1974 AMC 12 真题（P版）

1 答案 D

两边乘以 \(2xy\)，得 \(4y+6x=xy\)，所以 \(4y=x(y-6)\)。由于 \(y\ne6\)，\[ x=\frac{4y}{y-6}\text{。}\]

因此，正确答案是 D。

English

Multiplying by \(2xy\) gives \(4y+6x=xy,\) so \(4y=x(y-6).\) Since \(y\ne6,\) \[ x=\frac{4y}{y-6}. \]

Therefore, the correct answer is D.

2 答案 B

两个不同的数 \(x_1,x_2\) 是方程 \[ 3x^2-hx-b=0 \] 的两个根。由韦达定理，它们的和为 \(-\frac{(-h)}{3}=\frac{h}{3}\)。

因此，正确答案是 B。

English

The distinct numbers \(x_1,x_2\) are the two roots of \[ 3x^2-hx-b=0. \] By Vieta’s formulas, their sum is \(-\frac{(-h)}{3}=\frac{h}{3}.\)

Therefore, the correct answer is B.

3 答案 A

四个所选项的总次数为 \(7\) 的唯一方式，是选一次 \(2x\) 和三次 \(-x^2\)。贡献一次项的因子有 \(4\) 种选法，所以系数为 \[ 4(2)(-1)^3=-8\text{。}\]

因此，正确答案是 A。

English

The only way four selected terms can have total degree \(7\) is to choose \(2x\) once and \(-x^2\) three times. There are \(4\) choices for the first factor, so the coefficient is \[ 4(2)(-1)^3=-8. \]

Therefore, the correct answer is A.

4 答案 D

由多项式余式定理，余数为 \[ (-1)^{51}+51=-1+51=50\text{。}\]

因此，正确答案是 D。

English

By the polynomial remainder theorem, the remainder is \[ (-1)^{51}+51=-1+51=50. \]

Therefore, the correct answer is D.

5 答案 B

因为 \(ABCD\) 是圆内接四边形，所以 \(\angle ABC+\angle ADC=180^\circ\)。又因为 \(BE\) 是 \(BA\) 的延长线，所以也有 \(\angle EBC+\angle ABC=180^\circ\)。因此 \[ \angle EBC=\angle ADC=68^\circ\text{。}\]已知的 \(92^\circ\) 角无需使用。

因此，正确答案是 B。

English

Because \(ABCD\) is cyclic, \(\angle ABC+\angle ADC=180^\circ.\) Since \(BE\) extends \(BA,\) \(\angle EBC+\angle ABC=180^\circ\) as well. Hence \[ \angle EBC=\angle ADC=68^\circ. \] The given \(92^\circ\) angle is unnecessary.

Therefore, the correct answer is B.

6 答案 D

公式关于 \(x\)、\(y\) 对称，因此该运算满足交换律。此外，\[ \frac1{x*y}=\frac1x+\frac1y\text{。}\]所以 \((x*y)*z\) 和 \(x*(y*z)\) 的倒数都是 \(\frac1x+\frac1y+\frac1z\)，故两者相等。该运算也满足结合律。

因此，正确答案是 D。

English

Symmetry in \(x,\) \(y\) makes the operation commutative. Also, \[ \frac1{x*y}=\frac1x+\frac1y. \] Thus both \((x*y)*z\) and \(x*(y*z)\) have reciprocal \(\frac1x+\frac1y+\frac1z,\) so they are equal. The operation is associative too.

Therefore, the correct answer is D.

7 答案 D

设原有人口为 \(P\)。则 \[ 0.89(P+1200)=P-32\text{。}\]因此 \(0.11P=1100\)，得到 \(P=10{,}000\)。

因此，正确答案是 D。

English

Let the original population be \(P.\) Then \[ 0.89(P+1200)=P-32. \] Therefore \(0.11P=1100,\) giving \(P=10{,}000.\)

Therefore, the correct answer is D.

