1961 AMC 12 真题（P版）

1 答案 C

由于立方根是实数，\[ \left(-\frac1{125}\right)^{\frac{2}{3}} =\left(-\frac15\right)^2=\frac1{25}\text{。}\] 负指数表示取倒数，因此得到 \(25\)。

因此，正确答案是 C。

English

Because the cube root is real, \[ \left(-\frac1{125}\right)^{\frac{2}{3}} =\left(-\frac15\right)^2=\frac1{25}. \] The negative exponent takes the reciprocal, giving \(25.\)

Therefore, the correct answer is C.

2 答案 E

汽车的速度为每秒 \(\frac{a}{6r}\) 英尺。在 \(180\) 秒内，汽车行驶 \[ 180\cdot\frac{a}{6r}=\frac{30a}{r} \] 英尺，也就是 \(\frac{10a}{r}\) 码。

所以，正确答案是 E。

English

The speed is \(\frac{a}{6r}\) feet per second. In \(180\) seconds the automobile travels \[ 180\cdot\frac{a}{6r}=\frac{30a}{r} \] feet, or \(\frac{10a}{r}\) yards.

Thus, the correct answer is E.

3 答案 E

两条直线的斜率分别为 \(-\frac{1}{2}\) 和 \(-\frac{a}{3}\)。垂直条件要求 \[ \left(-\frac12\right)\left(-\frac a3\right)=-1\text{，}\] 因而 \(a=-6\)。

因此，正确答案是 E。

English

The two slopes are \(-\frac{1}{2}\) and \(-\frac{a}{3}.\) Perpendicularity requires \[ \left(-\frac12\right)\left(-\frac a3\right)=-1, \] so \(a=-6.\)

Therefore, the correct answer is E.

4 答案 B

对于正整数 \(m,n\)，\[ m^2n^2=(mn)^2\text{，}\] 它仍是一个正整数的平方。其他运算均不满足，例如 \(1+4=5,\frac{1}{4}\) 以及 \(\sqrt4=2\)。

所以，正确答案是 B。

English

For positive integers \(m,n,\) \[ m^2n^2=(mn)^2, \] which is again the square of a positive integer. The other operations fail, for example \(1+4=5,\frac{1}{4},\) and \(\sqrt4=2.\)

Thus, the correct answer is B.

5 答案 C

该式正是二项式展开 \[ ((x-1)+1)^4=x^4\text{。}\]

因此，正确答案是 C。

English

The expression is the binomial expansion \[ ((x-1)+1)^4=x^4. \]

Therefore, the correct answer is C.

6 答案 E

由于 \(\log(\frac{1}{8})=\log(8^{-1})=-\log8\)，\[ \frac{\log8}{\log(\frac{1}{8})}=-1\text{。}\]

所以，正确答案是 E。

English

Since \(\log(\frac{1}{8})=\log(8^{-1})=-\log8,\) \[ \frac{\log8}{\log(\frac{1}{8})}=-1. \]

Thus, the correct answer is E.

7 答案 A

第三项为 \[ \begin{aligned} &\binom62\left(\frac a{\sqrt x}\right)^4 \left(-\frac{\sqrt x}{a^2}\right)^2\\ &\qquad=15\cdot\frac{a^4}{x^2}\cdot\frac{x}{a^4} =\frac{15}{x} \end{aligned}\text{。}\]

因此，正确答案是 A。

English

The third term is \[ \begin{aligned} &\binom62\left(\frac a{\sqrt x}\right)^4 \left(-\frac{\sqrt x}{a^2}\right)^2\\ &\qquad=15\cdot\frac{a^4}{x^2}\cdot\frac{x}{a^4} =\frac{15}{x}. \end{aligned} \]

Therefore, the correct answer is A.

8 答案 B

由两个直角三角形可得 \[ \begin{aligned} A+C_1&=90^\circ,\\ B+C_2&=90^\circ \end{aligned}\text{。}\] 用第一个关系式减去第二个关系式，得到 \(C_1-C_2=B-A\)。

所以，正确答案是 B。

English

The two right triangles give \[ \begin{aligned} A+C_1&=90^\circ,\\ B+C_2&=90^\circ. \end{aligned} \] Subtracting the second relation from the first gives \(C_1-C_2=B-A.\)

Thus, the correct answer is B.

