2017A AMC 12 真题（P版）

1 答案 D

最便宜的冰棒来自 \(5\) 支装盒，单价为 \(\dfrac{\$3}{5}=\$0.60\)。即使按这个单价，\(14\) 支冰棒也要花 \(14\cdot\$0.60=\$8.40\)，超过 \(\$8\)。

因此 Pablo 最多能买 \(13\) 支，而且他可以用 \(\$6\) 买两盒 \(5\) 支装，再用 \(\$2\) 买一盒 \(3\) 支装，得到 \(2\cdot5+3=13\) 支冰棒。

所以正确答案是 D。

English

The cheapest popsicles come from the \(5\)-popsicle box, at \(\dfrac{\$3}{5}=\$0.60\) each. Even at that rate, \(14\) popsicles would cost \(14\cdot\$0.60=\$8.40,\) more than \(\$8.\)

So Pablo can buy at most \(13,\) and he achieves this with two \(5\)-boxes for \(\$6\) and one \(3\)-box for \(\$2,\) giving \(2\cdot5+3=13\) popsicles.

Thus, the correct answer is D.

2 答案 C

设这两个数为 \(x\) 和 \(y\)，则 \(x+y=4xy\)。

两边同时除以 \(xy\)，得到 \[ \dfrac{x+y}{xy}=4\text{，} \] 而左边正是 \(\dfrac{1}{y}+\dfrac{1}{x}\)。因此倒数之和为 \(4\)。

所以正确答案是 C。

English

Let the numbers be \(x\) and \(y,\) so \(x+y=4xy.\)

Dividing both sides by \(xy\) gives \[ \dfrac{x+y}{xy}=4, \] and the left side is exactly \(\dfrac{1}{y}+\dfrac{1}{x}.\) So the sum of the reciprocals is \(4.\)

Thus, the correct answer is C.

3 答案 B

这个承诺是“全部答对 \(\Rightarrow\) 得到 A”。一个蕴含命题只与它的逆否命题等价：“没有得到 A \(\Rightarrow\) 没有全部答对”。

“没有全部答对”表示至少有一道题答错，这正是选项 B。逆命题和否命题都不能推出，而“全部答错”比这个否定强得多。

所以正确答案是 B。

English

The promise is “all right \(\Rightarrow\) A.” An implication is equivalent only to its contrapositive: “not A \(\Rightarrow\) not all right.”

“Not all right” means at least one question was wrong, which is exactly statement B. The converse and inverse do not follow, and getting “all wrong” is a much stronger claim than the negation.

Thus, the correct answer is B.

4 答案 A

如果正方形边长为 \(x\)，Jerry 走 \(x+x=2x\) 而 Silvia 走对角线 \(\sqrt{x^2+x^2}=x\sqrt2\)。

Silvia 的路程短了的比例为 \[ \begin{aligned} &\dfrac{2x-x\sqrt2}{2x}=1-\dfrac{\sqrt2}{2} \\ &\approx 1-0.707=0.293 \end{aligned}\text{。}\]

这最接近 \(30\%\)。

所以正确答案是 A。

English

If the square has side \(x,\) Jerry walks \(x+x=2x,\) while Silvia walks the diagonal \(\sqrt{x^2+x^2}=x\sqrt2.\)

The fraction by which Silvia’s trip is shorter is \[ \begin{aligned} &\dfrac{2x-x\sqrt2}{2x}=1-\dfrac{\sqrt2}{2} \\ &\approx 1-0.707=0.293. \end{aligned} \]

This is closest to \(30\%.\)

Thus, the correct answer is A.

5 答案 B

那 \(20\) 个彼此认识的人只和 \(10\) 个陌生人握手。那 \(10\) 个陌生人中的每个人都和其他 \(29\) 人握手。

把握手次数相加并除以 \(2\)（每次握手涉及两个人），得到 \[ \begin{aligned} &\dfrac{1}{2}(20\cdot10+10\cdot29) \\ &=\dfrac{1}{2}(200+290)=245 \end{aligned}\text{。}\]

所以正确答案是 B。

English

Each of the \(20\) people who know each other shakes hands with only the \(10\) strangers. Each of the \(10\) strangers shakes hands with all \(29\) other people.

Summing handshake counts and dividing by \(2\) (each handshake involves two people) gives \[ \begin{aligned} &\dfrac{1}{2}(20\cdot10+10\cdot29) \\ &=\dfrac{1}{2}(200+290)=245. \end{aligned} \]

Thus, the correct answer is B.

