1960 AMC 12 真题(P版)
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
若 \(2\) 是方程 \(x^3+hx+10=0\) 的一个解(根),则 \(h\) 等于:
A\(10\) B\(9\) C\(2\) D\(-2\) E\(-9\)
English
If \(2\) is a solution (root) of \(x^3+hx+10=0,\) then \(h\) equals:
(A) \(10\)(B) \(9\)(C) \(2\)(D) \(-2\)(E) \(-9\)
2
一只钟从恰好 \(6{:}00\) 开始敲响 \(6\) 下,共用 \(5\) 秒。若每相邻两次敲响的时间间隔相同,那么敲响 \(12\) 下需要多少秒?
A\(9\dfrac13\) B\(10\) C\(11\) D\(14\dfrac23\) E以上都不是
English
It takes \(5\) seconds for a clock to strike \(6\) o’clock beginning at \(6{:}00\) o’clock precisely. If the strikings are uniformly spaced, how long, in seconds, does it take to strike \(12\) o’clock?
(A) \(9\dfrac13\)(B) \(10\)(C) \(11\)(D) \(14\dfrac23\)(E) none of these
3
对一张金额为 \(\$10{,}000\) 的账单,打 \(40\%\) 的折扣与先后打 \(36\%\) 和 \(4\%\) 的两次折扣之间的差额(以美元计)是:
A\(0\) B\(144\) C\(256\) D\(400\) E\(416\)
English
Applied to a bill for \(\$10{,}000\) the difference between a discount of \(40\%\) and two successive discounts of \(36\%\) and \(4\%,\) expressed in dollars, is:
(A) \(0\)(B) \(144\)(C) \(256\)(D) \(400\)(E) \(416\)
4
一个三角形的两个角均为 \(60^\circ\),它们的夹边长为 \(4\) 英寸。这个三角形的面积(以平方英寸计)是:
A\(8\sqrt3\) B\(8\) C\(4\sqrt3\) D\(4\) E\(2\sqrt3\)
English
Each of two angles of a triangle is \(60^\circ\) and the included side is \(4\) inches. The area of the triangle, in square inches, is:
(A) \(8\sqrt3\)(B) \(8\)(C) \(4\sqrt3\)(D) \(4\)(E) \(2\sqrt3\)
5
图像 \(x^2+y^2=9\) 与 \(y^2=9\) 的不同公共点共有:
A无穷多个 B四个 C两个 D一个 E没有
English
The number of distinct points common to the graphs of \(x^2+y^2=9\) and \(y^2=9\) is:
(A) infinitely many(B) four(C) two(D) one(E) none
6
一个圆的周长为 \(100\) 英寸。该圆内接正方形的边长(以英寸计)是:
A\(\dfrac{25\sqrt2}{\pi}\) B\(\dfrac{50\sqrt2}{\pi}\) C\(\dfrac{100}{\pi}\) D\(\dfrac{100\sqrt2}{\pi}\) E\(50\sqrt2\)
English
The circumference of a circle is \(100\) inches. The side of a square inscribed in this circle, expressed in inches, is:
(A) \(\dfrac{25\sqrt2}{\pi}\)(B) \(\dfrac{50\sqrt2}{\pi}\)(C) \(\dfrac{100}{\pi}\)(D) \(\dfrac{100\sqrt2}{\pi}\)(E) \(50\sqrt2\)
7
圆 I 经过圆 II 的圆心,并与圆 II 相切。圆 I 的面积为 \(4\) 平方英寸。那么圆 II 的面积(以平方英寸计)是:
A\(8\) B\(8\sqrt2\) C\(8\sqrt\pi\) D\(16\) E\(16\sqrt2\)
English
Circle I passes through the center of, and is tangent to, circle II. The area of circle I is \(4\) square inches. Then the area of circle II, in square inches, is:
(A) \(8\)(B) \(8\sqrt2\)(C) \(8\sqrt\pi\)(D) \(16\)(E) \(16\sqrt2\)
8
数 \(2.5252525\ldots\) 可以写成分数。将该分数约成最简分数后,其分子与分母之和是:
A\(7\) B\(29\) C\(141\) D\(349\) E以上都不是
English
The number \(2.5252525\ldots\) can be written as a fraction. When reduced to lowest terms the sum of the numerator and denominator of this fraction is:
(A) \(7\)(B) \(29\)(C) \(141\)(D) \(349\)(E) none of these
9
分式 \[ \frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac} \] 在 \(a\)、\(b\) 和 \(c\) 的取值满足适当限制时:
A不能约简 B可约简为 \(-1\) C可约简为一个三项式 D可约简为 \(\dfrac{a-b+c}{a+b-c}\) E可约简为 \(\dfrac{a+b-c}{a-b+c}\)
English
The fraction \[ \frac{a^2+b^2-c^2+2ab}{a^2+c^2-b^2+2ac} \] is (with suitable restrictions on the values of \(a,\) \(b,\) and \(c\)):
(A) irreducible(B) reducible to \(-1\)(C) reducible to a polynomial of three terms(D) reducible to \(\dfrac{a-b+c}{a+b-c}\)(E) reducible to \(\dfrac{a+b-c}{a-b+c}\)
10
给出以下六个陈述:
\((1)\) 所有女性都是好驾驶员。
\((2)\) 有些女性是好驾驶员。
\((3)\) 没有男性是好驾驶员。
\((4)\) 所有男性都是差驾驶员。
\((5)\) 至少有一名男性是差驾驶员。
\((6)\) 所有男性都是好驾驶员。
否定陈述 \((6)\) 的陈述是:
A\((1)\) B\((2)\) C\((3)\) D\((4)\) E\((5)\)
English
Given the following six statements:
\((1)\) All women are good drivers.
\((2)\) Some women are good drivers.
\((3)\) No men are good drivers.
\((4)\) All men are bad drivers.
\((5)\) At least one man is a bad driver.
\((6)\) All men are good drivers.
