1962 AMC 12 真题（P版）

1 答案 D

分子为 \(1\)，而 \[ 5^{-1}+3^{-1}=\frac15+\frac13=\frac8{15}\text{。}\]因此该表达式等于 \(\frac{1}{\frac{8}{15}}=\frac{15}{8}\)。

所以正确答案是 D。

English

The numerator is \(1,\) while \[ 5^{-1}+3^{-1}=\frac15+\frac13=\frac8{15}. \] Therefore the expression equals \(\frac{1}{\frac{8}{15}}=\frac{15}{8}.\)

Thus, the correct answer is D.

2 答案 A

有 \[ \frac2{\sqrt3}-\frac{\sqrt3}{2} =\frac{4-3}{2\sqrt3} =\frac{\sqrt3}{6}\text{。}\]

所以正确答案是 A。

English

We have \[ \frac2{\sqrt3}-\frac{\sqrt3}{2} =\frac{4-3}{2\sqrt3} =\frac{\sqrt3}{6}. \]

Therefore, the correct answer is A.

3 答案 B

由相邻项之差相等可得 \[ 2=(2x+3)-(x+1)=x+2\text{，}\]所以 \(x=0\)。

所以正确答案是 B。

English

Equality of consecutive differences gives \[ 2=(2x+3)-(x+1)=x+2, \] so \(x=0.\)

Thus, the correct answer is B.

4 答案 B

因为 \(8=2^3\) 且 \(32=2^5\)，原方程化为 \(2^{3x}=2^5\)。因此 \(3x=5\)，且 \(x=\frac{5}{3}\)。

所以正确答案是 B。

English

Since \(8=2^3\) and \(32=2^5,\) the equation becomes \(2^{3x}=2^5.\) Hence \(3x=5\) and \(x=\frac{5}{3}.\)

Therefore, the correct answer is B.

5 答案 C

对于任意圆，无论半径多大，周长除以直径都等于 \[ \frac{2\pi R}{2R}=\pi\text{。}\]

所以正确答案是 C。

English

For every circle, regardless of its radius, the circumference divided by the diameter is \[ \frac{2\pi R}{2R}=\pi. \]

Thus, the correct answer is C.

6 答案 D

若三角形的边长为 \(s\)，则 \[ \frac{\sqrt3}{4}s^2=9\sqrt3\text{，}\]所以 \(s=6\)。其周长为 \(18\)，故正方形边长为 \(\frac{18}{4}=\frac{9}{2}\)。因此正方形的对角线长为 \(\frac{9\sqrt2}{2}\)。

所以正确答案是 D。

English

If \(s\) is the triangle’s side, then \[ \frac{\sqrt3}{4}s^2=9\sqrt3, \] so \(s=6.\) Its perimeter is \(18,\) making the square’s side \(\frac{18}{4}=\frac{9}{2}.\) The square’s diagonal is therefore \(\frac{9\sqrt2}{2}.\)

Thus, the correct answer is D.

7 答案 C

三角形 \(BDC\) 在 \(B\) 和 \(C\) 处的角分别为 \(90^\circ-\frac{B}{2}\) 和 \(90^\circ-\frac{C}{2}\)。因此 \[ \begin{aligned} \angle BDC &=180^\circ-\left(90^\circ-\frac B2\right) \\ &\quad-\left(90^\circ-\frac C2\right)\\ &=\frac{B+C}{2} \\ &=\frac{180^\circ-A}{2} \end{aligned}\text{。}\]

所以正确答案是 C。

English

The angles of triangle \(BDC\) at \(B\) and \(C\) are \(90^\circ-\frac{B}{2}\) and \(90^\circ-\frac{C}{2}.\) Thus \[ \begin{aligned} \angle BDC &=180^\circ-\left(90^\circ-\frac B2\right) \\ &\quad-\left(90^\circ-\frac C2\right)\\ &=\frac{B+C}{2} \\ &=\frac{180^\circ-A}{2}. \end{aligned} \]

Therefore, the correct answer is C.

