1953 AMC 12 真题（P版）

1 答案 B

每个橙子的进价为 \(\frac{10}{3}\) 美分，售价为 \(4\) 美分，所以每个橙子的利润为 \(4-\frac{10}{3}=\frac{2}{3}\) 美分。要赚取 \(100\) 美分，他必须卖出 \[ \frac{100}{\frac{2}{3}}=150 \] 个橙子。

因此，正确答案是 B。

English

Each orange costs \(\frac{10}{3}\) cents and sells for \(4\) cents, so the profit is \(4-\frac{10}{3}=\frac{2}{3}\) cent per orange. To earn \(100\) cents, he must sell \[ \frac{100}{\frac{2}{3}}=150 \] oranges.

Thus, the correct answer is B.

2 答案 D

两次折扣后剩下原价的 \[ (0.80)(0.85)=0.68\text{。}\]因此，售价为 \(\$250.00\) 的 \(68\%\)，即 \(\$170.00\)。

因此，正确答案是 D。

English

The two discounts leave \[ (0.80)(0.85)=0.68 \] of the original price. The sale price is therefore \(68\%\) of \(\$250.00,\) or \(\$170.00.\)

Thus, the correct answer is D.

3 答案 D

因为 \(i^2=-1\)，所以 \[ \begin{aligned} (x+iy)(x-iy)&=x^2-(iy)^2\\ &=x^2+y^2 \end{aligned}\text{。}\]

因此，正确答案是 D。

English

Because \(i^2=-1,\) \[ \begin{aligned} (x+iy)(x-iy)&=x^2-(iy)^2\\ &=x^2+y^2. \end{aligned} \]

Thus, the correct answer is D.

4 答案 D

原方程因式分解为 \[ x(x+4)^2(4-x)=0\text{。}\]按重数计，根为 \(0\)、\(4\)、\(-4\) 和 \(-4\)。

因此，正确答案是 D。

English

The equation factors as \[ x(x+4)^2(4-x)=0. \] Its roots, counted with multiplicity, are \(0,\) \(4,\) \(-4,\) and \(-4.\)

Thus, the correct answer is D.

5 答案 C

由方程可得 \[ x=6^{2.5}=6^2\sqrt6=36\sqrt6\text{。}\]

因此，正确答案是 C。

English

The equation gives \[ x=6^{2.5}=6^2\sqrt6=36\sqrt6. \]

Thus, the correct answer is C.

6 答案 A

二十五美分硬币的枚数之差为 \[ (5q+1)-(q+5)=4q-4\text{。}\]每枚二十五美分硬币折合 \(\frac{5}{2}\) 枚十美分硬币，所以按此换算，得到 \[ \frac52(4q-4)=10(q-1)\text{。}\]

因此，正确答案是 A。

English

The difference is \[ (5q+1)-(q+5)=4q-4 \] quarters. Multiplying by \(\frac{5}{2}\) dimes per quarter gives \[ \frac52(4q-4)=10(q-1). \]

Thus, the correct answer is A.

7 答案 D

原分式的分子为 \[ \begin{aligned} &\frac{a^2+x^2-(x^2-a^2)} {\sqrt{a^2+x^2}}\\ &\qquad{}=\frac{2a^2}{\sqrt{a^2+x^2}} \end{aligned}\text{。}\]再除以 \(a^2+x^2\)，得到 \[ \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}\text{。}\]

因此，正确答案是 D。

English

The numerator is \[ \begin{aligned} &\frac{a^2+x^2-(x^2-a^2)} {\sqrt{a^2+x^2}}\\ &\qquad{}=\frac{2a^2}{\sqrt{a^2+x^2}}. \end{aligned} \] Dividing by \(a^2+x^2\) gives \[ \frac{2a^2}{(a^2+x^2)^{\frac{3}{2}}}. \]

Thus, the correct answer is D.

8 答案 C

代入后得到 \[ (2-x)(x^2+4)=8\text{。}\]展开并化简，得到 \[ x(x^2-2x+4)=0\text{。}\]二次因式的判别式为负，因此实数交点满足 \(x=0\)。

因此，正确答案是 C。

English

Substitution gives \[ (2-x)(x^2+4)=8. \] Expanding and simplifying, \[ x(x^2-2x+4)=0. \] The quadratic factor has negative discriminant, so the real intersection has \(x=0.\)

Thus, the correct answer is C.

