1965 AMC 12 真题（P版）

1 答案 C

由于底数为 \(2\)， 指数必须为零。因式分解得 \[ \begin{gathered} 2x^2-7x+5=0,\\ (2x-5)(x-1)=0 \end{gathered}\text{。}\] 因此 \(x=\frac{5}{2}\) 或 \(x=1\)， 共有两个实数解。

因此，正确答案是 C。

English

Since the base is \(2,\) the exponent must be zero. Factoring gives \[ \begin{gathered} 2x^2-7x+5=0,\\ (2x-5)(x-1)=0. \end{gathered} \] Thus \(x=\frac{5}{2}\) or \(x=1,\) giving two real values.

Therefore, the correct answer is C.

2 答案 D

若半径为 \(r\)， 则六边形的边长为 \(r\)。 该边所截劣弧的长度为 \(\frac{2\pi r}{6}=\frac{\pi r}{3}\)。 所求比为 \(r:(\frac{\pi r}{3})=3:\pi\)。

因此，正确答案是 D。

English

If the radius is \(r,\) the hexagon side is \(r.\) Its intercepted minor arc has length \(\frac{2\pi r}{6}=\frac{\pi r}{3}.\) The ratio is \(r:(\frac{\pi r}{3})=3:\pi.\)

Therefore, the correct answer is D.

3 答案 B

由于 \(2^{-2}=\frac{1}{4}\)， \[ \begin{aligned} 81^{-(2^{-2})} &=81^{-\frac{1}{4}}\\ &=(3^4)^{-\frac{1}{4}}=\frac13 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Since \(2^{-2}=\frac{1}{4},\) \[ \begin{aligned} 81^{-(2^{-2})} &=81^{-\frac{1}{4}}\\ &=(3^4)^{-\frac{1}{4}}=\frac13. \end{aligned} \]

Therefore, the correct answer is B.

4 答案 C

到平行线 \(l_1\) 和 \(l_3\) 距离相等的点位于两线之间唯一的平行中线上。到相交直线 \(l_1\) 和 \(l_2\) 距离相等的点位于它们的两条角平分线之一上。每条角平分线都与平行中线相交一次，因此共有两个点。

因此，正确答案是 C。

English

Points equidistant from parallel lines \(l_1\) and \(l_3\) lie on the unique line midway between them. Points equidistant from intersecting lines \(l_1\) and \(l_2\) lie on either of their two angle bisectors. Each angle bisector meets the midway parallel once, giving two points.

Therefore, the correct answer is C.

5 答案 A

令 \(z=0.363636\ldots\)。 则 \(100z-z=36\)， 所以 \(z=\frac{36}{99}=\frac{4}{11}\)。 所求的和为 \(4+11=15\)。

因此，正确答案是 A。

English

Let \(z=0.363636\ldots.\) Then \(100z-z=36,\) so \(z=\frac{36}{99}=\frac{4}{11}.\) The requested sum is \(4+11=15.\)

Therefore, the correct answer is A.

6 答案 B

指数函数与对数函数互相抵消，所以 \(10^{\log_{10}9}=9\)。 因此 \(8x+5=9\)， 从而 \(x=\frac{1}{2}\)。

因此，正确答案是 B。

English

The exponential and logarithm cancel, so \(10^{\log_{10}9}=9.\) Thus \(8x+5=9,\) giving \(x=\frac{1}{2}.\)

Therefore, the correct answer is B.

7 答案 E

根据韦达定理，\(r+s=-\frac{b}{a}\)，且 \(rs=\frac{c}{a}\)。 因此 \[ \begin{aligned} \frac1r+\frac1s &=\frac{r+s}{rs}\\ &=\frac{-\frac{b}{a}}{\frac{c}{a}}=-\frac bc \end{aligned}\text{。}\]

因此，正确答案是 E。

English

By Vieta’s formulas, \(r+s=-\frac{b}{a}\) and \(rs=\frac{c}{a}.\) Therefore \[ \begin{aligned} \frac1r+\frac1s &=\frac{r+s}{rs}\\ &=\frac{-\frac{b}{a}}{\frac{c}{a}}=-\frac bc. \end{aligned} \]

Thus, the correct answer is E.

