2012B AMC 12 真题（P版）

1 答案 C

每个教室里学生比兔子多 \(18-2=16\) 个。

全部 \(4\) 个教室中，学生比兔子多 \(4\cdot16=64\) 个。

因此正确答案是 C。

English

Each classroom has \(18-2=16\) more students than rabbits.

Across all \(4\) classrooms there are \(4\cdot16=64\) more students than rabbits.

Thus, the correct answer is C.

2 答案 E

圆内切于长方形，所以长方形的宽等于圆的直径，即 \(2\cdot5=10\)。

长方形的长为 \(2\cdot10=20\)，所以面积为 \(10\cdot20=200\)。

因此正确答案是 E。

English

The circle is inscribed, so the width of the rectangle equals the diameter, \(2\cdot5=10.\)

The length is then \(2\cdot10=20,\) so the area is \(10\cdot20=200.\)

Thus, the correct answer is E.

3 答案 D

设花栗鼠挖了 \(h\) 个洞。花栗鼠藏了 \(3h\) 颗橡子，松鼠藏了 \(4(h-4)\) 颗橡子。

因为它们藏的橡子数量相同，\(3h=4(h-4)\)，解得 \(h=16\)。

花栗鼠藏了 \(3\cdot16=48\) 颗橡子。

因此正确答案是 D。

English

Let \(h\) be the number of holes the chipmunk dug. The chipmunk hid \(3h\) acorns and the squirrel hid \(4(h-4)\) acorns.

Since they hid the same number, \(3h=4(h-4),\) which gives \(h=16.\)

The chipmunk hid \(3\cdot16=48\) acorns.

Thus, the correct answer is D.

4 答案 B

Étienne 的钱价值 \(400\cdot1.30=520\) 美元，而 Diana 有 \(500\) 美元。

Étienne 的钱的价值超过 Diana 的百分比为 \[\frac{520-500}{500}\cdot100\%=4\%\text{。}\]

因此正确答案是 B。

English

Étienne’s money is worth \(400\cdot1.30=520\) dollars, while Diana has \(500\) dollars.

The percent by which Étienne’s value exceeds Diana’s is \[\frac{520-500}{500}\cdot100\%=4\%.\]

Thus, the correct answer is B.

5 答案 A

三个连续的整数对的和分别是 \(26\)，\(41-26=15\)，和 \(57-41=16\)。

两个整数同奇偶时和为偶数，恰好一个是偶数时和为奇数。只有中间那一对的和为奇数，所以它必须至少含有一个偶数。

另外两对都可以全是奇数，所以最少可以只有 \(1\) 个偶数，例如 \(1,25,1,14,1,15\)。

因此正确答案是 A。

English

The three successive pairs have sums \(26,\) \(41-26=15,\) and \(57-41=16.\)

A pair sums to an even number when its two integers share parity, and to an odd number when exactly one is even. Only the middle pair sums to an odd number, so it must contain at least one even integer.

The other two pairs can be all odd, so as few as \(1\) even integer is possible, for example \(1,25,1,14,1,15.\)

Thus, the correct answer is A.

6 答案 A

设 \(d \gt 0\) 为那一小段量。Xiaoli 计算的是 \((x+d)-(y-d)\) \(=(x-y)+2d\)。

因为 \(2d \gt 0\)，她的估计值大于真实值 \(x-y\)。

因此正确答案是 A。

English

Let \(d \gt 0\) be the small amount. Xiaoli computes \((x+d)-(y-d)\) \(=(x-y)+2d.\)

Since \(2d \gt 0,\) her estimate exceeds the true value \(x-y.\)

Thus, the correct answer is A.

7 答案 E

灯泡按每 \(5\) 个一组重复，所以相邻两组的开头相距 \(5\cdot6=30\) 英寸，即 \(2.5\) 英尺。

每一组的开头都是一个奇数编号的红灯。第 \(3\) 个红灯是第 \(2\) 组的开头，第 \(21\) 个红灯是第 \(11\) 组的开头。

它们之间的距离是 \((11-2)\cdot2.5=22.5\) 英尺。

因此正确答案是 E。

English

The lights repeat in blocks of \(5,\) so consecutive blocks start \(5\cdot6=30\) inches, or \(2.5\) feet, apart.

Each block has one odd-numbered red light beginning it. The \(3\)rd red light begins the \(2\)nd block and the \(21\)st red light begins the \(11\)th block.

