2019B AMC 12 真题（P版）

1 答案 D

倒水前后水的体积相同，所以 \(\dfrac{5}{6}V_1=\dfrac{3}{4}V_2\)。

因此 \[ \dfrac{V_1}{V_2}=\dfrac{\frac{3}{4}}{\frac{5}{6}}=\dfrac{3}{4}\cdot\dfrac{6}{5}=\dfrac{9}{10}\text{。}\]

所以正确答案是 D。

English

The volume of water is the same before and after, so \(\dfrac{5}{6}V_1=\dfrac{3}{4}V_2.\)

Then \[ \dfrac{V_1}{V_2}=\dfrac{\frac{3}{4}}{\frac{5}{6}}=\dfrac{3}{4}\cdot\dfrac{6}{5}=\dfrac{9}{10}. \]

Thus, D is the correct answer.

2 答案 E

反例需要 \(n\) 不是质数（条件成立），且 \(n-2\) 不是质数（结论不成立）。

在选项中，\(27\) 不是质数，且 \(27-2=25\) 也不是质数。质数 \(11\) 和 \(19\) 不满足条件，而 \(15-2=13\)，\(21-2=19\) 都是质数。

所以正确答案是 E。

English

A counterexample needs \(n\) not prime (so the hypothesis holds) and \(n-2\) not prime (so the conclusion fails).

Among the choices, \(27\) is not prime and \(27-2=25\) is not prime. The primes \(11\) and \(19\) fail the hypothesis, and \(15-2=13,\) \(21-2=19\) are prime.

Thus, E is the correct answer.

3 答案 E

每个点都按 \((x,y)\to(-x,-y)\) 映射：确实有 \((-2,1)\to(2,-1)\) 且 \((-1,4)\to(1,-4)\)。

映射 \((x,y)\to(-x,-y)\) 是绕原点旋转 \(180^\circ\)。

所以正确答案是 E。

English

Each point maps by \((x,y)\to(-x,-y):\) indeed \((-2,1)\to(2,-1)\) and \((-1,4)\to(1,-4).\)

The map \((x,y)\to(-x,-y)\) is a \(180^\circ\) rotation about the origin.

Thus, E is the correct answer.

4 答案 C

分解左边：\[ \begin{gathered} (n+1)!+(n+2)! \\ =(n+1)!\,[1+(n+2)] \\ =(n+1)!\,(n+3) \end{gathered}\text{。}\]

两边同除以 \(n!\) 并使用 \((n+1)!=(n+1)\,n!\) 得到 \[ (n+1)(n+3)=440\text{。}\]

所以 \(n^2+4n-437=0\)，可分解为 \((n-19)(n+23)=0\)，得 \(n=19\)。它的数字和为 \(1+9=10\)。

所以正确答案是 C。

English

Factor the left side: \[ \begin{gathered} (n+1)!+(n+2)! \\ =(n+1)!\,[1+(n+2)] \\ =(n+1)!\,(n+3). \end{gathered} \]

Dividing both sides by \(n!\) and using \((n+1)!=(n+1)\,n!\) gives \[ (n+1)(n+3)=440. \]

So \(n^2+4n-437=0,\) which factors as \((n-19)(n+23)=0,\) giving \(n=19.\) Its digit sum is \(1+9=10.\)

Thus, C is the correct answer.

5 答案 B

设 Casper 有 \(M\) 美分。因为他可以刚好买 \(12\)、\(14\) 或 \(15\) 块价格为整数美分的糖果，所以 \(M\) 是 \(\operatorname{lcm}(12,14,15)=420\) 的倍数。

紫色糖果每块 \(20\) 美分，所以 \(n=\dfrac{M}{20}\)。最小的 \(M\) 是 \(420\)，得 \(n=\dfrac{420}{20}=21\)。

所以正确答案是 B。

English

Let \(M\) be Casper’s money in cents. Since he can exactly buy \(12,\) \(14,\) or \(15\) whole-cent pieces, \(M\) is a multiple of \(\operatorname{lcm}(12,14,15)=420.\)

Purple candy costs \(20\) cents, so \(n=\dfrac{M}{20}.\) The smallest \(M\) is \(420,\) giving \(n=\dfrac{420}{20}=21.\)

Thus, B is the correct answer.

