1956 AMC 12 真题（P版）

1 答案 A

代入 \(x=2\)，并先计算幂，得 \[ \begin{aligned} x+x(x^x)&=2+2(2^2)\\ &=2+8=10 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

Substituting \(x=2,\) and evaluating the exponent first, gives \[ \begin{aligned} x+x(x^x)&=2+2(2^2)\\ &=2+8=10. \end{aligned} \]

Thus, the correct answer is A.

2 答案 D

获利 \(20\%\) 的烟斗成本为 \[ \frac{\$1.20}{1.20}=\$1.00\text{，}\]而亏损 \(20\%\) 的烟斗成本为 \[ \frac{\$1.20}{0.80}=\$1.50\text{。}\]两根烟斗的总成本为 \(\$2.50\)，但总售价为 \(\$2.40\)，所以琼斯先生亏损 \(10\) 美分。

因此，正确答案是 D。

English

The pipe sold at a \(20\%\) profit cost \[ \frac{\$1.20}{1.20}=\$1.00, \] while the pipe sold at a \(20\%\) loss cost \[ \frac{\$1.20}{0.80}=\$1.50. \] Their combined cost was \(\$2.50,\) but they sold for \(\$2.40,\) so Mr. Jones lost \(10\) cents.

Thus, the correct answer is D.

3 答案 D

给定距离为 \(587\times10^{10}\) 英里。乘以 \(100=10^2\)，得 \[ (587\times10^{10})10^2=587\times10^{12} \] 英里。

因此，正确答案是 D。

English

The given distance is \(587\times10^{10}\) miles. Multiplying by \(100=10^2\) gives \[ (587\times10^{10})10^2=587\times10^{12} \] miles.

Thus, the correct answer is D.

4 答案 E

剩余可投资金额为 \[ 10000-4000-3500=2500\text{。}\]前两笔投资获得 \(0.05(4000)+0.04(3500)=340\) 美元，所以余款必须获得 \(500-340=160\) 美元。所需利率为 \[ \frac{160}{2500}=0.064=6.4\%\text{。}\]

因此，正确答案是 E。

English

The amount left to invest is \[ 10000-4000-3500=2500. \] The first two investments earn \(0.05(4000)+0.04(3500)=340\) dollars, so the remaining investment must earn \(500-340=160\) dollars. Its required rate is \[ \frac{160}{2500}=0.064=6.4\%. \]

Thus, the correct answer is E.

5 答案 C

中间硬币与任意两枚相邻硬币的圆心两两相距两个半径，所以它们构成等边三角形。因此，相邻两枚外圈硬币的圆心在中间硬币圆心处所成的角为 \(60^\circ\)。恰有 \[ \frac{360^\circ}{60^\circ}=6 \] 枚硬币可以围在它周围。

因此，正确答案是 C。

English

The centers of the middle nickel and any two neighboring nickels are pairwise two radii apart, so they form an equilateral triangle. Therefore consecutive outer centers subtend \(60^\circ\) at the center of the middle nickel. Exactly \[ \frac{360^\circ}{60^\circ}=6 \] nickels fit around it.

Thus, the correct answer is C.

6 答案 B

按每个头计算两条腿，正好计入每只鸡的两条腿，以及每头牛四条腿中的两条。因此，每头牛恰好贡献 \(14\) 条额外腿中的两条。所以牛的数量为 \[ \frac{14}{2}=7\text{。}\]

因此，正确答案是 B。

English

Counting two legs for every head accounts for both legs of each chicken and two of each cow’s four legs. Thus every cow contributes exactly two of the \(14\) extra legs. Hence the number of cows is \[ \frac{14}{2}=7. \]

Thus, the correct answer is B.

7 答案 C

由韦达定理，两根之积为 \(\frac{c}{a}\)。互为倒数的两根乘积为 \(1\)，所以 \[ \frac ca=1\text{，}\]这要求 \(c=a\)。

因此，正确答案是 C。

English

By Vieta’s formulas, the product of the two roots is \(\frac{c}{a}.\) Reciprocal roots have product \(1,\) so \[ \frac ca=1, \] which requires \(c=a.\)

Thus, the correct answer is C.

8 答案 B

当 \(y=-8\) 时，右边为 \(5^0=1\)。因此 \[ 8\cdot2^x=2^{x+3}=1=2^0\text{，}\]所以 \(x+3=0\)，且 \(x=-3\)。

因此，正确答案是 B。

English

When \(y=-8,\) the right side is \(5^0=1.\) Therefore \[ 8\cdot2^x=2^{x+3}=1=2^0, \] so \(x+3=0\) and \(x=-3.\)

Thus, the correct answer is B.

9 答案 D

对两个因式中的每一个，嵌套根式和外部四次方将 \(a\) 的指数乘以 \[ 9\cdot\frac16\cdot\frac13\cdot4=2\text{。}\]因此每个因式都是 \(a^2\)，两者之积为 \(a^2a^2=a^4\)。

因此，正确答案是 D。

English

For each of the two factors, the nested roots and outer fourth power multiply the exponent of \(a\) by \[ 9\cdot\frac16\cdot\frac13\cdot4=2. \] Thus each factor is \(a^2,\) and their product is \(a^2a^2=a^4.\)

Therefore, the correct answer is D.

