1989 AMC 12 真题（P版）

1 答案 C

由于 \(5^2=25\) 是奇数，\((-1)^{5^2}=-1\)。又有 \(1^{2^5}=1\)，所以和为 \(0\)。

所以正确答案是 C。

English

Because \(5^2=25\) is odd, \((-1)^{5^2}=-1.\) Also \(1^{2^5}=1,\) so the sum is \(0.\)

Thus the correct answer is C.

2 答案 D

有 \(\frac19+\frac1{16}=\frac{16+9}{144}=\frac{25}{144}\)。其正平方根为 \(\frac5{12}\)。

所以正确答案是 D。

English

We have \(\frac19+\frac1{16}=\frac{16+9}{144}=\frac{25}{144}.\) Its positive square root is \(\frac5{12}.\)

Thus the correct answer is D.

3 答案 D

设正方形边长为 \(s\)。每个矩形的长均为 \(s\)，且周长相同，所以三个宽相等，均为 \(\frac{s}{3}\)。因此 \[ 2\left(s+\frac{s}{3}\right)=24\text{，}\]故 \(s=9\)，正方形面积为 \(81\)。

所以正确答案是 D。

English

Let the square have side \(s.\) Since every rectangle has length \(s\) and the same perimeter, their three widths are equal and are each \(\frac{s}{3}.\) Thus \[ 2\left(s+\frac{s}{3}\right)=24, \] so \(s=9\) and the square’s area is \(81.\)

Thus the correct answer is D.

4 答案 D

作垂线 \(AG\) 和 \(BH\)，它们都垂直于 \(DF\)。在等腰梯形中，\(GH=AB=4\)，且 \[ DG=HC=\frac{DC-AB}{2}=3\text{。}\]由于 \(B\) 是 \(DE\) 的中点且 \(BH\parallel EF\)，中点定理说明 \(H\) 是 \(DF\) 的中点。因此 \(DH=DG+GH=7\)，故 \(DF=14\)，且 \(CF=14-10=4\)。

所以正确答案是 D。

English

Drop perpendiculars \(AG\) and \(BH\) to \(DF.\) In the isosceles trapezoid, \(GH=AB=4\) and \[ DG=HC=\frac{DC-AB}{2}=3. \] Since \(B\) is the midpoint of \(DE\) and \(BH\parallel EF,\) the midpoint theorem makes \(H\) the midpoint of \(DF.\) Hence \(DH=DG+GH=7,\) so \(DF=14\) and \(CF=14-10=4.\)

Thus the correct answer is D.

5 答案 E

有 \(11\) 条竖直网格线，每条含 \(20\) 根牙签；有 \(21\) 条水平网格线，每条含 \(10\) 根牙签。总数为 \[ 11\cdot20+21\cdot10=430\text{。}\]

所以正确答案是 E。

English

There are \(11\) vertical grid lines with \(20\) toothpicks each, and \(21\) horizontal grid lines with \(10\) toothpicks each. The total is \[ 11\cdot20+21\cdot10=430. \]

Thus the correct answer is E.

6 答案 A

两个截距分别为 \(\frac{6}{a}\) 和 \(\frac{6}{b}\)。因此三角形面积为 \[ \frac12\cdot\frac6a\cdot\frac6b=\frac{18}{ab}=6\text{，}\]所以 \(ab=3\)。

所以正确答案是 A。

English

The intercepts are \(\frac{6}{a}\) and \(\frac{6}{b}.\) Therefore the triangle’s area is \[ \frac12\cdot\frac6a\cdot\frac6b=\frac{18}{ab}=6, \] which gives \(ab=3.\)

Thus the correct answer is A.

