2021B AMC 12 真题（P版）

1 答案 D

因为 \(3\pi \approx 9.42\)，不等式 \(|x| \lt 3\pi\) 表示 \(-9.42 \lt x \lt 9.42\)。

这个范围内的整数从 \(-9\) 到 \(9\)，共有 \(19\) 个。

所以正确答案是 D。

English

Since \(3\pi \approx 9.42,\) the inequality \(|x| \lt 3\pi\) means \(-9.42 \lt x \lt 9.42.\)

The integers in this range run from \(-9\) to \(9,\) giving \(19\) values.

Thus, the correct answer is D.

2 答案 B

\(23\) 对全蓝组合用了 \(46\) 名蓝衬衫学生，还剩 \(57 - 46 = 11\) 名蓝衬衫学生。

这 \(11\) 名蓝衬衫学生每人都必须和一名黄衬衫学生配对，所以有 \(11\) 对混合组合，用掉 \(11\) 名黄衬衫学生。

剩下的 \(75 - 11 = 64\) 名黄衬衫学生组成 \(64 \div 2 = 32\) 对全黄组合。

所以正确答案是 B。

English

The \(23\) all-blue pairs account for \(46\) blue students, leaving \(57 - 46 = 11\) blue students.

Each of those \(11\) blue students must be paired with a yellow student, so there are \(11\) mixed pairs, using \(11\) yellow students.

The remaining \(75 - 11 = 64\) yellow students form \(64 \div 2 = 32\) all-yellow pairs.

Thus, the correct answer is B.

3 答案 A

从外向内计算，\(\dfrac{144}{53}-2=\dfrac{38}{53}\)，所以内层分式等于 \(\dfrac{38}{53}\)。

取倒数得 \(1+\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{53}{38}\)，所以 \(\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{15}{38}\)。

于是 \(2+\dfrac{2}{3+x}=\dfrac{38}{15}\)，所以 \(\dfrac{2}{3+x}=\dfrac{8}{15}\)，得到 \(3+x=\dfrac{15}{4}\)。

因此 \(x=\dfrac{15}{4}-3=\dfrac{3}{4}\)。

所以正确答案是 A。

English

Working from the outside in, \(\dfrac{144}{53}-2=\dfrac{38}{53},\) so the inner fraction equals \(\dfrac{38}{53}.\)

Its reciprocal gives \(1+\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{53}{38},\) so \(\cfrac{1}{2+\frac{2}{3+x}}=\dfrac{15}{38}.\)

Then \(2+\dfrac{2}{3+x}=\dfrac{38}{15},\) so \(\dfrac{2}{3+x}=\dfrac{8}{15},\) giving \(3+x=\dfrac{15}{4}.\)

Therefore \(x=\dfrac{15}{4}-3=\dfrac{3}{4}.\)

Thus, the correct answer is A.

4 答案 C

设上午班有 \(3\) 名学生，下午班有 \(4\) 名学生。

所有分数的总和是 \(3\cdot 84+4\cdot 70\) \(=252+280\) \(=532\)。

总平均数是 \(\dfrac{532}{7}=76\)。

所以正确答案是 C。

English

Suppose there are \(3\) students in the morning class and \(4\) in the afternoon class.

The total of all scores is \(3\cdot 84+4\cdot 70\) \(=252+280\) \(=532.\)

The overall mean is \(\dfrac{532}{7}=76.\)

Thus, the correct answer is C.

5 答案 D

绕 \((1,5)\) 逆时针旋转 \(90^\circ\) 会把 \((a,b)\) 变为 \((1-(b-5),\,5+(a-1))\) \(=(6-b,\,4+a)\)。

再关于 \(y=-x\) 反射（即把 \((x,y)\) 变为 \((-y,-x)\)），得到 \((-(4+a),\,-(6-b))\) \(=(-4-a,\,b-6)\)。

令它等于 \((-6,3)\)，得 \(-4-a=-6\) 且 \(b-6=3\)，所以 \(a=2\)，\(b=9\)。

因此 \(b-a=9-2=7\)。

所以正确答案是 D。

English

A \(90^\circ\) counterclockwise rotation about \((1,5)\) sends \((a,b)\) to \((1-(b-5),\,5+(a-1))\) \(=(6-b,\,4+a).\)

Reflecting that about \(y=-x\) (which maps \((x,y)\) to \((-y,-x)\)) gives \((-(4+a),\,-(6-b))\) \(=(-4-a,\,b-6).\)

Setting this equal to \((-6,3)\) gives \(-4-a=-6\) and \(b-6=3,\) so \(a=2\) and \(b=9.\)

Therefore \(b-a=9-2=7.\)

Thus, the correct answer is D.

6 答案 A

圆锥中的水量为 \(\dfrac13\pi(12)^2(18)=864\pi\) 立方厘米。

倒入圆柱后，水高为 \(h\)，满足 \(\pi(24)^2 h=864\pi\)。

于是 \(576h=864\)，所以 \(h=1.5\)。

所以正确答案是 A。

English

The cone holds \(\dfrac13\pi(12)^2(18)=864\pi\) cubic centimeters of water.