10 答案 B

由于 \(ABC\) 是等边三角形，圆心角 \(\angle ACB\) 为 \(60^\circ\)。圆周角 \(\angle ADB\) 所对的是同一条小弧 \(AB\)，所以其度数为圆心角的一半：\[ \angle ADB=\frac{60^\circ}{2}=30^\circ\text{。}\]

因此，正确答案是 B。

English

Since \(ABC\) is equilateral, the central angle \(\angle ACB\) is \(60^\circ.\) The inscribed angle \(\angle ADB\) intercepts the same minor arc \(AB,\) so its measure is half the central angle: \[ \angle ADB=\frac{60^\circ}{2}=30^\circ. \]

Thus, the correct answer is B.

11 答案 A

合并两个分式项，\[ \begin{gathered} -\frac1{1+\sqrt3}+\frac1{1-\sqrt3} \\ =\frac{(1+\sqrt3)-(1-\sqrt3)} {(1+\sqrt3)(1-\sqrt3)}\\ =\frac{2\sqrt3}{-2}\\ =-\sqrt3 \end{gathered}\text{。}\]再加上开头的 \(1\)，得 \(1-\sqrt3\)。

因此，正确答案是 A。

English

Combining the fractional terms, \[ \begin{gathered} -\frac1{1+\sqrt3}+\frac1{1-\sqrt3} \\ =\frac{(1+\sqrt3)-(1-\sqrt3)} {(1+\sqrt3)(1-\sqrt3)}\\ =\frac{2\sqrt3}{-2}\\ =-\sqrt3. \end{gathered} \] Adding the initial \(1\) gives \(1-\sqrt3.\)

Thus, the correct answer is A.

12 答案 E

在商有定义的地方，\[ \begin{aligned} \frac{x^{-1}-1}{x-1} &=\frac{\frac{1-x}{x}}{x-1}\\ &=\frac{-(x-1)}{x(x-1)}\\ &=-\frac1x \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Where the quotient is defined, \[ \begin{aligned} \frac{x^{-1}-1}{x-1} &=\frac{\frac{1-x}{x}}{x-1}\\ &=\frac{-(x-1)}{x(x-1)}\\ &=-\frac1x. \end{aligned} \]

Thus, the correct answer is E.

13 答案 C

两数之差为 \(y-x\)。以参照值 \(y\) 的比例表示，就是 \(\frac{y-x}{y}\)。乘以 \(100\) 将此分数化为百分比：\[ \frac{100(y-x)}y\text{。}\]

因此，正确答案是 C。

English

The difference is \(y-x.\) Measured as a fraction of the reference value \(y,\) this is \(\frac{y-x}{y}.\) Multiplying by \(100\) converts the fraction to a percent: \[ \frac{100(y-x)}y. \]

Thus, the correct answer is C.

14 答案 B

设 \(PB=t\)。由于 \(B\) 位于 \(P\) 与 \(C\) 之间，所以 \(PC=t+20\)。由切割线定理，\[ (10\sqrt3)^2=t(t+20)\text{。}\]因此 \(t^2+20t-300=0\)，即 \((t-10)(t+30)=0\)。长度为正，所以 \(t=10\)。

因此，正确答案是 B。

English

Let \(PB=t.\) Since \(B\) lies between \(P\) and \(C,\) we have \(PC=t+20.\) The tangent-secant theorem gives \[ (10\sqrt3)^2=t(t+20). \] Thus \(t^2+20t-300=0,\) or \((t-10)(t+30)=0.\) A length is positive, so \(t=10.\)

Therefore, the correct answer is B.

15 答案 A

当 \(x\ne\pm2\) 时，两边乘以 \(x^2-4\)，得 \[ 15-2(x+2)=x^2-4\text{。}\]因此 \(x^2+2x-15=0\)，所以 \[ (x+5)(x-3)=0\text{。}\]\(x=-5\) 和 \(x=3\) 都满足定义域限制。

因此，正确答案是 A。

English

For \(x\ne\pm2,\) multiplying by \(x^2-4\) gives \[ 15-2(x+2)=x^2-4. \] Hence \(x^2+2x-15=0,\) so \[ (x+5)(x-3)=0. \] Both \(x=-5\) and \(x=3\) satisfy the domain restriction.

Thus, the correct answer is A.

16 答案 C

第一个数与第二个数之比 \(2:3\)，以及第二个数与第三个数之比 \(5:8\)，合并后可得三个数之比为 \(10:15:24\)。\(49\) 份共等于 \(98\)，所以每份为 \(2\)。第二个数为 \(15\cdot2=30\)。

因此，正确答案是 C。

English

The first-to-second ratio \(2:3\) and second-to-third ratio \(5:8\) combine to give the three-number ratio \(10:15:24.\) The \(49\) total parts equal \(98,\) so each part is \(2.\) The second number is \(15\cdot2=30.\)

Thus, the correct answer is C.