The distance between them is \((11-2)\cdot2.5=22.5\) feet.

Thus, the correct answer is E.

8 答案 A

星期五固定为蛋糕。从星期五向两边计数。

另外六天（星期六，然后是星期四、星期三、星期二、星期一、星期日）都可以选择除已经确定的相邻那天甜点之外的任意一种甜点，因此各有 \(3\) 种选择。

菜单数为 \(3^6=729\)。

因此正确答案是 A。

English

Friday is fixed as cake. Work outward from Friday.

Each of the other six days (Saturday, then Thursday, Wednesday, Tuesday, Monday, Sunday) can be any dessert except the one served on the neighboring already-chosen day, giving \(3\) choices each.

The number of menus is \(3^6=729.\)

Thus, the correct answer is A.

9 答案 B

设 \(x\) 为 Clea 的步行速度，\(r\) 为自动扶梯速度，自动扶梯长度为 \(60x\)。在运行中的自动扶梯上步行给出 \(24(x+r)=60x\)，所以 \(r=\tfrac32x\)。

站着不动所需时间 \(t\) 满足 \(rt=60x\)，因此 \(\tfrac32xt=60x\) 得 \(t=40\) 秒。

因此正确答案是 B。

English

Let \(x\) be Clea’s walking rate and \(r\) the escalator’s rate, with the length equal to \(60x.\) Walking on the moving escalator gives \(24(x+r)=60x,\) so \(r=\tfrac32x.\)

Standing takes time \(t\) with \(rt=60x,\) so \(\tfrac32xt=60x\) and \(t=40\) seconds.

Thus, the correct answer is B.

10 答案 B

由 \(x^2+y^2=25\) 得 \(y^2=25-x^2\)。代入 \((x-4)^2+9y^2=81\) 得 \(x^2+x-20=0\)，所以 \(x=4\) 或 \(x=-5\)。

交点为 \((-5,0)\)，\((4,3)\)，和 \((4,-3)\)。

从 \((4,3)\) 到 \((4,-3)\) 的竖直边长为 \(6\)，到 \((-5,0)\) 的水平距离为 \(9\)，所以面积是 \(\tfrac12\cdot6\cdot9=27\)。

因此正确答案是 B。

English

From \(x^2+y^2=25\) we get \(y^2=25-x^2.\) Substituting into \((x-4)^2+9y^2=81\) gives \(x^2+x-20=0,\) so \(x=4\) or \(x=-5.\)

The intersection points are \((-5,0),\) \((4,3),\) and \((4,-3).\)

The vertical side from \((4,3)\) to \((4,-3)\) has length \(6,\) and the horizontal distance to \((-5,0)\) is \(9,\) so the area is \(\tfrac12\cdot6\cdot9=27.\)

Thus, the correct answer is B.

11 答案 C

展开各个数，\(132_A=A^2+3A+2\)，\(43_B=4B+3\)，且 \(69_{A+B}=6(A+B)+9\)。

当 \(B=A+1\) 时，方程变为 \(A^2+3A+2\) \(+4(A+1)+3\) \(=6(2A+1)+9\)，化简得 \((A-6)(A+1)=0\)。正整数解为 \(A=6\)，所以 \(B=7\)。

（当 \(B=A-1\) 时得到 \(A^2-5A-2=0\)，没有整数解。）

因此 \(A+B=13\)。

因此正确答案是 C。

English

Writing the numerals out, \(132_A=A^2+3A+2,\) \(43_B=4B+3,\) and \(69_{A+B}=6(A+B)+9.\)

With \(B=A+1,\) the equation becomes \(A^2+3A+2\) \(+4(A+1)+3\) \(=6(2A+1)+9,\) which simplifies to \((A-6)(A+1)=0.\) The positive solution is \(A=6,\) so \(B=7.\)

(The case \(B=A-1\) gives \(A^2-5A-2=0,\) which has no integer solution.)

Therefore \(A+B=13.\)

Thus, the correct answer is C.