8 答案 A

\(3^{11}\) 和 \(5^{13}\) 都是奇数，所以它们的和是偶数。因此，这个和能被最小的质数 \(2\) 整除。

因此，正确答案是 A。

English

Both \(3^{11}\) and \(5^{13}\) are odd, so their sum is even. Hence it is divisible by \(2,\) the smallest prime.

Therefore, the correct answer is A.

9 答案 B

每两行构成一个包含八个整数的重复区块。在每个区块中，其中的 \(8\) 的倍数都位于第二列，正如 \(8\) 和 \(16\) 一样。由于 \(1000=125\cdot8\)，它也位于第二列。

因此，正确答案是 B。

English

Each pair of rows is a repeating block of eight integers. In every such block, its multiple of \(8\) lies in the second column, as do \(8\) and \(16.\) Since \(1000=125\cdot8,\) it also lies in the second column.

Therefore, the correct answer is B.

10 答案 B

方程可写成 \((2k-1)x^2-8x+6=0\)。其判别式为 \[ 64-24(2k-1)=88-48k\text{。}\]当且仅当 \(k\gt\frac{11}{6}\) 时，判别式为负，所以满足条件的最小整数是 \(2\)。

因此，正确答案是 B。

English

The equation is \((2k-1)x^2-8x+6=0.\) Its discriminant is \[ 64-24(2k-1)=88-48k. \] This is negative exactly when \(k\gt\frac{11}{6},\) so the least integer possible is \(2.\)

Therefore, the correct answer is B.

11 答案 A

因为两点都在该直线上，所以 \(b-d=m(a-c)\)。因此，它们之间的距离为 \[ \begin{aligned} \ell&=\sqrt{(a-c)^2+(b-d)^2}\\ &=\sqrt{(a-c)^2(1+m^2)}\\ &=|a-c|\sqrt{1+m^2} \end{aligned}\text{。}\]

因此，正确答案是 A。

English

Since both points are on the line, \(b-d=m(a-c).\) Their distance is therefore \[ \begin{aligned} \ell&=\sqrt{(a-c)^2+(b-d)^2}\\ &=\sqrt{(a-c)^2(1+m^2)}\\ &=|a-c|\sqrt{1+m^2}. \end{aligned} \]

Therefore, the correct answer is A.

12 答案 B

若 \(g(x)=\frac{1}{2}\)，则 \(x^2=\frac{1}{2}\)，对应的两个 \(x\) 都非零。因此 \[ f(\frac{1}{2})=\frac{1-x^2}{x^2} =\frac{\frac{1}{2}}{\frac{1}{2}}=1\text{。}\]

因此，正确答案是 B。

English

If \(g(x)=\frac{1}{2},\) then \(x^2=\frac{1}{2},\) and either corresponding \(x\) is nonzero. Hence \[ f(\frac{1}{2})=\frac{1-x^2}{x^2} =\frac{\frac{1}{2}}{\frac{1}{2}}=1. \]

Therefore, the correct answer is B.

13 答案 D

“\(P\) 为真蕴含 \(Q\) 为假”的逆否命题是“\(Q\) 不为假蕴含 \(P\) 不为真”。这正是“若 \(Q\) 为真，则 \(P\) 为假”。

因此，正确答案是 D。

English

The contrapositive of “\(P\) true implies \(Q\) false” is “\(Q\) not false implies \(P\) not true.” This is exactly “If \(Q\) is true then \(P\) is false.”

Therefore, the correct answer is D.

14 答案 A

若 \(x\lt0\)，则 \(x^2\gt0\gt x\)，所以选项 A 恒成立。分别取 \(x=-1\)、\(x=-1\)、\(x=2\) 和 \(x=-1\)，即可否定 B、C、D 和 E。

因此，正确答案是 A。

English

If \(x\lt0,\) then \(x^2\gt0\gt x,\) so choice A always holds. The values \(x=-1,\) \(x=-1,\) \(x=2,\) and \(x=-1\) respectively disprove B, C, D, and E.

Therefore, the correct answer is A.

