2011B AMC 12 真题（P版）

1 答案 C

两个和分别是 \(2+4+6=12\) 与 \(1+3+5=9\)，所以原式等于 \(\dfrac{12}{9}-\dfrac{9}{12}=\dfrac{4}{3}-\dfrac{3}{4}\)。

通分得到 \[ \dfrac{16}{12}-\dfrac{9}{12}=\dfrac{7}{12}\text{。}\]

所以正确答案是 C。

English

The sums are \(2+4+6=12\) and \(1+3+5=9,\) so the expression equals \(\dfrac{12}{9}-\dfrac{9}{12}=\dfrac{4}{3}-\dfrac{3}{4}.\)

Over a common denominator this is \[ \dfrac{16}{12}-\dfrac{9}{12}=\dfrac{7}{12}. \]

Thus, the correct answer is C.

2 答案 E

前五次总分为 \(90+80+70+60+85=385\)，平均分为 \(77\)。目标是新的平均分至少为 \(80\)。

六次考试平均 \(80\) 分需要总分 \(6\cdot80=480\)，因此第六次至少需要 \(480-385=95\)。

所以正确答案是 E。

English

The five scores sum to \(90+80+70+60+85=385,\) giving an average of \(77.\) The goal is a new average of at least \(80.\)

Six tests averaging \(80\) must total \(6\cdot80=480,\) so the sixth score must be at least \(480-385=95.\)

Thus, the correct answer is E.

3 答案 C

总费用为 \(A+B\)，所以每个人的公平份额为 \(\dfrac{A+B}{2}\)。

LeRoy 已支付 \(A\)，少于他应承担的份额，因此他应给 Bernardo \[ \dfrac{A+B}{2}-A=\dfrac{B-A}{2}\text{。}\]

所以正确答案是 C。

English

The total cost is \(A+B,\) so each person’s fair share is \(\dfrac{A+B}{2}.\)

LeRoy paid \(A,\) which is less than his share, so he must give Bernardo \[ \dfrac{A+B}{2}-A=\dfrac{B-A}{2}. \]

Thus, the correct answer is C.

4 答案 E

因为 \(161=7\cdot23\)，唯一的两位数因数是 \(23\)。这必是 \(a\) 颠倒后的值，所以真正的 \(a\) 是 \(32\)，且 \(b=7\)。

正确乘积为 \[ 32\cdot7=224\text{。}\]

所以正确答案是 E。

English

Since \(161=7\cdot23,\) the only two-digit factor is \(23.\) This must be the reversed value of \(a,\) so the true value of \(a\) is \(32,\) and \(b=7.\)

The correct product is \[ 32\cdot7=224. \]

Thus, the correct answer is E.

5 答案 A

能被 \(1\) 到 \(6\) 每个整数整除的数必须是 \(\operatorname{lcm}(1,2,3,4,5,6)=60\) 的倍数。

\(60\) 的第二小正倍数是 \(120\)，其数字和为 \(1+2+0=3\)。

所以正确答案是 A。

English

A number divisible by every integer from \(1\) to \(6\) must be a multiple of \(\operatorname{lcm}(1,2,3,4,5,6)=60.\)

The second smallest positive multiple of \(60\) is \(120,\) whose digit sum is \(1+2+0=3.\)

Thus, the correct answer is A.

6 答案 C

设 \(O\) 为圆心。两段弧分别为 \(2x\) 和 \(3x\)，且 \(2x+3x=360^\circ\)，所以 \(x=72^\circ\)，小弧 \(BC\) 对应的圆心角为 \(\angle BOC=144^\circ\)。

到切点 \(B\) 和 \(C\) 的半径垂直于切线，所以 \(\angle ABO=\angle ACO=90^\circ\)。在四边形 \(ABOC\) 中，\[ \begin{gathered} \angle BAC=360^\circ-144^\circ-90^\circ \\ {}-90^\circ=36^\circ \end{gathered}\text{。}\]

所以正确答案是 C。

English

Let \(O\) be the center. The arcs measure \(2x\) and \(3x\) with \(2x+3x=360^\circ,\) so \(x=72^\circ\) and the minor arc \(BC\) gives central angle \(\angle BOC=144^\circ.\)

The radii to \(B\) and \(C\) are perpendicular to the tangents, so \(\angle ABO=\angle ACO=90^\circ.\) In quadrilateral \(ABOC,\) \[ \begin{gathered} \angle BAC=360^\circ-144^\circ-90^\circ \\ {}-90^\circ=36^\circ. \end{gathered} \]

Thus, the correct answer is C.