8 答案 D

这些数的总和为 \[ (n-1)+\left(1-\frac1n\right)=n-\frac1n\text{。}\]再除以 \(n\)，得到 \(1-\frac{1}{n^2}\)。

所以正确答案是 D。

English

The sum is \[ (n-1)+\left(1-\frac1n\right)=n-\frac1n. \] Dividing by \(n\) gives \(1-\frac{1}{n^2}.\)

Thus, the correct answer is D.

9 答案 B

在整数范围内分解得 \[ \begin{aligned} x^9-x &=x(x-1)(x+1)\\ &\quad\cdot(x^2+1)(x^4+1) \end{aligned}\text{。}\]最后两个非常数因式在整数范围内不可约，所以共有 \(5\) 个因式。

所以正确答案是 B。

English

Factoring over the integers gives \[ \begin{aligned} x^9-x &=x(x-1)(x+1)\\ &\quad\cdot(x^2+1)(x^4+1). \end{aligned} \] The last two nonconstant factors are irreducible over the integers, so there are \(5\) factors.

Thus, the correct answer is B.

10 答案 A

去程速度为每小时 \(\frac{150}{\frac{10}{3}}=45\) 英里。全程在 \[ 3\frac13+4\frac16=7\frac12 \] 小时内行驶 \(300\) 英里，所以 \(r=\frac{300}{\frac{15}{2}}=40\)。速度之差为 \(45-40=5\)。

所以正确答案是 A。

English

The outbound rate is \(\frac{150}{\frac{10}{3}}=45\) miles per hour. The entire trip covers \(300\) miles in \[ 3\frac13+4\frac16=7\frac12 \] hours, so \(r=\frac{300}{\frac{15}{2}}=40.\) The difference is \(45-40=5.\)

Thus, the correct answer is A.

11 答案 B

判别式为 \[ p^2-4\left(\frac{p^2-1}{4}\right)=1\text{。}\]两根为 \(\frac{p+1}{2}\) 和 \(\frac{p-1}{2}\)，两者之差为 \(1\)。

所以正确答案是 B。

English

The discriminant is \[ p^2-4\left(\frac{p^2-1}{4}\right)=1. \] The roots are \(\frac{p+1}{2}\) and \(\frac{p-1}{2},\) whose difference is \(1.\)

Therefore, the correct answer is B.

12 答案 C

最后三项的系数为 \[ \binom64,\quad-\binom65,\quad\binom66\text{，}\]所以它们的和为 \(15-6+1=10\)。

所以正确答案是 C。

English

The last three terms have coefficients \[ \binom64,\quad-\binom65,\quad\binom66, \] so their sum is \(15-6+1=10.\)

Thus, the correct answer is C.

13 答案 B

因为 \(R=\frac{kS}{T}\)，所以 \(S=\frac{RT}{k}\)。于是 \[ \begin{aligned} \frac{S_2}{S_1} &=\frac{R_2T_2}{R_1T_1}\\ &=\frac{\sqrt{48}\sqrt{75}} {(\frac{4}{3})(\frac{9}{14})}\\ &=\frac{60}{\frac{6}{7}}=70 \end{aligned}\text{。}\]因此 \(S_2=(\frac{3}{7})\cdot70=30\)。

所以正确答案是 B。

English

Because \(R=\frac{kS}{T},\) we have \(S=\frac{RT}{k}.\) Hence \[ \begin{aligned} \frac{S_2}{S_1} &=\frac{R_2T_2}{R_1T_1}\\ &=\frac{\sqrt{48}\sqrt{75}} {(\frac{4}{3})(\frac{9}{14})}\\ &=\frac{60}{\frac{6}{7}}=70. \end{aligned} \] Therefore \(S_2=(\frac{3}{7})\cdot70=30.\)

Thus, the correct answer is B.