15 答案 B

由于 \(x\lt-2\)，所以 \(1+x\lt0\)，从而 \(|1+x|=-1-x\)。于是 \(\bigl|1-|1+x|\bigr|\) 化为 \(|1-(-1-x)|\)，即 \(|2+x|=-2-x\)，因为 \(2+x\lt0\)。

因此，正确答案是 B。

English

Since \(x\lt-2,\) \(1+x\lt0,\) so \(|1+x|=-1-x.\) Thus \(\bigl|1-|1+x|\bigr|\) becomes \(|1-(-1-x)|,\) which is \(|2+x|=-2-x\) because \(2+x\lt0.\)

Therefore, the correct answer is B.

16 答案 A

两条直角边长均为 \(a\)，斜边长为 \(a\sqrt2\)，所以 \[ \begin{aligned} R&=\frac{a\sqrt2}{2},\\ r&=\frac{a+a-a\sqrt2}{2} \end{aligned}\text{。}\]因此 \[ \begin{aligned} \frac Rr&=\frac{\sqrt2}{2-\sqrt2}\\ &=\frac1{\sqrt2-1}=1+\sqrt2 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

With both legs of length \(a\) and hypotenuse \(a\sqrt2,\) \[ \begin{aligned} R&=\frac{a\sqrt2}{2},\\ r&=\frac{a+a-a\sqrt2}{2}. \end{aligned} \] Therefore \[ \begin{aligned} \frac Rr&=\frac{\sqrt2}{2-\sqrt2}\\ &=\frac1{\sqrt2-1}=1+\sqrt2. \end{aligned} \]

Therefore, the correct answer is A.

17 答案 C

由于 \((1+i)^2=2i\)，所以 \((1+i)^4=-4\)。同理，\((1-i)^2=-2i\)，且 \((1-i)^4=-4\)。因此，两个二十次方都等于 \((-4)^5=-1024\)，它们的差为 \(0\)。

因此，正确答案是 C。

English

Since \((1+i)^2=2i,\) we have \((1+i)^4=-4.\) Likewise \((1-i)^2=-2i\) and \((1-i)^4=-4.\) Hence each twentieth power equals \((-4)^5=-1024,\) and their difference is \(0.\)

Therefore, the correct answer is C.

18 答案 D

由 \(p=\log_8 3\) 可得 \(\log_2 3=3p\)。因此 \[ \log_2 5=(\log_2 3)(\log_3 5)=3pq\text{。}\]又因为 \(\log_2 10=1+\log_2 5=1+3pq\)，所以 \[ \log_{10}5=\frac{\log_2 5}{\log_2 10} =\frac{3pq}{1+3pq}\text{。}\]

因此，正确答案是 D。

English

From \(p=\log_8 3,\) \(\log_2 3=3p.\) Therefore \[ \log_2 5=(\log_2 3)(\log_3 5)=3pq. \] Since \(\log_2 10=1+\log_2 5=1+3pq,\) \[ \log_{10}5=\frac{\log_2 5}{\log_2 10} =\frac{3pq}{1+3pq}. \]

Therefore, the correct answer is D.

19 答案 A

正方形边长为 \(1\)。由于 \(CM=CN\)，比较它们长度的平方可得 \(DM=NB=x\)。此外，\[ \begin{aligned} CM^2&=1+x^2,\\ MN^2&=2(1-x)^2 \end{aligned}\text{。}\]因为 \(CMN\) 是等边三角形，令两式相等可得 \(x^2-4x+1=0\)。位于 \((0,1)\) 内的根是 \(x=2-\sqrt3\)。

三角形边长的平方为 \(1+x^2\)，所以其面积为 \[ \begin{aligned} A_\triangle&=\frac{\sqrt3}{4}(1+x^2)\\ &=\frac{\sqrt3}{4} \bigl(1+(2-\sqrt3)^2\bigr)\\ &=2\sqrt3-3 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

The square has side \(1.\) Since \(CM=CN,\) symmetry of their squared lengths gives \(DM=NB=x.\) Also, \[ \begin{aligned} CM^2&=1+x^2,\\ MN^2&=2(1-x)^2. \end{aligned} \] Equating these because \(CMN\) is equilateral gives \(x^2-4x+1=0.\) The root in \((0,1)\) is \(x=2-\sqrt3.\)

The triangle’s side has square \(1+x^2,\) so its area is \[ \begin{aligned} A_\triangle&=\frac{\sqrt3}{4}(1+x^2)\\ &=\frac{\sqrt3}{4} \bigl(1+(2-\sqrt3)^2\bigr)\\ &=2\sqrt3-3. \end{aligned} \]

Therefore, the correct answer is A.