6 答案 A

周长条件给出 \(CA+CB=50-10=40\)，所以 \(C\) 在以 \(A,B\) 为焦点、长轴 \(2a=40\) 的椭圆上。因此 \(a=20\) 且 \(c=5\)，所以半短轴为 \[ b=\sqrt{a^2-c^2}=\sqrt{375}\approx19.36\text{。}\]

若以 \(AB=10\) 为底且面积为 \(100\)，则从 \(C\) 到底边的高必须为 \(\dfrac{2\cdot100}{10}=20\)。但椭圆上的最大可能高度是 \(b\approx19.36\lt20\)，所以不存在这样的 \(C\)。

所以正确答案是 A。

English

The perimeter condition gives \(CA+CB=50-10=40,\) so \(C\) lies on an ellipse with foci \(A,B\) and major axis \(2a=40.\) Thus \(a=20\) and \(c=5,\) so the semi-minor axis is \[ b=\sqrt{a^2-c^2}=\sqrt{375}\approx19.36. \]

For area \(100\) with base \(AB=10,\) the height from \(C\) must be \(\dfrac{2\cdot100}{10}=20.\) But the greatest possible height on the ellipse is \(b\approx19.36\lt20,\) so no such \(C\) exists.

Thus, A is the correct answer.

7 答案 A

平均数为 \(\dfrac{4+6+8+17+x}{5}=\dfrac{35+x}{5}\)。

若 \(x\le6\)，中位数是 \(6\)，所以 \(\dfrac{35+x}{5}=6\)，得 \(x=-5\)，符合该范围。

若 \(6\lt x\lt8\)，中位数是 \(x\)，所以 \(\dfrac{35+x}{5}=x\)，得 \(x=8.75\)，不在范围内。若 \(x\ge8\)，中位数是 \(8\)，所以 \(\dfrac{35+x}{5}=8\)，得 \(x=5\)，不在范围内。

唯一解是 \(x=-5\)，因此和为 \(-5\)。

所以正确答案是 A。

English

The mean is \(\dfrac{4+6+8+17+x}{5}=\dfrac{35+x}{5}.\)

If \(x\le6,\) the median is \(6,\) so \(\dfrac{35+x}{5}=6\) gives \(x=-5,\) which is consistent.

If \(6\lt x\lt8,\) the median is \(x,\) so \(\dfrac{35+x}{5}=x\) gives \(x=8.75,\) not in range. If \(x\ge8,\) the median is \(8,\) so \(\dfrac{35+x}{5}=8\) gives \(x=5,\) not in range.

The only solution is \(x=-5,\) so the sum is \(-5.\)

Thus, A is the correct answer.

8 答案 A

因为 \(f(1-x)=(1-x)^2x^2=f(x)\)，所以 \(f\!\left(\tfrac{k}{2019}\right)=f\!\left(\tfrac{2019-k}{2019}\right)\)。

在这个和中，指标为 \(k\) 的项符号是 \((-1)^{k+1}\)，而指标为 \(2019-k\) 的项数值相等，但符号是 \((-1)^{2019-k+1}=(-1)^{k}\)，恰好相反。

每一项都与它的配对项抵消，所以总和为 \(0\)。

所以正确答案是 A。

English

Since \(f(1-x)=(1-x)^2x^2=f(x),\) we have \(f\!\left(\tfrac{k}{2019}\right)=f\!\left(\tfrac{2019-k}{2019}\right).\)

In the sum, the term with index \(k\) has sign \((-1)^{k+1},\) while the term with index \(2019-k\) equals it in value but has sign \((-1)^{2019-k+1}=(-1)^{k},\) the opposite.

Every term cancels with its partner, so the total is \(0.\)

Thus, A is the correct answer.

9 答案 B

令 \(t=\log_2 x\)，则 \(\log_4 x=\dfrac{t}{2}\)，三边为 \(t,\ \dfrac{t}{2},\ 3\)。

三角形不等式给出 \(t+\dfrac{t}{2}\gt3\)（所以 \(t\gt2\)）以及 \(\dfrac{t}{2}+3\gt t\)（所以 \(t\lt6\)）；第三个不等式自动成立。

因此 \(2\lt\log_2 x\lt6\)，即 \(4\lt x\lt64\)。整数 \(5,6,\ldots,63\) 共有 \(59\) 个。

所以正确答案是 B。

English

Let \(t=\log_2 x.\) Then \(\log_4 x=\dfrac{t}{2},\) and the sides are \(t,\ \dfrac{t}{2},\ 3.\)

The triangle inequalities give \(t+\dfrac{t}{2}\gt3\) (so \(t\gt2\)) and \(\dfrac{t}{2}+3\gt t\) (so \(t\lt6\)); the third inequality is automatic.

Thus \(2\lt\log_2 x\lt6,\) i.e. \(4\lt x\lt64.\) The integers \(5,6,\ldots,63\) number \(59.\)

Thus, B is the correct answer.

