1994 AMC 12 真题（P版）

1 答案 C

先合并同底数的幂，再将它们配对，\[ \begin{aligned} 4^4\cdot4^9\cdot9^4\cdot9^9 &=4^{13}9^{13}\\ &=(4\cdot9)^{13}\\ &=36^{13} \end{aligned}\text{。}\]因此正确答案是 C。

English

Combining like bases and then pairing them, \[ \begin{aligned} 4^4\cdot4^9\cdot9^4\cdot9^9 &=4^{13}9^{13}\\ &=(4\cdot9)^{13}\\ &=36^{13}. \end{aligned} \] Thus the correct answer is C.

2 答案 B

设两列的宽分别为 \(u,v\)，两行的高分别为 \(s,t\)。图中面积给出 \(us=6\)、\(vs=14\) 和 \(vt=35\)。因此缺少的面积为 \[ ut=\frac{(us)(vt)}{vs}=\frac{6\cdot35}{14}=15\text{。}\]所以正确答案是 B。

English

Let the column widths be \(u,v\) and the row heights be \(s,t.\) The displayed areas give \(us=6,\) \(vs=14,\) and \(vt=35.\) Therefore the missing area is \[ ut=\frac{(us)(vt)}{vs}=\frac{6\cdot35}{14}=15. \] Thus the correct answer is B.

3 答案 B

给定的和为 \(2x^x\)，所以式 I 恒等于它。式 II 在 \(x=1\) 时不成立，此时它的值为 \(1\)，而不是 \(2\)。式 III 在 \(x=2\) 时不成立，此时它的值为 \(16\)，而不是 \(8\)。式 IV 在 \(x=1\) 时不成立，此时它的值为 \(4\)，而不是 \(2\)。因此只有一个式子符合要求。所以正确答案是 B。

English

The given sum is \(2x^x,\) so expression I is always equal to it. Expression II fails at \(x=1,\) where its value is \(1\) instead of \(2.\) Expression III fails at \(x=2,\) where its value is \(16\) instead of \(8.\) Expression IV fails at \(x=1,\) where its value is \(4\) instead of \(2.\) Hence only one expression works. Thus the correct answer is B.

4 答案 A

圆心是 \((10,0)\)，半径是 \(15\)。因此 \[ (x-10)^2+15^2=15^2\text{，}\]所以 \(x=10\)。故正确答案是 A。

English

The center is \((10,0)\) and the radius is \(15.\) Thus \[ (x-10)^2+15^2=15^2, \] so \(x=10.\) Thus the correct answer is A.

5 答案 E

若起始数为 \(x\)，错误的计算给出 \[ \frac{x}{6}-14=16\text{，}\]所以 \(x=180\)。正确的计算应得到 \(6(180)+14=1094\)，它大于 \(1000\)。因此正确答案是 E。

English

If the starting number is \(x,\) the mistaken calculation gives \[ \frac{x}{6}-14=16, \] so \(x=180.\) The intended calculation would produce \(6(180)+14=1094,\) which is greater than \(1000.\) Thus the correct answer is E.

6 答案 A

向前推算，由 \(d+0=1\) 得 \(d=1\)，再由 \(c+d=0\) 得 \(c=-1\)，由 \(b+c=d\) 得 \(b=2\)。最后 \(a+b=c\)，所以 \(a=-1-2=-3\)。因此正确答案是 A。

English

Working backward, \(d+0=1\) gives \(d=1,\) then \(c+d=0\) gives \(c=-1,\) and \(b+c=d\) gives \(b=2.\) Finally \(a+b=c,\) so \(a=-1-2=-3.\) Thus the correct answer is A.

7 答案 E

每个正方形的面积都是 \(100\)。它们的重叠部分是 \(\triangle GAB\)，其底为 \(AB=10\)，从中心 \(G\) 到 \(\overline{AB}\) 的高为 \(5\)。其面积为 \(\frac12(10)(5)=25\)。因此并集面积为 \(100+100-25=175\)。所以正确答案是 E。

English

Each square has area \(100.\) Their overlap is \(\triangle GAB,\) whose base is \(AB=10\) and whose altitude from the center \(G\) to \(\overline{AB}\) is \(5.\) Its area is \(\frac12(10)(5)=25.\) Hence the union has area \(100+100-25=175.\) Thus the correct answer is E.