6 答案 B

四条长度能组成面积为正的四边形，当且仅当最长边严格小于另外三边之和。若第四根杆长为 \(n\)，则需要 \(15\lt 3+7+n\) 且 \(n\lt 3+7+15\)，所以 \[ 5\lt n\lt 25\text{。}\]

从 \(6\) 到 \(24\) 的整数共有 \(19\) 个，但长度为 \(7\) 和 \(15\) 的杆已经在桌上，剩下 \(19-2=17\) 种选择。

所以正确答案是 B。

English

Four lengths form a quadrilateral with positive area if and only if the longest is strictly less than the sum of the other three. With a fourth rod of length \(n,\) this requires \(15\lt 3+7+n\) and \(n\lt 3+7+15,\) so \[ 5\lt n\lt 25. \]

The integers from \(6\) to \(24\) give \(19\) values, but the rods of length \(7\) and \(15\) are already on the table, leaving \(19-2=17\) choices.

Thus, the correct answer is B.

7 答案 B

列出数值：\(f(1)=2\)，\(f(2)=f(1)+1=3\)，\(f(3)=f(1)+2=4\)，\(f(4)=f(3)+1=5\)，这说明可能有 \(f(n)=n+1\)。

两条规则都与 \(f(n)=n+1\) 一致：当 \(n\) 为偶数时，\((n-1)+1+1=n+1\)；当 \(n\) 为奇数时，\((n-2)+1+2=n+1\)。由于递归唯一确定 \(f\)，所以 \(f(2017)=2018\)。

所以正确答案是 B。

English

Listing values: \(f(1)=2,\) \(f(2)=f(1)+1=3,\) \(f(3)=f(1)+2=4,\) \(f(4)=f(3)+1=5,\) suggesting \(f(n)=n+1.\)

Both rules are consistent with \(f(n)=n+1:\) for even \(n,\) \((n-1)+1+1=n+1,\) and for odd \(n,\) \((n-2)+1+2=n+1.\) Since the recursion determines \(f\) uniquely, \(f(2017)=2018.\)

Thus, the correct answer is B.

8 答案 D

设 \(h=AB\)。 这个区域是一个半径为 \(3\)、高为 \(h\) 的圆柱，两端各接一个半径为 \(3\) 的半球。

圆柱体积为 \(\pi\cdot3^2\cdot h=9\pi h\)，两个半球合起来是一个球，体积为 \(\dfrac{4}{3}\pi\cdot3^3=36\pi\)。因此 \[ 9\pi h+36\pi=216\pi\text{，} \] 得 \(h=20\)。

所以正确答案是 D。

English

Let \(h=AB.\) The region is a cylinder of radius \(3\) and height \(h\) with a hemisphere of radius \(3\) on each end.

The cylinder has volume \(\pi\cdot3^2\cdot h=9\pi h,\) and the two hemispheres together form a sphere of volume \(\dfrac{4}{3}\pi\cdot3^3=36\pi.\) So \[ 9\pi h+36\pi=216\pi, \] giving \(h=20.\)

Thus, the correct answer is D.

9 答案 E

考虑 \(3\)，\(x+2\)，\(y-4\) 中哪两个是相等的较大值。

若 \(3=x+2\ge y-4\)，则 \(x=1\) 且 \(y\le7\)，得到一条从 \((1,7)\) 向下的射线。若 \(3=y-4\ge x+2\)，则 \(y=7\) 且 \(x\le1\)，得到一条从 \((1,7)\) 向左的射线。若 \(x+2=y-4\ge3\)，则 \(y=x+6\) 且 \(x\ge1\)，得到一条从 \((1,7)\) 向右上方的射线。

三条射线共享端点 \((1,7)\)，所以 \(S\) 是三条有共同端点的射线。

所以正确答案是 E。

English

Consider which two of \(3,\) \(x+2,\) \(y-4\) are the (equal) larger pair.

If \(3=x+2\ge y-4:\) then \(x=1\) and \(y\le7,\) a downward ray from \((1,7).\) If \(3=y-4\ge x+2:\) then \(y=7\) and \(x\le1,\) a leftward ray from \((1,7).\) If \(x+2=y-4\ge3:\) then \(y=x+6\) and \(x\ge1,\) a ray from \((1,7)\) going up and to the right.

All three rays share the endpoint \((1,7),\) so \(S\) is three rays with a common endpoint.

Thus, the correct answer is E.

