1952 AMC 12 真题（P版）

1 答案 B

若正有理数半径为 \(r\)，则圆的面积为 \(\pi r^2\)。因数 \(r^2\) 是非零有理数，而无理数 \(\pi\) 的非零有理数倍仍是无理数。

因此，正确答案是 B。

English

If the positive rational radius is \(r,\) then its area is \(\pi r^2.\) The factor \(r^2\) is a nonzero rational number, and a nonzero rational multiple of the irrational number \(\pi\) is irrational.

Thus, the correct answer is B.

2 答案 B

两个班的成绩总和为 \[ 20(80)+30(70)=3700 \] 个百分点，共有 \(50\) 名学生。因此，总平均成绩为 \(\frac{3700}{50}=74\%\)。

因此，正确答案是 B。

English

The two classes earned a combined total of \[ 20(80)+30(70)=3700 \] percentage points among \(50\) students. Their combined average is \(\frac{3700}{50}=74\%.\)

Thus, the correct answer is B.

3 答案 A

在 \(u=a\) 和 \(v=\frac{1}{a}\) 时使用立方差公式。因为 \(uv=1\)，所以 \[ \begin{aligned} a^3-\frac1{a^3} &=\left(a-\frac1a\right)\\ &\quad{}\cdot\left(a^2+1+\frac1{a^2}\right) \end{aligned}\text{。}\]

因此，正确答案是 A。

English

Apply the difference-of-cubes identity with \(u=a\) and \(v=\frac{1}{a}.\) Since \(uv=1,\) \[ \begin{aligned} a^3-\frac1{a^3} &=\left(a-\frac1a\right)\\ &\quad{}\cdot\left(a^2+1+\frac1{a^2}\right). \end{aligned} \]

Thus, the correct answer is A.

4 答案 C

第一磅收费 \(10\) 美分。其余 \(P-1\) 磅收费 \(3(P-1)\) 美分，所以 \[ C=10+3(P-1)\text{。}\]

因此，正确答案是 C。

English

The first pound costs \(10\) cents. The remaining \(P-1\) pounds cost \(3(P-1)\) cents, so \[ C=10+3(P-1). \]

Thus, the correct answer is C.

5 答案 A

斜率为 \[ \frac{12-(-6)}{6-0}=3\text{，}\]所以直线方程为 \(y=3x-6\)。代入 \(x=3\)，得到 \(y=3\)。

因此，正确答案是 A。

English

The slope is \[ \frac{12-(-6)}{6-0}=3, \] so the line is \(y=3x-6.\) Substituting \(x=3\) gives \(y=3.\)

Thus, the correct answer is A.

6 答案 E

两根为 \(\dfrac{7+\sqrt{49+36}}2\) 和 \(\dfrac{7-\sqrt{49+36}}2\)。因此，它们的正差为 \(\sqrt{85}\)。

因此，正确答案是 E。

English

The roots are \(\dfrac{7+\sqrt{49+36}}2\) and \(\dfrac{7-\sqrt{49+36}}2.\) Their positive difference is therefore \(\sqrt{85}.\)

Thus, the correct answer is E.

7 答案 B

通分后再取倒数，得到 \[ \begin{aligned} \left(\frac1x+\frac1y\right)^{-1} &=\left(\frac{x+y}{xy}\right)^{-1}\\ &=\frac{xy}{x+y} \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Combining the terms and then taking the reciprocal gives \[ \begin{aligned} \left(\frac1x+\frac1y\right)^{-1} &=\left(\frac{x+y}{xy}\right)^{-1}\\ &=\frac{xy}{x+y}. \end{aligned} \]

Thus, the correct answer is B.

8 答案 A

两个不同的等圆相交时有 \(2\) 条公切线，外切时有 \(3\) 条，相离时有 \(4\) 条。它们不能内切，因为内切要求圆心距等于两半径之差，即 \(0\)。圆心距为 \(0\) 时，两个等圆重合，并有无穷多条公切线。

因此，恰有 \(1\) 条公切线是不可能的，所以正确答案是 A。

English

Two distinct equal circles have \(2\) common tangents when they intersect, \(3\) when externally tangent, and \(4\) when disjoint. They cannot be internally tangent, because internal tangency requires the center distance to equal the difference of the radii, which is \(0.\) At center distance \(0\) the equal circles coincide and have infinitely many common tangents.

Thus exactly \(1\) common tangent is impossible, so the correct answer is A.

9 答案 D

消去分母，得到 \[ m(a-b)=cab\text{。}\]因此 \(ma=b(m+ca)\)，所以 \[ b=\frac{ma}{m+ca}\text{。}\]

因此，正确答案是 D。

English

Clearing the denominator gives \[ m(a-b)=cab. \] Hence \(ma=b(m+ca)\), so \[ b=\frac{ma}{m+ca}. \]

Thus, the correct answer is D.

10 答案 B

若单程为一英里，则总时间为 \[ \frac1{10}+\frac1{20}=\frac3{20} \] 小时。因此，平均速度为 \[ \frac{2}{\frac{3}{20}}=\frac{40}{3} =13\frac13\text{ 英里/小时}\text{。}\]

因此，正确答案是 B。

English

For one mile in each direction, the total time is \[ \frac1{10}+\frac1{20}=\frac3{20} \] hour. Thus the average speed is \[ \frac{2}{\frac{3}{20}}=\frac{40}{3} =13\frac13\text{ mph}. \]

Thus, the correct answer is B.