9 答案 D

剃须水中原有 \(9(0.50)=4.5\) 盎司酒精。若加入 \(w\) 盎司水，则 \[ \frac{4.5}{9+w}=0.30\text{。}\]所以 \(9+w=15\)，且 \(w=6\)。

因此，正确答案是 D。

English

The lotion initially contains \(9(0.50)=4.5\) ounces of alcohol. If \(w\) ounces of water are added, then \[ \frac{4.5}{9+w}=0.30. \] Hence \(9+w=15\) and \(w=6.\)

Thus, the correct answer is D.

10 答案 C

车轮的周长为 \(6\pi\) 英尺，所以所需圈数为 \[ \frac{5280}{6\pi}=\frac{880}{\pi}\text{。}\]

因此，正确答案是 C。

English

The wheel’s circumference is \(6\pi\) feet, so the required number of revolutions is \[ \frac{5280}{6\pi}=\frac{880}{\pi}. \]

Thus, the correct answer is C.

11 答案 C

若内圆半径为 \(r\)，则周长之差为 \[ 2\pi(r+10)-2\pi r=20\pi\approx62.8 \] 英尺，约为 \(60\) 英尺。

因此，正确答案是 C。

English

If the inner radius is \(r,\) the difference is \[ 2\pi(r+10)-2\pi r=20\pi\approx62.8 \] feet, which is about \(60\) feet.

Thus, the correct answer is C.

12 答案 B

面积之比等于直径之比的平方：\[ \left(\frac8{12}\right)^2=\left(\frac23\right)^2=\frac49\text{。}\]

因此，正确答案是 B。

English

The area ratio is the square of the diameter ratio: \[ \left(\frac8{12}\right)^2=\left(\frac23\right)^2=\frac49. \]

Thus, the correct answer is B.

13 答案 B

若公共的高为 \(h\)，则三角形的面积为 \[ \frac12(18)h=9h\text{。}\]梯形的面积等于其中位线乘以高，所以其中位线长必为 \(9\) 英寸。

因此，正确答案是 B。

English

If the common altitude is \(h,\) the triangle has area \[ \frac12(18)h=9h. \] A trapezoid’s area is its median times its altitude, so its median must be \(9\) inches.

Thus, the correct answer is B.

14 答案 E

两圆内切时有 \(PQ=p-q\)，外切时有 \(PQ=p+q\)。两圆相离时可以有 \(PQ\gt p+q\)，而许多相交或相离的情形满足 \(PQ\gt p-q\)。因此 A 至 D 中的每种情况都可能发生，没有一个说法是错误的。

因此，正确答案是 E。

English

Internal tangency realizes \(PQ=p-q,\) and external tangency realizes \(PQ=p+q.\) Disjoint circles can have \(PQ\gt p+q,\) and many intersecting or disjoint configurations have \(PQ\gt p-q.\) Thus each of A through D can occur, so none of them is false.

Therefore, the correct answer is E.

15 答案 B

设原正方形的边长为 \(s\)。其内切圆的直径为 \(s\)。这个直径就是从圆中切出的最大正方形的对角线，因此最后得到的正方形边长为 \(\frac{s}{\sqrt2}\)，面积为 \(\frac{s^2}{2}\)。两次切割后留下的是这个正方形，所以浪费的总面积为 \(s^2-\frac{s^2}{2}=\frac{s^2}{2}\)。

因此，正确答案是 B。

English

Let the original square have side \(s.\) The inscribed circle has diameter \(s.\) That diameter is the diagonal of the largest square cut from the circle, so the final square has side \(\frac{s}{\sqrt2}\) and area \(\frac{s^2}{2}.\) The material left after both cuts is this final square, so the total waste is \(s^2-\frac{s^2}{2}=\frac{s^2}{2}.\)

Thus, the correct answer is B.