8 答案 A

梯形占总面积的三分之一，所以较小相似三角形占三分之二。它的长度比为 \(\sqrt{\frac{2}{3}}\)。 因此平行线段的长度为 \[18\sqrt{\frac23}=6\sqrt6\text{。}\]

因此，正确答案是 A。

English

The trapezoid occupies one third of the area, so the smaller similar triangle occupies two thirds. Its linear scale factor is \(\sqrt{\frac{2}{3}}.\) Hence the parallel segment has length \[18\sqrt{\frac23}=6\sqrt6.\]

Therefore, the correct answer is A.

9 答案 E

配方得 \[y=(x-4)^2+c-16\text{。}\]顶点为 \((4,c-16)\)，所以当 \(c=16\) 时，顶点位于 \(x\) 轴上。

因此，正确答案是 E。

English

Completing the square gives \[y=(x-4)^2+c-16.\] The vertex is \((4,c-16),\) so it lies on the \(x\)-axis when \(c=16.\)

Therefore, the correct answer is E.

10 答案 A

有 \(x^2-x-6=(x-3)(x+2)\)。 该乘积恰在两根之间为负，因此 \(-2\lt x\lt3\)。

因此，正确答案是 A。

English

We have \(x^2-x-6=(x-3)(x+2).\) This product is negative exactly between its roots, so \(-2\lt x\lt3.\)

Therefore, the correct answer is A.

11 答案 B

命题 I 将 \(\sqrt a\sqrt b=\sqrt{ab}\) 错误地用于负被开方数；在实数范围内其左边没有定义，而取复数主值时左边等于 \(-8\)， 而不是 \(8\)。 命题 II 和 III 都能正确化简为 \(8\)。 因此只有 I 不正确。

因此，正确答案是 B。

English

Statement I improperly applies \(\sqrt a\sqrt b=\sqrt{ab}\) to negative radicands; over the reals its left side is undefined, and with principal complex roots it equals \(-8,\) not \(8.\) Statements II and III both correctly simplify to \(8.\) Thus only I is incorrect.

Therefore, the correct answer is B.

12 答案 D

以 \(A\) 为原点，并设从原点指向 \(B,C\) 的向量分别为 \(u,v\)。若菱形边长为 \(s\)，则其对角顶点为 \((\frac{s}{12})u+(\frac{s}{6})v\)。当这两个系数之和为 \(1\) 时，该点位于 \(BC\) 上。因此 \[ \frac{s}{12}+\frac{s}{6}=1\text{，}\]从而 \(s=4\)。

因此，正确答案是 D。

English

Use \(A\) as the origin and let the vectors to \(B,C\) be \(u,v.\) If the rhombus side is \(s,\) its opposite vertex is \((\frac{s}{12})u+(\frac{s}{6})v.\) A point lies on \(BC\) when these coefficients sum to \(1.\) Thus \[ \frac{s}{12}+\frac{s}{6}=1,\] giving \(s=4.\)

Therefore, the correct answer is D.

13 答案 E

该直线到原点的距离为 \(\frac{15}{\sqrt{(-3)^2+5^2}}=\frac{15}{\sqrt{34}}\lt4\)。 因此它穿过圆盘 \(x^2+y^2\leq16\) 的内部，交集是一条含有无穷多个点的线段。

因此，正确答案是 E。

English

The line’s distance from the origin is \(\frac{15}{\sqrt{(-3)^2+5^2}}=\frac{15}{\sqrt{34}}\lt4.\) It therefore crosses the interior of the disk \(x^2+y^2\leq16\) in a segment containing infinitely many points.

Thus, the correct answer is E.

14 答案 A

令 \(x=y=1\) 后，每个单项式都等于 \(1\)， 因此所得值就是数值系数之和。这个值为 \[(1-2+1)^7=0\text{。}\]

因此，正确答案是 A。

English

Setting \(x=y=1\) makes every monomial equal \(1,\) so the value is the sum of the numerical coefficients. It is \[(1-2+1)^7=0.\]

Therefore, the correct answer is A.

15 答案 B

用通常的记法，\(25_b=2b+5\)，且 \(52_b=5b+2\)。 因此 \[5b+2=2(2b+5)\text{，}\] 所以 \(b=8\)。

因此，正确答案是 B。

English

In ordinary notation, \(25_b=2b+5\) and \(52_b=5b+2.\) Thus \[5b+2=2(2b+5),\] so \(b=8.\)

Therefore, the correct answer is B.