17 答案 D

合并给出的部分分式，得 \[ \frac{A(2x-3)+B(x+2)}{(x+2)(2x-3)}\text{。}\]将其分子与 \(5x-11\) 比较，得 \[ \begin{aligned} 2A+B&=5,\\ -3A+2B&=-11 \end{aligned}\text{。}\]解得 \(A=3\)，且 \(B=-1\)。

因此，正确答案是 D。

English

Combining the proposed partial fractions gives \[ \frac{A(2x-3)+B(x+2)}{(x+2)(2x-3)}. \] Matching its numerator with \(5x-11\) yields \[ \begin{aligned} 2A+B&=5,\\ -3A+2B&=-11. \end{aligned} \] Solving gives \(A=3\) and \(B=-1.\)

Thus, the correct answer is D.

18 答案 E

我们有 \[ (10^y)^2=25\text{。}\]由于 \(10^y\) 为正数，所以 \(10^y=5\)。取倒数得 \(10^{-y}=\frac{1}{5}\)。

因此，正确答案是 E。

English

We have \[ (10^y)^2=25. \] Since \(10^y\) is positive, \(10^y=5.\) Taking the reciprocal gives \(10^{-y}=\frac{1}{5}.\)

Thus, the correct answer is E.

19 答案 D

设每根蜡烛的初始高度为 \(1\)。\(t\) 小时后，剩余高度分别为 \(1-\frac{t}{4}\) 和 \(1-\frac{t}{3}\)。所需条件为 \[ 1-\frac t4=2\left(1-\frac t3\right)\text{。}\]两边乘以 \(12\)，得 \(12-3t=24-8t\)，所以 \(5t=12\)，且 \(t=\frac{12}{5}=2\dfrac25\)。

因此，正确答案是 D。

English

Let each initial height be \(1.\) After \(t\) hours the remaining heights are \(1-\frac{t}{4}\) and \(1-\frac{t}{3}.\) The required condition is \[ 1-\frac t4=2\left(1-\frac t3\right). \] Multiplying by \(12\) gives \(12-3t=24-8t,\) so \(5t=12\) and \(t=\frac{12}{5}=2\dfrac25.\)

Thus, the correct answer is D.

20 答案 C

两边取常用对数，得 \[ x\log(0.2)=\log2\text{。}\]由于 \(\log(0.2)=\log2-1=-0.6990\)， \[ x=\frac{0.3010}{-0.6990}\approx-0.4306\text{。}\]精确到十分位为 \(-0.4\)。

因此，正确答案是 C。

English

Taking common logarithms gives \[ x\log(0.2)=\log2. \] Since \(\log(0.2)=\log2-1=-0.6990,\) \[ x=\frac{0.3010}{-0.6990}\approx-0.4306. \] To the nearest tenth, this is \(-0.4.\)

Thus, the correct answer is C.

21 答案 E

当直线方向平行于一条渐近线时，它可以与双曲线有一个有限交点；否则可以有两个有限交点。适当选取两条直线，使它们在双曲线上的有限交点互不重合，便可分别得到 \(1+1=2\)、\(1+2=3\) 或 \(2+2=4\) 个不同交点。若两条直线自身的交点恰好在双曲线上，则这个交点只计一次。

因此，\(2\)、\(3\) 或 \(4\) 都有可能，正确答案是 E。

English

A line can meet a hyperbola in either one finite point, when its direction is parallel to an asymptote, or two finite points. By choosing the two lines so that their finite intersection sets on the hyperbola are disjoint, they can therefore contribute \(1+1=2,\) \(1+2=3,\) or \(2+2=4\) distinct intersections. If their own intersection lies on the hyperbola, one point is shared instead.

Thus, \(2,\) \(3,\) or \(4\) are possible, and the correct answer is E.

22 答案 B

若第一次的速度为 \(v\)，则原旅行时间为 \(\frac{50}{v}\)。后一次的旅行时间为 \[ \frac{300}{3v}=\frac{100}{v}=2\left(\frac{50}{v}\right)\text{。}\]因此，新旅行时间是原来的两倍。

正确答案是 B。

English

If the first speed was \(v,\) the old time was \(\frac{50}{v}.\) The later time was \[ \frac{300}{3v}=\frac{100}{v}=2\left(\frac{50}{v}\right). \] Thus the new time was twice the old time.

The correct answer is B.

23 答案 C

实系数且判别式为零的二次方程有重根 \[ x=\frac{-(-2\sqrt2)}{2a}=\frac{\sqrt2}{a}\text{。}\]由于 \(a\) 是非零实数，此根为实数。它不一定总是有理数，也不一定总是无理数。

因此，两根必定相等且为实数，正确答案是 C。

English

A quadratic with real coefficients and discriminant zero has the repeated root \[ x=\frac{-(-2\sqrt2)}{2a}=\frac{\sqrt2}{a}. \] Since \(a\) is a nonzero real number, this root is real. It need not always be rational or always be irrational.