16 答案 D

利润和费用合计占销售额的 \(10\%+15\%=25\%\)，所以成本是售价的 \(75\%\)。加价额与成本之比为 \[ \frac{25\%}{75\%}=\frac13=33\frac13\%\text{。}\]

因此，正确答案是 D。

English

Profit and expenses are \(10\%+15\%=25\%\) of sales, so the cost is \(75\%\) of the selling price. The markup as a fraction of cost is \[ \frac{25\%}{75\%}=\frac13=33\frac13\%. \]

Thus, the correct answer is D.

17 答案 B

设按 \(4\%\) 投资的金额为 \(x\) 美元。由两笔收益相等可得 \[ 0.04x=0.06(4500-x)\text{，}\]所以 \(x=2700\)。每笔投资的收益都是 \(\$108\)，总收益为 \(\$216\)。平均利率为 \[ \frac{216}{4500}=0.048=4.8\%\text{。}\]

因此，正确答案是 B。

English

Let \(x\) dollars be invested at \(4\%.\) Equal returns give \[ 0.04x=0.06(4500-x), \] so \(x=2700.\) Each investment returns \(\$108,\) for a total of \(\$216.\) The average rate is \[ \frac{216}{4500}=0.048=4.8\%. \]

Thus, the correct answer is B.

18 答案 C

利用平方差公式， \[ \begin{gathered} x^4+4=(x^2+2)^2-(2x)^2\\ =(x^2-2x+2)(x^2+2x+2) \end{gathered}\text{。}\]

因此，正确答案是 C。

English

Using a difference of squares, \[ \begin{gathered} x^4+4=(x^2+2)^2-(2x)^2\\ =(x^2-2x+2)(x^2+2x+2). \end{gathered} \]

Thus, the correct answer is C.

19 答案 C

新的值为 \[ \left(\frac34x\right)\left(\frac34y\right)^2 =\frac{27}{64}xy^2\text{。}\]因此减少的量是原值的 \(1-\frac{27}{64}=\frac{37}{64}\)。

因此，正确答案是 C。

English

The new value is \[ \left(\frac34x\right)\left(\frac34y\right)^2 =\frac{27}{64}xy^2. \] Therefore the decrease is \(1-\frac{27}{64}=\frac{37}{64}\) of the original value.

Thus, the correct answer is C.

20 答案 D

因为 \(x\ne0\)，可将表达式写成 \[ \begin{aligned} x^2\Bigg[& \left(x^2+\frac1{x^2}\right)\\ &+\left(x+\frac1x\right)-4 \Bigg] \end{aligned}\text{。}\]由于 \(x^2+x^{-2}=y^2-2\)，可化为 \[ x^2(y^2+y-6)=0\text{。}\]

因此，正确答案是 D。

English

Because \(x\ne0,\) factor the expression as \[ \begin{aligned} x^2\Bigg[& \left(x^2+\frac1{x^2}\right)\\ &+\left(x+\frac1x\right)-4 \Bigg]. \end{aligned} \] Since \(x^2+x^{-2}=y^2-2,\) this becomes \[ x^2(y^2+y-6)=0. \]

Thus, the correct answer is D.

21 答案 D

该对数方程等价于 \[ x^2-3x+6=10\text{，}\]所以 \[ \begin{aligned} x^2-3x-4&=(x-4)(x+1)\\ &=0 \end{aligned}\text{。}\]因此 \(x=4\) 或 \(x=-1\)。

正确答案是 D。

English

The logarithmic equation is equivalent to \[ x^2-3x+6=10, \] so \[ \begin{aligned} x^2-3x-4&=(x-4)(x+1)\\ &=0. \end{aligned} \] Thus \(x=4\) or \(x=-1.\)

The correct answer is D.

22 答案 B

题目中的表达式是两个独立根式的乘积：\[ 27\sqrt[4]{9}\sqrt[3]{9} =3^{3+\frac{1}{2}+\frac{2}{3}}=3^{\frac{25}{6}}\text{。}\]因此它以 \(3\) 为底的对数是 \(\frac{25}{6}=4\dfrac16\)。

因此，正确答案是 B。

English

The printed expression is a product of two separate radicals: \[ 27\sqrt[4]{9}\sqrt[3]{9} =3^{3+\frac{1}{2}+\frac{2}{3}}=3^{\frac{25}{6}}. \] Its base-\(3\) logarithm is \(\frac{25}{6}=4\dfrac16.\)

Thus, the correct answer is B.

