1955 AMC 12 真题（P版）

1 答案 D

给定的小数为 \(3.75\times10^{-7}\)。选项 A 和 B 直接表示这个数，而 \[ 375\times10^{-9}=3.75\times10^{-7} \] 且 \(\frac38\times10^{-6}\) \(=0.375\times10^{-6}\) \(=3.75\times10^{-7}\)。但是 \[ \frac38\times10^{-7}=3.75\times10^{-8}\text{，}\]它小了十倍。

因此，正确答案是 D。

English

The given decimal is \(3.75\times10^{-7}.\) Choices A and B state this directly, while \[ 375\times10^{-9}=3.75\times10^{-7} \] and \(\frac38\times10^{-6}\) \(=0.375\times10^{-6}\) \(=3.75\times10^{-7}.\) But \[ \frac38\times10^{-7}=3.75\times10^{-8}, \] which is ten times smaller.

Thus, the correct answer is D.

2 答案 B

在 \(12{:}25\)，分针已从 \(12\) 点方向转过 \(25\cdot6^\circ=150^\circ\)，而时针转过 \(25\cdot0.5^\circ=12.5^\circ\)。两针之间的较小夹角为 \[ 150^\circ-12.5^\circ=137.5^\circ=137^\circ30'\text{。}\]

因此，正确答案是 B。

English

At \(12{:}25,\) the minute hand is \(25\cdot6^\circ=150^\circ\) past \(12,\) while the hour hand is \(25\cdot0.5^\circ=12.5^\circ.\) Their smaller separation is \[ 150^\circ-12.5^\circ=137.5^\circ=137^\circ30'. \]

Thus, the correct answer is B.

3 答案 B

若原来的总和为 \(S\)，则新的总和为 \(S+10\cdot20\)。因此新的平均数为 \[ \frac{S+200}{10}=\frac S{10}+20\text{。}\]所以平均数增加了 \(20\)。

因此，正确答案是 B。

English

If the original sum is \(S,\) the new sum is \(S+10\cdot20.\) Hence the new mean is \[ \frac{S+200}{10}=\frac S{10}+20. \] The mean is increased by \(20.\)

Thus, the correct answer is B.

4 答案 E

当 \(x\ne1,2\) 时，交叉相乘得 \[ x-2=2(x-1)=2x-2\text{，}\]所以 \(x=0\)。此值使原等式中的两个分母都不为零，并且满足等式。

因此，正确答案是 E。

English

For \(x\ne1,2,\) cross-multiplication gives \[ x-2=2(x-1)=2x-2, \] so \(x=0.\) This value makes both original denominators nonzero and satisfies the equality.

Thus, the correct answer is E.

5 答案 D

平方反比关系给出 \(y=\frac{k}{x^2}\)。由于 \(16=\frac{k}{1^2}\)，所以 \(k=16\)。当 \(x=8\) 时，\[ y=\frac{16}{8^2}=\frac14\text{。}\]

因此，正确答案是 D。

English

Inverse-square variation gives \(y=\frac{k}{x^2}.\) Since \(16=\frac{k}{1^2},\) we have \(k=16.\) At \(x=8,\) \[ y=\frac{16}{8^2}=\frac14. \]

Thus, the correct answer is D.

6 答案 B

假设他按每种价格各买 \(15\) 个橙子。前 \(15\) 个花费 \(50\)¢，后 \(15\) 个花费 \(60\)¢，总计 \(110\)¢。因此 \(30\) 个橙子必须卖 \(110\)¢，这等价于每 \(3\) 个卖 \(11\)¢。

因此，正确答案是 B。

English

Suppose he buys \(15\) oranges at each rate. The first \(15\) cost \(50\)¢ and the second \(15\) cost \(60\)¢, for \(110\)¢ total. Thus \(30\) oranges must sell for \(110\)¢, which is equivalent to \(3\) for \(11\)¢.

Therefore, the correct answer is B.

7 答案 B

若原工资为 \(W\)，则降薪后的工资为 \(0.8W\)。所需增加额为 \(0.2W\)，它占降薪后工资的比例为 \[ \frac{0.2W}{0.8W}=\frac14=25\%\text{。}\]

因此，正确答案是 B。

English

If the original wage is \(W,\) the reduced wage is \(0.8W.\) The needed increase is \(0.2W,\) which as a fraction of the reduced wage is \[ \frac{0.2W}{0.8W}=\frac14=25\%. \]

Thus, the correct answer is B.

8 答案 D

因式分解得 \[ x^2-4y^2=(x-2y)(x+2y)\text{。}\]因此图形是直线 \(x=2y\) 与 \(x=-2y\) 的并集。

正确答案是 D。

English

Factoring, \[ x^2-4y^2=(x-2y)(x+2y). \] Thus the graph is the union of the two straight lines \(x=2y\) and \(x=-2y.\)

The correct answer is D.