11 答案 C

该函数在 \(x=1\) 处无定义，因为此时分母为零，所以 \(f(1)=0\) 是错误的。其他说法均成立：\(f(0)=-2\)、\(f(-2)=0\)；解方程 \(y=\frac{x+2}{x-1}\) 得 \(x=\frac{y+2}{y-1}\)，同一公式也说明 \(f\) 是自身的反函数。

因此，正确答案是 C。

English

The function is undefined at \(x=1\) because its denominator is zero there, so \(f(1)=0\) is false. The other statements hold: \(f(0)=-2,\) \(f(-2)=0,\) solving \(y=\frac{x+2}{x-1}\) gives \(x=\frac{y+2}{y-1},\) and this same formula shows that \(f\) is its own inverse.

Thus, the correct answer is C.

12 答案 E

设首项为 \(a\)，公比为 \(r\)。由无穷和可得 \(a=6(1-r)\)。因此，前两项满足 \[ a(1+r)=6(1-r^2)=\frac92\text{。}\]所以 \(r^2=\frac{1}{4}\)：当 \(r=\frac{1}{2}\) 时 \(a=3\)，当 \(r=-\frac{1}{2}\) 时 \(a=9\)。

因此，正确答案是 E。

English

Let the first term be \(a\) and the common ratio be \(r.\) The infinite sum gives \(a=6(1-r).\) The first two terms therefore satisfy \[ a(1+r)=6(1-r^2)=\frac92. \] Thus \(r^2=\frac{1}{4},\) so \(r=\frac{1}{2}\) gives \(a=3,\) while \(r=-\frac{1}{2}\) gives \(a=9.\)

Thus, the correct answer is E.

13 答案 E

配方得到 \[ \begin{aligned} x^2+px+q &=\left(x+\frac p2\right)^2\\ &\quad{}+q-\frac{p^2}{4} \end{aligned}\text{。}\]当 \(x=-\frac{p}{2}\) 时，平方项最小。

因此，正确答案是 E。

English

Completing the square gives \[ \begin{aligned} x^2+px+q &=\left(x+\frac p2\right)^2\\ &\quad{}+q-\frac{p^2}{4}. \end{aligned} \] The square is least when \(x=-\frac{p}{2}.\)

Thus, the correct answer is E.

14 答案 B

房屋成本为 \(\frac{12000}{0.8}=15000\) 美元，商店成本为 \(\frac{12000}{1.2}=10000\) 美元。总成本为 \(25000\) 美元，总售价为 \(24000\) 美元，因此亏损 \(1000\) 美元。

因此，正确答案是 B。

English

The house cost \(\frac{12000}{0.8}=15000\) dollars, and the store cost \(\frac{12000}{1.2}=10000\) dollars. The combined cost was \(25000\) dollars, while the combined selling price was \(24000\) dollars, producing a loss of \(1000\) dollars.

Thus, the correct answer is B.

15 答案 C

该三角形的最大边为 \(9\)，且 \[ 9^2=81\lt 6^2+8^2=100\text{。}\]由勾股不等式的逆定理，边 \(9\) 所对的角为锐角。由于它是最大角，所以所有角都是锐角。

因此，正确答案是 C。

English

For a triangle with largest side \(9,\) \[ 9^2=81\lt 6^2+8^2=100. \] By the converse of the Pythagorean inequality, the angle opposite side \(9\) is acute. Since it is the largest angle, every angle is acute.

Thus, the correct answer is C.

16 答案 E

要保持面积不变，高应乘以 \(\frac{10}{11}\)。其减少的比例为 \[ 1-\frac{10}{11}=\frac1{11}\text{，}\]即 \(\frac{100}{11}\%=9\dfrac1{11}\%\)。

因此，正确答案是 E。

English

To keep the area fixed, the altitude is multiplied by \(\frac{10}{11}.\) Its fractional decrease is \[ 1-\frac{10}{11}=\frac1{11}, \] which is \(\frac{100}{11}\%=9\dfrac1{11}\%.\)

Thus, the correct answer is E.

17 答案 C

设原标价为 \(1\)。商人的成本为 \(0.8\)。若售价为 \(S\)，则 \(S-0.8=0.2S\)，所以 \(S=1\)。若所定价格为 \(M\)，打 \(20\%\) 折扣后有 \(0.8M=S=1\)，因此 \(M=1.25\)。

因此，所定价格应为原标价的 \(125\%\)，所以正确答案是 C。

English

Let the list price be \(1.\) The merchant’s cost is \(0.8.\) If the selling price is \(S,\) then \(S-0.8=0.2S,\) so \(S=1.\) If the marked price is \(M,\) the \(20\%\) discount gives \(0.8M=S=1,\) hence \(M=1.25.\)

The marked price must therefore be \(125\%\) of the list price, so the correct answer is C.

18 答案 D

利用对数乘积法则，原方程化为 \(\log(pq)=\log(p+q)\)，所以两个正真数必须满足 \[ pq=p+q\text{。}\]因此 \(p(q-1)=q\)，得到 \(p=\frac{q}{q-1}\)。允许的取值满足 \(q\gt1\)，这也保证所有对数的真数为正。

因此，正确答案是 D。

English

The logarithm product rule changes the equation to \(\log(pq)=\log(p+q),\) so the positive arguments must satisfy \[ pq=p+q. \] Therefore \(p(q-1)=q,\) giving \(p=\frac{q}{q-1}.\) Its admissible values have \(q\gt1,\) which also keeps all logarithm arguments positive.

