1984 AMC 12 真题（P版）

1 答案 C

分解分母得 \[ \begin{aligned} 252^2-248^2 &=(252+248)\\ &\quad{}\cdot(252-248)\\ &=500\cdot4=2000 \end{aligned}\text{。}\]因此原商为 \(\frac{1000^2}{2000}=500\)。所以正确答案是 C。

English

Factoring the denominator gives \[ \begin{aligned} 252^2-248^2 &=(252+248)\\ &\quad{}\cdot(252-248)\\ &=500\cdot4=2000. \end{aligned} \] Thus the given quotient is \(\frac{1000^2}{2000}=500.\) Therefore, the correct answer is C.

2 答案 B

有 \(x-\frac1y=\frac{xy-1}{y}\) 和 \(y-\frac1x=\frac{xy-1}{x}\)。已知条件保证所需各量不为零，所以它们的商为 \[ \frac{\frac{xy-1}{y}}{\frac{xy-1}{x}}=\frac{x}{y}\text{。}\]所以正确答案是 B。

English

We have \(x-\frac1y=\frac{xy-1}{y}\) and \(y-\frac1x=\frac{xy-1}{x}.\) The hypotheses make the needed quantities nonzero, so their quotient is \[ \frac{\frac{xy-1}{y}}{\frac{xy-1}{x}}=\frac{x}{y}. \] Therefore, the correct answer is B.

3 答案 C

\(n\) 的每个素因数都至少为 \(11\)。满足这一性质的最小合数是 \(11\cdot11=121\)，它位于 \(120\lt n\le130\) 内。

所以正确答案是 C。

English

Every prime factor of \(n\) is at least \(11.\) The smallest composite with that property is \(11\cdot11=121,\) which lies in \(120\lt n\le130.\)

Therefore, the correct answer is C.

4 答案 B

经过圆心并垂直于平行弦 \(BC\) 和 \(EF\) 的直线平分这两条弦。从长方形左边起测量，\(BC\) 的中点位置为 \(4+\frac52\)，而 \(EF\) 的中点位置为 \(3+\frac{EF}{2}\)。因此 \[ 4+\frac52=3+\frac{EF}{2}\text{，}\]得到 \(EF=7\)。

所以正确答案是 B。

English

The line through the circle’s center perpendicular to the parallel chords \(BC\) and \(EF\) bisects both. Measured from the rectangle’s left side, the midpoint of \(BC\) is \(4+\frac52,\) while the midpoint of \(EF\) is \(3+\frac{EF}{2}.\) Thus \[ 4+\frac52=3+\frac{EF}{2}, \] giving \(EF=7.\)

Therefore, the correct answer is B.

5 答案 D

对两边取正的一百分之一次方，得到 \(n^2\lt5^3=125\)。由于 \(11^2=121\lt125\)，而 \(12^2=144\gt125\)，所以最大可能整数为 \(11\)。

所以正确答案是 D。

English

Taking positive one-hundredth roots gives \(n^2\lt5^3=125.\) Since \(11^2=121\lt125\) but \(12^2=144\gt125,\) the largest possible integer is \(11.\)

Therefore, the correct answer is D.

6 答案 B

因为 \(b=3g\) 且 \(g=9t\)，所以 \(g=\frac{b}{3}\)，且 \(t=\frac{b}{27}\)。因此 \[ b+g+t=b+\frac b3+\frac b{27}=\frac{37}{27}b\text{。}\]所以正确答案是 B。

English

Because \(b=3g\) and \(g=9t,\) we have \(g=\frac{b}{3}\) and \(t=\frac{b}{27}.\) Hence \[ b+g+t=b+\frac b3+\frac b{27}=\frac{37}{27}b. \] Therefore, the correct answer is B.

7 答案 C

路线长为 \(16\cdot90\cdot75\) 厘米。杰克每分钟走 \(100\cdot60\) 厘米，所以所需时间为 \[ \frac{16\cdot90\cdot75}{100\cdot60}=18 \] 分钟。所以正确答案是 C。

English

The route is \(16\cdot90\cdot75\) centimeters long. Jack covers \(100\cdot60\) centimeters per minute, so his time is \[ \frac{16\cdot90\cdot75}{100\cdot60}=18 \] minutes. Therefore, the correct answer is C.