23 答案 B

两边乘以 \(\sqrt{x+10}\)，得到 \[ x+4=5\sqrt{x+10}\text{。}\]两边平方后得到 \[ \begin{gathered} x^2-17x-234\\ =(x-26)(x+9)\\ =0 \end{gathered}\text{。}\]\(x=26\) 满足原方程，而 \(x=-9\) 会使原方程左边等于 \(1-6=-5\)，而非 \(5\)。因此，介于 \(-10\) 与 \(-6\) 之间的 \(-9\) 是增根。

正确答案是 B。

English

Multiplying by \(\sqrt{x+10}\) gives \[ x+4=5\sqrt{x+10}. \] Squaring produces \[ \begin{gathered} x^2-17x-234\\ =(x-26)(x+9)\\ =0. \end{gathered} \] The value \(x=26\) satisfies the original equation, while \(x=-9\) makes its left side \(1-6=-5,\) not \(5.\) Thus \(-9,\) which lies between \(-10\) and \(-6,\) is extraneous.

The correct answer is B.

24 答案 A

展开后，从等式两边消去 \(100a^2+bc\)，得到 \[ 10a(b+c)=100a\text{。}\]因为 \(a\) 是正数，所以这等价于 \(b+c=10\)。

因此，正确答案是 A。

English

Expanding and canceling \(100a^2+bc\) from both sides leaves \[ 10a(b+c)=100a. \] Since \(a\) is positive, this is equivalent to \(b+c=10.\)

Thus, the correct answer is A.

25 答案 C

若某一项为 \(t\)，则后面两项为 \(tr\) 和 \(tr^2\)。因此 \[ t=tr+tr^2 \] 且 \(r^2+r-1=0\)。因为各项均为正数，所以 \[ r=\frac{\sqrt5-1}{2}\text{。}\]

因此，正确答案是 C。

English

If a term is \(t,\) the next two are \(tr\) and \(tr^2.\) Thus \[ t=tr+tr^2 \] and \(r^2+r-1=0.\) Because the terms are positive, \[ r=\frac{\sqrt5-1}{2}. \]

Thus, the correct answer is C.

26 答案 A

较靠近底边的平行线段在其上方围出一个较小三角形，其面积是整个三角形的 \(\frac{2}{3}\)。若该线段长为 \(x\)，由相似关系可得 \[ \frac{x^2}{15^2}=\frac23\text{。}\]因此 \(x^2=150\)，且 \(x=5\sqrt6\)。

因此，正确答案是 A。

English

The parallel closer to the base bounds a smaller triangle above it whose area is \(\frac{2}{3}\) of the whole triangle. If its length is \(x,\) similarity gives \[ \frac{x^2}{15^2}=\frac23. \] Hence \(x^2=150\) and \(x=5\sqrt6.\)

Thus, the correct answer is A.

27 答案 D

各圆面积依次为 \[ \pi,\ \frac{\pi}{4},\ \frac{\pi}{16},\ldots \] 因此它们的和为 \[ \frac{\pi}{1-\frac{1}{4}}=\frac{4\pi}{3}\text{。}\]

因此，正确答案是 D。

English

The areas are \[ \pi,\ \frac{\pi}{4},\ \frac{\pi}{16},\ldots \] Therefore their sum is \[ \frac{\pi}{1-\frac{1}{4}}=\frac{4\pi}{3}. \]

Thus, the correct answer is D.

28 答案 D

由角平分线定理可得 \[ \frac{y}{x}=\frac{c}{b}\text{，}\]即 \(\frac{y}{c}=\frac{x}{b}\)。因为 \(x+y=a\)，所以 \[ \frac yc=\frac xb=\frac{x+y}{b+c}=\frac a{b+c}\text{。}\]

因此，正确答案是 D。

English

The angle bisector theorem gives \[ \frac{y}{x}=\frac{c}{b}, \] or \(\frac{y}{c}=\frac{x}{b}.\) Since \(x+y=a,\) \[ \frac yc=\frac xb=\frac{x+y}{b+c}=\frac a{b+c}. \]

Thus, the correct answer is D.