Poured into the cylinder, this fills to height \(h\) where \(\pi(24)^2 h=864\pi.\)

Then \(576h=864,\) so \(h=1.5.\)

Thus, the correct answer is A.

7 答案 C

分解质因数，\(34=2\cdot 17\)，\(63=3^2\cdot 7\)，且 \(270=2\cdot 3^3\cdot 5\)，所以 \(N=2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2\)。

令 \(M\) 为奇数部分 \(3^5\cdot 5\cdot 7\cdot 17^2\)。所有因数之和为 \((1+2+4+8)\,\sigma(M)\) \(=15\,\sigma(M)\)。

奇因数之和为 \(\sigma(M)\)，所以偶因数之和为 \(15\,\sigma(M)-\sigma(M)=14\,\sigma(M)\)。

所求比为 \(\sigma(M):14\,\sigma(M)=1:14\)。

所以正确答案是 C。

English

Factoring, \(34=2\cdot 17,\) \(63=3^2\cdot 7,\) and \(270=2\cdot 3^3\cdot 5,\) so \(N=2^3\cdot 3^5\cdot 5\cdot 7\cdot 17^2.\)

Let \(M\) be the odd part \(3^5\cdot 5\cdot 7\cdot 17^2.\) The sum of all divisors is \((1+2+4+8)\,\sigma(M)\) \(=15\,\sigma(M).\)

The odd divisors sum to \(\sigma(M),\) so the even divisors sum to \(15\,\sigma(M)-\sigma(M)=14\,\sigma(M).\)

The ratio is \(\sigma(M):14\,\sigma(M)=1:14.\)

Thus, the correct answer is C.

8 答案 B

把圆心放在高度 \(0\)。两条等弦到圆心距离相等，所以三条等间距直线可放在高度 \(-\tfrac{d}{2},\tfrac{d}{2},\tfrac{3d}{2}\)，其中两条 \(38\) 的弦在 \(\pm\tfrac{d}{2}\)，\(34\) 的弦在 \(\tfrac{3d}{2}\)。

半弦关系给出 \(r^2-\left(\tfrac{d}{2}\right)^2=19^2\) 和 \(r^2-\left(\tfrac{3d}{2}\right)^2=17^2\)。

相减得 \(2d^2=19^2-17^2=72\)，所以 \(d^2=36\)，\(d=6\)。

所以正确答案是 B。

English

Place the center at height \(0.\) Two equal chords lie at equal distances from the center, so the three equally spaced lines are at heights \(-\tfrac{d}{2},\tfrac{d}{2},\tfrac{3d}{2},\) with the two \(38\)-chords at \(\pm\tfrac{d}{2}\) and the \(34\)-chord at \(\tfrac{3d}{2}.\)

Half-chord relations give \(r^2-\left(\tfrac{d}{2}\right)^2=19^2\) and \(r^2-\left(\tfrac{3d}{2}\right)^2=17^2.\)

Subtracting, \(2d^2=19^2-17^2=72,\) so \(d^2=36\) and \(d=6.\)

Thus, the correct answer is B.

9 答案 D

利用 \(\dfrac{1}{\log_{40}2}=\log_2 40\) 和 \(\dfrac{1}{\log_{20}2}=\log_2 20\)，原式变为 \((\log_2 80)(\log_2 40)\) \(-(\log_2 160)(\log_2 20)\)。

令 \(t=\log_2 5\)，则 \(\log_2 80=4+t\)，\(\log_2 40=3+t\)，\(\log_2 160=5+t\)，\(\log_2 20=2+t\)。

原式为 \((4+t)(3+t)\) \(-(5+t)(2+t)\) \(=(12+7t+t^2)\) \(-(10+7t+t^2)\) \(=2\)。

所以正确答案是 D。

English

Using \(\dfrac{1}{\log_{40}2}=\log_2 40\) and \(\dfrac{1}{\log_{20}2}=\log_2 20,\) the expression becomes \((\log_2 80)(\log_2 40)\) \(-(\log_2 160)(\log_2 20).\)

Let \(t=\log_2 5.\) Then \(\log_2 80=4+t,\) \(\log_2 40=3+t,\) \(\log_2 160=5+t,\) \(\log_2 20=2+t.\)

The value is \((4+t)(3+t)\) \(-(5+t)(2+t)\) \(=(12+7t+t^2)\) \(-(10+7t+t^2)\) \(=2.\)

Thus, the correct answer is D.

