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1968 AMC 12 真题(P版)

数学 1968 年 35 题 83 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

一个圆的直径增加 \(\pi\) 个单位后,周长增加 \(P\) 个单位。则 \(P\) 等于:

A\(\dfrac1\pi\) B\(\pi\) C\(\dfrac{\pi^2}{2}\) D\(\pi^2\) E\(2\pi\)

English

Let \(P\) units be the increase in the circumference of a circle resulting from an increase of \(\pi\) units in the diameter. Then \(P\) equals:

(A) \(\dfrac1\pi\)(B) \(\pi\)(C) \(\dfrac{\pi^2}{2}\)(D) \(\pi^2\)(E) \(2\pi\)

2

使 \(64^{x-1}\) 除以 \(4^{x-1}\) 等于 \(256^{2x}\) 的实数 \(x\) 为:

A\(-\dfrac23\) B\(-\dfrac13\) C\(0\) D\(\dfrac14\) E\(\dfrac38\)

English

The real value of \(x\) such that \(64^{x-1}\) divided by \(4^{x-1}\) equals \(256^{2x}\) is:

(A) \(-\dfrac23\)(B) \(-\dfrac13\)(C) \(0\)(D) \(\dfrac14\)(E) \(\dfrac38\)

3

一条过点 \((0,4)\) 的直线与直线 \(x-3y-7=0\) 垂直。它的方程是:

A\(y+3x-4=0\) B\(y+3x+4=0\) C\(y-3x-4=0\) D\(3y+x-12=0\) E\(3y-x-12=0\)

English

A straight line passing through the point \((0,4)\) is perpendicular to the line \(x-3y-7=0.\) Its equation is:

(A) \(y+3x-4=0\)(B) \(y+3x+4=0\)(C) \(y-3x-4=0\)(D) \(3y+x-12=0\)(E) \(3y-x-12=0\)

4

对正实数定义运算 \(*\):\(a*b=\dfrac{ab}{a+b}\)。则 \(4*(4*4)\) 等于:

A\(\dfrac34\) B\(1\) C\(\dfrac43\) D\(2\) E\(\dfrac{16}{3}\)

English

Define an operation \(*\) for positive real numbers by \(a*b=\dfrac{ab}{a+b}.\) Then \(4*(4*4)\) equals:

(A) \(\dfrac34\)(B) \(1\)(C) \(\dfrac43\)(D) \(2\)(E) \(\dfrac{16}{3}\)

5

若 \(f(n)=\dfrac13n(n+1)(n+2)\),则 \(f(r)-f(r-1)\) 等于:

A\(r(r+1)\) B\((r+1)(r+2)\) C\(\dfrac13r(r+1)\) D\(\dfrac13(r+1)(r+2)\) E\(\dfrac13r(r+1)(r+2)\)

English

If \(f(n)=\dfrac13n(n+1)(n+2),\) then \(f(r)-f(r-1)\) equals:

(A) \(r(r+1)\)(B) \((r+1)(r+2)\)(C) \(\dfrac13r(r+1)\)(D) \(\dfrac13(r+1)(r+2)\)(E) \(\dfrac13r(r+1)(r+2)\)

6

将凸四边形 \(ABCD\) 的边 \(AD\) 过 \(D\) 点延长,并将边 \(BC\) 过 \(C\) 点延长,两延长线交于点 \(E\)。设 \(S\) 表示角 \(CDE\) 与角 \(DCE\) 的度数之和,\(S'\) 表示角 \(BAD\) 与角 \(ABC\) 的度数之和。若 \(r=\frac{S}{S'}\),则:

A有时 \(r=1\),有时 \(r\gt1\) B有时 \(r=1\),有时 \(r\lt1\) C\(0\lt r\lt1\) D\(r\gt1\) E\(r=1\)

English

Let side \(AD\) of convex quadrilateral \(ABCD\) be extended through \(D,\) and let side \(BC\) be extended through \(C,\) to meet in point \(E.\) Let \(S\) represent the degree-sum of angles \(CDE\) and \(DCE,\) and let \(S'\) represent the degree-sum of angles \(BAD\) and \(ABC.\) If \(r=\frac{S}{S'},\) then:

(A) \(r=1\) sometimes, \(r\gt1\) sometimes(B) \(r=1\) sometimes, \(r\lt1\) sometimes(C) \(0\lt r\lt1\)(D) \(r\gt1\)(E) \(r=1\)

7

设 \(O\) 是三角形 \(ABC\) 的两条中线 \(AP\) 与 \(CQ\) 的交点。若 \(OQ\) 长 \(3\) 英寸,则 \(OP\) 的长度(英寸)为:

A\(3\) B\(\dfrac92\) C\(6\) D\(9\) E无法确定

English

Let \(O\) be the intersection point of medians \(AP\) and \(CQ\) of triangle \(ABC.\) If \(OQ\) is \(3\) inches, then \(OP,\) in inches, is:

(A) \(3\)(B) \(\dfrac92\)(C) \(6\)(D) \(9\)(E) undetermined

8

一个正数本应乘以 \(6\),却误被除以 \(6\)。以正确结果为基准,所产生的百分误差取最接近的整数为:

A\(100\) B\(97\) C\(83\) D\(17\) E\(3\)

English

A positive number is mistakenly divided by \(6\) instead of being multiplied by \(6.\) Based on the correct answer, the error thus committed, to the nearest percent, is:

(A) \(100\)(B) \(97\)(C) \(83\)(D) \(17\)(E) \(3\)

9

满足 \(\lvert x+2\rvert=2\lvert x-2\rvert\) 的所有实数 \(x\) 之和为:

A\(\dfrac13\) B\(\dfrac23\) C\(6\) D\(6\dfrac13\) E\(6\dfrac23\)

English

The sum of the real values of \(x\) satisfying \(\lvert x+2\rvert=2\lvert x-2\rvert\) is:

(A) \(\dfrac13\)(B) \(\dfrac23\)(C) \(6\)(D) \(6\dfrac13\)(E) \(6\dfrac23\)

10

假设对某所学校,下列陈述为真:

\(\mathrm{I}\):有些学生不诚实。

\(\mathrm{II}\):所有兄弟会成员都诚实。

必然可以得出的结论是:

A有些学生是兄弟会成员 B有些兄弟会成员不是学生 C有些学生不是兄弟会成员 D没有兄弟会成员是学生 E没有学生是兄弟会成员

English

Assume that, for a certain school, it is true that

\(\mathrm{I}:\) Some students are not honest.

