1959 AMC 12 真题（P版）

1 答案 B

若原棱长为 \(s\)，则新棱长为 \(1.5s\)。因此表面积变为原来的 \[ (1.5)^2=2.25 \] 倍。增加量是原表面积的 \(1.25\) 倍，即 \(125\%\)。

因此，正确答案是 B。

English

If the original edge length is \(s,\) the new edge length is \(1.5s.\) Thus the surface area is multiplied by \[ (1.5)^2=2.25. \] The increase is \(1.25\) times the original area, or \(125\%.\)

Thus, the correct answer is B.

2 答案 D

设所求距离为 \(x\)。平行线上方的三角形与原三角形相似，线性比为 \(1-x\)。其面积为原三角形的一半，所以 \[ (1-x)^2=\frac12\text{。}\]因为 \(0\lt x\lt1\)，所以 \(1-x=\frac{1}{\sqrt2}\)，从而 \[ x=1-\frac1{\sqrt2}=\frac{2-\sqrt2}{2}\text{。}\]

因此，正确答案是 D。

English

Let \(x\) be the requested distance. The triangle above the parallel line is similar to the original triangle, with linear ratio \(1-x.\) Its area is half the original area, so \[ (1-x)^2=\frac12. \] Since \(0\lt x\lt1,\) we have \(1-x=\frac{1}{\sqrt2},\) and therefore \[ x=1-\frac1{\sqrt2}=\frac{2-\sqrt2}{2}. \]

Therefore, the correct answer is D.

3 答案 E

筝形的对角线可以互相垂直，但它未必是菱形、矩形、正方形或等腰梯形。因此，仅有对角线垂直这一条件不能保证它属于前四类中的任何一类。

因此，正确答案是 E。

English

A kite can have perpendicular diagonals without being a rhombus, rectangle, square, or isosceles trapezoid. Therefore perpendicular diagonals alone do not force any of the first four classifications.

Thus, the correct answer is E.

4 答案 C

设三部分为 \(x,\frac{x}{3}\) 和 \(\frac{x}{6}\)。则 \[ x+\frac x3+\frac x6=\frac32x=78\text{，}\]所以 \(x=52\)。中间一部分为 \[ \frac{x}{3}=\frac{52}{3}=17\frac13\text{。}\]

因此，正确答案是 C。

English

Let the parts be \(x,\frac{x}{3},\) and \(\frac{x}{6}.\) Then \[ x+\frac x3+\frac x6=\frac32x=78, \] so \(x=52.\) The middle part is \[ \frac{x}{3}=\frac{52}{3}=17\frac13. \]

Thus, the correct answer is C.

5 答案 A

利用同底数幂的乘法法则，\[ \begin{aligned} 256^{0.16}\cdot256^{0.09} &=256^{0.25}\\ &=256^{\frac{1}{4}}=4 \end{aligned}\text{。}\]

因此，正确答案是 A。

English

Using the product rule for powers, \[ \begin{aligned} 256^{0.16}\cdot256^{0.09} &=256^{0.25}\\ &=256^{\frac{1}{4}}=4. \end{aligned} \]

Therefore, the correct answer is A.

6 答案 D

逆命题声称每个矩形都是正方形，这是假的。否命题声称每个不是正方形的四边形都不是矩形，这也是假的，因为非正方形的矩形就是一个反例。

因此两个命题都不为真，正确答案是 D。

English

The converse says that every rectangle is a square, which is false. The inverse says that every quadrilateral that is not a square is not a rectangle, which is also false because a nonsquare rectangle is a counterexample.

Thus neither statement is true, and the correct answer is D.

7 答案 D

由勾股定理得 \[ a^2+(a+d)^2=(a+2d)^2\text{。}\]化简得 \[ \begin{aligned} a^2-2ad-3d^2 &=0,\\ (a-3d)(a+d) &=0 \end{aligned}\text{。}\]因为二者为正，所以排除 \(a=-d\)，故 \(a=3d\)。

因此比为 \(3:1\)，正确答案是 D。

English

The Pythagorean theorem gives \[ a^2+(a+d)^2=(a+2d)^2. \] Simplifying, \[ \begin{aligned} a^2-2ad-3d^2 &=0,\\ (a-3d)(a+d) &=0. \end{aligned} \] Positivity excludes \(a=-d,\) so \(a=3d.\)

Thus the ratio is \(3:1,\) and the correct answer is D.