7 答案 B

由 \(\dfrac{x+y}{2}=60\)，得 \(x+y=120\)。要最大化 \(\dfrac{x}{y}\)，应让 \(y\) 尽可能小。

因为 \(x\le99\)，所以 \(y=120-x\ge21\)。取 \(x=99\)、\(y=21\) 可得最大值 \[ \dfrac{99}{21}=\dfrac{33}{7}\text{。}\]

所以正确答案是 B。

English

Since \(\dfrac{x+y}{2}=60,\) we have \(x+y=120.\) To maximize \(\dfrac{x}{y}\) we make \(y\) small.

Because \(x\le99,\) it follows that \(y=120-x\ge21.\) Taking \(x=99\) and \(y=21\) gives the maximum \[ \dfrac{99}{21}=\dfrac{33}{7}. \]

Thus, the correct answer is B.

8 答案 A

两条直线部分对两条路线相同，因此长度差只来自两个半圆端。设内侧半径为 \(r\)，则外侧多出的长度为 \[ 2\pi(r+6)-2\pi r=12\pi\text{。}\]

若速度为 \(x\) 米每秒，则多用的时间给出 \(36x=12\pi\)，所以 \(x=\dfrac{\pi}{3}\)。

所以正确答案是 A。

English

The straight sides are the same length for both paths, so the difference in length comes only from the two semicircular ends. If the inner radius is \(r,\) those ends combine into a full circle, and the extra length is \[ 2\pi(r+6)-2\pi r=12\pi. \]

If her speed is \(x\) meters per second, then the extra time gives \(36x=12\pi,\) so \(x=\dfrac{\pi}{3}.\)

Thus, the correct answer is A.

9 答案 D

区间长度为 \(30\)，其中负数部分长度为 \(20\)，正数部分长度为 \(10\)。因此每个数为正的概率为 \(\dfrac13\)，为负的概率为 \(\dfrac23\)。

乘积为正发生在两数都正或都负：\[ \left(\dfrac13\right)^2+\left(\dfrac23\right)^2=\dfrac19+\dfrac49=\dfrac59\text{。}\]

所以正确答案是 D。

English

The interval has length \(30,\) with \(20\) of it negative and \(10\) of it positive. So each number is positive with probability \(\dfrac13\) and negative with probability \(\dfrac23.\)

The product is positive when both are positive or both are negative: \[ \left(\dfrac13\right)^2+\left(\dfrac23\right)^2=\dfrac19+\dfrac49=\dfrac59. \]

Thus, the correct answer is D.

10 答案 E

因为 \(AB\parallel CD\)，所以 \(\angle CDM=\angle AMD\)。结合 \(\angle AMD=\angle CMD\)，可得 \(\angle CDM=\angle CMD\)，所以 \(\triangle CMD\) 是等腰三角形，且 \(CM=CD=6\)。

因此 \(\triangle MBC\) 在 \(B\) 处为直角，斜边 \(CM=6\)，直角边 \(BC=3\)，所以它是 \(30\)-\(60\)-\(90^\circ\) 三角形，且 \(\angle BMC=30^\circ\)。

最后 \(\angle AMD+\angle CMD\) \(+\angle BMC=180^\circ\)，所以 \(2\angle AMD+30^\circ=180^\circ\)，得 \(\angle AMD=75^\circ\)。

所以正确答案是 E。

English

Because \(AB\parallel CD,\) we have \(\angle CDM=\angle AMD.\) Combined with \(\angle AMD=\angle CMD,\) this gives \(\angle CDM=\angle CMD,\) so \(\triangle CMD\) is isosceles with \(CM=CD=6.\)

Then \(\triangle MBC\) is right-angled at \(B\) with hypotenuse \(CM=6\) and leg \(BC=3,\) so it is a \(30\)-\(60\)-\(90^\circ\) triangle with \(\angle BMC=30^\circ.\)

Finally, \(\angle AMD+\angle CMD\) \(+\angle BMC=180^\circ,\) so \(2\angle AMD+30^\circ=180^\circ,\) giving \(\angle AMD=75^\circ.\)

Thus, the correct answer is E.

11 答案 B

一次跳不行，因为 \((0,0)\) 与 \((1,0)\) 的距离只有 \(1\)。两次跳也不行：中间点必须同时距两端 \(5\)，只能在垂直平分线 \(x=\dfrac12\) 上，而这条线上没有格点。

三次跳可以做到，例如 \[ (0,0)\to(3,4)\to(6,0)\to(1,0)\text{，} \] 其中每一步长度都是 \(5\)。

所以正确答案是 B。

English

One jump cannot work, since \((0,0)\) and \((1,0)\) are only \(1\) apart. Two jumps also fail: the intermediate point would be at distance \(5\) from both, forcing it onto the perpendicular bisector \(x=\dfrac12,\) which contains no lattice points.

Three jumps suffice, for example \[ (0,0)\to(3,4)\to(6,0)\to(1,0), \] where each step has length \(5.\)

Thus, the correct answer is B.