\(\mathrm{II}:\) All fraternity members are honest.

A necessary conclusion is:

(A) Some students are fraternity members(B) Some fraternity members are not students(C) Some students are not fraternity members(D) No fraternity member is a student(E) No student is a fraternity member

11

若圆 \(\mathrm{I}\) 上一段 \(60^\circ\) 圆弧与圆 \(\mathrm{II}\) 上一段 \(45^\circ\) 圆弧长度相同,则圆 \(\mathrm{I}\) 与圆 \(\mathrm{II}\) 的面积之比为:

A\(16:9\) B\(9:16\) C\(4:3\) D\(3:4\) E以上皆非

English

If an arc of \(60^\circ\) on circle \(\mathrm{I}\) has the same length as an arc of \(45^\circ\) on circle \(\mathrm{II},\) the ratio of the area of circle \(\mathrm{I}\) to that of circle \(\mathrm{II}\) is:

(A) \(16:9\)(B) \(9:16\)(C) \(4:3\)(D) \(3:4\)(E) none of these

12

一个圆经过边长为 \(7\dfrac12\)、\(10\)、\(12\dfrac12\) 的三角形的三个顶点。该圆的半径为:

A\(\dfrac{15}{4}\) B\(5\) C\(\dfrac{25}{4}\) D\(\dfrac{35}{4}\) E\(\dfrac{15\sqrt2}{2}\)

English

A circle passes through the vertices of a triangle with side-lengths \(7\dfrac12,\) \(10,\) \(12\dfrac12.\) The radius of the circle is:

(A) \(\dfrac{15}{4}\)(B) \(5\)(C) \(\dfrac{25}{4}\)(D) \(\dfrac{35}{4}\)(E) \(\dfrac{15\sqrt2}{2}\)

13

若 \(m\) 和 \(n\) 是 \(x^2+mx+n=0\) 的两个根,且 \(m\ne0\)、\(n\ne0\),则两根之和为:

A\(-\dfrac12\) B\(-1\) C\(\dfrac12\) D\(1\) E无法确定

English

If \(m\) and \(n\) are the roots of \(x^2+mx+n=0,\) \(m\ne0,\) \(n\ne0,\) then the sum of the roots is:

(A) \(-\dfrac12\)(B) \(-1\)(C) \(\dfrac12\)(D) \(1\)(E) undetermined

14

若非零数 \(x\) 和 \(y\) 满足 \(x=1+\dfrac1y\) 与 \(y=1+\dfrac1x\),则 \(y\) 等于:

A\(x-1\) B\(1-x\) C\(1+x\) D\(-x\) E\(x\)

English

If \(x\) and \(y\) are nonzero numbers such that \(x=1+\dfrac1y\) and \(y=1+\dfrac1x,\) then \(y\) equals:

(A) \(x-1\)(B) \(1-x\)(C) \(1+x\)(D) \(-x\)(E) \(x\)

15

设 \(P\) 是任意三个连续正奇数的乘积。所有这样的 \(P\) 都能被整除的最大整数为:

A\(15\) B\(6\) C\(5\) D\(3\) E\(1\)

English

Let \(P\) be the product of any three consecutive positive odd integers. The largest integer dividing all such \(P\) is:

(A) \(15\)(B) \(6\)(C) \(5\)(D) \(3\)(E) \(1\)

16

若 \(x\) 满足 \(\dfrac1x\lt2\) 且 \(\dfrac1x\gt-3\),则:

A\(-\dfrac13\lt x\lt\dfrac12\) B\(-\dfrac12\lt x\lt3\) C\(x\gt\dfrac12\) D\(x\gt\dfrac12\) 或 \(-\dfrac13\lt x\lt0\) E\(x\gt\dfrac12\) 或 \(x\lt-\dfrac13\)

English

If \(x\) is such that \(\dfrac1x\lt2\) and \(\dfrac1x\gt-3,\) then:

(A) \(-\dfrac13\lt x\lt\dfrac12\)(B) \(-\dfrac12\lt x\lt3\)(C) \(x\gt\dfrac12\)(D) \(x\gt\dfrac12\) or \(-\dfrac13\lt x\lt0\)(E) \(x\gt\dfrac12\) or \(x\lt-\dfrac13\)

17

设 \(f(n)=\dfrac{x_1+x_2+\cdots+x_n}{n}\),其中 \(n\) 为正整数。若 \(x_k=(-1)^k\),其中 \(k=1\)、\(2\)、\(\ldots\)、\(n\),则 \(f(n)\) 的所有可能值组成的集合为:

A\(\{0\}\) B\(\left\{\dfrac1n\right\}\) C\(\left\{0,-\dfrac1n\right\}\) D\(\left\{0,\dfrac1n\right\}\) E\(\left\{1,\dfrac1n\right\}\)

English

Let \(f(n)=\dfrac{x_1+x_2+\cdots+x_n}{n},\) where \(n\) is a positive integer. If \(x_k=(-1)^k,\) \(k=1,\) \(2,\) \(\ldots,\) \(n,\) the set of possible values of \(f(n)\) is:

(A) \(\{0\}\)(B) \(\left\{\dfrac1n\right\}\)(C) \(\left\{0,-\dfrac1n\right\}\)(D) \(\left\{0,\dfrac1n\right\}\)(E) \(\left\{1,\dfrac1n\right\}\)

18

三角形 \(ABC\) 的边 \(AB\) 长 \(8\) 英寸。作直线 \(DEF\) 平行于 \(AB\),其中 \(D\) 在线段 \(AC\) 上,\(E\) 在线段 \(BC\) 上。直线 \(AE\) 的延长线平分角 \(FEC\)。若 \(DE\) 长 \(5\) 英寸,则 \(CE\) 的长度(英寸)为:

A\(\dfrac{51}{4}\) B\(13\) C\(\dfrac{53}{4}\) D\(\dfrac{40}{3}\) E\(\dfrac{27}{2}\)

English

Side \(AB\) of triangle \(ABC\) has length \(8\) inches. Line \(DEF\) is drawn parallel to \(AB\) so that \(D\) is on segment \(AC,\) and \(E\) is on segment \(BC.\) Line \(AE\) extended bisects angle \(FEC.\) If \(DE\) has length \(5\) inches, then the length of \(CE,\) in inches, is:

(A) \(\dfrac{51}{4}\)(B) \(13\)(C) \(\dfrac{53}{4}\)(D) \(\dfrac{40}{3}\)(E) \(\dfrac{27}{2}\)

19

将 \(10\) 美元兑换成一角硬币和二角五分硬币,且每种硬币至少使用一枚。设兑换方法数为 \(n\),则 \(n\) 等于:

A\(40\) B\(38\) C\(21\) D\(20\) E\(19\)

English

Let \(n\) be the number of ways that \(10\) dollars can be changed into dimes and quarters, with at least one of each coin being used. Then \(n\) equals:

(A) \(40\)(B) \(38\)(C) \(21\)(D) \(20\)(E) \(19\)

20

一个 \(n\) 边凸多边形的内角度数成等差数列。若公差为 \(5^\circ\),最大角为 \(160^\circ\),则 \(n\) 等于:

A\(9\) B\(10\) C\(12\) D\(16\) E\(32\)

English

The measures of the interior angles of a convex polygon of \(n\) sides are in arithmetic progression. If the common difference is \(5^\circ\) and the largest angle is \(160^\circ,\) then \(n\) equals:

(A) \(9\)(B) \(10\)(C) \(12\)(D) \(16\)(E) \(32\)

21

若 \(S=1!+2!+3!+\cdots+99!\),则 \(S\) 的个位数字为:

A\(9\) B\(8\) C\(5\) D\(3\) E\(0\)

English

If \(S=1!+2!+3!+\cdots+99!,\) then the units digit in the value of \(S\) is:

(A) \(9\)(B) \(8\)(C) \(5\)(D) \(3\)(E) \(0\)

22

将一条长为 \(1\) 的线段分成四段。以这四段为边能组成四边形,当且仅当每一段都:

A等于 \(\dfrac14\) B大于或等于 \(\dfrac18\),且小于 \(\dfrac12\) C大于 \(\dfrac18\),且小于 \(\dfrac12\) D大于 \(\dfrac18\),且小于 \(\dfrac14\) E小于 \(\dfrac12\)

English

A segment of length \(1\) is divided into four segments. Then there exists a quadrilateral with the four segments as sides if and only if each segment is:

(A) equal to \(\dfrac14\)(B) equal to or greater than \(\dfrac18\) and less than \(\dfrac12\)(C) greater than \(\dfrac18\) and less than \(\dfrac12\)(D) greater than \(\dfrac18\) and less than \(\dfrac14\)(E) less than \(\dfrac12\)

23

若所有对数均为实数,则等式

\[ \begin{aligned} \log(x+3)+\log(x-1)\\ =\log(x^2-2x-3) \end{aligned} \]

对下列哪种情形成立:

A所有实数 \(x\) B不存在实数 \(x\) C除 \(x=0\) 外的所有实数 \(x\) D除 \(x=0\) 外不存在实数 \(x\) E除 \(x=1\) 外的所有实数 \(x\)

English

If all the logarithms are real numbers, the equality

\[ \begin{aligned} \log(x+3)+\log(x-1)\\ =\log(x^2-2x-3) \end{aligned} \]

is satisfied for:

(A) all real values of \(x\)(B) no real values of \(x\)(C) all real values of \(x\) except \(x=0\)(D) no real values of \(x\) except \(x=0\)(E) all real values of \(x\) except \(x=1\)

24

将一幅 \(18''\times24''\) 的画装入木框,较长的一边竖直放置。上下两边的框宽是左右两边框宽的两倍。若木框面积等于画的面积,则装框后较短边与较长边的长度之比为:

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

English

A painting \(18''\times24''\) is to be placed into a wooden frame with the longer dimension vertical. The wood at the top and bottom is twice as wide as the wood on the sides. If the frame area equals that of the painting itself, the ratio of the smaller to the larger dimension of the framed painting is:

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

25

艾斯以恒定速度跑步,弗拉什的速度是他的 \(x\) 倍,其中 \(x\gt1\)。弗拉什让艾斯领先 \(y\) 码,发出信号后两人同向出发。则弗拉什追上艾斯前必须跑的码数为:

A\(xy\) B\(\dfrac{y}{x+y}\) C\(\dfrac{xy}{x-1}\) D\(\dfrac{x+y}{x+1}\) E\(\dfrac{x+y}{x-1}\)

English

Ace runs with constant speed and Flash runs \(x\) times as fast, \(x\gt1.\) Flash gives Ace a head start of \(y\) yards, and, at a given signal, they start off in the same direction. Then the number of yards Flash must run to catch Ace is:

(A) \(xy\)(B) \(\dfrac{y}{x+y}\)(C) \(\dfrac{xy}{x-1}\)(D) \(\dfrac{x+y}{x+1}\)(E) \(\dfrac{x+y}{x-1}\)

26

设 \(S=2+4+6+\cdots+2N\),其中 \(N\) 是使 \(S\gt1{,}000{,}000\) 的最小正整数。则 \(N\) 的各位数字之和为:

A\(27\) B\(12\) C\(6\) D\(2\) E\(1\)

English

Let \(S=2+4+6+\cdots+2N,\) where \(N\) is the smallest positive integer such that \(S\gt1{,}000{,}000.\) Then the sum of the digits of \(N\) is:

(A) \(27\)(B) \(12\)(C) \(6\)(D) \(2\)(E) \(1\)

27

设

\[ \begin{aligned} S_n&=1-2+3-4+\cdots\\ &\quad+(-1)^{n-1}n\text{。} \end{aligned} \]

其中 \(n=1\)、\(2\)、\(\ldots\)。则 \(S_{17}+S_{33}+S_{50}\) 等于:

A\(0\) B\(1\) C\(2\) D\(-1\) E\(-2\)

English

Let

\[ \begin{aligned} S_n&=1-2+3-4+\cdots\\ &\quad+(-1)^{n-1}n. \end{aligned} \]

Here \(n=1,\) \(2,\) \(\ldots.\) Then \(S_{17}+S_{33}+S_{50}\) equals:

(A) \(0\)(B) \(1\)(C) \(2\)(D) \(-1\)(E) \(-2\)

28

若 \(a\) 与 \(b\) 的算术平均数是其几何平均数的两倍,且 \(a\gt b\gt0\),则 \(\frac{a}{b}\) 取到最接近整数后的一个可能值为:

A\(5\) B\(8\) C\(11\) D\(14\) E以上皆非

English

If the arithmetic mean of \(a\) and \(b\) is double their geometric mean, with \(a\gt b\gt0,\) then a possible value for the ratio \(\frac{a}{b},\) to the nearest integer, is:

(A) \(5\)(B) \(8\)(C) \(11\)(D) \(14\)(E) none of these

29

已知三个数 \(x\)、\(y=x^x\)、\(z=x^{(x^x)}\),其中 \(0.9\lt x\lt1.0\)。按从小到大排列为:

A\(x\)、\(z\)、\(y\) B\(x\)、\(y\)、\(z\) C\(y\)、\(x\)、\(z\) D\(y\)、\(z\)、\(x\) E\(z\)、\(x\)、\(y\)

English

Given the three numbers \(x,\) \(y=x^x,\) \(z=x^{(x^x)},\) with \(0.9\lt x\lt1.0.\) Arranged in order of increasing magnitude, they are:

(A) \(x,\) \(z,\) \(y\)(B) \(x,\) \(y,\) \(z\)(C) \(y,\) \(x,\) \(z\)(D) \(y,\) \(z,\) \(x\)(E) \(z,\) \(x,\) \(y\)

30

在同一平面内画有凸多边形 \(P_1\) 和 \(P_2\),其边数分别为 \(n_1\) 和 \(n_2\),且 \(n_1\leq n_2\)。若 \(P_1\) 与 \(P_2\) 没有任何公共线段,则 \(P_1\) 与 \(P_2\) 交点数的最大值为:

A\(2n_1\) B\(2n_2\) C\(n_1n_2\) D\(n_1+n_2\) E以上皆非

English

Convex polygons \(P_1\) and \(P_2\) are drawn in the same plane with \(n_1\) and \(n_2\) sides, respectively, \(n_1\leq n_2.\) If \(P_1\) and \(P_2\) do not have any line segment in common, then the maximum number of intersections of \(P_1\) and \(P_2\) is:

(A) \(2n_1\)(B) \(2n_2\)(C) \(n_1n_2\)(D) \(n_1+n_2\)(E) none of these

31

在这幅未按比例绘制的图中,图形 \(\mathrm{I}\) 和 \(\mathrm{III}\) 是面积分别为 \(32\sqrt3\) 与 \(8\sqrt3\) 平方英寸的等边三角形区域。图形 \(\mathrm{II}\) 是面积为 \(32\) 平方英寸的正方形区域。将线段 \(AD\) 的长度减少其自身的 \(12\dfrac12\%\),而 \(AB\) 与 \(CD\) 的长度保持不变。正方形面积减少的百分比为:

题图

A\(12\dfrac12\) B\(25\) C\(50\) D\(75\) E\(87\dfrac12\)

English

In this diagram, not drawn to scale, figures \(\mathrm{I}\) and \(\mathrm{III}\) are equilateral triangular regions with respective areas of \(32\sqrt3\) and \(8\sqrt3\) square inches. Figure \(\mathrm{II}\) is a square region with area \(32\) square inches. Let the length of segment \(AD\) be decreased by \(12\dfrac12\%\) of itself, while the lengths of \(AB\) and \(CD\) remain unchanged. The percent decrease in the area of the square is:

题图

(A) \(12\dfrac12\)(B) \(25\)(C) \(50\)(D) \(75\)(E) \(87\dfrac12\)

32

\(A\) 和 \(B\) 分别沿两条在点 \(O\) 垂直相交的直线路径做匀速运动。当 \(A\) 位于 \(O\) 时,\(B\) 距 \(O\) 尚有 \(500\) 码。经过 \(2\) 分钟,两者到 \(O\) 的距离相等;再过 \(8\) 分钟,两者到 \(O\) 的距离又一次相等。则 \(A\) 与 \(B\) 的速度之比为:

A\(4:5\) B\(5:6\) C\(2:3\) D\(5:8\) E\(1:2\)

English

\(A\) and \(B\) move uniformly along two straight paths intersecting at right angles in point \(O.\) When \(A\) is at \(O,\) \(B\) is \(500\) yards short of \(O.\) In \(2\) minutes they are equidistant from \(O,\) and in \(8\) minutes more they are again equidistant from \(O.\) Then the ratio of \(A\)’s speed to \(B\)’s speed is:

(A) \(4:5\)(B) \(5:6\)(C) \(2:3\)(D) \(5:8\)(E) \(1:2\)

33

数 \(N\) 用 \(7\) 进制表示时有三位。当 \(N\) 用 \(9\) 进制表示时,三个数字的顺序恰好颠倒。则中间一位数字为:

A\(0\) B\(1\) C\(3\) D\(4\) E\(5\)

English

A number \(N\) has three digits when expressed in base \(7.\) When \(N\) is expressed in base \(9\) the digits are reversed. Then the middle digit is:

(A) \(0\)(B) \(1\)(C) \(3\)(D) \(4\)(E) \(5\)

34

众议院共有 \(400\) 名议员投票,一项法案未获通过。由同一批议员重新投票后,法案以原先落败票差两倍的票差通过。重新投票时的赞成票数是原先反对票数的 \(\dfrac{12}{11}\)。第二次投票的赞成票比第一次多多少票?

A\(75\) B\(60\) C\(50\) D\(45\) E\(20\)

English

With \(400\) members voting, the House of Representatives defeated a bill. A re-vote, with the same members voting, resulted in passage of the bill by twice the margin by which it was originally defeated. The number voting for the bill on the re-vote was \(\dfrac{12}{11}\) of the number voting against it originally. How many more members voted for the bill the second time than voted for it the first time?

