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1963 AMC 12 真题(P版)

数学 1963 年 40 题 143 KB 含答案解析

中英双语:每题先中文,后附英文原题;解答在同页下方「查看答案解析与下载」里。

1

以下哪一点不在函数 \(y=\dfrac{x}{x+1}\) 的图像上?

A\((0,0)\) B\(\left(-\dfrac12,-1\right)\) C\(\left(\dfrac12,\dfrac13\right)\) D\((-1,1)\) E\((-2,2)\)

English

Which one of the following points is not on the graph of \(y=\dfrac{x}{x+1}?\)

(A) \((0,0)\)(B) \(\left(-\dfrac12,-1\right)\)(C) \(\left(\dfrac12,\dfrac13\right)\)(D) \((-1,1)\)(E) \((-2,2)\)

2

设 \(n=x-y^{x-y}\)。求 \(n\) 在 \(x=2\) 且 \(y=-2\) 时的值。

A\(-14\) B\(0\) C\(1\) D\(18\) E\(256\)

English

Let \(n=x-y^{x-y}.\) Find \(n\) when \(x=2\) and \(y=-2.\)

(A) \(-14\)(B) \(0\)(C) \(1\)(D) \(18\)(E) \(256\)

3

若 \(x+1\) 的倒数是 \(x-1\),则 \(x\) 等于:

A\(0\) B\(1\) C\(-1\) D\(\pm1\) E以上都不是

English

If the reciprocal of \(x+1\) is \(x-1,\) then \(x\) equals:

(A) \(0\)(B) \(1\)(C) \(-1\)(D) \(\pm1\)(E) none of these

4

当 \(k\) 取何值时,方程 \(y=x^2\) 与 \(y=3x+k\) 有两个相同的解?

A\(\dfrac49\) B\(-\dfrac49\) C\(\dfrac94\) D\(-\dfrac94\) E\(\pm\dfrac94\)

English

For what value(s) of \(k\) does the pair of equations \(y=x^2\) and \(y=3x+k\) have two identical solutions?

(A) \(\dfrac49\)(B) \(-\dfrac49\)(C) \(\dfrac94\)(D) \(-\dfrac94\)(E) \(\pm\dfrac94\)

5

若 \(x\) 和 \(\log_{10}x\) 都是实数,并且 \(\log_{10}x\lt0\),则:

A\(x\lt0\) B\(-1\lt x\lt1\) C\(0\lt x\leq1\) D\(-1\lt x\lt0\) E\(0\lt x\lt1\)

English

If \(x\) and \(\log_{10}x\) are real numbers and \(\log_{10}x\lt0,\) then:

(A) \(x\lt0\)(B) \(-1\lt x\lt1\)(C) \(0\lt x\leq1\)(D) \(-1\lt x\lt0\)(E) \(0\lt x\lt1\)

6

三角形 \(BAD\) 在 \(B\) 点为直角。线段 \(AD\) 上有一点 \(C\),满足 \(AC=CD\) 且 \(AB=BC\)。角 \(DAB\) 的度数是:

A\(67\dfrac12\) B\(60\) C\(45\) D\(30\) E\(22\dfrac12\)

English

Triangle \(BAD\) is right-angled at \(B.\) On \(AD\) there is a point \(C\) for which \(AC=CD\) and \(AB=BC.\) The magnitude of angle \(DAB,\) in degrees, is:

(A) \(67\dfrac12\)(B) \(60\)(C) \(45\)(D) \(30\)(E) \(22\dfrac12\)

7

给出以下四个方程:

\[ \begin{aligned} (1)\quad&3y-2x=12,\\ (2)\quad&-2x-3y=10,\\ (3)\quad&3y+2x=12,\\ (4)\quad&2y+3x=10 \end{aligned}\text{。}\]

其中表示两条垂直直线的一组是:

A\((1)\) 和 \((4)\) B\((1)\) 和 \((3)\) C\((1)\) 和 \((2)\) D\((2)\) 和 \((4)\) E\((2)\) 和 \((3)\)

English

Given the four equations:

\[ \begin{aligned} (1)\quad&3y-2x=12,\\ (2)\quad&-2x-3y=10,\\ (3)\quad&3y+2x=12,\\ (4)\quad&2y+3x=10. \end{aligned} \]

The pair representing perpendicular lines is:

(A) \((1)\) and \((4)\)(B) \((1)\) and \((3)\)(C) \((1)\) and \((2)\)(D) \((2)\) and \((4)\)(E) \((2)\) and \((3)\)

8

最小正整数 \(x\) 是多少,能使 \(1260x=N^3\) 成立?其中 \(N\) 是整数。

A\(1050\) B\(1260\) C\(1260^2\) D\(7350\) E\(44{,}100\)

English

The smallest positive integer \(x\) for which \(1260x=N^3,\) where \(N\) is an integer, is:

(A) \(1050\)(B) \(1260\)(C) \(1260^2\)(D) \(7350\)(E) \(44{,}100\)

9

在 \(\left(a-\dfrac1{\sqrt a}\right)^7\) 的展开式中,\(a^{-\frac{1}{2}}\) 的系数是:

A\(-7\) B\(7\) C\(-21\) D\(21\) E\(35\)

English

In the expansion of \(\left(a-\dfrac1{\sqrt a}\right)^7\) the coefficient of \(a^{-\frac{1}{2}}\) is:

(A) \(-7\)(B) \(7\)(C) \(-21\)(D) \(21\)(E) \(35\)

10

在正方形内部取一点 \(P\),该正方形的边长为 \(a\),所取点到两个相邻顶点的距离以及到这两个顶点对边的距离都相等。若 \(d\) 表示这个公共距离,则 \(d\) 等于:

A\(\dfrac{3a}{5}\) B\(\dfrac{5a}{8}\) C\(\dfrac{3a}{8}\) D\(\dfrac{a\sqrt2}{2}\) E\(\dfrac a2\)

English

Point \(P\) is taken interior to a square with side-length \(a\) and such that it is equally distant from two consecutive vertices and from the side opposite these vertices. If \(d\) represents the common distance, then \(d\) equals:

(A) \(\dfrac{3a}{5}\)(B) \(\dfrac{5a}{8}\)(C) \(\dfrac{3a}{8}\)(D) \(\dfrac{a\sqrt2}{2}\)(E) \(\dfrac a2\)

11

一组 \(50\) 个数的算术平均数是 \(38\)。若从中去掉 \(45\) 和 \(55\) 这两个数,剩余各数的算术平均数是:

A\(38.5\) B\(37.5\) C\(37\) D\(36.5\) E\(36\)

English

The arithmetic mean of a set of \(50\) numbers is \(38.\) If two numbers of the set, namely \(45\) and \(55,\) are discarded, the arithmetic mean of the remaining set of numbers is:

(A) \(38.5\)(B) \(37.5\)(C) \(37\)(D) \(36.5\)(E) \(36\)

12

平行四边形 \(PQRS\) 的三个顶点是 \(P(-3,-2)\)、\(Q(1,-5)\)、\(R(9,1)\),其中 \(P\) 与 \(R\) 是对角顶点。顶点 \(S\) 的两个坐标之和是:

A\(13\) B\(12\) C\(11\) D\(10\) E\(9\)

English

Three vertices of parallelogram \(PQRS\) are \(P(-3,-2),\) \(Q(1,-5),\) \(R(9,1)\) with \(P\) and \(R\) diagonally opposite. The sum of the coordinates of vertex \(S\) is:

(A) \(13\)(B) \(12\)(C) \(11\)(D) \(10\)(E) \(9\)

13

若 \(2^a+2^b=3^c+3^d\),则整数 \(a\)、\(b\)、\(c\)、\(d\) 中最多有几个可以是负数?