Thus, the roots are necessarily equal and real, so the correct answer is C.

24 答案 D

设 \(\angle DAE=\alpha\)。由于 \(AB=AC\) 且 \(AE=AD\)， \[ \begin{aligned} \angle ACB&=75^\circ-\frac\alpha2,\\ \angle ADE&=90^\circ-\frac\alpha2 \end{aligned}\text{。}\]在三角形 \(ACD\) 中，\(\angle ADC=180^\circ-\alpha-\angle ACB\)，所以 \(\angle ADC=105^\circ-\frac\alpha2\)。由于 \(\angle ADC=\angle ADE+x\)， \[ \begin{aligned} x&=\left(105^\circ-\frac\alpha2\right) -\left(90^\circ-\frac\alpha2\right)\\ &=15^\circ \end{aligned}\text{。}\]

因此，正确答案是 D。

English

Let \(\angle DAE=\alpha.\) Since \(AB=AC\) and \(AE=AD,\) \[ \begin{aligned} \angle ACB&=75^\circ-\frac\alpha2,\\ \angle ADE&=90^\circ-\frac\alpha2. \end{aligned} \] Triangle \(ACD\) gives \(\angle ADC=180^\circ-\alpha-\angle ACB,\) so \(\angle ADC=105^\circ-\frac\alpha2.\) Since \(\angle ADC=\angle ADE+x,\) \[ \begin{aligned} x&=\left(105^\circ-\frac\alpha2\right) -\left(90^\circ-\frac\alpha2\right)\\ &=15^\circ. \end{aligned} \]

Thus, the correct answer is D.

25 答案 C

计算得 \[ \begin{aligned} \sum_{k=1}^n(2k+1) &=2\sum_{k=1}^n k+\sum_{k=1}^n1\\ &=n(n+1)+n\\ &=n(n+2) \end{aligned}\text{。}\]

因此，正确答案是 C。

English

We compute \[ \begin{aligned} \sum_{k=1}^n(2k+1) &=2\sum_{k=1}^n k+\sum_{k=1}^n1\\ &=n(n+1)+n\\ &=n(n+2). \end{aligned} \]

Thus, the correct answer is C.

26 答案 A

在等腰三角形中，一个底角和顶角可以确定三个角，但没有指定任何边长。因此，任意大小的相似三角形都满足同样的已知条件，三角形不能唯一确定。其余每个选项都直接给出或通过半径给出了足够的长度信息，可以同时确定大小和形状。

因此，正确答案是 A。

English

In an isosceles triangle, a base angle and the vertex angle determine all three angles, but no side length is specified. Therefore triangles of every scale have the same given data, so the triangle is not determined. Each other choice includes enough length information, directly or through a radius, to fix the scale as well as the shape.

Thus, the correct answer is A.

27 答案 C

若两条夹边为 \(a\) 和 \(b\)，保持不变的夹角为 \(C\)，则原面积为 \(\frac12ab\sin C\)。两条边都加倍后，面积为 \[ \begin{gathered} \frac12(2a)(2b)\sin C \\ =2ab\sin C\\ =4\left(\frac12ab\sin C\right) \end{gathered}\text{。}\]

因此，正确答案是 C。

English

If the included sides are \(a\) and \(b,\) and their unchanged angle is \(C,\) the original area is \(\frac12ab\sin C.\) After both sides are doubled, the area is \[ \begin{gathered} \frac12(2a)(2b)\sin C \\ =2ab\sin C\\ =4\left(\frac12ab\sin C\right). \end{gathered} \]

Thus, the correct answer is C.

28 答案 D

设女儿得到 \(4x\)，儿子得到 \(3x\)。两人共得到 \(7x\)，即遗产的一半。妻子得到 \(6x\)，所以由另一半可得 \[ 7x=6x+500\text{。}\]因此 \(x=500\)，全部遗产为 \(2(7x)=14(500)=7000\) 美元。

因此，正确答案是 D。

English

Let the daughter receive \(4x\) and the son \(3x.\) Together they receive \(7x,\) which is half the estate. The wife receives \(6x,\) so the other half gives \[ 7x=6x+500. \] Thus \(x=500,\) and the whole estate is \(2(7x)=14(500)=7000\) dollars.

Therefore, the correct answer is D.