8 答案 D

从 \(A\) 和 \(B\) 向 \(DC\) 作垂线。由于 \(BC=3\sqrt2\)，且 \(C\) 处的角为 \(45^\circ\)，高和右侧的水平偏移量都为 \(3\)。左侧的 \(30^\circ\)-\(60^\circ\)-\(90^\circ\) 三角形高为 \(3\)，所以其水平偏移量为 \(\sqrt3\)。因此 \[ DC=\sqrt3+AB+3=8+\sqrt3\text{。}\]正确答案是 D。

English

Drop perpendiculars from \(A\) and \(B\) to \(DC.\) Since \(BC=3\sqrt2\) and the angle at \(C\) is \(45^\circ,\) both the height and the right horizontal offset are \(3.\) The left \(30^\circ\)-\(60^\circ\)-\(90^\circ\) triangle has height \(3,\) so its horizontal offset is \(\sqrt3.\) Therefore \[ DC=\sqrt3+AB+3=8+\sqrt3. \] The correct answer is D.

9 答案 D

有 \[ \begin{aligned} 4^{16}5^{25} &=2^{32}5^{25}\\ &=2^7\cdot10^{25}\\ &=128\cdot10^{25} \end{aligned}\text{。}\]这是 \(128\) 后面跟着 \(25\) 个零，所以共有 \(3+25=28\) 位。正确答案是 D。

English

We have \[ \begin{aligned} 4^{16}5^{25} &=2^{32}5^{25}\\ &=2^7\cdot10^{25}\\ &=128\cdot10^{25}. \end{aligned} \] This is \(128\) followed by \(25\) zeros, so it has \(3+25=28\) digits. Therefore, the correct answer is D.

10 答案 B

点 \(1+2i\) 和 \(-1-2i\) 是相对的顶点，所以它们的中点 \(0\) 是正方形的中心。因此第四个顶点是 \(-2+i\) 关于原点的对称点，即 \(2-i\)。

所以正确答案是 B。

English

The points \(1+2i\) and \(-1-2i\) are opposite vertices, so their midpoint \(0\) is the center of the square. The fourth vertex is therefore the reflection of \(-2+i\) through the origin, namely \(2-i.\)

Therefore, the correct answer is B.

11 答案 A

将显示的数写成 \(x^e\)。平方使 \(e\) 变为 \(2e\)，随后取倒数使它变为 \(-2e\)。从 \(e=1\) 开始，经过 \(n\) 组这样的操作后，指数为 \((-2)^n\)。因此 \(y=x^{(-2)^n}\)。

所以正确答案是 A。

English

Write the displayed value as \(x^e.\) Squaring changes \(e\) to \(2e,\) and then reciprocating changes it to \(-2e.\) Starting from \(e=1,\) after \(n\) such pairs the exponent is \((-2)^n.\) Thus \(y=x^{(-2)^n}.\)

Therefore, the correct answer is A.

12 答案 B

将递推式逐项相消，得到 \[ \begin{aligned} a_{100} &=a_1+2\sum_{n=1}^{99}n\\ &=2+2\cdot\frac{99\cdot100}{2}\\ &=9902 \end{aligned}\text{。}\]所以正确答案是 B。

English

Telescoping the recurrence gives \[ \begin{aligned} a_{100} &=a_1+2\sum_{n=1}^{99}n\\ &=2+2\cdot\frac{99\cdot100}{2}\\ &=9902. \end{aligned} \] Therefore, the correct answer is B.