8 答案 A

配方得 \[ x^2-6x+13=(x-3)^2+4\text{。}\]此式至少为 \(4\)，并在 \(x=3\) 时取等号。

因此，正确答案是 A。

English

Completing the square, \[ x^2-6x+13=(x-3)^2+4. \] This expression is at least \(4,\) with equality at \(x=3.\)

Thus, the correct answer is A.

9 答案 C

前三个儿子共得到牛群的 \[ \frac12+\frac14+\frac15=\frac{19}{20}\text{。}\]因此剩余的 \(\frac{1}{20}\) 是 \(7\) 头牛，所以 \(n=140\)。

因此，正确答案是 C。

English

The first three sons receive \[ \frac12+\frac14+\frac15=\frac{19}{20} \] of the herd. Thus the remaining \(\frac{1}{20}\) of the herd is \(7,\) so \(n=140.\)

Therefore, the correct answer is C.

10 答案 D

三角形 \(AED\) 和 \(ABC\) 的底边位于同一直线 \(AC\) 上。由于 \[ \frac{AD}{AB}=\frac{1.2}{3.6}=\frac13\text{，}\]点 \(D\) 到直线 \(AC\) 的垂直距离是点 \(B\) 到该直线相应距离的三分之一。要使面积相等，必须有 \[ AE\left(\frac h3\right)=AC(h)\text{，}\]因此 \(AE=3AC=10.8\)。

因此，正确答案是 D。

English

Triangles \(AED\) and \(ABC\) use bases on the same line \(AC.\) Since \[ \frac{AD}{AB}=\frac{1.2}{3.6}=\frac13, \] the perpendicular distance from \(D\) to line \(AC\) is one-third the corresponding distance from \(B.\) Equality of areas therefore requires \[ AE\left(\frac h3\right)=AC(h), \] so \(AE=3AC=10.8.\)

Thus, the correct answer is D.

11 答案 D

因为 \[ 0.0625=\frac1{16}=2^{-4}\text{，}\]所以 \(\log_2(0.0625)=-4\)。

因此，正确答案是 D。

English

Since \[ 0.0625=\frac1{16}=2^{-4}, \] we have \(\log_2(0.0625)=-4.\)

Therefore, the correct answer is D.

12 答案 A

等比数列的条件为 \[ (50+c)^2=(20+c)(100+c)\text{。}\]展开并消去 \(c^2\) 得 \(500=20c\)，所以 \(c=25\)。公比为 \[ \frac{50+25}{20+25}=\frac{75}{45}=\frac53\text{。}\]

因此，正确答案是 A。

English

The geometric-progression condition is \[ (50+c)^2=(20+c)(100+c). \] Expanding and cancelling \(c^2\) gives \(500=20c,\) so \(c=25.\) The common ratio is \[ \frac{50+25}{20+25}=\frac{75}{45}=\frac53. \]

Thus, the correct answer is A.

13 答案 D

原来的总和为 \(50\cdot38=1900\)。去掉 \(45\) 和 \(55\) 后，剩余总和为 \(1800\)，所以新的平均数为 \[ \frac{1800}{48}=37.5\text{。}\]

因此，正确答案是 D。

English

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

Therefore, the correct answer is D.

14 答案 D

偶数在加法、减法和乘法下封闭。除法不成立，因为 \(\frac{2}{4}=\frac{1}{2}\notin S\)。求平均数也不成立，因为 \(0\) 与 \(2\) 的平均数是 \(1\notin S\)。

因此只有运算 \(1\)、\(2\) 和 \(3\) 总是成立，正确答案是 D。

English

Even integers are closed under addition, subtraction, and multiplication. Division fails, since \(\frac{2}{4}=\frac{1}{2}\notin S.\) Averaging fails, since the mean of \(0\) and \(2\) is \(1\notin S.\)

Thus only operations \(1,\) \(2,\) and \(3\) always work, and the correct answer is D.

15 答案 C

若两条直角边为 \(a,b\)，斜边为 \(c\)，则 \[ a^2+b^2=c^2=2ab\text{。}\]因此 \((a-b)^2=0\)，所以两条直角边相等。这个直角三角形是等腰三角形，其锐角均为 \(45^\circ\)。

因此，正确答案是 C。

English

If the legs are \(a,b\) and the hypotenuse is \(c,\) then \[ a^2+b^2=c^2=2ab. \] Hence \((a-b)^2=0,\) so the legs are equal. The right triangle is isosceles and its acute angles are \(45^\circ.\)

Thus, the correct answer is C.