9 答案 C

有 \[ r=(2a)^{2b}=(4a^2)^b \] 同时 \(r=a^bx^b=(ax)^b\)。因此 \(ax=4a^2\)，所以 \(x=4a\)。

因此，正确答案是 C。

English

We have \[ r=(2a)^{2b}=(4a^2)^b \] and also \(r=a^bx^b=(ax)^b.\) Thus \(ax=4a^2,\) so \(x=4a.\)

Therefore, the correct answer is C.

10 答案 D

在等边三角形中，\(BD=6\)，且 \(AD=6\sqrt3\)。因此 \(DE=3\sqrt3\)，再由直角三角形 \(BDE\) 得 \[ \begin{aligned} BE^2&=6^2+(3\sqrt3)^2\\ &=36+27=63 \end{aligned}\text{。}\]所以 \(BE=\sqrt{63}\)。

所以，正确答案是 D。

English

In the equilateral triangle, \(BD=6\) and \(AD=6\sqrt3.\) Hence \(DE=3\sqrt3,\) and right triangle \(BDE\) gives \[ \begin{aligned} BE^2&=6^2+(3\sqrt3)^2\\ &=36+27=63. \end{aligned} \] Therefore \(BE=\sqrt{63}.\)

Thus, the correct answer is D.

11 答案 C

由切线段相等可得 \(PQ=PB\) 且 \(RQ=RC\)。因此 \[ \begin{aligned} AP+PR+RA &=AP+PQ\\ &\quad+RQ+RA\\ &=AP+PB\\ &\quad+RC+RA\\ &=AB+AC \end{aligned}\text{。}\]又因为 \(AB=AC=20\)，所以周长为 \(40\)。

因此，正确答案是 C。

English

Equal tangent segments give \(PQ=PB\) and \(RQ=RC.\) Thus \[ \begin{aligned} AP+PR+RA &=AP+PQ\\ &\quad+RQ+RA\\ &=AP+PB\\ &\quad+RC+RA\\ &=AB+AC. \end{aligned} \] Also \(AB=AC=20,\) so the perimeter is \(40.\)

Therefore, the correct answer is C.

12 答案 A

各项的指数为 \(\frac{1}{2}\)、\(\frac{1}{3}\)、\(\frac{1}{6}\)。每一项都是前一项乘以 \(2^{-\frac{1}{6}}\) 得到的，所以后一项的指数为 \(\frac{1}{6}-\frac{1}{6}=0\)。第四项是 \(2^0=1\)。

所以，正确答案是 A。

English

The exponents are \(\frac{1}{2},\) \(\frac{1}{3},\) and \(\frac{1}{6}.\) Each term is obtained by multiplying by \(2^{-\frac{1}{6}},\) so the next exponent is \(\frac{1}{6}-\frac{1}{6}=0.\) The fourth term is \(2^0=1.\)

Thus, the correct answer is A.

13 答案 E

在根号内因式分解可得 \[ \begin{aligned} \sqrt{t^2(t^2+1)} &=\sqrt{t^2}\sqrt{t^2+1}\\ &=|t|\sqrt{1+t^2} \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Factoring inside the radical gives \[ \begin{aligned} \sqrt{t^2(t^2+1)} &=\sqrt{t^2}\sqrt{t^2+1}\\ &=|t|\sqrt{1+t^2}. \end{aligned} \]

Therefore, the correct answer is E.

14 答案 E

设两条对角线为 \(d\) 和 \(2d\)。则 \(K=\frac{d(2d)}{2}=d^2\)。一条边的长度为 \[ \begin{aligned} \sqrt{\left(\frac d2\right)^2+d^2} &=\frac{\sqrt5}{2}d\\ &=\frac12\sqrt{5K} \end{aligned}\text{。}\] 前四个选项中没有这个结果。

所以，正确答案是 E。

English

Let the diagonals be \(d\) and \(2d.\) Then \(K=\frac{d(2d)}{2}=d^2.\) A side has length \[ \begin{aligned} \sqrt{\left(\frac d2\right)^2+d^2} &=\frac{\sqrt5}{2}d\\ &=\frac12\sqrt{5K}. \end{aligned} \] This is not among the first four choices.

Thus, the correct answer is E.