13 答案 A

注意 \[ \begin{aligned} &(\sqrt2+\sqrt3+\sqrt5)\\ &\quad{}\cdot(\sqrt2+\sqrt3-\sqrt5)\\ &=(\sqrt2+\sqrt3)^2-5\\ &=2\sqrt6 \end{aligned}\text{。}\]因此，该商为 \(\sqrt2+\sqrt3-\sqrt5\)。

所以正确答案是 A。

English

Observe that \[ \begin{aligned} &(\sqrt2+\sqrt3+\sqrt5)\\ &\quad{}\cdot(\sqrt2+\sqrt3-\sqrt5)\\ &=(\sqrt2+\sqrt3)^2-5\\ &=2\sqrt6. \end{aligned} \] Hence the quotient is \(\sqrt2+\sqrt3-\sqrt5.\)

Therefore, the correct answer is A.

14 答案 A

对数要求 \(x\gt0\)。等式两边取以 \(10\) 为底的对数，得到 \[ (\log_{10}x)^2=1\text{。}\]因此 \(\log_{10}x=\pm1\)，所以两个根为 \(10\) 和 \(10^{-1}\)，它们的积为 \(1\)。

所以正确答案是 A。

English

The logarithm requires \(x\gt0.\) Taking base-\(10\) logarithms gives \[ (\log_{10}x)^2=1. \] Thus \(\log_{10}x=\pm1,\) so the two roots are \(10\) and \(10^{-1},\) whose product is \(1.\)

Therefore, the correct answer is A.

15 答案 A

原方程等价于 \[ \begin{aligned} &\cos2x\cos3x\\ &\quad-\sin2x\sin3x\\ &=\cos5x=0 \end{aligned}\text{。}\]因此 \(5x=90^\circ+180^\circ k\)。取 \(k=0\)，得到题目所列的值 \(x=18^\circ\)。

所以正确答案是 A。

English

The equation is equivalent to \[ \begin{aligned} &\cos2x\cos3x\\ &\quad-\sin2x\sin3x\\ &=\cos5x=0. \end{aligned} \] Therefore \(5x=90^\circ+180^\circ k.\) Taking \(k=0\) gives \(x=18^\circ,\) the listed value.

Therefore, the correct answer is A.

16 答案 E

该关系说明，每个根 \(2+r\) 都与根 \(2-r\) 配对。因此，四个互不相同的根组成两对，每对的和为 \(4\)。四个根之和为 \(2\cdot4=8\)。

所以正确答案是 E。

English

The relation shows that every root \(2+r\) is paired with \(2-r.\) Four distinct roots therefore form two such pairs, and each pair has sum \(4.\) The sum of all four roots is \(2\cdot4=8.\)

Therefore, the correct answer is E.

17 答案 C

令 \(AH=h\)。由直角三角形中的相似关系可得 \[ \begin{aligned} AC^2&=AH\cdot AB,\\ 225&=h(h+16) \end{aligned}\text{，}\]所以 \(h=9\)。因此 \(AB=25\)。又有 \(CH=\sqrt{AH\cdot HB}\)，所以 \(CH=\sqrt{9\cdot16}=12\)。面积为 \(\frac12\cdot25\cdot12=150\)。

所以正确答案是 C。

English

Let \(AH=h.\) Similarity in the right triangle gives \[ \begin{aligned} AC^2&=AH\cdot AB,\\ 225&=h(h+16), \end{aligned} \] so \(h=9.\) Hence \(AB=25.\) Also \(CH=\sqrt{AH\cdot HB},\) so \(CH=\sqrt{9\cdot16}=12.\) The area is \(\frac12\cdot25\cdot12=150.\)

Therefore, the correct answer is C.

18 答案 E

到两坐标轴距离相等给出 \(y=x\) 或 \(y=-x\)。在 \(y=-x\) 上，到直线 \(x+y=2\) 的距离为 \(\sqrt2\)，所以 \((\sqrt2,-\sqrt2)\) 和 \((-\sqrt2,\sqrt2)\) 都满足三个距离条件。它们的 \(x\) 坐标不同，所以 \(x\) 不能唯一确定。

所以正确答案是 E。

English

Equal distance from the axes gives \(y=x\) or \(y=-x.\) On \(y=-x,\) the distance to the line \(x+y=2\) is \(\sqrt2,\) so both \((\sqrt2,-\sqrt2)\) and \((-\sqrt2,\sqrt2)\) satisfy all three distance conditions. Their \(x\)-coordinates differ, so \(x\) is not uniquely determined.