9 答案 D

因为 \(8^2+15^2=17^2\)，所以这个三角形是直角三角形。其内切圆半径为 \[ r=\frac{8+15-17}{2}=3\text{。}\]

因此，正确答案是 D。

English

Because \(8^2+15^2=17^2,\) the triangle is right. Its inradius is \[ r=\frac{8+15-17}{2}=3. \]

Thus, the correct answer is D.

10 答案 A

火车行驶 \(\frac{a}{40}\) 小时，停靠 \(\frac{mn}{60}\) 小时。因此总时间为 \[ \frac a{40}+\frac{mn}{60} =\frac{3a+2mn}{120}\text{。}\]

因此，正确答案是 A。

English

The train moves for \(\frac{a}{40}\) hours and is stopped for \(\frac{mn}{60}\) hours. Hence the total time is \[ \frac a{40}+\frac{mn}{60} =\frac{3a+2mn}{120}. \]

Thus, the correct answer is A.

11 答案 C

原命题表示所有学习迟缓的学生都不在这所学校就读。它的否定是至少有一名学习迟缓的学生在这所学校就读。

因此，“有些学习迟缓的学生在这所学校就读”是正确的，所以答案是 C。

English

The original statement says that every slow learner is absent from the school. Its negation is that at least one slow learner attends the school.

Thus, “Some slow learners attend this school” is correct, so the answer is C.

12 答案 D

定义域要求 \(x\ge1\)。当 \(x=1\) 时，左边为 \(2+0=2\)，所以 \(x=1\) 是一个解。要说明没有其他解，注意当 \(x\ge1\) 时，两个根式都不减，而 \(\sqrt{5x-1}\) 严格递增。因此，对每个 \(x>1\)，两个根式之和都大于 \(2\)。

因此，正确答案是 D。

English

The domain requires \(x\ge1.\) At \(x=1,\) the left side is \(2+0=2,\) so \(x=1\) works. To see that there is no other solution, note that both radicals are nondecreasing for \(x\ge1,\) and \(\sqrt{5x-1}\) is strictly increasing. Therefore the sum exceeds \(2\) for every \(x>1.\)

Thus, the correct answer is D.

13 答案 C

将分子因式分解：\[ \begin{aligned} a^{-4}-b^{-4} &=(a^{-2}-b^{-2})\\ &\quad\cdot(a^{-2}+b^{-2}) \end{aligned}\text{。}\]在原分式有定义的情况下，约去公因式后得到 \(a^{-2}+b^{-2}\)。

因此，正确答案是 C。

English

Factoring the numerator, \[ \begin{aligned} a^{-4}-b^{-4} &=(a^{-2}-b^{-2})\\ &\quad\cdot(a^{-2}+b^{-2}). \end{aligned} \] Where the original fraction is defined, canceling the common factor leaves \(a^{-2}+b^{-2}.\)

Thus, the correct answer is C.

14 答案 A

若正方形的边长为 \(s\)，则矩形的长和宽分别为 \(1.1s\) 和 \(0.9s\)。因此 \[ \begin{aligned} \frac{[R]}{[S]} &=\frac{(1.1s)(0.9s)}{s^2}\\ &=0.99=\frac{99}{100} \end{aligned}\text{。}\]

因此，正确答案是 A。

English

If the square side is \(s,\) then the rectangle has dimensions \(1.1s\) and \(0.9s.\) Therefore \[ \begin{aligned} \frac{[R]}{[S]} &=\frac{(1.1s)(0.9s)}{s^2}\\ &=0.99=\frac{99}{100}. \end{aligned} \]

Thus, the correct answer is A.

15 答案 D

若较大圆的半径为 \(R\)，则 \[ \frac{\pi r^2}{\pi R^2}=\frac13\text{，}\]所以 \(R=r\sqrt3\)。两半径之差为 \[ R-r=(\sqrt3-1)r\approx0.732r\text{。}\]

因此，最佳近似值为 \(0.73r\)，正确答案是 D。

English

If the larger radius is \(R,\) then \[ \frac{\pi r^2}{\pi R^2}=\frac13, \] so \(R=r\sqrt3.\) The difference is \[ R-r=(\sqrt3-1)r\approx0.732r. \]

Thus, the best approximation is \(0.73r,\) and the correct answer is D.

16 答案 E

代入得 \(a+b=4+(-4)=0\)。因此原式变为 \(\frac{3}{0}\)，它没有定义。

因此，这个式子无意义，正确答案是 E。

English

Substitution gives \(a+b=4+(-4)=0.\) Hence the expression becomes \(\frac{3}{0},\) which is undefined.

Thus, the expression is meaningless and the correct answer is E.

17 答案 C

使用常用对数，\[ \begin{aligned} \log x&=5\log3-2\\ &=\log(3^5)-\log100\\ &=\log\left(\frac{243}{100}\right) \end{aligned}\text{。}\]因此 \(x=2.43\)。

正确答案是 C。

English

Using common logarithms, \[ \begin{aligned} \log x&=5\log3-2\\ &=\log(3^5)-\log100\\ &=\log\left(\frac{243}{100}\right). \end{aligned} \] Therefore \(x=2.43.\)

The correct answer is C.