10 答案 E

总和为 \(1+2+\cdots+37=703\)。若选出的两个数为 \(a\) 和 \(b\)，则 \(703-a-b=ab\)。

所以 \(ab+a+b=703\)，两边加 \(1\) 得 \((a+1)(b+1)=704=2^6\cdot 11\)。

我们需要 \(a+1,b+1\) 是 \(2\) 到 \(38\) 之间的因数。因数对 \(22\cdot 32=704\) 可行，得到 \(a=21\)，\(b=31\)。

它们的差为 \(31-21=10\)。

所以正确答案是 E。

English

The sum \(1+2+\cdots+37=703.\) If the chosen numbers are \(a\) and \(b,\) then \(703-a-b=ab.\)

So \(ab+a+b=703,\) and adding \(1\) gives \((a+1)(b+1)=704=2^6\cdot 11.\)

We need factors \(a+1,b+1\) between \(2\) and \(38.\) The pair \(22\cdot 32=704\) works, giving \(a=21,\) \(b=31.\)

Their difference is \(31-21=10.\)

Thus, the correct answer is E.

11 答案 D

令 \(A=(0,0)\)，\(C=(15,0)\)。则 \(B=\left(\tfrac{33}{5},\tfrac{56}{5}\right)\)，且因为 \(PC=10\)，\(P=(5,0)\)。直线 \(BP\) 的斜率为 \(7\)，所以其方程为 \(y=7(x-5)\)。

要使 \(ABCD\) 成为梯形且 \(D\) 在直线 \(BP\) 上，取 \(CD\parallel AB\)。过 \(C\) 作平行于 \(AB\) 的直线，与 \(BP\) 相交于 \((\tfrac95,-\tfrac{112}{5})\)。

对 \(ABCE\)，其中 \(E\) 在直线 \(BP\) 上，取 \(AE\parallel BC\)。过 \(A\) 作平行于 \(BC\) 的直线，与 \(BP\) 相交于 \((\tfrac{21}{5},-\tfrac{28}{5})\)。

两点横坐标之差为 \(\tfrac{12}{5}\)，纵坐标之差为 \(\tfrac{84}{5}\)，所以 \(DE=\tfrac15\sqrt{12^2+84^2}=12\sqrt2\)。

所以正确答案是 D。

English

Place \(A=(0,0)\) and \(C=(15,0).\) Then \(B=\left(\tfrac{33}{5},\tfrac{56}{5}\right),\) and since \(PC=10,\) \(P=(5,0).\) Line \(BP\) has slope \(7,\) so it is \(y=7(x-5).\)

For \(ABCD\) to be a trapezoid with \(D\) on line \(BP,\) take \(CD\parallel AB.\) The line through \(C\) parallel to \(AB\) meets line \(BP\) at \((\tfrac95,-\tfrac{112}{5}).\)

For \(ABCE\) with \(E\) on line \(BP,\) take \(AE\parallel BC.\) The line through \(A\) parallel to \(BC\) meets line \(BP\) at \((\tfrac{21}{5},-\tfrac{28}{5}).\)

Their coordinate differences are \(\tfrac{12}{5}\) and \(\tfrac{84}{5},\) so \(DE=\tfrac15\sqrt{12^2+84^2}=12\sqrt2.\)

Thus, the correct answer is D.

12 答案 D

令 \(n=|S|\)，\(T\) 为总和，\(M\) 为最大数，\(L\) 为最小数。则 \(\dfrac{T-M}{n-1}=32\)，\(\dfrac{T-M-L}{n-2}=35\)，且 \(\dfrac{T-L}{n-1}=40\)。

用第三个方程减第一个方程：\(\dfrac{M-L}{n-1}=8\)。因为 \(M-L=72\)，得 \(n-1=9\)，所以 \(n=10\)。

于是 \(T-M=288\)，且 \(T-L=360\)。中间的方程给出 \(T-M-L=35\cdot 8=280\)，所以 \(L=288-280=8\)，\(M=80\)。

因此 \(T=288+80=368\)，平均值为 \(\dfrac{368}{10}=36.8\)。

所以正确答案是 D。

English

Let \(n=|S|,\) let \(T\) be the total, \(M\) the greatest, and \(L\) the least. Then \(\dfrac{T-M}{n-1}=32,\) \(\dfrac{T-M-L}{n-2}=35,\) and \(\dfrac{T-L}{n-1}=40.\)

Subtracting the first from the third: \(\dfrac{M-L}{n-1}=8.\) Since \(M-L=72,\) we get \(n-1=9,\) so \(n=10.\)

Then \(T-M=288\) and \(T-L=360.\) The middle equation gives \(T-M-L=35\cdot 8=280,\) so \(L=288-280=8\) and \(M=80.\)

Thus \(T=288+80=368,\) and the average is \(\dfrac{368}{10}=36.8.\)

Thus, the correct answer is D.