16 答案 C

取 \(C=(0,0),\ A=(0,30)\) 和 \(E=(30,0)\)。则 \(B=(0,15)\)，且 \(D=(15,0)\)。直线 \(AD\) 和 \(EB\) 是三角形 \(ACE\) 的中线，所以它们交于重心 \(F=(10,10)\)。线段 \(DE\) 长为 \(15\)，点 \(F\) 到其所在直线的距离为 \(10\)。因此 \[[DFE]=\frac12(15)(10)=75\text{。}\]

因此，正确答案是 C。

English

Put \(C=(0,0),\ A=(0,30),\) and \(E=(30,0).\) Then \(B=(0,15)\) and \(D=(15,0).\) Lines \(AD\) and \(EB\) are medians of triangle \(ACE,\) so they meet at the centroid \(F=(10,10).\) Segment \(DE\) has length \(15\) and \(F\) is \(10\) units above its line. Therefore \[[DFE]=\frac12(15)(10)=75.\]

Thus, the correct answer is C.

17 答案 E

该命题的含义是：如果野餐没有举行，那么天气不好。它的逆否命题是：如果天气晴朗，那么野餐会举行。

因此，正确答案是 E。

English

The statement says: if the picnic is not held, then the weather is not fair. Its contrapositive is: if the weather is fair, then the picnic will be held.

Therefore, the correct answer is E.

18 答案 B

误差为 \[\frac1{1+y}-(1-y)=\frac{y^2}{1+y}\text{。}\] 再除以正确值 \(\frac{1}{1+y}\)，所得比值为 \(y^2\)。

因此，正确答案是 B。

English

The error is \[\frac1{1+y}-(1-y)=\frac{y^2}{1+y}.\] Dividing by the correct value \(\frac{1}{1+y}\) gives the ratio \(y^2.\)

Therefore, the correct answer is B.

19 答案 C

由首项可知商必为 \(x+1\)。 展开得 \[ \begin{aligned} &(x+1)(x^3+3x^2+9x+3)\\ &\quad=x^4+4x^3+12x^2+12x+3 \end{aligned}\text{。}\] 因此 \(6p=12,\ 4q=12,\ r=3\)， 所以 \(p=2,\ q=3\)，且 \((p+q)r=15\)。

因此，正确答案是 C。

English

The leading terms force quotient \(x+1.\) Expanding, \[ \begin{aligned} &(x+1)(x^3+3x^2+9x+3)\\ &\quad=x^4+4x^3+12x^2+12x+3. \end{aligned} \] Thus \(6p=12,\ 4q=12,\ r=3,\) so \(p=2,\ q=3\) and \((p+q)r=15.\)

Therefore, the correct answer is C.

20 答案 C

第 \(r\) 项等于相邻两个部分和之差。代入并化简得 \[S_r-S_{r-1}=6r-1\text{。}\]

因此，正确答案是 C。

English

The \(r\)th term is the difference of consecutive partial sums. Substitution and simplification give \[S_r-S_{r-1}=6r-1.\]

Therefore, the correct answer is C.

21 答案 D

该表达式等于 \[ \begin{gathered} \log_{10}\left(\frac{x^2+3}{x^2}\right)\\ =\log_{10}\left(1+\frac3{x^2}\right) \end{gathered}\text{。}\] 它始终为正，但随着 \(x\) 增大而趋近于 \(0\)。因此可以使它小于任意给定的正数。

因此，正确答案是 D。

English

The expression is \[ \begin{gathered} \log_{10}\left(\frac{x^2+3}{x^2}\right)\\ =\log_{10}\left(1+\frac3{x^2}\right). \end{gathered} \] It is always positive, but tends to \(0\) as \(x\) grows. Therefore it can be made smaller than any specified positive number.

Thus, the correct answer is D.

22 答案 A

若 \(a_0\ne0\)，则 \(rs=\frac{a_0}{a_2}\ne0\)，并且 \[ \begin{gathered} a_0\left(1-\frac xr\right) \left(1-\frac xs\right)\\ =\frac{a_0}{rs}(x-r)(x-s)\\ =a_2(x-r)(x-s) \end{gathered} \] 对每个 \(x\) 都等于原多项式。若 \(a_0=0\)，则右边至少有一个分母为零。

因此，正确答案是 A。

English

If \(a_0\ne0,\) then \(rs=\frac{a_0}{a_2}\ne0\) and \[ \begin{gathered} a_0\left(1-\frac xr\right) \left(1-\frac xs\right)\\ =\frac{a_0}{rs}(x-r)(x-s)\\ =a_2(x-r)(x-s), \end{gathered} \] which is the original polynomial for every \(x.\) If \(a_0=0,\) at least one denominator on the right is zero.