A\(4\) B\(3\) C\(2\) D\(1\) E\(0\)

English

If \(2^a+2^b=3^c+3^d,\) the number of integers \(a,\) \(b,\) \(c,\) \(d\) which can possibly be negative is, at most:

(A) \(4\)(B) \(3\)(C) \(2\)(D) \(1\)(E) \(0\)

14

给定方程 \(x^2+kx+6=0\) 和 \(x^2-kx+6=0\)。如果适当排列两方程的根后,第二个方程的每个根都比第一个方程的对应根大 \(5\),那么 \(k\) 等于:

A\(5\) B\(-5\) C\(7\) D\(-7\) E以上都不是

English

Given the equations \(x^2+kx+6=0\) and \(x^2-kx+6=0.\) If, when the roots of the equations are suitably listed, each root of the second equation is \(5\) more than the corresponding root of the first equation, then \(k\) equals:

(A) \(5\)(B) \(-5\)(C) \(7\)(D) \(-7\)(E) none of these

15

一个圆内切于等边三角形,一个正方形内接于该圆。三角形面积与正方形面积之比是:

A\(\sqrt3:1\) B\(\sqrt3:\sqrt2\) C\(3\sqrt3:2\) D\(3:\sqrt2\) E\(3:2\sqrt2\)

English

A circle is inscribed in an equilateral triangle, and a square is inscribed in the circle. The ratio of the area of the triangle to the area of the square is:

(A) \(\sqrt3:1\)(B) \(\sqrt3:\sqrt2\)(C) \(3\sqrt3:2\)(D) \(3:\sqrt2\)(E) \(3:2\sqrt2\)

16

三个非零数 \(a\)、\(b\)、\(c\) 成等差数列。把 \(a\) 增加 \(1\) 或把 \(c\) 增加 \(2\),所得三个数都成等比数列。则 \(b\) 等于:

A\(16\) B\(14\) C\(12\) D\(10\) E\(8\)

English

Three numbers \(a,\) \(b,\) \(c,\) none zero, form an arithmetic progression. Increasing \(a\) by \(1\) or increasing \(c\) by \(2\) results in a geometric progression. Then \(b\) equals:

(A) \(16\)(B) \(14\)(C) \(12\)(D) \(10\)(E) \(8\)

17

表达式

\[ \frac{\dfrac{a}{a+y}+\dfrac{y}{a-y}} {\dfrac{y}{a+y}-\dfrac{a}{a-y}}\text{,}\]

其中 \(a\) 为实数且 \(a\ne0\),对于下列哪些取值,表达式的值为 \(-1\)?

A除两个实数值外的所有 \(y\) B仅两个实数值的 \(y\) C所有实数值的 \(y\) D仅一个实数值的 \(y\) E没有实数值的 \(y\)

English

The expression

\[ \frac{\dfrac{a}{a+y}+\dfrac{y}{a-y}} {\dfrac{y}{a+y}-\dfrac{a}{a-y}}, \]

\(a\) real, \(a\ne0,\) has the value \(-1\) for:

(A) all but two real values of \(y\)(B) only two real values of \(y\)(C) all real values of \(y\)(D) only one real value of \(y\)(E) no real values of \(y\)

18

弦 \(EF\) 是弦 \(BC\) 的垂直平分线,并与其交于 \(M\)。在 \(B\) 与 \(M\) 之间取一点 \(U\),延长 \(EU\) 与圆交于 \(A\)。对于任意如此选取的 \(U\),三角形 \(EUM\) 与下列哪个三角形相似?

题图

A\(EFA\) B\(EFC\) C\(ABM\) D\(ABU\) E\(FMC\)

English

Chord \(EF\) is the perpendicular bisector of chord \(BC,\) intersecting it in \(M.\) Between \(B\) and \(M\) point \(U\) is taken, and \(EU\) extended meets the circle in \(A.\) Then, for any selection of \(U,\) as described, triangle \(EUM\) is similar to triangle:

题图

(A) \(EFA\)(B) \(EFC\)(C) \(ABM\)(D) \(ABU\)(E) \(FMC\)

19

清点 \(n\) 个彩球,其中一些是红球,另一些是黑球。最先清点的 \(50\) 个球中有 \(49\) 个红球;此后每清点 \(8\) 个球,就有 \(7\) 个红球。如果清点的所有球中至少有 \(90\%\) 是红球,那么 \(n\) 的最大值是:

A\(225\) B\(210\) C\(200\) D\(180\) E\(175\)

English

In counting \(n\) colored balls, some red and some black, it was found that \(49\) of the first \(50\) counted were red. Thereafter, \(7\) out of every \(8\) counted were red. If, in all, \(90\%\) or more of the balls counted were red, the maximum value of \(n\) is:

(A) \(225\)(B) \(210\)(C) \(200\)(D) \(180\)(E) \(175\)

20

\(R\) 点与 \(S\) 点相距 \(76\) 英里。两人分别从这两点同时出发,相向而行。从 \(R\) 点出发的人始终以每小时 \(4\dfrac12\) 英里的速度行走;从 \(S\) 点出发的人第一小时以每小时 \(3\dfrac14\) 英里的速度行走,第二小时以每小时 \(3\dfrac34\) 英里的速度行走,之后各小时的速度依次成等差数列。若两人在整数小时后相遇,且相遇点到 \(R\) 点比到 \(S\) 点近 \(x\) 英里,则 \(x\) 是:

A\(10\) B\(8\) C\(6\) D\(4\) E\(2\)

English

Two men at points \(R\) and \(S,\) \(76\) miles apart, set out at the same time to walk towards each other. The man at \(R\) walks uniformly at \(4\dfrac12\) miles per hour; the man at \(S\) walks at \(3\dfrac14\) miles per hour for the first hour, at \(3\dfrac34\) miles per hour for the second hour, and so on, in arithmetic progression. If the men meet \(x\) miles nearer \(R\) than \(S\) in an integral number of hours, then \(x\) is:

(A) \(10\)(B) \(8\)(C) \(6\)(D) \(4\)(E) \(2\)

21

表达式 \(x^2-y^2-z^2+2yz+x+y-z\) 具有:

A没有系数和指数均为整数的一次因式 B因式 \(-x+y+z\) C因式 \(x-y-z+1\) D因式 \(x+y-z+1\) E因式 \(x-y+z+1\)

English

The expression \(x^2-y^2-z^2+2yz+x+y-z\) has:

(A) no linear factor with integer coefficients and integer exponents(B) the factor \(-x+y+z\)(C) the factor \(x-y-z+1\)(D) the factor \(x+y-z+1\)(E) the factor \(x-y+z+1\)

22

锐角三角形 \(ABC\) 内接于以 \(O\) 为圆心的圆,且 \(\overset{\frown}{AB}=120^\circ\)、\(\overset{\frown}{BC}=72^\circ\)。取一点 \(E\) 于劣弧 \(AC\) 上,使 \(OE\) 垂直于 \(AC\)。角 \(OBE\) 与角 \(BAC\) 的度数之比是:

A\(\dfrac5{18}\) B\(\dfrac29\) C\(\dfrac14\) D\(\dfrac13\) E\(\dfrac49\)

English

Acute-angled triangle \(ABC\) is inscribed in a circle with center at \(O;\) \(\overset{\frown}{AB}=120^\circ\) and \(\overset{\frown}{BC}=72^\circ.\) A point \(E\) is taken in minor arc \(AC\) such that \(OE\) is perpendicular to \(AC.\) Then the ratio of the magnitudes of angles \(OBE\) and \(BAC\) is:

(A) \(\dfrac5{18}\)(B) \(\dfrac29\)(C) \(\dfrac14\)(D) \(\dfrac13\)(E) \(\dfrac49\)

23

\(A\) 先给 \(B\) 与 \(B\) 原有钱数相同的美分,再给 \(C\) 与 \(C\) 原有钱数相同的美分。接着,\(B\) 同样分别给 \(A\) 和 \(C\) 与他们当时各自钱数相同的美分。最后,\(C\) 也同样分别给 \(A\) 和 \(B\) 与他们当时各自钱数相同的美分。如果最终每人都有 \(16\) 美分,那么 \(A\) 最初有多少美分?

A\(24\) B\(26\) C\(28\) D\(30\) E\(32\)

English

\(A\) gives \(B\) as many cents as \(B\) has and \(C\) as many cents as \(C\) has. Similarly, \(B\) then gives \(A\) and \(C\) as many cents as each then has. \(C\), similarly, then gives \(A\) and \(B\) as many cents as each then has. If each finally has \(16\) cents, with how many cents does \(A\) start?

(A) \(24\)(B) \(26\)(C) \(28\)(D) \(30\)(E) \(32\)

24

考虑形如 \(x^2+bx+c=0\) 的方程。若系数 \(b\) 和 \(c\) 都从整数集合 \(\{1,2,3,4,5,6\}\) 中选取,其中有多少个方程具有实根?

A\(20\) B\(19\) C\(18\) D\(17\) E\(16\)

English

Consider equations of the form \(x^2+bx+c=0.\) How many such equations have real roots and have coefficients \(b\) and \(c\) selected from the set of integers \(\{1,2,3,4,5,6\}?\)

(A) \(20\)(B) \(19\)(C) \(18\)(D) \(17\)(E) \(16\)

25

在正方形 \(ABCD\) 的边 \(AD\) 上取一点 \(F\)。过 \(C\) 作 \(CF\) 的垂线,与 \(AB\) 的延长线交于 \(E\)。\(ABCD\) 的面积为 \(256\) 平方英寸,三角形 \(CEF\) 的面积为 \(200\) 平方英寸。则 \(BE\) 的长度为多少英寸?

题图

A\(12\) B\(14\) C\(15\) D\(16\) E\(20\)

English

Point \(F\) is taken in side \(AD\) of square \(ABCD.\) At \(C\) a perpendicular is drawn to \(CF,\) meeting \(AB\) extended at \(E.\) The area of \(ABCD\) is \(256\) square inches and the area of triangle \(CEF\) is \(200\) square inches. Then the number of inches in \(BE\) is:

题图

(A) \(12\)(B) \(14\)(C) \(15\)(D) \(16\)(E) \(20\)

26

形式 I。考虑以下命题:

\[ \begin{aligned} (1)\quad&p\land\neg q\land r,\\ (2)\quad&\neg p\land\neg q\land r,\\ (3)\quad&p\land\neg q\land\neg r,\\ (4)\quad&\neg p\land q\land r \end{aligned}\text{,}\]

其中 \(p\)、\(q\)、\(r\) 是命题。以上有多少个能推出 \((p\to q)\to r\) 为真?

形式 II。考虑以下命题:\((1)\) \(p\) 和 \(r\) 为真且 \(q\) 为假;\((2)\) \(r\) 为真且 \(p\) 和 \(q\) 为假;\((3)\) \(p\) 为真且 \(q\) 和 \(r\) 为假;\((4)\) \(q\) 和 \(r\) 为真且 \(p\) 为假。以上有多少个能推出命题“\(r\) 可由‘\(p\) 蕴含 \(q\)’推出”为真?

A\(0\) B\(1\) C\(2\) D\(3\) E\(4\)

English

Form I. Consider the statements

\[ \begin{aligned} (1)\quad&p\land\neg q\land r,\\ (2)\quad&\neg p\land\neg q\land r,\\ (3)\quad&p\land\neg q\land\neg r,\\ (4)\quad&\neg p\land q\land r, \end{aligned} \]

where \(p,\) \(q,\) \(r\) are propositions. How many of these imply the truth of \((p\to q)\to r?\)

Form II. Consider the statements \((1)\) \(p\) and \(r\) are true and \(q\) is false, \((2)\) \(r\) is true and \(p\) and \(q\) are false, \((3)\) \(p\) is true and \(q\) and \(r\) are false, \((4)\) \(q\) and \(r\) are true and \(p\) is false. How many of these imply the truth of the statement “\(r\) is implied by the statement that \(p\) implies \(q\)”?

(A) \(0\)(B) \(1\)(C) \(2\)(D) \(3\)(E) \(4\)

27

在平面上画六条直线,其中任意两条都不平行,任意三条都不共点。它们把平面分成的区域数是:

A\(16\) B\(20\) C\(22\) D\(24\) E\(26\)

English

Six straight lines are drawn in a plane with no two parallel and no three concurrent. The number of regions into which they divide the plane is:

(A) \(16\)(B) \(20\)(C) \(22\)(D) \(24\)(E) \(26\)

28

已知方程 \(3x^2-4x+k=0\) 有实根。使该方程两根之积最大的 \(k\) 值是:

A\(\dfrac{16}{9}\) B\(\dfrac{16}{3}\) C\(\dfrac49\) D\(\dfrac43\) E\(-\dfrac43\)

English

Given the equation \(3x^2-4x+k=0\) with real roots. The value of \(k\) for which the product of the roots of the equation is a maximum is:

(A) \(\dfrac{16}{9}\)(B) \(\dfrac{16}{3}\)(C) \(\dfrac49\)(D) \(\dfrac43\)(E) \(-\dfrac43\)

29

一个竖直向上发射的质点在 \(t\) 秒末到达 \(s\) 英尺的高度,其中 \(s=160t-16t^2\)。它能达到的最大高度是:

A\(800\) B\(640\) C\(400\) D\(320\) E\(160\)

English

A particle projected vertically upward reaches, at the end of \(t\) seconds, an elevation of \(s\) feet where \(s=160t-16t^2.\) The highest elevation is:

(A) \(800\)(B) \(640\)(C) \(400\)(D) \(320\)(E) \(160\)

30

设

\[ F=\log\frac{1+x}{1-x}\text{。}\]

求新函数 \(G\):把 \(x\) 在 \(F\) 中的每一次出现都替换为

\[ \frac{3x+x^3}{1+3x^2}\text{,}\]

并化简。化简后的 \(G\) 等于:

A\(-F\) B\(F\) C\(3F\) D\(F^3\) E\(F^3-F\)

English

Let

\[ F=\log\frac{1+x}{1-x}. \]

Form a new function \(G\) by replacing each \(x\) in \(F\) by

\[ \frac{3x+x^3}{1+3x^2}, \]

and simplify. The simplified expression \(G\) is equal to:

(A) \(-F\)(B) \(F\)(C) \(3F\)(D) \(F^3\)(E) \(F^3-F\)

31

方程 \(2x+3y=763\) 的正整数解共有:

A\(255\) B\(254\) C\(128\) D\(127\) E\(0\)

English

The number of solutions in positive integers of \(2x+3y=763\) is:

(A) \(255\)(B) \(254\)(C) \(128\)(D) \(127\)(E) \(0\)

32

矩形 \(R\) 的边长为 \(a\) 和 \(b\),其中 \(a\lt b\)。现要得到一个边长为 \(x\) 和 \(y\) 的矩形,其中 \(x\lt a\)、\(y\lt a\),使它的周长是 \(R\) 的三分之一,面积也是 \(R\) 的三分之一。这样的不同矩形共有:

A\(0\) B\(1\) C\(2\) D\(4\) E无穷多个

English

The dimensions of a rectangle \(R\) are \(a\) and \(b,\) \(a\lt b.\) It is required to obtain a rectangle with dimensions \(x\) and \(y,\) \(x\lt a,\) \(y\lt a,\) so that its perimeter is one-third that of \(R,\) and its area is one-third that of \(R.\) The number of such (different) rectangles is:

(A) \(0\)(B) \(1\)(C) \(2\)(D) \(4\)(E) infinitely many

33

已知直线 \(y=\dfrac34x+6\),另有直线 \(L\) 与它平行且相距 \(4\) 个单位。下列哪一个可能是 \(L\) 的方程?

A\(y=\dfrac34x+1\) B\(y=\dfrac34x\) C\(y=\dfrac34x-\dfrac23\) D\(y=\dfrac34x-1\) E\(y=\dfrac34x+2\)

English

Given the line \(y=\dfrac34x+6\) and a line \(L\) parallel to the given line and \(4\) units from it. A possible equation for \(L\) is:

(A) \(y=\dfrac34x+1\)(B) \(y=\dfrac34x\)(C) \(y=\dfrac34x-\dfrac23\)(D) \(y=\dfrac34x-1\)(E) \(y=\dfrac34x+2\)

34

在三角形 \(ABC\) 中,边 \(a=\sqrt3\),边 \(b=\sqrt3\),且边 \(c\gt3\)。设 \(x\) 是满足“边 \(c\) 所对角的度数必大于 \(x\)”的最大数,则 \(x\) 等于:

A\(150\) B\(120\) C\(105\) D\(90\) E\(60\)

English

In triangle \(ABC,\) side \(a=\sqrt3,\) side \(b=\sqrt3,\) and side \(c\gt3.\) Let \(x\) be the largest number such that the magnitude, in degrees, of the angle opposite side \(c\) exceeds \(x.\) Then \(x\) equals:

(A) \(150\)(B) \(120\)(C) \(105\)(D) \(90\)(E) \(60\)

35

一个三角形的三边长都是整数,面积也是整数。其中一边长为 \(21\),周长为 \(48\)。最短边的长度是:

A\(8\) B\(10\) C\(12\) D\(14\) E\(16\)

English

The lengths of the sides of a triangle are integers, and its area is also an integer. One side is \(21\) and the perimeter is \(48.\) The shortest side is:

(A) \(8\)(B) \(10\)(C) \(12\)(D) \(14\)(E) \(16\)

36

某人起初有 \(64\) 美分,共下注 \(6\) 次,赢三次、输三次,输赢的次序随机,每次获胜和失败的概率相同。每次下注金额都是下注时剩余钱数的一半,则最终结果是:

A损失 \(27\) 美分 B获利 \(27\) 美分 C损失 \(37\) 美分 D既没有获利也没有损失 E获利或损失取决于输赢的次序

English

A person starting with \(64\) cents and making \(6\) bets, wins three times and loses three times, the wins and losses occurring in random order. The chance for a win is equal to the chance for a loss. If each wager is for half the money remaining at the time of the bet, then the final result is:

(A) a loss of \(27\)¢(B) a gain of \(27\)¢(C) a loss of \(37\)¢(D) neither a gain nor a loss(E) a gain or a loss depending upon the order in which the wins and losses occur

37

直线上的点 \(P_1\)、\(P_2\)、\(\ldots\)、\(P_7\) 按此顺序排列,但间距不一定相等。在该直线上任取一点 \(P\),令 \(s\) 为以下无向线段长度之和:

\[ PP_1,\ PP_2,\ \ldots,\ PP_7\text{。}\]

则 \(s\) 取得最小值的充要条件是点 \(P\) 位于:

A\(P_1\) 与 \(P_7\) 的中点 B\(P_2\) 与 \(P_6\) 的中点 C\(P_3\) 与 \(P_5\) 的中点 D\(P_4\) 处 E\(P_1\) 处

English

Given points \(P_1,\) \(P_2,\) \(\ldots,\) \(P_7\) on a straight line, in the order stated (not necessarily evenly spaced). Let \(P\) be an arbitrarily selected point on the line and let \(s\) be the sum of the undirected lengths

\[ PP_1,\ PP_2,\ \ldots,\ PP_7. \]

Then \(s\) is smallest if and only if the point \(P\) is:

(A) midway between \(P_1\) and \(P_7\)(B) midway between \(P_2\) and \(P_6\)(C) midway between \(P_3\) and \(P_5\)(D) at \(P_4\)(E) at \(P_1\)

38

点 \(F\) 位于平行四边形 \(ABCD\) 的边 \(AD\) 的延长线上。\(BF\) 与对角线 \(AC\) 交于 \(E\),与边 \(DC\) 交于 \(G\)。若 \(EF=32\) 且 \(GF=24\),则 \(BE\) 等于:

题图

A\(4\) B\(8\) C\(10\) D\(12\) E\(16\)

English

Point \(F\) is taken on the extension of side \(AD\) of parallelogram \(ABCD.\) \(BF\) intersects diagonal \(AC\) at \(E\) and side \(DC\) at \(G.\) If \(EF=32\) and \(GF=24,\) then \(BE\) equals:

题图

(A) \(4\)(B) \(8\)(C) \(10\)(D) \(12\)(E) \(16\)

39

在三角形 \(ABC\) 中,作线段 \(CE\) 和 \(AD\),使 \(\dfrac{CD}{DB}=\dfrac31\) 且 \(\dfrac{AE}{EB}=\dfrac32\)。

设 \(r=\dfrac{CP}{PE}\),其中 \(P\) 是 \(CE\) 与 \(AD\) 的交点。则 \(r\) 等于:

题图

A\(3\) B\(\dfrac32\) C\(4\) D\(5\) E\(\dfrac52\)

English

In triangle \(ABC\) lines \(CE\) and \(AD\) are drawn so that \(\dfrac{CD}{DB}=\dfrac31\) and \(\dfrac{AE}{EB}=\dfrac32.\)

Let \(r=\dfrac{CP}{PE},\) where \(P\) is the intersection point of \(CE\) and \(AD.\) Then \(r\) equals:

题图

(A) \(3\)(B) \(\dfrac32\)(C) \(4\)(D) \(5\)(E) \(\dfrac52\)

40

若数 \(x\) 满足方程 \(\sqrt[3]{x+9}-\sqrt[3]{x-9}=3\),则 \(x^2\) 介于:

A\(55\) 与 \(65\) 之间 B\(65\) 与 \(75\) 之间 C\(75\) 与 \(85\) 之间 D\(85\) 与 \(95\) 之间 E\(95\) 与 \(105\) 之间

English

If \(x\) is a number satisfying the equation \(\sqrt[3]{x+9}-\sqrt[3]{x-9}=3,\) then \(x^2\) is between:

(A) \(55\) and \(65\)(B) \(65\) and \(75\)(C) \(75\) and \(85\)(D) \(85\) and \(95\)(E) \(95\) and \(105\)

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逐题解析

1 答案 D

表达式 \(\frac{x}{x+1}\) 在 \(x=-1\) 时没有定义。因此,横坐标为 \(-1\) 的点不可能在其图像上。

所以正确答案是 D。

English

The expression \(\frac{x}{x+1}\) is undefined when \(x=-1.\) Therefore no point whose first coordinate is \(-1\) can lie on its graph.

