1998 AMC 12 真题（P版）

1 答案 E

位置 \(II\) 中长方形的左、右标号必须分别能在另两个长方形的右、左边找到。只有 \(D\) 能同时满足，因为它的两个侧边标号为 \(7\) 和 \(4\)：\(E\) 的右边标号为 \(7\)，而 \(A\) 的左边标号为 \(4\)。因此位置 \(I\) 中是 \(E\)，正确答案是 E。

English

Position \(II\) needs both its left and right labels to occur as right and left labels, respectively, on two other rectangles. Only \(D,\) with side labels \(7\) and \(4,\) permits both matches: \(E\) has right label \(7,\) and \(A\) has left label \(4.\) Hence position \(I\) contains \(E,\) so the correct answer is E.

2 答案 E

最小的正分母是 \(0+1=1\)，再用另两个数字组成的最大分子是 \(8+9=17\)。它们的比是整数 \(17\)，显然已经最大。因此 \(A+B=17\)，正确答案是 E。

English

The least positive denominator is \(0+1=1,\) and the greatest numerator using two other digits is \(8+9=17.\) Their ratio is the integer \(17,\) which is plainly maximal. Thus \(A+B=17,\) and E is correct.

3 答案 D

个位给出 \(12-b=3\)，所以 \(b=9\)。十位中，\(a-1\) 必须再借位，而 \(a-1+10-8=7\)，所以 \(a=6\)。百位于是给出 \(7-1-4=c=2\)。因此 \(a+b+c=6+9+2=17\)，正确答案是 D。

English

The units column gives \(12-b=3,\) so \(b=9.\) In the tens column, \(a-1\) must borrow, and \(a-1+10-8=7,\) giving \(a=6.\) The hundreds column then gives \(7-1-4=c=2.\) Therefore \(a+b+c=6+9+2=17,\) so D is correct.

4 答案 E

三个内层值都是 \(1\)：\(\frac{90}{90}=1,\frac{3}{3}=1\)，且 \(\frac{15}{15}=1\)。所以外层表达式为 \([1,1,1]=\frac{1+1}{1}=2\)。正确答案是 E。

English

Each inner value is \(1:\) \(\frac{90}{90}=1,\frac{3}{3}=1,\) and \(\frac{15}{15}=1.\) Thus the outer expression is \([1,1,1]=\frac{1+1}{1}=2.\) The correct answer is E.

5 答案 C

因式分解得 \(2^{1995}(8-4-2+1)=3\cdot2^{1995}\)。因此 \(k=3\)，正确答案是 C。

English

Factoring gives \(2^{1995}(8-4-2+1)=3\cdot2^{1995}.\) Hence \(k=3,\) so C is correct.

6 答案 C

因为 \(1998=2\cdot3^3\cdot37\)，最接近 \(\sqrt{1998}\) 的因数对是 \(37\cdot54\)。它们的差为 \(54-37=17\)，正确答案是 C。

English

Since \(1998=2\cdot3^3\cdot37,\) its factor pair nearest \(\sqrt{1998}\) is \(37\cdot54.\) Their difference is \(54-37=17,\) so C is correct.

7 答案 D

最内层根式为 \(N^{\frac{1}{3}}\)。下一层为 \((N\cdot N^{\frac{1}{3}})^{\frac{1}{3}}=N^{\frac{4}{9}}\)。最外层为 \((N\cdot N^{\frac{4}{9}})^{\frac{1}{3}}=N^{\frac{13}{27}}\)。所以正确答案是 D。

English

The innermost radical is \(N^{\frac{1}{3}}.\) The next is \((N\cdot N^{\frac{1}{3}})^{\frac{1}{3}}=N^{\frac{4}{9}}.\) The outermost is \((N\cdot N^{\frac{4}{9}})^{\frac{1}{3}}=N^{\frac{13}{27}}.\) Thus D is correct.

8 答案 D

每个梯形的面积为 \(\frac{1}{3}\)。它的高为 \(\frac{1}{2}\)，两条平行边长为 \(x\) 和 \(\frac{1}{2}\)。因此 \[ \frac12\cdot\frac12\left(x+\frac12\right)=\frac13\text{，}\]解得 \(x=\frac{5}{6}\)。所以正确答案是 D。

English

Each trapezoid has area \(\frac{1}{3}.\) Its height is \(\frac{1}{2},\) and its parallel sides have lengths \(x\) and \(\frac{1}{2}.\) Therefore \[ \frac12\cdot\frac12\left(x+\frac12\right)=\frac13, \] which gives \(x=\frac{5}{6}.\) Thus D is correct.