7 答案 C

由于 \(M\) 是 \(AC\) 的中点，从 \(H\) 到 \(M\) 的水平变化为 \(\frac{HC}{2}\)，竖直变化为 \(\frac{AH}{2}\)。因此 \[ \begin{aligned} \tan\angle MHC &=\frac{\frac{AH}{2}}{\frac{HC}{2}}\\ &=\frac{AH}{HC}\\ &=\tan\angle ACH \end{aligned}\text{。}\]所以 \(\angle MHC=\angle C=30^\circ\)。

所以正确答案是 C。

English

Because \(M\) is the midpoint of \(AC,\) the horizontal change from \(H\) to \(M\) is \(\frac{HC}{2},\) while the vertical change is \(\frac{AH}{2}.\) Thus \[ \begin{aligned} \tan\angle MHC &=\frac{\frac{AH}{2}}{\frac{HC}{2}}\\ &=\frac{AH}{HC}\\ &=\tan\angle ACH. \end{aligned} \] Hence \(\angle MHC=\angle C=30^\circ.\)

Thus the correct answer is C.

8 答案 D

整系数分解必为 \((x-k)(x+k+1)\)，其中 \(k\) 为正整数。展开得 \(x^2+x-k(k+1)\)，所以 \(n=k(k+1)\)。\(k=1,2,\ldots,9\) 时 \(n\le100\)，而 \(10\cdot11\gt100\)。因此共有 \(9\) 个值。

所以正确答案是 D。

English

An integer factorization must have the form \((x-k)(x+k+1)\) for some positive integer \(k.\) Expanding gives \(x^2+x-k(k+1),\) so \(n=k(k+1).\) The values \(k=1,2,\ldots,9\) give \(n\le100,\) while \(10\cdot11\gt100.\) There are \(9\) values.

Thus the correct answer is D.

9 答案 B

前两个首字母必须从 \(A\) 到 \(Y\) 中选取两个不同字母。一旦选定，顺序也随之确定。因此首字母组合数为 \[ \binom{25}{2}=300\text{。}\]

所以正确答案是 B。

English

The first two initials must be two distinct letters chosen from \(A\) through \(Y.\) Once chosen, their order is forced. Therefore the number of monograms is \[ \binom{25}{2}=300. \]

Thus the correct answer is B.

10 答案 C

直接代入得 \[ \begin{aligned} u_2&=-\frac1{a+1},\\ u_3&=-\frac{a+1}{a},\\ u_4&=a \end{aligned}\text{。}\]因此数列每 \(3\) 项重复一次，而且 \(u_n=a\) 恰在 \(n\equiv1\pmod3\) 时成立。选项中只有 \(16\) 符合。

所以正确答案是 C。

English

Direct substitution gives \[ \begin{aligned} u_2&=-\frac1{a+1},\\ u_3&=-\frac{a+1}{a},\\ u_4&=a. \end{aligned} \] The sequence therefore repeats every \(3\) terms, so \(u_n=a\) whenever \(n\equiv1\pmod3.\) Among the choices, only \(16\) has that form.

Thus the correct answer is C.

11 答案 A

各量的最大可能值依次为 \(d=99,\ c=395,\ b=1184\)，最后 \(a=2367\)。每个值都满足相应严格不等式，所以该最大值可以达到。

所以正确答案是 A。

English

The largest possible values are \(d=99,\ c=395,\ b=1184,\) and finally \(a=2367.\) Each value satisfies its strict inequality, so this maximum is attainable.

Thus the correct answer is A.

12 答案 C

相对速度为每小时 \(120\) 英里，所以五分钟内司机相对行驶 \(10\) 英里。在这段距离内经过 \(20\) 辆等距车辆，间距约为 \(\frac{10}{20}=\frac{1}{2}\) 英里。因此一段 \(100\) 英里的公路约有 \(\frac{100}{\frac{1}{2}}=200\) 辆车。

所以正确答案是 C。

English

The relative speed is \(120\) miles per hour, so in five minutes the driver covers \(10\) relative miles. Passing \(20\) equally spaced vehicles in that distance means the spacing is about \(\frac{10}{20}=\frac{1}{2}\) mile. Thus a \(100\)-mile section contains about \(\frac{100}{\frac{1}{2}}=200\) vehicles.

Thus the correct answer is C.