18 答案 A

由于判别式为零，求根公式两次都给出 \[ x=\frac{-2\sqrt3}{2}=-\sqrt3\text{。}\]这个根是无理数，但题目所列的适用描述是“相等的实数”。

因此，正确答案是 A。

English

The quadratic formula gives \[ x=\frac{-2\sqrt3}{2}=-\sqrt3 \] twice because the discriminant is zero. The root is irrational, but the requested description that applies is “real and equal.”

Thus, the correct answer is A.

19 答案 B

令前者为较大的数，则 \[ u+v=6,\qquad u-v=8 \] 给出 \(u=7\) 和 \(v=-1\)。两数之积为 \(-7\)，所以以它们为根的首一方程为 \[ x^2-6x-7=0\text{。}\]

因此，正确答案是 B。

English

Taking the larger number first, \[ u+v=6,\qquad u-v=8 \] gives \(u=7\) and \(v=-1.\) Their product is \(-7,\) so the monic equation with these roots is \[ x^2-6x-7=0. \]

Thus, the correct answer is B.

20 答案 A

要使表达式为零，必须有 \[ \sqrt{25-t^2}=-5\text{。}\]根号表示主平方根。当被开方数为非负实数时，主平方根非负；在复数范围内，则选取实部非负的平方根。因此它不可能等于 \(-5\)。两边平方会引入增根 \(t=0\)，此时原表达式等于 \(10\)。

因此，没有任何实数或虚数满足条件，正确答案是 A。

English

For the expression to vanish, one would need \[ \sqrt{25-t^2}=-5. \] The radical sign denotes the principal square root. It is nonnegative for a nonnegative real radicand and, over the complex numbers, is chosen with nonnegative real part; it therefore cannot equal \(-5.\) Squaring would introduce the extraneous candidate \(t=0,\) for which the original expression is \(10.\)

Thus, no real or imaginary value works, and the correct answer is A.

21 答案 B

以斜边为底，并用 \(h\) 表示其上的高，则 \[ A=\frac12ch\text{。}\]解得 \(h=\frac{2A}{c}\)。

因此，正确答案是 B。

English

Taking the hypotenuse as the base and writing its altitude as \(h,\) \[ A=\frac12ch. \] Solving gives \(h=\frac{2A}{c}.\)

Thus, the correct answer is B.

22 答案 D

第一种优惠的实付价格是原价的 \[ 0.8\cdot0.8\cdot0.9=0.576\text{。}\]第二种优惠的实付价格是原价的 \[ 0.6\cdot0.95^2=0.5415\text{。}\]第二种比第一种少付订单金额的 \(0.576-0.5415=0.0345\)，即 \[ 0.0345(10000)=345 \] 美元。

因此，正确答案是 D。

English

The first offer leaves a fraction \[ 0.8\cdot0.8\cdot0.9=0.576 \] of the price. The second leaves \[ 0.6\cdot0.95^2=0.5415. \] The second is cheaper by \(0.576-0.5415=0.0345\) of the order, or \[ 0.0345(10000)=345 \] dollars.

Thus, the correct answer is D.

23 答案 C

把 \(x\) 枚五美分硬币当作二十五美分硬币，使总额多算 \(20x\)¢。把 \(x\) 枚十美分硬币当作一美分硬币，使总额少算 \(9x\)¢。因此净多算了 \[ 20x-9x=11x \] 美分，必须减去这个数额。

因此，正确答案是 C。

English

The \(x\) nickels counted as quarters overstate the total by \(20x\)¢. The \(x\) dimes counted as cents understate it by \(9x\)¢. The net overstatement is \[ 20x-9x=11x \] cents, so that amount must be subtracted.

Thus, the correct answer is C.

24 答案 E

配方得 \[ \begin{aligned} 4x^2-12x-1 &=4\left(x-\frac32\right)^2\\ &\quad{}-10 \end{aligned}\text{。}\]平方项非负，所以最小值为 \(-10\)，并在 \(x=\frac{3}{2}\) 时取得。

因此，正确答案是 E。

English

Completing the square, \[ \begin{aligned} 4x^2-12x-1 &=4\left(x-\frac32\right)^2\\ &\quad{}-10. \end{aligned} \] The squared term is nonnegative, so the minimum value is \(-10,\) attained at \(x=\frac{3}{2}.\)

Thus, the correct answer is E.

25 答案 E

我们有 \[ \begin{aligned} x^4+2x^2+9 &=(x^2+3)^2\\ &\quad{}-(2x)^2\\ &=(x^2-2x+3)\\ &\quad\cdot(x^2+2x+3) \end{aligned}\text{。}\]这两个因式都不在选项 A 至 D 中。

因此，正确答案是 E。

English

We have \[ \begin{aligned} x^4+2x^2+9 &=(x^2+3)^2\\ &\quad{}-(2x)^2\\ &=(x^2-2x+3)\\ &\quad\cdot(x^2+2x+3). \end{aligned} \] Neither factor appears among choices A through D.

Thus, the correct answer is E.