The statement that negates statement \((6)\) is:
(A) \((1)\)(B) \((2)\)(C) \((3)\)(D) \((4)\)(E) \((5)\)
11
对于某个给定的 \(k\),方程 \[ x^2-3kx+2k^2-1=0 \] 的两根之积为 \(7\)。这两个根可以描述为:
A正整数 B负整数 C有理数,但不是整数 D无理数 E虚数
English
For a given value of \(k,\) the product of the roots of \[ x^2-3kx+2k^2-1=0 \] is \(7.\) The roots may be characterized as:
(A) integral and positive(B) integral and negative(C) rational, but not integral(D) irrational(E) imaginary
12
在同一平面内,所有半径为给定值 \(a\),且经过一个定点的圆,其圆心的轨迹是:
A一个点 B一条直线 C两条直线 D一个圆 E两个圆
English
The locus of the centers of all circles of given radius \(a,\) in the same plane, passing through a fixed point, is:
(A) a point(B) a straight line(C) two straight lines(D) a circle(E) two circles
13
由 \(y=3x+2\)、\(y=-3x+2\) 和 \(y=-2\) 围成的多边形是:
A等边三角形 B等腰三角形 C直角三角形 D一个三角形和一个梯形 E四边形
English
The polygon(s) formed by \(y=3x+2,\) \(y=-3x+2,\) and \(y=-2,\) is (are):
(A) an equilateral triangle(B) an isosceles triangle(C) a right triangle(D) a triangle and a trapezoid(E) a quadrilateral
14
若 \(a\) 和 \(b\) 是实数,则方程 \(3x-5+a=bx+1\) 有唯一解 \(x\) 的条件是[符号 \(a\ne0\) 表示 \(a\) 不等于零]:
A对所有 \(a\) 和 \(b\) B当 \(a\ne2b\) 时 C当 \(a\ne6\) 时 D当 \(b\ne0\) 时 E当 \(b\ne3\) 时
English
If \(a\) and \(b\) are real numbers, the equation \(3x-5+a=bx+1\) has a unique solution \(x\) [the symbol \(a\ne0\) means that \(a\) is different from zero]:
(A) for all \(a\) and \(b\)(B) if \(a\ne2b\)(C) if \(a\ne6\)(D) if \(b\ne0\)(E) if \(b\ne3\)
15
三角形 I 是等边三角形,边长为 \(A\)、周长为 \(P\)、面积为 \(K\)、外接圆半径为 \(R\)。三角形 II 也是等边三角形,边长为 \(a\)、周长为 \(p\)、面积为 \(k\)、外接圆半径为 \(r\)。若 \(A\) 不同于 \(a\),则:
A\(P:p=R:r\) 仅在某些情况下成立 B\(P:p=R:r\) 总是成立 C\(P:p=K:k\) 仅在某些情况下成立 D\(R:r=K:k\) 总是成立 E\(R:r=K:k\) 仅在某些情况下成立
English
Triangle I is equilateral with side \(A,\) perimeter \(P,\) area \(K,\) and circumradius \(R\) (radius of the circumscribed circle). Triangle II is equilateral with side \(a,\) perimeter \(p,\) area \(k,\) and circumradius \(r.\) If \(A\) is different from \(a,\) then:
(A) \(P:p=R:r\) only sometimes(B) \(P:p=R:r\) always(C) \(P:p=K:k\) only sometimes(D) \(R:r=K:k\) always(E) \(R:r=K:k\) only sometimes
16
在以 \(5\) 为底的计数系统中,数数方式如下:\(1\)、\(2\)、\(3\)、\(4\)、\(10\)、\(11\)、\(12\)、\(13\)、\(14\)、\(20\)、\(\ldots\)。在十进位系统中表示为 \(69\) 的数,用 \(5\) 进制表示时,其数位是:
A两个连续的数字 B两个不连续的数字 C三个连续的数字 D三个不连续的数字 E四个数字
English
In the numeration system with base \(5,\) counting is as follows: \(1,\) \(2,\) \(3,\) \(4,\) \(10,\) \(11,\) \(12,\) \(13,\) \(14,\) \(20,\) \(\ldots\) The number whose description in the decimal system is \(69,\) when described in the base \(5\) system, is a number with:
(A) two consecutive digits(B) two non-consecutive digits(C) three consecutive digits(D) three non-consecutive digits(E) four digits
17
对某一群体,公式 \(N=8\cdot10^8\cdot x^{-\frac{3}{2}}\) 给出了收入超过 \(x\) 美元的人数。收入最高的 \(800\) 人中,最低收入至少为多少美元?
A\(10^4\) B\(10^6\) C\(10^8\) D\(10^{12}\) E\(10^{16}\)
English
The formula \(N=8\cdot10^8\cdot x^{-\frac{3}{2}}\) gives, for a certain group, the number of individuals whose income exceeds \(x\) dollars. The lowest income, in dollars, of the wealthiest \(800\) individuals is at least:
(A) \(10^4\)(B) \(10^6\)(C) \(10^8\)(D) \(10^{12}\)(E) \(10^{16}\)
18
方程组 \(3^{x+y}=81\) 和 \(81^{x-y}=3\):
A没有公共解 B解为 \(x=2\)、\(y=2\) C解为 \(x=2\dfrac12\)、\(y=1\dfrac12\) D有由一正一负两个整数组成的公共解 E以上都不是
English
The pair of equations \(3^{x+y}=81\) and \(81^{x-y}=3\) has:
(A) no common solution(B) the solution \(x=2,\) \(y=2\)(C) the solution \(x=2\dfrac12,\) \(y=1\dfrac12\)(D) a common solution in positive and negative integers(E) none of these
19
考虑方程 I:\(x+y+z=46\),其中 \(x\)、\(y\) 和 \(z\) 为正整数;以及方程 II:\(x+y+z+w=46\),其中 \(x\)、\(y\)、\(z\) 和 \(w\) 为正整数。则:
A\(\mathrm{I}\) 可由连续整数满足 B\(\mathrm{I}\) 可由连续偶数满足 C\(\mathrm{II}\) 可由连续整数满足 D\(\mathrm{II}\) 可由连续偶数满足 E\(\mathrm{II}\) 可由连续奇数满足
English
Consider equation I: \(x+y+z=46,\) where \(x,\) \(y,\) and \(z\) are positive integers, and equation II: \(x+y+z+w=46,\) where \(x,\) \(y,\) \(z,\) and \(w\) are positive integers. Then:
(A) \(\mathrm{I}\) can be solved in consecutive integers(B) \(\mathrm{I}\) can be solved in consecutive even integers(C) \(\mathrm{II}\) can be solved in consecutive integers(D) \(\mathrm{II}\) can be solved in consecutive even integers(E) \(\mathrm{II}\) can be solved in consecutive odd integers
20
在 \[ \left(\frac{x^2}{2}-\frac2x\right)^8 \] 的展开式中,\(x^7\) 的系数是:
A\(56\) B\(-56\) C\(14\) D\(-14\) E\(0\)
English
The coefficient of \(x^7\) in the expansion of \[ \left(\frac{x^2}{2}-\frac2x\right)^8 \] is:
(A) \(56\)(B) \(-56\)(C) \(14\)(D) \(-14\)(E) \(0\)
21
正方形 I 的对角线长为 \(a+b\)。面积为正方形 I 两倍的正方形 II,其周长是:
A\((a+b)^2\) B\(\sqrt2(a+b)^2\) C\(2(a+b)\) D\(\sqrt8(a+b)\) E\(4(a+b)\)
English
The diagonal of square I is \(a+b.\) The perimeter of square II with twice the area of I is:
(A) \((a+b)^2\)(B) \(\sqrt2(a+b)^2\)(C) \(2(a+b)\)(D) \(\sqrt8(a+b)\)(E) \(4(a+b)\)
22
在等式 \((x+m)^2-(x+n)^2=(m-n)^2\) 中,\(m\) 和 \(n\) 是不相等的非零常数。若 \(x=am+bn\) 满足该等式,则:
A\(a=0\),且 \(b\) 有唯一的非零值 B\(a=0\),且 \(b\) 有两个非零值 C\(b=0\),且 \(a\) 有唯一的非零值 D\(b=0\),且 \(a\) 有两个非零值 E\(a\) 和 \(b\) 各有唯一的非零值
English
The equality \((x+m)^2-(x+n)^2=(m-n)^2,\) where \(m\) and \(n\) are unequal non-zero constants, is satisfied by \(x=am+bn\) where:
(A) \(a=0,\) \(b\) has a unique non-zero value(B) \(a=0,\) \(b\) has two non-zero values(C) \(b=0,\) \(a\) has a unique non-zero value(D) \(b=0,\) \(a\) has two non-zero values(E) \(a\) and \(b\) each have a unique non-zero value
23
一个圆柱形盒子的半径 \(R\) 为 \(8\) 英寸,高 \(H\) 为 \(3\) 英寸。体积为 \(V=\pi R^2H\)。将 \(R\) 增加 \(x\) 英寸与将 \(H\) 增加 \(x\) 英寸时,体积要增加同一个固定的正数。满足这一条件的是:
A\(x\) 没有实数取值 B\(x\) 有一个整数取值 C\(x\) 有一个有理数但非整数的取值 D\(x\) 有一个无理数取值 E\(x\) 有两个实数取值
English
The radius \(R\) of a cylindrical box is \(8\) inches, the height \(H\) is \(3\) inches. The volume \(V=\pi R^2H\) is to be increased by the same fixed positive amount when \(R\) is increased by \(x\) inches as when \(H\) is increased by \(x\) inches. This condition is satisfied by:
(A) no real value of \(x\)(B) one integral value of \(x\)(C) one rational, but not integral, value of \(x\)(D) one irrational value of \(x\)(E) two real values of \(x\)
24
若 \(\log_{2x}216=x\),其中 \(x\) 为实数,则 \(x\) 是:
A既不是完全平方数也不是完全立方数的整数 B既不是完全平方数也不是完全立方数的非整数有理数 C无理数 D完全平方数 E完全立方数
English
If \(\log_{2x}216=x,\) where \(x\) is real, then \(x\) is:
(A) a non-square, non-cube integer(B) a non-square, non-cube, non-integral rational number(C) an irrational number(D) a perfect square(E) a perfect cube
25
设 \(m\) 和 \(n\) 为任意两个奇数,且 \(n\) 小于 \(m\)。能整除所有可能的 \(m^2-n^2\) 的最大整数是:
A\(2\) B\(4\) C\(6\) D\(8\) E\(16\)
English
Let \(m\) and \(n\) be any two odd numbers, with \(n\) less than \(m.\) The largest integer which divides all possible numbers of the form \(m^2-n^2\) is:
(A) \(2\)(B) \(4\)(C) \(6\)(D) \(8\)(E) \(16\)
26
求满足下列不等式的所有 \(x\) 值:\[ \left|\frac{5-x}{3}\right|\lt2 \] [符号 \(|a|\) 表示:若 \(a\) 为正,则取 \(+a\);若 \(a\) 为负,则取 \(-a\);若 \(a\) 为零,则取 \(0\)。记号 \(1\lt a\lt2\) 表示 \(a\) 可以取 \(1\) 与 \(2\) 之间的任意值,但不包括 \(1\) 和 \(2\)。]
A\(1\lt x\lt11\) B\(-1\lt x\lt11\) C\(x\lt11\) D\(x\gt11\) E\(|x|\lt6\)
English
Find the set of \(x\)-values satisfying the inequality \[ \left|\frac{5-x}{3}\right|\lt2. \] [The symbol \(|a|\) means \(+a\) if \(a\) is positive, \(-a\) if \(a\) is negative, \(0\) if \(a\) is zero. The notation \(1\lt a\lt2\) means that \(a\) can have any value between \(1\) and \(2,\) excluding \(1\) and \(2.\)]
(A) \(1\lt x\lt11\)(B) \(-1\lt x\lt11\)(C) \(x\lt11\)(D) \(x\gt11\)(E) \(|x|\lt6\)
27
设多边形 \(P\) 的内角和为 \(S\),且每个内角都是同一顶点处外角的 \(7\dfrac12\) 倍。则:
A\(S=2660^\circ\),且 \(P\) 可能是正多边形 B\(S=2660^\circ\),且 \(P\) 不是正多边形 C\(S=2700^\circ\),且 \(P\) 是正多边形 D\(S=2700^\circ\),且 \(P\) 不是正多边形 E\(S=2700^\circ\),且 \(P\) 可能是正多边形,也可能不是
English
Let \(S\) be the sum of the interior angles of a polygon \(P\) for which each interior angle is \(7\dfrac12\) times the exterior angle at the same vertex. Then:
(A) \(S=2660^\circ\) and \(P\) may be regular(B) \(S=2660^\circ\) and \(P\) is not regular(C) \(S=2700^\circ\) and \(P\) is regular(D) \(S=2700^\circ\) and \(P\) is not regular(E) \(S=2700^\circ\) and \(P\) may or may not be regular
28
方程 \[ x-\frac7{x-3}=3-\frac7{x-3} \] 的根的情况是:
A有无穷多个整数根 B无根 C有一个整数根 D有两个相等的整数根 E有两个相等的非整数根
English
The equation \[ x-\frac7{x-3}=3-\frac7{x-3} \] has:
(A) infinitely many integral roots(B) no root(C) one integral root(D) two equal integral roots(E) two equal non-integral roots
29
\(A\) 的钱的五倍加上 \(B\) 的钱多于 \(\$51.00\)。\(A\) 的钱的三倍减去 \(B\) 的钱等于 \(\$21.00\)。若 \(a\) 表示 \(A\) 的钱(美元),\(b\) 表示 \(B\) 的钱(美元),则:
A\(a\gt9\),\(b\gt6\) B\(a\gt9\),\(b\lt6\) C\(a\gt9\),\(b=6\) D\(a\gt9\),但无法给 \(b\) 确定界限 E\(2a=3b\)
English
Five times \(A\)’s money added to \(B\)’s money is more than \(\$51.00.\) Three times \(A\)’s money minus \(B\)’s money is \(\$21.00.\) If \(a\) represents \(A\)’s money in dollars and \(b\) represents \(B\)’s money in dollars, then:
(A) \(a\gt9,\) \(b\gt6\)(B) \(a\gt9,\) \(b\lt6\)(C) \(a\gt9,\) \(b=6\)(D) \(a\gt9,\) but we can put no bounds on \(b\)(E) \(2a=3b\)
30
给定直线 \(3x+5y=15\),考察该直线上到两坐标轴距离相等的点。这样的点存在于:
A任何象限都没有 B仅第 \(\mathrm{I}\) 象限 C仅第 \(\mathrm{I}\)、\(\mathrm{II}\) 象限 D仅第 \(\mathrm{I}\)、\(\mathrm{II}\)、\(\mathrm{III}\) 象限 E每个象限
English
Given the line \(3x+5y=15\) and a point on this line equidistant from the coordinate axes. Such a point exists in:
(A) none of the quadrants(B) quadrant \(\mathrm{I}\) only(C) quadrants \(\mathrm{I},\) \(\mathrm{II}\) only(D) quadrants \(\mathrm{I},\) \(\mathrm{II},\) \(\mathrm{III}\) only(E) each of the quadrants