13 答案 B

交叠部分是平行四边形。取倾斜的一边为底，其垂直高为斜带的宽 \(1\)。若底长为 \(L\)，其竖直分量等于水平带的宽，所以 \(L\sin\alpha=1\)。因此 \(L=\frac{1}{\sin\alpha}\)，面积为 \(L\cdot1=\frac{1}{\sin\alpha}\)。

所以正确答案是 B。

English

The overlap is a parallelogram. Taking a slanted side as its base, the perpendicular height is the width \(1\) of the slanted strip. If that base has length \(L,\) its vertical component is the width of the horizontal strip, so \(L\sin\alpha=1.\) Hence \(L=\frac{1}{\sin\alpha},\) and the area is \(L\cdot1=\frac{1}{\sin\alpha}.\)

Thus the correct answer is B.

14 答案 B

利用 \(\sin10=2\sin5\cos5\)，\[ \begin{aligned} \cot10+\tan5 &=\frac{\cos10}{\sin10}\\ &\quad+\frac{\sin5}{\cos5} \end{aligned}\text{。}\]通分后的分子为 \(\cos10+2\sin^25\)。由于 \(\cos10=\cos^25-\sin^25\)，该分子为 \(1\)，原式为 \(\csc10\)。

所以正确答案是 B。

English

Using \(\sin10=2\sin5\cos5,\) \[ \begin{aligned} \cot10+\tan5 &=\frac{\cos10}{\sin10}\\ &\quad+\frac{\sin5}{\cos5}. \end{aligned} \] Combining the fractions gives numerator \(\cos10+2\sin^25.\) Since \(\cos10=\cos^25-\sin^25,\) this numerator is \(1,\) and the expression is \(\csc10.\)

Thus the correct answer is B.

15 答案 E

令 \(AD=m\)、\(DC=9-m\)。斯图尔特定理给出 \[ \begin{gathered} 7^2m+5^2(9-m) \\ =9\bigl(5^2+m(9-m)\bigr) \end{gathered}\text{。}\]化简得 \(9m^2-57m=0\)，所以 \(m=\frac{19}{3}\)。于是 \(DC=\frac{8}{3}\)，且 \(AD:DC=19:8\)。

所以正确答案是 E。

English

Let \(AD=m\) and \(DC=9-m.\) Stewart’s Theorem gives \[ \begin{gathered} 7^2m+5^2(9-m) \\ =9\bigl(5^2+m(9-m)\bigr). \end{gathered} \] Simplifying yields \(9m^2-57m=0,\) so \(m=\frac{19}{3}.\) Then \(DC=\frac{8}{3},\) and \(AD:DC=19:8.\)

Thus the correct answer is E.

16 答案 B

横坐标与纵坐标之差分别为 \(45\) 和 \(264\)，它们的最大公因数为 \(3\)。因此，包括两个端点在内，格点数为 \(\gcd(45,264)+1=4\)。

所以正确答案是 B。

English

The coordinate differences are \(45\) and \(264,\) whose greatest common divisor is \(3.\) The number of lattice points, including both endpoints, is therefore \(\gcd(45,264)+1=4.\)

Thus the correct answer is B.

17 答案 D

若正方形的边长为 \(s\)，则三角形的边长为 \(s+d\)。周长条件给出 \[ 3(s+d)-4s=1989\text{，}\]所以 \(s=3d-1989\)。条件 \(s\gt0\) 等价于 \(d\gt663\)。因此，正整数 \(1\) 至 \(663\) 恰好是不可能的值。

所以正确答案是 D。

English

If the square has side \(s,\) the triangle has side \(s+d.\) The perimeter condition gives \[ 3(s+d)-4s=1989, \] so \(s=3d-1989.\) The condition \(s\gt0\) is equivalent to \(d\gt663.\) Thus the positive integers \(1\) through \(663\) are precisely the impossible values.

Thus the correct answer is D.

18 答案 B

分母有理化可得 \[ \frac1{x+\sqrt{x^2+1}}=\sqrt{x^2+1}-x\text{。}\]因此整个式子等于 \(2x\)。它是有理数，当且仅当 \(x\) 是有理数。

所以正确答案是 B。

English

Rationalizing gives \[ \frac1{x+\sqrt{x^2+1}}=\sqrt{x^2+1}-x. \] The entire expression is therefore \(2x.\) It is rational exactly when \(x\) is rational.

Thus the correct answer is B.