15 答案 B

第一组用 \(x^3\) 个工时生产 \(x\) 件产品，所以每工时的产量为 \(\frac{1}{x^2}\) 件。第二组提供 \(y^3\) 个工时，因此生产 \(\frac{y^3}{x^2}\) 件产品。

所以，正确答案是 B。

English

The first group uses \(x^3\) man-hours to make \(x\) articles, so the rate is \(\frac{1}{x^2}\) article per man-hour. The second group supplies \(y^3\) man-hours and therefore makes \(\frac{y^3}{x^2}\) articles.

Thus, the correct answer is B.

16 答案 E

若减去 \(t\)，新面积应满足 \[ \frac12(b-t)(h+m)=\frac12\left(\frac12bh\right)\text{。}\] 因此 \(b-t=\frac{bh}{2(h+m)}\)，所以 \[ \begin{aligned} t&=b-\frac{bh}{2(h+m)}\\ &=\frac{b(2m+h)}{2(h+m)} \end{aligned}\text{。}\]

因此，正确答案是 E。

English

If \(t\) is removed, the new area condition is \[ \frac12(b-t)(h+m)=\frac12\left(\frac12bh\right). \] Hence \(b-t=\frac{bh}{2(h+m)},\) so \[ \begin{aligned} t&=b-\frac{bh}{2(h+m)}\\ &=\frac{b(2m+h)}{2(h+m)}. \end{aligned} \]

Therefore, the correct answer is E.

17 答案 D

用通常的记法，这笔交易表示为 \[ r^3-(4r^2+4r)=3r^2+4r\text{。}\] 因此 \(r(r^2-7r-8)=0\)，所以 \(r=8\) 或 \(r=-1\)。进制的底必须为正且大于 \(4\)，故 \(r=8\)。

因此，正确答案是 D。

English

In ordinary notation the transaction says \[ r^3-(4r^2+4r)=3r^2+4r. \] Thus \(r(r^2-7r-8)=0,\) so \(r=8\) or \(r=-1.\) A numeral base must be positive and exceed \(4,\) hence \(r=8.\)

Therefore, the correct answer is D.

18 答案 A

最终人口相当于原人口乘以 \[ (1.25)^2(0.75)^2=\frac{225}{256}\approx0.879\text{。}\] 这表示约减少 \(12.1\%\)，取整后为 \(12\%\)。

所以，正确答案是 A。

English

The final population is multiplied by \[ (1.25)^2(0.75)^2=\frac{225}{256}\approx0.879. \] This is a decrease of about \(12.1\%,\) which rounds to \(12\%.\)

Thus, the correct answer is A.

19 答案 B

在定义域 \(x>0\) 上，两式相等要求 \[ \log(x^2)=\log(2x)\text{，}\] 因而 \(x^2=2x\)。唯一的正解是 \(x=2\)，所以恰有一个交点。

因此，正确答案是 B。

English

On the domain \(x>0,\) equality requires \[ \log(x^2)=\log(2x), \] so \(x^2=2x.\) The only positive solution is \(x=2,\) giving exactly one intersection.

Therefore, the correct answer is B.

20 答案 A

若 \(x\ge0\)，则 \(y>2x\ge0\)。若 \(x<0\)，则 \(y>4-x>0\)。因此每个满足条件的点都位于 \(x\) 轴上方，同时正、负 \(x\) 值都能出现。该区域位于第 \(\mathrm{I}\)、\(\mathrm{II}\) 象限。

因此，正确答案是 A。

English

If \(x\ge0,\) then \(y>2x\ge0.\) If \(x<0,\) then \(y>4-x>0.\) Thus every feasible point is above the \(x\)-axis, while both positive and negative \(x\)-values occur. The region lies in quadrants \(\mathrm{I}\) and \(\mathrm{II}.\)

Therefore, the correct answer is A.

21 答案 D

令 \(A=\mathbf0\)、\(B=\mathbf b\)、\(C=\mathbf c\)。则 \[ \begin{aligned} E&=\frac{\mathbf b}{2},\\ N&=\frac{\mathbf b}{4},\\ M&=\frac{\mathbf b+\mathbf c}{3} \end{aligned}\text{。}\]计算行列式得 \[ [MNE]=\frac1{24} \left|\det(\mathbf b,\mathbf c)\right|\text{，}\]而 \[ [ABC]=\frac12 \left|\det(\mathbf b,\mathbf c)\right|\text{。}\]两者之比为 \(\frac{1}{12}\)。