31
要使 \(x^2+2x+5\) 成为 \(x^4+px^2+q\) 的因式,\(p\) 和 \(q\) 的值必须依次为:
A\(-2\),\(5\) B\(5\),\(25\) C\(10\),\(20\) D\(6\),\(25\) E\(14\),\(25\)
English
For \(x^2+2x+5\) to be a factor of \(x^4+px^2+q,\) the values of \(p\) and \(q\) must be, respectively:
(A) \(-2,\) \(5\)(B) \(5,\) \(25\)(C) \(10,\) \(20\)(D) \(6,\) \(25\)(E) \(14,\) \(25\)
32
在图中,圆心为 \(O\)。\(\overline{AB}\perp\overline{BC}\),\(ADOE\) 是一条直线,\(\overline{AP}=\overline{AD}\),且 \(\overline{AB}\) 的长度是半径的两倍。则:
A\(\overline{AP}^{\,2}=\overline{PB}\cdot\overline{AB}\) B\(\overline{AP}\cdot\overline{DO}=\overline{PB}\cdot\overline{AD}\) C\(\overline{AB}^{\,2}=\overline{AD}\cdot\overline{DE}\) D\(\overline{AB}\cdot\overline{AD}=\overline{OB}\cdot\overline{AO}\) E以上都不是
English
In this figure the center of the circle is \(O.\) \(\overline{AB}\perp\overline{BC},\) \(ADOE\) is a straight line, \(\overline{AP}=\overline{AD},\) and \(\overline{AB}\) has a length twice the radius. Then:
(A) \(\overline{AP}^{\,2}=\overline{PB}\cdot\overline{AB}\)(B) \(\overline{AP}\cdot\overline{DO}=\overline{PB}\cdot\overline{AD}\)(C) \(\overline{AB}^{\,2}=\overline{AD}\cdot\overline{DE}\)(D) \(\overline{AB}\cdot\overline{AD}=\overline{OB}\cdot\overline{AO}\)(E) none of these
33
给定一个含 \(58\) 项的数列;每一项都形如 \(P+n\),其中 \(P\) 表示所有不超过 \(61\) 的质数的乘积 \(2\cdot3\cdot5\cdots61\),而 \(n\) 依次取值 \(2\)、\(3\)、\(4\)、\(\ldots\)、\(59\)。设该数列中质数的个数为 \(N\)。则 \(N\) 是:
A\(0\) B\(16\) C\(17\) D\(57\) E\(58\)
English
You are given a sequence of \(58\) terms; each term has the form \(P+n\) where \(P\) stands for the product \(2\cdot3\cdot5\cdots61\) of all prime numbers less than or equal to \(61,\) and \(n\) takes, successively, the values \(2,\) \(3,\) \(4,\) \(\ldots,\) \(59.\) Let \(N\) be the number of primes appearing in this sequence. Then \(N\) is:
(A) \(0\)(B) \(16\)(C) \(17\)(D) \(57\)(E) \(58\)
34
两名游泳者分别位于一个长 \(90\) 英尺的泳池两端,同时开始沿泳池长度方向游泳,一人的速度为每秒 \(3\) 英尺,另一人的速度为每秒 \(2\) 英尺。他们来回游了 \(12\) 分钟。假设转身不耗费时间,求他们相互经过的次数。
A\(24\) B\(21\) C\(20\) D\(19\) E\(18\)
English
Two swimmers, at opposite ends of a \(90\)-foot pool, start to swim the length of the pool, one at the rate of \(3\) feet per second, the other at \(2\) feet per second. They swim back and forth for \(12\) minutes. Allowing no loss of time at the turns, find the number of times they pass each other.
(A) \(24\)(B) \(21\)(C) \(20\)(D) \(19\)(E) \(18\)
35
从周长为 \(10\) 个单位的圆外一点 \(P\) 作圆的切线;又从 \(P\) 作一条割线,将圆分成长度分别为 \(m\) 和 \(n\) 的两段不等弧。已知切线长 \(t\) 是 \(m\) 与 \(n\) 的比例中项。若 \(m\) 和 \(t\) 都是整数,则 \(t\) 可能取值的个数为:
A零个 B一个 C两个 D三个 E无穷多个
English
From point \(P\) outside a circle, with a circumference of \(10\) units, a tangent is drawn. Also from \(P\) a secant is drawn dividing the circle into unequal arcs with lengths \(m\) and \(n.\) It is found that \(t,\) the length of the tangent, is the mean proportional between \(m\) and \(n.\) If \(m\) and \(t\) are integers, then \(t\) may have the following number of values:
(A) zero(B) one(C) two(D) three(E) infinitely many
36
在同一个首项为 \(a\)、公差为 \(d\) 的等差数列中,设前 \(n\)、\(2n\)、\(3n\) 项的和依次为 \(s_1\)、\(s_2\)、\(s_3\)。令 \(R=s_3-s_2-s_1\)。则 \(R\) 取决于:
A\(a\) 和 \(d\) B\(d\) 和 \(n\) C\(a\) 和 \(n\) D\(a\)、\(d\) 和 \(n\) E不取决于 \(a\)、\(d\) 或 \(n\) 中的任何一个
English
Let \(s_1,\) \(s_2,\) \(s_3\) be the respective sums of \(n,\) \(2n,\) \(3n\) terms of the same arithmetic progression with \(a\) as the first term and \(d\) as the common difference. Let \(R=s_3-s_2-s_1.\) Then \(R\) is dependent on:
(A) \(a\) and \(d\)(B) \(d\) and \(n\)(C) \(a\) and \(n\)(D) \(a,\) \(d,\) and \(n\)(E) neither \(a\) nor \(d\) nor \(n\)
37
一个三角形的底长为 \(b\),高为 \(h\)。在该三角形内接一个高为 \(x\) 的矩形,且矩形的底边在三角形的底边上。该矩形的面积是:
A\(\dfrac{bx}{h}(h-x)\) B\(\dfrac{hx}{b}(b-x)\) C\(\dfrac{bx}{h}(h-2x)\) D\(x(b-x)\) E\(x(h-x)\)
English
The base of a triangle is of length \(b,\) and the altitude is of length \(h.\) A rectangle of height \(x\) is inscribed in the triangle with the base of the rectangle in the base of the triangle. The area of the rectangle is:
(A) \(\dfrac{bx}{h}(h-x)\)(B) \(\dfrac{hx}{b}(b-x)\)(C) \(\dfrac{bx}{h}(h-2x)\)(D) \(x(b-x)\)(E) \(x(h-x)\)
38
在图中,\(\overline{AB}\) 和 \(\overline{AC}\) 是等腰三角形 \(ABC\) 的两条等长边,其中内接等边三角形 \(DEF\)。记角 \(BFD\) 为 \(a\)、角 \(ADE\) 为 \(b\)、角 \(FEC\) 为 \(c\)。则:
A\(b=\dfrac{a+c}{2}\) B\(b=\dfrac{a-c}{2}\) C\(a=\dfrac{b-c}{2}\) D\(a=\dfrac{b+c}{2}\) E以上都不是
English
In this diagram \(\overline{AB}\) and \(\overline{AC}\) are the equal sides of an isosceles triangle \(ABC,\) in which is inscribed equilateral triangle \(DEF.\) Designate angle \(BFD\) by \(a,\) angle \(ADE\) by \(b,\) and angle \(FEC\) by \(c.\) Then:
(A) \(b=\dfrac{a+c}{2}\)(B) \(b=\dfrac{a-c}{2}\)(C) \(a=\dfrac{b-c}{2}\)(D) \(a=\dfrac{b+c}{2}\)(E) none of these
39
