1991 AMC 12 真题（P版）

1 答案 E

直接代入可得 \[ \begin{aligned} \boxed{1,-2,-3} &=\frac{-3+1}{-3-(-2)}\\ &=\frac{-2}{-1}=2 \end{aligned}\text{。}\]

因此正确答案为 E。

English

Direct substitution gives \[ \begin{aligned} \boxed{1,-2,-3} &=\frac{-3+1}{-3-(-2)}\\ &=\frac{-2}{-1}=2. \end{aligned} \]

Thus the correct answer is E.

2 答案 E

由于 \(\pi\gt3\)，\(3-\pi\) 为负数。因此 \(|3-\pi|=-(3-\pi)=\pi-3\)。

因此正确答案为 E。

English

Since \(\pi\gt3,\) the quantity \(3-\pi\) is negative. Therefore \(|3-\pi|=-(3-\pi)=\pi-3.\)

Thus the correct answer is E.

3 答案 A

有 \(4^{-1}-3^{-1}=\frac14-\frac13=-\frac1{12}\)。再取 \(-1\) 次幂就是求倒数，得到 \(-12\)。

因此正确答案为 A。

English

We have \(4^{-1}-3^{-1}=\frac14-\frac13=-\frac1{12}.\) Raising this result to the power \(-1\) takes its reciprocal, giving \(-12.\)

Thus the correct answer is A.

4 答案 C

直角三角形有一个 \(90^\circ\) 角，另外两个正角之和只有 \(90^\circ\)。所以另外两个角都不可能是钝角，三角形不可能既是直角三角形又是钝角三角形。

因此正确答案为 C。

English

A right triangle has one \(90^\circ\) angle, leaving only \(90^\circ\) for its other two positive angles together. Neither remaining angle can therefore be obtuse. Hence a triangle cannot be both right and obtuse.

Thus the correct answer is C.

5 答案 E

箭杆 \(CDEF\) 是一个 \(20\)×\(10\) 的矩形，所以面积为 \(200\)。另外，\(BG=BC+CF+FG\) \(=5+10+5=20\)。由于 \(\angle BAG=90^\circ\) 且 \(AB=AG\)，三角形 \(ABG\) 是斜边长为 \(20\) 的等腰直角三角形。两条直角边长均为 \(10\sqrt2\)，所以其面积为 \(\frac12(10\sqrt2)^2=100\)。总面积为 \(200+100=300\)。

因此正确答案为 E。

English

The shaft \(CDEF\) is a \(20\)-by-\(10\) rectangle, so its area is \(200.\) Also \(BG=BC+CF+FG\) \(=5+10+5=20.\) Because \(\angle BAG=90^\circ\) and \(AB=AG,\) triangle \(ABG\) is an isosceles right triangle with hypotenuse \(20.\) Its legs are \(10\sqrt2,\) so its area is \(\frac12(10\sqrt2)^2=100.\) The total area is \(200+100=300.\)

Thus the correct answer is E.

6 答案 E

从内向外使用分数指数，得到 \[ \begin{aligned} \sqrt{x\sqrt{x\sqrt{x}}} &=\sqrt{x\cdot x^{\frac{3}{4}}}\\ &=x^{\frac{7}{8}} =\sqrt[8]{x^7} \end{aligned}\text{。}\]

因此正确答案为 E。

English

Starting inside and using fractional exponents, \[ \begin{aligned} \sqrt{x\sqrt{x\sqrt{x}}} &=\sqrt{x\cdot x^{\frac{3}{4}}}\\ &=x^{\frac{7}{8}} =\sqrt[8]{x^7}. \end{aligned} \]

Thus the correct answer is E.

7 答案 B

分子和分母同时除以非零数 \(b\)，可得 \[ \frac{a+b}{a-b}=\frac{\frac{a}{b}+1}{\frac{a}{b}-1}=\frac{x+1}{x-1}\text{。}\]

因此正确答案为 B。

English

Dividing both numerator and denominator by the nonzero number \(b\) gives \[ \frac{a+b}{a-b}=\frac{\frac{a}{b}+1}{\frac{a}{b}-1}=\frac{x+1}{x-1}. \]

Thus the correct answer is B.