12 答案 A

设八边形边长为 \(1\)。四个角上的三角形是直角等腰三角形，直角边长为 \(\dfrac{\sqrt2}{2}\)，每个面积为 \(\dfrac14\)。四个长方形尺寸为 \(1\) 乘 \(\dfrac{\sqrt2}{2}\)，每个面积为 \(\dfrac{\sqrt2}{2}\)，中心正方形面积为 \(1\)。

总面积为 \[ 4\cdot\dfrac14+4\cdot\dfrac{\sqrt2}{2}+1=2+2\sqrt2\text{。} \] 因此击中中心正方形的概率为 \[ \dfrac{1}{2+2\sqrt2}=\dfrac{\sqrt2-1}{2}\text{。} \]

所以正确答案是 A。

English

Assume the octagon has edge length \(1.\) The four corner triangles are right isosceles with legs \(\dfrac{\sqrt2}{2}\) and area \(\dfrac14\) each. The four rectangles are \(1\) by \(\dfrac{\sqrt2}{2}\) with area \(\dfrac{\sqrt2}{2}\) each, and the center square has area \(1.\)

The total area is \[ 4\cdot\dfrac14+4\cdot\dfrac{\sqrt2}{2}+1=2+2\sqrt2. \] The probability of hitting the center square is \[ \dfrac{1}{2+2\sqrt2}=\dfrac{\sqrt2-1}{2}. \]

Thus, the correct answer is A.

13 答案 B

最大差为 \(w-z=9\)。对任意一个中间数 \(n\)，都有 \(9=(w-n)+(n-z)\)。所列差值中，和为 \(9\) 的数对只有 \(3+6\) 和 \(4+5\)，所以剩下的差必为 \(x-y=1\)。

第二大的差 \(6\) 只能是 \(w-y\) 或 \(x-z\)。若 \(w-y=6\)，四个数是 \(\{w,w-5,w-6,w-9\}\)，所以 \(4w-20=44\)，得到 \(w=16\)。若 \(x-z=6\)，四个数是 \(\{w,w-3,w-4,w-9\}\)，所以 \(4w-16=44\)，得到 \(w=15\)。

可能的值是 \(16\) 和 \(15\)，它们的和为 \(31\)。

因此，正确答案是 B。

English

The largest difference is \(w-z=9.\) For either interior number \(n,\) we have \(9=(w-n)+(n-z).\) The only pairs among the listed differences that sum to \(9\) are \(3+6\) and \(4+5,\) so the remaining difference must be \(x-y=1.\)

The second largest difference \(6\) is either \(w-y\) or \(x-z.\) If \(w-y=6,\) the numbers are \(\{w,w-5,w-6,w-9\},\) so \(4w-20=44\) and \(w=16.\) If \(x-z=6,\) the numbers are \(\{w,w-3,w-4,w-9\},\) so \(4w-16=44\) and \(w=15.\)

The possible values are \(16\) and \(15,\) which sum to \(31.\)

Thus, the correct answer is B.

14 答案 D

令 \(p=FV\)，准线为 \(\ell\)。把 \(F\) 与 \(B\) 投影到 \(\ell\) 上，由焦点与准线的性质可得 \(FB=2p\)，即 \(B\) 到 \(\ell\) 的距离。再由勾股定理，\[ \begin{aligned} VB&=\sqrt{FV^2+FB^2} \\ &=\sqrt{p^2+4p^2}=\sqrt5\,p \end{aligned}\text{。}\]

于是 \(\cos(\angle FVB)=\dfrac{FV}{VB}\) \(=\dfrac{p}{\sqrt5\,p}\) \(=\dfrac{1}{\sqrt5}\)。因为 \(\angle AVB=2\angle FVB\)，所以 \[ \begin{aligned} \cos(\angle AVB) &=2\cos^2(\angle FVB) \\ &\quad {}-1 \\ &=2\cdot\dfrac15-1 \\ &=-\dfrac35 \end{aligned}\text{。}\]

所以正确答案是 D。

English

Let \(p=FV\) and let the directrix be \(\ell.\) Projecting \(F\) and \(B\) onto \(\ell,\) the focus-directrix property gives \(FB=2p\) (the distance from \(B\) to \(\ell\)), and by the Pythagorean Theorem \[ \begin{aligned} VB&=\sqrt{FV^2+FB^2} \\ &=\sqrt{p^2+4p^2}=\sqrt5\,p. \end{aligned} \]

Then \(\cos(\angle FVB)=\dfrac{FV}{VB}\) \(=\dfrac{p}{\sqrt5\,p}\) \(=\dfrac{1}{\sqrt5}.\) Since \(\angle AVB=2\angle FVB,\) \[ \begin{aligned} \cos(\angle AVB) &=2\cos^2(\angle FVB) \\ &\quad {}-1 \\ &=2\cdot\dfrac15-1 \\ &=-\dfrac35. \end{aligned} \]

Thus, the correct answer is D.