12 答案 E

设 \(A\) 为所有零连续的序列，\(B\) 为所有一连续的序列。

对于 \(A\)，有一个全是一的序列，\(20\) 个恰有一个零的序列，以及 \(\binom{20}{2}=190\) 个含有至少两个零的序列（选择第一个和最后一个零的位置）。所以 \(|A|=1+20+190=211\)，由对称性 \(|B|=211\)。

\(A\cap B\) 中的序列是一个零的块后接一个一的块，或顺序相反；这样的序列有 \(|A\cap B|=40\) 个。

因此 \(|A\cup B|=211+211-40\) \(=382\)。

因此正确答案是 E。

English

Let \(A\) be the sequences in which all zeros are consecutive and \(B\) those in which all ones are consecutive.

For \(A,\) there is one all-ones sequence, \(20\) sequences with exactly one zero, and \(\binom{20}{2}=190\) sequences with two or more zeros (choose the first and last zero position). So \(|A|=1+20+190=211,\) and by symmetry \(|B|=211.\)

A sequence in \(A\cap B\) is a block of zeros followed by a block of ones, or the reverse; there are \(|A\cap B|=40\) of these.

Therefore \(|A\cup B|=211+211-40\) \(=382.\)

Thus, the correct answer is E.

13 答案 D

抛物线相交处满足 \(x^2+ax+b=x^2+cx+d\)，即 \(ax+b=cx+d\)。这个方程无解，恰好发生在两条直线平行且不重合的时候，也就是 \(a=c\) 且 \(b\neq d\)。

\(a=c\) 的概率是 \(\tfrac16\)，且 \(b\neq d\) 的概率是 \(\tfrac56\)，所以没有公共点的概率为 \(\tfrac16\cdot\tfrac56=\tfrac5{36}\)。

至少有一个公共点的概率为 \(1-\tfrac5{36}=\tfrac{31}{36}\)。

因此正确答案是 D。

English

The parabolas meet where \(x^2+ax+b=x^2+cx+d,\) i.e. \(ax+b=cx+d.\) This has no solution exactly when the lines are parallel and distinct: \(a=c\) and \(b\neq d.\)

The probability that \(a=c\) is \(\tfrac16,\) and the probability that \(b\neq d\) is \(\tfrac56,\) so the probability of no common point is \(\tfrac16\cdot\tfrac56=\tfrac5{36}.\)

The probability of at least one common point is \(1-\tfrac5{36}=\tfrac{31}{36}.\)

Thus, the correct answer is D.

14 答案 A

当 Bernardo 在一轮后给出的加倍数 \(2n+50\ge1000\) 且之前的数都小于 \(1000\) 时，他获胜。满足 \(2n+50\ge1000\) 的最小 \(n\) 是 \(475\)。

倒推，经过二、三、四轮后导致胜利的最小初始值分别是满足 \(2n+50\ge475\)，\(\ge213\)，和 \(\ge82\) 的最小整数，即 \(213\)，\(82\)，和 \(16\)。不会有超过四轮后才获胜的初始值。

所以 \(N=16\)，各位数字之和为 \(1+6=7\)。

因此正确答案是 A。

English

Bernardo wins after a round when his doubled number \(2n+50\ge1000\) but the previous numbers stayed below \(1000.\) The smallest \(n\) with \(2n+50\ge1000\) is \(475.\)

Working backwards, the smallest starting values that lead to a win after two, three, and four rounds are the smallest integers with \(2n+50\ge475,\) \(\ge213,\) and \(\ge82,\) namely \(213,\) \(82,\) and \(16.\) No start wins after more than four rounds.

So \(N=16,\) and the sum of its digits is \(1+6=7.\)

Thus, the correct answer is A.

15 答案 C

每个扇形形成的圆锥母线长为 \(12\)。较小扇形的弧长为 \(\tfrac{120}{360}\cdot2\pi\cdot12=8\pi\)，因此底面半径为 \(4\)，高为 \(\sqrt{12^2-4^2}=8\sqrt2\)。

较大扇形（圆心角 \(240^\circ\)）的弧长为 \(16\pi\)，底面半径为 \(8\)，高为 \(\sqrt{12^2-8^2}=4\sqrt5\)。

体积之比为 \[\frac{\tfrac13\pi\cdot4^2\cdot8\sqrt2}{\tfrac13\pi\cdot8^2\cdot4\sqrt5} =\frac{\sqrt{10}}{10}\text{。}\]

因此正确答案是 C。

English

Each sector forms a cone with slant height \(12.\) The smaller sector’s arc length is \(\tfrac{120}{360}\cdot2\pi\cdot12=8\pi,\) so its base radius is \(4\) and its height is \(\sqrt{12^2-4^2}=8\sqrt2.\)

The larger sector (central angle \(240^\circ\)) has arc length \(16\pi,\) base radius \(8,\) and height \(\sqrt{12^2-8^2}=4\sqrt5.\)

The ratio of volumes is \[\frac{\tfrac13\pi\cdot4^2\cdot8\sqrt2}{\tfrac13\pi\cdot8^2\cdot4\sqrt5} =\frac{\sqrt{10}}{10}.\]

Thus, the correct answer is C.