10 答案 E

将上排四座城市依次命名为 \(A,B,C,D\)，中排为 \(E,F,G,H\)，下排为 \(I,J,K,L\)。一条使用 \(13\) 条道路的路线是一条开放欧拉迹，所以在所用道路构成的图中，只有 \(A\) 和 \(L\) 的度数为奇数。

在完整地图中，需要改变度数奇偶性的顶点是 \(A,B,C,E,H,J,K,L\)。因为只删除 \(4\) 条道路，这四条道路必须将这 \(8\) 个顶点两两配对。其中 \(E\) 只与 \(A\) 相邻，所以必须删除 \(AE\)；随后 \(BC\) 也被迫删除。同理，\(H\) 只与 \(L\) 相邻，所以必须删除 \(HL\)，随后删除 \(JK\)。

剩余图是一条链 \[ \begin{gathered} A-B-F,\\ F-E-I-J-F,\\ F-G,\\ G-C-D-H-G,\\ G-K-L \end{gathered}\text{。}\]两个 \(4\) 边形环各可沿两个方向遍历，其余部分都被迫确定。因此共有 \(2\cdot2=4\) 条路线。

所以 E 是正确答案。

English

Name the four cities in the top row \(A,B,C,D,\) those in the middle row \(E,F,G,H,\) and those in the bottom row \(I,J,K,L.\) A route using \(13\) roads is an open Euler trail, so in the used graph exactly \(A\) and \(L\) have odd degree.

In the full map, the vertices whose degree parity must change are \(A,B,C,E,H,J,K,L.\) Because only \(4\) roads are removed, those roads must pair these \(8\) vertices. Among them, \(E\) is adjacent only to \(A,\) forcing \(AE\) to be removed; then \(BC\) is forced. Similarly \(H\) is adjacent only to \(L,\) forcing \(HL,\) and then \(JK.\)

The remaining graph is a chain \[ \begin{gathered} A-B-F,\\ F-E-I-J-F,\\ F-G,\\ G-C-D-H-G,\\ G-K-L. \end{gathered} \] Each of the two \(4\)-cycles can be traversed in either direction, and everything else is forced. Hence there are \(2\cdot2=4\) routes.

Thus, E is the correct answer.

11 答案 D

两条棱恰好在共面时能确定一个平面，也就是它们平行或相交。

\(12\) 条棱分成 \(3\) 个方向，每个方向有 \(4\) 条平行棱，得到 \(3\binom{4}{2}=18\) 对平行棱。共顶点的棱对有 \(8\binom{3}{2}=24\) 对。

总数为 \(18+24=42\)。

所以正确答案是 D。

English

Two edges determine a plane exactly when they are coplanar, that is, parallel or intersecting.

The \(12\) edges split into \(3\) directions of \(4\) parallel edges, giving \(3\binom{4}{2}=18\) parallel pairs. Edges sharing a vertex give \(8\binom{3}{2}=24\) intersecting pairs.

The total is \(18+24=42.\)

Thus, D is the correct answer.

12 答案 D

三角形 \(ABC\) 的周长为 \(1+1+\sqrt2=2+\sqrt2\)，且 \(AC=\sqrt2\)。在 \(\triangle ACD\) 中设 \(CD=d\)，则 \(AD=\sqrt{2+d^2}\)，周长相等给出 \[ \sqrt2+d+\sqrt{2+d^2}=2+\sqrt2\text{。}\]

因此 \(\sqrt{2+d^2}=2-d\)，所以 \(2+d^2=4-4d+d^2\)，得 \(d=\dfrac12\) 且 \(AD=\dfrac32\)。

因为 \(\angle BAC=45^\circ\)，令 \(\theta=\angle CAD\) 则 \(2\angle BAD=90^\circ+2\theta\)，所以 \(\sin(2\angle BAD)=\cos 2\theta\)。又 \(\tan\theta=\dfrac{CD}{AC}=\dfrac{1}{2\sqrt2}\)，因此 \[ \begin{gathered} \cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta} \\ =\dfrac{1-\tfrac18}{1+\tfrac18}=\dfrac{7}{9} \end{gathered}\text{。}\]

所以正确答案是 D。

English

Triangle \(ABC\) has perimeter \(1+1+\sqrt2=2+\sqrt2\) and \(AC=\sqrt2.\) In \(\triangle ACD\) let \(CD=d,\) so \(AD=\sqrt{2+d^2}\) and equal perimeters give \[ \sqrt2+d+\sqrt{2+d^2}=2+\sqrt2. \]

Then \(\sqrt{2+d^2}=2-d,\) so \(2+d^2=4-4d+d^2,\) giving \(d=\dfrac12\) and \(AD=\dfrac32.\)

Since \(\angle BAC=45^\circ,\) writing \(\theta=\angle CAD\) gives \(2\angle BAD=90^\circ+2\theta,\) so \(\sin(2\angle BAD)=\cos 2\theta.\) With \(\tan\theta=\dfrac{CD}{AC}=\dfrac{1}{2\sqrt2},\) we get \[ \begin{gathered} \cos2\theta=\dfrac{1-\tan^2\theta}{1+\tan^2\theta} \\ =\dfrac{1-\tfrac18}{1+\tfrac18}=\dfrac{7}{9}. \end{gathered} \]

Thus, D is the correct answer.

