比上一行斜上方两个项之和大 \(1\)，对整行求和得到递推 \[ S_n=2S_{n-1}+(n-2)\text{。}\]

由 \(S_1=1\)，可解得 \(S_n=2^n-n\)（检验：\(S_5=32-5=27\) \(=1+7+11+7+1\)）。

所以 \(S_{2023}=2^{2023}-2023\)。\(2\) 的幂的个位数字按 \(2,4,8,6\) 循环，且 \(2023\equiv 3\pmod 4\)，所以 \(2^{2023}\) 的个位是 \(8\)。于是 \(8-3=5\) 个位数字为 \(5\)。

所以正确答案是 C。

English

Let \(S_n\) be the sum of row \(n.\) Each interior entry is \(1\) more than the sum of the two entries above it, and summing over the row gives the recurrence \[ S_n=2S_{n-1}+(n-2). \]

With \(S_1=1,\) this solves to \(S_n=2^n-n\) (check: \(S_5=32-5=27\) \(=1+7+11+7+1\)).

So \(S_{2023}=2^{2023}-2023.\) Since powers of \(2\) cycle with units digits \(2,4,8,6\) and \(2023\equiv 3\pmod 4,\) \(2^{2023}\) ends in \(8.\) Then \(8-3=5\) gives units digit \(5.\)

Thus, the correct answer is C.

21 答案 A

固定 \(R\)。正二十面体的其他 \(11\) 个顶点中，有 \(5\) 个到该点的距离为 \(1\)，\(5\) 个距离为 \(2\)，还有 \(1\) 个（对顶点）距离为 \(3\)。

有序选取互不相同的 \(Q,S\) 时，\(d(Q,R)=d(R,S)\) 的概率为 \[ \dfrac{5\cdot 4+5\cdot 4}{11\cdot 10}=\dfrac{40}{110}=\dfrac{4}{11}\text{。}\]

由 \(Q\) 和 \(S\) 的对称性，\[ \begin{gathered} P(d(Q,R)\gt d(R,S))\\ {}=\dfrac{1-\tfrac{4}{11}}{2}\\ {}=\dfrac{7}{22} \end{gathered}\text{。}\]

所以正确答案是 A。

English

Fix \(R.\) Among the other \(11\) vertices of the icosahedron, \(5\) are at distance \(1,\) \(5\) are at distance \(2,\) and \(1\) (the antipode) is at distance \(3.\)

Choosing ordered distinct \(Q,S,\) the probability that \(d(Q,R)=d(R,S)\) is \[ \dfrac{5\cdot 4+5\cdot 4}{11\cdot 10}=\dfrac{40}{110}=\dfrac{4}{11}. \]

By the symmetry between \(Q\) and \(S,\) \[ \begin{gathered} P(d(Q,R)\gt d(R,S))\\ {}=\dfrac{1-\tfrac{4}{11}}{2}\\ {}=\dfrac{7}{22}. \end{gathered} \]

Thus, the correct answer is A.

22 答案 B

令 \(n=1\)，得 \(f(1)=1\)。对素数 \(p\)，令 \(n=p\)，得 \(f(p)+p\cdot f(1)=1\)，所以 \(f(p)=1-p\)。令 \(n=p^2\)，由 \(f(p^2)+p\,f(p)+p^2 f(1)=1\) 得 \(f(p^2)=1-p\)。

由于定义关系是积性函数的 Dirichlet 卷积，\(f\) 也是积性的。又 \(2023=7\cdot 17^2\)，\[ \begin{gathered} f(2023)=f(7)\cdot f(17^2)\\ {}=(1-7)(1-17)\\ {}=(-6)(-16)\\ {}=96 \end{gathered}\text{。}\]

所以正确答案是 B。

English

Setting \(n=1\) gives \(f(1)=1.\) For a prime \(p,\) \(n=p\) gives \(f(p)+p\cdot f(1)=1,\) so \(f(p)=1-p.\) For \(n=p^2,\) \(f(p^2)+p\,f(p)+p^2 f(1)=1\) gives \(f(p^2)=1-p.\)

Since the defining relation is a Dirichlet convolution of multiplicative functions, \(f\) is multiplicative. With \(2023=7\cdot 17^2,\) \[ \begin{gathered} f(2023)=f(7)\cdot f(17^2)\\ {}=(1-7)(1-17)\\ {}=(-6)(-16)\\ {}=96. \end{gathered} \]

Thus, the correct answer is B.