10 答案 C

Laurent 的数有概率 \(\dfrac{1}{2}\) 落在 \([2017,4034]\)，这会大于 Chloé 可能选择的任何数，因此他必胜。

另有概率 \(\dfrac{1}{2}\)，Laurent 的数落在 \([0,2017]\)，与 Chloé 的区间相同；由对称性，他较大的概率是一半。总概率为 \[ \dfrac{1}{2}\cdot1+\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\text{。}\]

所以正确答案是 C。

English

With probability \(\dfrac{1}{2},\) Laurent’s number lies in \([2017,4034],\) which exceeds any number Chloé could choose, so he wins for certain.

With the other probability \(\dfrac{1}{2},\) Laurent’s number lies in \([0,2017],\) matching Chloé’s interval; by symmetry he is larger half the time. The total probability is \[ \dfrac{1}{2}\cdot1+\dfrac{1}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}. \]

Thus, the correct answer is C.

11 答案 D

若多边形有 \(n\) 条边，漏掉的角为 \(\alpha\)，则 \((n-2)180=2017+\alpha\)。因为 \(0\lt\alpha\lt180\)，所以 \[ 2017\lt(n-2)180\lt2197\text{。}\]

这个范围内唯一的 \(180\) 的倍数是 \(2160=(14-2)180\)，所以 \(n=14\)，且 \[ \alpha=2160-2017=143\text{。}\]

所以正确答案是 D。

English

If the polygon has \(n\) sides and the forgotten angle is \(\alpha,\) then \((n-2)180=2017+\alpha.\) Since \(0\lt\alpha\lt180,\) \[ 2017\lt(n-2)180\lt2197. \]

The only multiple of \(180\) in this range is \(2160=(14-2)180,\) so \(n=14\) and \[ \alpha=2160-2017=143. \]

Thus, the correct answer is D.

12 答案 B

Horse \(k\) 在时刻 \(t\) 位于起点，恰好当 \(t\) 是 \(k\) 的倍数。因此我们需要最小的 \(t\)，使它在 \(1,2,\ldots,10\) 中至少有 \(5\) 个因数。

小于 \(12\) 的正整数至多有 \(4\) 个这样的因数，而 \(12\) 可被 \(1,2,3,4\) 和 \(6\) 整除。因此 \(T=12\)，它的各位数字之和是 \(1+2=3\)。

所以正确答案是 B。

English

Horse \(k\) is at the starting point at time \(t\) precisely when \(t\) is a multiple of \(k.\) So we want the smallest \(t\) with at least \(5\) divisors among \(1,2,\ldots,10.\)

The positive integers below \(12\) have at most \(4\) divisors, while \(12\) is divisible by \(1,2,3,4,\) and \(6.\) Thus \(T=12,\) and the sum of its digits is \(1+2=3.\)

Thus, the correct answer is B.

13 答案 B

设距离为 \(d\) 英里，通常速度为每小时 \(r\) 英里。由于通常行程为 \(3\) 小时，\(d=3r\)。

前 \(\dfrac{1}{3}\) 路程以速度 \(r\) 行驶，用时 \(\dfrac{1}{3}\cdot180=60\) 分钟，所以剩下 \(\dfrac{2}{3}\) 路程用时 \(276-60=216\) 分钟，\(=\dfrac{18}{5}\) 小时，速度为 \(r-20\)。

最后一段距离为 \(\dfrac{2}{3}d=2r\) 英里，所以 \[ 2r=(r-20)\cdot\dfrac{18}{5}\text{。} \] 解得 \(10r=18r-360\)，因此 \(r=45\)，\(d=3\cdot45=135\)。

所以正确答案是 B。

English

Let the distance be \(d\) miles and the usual speed \(r\) mph. Since the usual trip is \(3\) hours, \(d=3r.\)

The first \(\dfrac{1}{3}\) of the drive takes \(\dfrac{1}{3}\cdot180=60\) minutes at speed \(r,\) so the remaining \(\dfrac{2}{3}\) takes \(276-60=216\) minutes \(=\dfrac{18}{5}\) hours at speed \(r-20.\)

That final portion covers \(\dfrac{2}{3}d=2r\) miles, so \[ 2r=(r-20)\cdot\dfrac{18}{5}. \] Solving gives \(10r=18r-360,\) so \(r=45\) and \(d=3\cdot45=135.\)

Thus, the correct answer is B.