14 答案 B

首项为 \(4\)，公比为 \(-\frac{2}{3}\)。因此 \[ s=\frac4{1-(-\frac{2}{3})}=\frac{12}{5}=2.4\text{。}\]

所以正确答案是 B。

English

The first term is \(4\) and the common ratio is \(-\frac{2}{3}.\) Thus \[ s=\frac4{1-(-\frac{2}{3})}=\frac{12}{5}=2.4. \]

Therefore, the correct answer is B.

15 答案 D

也用 \(A\)、\(B\) 和 \(C\) 表示相应的位置向量。重心为 \[ G=\frac{A+B+C}{3}\text{。}\]因为 \(A\) 和 \(B\) 固定，所以这是将 \(C\) 的位置按 \(\frac{1}{3}\) 缩放后再平移。因此直线轨迹仍映射成直线轨迹。

所以正确答案是 D。

English

Let \(A,\) \(B,\) and \(C\) also denote their position vectors. The centroid is \[ G=\frac{A+B+C}{3}. \] Since \(A\) and \(B\) are fixed, this is a translation and scaling by \(\frac{1}{3}\) of the position of \(C.\) A straight-line locus therefore maps to a straight-line locus.

Thus, the correct answer is D.

16 答案 C

\(R_1\) 的对角线长为 \(\sqrt{2^2+6^2}=2\sqrt{10}\)。因此 \[ \frac{[R_2]}{[R_1]} =\left(\frac{15}{2\sqrt{10}}\right)^2 =\frac{45}{8}\text{。}\]所以 \([R_2]=12(\frac{45}{8})=\frac{135}{2}\)。

所以正确答案是 C。

English

The diagonal of \(R_1\) is \(\sqrt{2^2+6^2}=2\sqrt{10}.\) Therefore \[ \frac{[R_2]}{[R_1]} =\left(\frac{15}{2\sqrt{10}}\right)^2 =\frac{45}{8}. \] Thus \([R_2]=12(\frac{45}{8})=\frac{135}{2}.\)

Therefore, the correct answer is C.

17 答案 B

换成以 \(2\) 为底，得到 \[ \begin{aligned} a&=\frac{\log_2(15^2)}{\log_2(2^3)}\\ &=\frac{2\log_2 15}{3} =\frac{2b}{3} \end{aligned}\text{。}\]

所以正确答案是 B。

English

Changing to base \(2\) gives \[ \begin{aligned} a&=\frac{\log_2(15^2)}{\log_2(2^3)}\\ &=\frac{2\log_2 15}{3} =\frac{2b}{3}. \end{aligned} \]

Thus, the correct answer is B.

18 答案 A

这 \(12\) 个以圆心为顶点的三角形，每个面积为 \[ \frac12r^2\sin30^\circ=\frac{r^2}{4}\text{。}\]因此总面积为 \(12(\frac{r^2}{4})=3r^2\)。

所以正确答案是 A。

English

Each of the \(12\) central triangles has area \[ \frac12r^2\sin30^\circ=\frac{r^2}{4}. \] Their total area is \(12(\frac{r^2}{4})=3r^2.\)

Thus, the correct answer is A.

19 答案 C

由点 \((0,5)\) 得 \(c=5\)。另外两个点给出 \[ a-b=7,\qquad 4a+2b=-8\text{。}\]解得 \(a=1\)、\(b=-6\)。因此 \(a+b+c=1-6+5=0\)。

所以正确答案是 C。

English

The point \((0,5)\) gives \(c=5.\) The other two points give \[ a-b=7,\qquad 4a+2b=-8. \] Solving yields \(a=1\) and \(b=-6.\) Hence \(a+b+c=1-6+5=0.\)

Thus, the correct answer is C.

20 答案 A

五个角的平均数为 \(\frac{540^\circ}{5}=108^\circ\)。对于五个成等差数列的数，中间项等于平均数，所以必有一个角为 \(108^\circ\)。

所以正确答案是 A。

English

The average of the five angles is \(\frac{540^\circ}{5}=108^\circ.\) For five terms in arithmetic progression, the middle term equals the average, so one angle must be \(108^\circ.\)

Thus, the correct answer is A.