16 答案 D

因式分解并乘以倒数得 \[ \begin{aligned} &\frac{(x-1)(x-2)}{(x-2)(x-3)}\\ &\quad{}\cdot \frac{(x-3)(x-4)}{(x-1)(x-4)}=1 \end{aligned} \] 对原表达式定义域内的每个值都成立。

因此，正确答案是 D。

English

Factoring and multiplying by the reciprocal gives \[ \begin{aligned} &\frac{(x-1)(x-2)}{(x-2)(x-3)}\\ &\quad{}\cdot \frac{(x-3)(x-4)}{(x-1)(x-4)}=1 \end{aligned} \] for every value in the domain of the original expression.

Therefore, the correct answer is D.

17 答案 E

两个条件给出 \[ a-b=1,\qquad a-\frac b5=5\text{。}\]两式相减得 \(\frac{4b}{5}=4\)，所以 \(b=5\) 且 \(a=6\)。因此 \(a+b=11\)。

因此，正确答案是 E。

English

The two conditions give \[ a-b=1,\qquad a-\frac b5=5. \] Subtracting yields \(\frac{4b}{5}=4,\) so \(b=5\) and \(a=6.\) Therefore \(a+b=11.\)

Thus, the correct answer is E.

18 答案 E

前 \(n\) 个正整数之和为 \(\frac{n(n+1)}{2}\)。除以 \(n\) 得平均数 \[ \frac{n+1}{2}\text{。}\]

因此，正确答案是 E。

English

The sum of the first \(n\) positive integers is \(\frac{n(n+1)}{2}.\) Dividing by \(n\) gives the mean \[ \frac{n+1}{2}. \]

Therefore, the correct answer is E.

19 答案 B

对砝码 \(1,3,9\) 使用系数 \(-1,0,1\)，平衡三进制可以表示从 \(1\) 到 \[ 1+3+9=13 \] 的每个整数重量。因此可以称量 \(13\) 种不同的正重量。

因此，正确答案是 B。

English

Using coefficients \(-1,0,1\) for the weights \(1,3,9,\) balanced ternary represents every integer weight from \(1\) through \[ 1+3+9=13. \] Thus there are \(13\) different positive object weights that can be measured.

Therefore, the correct answer is B.

20 答案 B

因为 \(x=\frac{ky}{z^2}\)，所以 \[ \frac{x_{\rm new}}{10} =\frac{16}{4}\left(\frac{14}{7}\right)^2 =4\cdot4=16\text{。}\]因此 \(x_{\rm new}=160\)。

因此，正确答案是 B。

English

Because \(x=\frac{ky}{z^2},\) \[ \frac{x_{\rm new}}{10} =\frac{16}{4}\left(\frac{14}{7}\right)^2 =4\cdot4=16. \] Hence \(x_{\rm new}=160.\)

Thus, the correct answer is B.

21 答案 C

边长为 \(\frac{p}{3}\)，所以外接圆半径为 \[ R=\frac{\frac{p}{3}}{\sqrt3}=\frac{p}{3\sqrt3}\text{。}\]因此圆的面积为 \[ \pi R^2=\frac{\pi p^2}{27}\text{。}\]

因此，正确答案是 C。

English

The side length is \(\frac{p}{3},\) so the circumradius is \[ R=\frac{\frac{p}{3}}{\sqrt3}=\frac{p}{3\sqrt3}. \] Therefore the circle’s area is \[ \pi R^2=\frac{\pi p^2}{27}. \]

Thus, the correct answer is C.

22 答案 C

若较短底边为 \(b\)，则对角线中点连线的长度为两底边之差的一半：\[ \frac{97-b}{2}=3\text{。}\]因此 \(97-b=6\)，且 \(b=91\)。

因此，正确答案是 C。

English

If the shorter base is \(b,\) the diagonal-midpoint segment has length half the difference of the bases: \[ \frac{97-b}{2}=3. \] Thus \(97-b=6\) and \(b=91.\)

Therefore, the correct answer is C.

23 答案 A

该方程等价于 \[ a^2-15a=100\text{，}\]即 \((a-20)(a+5)=0\)。因此 \(a=20\) 或 \(a=-5\)。两种情况下对数的真数都是 \(100\)，所以两个解都是有效的整数。

因此，正确答案是 A。

English

The equation is equivalent to \[ a^2-15a=100, \] or \((a-20)(a+5)=0.\) Thus \(a=20\) or \(a=-5.\) In both cases the logarithm’s argument is \(100,\) so both solutions are valid integers.