9 答案 D

平均时间为 \(0.20(60)+0.10(0)\) \(+0.35(20)+0.35(40)\) \(=12+7+14=33\) 分钟。所以正确答案是 D。

English

The average is \(0.20(60)+0.10(0)\) \(+0.35(20)+0.35(40)\) \(=12+7+14=33\) minutes. Thus D is correct.

10 答案 A

若长方形的边长为 \(x,y\)，则 \(2x+2y=14\)，所以 \(x+y=7\)。大正方形的边长为 \(x+y\)，故其面积为 \(7^2=49\)。正确答案是 A。

English

If a rectangle has sides \(x,y,\) then \(2x+2y=14,\) so \(x+y=7.\) The large square has side \(x+y,\) hence area \(7^2=49.\) The correct answer is A.

11 答案 D

四条边分别确定四个不同的直径圆。两条对角线等长且中点相同，所以它们确定同一个、也就是第五个圆。因此共有 \(5\) 个，正确答案是 D。

English

The four sides determine four distinct diameter circles. The two diagonals are equal and share a midpoint, so they determine the same fifth circle. Hence there are \(5,\) and D is correct.

12 答案 A

依次取幂得到 \[ N=7^{\,5^{\,3^{\,2^{11}}}}\text{。}\]因此 \(7\) 是 \(N\) 唯一的质因数，正确答案是 A。

English

Successively exponentiating gives \[ N=7^{\,5^{\,3^{\,2^{11}}}}. \] Thus \(7\) is the only prime factor of \(N,\) so the correct answer is A.

13 答案 E

可行的结果包括：\((4,4,3,3)\) 的和为 \(14\)，\((6,4,3,2)\) 的和为 \(15\)，\((6,6,2,2)\) 的和为 \(16\)，而 \((6,6,4,1)\) 的和为 \(17\)。若出现两个 \(6\)，另外两个面的乘积为 \(4\)，其和至多为 \(5\)，故总和至多为 \(17\)；出现不足两个 \(6\) 时，也不可能使总和为 \(18\) 且乘积仍为 \(144\)。所以 \(18\) 不可能出现，正确答案是 E。

English

Valid rolls include \((4,4,3,3)\) with sum \(14,\) \((6,4,3,2)\) with sum \(15,\) \((6,6,2,2)\) with sum \(16,\) and \((6,6,4,1)\) with sum \(17.\) If two \(6\)s occur, the other two faces have product \(4\) and sum at most \(5,\) giving total at most \(17;\) the cases with fewer \(6\)s also cannot total \(18\) while retaining product \(144.\) Thus \(18\) is impossible, and E is correct.

14 答案 A

抛物线开口向上，所以 \(a\gt0\)。两根异号，因此 \(\frac{c}{a}\lt0\)，从而 \(c\lt0\)。另外，\(-\frac{b}{2a}=4\gt0\)，所以 \(b\lt0\)。因此只有 \(a\) 一定为正，正确答案是 A。

English

The parabola opens upward, so \(a\gt0.\) Its roots have opposite signs, so \(\frac{c}{a}\lt0\) and hence \(c\lt0.\) Also \(-\frac{b}{2a}=4\gt0,\) so \(b\lt0.\) Therefore only \(a\) must be positive, and A is correct.

15 答案 C

边长为 \(s\) 的正六边形由六个边长为 \(s\) 的等边三角形组成。若面积相等的等边三角形边长为 \(t\)，则 \(t^2=6s^2\)，所以 \(\frac{t}{s}=\sqrt6\)。正确答案是 C。

English

A regular hexagon of side \(s\) is six equilateral triangles of side \(s.\) An equal-area equilateral triangle of side \(t\) therefore satisfies \(t^2=6s^2,\) so \(\frac{t}{s}=\sqrt6.\) The correct answer is C.

16 答案 B

阴影面积为 \[ \begin{aligned} &\frac{\pi}{2}\left(\frac{a+b}{2}\right)^2 +\frac{\pi}{2}\left(\frac a2\right)^2\\ &\qquad-\frac{\pi}{2}\left(\frac b2\right)^2 =\frac{\pi a(a+b)}4 \end{aligned}\text{。}\]同理，非阴影面积为 \(\frac{\pi b(a+b)}{4}\)。二者之比为 \(\frac{a}{b}\)，正确答案是 B。

English

The shaded area is \[ \begin{aligned} &\frac{\pi}{2}\left(\frac{a+b}{2}\right)^2 +\frac{\pi}{2}\left(\frac a2\right)^2\\ &\qquad-\frac{\pi}{2}\left(\frac b2\right)^2 =\frac{\pi a(a+b)}4. \end{aligned} \] Similarly, the unshaded area is \(\frac{\pi b(a+b)}{4}.\) Their ratio is \(\frac{a}{b},\) so B is correct.