29 答案 D

给出的面积表示真实面积满足 \[ 1.10245\le A\lt1.10255\text{。}\]两边开平方，近似得到 \[ 1.049976\le\sqrt A\lt1.050024\text{。}\]因此，任何可能的边长精确到万分之一时都舍入为 \(1.0500\)。小数点后的末尾零是有效数字，所以 \(1.0500\) 有五位有效数字。

因此，正确答案是 D。

English

The reported area means the true area lies in \[ 1.10245\le A\lt1.10255. \] Taking square roots gives approximately \[ 1.049976\le\sqrt A\lt1.050024. \] Every possible side length therefore rounds to \(1.0500\) to the nearest ten-thousandth. The trailing zeros after the decimal are significant, so \(1.0500\) has five significant digits.

Thus, the correct answer is D.

30 答案 D

\(B\) 先生先支付 \[ 9000(0.90)=8100 \] 美元，然后以 \(8100(1.10)=8910\) 美元的价格把房屋卖回。\(A\) 先生收到 \(\$8100\)，又支付 \(\$8910\)，所以亏损 \(\$810\)。

因此，正确答案是 D。

English

Mr. \(B\) first pays \[ 9000(0.90)=8100 \] dollars. He then sells the house back for \(8100(1.10)=8910\) dollars. Mr. \(A\) receives \(\$8100\) and pays \(\$8910,\) so he loses \(\$810.\)

Thus, the correct answer is D.

31 答案 A

\(v\) 英里每小时等于 \(\frac{22v}{15}\) 英尺每秒。每次咔嗒声代表行驶 \(30\) 英尺，所以每秒的咔嗒声次数为 \[ \frac{\frac{22v}{15}}{30}=\frac{11v}{225}\text{。}\]在 \(\frac{225}{11}\approx20.45\) 秒内，咔嗒声次数为 \(v\)。最接近的选项是 \(20\) 秒。

因此，正确答案是 A。

English

A speed of \(v\) miles per hour is \(\frac{22v}{15}\) feet per second. Since each click represents \(30\) feet, the click rate is \[ \frac{\frac{22v}{15}}{30}=\frac{11v}{225} \] clicks per second. In \(\frac{225}{11}\approx20.45\) seconds, the number of clicks is \(v.\) The closest listed interval is \(20\) seconds.

Thus, the correct answer is A.

32 答案 D

对于矩形的每一条边，与它相邻的两条三等分线在该边的垂直平分线上相交。这样得到的相对交点关于矩形中心对称，因此所得四边形的两条对角线互相平分。一条对角线位于水平对称轴上，另一条位于竖直对称轴上，所以它们互相垂直。

对角线互相平分的四边形是平行四边形；若其对角线还互相垂直，则四条边相等。因此这个四边形是菱形。由于两条对角线不一定等长，它不一定是正方形。

因此，正确答案是 D。

English

For each side, the two trisectors adjacent to it meet on that side’s perpendicular bisector. Opposite such intersection points are reflections across the center of the rectangle, so the diagonals of the resulting quadrilateral bisect each other. One diagonal lies on the horizontal symmetry axis and the other on the vertical symmetry axis, so they are perpendicular.

A quadrilateral whose diagonals bisect each other is a parallelogram; if those diagonals are perpendicular, its four sides are equal. Hence the quadrilateral is a rhombus. It need not be a square because the two diagonals need not have equal length.

Thus, the correct answer is D.

33 答案 C

若两条直角边的长度均为 \(s\)，则 \[ s(2+\sqrt2)=2p\text{，}\]所以 \(s=p(2-\sqrt2)\)。因此面积为 \[ \begin{gathered} \frac{s^2}{2} =\frac{p^2(2-\sqrt2)^2}{2}\\ =(3-2\sqrt2)p^2 \end{gathered}\text{。}\]

因此，正确答案是 C。

English

If each leg is \(s,\) then \[ s(2+\sqrt2)=2p, \] so \(s=p(2-\sqrt2).\) Therefore the area is \[ \begin{gathered} \frac{s^2}{2} =\frac{p^2(2-\sqrt2)^2}{2}\\ =(3-2\sqrt2)p^2. \end{gathered} \]

Thus, the correct answer is C.