13 答案 D

令 \(f(\theta)=1-3\sin\theta+5\cos 3\theta\)。在 \(\theta=0,\tfrac\pi3,\tfrac{2\pi}3,\ldots,2\pi\) 处，其符号依次为 \(+,-,+,-,+,-,+\)。因此，相邻两点之间的六个区间各至少有一个根。

在任意根处，\(5\cos3\theta=3\sin\theta-1\)。于是 \[ \begin{aligned} &25\sin^2 3\theta-\cos^2\theta \\ &\quad =23+6\sin\theta-8\sin^2\theta>0 \end{aligned}\text{。}\]所以 \(15|\sin3\theta|>3|\cos\theta|\)。在区间 \((\tfrac{k\pi}{3},\tfrac{(k+1)\pi}{3})\) 中，每个根处的 \(f'(\theta)=-3\cos\theta-15\sin3\theta\) 的符号都是 \((-1)^{k+1}\)。

因此，同一区间内的每个根都以相同方向穿过横轴。若有两个这样的根，它们之间必有一个反方向的穿越，产生矛盾。所以每个区间恰有一个根，共有 \(6\) 个解。

所以正确答案是 D。

English

Let \(f(\theta)=1-3\sin\theta+5\cos 3\theta.\) At \(\theta=0,\tfrac\pi3,\tfrac{2\pi}3,\ldots,2\pi,\) its signs alternate \(+,-,+,-,+,-,+.\) Therefore there is at least one root in each of the six intervening intervals.

At any root, \(5\cos3\theta=3\sin\theta-1.\) Consequently \[ \begin{aligned} &25\sin^2 3\theta-\cos^2\theta \\ &\quad =23+6\sin\theta-8\sin^2\theta>0. \end{aligned} \] Hence \(15|\sin3\theta|>3|\cos\theta|.\) In the interval \((\tfrac{k\pi}{3},\tfrac{(k+1)\pi}{3}),\) the sign of \(f'(\theta)=-3\cos\theta-15\sin3\theta\) is therefore \((-1)^{k+1}\) at every root.

Thus every root in a given interval crosses the axis in the same direction. Two such roots would require an intervening crossing in the opposite direction, so each interval has exactly one root. There are \(6\) solutions in all.

Thus, the correct answer is D.

14 答案 A

把 \(D\) 放在原点，令 \(A\)，\(C\) 沿矩形的两条边，且 \(M\) 正好在 \(D\) 的上方。则 \(MA^2=AD^2+DM^2\)，\(MC^2=CD^2+DM^2\)，且 \(MB^2=AD^2+CD^2+DM^2\)。

因此 \(MB^2=MA^2+MC^2-DM^2\)。写成 \(MA,MC,MB=k,k+2,k+4\)，得 \(DM^2=k^2+(k+2)^2\) \(-(k+4)^2\) \(=k^2-4k-12\)。

若 \(DM=t\) 是正整数，则 \((k-2)^2-t^2=16\)，所以 \((k-2-t)(k-2+t)=16\)。能使 \(t>0\) 的同奇偶正因数对只有 \((2,8)\)，由此得到 \(k=7\) 和 \(t=3\)。于是 \(AD^2=49-9=40\)，且 \(CD^2=81-9=72\)。

底面积为 \(AD\cdot CD=\sqrt{40}\cdot\sqrt{72}\) \(=\sqrt{2880}=24\sqrt5\)，体积为 \(\tfrac13\cdot 24\sqrt5\cdot 3=24\sqrt5\)。

所以正确答案是 A。

English

Place \(D\) at the origin with \(A,\) \(C\) along the rectangle’s edges and \(M\) directly above \(D.\) Then \(MA^2=AD^2+DM^2,\) \(MC^2=CD^2+DM^2,\) and \(MB^2=AD^2+CD^2+DM^2.\)

Thus \(MB^2=MA^2+MC^2-DM^2.\) Writing \(MA,MC,MB=k,k+2,k+4,\) we get \(DM^2=k^2+(k+2)^2\) \(-(k+4)^2\) \(=k^2-4k-12.\)

If \(DM=t\) is a positive integer, then \((k-2)^2-t^2=16,\) so \((k-2-t)(k-2+t)=16.\) The only positive same-parity factor pair giving \(t>0\) is \((2,8),\) which yields \(k=7\) and \(t=3.\) Thus \(AD^2=49-9=40\) and \(CD^2=81-9=72.\)

The base area is \(AD\cdot CD=\sqrt{40}\cdot\sqrt{72}\) \(=\sqrt{2880}=24\sqrt5,\) and the volume is \(\tfrac13\cdot 24\sqrt5\cdot 3=24\sqrt5.\)

Thus, the correct answer is A.

15 答案 D

右侧内部点到 \(A\)、\(B\) 和 \(C\) 的距离都是 \(2\)，所以这三点位于一个半径为 \(2\) 的圆上。弦 \(AB\) 和 \(BC\) 的长度也都是 \(2\)，所以它们各自在圆心处所对的角为 \(60^\circ\)。因此 \(AC=2\sqrt3\)。另一半同样是它的镜像。

取 \(C=(-1,0)\)，\(D=(1,0)\)，和 \(A=(0,\sqrt{11})\)；则 \(AC=AD=2\sqrt3\)。以 \(A\) 和 \(C\) 为圆心的两个半径为 \(2\) 的圆的另一个交点是 \[ B=\left(-\tfrac12-\tfrac{\sqrt{11}}{2\sqrt3}, \tfrac{\sqrt{11}}2+\tfrac{1}{2\sqrt3}\right)\text{，}\]而 \(E\) 是它关于 \(y\) 轴的反射。

对 \(A,B,C,D,E\) 使用鞋带公式，得到 \([ABCDE]=\sqrt{11}+2\sqrt3\)，等于 \(\sqrt{11}+\sqrt{12}\)。

所以 \(m+n=11+12=23\)。

因此，正确答案是 D。

English

The right interior point is distance \(2\) from \(A,\) \(B,\) and \(C,\) so these three points lie on a circle of radius \(2.\) Chords \(AB\) and \(BC\) also have length \(2,\) so each subtends \(60^\circ\) at the center. Thus \(AC=2\sqrt3.\) Similarly, the other half is its mirror image.