Thus, the correct answer is D.

2 答案 A

这里 \(x-y=2-(-2)=4\),所以 \[n=2-(-2)^4=2-16=-14\text{。}\]

所以正确答案是 A。

English

Here \(x-y=2-(-2)=4,\) so \[n=2-(-2)^4=2-16=-14.\]

Therefore, the correct answer is A.

3 答案 E

方程为 \(\frac1{x+1}=x-1\)。因此 \(1=x^2-1\),所以 \(x=\pm\sqrt2\)。这两个值都不在选项中。

所以正确答案是 E。

English

The equation is \(\frac1{x+1}=x-1.\) Hence \(1=x^2-1,\) so \(x=\pm\sqrt2.\) Neither value is listed.

Thus, the correct answer is E.

4 答案 D

交点满足 \(x^2-3x-k=0\)。要有两个相同的解,必须满足 \[(-3)^2-4(1)(-k)=9+4k=0\text{,}\]从而 \(k=-\frac{9}{4}\)。

所以正确答案是 D。

English

Intersections satisfy \(x^2-3x-k=0.\) Two identical solutions require \[(-3)^2-4(1)(-k)=9+4k=0,\] giving \(k=-\frac{9}{4}.\)

Therefore, the correct answer is D.

5 答案 E

对数为实数要求 \(x\gt0\)。由于底数 \(10\) 大于 \(1\),\(\log_{10}x\lt0=\log_{10}1\) 推出 \(x\lt1\)。

所以正确答案是 E。

English

The logarithm is real only for \(x\gt0.\) Since base \(10\) is greater than \(1,\) \(\log_{10}x\lt0=\log_{10}1\) implies \(x\lt1.\)

Thus, the correct answer is E.

6 答案 B

因为 \(C\) 是斜边 \(AD\) 的中点,所以 \(AC=BC=CD\)。又因为 \(AB=BC\),三角形 \(ABC\) 是等边三角形。因此 \(\angle DAB=\angle CAB=60^\circ\)。

所以正确答案是 B。

English

Since \(C\) is the midpoint of hypotenuse \(AD,\) \(AC=BC=CD.\) Given \(AB=BC,\) triangle \(ABC\) is equilateral. Thus \(\angle DAB=\angle CAB=60^\circ.\)

Therefore, the correct answer is B.

7 答案 A

直线 \((1)\) 的斜率是 \(\frac{2}{3}\),直线 \((4)\) 的斜率是 \(-\frac{3}{2}\)。它们的积为 \(-1\),所以这两条直线互相垂直。

所以正确答案是 A。

English

Line \((1)\) has slope \(\frac{2}{3},\) and line \((4)\) has slope \(-\frac{3}{2}.\) Their product is \(-1,\) so these two lines are perpendicular.

Thus, the correct answer is A.

8 答案 D

因为 \(1260=2^2\cdot3^2\cdot5\cdot7\),要使每个指数都能被 \(3\) 整除,最小的乘数是 \[2\cdot3\cdot5^2\cdot7^2=7350\text{。}\]

所以正确答案是 D。

English

Since \(1260=2^2\cdot3^2\cdot5\cdot7,\) the least multiplier making every exponent divisible by \(3\) is \[2\cdot3\cdot5^2\cdot7^2=7350.\]

Thus, the correct answer is D.

9 答案 C

恰好选取 \(-a^{-\frac{1}{2}}\) 共 \(r\) 次所得项的指数为 \(7-r-\frac{r}{2}=7-\frac{3r}{2}\)。令它等于 \(-\frac{1}{2}\),得到 \(r=5\)。该项的系数是 \(\binom75(-1)^5=-21\)。

所以正确答案是 C。

English

The term using \(-a^{-\frac{1}{2}}\) exactly \(r\) times has exponent \(7-r-\frac{r}{2}=7-\frac{3r}{2}.\) Setting this to \(-\frac{1}{2}\) gives \(r=5.\) Its coefficient is \(\binom75(-1)^5=-21.\)

Therefore, the correct answer is C.

10 答案 B

令对边的高度为 \(0\),两个顶点的高度为 \(a\)。则 \(P=(\frac{a}{2},d)\)。顶点距离与 \(d\) 相等,故 \[\left(\frac a2\right)^2+(a-d)^2=d^2\text{,}\]所以 \(\frac{5a^2}{4}=2ad\),从而 \(d=\frac{5a}{8}\)。

所以正确答案是 B。

English

Place the opposite side at height \(0\) and the two vertices at height \(a.\) Then \(P=(\frac{a}{2},d).\) Equality of the vertex distance and \(d\) gives \[\left(\frac a2\right)^2+(a-d)^2=d^2,\] so \(\frac{5a^2}{4}=2ad\) and \(d=\frac{5a}{8}.\)

Thus, the correct answer is B.

11 答案 B

原来的总和是 \(50\cdot38=1900\)。去掉 \(45+55=100\) 后,剩余各数之和为 \(1800\),所以新的平均数为 \(\frac{1800}{48}=37.5\)。

所以正确答案是 B。

English

The original sum is \(50\cdot38=1900.\) After removing \(45+55=100,\) the remaining sum is \(1800,\) so the new mean is \(\frac{1800}{48}=37.5.\)

Thus, the correct answer is B.

12 答案 E

因为 \(P+R=Q+S\),所以 \[ \begin{aligned} S&=P+R-Q\\ &=(-3,-2)+(9,1)-(1,-5)\\ &=(5,4) \end{aligned}\text{。}\]坐标之和为 \(5+4=9\)。

所以正确答案是 E。

English

Since \(P+R=Q+S,\) \[ \begin{aligned} S&=P+R-Q\\ &=(-3,-2)+(9,1)-(1,-5)\\ &=(5,4). \end{aligned} \] The coordinate sum is \(5+4=9.\)

Therefore, the correct answer is E.

13 答案 E

化为最简分数后,左边的分母只能是 \(2\) 的幂,右边的分母只能是 \(3\) 的幂。两边相等就迫使两边都是整数。若 \(c\) 或 \(d\) 为负数,\(3^c+3^d\) 的分母中仍会含因数 \(3\)。因此 \(c,d\geq0\)。

同理,\(a\) 或 \(b\) 为负数时,和为整数的唯一特殊情形是 \(a=b=-1\),此时和为 \(1\)。但 \(3^c+3^d\geq2\)。所以也有 \(a,b\geq0\),没有一个可以是负数。

所以正确答案是 E。

English

In lowest terms, the left side can have only a power of \(2\) in its denominator, while the right side can have only a power of \(3.\) Equality therefore forces both sides to be integers. If either \(c\) or \(d\) were negative, \(3^c+3^d\) would retain a factor \(3\) in its denominator. Thus \(c,d\geq0.\)

Likewise, negative \(a\) or \(b\) can give an integer only in the exceptional case \(a=b=-1,\) whose sum is \(1.\) But \(3^c+3^d\geq2.\) Hence \(a,b\geq0\) as well, so none can be negative.