20 答案 D

将五项的分母有理化，得 \[ \begin{aligned} T={}&(3+\sqrt8)-(\sqrt8+\sqrt7)\\ &+(\sqrt7+\sqrt6)-(\sqrt6+\sqrt5)\\ &+(\sqrt5+2)=5 \end{aligned}\text{。}\]因此 \(T\gt2\)。

因此，正确答案是 D。

English

Rationalizing the five terms gives \[ \begin{aligned} T={}&(3+\sqrt8)-(\sqrt8+\sqrt7)\\ &+(\sqrt7+\sqrt6)-(\sqrt6+\sqrt5)\\ &+(\sqrt5+2)=5. \end{aligned} \] Thus \(T\gt2.\)

Therefore, the correct answer is D.

21 答案 B

设首项为 \(a\)，公比为 \(r\)。由条件可得 \[ \begin{aligned} ar^3(r-1)&=576,\\ a(r-1)&=9 \end{aligned}\text{。}\]两式相除得 \(r^3=64\)，所以 \(r=4\)。于是 \(3a=9\)，从而 \(a=3\)。所求和为 \[ \begin{aligned} S&=3(1+4+16+64+256)\\ &=3\cdot341\\ &=1023 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Let the first term be \(a\) and the ratio be \(r.\) The conditions give \[ \begin{aligned} ar^3(r-1)&=576,\\ a(r-1)&=9. \end{aligned} \] Their quotient gives \(r^3=64,\) so \(r=4.\) Then \(3a=9,\) hence \(a=3.\) The sum is \[ \begin{aligned} S&=3(1+4+16+64+256)\\ &=3\cdot341\\ &=1023. \end{aligned} \]

Therefore, the correct answer is B.

22 答案 E

令 \(u=\frac{A}{2}\)。则 \(\sin u-\sqrt3\cos u\) 等于 \(2\sin(u-60^\circ)\)。当 \(u-60^\circ=270^\circ+360^\circ n\) 时取得最小值，所以 \[ A=660^\circ+720^\circ n \] 其中 \(n\) 为整数。前四个选项都不具有这种形式。

因此，正确答案是 E。

English

Let \(u=\frac{A}{2}.\) Then \(\sin u-\sqrt3\cos u\) equals \(2\sin(u-60^\circ).\) Its minimum occurs when \(u-60^\circ=270^\circ+360^\circ n,\) so \[ A=660^\circ+720^\circ n \] for an integer \(n.\) None of the first four choices has this form.

Therefore, the correct answer is E.

23 答案 B

从每个外点出发，到切点的半径构成全等的直角三角形，所以 \(OP\) 与 \(OQ\) 分别平分两个切线夹角。由于另外两条切线平行，这两个半角之和为 \(90^\circ\)，因此 \(\angle POQ=90^\circ\)。

因此 \(OT''=r\) 是斜边 \(PQ\) 上的高。由同一外点引出的切线段相等，得 \(PT''=PT=4\) 和 \(QT''=QT'=9\)。由斜边上高的定理，\[ r^2=(PT'')(QT'')=4\cdot9=36\text{，}\]所以 \(r=6\)。

因此，正确答案是 B。

English

The radii to the tangency points make congruent right triangles from each external point, so \(OP\) and \(OQ\) bisect the two tangent angles. Because the other two tangents are parallel, these half-angles add to \(90^\circ;\) hence \(\angle POQ=90^\circ.\)

Thus \(OT''=r\) is the altitude to the hypotenuse \(PQ.\) Equal tangent segments give \(PT''=PT=4\) and \(QT''=QT'=9.\) The altitude theorem yields \[ r^2=(PT'')(QT'')=4\cdot9=36, \] so \(r=6.\)

Therefore, the correct answer is B.