Therefore, the correct answer is E.

19 答案 D

六个所选球的数字之和为奇数，要求其中编号为奇数的球有奇数个。有 \(6\) 个奇数球和 \(5\) 个偶数球，所以有利选法数为 \[ \begin{aligned} &\binom61\binom55+\binom63\binom53\\ &\quad+\binom65\binom51\\ &=6+200+30=236 \end{aligned}\text{。}\]全部 \(\binom{11}{6}=462\) 种选法中，所求概率为 \(\frac{236}{462}=\frac{118}{231}\)。

所以正确答案是 D。

English

An odd sum requires an odd number of the six selected balls to be odd. There are \(6\) odd and \(5\) even balls, so the favorable count is \[ \begin{aligned} &\binom61\binom55+\binom63\binom53\\ &\quad+\binom65\binom51\\ &=6+200+30=236. \end{aligned} \] Of the \(\binom{11}{6}=462\) selections, the desired probability is \(\frac{236}{462}=\frac{118}{231}.\)

Therefore, the correct answer is D.

20 答案 C

若 \(x-|2x+1|=3\)，则 \(|2x+1|=x-3\)，这要求 \(x\ge3\)，但求解后没有符合条件的解。若 \(x-|2x+1|=-3\)，则 \(|2x+1|=x+3\)。分别解两个线性方程，得到 \(x=2\) 和 \(x=-\frac{4}{3}\)，二者都符合条件。因此，共有 \(2\) 个不同的解。

所以正确答案是 C。

English

If \(x-|2x+1|=3,\) then \(|2x+1|=x-3,\) which requires \(x\ge3\) and yields no solution. If \(x-|2x+1|=-3,\) then \(|2x+1|=x+3.\) Its two linear cases give \(x=2\) and \(x=-\frac{4}{3},\) both valid. Thus there are \(2\) distinct solutions.

Therefore, the correct answer is C.

21 答案 C

由于 \(c(a+b)=23\)，且所有变量都是正整数，所以 \(c=1\)，且 \(a+b=23\)。将 \(b=23-a\) 代入 \(ab+b=44\)。于是 \[ (23-a)(a+1)=44\text{，}\]即 \(a^2-22a+21=0\)，所以 \(a=1\) 或 \(21\)。每种情形都会得到正整数 \(b\)，因此恰有两个三元组。

所以正确答案是 C。

English

Since \(c(a+b)=23\) and all variables are positive integers, \(c=1\) and \(a+b=23.\) Put \(b=23-a\) in \(ab+b=44.\) Then \[ (23-a)(a+1)=44, \] or \(a^2-22a+21=0,\) so \(a=1\) or \(21.\) Each gives a positive \(b,\) producing exactly two triples.

Therefore, the correct answer is C.

22 答案 B

顶点坐标为 \[ x_t=-\frac{t}{2a},\qquad y_t=c-\frac{t^2}{4a}\text{。}\]由于 \(t=-2ax_t\)，消去 \(t\) 得到 \(y_t=c-ax_t^2\)。当 \(t\) 遍历所有实数时，\(x_t\) 也遍历所有实数，因此描出了整条抛物线。

所以正确答案是 B。

English

The vertex coordinates are \[ x_t=-\frac{t}{2a},\qquad y_t=c-\frac{t^2}{4a}. \] Since \(t=-2ax_t,\) eliminating \(t\) gives \(y_t=c-ax_t^2.\) As \(t\) ranges over all reals, so does \(x_t,\) so the entire parabola is traced.

Therefore, the correct answer is B.

23 答案 D

设 \(R\) 表示所给比值。由和差化积公式，\[ \begin{aligned} R&=\frac{\sin10^\circ+\sin20^\circ} {\cos10^\circ+\cos20^\circ},\\ R&=\frac{2\sin15^\circ\cos5^\circ} {2\cos15^\circ\cos5^\circ}\\ &=\tan15^\circ \end{aligned}\text{。}\]所以正确答案是 D。

English

Let \(R\) denote the given ratio. By the sum-to-product identities, \[ \begin{aligned} R&=\frac{\sin10^\circ+\sin20^\circ} {\cos10^\circ+\cos20^\circ},\\ R&=\frac{2\sin15^\circ\cos5^\circ} {2\cos15^\circ\cos5^\circ}\\ &=\tan15^\circ. \end{aligned} \] Therefore, the correct answer is D.