19 答案 E

半径为 \(R=\frac{12}{2\pi}=\frac{6}{\pi}\)。弧长 \(3,4,5\) 对应的圆心角为 \(\frac{\pi}{2},\frac{2\pi}{3},\frac{5\pi}{6}\)。这些角的正弦之和为 \(1+\frac{\sqrt3}{2}+\frac{1}{2}=\frac{3+\sqrt3}{2}\)。从圆心将三角形分割后，其面积为 \[ \begin{aligned} K&=\frac12\cdot\frac{36}{\pi^2} \cdot\frac{3+\sqrt3}{2}\\ &=\frac9{\pi^2}(\sqrt3+3) \end{aligned}\text{。}\]

所以正确答案是 E。

English

The radius is \(R=\frac{12}{2\pi}=\frac{6}{\pi}.\) The arc lengths \(3,4,5\) give central angles \(\frac{\pi}{2},\frac{2\pi}{3},\frac{5\pi}{6}.\) Their sines sum to \(1+\frac{\sqrt3}{2}+\frac{1}{2}=\frac{3+\sqrt3}{2}.\) Splitting the triangle at the center, its area is \[ \begin{aligned} K&=\frac12\cdot\frac{36}{\pi^2} \cdot\frac{3+\sqrt3}{2}\\ &=\frac9{\pi^2}(\sqrt3+3). \end{aligned} \]

Thus the correct answer is E.

20 答案 B

已知条件等价于 \(144\le x\lt169\)，该区间的长度为 \(25\)。所求条件等价于 \[ 120^2\le100x\lt121^2\text{，}\]即 \(144\le x\lt146.41\)。这个区间的长度为 \(2.41=\frac{241}{100}\)，所以条件概率为 \[ \frac{\frac{241}{100}}{25}=\frac{241}{2500}\text{。}\]

所以正确答案是 B。

English

The given condition is \(144\le x\lt169,\) an interval of length \(25.\) The desired condition is \[ 120^2\le100x\lt121^2, \] or \(144\le x\lt146.41.\) Its length is \(2.41=\frac{241}{100},\) so the conditional probability is \[ \frac{\frac{241}{100}}{25}=\frac{241}{2500}. \]

Thus the correct answer is B.

21 答案 C

将旗帜缩放为单位正方形，并设其中心到蓝色正方形一个顶点的水平距离为 \(h\)。四个全等的白色角三角形的总面积为 \((1-2h)^2\)。因为十字占 \(0.36\)，\[ (1-2h)^2=0.64\text{。}\]由 \(0\lt h\lt\frac{1}{2}\)，可得 \(h=0.1\)。蓝色正方形的两条互相垂直的对角线长度分别为 \(2h\) 和 \(2h\)，所以其面积为 \(\frac12(2h)^2=2h^2=0.02\)，即旗帜面积的 \(2\%\)。

所以正确答案是 C。

English

Scale the flag to a unit square, and let \(h\) be the horizontal distance from its center to a vertex of the blue square. The four congruent white corner triangles have total area \((1-2h)^2.\) Since the cross occupies \(0.36,\) \[ (1-2h)^2=0.64. \] With \(0\lt h\lt\frac{1}{2},\) this gives \(h=0.1.\) The blue square has perpendicular diagonals \(2h\) and \(2h,\) so its area is \(\frac12(2h)^2=2h^2=0.02,\) or \(2\%\) of the flag.

Thus the correct answer is C.

22 答案 A

材料、大小、颜色和形状各自的其他选择数依次为 \(1,2,3,3\)。恰好改变两个属性，共有 \[ \begin{aligned} &1\cdot2+1\cdot3+1\cdot3\\ &\qquad+2\cdot3+2\cdot3+3\cdot3=29 \end{aligned}\text{。}\]

所以正确答案是 A。

English

The numbers of alternatives for material, size, color, and shape are \(1,2,3,3.\) Changing exactly two attributes gives \[ \begin{aligned} &1\cdot2+1\cdot3+1\cdot3\\ &\qquad+2\cdot3+2\cdot3+3\cdot3=29. \end{aligned} \]

Thus the correct answer is A.