所以，正确答案是 D。

English

Set \(A=\mathbf0,\) \(B=\mathbf b,\) and \(C=\mathbf c.\) Then \[ \begin{aligned} E&=\frac{\mathbf b}{2},\\ N&=\frac{\mathbf b}{4},\\ M&=\frac{\mathbf b+\mathbf c}{3}. \end{aligned} \] A determinant calculation gives \[ [MNE]=\frac1{24} \left|\det(\mathbf b,\mathbf c)\right|, \] while \[ [ABC]=\frac12 \left|\det(\mathbf b,\mathbf c)\right|. \] Their ratio is \(\frac{1}{12}.\)

Thus, the correct answer is D.

22 答案 C

令 \(P(x)=3x^3-9x^2+kx-12\)。由于 \(x-3\) 是因式，\[ P(3)=81-81+3k-12=0\text{，}\]所以 \(k=4\)。直接相除得 \[ P(x)=(x-3)(3x^2+4)\text{。}\]

因此，正确答案是 C。

English

Let \(P(x)=3x^3-9x^2+kx-12.\) Since \(x-3\) is a factor, \[ P(3)=81-81+3k-12=0, \] so \(k=4.\) Direct division gives \[ P(x)=(x-3)(3x^2+4). \]

Therefore, the correct answer is C.

23 答案 B

由分点比得 \(\frac{AP}{AB}=\frac25\) 和 \(\frac{AQ}{AB}=\frac37\)。因此 \(PQ=(\frac{3}{7}-\frac{2}{5})AB=\frac{AB}{35}\)。因为 \(PQ=2\)，所以 \(AB=70\)。

所以，正确答案是 B。

English

The division ratios give \(\frac{AP}{AB}=\frac25\) and \(\frac{AQ}{AB}=\frac37.\) Hence \(PQ=(\frac{3}{7}-\frac{2}{5})AB=\frac{AB}{35}.\) Since \(PQ=2,\) we obtain \(AB=70.\)

Thus, the correct answer is B.

24 答案 A

各书价格为 \(p\)、\(p+2\)、\(\ldots\)、\(p+60\)，中间一本的价格为 \(p+30\)。若取右侧相邻书，则 \[ (p+30)+(p+32)=p+60\text{，}\]得 \(p=-2\)，不可能是售价。取左侧相邻书时 \[ (p+30)+(p+28)=p+60\text{，}\]所以 \(p=2\)。因此相邻书必须是左侧那一本。

因此，正确答案是 A。

English

The prices are \(p,\) \(p+2,\) \(\ldots,\) \(p+60,\) and the middle price is \(p+30.\) If the right neighbor is used, then \[ (p+30)+(p+32)=p+60, \] which gives \(p=-2,\) impossible for a selling price. The left neighbor gives \[ (p+30)+(p+28)=p+60, \] so \(p=2.\) Thus the adjacent book must be the one to the left.

Therefore, the correct answer is A.

25 答案 A

令 \(\angle B=m\)。由 \(PQ=QB\)，三角形 \(BQP\) 给出 \(\angle QPB=m\) 和 \(\angle AQP=2m\)。由 \(AP=PQ\)，三角形 \(AQP\) 给出 \(\angle QAP=2m\)，故 \(\angle QPA=180^\circ-4m\)。由于 \(B,P,C\) 共线，\(\angle APC=3m\)。再由 \(AP=AC\) 得 \(\angle ACP=3m\)。最后，\(AB=BC\) 使 \(ABC\) 的两个底角均为 \(3m\)，且 \[ m+3m+3m=180^\circ\text{。}\]因此 \(m=\frac{180^\circ}{7}=25\dfrac57^\circ\)。

所以，正确答案是 A。

English

Let \(\angle B=m.\) Since \(PQ=QB,\) triangle \(BQP\) gives \(\angle QPB=m\) and \(\angle AQP=2m.\) Since \(AP=PQ,\) triangle \(AQP\) gives \(\angle QAP=2m,\) so \(\angle QPA=180^\circ-4m.\) As \(B,P,C\) are collinear, \(\angle APC=3m.\) Now \(AP=AC\) gives \(\angle ACP=3m.\) Finally \(AB=BC\) makes both base angles of \(ABC\) equal to \(3m,\) and \[ m+3m+3m=180^\circ. \] Therefore \(m=\frac{180^\circ}{7}=25\dfrac57^\circ.\)

Thus, the correct answer is A.