21 答案 E

另一个根为 \(3-2i\)。两根之积为 \[ (3+2i)(3-2i)=9+4=13\text{。}\]由韦达定理得 \(\frac{s}{2}=13\)，所以 \(s=26\)。

所以正确答案是 E。

English

The other root is \(3-2i.\) Their product is \[ (3+2i)(3-2i)=9+4=13. \] Vieta’s formula gives \(\frac{s}{2}=13,\) so \(s=26.\)

Therefore, the correct answer is E.

22 答案 D

用通常的记数法， \[ 121_b=b^2+2b+1=(b+1)^2\text{。}\]对每个允许的进制，这都是一个平方数。由于出现了数字 \(2\)，恰好只有满足 \(b\gt2\) 的整数进制才可用。

所以正确答案是 D。

English

In ordinary notation, \[ 121_b=b^2+2b+1=(b+1)^2. \] This is a square for every allowable base. Since digit \(2\) occurs, precisely the integral bases \(b\gt2\) are allowable.

Thus, the correct answer is D.

23 答案 E

设 \(AB=c\)、\(CD=h\)，且 \(AE=e\)。令两个面积公式相等，得到 \[ \frac12ch=\frac12(BC)e\text{，}\]所以 \(BC=\frac{ch}{e}\)。由于三角形 \(BCD\) 在 \(D\) 处为直角， \[ DB=\sqrt{BC^2-CD^2}\text{。}\]无论原三角形是锐角、直角还是钝角三角形，这都能确定 \(DB\)，所以前四个选项均不正确。

所以正确答案是 E。

English

Let \(AB=c,\) \(CD=h,\) and \(AE=e.\) Equating two area formulas gives \[ \frac12ch=\frac12(BC)e, \] so \(BC=\frac{ch}{e}.\) Since triangle \(BCD\) is right at \(D,\) \[ DB=\sqrt{BC^2-CD^2}. \] This determines \(DB\) whether the original triangle is acute, right, or obtuse, so none of the first four choices is correct.

Thus, the correct answer is E.

24 答案 A

工作速度满足 \[ \frac1{x+6}+\frac1{x+1}+\frac1{2x}=\frac1x\text{。}\]因此 \[ \frac1{x+6}+\frac1{x+1}=\frac1{2x}\text{，}\]化简得 \(3x^2+7x-6=0\)，即 \((3x-2)(x+3)=0\)。时间必须为正，所以 \(x=\frac{2}{3}\)。

所以正确答案是 A。

English

The rates satisfy \[ \frac1{x+6}+\frac1{x+1}+\frac1{2x}=\frac1x. \] Thus \[ \frac1{x+6}+\frac1{x+1}=\frac1{2x}, \] which simplifies to \(3x^2+7x-6=0,\) or \((3x-2)(x+3)=0.\) A time must be positive, so \(x=\frac{2}{3}.\)

Therefore, the correct answer is A.

25 答案 C

设圆心为 \((h,4)\)。由于圆经过 \(A\)， \[ r^2=h^2+16\text{。}\]与 \(x=8\) 相切给出 \(r=8-h\)。因此 \(h^2+16=(8-h)^2\)，所以 \(h=3\)，且 \(r=5\)。

所以正确答案是 C。

English

Place the center at \((h,4).\) Since the circle passes through \(A,\) \[ r^2=h^2+16. \] Tangency to \(x=8\) gives \(r=8-h.\) Therefore \(h^2+16=(8-h)^2,\) so \(h=3\) and \(r=5.\)

Thus, the correct answer is C.

26 答案 E

配方得 \[ 8x-3x^2 =\frac{16}{3}-3\left(x-\frac43\right)^2\text{。}\]平方项非负，所以最大值为 \(\frac{16}{3}\)。

所以正确答案是 E。

English

Completing the square, \[ 8x-3x^2 =\frac{16}{3}-3\left(x-\frac43\right)^2. \] The square term is nonnegative, so the maximum is \(\frac{16}{3}.\)

Thus, the correct answer is E.