13 答案 C

两球落入同一箱子的概率为 \[ \begin{gathered} \sum_{k=1}^\infty \left(2^{-k}\right)^2=\sum_{k=1}^\infty 4^{-k} \\ =\dfrac{\frac{1}{4}}{1-\frac{1}{4}}=\dfrac13 \end{gathered}\text{。}\]

由对称性，红球编号更大和绿球编号更大的概率相等，所以各自概率为 \[ \dfrac{1-\tfrac13}{2}=\dfrac13\text{。}\]

所以正确答案是 C。

English

The probability the balls land in the same bin is \[ \begin{gathered} \sum_{k=1}^\infty \left(2^{-k}\right)^2=\sum_{k=1}^\infty 4^{-k} \\ =\dfrac{\frac{1}{4}}{1-\frac{1}{4}}=\dfrac13. \end{gathered} \]

By symmetry, the red ball being higher and the green ball being higher are equally likely, so each has probability \[ \dfrac{1-\tfrac13}{2}=\dfrac13. \]

Thus, C is the correct answer.

14 答案 C

因为 \(100{,}000=2^5\cdot5^5\)，每个因数都形如 \(2^a5^b\)，其中 \(0\le a,b\le5\)。两个因数的乘积形如 \(2^x5^y\)，其中 \(0\le x,y\le10\)，且每个这样的 \((x,y)\) 都能达到，得到 \(11\cdot11=121\) 个值。

我们需要两个不同的因数。一个值 \(2^x5^y\) 只能表示为某个因数乘以自身，恰好发生在 \(x\) 和 \(y\) 都只有一种拆分方式时，也就是 \(x,y\in\{0,10\}\)。这 \(4\) 个角上的值（\(1,\ 2^{10},\ 5^{10},\ 2^{10}5^{10}\)）不能使用两个不同因数表示。

数量为 \(121-4=117\)。

所以正确答案是 C。

English

Since \(100{,}000=2^5\cdot5^5,\) every divisor is \(2^a5^b\) with \(0\le a,b\le5.\) A product of two divisors is \(2^x5^y\) with \(0\le x,y\le10,\) and every such pair \((x,y)\) is attainable, giving \(11\cdot11=121\) values.

We need two distinct divisors. A value \(2^x5^y\) is forced to be a divisor times itself only when both \(x\) and \(y\) have a unique split, which happens exactly when \(x,y\in\{0,10\}.\) Those \(4\) corner values (\(1,\ 2^{10},\ 5^{10},\ 2^{10}5^{10}\)) cannot use two distinct divisors.

The count is \(121-4=117.\)

Thus, C is the correct answer.

15 答案 E

设 \(A=(0,0)\)，\(\ B=(2,0)\)，\(\ C=(4,0)\)，\(\ D=(6,0)\)，则三个半圆的圆心为 \((1,0),(3,0),(5,0)\)，顶点为 \(E=(1,1)\)，\(\ F=(3,1)\)，\(\ G=(5,1)\)。该圆的圆心为 \(F=(3,1)\)，半径为 \(2\)，所以它经过 \(E\) 和 \(G\)，面积为 \(4\pi\)。

中间半圆 \(BFC\) 完全落在该圆内，去掉面积 \(\dfrac{\pi}{2}\)。对于左边的半圆，重叠部分从 \(E\) 开始，包含它右侧的四分之一圆，但要除去落在大圆下方的那块区域。大圆与 \(x\) 轴相交于 \(x=3-\sqrt3\)。被除去的面积为\[ \begin{aligned} I &=\int_1^{3-\sqrt3}1\,dx\\ &\quad-\int_1^{3-\sqrt3} \sqrt{4-(x-3)^2}\,dx\\ &=2-\dfrac{\sqrt3}{2}-\dfrac{\pi}{3} \end{aligned}\text{。}\] 因此重叠部分的面积为\[ \begin{aligned} R&=\dfrac{\pi}{4}-I\\ &=\dfrac{7\pi}{12}-2+\dfrac{\sqrt3}{2} \end{aligned}\text{。}\] 由对称性，右边的半圆贡献同样大的重叠面积。

阴影面积为\[ 4\pi-\dfrac{\pi}{2}-2R=\dfrac{7}{3}\pi-\sqrt3+4\text{。}\] 因此 \(a=7,\ b=3,\ c=3,\ d=4\)，所以 \(a+b+c+d=17\)。

所以正确答案是 E。

English

Put \(A=(0,0),\) \(\ B=(2,0),\) \(\ C=(4,0),\) \(\ D=(6,0),\) so the semicircles are centered at \((1,0),(3,0),(5,0)\) and their tops are \(E=(1,1),\) \(\ F=(3,1),\) \(\ G=(5,1).\) The circle has center \(F=(3,1)\) and radius \(2,\) so it passes through \(E\) and \(G,\) and has area \(4\pi.\)