Thus, the correct answer is D.

19 答案 D

三角形 \(ABD\) 与 \(EBC\) 的底边 \(AD\) 和 \(EC\) 位于同一直线 \(AC\) 上，所以从 \(B\) 到该直线的公共高给出 \[ \frac{[ABD]}{[EBC]}=\frac{AD}{EC}\text{。}\]又有 \(\angle ABD=\angle EBC\)，因为两者都是角 \(B\) 的三分之一。再用两边及其夹角表示同一个面积比，得到 \[ \frac{[ABD]}{[EBC]} =\frac{AB\cdot BD}{BE\cdot BC}\text{。}\]

因此，正确答案是 D。

English

Triangles \(ABD\) and \(EBC\) have bases \(AD\) and \(EC\) on the same line \(AC,\) so their common altitude from \(B\) gives \[ \frac{[ABD]}{[EBC]}=\frac{AD}{EC}. \] Also, \(\angle ABD=\angle EBC\) because both are one-third of angle \(B.\) Using two sides and the included angle for the same area ratio gives \[ \frac{[ABD]}{[EBC]} =\frac{AB\cdot BD}{BE\cdot BC}. \]

Thus, the correct answer is D.

20 答案 E

设 \(x=3k\)，\(y=4k\)。前四个表达式依次为 \(\frac{7}{4}\)、\(4\)、\(\frac{11}{3}\) 和 \(\frac{3}{8}\)。但 \[ \frac{x-y}{y}=\frac{3k-4k}{4k}=-\frac14\text{，}\]而不是 \(\frac{1}{4}\)。

因此，正确答案是 E。

English

Set \(x=3k\) and \(y=4k.\) The first four expressions become \(\frac{7}{4},\) \(4,\) \(\frac{11}{3},\) and \(\frac{3}{8},\) respectively. But \[ \frac{x-y}{y}=\frac{3k-4k}{4k}=-\frac14, \] not \(\frac{1}{4}.\)

Thus, the correct answer is E.

21 答案 B

正 \(n\) 边形的每个外角为 \(\frac{360}{n}\) 度。形成星形一个尖角的两条边与它们之间的多边形边组成一个三角形，其两个底角就是这两个外角。因此，该尖角为 \[ 180-2\left(\frac{360}{n}\right) =\frac{(n-4)180}{n}\text{。}\]

因此，正确答案是 B。

English

Each exterior angle of a regular \(n\)-gon is \(\frac{360}{n}\) degrees. The two sides forming a point of the star, together with the intervening polygon side, make a triangle whose two base angles are those exterior angles. Hence its point angle is \[ 180-2\left(\frac{360}{n}\right) =\frac{(n-4)180}{n}. \]

Thus, the correct answer is B.

22 答案 B

在第一个直角三角形中，\[ AB^2=AC^2+BC^2=b^2+1\text{。}\]在第二个直角三角形中，\(AB\) 是斜边且 \(AD=2\)，所以 \[ \begin{aligned} BD^2&=AB^2-AD^2\\ &=b^2+1-4=b^2-3 \end{aligned}\text{。}\]因此 \(BD=\sqrt{b^2-3}\)。

因此，正确答案是 B。

English

In the first right triangle, \[ AB^2=AC^2+BC^2=b^2+1. \] In the second right triangle, \(AB\) is the hypotenuse and \(AD=2,\) so \[ \begin{aligned} BD^2&=AB^2-AD^2\\ &=b^2+1-4=b^2-3. \end{aligned} \] Therefore \(BD=\sqrt{b^2-3}.\)

Thus, the correct answer is B.

23 答案 A

交叉相乘，得到 \[ \begin{aligned} (m+1)x^2 &-(m+1)bx\\ &-(m-1)ax\\ &\quad{}+(m-1)c=0 \end{aligned}\text{。}\]绝对值相等、符号相反的两根之和为 \(0\)，所以 \(x\) 的系数为 \(0\)。因此 \[ m(a+b)+(b-a)=0\text{，}\]并且 \[ m=\frac{a-b}{a+b}\text{。}\]

因此，正确答案是 A。

English

Cross-multiplication gives \[ \begin{aligned} (m+1)x^2 &-(m+1)bx\\ &-(m-1)ax\\ &\quad{}+(m-1)c=0. \end{aligned} \] Roots that are equal in magnitude and opposite in sign have sum \(0,\) so the coefficient of \(x\) is \(0.\) Thus \[ m(a+b)+(b-a)=0, \] and \[ m=\frac{a-b}{a+b}. \]

Thus, the correct answer is A.