15 答案 D

分解得 \[ \begin{aligned} 2^{24}-1 &=(2^{12}-1)(2^{12}+1) \\ &=(2^6-1)(2^6+1) \\ &\quad {}\cdot(2^4+1)(2^8-2^4+1)\text{，} \end{aligned} \] 即 \(63\cdot65\cdot17\cdot241\) \(=3^2\cdot5\cdot7\cdot13\cdot17\cdot241\)。

因为 \(241\) 是三位质数，两位数因数只来自 \(3^2\cdot5\cdot7\cdot13\cdot17\)。它们是 \[ \begin{gathered} 13,15,17,21,35,39,45,51, \\ 63,65,85,91\text{，} \end{gathered} \] 共 \(12\) 个。

所以正确答案是 D。

English

Factoring, \[ \begin{aligned} 2^{24}-1 &=(2^{12}-1)(2^{12}+1) \\ &=(2^6-1)(2^6+1) \\ &\quad {}\cdot(2^4+1)(2^8-2^4+1), \end{aligned} \] which equals \(63\cdot65\cdot17\cdot241\) \(=3^2\cdot5\cdot7\cdot13\cdot17\cdot241.\)

Since \(241\) is a three-digit prime, the two-digit factors come from \(3^2\cdot5\cdot7\cdot13\cdot17.\) They are \[ \begin{gathered} 13,15,17,21,35,39,45,51, \\ 63,65,85,91, \end{gathered} \] for a total of \(12.\)

Thus, the correct answer is D.

16 答案 C

设 \(E\) 和 \(H\) 分别为 \(AB\) 与 \(BC\) 的中点。\(AB\) 的垂直平分线经过 \(E\) 与对角线 \(AC\) 交于 \(F\)；\(BC\) 的垂直平分线经过 \(H\) 与 \(AC\) 交于 \(G\)。区域 \(R\) 是五边形 \(BEFGH\)。

三角形 \(AFE\) 是 \(30\)-\(60\)-\(90^\circ\) 三角形，且 \(AE=1\)，所以面积为 \(\dfrac12\cdot1\cdot\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{6}\)。三角形 \(BFE\) 和 \(BGH\) 与它全等，而 \(\triangle FBG\) 是等边三角形，可分成另外两个相同的小三角形。

因此 \(R\) 由四个全等三角形组成，面积为 \(4\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}\)。

所以正确答案是 C。

English

Let \(E\) and \(H\) be the midpoints of \(AB\) and \(BC.\) The perpendicular bisector of \(AB\) through \(E\) meets diagonal \(AC\) at \(F,\) and the perpendicular bisector of \(BC\) through \(H\) meets \(AC\) at \(G.\) The region \(R\) is the pentagon \(BEFGH.\)

Triangle \(AFE\) is a \(30\)-\(60\)-\(90^\circ\) triangle with \(AE=1,\) so its area is \(\dfrac12\cdot1\cdot\dfrac{1}{\sqrt3}=\dfrac{\sqrt3}{6}.\) Triangles \(BFE\) and \(BGH\) are congruent to it, and \(\triangle FBG\) is equilateral, splitting into two more copies.

Hence \(R\) consists of four congruent triangles, giving area \(4\cdot\dfrac{\sqrt3}{6}=\dfrac{2\sqrt3}{3}.\)

Thus, the correct answer is C.

17 答案 B

首先，\[ \begin{aligned} h_1(x) &=\log_{10}\!\left(\dfrac{10^{10x}}{10}\right) \\ &=\log_{10}\!\left(10^{10x-1}\right) \\ &=10x-1 \end{aligned}\text{。}\]

迭代可得 \(h_n(x)=10^n x\) \(-(1+10+\cdots+10^{n-1})\)。因此 \(h_n(1)\) 是一个 \(n\) 位整数，个位为 \(9\)，其余各位都是 \(8\)。

当 \(n=2011\) 时，数字和为 \[ 8\cdot2010+9=16{,}089\text{。}\]

所以正确答案是 B。

English

First, \[ \begin{aligned} h_1(x) &=\log_{10}\!\left(\dfrac{10^{10x}}{10}\right) \\ &=\log_{10}\!\left(10^{10x-1}\right) \\ &=10x-1. \end{aligned} \]

Iterating, \(h_n(x)=10^n x\) \(-(1+10+\cdots+10^{n-1}).\) Therefore \(h_n(1)\) is an \(n\)-digit integer whose units digit is \(9\) and all of whose other digits are \(8.\)

For \(n=2011,\) the digit sum is \[ 8\cdot2010+9=16{,}089. \]

Thus, the correct answer is B.