Put \(C=(-1,0),\) \(D=(1,0),\) and \(A=(0,\sqrt{11});\) then \(AC=AD=2\sqrt3.\) The other intersection of the radius-\(2\) circles centered at \(A\) and \(C\) is \[ B=\left(-\tfrac12-\tfrac{\sqrt{11}}{2\sqrt3}, \tfrac{\sqrt{11}}2+\tfrac{1}{2\sqrt3}\right), \] and \(E\) is its reflection across the \(y\)-axis.

Applying the shoelace formula to \(A,B,C,D,E\) gives \([ABCDE]=\sqrt{11}+2\sqrt3,\) which equals \(\sqrt{11}+\sqrt{12}.\)

So \(m+n=11+12=23.\)

Thus, the correct answer is D.

16 答案 A

设 \(f\) 的根为 \(r,s,t\)。因为 \(g\) 是首项系数为一、根为 \(\tfrac1r,\tfrac1s,\tfrac1t\) 的多项式，\[ \begin{aligned} g(1) &= \left(1-\tfrac1r\right)\left(1-\tfrac1s\right) \\ &\quad {}\cdot \left(1-\tfrac1t\right) \\ &= \dfrac{(r-1)(s-1)(t-1)}{rst} \end{aligned}\text{。}\]

现在 \(f(1)=(1-r)(1-s)(1-t)\) \(=1+a+b+c\)，所以 \((r-1)(s-1)(t-1)\) \(=-(1+a+b+c)\)。另外 \(rst=-c\)。

因此 \(g(1)\) \(=\dfrac{-(1+a+b+c)}{-c}\) \(=\dfrac{1+a+b+c}{c}\)。

所以正确答案是 A。

English

Let \(f\) have roots \(r,s,t.\) Since \(g\) is monic with roots \(\tfrac1r,\tfrac1s,\tfrac1t,\) \[ \begin{aligned} g(1) &= \left(1-\tfrac1r\right)\left(1-\tfrac1s\right) \\ &\quad {}\cdot \left(1-\tfrac1t\right) \\ &= \dfrac{(r-1)(s-1)(t-1)}{rst}. \end{aligned} \]

Now \(f(1)=(1-r)(1-s)(1-t)\) \(=1+a+b+c,\) so \((r-1)(s-1)(t-1)\) \(=-(1+a+b+c).\) Also \(rst=-c.\)

Therefore \(g(1)\) \(=\dfrac{-(1+a+b+c)}{-c}\) \(=\dfrac{1+a+b+c}{c}.\)

Thus, the correct answer is A.

17 答案 B

令 \(AB=a\)、\(CD=b\)，并设内点到 \(AB\) 和 \(CD\) 的高度分别为 \(h_a\) 与 \(h_b\)。底边三角形给出 \(\tfrac12 a h_a=4\) 和 \(\tfrac12 b h_b=2\)，所以 \(a h_a=8\)，\(b h_b=4\)。

总面积为 \(2+3+4+5=14\) \(=\tfrac12(a+b)(h_a+h_b)\)，所以 \((a+b)(h_a+h_b)=28\)。展开得 \(a h_a+b h_b+a h_b+b h_a=28\)，因此 \(a h_b+b h_a=16\)。

令 \(u=a h_b\)，\(v=b h_a\)，则 \(u+v=16\)，且 \(uv=(a h_a)(b h_b)=32\)，所以 \(u,v=8\pm 4\sqrt2\)。

最后 \(\dfrac{AB}{CD}\) \(=\dfrac{a}{b}\) \(=\dfrac{a h_b}{b h_b}\) \(=\dfrac{u}{4}\) \(=\dfrac{8+4\sqrt2}{4}\) \(=2+\sqrt2\)。

所以正确答案是 B。

English

Let \(AB=a,\) \(CD=b,\) and let the interior point be at heights \(h_a\) from \(AB\) and \(h_b\) from \(CD.\) The base triangles give \(\tfrac12 a h_a=4\) and \(\tfrac12 b h_b=2,\) so \(a h_a=8\) and \(b h_b=4.\)

The total area is \(2+3+4+5=14\) \(=\tfrac12(a+b)(h_a+h_b),\) so \((a+b)(h_a+h_b)=28.\) Expanding, \(a h_a+b h_b+a h_b+b h_a=28,\) giving \(a h_b+b h_a=16.\)

Let \(u=a h_b\) and \(v=b h_a.\) Then \(u+v=16\) and \(uv=(a h_a)(b h_b)=32,\) so \(u,v=8\pm 4\sqrt2.\)

Finally \(\dfrac{AB}{CD}\) \(=\dfrac{a}{b}\) \(=\dfrac{a h_b}{b h_b}\) \(=\dfrac{u}{4}\) \(=\dfrac{8+4\sqrt2}{4}\) \(=2+\sqrt2.\)

Thus, the correct answer is B.