24 答案 B

因为 \(AB=20\)、\(AC=12\)，且角 \(C\) 为直角，所以 \(BC=16\)。令 \(A=(0,0)\)、\(B=(20,0)\)。解方程组 \[ \begin{aligned} x_C^2+y_C^2&=144,\\ (x_C-20)^2+y_C^2&=256 \end{aligned} \] 得 \(C=(\frac{36}{5},\frac{48}{5})\)。另有 \(D=(10,0)\)。

直线 \(CB\) 与 \(x=10\) 相交于 \(E=(10,\frac{15}{2})\)。现在对 \(A,D,E,C\) 使用鞋带公式，得到 \[ \begin{aligned} [ADEC] &=\frac12\left( 10\cdot\frac{15}{2} +10\cdot\frac{48}{5}\right.\\ &\qquad\left.{} -\frac{15}{2}\cdot\frac{36}{5}\right)\\ &=\frac{117}{2}=58\frac12 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Since \(AB=20,\) \(AC=12,\) and angle \(C\) is right, \(BC=16.\) Put \(A=(0,0)\) and \(B=(20,0).\) Solving \[ \begin{aligned} x_C^2+y_C^2&=144,\\ (x_C-20)^2+y_C^2&=256 \end{aligned} \] gives \(C=(\frac{36}{5},\frac{48}{5}).\) Also \(D=(10,0).\)

The line \(CB\) meets \(x=10\) at \(E=(10,\frac{15}{2}).\) The shoelace formula for \(A,D,E,C\) now gives \[ \begin{aligned} [ADEC] &=\frac12\left( 10\cdot\frac{15}{2} +10\cdot\frac{48}{5}\right.\\ &\qquad\left.{} -\frac{15}{2}\cdot\frac{36}{5}\right)\\ &=\frac{117}{2}=58\frac12. \end{aligned} \]

Thus, the correct answer is B.

25 答案 D

设总时间为 \(T\) 秒。爆破工与爆炸地点相距 \(8T\) 码，即 \(24T\) 英尺。因为声音在 \(30\) 秒后才开始传播，所以 \[ 1080(T-30)=24T\text{。}\]因此 \(T=\frac{1350}{44}\) 秒，他跑出的距离为 \[ 8T=\frac{2700}{11}\approx245.45\text{ 码}\text{。}\]

选项中与之近似的是 \(245\) 码，所以正确答案是 D。

English

Let \(T\) be the total time in seconds. The powderman is \(8T\) yards, or \(24T\) feet, from the blast. Because the sound starts after \(30\) seconds, \[ 1080(T-30)=24T. \] Thus \(T=\frac{1350}{44}\) seconds, and the distance he ran is \[ 8T=\frac{2700}{11}\approx245.45\text{ yards}. \]

Approximately \(245\) yards is the listed value, so the correct answer is D.

26 答案 C

设 \(u=r+\frac{1}{r}\)。展开 \(u^3\)，得到 \[ r^3+\frac1{r^3}=u^3-3u\text{。}\]因为 \(u^2=3\)，该式变为 \(u(u^2-3)=0\)。

因此，所求值为 \(0\)，所以正确答案是 C。

English

Let \(u=r+\frac{1}{r}.\) Expanding \(u^3\) gives \[ r^3+\frac1{r^3}=u^3-3u. \] Since \(u^2=3,\) this becomes \(u(u^2-3)=0.\)

Thus the value is \(0,\) so the correct answer is C.

27 答案 E

边长为 \(s\) 的等边三角形，其高为 \(\frac{s\sqrt3}{2}\)。因此，第一个三角形的边长为 \(\frac{2R}{\sqrt3}\)，周长为 \(2\sqrt3R\)。内接于半径为 \(R\) 的圆的等边三角形边长为 \(\sqrt3R\)，周长为 \(3\sqrt3R\)。

因此，周长之比为 \(2:3\)，所以正确答案是 E。

English

An equilateral triangle with side \(s\) has altitude \(\frac{s\sqrt3}{2}.\) Thus the first triangle has side \(\frac{2R}{\sqrt3}\) and perimeter \(2\sqrt3R.\) An equilateral triangle inscribed in a circle of radius \(R\) has side \(\sqrt3R\) and perimeter \(3\sqrt3R.\)

The perimeter ratio is therefore \(2:3,\) so the correct answer is E.

28 答案 C

\(y\) 值的相邻差为 \(4,6,8,10\)，这些数的相邻差都为 \(2\)。因此，数据符合首项系数为一的二次式。代入可知，\[ y=x^2+x+1 \] 依次给出 \(3,7,13,21,31\)，对应于 \(x=1,2,3,4,5\)。

因此，正确答案是 C。

English

The successive differences of the \(y\)-values are \(4,6,8,10,\) whose successive differences are all \(2.\) Thus the data fit a monic quadratic. Substitution shows \[ y=x^2+x+1 \] gives \(3,7,13,21,31\) at \(x=1,2,3,4,5,\) respectively.

Thus, the correct answer is C.

29 答案 A

将圆心置于原点，取 \(C=(0,5)\)，并设 \(H=(x,y)\)。方程组 \[ \begin{aligned} x^2+y^2&=25,\\ x^2+(y-5)^2&=64 \end{aligned} \] 给出 \(y=-\frac{7}{5}\) 和 \(x=\pm\frac{24}{5}\)。取 \(x=\frac{24}{5}\) 不影响两段的长度。

从 \(C=(0,5)\) 到 \(H=(\frac{24}{5},-\frac{7}{5})\) 的直线与 \(y=0\) 相交于 \(x=\frac{15}{4}\) 处。因此，该交点到端点 \((-5,0)\) 和 \((5,0)\) 的距离如下；这两个端点构成直径 \(AB\)：\[ \begin{aligned} 5+\frac{15}{4}&=\frac{35}{4}=8.75,\\ 5-\frac{15}{4}&=\frac54=1.25 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

Put the circle at the origin with \(C=(0,5)\) and let \(H=(x,y).\) The equations \[ \begin{aligned} x^2+y^2&=25,\\ x^2+(y-5)^2&=64 \end{aligned} \] give \(y=-\frac{7}{5}\) and \(x=\pm\frac{24}{5}.\) Taking \(x=\frac{24}{5}\) does not change the segment lengths.