17 答案 E

令 \(y=0\)，得到 \(f(x)=x+f(0)=x+2\)。因此 \(f(1998)=2000\)，正确答案是 E。

English

Taking \(y=0\) gives \(f(x)=x+f(0)=x+2.\) Hence \(f(1998)=2000,\) so E is correct.

18 答案 A

三个体积分别为 \(A=\frac23\pi r^3\)、\(M=2\pi r^3\) 和 \(C=\frac43\pi r^3\)。因此 \(A-M+C\) \(=(\frac{2}{3}-2+\frac{4}{3})\pi r^3=0\)，正确答案是 A。

English

The volumes are \(A=\frac23\pi r^3,\) \(M=2\pi r^3,\) and \(C=\frac43\pi r^3.\) Therefore \(A-M+C\) \(=(\frac{2}{3}-2+\frac{4}{3})\pi r^3=0,\) so A is correct.

19 答案 C

面积为 \(\frac12(10)|5\sin\theta|=25|\sin\theta|\)。所以 \(|\sin\theta|=\frac{2}{5}\)。圆上有四个相应的点，每个点都给出一个不同的三角形。因此答案为 \(4\)，正确答案是 C。

English

The area is \(\frac12(10)|5\sin\theta|=25|\sin\theta|.\) Thus \(|\sin\theta|=\frac{2}{5}.\) There are four corresponding points on the circle, and each gives a distinct triangle. Hence the answer is \(4,\) making C correct.

20 答案 C

八个三元组为 \[ \begin{gathered} (1,2,10),(1,3,9),\\ (1,4,8),(1,5,7),\\ (2,3,8),(2,4,7),\\ (2,5,6),(3,4,6) \end{gathered}\text{。}\]凯西的话排除 \((3,4,6)\)。得知这一点后，特蕾西的话排除最右端为 \(10,9\) 或 \(6\) 的情形。剩下的三元组为 \((1,4,8),(2,3,8)\) 和 \((1,5,7),(2,4,7)\)。若中间数为 \(3\) 或 \(5\)，斯泰西此时就能确定三元组，所以她的话迫使中间数为 \(4\)。因此正确答案是 C。

English

The eight triples are \[ \begin{gathered} (1,2,10),(1,3,9),\\ (1,4,8),(1,5,7),\\ (2,3,8),(2,4,7),\\ (2,5,6),(3,4,6). \end{gathered} \] Casey’s statement excludes \((3,4,6).\) With that known, Tracy’s statement excludes rightmost values \(10,9,\) and \(6.\) The remaining triples are \((1,4,8),(2,3,8)\) and \((1,5,7),(2,4,7).\) A middle value \(3\) or \(5\) would now identify the triple, so Stacy’s statement forces the middle value \(4.\) Thus C is correct.

21 答案 C

若桑妮的速度为 \(r\)，则温迪的速度为 \(\frac{r(h-d)}{h}\)。第二次比赛中，桑妮需要 \(\frac{h+d}{r}\) 的时间，在此期间温迪跑了 \[ \frac{r(h-d)}h\cdot\frac{h+d}{r} =h-\frac{d^2}{h}\text{。}\]桑妮到达位置 \(h\) 时完成比赛，所以领先距离为 \(\frac{d^2}{h}\)。正确答案是 C。

English

If Sunny’s speed is \(r,\) Windy’s is \(\frac{r(h-d)}{h}.\) Sunny needs time \(\frac{h+d}{r}\) in the second race, during which Windy runs \[ \frac{r(h-d)}h\cdot\frac{h+d}{r} =h-\frac{d^2}{h}. \] Sunny finishes at position \(h,\) so the lead is \(\frac{d^2}{h}.\) Thus C is correct.

22 答案 C

每一项都等于 \(\log_{100!}b\)。因此总和为 \[ \begin{aligned} &\log_{100!}(2\cdot3\cdots100)\\ &\qquad=\log_{100!}(100!)=1 \end{aligned}\text{。}\]所以正确答案是 C。

English

Each term equals \(\log_{100!}b.\) Therefore the sum is \[ \begin{aligned} &\log_{100!}(2\cdot3\cdots100)\\ &\qquad=\log_{100!}(100!)=1. \end{aligned} \] Thus C is correct.