26 答案 C

两个和给出 \[ \begin{aligned} 25(2a+49d)&=200,\\ 25(2a+149d)&=2700 \end{aligned}\text{。}\] 因此 \(2a+49d=8\) 且 \(2a+149d=108\)。相减得 \(d=1\)，继而 \(2a=-41\)，所以 \(a=-20.5\)。

因此，正确答案是 C。

English

The two sums give \[ \begin{aligned} 25(2a+49d)&=200,\\ 25(2a+149d)&=2700. \end{aligned} \] Thus \(2a+49d=8\) and \(2a+149d=108.\) Subtraction gives \(d=1,\) and then \(2a=-41,\) so \(a=-20.5.\)

Therefore, the correct answer is C.

27 答案 D

每个多边形内角至少为 \(60^\circ\)，每个凸多边形内角小于 \(180^\circ\)。由于 \(k\ge2\) 且 \(kx<180^\circ\)，必须有 \(x<90^\circ\)。区间 \([60^\circ,90^\circ)\) 内唯一可能的等角多边形内角是三角形的 \(x=60^\circ\)。此时 \(k=2\) 给出 \(kx=120^\circ\)，即正六边形的内角；更大的 \(k\) 不可能。因此只有一对。

所以，正确答案是 D。

English

Every polygon angle is at least \(60^\circ,\) and every convex polygon angle is less than \(180^\circ.\) Since \(k\ge2\) and \(kx<180^\circ,\) we need \(x<90^\circ.\) The only possible equiangular polygon angle in \([60^\circ,90^\circ)\) is the triangle angle \(x=60^\circ.\) Then \(k=2\) gives \(kx=120^\circ,\) the angle of a regular hexagon; larger \(k\) is impossible. Hence there is one pair.

Thus, the correct answer is D.

28 答案 D

\(7\) 的各次幂个位数字按 \(7\)、\(9\)、\(3\)、\(1\) 循环，周期为 \(4\)。由于 \(753\equiv1\pmod4\)，\(2137^{753}\) 的个位数字为 \(7\)。

因此，正确答案是 D。

English

Powers of \(7\) have units digits \(7,\) \(9,\) \(3,\) \(1\) in a cycle of length \(4.\) Since \(753\equiv1\pmod4,\) the units digit of \(2137^{753}\) is \(7.\)

Therefore, the correct answer is D.

29 答案 B

变换后的两根之和为 \[ a(r+s)+2b=-b+2b=b \] 乘积为 \[ \begin{gathered} (ar+b)(as+b)\\ =a^2rs+ab(r+s)+b^2\\ =ac \end{gathered}\text{。}\] 因此以它们为根的首一方程为 \(x^2-bx+ac=0\)。

所以，正确答案是 B。

English

The transformed roots have sum \[ a(r+s)+2b=-b+2b=b \] and product \[ \begin{gathered} (ar+b)(as+b)\\ =a^2rs+ab(r+s)+b^2\\ =ac. \end{gathered} \] Therefore their monic equation is \(x^2-bx+ac=0.\)

Thus, the correct answer is B.

30 答案 D

由换底公式，\[ \begin{aligned} \log_5 12 &=\frac{\log12}{\log5}\\ &=\frac{2\log2+\log3} {\log10-\log2}\\ &=\frac{2a+b}{1-a} \end{aligned}\text{。}\]

因此，正确答案是 D。

English

By change of base, \[ \begin{aligned} \log_5 12 &=\frac{\log12}{\log5}\\ &=\frac{2\log2+\log3} {\log10-\log2}\\ &=\frac{2a+b}{1-a}. \end{aligned} \]

Therefore, the correct answer is D.

31 答案 D

由外角平分线定理，\[ \frac{PA}{PB}=\frac{AC}{CB}=\frac34\text{。}\] 因为 \(PB=PA+AB\)，令 \(PA=3t\)、\(PB=4t\)。则 \(AB=t\)，所以 \(PA:AB=3:1\)。

所以，正确答案是 D。

English

The exterior angle bisector theorem gives \[ \frac{PA}{PB}=\frac{AC}{CB}=\frac34. \] Since \(PB=PA+AB,\) write \(PA=3t,\) \(PB=4t.\) Then \(AB=t,\) so \(PA:AB=3:1.\)

Thus, the correct answer is D.