23 答案 B

由 AM-GM，\(1+2a\ge 2\sqrt{2a}\)，\(2+2b\ge 4\sqrt{b}\)，且 \(2a+b\ge 2\sqrt{2ab}\)。相乘得 \[ \begin{gathered} (1+2a)(2+2b)(2a+b)\\ {}\ge 16\sqrt{2a}\cdot\sqrt{b}\cdot\sqrt{2ab}\\ {}=32ab \end{gathered}\text{。}\]

等号要求 \(1=2a\)、\(2=2b\) 和 \(2a=b\) 同时成立。这给出 \(a=\tfrac12\)、\(b=1\)，且三者相容，所以恰有一个解。

所以正确答案是 B。

English

By AM-GM, \(1+2a\ge 2\sqrt{2a},\) \(2+2b\ge 4\sqrt{b},\) and \(2a+b\ge 2\sqrt{2ab}.\) Multiplying, \[ \begin{gathered} (1+2a)(2+2b)(2a+b)\\ {}\ge 16\sqrt{2a}\cdot\sqrt{b}\cdot\sqrt{2ab}\\ {}=32ab. \end{gathered} \]

Equality requires \(1=2a,\) \(2=2b,\) and \(2a=b\) simultaneously. These give \(a=\tfrac12,\) \(b=1,\) which are consistent, so there is exactly one solution.

Thus, the correct answer is B.

24 答案 C

对固定长度 \(n\)，\(\{1,\ldots,10\}\) 中的每个元素可以独立选择永不出现，或第一次出现在 \(A_1,\ldots,A_n\) 中的某一个，因此每个元素有 \(n+1\) 种选择。于是长度为 \(n\) 的链有 \((n+1)^{10}\) 条。

因此 \[ K=\sum_{n=1}^{10}(n+1)^{10}=\sum_{k=2}^{11}k^{10}\text{。} \] 模 \(10\) 时，\(k=2,\ldots,11\) 各项化为 \(4,9,6,5,6,9,4,1,0,1\)，和为 \(45\equiv 5\)。

所以正确答案是 C。

English

For a fixed length \(n,\) each element of \(\{1,\ldots,10\}\) independently either never appears or first appears in one of \(A_1,\ldots,A_n,\) giving \(n+1\) choices. Hence there are \((n+1)^{10}\) chains of length \(n.\)

Summing, \[ K=\sum_{n=1}^{10}(n+1)^{10}=\sum_{k=2}^{11}k^{10}. \] Modulo \(10,\) the terms \(k=2,\ldots,11\) reduce to \(4,9,6,5,6,9,4,1,0,1,\) which sum to \(45\equiv 5.\)

Thus, the correct answer is C.

25 答案 C

由 De Moivre 公式，\((\cos x+i\sin x)^{2023}\) \(=\cos 2023x+i\sin 2023x\)。展开左边并取虚部与实部之比，再把分子、分母同除以 \(\cos^{2023}x\)，即可把 \(\tan 2023x\) 写成题中关于 \(\tan x\) 的有理函数。

系数 \(a_{2023}\) 是分子中 \(\tan^{2023}x\) 的系数，来自 \(k=2023\) 项：\[ \begin{gathered} a_{2023}=(-1)^{\frac{2023-1}{2}}\binom{2023}{2023}\\ {}=(-1)^{1011}\\ {}=-1 \end{gathered}\text{。}\]

所以正确答案是 C。

English

By De Moivre, \((\cos x+i\sin x)^{2023}\) \(=\cos 2023x+i\sin 2023x.\) Expanding the left side and taking the ratio of imaginary to real parts gives \(\tan 2023x\) as the stated rational function of \(\tan x\) after dividing numerator and denominator by \(\cos^{2023}x.\)

The coefficient \(a_{2023}\) is the coefficient of \(\tan^{2023}x\) in the numerator, which comes from the \(k=2023\) term: \[ \begin{gathered} a_{2023}=(-1)^{\frac{2023-1}{2}}\binom{2023}{2023}\\ {}=(-1)^{1011}\\ {}=-1. \end{gathered} \]

Thus, the correct answer is C.