29 答案 D

在交点处，\[ \begin{aligned} (x+y)^2&=x^2+y^2+2xy\\ &=25+24=49 \end{aligned}\text{，}\]所以 \(x+y=\pm7\)。再结合 \(xy=12\)，得到四个点 \[ \begin{aligned} &(3,4),\ (4,3),\\ &(-3,-4),\ (-4,-3) \end{aligned}\text{。}\]按它们在圆周上的顺序连接，相邻边向量互相垂直，所以四边形是矩形。相邻边长不相等，因此它不是正方形。

因此，正确答案是 D。

English

At an intersection, \[ \begin{aligned} (x+y)^2&=x^2+y^2+2xy\\ &=25+24=49, \end{aligned} \] so \(x+y=\pm7.\) Together with \(xy=12,\) this gives the four points \[ \begin{aligned} &(3,4),\ (4,3),\\ &(-3,-4),\ (-4,-3). \end{aligned} \] In their cyclic order around the circle, adjacent side vectors are perpendicular, so the quadrilateral is a rectangle. Its adjacent side lengths are unequal, so it is not a square.

Thus, the correct answer is D.

30 答案 B

设边长为 \(s\)。由 \[ \frac{s\sqrt3}{2}=\sqrt6\text{，}\]得 \(s=2\sqrt2\)。因此面积为 \[ \frac12s\sqrt6 =\frac12(2\sqrt2)(\sqrt6)=2\sqrt3\text{。}\]

因此，正确答案是 B。

English

Let the side length be \(s.\) From \[ \frac{s\sqrt3}{2}=\sqrt6, \] we get \(s=2\sqrt2.\) Hence the area is \[ \frac12s\sqrt6 =\frac12(2\sqrt2)(\sqrt6)=2\sqrt3. \]

Thus, the correct answer is B.

31 答案 E

第二十个正整数的通常数值为 \(20\)。由于 \[ 20=1\cdot4^2+1\cdot4+0\text{，}\]它的四进制表示为 \(110\)。

因此，正确答案是 E。

English

The twentieth positive integer has ordinary value \(20.\) Since \[ 20=1\cdot4^2+1\cdot4+0, \] its base-four representation is \(110.\)

Thus, the correct answer is E.

32 答案 B

因为他们从两端同时出发并在中央第一次相遇，所以两人的速度相同。每个人都在 \(3\) 分钟后到达对岸。他们立即转身，各自再用 \(1.5\) 分钟回到中央并再次相遇。第二次相遇发生在出发 \[ 3+1.5=4.5=4\frac12 \] 分钟后。

因此，正确答案是 B。

English

Because they started simultaneously from opposite ends and first met at the center, their speeds are equal. Each reaches the opposite end after \(3\) minutes. They turn immediately, and each then needs another \(1.5\) minutes to return to the center, where they meet again. The second meeting occurs after \[ 3+1.5=4.5=4\frac12 \] minutes.

Thus, the correct answer is B.

33 答案 E

数 \(\sqrt2\) 是无理数。有理数的小数展开要么终止，要么最终循环，而无理数的小数展开则是无限不循环的。有限小数 \(1.41421\) 只是一个近似值。

因此，正确答案是 E。

English

The number \(\sqrt2\) is irrational. A rational number has a decimal expansion that either terminates or eventually repeats, whereas an irrational number has an infinite non-repeating decimal expansion. The finite decimal \(1.41421\) is only an approximation.

Thus, the correct answer is E.

34 答案 A

我们有 \[ n^2(n^2-1)=n^2(n-1)(n+1)\text{。}\]在 \(n-1\)、\(n\)、\(n+1\) 中，有一个能被 \(3\) 整除。若 \(n\) 为偶数，则 \(n^2\) 能被 \(4\) 整除；若 \(n\) 为奇数，则 \(n-1\) 和 \(n+1\) 都是偶数，所以它们的乘积能被 \(4\) 整除。因此，这个式子一定能被 \(12\) 整除。它不一定能被 \(24\) 整除，因为 \(n=2\) 时式子的值为 \(12\)。

因此，正确答案是 A。

English

We have \[ n^2(n^2-1)=n^2(n-1)(n+1). \] Among \(n-1,\) \(n,\) and \(n+1,\) one is divisible by \(3.\) If \(n\) is even, \(n^2\) is divisible by \(4;\) if \(n\) is odd, both \(n-1\) and \(n+1\) are even, so their product is divisible by \(4.\) Thus the expression is always divisible by \(12.\) It is not always divisible by \(24,\) since \(n=2\) gives \(12.\)

Therefore, the correct answer is A.

35 答案 B

菱形的四条边都等于圆的半径 \(16\)。一条长度等于半径的弦与连接其两个端点的两条半径组成等边三角形，所以所夹的圆心角为 \(60^\circ\)。因此，菱形的面积为 \[ \begin{aligned} 16^2\sin60^\circ &=256\cdot\frac{\sqrt3}{2}\\ &=128\sqrt3 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

All four sides of the rhombus equal the circle’s radius, \(16.\) A chord of length equal to the radius forms an equilateral triangle with the two radii to its endpoints, so the included central angle is \(60^\circ.\) Therefore the rhombus has area \[ \begin{aligned} 16^2\sin60^\circ &=256\cdot\frac{\sqrt3}{2}\\ &=128\sqrt3. \end{aligned} \]

Thus, the correct answer is B.