The middle semicircle \(BFC\) lies entirely inside the circle, removing area \(\dfrac{\pi}{2}.\) For the left semicircle, the overlap starts at \(E\) and includes its right-hand quarter-circle, except for the region below the large circle. The large circle meets the \(x\)-axis at \(x=3-\sqrt3.\) The excluded area is \[ \begin{aligned} I &=\int_1^{3-\sqrt3}1\,dx\\ &\quad-\int_1^{3-\sqrt3} \sqrt{4-(x-3)^2}\,dx\\ &=2-\dfrac{\sqrt3}{2}-\dfrac{\pi}{3}. \end{aligned} \] Therefore the overlap has area \[ \begin{aligned} R&=\dfrac{\pi}{4}-I\\ &=\dfrac{7\pi}{12}-2+\dfrac{\sqrt3}{2}. \end{aligned} \] By symmetry, the right semicircle contributes the same overlap.

The shaded area is \[ 4\pi-\dfrac{\pi}{2}-2R=\dfrac{7}{3}\pi-\sqrt3+4. \] Hence \(a=7,\ b=3,\ c=3,\ d=4,\) so \(a+b+c+d=17.\)

Thus, E is the correct answer.

16 答案 A

设 \(p(n)\) 为在之前没有落到第 \(3\) 号或第 \(6\) 号的情况下落到第 \(n\) 号睡莲叶的概率。每片睡莲叶把概率 \(\dfrac12\) 传到下一片，另一个 \(\dfrac12\) 传到再下一片，而第 \(3\) 号和第 \(6\) 号不再向外传递概率。

因此 \(p(0)=1,\ p(1)=\dfrac12,\ p(2)=\dfrac34\)，并且（跳过 \(3\)） \(p(4)=\dfrac38,\ p(5)=\dfrac{3}{16}\)，然后（跳过 \(6\)） \(p(7)=\dfrac{3}{32}\)，\(\ p(8)=\dfrac{3}{64}\)，\(\ p(9)=\dfrac{9}{128}\)。

最后 \[ \begin{gathered} p(10)=\dfrac12 p(8)+\dfrac12 p(9) \\ =\dfrac{3}{128}+\dfrac{9}{256} \\ =\dfrac{15}{256} \end{gathered}\text{。}\]

所以正确答案是 A。

English

Let \(p(n)\) be the probability of landing on pad \(n\) without first landing on pad \(3\) or \(6.\) Each pad sends probability \(\dfrac12\) to the next pad and \(\dfrac12\) two pads ahead, and pads \(3\) and \(6\) pass nothing on.

Then \(p(0)=1,\ p(1)=\dfrac12,\ p(2)=\dfrac34,\) and (skipping \(3\)) \(p(4)=\dfrac38,\ p(5)=\dfrac{3}{16},\) then (skipping \(6\)) \(p(7)=\dfrac{3}{32},\) \(\ p(8)=\dfrac{3}{64},\) \(\ p(9)=\dfrac{9}{128}.\)

Finally \[ \begin{gathered} p(10)=\dfrac12 p(8)+\dfrac12 p(9) \\ =\dfrac{3}{128}+\dfrac{9}{256} \\ =\dfrac{15}{256}. \end{gathered} \]

Thus, A is the correct answer.

17 答案 D

这三个点构成等边三角形，当且仅当 \(|z|=|z^3|=|z^3-z|\)。由 \(|z|=|z^3|=|z|^3\) 得 \(|z|=1\)。

于是 \(|z^3-z|=|z|\,|z^2-1|=|z^2-1|\)，所以需要 \(|z^2-1|=1\)。写 \(z=e^{i\theta}\)，则 \(|z^2-1|=2|\sin\theta|=1\)，所以 \(|\sin\theta|=\dfrac12\)。

因此 \(\theta=30^\circ,150^\circ,210^\circ,330^\circ\)，给出四个 \(z\) 的值，并且所得顶点互不相同。

所以正确答案是 D。

English

The three points form an equilateral triangle iff \(|z|=|z^3|=|z^3-z|.\) From \(|z|=|z^3|=|z|^3\) we get \(|z|=1.\)

Then \(|z^3-z|=|z|\,|z^2-1|=|z^2-1|,\) so we need \(|z^2-1|=1.\) Writing \(z=e^{i\theta},\) \(|z^2-1|=2|\sin\theta|=1,\) so \(|\sin\theta|=\dfrac12.\)

This gives \(\theta=30^\circ,150^\circ,210^\circ,330^\circ,\) four values of \(z,\) all yielding distinct vertices.

Thus, D is the correct answer.