要满足方程 \[ \frac{a+b}{a}=\frac{b}{a+b}\text{,}\] \(a\) 和 \(b\) 必须:
A都是有理数 B都是实数但不是有理数 C都不是实数 D一个是实数,另一个不是实数 E一个是实数而另一个不是,或者两者都不是实数
English
To satisfy the equation \[ \frac{a+b}{a}=\frac{b}{a+b}, \] \(a\) and \(b\) must be:
(A) both rational(B) both real but not rational(C) both not real(D) one real, one not real(E) one real, one not real or both not real
40
给定直角三角形 \(ABC\),两直角边为 \(\overline{BC}=3\)、\(\overline{AC}=4\)。求从 \(C\) 引向斜边的两条三等分角线中较短一条的长度:
A\(\dfrac{32\sqrt3-24}{13}\) B\(\dfrac{12\sqrt3-9}{13}\) C\(6\sqrt3-8\) D\(\dfrac{5\sqrt{10}}6\) E\(\dfrac{25}{12}\)
English
Given right triangle \(ABC\) with legs \(\overline{BC}=3,\) \(\overline{AC}=4.\) Find the length of the shorter angle trisector from \(C\) to the hypotenuse:
(A) \(\dfrac{32\sqrt3-24}{13}\)(B) \(\dfrac{12\sqrt3-9}{13}\)(C) \(6\sqrt3-8\)(D) \(\dfrac{5\sqrt{10}}6\)(E) \(\dfrac{25}{12}\)
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答案速查
| 题号 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 答案 | E | C | B | C | C | B | D | D | E |
| 题号 | 10 | 11 | 12 | 13 | 14 | 15 | 16 | 17 | 18 |
| 答案 | E | D | D | B | E | B | C | A | E |
| 题号 | 19 | 20 | 21 | 22 | 23 | 24 | 25 | 26 | 27 |
| 答案 | C | D | E | A | C | A | D | B | E |
| 题号 | 28 | 29 | 30 | 31 | 32 | 33 | 34 | 35 | 36 |
| 答案 | B | A | C | D | A | A | C | C | B |
| 题号 | 37 | 38 | 39 | 40 |
| 答案 | A | D | E | A |
答案串:ECBCCBDDEEDDBEBCAECDEACADBEBACDAACCBADEA
逐题解析
1 答案 E
代入给定的根,得到 \[ 2^3+2h+10=0\text{。}\]因此 \(2h=-18\),所以 \(h=-9\)。
因此,正确答案是 E。
English
Substituting the given root gives \[ 2^3+2h+10=0. \] Thus \(2h=-18,\) so \(h=-9.\)
Therefore, the correct answer is E.
2 答案 C
从 \(6\) 下中的第一下到最后一下共有 \(5\) 个相等的时间间隔,所以每个间隔为 \(1\) 秒。敲响十二下共有 \(11\) 个这样的间隔,因此需要 \(11\) 秒。
因此,正确答案是 C。
English
There are \(5\) equal intervals from the first of \(6\) strikes to the last, so each interval lasts \(1\) second. Twelve strikes contain \(11\) such intervals and therefore take \(11\) seconds.
Thus, the correct answer is C.
3 答案 B
\(40\%\) 的折扣为 \(\$4000\)。两次连续折扣的总额为 \[ 3600+0.04(6400)=3856\text{。}\]两者之差为 \(4000-3856=144\) 美元。
因此,正确答案是 B。
English
A \(40\%\) discount is \(\$4000.\) The successive discounts total \[ 3600+0.04(6400)=3856. \] Their difference is \(4000-3856=144\) dollars.
Therefore, the correct answer is B.
4 答案 C
第三个角也是 \(60^\circ\),所以该三角形是边长为 \(4\) 的等边三角形。其面积为 \[ \frac{\sqrt3}{4}(4)^2=4\sqrt3\text{。}\]
因此,正确答案是 C。
English
The third angle is also \(60^\circ,\) so the triangle is equilateral with side \(4.\) Its area is \[ \frac{\sqrt3}{4}(4)^2=4\sqrt3. \]
Thus, the correct answer is C.
5 答案 C
代入得 \(x^2+9=9\),所以 \(x=0\)。再由 \(y^2=9\) 得 \(y=3\) 或 \(y=-3\)。因此共有两个公共点。
因此,正确答案是 C。
English
Substitution gives \(x^2+9=9,\) so \(x=0.\) Then \(y^2=9\) gives \(y=3\) or \(y=-3.\) Hence there are two common points.
Therefore, the correct answer is C.
6 答案 B
直径为 \(\frac{100}{\pi}\)。若正方形的边长为 \(s\),则其对角线为 \(s\sqrt2\),所以 \[ s\sqrt2=\frac{100}{\pi}\text{。}\]因此 \(s=\frac{50\sqrt2}{\pi}\)。
因此,正确答案是 B。
English
The diameter is \(\frac{100}{\pi}.\) If the square has side \(s,\) its diagonal is \(s\sqrt2,\) so \[ s\sqrt2=\frac{100}{\pi}. \] Therefore \(s=\frac{50\sqrt2}{\pi}.\)
Thus, the correct answer is B.
7 答案 D
因为圆 I 经过圆 II 的圆心,所以两圆心之间的距离等于圆 I 的半径 \(r\)。内切又使该距离等于 \(R-r\),其中 \(R\) 是圆 II 的半径。因此 \(R=2r\),面积便乘以 \(2^2=4\)。所以圆 II 的面积为 \(4\cdot4=16\)。
因此,正确答案是 D。
English
Because circle I passes through the center of circle II, the distance between the centers equals the radius \(r\) of circle I. Internal tangency also makes that distance \(R-r,\) where \(R\) is the radius of circle II. Thus \(R=2r,\) and the area is multiplied by \(2^2=4.\) It is therefore \(4\cdot4=16.\)
Thus, the correct answer is D.
8 答案 D
有 \[ 2.525252\ldots=2+\frac{52}{99}=\frac{250}{99}\text{。}\]这个分数已经是最简分数,并且 \(250+99=349\)。
因此,正确答案是 D。
English
We have \[ 2.525252\ldots=2+\frac{52}{99}=\frac{250}{99}. \] This fraction is already in lowest terms, and \(250+99=349.\)
Therefore, the correct answer is D.
9 答案 E
因式分解得 \[ \begin{gathered} a^2+b^2-c^2+2ab\\ {}=(a+b-c)(a+b+c),\\ a^2+c^2-b^2+2ac\\ {}=(a-b+c)(a+b+c) \end{gathered}\text{。}\]在题述的适当限制下约去公因式,得到 \[ \frac{a+b-c}{a-b+c}\text{。}\]
因此,正确答案是 E。
English
Factoring gives \[ \begin{gathered} a^2+b^2-c^2+2ab\\ {}=(a+b-c)(a+b+c),\\ a^2+c^2-b^2+2ac\\ {}=(a-b+c)(a+b+c). \end{gathered} \] Cancelling the common factor, under the stated suitable restrictions, leaves \[ \frac{a+b-c}{a-b+c}. \]
Thus, the correct answer is E.
10 答案 E
“所有男性都是好驾驶员”的否定是“至少有一名男性不是好驾驶员”。按照选项中的说法,这就是陈述 \((5)\):“至少有一名男性是差驾驶员。”
因此,正确答案是 E。
English
The negation of “All men are good drivers” is “At least one man is not a good driver.” In the terminology of the choices, that is statement \((5),\) “At least one man is a bad driver.”
Therefore, the correct answer is E.