8 答案 C

液体的体积为 \(6\cdot3\cdot12=216\) 立方厘米。因此 \[ 0.1\pi r^2=216\text{，}\]所以 \(r^2=\frac{2160}{\pi}\)，并且 \(r=\sqrt{\frac{2160}{\pi}}\)。

因此正确答案为 C。

English

The liquid volume is \(6\cdot3\cdot12=216\) cubic centimeters. Thus \[ 0.1\pi r^2=216, \] so \(r^2=\frac{2160}{\pi}\) and \(r=\sqrt{\frac{2160}{\pi}}.\)

Thus the correct answer is C.

9 答案 D

总增长因子为 \[ \begin{aligned} &\left(1+\frac{i}{100}\right) \left(1+\frac{j}{100}\right)\\ &\qquad=1+\frac{i+j}{100} +\frac{ij}{10000} \end{aligned}\text{。}\]因此，增长的百分数为 \(i+j+\frac{ij}{100}\)。

因此正确答案为 D。

English

The combined growth factor is \[ \begin{aligned} &\left(1+\frac{i}{100}\right) \left(1+\frac{j}{100}\right)\\ &\qquad=1+\frac{i+j}{100} +\frac{ij}{10000}. \end{aligned} \] Hence the percent increase is \(i+j+\frac{ij}{100}.\)

Thus the correct answer is D.

10 答案 B

经过 \(P\) 的最长弦是长为 \(30\) 的直径。最短弦垂直于经过 \(P\) 的半径，其长度为 \[ 2\sqrt{15^2-9^2}=24\text{。}\]当弦在这两个位置之间转动时，其长度连续地从 \(24\) 变到 \(30\)。两个端点长度各出现一次，而中间的五个整数长度 \(25,26,27,28,29\) 各出现两次。因此总数为 \(1+2(5)+1=12\)。

因此正确答案为 B。

English

The longest chord through \(P\) is a diameter of length \(30.\) The shortest is perpendicular to the radius through \(P,\) and has length \[ 2\sqrt{15^2-9^2}=24. \] As the chord rotates between these positions, its length varies continuously from \(24\) to \(30.\) The endpoint lengths each occur once, while each of the five interior integer lengths \(25,26,27,28,29\) occurs twice. Thus the count is \(1+2(5)+1=12.\)

Thus the correct answer is B.

11 答案 B

杰克在 \(\frac{5}{15}=\frac{1}{3}\) 小时后到达山顶。此时吉尔已经跑了 \(\frac{1}{3}-\frac{1}{6}=\frac{1}{6}\) 小时，上坡跑了 \(\frac{16}{6}=\frac{8}{3}\) 千米，所以两人相距 \(5-\frac{8}{3}=\frac{7}{3}\) 千米。他们的相对接近速度为 \(20+16=36\) 千米/小时，因此经过 \(\frac{\frac{7}{3}}{36}=\frac{7}{108}\) 小时相遇。杰克下坡的路程为 \[ 20\left(\frac7{108}\right)=\frac{35}{27} \] 千米。

因此正确答案为 B。

English

Jack reaches the top in \(\frac{5}{15}=\frac{1}{3}\) hour. Jill has then run for \(\frac{1}{3}-\frac{1}{6}=\frac{1}{6}\) hour and is \(\frac{16}{6}=\frac{8}{3}\) km uphill, so they are \(5-\frac{8}{3}=\frac{7}{3}\) km apart. Their closing speed is \(20+16=36\) km/hr, so they meet after \(\frac{\frac{7}{3}}{36}=\frac{7}{108}\) hour. Jack descends \[ 20\left(\frac7{108}\right)=\frac{35}{27} \] kilometers.

Thus the correct answer is B.