8 答案 C

每条边长为 \(2\)。以边长为单位，七个水平条带的宽度为 \[ 1,3,5,7,5,3,1\text{，}\]总计 \(25\) 个单位正方形。每个正方形的面积为 \(2^2=4\)，所以多边形的面积为 \(25\cdot4=100\)。因此正确答案是 C。

English

Each side has length \(2.\) Measured in side-length units, the seven horizontal strips have widths \[ 1,3,5,7,5,3,1, \] totaling \(25\) unit squares. Each such square has area \(2^2=4,\) so the polygon’s area is \(25\cdot4=100.\) Thus the correct answer is C.

9 答案 D

设 \(\angle B=x\)，则 \(\angle A=4x\)。余角条件给出 \[ 90-x=4(90-4x)\text{。}\]因此 \(15x=270\)，且 \(x=18^\circ\)。所以正确答案是 D。

English

Let \(\angle B=x,\) so \(\angle A=4x.\) The complement condition gives \[ 90-x=4(90-4x). \] Hence \(15x=270\) and \(x=18^\circ.\) Thus the correct answer is D.

10 答案 B

由已知顺序，\(m(b,c)=b\)，所以 \(M(a,m(b,c))=M(a,b)=b\)。此外，\(m(a,e)=a\)，因此 \(m(d,m(a,e))=m(d,a)=a\)。最外层表达式为 \(M(b,a)=b\)。所以正确答案是 B。

English

From the ordering, \(m(b,c)=b,\) so \(M(a,m(b,c))=M(a,b)=b.\) Also \(m(a,e)=a,\) hence \(m(d,m(a,e))=m(d,a)=a.\) The outer expression is therefore \(M(b,a)=b.\) Thus the correct answer is B.

11 答案 D

粘合前的总表面积为 \[ 6(1^2+2^2+3^2)=84\text{。}\]边长为 \(2\) 的立方体可以共用面积为 \(4\) 的面与边长为 \(3\) 的立方体粘合；把边长为 \(1\) 的立方体放在它们的公共棱处，它就能与两个较大的立方体各共用面积为 \(1\) 的面。因此总接触面积为 \(6\)。每块粘合面积会去掉两份外露面，得到 \(84-2(6)=72\)。任意两个立方体共用的面积都不可能超过较小立方体的一个面，所以这里的接触面积已达到最大，表面积也达到最小。因此正确答案是 D。

English

Before gluing, the total surface area is \[ 6(1^2+2^2+3^2)=84. \] The side-\(2\) cube can share area \(4\) with the side-\(3\) cube, while the side-\(1\) cube is placed at their common edge so that it shares area \(1\) with each larger cube. Thus the total contact area is \(6.\) Each glued area removes two exposed copies, giving \(84-2(6)=72.\) No pair can share more than the smaller face, so this is maximal contact and minimal surface area. Thus the correct answer is D.

12 答案 D

因为 \(i^{-1}=-i\)，\[ \begin{aligned} \left(i-i^{-1}\right)^{-1} &=(i+i)^{-1}\\ &=\frac1{2i}\\ &=-\frac i2 \end{aligned}\text{。}\]所以正确答案是 D。

English

Because \(i^{-1}=-i,\) \[ \begin{aligned} \left(i-i^{-1}\right)^{-1} &=(i+i)^{-1}\\ &=\frac1{2i}\\ &=-\frac i2. \end{aligned} \] Thus the correct answer is D.

13 答案 B

设 \(\angle A=\theta\)。由于 \(AP=PC\)，三角形 \(APC\) 中有 \(\angle PAC=\angle PCA=\theta\)。由于 \(AB=AC\)，\(\triangle ABC\) 的每个底角都是 \(\frac{180^\circ-\theta}{2}\)。又因为 \(PC=CB\)，所以 \(\triangle PCB\) 在 \(P\) 和 \(B\) 处的底角相等；其中 \(B\) 处的角为 \(\frac{180^\circ-\theta}{2}\)。因此 \(\angle PCB=\theta\)。在 \(C\) 处，\[ 2\theta=\frac{180^\circ-\theta}{2}\text{，}\]从而 \(5\theta=180^\circ\)，且 \(\theta=36^\circ\)。所以正确答案是 B。