Thus, the correct answer is E.

14 答案 A

第一个方程的两根之和为 \(-k\),第二个方程的两根之和为 \(k\)。因为两个根都增加了 \(5\),所以 \(k=-k+10\),从而 \(k=5\)。

所以正确答案是 A。

English

The first pair of roots has sum \(-k,\) while the second has sum \(k.\) Since both roots increase by \(5,\) \(k=-k+10,\) so \(k=5.\)

Therefore, the correct answer is A.

15 答案 C

若圆的半径为 \(r\),则等边三角形的边长为 \(2\sqrt3r\),面积为 \(3\sqrt3r^2\)。正方形的边长为 \(r\sqrt2\),面积为 \(2r^2\)。所求比为 \(3\sqrt3:2\)。

所以正确答案是 C。

English

If the circle has radius \(r,\) the equilateral triangle has side \(2\sqrt3r\) and area \(3\sqrt3r^2.\) The square has side \(r\sqrt2\) and area \(2r^2.\) The ratio is \(3\sqrt3:2.\)

Thus, the correct answer is C.

16 答案 C

这些条件给出 \[ \begin{gathered} 2b=a+c,\\ b^2=(a+1)c,\qquad b^2=a(c+2) \end{gathered}\text{。}\]比较后两个等式,得 \(c=2a\)。因此 \(b=\frac{3a}{2}\),并且 \(\frac{9a^2}{4}=2a(a+1)\)。由于 \(a\ne0\),得 \(a=8\),所以 \(b=12\)。

所以正确答案是 C。

English

The conditions give \[ \begin{gathered} 2b=a+c,\\ b^2=(a+1)c,\qquad b^2=a(c+2). \end{gathered} \] Comparing the last two yields \(c=2a.\) Thus \(b=\frac{3a}{2},\) and \(\frac{9a^2}{4}=2a(a+1).\) Since \(a\ne0,\) \(a=8,\) so \(b=12.\)

Therefore, the correct answer is C.

17 答案 A

当 \(y\ne\pm a\) 时,分子化简为 \(\frac{a^2+y^2}{a^2-y^2}\),分母恰为它的相反数。因此商为 \(-1\)。当 \(y=a\) 或 \(y=-a\) 时,原式没有定义。

所以正确答案是 A。

English

For \(y\ne\pm a,\) the numerator simplifies to \(\frac{a^2+y^2}{a^2-y^2}\) and the denominator to its negative. Their quotient is therefore \(-1.\) The two values \(y=a\) and \(y=-a\) are undefined.

Thus, the correct answer is A.

18 答案 A

因为 \(EF\) 是直径,所以 \(\angle EAF=90^\circ\)。又因为 \(EF\perp BC\),所以 \(\angle EMU=90^\circ\)。由于 \(E,U,A\) 共线,两个三角形在 \(E\) 点有同一个锐角。因此 \(\triangle EUM\sim\triangle EFA\)。

所以正确答案是 A。

English

Since \(EF\) is a diameter, \(\angle EAF=90^\circ.\) Also \(EF\perp BC,\) so \(\angle EMU=90^\circ.\) Because \(E,U,A\) are collinear, the two triangles share the same acute angle at \(E.\) Hence \(\triangle EUM\sim\triangle EFA.\)

Therefore, the correct answer is A.

19 答案 B

条件为 \[49+\frac78(n-50)\geq\frac9{10}n\text{。}\]化简得 \(\frac{21}{4}\geq \frac{n}{40}\),所以 \(n\leq210\)。最大值为 \(210\)。

所以正确答案是 B。

English

The condition is \[49+\frac78(n-50)\geq\frac9{10}n.\] Simplifying gives \(\frac{21}{4}\geq \frac{n}{40},\) so \(n\leq210.\) The maximum is \(210.\)

Thus, the correct answer is B.

20 答案 D

在 \(h\) 小时内,第一个人走了 \(\frac{9h}{2}\) 英里。第二个人走过的路程是等差数列之和 \[ \begin{aligned} &\frac h2\left(2\cdot\frac{13}{4} +(h-1)\frac12\right)\\ &\qquad=3h+\frac{h^2}{4} \end{aligned}\text{。}\]两人的路程之和为 \(76\),得到 \(h^2+30h-304=0\),其正根为 \(h=8\)。他们分别走了 \(36\) 英里和 \(40\) 英里,因此相遇点到 \(R\) 点比到 \(S\) 点近 \(4\) 英里。

所以正确答案是 D。

English

In \(h\) hours the first man walks \(\frac{9h}{2}.\) The second walks an arithmetic-series total \[ \begin{aligned} &\frac h2\left(2\cdot\frac{13}{4} +(h-1)\frac12\right)\\ &\qquad=3h+\frac{h^2}{4}. \end{aligned} \] Their distances sum to \(76,\) giving \(h^2+30h-304=0,\) whose positive root is \(h=8.\) They walk \(36\) and \(40\) miles, so the meeting point is \(4\) miles nearer \(R\) than \(S.\)

Thus, the correct answer is D.

21 答案 E

令 \(u=y-z\)。两部分可分解为 \[ \begin{gathered} x^2-u^2=(x+u)(x-u),\\ x+u=(x+u)\cdot1 \end{gathered}\text{。}\]因此整个表达式是 \((x+u)(x-u+1)\),它的一个因式是 \(x-y+z+1\)。

所以正确答案是 E。

English

Let \(u=y-z.\) The two groups factor as \[ \begin{gathered} x^2-u^2=(x+u)(x-u),\\ x+u=(x+u)\cdot1. \end{gathered} \] Thus the whole expression is \((x+u)(x-u+1),\) and one factor is \(x-y+z+1.\)

Therefore, the correct answer is E.

22 答案 D

劣弧 \(AC\) 的度数为 \(360^\circ-120^\circ-72^\circ=168^\circ\)。因为 \(OE\perp AC\),所以 \(E\) 平分该弧,故 \(\angle BOE=72^\circ+84^\circ=156^\circ\)。等腰三角形 \(OBE\) 给出 \(\angle OBE=12^\circ\),而 \(\angle BAC=\frac{72^\circ}{2}=36^\circ\)。二者之比为 \(\frac{1}{3}\)。

所以正确答案是 D。

English

Minor arc \(AC\) has measure \(360^\circ-120^\circ-72^\circ=168^\circ.\) Since \(OE\perp AC,\) \(E\) bisects that arc, so \(\angle BOE=72^\circ+84^\circ=156^\circ.\) Isosceles triangle \(OBE\) gives \(\angle OBE=12^\circ,\) while \(\angle BAC=\frac{72^\circ}{2}=36^\circ.\) Their ratio is \(\frac{1}{3}.\)

Thus, the correct answer is D.