The line from \(C=(0,5)\) to \(H=(\frac{24}{5},-\frac{7}{5})\) crosses \(y=0\) at \(x=\frac{15}{4}.\) Its distances to the endpoints \((-5,0)\) and \((5,0)\) of diameter \(AB\) are therefore \[ \begin{aligned} 5+\frac{15}{4}&=\frac{35}{4}=8.75,\\ 5-\frac{15}{4}&=\frac54=1.25. \end{aligned} \]

Thus, the correct answer is A.

30 答案 A

求和公式给出 \[ \begin{aligned} S_{10}&=5(2a+9d),\\ S_5&=\frac52(2a+4d) \end{aligned}\text{。}\]方程 \(S_{10}=4S_5\) 化为 \[ 2a+9d=2(2a+4d)\text{，}\]所以 \(d=2a\)。因此 \(a:d=1:2\)。

因此，正确答案是 A。

English

The sum formula gives \[ \begin{aligned} S_{10}&=5(2a+9d),\\ S_5&=\frac52(2a+4d). \end{aligned} \] The equation \(S_{10}=4S_5\) becomes \[ 2a+9d=2(2a+4d), \] so \(d=2a.\) Hence \(a:d=1:2.\)

Thus, the correct answer is A.

31 答案 D

每一对点确定一条直线；又因为任意三点不共线，所以没有一条直线会由多于一对点重复计数。因此，直线条数为 \[ \binom{12}{2}=\frac{12\cdot11}{2}=66\text{。}\]

因此，正确答案是 D。

English

Every pair of points determines one line, and no line is counted by more than one pair because no three points are collinear. Therefore the number of lines is \[ \binom{12}{2}=\frac{12\cdot11}{2}=66. \]

Thus, the correct answer is D.

32 答案 D

对 \(K\) 而言，路程为 \(30\) 英里，速度为每小时 \(x\) 英里。因此 \[ \text{时间}=\frac{\text{路程}}{\text{速度}} =\frac{30}{x}\text{ 小时}\text{。}\]写出这个表达式并不需要利用与 \(M\) 的比较。

因此，正确答案是 D。

English

For \(K,\) the distance is \(30\) miles and the rate is \(x\) miles per hour. Thus \[ \text{time}=\frac{\text{distance}}{\text{rate}} =\frac{30}{x}\text{ hours}. \] The comparison with \(M\) is not needed for this expression.

Thus, the correct answer is D.

33 答案 B

公共周长为 \(P\) 时，圆的半径为 \(\frac{P}{2\pi}\)，所以面积为 \(\frac{P^2}{4\pi}\)。正方形边长为 \(\frac{P}{4}\)，所以面积为 \(\frac{P^2}{16}\)。因为 \(4\pi\lt16\)，所以 \[ \frac{P^2}{4\pi}\gt\frac{P^2}{16}\text{。}\]

圆的面积较大，所以正确答案是 B。

English

With common perimeter \(P,\) the circle’s radius is \(\frac{P}{2\pi}\), so its area is \(\frac{P^2}{4\pi}.\) The square’s side is \(\frac{P}{4},\) so its area is \(\frac{P^2}{16}.\) Since \(4\pi\lt16,\) \[ \frac{P^2}{4\pi}\gt\frac{P^2}{16}. \]

The circle has the greater area, so the correct answer is B.

34 答案 E

若原价为 \(P\) 美元，则两次变价后的价格为 \[ \begin{aligned} &P\left(1+\frac p{100}\right) \left(1-\frac p{100}\right)\\ &\qquad=P\left(1-\frac{p^2}{10{,}000}\right) \end{aligned}\text{。}\]令该式等于 \(1\)，得到 \[ P=\frac{10{,}000}{10{,}000-p^2}\text{。}\]

因此，正确答案是 E。

English

If the original price is \(P\) dollars, then after both changes its price is \[ \begin{aligned} &P\left(1+\frac p{100}\right) \left(1-\frac p{100}\right)\\ &\qquad=P\left(1-\frac{p^2}{10{,}000}\right). \end{aligned} \] Setting this equal to \(1\) gives \[ P=\frac{10{,}000}{10{,}000-p^2}. \]

Thus, the correct answer is E.