36 答案 E

题设条件为 \[ \frac{K(K+1)}2=N^2\text{，}\]即 \[ (2K+1)^2-8N^2=1\text{。}\]这个佩尔方程的正整数解由基本解 \(3+\sqrt8\) 生成。最前面的几组 \((2K+1,N)\) 为 \[ \begin{aligned} &(3,1),\ (17,6),\\ &(99,35),\ (577,204),\ldots \end{aligned} \] 因此，满足 \(N\lt100\) 的值为 \(K=1,8,49\)。

因此，正确答案是 E。

English

The condition is \[ \frac{K(K+1)}2=N^2, \] or \[ (2K+1)^2-8N^2=1. \] The positive solutions of this Pell equation are generated from the fundamental solution \(3+\sqrt8.\) Their first pairs \((2K+1,N)\) are \[ \begin{aligned} &(3,1),\ (17,6),\\ &(99,35),\ (577,204),\ldots \end{aligned} \] Thus the values with \(N\lt100\) are \(K=1,8,49.\)

Therefore, the correct answer is E.

37 答案 E

\(60^\circ\) 角所对的对角线把菱形分成两个等边三角形，所以它的长度等于菱形的边长。地图的比例尺为每英寸 \(\frac{800}{3}\) 英里，因此实际边长为 \[ \frac3{16}\cdot\frac{800}{3}=50 \] 英里。所以菱形的面积为 \[ \begin{aligned} 50^2\sin60^\circ &=2500\cdot\frac{\sqrt3}{2}\\ &=1250\sqrt3 \end{aligned}\text{。}\]

因此，正确答案是 E。

English

The diagonal opposite the \(60^\circ\) angle divides the rhombus into two equilateral triangles, so its length equals the rhombus side. The map scale is \(\frac{800}{3}\) miles per inch, hence the actual side length is \[ \frac3{16}\cdot\frac{800}{3}=50 \] miles. Therefore the rhombus area is \[ \begin{aligned} 50^2\sin60^\circ &=2500\cdot\frac{\sqrt3}{2}\\ &=1250\sqrt3. \end{aligned} \]

Thus, the correct answer is E.

38 答案 D

令两种面积公式相等，得到 \[ \frac12ab=\frac12cx\text{，}\]所以 \(ab=cx\)。两边平方并利用 \(c^2=a^2+b^2\)，得到 \[ a^2b^2=x^2(a^2+b^2)\text{。}\]两边除以 \(a^2b^2x^2\)，得到 \[ \frac1{x^2}=\frac1{a^2}+\frac1{b^2}\text{。}\]

因此，正确答案是 D。

English

Equating two area formulas gives \[ \frac12ab=\frac12cx, \] so \(ab=cx.\) Squaring and using \(c^2=a^2+b^2\) yields \[ a^2b^2=x^2(a^2+b^2). \] Dividing by \(a^2b^2x^2\) gives \[ \frac1{x^2}=\frac1{a^2}+\frac1{b^2}. \]

Thus, the correct answer is D.

39 答案 C

由勾股定理，\[ b^2=c^2-a^2=(c-a)(c+a)\text{。}\]因为 \(c\) 与 \(a\) 是相邻整数且 \(c\gt a\)，所以 \(c-a=1\)。因此 \(b^2=c+a\)。

因此，正确答案是 C。

English

By the Pythagorean theorem, \[ b^2=c^2-a^2=(c-a)(c+a). \] Since \(c\) and \(a\) are consecutive and \(c\gt a,\) we have \(c-a=1.\) Hence \(b^2=c+a.\)

Thus, the correct answer is C.

40 答案 A

由第一个方程可知 \(gt=V-V_0\)。因此 \[ \begin{aligned} S&=\frac12(gt)t+V_0t\\ &=\frac12(V-V_0)t+V_0t\\ &=\frac12(V+V_0)t \end{aligned}\text{。}\]解得 \(t=\frac{2S}{V+V_0}\)。

因此，正确答案是 A。

English

From the first equation, \(gt=V-V_0.\) Therefore \[ \begin{aligned} S&=\frac12(gt)t+V_0t\\ &=\frac12(V-V_0)t+V_0t\\ &=\frac12(V+V_0)t. \end{aligned} \] Solving gives \(t=\frac{2S}{V+V_0}.\)

Thus, the correct answer is A.

41 答案 C

代入 \(y=2x\)，得到 \[ 12x^2+2x+4=12x^2+4x+4\text{。}\]消去两边相同的二次项和常数项后，剩下 \(2x=4x\)，所以 \(x=0\)。这个值满足两个方程。

因此，正确答案是 C。

English

Substituting \(y=2x\) gives \[ 12x^2+2x+4=12x^2+4x+4. \] Cancelling the common quadratic and constant terms leaves \(2x=4x,\) so \(x=0.\) This value satisfies both equations.

Thus, the correct answer is C.