16 答案 B

每首歌恰好被三对中的一对喜欢，或被单个女生喜欢，或无人喜欢。每一对都必须被表示出来。

情况 \(1\)： 每首歌都被一对女生喜欢。某一对得到四首歌中的两首（\(\binom42=6\) 种，且有 \(3\) 种选择是哪一对），另外两对各得到一首歌（\(2\) 种）。共有 \(3\cdot6\cdot2=36\)。

情况 \(2\)： 三首歌分别对应三对女生（各一首），第四首歌被单个女生喜欢或无人喜欢。把四首歌分配到这四个角色有 \(4!=24\) 种，剩余角色有 \(4\) 种选择（Amy、Beth、Jo 或无人）：\(24\cdot4=96\)。

总数为 \(36+96=132\)。

因此正确答案是 B。

English

Each song is liked by exactly one of the three pairs, by a single girl, or by no one. Every pair must be represented.

Case \(1:\) every song is liked by a pair. One pair gets two of the four songs (\(\binom42=6\) ways, and \(3\) choices for which pair), and the other two pairs get one song each (\(2\) ways). This gives \(3\cdot6\cdot2=36.\)

Case \(2:\) three songs go to the three pairs (one each) and the fourth song is liked by a single girl or no one. Assigning the four songs to these four roles gives \(4!=24\) ways, and the leftover role has \(4\) options (Amy, Beth, Jo, or no one): \(24\cdot4=96.\)

The total is \(36+96=132.\)

Thus, the correct answer is B.

17 答案 C

设 \(\theta\) 为直线 \(PQ\) 与 \(x\)-轴所成的锐角。边 \(SR=PQ\) 在从 \((3,0)\) 到 \((5,0)\) 的线段上投影为 \(2\cos\theta\)，而边 \(SP=QR\) 在从 \((7,0)\) 到 \((13,0)\) 的线段上投影为 \(6\sin\theta\)。

因为正方形的边长相等，\(2\cos\theta=6\sin\theta\)， 所以 \(\tan\theta=\tfrac13\)。 因此直线 \(SP,RQ\) 的斜率为 \(3\)，直线 \(SR,PQ\) 的斜率为 \(-\tfrac13\)。

中心位于过 \((4,0)\) 且斜率为 \(3\) 的直线上，也位于过 \((10,0)\) 且斜率为 \(-\tfrac13\) 的直线上： \[y=3(x-4)\text{，}\] \[y=-\tfrac13(x-10)\text{。}\] 两线交于 \((4.6,1.8)\)。

坐标之和为 \(4.6+1.8=6.4\)。

因此正确答案是 C。

English

Let \(\theta\) be the acute angle line \(PQ\) makes with the \(x\)-axis. Sides \(SR=PQ\) span the segment from \((3,0)\) to \((5,0)\) as \(2\cos\theta,\) while \(SP=QR\) span the segment from \((7,0)\) to \((13,0)\) as \(6\sin\theta.\)

Since the square has equal sides, \(2\cos\theta=6\sin\theta,\) so \(\tan\theta=\tfrac13.\) Thus lines \(SP,RQ\) have slope \(3\) and lines \(SR,PQ\) have slope \(-\tfrac13.\)

The center lies on the line through \((4,0)\) with slope \(3\) and the line through \((10,0)\) with slope \(-\tfrac13:\) \[y=3(x-4),\] \[y=-\tfrac13(x-10).\] These meet at \((4.6,1.8).\)

The sum of the coordinates is \(4.6+1.8=6.4.\)

Thus, the correct answer is C.