23 答案 B

逆向还原这些交易。\(C\) 给钱之前,三人的钱数为 \((8,8,32)\)。\(B\) 给钱之前,钱数为 \((4,28,16)\)。\(A\) 给钱之前,\(B\) 和 \(C\) 必须分别有 \(14\) 和 \(8\) 美分,而 \(A\) 有 \(4+14+8=26\) 美分。

所以正确答案是 B。

English

Reverse the transactions. Before \(C\) gives, the holdings are \((8,8,32).\) Before \(B\) gives, they are \((4,28,16).\) Before \(A\) gives, \(B\) and \(C\) must have had \(14\) and \(8,\) while \(A\) had \(4+14+8=26.\)

Therefore, the correct answer is B.

24 答案 B

当 \(c\) 依次等于 \(1\)、\(2\)、\(3\)、\(4\)、\(5\) 和 \(6\) 时,条件 \(b^2\geq4c\) 相应地允许 \(5\)、\(4\)、\(3\)、\(3\)、\(2\) 和 \(2\) 个给定集合中的 \(b\) 值。它们的总和为 \(5+4+3+3+2+2=19\)。

所以正确答案是 B。

English

For \(c\) equal to \(1,\) \(2,\) \(3,\) \(4,\) \(5,\) and \(6,\) the condition \(b^2\geq4c\) permits respectively \(5,\) \(4,\) \(3,\) \(3,\) \(2,\) and \(2\) values of \(b\) in the given set. Their sum is \(5+4+3+3+2+2=19.\)

Thus, the correct answer is B.

25 答案 A

互余的锐角和正方形的等边给出 \(\triangle CDF\cong\triangle CBE\),从而 \(CF=CE\)。由于 \(CF\perp CE\),\[ \begin{aligned} 200=[CEF] &=\frac12 CE\cdot CF\\ &=\frac12CE^2 \end{aligned}\text{,}\]所以 \(CE=20\)。正方形边长为 \(16\),在直角三角形 \(CBE\) 中,\(BE=\sqrt{20^2-16^2}=12\)。

所以正确答案是 A。

English

The complementary acute angles and the equal square sides give \(\triangle CDF\cong\triangle CBE,\) hence \(CF=CE.\) Since \(CF\perp CE,\) \[ \begin{aligned} 200=[CEF] &=\frac12 CE\cdot CF\\ &=\frac12CE^2, \end{aligned} \] so \(CE=20.\) The square side is \(16,\) and right triangle \(CBE\) gives \(BE=\sqrt{20^2-16^2}=12.\)

Therefore, the correct answer is A.

26 答案 E

在情形 \((1)\) 和 \((3)\) 中,\(p\to q\) 为假,所以外层蕴含为真。在情形 \((2)\) 和 \((4)\) 中,\(p\to q\) 为真,而 \(r\) 也为真,所以外层蕴含仍为真。因此四个命题都能推出它。

所以正确答案是 E。

English

In cases \((1)\) and \((3),\) \(p\to q\) is false, so the outer implication is true. In cases \((2)\) and \((4),\) \(p\to q\) is true, but \(r\) is also true, so the outer implication is again true. All four statements imply it.

Thus, the correct answer is E.

27 答案 C

依次画出的直线分别新增 \(1,2,\ldots,6\) 个区域。因此区域总数为 \[ 1+\sum_{k=1}^{6}k=1+21=22\text{。}\]

所以正确答案是 C。

English

Successive lines create \(1,2,\ldots,6\) new regions. Thus the total is \[ 1+\sum_{k=1}^{6}k=1+21=22. \]

Therefore, the correct answer is C.

28 答案 D

有实根要求 \(16-12k\geq0\),所以 \(k\leq\frac{4}{3}\)。两根之积为 \(\frac{k}{3}\),它随 \(k\) 增大。因此在 \(k=\frac{4}{3}\) 时达到最大。

所以正确答案是 D。

English

Real roots require \(16-12k\geq0,\) so \(k\leq\frac{4}{3}.\) The root product is \(\frac{k}{3},\) which increases with \(k.\) It is therefore largest at \(k=\frac{4}{3}.\)

Thus, the correct answer is D.

29 答案 C

顶点对应的时刻为 \(t=-\frac{160}{2\cdot-16}=5\)。此时 \[s(5)=160(5)-16(25)=400\text{。}\]

所以正确答案是 C。

English

The vertex occurs at \(t=-\frac{160}{2\cdot-16}=5.\) Then \[s(5)=160(5)-16(25)=400.\]

Thus, the correct answer is C.

30 答案 C

令 \(u=\frac{3x+x^3}{1+3x^2}\),则 \[ \begin{aligned} \frac{1+u}{1-u} &=\frac{1+3x+3x^2+x^3} {1-3x+3x^2-x^3}\\ &=\left(\frac{1+x}{1-x}\right)^3 \end{aligned}\text{。}\]因此 \(G=\log(\frac{1+x}{1-x})^3\),所以 \(G=3F\)。

所以正确答案是 C。

English

For \(u=\frac{3x+x^3}{1+3x^2},\) \[ \begin{aligned} \frac{1+u}{1-u} &=\frac{1+3x+3x^2+x^3} {1-3x+3x^2-x^3}\\ &=\left(\frac{1+x}{1-x}\right)^3. \end{aligned} \] Therefore \(G=\log(\frac{1+x}{1-x})^3,\) so \(G=3F.\)

Thus, the correct answer is C.

31 答案 D

\(y\) 必须为奇数,才能使 \(763-3y\) 为偶数;正数条件给出 \(1\leq y\leq253\)。奇数 \(1\)、\(3\)、\(\ldots\) 和 \(253\) 共有 \(\frac{253+1}{2}=127\) 个。

所以正确答案是 D。

English

We need \(y\) odd so that \(763-3y\) is even, and positivity gives \(1\leq y\leq253.\) The odd values \(1,\) \(3,\) \(\ldots,\) and \(253\) number \(\frac{253+1}{2}=127.\)

Therefore, the correct answer is D.

32 答案 A

这些条件给出 \(3(x+y)=a+b\) 和 \(3xy=ab\)。相除得到 \[\frac1x+\frac1y=\frac1a+\frac1b\text{。}\]但是 \(x\lt a\) 且 \(y\lt a\) 使左边大于 \(\frac{2}{a}\),而 \(a\lt b\) 使右边小于 \(\frac{2}{a}\)。这不可能成立。

所以正确答案是 A。

English

The conditions give \(3(x+y)=a+b\) and \(3xy=ab.\) Dividing yields \[\frac1x+\frac1y=\frac1a+\frac1b.\] But \(x\lt a\) and \(y\lt a\) make the left side greater than \(\frac{2}{a},\) while \(a\lt b\) makes the right side less than \(\frac{2}{a}.\) This is impossible.

Thus, the correct answer is A.

33 答案 A

已知直线可写成 \(3x-4y+24=0\)。平行直线 \(y=\frac{3x}{4}+b\) 可写成 \(3x-4y+4b=0\),所以两直线间的距离为 \(\frac{|24-4b|}{5}\)。令它等于 \(4\),得到 \(|6-b|=5\),从而 \(b=1\) 或 \(11\)。选项 A 是可能的。

所以正确答案是 A。

English

The given line is \(3x-4y+24=0.\) A parallel line \(y=\frac{3x}{4}+b\) is \(3x-4y+4b=0,\) so the distance is \(\frac{|24-4b|}{5}.\) Setting this equal to \(4\) gives \(|6-b|=5,\) hence \(b=1\) or \(11.\) Choice A is possible.