18 答案 A

令 \(p=|z|^2=z\bar z\)，\(s=z+\bar z\)。则 \(|z+2|^2=p+2s+4\)，且 \(|z^2+1|^2\) \(=p^2+(z^2+\bar z^2)+1\) \(=p^2+(s^2-2p)+1\)。

代入得 \(12p=2(p+2s+4)\) \(+p^2+s^2-2p+1+31\)，化简为 \(p^2-12p+s^2+4s+40=0\)。

配方得 \((p-6)^2+(s+2)^2=0\)，所以 \(p=6\)，\(s=-2\)。

于是 \(z+\dfrac{6}{z}=z+\dfrac{6\bar z}{|z|^2}=z+\bar z=-2\)。

所以正确答案是 A。

English

Let \(p=|z|^2=z\bar z\) and \(s=z+\bar z.\) Then \(|z+2|^2=p+2s+4,\) and \(|z^2+1|^2\) \(=p^2+(z^2+\bar z^2)+1\) \(=p^2+(s^2-2p)+1.\)

Substituting, \(12p=2(p+2s+4)\) \(+p^2+s^2-2p+1+31,\) which simplifies to \(p^2-12p+s^2+4s+40=0.\)

Completing the square gives \((p-6)^2+(s+2)^2=0,\) so \(p=6\) and \(s=-2.\)

Then \(z+\dfrac{6}{z}=z+\dfrac{6\bar z}{|z|^2}=z+\bar z=-2.\)

Thus, the correct answer is A.

19 答案 B

设两个骰子的面数为 \(a\le b\)。因为两个骰子都至少有 \(6\) 个面，和为 \(7\) 恰有 \(6\) 种方式，所以和为 \(10\) 有 \(6\div\tfrac34=8\) 种方式。

掷出 \(10\) 的方式数为 \(\min(a,9)\) \(-\max(1,10-b)+1=8\)。和为 \(12\) 的概率为 \(\tfrac{1}{12}\)，所以它有 \(\tfrac{ab}{12}\) 种方式。

和为 \(10\) 有 \(8\) 种结果，要求 \(a\ge8\) 且 \(b\ge9\)，所以必有 \(a+b\ge17\)。当 \((a,b)=(8,9)\) 时，和为 \(10\) 恰有 \(8\) 种结果，而和为 \(12\) 恰有第一枚骰子掷出 \(3,4,\ldots,8\) 时的 \(6\) 种结果。因为 \(\frac{6}{8\cdot9}=\frac{1}{12}\)，两个条件都成立，从而下界 \(17\) 可以达到。

所以正确答案是 B。

English

Let the dice have \(a\le b\) faces. Since both have at least \(6\) faces, a sum of \(7\) occurs in exactly \(6\) ways, so a sum of \(10\) occurs in \(6\div\tfrac34=8\) ways.

The number of ways to roll \(10\) is \(\min(a,9)\) \(-\max(1,10-b)+1=8.\) A sum of \(12\) has probability \(\tfrac{1}{12},\) so it occurs in \(\tfrac{ab}{12}\) ways.

Having \(8\) outcomes for a sum of \(10\) requires \(a\ge8\) and \(b\ge9,\) so necessarily \(a+b\ge17.\) For \((a,b)=(8,9),\) a sum of \(10\) has \(8\) outcomes, while a sum of \(12\) has the \(6\) outcomes with the first die showing \(3,4,\ldots,8.\) Since \(\frac{6}{8\cdot9}=\frac{1}{12},\) both conditions hold and the lower bound \(17\) is attained.

Thus, the correct answer is B.

20 答案 A

因为 \(z^3\equiv 1\pmod{z^2+z+1}\) 且 \(2021=3\cdot 673+2\)，所以 \(z^{2021}\equiv z^2\)。

因此 \(z^{2021}+1\equiv z^2+1\)。再用 \(z^2\equiv -z-1\)，化简，得到 \(-z-1+1=-z\)。

所以 \(R(z)=-z\)。

所以正确答案是 A。

English

Since \(z^3\equiv 1\pmod{z^2+z+1}\) and \(2021=3\cdot 673+2,\) we have \(z^{2021}\equiv z^2.\)

So \(z^{2021}+1\equiv z^2+1.\) Reducing further with \(z^2\equiv -z-1,\) this is \(-z-1+1=-z.\)

Therefore \(R(z)=-z.\)

Thus, the correct answer is A.