35 答案 A

先乘以共轭式 \(\sqrt2+\sqrt3+\sqrt5\)，得到 \[ \begin{aligned} &\frac{\sqrt2(\sqrt2+\sqrt3+\sqrt5)} {(\sqrt2+\sqrt3)^2-5}\\ &\qquad=\frac{2+\sqrt6+\sqrt{10}}{2\sqrt6} \end{aligned}\text{。}\]再将分子、分母同乘 \(\sqrt6\)，得到 \[ \begin{gathered} \frac{2\sqrt6+6+\sqrt{60}}{12} \\ =\frac{3+\sqrt6+\sqrt{15}}6 \end{gathered}\text{。}\]

因此，正确答案是 A。

English

Multiply first by the conjugate \(\sqrt2+\sqrt3+\sqrt5.\) This gives \[ \begin{aligned} &\frac{\sqrt2(\sqrt2+\sqrt3+\sqrt5)} {(\sqrt2+\sqrt3)^2-5}\\ &\qquad=\frac{2+\sqrt6+\sqrt{10}}{2\sqrt6}. \end{aligned} \] Multiplying numerator and denominator by \(\sqrt6\) yields \[ \begin{gathered} \frac{2\sqrt6+6+\sqrt{60}}{12} \\ =\frac{3+\sqrt6+\sqrt{15}}6. \end{gathered} \]

Thus, the correct answer is A.

36 答案 E

当 \(x\ne-1\) 时，\[ \begin{aligned} \frac{x^3+1}{x^2-1} &=\frac{(x+1)(x^2-x+1)} {(x+1)(x-1)}\\ &=\frac{x^2-x+1}{x-1} \end{aligned}\text{。}\]它在 \(x=-1\) 处的极限为 \[ \frac{1+1+1}{-2}=-\frac32\text{。}\]将函数在该点定义为此值即可消除间断。

因此，正确答案是 E。

English

For \(x\ne-1,\) \[ \begin{aligned} \frac{x^3+1}{x^2-1} &=\frac{(x+1)(x^2-x+1)} {(x+1)(x-1)}\\ &=\frac{x^2-x+1}{x-1}. \end{aligned} \] Its limit at \(x=-1\) is \[ \frac{1+1+1}{-2}=-\frac32. \] Assigning this value removes the discontinuity.

Thus, the correct answer is E.

37 答案 B

两条等弦关于圆心对称，各距圆心 \(4\) 英寸。对任一弦，圆心处的半角满足 \(\cos\theta=\frac{4}{8}=\frac{1}{2}\)，所以 \(\theta=60^\circ\)。每个外侧弓形的面积等于一个 \(120^\circ\) 扇形面积减去等腰三角形面积：\[ \begin{aligned} &\frac{120}{360}\pi(8)^2 -\frac12(8)^2\sin120^\circ\\ &\qquad=\frac{64\pi}{3}-16\sqrt3 \end{aligned}\text{。}\]因此，两弦之间的面积为 \[ \begin{aligned} &64\pi-2\left(\frac{64\pi}{3}-16\sqrt3\right)\\ &\qquad=\frac{64\pi}{3}+32\sqrt3\\ &\qquad=21\frac13\pi+32\sqrt3 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

The equal chords are symmetrically \(4\) inches from the center. For either chord, the half-angle at the center satisfies \(\cos\theta=\frac{4}{8}=\frac{1}{2},\) so \(\theta=60^\circ.\) Each outer segment is a \(120^\circ\) sector minus the isosceles triangle: \[ \begin{aligned} &\frac{120}{360}\pi(8)^2 -\frac12(8)^2\sin120^\circ\\ &\qquad=\frac{64\pi}{3}-16\sqrt3. \end{aligned} \] The area between the chords is therefore \[ \begin{aligned} &64\pi-2\left(\frac{64\pi}{3}-16\sqrt3\right)\\ &\qquad=\frac{64\pi}{3}+32\sqrt3\\ &\qquad=21\frac13\pi+32\sqrt3. \end{aligned} \]

Thus, the correct answer is B.

38 答案 D

若两条底边为 \(b_1,b_2\)，则 \[ 1400=\frac12(b_1+b_2)(50)\text{，}\]所以 \(b_1+b_2=56\)。由正的 \(8\) 的倍数组成的无序对为 \[ (8,48),\quad(16,40),\quad(24,32)\text{。}\]因此共有三组解。

因此，正确答案是 D。

English

If the bases are \(b_1,b_2,\) then \[ 1400=\frac12(b_1+b_2)(50), \] so \(b_1+b_2=56.\) The unordered positive pairs of multiples of \(8\) are \[ (8,48),\quad(16,40),\quad(24,32). \] There are three solutions.

Thus, the correct answer is D.

39 答案 A

由周长和对角线可得 \[ \begin{aligned} L+W&=\frac p2,\\ L^2+W^2&=d^2 \end{aligned}\text{。}\]因此 \[ \begin{aligned} (L-W)^2 &=2(L^2+W^2)\\ &\quad{}-(L+W)^2\\ &=2d^2-\frac{p^2}{4}\\ &=\frac{8d^2-p^2}{4} \end{aligned}\text{。}\]取非负平方根，得到 \[ L-W=\frac{\sqrt{8d^2-p^2}}2\text{。}\]

因此，正确答案是 A。

English

The perimeter and diagonal give \[ \begin{aligned} L+W&=\frac p2,\\ L^2+W^2&=d^2. \end{aligned} \] Hence \[ \begin{aligned} (L-W)^2 &=2(L^2+W^2)\\ &\quad{}-(L+W)^2\\ &=2d^2-\frac{p^2}{4}\\ &=\frac{8d^2-p^2}{4}. \end{aligned} \] Taking the nonnegative square root gives \[ L-W=\frac{\sqrt{8d^2-p^2}}2. \]

Thus, the correct answer is A.