Therefore, the correct answer is A.

24 答案 C

设加入 \(x\) 盎司盐。浓度方程为 \[ \frac{\frac{m^2}{100}+x}{m+x}=\frac{2m}{100}\text{。}\]两边同乘分母并合并含 \(x\) 的项，得 \[ x(100-2m)=m^2\text{，}\]所以 \(x=\frac{m^2}{100-2m}\)。

因此，正确答案是 C。

English

Let \(x\) ounces of salt be added. The concentration equation is \[ \frac{\frac{m^2}{100}+x}{m+x}=\frac{2m}{100}. \] Multiplying through and collecting the \(x\)-terms gives \[ x(100-2m)=m^2, \] so \(x=\frac{m^2}{100-2m}.\)

Therefore, the correct answer is C.

25 答案 D

该绝对值不等式等价于 \[ -4\lt3-x\lt4\text{。}\]减去 \(3\) 后再乘以 \(-1\)，不等号方向反转，得到 \[ -1\lt x\lt7\text{。}\]

因此，正确答案是 D。

English

The absolute-value inequality is equivalent to \[ -4\lt3-x\lt4. \] Subtracting \(3\) and multiplying by \(-1\) reverses the inequalities, giving \[ -1\lt x\lt7. \]

Thus, the correct answer is D.

26 答案 A

将底边两端点置于 \(A=(-\frac{\sqrt2}{2},0)\) 和 \(B=(\frac{\sqrt2}{2},0)\)，并设第三个顶点为 \(C=(0,h)\)。从 \(A\) 出发的中线方向为 \[ \left(\frac{3\sqrt2}{4},\frac h2\right)\text{，}\]从 \(B\) 出发的中线方向为 \[ \left(-\frac{3\sqrt2}{4},\frac h2\right)\text{。}\]两者点积为零，所以 \[ -\frac98+\frac{h^2}{4}=0\text{，}\]且 \(h=\frac{3\sqrt2}{2}\)。因此面积为 \[ \frac12\cdot\sqrt2\cdot\frac{3\sqrt2}{2}=\frac32\text{。}\]

因此，正确答案是 A。

English

Put the base endpoints at \(A=(-\frac{\sqrt2}{2},0)\) and \(B=(\frac{\sqrt2}{2},0),\) and let the third vertex be \(C=(0,h).\) The median from \(A\) has direction \[ \left(\frac{3\sqrt2}{4},\frac h2\right), \] while the median from \(B\) has direction \[ \left(-\frac{3\sqrt2}{4},\frac h2\right). \] Their dot product is zero, so \[ -\frac98+\frac{h^2}{4}=0, \] and \(h=\frac{3\sqrt2}{2}.\) Thus the area is \[ \frac12\cdot\sqrt2\cdot\frac{3\sqrt2}{2}=\frac32. \]

Therefore, the correct answer is A.

27 答案 A

由韦达定理，两根之和为 \[ -\frac{-1}{i}=\frac1i=-i\text{，}\]而不是 \(2\)。判别式为 \[ (-1)^2-4(i)(2i)=1-8i^2=9\text{。}\]求根公式给出虚根 \(i\) 和 \(-2i\)，所以其余陈述都正确。

因此，正确答案是 A。

English

By Vieta’s formula, the sum of the roots is \[ -\frac{-1}{i}=\frac1i=-i, \] not \(2.\) Also the discriminant is \[ (-1)^2-4(i)(2i)=1-8i^2=9. \] The quadratic formula gives the imaginary roots \(i\) and \(-2i,\) so the remaining statements are true.

Thus, the correct answer is A.

28 答案 E

由角平分线定理，\[ \begin{aligned} \frac{AM}{MB}&=\frac{AC}{CB}=\frac ba,\\ \frac{CL}{LB}&=\frac{AC}{AB}=\frac bc \end{aligned}\text{。}\]因此 \[ k=\frac{\frac{b}{a}}{\frac{b}{c}}=\frac ca\text{。}\]

因此，正确答案是 E。

English

By the angle bisector theorem, \[ \begin{aligned} \frac{AM}{MB}&=\frac{AC}{CB}=\frac ba,\\ \frac{CL}{LB}&=\frac{AC}{AB}=\frac bc. \end{aligned} \] Therefore \[ k=\frac{\frac{b}{a}}{\frac{b}{c}}=\frac ca. \]

Thus, the correct answer is E.