14 答案 C

令 \(X\)，\(Y\)，\(Z\) 分别表示 Alice-Bob、Alice-Carla、Derek-Eric 相邻的坐法。所求为 \(5!-|X\cup Y\cup Z|\)。

把一个被禁止的配对看作一个块，得到 \(|X|=|Y|=|Z|=2\cdot4!=48\)。对交集，有 \(|X\cap Y|=2\cdot3!=12\)（Alice 在 Bob 和 Carla 中间），\(|X\cap Z|=|Y\cap Z|\) \(=2\cdot2\cdot3!=24\)，且 \(|X\cap Y\cap Z|=2\cdot2\cdot2!=8\)。

由容斥，\(|X\cup Y\cup Z|=(48\cdot3)\) \(-(12+24+24)\) \(+8=92\)，所以答案是 \(120-92=28\)。

所以正确答案是 C。

English

Let \(X,\) \(Y,\) \(Z\) be the seatings where Alice-Bob, Alice-Carla, and Derek-Eric are adjacent, respectively. The answer is \(5!-|X\cup Y\cup Z|.\)

Treating a forbidden pair as a block gives \(|X|=|Y|=|Z|=2\cdot4!=48.\) For intersections, \(|X\cap Y|=2\cdot3!=12\) (Alice between Bob and Carla), \(|X\cap Z|=|Y\cap Z|\) \(=2\cdot2\cdot3!=24,\) and \(|X\cap Y\cap Z|=2\cdot2\cdot2!=8.\)

By inclusion-exclusion, \(|X\cup Y\cup Z|=(48\cdot3)\) \(-(12+24+24)\) \(+8=92,\) so the answer is \(120-92=28.\)

Thus, the correct answer is C.

15 答案 D

对 \(0\lt x\lt\dfrac{\pi}{2}\)，三项都为正，所以 \(f(x)\gt0\)。对 \(\dfrac{\pi}{2}\lt x\lt\pi\)，令 \(s=\sin x\gt0\)，\(t=-\cos x\gt0\)。因为 \(0\lt t\le1\)，\[ \begin{aligned} f(x)&=s-2t-\dfrac{3s}{t} \\ &=-2t-s\left(\dfrac{3}{t}-1\right)\lt0 \end{aligned}\text{。}\] 因此在 \(x=\pi\) 之前没有根。

当 \(x=\pi\)，\(f(\pi)=0+2(-1)+0=-2\lt0\)。当 \(x=\dfrac{5\pi}{4}\)，\(\tan x=1\)，所以 \(f=-\dfrac{\sqrt2}{2}+3\gt0\)。由介值定理，最小正根位于 \(\left(\pi,\dfrac{5\pi}{4}\right)\)。

因为 \(\pi\gt3\) 且 \(\dfrac{5\pi}{4}\lt4\)，这个区间包含在 \((3,4)\) 中。

所以正确答案是 D。

English

For \(0\lt x\lt\dfrac{\pi}{2}\) all three terms are positive, so \(f(x)\gt0.\) For \(\dfrac{\pi}{2}\lt x\lt\pi,\) set \(s=\sin x\gt0\) and \(t=-\cos x\gt0.\) Since \(0\lt t\le1,\) \[ \begin{aligned} f(x)&=s-2t-\dfrac{3s}{t} \\ &=-2t-s\left(\dfrac{3}{t}-1\right)\lt0. \end{aligned} \] Thus no root occurs before \(x=\pi.\)

At \(x=\pi,\) \(f(\pi)=0+2(-1)+0=-2\lt0.\) At \(x=\dfrac{5\pi}{4},\) \(\tan x=1\) so \(f=-\dfrac{\sqrt2}{2}+3\gt0.\) By the intermediate value theorem the smallest positive root lies in \(\left(\pi,\dfrac{5\pi}{4}\right).\)

Since \(\pi\gt3\) and \(\dfrac{5\pi}{4}\lt4,\) this interval sits inside \((3,4).\)

Thus, the correct answer is D.

16 答案 B

大半圆半径为 \(3\)，圆心 \(C\) 是 \(\overline{JK}\) 的中点。把 \(J\) 放在原点，则沿底边有 \(A=2\)、\(B=5\)、\(C=3\)、\(K=6\)。设 \(P\) 处圆的半径为 \(r\)。

由相切可知 \(PA=2+r\)，\(PB=1+r\)，且 \(PC=3-r\)。从 \(P\) 向底边作垂线，垂足的水平位置为 \(3+x\)，高度为 \(h\)，由勾股定理得 \[ \begin{aligned} h^2 &=(2+r)^2-(1+x)^2 \\ &=(3-r)^2-x^2 \\ &=(1+r)^2-(2-x)^2 \end{aligned}\text{。}\]

令第一个表达式与中间那个相等，得到 \(5r-x=3\)；令最后一个表达式与中间那个相等，得到 \(2r+x=3\)。两式相加得 \(7r=6\)，所以 \(r=\dfrac{6}{7}\)。