23 答案 D

在时刻 \(n^2\)，粒子位于 \((n,0)\)（当 \(n\) 为奇数）或 \((0,n)\)（当 \(n\) 为偶数）。因此，在时刻 \(44^2=1936\)，粒子位于 \((0,44)\)。接着它向右移动 \(44\) 个单位，到达 \((44,44)\) 时是时刻 \(1980\)，然后在接下来的 \(9\) 分钟内向下移动。因此，在时刻 \(1989\)，它位于 \((44,35)\)。

所以正确答案是 D。

English

At time \(n^2,\) the particle is at \((n,0)\) for odd \(n\) and at \((0,n)\) for even \(n.\) Thus at time \(44^2=1936\) it is at \((0,44).\) It then moves \(44\) units right, reaching \((44,44)\) at time \(1980,\) and moves downward for the next \(9\) minutes. At time \(1989\) it is therefore at \((44,35).\)

Thus the correct answer is D.

24 答案 B

当女性人数分别为 \(0,1,2\) 时，可能的数对依次为 \[ \begin{aligned} &(0,5),\ (2,5),\\ &(4,5),\ (3,4) \end{aligned}\text{，}\]其中有两名女性的两种情形分别对应她们相邻和不相邻。交换两种性别，可得 \[ \begin{aligned} &(5,0),\ (5,2),\\ &(5,4),\ (4,3) \end{aligned}\text{。}\]这些是 \(8\) 个不同的有序数对。

所以正确答案是 B。

English

For \(0,1,2\) females, the possible pairs are respectively \[ \begin{aligned} &(0,5),\ (2,5),\\ &(4,5),\ (3,4), \end{aligned} \] where the two-female cases distinguish adjacent from nonadjacent females. Swapping the sexes gives \[ \begin{aligned} &(5,0),\ (5,2),\\ &(5,4),\ (4,3). \end{aligned} \] These are \(8\) distinct ordered pairs.

Thus the correct answer is B.

25 答案 B

两队分数之和为 \(55\)，所以获胜分数至多为 \(27\)。其最小值为 \(1+2+3+4+5=15\)。从 \(15\) 到 \(20\) 的每个分数都可由 \(\{1,2,3,4,k\}\) 取得，其中 \(k=5,\ldots,10\)。从 \(21\) 到 \(25\) 的分数可由 \(\{1,2,3,k,10\}\) 取得，其中 \(k=5,\ldots,9\)；而 \(26,27\) 分别可由 \(\{1,2,4,9,10\}\) 和 \(\{1,2,5,9,10\}\) 取得。因此，一共有 \(13\) 个可能的整数，即从 \(15\) 到 \(27\)。

所以正确答案是 B。

English

The two scores sum to \(55,\) so the winning score is at most \(27.\) Its minimum is \(1+2+3+4+5=15.\) Every score from \(15\) through \(20\) is obtained by \(\{1,2,3,4,k\}\) for \(k=5,\ldots,10.\) Scores \(21\) through \(25\) use \(\{1,2,3,k,10\}\) for \(k=5,\ldots,9,\) and \(26,27\) use \(\{1,2,4,9,10\}\) and \(\{1,2,5,9,10\}.\) Thus all \(13\) integers from \(15\) through \(27\) are possible.

Thus the correct answer is B.

26 答案 C

取边长为 \(2\) 的立方体，则其体积为 \(8\)。各面的中心为 \((\pm1,0,0),(0,\pm1,0),(0,0,\pm1)\)。在每个卦限中，八面体截出一个三条互相垂直的棱长均为单位长度的四面体，其体积为 \(\frac{1}{6}\)。所以八面体的体积为 \(8(\frac{1}{6})=\frac{4}{3}\)，所求体积比为 \(\frac{\frac{4}{3}}{8}=\frac{1}{6}\)。

所以正确答案是 C。

English

Take a cube of side \(2,\) so its volume is \(8.\) The face centers are \((\pm1,0,0),(0,\pm1,0),(0,0,\pm1).\) In each octant, the octahedron cuts out a tetrahedron with three perpendicular unit edges and volume \(\frac{1}{6}.\) Thus its volume is \(8(\frac{1}{6})=\frac{4}{3},\) and the ratio is \(\frac{\frac{4}{3}}{8}=\frac{1}{6}.\)

Thus the correct answer is C.