26 答案 E

\(A\) 先生先收到 \(1.10(10000)=11000\) 美元。然后 \(B\) 先生以其 \(11000\) 美元成本的 \(10\%\) 亏损卖出，所以 \(A\) 先生以 \[ 0.90(11000)=9900 \] 美元买回房子。\(A\) 先生重新拥有房子，并赚得 \(11000-9900=1100\) 美元；\(B\) 先生亏损 \(1100\) 美元。题中列出的金额都不正确。

因此，正确答案是 E。

English

Mr. \(A\) first receives \(1.10(10000)=11000\) dollars. Mr. \(B\) then sells at a loss of \(10\%\) of his \(11000\)-dollar cost, so Mr. \(A\) buys the house back for \[ 0.90(11000)=9900 \] dollars. Mr. \(A\) again owns the house and has gained \(11000-9900=1100\) dollars; Mr. \(B\) has lost \(1100\) dollars. None of the stated amounts is correct.

Thus, the correct answer is E.

27 答案 B

由韦达定理，\(r+s=p\)，且 \(rs=q\)。因此 \[ \begin{aligned} r^2+s^2 &=(r+s)^2-2rs\\ &=p^2-2q \end{aligned}\text{。}\]

因此，正确答案是 B。

English

By Vieta’s formulas, \(r+s=p\) and \(rs=q.\) Therefore \[ \begin{aligned} r^2+s^2 &=(r+s)^2-2rs\\ &=p^2-2q. \end{aligned} \]

Thus, the correct answer is B.

28 答案 E

用 \(-x\) 替换 \(x\)，得 \(y=ax^2-bx+c\)。在交点处，\[ ax^2+bx+c=ax^2-bx+c\text{，}\]所以 \(2bx=0\)。由于 \(b\ne0\)，必有 \(x=0\)，进而 \(y=c\ne0\)。两图像恰有一个交点 \((0,c)\)，且它在 \(y\) 轴上。

因此，正确答案是 E。

English

Replacing \(x\) by \(-x\) gives \(y=ax^2-bx+c.\) At an intersection, \[ ax^2+bx+c=ax^2-bx+c, \] so \(2bx=0.\) Since \(b\ne0,\) we must have \(x=0,\) and then \(y=c\ne0.\) There is exactly one intersection, \((0,c),\) on the \(y\)-axis.

Thus, the correct answer is E.

29 答案 E

直线 \(PR\) 在两半圆的公共端点 \(R\) 处与它们都相切。对较大的圆，两切线夹角定理给出 \[ \angle APR=180^\circ-a=c\text{。}\]对较小的圆，同一定理给出 \[ \angle RPB=180^\circ-b=d\text{。}\]因此，从 \(\overline{PA}\) 经 \(\overline{PR}\) 到 \(\overline{PB}\) 的优角为 \(c+d\)。所以两条切线之间的另一个角为 \[ \begin{aligned} 360^\circ-(c+d) &=(180^\circ-c)\\ &\quad{}+(180^\circ-d)\\ &=a+b \end{aligned}\text{，}\]因为每个上半圆的度数都是 \(180^\circ\)。

因此，正确答案是 E。

English

The line \(PR\) is tangent to both semicircles at their common endpoint \(R.\) For the larger circle, the tangent-tangent angle theorem gives \[ \angle APR=180^\circ-a=c. \] For the smaller circle, the same theorem gives \[ \angle RPB=180^\circ-b=d. \] Thus the reflex angle from \(\overline{PA}\) to \(\overline{PB}\) through \(\overline{PR}\) has measure \(c+d.\) Hence the other angle between the tangents is \[ \begin{aligned} 360^\circ-(c+d) &=(180^\circ-c)\\ &\quad{}+(180^\circ-d)\\ &=a+b, \end{aligned} \] because each upper semicircle has measure \(180^\circ.\)

Thus, the correct answer is E.

30 答案 B

第一个方程给出 \(x=\pm3\)。对第二个方程作平方差因式分解，得到 \[ \begin{aligned} &[(2x-1)-(x-1)]\\ &\quad\cdot[(2x-1)+(x-1)]\\ &\qquad=x(3x-2)=0 \end{aligned}\text{，}\]所以 \(x=0\) 或 \(\frac{2}{3}\)。将第三个方程两边平方，得到 \(x^2-x-6=0\)，候选值为 \(3\) 和 \(-2\)；只有 \(3\) 满足实数定义域的限制。所得的每个根都不大于 \(3\)。

因此，每个方程都没有大于 \(3\) 的根，正确答案是 B。

English

The first equation gives \(x=\pm3.\) Factoring the difference of squares in the second gives \[ \begin{aligned} &[(2x-1)-(x-1)]\\ &\quad\cdot[(2x-1)+(x-1)]\\ &\qquad=x(3x-2)=0, \end{aligned} \] so \(x=0\) or \(\frac{2}{3}.\) Squaring the third gives \(x^2-x-6=0,\) with candidates \(3\) and \(-2;\) only \(3\) satisfies the real-domain restrictions. Every root obtained is at most \(3.\)

Thus, each equation has no root greater than \(3,\) and the correct answer is B.