(A) \(75\)(B) \(60\)(C) \(50\)(D) \(45\)(E) \(20\)

35

在图中,圆心为 \(O\),半径为 \(a\) 英寸,弦 \(EF\) 平行于弦 \(CD\),\(O\)、\(G\)、\(H\)、\(J\) 共线,且 \(G\) 是 \(CD\) 的中点。设 \(K\)(平方英寸)表示梯形 \(CDFE\) 的面积,\(R\)(平方英寸)表示矩形 \(ELMF\) 的面积。将 \(CD\) 与 \(EF\) 向上平移,使 \(OG\) 增大并趋近于 \(a\),同时始终保持 \(JH\) 等于 \(HG\),则比值 \(K:R\) 可以任意接近:

题图

A\(0\) B\(1\) C\(\sqrt2\) D\(\dfrac1{\sqrt2}+\dfrac12\) E\(\dfrac1{\sqrt2}+1\)

English

In this diagram the center of the circle is \(O,\) the radius is \(a\) inches, chord \(EF\) is parallel to chord \(CD,\) \(O,\) \(G,\) \(H,\) \(J\) are collinear, and \(G\) is the midpoint of \(CD.\) Let \(K\) (square inches) represent the area of trapezoid \(CDFE\) and let \(R\) (square inches) represent the area of rectangle \(ELMF.\) Then, as \(CD\) and \(EF\) are translated upward so that \(OG\) increases toward the value \(a,\) while \(JH\) always equals \(HG,\) the ratio \(K:R\) becomes arbitrarily close to:

题图

(A) \(0\)(B) \(1\)(C) \(\sqrt2\)(D) \(\dfrac1{\sqrt2}+\dfrac12\)(E) \(\dfrac1{\sqrt2}+1\)

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逐题解析

1 答案 D

原周长为 \(\pi d\)。直径增加 \(\pi\) 后,周长为 \(\pi(d+\pi)\)。因此增加量为 \(\pi(d+\pi)-\pi d=\pi^2\)。

因此,正确答案是 D。

English

The original circumference is \(\pi d.\) After the diameter increases by \(\pi,\) it is \(\pi(d+\pi).\) The increase is therefore \(\pi(d+\pi)-\pi d=\pi^2.\)

Therefore, the correct answer is D.

2 答案 B

左边为 \((\frac{64}{4})^{x-1}=16^{x-1}\),而 \(256^{2x}=(16^2)^{2x}=16^{4x}\)。因此 \(x-1=4x\),所以 \(x=-\frac{1}{3}\)。

因此,正确答案是 B。

English

The left side is \((\frac{64}{4})^{x-1}=16^{x-1},\) while \(256^{2x}=(16^2)^{2x}=16^{4x}.\) Hence \(x-1=4x,\) so \(x=-\frac{1}{3}.\)

Therefore, the correct answer is B.

3 答案 A

已知直线的斜率为 \(\frac{1}{3}\),所以垂线的斜率为 \(-3\)。过点 \((0,4)\) 的垂线方程为 \(y=-3x+4\),即 \(y+3x-4=0\)。

因此,正确答案是 A。

English

The given line has slope \(\frac{1}{3},\) so a perpendicular line has slope \(-3.\) Through \((0,4)\) its equation is \(y=-3x+4,\) or \(y+3x-4=0.\)

Therefore, the correct answer is A.

4 答案 C

首先,\(4*4=\frac{16}{8}=2\)。因此 \(4*(4*4)=4*2\),其值为 \(\frac{8}{4+2}=\frac{4}{3}\)。

因此,正确答案是 C。

English

First, \(4*4=\frac{16}{8}=2.\) Therefore \(4*(4*4)=4*2,\) which equals \(\frac{8}{4+2}=\frac{4}{3}.\)

Therefore, the correct answer is C.

5 答案 A

有 \(f(r)=\frac13r(r+1)(r+2)\) 和 \(f(r-1)=\frac13(r-1)r(r+1)\)。两者之差可写成 \(\frac13r(r+1)\) 乘以 \((r+2)-(r-1)=3\)。结果为 \(r(r+1)\)。

因此,正确答案是 A。

English

We have \(f(r)=\frac13r(r+1)(r+2)\) and \(f(r-1)=\frac13(r-1)r(r+1).\) Factor their difference as \(\frac13r(r+1)\) times \((r+2)-(r-1)=3.\) The result is \(r(r+1).\)

Therefore, the correct answer is A.

6 答案 E

在三角形 \(ECD\) 中,\(S=180^\circ-\angle E\)。由于 \(D\) 在 \(AE\) 上,\(C\) 在 \(BE\) 上,由三角形 \(EAB\) 同样得到 \(S'=180^\circ-\angle E\)。因此 \(S=S'\),且 \(r=1\)。

因此,正确答案是 E。

English

In triangle \(ECD,\) \(S=180^\circ-\angle E.\) Since \(D\) lies on \(AE\) and \(C\) lies on \(BE,\) triangle \(EAB\) gives \(S'=180^\circ-\angle E\) as well. Thus \(S=S'\) and \(r=1.\)

Therefore, the correct answer is E.

7 答案 E

重心按 \(2:1\) 分割每条中线,所以 \(OQ=3\) 可得 \(CQ=9\)。但这并未给出两条不同中线 \(CQ\) 与 \(AP\) 的长度关系。满足 \(CQ=9\) 的三角形可以有不同的 \(AP\),所以 \(OP\) 无法确定。

因此,正确答案是 E。

English

The centroid divides each median in a \(2:1\) ratio, so \(OQ=3\) determines \(CQ=9.\) It gives no relation between the lengths of the two different medians \(CQ\) and \(AP.\) Triangles with \(CQ=9\) can have different \(AP,\) so \(OP\) is undetermined.

Therefore, the correct answer is E.

8 答案 B

正确结果为 \(6N\),误算结果为 \(\frac{N}{6}\)。以正确结果为基准,误差为 \(\frac{6N-\frac{N}{6}}{6N}=\frac{35}{36}\approx97.2\%\)。

因此,正确答案是 B。

English

The correct result is \(6N,\) while the mistaken result is \(\frac{N}{6}.\) Relative to the correct result, the error is \(\frac{6N-\frac{N}{6}}{6N}=\frac{35}{36}\approx97.2\%.\)

Therefore, the correct answer is B.

9 答案 E

平方后得到 \((x+2)^2=4(x-2)^2\),即 \(3x^2-20x+12=0\)。两个根都满足原绝对值方程,它们的和为 \(\frac{20}{3}=6\frac23\)。

因此,正确答案是 E。

English

Squaring gives \((x+2)^2=4(x-2)^2,\) or \(3x^2-20x+12=0.\) Both roots satisfy the original absolute-value equation, and their sum is \(\frac{20}{3}=6\frac23.\)

Therefore, the correct answer is E.

10 答案 C

陈述 \(\mathrm{I}\) 保证存在一名不诚实的学生。陈述 \(\mathrm{II}\) 说明每位兄弟会成员都诚实,所以这名不诚实的学生不可能是兄弟会成员。因此,有些学生不是兄弟会成员。

因此,正确答案是 C。

English

Statement \(\mathrm{I}\) guarantees a dishonest student. Statement \(\mathrm{II}\) says every fraternity member is honest, so that dishonest student cannot be a fraternity member. Hence some student is not a fraternity member.