24 答案 A

把掷出 \(5\) 或 \(6\) 记为一次成功，其概率为 \(\frac{1}{3}\)。所求概率为 \[ \begin{aligned} &\binom65\left(\frac13\right)^5 \left(\frac23\right) +\left(\frac13\right)^6\\ &\qquad=\frac{12}{729}+\frac1{729}\\ &\qquad=\frac{13}{729} \end{aligned}\text{。}\]

因此，正确答案是 A。

English

Let a success be a \(5\) or \(6,\) so its probability is \(\frac{1}{3}.\) The desired probability is \[ \begin{aligned} &\binom65\left(\frac13\right)^5 \left(\frac23\right) +\left(\frac13\right)^6\\ &\qquad=\frac{12}{729}+\frac1{729}\\ &\qquad=\frac{13}{729}. \end{aligned} \]

Therefore, the correct answer is A.

25 答案 C

选取坐标 \(A=(0,0)\)、\(B=(1,0)\)、\(D=(u,v)\) 和 \(C=(u+1,v)\)，于是 \(k=v\)。两个中点为 \[ \begin{aligned} M&=(\frac{u}{2},\frac{v}{2}),\\ N&=(1+\frac{u}{2},\frac{v}{2}) \end{aligned}\text{。}\]将 \(CM\) 和 \(DN\) 延长至 \(y=0\)，可得 \(Q=(-1,0)\) 和 \(P=(2,0)\)。

将 \(DP\) 与 \(CQ\) 参数化，可得它们的交点为 \(O=(\frac{3u}{4}+\frac{1}{2},\frac{3v}{4})\)。因此 \(QP=3\)，从 \(O\) 到基线的高为 \(\frac{3v}{4}\)，所以 \[ \begin{aligned} [QPO]&=\frac12(3)\left(\frac{3v}{4}\right)\\ &=\frac{9v}{8}=\frac{9k}{8} \end{aligned}\text{。}\]

因此，正确答案是 C。

English

Choose coordinates \(A=(0,0),\) \(B=(1,0),\) \(D=(u,v),\) and \(C=(u+1,v),\) so \(k=v.\) The midpoints are \[ \begin{aligned} M&=(\frac{u}{2},\frac{v}{2}),\\ N&=(1+\frac{u}{2},\frac{v}{2}). \end{aligned} \] Extending \(CM\) and \(DN\) to \(y=0\) gives \(Q=(-1,0)\) and \(P=(2,0).\)

Parameterizing \(DP\) and \(CQ\) shows their intersection is \(O=(\frac{3u}{4}+\frac{1}{2},\frac{3v}{4}).\) Thus \(QP=3\) and the height from \(O\) is \(\frac{3v}{4},\) so \[ \begin{aligned} [QPO]&=\frac12(3)\left(\frac{3v}{4}\right)\\ &=\frac{9v}{8}=\frac{9k}{8}. \end{aligned} \]

Therefore, the correct answer is C.

26 答案 C

因为 \[ 30^4=2^4\cdot3^4\cdot5^4\text{，}\]每个因数都可独立选择三个指数，每个指数取自 \(0\)、\(1\)、\(2\)、\(3\)、\(4\)。共有 \(5^3=125\) 个因数。除去 \(1\) 和 \(30^4\)，还剩 \(125-2=123\) 个。

因此，正确答案是 C。

English

Because \[ 30^4=2^4\cdot3^4\cdot5^4, \] a divisor independently chooses each of three exponents from \(0,\) \(1,\) \(2,\) \(3,\) \(4.\) There are \(5^3=125\) divisors. Excluding \(1\) and \(30^4\) leaves \(125-2=123.\)

Therefore, the correct answer is C.