24 答案 E

两个判别式给出 \(a^2\ge8b\) 和 \(b^2\ge a\)。因此 \[ a^4\ge64b^2\ge64a\text{。}\]由于 \(a\gt0\)，可得 \(a\ge4\)；进而由 \(b^2\ge a\ge4\) 得 \(b\ge2\)。所以 \(a+b\ge6\)。当 \(a=4,b=2\) 时取等号，此时两个判别式都为零。

所以正确答案是 E。

English

The two discriminants give \(a^2\ge8b\) and \(b^2\ge a.\) Therefore \[ a^4\ge64b^2\ge64a. \] Since \(a\gt0,\) this yields \(a\ge4,\) and then \(b^2\ge a\ge4\) gives \(b\ge2.\) Thus \(a+b\ge6.\) Equality occurs at \(a=4,b=2,\) for which both discriminants are zero.

Therefore, the correct answer is E.

25 答案 D

条件给出 \(2(xy+xz+yz)=22\) 和 \(4(x+y+z)=24\)，所以 \(xy+xz+yz=11\)，且 \(x+y+z=6\)。因此，一条体对角线长度的平方为 \[ x^2+y^2+z^2=6^2-2(11)=14\text{。}\]其长度为 \(\sqrt{14}\)。

所以正确答案是 D。

English

The conditions give \(2(xy+xz+yz)=22\) and \(4(x+y+z)=24,\) so \(xy+xz+yz=11\) and \(x+y+z=6.\) Hence the square of an internal diagonal is \[ x^2+y^2+z^2=6^2-2(11)=14. \] Its length is \(\sqrt{14}.\)

Therefore, the correct answer is D.

26 答案 B

连接 \(MC\)。由于 \(EC\parallel MD\)，点 \(E\) 和 \(C\) 到 \(MD\) 的垂直距离相等，所以 \([DME]=[DMC]\)。因此 \[ \begin{aligned} [BED]&=[BMD]+[DME]\\ &=[BMD]+[DMC]\\ &=[BMC] \end{aligned}\text{。}\]因为 \(M\) 是 \(AB\) 的中点，所以三角形 \(BMC\) 的面积是 \(ABC\) 面积的一半。因此 \([BED]=12\)。

所以正确答案是 B。

English

Draw \(MC.\) Since \(EC\parallel MD,\) points \(E\) and \(C\) have equal perpendicular distances from \(MD,\) so \([DME]=[DMC].\) Therefore \[ \begin{aligned} [BED]&=[BMD]+[DME]\\ &=[BMD]+[DMC]\\ &=[BMC]. \end{aligned} \] Because \(M\) is the midpoint of \(AB,\) triangle \(BMC\) has half the area of \(ABC.\) Thus \([BED]=12.\)

Therefore, the correct answer is B.

27 答案 C

令 \(AC=s\)。由直角三角形的射影定理可得 \[ AC^2=FC\cdot BC\text{，}\]所以 \(BC=s^2\)。由于 \(BD=DC=1\)，余弦定理中的分子和分母分别满足 \[ \begin{aligned} BC^2+DC^2-BD^2&=s^4,\\ 2(BC)(DC)&=2s^2 \end{aligned}\text{。}\]因此 \(\cos\angle BCD=\frac{s^2}{2}\)。但 \(D\) 在 \(AC\) 上，并且在直角三角形 \(ABC\) 中，\(\cos\angle BCA=\frac{AC}{BC}=\frac{1}{s}\)。所以 \(\frac{s^2}{2}=\frac{1}{s}\)，从而 \(s^3=2\)，且 \(AC=\sqrt[3]{2}\)。