18 答案 C

取 \(A=(0,0,0)\)，\(\ B=(3,0,0)\)，\(\ C=(3,3,0)\)，\(\ D=(0,3,0)\)，\(\ E=(0,0,6)\)。则 \[ \begin{gathered} P=(2,0,2), \\ Q=(0,2,2), \\ R=(1,1,4) \end{gathered}\text{。}\]

所以 \(\vec{PQ}=(-2,2,0)\)，\(\vec{PR}=(-1,1,2)\)，从而 \(\vec{PQ}\times\vec{PR}=(4,4,0)\)。

面积为 \(\dfrac12\bigl|(4,4,0)\bigr|=\dfrac12\cdot4\sqrt2=2\sqrt2\)。

所以正确答案是 C。

English

Place \(A=(0,0,0),\) \(\ B=(3,0,0),\) \(\ C=(3,3,0),\) \(\ D=(0,3,0),\) \(\ E=(0,0,6).\) Then \[ \begin{gathered} P=(2,0,2), \\ Q=(0,2,2), \\ R=(1,1,4). \end{gathered} \]

So \(\vec{PQ}=(-2,2,0)\) and \(\vec{PR}=(-1,1,2),\) giving \(\vec{PQ}\times\vec{PR}=(4,4,0).\)

The area is \(\dfrac12\bigl|(4,4,0)\bigr|=\dfrac12\cdot4\sqrt2=2\sqrt2.\)

Thus, C is the correct answer.

19 答案 B

从 \((1,1,1)\) 出发，三名玩家各自把钱给另外两人之一，所以有 \(8\) 个等可能结果；只有 \(2\) 种循环赠送模式会回到 \((1,1,1)\)，概率为 \(\dfrac14\)。

从 \((2,1,0)\) 状态出发，没钱的玩家不给钱；检查另外两人的 \(4\) 个等可能选择，恰好一个会得到 \((1,1,1)\)，概率仍为 \(\dfrac14\)。

因此任意一次铃响后，状态为 \((1,1,1)\) 的概率都是 \(\dfrac14\)，包括铃响 \(2019\) 次之后。

所以正确答案是 B。

English

From \((1,1,1),\) each of the three players gives to one of two others, so there are \(8\) equally likely outcomes; only the \(2\) cyclic gift patterns return to \((1,1,1),\) a probability of \(\dfrac14.\)

From a \((2,1,0)\) state the broke player gives nothing, and checking the \(4\) equally likely choices of the other two shows exactly one yields \((1,1,1),\) again probability \(\dfrac14.\)

So after any ring the probability of \((1,1,1)\) is \(\dfrac14,\) including after \(2019\) rings.

Thus, B is the correct answer.

20 答案 C

设切线交点为 \(P=(x,0)\)。相等切线长给出 \(PA=PB\)，所以 \((x-6)^2+13^2\) \(=(x-12)^2+11^2\)，得 \(x=5\)，即 \(P=(5,0)\)。

圆心 \(O=(h,k)\) 满足 \(OA\perp PA\) 和 \(OB\perp PB\)。因为 \(PA=(1,13)\) 且 \(PB=(7,11)\)，得到 \(h+13k=175\) 和 \(7h+11k=205\)，所以 \(O=\left(\dfrac{37}{4},\dfrac{51}{4}\right)\)。

于是 \(r^2=OA^2\) \(=\left(\dfrac{13}{4}\right)^2+\left(\dfrac14\right)^2\) \(=\dfrac{170}{16}=\dfrac{85}{8}\)，所以面积为 \(\dfrac{85\pi}{8}\)。

所以正确答案是 C。

English

Let \(P=(x,0)\) be the intersection. Equal tangent lengths give \(PA=PB,\) so \((x-6)^2+13^2\) \(=(x-12)^2+11^2,\) yielding \(x=5\) and \(P=(5,0).\)

The center \(O=(h,k)\) satisfies \(OA\perp PA\) and \(OB\perp PB.\) With \(PA=(1,13)\) and \(PB=(7,11),\) these give \(h+13k=175\) and \(7h+11k=205,\) so \(O=\left(\dfrac{37}{4},\dfrac{51}{4}\right).\)

Then \(r^2=OA^2\) \(=\left(\dfrac{13}{4}\right)^2+\left(\dfrac14\right)^2\) \(=\dfrac{170}{16}=\dfrac{85}{8},\) so the area is \(\dfrac{85\pi}{8}.\)

Thus, C is the correct answer.

21 答案 B

如果三个系数都等于同一个值 \(u\)，多项式就是 \(u(x^2+x+1)\)，其根不等于 \(u\)。因此系数集合和根集合都有两个不同的值，所以恰有两个系数相同。由韦达定理，\(r+s=-\dfrac{b}{a}\) 且 \(rs=\dfrac{c}{a}\)。

先设 \(a=b=u\) 且 \(c=v\)。根为 \(u,v\)，所以韦达定理给出 \(u+v=-1\) 且 \(uv=\frac{v}{u}\)。因而 \(v(u^2-1)=0\)，得到 \(x^2+x-2\) 和 \(-x^2-x\)。