40 答案 E

这些数值应当是连续的平方数 \[ \begin{gathered} 62^2,63^2,64^2,65^2,\\ 66^2,67^2,68^2,69^2 \end{gathered}\text{。}\]即 \(3844,3969,4096,4225\)、\(4356,4489,4624,4761\)。因此，表中错误的数值是 \(4227\)，但它不在选项 A–D 中。

所以，列出的四个数都不是错误值，正确答案是 E。

English

The values are intended to be the consecutive squares \[ \begin{gathered} 62^2,63^2,64^2,65^2,\\ 66^2,67^2,68^2,69^2. \end{gathered} \] These are \(3844,3969,4096,4225,\) \(4356,4489,4624,4761.\) Thus the incorrect table entry is \(4227,\) but it is not one of choices A-D.

Consequently none of the four listed numbers is incorrect, so the correct answer is E.

41 答案 C

原高为 \(2\)、原半径为 \(r\) 时，增加半径所增加的体积为 \[ 2\pi\bigl((r+6)^2-r^2\bigr)=24\pi r+72\pi \] 个立方单位。将高增加 \(6\) 所增加的体积为 \(6\pi r^2\) 个立方单位。令二者相等，得到 \[ 6r^2=24r+72\text{，}\]即 \(r^2-4r-12=0\)。其正根为 \(r=6\)。

因此，正确答案是 C。

English

With original height \(2\) and radius \(r,\) increasing the radius adds \[ 2\pi\bigl((r+6)^2-r^2\bigr)=24\pi r+72\pi \] cubic units. Increasing the height by \(6\) adds \(6\pi r^2\) cubic units. Equating these gives \[ 6r^2=24r+72, \] or \(r^2-4r-12=0.\) Its positive root is \(r=6.\)

Thus, the correct answer is C.

42 答案 D

小数点越过非循环部分后，\[ 10^rD=P.QQQ\ldots\text{，}\]再越过一组循环部分后，\[ 10^{r+s}D=PQ.QQQ\ldots\text{。}\]两式相减，得到标准关系 \[ \begin{gathered} 10^r(10^s-1)D \\ =P(10^s-1)+Q \end{gathered}\text{。}\]这并非选项 D 所写的 \(Q(P-1)\)。

因此，不正确的表达式是选项 D，这也是本题的正确答案。

English

After shifting past the nonrepeating block, \[ 10^rD=P.QQQ\ldots, \] and after shifting one repeating block farther, \[ 10^{r+s}D=PQ.QQQ\ldots. \] Subtracting yields the standard relation \[ \begin{gathered} 10^r(10^s-1)D \\ =P(10^s-1)+Q. \end{gathered} \] This is not \(Q(P-1),\) as stated in choice D.

Thus, the incorrect expression, and hence the correct answer, is D.

43 答案 A

每个小半圆的直径为 \(\frac{d}{n}\)，所以弧长为 \(\frac{\pi d}{2n}\)。全部 \(n\) 条弧的长度之和为 \[ n\cdot\frac{\pi d}{2n}=\frac{\pi d}{2}\text{，}\]恰好等于原圆周长的一半。这个等式对每个正 \(n\) 都成立，而不仅仅在极限情况下成立。

因此，正确答案是 A。

English

Each small semicircle has diameter \(\frac{d}{n},\) so its arc length is \(\frac{\pi d}{2n}.\) The sum of all \(n\) arc lengths is \[ n\cdot\frac{\pi d}{2n}=\frac{\pi d}{2}, \] exactly the semi-circumference of the original circle. This equality holds for every positive \(n,\) not merely in the limit.

Thus, the correct answer is A.

44 答案 C

设原数为 \(10a+b=k(a+b)\)。倒序数为 \(10b+a\)。因为 \[ \begin{gathered} (10a+b)+(10b+a) \\ =11(a+b) \end{gathered}\text{，}\]所以倒序数等于 \[ \begin{aligned} &11(a+b)-k(a+b)\\ &\qquad=(11-k)(a+b) \end{aligned}\text{。}\]

因此，正确答案是 C。

English

Let the original number be \(10a+b=k(a+b).\) The reversed number is \(10b+a.\) Since \[ \begin{gathered} (10a+b)+(10b+a) \\ =11(a+b), \end{gathered} \] the reversed number equals \[ \begin{aligned} &11(a+b)-k(a+b)\\ &\qquad=(11-k)(a+b). \end{aligned} \]

Thus, the correct answer is C.

45 答案 E

对不相等的正数 \(a,b\)，算术平均值与几何平均值不等式给出 \[ \frac{a+b}{2}\gt\sqrt{ab}\text{。}\]对 \(\frac{1}{a}\) 与 \(\frac{1}{b}\) 使用同一不等式，得到 \[ \sqrt{ab}\gt\frac{2ab}{a+b}\text{。}\]因此，从大到小依次为算术平均数、几何平均数、调和平均数。

因此，正确答案是 E。

English

The arithmetic-geometric mean inequality gives \[ \frac{a+b}{2}\gt\sqrt{ab} \] for unequal positive \(a,b.\) Applying the same inequality to \(\frac{1}{a}\) and \(\frac{1}{b}\) gives \[ \sqrt{ab}\gt\frac{2ab}{a+b}. \] Therefore the decreasing order is arithmetic mean, geometric mean, harmonic mean.

Thus, the correct answer is E.