Thus, the correct answer is A.

34 答案 B

余弦定理给出 \[\cos C=\frac{3+3-c^2}{2\sqrt3\sqrt3}=\frac{6-c^2}{6}\text{。}\]因为 \(c\gt3\),所以 \(\cos C\lt-\frac{1}{2}\),从而 \(C\gt120^\circ\)。角度可以趋近 \(120^\circ\),只要 \(c\) 从右侧趋近 \(3\),所以能保证的最大下界是 \(120\)。

所以正确答案是 B。

English

The law of cosines gives \[\cos C=\frac{3+3-c^2}{2\sqrt3\sqrt3}=\frac{6-c^2}{6}.\] Since \(c\gt3,\) \(\cos C\lt-\frac{1}{2},\) so \(C\gt120^\circ.\) Values can approach \(120^\circ\) as \(c\) approaches \(3\) from above, so the largest guaranteed bound is \(120.\)

Therefore, the correct answer is B.

35 答案 B

设另外两边为 \(x\) 和 \(27-x\),并令 \(x\leq27-x\)。半周长是 \(24\),所以海伦公式给出 \[ \begin{aligned} K^2 &=24\cdot3(24-x)(x-3)\\ &=72(24-x)(x-3) \end{aligned}\text{。}\]三角形不等式给出 \(4\leq x\leq13\)。逐一检查这些整数,当 \(x=4,\ldots,9\) 时该表达式都不是完全平方数;当 \(x=10\) 时,它等于 \(72\cdot14\cdot7=7056=84^2\)。因此最短边长为 \(10\)。

所以正确答案是 B。

English

Let the other sides be \(x\) and \(27-x,\) with \(x\leq27-x.\) The semiperimeter is \(24,\) so Heron’s formula gives \[ \begin{aligned} K^2 &=24\cdot3(24-x)(x-3)\\ &=72(24-x)(x-3). \end{aligned} \] The triangle inequalities give \(4\leq x\leq13.\) Checking these integers, the expression is not a square for \(x=4,\ldots,9,\) while at \(x=10\) it is \(72\cdot14\cdot7=7056=84^2.\) Thus the shortest side is \(10.\)

Therefore, the correct answer is B.

36 答案 C

三胜三负之后,剩余钱数(单位:美分)为 \[ \begin{aligned} 64\left(\frac32\right)^3 \left(\frac12\right)^3 &=64\cdot\frac{27}{64}\\ &=27 \end{aligned}\text{。}\]因此无论次序如何,损失都是 \(64-27=37\) 美分。

所以正确答案是 C。

English

After three wins and three losses, the amount, in cents, is \[ \begin{aligned} 64\left(\frac32\right)^3 \left(\frac12\right)^3 &=64\cdot\frac{27}{64}\\ &=27. \end{aligned} \] The loss is \(64-27=37\) cents, regardless of order.

Thus, the correct answer is C.

37 答案 D

对任意一对 \(P_i,P_{8-i}\),距离之和 \(PP_i+PP_{8-i}\) 在 \(P\) 位于它们之间时最小。这三个区间都包含 \(P_4\)。剩余的一项 \(PP_4\) 仅在 \(P=P_4\) 时达到最小。

所以正确答案是 D。

English

For any pair \(P_i,P_{8-i},\) the sum \(PP_i+PP_{8-i}\) is minimized when \(P\) lies between them. The three such intervals all contain \(P_4.\) The remaining term \(PP_4\) is uniquely minimized at \(P=P_4.\)

Thus, the correct answer is D.

38 答案 E

取仿射坐标 \(A=0\)、\(B=u\)、\(D=v\)、\(C=u+v\),以及 \(F=tv\)。\(FB\) 上一点可写为 \(su+t(1-s)v\)。与 \(DC\) 相交得到 \(s_G=\frac{t-1}{t}\),与 \(AC\) 相交得到 \(s_E=\frac{t}{t+1}\)。因此 \[ \begin{aligned} \frac{GF}{EF} &=\frac{s_G}{s_E} =\frac{t^2-1}{t^2}\\ &=\frac{24}{32}=\frac34 \end{aligned}\text{,}\]所以 \(t=2\),且 \(s_E=\frac{2}{3}\)。于是 \(EF=(\frac{2}{3})FB=32\),从而 \(FB=48\),\(BE=16\)。

所以正确答案是 E。

English

Use affine coordinates \(A=0,\) \(B=u,\) \(D=v,\) \(C=u+v,\) and \(F=tv.\) A point on \(FB\) is \(su+t(1-s)v.\) Intersecting \(DC\) gives \(s_G=\frac{t-1}{t},\) while intersecting \(AC\) gives \(s_E=\frac{t}{t+1}.\) Therefore \[ \begin{aligned} \frac{GF}{EF} &=\frac{s_G}{s_E} =\frac{t^2-1}{t^2}\\ &=\frac{24}{32}=\frac34, \end{aligned} \] so \(t=2\) and \(s_E=\frac{2}{3}.\) Thus \(EF=(\frac{2}{3})FB=32,\) making \(FB=48\) and \(BE=16.\)

Therefore, the correct answer is E.

39 答案 D

比 \(AE:EB=3:2\) 对应质量 \(m_A=2\) 和 \(m_B=3\)。比 \(CD:DB=3:1\) 随即给出 \(m_C=1\)。点 \(E\) 的质量为 \(m_A+m_B=5\),所以在塞瓦线 \(CE\) 上,\[\frac{CP}{PE}=\frac{m_E}{m_C}=\frac51=5\text{。}\]

所以正确答案是 D。

English

The ratio \(AE:EB=3:2\) is represented by masses \(m_A=2\) and \(m_B=3.\) The ratio \(CD:DB=3:1\) then gives \(m_C=1.\) Point \(E\) has mass \(m_A+m_B=5,\) so along cevian \(CE,\) \[\frac{CP}{PE}=\frac{m_E}{m_C}=\frac51=5.\]

Thus, the correct answer is D.

40 答案 C

令 \(u=\sqrt[3]{x+9}\),\(v=\sqrt[3]{x-9}\)。则 \(u-v=3\),并且 \[ \begin{aligned} 18&=u^3-v^3\\ &=(u-v)(u^2+uv+v^2) \end{aligned}\text{,}\]所以 \(u^2+uv+v^2=6\)。与 \((u-v)^2=9\) 比较,得 \(uv=-1\)。因此 \((u+v)^2=5\)。对 \(2u=3\pm\sqrt5\) 两边取立方,得到 \(x=\pm4\sqrt5\);无论哪种情形,\(x^2=80\),介于 \(75\) 与 \(85\) 之间。

所以正确答案是 C。

English

Let \(u=\sqrt[3]{x+9}\) and \(v=\sqrt[3]{x-9}.\) Then \(u-v=3\) and \[ \begin{aligned} 18&=u^3-v^3\\ &=(u-v)(u^2+uv+v^2), \end{aligned} \] so \(u^2+uv+v^2=6.\) Comparing with \((u-v)^2=9\) gives \(uv=-1.\) Hence \((u+v)^2=5.\) Cubing \(2u=3\pm\sqrt5\) gives \(x=\pm4\sqrt5,\) and either way \(x^2=80,\) between \(75\) and \(85.\)

Therefore, the correct answer is C.

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