18 答案 B

一旦固定 \(a_1=k\)，数 \(k,k+1,\ldots,10\) 必须从左到右按递增顺序出现，而数 \(1,\ldots,k-1\) 必须从右到左按递增顺序出现（这样每个新出现的较小数都有它的后继已经放好）。

对每个 \(k\)，只需选择第一个位置之后的 \(9\) 个位置中哪些放小于 \(k\) 的数，因此有 \(\binom{9}{k-1}\) 个排列。

求和得 \[ \begin{aligned} \sum_{k=1}^{10}\binom{9}{k-1} &= \sum_{j=0}^{9}\binom{9}{j} \\ &= 2^9=512 \end{aligned}\text{。}\]

因此正确答案是 B。

English

Once \(a_1=k\) is fixed, the numbers \(k,k+1,\ldots,10\) must appear left to right in increasing order, and the numbers \(1,\ldots,k-1\) must appear from right to left in increasing order (so each new small number has its successor already placed).

For each \(k,\) the list is determined by choosing which of the \(9\) positions after the first hold the numbers below \(k,\) giving \(\binom{9}{k-1}\) lists.

Summing, \[ \begin{aligned} \sum_{k=1}^{10}\binom{9}{k-1} &= \sum_{j=0}^{9}\binom{9}{j} \\ &= 2^9=512. \end{aligned} \]

Thus, the correct answer is B.

19 答案 A

把 \(P_1\) 放在原点，使三条棱沿坐标轴，并设靠近 \(P_1\) 的三个八面体顶点都与 \(P_1\) 相距 \(t\)。由对称性，靠近 \(P_1'\) 的三个顶点也都与 \(P_1'\) 相距 \(t\)。

两个同靠近 \(P_1\) 的顶点，例如 \((t,0,0)\) 和 \((0,t,0)\)，相距 \(t\sqrt2\)。靠近 \(P_1\) 的一个顶点，例如 \((t,0,0)\)，与靠近 \(P_1'\) 的相应顶点，例如 \((1,1-t,1)\)，之间也必须有相同距离。

令这两种边长的平方相等，并使用单位立方体的棱长，得到 \(t=\tfrac34\)，因此边长为 \(t\sqrt2=\dfrac{3\sqrt2}{4}\)。

因此正确答案是 A。

English

Place \(P_1\) at the origin with edges along the axes, and let each of the three octahedron vertices near \(P_1\) be a distance \(t\) from \(P_1;\) by symmetry the three near \(P_1'\) are also a distance \(t\) from \(P_1'.\)

Two vertices sharing \(P_1,\) such as \((t,0,0)\) and \((0,t,0),\) are a distance \(t\sqrt2\) apart. A vertex near \(P_1,\) say \((t,0,0),\) and the appropriate vertex near \(P_1',\) say \((1,1-t,1),\) must be the same distance apart.

Setting the two squared side lengths equal and using the cube’s unit edges yields \(t=\tfrac34,\) so the side length is \(t\sqrt2=\dfrac{3\sqrt2}{4}.\)

Thus, the correct answer is A.

20 答案 D

对于平行边为 \(a\lt c\)、两腰为 \(b,d\) 的梯形，平移一条腰会形成边长为 \(b\)，\(d\)，和 \(c-a\) 的三角形。三角形不等式迫使较长的平行边为 \(c=11\)。

若 \(a=3\)，三角形边长为 \(5,7,8\)，面积为 \(10\sqrt3\)，梯形面积为 \(\tfrac{35}{2}\sqrt3\)。若 \(a=5\)，三角形边长为 \(3,6,7\)，面积为 \(4\sqrt5\)，梯形面积为 \(\tfrac{32}{3}\sqrt5\)。若 \(a=7\)，三角形边长为 \(3,4,5\)，是直角三角形，梯形面积为 \(27\)。

总和为 \(\tfrac{35}{2}\sqrt3+\tfrac{32}{3}\sqrt5+27\)，所以 \[ \begin{gathered} r_1+r_2+r_3+n_1+n_2 \\ = \tfrac{35}{2}+\tfrac{32}{3}+27+3+5 \\ = 63+\tfrac16 \end{gathered}\text{。}\]

小于或等于这个值的最大整数是 \(63\)。

因此正确答案是 D。

English

For a trapezoid with parallel sides \(a\lt c\) and legs \(b,d,\) translating a leg forms a triangle with sides \(b,\) \(d,\) and \(c-a.\) The triangle inequality forces the longer parallel side to be \(c=11.\)