34 答案 C

由正弦定理的推广形式，外接圆直径为 \[ d=\frac{12}{\sin30^\circ}=\frac{12}{\frac{1}{2}}=24 \] 英寸。

因此，正确答案是 C。

English

By the extended law of sines, the circumdiameter is \[ d=\frac{12}{\sin30^\circ}=\frac{12}{\frac{1}{2}}=24 \] inches.

Thus, the correct answer is C.

35 答案 E

直接代入可得 \[ f(x+2)=\frac{(x+2)(x+1)}2\text{。}\]又因为 \(f(x+1)=\frac{x(x+1)}{2}\)，所以 \[ \begin{gathered} \frac{(x+2)f(x+1)}x\\ =\frac{(x+2)(x+1)}2 \end{gathered}\text{。}\]

因此，正确答案是 E。

English

Directly, \[ f(x+2)=\frac{(x+2)(x+1)}2. \] Also \(f(x+1)=\frac{x(x+1)}{2},\) so \[ \begin{gathered} \frac{(x+2)f(x+1)}x\\ =\frac{(x+2)(x+1)}2. \end{gathered} \]

Thus, the correct answer is E.

36 答案 C

要能被 \(x-3\) 整除，必须满足 \[ 4(3)^2-6(3)+m=0\text{，}\]所以 \(18+m=0\)，且 \(m=-18\)。在所列各数中，只有 \(36\) 是 \(-18\) 的整数倍。

因此，正确答案是 C。

English

Divisibility by \(x-3\) requires \[ 4(3)^2-6(3)+m=0, \] so \(18+m=0\) and \(m=-18.\) Of the listed numbers, only \(36\) is an exact multiple of \(-18.\)

Thus, the correct answer is C.

37 答案 E

高为 \[ \sqrt{12^2-3^2}=3\sqrt{15}\text{，}\]所以面积为 \(K=\tfrac12(6)(3\sqrt{15})=9\sqrt{15}\)。因此外接圆半径为 \[ R=\frac{(12)(12)(6)}{4(9\sqrt{15})} =\frac{8\sqrt{15}}5\text{。}\]这个值不在选项中。

因此，正确答案是 E。

English

The altitude is \[ \sqrt{12^2-3^2}=3\sqrt{15}, \] so the area is \(K=\tfrac12(6)(3\sqrt{15})=9\sqrt{15}.\) Hence the circumradius is \[ R=\frac{(12)(12)(6)}{4(9\sqrt{15})} =\frac{8\sqrt{15}}5. \] This value is not listed.

Thus, the correct answer is E.

38 答案 C

首先，\(f(4)=4-2=2\)。因此 \[ \begin{aligned} F[3,f(4)]&=F(3,2)\\ &=2^2+3=7 \end{aligned}\text{。}\]

因此，正确答案是 C。

English

First \(f(4)=4-2=2.\) Therefore \[ \begin{aligned} F[3,f(4)]&=F(3,2)\\ &=2^2+3=7. \end{aligned} \]

Thus, the correct answer is C.

39 答案 A

当底数和真数均满足定义条件时， \[ \begin{aligned} \log_a b\cdot\log_b a &=\frac{\log b}{\log a}\\ &\quad{}\cdot\frac{\log a}{\log b}\\ &=1 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

For permissible bases and arguments, \[ \begin{aligned} \log_a b\cdot\log_b a &=\frac{\log b}{\log a}\\ &\quad{}\cdot\frac{\log a}{\log b}\\ &=1. \end{aligned} \]

Thus, the correct answer is A.

40 答案 C

“每个人都诚实”的否定是“存在一个不诚实的人”。用选项中的说法，就是有些人不诚实。

因此，正确答案是 C。

English

The negation of “every man is honest” is “there exists a man who is not honest.” In the language of the choices, some men are dishonest.

Thus, the correct answer is C.