32 答案 C

多边形的面积为 \[ \frac n2R^2\sin\frac{360^\circ}{n}\text{。}\] 当 \(n=12\) 时，它变为 \[ 6R^2\sin30^\circ=3R^2\text{，}\] 符合要求。

因此，正确答案是 C。

English

The polygon’s area is \[ \frac n2R^2\sin\frac{360^\circ}{n}. \] For \(n=12,\) this becomes \[ 6R^2\sin30^\circ=3R^2, \] as required.

Therefore, the correct answer is C.

33 答案 B

因式分解得 \[ (2^x-3^y)(2^x+3^y)=55\text{。}\]首先，\(x>0\)。若 \(y\le0\)，则 \(55<4^x\le56\)，而 \(4\) 的整数次幂不可能落在这个范围内。因此 \(y>0\)，两个因数都是正奇整数。因数对 \(5\)、\(11\) 给出 \(2^{x+1}=16\) 和 \(2\cdot3^y=6\)，所以 \((x,y)=(3,1)\)。因数对 \(1\)、\(55\) 则要求 \(2^x=28\)，不可能。因此恰有一个解。

所以，正确答案是 B。

English

Factor: \[ (2^x-3^y)(2^x+3^y)=55. \] First, \(x>0.\) If \(y\le0,\) then \(55<4^x\le56,\) which is impossible for an integral power of \(4.\) Hence \(y>0,\) so both factors are positive odd integers. The factor pair \(5,\) \(11\) gives \(2^{x+1}=16\) and \(2\cdot3^y=6,\) so \((x,y)=(3,1).\) The factor pair \(1,\) \(55\) would require \(2^x=28,\) impossible. Hence there is exactly one solution.

Thus, the correct answer is B.

34 答案 A

有 \[ \frac{2x+3}{x+2}=2-\frac1{x+2}\text{。}\] 当 \(x\ge0\) 时，该值从 \(\frac{3}{2}\) 递增并趋近 \(2\)，但不等于 \(2\)。因此 \(S=[\frac32,2)\)，其最大下界 \(m=\frac{3}{2}\) 属于 \(S\)，而最小上界 \(M=2\) 不属于其中。

因此，正确答案是 A。

English

We have \[ \frac{2x+3}{x+2}=2-\frac1{x+2}. \] For \(x\ge0,\) this increases from \(\frac{3}{2}\) toward \(2\) without reaching \(2.\) Thus \(S=[\frac32,2),\) so its greatest lower bound \(m=\frac{3}{2}\) belongs to \(S,\) while its least upper bound \(M=2\) does not.

Therefore, the correct answer is A.

35 答案 D

因为 \(5!=120\)，\[ 695=5\cdot120+95\text{。}\] 接着 \(4!=24\)，且 \(95=3\cdot24+23\)。因此系数 \(a_4\) 为 \(3\)。

所以，正确答案是 D。

English

Since \(5!=120,\) \[ 695=5\cdot120+95. \] Next \(4!=24,\) and \(95=3\cdot24+23.\) Therefore the coefficient \(a_4\) is \(3.\)

Thus, the correct answer is D.

36 答案 B

令 \(A=\mathbf0\)、\(B=\mathbf b\)、\(C=\mathbf c\)，其中 \(|\mathbf b|=\ell\)、\(|\mathbf c|=6\)，且 \(|\mathbf c-\mathbf b|=7\)。从 \(A\) 和 \(B\) 引出的中线方向分别为 \(\frac{\mathbf b+\mathbf c}{2}\) 和 \(\frac{\mathbf c}{2}-\mathbf b\)。其点积为零，故 \[ (\mathbf b+\mathbf c)\mathbin{\cdot}(\mathbf c-2\mathbf b)=0\text{。}\]使用 \(\mathbf b\mathbin{\cdot}\mathbf c=\frac{\ell^2+36-49}{2}\)，化简得 \(\ell^2=17\)。所以 \(AB=\sqrt{17}\)。

因此，正确答案是 B。

English

Let \(A=\mathbf0,\) \(B=\mathbf b,\) and \(C=\mathbf c,\) with \(|\mathbf b|=\ell,\) \(|\mathbf c|=6,\) and \(|\mathbf c-\mathbf b|=7.\) The median directions are \(\frac{\mathbf b+\mathbf c}{2}\) from \(A\) and \(\frac{\mathbf c}{2}-\mathbf b\) from \(B.\) Their dot product is zero, so \[ (\mathbf b+\mathbf c)\mathbin{\cdot}(\mathbf c-2\mathbf b)=0. \] Using \(\mathbf b\mathbin{\cdot}\mathbf c=\frac{\ell^2+36-49}{2},\) this simplifies to \(\ell^2=17.\) Hence \(AB=\sqrt{17}.\)

Therefore, the correct answer is B.