所以正确答案是 B。

English

The large semicircle has radius \(3\) and center \(C,\) the midpoint of \(\overline{JK}.\) Placing \(J\) at the origin, \(A=2,\) \(B=5,\) \(C=3,\) \(K=6\) along the base. Let \(r\) be the radius of the circle at \(P.\)

By tangency, \(PA=2+r,\) \(PB=1+r,\) and \(PC=3-r.\) Dropping a perpendicular from \(P\) to the base at horizontal position \(3+x\) with height \(h,\) the Pythagorean theorem gives \[ \begin{aligned} h^2 &=(2+r)^2-(1+x)^2 \\ &=(3-r)^2-x^2 \\ &=(1+r)^2-(2-x)^2. \end{aligned} \]

Equating the first expression with the middle one gives \(5r-x=3,\) while equating the last expression with the middle one gives \(2r+x=3.\) Adding yields \(7r=6,\) so \(r=\dfrac{6}{7}.\)

Thus, the correct answer is B.

17 答案 D

这 \(24\) 个解是 \(24\) 次单位根，即 \(z=e^{\frac{\pi i k}{12}}\)，其中 \(k=0,1,\ldots,23\)。

于是 \(z^6=e^{\frac{\pi i k}{2}}=\cos\dfrac{k\pi}{2}+i\sin\dfrac{k\pi}{2}\)，它为实数当且仅当 \(\sin\dfrac{k\pi}{2}=0\)，即 \(k\) 为偶数。这个范围内有 \(12\) 个偶数 \(k\)。

所以正确答案是 D。

English

The \(24\) solutions are the \(24\)th roots of unity, \(z=e^{\frac{\pi i k}{12}}\) for \(k=0,1,\ldots,23.\)

Then \(z^6=e^{\frac{\pi i k}{2}}=\cos\dfrac{k\pi}{2}+i\sin\dfrac{k\pi}{2},\) which is real exactly when \(\sin\dfrac{k\pi}{2}=0,\) i.e. when \(k\) is even. There are \(12\) even values of \(k\) in the range.

Thus, the correct answer is D.

18 答案 D

给 \(n\) 加 \(1\) 通常会使数字和增加 \(1\)，但末尾的每个 \(9\) 都会变成 \(0\)，使数字和减少 \(9\)。如果 \(n\) 恰好以 \(k\) 个九结尾，则 \(S(n+1)=S(n)+1-9k\) \(=1275-9k\)。

因此可能值为 \(1275,1266,1257,\ldots\) 在选项中，只有 \(1239=1275-9\cdot4\) 符合（例如，\(n\) 可以以四个 \(9\) 结尾，前面有足够多的 \(1\)）。

所以正确答案是 D。

English

Adding \(1\) to \(n\) increases the digit sum by \(1,\) except that each trailing \(9\) turns into a \(0,\) losing \(9.\) If \(n\) ends in exactly \(k\) nines, then \(S(n+1)=S(n)+1-9k\) \(=1275-9k.\)

So the possible values are \(1275,1266,1257,\ldots\) Among the choices, only \(1239=1275-9\cdot4\) fits (for example, \(n\) ending in four \(9\)s preceded by enough \(1\)s).

Thus, the correct answer is D.

19 答案 D

对第一个正方形，它截出的两个小三角形与整个三角形相似，得到 \(\dfrac{x}{3-x}=\dfrac{4-x}{x}\)，所以 \(x=\dfrac{12}{7}\)。（等价地，直角处内接正方形的边长为 \(\dfrac{3\cdot4}{3+4}\)。）

对第二个正方形，以长度为 \(5\) 的斜边作底；到斜边的高为 \(h=\dfrac{3\cdot4}{5}=\dfrac{12}{5}\)。若正方形一边在底边 \(b\) 上，高为 \(h\)，则正方形边长为 \(\dfrac{bh}{b+h}\)，所以 \[ y=\dfrac{5\cdot\tfrac{12}{5}}{5+\tfrac{12}{5}}=\dfrac{12}{\tfrac{37}{5}}=\dfrac{60}{37}\text{。}\]

因此 \(\dfrac{x}{y}=\dfrac{12}{7}\cdot\dfrac{37}{60}=\dfrac{37}{35}\)。

所以正确答案是 D。

English

For the first square, the two smaller triangles it cuts off are similar to the whole triangle, giving \(\dfrac{x}{3-x}=\dfrac{4-x}{x},\) so \(x=\dfrac{12}{7}.\) (Equivalently, a square in the right angle has side \(\dfrac{3\cdot4}{3+4}.\))

For the second square, take the hypotenuse of length \(5\) as base; the altitude to it is \(h=\dfrac{3\cdot4}{5}=\dfrac{12}{5}.\) A square with a side on a base \(b\) and height \(h\) has side \(\dfrac{bh}{b+h},\) so \[ y=\dfrac{5\cdot\tfrac{12}{5}}{5+\tfrac{12}{5}}=\dfrac{12}{\tfrac{37}{5}}=\dfrac{60}{37}. \]

Therefore \(\dfrac{x}{y}=\dfrac{12}{7}\cdot\dfrac{37}{60}=\dfrac{37}{35}.\)

Thus, the correct answer is D.