23 答案 D

两个圆分别为 \[ (x-6)^2+(y-3)^2=49\text{，}\]\[ (x-2)^2+(y-6)^2=k+40\text{。}\]两圆圆心距为 \(5\)。设第二个圆的半径为 \(r=\sqrt{k+40}\)，相交条件为 \(|7-r|\le5\le7+r\)，即 \(2\le r\le12\)。所以 \(-36\le k\le104\)，且 \(b-a=140\)。正确答案是 D。

English

The circles are \[ (x-6)^2+(y-3)^2=49, \] \[ (x-2)^2+(y-6)^2=k+40. \] Their centers are \(5\) units apart. With second radius \(r=\sqrt{k+40},\) intersection requires \(|7-r|\le5\le7+r,\) or \(2\le r\le12.\) Thus \(-36\le k\le104,\) and \(b-a=140.\) The correct answer is D.

24 答案 C

每个匹配条件都给出 \(10^4=10,000\) 个号码：选择三个前缀数字和剩余一个不受限制的数字。若两个条件同时成立，则 \(d_1d_2d_3=d_4d_5d_6=d_5d_6d_7\)，迫使所有数字相同，所以交集有 \(10\) 个。由容斥原理，答案为 \(10,000+10,000-10=19,990\)，正确答案是 C。

English

Each matching condition gives \(10^4=10,000\) numbers: choose the three prefix digits and the one unconstrained remaining digit. If both hold, then \(d_1d_2d_3=d_4d_5d_6=d_5d_6d_7,\) forcing all digits equal, so there are \(10\) overlaps. Inclusion-exclusion gives \(10,000+10,000-10=19,990,\) so C is correct.

25 答案 B

折痕经过 \((2,1)\)，斜率为 \(2\)，所以其方程为 \(y=2x-3\)。过 \((7,3)\) 的垂线斜率为 \(-\frac{1}{2}\)，与折痕交于 \((\frac{19}{5},\frac{23}{5})\)。该交点是 \((7,3)\) 与 \((m,n)\) 的中点，所以 \(m=\frac{3}{5},\ n=\frac{31}{5}\)。因此 \(m+n=\frac{34}{5}=6.8\)，正确答案是 B。

English

The crease passes through \((2,1)\) with slope \(2,\) so it is \(y=2x-3.\) The perpendicular through \((7,3)\) has slope \(-\frac{1}{2}\) and meets the crease at \((\frac{19}{5},\frac{23}{5}).\) This intersection is the midpoint of \((7,3)\) and \((m,n),\) giving \(m=\frac{3}{5},\ n=\frac{31}{5}.\) Hence \(m+n=\frac{34}{5}=6.8,\) so B is correct.

26 答案 B

因为 \(\angle B+\angle D=180^\circ\)，四边形为圆内接四边形，且 \(AC\) 是其直径。在三角形 \(ABD\) 中，\[ \begin{aligned} BD^2&=13^2+46^2\\ &\quad-2(13)(46)\cos120^\circ\\ &=2883=(31\sqrt3)^2 \end{aligned}\text{。}\]由正弦定理的推广形式，\(AC=\frac{BD}{\sin120^\circ}\) \(=\frac{31\sqrt3}{\frac{\sqrt3}{2}}=62\)。所以正确答案是 B。

English

Because \(\angle B+\angle D=180^\circ,\) the quadrilateral is cyclic, and \(AC\) is its diameter. In triangle \(ABD,\) \[ \begin{aligned} BD^2&=13^2+46^2\\ &\quad-2(13)(46)\cos120^\circ\\ &=2883=(31\sqrt3)^2. \end{aligned} \] By the extended law of sines, \(AC=\frac{BD}{\sin120^\circ}\) \(=\frac{31\sqrt3}{\frac{\sqrt3}{2}}=62.\) Thus B is correct.

27 答案 E

第一步之后，\(8\) 个角块各贡献 \(27\) 个外露单位面，\(12\) 个棱块各贡献 \(36\) 个。给一个角块打隧道会移除 \(3\) 个外露单位正方形，并增加 \(24\) 个隧道内壁正方形；给一个棱块打隧道会移除 \(4\) 个外露面，也增加 \(24\) 个内壁面。因此最终面积为 \[ \begin{aligned} &8(27-3+24)\\ &\qquad+12(36-4+24)\\ &=384+672=1056 \end{aligned}\text{。}\]所以正确答案是 E。

English

After the first stage, \(8\) corner subcubes contribute \(27\) exposed units each, and \(12\) edge subcubes contribute \(36\) each. Tunneling a corner subcube removes \(3\) exposed unit squares and adds \(24\) tunnel-wall squares; tunneling an edge subcube removes \(4\) and also adds \(24.\) Hence the final area is \[ \begin{aligned} &8(27-3+24)\\ &\qquad+12(36-4+24)\\ &=384+672=1056. \end{aligned} \] Thus E is correct.