41 答案 E

将女生营地到道路的垂足设为 \((0,0)\)。两个营地可分别设为 \(G=(0,300)\) 和 \(B=(400,0)\)。若食堂为 \(C=(t,0)\)，由到两营地距离相等可得 \[ t^2+300^2=(400-t)^2\text{，}\]所以 \(t=87.5\)。公共距离为 \[ BC=400-87.5=312.5 \] 杆，不在选项中。

因此，正确答案是 E。

English

Let the foot of the perpendicular from the girls’ camp be \((0,0).\) The camps can be placed at \(G=(0,300)\) and \(B=(400,0).\) If the canteen is \(C=(t,0),\) equidistance gives \[ t^2+300^2=(400-t)^2, \] so \(t=87.5.\) The common distance is \[ BC=400-87.5=312.5 \] rods, which is not listed.

Thus, the correct answer is E.

42 答案 E

两圆心的连线、公切线段以及长度为 \(4+5=9\) 的垂直线段构成直角三角形。因此公切线段长为 \[ \sqrt{41^2-9^2}=\sqrt{1600}=40 \] 英寸。

因此，正确答案是 E。

English

The center segment, the tangent segment, and a perpendicular leg of length \(4+5=9\) form a right triangle. Thus the tangent length is \[ \sqrt{41^2-9^2}=\sqrt{1600}=40 \] inches.

Thus, the correct answer is E.

43 答案 C

将百分率写成相对于原来数量的比例，收入相同要求 \[ (1+p)(1-d)=1\text{。}\]因此 \(p-d-pd=0\)，所以 \[ d=\frac p{1+p}\text{。}\]

因此，正确答案是 C。

English

Writing the percentage rates as fractions of the original quantities, equal income requires \[ (1+p)(1-d)=1. \] Hence \(p-d-pd=0,\) so \[ d=\frac p{1+p}. \]

Thus, the correct answer is C.

44 答案 A

第一名学生只改错了常数项，所以正确的 \(x\) 系数应与根为 \(8\) 和 \(2\) 的首一二次方程相同，即 \(-(8+2)=-10\)。第二名学生只改错了这个一次项系数，所以正确的常数项为 \((-9)(-1)=9\)。因此正确方程是 \[ x^2-10x+9=0\text{。}\]

因此，正确答案是 A。

English

The first student changed only the constant term, so the correct coefficient of \(x\) is the one in the monic quadratic with roots \(8\) and \(2:\) it is \(-(8+2)=-10.\) The second student changed only that linear coefficient, so the correct constant is \((-9)(-1)=9.\) Therefore the correct equation is \[ x^2-10x+9=0. \]

Thus, the correct answer is A.

45 答案 E

因为 \[ (a-b)^2\ge0\text{，}\]所以 \((a+b)^2\ge4ab\)。两边都非负，因此 \[ \frac{a+b}{2}\ge\sqrt{ab}\text{。}\]当 \(a=b\) 时等号成立。

因此，正确答案是 E。

English

Since \[ (a-b)^2\ge0, \] we have \((a+b)^2\ge4ab.\) Both sides are nonnegative, so \[ \frac{a+b}{2}\ge\sqrt{ab}. \] Equality occurs when \(a=b.\)

Thus, the correct answer is E.

46 答案 D

由节省路程的条件可得 \[ \sqrt{L^2+W^2}=\frac L2+W\text{。}\]两边平方并消去 \(W^2\)，得到 \[ L^2=\frac{L^2}{4}+LW\text{，}\]所以 \(\frac{W}{L}=\frac{3}{4}\)。

因此，正确答案是 D。

English

The saving condition gives \[ \sqrt{L^2+W^2}=\frac L2+W. \] Squaring and canceling \(W^2\) yields \[ L^2=\frac{L^2}{4}+LW, \] so \(\frac{W}{L}=\frac{3}{4}.\)

Thus, the correct answer is D.