Therefore, the correct answer is C.

11 答案 B

弧长相等给出 \(60r_1=45r_2\),所以 \(\frac{r_1}{r_2}=\frac{3}{4}\)。圆的面积与半径的平方成正比,故 \(A_1:A_2=9:16\)。

因此,正确答案是 B。

English

Equal arc lengths give \(60r_1=45r_2,\) so \(\frac{r_1}{r_2}=\frac{3}{4}.\) Circle areas scale as the squares of the radii, giving \(A_1:A_2=9:16.\)

Therefore, the correct answer is B.

12 答案 C

三边为 \(\frac52(3,4,5)\),所以该三角形为直角三角形,斜边长 \(\frac{25}{2}\)。斜边就是外接圆直径,因此半径为 \(\frac{25}{4}\)。

因此,正确答案是 C。

English

The sides are \(\frac52(3,4,5),\) so the triangle is right with hypotenuse \(\frac{25}{2}.\) The hypotenuse is the circumdiameter, making the radius \(\frac{25}{4}.\)

Therefore, the correct answer is C.

13 答案 B

由于两根为 \(m\)、\(n\),韦达定理给出 \(m+n=-m\) 和 \(mn=n\)。因为 \(n\ne0\),第二个等式给出 \(m=1\),进而 \(n=-2\)。两根之和为 \(-1\)。

因此,正确答案是 B。

English

Because the roots are \(m,\) \(n,\) Vieta gives \(m+n=-m\) and \(mn=n.\) Since \(n\ne0,\) the second equation gives \(m=1,\) and then \(n=-2.\) Their sum is \(-1.\)

Therefore, the correct answer is B.

14 答案 E

两个等式分别给出 \(xy=y+1\) 和 \(xy=x+1\)。因此 \(x+1=y+1\),所以 \(y=x\)。

因此,正确答案是 E。

English

The equations give \(xy=y+1\) and \(xy=x+1.\) Therefore \(x+1=y+1,\) so \(y=x.\)

Therefore, the correct answer is E.

15 答案 D

三个连续奇数模 \(3\) 的余数依次连续,所以其中一个能被 \(3\) 整除。因此每个这样的乘积都能被 \(3\) 整除。两个乘积 \(1\cdot3\cdot5=15\) 与 \(7\cdot9\cdot11=693\) 的最大公因数为 \(3\),所以不存在总能整除该乘积的更大整数。

因此,正确答案是 D。

English

Three consecutive odd integers occupy three consecutive residues modulo \(3,\) so one is divisible by \(3.\) Thus every product is divisible by \(3.\) The products \(1\cdot3\cdot5=15\) and \(7\cdot9\cdot11=693\) have greatest common divisor \(3,\) so no larger integer always divides the product.

Therefore, the correct answer is D.

16 答案 E

当 \(x\gt0\) 时,\(\frac{1}{x}\lt2\) 给出 \(x\gt\frac{1}{2}\),另一个不等式自动成立。当 \(x\lt0\) 时,\(\frac{1}{x}\gt-3\) 给出 \(x\lt-\frac{1}{3}\),第一个不等式自动成立。因此 \(x\gt\frac{1}{2}\) 或 \(x\lt-\frac{1}{3}\)。

因此,正确答案是 E。

English

For \(x\gt0,\) \(\frac{1}{x}\lt2\) gives \(x\gt\frac{1}{2},\) while the other inequality is automatic. For \(x\lt0,\) \(\frac{1}{x}\gt-3\) gives \(x\lt-\frac{1}{3},\) while the first is automatic. Thus \(x\gt\frac{1}{2}\) or \(x\lt-\frac{1}{3}.\)

Therefore, the correct answer is E.

17 答案 C

若 \(n\) 为偶数,各项两两配对后和为 \(0\),所以 \(f(n)=0\)。若 \(n\) 为奇数,最后还剩一个 \(-1\),所以 \(f(n)=-\frac{1}{n}\)。因此所有可能值为 \(\{0,-\frac{1}{n}\}\)。

因此,正确答案是 C。

English

If \(n\) is even, the terms pair to give sum \(0,\) so \(f(n)=0.\) If \(n\) is odd, one final \(-1\) remains, so \(f(n)=-\frac{1}{n}.\) The possible values are therefore \(\{0,-\frac{1}{n}\}.\)

Therefore, the correct answer is C.

18 答案 D

在 \(AE\) 过 \(E\) 的延长线上取点 \(G\)。由角平分线条件,\(\angle FEG=\angle GEC\)。由于 \(FE\parallel AB\),有 \(\angle FEG=\angle BAE\)。又因为射线 \(EG\) 与 \(EA\) 方向相反,射线 \(EC\) 与 \(EB\) 方向也相反,所以 \(\angle GEC=\angle AEB\)。因此三角形 \(ABE\) 为等腰三角形,且 \(BE=AB=8\)。由相似三角形 \(CDE\) 与 \(CAB\) 得 \(\frac{CE}{CE+8}=\frac58\)。因此 \(8CE=5CE+40\),所以 \(CE=\frac{40}{3}\)。

因此,正确答案是 D。

English

Put \(G\) on the extension of \(AE\) beyond \(E.\) The bisector condition gives \(\angle FEG=\angle GEC.\) Since \(FE\parallel AB,\) we have \(\angle FEG=\angle BAE.\) Also, rays \(EG\) and \(EA\) are opposite, as are rays \(EC\) and \(EB,\) so \(\angle GEC=\angle AEB.\) Thus triangle \(ABE\) is isosceles and \(BE=AB=8.\) Similar triangles \(CDE\) and \(CAB\) give \(\frac{CE}{CE+8}=\frac58.\) Hence \(8CE=5CE+40,\) so \(CE=\frac{40}{3}.\)

Therefore, the correct answer is D.

19 答案 E

方程 \(10d+25q=1000\) 化简为 \(2d+5q=200\)。因此 \(q\) 必须为偶数;令 \(q=2k\),则 \(d=100-5k\)。由两者均为正数可得 \(k=1\)、\(2\)、\(\ldots\)、\(19\),所以共有 \(19\) 种方法。

因此,正确答案是 E。

English

The equation \(10d+25q=1000\) reduces to \(2d+5q=200.\) Thus \(q\) must be even; write \(q=2k,\) giving \(d=100-5k.\) Positivity requires \(k=1,\) \(2,\) \(\ldots,\) \(19,\) so there are \(19\) ways.