20 答案 E

令 \(u=\log_b a\)。因为 \(\log_b(a^{2017})=2017\log_b a\)，方程为 \(u^{2017}=2017u\)，所以 \(u=0\) 或 \(u^{2016}=2017\)。

若 \(u=0\) 则 \(a=1\)，对所有 \(199\) 个底数都有效。若 \(u^{2016}=2017\) 则 \(u=\pm2017^{\frac{1}{2016}}\)，对每个底数给出 \(2\) 个 \(a\) 值，也就是 \(2\cdot199=398\) 个有序对。

总共有 \(199+398=597\) 个有序对。

所以正确答案是 E。

English

Let \(u=\log_b a.\) Since \(\log_b(a^{2017})=2017\log_b a,\) the equation is \(u^{2017}=2017u,\) so \(u=0\) or \(u^{2016}=2017.\)

If \(u=0,\) then \(a=1,\) valid for every one of the \(199\) bases. If \(u^{2016}=2017,\) then \(u=\pm2017^{\frac{1}{2016}},\) giving \(2\) values of \(a\) for each base, i.e. \(2\cdot199=398\) pairs.

In total there are \(199+398=597\) ordered pairs.

Thus, the correct answer is E.

21 答案 D

使用多项式 \(10x+10\)，根 \(-1\) 进入 \(S\)。然后 \(1\) 作为 \(-x^{10}-x^9-\cdots-x+10\) 的根进入，而 \(-10\) 由 \(x+10\) 得到。

现在 \(x^3+x-10\) 有根 \(2\)，而 \(x+2\) 给出 \(-2\)；接着 \(2x-10\) 和 \(2x+10\) 给出 \(\pm5\)。此时 \(S=\{0,\pm1,\pm2,\pm5,\pm10\}\)。

不会再出现其他整数。用归纳法，\(S\) 中每个非零元素都整除 \(10\)。如果规则中使用的多项式常数项为 \(0\)，先提出最高可能次幂的 \(x\)；任何非零根就成为一个新多项式的根，而其常数项是原多项式的第一个非零系数。由有理根定理，该根整除这个系数；归纳假设说明该系数整除 \(10\)。因此 \(S\) 有 \(9\) 个元素。

因此，正确答案是 D。

English

Using \(10x+10,\) the root \(-1\) enters \(S.\) Then \(1\) enters as a root of \(-x^{10}-x^9-\cdots-x+10,\) and \(-10\) enters from \(x+10.\)

Now \(x^3+x-10\) has root \(2,\) and \(x+2\) gives \(-2;\) then \(2x-10\) and \(2x+10\) give \(\pm5.\) At this point \(S=\{0,\pm1,\pm2,\pm5,\pm10\}.\)

No further integer can appear. Inductively, every nonzero member of \(S\) divides \(10.\) If a polynomial used in the rule has constant term \(0,\) factor out the largest possible power of \(x;\) any nonzero root is then a root of a polynomial whose constant term is the first nonzero original coefficient. The Rational Root Theorem shows that the root divides this coefficient, which by the inductive hypothesis divides \(10.\) So \(S\) has \(9\) elements.

Thus, the correct answer is D.

22 答案 E

由对称性，把相关内部点分成三类：\(C=\{(0,0)\}\)、“轴向”点 \(A=\{(\pm1,0),(0,\pm1)\}\) 和“对角”点 \(I=\{(\pm1,\pm1)\}\)。设从 \(A,C,I\) 三类点出发最终碰到顶点的概率分别为 \(a,c,i\)。

读出转移概率（例如，\(A\) 中一点以概率 \(\tfrac28\) 到 \(A\)，以 \(\tfrac18\) 到 \(C\)，以 \(\tfrac28\) 到 \(I\)，并以 \(\tfrac38\) 到边内部，等等），得到 \[ \begin{aligned} &a=\tfrac28 a+\tfrac18 c+\tfrac28 i,\quad \\ &c=\tfrac48 a+\tfrac48 i,\quad \\ &i=\tfrac28 a+\tfrac18 c+\tfrac18 \end{aligned}\text{。}\]