English

Let \(\angle A=\theta.\) Since \(AP=PC,\) triangle \(APC\) has \(\angle PAC=\angle PCA=\theta.\) Since \(AB=AC,\) each base angle of \(\triangle ABC\) is \(\frac{180^\circ-\theta}{2}.\) Also \(PC=CB,\) so the base angles of \(\triangle PCB\) at \(P\) and \(B\) are equal; the one at \(B\) is \(\frac{180^\circ-\theta}{2}.\) Hence \(\angle PCB=\theta.\) At \(C,\) \[ 2\theta=\frac{180^\circ-\theta}{2}, \] giving \(5\theta=180^\circ\) and \(\theta=36^\circ.\) Thus the correct answer is B.

14 答案 B

项数为 \[ \frac{40-20}{\frac{1}{5}}+1=101\text{。}\]各项的平均数是 \(\frac{20+40}{2}=30\)，所以总和是 \(101\cdot30=3030\)。因此正确答案是 B。

English

The number of terms is \[ \frac{40-20}{\frac{1}{5}}+1=101. \] Their average is \(\frac{20+40}{2}=30,\) so the sum is \(101\cdot30=3030.\) Thus the correct answer is B.

15 答案 B

写成 \(n=10a+b\)。模 \(100\) 意义下，\[ n^2\equiv20ab+b^2\text{。}\]项 \(20ab\) 使十位数字改变一个偶数，所以只需考虑 \(b^2\) 的十位数字。在 \(b=0,1,\ldots,9\) 中，只有 \(b=4\) 和 \(b=6\) 时该数字为奇数。从 \(1\) 到 \(100\) 中，每种个位数字都出现 \(10\) 次，所以共有 \(2\cdot10=20\) 个。因此正确答案是 B。

English

Write \(n=10a+b.\) Modulo \(100,\) \[ n^2\equiv20ab+b^2. \] The term \(20ab\) changes the tens digit by an even amount, so only the tens digit of \(b^2\) matters. Among \(b=0,1,\ldots,9,\) it is odd only for \(b=4\) and \(b=6.\) Each units digit occurs \(10\) times from \(1\) through \(100,\) giving \(2\cdot10=20.\) Thus the correct answer is B.

16 答案 B

两个条件给出 \(\frac{R-1}{T-1}=\frac17\) 和 \(\frac{R}{T-2}=\frac15\)。因此 \(T=7R-6\) 且 \(T=5R+2\)。于是 \(R=4\)，\(T=22\)。所以正确答案是 B。

English

The two conditions give \(\frac{R-1}{T-1}=\frac17\) and \(\frac{R}{T-2}=\frac15.\) Thus \(T=7R-6\) and \(T=5R+2.\) Hence \(R=4\) and \(T=22.\) Thus the correct answer is B.

17 答案 D

长方形比圆宽，所以公共区域是圆去掉与圆心距离超过 \(\sqrt2\) 的上、下两个弓形。一个弓形的面积为 \[ \begin{aligned} 2^2\left(\frac{\pi}{4}\right) &-\sqrt2\sqrt{2^2-(\sqrt2)^2}\\ &=\pi-2 \end{aligned}\text{。}\]因此公共区域面积为 \(4\pi-2(\pi-2)=2\pi+4\)。所以正确答案是 D。

English

The rectangle is wider than the circle, so the common region is the circle with the top and bottom caps beyond distance \(\sqrt2\) from the center removed. One cap has area \[ \begin{aligned} 2^2\left(\frac{\pi}{4}\right) &-\sqrt2\sqrt{2^2-(\sqrt2)^2}\\ &=\pi-2. \end{aligned} \] Therefore the common area is \(4\pi-2(\pi-2)=2\pi+4.\) Thus the correct answer is D.