42 答案 A

要有实数解，必须满足 \(x\ge3\)。此时 \(x+4>x-3\)，所以 \[ \sqrt{x+4}>\sqrt{x-3}\text{。}\]因此 \(\sqrt{x+4}-\sqrt{x-3}+1>1\)，不可能等于零。原方程是实数范围内的根式方程，所以非实数根不在其定义域内。

正确答案是 A。

English

For a real solution, \(x\ge3.\) Then \(x+4>x-3,\) so \[ \sqrt{x+4}>\sqrt{x-3}. \] Consequently \(\sqrt{x+4}-\sqrt{x-3}+1>1,\) and it cannot equal zero. The original equation is a real radical expression, so nonreal roots are outside its domain.

The correct answer is A.

43 答案 C

将互不相同的整数边长按递增顺序排列。在三角形不等式和周长限制下检查较小的可能，得到 \[ \begin{aligned} &(2,3,4),\ (2,4,5),\\ &(3,4,5) \end{aligned}\text{。}\]当 \(c\ge6\) 时，满足 \(a+b\gt c\) 的最小新不等边候选为 \((3,4,6)\)，其周长已经是 \(13\)，更大的选择都不合要求。因此共有 \(3\) 个三角形。

正确答案是 C。

English

Write the distinct integer sides in increasing order. Checking the small possibilities under the triangle inequality and perimeter bound gives \[ \begin{aligned} &(2,3,4),\ (2,4,5),\\ &(3,4,5). \end{aligned} \] For \(c\ge6,\) the smallest new scalene candidate satisfying \(a+b\gt c\) is \((3,4,6),\) whose perimeter is already \(13,\) and larger choices cannot qualify. Thus there are \(3\) triangles.

The correct answer is C.

44 答案 B

因为 \(x\lt a\lt0\)，将 \(x\lt a\) 乘以负数 \(x\) 时不等号反向，得到 \(x^2\gt ax\)。将同一个不等式乘以负数 \(a\)，得到 \(ax\gt a^2\)。因此 \[ x^2\gt ax\gt a^2\text{。}\]

因此，正确答案是 B。

English

Since \(x\lt a\lt0,\) multiplying \(x\lt a\) by the negative number \(x\) reverses the inequality and gives \(x^2\gt ax.\) Multiplying the same inequality by the negative number \(a\) gives \(ax\gt a^2.\) Therefore \[ x^2\gt ax\gt a^2. \]

Thus, the correct answer is B.

45 答案 A

原半径为 \(12.5\) 英寸，新半径为 \(12.25\) 英寸。对于固定距离，转数与半径成反比，所以相对增幅为 \[ \begin{aligned} \frac{12.5}{12.25}-1 &=\frac{50}{49}-1\\ &=\frac1{49}\\ &\approx2.04\% \end{aligned}\text{。}\]这约为 \(2\%\)。

因此，正确答案是 A。

English

The original radius is \(12.5\) inches and the new radius is \(12.25\) inches. For a fixed distance, the number of revolutions varies inversely with radius, so the relative increase is \[ \begin{aligned} \frac{12.5}{12.25}-1 &=\frac{50}{49}-1\\ &=\frac1{49}\\ &\approx2.04\%. \end{aligned} \] This is about \(2\%.\)

Thus, the correct answer is A.

46 答案 A

交叉相乘得到 \[ N(1+x)=(N+1)(1-x)\text{，}\]所以 \(x(2N+1)=1\)，并且 \[ x=\frac1{2N+1}\text{。}\]每个正数 \(N\) 都给出 \(0\lt x\lt1\)。反过来，对任意 \(0\lt x\lt1\)，有 \[ N=\frac{1-x}{2x}>0\text{，}\]因此存在这样的 \(N\)。

因此，\(x\) 可以是任意小于 \(1\) 的正数，正确答案是 A。

English

Cross-multiplication gives \[ N(1+x)=(N+1)(1-x), \] so \(x(2N+1)=1\) and \[ x=\frac1{2N+1}. \] Every positive \(N\) gives \(0\lt x\lt1.\) Conversely, for any \(0\lt x\lt1,\) \[ N=\frac{1-x}{2x}>0, \] so such an \(N\) exists.

Thus, \(x\) may be any positive value less than \(1,\) and the correct answer is A.

47 答案 D

若现有 \(m\) 台机器，则这项工程需要 \(3m\) 个机器·天。增加三台机器后，需要 \(2(m+3)\) 个机器·天，所以 \[ 3m=2(m+3)\text{，}\]解得 \(m=6\)。因此，这项工程需要 \(3m=18\) 个机器·天，一台机器单独完成需要 \(18\) 天。

因此，正确答案是 D。

English

If there are presently \(m\) machines, the job requires \(3m\) machine-days. With three more machines it requires \(2(m+3)\) machine-days, so \[ 3m=2(m+3), \] giving \(m=6.\) The job therefore requires \(3m=18\) machine-days, so one machine would take \(18\) days.

Thus, the correct answer is D.