29 答案 D

得分条件给出 \[ 15+\frac{n-20}{3}=\frac n2\text{。}\]两边乘以 \(6\) 得 \(90+2n-40=3n\)，所以 \(n=50\)。此值也使其余题目中答对的题数为整数。因此 \(n\) 恰有一个可能值。

因此，正确答案是 D。

English

The score condition gives \[ 15+\frac{n-20}{3}=\frac n2. \] Multiplying by \(6\) yields \(90+2n-40=3n,\) so \(n=50.\) This value also makes the number of remaining correct answers an integer. Hence there is exactly one possible value of \(n.\)

Therefore, the correct answer is D.

30 答案 B

若 \(B\) 跑一圈的时间为 \(t\)，则两人的相对速度为 \[ \frac1{40}+\frac1t=\frac1{15}\text{。}\]因此 \(\frac{1}{t}=\frac{1}{24}\)，所以 \(t=24\) 秒。

因此，正确答案是 B。

English

If \(B\)’s lap time is \(t,\) their relative speed is \[ \frac1{40}+\frac1t=\frac1{15}. \] Thus \(\frac{1}{t}=\frac{1}{24},\) so \(t=24\) seconds.

Therefore, the correct answer is B.

31 答案 B

设半圆半径为 \(r\)，正方形边长为 \(s\)，其中 \(s^2=40\)。正方形一个上顶点相对圆心的水平距离为 \(\frac{s}{2}\)，竖直距离为 \(s\)，所以 \[ r^2=s^2+\left(\frac s2\right)^2=\frac54s^2=50\text{。}\]内接于整圆的正方形对角线长为 \(2r\)，故面积为 \(2r^2=100\)。

因此，正确答案是 B。

English

Let the semicircle have radius \(r\) and the square have side \(s,\) where \(s^2=40.\) A top vertex of the square is \(\frac{s}{2}\) horizontally and \(s\) vertically from the center, so \[ r^2=s^2+\left(\frac s2\right)^2=\frac54s^2=50. \] A square inscribed in the full circle has diagonal \(2r,\) hence area \(2r^2=100.\)

Thus, the correct answer is B.

32 答案 C

若 \(O\) 为圆心，\(T\) 为切点，则 \(OT\perp AT\)。因此 \[ \begin{aligned} AO &=\sqrt{r^2+l^2}\\ &=\sqrt{r^2+\frac{16r^2}{9}}\\ &=\frac{5r}{3} \end{aligned}\text{。}\]点 \(A\) 到圆的最短距离为 \(AO-r=\frac{2r}{3}\)。由于 \(l=\frac{4r}{3}\)，此距离等于 \(\frac{l}{2}\)。

因此，正确答案是 C。

English

If \(O\) is the center and \(T\) the point of tangency, then \(OT\perp AT.\) Hence \[ \begin{aligned} AO &=\sqrt{r^2+l^2}\\ &=\sqrt{r^2+\frac{16r^2}{9}}\\ &=\frac{5r}{3}. \end{aligned} \] The shortest distance from \(A\) to the circle is \(AO-r=\frac{2r}{3}.\) Since \(l=\frac{4r}{3},\) this distance equals \(\frac{l}{2}.\)

Therefore, the correct answer is C.

33 答案 B

各项的倒数开始为 \[ \frac13,\quad\frac14,\quad\frac16\text{，}\]公差为 \(-\frac{1}{12}\)。下一个倒数是 \(\frac{1}{12}\)，所以第四项为 \(12\)。因此 \[ S_4=3+4+6+12=25\text{。}\]

因此，正确答案是 B。

English

The reciprocals begin \[ \frac13,\quad\frac14,\quad\frac16, \] with common difference \(-\frac{1}{12}.\) The next reciprocal is \(\frac{1}{12},\) so the fourth term is \(12.\) Therefore \[ S_4=3+4+6+12=25. \]

Thus, the correct answer is B.

34 答案 A

由韦达定理，\(r+s=3\) 且 \(rs=1\)。因此 \[ \begin{aligned} r^2+s^2 &=(r+s)^2-2rs\\ &=9-2=7 \end{aligned}\text{，}\]它是正整数。

因此，正确答案是 A。

English

Vieta’s formulas give \(r+s=3\) and \(rs=1.\) Therefore \[ \begin{aligned} r^2+s^2 &=(r+s)^2-2rs\\ &=9-2=7, \end{aligned} \] which is a positive integer.

Thus, the correct answer is A.