21 答案 D

两边取 \(\log_2\)，方程变为 \(2^{\sqrt2}\log_2 x=2^{x-1}\)。代入 \(x=\sqrt2\) 得 \(2^{\sqrt2}\cdot\tfrac12=2^{\sqrt2-1}\) 成立，所以 \(x=\sqrt2\) 是一个解。

令 \(f(x)=2^{x-1}-2^{\sqrt2}\log_2 x\)。则 \(f(1)\gt 0\)、\(f(\sqrt2)=0\)、\(f(2)\lt 0\)，且 \(f(4)\gt 0\)，所以在 \(2\) 与 \(4\) 之间有第二个根 \(x_0\)。

为证明没有其他解，将方程除以 \(2^x\)，并在 \(x\gt1\) 时考察 \(h(x)=\frac{\ln x}{2^x}\)。它的导数符号与 \(\frac{1}{x}-(\ln2)(\ln x)\) 相同，而后者严格递减。因此，\(h\) 先增后减一次，任意水平线与其图像至多相交两次。前面找到的两个根就是全部解。因为 \(2\lt x_0\lt4\)，所以 \(2\lt S=\sqrt2+x_0\lt6\)。

所以正确答案是 D。

English

Taking \(\log_2,\) the equation becomes \(2^{\sqrt2}\log_2 x=2^{x-1}.\) Substituting \(x=\sqrt2\) gives \(2^{\sqrt2}\cdot\tfrac12=2^{\sqrt2-1},\) which holds, so \(x=\sqrt2\) is a solution.

Let \(f(x)=2^{x-1}-2^{\sqrt2}\log_2 x.\) Then \(f(1)\gt 0,\) \(f(\sqrt2)=0,\) \(f(2)\lt 0,\) and \(f(4)\gt 0,\) so there is a second root \(x_0\) between \(2\) and \(4.\)

To prove there are no others, divide the equation by \(2^x\) and consider \(h(x)=\frac{\ln x}{2^x}\) for \(x\gt1.\) Its derivative has the sign of \(\frac{1}{x}-(\ln2)(\ln x),\) a strictly decreasing expression. Thus \(h\) increases once and then decreases, so a horizontal line meets its graph at most twice. The two roots already found are all the solutions. Since \(2\lt x_0\lt4,\) we have \(2\lt S=\sqrt2+x_0\lt6.\)

Thus, the correct answer is D.

22 答案 B

把每堵砖墙看作一个类似 Nim 的堆，并赋予 Grundy 值。一步移走 \(1\) 块或 \(2\) 块相邻砖，可能把一堵墙分成长度为 \(a,b\) 的两堵。因此，\(g(n)\) 是所有满足 \(a+b=n-1\) 或 \(a+b=n-2\) 的 \(g(a)\oplus g(b)\) 的最小未出现非负整数。

从 \(g(0)=0\) 开始，这个递推给出 \(g(1),g(2),\ldots,g(6)\) 依次为 \(1,2,3,1,4,3\)。

后手 Beth 获胜当且仅当各墙 Grundy 值的异或为 \(0\)。检查每个选项，只有 \((6,2,1)\) 给出 \(g(6)\oplus g(2)\oplus g(1)\) \(=3\oplus 2\oplus 1=0\)。

所以正确答案是 B。

English

Treat each wall as a Nim-like heap with a Grundy value. A move removes \(1\) or \(2\) adjacent bricks, possibly splitting a wall into lengths \(a,b.\) Thus \(g(n)\) is the mex of \(g(a)\oplus g(b)\) over \(a+b=n-1\) or \(a+b=n-2.\)

Starting with \(g(0)=0,\) this recurrence gives \(g(1),g(2),\ldots,g(6)\) equal to \(1,2,3,1,4,3,\) respectively.

The second player Beth wins exactly when the XOR of the walls’ Grundy values is \(0.\) Checking each option, only \((6,2,1)\) gives \(g(6)\oplus g(2)\oplus g(1)\) \(=3\oplus 2\oplus 1=0.\)

Thus, the correct answer is B.

23 答案 A

不同且等间距的箱子形成等差数列 \(n,n+d,n+2d\)，其中 \(n,d\ge 1\)。三个编号之和为 \(3(n+d)\)，因此把球固定分配到这三个箱子的概率为 \(2^{-3(n+d)}\)。

三个球可以按 \(3!=6\) 种方式对应到这些箱子，所以总概率为 \(6\sum_{n\ge 1}\sum_{d\ge 1}2^{-3(n+d)}\) \(=6\left(\sum_{n\ge 1}\tfrac{1}{8^n}\right)^2\) \(=6\cdot\tfrac17\cdot\tfrac17=\tfrac{6}{49}\)。

因为 \(\gcd(6,49)=1\)，得 \(p+q=6+49=55\)。

所以正确答案是 A。

English

Evenly spaced distinct bins form an arithmetic progression \(n,n+d,n+2d\) with \(n,d\ge 1.\) The three labels sum to \(3(n+d),\) so a fixed assignment of balls to these bins has probability \(2^{-3(n+d)}.\)

The three balls can be ordered in \(3!=6\) ways, so the total probability is \(6\sum_{n\ge 1}\sum_{d\ge 1}2^{-3(n+d)}\) \(=6\left(\sum_{n\ge 1}\tfrac{1}{8^n}\right)^2\) \(=6\cdot\tfrac17\cdot\tfrac17=\tfrac{6}{49}.\)

Since \(\gcd(6,49)=1,\) we get \(p+q=6+49=55.\)

Thus, the correct answer is A.