11 答案 D
由韦达定理,\(2k^2-1=7\),所以 \(k=\pm2\)。当 \(k=2\) 时,两根为 \(3\pm\sqrt2\);当 \(k=-2\) 时,两根为 \(-3\pm\sqrt2\)。无论哪种情况,两根都是无理数。
因此,正确答案是 D。
English
By Vieta’s formulas, \(2k^2-1=7,\) so \(k=\pm2.\) The roots are \(3\pm\sqrt2\) when \(k=2,\) and \(-3\pm\sqrt2\) when \(k=-2.\) In either case both roots are irrational.
Thus, the correct answer is D.
12 答案 D
半径为 \(a\) 的圆经过该定点,当且仅当其圆心到该点的距离为 \(a\)。所有这些圆心的轨迹是一个半径为 \(a\) 的圆。
因此,正确答案是 D。
English
A circle of radius \(a\) passes through the fixed point exactly when its center is distance \(a\) from that point. The locus of all such centers is a circle of radius \(a.\)
Therefore, the correct answer is D.
13 答案 B
两条斜线交于 \((0,2)\),而它们与 \(y=-2\) 的交点分别为 \((-\frac{4}{3},-2)\) 和 \((\frac{4}{3},-2)\)。后两个点关于 \(y\) 轴对称,所以两条斜边等长。该图形是等腰三角形。
因此,正确答案是 B。
English
The slanted lines meet at \((0,2),\) while their intersections with \(y=-2\) are \((-\frac{4}{3},-2)\) and \((\frac{4}{3},-2).\) The latter two points are symmetric about the \(y\)-axis, so the two slanted sides have equal length. The figure is an isosceles triangle.
Thus, the correct answer is B.
14 答案 E
整理得 \[ (3-b)x=6-a\text{。}\]当且仅当 \(3-b\ne0\),即 \(b\ne3\) 时,该方程有唯一解。
因此,正确答案是 E。
English
Rearranging gives \[ (3-b)x=6-a. \] This has a unique solution exactly when \(3-b\ne0,\) or \(b\ne3.\)
Thus, the correct answer is E.
15 答案 B
对于等边三角形,\(P=3A\),且 \(R=\frac{A}{\sqrt3}\);\(p\) 和 \(r\) 的关系同理。因此 \[ P:p=A:a=R:r \] 对任意这样的一对三角形都成立。
因此,正确答案是 B。
English
For equilateral triangles, \(P=3A\) and \(R=\frac{A}{\sqrt3},\) with analogous formulas for \(p\) and \(r.\) Therefore \[ P:p=A:a=R:r \] for every such pair of triangles.
Thus, the correct answer is B.
16 答案 C
因为 \[ 69=2\cdot25+3\cdot5+4\text{,}\]所以它的 \(5\) 进制表示是 \(234_5\)。其中三个数字 \(2\)、\(3\)、\(4\) 是连续的。
因此,正确答案是 C。
English
Since \[ 69=2\cdot25+3\cdot5+4, \] its base-\(5\) representation is \(234_5.\) Its three digits \(2,\) \(3,\) and \(4\) are consecutive.
Therefore, the correct answer is C.
17 答案 A
在分界点,\[ 800=8\cdot10^8x^{-\frac{3}{2}}\text{。}\]因此 \(x^{\frac{3}{2}}=10^6\),将两边都取 \(\frac{2}{3}\) 次幂,得到 \(x=10^4\)。
因此,正确答案是 A。
English
At the cutoff, \[ 800=8\cdot10^8x^{-\frac{3}{2}}. \] Hence \(x^{\frac{3}{2}}=10^6,\) and raising both sides to the \(\frac{2}{3}\) power gives \(x=10^4.\)
Thus, the correct answer is A.
18 答案 E
第一个方程给出 \(x+y=4\)。第二个方程给出 \[ 3^{4(x-y)}=3\text{,}\]所以 \(x-y=\frac{1}{4}\)。因此 \(x=\frac{17}{8}\),且 \(y=\frac{15}{8}\)。这组解不在前四个选项中。
因此,正确答案是 E。
English
The first equation gives \(x+y=4.\) The second gives \[ 3^{4(x-y)}=3, \] so \(x-y=\frac{1}{4}.\) Therefore \(x=\frac{17}{8}\) and \(y=\frac{15}{8}.\) This pair is not listed among the first four choices.
Thus, the correct answer is E.
19 答案 C
三个连续整数之和能被 \(3\) 整除,所以不可能是 \(46\)。从 \(n\) 开始的四个连续整数之和为 \(4n+6\)。令 \(4n+6=46\),得 \(n=10\),并且确实有 \[ 10+11+12+13=46\text{。}\]
因此,正确答案是 C。
English
The sum of three consecutive integers is divisible by \(3,\) so it cannot be \(46.\) Four consecutive integers beginning with \(n\) have sum \(4n+6.\) Setting \(4n+6=46\) gives \(n=10,\) and indeed \[ 10+11+12+13=46. \]
Therefore, the correct answer is C.
20 答案 D
选用 \(k\) 个 \(-\frac{2}{x}\) 因子的项,其指数为 \[ 2(8-k)-k=16-3k\text{。}\]要得到 \(x^7\),需要 \(k=3\)。其系数为 \[ \binom83\left(\frac12\right)^5(-2)^3=-14\text{。}\]
因此,正确答案是 D。
English
The term using \(k\) factors of \(-\frac{2}{x}\) has exponent \[ 2(8-k)-k=16-3k. \] To obtain \(x^7,\) we need \(k=3.\) Its coefficient is \[ \binom83\left(\frac12\right)^5(-2)^3=-14. \]
Thus, the correct answer is D.
21 答案 E
对角线长为 \(a+b\) 的正方形,其面积为 \(\frac{(a+b)^2}{2}\)。因此正方形 II 的面积为 \((a+b)^2\),所以它的边长为 \(a+b\),周长为 \(4(a+b)\)。
因此,正确答案是 E。
English
A square with diagonal \(a+b\) has area \(\frac{(a+b)^2}{2}.\) Square II therefore has area \((a+b)^2,\) so its side is \(a+b\) and its perimeter is \(4(a+b).\)
Thus, the correct answer is E.
22 答案 A
利用 \(m\ne n\) 并因式分解,得 \[ \begin{gathered} (m-n)(2x+m+n)\\ {}=(m-n)^2 \end{gathered}\text{。}\]因此 \(2x+m+n=m-n\),所以 \(x=-n\)。于是 \(a=0\),且 \(b=-1\),这是唯一的非零值。
因此,正确答案是 A。
English
Factoring and using \(m\ne n\) gives \[ \begin{gathered} (m-n)(2x+m+n)\\ {}=(m-n)^2. \end{gathered} \] Hence \(2x+m+n=m-n,\) so \(x=-n.\) Thus \(a=0\) and \(b=-1,\) a unique nonzero value.
Therefore, the correct answer is A.
23 答案 C
增加半径时,体积增加 \[ 3\pi\big((8+x)^2-8^2\big)\text{,}\]而增加高度时,体积增加 \(64\pi x\)。令两者相等,得 \[ 48x+3x^2=64x\text{,}\]所以 \(x=0\) 或 \(x=\frac{16}{3}\)。增加量必须为正,因此只剩一个有理数但非整数的取值。
因此,正确答案是 C。
English
Increasing the radius changes the volume by \[ 3\pi\big((8+x)^2-8^2\big), \] while increasing the height changes it by \(64\pi x.\) Equating these gives \[ 48x+3x^2=64x, \] so \(x=0\) or \(x=\frac{16}{3}.\) The increase must be positive, leaving one rational but nonintegral value.