18 答案 A

设顶点为 \(A\)，底面正方形为 \(BCDE\)。于是 \(AB=AD=1\)，且 \(BD=\sqrt2\)，所以 \(\triangle BAD\) 是等腰直角三角形。

设立方体边长为 \(x\)。它与 \(\triangle BAD\) 所在平面的交线是一个高为 \(x\)、宽为 \(\sqrt2\,x\) 的长方形，其上方两个顶点在 \(AB\) 与 \(AD\) 上。因为 \(AB\) 与 \(AD\) 与底边成 \(45^\circ\)，长方形外侧的两段 \(BD\) 各长 \(x\)，所以 \[ \sqrt2=BD=\sqrt2\,x+2x\text{，} \] 得 \(x=\dfrac{\sqrt2}{2+\sqrt2}=\sqrt2-1\)。

体积为 \[ (\sqrt2-1)^3=5\sqrt2-7\text{。} \]

所以正确答案是 A。

English

Let the apex be \(A\) and the base be square \(BCDE.\) Then \(AB=AD=1\) and \(BD=\sqrt2,\) so \(\triangle BAD\) is an isosceles right triangle.

Let the cube have edge length \(x.\) Its intersection with the plane of \(\triangle BAD\) is a rectangle of height \(x\) and width \(\sqrt2\,x,\) whose top corners lie on \(AB\) and \(AD.\) Because the legs \(AB\) and \(AD\) meet the base at \(45^\circ,\) each portion of \(BD\) outside the rectangle has length \(x,\) so \[ \sqrt2=BD=\sqrt2\,x+2x, \] which reduces to \(x=\dfrac{\sqrt2}{2+\sqrt2}=\sqrt2-1.\)

The volume is \[ (\sqrt2-1)^3=5\sqrt2-7. \]

Thus, the correct answer is A.

19 答案 B

对 \(0\lt x\le100\)，在直线 \(y=\tfrac12x+2\) 上方最近的格点为：当 \(x\) 为偶数时是 \(\left(x,\tfrac12x+3\right)\)；当 \(x\) 为奇数时是 \(\left(x,\tfrac12x+\tfrac52\right)\)。

从 \((0,2)\) 到该点的斜率，偶数 \(x\) 时为 \(\dfrac12+\dfrac1x\)，奇数 \(x\) 时为 \(\dfrac12+\dfrac{1}{2x}\)。这些斜率的最小值在偶数 \(x\) 时为 \(\dfrac{51}{100}\)，在奇数 \(x\) 时为 \(\dfrac{50}{99}\)。

因为 \(\dfrac{50}{99}\lt\dfrac{51}{100}\)，直线恰在 \(\dfrac12\lt m\lt\dfrac{50}{99}\) 时避开所有这些格点，所以最大值为 \(a=\dfrac{50}{99}\)。

所以正确答案是 B。

English

For \(0\lt x\le100,\) the nearest lattice point above the line \(y=\tfrac12x+2\) is \(\left(x,\tfrac12x+3\right)\) if \(x\) is even and \(\left(x,\tfrac12x+\tfrac52\right)\) if \(x\) is odd.

The slope from \((0,2)\) to that point is \(\dfrac12+\dfrac1x\) for even \(x\) and \(\dfrac12+\dfrac{1}{2x}\) for odd \(x.\) The minimum such slope is \(\dfrac{51}{100}\) for even \(x\) and \(\dfrac{50}{99}\) for odd \(x.\)

Since \(\dfrac{50}{99}\lt\dfrac{51}{100},\) the line avoids all these lattice points exactly when \(\dfrac12\lt m\lt\dfrac{50}{99},\) so the maximum is \(a=\dfrac{50}{99}.\)

Thus, the correct answer is B.

20 答案 C

因为 \(DE\parallel AC\) 且 \(EF\parallel AB\)，所以 \(\angle BDE=\angle BAC=\angle EFC\)。由圆周角定理，\(\angle BXE=\angle BDE\) 且 \(\angle EXC=\angle EFC\)，从而 \(\angle BXE=\angle EXC\)。再由 \(BE=EC\)，可得 \(XB=XC\)。

此外，\(\angle BXC=2\angle BAC\)。在 \(\triangle BXC\) 中使用弦长公式，得到 \(BC=2XB\sin(\angle BAC)\)，在 \(\triangle ABC\) 中使用同一公式，得到 \(BC=2R\sin(\angle BAC)\)。因此 \(XB=XC=R\)。在 \(BC\) 的这一侧，到 \(B\) 和 \(C\) 的距离都等于 \(R\) 的点就是 \(\triangle ABC\) 的外心，所以 \(XA=XB=XC=R\)。