31 答案 D

小三角形的面积也等于原三角形面积的一半。若小三角形中平行于原底边的边长为 \(b\)，由相似关系 \[ \left(\frac b2\right)^2=\frac12\text{，}\]所以 \(b=\sqrt2\)。梯形两条平行边的长度为 \(2\) 和 \(\sqrt2\)，因此其中位线长为 \[ \frac{2+\sqrt2}{2}\text{。}\]

因此，正确答案是 D。

English

The small triangle also has half the original area. If its side parallel to the original base has length \(b,\) similarity gives \[ \left(\frac b2\right)^2=\frac12, \] so \(b=\sqrt2.\) The parallel sides of the trapezoid have lengths \(2\) and \(\sqrt2,\) so its median has length \[ \frac{2+\sqrt2}{2}. \]

Thus, the correct answer is D.

32 答案 B

判别式为零给出 \[ (2b)^2-4ac=0\text{，}\]因而 \(b^2=ac\)。等价地，在比值有定义时，\(\frac{a}{b}=\frac{b}{c}\)，这正是 \(a,b,c\) 构成等比数列的条件。

因此，正确答案是 B。

English

A zero discriminant gives \[ (2b)^2-4ac=0, \] hence \(b^2=ac.\) Equivalently, \(\frac{a}{b}=\frac{b}{c}\) where the ratios are defined, which is the defining relation for \(a,b,c\) to form a geometric progression.

Thus, the correct answer is B.

33 答案 A

在 \(8{:}00\) 时，时针领先 \(240^\circ\)。分针以每分钟 \(5.5^\circ\) 的相对速度追赶，所以两针在 \(8\) 点后的 \(\frac{240}{5.5}=\frac{480}{11}\) 分钟重合。在 \(2{:}00\) 时，时针领先 \(60^\circ\)。要使两针沿相关方向相差 \(180^\circ\)，分针必须追赶 \(240^\circ\)，同样需要 \(\frac{480}{11}\) 分钟。因此，出发和到达时刻在各自整点后的分钟数相同，恰好相隔六小时。

因此，正确答案是 A。

English

At \(8{:}00,\) the hour hand is \(240^\circ\) ahead. The minute hand gains at \(5.5^\circ\) per minute, so the hands coincide \(\frac{240}{5.5}=\frac{480}{11}\) minutes after \(8.\) At \(2{:}00,\) the hour hand is \(60^\circ\) ahead. For the hands to be \(180^\circ\) apart in the relevant direction, the minute hand must gain \(240^\circ,\) again taking \(\frac{480}{11}\) minutes. The start and finish therefore have the same minute offset within their hours, exactly six hours apart.

Thus, the correct answer is A.

34 答案 C

两圆心相距 \(12\) 英寸，半径相差 \(6\)。因此每条公外切线段长 \[ \sqrt{12^2-6^2}=6\sqrt3\text{，}\]两条共长 \(12\sqrt3\)。由切线的几何关系，半径为 \(3\) 的圆上有一段 \(120^\circ\) 的外露弧，半径为 \(9\) 的圆上有一段 \(240^\circ\) 的外露弧。两段弧长之和为 \[ \begin{aligned} &\frac{120}{360}(2\pi\cdot3)\\ &\quad{}+\frac{240}{360}(2\pi\cdot9)\\ &\qquad=2\pi+12\pi\\ &\qquad=14\pi \end{aligned}\text{。}\]因此铁丝长为 \(12\sqrt3+14\pi\)。

因此，正确答案是 C。

English

The centers are \(12\) inches apart and their radii differ by \(6.\) Thus each common external tangent segment has length \[ \sqrt{12^2-6^2}=6\sqrt3, \] contributing \(12\sqrt3\) in all. The geometry of the tangent lines leaves a \(120^\circ\) exposed arc on the radius-\(3\) circle and a \(240^\circ\) exposed arc on the radius-\(9\) circle. Their lengths total \[ \begin{aligned} &\frac{120}{360}(2\pi\cdot3)\\ &\quad{}+\frac{240}{360}(2\pi\cdot9)\\ &\qquad=2\pi+12\pi\\ &\qquad=14\pi. \end{aligned} \] Hence the wire length is \(12\sqrt3+14\pi.\)

Thus, the correct answer is C.

35 答案 B

第一个男孩拿走 \(\frac{n}{2}+1\)，剩下 \(\frac{n}{2}-1\)。第二个男孩拿走余数的三分之一，第三个男孩得到另外三分之二，自动是第二个男孩所得的两倍。因此分配条件不能唯一确定 \(n\)。它只要求 \(\frac{n}{2}-1\) 是 \(3\) 的非负倍数，所以 \(8,14,20,26,\ldots\) 等许多数值都可行。

因此，原有弹珠数无法确定，正确答案是 B。

English

The first boy takes \(\frac{n}{2}+1,\) leaving \(\frac{n}{2}-1.\) The second takes one third of that remainder, and the third receives the other two thirds, automatically twice the second boy’s share. Thus the share condition imposes no unique value of \(n.\) It only requires \(\frac{n}{2}-1\) to be a nonnegative multiple of \(3,\) so many values such as \(8,14,20,26,\ldots\) work.