Thus, the correct answer is C.
24 答案 A
该方程等价于 \[ (2x)^x=216\text{。}\]由于 \(216=6^3\),\(x=3\) 满足方程。为说明它是唯一允许的实数解,注意对数的底要求 \(x>0\) 且 \(2x\ne1\);当 \(0\lt x\lt\frac{1}{2}\) 时,\(x\ln(2x)\lt0\);而当 \(x>\frac{1}{2}\) 时,函数 \(x\ln(2x)\) 严格递增。因此 \(x=3\) 是一个既非完全平方数又非完全立方数的整数。
因此,正确答案是 A。
English
The equation is equivalent to \[ (2x)^x=216. \] Since \(216=6^3,\) \(x=3\) satisfies the equation. To see it is the only admissible real solution, note that the base requires \(x>0\) and \(2x\ne1;\) on \(0\lt x\lt\frac{1}{2},\) \(x\ln(2x)\lt0,\) while for \(x>\frac{1}{2}\) the function \(x\ln(2x)\) is strictly increasing. Thus \(x=3,\) a non-square, non-cube integer.
Therefore, the correct answer is A.
25 答案 D
有 \[ m^2-n^2=(m-n)(m+n)\text{。}\]两个因子都是偶数,其中一个能被 \(4\) 整除,所以每个这样的差都能被 \(8\) 整除。取 \(m=3,n=1\),得到 \(m^2-n^2=8\),这说明不存在总能整除它的更大整数。
因此,正确答案是 D。
English
We have \[ m^2-n^2=(m-n)(m+n). \] Both factors are even, and one is divisible by \(4,\) so every such difference is divisible by \(8.\) Taking \(m=3,n=1\) gives \(m^2-n^2=8,\) proving that no larger integer always divides it.
Thus, the correct answer is D.
26 答案 B
该不等式等价于 \[ |5-x|\lt6\text{,}\]所以 \(-6\lt5-x\lt6\)。减去 \(5\) 并按需要变号,得到 \[ -1\lt x\lt11\text{。}\]
因此,正确答案是 B。
English
The inequality is equivalent to \[ |5-x|\lt6, \] so \(-6\lt5-x\lt6.\) Subtracting \(5\) and reversing signs as needed gives \[ -1\lt x\lt11. \]
Therefore, the correct answer is B.
27 答案 E
若一个外角为 \(e\),则 \[ e+\frac{15}{2}e=180^\circ\text{,}\]所以 \(e=\frac{360^\circ}{17}\)。由于外角和为 \(360^\circ\),多边形有 \(17\) 个顶点,并且 \[ S=(17-2)180^\circ=2700^\circ\text{。}\]它的所有角都相等,但各边不一定相等,所以它可能是正多边形,也可能不是。
因此,正确答案是 E。
English
If an exterior angle is \(e,\) then \[ e+\frac{15}{2}e=180^\circ, \] so \(e=\frac{360^\circ}{17}.\) Since the exterior angles total \(360^\circ,\) the polygon has \(17\) vertices and \[ S=(17-2)180^\circ=2700^\circ. \] All its angles are equal, but its sides need not be equal, so it may or may not be regular.
Thus, the correct answer is E.
28 答案 B
当 \(x\ne3\) 时,从两边减去相同的分式后只剩 \(x=3\)。但 \(x=3\) 会使原式的分母为零,所以它不在定义域内。该方程无根。
因此,正确答案是 B。
English
For \(x\ne3,\) subtracting the identical fractions from both sides leaves \(x=3.\) But \(x=3\) makes the original denominators zero, so it is not in the domain. The equation has no root.
Therefore, the correct answer is B.
29 答案 A
由 \(3a-b=21\),得 \(b=3a-21\)。代入不等式,得到 \[ 5a+3a-21>51\text{,}\]所以 \(a>9\)。进而 \(b=3a-21>6\)。
因此,正确答案是 A。
English
From \(3a-b=21,\) we have \(b=3a-21.\) Substitution into the inequality gives \[ 5a+3a-21>51, \] so \(a>9.\) It follows that \(b=3a-21>6.\)
Thus, the correct answer is A.
30 答案 C
当 \(y=x\) 时,直线给出 \(8x=15\),所得点在第一象限。当 \(y=-x\) 时,得到 \(-2x=15\),所以 \(x\lt0\) 且 \(y>0\),所得点在第二象限。没有其他可能。
因此,正确答案是 C。
English
For \(y=x,\) the line gives \(8x=15,\) producing a point in quadrant I. For \(y=-x,\) it gives \(-2x=15,\) so \(x\lt0\) and \(y>0,\) producing a point in quadrant II. There are no other possibilities.
Therefore, the correct answer is C.
31 答案 D
对 \(x^2+2x+5\) 取模,有 \(x^2=-2x-5\)。于是 \[ x^3=10-x,\qquad x^4=12x+5\text{。}\]因此 \[ \begin{gathered} x^4+px^2+q\\ {}\equiv(12-2p)x+(5-5p+q) \end{gathered}\text{。}\]当 \(p=6\) 且 \(q=25\) 时,两个系数都为零。
因此,正确答案是 D。
English
Modulo \(x^2+2x+5,\) we have \(x^2=-2x-5.\) Then \[ x^3=10-x,\qquad x^4=12x+5. \] Thus \[ \begin{gathered} x^4+px^2+q\\ {}\equiv(12-2p)x+(5-5p+q). \end{gathered} \] Both coefficients vanish when \(p=6\) and \(q=25.\)
Therefore, the correct answer is D.
32 答案 A
设半径为 \(r\)。因为 \(AB\) 在 \(B\) 点处与圆相切,所以 \[ AB^2=AD\cdot AE\text{。}\]令 \(AD=t\)。由于割线经过圆心,\(AE=t+2r\),而 \(AB=2r\)。因此 \[ t(t+2r)=4r^2\text{。}\]另外,\(AP=AD=t\),且 \(PB=AB-AP=2r-t\)。将上式整理为 \[ t^2=2r(2r-t)=PB\cdot AB\text{。}\]因此 \(AP^2=PB\cdot AB\)。
因此,正确答案是 A。
English
Let the radius be \(r.\) Since \(AB\) is tangent at \(B,\) \[ AB^2=AD\cdot AE. \] Write \(AD=t.\) Because the secant passes through the center, \(AE=t+2r,\) while \(AB=2r.\) Hence \[ t(t+2r)=4r^2. \] Also \(AP=AD=t\) and \(PB=AB-AP=2r-t.\) The displayed equation rearranges to \[ t^2=2r(2r-t)=PB\cdot AB. \] Therefore \(AP^2=PB\cdot AB.\)
Thus, the correct answer is A.
33 答案 A
对给定范围内的每个 \(n\),选择 \(n\) 的一个质因子 \(p\)。由于 \(p\le59\),它是 \(P\) 的一个因子。因此 \(p\) 整除 \(P+n\)。又因为 \(P+n>p\),所以 \(P+n\) 是合数。没有一项是质数,故 \(N=0\)。
因此,正确答案是 A。
English
For each \(n\) in the given range, choose a prime divisor \(p\) of \(n.\) Since \(p\le59,\) it is one of the factors of \(P.\) Therefore \(p\) divides \(P+n.\) Moreover \(P+n>p,\) so \(P+n\) is composite. No term is prime, and \(N=0.\)
Thus, the correct answer is A.