12 答案 D

内角和为 \(720^\circ\)，所以 \[ \begin{aligned} 6a+15d&=720,\\ 2a+5d&=240 \end{aligned}\text{。}\]因此 \(d\) 为偶数。最大角为 \(a+5d=120+\frac52d\)，且必须小于 \(180\)。所以 \(d\lt24\)，允许的最大偶数为 \(d=22\)。此时 \(a=65\)，且 \(m=65+5(22)=175\)。

因此正确答案为 D。

English

The angle sum is \(720^\circ,\) so \[ \begin{aligned} 6a+15d&=720,\\ 2a+5d&=240. \end{aligned} \] Hence \(d\) is even. The largest angle is \(a+5d=120+\frac52d,\) which must be less than \(180.\) Thus \(d\lt24,\) and the largest allowable even value is \(d=22.\) It gives \(a=65\) and \(m=65+5(22)=175.\)

Thus the correct answer is D.

13 答案 D

已知概率为 \(P(X)=\frac14\) 和 \(P(Y)=\frac35\)。因此 \[ P(Z)=1-\frac14-\frac35=\frac3{20}\text{。}\]\(Z\) 失败的概率为 \(\frac{17}{20}\)，所以 \(Z\) 获胜的反向赔率为 \(17\) 比 \(3\)。

因此正确答案为 D。

English

The given probabilities are \(P(X)=\frac14\) and \(P(Y)=\frac35.\) Therefore \[ P(Z)=1-\frac14-\frac35=\frac3{20}. \] The probability that \(Z\) loses is \(\frac{17}{20},\) so the odds against \(Z\) are \(17\)-to-\(3.\)

Thus the correct answer is D.

14 答案 C

对任意质数 \(p\)，取 \(x=p^{201}=(p^{67})^3\)。这是一个立方数，它的正因数为 \(1,p,p^2,\ldots,p^{201}\)，共有 \(202\) 个。因此 \(202\) 可以出现。

因此正确答案为 C。

English

For any prime \(p,\) take \(x=p^{201}=(p^{67})^3.\) This is a cube, and its positive divisors are \(1,p,p^2,\ldots,p^{201},\) a total of \(202.\) Thus \(202\) can occur.

Thus the correct answer is C.

15 答案 B

条件说明每把空椅子都与一把已占椅子相邻，所以不可能有三把连续的椅子都空着。因此，每把已占椅子至多可以对应它自己以及其后的两把空椅子，从而 \(60\le3N\)，即 \(N\ge20\)。沿圆桌重复“已占、空、空”的排列即可达到 \(N=20\)。

因此正确答案为 B。

English

The condition says that every empty chair has an occupied neighbor, so no three consecutive chairs may all be empty. Therefore each occupied chair can account for at most itself and the two empty chairs following it, giving \(60\le3N\) and \(N\ge20.\) Repeating the pattern occupied-empty-empty around the table attains \(N=20.\)

Thus the correct answer is B.

16 答案 D

若有 \(s\) 名毕业班学生，则有 \(1.5s\) 名非毕业班学生，所以 \(2.5s=100\)，从而 \(s=40\)。设非毕业班学生的平均分为 \(x\)，则毕业班学生的平均分为 \(1.5x\)。总分方程为 \[ 60x+40(1.5x)=100(100)\text{，}\]所以 \(120x=10000\)，毕业班学生的平均分为 \(1.5x=125\)。

因此正确答案为 D。

English

If there are \(s\) seniors, then there are \(1.5s\) non-seniors, so \(2.5s=100\) and \(s=40.\) Let the non-senior mean be \(x;\) the senior mean is \(1.5x.\) The total score equation is \[ 60x+40(1.5x)=100(100), \] so \(120x=10000,\) and the senior mean is \(1.5x=125.\)

Thus the correct answer is D.