Therefore, the correct answer is E.

20 答案 A

最小角为 \(160-5(n-1)\)。内角和为 \(\frac n2\) 乘以 \(320-5(n-1)\),并且等于 \(180(n-2)\)。化简得 \[ \begin{aligned} n^2+7n-144&=0,\\ (n-9)(n+16)&=0\text{。} \end{aligned} \] 因此 \(n=9\)。

因此,正确答案是 A。

English

The smallest angle is \(160-5(n-1).\) The angle sum is \(\frac n2\) times \(320-5(n-1),\) and it equals \(180(n-2).\) This simplifies to \[ \begin{aligned} n^2+7n-144&=0,\\ (n-9)(n+16)&=0. \end{aligned} \] Thus \(n=9.\)

Therefore, the correct answer is A.

21 答案 D

当 \(k\ge5\) 时,\(k!\) 能被 \(10\) 整除。只有前四项会影响个位。它们的和为 \(1+2+6+24=33\),个位数字是 \(3\)。

因此,正确答案是 D。

English

For \(k\ge5,\) \(k!\) is divisible by \(10.\) Only the first four terms matter. Their sum is \(1+2+6+24=33,\) whose units digit is \(3.\)

Therefore, the correct answer is D.

22 答案 E

能组成简单非退化四边形,当且仅当每一边都小于其余三边之和。由于总长为 \(1\),对每一段而言,这个条件是 \(s\lt1-s\),即 \(s\lt\frac{1}{2}\)。

因此,正确答案是 E。

English

A simple nondegenerate quadrilateral exists exactly when each side is less than the sum of the other three. Since the total is \(1,\) this condition is \(s\lt1-s,\) or \(s\lt\frac{1}{2},\) for every segment.

Therefore, the correct answer is E.

23 答案 B

合并对数要求 \((x+3)(x-1)=x^2-2x-3\)。由此得到 \(x=0\)。但此时 \(x-1=-1\),所以左边的对数不是实数。因此没有实数解。

因此,正确答案是 B。

English

Combining logarithms would require \((x+3)(x-1)=x^2-2x-3.\) This gives \(x=0.\) But then \(x-1=-1,\) so the logarithm on the left is not real. Hence there are no real solutions.

Therefore, the correct answer is B.

24 答案 C

外部尺寸为 \(18+2x\) 和 \(24+4x\)。由于木框面积等于画的面积,\((18+2x)(24+4x)=2(18)(24)\)。由此得到 \(x^2+15x-54=0\),所以 \(x=3\)。装框后的尺寸为 \(24\) 和 \(36\),其比为 \(2:3\)。

因此,正确答案是 C。

English

The outside dimensions are \(18+2x\) and \(24+4x.\) Since the frame area equals the painting area, \((18+2x)(24+4x)=2(18)(24).\) This gives \(x^2+15x-54=0,\) so \(x=3.\) The dimensions are \(24\) and \(36,\) with ratio \(2:3.\)

Therefore, the correct answer is C.

25 答案 C

若艾斯的速度为 \(v\),弗拉什的速度为 \(xv\),所以两人的相对速度为 \((x-1)v\)。追赶时间为 \(\frac{y}{(x-1)v}\),在此期间弗拉什跑了 \(\frac{xv\cdot y}{(x-1)v}=\frac{xy}{x-1}\)。

因此,正确答案是 C。

English

If Ace runs at speed \(v,\) Flash runs at \(xv,\) so their closing speed is \((x-1)v.\) The catch-up time is \(\frac{y}{(x-1)v},\) during which Flash runs \(\frac{xv\cdot y}{(x-1)v}=\frac{xy}{x-1}.\)

Therefore, the correct answer is C.

26 答案 E

有 \(S=N(N+1)\)。当 \(N=999\) 时,\(999{,}000\lt1{,}000{,}000\);当 \(N=1000\) 时,结果为 \(1{,}001{,}000\)。因此 \(N=1000\),其各位数字之和为 \(1\)。

因此,正确答案是 E。

English

We have \(S=N(N+1).\) For \(N=999,\) this is \(999{,}000\lt1{,}000{,}000,\) while for \(N=1000\) it is \(1{,}001{,}000.\) Thus \(N=1000,\) whose digit sum is \(1.\)

Therefore, the correct answer is E.

27 答案 B

配对后可得:当 \(n\) 为偶数时,\(S_n=-\frac{n}{2}\);当 \(n\) 为奇数时,\(S_n=\frac{n+1}{2}\)。因此 \(S_{17}=9\)、\(S_{33}=17\)、\(S_{50}=-25\),三者之和为 \(1\)。

因此,正确答案是 B。

English

Pairing gives \(S_n=-\frac{n}{2}\) for even \(n\) and \(S_n=\frac{n+1}{2}\) for odd \(n.\) Hence \(S_{17}=9,\) \(S_{33}=17,\) \(S_{50}=-25,\) and their sum is \(1.\)

Therefore, the correct answer is B.

28 答案 D

令 \(t=\frac{a}{b}\gt1\)。由条件得 \(\frac{t+1}{2}=2\sqrt t\),所以 \(t^2-14t+1=0\)。大于 \(1\) 的根为 \(t=7+4\sqrt3\approx13.93\),最接近的整数为 \(14\)。

因此,正确答案是 D。

English

Let \(t=\frac{a}{b}\gt1.\) The condition gives \(\frac{t+1}{2}=2\sqrt t,\) so \(t^2-14t+1=0.\) The root greater than \(1\) is \(t=7+4\sqrt3\approx13.93,\) whose nearest integer is \(14.\)

Therefore, the correct answer is D.

29 答案 A

因为 \(0\lt x\lt1\) 且 \(x\lt1\),将底数 \(x\) 提到指数 \(x\) 可得 \(x\lt y=x^x\lt1\)。当底数介于 \(0\) 与 \(1\) 之间时,指数越大,幂值越小。由于 \(x\lt y\lt1\),有 \(x=x^1\lt x^y=z\lt x^x=y\)。所以 \(x\lt z\lt y\)。

因此,正确答案是 A。

English

Since \(0\lt x\lt1\) and \(x\lt1,\) raising \(x\) to the exponent \(x\) gives \(x\lt y=x^x\lt1.\) For a base between \(0\) and \(1,\) a larger exponent gives a smaller value. Because \(x\lt y\lt1,\) we have \(x=x^1\lt x^y=z\lt x^x=y.\) Thus \(x\lt z\lt y.\)

Therefore, the correct answer is A.