48 答案 B

令 \(q=2p-5\)。商为正数要求 \(q>0\)，而 \(q\) 是奇数。由于 \(p=\frac{q+5}{2}\)，\[ \begin{aligned} \frac{3p+25}{2p-5} &=\frac{3q+65}{2q}\\ &=\frac{3+\frac{65}{q}}{2} \end{aligned}\text{。}\]所以 \(q\) 必须是 \(65\) 的正因数。四种可能 \(q=1\)、\(q=5\)、\(q=13\)、\(q=65\) 分别给出 \(p=3\)、\(p=5\)、\(p=9\)、\(p=35\)，且每个都符合。不存在其他符合条件的正整数 \(p\)。

因此，数学上正确的集合是 \(\{3,5,9,35\}\)，即调整后的选项 B。

English

Set \(q=2p-5.\) A positive quotient requires \(q>0,\) and \(q\) is odd. Since \(p=\frac{q+5}{2},\) \[ \begin{aligned} \frac{3p+25}{2p-5} &=\frac{3q+65}{2q}\\ &=\frac{3+\frac{65}{q}}{2}. \end{aligned} \] Thus \(q\) must be a positive divisor of \(65.\) The possibilities \(q=1,\) \(q=5,\) \(q=13,\) and \(q=65\) give \(p=3,\) \(p=5,\) \(p=9,\) and \(p=35,\) respectively, and each works. No other positive integer \(p\) works.

Therefore the mathematically correct set is \(\{3,5,9,35\},\) shown in the adjusted choice B.

49 答案 E

圆位于边 \(AB\) 的另一侧，与顶点 \(P\) 相对，所以 \(O\) 是 \(P\) 所对的旁心。因此，\(AO\) 和 \(BO\) 分别平分 \(A\) 与 \(B\) 处的外角。若 \(A\) 与 \(B\) 处的内角分别为 \(\alpha\) 和 \(\beta\)，则三角形 \(AOB\) 在 \(A\) 与 \(B\) 处的角分别为 \(90^\circ-\frac{\alpha}{2}\) 和 \(90^\circ-\frac{\beta}{2}\)。因此 \[ \begin{aligned} \angle AOB &=180^\circ-\left(90^\circ-\frac\alpha2\right)\\ &\mathrel{\phantom{=}}-\left(90^\circ-\frac\beta2\right)\\ &=\frac{\alpha+\beta}{2} =\frac{180^\circ-40^\circ}{2}\\ &=70^\circ \end{aligned}\text{。}\]

因此，正确答案是 E。

English

The circle lies opposite \(P\) across side \(AB,\) so \(O\) is the excenter opposite \(P.\) Thus \(AO\) and \(BO\) bisect the exterior angles at \(A\) and \(B.\) If the interior angles at \(A\) and \(B\) are \(\alpha\) and \(\beta,\) then triangle \(AOB\) has angles \(90^\circ-\frac{\alpha}{2}\) and \(90^\circ-\frac{\beta}{2}\) at \(A\) and \(B.\) Hence \[ \begin{aligned} \angle AOB &=180^\circ-\left(90^\circ-\frac\alpha2\right)\\ &\mathrel{\phantom{=}}-\left(90^\circ-\frac\beta2\right)\\ &=\frac{\alpha+\beta}{2} =\frac{180^\circ-40^\circ}{2}\\ &=70^\circ. \end{aligned} \]

Thus, the correct answer is E.

50 答案 B

将两条相等的边缩放为 \(1\)，并令 \[ \begin{aligned} C&=(0,0),\qquad B=(1,0),\\ A&=(\cos\theta,\sin\theta) \end{aligned}\text{。}\]因为正方形 \(BCDE\) 作在三角形外部，所以 \(D=(0,-1)\)。令 \(c=\cos\theta\) 且 \(s=\sin\theta\)。则 \[ \begin{aligned} \overrightarrow{AB}&=(1-c,-s),\\ \overrightarrow{AD}&=(-c,-1-s) \end{aligned}\text{。}\]两个向量的点积为 \(1+s-c\)，二维叉积的绝对值也为 \(1+s-c\)。因此 \[ \tan\angle DAB=1\text{，}\]所以 \(\angle DAB=45^\circ\)，它与 \(\theta\) 无关，因而也与三角形的形状无关。

因此，正确答案是 B。

English

Scale the equal sides to \(1\) and place \[ \begin{aligned} C&=(0,0),\qquad B=(1,0),\\ A&=(\cos\theta,\sin\theta). \end{aligned} \] Since square \(BCDE\) is constructed away from the triangle, \(D=(0,-1).\) Put \(c=\cos\theta\) and \(s=\sin\theta.\) Then \[ \begin{aligned} \overrightarrow{AB}&=(1-c,-s),\\ \overrightarrow{AD}&=(-c,-1-s). \end{aligned} \] Their dot product is \(1+s-c,\) while the absolute value of their two-dimensional cross product is also \(1+s-c.\) Therefore \[ \tan\angle DAB=1, \] so \(\angle DAB=45^\circ,\) independent of \(\theta\) and hence independent of the triangle’s shape.

Thus, the correct answer is B.