如果 \(b=c=v\) 且 \(a=u\)，同一个乘积方程给出 \(v(u^2-1)=0\)，而和的方程只留下 \(u=1,\ v=-\tfrac12\)。这给出 \(x^2-\tfrac12x-\tfrac12\)。

最后，如果 \(a=c=u\) 且 \(b=v\)，乘积方程给出 \(uv=1\)，所以 \(v=\frac{1}{u}\)。和的方程变为 \(u^3+u+1=0\)。这个严格递增的三次函数有唯一实根 \(u\)，因此恰好再得到一个多项式 \(ux^2+\dfrac1u x+u\)。所以共有 \(4\) 个多项式。

所以 B 是正确答案。

English

If all three coefficients had one value \(u,\) the polynomial would be \(u(x^2+x+1),\) whose roots do not equal \(u.\) Thus the coefficient and root sets both have two distinct values, so exactly two coefficients coincide. By Vieta’s formulas, \(r+s=-\dfrac{b}{a}\) and \(rs=\dfrac{c}{a}.\)

First suppose \(a=b=u\) and \(c=v.\) The roots are \(u,v,\) so Vieta gives \(u+v=-1\) and \(uv=\frac{v}{u}.\) Hence \(v(u^2-1)=0,\) producing \(x^2+x-2\) and \(-x^2-x.\)

If \(b=c=v\) and \(a=u,\) the same product equation gives \(v(u^2-1)=0,\) while the sum equation leaves only \(u=1,\ v=-\tfrac12.\) This gives \(x^2-\tfrac12x-\tfrac12.\)

Finally, if \(a=c=u\) and \(b=v,\) the product equation gives \(uv=1,\) so \(v=\frac{1}{u}.\) The sum equation becomes \(u^3+u+1=0.\) This strictly increasing cubic has one real root \(u,\) producing exactly one more polynomial, \(ux^2+\dfrac1u x+u.\) Therefore there are \(4\) polynomials.

Thus, B is the correct answer.

22 答案 C

令 \(a_n=x_n-4\)。简单计算得到 \[ \begin{gathered} a_{n+1}=x_{n+1}-4 \\ =\dfrac{(x_n+5)(x_n-4)}{x_n+6} \\ =a_n\cdot\dfrac{x_n+5}{x_n+6} \end{gathered}\text{。}\]

从 \(a_0=1\) 开始，各项保持为正并递减。因为 \(x_n\) 从 \(5\) 递减趋向 \(4\)，每个比值 \(\dfrac{x_n+5}{x_n+6}\) 都严格介于 \(\dfrac{9}{10}\) 和 \(\dfrac{10}{11}\) 之间。

因此 \(a_m\) 被夹在 \(\left(\dfrac{9}{10}\right)^m\) 与 \(\left(\dfrac{10}{11}\right)^m\) 之间。求解 \(a_m\le2^{-20}\) 可知 \(m\) 大约在 \(132\) 到 \(146\) 之间，落在 \([81,242]\) 中。

所以正确答案是 C。

English

Let \(a_n=x_n-4.\) A short computation gives \[ \begin{gathered} a_{n+1}=x_{n+1}-4 \\ =\dfrac{(x_n+5)(x_n-4)}{x_n+6} \\ =a_n\cdot\dfrac{x_n+5}{x_n+6}. \end{gathered} \]

Starting from \(a_0=1,\) the terms stay positive and decrease. Because \(x_n\) decreases from \(5\) toward \(4,\) each ratio \(\dfrac{x_n+5}{x_n+6}\) lies strictly between \(\dfrac{9}{10}\) and \(\dfrac{10}{11}.\)

Hence \(a_m\) is squeezed between \(\left(\dfrac{9}{10}\right)^m\) and \(\left(\dfrac{10}{11}\right)^m.\) Solving \(a_m\le2^{-20}\) puts \(m\) between about \(132\) and \(146,\) which lies in \([81,242].\)

Thus, C is the correct answer.

23 答案 C

没有两个 \(0\) 相邻，所以 \(0\) 之间由若干 \(1\) 的块隔开，每块大小为 \(1\) 或 \(2\) （不能为 \(3\)）。若有 \(k\) 个零，则有 \(k-1\) 个这样的块，合计 \(19-k\) 个一。

大小为 \(2\) 的块数为 \((19-k)-(k-1)=20-2k\)，它必须满足 \(0\le20-2k\le k-1\)，即 \(7\le k\le10\)。

对 \(k=7,8,9,10\) 求和 \(\binom{k-1}{20-2k}\) 得到 \(\binom66+\binom74+\binom82+\binom90\) \(=1+35+28+1\) \(=65\)。

所以正确答案是 C。

English

No two \(0\)s are adjacent, so the \(0\)s are separated by blocks of \(1\)s, each of size \(1\) or \(2\) (never \(3\)). If there are \(k\) zeros, there are \(k-1\) such blocks summing to \(19-k\) ones.