28 答案 B

设 \(\angle DAB=\alpha\)。因为 \(\frac{AC}{AD}=\cos2\alpha=\frac{2}{3}\)，由 \(\cos2\alpha\) 的恒等式可得 \(\tan\alpha=\frac{1}{\sqrt5}\)。取 \(AC=1\)，则 \[ \begin{aligned} CD&=\tan2\alpha=\frac{\sqrt5}{2},\\ CB&=\tan3\alpha=\frac{7\sqrt5}{5} \end{aligned}\text{。}\]因此 \(BD=CB-CD=\frac{9\sqrt5}{10}\)，且 \(\frac{CD}{BD}=\frac{5}{9}\)。所以 \(m+n=14\)，正确答案是 B。

English

Let \(\angle DAB=\alpha.\) Since \(\frac{AC}{AD}=\cos2\alpha=\frac{2}{3},\) the identity for \(\cos2\alpha\) gives \(\tan\alpha=\frac{1}{\sqrt5}.\) Taking \(AC=1,\) \[ \begin{aligned} CD&=\tan2\alpha=\frac{\sqrt5}{2},\\ CB&=\tan3\alpha=\frac{7\sqrt5}{5}. \end{aligned} \] Thus \(BD=CB-CD=\frac{9\sqrt5}{10},\) and \(\frac{CD}{BD}=\frac{5}{9}.\) Therefore \(m+n=14,\) so B is correct.

29 答案 D

只需包围三个非共线格点 \((0,0),(0,1),(1,0)\)。在最大位置，相对的两条边被邻近格点卡住；最大可能距离为 \(\sqrt5\)，对应的两条平行线分别经过 \((1,1)\) 与 \((0,-1)\)。因此面积至多为 \(5\)。由下列四条直线围成的正方形达到这个上界：\[ \begin{aligned} y&=2x+2,\\ y&=2x-3,\\ 2y+x&=3,\\ 2y+x&=-2 \end{aligned}\text{。}\]其面积为 \(5\)，且内部恰好包含上述三个格点。所以正确答案是 D。

English

It suffices to enclose the three noncollinear lattice points \((0,0),(0,1),(1,0).\) In a maximal placement, two opposite sides are pinned by neighboring lattice points; the greatest possible separation is the distance \(\sqrt5\) between parallel lines through \((1,1)\) and \((0,-1).\) Hence the area is at most \(5.\) This is attained by the square bounded by \[ \begin{aligned} y&=2x+2,\\ y&=2x-3,\\ 2y+x&=3,\\ 2y+x&=-2. \end{aligned} \] Its area is \(5\) and exactly the three stated lattice points lie inside. Thus D is correct.

30 答案 E

任意十个连续整数中的五个偶数至少贡献 \(2^8\)。所以只有当数块至少含八个因数 \(5\) 时，最右端非零数字才可能为奇数；第一次可能出现这种情形，是数块包含 \(5^7=78125\) 时。对于 \(n=5^7-9\)，数块中 \(v_2=9\)、\(v_5=8\)，所以该数字仍为偶数。对于 \(n=5^7-8=78117\)，两个指数都为 \(8\)。约去 \(2^8 5^8\)，再将剩余奇数的个位数字相乘，得到 \[ \begin{aligned} &7\cdot9\cdot1\cdot3\cdot9\cdot3\\ &\qquad\cdot1\cdot1\cdot3 \equiv9\pmod{10} \end{aligned}\text{。}\]因此第一个为奇数的最右端非零数字是 \(9\)，正确答案是 E。

English

The five even terms in any ten consecutive integers contribute at least \(2^8.\) Thus the rightmost nonzero digit can be odd only when the block contains at least eight factors of \(5,\) first possible when it contains \(5^7=78125.\) For \(n=5^7-9,\) the block has \(v_2=9\) and \(v_5=8,\) so the digit remains even. For \(n=5^7-8=78117,\) both valuations are \(8.\) Cancelling \(2^8 5^8\) and multiplying the remaining odd unit digits gives \[ \begin{aligned} &7\cdot9\cdot1\cdot3\cdot9\cdot3\\ &\qquad\cdot1\cdot1\cdot3 \equiv9\pmod{10}. \end{aligned} \] Hence the first odd rightmost nonzero digit is \(9,\) and E is correct.