解得 \(a=\dfrac{1}{14}\)，\(c=\dfrac{4}{35}\)，\(i=\dfrac{11}{70}\)。所求概率是 \(c=\dfrac{4}{35}\)，所以 \(m+n=4+35=39\)。

所以正确答案是 E。

English

By symmetry, group the relevant interior points into three types: \(C=\{(0,0)\},\) the “axis” points \(A=\{(\pm1,0),(0,\pm1)\},\) and the “diagonal” points \(I=\{(\pm1,\pm1)\}.\) Let \(a,c,i\) be the probabilities of eventually hitting a corner starting from a point of type \(A,C,I.\)

Reading off the transition probabilities (a point in \(A\) goes to \(A\) with prob \(\tfrac28,\) to \(C\) with \(\tfrac18,\) to \(I\) with \(\tfrac28,\) and to a side interior with \(\tfrac38,\) etc.) gives \[ \begin{aligned} &a=\tfrac28 a+\tfrac18 c+\tfrac28 i,\quad \\ &c=\tfrac48 a+\tfrac48 i,\quad \\ &i=\tfrac28 a+\tfrac18 c+\tfrac18. \end{aligned} \]

Solving yields \(a=\dfrac{1}{14},\) \(c=\dfrac{4}{35},\) \(i=\dfrac{11}{70}.\) The required probability is \(c=\dfrac{4}{35},\) so \(m+n=4+35=39.\)

Thus, the correct answer is E.

23 答案 C

因为 \(g\) 有三个不同的根，且都被四次多项式 \(f\) 共享，所以可写成 \(f(x)=(x-q)g(x)\)，其中 \(q\) 是剩下的根。展开得 \[ \begin{aligned} &f(x)=x^4+(a-q)x^3 \\ &\quad {}+(1-qa)x^2 \\ &\quad {}+(10-q)x-10q \end{aligned}\text{。}\]

比较 \(x\) 的系数，\(10-q=100\)，所以 \(q=-90\)。比较 \(x^3\) 的系数，\(a-q=1\)，所以 \(a=-89\)。

因此 \(g(1)=1+a+1+10\) \(=12-89=-77\)，且 \(1-q=91\)，所以 \[ \begin{aligned} &f(1)=(1-q)g(1) \\ &=91\cdot(-77)=-7007 \end{aligned}\text{。}\]

所以正确答案是 C。

English

Since \(g\) has three distinct roots all shared by the quartic \(f,\) we can write \(f(x)=(x-q)g(x)\) for some remaining root \(q.\) Expanding, \[ \begin{aligned} &f(x)=x^4+(a-q)x^3 \\ &\quad {}+(1-qa)x^2 \\ &\quad {}+(10-q)x-10q. \end{aligned} \]

Matching the \(x\) coefficient, \(10-q=100,\) so \(q=-90.\) Matching the \(x^3\) coefficient, \(a-q=1,\) so \(a=-89.\)

Then \(g(1)=1+a+1+10\) \(=12-89=-77\) and \(1-q=91,\) so \[ \begin{aligned} &f(1)=(1-q)g(1) \\ &=91\cdot(-77)=-7007. \end{aligned} \]

Thus, the correct answer is C.

24 答案 A

因为 \(YE\parallel AD\) 且 \(EF\parallel AC\)，可得到 \(\triangle XEY\sim\triangle XAD\) 和 \(\triangle XEF\sim\triangle XAC\)，所以 \(\dfrac{XY}{XE}=\dfrac{XD}{XA}\) 且 \(\dfrac{XF}{XE}=\dfrac{XC}{XA}\)。因而 \(\dfrac{XC}{XD}=\dfrac{XF}{XY}\)，所以 \(XF\cdot XD=XC\cdot XY\)。

在点 \(X\) 使用点的幂，\(XC\cdot XG=XD\cdot XB\)，合并可得 \(XF\cdot XG=XB\cdot XY\)。令 \(d=BD\)，则 \(DX=\dfrac14 d\) 且 \(BY=\dfrac{11}{36}d\)，所以 \[ \begin{aligned} XF\cdot XG &=\left(d-\tfrac14 d\right) \\ &\quad {}\cdot\left(d-\tfrac14 d-\tfrac{11}{36}d\right) \\ &=\dfrac34 d\cdot\dfrac49 d=\dfrac{d^2}{3} \end{aligned}\text{。}\]

由于 \(ABCD\) 是圆内接四边形，\(\angle BAD\) 与 \(\angle BCD\) 互补。在 \(\triangle ABD\) 和 \(\triangle CBD\) 中使用余弦定理，得到 \(\dfrac{73-d^2}{48}=\dfrac{d^2-40}{24}\)，所以 \(d^2=51\)。因此 \(XF\cdot XG=\dfrac{51}{3}=17\)。