30 答案 A

\(P_1\) 的每条边都位于一条直线上,它与凸区域 \(P_2\) 的交集要么是一条线段,要么为空。因此该边与 \(P_2\) 的边界至多相交两次。对 \(n_1\) 条边合计,交点数至多为 \(2n_1\);取适当的狭长凸 \(n_1\) 边形穿过一个凸 \(n_2\) 边形即可达到此上界。

因此,正确答案是 A。

English

Each side of \(P_1\) lies on a line, and its intersection with the convex region \(P_2\) is a single segment or empty. Thus that side crosses the boundary of \(P_2\) at most twice. Across \(n_1\) sides there are at most \(2n_1,\) and a suitable thin convex \(n_1\)-gon crossing a convex \(n_2\)-gon attains this bound.

Therefore, the correct answer is A.

31 答案 D

由 \((\frac{\sqrt3}{4})s^2=32\sqrt3\) 得 \(AB=8\sqrt2\)。正方形的边长 \(BC=4\sqrt2\),较小等边三角形的边长也有 \(CD=4\sqrt2\)。因此 \(AD=16\sqrt2\)。将它减少 \(\frac{1}{8}\),会使 \(BC\) 减少 \(2\sqrt2\),其新长度为 \(2\sqrt2\)。正方形面积从 \(32\) 降至 \(8\),减少了 \(75\%\)。

因此,正确答案是 D。

English

From \((\frac{\sqrt3}{4})s^2=32\sqrt3,\) \(AB=8\sqrt2.\) The square has \(BC=4\sqrt2,\) and the smaller equilateral triangle also has \(CD=4\sqrt2.\) Thus \(AD=16\sqrt2.\) Decreasing it by \(\frac{1}{8}\) removes \(2\sqrt2\) from \(BC,\) whose new length is \(2\sqrt2.\) The square’s area falls from \(32\) to \(8,\) a \(75\%\) decrease.

Therefore, the correct answer is D.

32 答案 C

设 \(A\)、\(B\) 的速度分别为 \(u\)、\(v\)。当 \(t=2\) 时,\(2u=500-2v\),所以 \(u+v=250\)。当 \(t=10\) 时,\(B\) 已通过 \(O\),且 \(10u=10v-500\),所以 \(v-u=50\)。于是 \(u=100\)、\(v=150\),且 \(u:v=2:3\)。

因此,正确答案是 C。

English

Let the speeds of \(A,\) \(B\) be \(u,\) \(v.\) At \(t=2,\) \(2u=500-2v,\) so \(u+v=250.\) At \(t=10,\) \(B\) has passed \(O,\) and \(10u=10v-500,\) so \(v-u=50.\) Hence \(u=100,\) \(v=150,\) and \(u:v=2:3.\)

Therefore, the correct answer is C.

33 答案 A

设 \(N=(abc)_7=(cba)_9\)。则 \(49a+7b+c=81c+9b+a\),所以 \(24a-b=40c\)。在 \(1\leq a\leq6\)、\(0\leq b\leq6\) 且 \(1\leq c\leq6\) 的条件下,唯一可能是 \(a=5\)、\(c=3\)、\(b=0\)。事实上,\((503)_7=(305)_9=248\)。

因此,正确答案是 A。

English

Let \(N=(abc)_7=(cba)_9.\) Then \(49a+7b+c=81c+9b+a,\) so \(24a-b=40c.\) With \(1\leq a\leq6,\) \(0\leq b\leq6,\) and \(1\leq c\leq6,\) the only possibility is \(a=5,\) \(c=3,\) \(b=0.\) Indeed, \((503)_7=(305)_9=248.\)

Therefore, the correct answer is A.

34 答案 B

设第一次和第二次投票的赞成票数为 \(F\)、\(G\)。由比例条件得 \(G=\frac{12}{11}(400-F)\)。通过时的票差是原先落败票差的两倍,所以 \(2G-400=2(400-2F)\),即 \(G+2F=600\)。解得 \(F=180\)、\(G=240\),所以第二次多了 \(60\) 张赞成票。

因此,正确答案是 B。

English

Let \(F,\) \(G\) be the original and second numbers voting for the bill. The ratio condition gives \(G=\frac{12}{11}(400-F).\) The passage margin is twice the original defeat margin, so \(2G-400=2(400-2F),\) or \(G+2F=600.\) Solving gives \(F=180,\) \(G=240,\) so \(60\) more members voted for it.

Therefore, the correct answer is B.

35 答案 D

令 \(JH=HG=x\)。则 \(OG=a-2x\),且 \(OH=a-x\)。两条半弦为 \[ \begin{aligned} GD&=2\sqrt{x(a-x)},\\ HF&=\sqrt{x(2a-x)}\text{。} \end{aligned} \] 由于两个图形的高均为 \(x\),\[ \begin{aligned} \frac KR&=\frac{x(GD+HF)}{2x(HF)}\\ &=\frac12+\frac{GD}{2HF}\text{。} \end{aligned} \] 当 \(x\) 趋近于 \(0\) 时,后一个分数趋近于 \[ \frac{2\sqrt{ax}}{2\sqrt{2ax}}=\frac1{\sqrt2}\text{。} \] 因此 \(K:R\) 趋近于 \(\frac{1}{\sqrt2}+\frac{1}{2}\)。

因此,正确答案是 D。

English

Let \(JH=HG=x.\) Then \(OG=a-2x\) and \(OH=a-x.\) The half-chords are \[ \begin{aligned} GD&=2\sqrt{x(a-x)},\\ HF&=\sqrt{x(2a-x)}. \end{aligned} \] Since both figures have height \(x,\) \[ \begin{aligned} \frac KR&=\frac{x(GD+HF)}{2x(HF)}\\ &=\frac12+\frac{GD}{2HF}. \end{aligned} \] As \(x\) approaches \(0,\) the latter fraction approaches \[ \frac{2\sqrt{ax}}{2\sqrt{2ax}}=\frac1{\sqrt2}. \] Thus \(K:R\) approaches \(\frac{1}{\sqrt2}+\frac{1}{2}.\)

Therefore, the correct answer is D.

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