27 答案 E

取最大值满足交换律和结合律，所以 \((1)\) 与 \((2)\) 成立。法则 \((3)\) 是分配恒等式 \[ \begin{aligned} &\min(a,\max(b,c))\\ &\quad=\max(\min(a,b),\min(a,c)) \end{aligned}\text{。}\]若 \(a\ge\max(b,c)\)，两边都等于 \(\max(b,c)\)；若 \(a\lt\max(b,c)\)，两边都等于 \(a\)。因此 \((3)\) 也成立。

所以正确答案是 E。

English

Maximum is commutative and associative, so \((1)\) and \((2)\) hold. Rule \((3)\) is the distributive identity \[ \begin{aligned} &\min(a,\max(b,c))\\ &\quad=\max(\min(a,b),\min(a,c)). \end{aligned} \] If \(a\ge\max(b,c),\) both sides equal \(\max(b,c);\) if \(a\lt\max(b,c),\) both sides equal \(a.\) Thus \((3)\) also holds.

Therefore, the correct answer is E.

28 答案 D

令 \(y=\log_{10}x\)，则 \(x=10^y\)。两边取以 \(10\) 为底的对数，得到 \[ y^2=3y-2\text{。}\]因此 \((y-1)(y-2)=0\)，所以 \(x=10\) 或 \(x=100\)。

所以正确答案是 D。

English

Set \(y=\log_{10}x,\) so \(x=10^y.\) Taking base-\(10\) logarithms gives \[ y^2=3y-2. \] Hence \((y-1)(y-2)=0,\) so \(x=10\) or \(x=100.\)

Thus, the correct answer is D.

29 答案 A

该不等式为 \(2x^2+x-6\lt0\)。分解得 \((2x-3)(x+2)\lt0\)。乘积在两根之间为负，所以 \(-2\lt x\lt\frac{3}{2}\)。

所以正确答案是 A。

English

The inequality is \(2x^2+x-6\lt0.\) Factoring gives \((2x-3)(x+2)\lt0.\) The product is negative between its roots, so \(-2\lt x\lt\frac{3}{2}.\)

Thus, the correct answer is A.

30 答案 D

只要 \(p\) 和 \(q\) 中至少一个为假，“\(p\) 和 \(q\) 都为真”的否定就成立。这发生在情形 \((2)\)、\((3)\) 和 \((4)\) 中，共有 \(3\) 种。

所以正确答案是 D。

English

The negation of “\(p\) and \(q\) are both true” holds whenever at least one of \(p\) and \(q\) is false. This occurs in cases \((2),\) \((3),\) and \((4),\) for a total of \(3.\)

Thus, the correct answer is D.

31 答案 C

设较小和较大的多边形分别有 \(n\) 和 \(N\) 条边。则 \[ \frac{\frac{N-2}{N}}{\frac{n-2}{n}}=\frac32\text{，}\]解得 \(N=\frac{4n}{6-n}\)。由正数条件及 \(N\gt n\)，可得 \(n=3\)、\(n=4\) 或 \(n=5\)。它们分别给出 \(N=4\)、\(N=8\) 和 \(N=20\)。因此共有 \(3\) 组。

所以正确答案是 C。

English

Let the smaller and larger polygons have \(n\) and \(N\) sides. Then \[ \frac{\frac{N-2}{N}}{\frac{n-2}{n}}=\frac32, \] which gives \(N=\frac{4n}{6-n}.\) Positivity and \(N\gt n\) require \(n=3,\) \(n=4,\) or \(n=5.\) These give \(N=4,\) \(N=8,\) and \(N=20,\) respectively. Thus there are \(3\) pairs.

Therefore, the correct answer is C.