35 答案 E

将左边因式分解得 \[ \begin{aligned} &(n-m)(2x-m-n)\\ &\qquad=(m-n)^2 \end{aligned}\text{。}\]因为 \(m\ne n\)，除以 \(n-m\) 得 \(2x-m-n=n-m\)，所以 \(x=n\)。这个唯一的负值不属于选项 A 至 D 中的任何一种情况。

因此，正确答案是 E。

English

Factoring the left side gives \[ \begin{aligned} &(n-m)(2x-m-n)\\ &\qquad=(m-n)^2. \end{aligned} \] Since \(m\ne n,\) division by \(n-m\) yields \(2x-m-n=n-m,\) so \(x=n.\) This single negative value is not any of choices A through D.

Therefore, the correct answer is E.

36 答案 D

设与 \(60^\circ\) 底角相邻的边长为 \(x\)，对边长为 \(90-x\)。由余弦定理 \[ \begin{aligned} (90-x)^2 &=80^2+x^2\\ &\quad{}-2(80)(x)\cos60^\circ \end{aligned}\text{。}\]化简得 \(8100-180x=6400-80x\)，所以 \(x=17\)。三边长为 \(17,\ 73\) 和 \(80\)，最短边为 \(17\)。

因此，正确答案是 D。

English

Let the side adjacent to the \(60^\circ\) base angle be \(x,\) and let the opposite side be \(90-x.\) The law of cosines gives \[ \begin{aligned} (90-x)^2 &=80^2+x^2\\ &\quad{}-2(80)(x)\cos60^\circ. \end{aligned} \] Simplifying yields \(8100-180x=6400-80x,\) so \(x=17.\) The side lengths are \(17,\ 73,\) and \(80,\) and the shortest is \(17.\)

Thus, the correct answer is D.

37 答案 B

该乘积逐项消去：\[ \frac23\cdot\frac34\cdot\frac45\cdots\frac{n-1}{n} =\frac2n\text{。}\]

因此，正确答案是 B。

English

The product telescopes: \[ \frac23\cdot\frac34\cdot\frac45\cdots\frac{n-1}{n} =\frac2n. \]

Therefore, the correct answer is B.

38 答案 B

令 \(u=\sqrt{2x}\ge0\)。由于 \(x=\frac{u^2}{2}\)，方程变为 \[ 2u^2+u=1\text{，}\]即 \((2u-1)(u+1)=0\)。因此 \(u=\frac{1}{2}\) 且 \(x=\frac{u^2}{2}=\frac{1}{8}\)。另一个二次方程根因 \(u\ge0\) 而不合条件。故 \(x\) 是分数。

因此，正确答案是 B。

English

Let \(u=\sqrt{2x}\ge0.\) Since \(x=\frac{u^2}{2},\) the equation becomes \[ 2u^2+u=1, \] or \((2u-1)(u+1)=0.\) Thus \(u=\frac{1}{2}\) and \(x=\frac{u^2}{2}=\frac{1}{8}.\) The other quadratic root is inadmissible because \(u\ge0.\) Hence \(x\) is fractional.

Therefore, the correct answer is B.

39 答案 D

将前九项相加，得 \[ \begin{aligned} S={}&(x+x^2+\cdots+x^9)\\ &{}+(1+2+\cdots+9)a \end{aligned}\text{。}\]因此 \[ S=\frac{x^{10}-x}{x-1}+45a\text{。}\]当 \(x=1\) 时按连续性理解此表达式，此时两种形式都等于 \(9+45a\)。

因此，正确答案是 D。

English

Adding the first nine terms gives \[ \begin{aligned} S={}&(x+x^2+\cdots+x^9)\\ &{}+(1+2+\cdots+9)a. \end{aligned} \] Therefore \[ S=\frac{x^{10}-x}{x-1}+45a. \] The expression is understood by continuity at \(x=1,\) where both forms equal \(9+45a.\)

Thus, the correct answer is D.

40 答案 C

使用向量，令 \(A=\mathbf0,\ B=\mathbf b\)，且 \(C=\mathbf c\)。因为 \(D=\frac{\mathbf c}{2}\)，且 \(E\) 是 \(BD\) 的中点，所以 \[ E=\frac12\mathbf b+\frac14\mathbf c\text{。}\]\(AB\) 上一点 \(F\) 可写成 \(F=t\mathbf b\)。由于 \(E\) 位于 \(CF\) 上，比较 \(\mathbf c\) 的系数可得 \(E=\tfrac14C+\tfrac34F\)。于是 \[ \frac34t=\frac12\text{，}\]所以 \(t=\frac{2}{3}\)。因此 \(\frac{AF}{AB}=\frac{2}{3}\)，且 \(\frac{BF}{BA}=\frac{1}{3}\)。由于 \(BF=5\)，所以 \(BA=15\)。