46 答案 C

设较长边和较短边的长度分别为 \(L\) 和 \(W\)，对角线长为 \(d\)。新长方形的面积为 \[ (d+L)(d-L)=d^2-L^2\text{。}\]由勾股定理，\(d^2=L^2+W^2\)，所以新面积为 \(W^2\)。这正是以原长方形较短边为边长的正方形面积。

因此，正确答案是 C。

English

Let the greater and smaller side lengths be \(L\) and \(W,\) and let the diagonal be \(d.\) The new rectangle’s area is \[ (d+L)(d-L)=d^2-L^2. \] The Pythagorean theorem gives \(d^2=L^2+W^2,\) so the new area is \(W^2.\) This is the area of a square whose side equals the smaller side of the original rectangle.

Thus, the correct answer is C.

47 答案 D

对于题目预期的正整数解，第一个方程给出 \(z=y^2\)。第二个方程为 \[ 2^z=2^{2x+1}\text{，}\]所以 \(z=2x+1\)，且 \(x=\frac{y^2-1}{2}\)。代入 \(x+y+z=16\)，得到 \[ \frac{y^2-1}{2}+y+y^2=16\text{。}\]正整数解为 \(y=3\)，从而 \(x=4\)、\(z=9\)。直接代入即可验证三个方程均成立。

因此，题目各选项中预期的答案是 D。

English

For the intended positive integral solution, the first equation gives \(z=y^2.\) The second equation is \[ 2^z=2^{2x+1}, \] so \(z=2x+1\) and \(x=\frac{y^2-1}{2}.\) Substituting into \(x+y+z=16\) gives \[ \frac{y^2-1}{2}+y+y^2=16. \] The positive integral solution is \(y=3,\) which gives \(x=4\) and \(z=9.\) Direct substitution verifies all three displayed equations.

Thus, the intended listed answer is D.

48 答案 A

设两人的速度满足 \(u\gt v\)。两个相遇条件给出 \[ u-v=\frac kr,\qquad u+v=\frac kt\text{。}\]将两式相加、相减，得到 \[ \begin{aligned} u&=\frac k2\left(\frac1r+\frac1t\right),\\ v&=\frac k2\left(\frac1t-\frac1r\right) \end{aligned}\text{。}\]因此 \[ \frac uv=\frac{r+t}{r-t}\text{。}\]

因此，正确答案是 A。

English

Let the speeds be \(u\gt v.\) The two meeting conditions give \[ u-v=\frac kr,\qquad u+v=\frac kt. \] Adding and subtracting these equations yields \[ \begin{aligned} u&=\frac k2\left(\frac1r+\frac1t\right),\\ v&=\frac k2\left(\frac1t-\frac1r\right). \end{aligned} \] Therefore \[ \frac uv=\frac{r+t}{r-t}. \]

Thus, the correct answer is A.

49 答案 C

取 \(A=(0,1)\)、\(B=(0,0)\)、\(C=(1,0)\)。则 \[ \begin{aligned} D&=\left(\frac23,0\right),\\ E&=\left(\frac13,\frac23\right),\\ F&=\left(0,\frac13\right) \end{aligned}\text{。}\]分别求 \(AD,BE,CF\) 两两相交的交点，得到 \[ \begin{aligned} N_1&=\left(\frac47,\frac17\right),\\ N_2&=\left(\frac27,\frac47\right),\\ N_3&=\left(\frac17,\frac27\right) \end{aligned}\text{。}\]行列式面积公式给出 \[ \begin{aligned} [N_1N_2N_3]&=\frac1{14},\\ [ABC]&=\frac12 \end{aligned}\text{。}\]因此 \(\frac{[N_1N_2N_3]}{[ABC]}=\frac{1}{7}\)。

因此，正确答案是 C。

English

Use \(A=(0,1),\) \(B=(0,0),\) \(C=(1,0).\) Then \[ \begin{aligned} D&=\left(\frac23,0\right),\\ E&=\left(\frac13,\frac23\right),\\ F&=\left(0,\frac13\right). \end{aligned} \] Intersecting \(AD,BE,CF\) in pairs gives \[ \begin{aligned} N_1&=\left(\frac47,\frac17\right),\\ N_2&=\left(\frac27,\frac47\right),\\ N_3&=\left(\frac17,\frac27\right). \end{aligned} \] The determinant area formula gives \[ \begin{aligned} [N_1N_2N_3]&=\frac1{14},\\ [ABC]&=\frac12. \end{aligned} \] Hence \(\frac{[N_1N_2N_3]}{[ABC]}=\frac{1}{7}.\)

Thus, the correct answer is C.

50 答案 D

在初始项之后，每个幂 \(4^{-k}\) 分别单独出现一次，并乘以 \(\sqrt2\) 出现一次。因为 \(\sum_{k=1}^{\infty}4^{-k}=\frac{1}{3}\)，所以极限为 \[ 1+\frac{1+\sqrt2}{3}=\frac{4+\sqrt2}{3}\text{。}\]

因此，正确答案是 D。

English

After the initial term, each power \(4^{-k}\) appears once by itself and once multiplied by \(\sqrt2.\) Since \(\sum_{k=1}^{\infty}4^{-k}=\frac{1}{3},\) the limit is \[ 1+\frac{1+\sqrt2}{3}=\frac{4+\sqrt2}{3}. \]

Thus, the correct answer is D.