18 答案 C

由内角和得 \(9\angle A=180^\circ\)，所以 \(\angle A=20^\circ\)。因此弦 \(\overline{BC}\) 所对的圆心角为 \(40^\circ\)。正 \(n\) 边形的相邻顶点所对圆心角为 \(\frac{360^\circ}{n}\)，所以 \[ \frac{360^\circ}{n}=40^\circ\text{，}\]从而 \(n=9\)。因此正确答案是 C。

English

The angle sum gives \(9\angle A=180^\circ,\) so \(\angle A=20^\circ.\) The central angle subtending chord \(\overline{BC}\) is therefore \(40^\circ.\) Adjacent vertices of a regular \(n\)-gon subtend \(\frac{360^\circ}{n},\) so \[ \frac{360^\circ}{n}=40^\circ, \] and \(n=9.\) Thus the correct answer is C.

19 答案 C

为了避免出现十张标号相同的圆片，可以抽出 \[ 1+2+\cdots+9=45 \] 张标号小于 \(10\) 的所有圆片，并从每种标号中至多抽 \(9\) 张；这样的标号共有 \(41\) 种，从 \(10\) 到 \(50\)。因此，在尚未被迫得到十张相同标号的圆片前，最多可抽 \(45+9(41)=414\) 张；再抽一张就能保证达到要求。最小数量为 \(415\)。所以正确答案是 C。

English

To avoid ten equal labels, one may draw all \[ 1+2+\cdots+9=45 \] disks with labels below \(10,\) and at most \(9\) from each of the \(41\) labels \(10\) through \(50.\) Thus \(45+9(41)=414\) disks can be drawn without forcing ten alike, and the next draw guarantees them. The minimum is \(415.\) Thus the correct answer is C.

20 答案 B

因为 \(y=xr\) 且 \(z=xr^2\)，等差数列条件给出 \[ 4xr=x+3xr^2\text{。}\]这里 \(x\ne0\)，所以 \(3r^2-4r+1=0\)，得到 \(r=1\) 或 \(r=\frac{1}{3}\)。条件 \(x\ne y\) 排除 \(r=1\)，只剩 \(r=\frac{1}{3}\)。因此正确答案是 B。

English

Because \(y=xr\) and \(z=xr^2,\) the arithmetic-sequence condition gives \[ 4xr=x+3xr^2. \] Here \(x\ne0,\) so \(3r^2-4r+1=0,\) yielding \(r=1\) or \(r=\frac{1}{3}.\) The condition \(x\ne y\) excludes \(r=1,\) leaving \(r=\frac{1}{3}.\) Thus the correct answer is B.

21 答案 C

可能的奇数是 \(13,31,121,211\) 和 \(1111\)。数 \(13,31\) 和 \(211\) 是质数，而 \(121=11^2\)，且 \(1111=11\cdot101\)。所以有 \(2\) 个反例。因此正确答案是 C。

English

The odd possibilities are \(13,31,121,211,\) and \(1111.\) The numbers \(13,31,\) and \(211\) are prime, while \(121=11^2\) and \(1111=11\cdot101.\) Thus there are \(2\) counterexamples. Thus the correct answer is C.

22 答案 C

教授必须占据第 \(2,3,\ldots,8\) 把椅子中的三个互不相邻的位置。这种 \(3\) 元子集从连续 \(7\) 个位置中选取，其数量为 \[ \binom{7-3+1}{3}=\binom53=10\text{。}\]三位不同的教授可以用 \(3!=6\) 种方式安排到所选椅子上，共有 \(10\cdot6=60\) 种。因此正确答案是 C。

English

The professors must occupy three nonconsecutive positions among chairs \(2,3,\ldots,8.\) The number of such \(3\)-subsets of \(7\) consecutive positions is \[ \binom{7-3+1}{3}=\binom53=10. \] The three distinct professors can be assigned to the selected chairs in \(3!=6\) ways, for \(10\cdot6=60.\) Thus the correct answer is C.

23 答案 E

该区域的面积为 \(3\cdot3+2\cdot1=11\)，所以每一半的面积为 \(\frac{11}{2}\)。若直线为 \(y=mx\)，则直线下方包含面积为 \(2\) 的整个长方形（从 \(x=3\) 到 \(x=5\)），以及一个底为 \(3\)、高为 \(3m\) 的三角形。因此 \[ 2+\frac12(3)(3m)=\frac{11}{2}\text{。}\]由此得 \(m=\frac{7}{9}\)，它确实位于所假设的范围内。因此正确答案是 E。

English

The region has area \(3\cdot3+2\cdot1=11,\) so each half has area \(\frac{11}{2}.\) If the line is \(y=mx,\) the part below it consists of the entire \(2\)-unit rectangle from \(x=3\) to \(x=5,\) together with a triangle of base \(3\) and height \(3m.\) Therefore \[ 2+\frac12(3)(3m)=\frac{11}{2}. \] This gives \(m=\frac{7}{9},\) which indeed lies in the assumed range. Thus the correct answer is E.