If \(a=3,\) the triangle has sides \(5,7,8\) with area \(10\sqrt3,\) and the trapezoid has area \(\tfrac{35}{2}\sqrt3.\) If \(a=5,\) the triangle has sides \(3,6,7\) with area \(4\sqrt5,\) giving trapezoid area \(\tfrac{32}{3}\sqrt5.\) If \(a=7,\) the triangle has sides \(3,4,5,\) a right triangle, giving trapezoid area \(27.\)

The total is \(\tfrac{35}{2}\sqrt3+\tfrac{32}{3}\sqrt5+27,\) so \[ \begin{gathered} r_1+r_2+r_3+n_1+n_2 \\ = \tfrac{35}{2}+\tfrac{32}{3}+27+3+5 \\ = 63+\tfrac16. \end{gathered} \]

The greatest integer at most this value is \(63.\)

Thus, the correct answer is D.

21 答案 A

延长 \(EF\) 和 \(CB\)，并过 \(A\) 作一条同时垂直于它们的直线，交点分别为 \(H\) 和 \(J\)。 因为 \(\angle ABJ=60^\circ\) 有 \(BJ=20\) 且 \(AJ=20\sqrt3\)。 令 \(u=BX\)， 勾股定理给出 \(s^2=(20+u)^2+(20\sqrt3)^2\)。

等角条件使四个角部三角形全等，沿 \(EF\) 追踪相等线段可得 \[ \begin{aligned} u+20\sqrt3 &= 41(\sqrt3-1) \\ &\quad {}+\frac{20+u}{\sqrt3}\text{，} \end{aligned} \] 所以 \(u=21\sqrt3-20\)。

因为 \(20+u=21\sqrt3\)，所以 \[ \begin{aligned} s^2 &= (21\sqrt3)^2+(20\sqrt3)^2 \\ &= 3(441+400)=3\cdot29^2\text{，} \end{aligned} \] 因此 \(s=29\sqrt3\)。

因此正确答案是 A。

English

Extend \(EF\) and \(CB\) to a line through \(A\) perpendicular to both, meeting them at \(H\) and \(J.\) Since \(\angle ABJ=60^\circ,\) we have \(BJ=20\) and \(AJ=20\sqrt3.\) With \(u=BX,\) the Pythagorean theorem gives \(s^2=(20+u)^2+(20\sqrt3)^2.\)

The equiangular angles make the four corner triangles congruent, and chasing the equal segments along \(EF\) yields \[ \begin{aligned} u+20\sqrt3 &= 41(\sqrt3-1) \\ &\quad {}+\frac{20+u}{\sqrt3}, \end{aligned} \] so \(u=21\sqrt3-20.\)

Since \(20+u=21\sqrt3,\) we get \[ \begin{aligned} s^2 &= (21\sqrt3)^2+(20\sqrt3)^2 \\ &= 3(441+400)=3\cdot29^2, \end{aligned} \] giving \(s=29\sqrt3.\)

Thus, the correct answer is A.

22 答案 E

给七列向前（向右）线段编号；若路径不使用返回线段，它只是在每一列选择一条向前线段。选择数为 \(2,2,4,4,4,2,2\)，因此有 \(2^{10}\) 条路径。

设 \(s_1,s_2,s_3\) 为三条向左的返回线段（位于第 \(2,4,6\) 列）。分析一旦走过一条返回线段，哪些列会被迫确定，可得：对 \(\{s_1\}\) 和 \(\{s_3\}\)，各有 \(2^8\) 条路径，对 \(\{s_1,s_3\}\) 有 \(2^6\) 条，对 \(\{s_2\}\) 有 \(2^9\) 条，对 \(\{s_1,s_2\}\) 和 \(\{s_2,s_3\}\)，各有 \(2^7\) 条，对 \(\{s_1,s_2,s_3\}\) 有 \(2^5\) 条。

相加得 \[ \begin{aligned} &2^{10}+2\cdot2^8+2^6 \\ &\quad {}+2^9+2\cdot2^7+2^5=2400 \end{aligned}\text{。}\]

因此正确答案是 E。

English

Label the seven columns of forward (rightward) segments; a path with no back segment simply chooses one forward segment in each column. The numbers of choices are \(2,2,4,4,4,2,2,\) giving \(2^{10}\) paths.