所以正确答案是 C。

English

Let \(AC=s.\) The right-triangle projection relation gives \[ AC^2=FC\cdot BC, \] so \(BC=s^2.\) Since \(BD=DC=1,\) the numerator and denominator in the Law of Cosines satisfy \[ \begin{aligned} BC^2+DC^2-BD^2&=s^4,\\ 2(BC)(DC)&=2s^2. \end{aligned} \] Hence \(\cos\angle BCD=\frac{s^2}{2}.\) But \(D\) lies on \(AC,\) and in right triangle \(ABC,\) \(\cos\angle BCA=\frac{AC}{BC}=\frac{1}{s}.\) Thus \(\frac{s^2}{2}=\frac{1}{s},\) so \(s^3=2\) and \(AC=\sqrt[3]{2}.\)

Therefore, the correct answer is C.

28 答案 D

因为 \(\sqrt{x}+\sqrt{y}=8\sqrt{31}\)，两个根式必须具有相同的无平方因子部分 \(31\)。写成 \(x=31u^2\) 和 \(y=31v^2\)，其中 \(u,v\) 为正整数。于是 \[ u+v=8,\qquad u\lt v\text{。}\]可能的情况为 \((u,v)=(1,7)\)、\((2,6)\)、\((3,5)\)，得到三个不同的整数对 \((x,y)\)。

所以正确答案是 D。

English

Because \(\sqrt{x}+\sqrt{y}=8\sqrt{31},\) the two radicals must have common squarefree part \(31.\) Write \(x=31u^2\) and \(y=31v^2\) for positive integers \(u,v.\) Then \[ u+v=8,\qquad u\lt v. \] The possibilities are \((u,v)=(1,7),\) \((2,6),\) and \((3,5),\) producing three distinct pairs \((x,y).\)

Therefore, the correct answer is D.

29 答案 A

该圆全部位于 \(x\gt0\) 的区域内，所以 \(\frac{y}{x}\) 是斜率 \(m\)，相应直线 \(y=mx\) 经过圆上一点。取极值时，这条直线与圆相切。圆心 \((3,3)\) 到直线 \(mx-y=0\) 的距离必须等于 \(\sqrt6\)，因此 \[ \frac{|3m-3|}{\sqrt{m^2+1}}=\sqrt6\text{。}\]两边平方并化简，得到 \(m^2-6m+1=0\)，所以 \(m=3\pm2\sqrt2\)。较大的值为 \(3+2\sqrt2\)。

所以正确答案是 A。

English

The circle lies entirely where \(x\gt0,\) so \(\frac{y}{x}\) is the slope \(m\) of the line \(y=mx\) through a point on it. At an extreme, this line is tangent. The distance from the center \((3,3)\) to \(mx-y=0\) must equal \(\sqrt6,\) so \[ \frac{|3m-3|}{\sqrt{m^2+1}}=\sqrt6. \] Squaring and simplifying gives \(m^2-6m+1=0,\) hence \(m=3\pm2\sqrt2.\) The larger value is \(3+2\sqrt2.\)

Therefore, the correct answer is A.

30 答案 B

令 \(S=\sum_{k=1}^9kw^k\)。由于 \(w^9=1\)，且 \(w^{10}=w\)，\[ \begin{aligned} (1-w)S &=(w+w^2+\cdots+w^9)\\ &\quad-9w^{10}\\ &=-9w \end{aligned}\text{。}\]所以 \(|S|=\frac{9}{|1-w|}\)，这是因为 \(|w|=1\)。因此 \[ |S|^{-1}=\frac{|1-w|}{9} =\frac{2}{9}\sin20^\circ\text{。}\]所以正确答案是 B。

English

Let \(S=\sum_{k=1}^9kw^k.\) Since \(w^9=1\) and \(w^{10}=w,\) \[ \begin{aligned} (1-w)S &=(w+w^2+\cdots+w^9)\\ &\quad-9w^{10}\\ &=-9w. \end{aligned} \] Therefore \(|S|=\frac{9}{|1-w|},\) because \(|w|=1.\) Thus \[ |S|^{-1}=\frac{|1-w|}{9} =\frac{2}{9}\sin20^\circ. \] Therefore, the correct answer is B.