Therefore the original number cannot be determined, and the correct answer is B.

36 答案 E

矩形油面的长为 \(10\)，所以它在圆形截面中的弦宽为 \(\frac{40}{10}=4\)。半弦长为 \(2\)，半径为 \(3\)，因此圆心到弦的距离为 \[ \sqrt{3^2-2^2}=\sqrt5\text{。}\]这样长的弦可以位于圆心下方，也可以位于圆心上方。从油罐底部量起，相应的深度为 \(3-\sqrt5\) 和 \(3+\sqrt5\)。

因此，正确答案是 E。

English

The rectangular surface has length \(10,\) so its chord width in the circular cross-section is \(\frac{40}{10}=4.\) Half the chord is \(2,\) and the radius is \(3,\) so the distance from the center to the chord is \[ \sqrt{3^2-2^2}=\sqrt5. \] A chord of this length can lie either below or above the center. Measured from the bottom of the tank, the corresponding depths are \(3-\sqrt5\) and \(3+\sqrt5.\)

Thus, the correct answer is E.

37 答案 B

差为 \(99(h-u)\)。由于 \(h-u\) 是从 \(1\) 到 \(9\) 的整数，并且个位数字是 \(4\)，所以需要 \(9(h-u)\equiv4\pmod{10}\)。由此得 \(h-u=6\)，因而差为 \[ 99\cdot6=594\text{。}\]从右向左，在个位数字 \(4\) 之后的两个数字是 \(9\) 和 \(5\)。

因此，正确答案是 B。

English

The difference is \(99(h-u).\) Since \(h-u\) is an integer from \(1\) through \(9,\) and the units digit is \(4,\) we need \(9(h-u)\equiv4\pmod{10}.\) This gives \(h-u=6,\) so the difference is \[ 99\cdot6=594. \] Moving from right to left after the units digit \(4,\) the next digits are \(9\) and \(5.\)

Thus, the correct answer is B.

38 答案 B

若原来的四个整数之和为 \(S\)，则与单独取出的整数 \(x\) 对应的结果为 \[ x+\frac{S-x}{3}=\frac{S+2x}{3}\text{。}\]将四个给出的结果相加会把总和计为 \(2S\)，所以 \[ S=\frac{29+23+21+17}{2}=45\text{。}\]因此结果 \(29\) 来自 \(x=\dfrac{3(29)-45}{2}=21\)。

因此，原来的一个整数是 \(21\)，正确答案是 B。

English

If the original integers sum to \(S,\) the result associated with singled-out integer \(x\) is \[ x+\frac{S-x}{3}=\frac{S+2x}{3}. \] Summing all four reported results counts the total as \(2S,\) so \[ S=\frac{29+23+21+17}{2}=45. \] The result \(29\) therefore comes from \(x=\dfrac{3(29)-45}{2}=21.\)

Thus, one original integer is \(21,\) and the correct answer is B.

39 答案 B

配方得 \[ y=\left(x+\frac p2\right)^2+q-\frac{p^2}{4}\text{。}\]它的最小值为 \(q-\frac{p^2}{4}\)。令其等于零，得 \(q=\frac{p^2}{4}\)。

因此，正确答案是 B。

English

Completing the square, \[ y=\left(x+\frac p2\right)^2+q-\frac{p^2}{4}. \] Its least value is \(q-\frac{p^2}{4}.\) Setting this equal to zero gives \(q=\frac{p^2}{4}.\)

Thus, the correct answer is B.

40 答案 A

在两个分式都有定义的地方，若它们相等，则必须有 \[ d(ax+b)=b(cx+d)\text{，}\]化简为 \(x(ad-bc)=0\)。在选项 A 中，\(a=c=1\) 给出 \(ad-bc=d-b\ne0\)，且 \(x\ne0\)，所以两分式不可能相等。选项 B 至 E 中的每一个反而都直接保证两式相等。

因此，在选项 A 的条件下，两分式不相等。

English

Where the fractions are defined, equality would require \[ d(ax+b)=b(cx+d), \] which simplifies to \(x(ad-bc)=0.\) Under choice A, \(a=c=1\) gives \(ad-bc=d-b\ne0,\) and \(x\ne0,\) so equality is impossible. Each of B through E instead forces equality directly.

Thus, the fractions are unequal under choice A.