由海伦公式，\(13\)-\(14\)-\(15\) 三角形的面积是 \(84\)，所以 \[ R=\dfrac{13\cdot14\cdot15}{4\cdot84}=\dfrac{65}{8}\text{，}\]并且 \(XA+XB+XC=3R=\dfrac{195}{8}\)。

因此，正确答案是 C。

English

Since \(DE\parallel AC\) and \(EF\parallel AB,\) we get \(\angle BDE=\angle BAC=\angle EFC.\) By the Inscribed Angle Theorem, \(\angle BXE=\angle BDE\) and \(\angle EXC=\angle EFC,\) so \(\angle BXE=\angle EXC.\) With \(BE=EC,\) this forces \(XB=XC.\)

Also \(\angle BXC=2\angle BAC.\) The chord formula in \(\triangle BXC\) gives \(BC=2XB\sin(\angle BAC),\) while the same formula in \(\triangle ABC\) gives \(BC=2R\sin(\angle BAC).\) Thus \(XB=XC=R.\) The point at distance \(R\) from both \(B\) and \(C\) on this side of \(BC\) is the circumcenter of \(\triangle ABC,\) so \(XA=XB=XC=R.\)

The area of the \(13\)-\(14\)-\(15\) triangle is \(84\) by Heron’s formula, so \[ R=\dfrac{13\cdot14\cdot15}{4\cdot84}=\dfrac{65}{8}, \] and \(XA+XB+XC=3R=\dfrac{195}{8}.\)

Thus, the correct answer is C.

21 答案 D

设算术平均数为 \(10a+b\)，几何平均数为 \(10b+a\)，则 \(x+y=2(10a+b)\)，且 \(xy=(10b+a)^2\)。

因此 \[ \begin{aligned} (x-y)^2 &=(x+y)^2-4xy \\ &=396(a^2-b^2) \\ &=11\cdot6^2 \\ &\quad {}\cdot(a+b)(a-b) \end{aligned}\text{。}\] 由于对两个不相等的正数，算术平均数大于几何平均数，所以 \(a>b\)。令 \(u=a-b\)，\(v=a+b\)。于是 \(1\le u\le9\) 且 \(1\le v\le17\)。要使 \(11uv\) 为完全平方数，\(uv\) 中必须含有奇数次幂的 \(11\)。因此 \(v=11\)，这是因为 \(u<11\) 且 \(v<22\)。此时 \(11uv=121u\) 是完全平方数，所以 \(u\) 也是完全平方数。又因为 \(u\) 与 \(v\) 的奇偶性相同，只剩下 \(u=1\) 或 \(9\)。后者给出 \(a=10\)，它不是一位数字，所以 \(u=1\)，从而 \((a,b)=(6,5)\)。

于是 \((x-y)^2=11\cdot6^2\cdot11=66^2\)，所以 \(|x-y|=66\)。（确实有 \(\{x,y\}=\{32,98\}\)。）

所以正确答案是 D。

English

Let the arithmetic mean be \(10a+b\) and the geometric mean be \(10b+a.\) Then \(x+y=2(10a+b)\) and \(xy=(10b+a)^2.\)

Therefore \[ \begin{aligned} (x-y)^2 &=(x+y)^2-4xy \\ &=396(a^2-b^2) \\ &=11\cdot6^2 \\ &\quad {}\cdot(a+b)(a-b). \end{aligned} \] Since the arithmetic mean exceeds the geometric mean for distinct positive numbers, \(a>b.\) Put \(u=a-b\) and \(v=a+b.\) Then \(1\le u\le9\) and \(1\le v\le17.\) For \(11uv\) to be a square, \(uv\) must contain an odd power of \(11.\) Therefore \(v=11,\) because \(u<11\) and \(v<22.\) Now \(11uv=121u\) is a square, so \(u\) is a square. Also \(u\) and \(v\) have the same parity, leaving \(u=1\) or \(9.\) The latter gives \(a=10,\) not a digit, so \(u=1\) and \((a,b)=(6,5).\)

Then \((x-y)^2=11\cdot6^2\cdot11=66^2,\) so \(|x-y|=66.\) (Indeed \(\{x,y\}=\{32,98\}.\))

Thus, the correct answer is D.