Let \(s_1,s_2,s_3\) be the three left-pointing back segments (in columns \(2,4,6\)). Analyzing which columns become forced once a back segment is traversed gives \(2^8\) paths for each of \(\{s_1\}\) and \(\{s_3\},\) \(2^6\) for \(\{s_1,s_3\},\) \(2^9\) for \(\{s_2\},\) \(2^7\) for each of \(\{s_1,s_2\}\) and \(\{s_2,s_3\},\) and \(2^5\) for \(\{s_1,s_2,s_3\}.\)

Adding, \[ \begin{aligned} &2^{10}+2\cdot2^8+2^6 \\ &\quad {}+2^9+2\cdot2^7+2^5=2400. \end{aligned} \]

Thus, the correct answer is E.

23 答案 B

因为 \(|z_0|=1\)，对恒等式 \[ \begin{aligned} &4z_0^5-(z_0-1)P(z_0) \\ &\quad =z_0^4(4-a)+z_0^3(a-b) \\ &\quad {}+z_0^2(b-c)+z_0(c-d)+d \end{aligned} \] 使用三角不等式时，左边的绝对值是 \(4\)。右边非负的系数差之和也是 \(4\)，所以三角不等式取等号，所有非零复数项都指向同一方向。

如果有两个系数差非零，取它们的商可知，对某个 \(1\le k\le4\)，\(z_0^k\) 是正实数。由于 \(|z_0|=1\)，这意味着 \(z_0^k=1\)。当 \(k=2,3,4\) 时，分别得到 \(a=4,b=c,d=0\)；\(a=b=4,c=d=0\)；以及一个已经属于第一族的多项式。若不存在这样的幂，等号条件迫使恰有一个非零系数差。常数项的差给出 \(a=b=c=d=4\)；其他位置的差也会迫使 \(z_0^{5-j}=1\)，从而回到上述情形。因此，多项式恰好是 \(4z^4+4z^3+bz^2+bz\)，其中 \(0\le b\le4\)，再加上 \(4z^4+4z^3+4z^2\) 和 \(4z^4+4z^3+4z^2+4z+4\)。

它们在 \(1\) 处的值分别为 \(20\)，\(12\)，以及 \(8+2b\)；总和为 \[ \begin{gathered} 20+12+\sum_{b=0}^{4}(8+2b) \\ = 32+40+20 \\ = 92 \end{gathered}\text{。}\]

因此，正确答案是 B。

English

Because \(|z_0|=1,\) applying the triangle inequality to the identity \[ \begin{aligned} &4z_0^5-(z_0-1)P(z_0) \\ &\quad =z_0^4(4-a)+z_0^3(a-b) \\ &\quad {}+z_0^2(b-c)+z_0(c-d)+d \end{aligned} \] has left side of absolute value \(4.\) The nonnegative coefficient jumps on the right sum to \(4,\) so the triangle inequality is an equality and all its nonzero complex summands point in the same direction.

If two jumps are nonzero, their quotient shows that \(z_0^k\) is a positive real for some \(1\le k\le4.\) Since \(|z_0|=1,\) this means \(z_0^k=1.\) The cases \(k=2,3,4\) give, respectively, \(a=4,b=c,d=0;\) \(a=b=4,c=d=0;\) and a polynomial already in the first family. If no such power exists, equality forces exactly one nonzero jump. The constant jump gives \(a=b=c=d=4;\) any other jump again forces \(z_0^{5-j}=1\) and returns to the cases just listed. Hence the polynomials are exactly \(4z^4+4z^3+bz^2+bz\) for \(0\le b\le4,\) together with \(4z^4+4z^3+4z^2\) and \(4z^4+4z^3+4z^2+4z+4.\)

Their values at \(1\) are \(20,\) \(12,\) and \(8+2b;\) summing gives \[ \begin{gathered} 20+12+\sum_{b=0}^{4}(8+2b) \\ = 32+40+20 \\ = 92. \end{gathered} \]

Thus, the correct answer is B.

24 答案 D

若 \(N_2\) 是 \(N_1\) 的倍数，则 \(f_1(N_2)\) 是 \(f_1(N_1)\) 的倍数，所以若 \(S_{N_1}\) 无界，则 \(S_{N_2}\) 也无界。称 \(N\) 为本质数，如果它无界而任何真因数都不无界。本质数 \(N\) 的所有指数都至少为 \(2\)，并且 \((p_1\cdots p_k)^2\le400\) 迫使它至多含两个质数。