因此，正确答案是 C。

English

Use vectors with \(A=\mathbf0,\ B=\mathbf b,\) and \(C=\mathbf c.\) Since \(D=\frac{\mathbf c}{2}\) and \(E\) is the midpoint of \(BD,\) \[ E=\frac12\mathbf b+\frac14\mathbf c. \] A point \(F\) on \(AB\) has the form \(F=t\mathbf b.\) Since \(E\) lies on \(CF,\) comparison of the \(\mathbf c\)-coefficient shows that \(E=\tfrac14C+\tfrac34F.\) Hence \[ \frac34t=\frac12, \] so \(t=\frac{2}{3}.\) Thus \(\frac{AF}{AB}=\frac{2}{3},\) and \(\frac{BF}{BA}=\frac{1}{3}.\) Since \(BF=5,\) \(BA=15.\)

Therefore, the correct answer is C.

41 答案 D

设两个全等圆的半径为 \(R\)。它们的圆心相距 \(2R\)，所以半径为 \(4\) 的圆的圆心位于两者正中间。小圆圆心与任一大圆圆心的水平距离和竖直距离分别为 \(R\) 和 \(R-4\)。由相切条件得 \[ R^2+(R-4)^2=(R+4)^2\text{。}\]化简得 \(R^2-16R=0\)，由正性知 \(R=16\)。

因此，正确答案是 D。

English

Let the equal circles have radius \(R.\) Their centers are \(2R\) apart, so the center of the radius-\(4\) circle lies midway between them. The horizontal and vertical separations between its center and either large center are \(R\) and \(R-4.\) Tangency gives \[ R^2+(R-4)^2=(R+4)^2. \] Simplifying yields \(R^2-16R=0,\) and positivity gives \(R=16.\)

Thus, the correct answer is D.

42 答案 E

对任一质数，设它在 \(a,b,c\) 中的指数为 \(u\le v\le w\)。它在 \(MD\) 中的指数为 \(u+w\)，而在 \(abc\) 中的指数为 \(u+v+w\)。因此 \(MD\le abc\)，证明了陈述 \((2)\)。对每个质数都恰有 \(v=0\) 时等号成立，这意味着没有任何质数同时整除 \(a,b,c\) 中的两个数。这正是两两互质，证明了陈述 \((4)\)。

因此，正确答案是 E。

English

For any prime, let its exponents in \(a,b,c\) be \(u\le v\le w.\) Its exponent in \(MD\) is \(u+w,\) while its exponent in \(abc\) is \(u+v+w.\) Thus \(MD\le abc,\) proving statement \((2).\) Equality holds exactly when \(v=0\) for every prime, meaning no prime divides two of \(a,b,c.\) That is precisely pairwise relative primality, proving statement \((4).\)

Therefore, the correct answer is E.

43 答案 B

半周长为 \(52\)，所以由海伦公式 \[ K=\sqrt{52\cdot27\cdot13\cdot12}=468\text{。}\]若 \(R\) 为外接圆半径，则 \[ R=\frac{25\cdot39\cdot40}{4\cdot468}=\frac{125}{6}\text{。}\]因此直径为 \(2R=\frac{125}{3}\)。

因此，正确答案是 B。

English

The semiperimeter is \(52,\) so Heron’s formula gives \[ K=\sqrt{52\cdot27\cdot13\cdot12}=468. \] If \(R\) is the circumradius, then \[ R=\frac{25\cdot39\cdot40}{4\cdot468}=\frac{125}{6}. \] Hence the diameter is \(2R=\frac{125}{3}.\)

Thus, the correct answer is B.

44 答案 C

若两根为 \(u,v>1\)，则 \(b=-(u+v)\) 且 \(c=uv\)。因此 \[ \begin{aligned} s&=1-u-v+uv\\ &=(u-1)(v-1)>0 \end{aligned}\text{。}\]

因此 \(s\) 必大于零，正确答案是 C。

English

If the roots are \(u,v>1,\) then \(b=-(u+v)\) and \(c=uv.\) Therefore \[ \begin{aligned} s&=1-u-v+uv\\ &=(u-1)(v-1)>0. \end{aligned} \]

Thus \(s\) must be greater than zero, and the correct answer is C.