22 答案 D

对边长为 \(a,b,c\) 的三角形，切线段长度为 \(AD=\tfrac12(b+c-a)\)、\(BE=\tfrac12(a+c-b)\)、\(CF=\tfrac12(a+b-c)\)。若 \(T_n\) 的边长为 \((y-1,y,y+1)\)，则 \(T_{n+1}\) 的边长为 \(\left(\tfrac{y}{2}-1,\tfrac{y}{2},\tfrac{y}{2}+1\right)\)。

从中间边为 \(2012\) 的 \(T_1\) 开始，中间边每步减半，且 \(T_{n+1}\) 的周长是 \(T_n\) 周长的 \(\tfrac12\)。这种形式的三角形存在当且仅当中间边大于 \(2\)。

\(T_n\) 的中间边为 \(\dfrac{2012}{2^{n-1}}\)。它在 \(n=11\) 时第一次不超过 \(2\)，所以最后一个有效三角形是 \(T_{10}\)，其中间边为 \(\dfrac{2012}{2^9}\)，周长为 \[ 3\cdot\dfrac{2012}{2^9}=\dfrac{6036}{512}=\dfrac{1509}{128}\text{。}\]

所以正确答案是 D。

English

For a triangle with sides \(a,b,c,\) the tangent lengths are \(AD=\tfrac12(b+c-a),\) \(BE=\tfrac12(a+c-b),\) and \(CF=\tfrac12(a+b-c).\) If \(T_n\) has sides \((y-1,y,y+1),\) then \(T_{n+1}\) has sides \(\left(\tfrac{y}{2}-1,\tfrac{y}{2},\tfrac{y}{2}+1\right).\)

Starting from \(T_1\) with middle side \(2012,\) the middle side halves each step and the perimeter of \(T_{n+1}\) is \(\tfrac12\) the perimeter of \(T_n.\) A triangle of this form exists only while its middle side exceeds \(2.\)

The middle side of \(T_n\) is \(\dfrac{2012}{2^{n-1}}.\) This first drops to \(2\) or below at \(n=11,\) so the last valid triangle is \(T_{10},\) whose middle side is \(\dfrac{2012}{2^9}\) and whose perimeter is \[ 3\cdot\dfrac{2012}{2^9}=\dfrac{6036}{512}=\dfrac{1509}{128}. \]

Thus, the correct answer is D.

23 答案 C

格点 \(X=(x,y)\) 位于某条路径上，当且仅当 \[ \begin{aligned} d&=|x-3|+|x+3| \\ &\quad {}+|y-2|+|y+2|\le20 \end{aligned}\text{。}\] 该式在 \(x\to-x\) 或 \(y\to-y\) 时不变，所以先数 \(x\ge0\)、\(y\ge0\) 的点，再乘以 \(4\)，并校正坐标轴上的重复计数。

若 \(0\le x\le3\) 且 \(0\le y\le2\)，则这 \(4\cdot3=12\) 个点全都满足条件。若 \(0\le x\le3\) 且 \(y\ge3\)，则 \(y\le7\)，共给出 \(4\cdot5=20\) 个点。若 \(x\ge4\) 且 \(0\le y\le2\)，则 \(x\le8\)，共给出 \(5\cdot3=15\) 个点。最后，当 \(x\ge4\) 且 \(y\ge3\) 时，条件化为 \(x+y\le10\)，共给出 \(4+3+2+1=10\) 个点。于是第一象限中共有 \(57\) 个点，其中 \(15\) 个位于非负坐标轴上。由对称性，总数为 \[ 4\cdot57-2\cdot15-3=195\text{。}\]

所以正确答案是 C。

English

A lattice point \(X=(x,y)\) lies on some path exactly when \[ \begin{aligned} d&=|x-3|+|x+3| \\ &\quad {}+|y-2|+|y+2|\le20. \end{aligned} \] This expression is unchanged when \(x\to-x\) or \(y\to-y,\) so we count points with \(x\ge0,\) \(y\ge0,\) multiply by \(4,\) and correct for the axes.

If \(0\le x\le3\) and \(0\le y\le2,\) all \(4\cdot3=12\) points work. If \(0\le x\le3\) and \(y\ge3,\) then \(y\le7,\) giving \(4\cdot5=20\) points. If \(x\ge4\) and \(0\le y\le2,\) then \(x\le8,\) giving \(5\cdot3=15\) points. Finally, for \(x\ge4\) and \(y\ge3,\) the condition is \(x+y\le10,\) giving \(4+3+2+1=10\) points. Thus there are \(57\) in the first quadrant, including \(15\) on the nonnegative axes. By symmetry the total is \[ 4\cdot57-2\cdot15-3=195. \]

Thus, the correct answer is C.