对于 \(n=2^a3^b\)，两次迭代把指数对变成 \((2a-4,2b-3)\)。因此轨道无界，当且仅当 \(a\ge5\) 或 \(b\ge4\)，由此得到本质数 \(2^5=32\) 和 \(3^4=81\)。对于只含另一个质数的情形，上界只留下 \(5^2,5^3,7^2,7^3,11^2,13^2,17^2,19^2\)；直接应用 \(f_1\) 后只有 \(7^3=343\) 是本质数。对于两个质数，\(p_1p_2\le20\) 除 \((2,3)\) 外只留下 \((2,5),(2,7),(3,5)\)。对相应的平方因子乘积应用 \(f_1\)，只剩 \(2^4\cdot5^2=400\)。

它们在 \(400\) 以内的倍数个数分别为 \(\lfloor\frac{400}{32}\rfloor=12\)，\(\lfloor\frac{400}{81}\rfloor=4\)，\(\lfloor\frac{400}{343}\rfloor=1\)，和 \(\lfloor\frac{400}{400}\rfloor=1\)，且没有重叠，总数为 \(12+4+1+1=18\)。

因此，正确答案是 D。

English

If \(N_2\) is a multiple of \(N_1\) then \(f_1(N_2)\) is a multiple of \(f_1(N_1),\) so if \(S_{N_1}\) is unbounded so is \(S_{N_2}.\) Call \(N\) essential if it is unbounded but no proper divisor is. An essential \(N\) must have all exponents at least \(2,\) and \((p_1\cdots p_k)^2\le400\) forces at most two primes.

For \(n=2^a3^b,\) two iterations send the exponent pair to \((2a-4,2b-3).\) Thus the orbit is unbounded exactly when \(a\ge5\) or \(b\ge4,\) producing the essential values \(2^5=32\) and \(3^4=81.\) For a single other prime, the bound leaves only \(5^2,5^3,7^2,7^3,11^2,13^2,17^2,19^2;\) direct application of \(f_1\) leaves only \(7^3=343\) essential. With two primes, \(p_1p_2\le20\) leaves the pairs \((2,5),(2,7),(3,5)\) besides \((2,3).\) Applying \(f_1\) to their possible squareful products leaves only \(2^4\cdot5^2=400.\)

Their multiples up to \(400\) number \(\lfloor\frac{400}{32}\rfloor=12,\) \(\lfloor\frac{400}{81}\rfloor=4,\) \(\lfloor\frac{400}{343}\rfloor=1,\) and \(\lfloor\frac{400}{400}\rfloor=1,\) with no overlaps, for a total of \(12+4+1+1=18.\)

Thus, the correct answer is D.

25 答案 B

等腰直角三角形贡献 \(f(t)=1\)。 对于三边互不相等的直角三角形，关于合适的直线反射可把它与三角形 \(t_1\) 配对，使得 \(f(t)f(t_1)\) \(=\tan(\angle CBA)\tan(\angle ACB)\) \(=1\)。

连续反射（关于 \(x=2\)， 再关于 \(x=y\)， 再关于 \(y=\tfrac52\)）把乘积化简为仅六个形如 \(OYZ\) 且 \(Y\) 在最上排的三角形对应乘积的倒数。

这六个三角形给出 \[ \begin{aligned} &\frac15\cdot\frac25\cdot\frac35\cdot\frac45\cdot\frac{3\sqrt2}{\sqrt2} \\ &\quad {}\cdot\frac{4\sqrt2}{2\sqrt2}=\frac{144}{625} \end{aligned}\text{，}\] 因此所求乘积是其倒数 \(\dfrac{625}{144}\)。

因此正确答案是 B。

English

Isosceles right triangles contribute \(f(t)=1.\) For a scalene right triangle, reflecting across a suitable line pairs it with a triangle \(t_1\) so that \(f(t)f(t_1)\) \(=\tan(\angle CBA)\tan(\angle ACB)\) \(=1.\)

Successive reflections (across \(x=2,\) then \(x=y,\) then \(y=\tfrac52\)) reduce the product to the reciprocal of the product over just six triangles of the form \(OYZ\) with \(Y\) on the top row.

Those six give \[ \begin{aligned} &\frac15\cdot\frac25\cdot\frac35\cdot\frac45\cdot\frac{3\sqrt2}{\sqrt2} \\ &\quad {}\cdot\frac{4\sqrt2}{2\sqrt2}=\frac{144}{625}, \end{aligned} \] so the required product is its reciprocal, \(\dfrac{625}{144}.\)

Thus, the correct answer is B.