45 答案 B

这些对数逐项消去：\[ \begin{aligned} &(\log_3x)(\log_x2x)(\log_{2x}y)\\ &\qquad=\log_3y \end{aligned}\text{。}\]又因为 \(\log_xx^2=2\)，所以 \(\log_3y=2\)，且 \(y=3^2=9\)。

因此，正确答案是 B。

English

The logarithms telescope: \[ \begin{aligned} &(\log_3x)(\log_x2x)(\log_{2x}y)\\ &\qquad=\log_3y. \end{aligned} \] Also \(\log_xx^2=2,\) so \(\log_3y=2\) and \(y=3^2=9.\)

Therefore, the correct answer is B.

46 答案 B

下雨的上午有 \(d-6\) 个，下雨的下午有 \(d-5\) 个。由条件 \((2)\)，可知这些下雨时段互不重合，总数为 \(7\)。因此 \[ (d-6)+(d-5)=7\text{，}\]所以 \(2d=18\)，且 \(d=9\)。

因此，正确答案是 B。

English

There are \(d-6\) rainy mornings and \(d-5\) rainy afternoons. By condition \((2),\) these are distinct rainy occasions, and their total is \(7.\) Hence \[ (d-6)+(d-5)=7, \] so \(2d=18\) and \(d=9.\)

Thus, the correct answer is B.

47 答案 A

陈述 III 保证至少存在一名会思考的学生。由陈述 II，该学生是人，所以有些人会思考，从而推出陈述 \((2)\)。题设没有说明新生与学生的关系，也没有说明新生是否会思考，更不能保证存在不属于学生的、会思考的人。因此 \((1)\)、\((3)\) 或 \((4)\) 都不能推出。

因此，正确答案是 A。

English

Statement III supplies at least one student who thinks. By statement II, that student is human, so some human thinks and statement \((2)\) follows. Nothing relates freshmen to students, says whether freshmen think, or guarantees a thinking human outside the students. Thus none of \((1),\) \((3),\) or \((4)\) follows.

Therefore, the correct answer is A.

48 答案 B

按次数分类计数。若 \(n=0\)，则 \(a_0=3\)，得到一个多项式。若 \(n=1\)，则 \[ a_0+|a_1|=2\text{。}\]此时 \(a_0=2,a_1=0\) 或 \(a_0=1,a_1=\pm1\)，共有三个多项式。若 \(n=2\)，则 \(a_0=1\) 且 \(a_1=a_2=0\)，再得一个。次数不可能更高。总数为 \(1+3+1=5\)。

因此，正确答案是 B。

English

We count by degree. If \(n=0,\) then \(a_0=3,\) giving one polynomial. If \(n=1,\) then \[ a_0+|a_1|=2. \] This gives \(a_0=2,a_1=0\) or \(a_0=1,a_1=\pm1,\) for three polynomials. If \(n=2,\) then \(a_0=1\) and \(a_1=a_2=0,\) giving one more. No higher degree is possible. The total is \(1+3+1=5.\)

Thus, the correct answer is B.

49 答案 B

将级数分组为 \[ \begin{aligned} S={}&\left(1-\frac12-\frac14\right)\\ &+\left(\frac18-\frac1{16}-\frac1{32}\right) +\cdots \end{aligned}\text{。}\]第一组为 \(\frac{1}{4}\)，各组构成公比为 \(\frac{1}{8}\) 的等比级数。因此 \[ S=\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac27\text{。}\]

因此，正确答案是 B。

English

Group the series as \[ \begin{aligned} S={}&\left(1-\frac12-\frac14\right)\\ &+\left(\frac18-\frac1{16}-\frac1{32}\right) +\cdots. \end{aligned} \] The first block is \(\frac{1}{4},\) and successive blocks form a geometric series with ratio \(\frac{1}{8}.\) Hence \[ S=\frac{\frac{1}{4}}{1-\frac{1}{8}}=\frac27. \]

Therefore, the correct answer is B.

50 答案 D

每名成员恰好对应一对无序的委员会。反之，每对委员会恰有一名共同成员。因此成员与四个委员会的两两组合一一对应，所以 \[ x=\binom42=6\text{。}\]这是 \(4\) 与 \(8\) 之间的唯一值。

因此，正确答案是 D。

English

Each member belongs to exactly one unordered pair of committees. Conversely, each pair of committees has exactly one common member. Thus the members are in one-to-one correspondence with the pairs of four committees, so \[ x=\binom42=6. \] This is a single value between \(4\) and \(8.\)

Therefore, the correct answer is D.