24 答案 B

按 \(z^4\) 分解： \[ \begin{aligned} P(z) &=(z^4-1) \\ &\quad {}\cdot\big(z^4+(4\sqrt3+7)\big)\text{。} \end{aligned} \] 第一因子给出根 \(1,-1,i,-i\)。又因为 \(4\sqrt3+7=(\sqrt3+2)^2\)，且 \(2(\sqrt3+2)=(\sqrt3+1)^2\)，令 \(w=\tfrac12(\sqrt3+1)\)，另四个根为 \(w(\pm1\pm i)\)。

这八个根关于原点对称，并具有 \(4\) 重旋转对称性；任意两个根之间的线段长度至少为 \(\sqrt2\)。因此任何这样的八边形周长至少为 \(8\sqrt2\)，而依次取顶点 \(1\)、\(w(1+i)\)、\(i\)、\(w(-1+i)\)、\(-1\)、\(w(-1-i)\)、\(-i\)、\(w(1-i)\) 时可达到该下界。

所以正确答案是 B。

English

Factoring in \(z^4,\) \[ \begin{aligned} P(z) &=(z^4-1) \\ &\quad {}\cdot\big(z^4+(4\sqrt3+7)\big). \end{aligned} \] The first factor gives the roots \(1,-1,i,-i.\) Since \(4\sqrt3+7=(\sqrt3+2)^2\) and \(2(\sqrt3+2)=(\sqrt3+1)^2,\) writing \(w=\tfrac12(\sqrt3+1)\) the other four roots are \(w(\pm1\pm i).\)

The eight roots are symmetric about the origin with \(4\)-fold symmetry, and every segment joining two of them has length at least \(\sqrt2.\) Thus any such polygon has perimeter at least \(8\sqrt2,\) and the polygon with vertices \(1,\) \(w(1+i),\) \(i,\) \(w(-1+i),\) \(-1,\) \(w(-1-i),\) \(-i,\) \(w(1-i)\) achieves it.

Thus, the correct answer is B.

25 答案 D

因为 \(\left[\dfrac{n+mk}{k}\right]=\left[\dfrac{n}{k}\right]+m\)，所以 \(n\) 是否满足恒等式只取决于 \(n\bmod k\)。由于 \(99!\) 能被 \(k\) 整除对 \(1\le k\le99\) 成立，每个剩余类等可能。

写 \(100=qk+r\) 和 \(n=q_1k+r_1\)，其中两个余数都取自 \([-\frac{(k-1)}{2},\frac{k-1}{2}]\)。若 \(r\ge0\)，则不发生进位当且仅当 \(r-\frac{k-1}{2}\le r_1\le\frac{k-1}{2}\)；这给出 \(k-r\) 个剩余类。\(r<0\) 的情形同样给出 \(k+r\) 个剩余类。因此两种情形下都有 \[ P(k)=1-\dfrac{|r|}{k}\text{。}\]

要最小化 \(P(k)\)，就要最大化 \(|r|/k\)。若 \(r=\frac{k-1}{2}\)，则 \(201=k(2q+1)\)，而满足 \(k\le99\) 的最大取值是 \(201\) 的因数 \(67\)。若 \(r=-\frac{(k-1)}{2}\)，则 \(199=k(2q-1)\)；因为 \(199\) 是质数，只有 \(k=1\) 可能。在其余所有情形中都有 \(|r|\le\frac{k-3}{2}\)，所以 \[ \begin{aligned} P(k)&\ge\dfrac12+\dfrac{3}{2k} \\ &\ge\dfrac12+\dfrac{3}{198} \\ &>\dfrac{34}{67} \end{aligned}\text{。}\] 而当 \(k=67\) 时， \[ P(67)=\dfrac12+\dfrac{1}{2\cdot67}=\dfrac{34}{67}\text{。}\]

所以正确答案是 D。

English

Because \(\left[\dfrac{n+mk}{k}\right]=\left[\dfrac{n}{k}\right]+m,\) whether \(n\) satisfies the identity depends only on \(n\bmod k.\) Since \(99!\) is divisible by \(k\) for \(1\le k\le99,\) every residue class is equally likely.

Write \(100=qk+r\) and \(n=q_1k+r_1,\) choosing both remainders in \([-\frac{(k-1)}{2},\frac{k-1}{2}].\) If \(r\ge0,\) no carry occurs precisely when \(r-\frac{k-1}{2}\le r_1\le\frac{k-1}{2};\) this gives \(k-r\) residue classes. The case \(r<0\) similarly gives \(k+r\) classes. Hence in both cases \[ P(k)=1-\dfrac{|r|}{k}. \]

To minimize \(P(k)\) we maximize \(|r|/k.\) If \(r=\frac{k-1}{2},\) then \(201=k(2q+1),\) and the largest possible \(k\le99\) is the divisor \(67\) of \(201.\) If \(r=-\frac{(k-1)}{2},\) then \(199=k(2q-1);\) because \(199\) is prime, only \(k=1\) is possible. In every remaining case \(|r|\le\frac{k-3}{2},\) so \[ \begin{aligned} P(k)&\ge\dfrac12+\dfrac{3}{2k} \\ &\ge\dfrac12+\dfrac{3}{198} \\ &>\dfrac{34}{67}. \end{aligned} \] For \(k=67,\) \[ P(67)=\dfrac12+\dfrac{1}{2\cdot67}=\dfrac{34}{67}. \]

Thus, the correct answer is D.