所以正确答案是 A。

English

Because \(YE\parallel AD\) and \(EF\parallel AC,\) we get \(\triangle XEY\sim\triangle XAD\) and \(\triangle XEF\sim\triangle XAC,\) giving \(\dfrac{XY}{XE}=\dfrac{XD}{XA}\) and \(\dfrac{XF}{XE}=\dfrac{XC}{XA}.\) Hence \(\dfrac{XC}{XD}=\dfrac{XF}{XY},\) so \(XF\cdot XD=XC\cdot XY.\)

Power of a Point at \(X\) gives \(XC\cdot XG=XD\cdot XB,\) and combining yields \(XF\cdot XG=XB\cdot XY.\) With \(d=BD,\) \(DX=\dfrac14 d\) and \(BY=\dfrac{11}{36}d,\) so \[ \begin{aligned} XF\cdot XG &=\left(d-\tfrac14 d\right) \\ &\quad {}\cdot\left(d-\tfrac14 d-\tfrac{11}{36}d\right) \\ &=\dfrac34 d\cdot\dfrac49 d=\dfrac{d^2}{3}. \end{aligned} \]

Since \(ABCD\) is cyclic, \(\angle BAD\) and \(\angle BCD\) are supplementary. The Law of Cosines on \(\triangle ABD\) and \(\triangle CBD\) gives \(\dfrac{73-d^2}{48}=\dfrac{d^2-40}{24},\) so \(d^2=51.\) Therefore \(XF\cdot XG=\dfrac{51}{3}=17.\)

Thus, the correct answer is A.

25 答案 E

令 \(A=\{\sqrt2 i,-\sqrt2 i\}\)（每个模为 \(\sqrt2\)），令 \(B\) 为另外四个元素 （每个模为 \(\dfrac12\)）。由于 \(|P|=(\sqrt2)^{\#A}\left(\tfrac12\right)^{\#B}=1\) 强制 \(\#A=8\) 且 \(\#B=4\)，必须恰有 \(8\) 个因子来自 \(A\)，\(4\) 个来自 \(B\)。

\(8\) 个 \(A\) 中元素的乘积等于 \(\pm16\)（实数），而 \(4\) 个 \(B\) 中元素的乘积等于 \(\pm\tfrac{1}{16},\pm\tfrac{i}{16}\) 中的一个。它们的乘积是 \(\pm1,\pm i\) 中的一个，且四种结果等可能，所以这些配置中恰有 \(\tfrac14\) 给出 \(P=-1\)。

出现 \(8\) 个来自 \(A\)、\(4\) 个来自 \(B\) 这种模式的概率为 \(\binom{12}{4}\left(\tfrac13\right)^8\left(\tfrac23\right)^4=\dfrac{880}{3^{10}}\)。再乘以 \(\tfrac14\) 得到 \[ \begin{aligned} P&=\dfrac{1}{4}\cdot\dfrac{880}{3^{10}} \\ &=\dfrac{220}{3^{10}}=\dfrac{2^2\cdot5\cdot11}{3^{10}} \end{aligned}\text{。}\]

所以正确答案是 E。

English

Let \(A=\{\sqrt2 i,-\sqrt2 i\}\) (each of magnitude \(\sqrt2\)) and \(B\) be the other four elements (each of magnitude \(\dfrac12\)). Since \(|P|=(\sqrt2)^{\#A}\left(\tfrac12\right)^{\#B}=1\) forces \(\#A=8\) and \(\#B=4,\) exactly \(8\) factors must come from \(A\) and \(4\) from \(B.\)

A product of \(8\) elements of \(A\) equals \(\pm16\) (real), and a product of \(4\) elements of \(B\) equals one of \(\pm\tfrac{1}{16},\pm\tfrac{i}{16}.\) Their product is one of \(\pm1,\pm i,\) each equally likely, so exactly \(\tfrac14\) of these configurations give \(P=-1.\)

The chance of landing in the \(8\)-from-\(A\), \(4\)-from-\(B\) pattern is \(\binom{12}{4}\left(\tfrac13\right)^8\left(\tfrac23\right)^4=\dfrac{880}{3^{10}}.\) Multiplying by \(\tfrac14\) gives \[ \begin{aligned} P&=\dfrac{1}{4}\cdot\dfrac{880}{3^{10}} \\ &=\dfrac{220}{3^{10}}=\dfrac{2^2\cdot5\cdot11}{3^{10}}. \end{aligned} \]

Thus, the correct answer is E.