17 答案 D

区间内的回文数形如 \(1dd1\)，其中 \(d\) 是一个数字。代数上，\[ \begin{aligned} 1dd1&=1001+110d\\ &=11(91+10d) \end{aligned}\text{。}\]因此，两位回文质数只能是 \(11\)，且 \(91+10d\) 必须是一个三位回文质数。当 \(d=0,1,\ldots,9\) 时，\(d=1,4,6,9\) 给出质数 \(101,131,151,181\)。所以共有 \(4\) 个年份。

因此正确答案为 D。

English

The palindromes in the interval are \(1dd1,\) where \(d\) is a digit. Algebraically, \[ \begin{aligned} 1dd1&=1001+110d\\ &=11(91+10d). \end{aligned} \] The two-digit prime palindrome must therefore be \(11,\) and \(91+10d\) must be a three-digit prime palindrome. For \(d=0,1,\ldots,9,\) the prime values occur for \(d=1,4,6,9,\) giving \(101,131,151,181.\) Thus there are \(4\) years.

Thus the correct answer is D.

18 答案 D

令 \(z=x+yi\)，则 \[ \begin{aligned} (3+4i)z &=(3x-4y)\\ &\quad +(4x+3y)i \end{aligned}\text{。}\]它恰在 \(4x+3y=0\) 时为实数，而这是一条经过原点的直线的方程。

因此正确答案为 D。

English

Writing \(z=x+yi,\) \[ \begin{aligned} (3+4i)z &=(3x-4y)\\ &\quad +(4x+3y)i. \end{aligned} \] This is real exactly when \(4x+3y=0,\) which is the equation of a line through the origin.

Thus the correct answer is D.

19 答案 B

设 \(C=(0,0)\)、\(A=(0,3)\)、\(B=(4,0)\)。由于 \(\overrightarrow{AB}=(4,-3)\)，长度为 \(12\)、垂直于 \(\overrightarrow{AB}\) 且背离 \(C\) 的向量为 \((\frac{36}{5},\frac{48}{5})\)。因此 \(D=(\frac{36}{5},\frac{63}{5})\)。又因 \(DE\parallel AC\)，有 \(E=(\frac{36}{5},0)\)，所以 \(DE=\frac{63}{5}\)，\(DB=13\)。于是 \[ \frac{DE}{DB}=\frac{\frac{63}{5}}{13}=\frac{63}{65}\text{。}\]因此 \(m+n=63+65=128\)。

因此正确答案为 B。

English

Place \(C=(0,0),\) \(A=(0,3),\) and \(B=(4,0).\) Since \(\overrightarrow{AB}=(4,-3),\) the length-\(12\) vector perpendicular to \(\overrightarrow{AB}\) and directed away from \(C\) is \((\frac{36}{5},\frac{48}{5}).\) Hence \(D=(\frac{36}{5},\frac{63}{5}).\) Since \(DE\parallel AC,\) we have \(E=(\frac{36}{5},0),\) so \(DE=\frac{63}{5}\) and \(DB=13.\) Therefore \[ \frac{DE}{DB}=\frac{\frac{63}{5}}{13}=\frac{63}{65}. \] Thus \(m+n=63+65=128.\)

Thus the correct answer is B.

20 答案 E

令 \(a=2^x-4\)，\(b=4^x-2\)。右边为 \((a+b)^3\)，所以 \[ a^3+b^3=(a+b)^3 \] 等价于 \(ab(a+b)=0\)。三种情形分别给出 \(2^x=4\)、\(4^x=2\) 或 \(4^x+2^x=6\)，对应的实数解分别为 \(x=2,\frac12\) 和 \(1\)。它们的和为 \(2+\frac12+1=\frac72\)。

因此正确答案为 E。

English

Let \(a=2^x-4\) and \(b=4^x-2.\) The right side is \((a+b)^3,\) so \[ a^3+b^3=(a+b)^3 \] is equivalent to \(ab(a+b)=0.\) The cases give \(2^x=4,\) \(4^x=2,\) or \(4^x+2^x=6,\) whose real solutions are \(x=2,\frac12,\) and \(1,\) respectively. Their sum is \(2+\frac12+1=\frac72.\)

Thus the correct answer is E.