24 答案 C

样本 \(7,7,12,12,12\) 的总和为 \(50\)，中位数为 \(12\)，极差为 \(5\)，所以极差 \(5\) 可以达到。假设极差至多为 \(4\)。若最大值至多为 \(12\)，则最大的三个值都必须是 \(12\)，余下两个较小值之和为 \(14\)；但它们都至少为 \(8\)，产生矛盾。若最大值大于 \(12\)，则最小值大于 \(8\)，使总和大于 \(8+8+12+12+12=52\)。因此更小的极差不可能达到。正确答案是 C。

English

The sample \(7,7,12,12,12\) has sum \(50,\) median \(12,\) and range \(5,\) so range \(5\) is possible. Suppose the range were at most \(4.\) If the largest value were at most \(12,\) then the top three values would all be \(12,\) leaving the bottom two with sum \(14;\) yet each would be at least \(8,\) a contradiction. If the largest value exceeded \(12,\) then the smallest would exceed \(8,\) making the total exceed \(8+8+12+12+12=52.\) Thus no smaller range works. The correct answer is C.

25 答案 A

若 \(x\gt0\)，第二个方程给出 \(y=-x^2\)，而第一个方程给出 \(y=3-x\)；这要求 \(x^2-x+3=0\)，但它没有实根。因此 \(x\lt0\)。令 \(t=-x\gt0\)。两个方程变为 \(t+y=3\) 和 \(ty-t^3=0\)，所以 \(y=t^2\)。因此 \(x-y=-t-t^2=-3\)，它本身就是整数。所以正确答案是 A。

English

If \(x\gt0,\) the second equation gives \(y=-x^2,\) while the first gives \(y=3-x;\) these would require \(x^2-x+3=0,\) which has no real root. Hence \(x\lt0.\) Put \(t=-x\gt0.\) Then the equations become \(t+y=3\) and \(ty-t^3=0,\) so \(y=t^2.\) Therefore \(x-y=-t-t^2=-3,\) already an integer. Thus the correct answer is A.

26 答案 A

在内侧多边形的一个顶点处，它的内角与周围多边形的两个内角之和为 \(360^\circ\)。因此 \[ \begin{gathered} 180^\circ\left(1-\frac2m\right) \\ {}+2\cdot180^\circ\left(1-\frac2n\right)\\ {}=360^\circ \end{gathered}\text{。}\]化简得 \(\frac{1}{m}+\frac{2}{n}=\frac{1}{2}\)。代入 \(m=10\)，得到 \(\frac{2}{n}=\frac{2}{5}\)，所以 \(n=5\)。因此正确答案是 A。

English

At a vertex of the inner polygon, its interior angle and two angles from the surrounding polygons total \(360^\circ.\) Thus \[ \begin{gathered} 180^\circ\left(1-\frac2m\right) \\ {}+2\cdot180^\circ\left(1-\frac2n\right)\\ {}=360^\circ. \end{gathered} \] This simplifies to \(\frac{1}{m}+\frac{2}{n}=\frac{1}{2}.\) With \(m=10,\) we get \(\frac{2}{n}=\frac{2}{5},\) so \(n=5.\) Thus the correct answer is A.

27 答案 D

选中一粒会爆开的玉米的两种概率为 \[ \begin{gathered} P(W\text{ 且爆开})=\frac23\cdot\frac12=\frac13, \\ P(Y\text{ 且爆开})=\frac13\cdot\frac23=\frac29 \end{gathered}\text{。}\]因此 \(P(\text{爆开})=\frac{5}{9}\)，且 \[ P(W\mid\text{爆开})=\frac{\frac{1}{3}}{\frac{5}{9}}=\frac35\text{。}\]所以正确答案是 D。

English

The probabilities of selecting a kernel that pops are \[ \begin{gathered} P(W\text{ and pop})=\frac23\cdot\frac12=\frac13, \\ P(Y\text{ and pop})=\frac13\cdot\frac23=\frac29. \end{gathered} \] Hence \(P(\text{pop})=\frac{5}{9},\) and \[ P(W\mid\text{pop})=\frac{\frac{1}{3}}{\frac{5}{9}}=\frac35. \] Thus the correct answer is D.