24 答案 A

设两条对角线交于 \(O\)，夹角为 \(\theta\)。从 \(A\) 和 \(C\) 到 \(BD\) 的垂足关于 \(O\) 对称，所以 \(PQ=AC\cos\theta=6\)；同理 \(RS=BD\cos\theta=8\)。

平行四边形面积为 \(\tfrac12\cdot AC\cdot BD\sin\theta=15\)，所以 \(AC\cdot BD\sin\theta=30\)。于是 \(\dfrac{48\sin\theta}{\cos^2\theta}=30\)，得 \(\dfrac{\sin\theta}{\cos^2\theta}=\dfrac58\)。

令 \(s=\sin\theta\)，则 \(8s=5(1-s^2)\)，解得 \(s=\dfrac{-4+\sqrt{41}}{5}\)，所以 \(\cos^2\theta=1-s^2=\dfrac{8(\sqrt{41}-4)}{25}\)。

因此 \(d^2=BD^2\) \(=\dfrac{64}{\cos^2\theta}\) \(=8(\sqrt{41}+4)\) \(=32+8\sqrt{41}\)，所以 \(m+n+p=32+8+41=81\)。

所以正确答案是 A。

English

Let the diagonals meet at \(O\) at angle \(\theta.\) The feet of the perpendiculars from \(A\) and \(C\) to \(BD\) are symmetric about \(O,\) so \(PQ=AC\cos\theta=6;\) likewise \(RS=BD\cos\theta=8.\)

The parallelogram’s area is \(\tfrac12\cdot AC\cdot BD\sin\theta=15,\) so \(AC\cdot BD\sin\theta=30.\) Then \(\dfrac{48\sin\theta}{\cos^2\theta}=30,\) giving \(\dfrac{\sin\theta}{\cos^2\theta}=\dfrac58.\)

Writing \(s=\sin\theta,\) \(8s=5(1-s^2)\) gives \(s=\dfrac{-4+\sqrt{41}}{5},\) so \(\cos^2\theta=1-s^2=\dfrac{8(\sqrt{41}-4)}{25}.\)

Then \(d^2=BD^2\) \(=\dfrac{64}{\cos^2\theta}\) \(=8(\sqrt{41}+4)\) \(=32+8\sqrt{41},\) so \(m+n+p=32+8+41=81.\)

Thus, the correct answer is A.

25 答案 E

对斜率 \(m\)，第 \(x\) 列（其中 \(1\le x\le 30\)）贡献 \(\min(30,\lfloor mx\rfloor)\) 个在直线 \(y=mx\) 上或其下方的点，我们需要总数等于 \(300\)。

在 \(m=\tfrac23\) 时，\(30\) 的上限尚未起作用，并且 \(\sum_{x=1}^{30}\lfloor \frac{2x}{3}\rfloor=300\)。计数会保持不变，直到遇到下一个更大的斜率 \(\frac{y}{x}\)，其中 \(x\le30\)。

若 \(\frac{y}{x}>\frac{2}{3}\)，则 \(3y-2x\) 是正整数。最接近的情形满足 \(3y-2x=1\)；在约束 \(x\le30\) 下取满足这个同余式的最大横坐标，得到 \((y,x)=(19,28)\)。若这个正整数至少为 \(2\)，斜率差会更大。因此，所求区间是 \([\tfrac23,\tfrac{19}{28})\)，长度为 \(\tfrac{19}{28}-\tfrac23=\tfrac1{84}\)。

因为 \(\gcd(1,84)=1\)，所以 \(a+b=1+84=85\)。

所以正确答案是 E。

English

For slope \(m,\) column \(x\) (with \(1\le x\le 30\)) contributes \(\min(30,\lfloor mx\rfloor)\) points on or below \(y=mx,\) and we need the total to equal \(300.\)

At \(m=\tfrac23,\) the cap at \(30\) is inactive and \(\sum_{x=1}^{30}\lfloor \frac{2x}{3}\rfloor=300.\) The count remains fixed until the next larger slope \(\frac{y}{x}\) with \(x\le30.\)

If \(\frac{y}{x}>\frac{2}{3},\) then \(3y-2x\) is a positive integer. The closest possibility has \(3y-2x=1;\) maximizing \(x\le30\) in this congruence gives \((y,x)=(19,28).\) Any numerator at least \(2\) gives a larger gap. Hence the interval is \([\tfrac23,\tfrac{19}{28}),\) whose length is \(\tfrac{19}{28}-\tfrac23=\tfrac1{84}.\)

Since \(\gcd(1,84)=1,\) \(a+b=1+84=85.\)

Thus, the correct answer is E.