27 答案 D

对于 \(s=x+y\ge2\)，共有 \(s-1\) 个正整数有序数对 \((x,y)\)；而 \(z=n-2s\) 在 \(s\le\lfloor\frac{n-1}{2}\rfloor\) 时为正数。令 \(m=\lfloor\frac{n-1}{2}\rfloor\)，解的个数为 \[ \sum_{s=2}^{m}(s-1)=\frac{m(m-1)}2\text{。}\]令它等于 \(28\)，得到 \(m=8\)。因此 \(\lfloor\frac{n-1}{2}\rfloor=8\)，所以 \(n=17\) 或 \(18\)。

所以正确答案是 D。

English

For \(s=x+y\ge2,\) there are \(s-1\) positive ordered pairs \((x,y),\) and \(z=n-2s\) is positive when \(s\le\lfloor\frac{n-1}{2}\rfloor.\) Writing \(m=\lfloor\frac{n-1}{2}\rfloor,\) the number of solutions is \[ \sum_{s=2}^{m}(s-1)=\frac{m(m-1)}2. \] Setting this equal to \(28\) gives \(m=8.\) Hence \(\lfloor\frac{n-1}{2}\rfloor=8,\) so \(n=17\) or \(18.\)

Thus the correct answer is D.

28 答案 D

方程 \(t^2-9t+1=0\) 的两个根 \(r,s\) 都是正数，并且满足 \(rs=1\)。因此，它们的锐角反正切之和为 \(\frac{\pi}{2}\)。在 \(0\) 与 \(2\pi\) 之间，每个正切值都出现两次，第二次比第一次增加 \(\pi\)。全部四个根之和为 \[ 2\left(\frac\pi2\right)+2\pi=3\pi\text{。}\]

所以正确答案是 D。

English

The two roots \(r,s\) of \(t^2-9t+1=0\) are positive and satisfy \(rs=1.\) Therefore their acute arctangents add to \(\frac{\pi}{2}.\) Each tangent value occurs twice between \(0\) and \(2\pi,\) with the second occurrence shifted by \(\pi.\) The sum of all four roots is \[ 2\left(\frac\pi2\right)+2\pi=3\pi. \]

Thus the correct answer is D.

29 答案 B

所求和是 \[ (1+i)^{99}=2^{\frac{99}{2}}e^{\frac{99\pi i}{4}} \] 的实部。由于 \(99\equiv3\pmod8\)，其实部为 \[ \begin{aligned} 2^{\frac{99}{2}}\cos\frac{3\pi}{4} &=2^{\frac{99}{2}}\left(-\frac{\sqrt2}{2}\right)\\ &=-2^{49} \end{aligned}\text{。}\]

所以正确答案是 B。

English

The sum is the real part of \[ (1+i)^{99}=2^{\frac{99}{2}}e^{\frac{99\pi i}{4}}. \] Since \(99\equiv3\pmod8,\) its real part is \[ \begin{aligned} 2^{\frac{99}{2}}\cos\frac{3\pi}{4} &=2^{\frac{99}{2}}\left(-\frac{\sqrt2}{2}\right)\\ &=-2^{49}. \end{aligned} \]

Thus the correct answer is B.

30 答案 A

对于 \(19\) 对相邻位置中的每一对，两人性别不同的概率为 \[ \frac7{20}\cdot\frac{13}{19} +\frac{13}{20}\cdot\frac7{19}\text{。}\]由期望的线性性质，\[ \begin{aligned} \mathbb E[S] &=19\cdot\frac{2\cdot7\cdot13}{20\cdot19}\\ &=\frac{91}{10}=9.1 \end{aligned}\text{，}\]最接近 \(9\)。

所以正确答案是 A。

English

For each of the \(19\) adjacent position pairs, the probability of mixed sexes is \[ \frac7{20}\cdot\frac{13}{19} +\frac{13}{20}\cdot\frac7{19}. \] By linearity of expectation, \[ \begin{aligned} \mathbb E[S] &=19\cdot\frac{2\cdot7\cdot13}{20\cdot19}\\ &=\frac{91}{10}=9.1, \end{aligned} \] which is closest to \(9.\)

Thus the correct answer is A.