47 答案 D

当 \(x\gt0\) 时，由标准指数不等式可得 \(1+x\lt e^x\)。两边取自然对数，得到 \[ \ln(1+x)\lt x\text{。}\]若 \(\log\) 表示常用对数，则 \[ \begin{aligned} \log(1+x)&=\frac{\ln(1+x)}{\ln10}\\ &\lt\ln(1+x) \end{aligned}\text{，}\]因而选项中的同一不等式成立。

因此，正确答案是 D。

English

For \(x\gt0,\) the standard exponential inequality gives \(1+x\lt e^x.\) Taking natural logarithms yields \[ \ln(1+x)\lt x. \] If \(\log\) denotes the common logarithm, then \[ \begin{aligned} \log(1+x)&=\frac{\ln(1+x)}{\ln10}\\ &\lt\ln(1+x), \end{aligned} \] so the same listed inequality holds.

Thus, the correct answer is D.

48 答案 D

设长底为 \(1\)，短底和高均为 \(r\)。一条对角线的水平投影为 \(\frac{1+r}{2}\)，长度为 \(1\)。因此 \[ 1=r^2+\left(\frac{1+r}{2}\right)^2\text{。}\]化简得 \[ \begin{aligned} 5r^2+2r-3&=(5r-3)(r+1)\\ &=0 \end{aligned}\text{。}\]正的比值为 \(r=\frac{3}{5}\)。

因此，正确答案是 D。

English

Let the larger base be \(1,\) and let the smaller base and altitude both be \(r.\) A diagonal has horizontal projection \(\frac{1+r}{2}\) and length \(1.\) Thus \[ 1=r^2+\left(\frac{1+r}{2}\right)^2. \] This simplifies to \[ \begin{aligned} 5r^2+2r-3&=(5r-3)(r+1)\\ &=0. \end{aligned} \] The positive ratio is \(r=\frac{3}{5}.\)

Thus, the correct answer is D.

49 答案 E

将 \(B=(2,1)\) 关于 \(y\) 轴反射到 \(B'=(-2,1)\)。对于位于 \(y\) 轴上的 \(C\)，有 \(BC=B'C\)，所以当 \(A,C,B'\) 共线时，\(AC+BC\) 最小。从 \(A=(5,5)\) 到 \(B'=(-2,1)\) 的直线斜率为 \(\frac{4}{7}\)。当 \(x=0\) 时，其纵坐标为 \[ 5-\frac47(5)=\frac{15}{7}=2\frac17\text{。}\]因此 \(k=\frac{15}{7}\)。

因此，正确答案是 E。

English

Reflect \(B=(2,1)\) across the \(y\)-axis to \(B'=(-2,1).\) For \(C\) on the \(y\)-axis, \(BC=B'C,\) so \(AC+BC\) is minimized when \(A,C,B'\) are collinear. The line from \(A=(5,5)\) to \(B'=(-2,1)\) has slope \(\frac{4}{7}.\) At \(x=0,\) its height is \[ 5-\frac47(5)=\frac{15}{7}=2\frac17. \] Hence \(k=\frac{15}{7}.\)

Thus, the correct answer is E.

50 答案 B

设从第三个顶点引出的两条切线段长均为 \(z\)。由同一点引出的切线段相等，所以三边长为 \[ 14,\qquad 6+z,\qquad 8+z\text{，}\]半周长为 \(s=14+z\)。因为内切圆半径为 \(4\)， \[ K=rs=4(14+z)\text{。}\]由海伦公式， \[ \begin{aligned} K^2 &=(14+z)z\cdot8\cdot6\\ &=48z(14+z) \end{aligned}\text{。}\]将两种面积表达式平方后相等，得到 \(16(14+z)=48z\)，所以 \(z=7\)。三边长为 \(14\)、\(13\) 和 \(15\)，最短边长为 \(13\)。

因此，正确答案是 B。

English

Let the two tangent segments from the third vertex each have length \(z.\) Equal tangents from a common vertex make the side lengths \[ 14,\qquad 6+z,\qquad 8+z, \] with semiperimeter \(s=14+z.\) Since the inradius is \(4,\) \[ K=rs=4(14+z). \] Heron’s formula gives \[ \begin{aligned} K^2 &=(14+z)z\cdot8\cdot6\\ &=48z(14+z). \end{aligned} \] Equating the squares yields \(16(14+z)=48z,\) so \(z=7.\) The sides are \(14,\) \(13,\) and \(15,\) and the shortest is \(13.\)

Thus, the correct answer is B.