41 答案 A

设通常速度为每小时 \(R\) 英里，全程距离为 \(D\)。行驶第一小时后，剩余路程按正常速度所需时间为 \(\frac{D}{R}-1\)。以 \(\frac{4R}{5}\) 的速度行驶会多用这段正常时间的四分之一，所以 \[ \frac12+\frac14\left(\frac DR-1\right)=2\text{。}\]若事故在多行驶 \(80\) 英里后发生，则 \[ \frac12+\frac14\left(\frac DR-1-\frac{80}{R}\right)=1\text{。}\]用第一个方程减去第二个方程，得 \(\frac{20}{R}=1\)，因此 \(R=20\) 英里/小时。

因此，正确答案是 A。

English

Let the usual rate be \(R\) mph and the total distance be \(D.\) After the first hour, the normal time for the remaining distance is \(\frac{D}{R}-1.\) Traveling it at \(\frac{4R}{5}\) adds one-fourth of that time, so \[ \frac12+\frac14\left(\frac DR-1\right)=2. \] If the accident occurs \(80\) miles later, \[ \frac12+\frac14\left(\frac DR-1-\frac{80}{R}\right)=1. \] Subtracting the second equation from the first gives \(\frac{20}{R}=1,\) hence \(R=20\) mph.

Thus, the correct answer is A.

42 答案 C

由于所有量均为正数，平方是等价变形：\[ a+\frac bc=a^2\frac bc\text{。}\]两边乘以 \(c\)，得 \(ac+b=a^2b\)，所以 \[ c=\frac{b(a^2-1)}a\text{。}\]

因此，正确答案是 C。

English

Because all quantities are positive, squaring is reversible: \[ a+\frac bc=a^2\frac bc. \] Multiplying by \(c\) gives \(ac+b=a^2b,\) so \[ c=\frac{b(a^2-1)}a. \]

Thus, the correct answer is C.

43 答案 E

代入 \(y(x+1)=1\)，得 \[ (x+1)^3=1\text{。}\]\(1\) 的三个立方根包括一个实根和两个非实根。每个根都恰好确定一个对应的 \(y=(x+1)^2\)。因此有一对实数解和两对虚数解。

因此，正确答案是 E。

English

Substitution into \(y(x+1)=1\) gives \[ (x+1)^3=1. \] The three cube roots of \(1\) consist of one real root and two nonreal roots. Each determines exactly one corresponding value \(y=(x+1)^2.\) Therefore there is one real pair and two imaginary pairs.

Thus, the correct answer is E.

44 答案 A

因为 \(OB=BC\)，所以三角形 \(OBC\) 是等腰三角形，故 \(\angle BOC=\angle BCO=y\)。因此它在 \(B\) 点处的外角为 \(\angle ABO=2y\)。又因为 \(OA=OB\)，三角形 \(AOB\) 也是等腰三角形，且 \(\angle OAB=2y\)。在三角形 \(AOC\) 中，\(A\) 点和 \(C\) 点处的角分别为 \(2y\) 和 \(y\)，所以 \(\angle AOC=180^\circ-3y\)。角 \(x=\angle AOD\) 与 \(\angle AOC\) 互为补角，因此 \(x=3y\)。

因此，正确答案是 A。

English

Because \(OB=BC,\) triangle \(OBC\) is isosceles, so \(\angle BOC=\angle BCO=y.\) Its exterior angle at \(B\) is therefore \(\angle ABO=2y.\) Since \(OA=OB,\) triangle \(AOB\) is also isosceles and \(\angle OAB=2y.\) In triangle \(AOC,\) the angles at \(A\) and \(C\) are \(2y\) and \(y,\) so \(\angle AOC=180^\circ-3y.\) Angle \(x=\angle AOD\) is supplementary to \(\angle AOC,\) hence \(x=3y.\)

Thus, the correct answer is A.

45 答案 A

前三个对应项之和给出 \[ \begin{aligned} a&=1,\\ r+d&=1,\\ r^2+2d&=2 \end{aligned}\text{。}\]代入 \(d=1-r\)，得 \(r(r-2)=0\)。由于 \(r\ne0\)，所以 \(r=2\)，且 \(d=-1\)。等比数列前十项之和为 \(2^{10}-1=1023\)，等差数列前十项之和为 \(0-1-\cdots-9=-45\)。两者之和为 \(1023-45=978\)。

因此，正确答案是 A。

English

The first three termwise sums give \[ \begin{aligned} a&=1,\\ r+d&=1,\\ r^2+2d&=2. \end{aligned} \] Substituting \(d=1-r\) yields \(r(r-2)=0.\) Since \(r\ne0,\) we have \(r=2\) and \(d=-1.\) The first ten geometric terms sum to \(2^{10}-1=1023,\) while the first ten arithmetic terms sum to \(0-1-\cdots-9=-45.\) Their combined sum is \(1023-45=978.\)

Thus, the correct answer is A.

46 答案 B

前两个方程相加得 \(6x-12=0\)，所以 \(x=2\)。代入得 \(3y=2\)，因而 \(y=\frac{2}{3}\)。此点也在后面给出的两条直线上。因此，四个图像仅有一个公共点 \((2,\frac{2}{3})\)。

因此，正确答案是 B。

English

Adding the first two equations gives \(6x-12=0,\) so \(x=2.\) Substitution gives \(3y=2,\) hence \(y=\frac{2}{3}.\) This point also lies on the last two given lines. Therefore all four graphs have the single common point \((2,\frac{2}{3}).\)

Thus, the correct answer is B.