34 答案 C
速度较快者的位置每 \(60\) 秒重复一次,速度较慢者的位置每 \(90\) 秒重复一次。因此,两人的位置变化共同以 \(180\) 秒为周期。在一个周期内,他们的位置在 \[ t=18,54,90,126,162 \] 秒时重合。原题答案采用的计数方式把 \(t=90\) 秒时两人在端点同时转身也计作一次相遇。因此每个周期计 \(5\) 次,\(12\) 分钟内共有 \(4\) 个周期,得到 \(5\cdot4=20\)。
因此,正确答案是 C。
English
The faster swimmer’s position repeats every \(60\) seconds, and the slower swimmer’s position repeats every \(90\) seconds. Thus their joint position pattern repeats every \(180\) seconds. During one such cycle, their positions coincide at \[ t=18,54,90,126,162 \] seconds. The keyed interpretation counts the simultaneous turn at the endpoint when \(t=90\) as one of these encounters. Thus there are \(5\) counted encounters per cycle and \(4\) cycles in \(12\) minutes, giving \(5\cdot4=20.\)
Therefore, the correct answer is C.
35 答案 C
弧长满足 \(m+n=10\),而比例中项条件给出 \[ t^2=mn=m(10-m)\text{。}\]当 \(m\) 取从 \(1\) 到 \(9\) 的整数时,不计对称重复,可能的乘积为 \(9\)、\(16\)、\(21\)、\(24\) 和 \(25\)。最后一个来自相等的两弧 \(m=n=5\),应排除。其余完全平方数给出 \(t=3\) 和 \(t=4\),共两个值。
因此,正确答案是 C。
English
The arc lengths satisfy \(m+n=10,\) while the mean-proportional condition gives \[ t^2=mn=m(10-m). \] For integral \(m\) from \(1\) through \(9,\) the possible products, up to symmetry, are \(9,\) \(16,\) \(21,\) \(24,\) and \(25.\) The last comes from equal arcs \(m=n=5\) and is excluded. The remaining squares give \(t=3\) and \(t=4,\) two values.
Thus, the correct answer is C.
36 答案 B
使用等差数列求和公式,\[ s_j=\frac{jn}{2}\bigl(2a+(jn-1)d\bigr)\text{。}\]在 \(s_3-s_2-s_1\) 中,\(a\) 的系数为 \(3n-2n-n=0\)。化简其余各项,得到 \[ R=2n^2d\text{。}\]因此 \(R\) 取决于 \(d\) 和 \(n\),但不取决于 \(a\)。
因此,正确答案是 B。
English
Using the arithmetic-series formula, \[ s_j=\frac{jn}{2}\bigl(2a+(jn-1)d\bigr). \] In \(s_3-s_2-s_1,\) the coefficient of \(a\) is \(3n-2n-n=0.\) Simplifying the remaining terms gives \[ R=2n^2d. \] Thus \(R\) depends on \(d\) and \(n,\) but not on \(a.\)
Therefore, the correct answer is B.
37 答案 A
由相似关系,高度 \(x\) 处平行于底边的截线长度为 \[ b\left(1-\frac{x}{h}\right)=\frac{b(h-x)}{h}\text{。}\]将它乘以矩形的高 \(x\),得到 \[ \frac{bx}{h}(h-x)\text{。}\]
因此,正确答案是 A。
English
The cross-section parallel to the base at height \(x\) has width \[ b\left(1-\frac{x}{h}\right)=\frac{b(h-x)}{h} \] by similarity. Multiplying by the rectangle’s height \(x\) gives \[ \frac{bx}{h}(h-x). \]
Thus, the correct answer is A.
38 答案 D
设 \(ABC\) 的每个底角为 \(\theta\)。从 \(BC\) 的方向量起,直线 \(FD\) 的方向角为 \(180^\circ-a\),所以由等边三角形条件,\(DE\) 的方向角为 \(60^\circ-a\)。在 \(D\) 点处,\[ b=\theta-(60^\circ-a)=\theta-60^\circ+a\text{。}\]同样,在 \(E\) 点追角可得 \[ c=60^\circ+a-\theta\text{。}\]两式相加,得到 \(b+c=2a\)。
因此 \(a=\frac{b+c}{2}\),正确答案是 D。
English
Let each base angle of \(ABC\) be \(\theta.\) Measured from the direction \(BC,\) the line \(FD\) has direction \(180^\circ-a,\) so the equilateral condition makes \(DE\) have direction \(60^\circ-a.\) At \(D,\) \[ b=\theta-(60^\circ-a)=\theta-60^\circ+a. \] Similarly, angle chasing at \(E\) gives \[ c=60^\circ+a-\theta. \] Adding these equations yields \(b+c=2a.\)
Therefore \(a=\frac{b+c}{2},\) and the correct answer is D.
39 答案 E
由分母可知 \(a\ne0\) 且 \(a+b\ne0\)。令 \(t=\frac{b}{a}\) 并交叉相乘,得到 \[ (1+t)^2=t\text{,}\]即 \(t^2+t+1=0\)。其判别式为 \(-3\),所以 \(t\) 不是实数。因此 \(a\) 和 \(b\) 不可能都是实数。根据为 \(a\) 选择的非零复数倍数,可以一个为实数而另一个为非实数,也可以两者都是非实数。
因此,正确答案是 E。
English
The denominators require \(a\ne0\) and \(a+b\ne0.\) Setting \(t=\frac{b}{a}\) and cross-multiplying gives \[ (1+t)^2=t, \] or \(t^2+t+1=0.\) Its discriminant is \(-3,\) so \(t\) is not real. Therefore \(a\) and \(b\) cannot both be real. Depending on the nonzero complex scale chosen for \(a,\) one can be real and the other nonreal, or both can be nonreal.
Thus, the correct answer is E.
40 答案 A
设 \(C=(0,0)\)、\(B=(3,0)\)、\(A=(0,4)\)。斜边的方程为 \(\frac{x}{3}+\frac{y}{4}=1\)。与水平边成 \(30^\circ\) 的三等分角射线为 \(y=\frac{x}{\sqrt3}\)。代入得 \[ x=\frac{12\sqrt3}{4\sqrt3+3}\text{。}\]其长度为 \(\frac{x}{\cos30^\circ}=\frac{2x}{\sqrt3}\),即 \[ \frac{24}{4\sqrt3+3} =\frac{32\sqrt3-24}{13}\text{。}\]与水平边成 \(60^\circ\) 的射线给出较长的三等分角线。
因此,正确答案是 A。
English
Place \(C=(0,0),\) \(B=(3,0),\) \(A=(0,4).\) The hypotenuse has equation \(\frac{x}{3}+\frac{y}{4}=1.\) The \(30^\circ\) trisector ray is \(y=\frac{x}{\sqrt3}.\) Substitution gives \[ x=\frac{12\sqrt3}{4\sqrt3+3}. \] Its length is \(\frac{x}{\cos30^\circ}=\frac{2x}{\sqrt3},\) hence \[ \frac{24}{4\sqrt3+3} =\frac{32\sqrt3-24}{13}. \] The \(60^\circ\) ray gives the longer trisector.
Thus, the correct answer is A.