32 答案 E

有 \(x_n=1+\frac{n-1}{2}=\frac{n+1}{2}\)。因此 \[ \begin{aligned} x_1+\cdots+x_n &=\frac n2\left(1+\frac{n+1}{2}\right)\\ &=\frac{n^2+3n}{4} \end{aligned}\text{。}\]

所以正确答案是 E。

English

We have \(x_n=1+\frac{n-1}{2}=\frac{n+1}{2}.\) Therefore \[ \begin{aligned} x_1+\cdots+x_n &=\frac n2\left(1+\frac{n+1}{2}\right)\\ &=\frac{n^2+3n}{4}. \end{aligned} \]

Thus, the correct answer is E.

33 答案 A

当 \(x-1\ge0\) 时，界限给出 \(3\le x\le6\)。当 \(x-1\le0\) 时，给出 \(-4\le x\le-1\)。解集是这两个区间的并集。

所以正确答案是 A。

English

For \(x-1\ge0,\) the bounds give \(3\le x\le6.\) For \(x-1\le0,\) they give \(-4\le x\le-1.\) The solution is the union of these two intervals.

Thus, the correct answer is A.

34 答案 E

整理得 \[ K^2x^2-(3K^2+1)x+2K^2=0\text{。}\]对每个实数 \(K\)，其判别式均满足 \[ \begin{aligned} &(3K^2+1)^2-8K^4\\ &\qquad=K^4+6K^2+1\gt0 \end{aligned}\text{。}\]这也包括 \(K=0\) 的情形，此时原方程给出 \(x=0\)。

所以正确答案是 E。

English

Rearranging gives \[ K^2x^2-(3K^2+1)x+2K^2=0. \] Its discriminant is \[ \begin{aligned} &(3K^2+1)^2-8K^4\\ &\qquad=K^4+6K^2+1\gt0 \end{aligned} \] for every real \(K.\) This also covers \(K=0,\) when the original equation gives \(x=0.\)

Therefore, the correct answer is E.

35 答案 B

两次观察分别位于两针重合时刻的前后两侧。带符号的角度差变化了 \(220^\circ\)。由于分针每分钟比时针多转 \(6-0.5=5.5^\circ\)，所以经过的时间为 \[ \frac{220}{5.5}=40 \] 分钟。

所以正确答案是 B。

English

The two observations lie on opposite sides of the instant when the hands coincide. Their signed angular separation changes by \(220^\circ.\) Since the minute hand gains on the hour hand at \(6-0.5=5.5^\circ\) per minute, the elapsed time is \[ \frac{220}{5.5}=40 \] minutes.

Thus, the correct answer is B.

36 答案 C

令 \(n=x-9\)。则 \[ 2^y=n^2-1=(n-1)(n+1)\text{。}\]由于乘积是 \(2\) 的幂，两个因式都不能含有奇素因子。相差 \(2\) 且均为带符号的 \(2\) 的幂的相邻偶数，只有 \((-4,-2)\) 和 \((2,4)\)。因此 \(n=\pm3\) 且 \(y=3\)，给出 \(x=6\) 或 \(12\)。共有 \(2\) 个有序数对 \((x,y)\)。

所以正确答案是 C。

English

Put \(n=x-9.\) Then \[ 2^y=n^2-1=(n-1)(n+1). \] Since the product is a power of \(2,\) both factors must have no odd prime divisor. The only consecutive even integers differing by \(2\) that are both signed powers of \(2\) are \((-4,-2)\) and \((2,4).\) Thus \(n=\pm3\) and \(y=3,\) giving \(x=6\) or \(12.\) There are \(2\) ordered pairs \((x,y).\)

Therefore, the correct answer is C.