27 答案 A

有 \[ |f(x)+4|=|3x+6|=3|x+2|\text{。}\]若 \(|x+2|\lt b\)，且 \(b\le \frac{a}{3}\)，则 \(3|x+2|\lt3b\le a\)，符合要求。若 \(b\gt \frac{a}{3}\)，则可选取 \(|x+2|\)，使它严格介于 \(\frac{a}{3}\) 与 \(b\) 之间，命题便不成立。因此，确切条件是 \(b\le \frac{a}{3}\)。

因此，正确答案是 A。

English

We have \[ |f(x)+4|=|3x+6|=3|x+2|. \] If \(|x+2|\lt b\) and \(b\le \frac{a}{3},\) then \(3|x+2|\lt3b\le a,\) as required. If \(b\gt \frac{a}{3},\) one can choose \(|x+2|\) strictly between \(\frac{a}{3}\) and \(b,\) so the statement fails. Thus the exact condition is \(b\le \frac{a}{3}.\)

Therefore, the correct answer is A.

28 答案 D

若 \(a_1=0\)，则 \[ 0\le x\le\sum_{n=2}^{25}\frac2{3^n} \lt\sum_{n=2}^{\infty}\frac2{3^n}=\frac13\text{。}\]若 \(a_1=2\)，则 \(x\ge\frac{2}{3}\)，同时 \[ x\le\sum_{n=1}^{25}\frac2{3^n} \lt\sum_{n=1}^{\infty}\frac2{3^n}=1\text{。}\]因此，每个这样的 \(x\) 都位于所述两个外侧三等分区间之一。

因此，正确答案是 D。

English

If \(a_1=0,\) then \[ 0\le x\le\sum_{n=2}^{25}\frac2{3^n} \lt\sum_{n=2}^{\infty}\frac2{3^n}=\frac13. \] If \(a_1=2,\) then \(x\ge\frac{2}{3},\) while \[ x\le\sum_{n=1}^{25}\frac2{3^n} \lt\sum_{n=1}^{\infty}\frac2{3^n}=1. \] Hence every such \(x\) lies in one of the two stated outer thirds.

Therefore, the correct answer is D.

29 答案 B

对每个 \(p\)，\[ \begin{aligned} S_p&=\frac{40}{2} \bigl(2p+39(2p-1)\bigr)\\ &=1600p-780 \end{aligned}\text{。}\]由于 \(1+2+\cdots+10=55\)，所以 \[ \begin{aligned} \sum_{p=1}^{10}S_p &=1600(55)-7800\\ &=80{,}200 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

For each \(p,\) \[ \begin{aligned} S_p&=\frac{40}{2} \bigl(2p+39(2p-1)\bigr)\\ &=1600p-780. \end{aligned} \] Since \(1+2+\cdots+10=55,\) therefore \[ \begin{aligned} \sum_{p=1}^{10}S_p &=1600(55)-7800\\ &=80{,}200. \end{aligned} \]

Therefore, the correct answer is B.

30 答案 A

将较长部分缩放为 \(1\)，则较短部分为 \(R\)。由条件得 \(R=\frac{1}{1+R}\)，所以 \[ R^2+R=1,\qquad R^{-1}=R+1\text{。}\]因此 \[ R^2+R^{-1}=R^2+R+1=2\text{。}\]从内向外化简，最外层 \(R\) 的指数变为 \[ \begin{aligned} R^{\,R^2+R^{-1}}+R^{-1} &=R^2+R^{-1}\\ &=2 \end{aligned}\text{。}\]所以整个表达式为 \(R^2+R^{-1}=2\)。

因此，正确答案是 A。

English

Scale the greater part to \(1,\) making the lesser part \(R.\) The condition gives \(R=\frac{1}{1+R},\) so \[ R^2+R=1,\qquad R^{-1}=R+1. \] Consequently \[ R^2+R^{-1}=R^2+R+1=2. \] Working outward, the exponent of the outer \(R\) becomes \[ \begin{aligned} R^{\,R^2+R^{-1}}+R^{-1} &=R^2+R^{-1}\\ &=2. \end{aligned} \] The whole expression is therefore \(R^2+R^{-1}=2.\)

Therefore, the correct answer is A.