21 答案 A

由于 \[ \sec^2\theta=\frac1{\cos^2\theta} =\frac{\csc^2\theta}{\csc^2\theta-1}\text{，}\]在定义中取 \(x=\csc^2\theta\)。于是 \[ f(\sec^2\theta)=\frac1{\csc^2\theta}=\sin^2\theta\text{。}\]

因此正确答案为 A。

English

Since \[ \sec^2\theta=\frac1{\cos^2\theta} =\frac{\csc^2\theta}{\csc^2\theta-1}, \] choose \(x=\csc^2\theta\) in the definition. Then \[ f(\sec^2\theta)=\frac1{\csc^2\theta}=\sin^2\theta. \]

Thus the correct answer is A.

22 答案 B

由于 \(PA=4\) 且 \(AB=4\)，有 \(PB=8\)。因此，以 \(P\) 为中心、把小圆映到大圆的位似比为 \(2\)。若小圆半径为 \(r\)，则 \(3r\) 是其圆心到 \(P\) 的距离：两个圆心位于同一条射线上，它们到 \(P\) 的距离之比为 \(2\)，且两圆心间的距离为 \(r+2r=3r\)。对由 \(P\)、小圆圆心和 \(A\) 构成的直角三角形应用勾股定理，\[ PA^2=(3r)^2-r^2=8r^2\text{。}\]所以 \(16=8r^2\)，从而 \(r^2=2\)，面积为 \(2\pi\)。

因此正确答案为 B。

English

Since \(PA=4\) and \(AB=4,\) we have \(PB=8.\) The homothety centered at \(P\) taking the smaller circle to the larger therefore has ratio \(2.\) If the smaller radius is \(r,\) its center is \(3r\) from \(P:\) the centers are on the same ray, their distances from \(P\) have ratio \(2,\) and their difference is \(r+2r=3r.\) Applying the right triangle formed by \(P,\) the small center, and \(A\), \[ PA^2=(3r)^2-r^2=8r^2. \] Thus \(16=8r^2,\) so \(r^2=2\) and the area is \(2\pi.\)

Thus the correct answer is B.

23 答案 C

设 \(B=(0,0)\)、\(C=(2,0)\)、\(D=(2,2)\)、\(A=(0,2)\)。则 \(E=(0,1)\)、\(F=(1,0)\)，且直线交点为 \[ \begin{aligned} I&=AF\cap DE =\left(\frac25,\frac65\right),\\ H&=AF\cap BD =\left(\frac23,\frac23\right) \end{aligned}\text{。}\]鞋带公式给出 \[ \begin{aligned} [BEIH] &=\frac12\left| \frac4{15}-\frac25-\frac45 \right|\\ &=\frac7{15} \end{aligned}\text{。}\]

因此正确答案为 C。

English

Set \(B=(0,0),\) \(C=(2,0),\) \(D=(2,2),\) and \(A=(0,2).\) Then \(E=(0,1),\) \(F=(1,0),\) and the line intersections are \[ \begin{aligned} I&=AF\cap DE =\left(\frac25,\frac65\right),\\ H&=AF\cap BD =\left(\frac23,\frac23\right). \end{aligned} \] The shoelace formula gives \[ \begin{aligned} [BEIH] &=\frac12\left| \frac4{15}-\frac25-\frac45 \right|\\ &=\frac7{15}. \end{aligned} \]

Thus the correct answer is C.

24 答案 D

点 \((u,\log_{10}u)\) 旋转后变为 \[ (x,y)=(-\log_{10}u,u)\text{。}\]因此 \(x=-\log_{10}y\)，所以 \(\log_{10}y=-x\)，且 \(y=10^{-x}\)。

因此正确答案为 D。

English

A point \((u,\log_{10}u)\) rotates to \[ (x,y)=(-\log_{10}u,u). \] Hence \(x=-\log_{10}y,\) so \(\log_{10}y=-x\) and \(y=10^{-x}.\)

Thus the correct answer is D.