37 答案 D

令 \(A=(0,0)\)、\(B=(1,0)\)、\(C=(1,1)\) 且 \(D=(0,1)\)。则 \(E=(t,0)\)，且 \(F=(0,t)\)。由鞋带公式得 \[ \begin{aligned} [CDFE]&=\frac{1+t-t^2}{2}\\ &=\frac58-\frac12 \left(t-\frac12\right)^2 \end{aligned}\text{。}\]其最大值为 \(\frac{5}{8}\)。

所以正确答案是 D。

English

Set \(A=(0,0),\) \(B=(1,0),\) \(C=(1,1),\) and \(D=(0,1).\) Then \(E=(t,0)\) and \(F=(0,t).\) The shoelace formula gives \[ \begin{aligned} [CDFE]&=\frac{1+t-t^2}{2}\\ &=\frac58-\frac12 \left(t-\frac12\right)^2. \end{aligned} \] Its maximum is \(\frac{5}{8}.\)

Thus, the correct answer is D.

38 答案 B

设原来的人口为 \(a^2\)。则 \[ \begin{aligned} b^2-a^2&=99,\\ c^2-a^2&=200\text{。} \end{aligned} \] 由 \((b-a)(b+a)=99\)，\(a\) 的正数可能值为 \(49\)、\(15\) 和 \(1\)。检验第二个条件，只有 \(a=49\) 可行，因为 \[ 49^2+200=2601=51^2\text{。}\]人口为 \(49^2=7^4\)，是 \(7\) 的倍数。

所以正确答案是 B。

English

Let the original population be \(a^2.\) Then \[ \begin{aligned} b^2-a^2&=99,\\ c^2-a^2&=200. \end{aligned} \] From \((b-a)(b+a)=99,\) the positive possibilities for \(a\) are \(49,\) \(15,\) and \(1.\) Checking the second condition, only \(a=49\) works, since \[ 49^2+200=2601=51^2. \] The population is \(49^2=7^4,\) a multiple of \(7.\)

Thus, the correct answer is B.

39 答案 C

以三条中线为边所组成的三角形，其面积为 \[ \frac34\left(3\sqrt{15}\right)=\frac{9\sqrt{15}}4\text{。}\]若 \(\theta\) 是边长 \(3\) 与 \(6\) 的夹角，则 \[ 9\sin\theta=\frac{9\sqrt{15}}4\text{，}\]所以 \(\cos\theta=\pm\frac{1}{4}\)。由余弦定理，第三条中线 \(m\) 满足 \[ m^2=3^2+6^2-2(3)(6)\cos\theta\text{，}\]得到 \(m^2=36\) 或 \(54\)。若 \(m=6\)，则两条中线相等，从而两条边也相等。由于原三角形各边不等，故 \(m=\sqrt{54}=3\sqrt6\)。

所以正确答案是 C。

English

The triangle whose sides are the three medians has area \[ \frac34\left(3\sqrt{15}\right)=\frac{9\sqrt{15}}4. \] If \(\theta\) is the included angle between its sides \(3\) and \(6,\) then \[ 9\sin\theta=\frac{9\sqrt{15}}4, \] so \(\cos\theta=\pm\frac{1}{4}.\) By the law of cosines, the third median \(m\) satisfies \[ m^2=3^2+6^2-2(3)(6)\cos\theta, \] giving \(m^2=36\) or \(54.\) The value \(m=6\) would make two medians, and hence two sides, equal. Because the triangle has unequal sides, \(m=\sqrt{54}=3\sqrt6.\)

Therefore, the correct answer is C.

40 答案 B

当 \(|x|\lt1\) 时， \[ \sum_{n=1}^{\infty}nx^n=\frac{x}{(1-x)^2}\text{。}\]取 \(x=\frac{1}{10}\)，得到 \[ \frac{\frac{1}{10}}{(\frac{9}{10})^2}=\frac{10}{81}\text{。}\]

所以正确答案是 B。

English

For \(|x|\lt1,\) \[ \sum_{n=1}^{\infty}nx^n=\frac{x}{(1-x)^2}. \] Taking \(x=\frac{1}{10}\) gives \[ \frac{\frac{1}{10}}{(\frac{9}{10})^2}=\frac{10}{81}. \]

Thus, the correct answer is B.