47 答案 C

两式之差为 \[ \begin{aligned} &(a+b)(a+c)-(a+bc)\\ &\qquad=a^2+ab+ac-a\\ &\qquad=a(a+b+c-1) \end{aligned}\text{。}\]特别地，每当 \(a+b+c=1\) 时，这个差为零，两式相等。

因此，正确答案是 C。

English

The difference is \[ \begin{aligned} &(a+b)(a+c)-(a+bc)\\ &\qquad=a^2+ab+ac-a\\ &\qquad=a(a+b+c-1). \end{aligned} \] In particular, whenever \(a+b+c=1,\) this difference is zero and the expressions are equal.

Thus, the correct answer is C.

48 答案 B

设 \(A=(0,0)\)、\(B=(2,0)\)、\(C=(u,v)\)。则 \[ \begin{aligned} D&=(1,0),\\ E&=\left(\frac{u+2}{2},\frac v2\right),\\ F&=\left(\frac u2,\frac v2\right) \end{aligned}\text{。}\]由 \(\overrightarrow{FH}=\overrightarrow{AE}\)，得 \(H=(u+1,v)\)。水平直线 \(FE\) 在从 \(B\) 到 \(H\) 的中点处与 \(BH\) 相交，即 \(G=(\frac{u+3}{2},\frac{v}{2})\)。这些坐标验证了 \(AEHF\) 是平行四边形、\(\overrightarrow{BH}=\overrightarrow{DC}\)、\(FG=\frac{3AB}{4}\)，且 \(G\) 是 \(BH\) 的中点，所以 \(FG\) 是一条中线。但是 \[ \begin{aligned} HE&=\frac12\sqrt{u^2+v^2},\\ HG&=\frac12\sqrt{(u-1)^2+v^2} \end{aligned}\text{，}\]它们通常并不相等。

因此，选项 B 不一定正确。

English

Set \(A=(0,0),\) \(B=(2,0),\) \(C=(u,v).\) Then \[ \begin{aligned} D&=(1,0),\\ E&=\left(\frac{u+2}{2},\frac v2\right),\\ F&=\left(\frac u2,\frac v2\right). \end{aligned} \] Since \(\overrightarrow{FH}=\overrightarrow{AE},\) we get \(H=(u+1,v).\) The horizontal line \(FE\) meets \(BH\) halfway from \(B\) to \(H,\) at \(G=(\frac{u+3}{2},\frac{v}{2}).\) These coordinates verify that \(AEHF\) is a parallelogram, \(\overrightarrow{BH}=\overrightarrow{DC},\) \(FG=\frac{3AB}{4},\) and \(G\) is the midpoint of \(BH,\) making \(FG\) a median. But \[ \begin{aligned} HE&=\frac12\sqrt{u^2+v^2},\\ HG&=\frac12\sqrt{(u-1)^2+v^2}, \end{aligned} \] which are not generally equal.

Thus, statement B is not necessarily correct.

49 答案 C

当 \(x\ne2\) 时，\[ \frac{x^2-4}{x-2}=x+2\text{。}\]直线 \(y=x+2\) 与 \(y=2x\) 原本会在 \(x=2,\ y=4\) 处相交。但是 \(x=2\) 不在原有理函数的定义域内，所以该点是一个空点，两图像没有交点。

因此，正确答案是 C。

English

For \(x\ne2,\) \[ \frac{x^2-4}{x-2}=x+2. \] The lines \(y=x+2\) and \(y=2x\) would meet at \(x=2,\ y=4.\) But \(x=2\) is excluded from the original rational function, so that point is a hole and there is no intersection.

Thus, the correct answer is C.

50 答案 C

设 \(v\) 为 \(A\) 的新速度。相对于 \(B\)，\(A\) 以每小时 \(v-40\) 英里的速度追赶，因此超车时间与 \(\frac{30}{v-40}\) 成正比。\(A\) 与 \(C\) 以每小时 \(v+50\) 英里的相对速度缩短 \(210\) 英尺的间距，因此相遇时间与 \(\frac{210}{v+50}\) 成正比。在恰好能够安全超车的临界时刻，\[ \frac{30}{v-40}=\frac{210}{v+50}\text{。}\]所以 \(30v+1500=210v-8400\)，得 \(v=55\) 英里/小时。\(A\) 必须把原速度提高 \(5\) 英里/小时。

因此，正确答案是 C。

English

Let \(v\) be \(A\)’s new speed. Relative to \(B,\) \(A\) gains at \(v-40\) mph, so the passing time is proportional to \(\frac{30}{v-40}.\) \(A\) and \(C\) close their \(210\)-foot separation at \(v+50\) mph, so their meeting time is proportional to \(\frac{210}{v+50}.\) At the limiting safe time, \[ \frac{30}{v-40}=\frac{210}{v+50}. \] Thus \(30v+1500=210v-8400,\) giving \(v=55\) mph. \(A\) must increase his original speed by \(5\) mph.

Therefore, the correct answer is C.