37 答案 C

由领先距离得 \[ \begin{aligned} \frac{v_B}{v_A}&=\frac{d-20}{d},\\ \frac{v_C}{v_B}&=\frac{d-10}{d},\\ \frac{v_C}{v_A}&=\frac{d-28}{d} \end{aligned}\text{。}\] 因此 \[ \frac{(d-20)(d-10)}{d^2}=\frac{d-28}{d}\text{。}\] 展开并化简得 \(2d=200\)，所以 \(d=100\)。

所以，正确答案是 C。

English

The margins give \[ \begin{aligned} \frac{v_B}{v_A}&=\frac{d-20}{d},\\ \frac{v_C}{v_B}&=\frac{d-10}{d},\\ \frac{v_C}{v_A}&=\frac{d-28}{d}. \end{aligned} \] Therefore \[ \frac{(d-20)(d-10)}{d^2}=\frac{d-28}{d}. \] Expanding and simplifying yields \(2d=200,\) so \(d=100.\)

Thus, the correct answer is C.

38 答案 A

由泰勒斯定理，\(ABC\) 在 \(C\) 处为直角。令 \(p=AC\)、\(q=BC\)。则 \[ p^2+q^2=AB^2=4r^2\text{。}\] 因为 \((p-q)^2\ge0\)，所以 \(2pq\le p^2+q^2\)。因此 \[ s^2=(p+q)^2\le2(p^2+q^2)=8r^2\text{。}\]

所以，正确答案是 A。

English

By Thales’ theorem, \(ABC\) is right at \(C.\) Put \(p=AC\) and \(q=BC.\) Then \[ p^2+q^2=AB^2=4r^2. \] Since \((p-q)^2\ge0,\) we have \(2pq\le p^2+q^2.\) Therefore \[ s^2=(p+q)^2\le2(p^2+q^2)=8r^2. \]

Thus, the correct answer is A.

39 答案 B

将正方形分成四个边长为 \(\frac{1}{2}\) 的小正方形。五点中有两点落在同一小正方形内，所以它们的距离至多为其对角线 \(\frac{\sqrt2}{2}\)。在四个顶点和中心各放一点可以达到此界：此时最短距离为 \(\frac{\sqrt2}{2}\)。因此所求最小保证值是 \(\frac{\sqrt2}{2}\)。

因此，正确答案是 B。

English

Divide the square into four squares of side \(\frac{1}{2}.\) Two of the five points lie in the same small square, so their distance is at most its diagonal, \(\frac{\sqrt2}{2}.\) This bound is attainable by placing points at the four corners and the center: the shortest distance is then \(\frac{\sqrt2}{2}.\) Hence the least guaranteed value is \(\frac{\sqrt2}{2}.\)

Therefore, the correct answer is B.

40 答案 A

由柯西-施瓦茨不等式，\[ \begin{aligned} 60&=5x+12y\\ &\le\sqrt{5^2+12^2}\\ &\qquad\cdot\sqrt{x^2+y^2}\\ &=13\sqrt{x^2+y^2} \end{aligned}\text{。}\] 因此该距离至少为 \(\frac{60}{13}\)。当 \((x,y)\) 与 \((5,12)\) 成比例时取等号，所以最小值为 \(\frac{60}{13}\)。

因此，正确答案是 A。

English

By Cauchy-Schwarz, \[ \begin{aligned} 60&=5x+12y\\ &\le\sqrt{5^2+12^2}\\ &\qquad\cdot\sqrt{x^2+y^2}\\ &=13\sqrt{x^2+y^2}. \end{aligned} \] Thus the distance is at least \(\frac{60}{13}.\) Equality occurs when \((x,y)\) is proportional to \((5,12),\) so the minimum is \(\frac{60}{13}.\)

Therefore, the correct answer is A.