The number of size-\(2\) blocks is \((19-k)-(k-1)=20-2k,\) which must satisfy \(0\le20-2k\le k-1,\) i.e. \(7\le k\le10.\)

Summing \(\binom{k-1}{20-2k}\) over \(k=7,8,9,10\) gives \(\binom66+\binom74+\binom82+\binom90\) \(=1+35+28+1\) \(=65.\)

Thus, C is the correct answer.

24 答案 C

当 \(a,b,c\) 在 \([0,1]\) 中变化时，集合 \(S\) 是沿向量 \(v_1=1=(1,0)\)，\(\ v_2=\omega=\left(-\dfrac12,\dfrac{\sqrt3}{2}\right)\)，\(\ v_3=\omega^2=\left(-\dfrac12,-\dfrac{\sqrt3}{2}\right)\) 的三个单位线段的闵可夫斯基和。

这是一个带状多边形，其面积等于所有向量对的叉积大小之和。每一对都给出 \(|v_i\times v_j|=\dfrac{\sqrt3}{2}\)。

因此面积为 \(3\cdot\dfrac{\sqrt3}{2}=\dfrac{3\sqrt3}{2}\)。

所以正确答案是 C。

English

As \(a,b,c\) range over \([0,1],\) the set \(S\) is the Minkowski sum of the three unit segments along \(v_1=1=(1,0),\) \(\ v_2=\omega=\left(-\dfrac12,\dfrac{\sqrt3}{2}\right),\) \(\ v_3=\omega^2=\left(-\dfrac12,-\dfrac{\sqrt3}{2}\right).\)

This is a zonogon whose area is the sum of the cross-product magnitudes over pairs. Each pair gives \(|v_i\times v_j|=\dfrac{\sqrt3}{2}.\)

Therefore the area is \(3\cdot\dfrac{\sqrt3}{2}=\dfrac{3\sqrt3}{2}.\)

Thus, C is the correct answer.

25 答案 C

三个重心分别为 \(\dfrac{A+B+C}{3}\)，\(\ \dfrac{B+C+D}{3}\)，\(\ \dfrac{A+C+D}{3}\)。它们两两之差为 \(\dfrac{A-D}{3},\ \dfrac{B-A}{3},\ \dfrac{B-D}{3}\)，所以重心三角形为等边三角形迫使 \(AB=BD=DA\)；也就是说，\(\triangle ABD\) 是边长为 \(s=BD\) 的等边三角形。

沿 \(BD\) 分割，\[ \begin{gathered} [ABCD]=[ABD]+[BCD] \\ =\dfrac{\sqrt3}{4}s^2 \\ {}+\dfrac12\cdot2\cdot6\sin C \end{gathered}\text{，}\]其中 \(C=\angle BCD\)。由余弦定理，\(s^2=40-24\cos C\)，所以 \[ \begin{gathered} [ABCD]=10\sqrt3 \\ {}-6\sqrt3\cos C+6\sin C \end{gathered}\text{。}\]

表达式 \(6\sin C-6\sqrt3\cos C\) 的最大值为 \(\sqrt{6^2+(6\sqrt3)^2}=12\)，所以最大面积是 \(10\sqrt3+12=12+10\sqrt3\)。当 \(C=150^\circ\) 时取等号；构造具有这个角的 \(\triangle BCD\)，并在 \(\overline{BD}\) 另一侧作等边 \(\triangle ABD\)，可以得到凸四边形，所以最大值能够达到。

所以 C 是正确答案。

English

The centroids are \(\dfrac{A+B+C}{3},\) \(\ \dfrac{B+C+D}{3},\) \(\ \dfrac{A+C+D}{3}.\) Their pairwise differences are \(\dfrac{A-D}{3},\ \dfrac{B-A}{3},\ \dfrac{B-D}{3},\) so an equilateral centroid triangle forces \(AB=BD=DA;\) that is, \(\triangle ABD\) is equilateral with side \(s=BD.\)

Splitting along \(BD,\) \[ \begin{gathered} [ABCD]=[ABD]+[BCD] \\ =\dfrac{\sqrt3}{4}s^2 \\ {}+\dfrac12\cdot2\cdot6\sin C, \end{gathered} \] where \(C=\angle BCD.\) By the Law of Cosines \(s^2=40-24\cos C,\) so \[ \begin{gathered} [ABCD]=10\sqrt3 \\ {}-6\sqrt3\cos C+6\sin C. \end{gathered} \]

The expression \(6\sin C-6\sqrt3\cos C\) has maximum \(\sqrt{6^2+(6\sqrt3)^2}=12,\) so the greatest area is \(10\sqrt3+12=12+10\sqrt3.\) Equality occurs at \(C=150^\circ;\) constructing \(\triangle BCD\) with that angle and placing equilateral \(\triangle ABD\) on the opposite side of \(\overline{BD}\) produces a convex quadrilateral, so the maximum is attainable.

Thus, C is the correct answer.