25 答案 D

由于 \(T_k=\frac{k(k+1)}{2}\) 且 \(T_k-1=\frac{(k-1)(k+2)}{2}\)，\[ \frac{T_k}{T_k-1} =\frac{k}{k-1}\cdot\frac{k+1}{k+2}\text{。}\]连乘后大量因子相消：\[ \begin{aligned} P_n &=\left(\prod_{k=2}^n\frac{k}{k-1}\right) \left(\prod_{k=2}^n\frac{k+1}{k+2}\right)\\ &=n\cdot\frac3{n+2}\\ &=\frac{3n}{n+2} \end{aligned}\text{。}\]当 \(n=1991\) 时，该值略小于 \(3\)，最接近 \(2.9\)。

因此正确答案为 D。

English

Since \(T_k=\frac{k(k+1)}{2}\) and \(T_k-1=\frac{(k-1)(k+2)}{2},\) \[ \frac{T_k}{T_k-1} =\frac{k}{k-1}\cdot\frac{k+1}{k+2}. \] The product telescopes: \[ \begin{aligned} P_n &=\left(\prod_{k=2}^n\frac{k}{k-1}\right) \left(\prod_{k=2}^n\frac{k+1}{k+2}\right)\\ &=n\cdot\frac3{n+2}\\ &=\frac{3n}{n+2}. \end{aligned} \] For \(n=1991,\) this is just under \(3\) and is closest to \(2.9.\)

Thus the correct answer is D.

26 答案 C

能被 \(5\) 整除迫使第五位为 \(5\)。能被 \(2\) 和 \(6\) 整除迫使第二位和第六位为偶数。接着依次应用整除规则：前三位的数字和必须能被 \(3\) 整除，由第三位和第四位组成的两位数必须能被 \(4\) 整除。检查集合 \(\{1,2,3,4,6\}\) 中余下数字的排列，只剩 \(123654\) 和 \(321654\)。直接检验可知，这两个数都满足全部六个前缀整除条件，所以共有 \(2\) 个。

因此正确答案为 C。

English

Divisibility by \(5\) forces the fifth digit to be \(5.\) Divisibility by \(2\) and \(6\) forces the second and sixth digits to be even. Now apply the divisibility tests successively: the first three digits must have sum divisible by \(3,\) and the two-digit number formed by the third and fourth digits must be divisible by \(4.\) Checking the remaining choices from \(\{1,2,3,4,6\}\) leaves \(123654\) and \(321654.\) Each number directly satisfies all six prefix divisibility conditions, so there are \(2.\)

Thus the correct answer is C.

27 答案 C

因为 \[ \frac1{x-\sqrt{x^2-1}}=x+\sqrt{x^2-1}\text{，}\]已知方程给出 \(x+\sqrt{x^2-1}=10\)。其倒数为 \(x-\sqrt{x^2-1}=\frac{1}{10}\)，相加得到 \(2x=10.1\)，所以 \(x=5.05\)。另外，\[ \frac1{x^2+\sqrt{x^4-1}}=x^2-\sqrt{x^4-1}\text{。}\]因此所求式为 \(2x^2=2(5.05)^2=51.005\)。

因此正确答案为 C。

English

Because \[ \frac1{x-\sqrt{x^2-1}}=x+\sqrt{x^2-1}, \] the given equation implies \(x+\sqrt{x^2-1}=10.\) Its reciprocal is \(x-\sqrt{x^2-1}=\frac{1}{10},\) so adding yields \(2x=10.1\) and \(x=5.05.\) Also \[ \frac1{x^2+\sqrt{x^4-1}}=x^2-\sqrt{x^4-1}. \] Therefore the requested expression is \(2x^2=2(5.05)^2=51.005.\)

Thus the correct answer is C.