28 答案 C

设 \(x\) 轴截距为质数 \(p\)，\(y\) 轴截距为正整数 \(q\)。截距式给出 \[ \begin{gathered} \frac4p+\frac3q=1, \\ q=\frac{3p}{p-4}=3+\frac{12}{p-4} \end{gathered}\text{。}\]因此 \(p-4\) 是 \(12\) 的正因数。检验 \(1,2,3,4,6,12\)，只有质数 \(p\) 对应 \(p=5\) 和 \(p=7\)。所以共有 \(2\) 条直线。因此正确答案是 C。

English

Let the \(x\)-intercept be the prime \(p\) and the \(y\)-intercept be the positive integer \(q.\) Intercept form gives \[ \begin{gathered} \frac4p+\frac3q=1, \\ q=\frac{3p}{p-4}=3+\frac{12}{p-4}. \end{gathered} \] Thus \(p-4\) is a positive divisor of \(12.\) Testing \(1,2,3,4,6,12\) gives prime \(p\) only for \(p=5\) and \(p=7.\) Therefore there are \(2\) lines. Thus the correct answer is C.

29 答案 A

小弧 \(\overset{\frown}{BC}\) 所对的圆心角为 \(\frac{r}{r}=1\)，所以 \(BC=2r\sin\frac12\)。由于 \(AB=AC\) 且 \(AB\gt r\)，点 \(A\) 是大弧 \(BC\) 的中点。从 \(A\) 到 \(B\) 的较小圆心角为 \(\pi-\frac12\)，所以 \(AB=2r\) \(\sin\left(\frac{\pi-\frac{1}{2}}{2}\right)=2r\cos\frac14\)。因此 \(\frac{AB}{BC}=\frac{\cos(\frac{1}{4})}{\sin(\frac{1}{2})}\) \(=\frac1{2\sin(\frac{1}{4})}=\frac12\csc\frac14\)。所以正确答案是 A。

English

The central angle subtending minor arc \(\overset{\frown}{BC}\) is \(\frac{r}{r}=1,\) so \(BC=2r\sin\frac12.\) Since \(AB=AC\) and \(AB\gt r,\) point \(A\) is the midpoint of the major arc \(BC.\) The minor central angle from \(A\) to \(B\) is \(\pi-\frac12,\) so \(AB=2r\)\(\sin\left(\frac{\pi-\frac{1}{2}}{2}\right)=2r\cos\frac14.\) Therefore \(\frac{AB}{BC}=\frac{\cos(\frac{1}{4})}{\sin(\frac{1}{2})}\)\(=\frac1{2\sin(\frac{1}{4})}=\frac12\csc\frac14.\) Thus the correct answer is A.

30 答案 C

可行的最少骰子数为 \[ n=\left\lceil\frac{1994}{6}\right\rceil=333\text{。}\]把每个骰子的结果 \(d\) 替换为 \(7-d\)，就会在点数和为 \(T\) 与点数和为 \(7n-T\) 的结果之间建立一一对应。因此当 \(n=333\) 时，与 \(1994\) 配对的点数和为 \[ S=7(333)-1994=337\text{。}\]从最小点数和到平均值，骰子点数和的出现次数严格递增，所以对任意可行的 \(n\)，都不会有小于 \(337\) 的点数和与 \(1994\) 的出现次数相同。因此最小值为 \(337\)，正确答案是 C。

English

The smallest feasible number of dice is \[ n=\left\lceil\frac{1994}{6}\right\rceil=333. \] Replacing every die result \(d\) by \(7-d\) is a bijection between outcomes of sum \(T\) and outcomes of sum \(7n-T.\) Thus for \(n=333,\) the sum paired with \(1994\) is \[ S=7(333)-1994=337. \] Dice-sum counts increase strictly from the minimum sum up to the mean, so for any feasible \(n\) no sum below \(337\) can match the count at \(1994.\) Thus the smallest value is \(337,\) and the correct answer is C.