28 答案 B

白弹珠数量的奇偶性始终不变，所以含一颗白弹珠的选项不可能出现。此外，最后一次操作从三颗弹珠开始时，可能留下 \(1\) 颗黑弹珠、\(1\) 颗黑弹珠和 \(1\) 颗白弹珠，或 \(2\) 颗白弹珠，但绝不会留下 \(2\) 颗黑弹珠。留下 \(2\) 颗白弹珠的状态确实可以达到：不断把 \(3\) 颗黑弹珠替换为 \(1\) 颗黑弹珠，直到剩下 \(2\) 颗黑弹珠；再两次使用“\(1\) 黑 \(2\) 白”的规则，留下 \(100\) 颗白弹珠；然后交替使用“\(3\) 白”规则和“\(1\) 黑 \(2\) 白”规则，每两次操作把白弹珠数减少 \(2\)，直到剩下 \(2\) 颗。

因此正确答案为 B。

English

The parity of the number of white marbles never changes, so choices with one white marble are impossible. Also a final operation starting with three marbles can leave \(1\) black, \(1\) black and \(1\) white, or \(2\) white, but never \(2\) black. The state with \(2\) white is attainable: repeatedly replace \(3\) black by \(1\) black until \(2\) black remain; twice use the \(1\)-black-\(2\)-white rule, leaving \(100\) white; then alternate the \(3\)-white rule with the \(1\)-black-\(2\)-white rule, reducing the number of white marbles by \(2\) per pair until \(2\) remain.

Thus the correct answer is B.

29 答案 B

使用提示中的坐标，点 \(X=(x,y)\) 的垂直平分线条件为 \(|X-A|=|X-A'|\)，化简得 \[ x+3\sqrt3\,y=8\text{。}\]该直线与 \(AB,\ y=\sqrt3x\) 相交于 \(P=(\frac{4}{5},\frac{4\sqrt3}{5})\)，与 \(AC,\ y=\sqrt3(3-x)\) 相交于 \(Q=(\frac{19}{8},\frac{5\sqrt3}{8})\)。因此 \[ \begin{aligned} PQ &=\sqrt{\left(\frac{63}{40}\right)^2 +\left(\frac{7\sqrt3}{40}\right)^2}\\ &=\frac7{20}\sqrt{21} \end{aligned}\text{。}\]

因此正确答案为 B。

English

With the coordinates in the hint, the perpendicular-bisector condition for \(X=(x,y)\) is \(|X-A|=|X-A'|,\) which simplifies to \[ x+3\sqrt3\,y=8. \] Intersecting this line with \(AB,\ y=\sqrt3x,\) gives \(P=(\frac{4}{5},\frac{4\sqrt3}{5}).\) Intersecting it with \(AC,\ y=\sqrt3(3-x),\) gives \(Q=(\frac{19}{8},\frac{5\sqrt3}{8}).\) Hence \[ \begin{aligned} PQ &=\sqrt{\left(\frac{63}{40}\right)^2 +\left(\frac{7\sqrt3}{40}\right)^2}\\ &=\frac7{20}\sqrt{21}. \end{aligned} \]

Thus the correct answer is B.

30 答案 B

令 \(c=|C|\)，\(u=|A\cup B\cup C|\)。由于 \(n(S)=2^{|S|}\)，方程变为 \(2^{101}+2^c=2^u\)。两个加数必须相等，所以 \(c=101\)，\(u=102\)。在并集中，\(A\)、\(B\) 和 \(C\) 分别缺少 \(2\)、\(2\) 和 \(1\) 个元素。因此，三重交集中最多缺少五个不同的元素，从而 \(|A\cap B\cap C|\ge97\)。令这些缺少的元素彼此不同即可取到等号。

因此正确答案为 B。

English

Let \(c=|C|\) and \(u=|A\cup B\cup C|.\) Since \(n(S)=2^{|S|},\) the equation becomes \(2^{101}+2^c=2^u.\) The two summands must be equal, so \(c=101\) and \(u=102.\) Within the union, \(A,\) \(B,\) and \(C\) omit \(2,\) \(2,\) and \(1\) elements. Thus at most five distinct elements are absent from the triple intersection, giving \(|A\cap B\cap C|\ge97.\) Equality is attained by making those omissions distinct.

Thus the correct answer is B.