Therefore, the correct answer is A.

23 答案 D

当 \(1.99\lt x\lt2.01\) 时，\[ \begin{gathered} |x^2-4|=|x-2||x+2|\\ \lt(0.01)(4.01)=0.0401 \end{gathered}\text{。}\]当 \(x\) 任意接近 \(2.01\) 时，该表达式也任意接近 \(0.0401\)，因此任何更小的上界都不成立。

因此，正确答案是 D。

English

For \(1.99\lt x\lt2.01,\) \[ \begin{gathered} |x^2-4|=|x-2||x+2|\\ \lt(0.01)(4.01)=0.0401. \end{gathered} \] Values with \(x\) arbitrarily close to \(2.01\) make the expression arbitrarily close to \(0.0401,\) so no smaller bound works.

Therefore, the correct answer is D.

24 答案 E

该乘积为 \[ 10^{\frac{1+2+\cdots+n}{11}} =10^{\frac{n(n+1)}{22}}\text{。}\] 要大于 \(100{,}000=10^5\)， 需有 \(n(n+1)\gt110\)。 当 \(n=10\) 时恰好相等，而 \(n=11\) 时超过它。

因此，正确答案是 E。

English

The product is \[ 10^{\frac{1+2+\cdots+n}{11}} =10^{\frac{n(n+1)}{22}}. \] To exceed \(100{,}000=10^5,\) we need \(n(n+1)\gt110.\) At \(n=10\) equality holds, while \(n=11\) exceeds it.

Therefore, the correct answer is E.

25 答案 D

若 \(\angle ACE=90^\circ\)，则 \(AE\) 是直角三角形 \(ACE\) 的斜边。斜边中点 \(B\) 到 \(A,C,E\) 的距离相等。因此 \(AB=BC\)，所以四边形 \(ABCD\) 必定至少有一对相等的边。无法推出任何关于 \(D\) 的条件。

因此，正确答案是 D。

English

If \(\angle ACE=90^\circ,\) then \(AE\) is the hypotenuse of right triangle \(ACE.\) Its midpoint \(B\) is equidistant from \(A,C,E.\) Thus \(AB=BC,\) so quadrilateral \(ABCD\) necessarily has at least one pair of equal sides. No condition involving \(D\) follows.

Therefore, the correct answer is D.

26 答案 E

有 \[m=\frac{2k+3l}{5}\] 且 \[p=\frac{k+l}{2}\text{，}\]所以 \(m-p=\frac{l-k}{10}\)。该差值可能为正、为零或为负，取决于所选各数。因此前四种关系没有一种恒成立。

因此，正确答案是 E。

English

We have \[m=\frac{2k+3l}{5}\] and \[p=\frac{k+l}{2},\] so \(m-p=\frac{l-k}{10}.\) This difference may be positive, zero, or negative depending on the chosen numbers. None of the first four relations always holds.

Therefore, the correct answer is E.

27 答案 A

由余式定理得 \(R_1=1+m+2=m+3\)，且 \(R_2=1-m+2=3-m\)。 两者相等可得 \(m=0\)。

因此，正确答案是 A。

English

The remainder theorem gives \(R_1=1+m+2=m+3\) and \(R_2=1-m+2=3-m.\) Equality implies \(m=0.\)

Therefore, the correct answer is A.

28 答案 B

设 \(Z\) 每分钟走 \(v\) 步，则 \(A\) 每分钟走 \(2v\) 步，并设扶梯每分钟贡献 \(e\) 级台阶。他们的行进时间分别为 \(\frac{27}{2v}\) 和 \(\frac{18}{v}\)。因此 \[ n=27+\frac{27e}{2v} =18+\frac{18e}{v}\text{。}\]解得 \(\frac{e}{v}=2\)，从而 \(n=18+18(2)=54\)。

因此，正确答案是 B。

English

Let \(Z\) take \(v\) steps per minute, so \(A\) takes \(2v,\) and let the escalator contribute \(e\) steps per minute. Their travel times are \(\frac{27}{2v}\) and \(\frac{18}{v}.\) Thus \[ n=27+\frac{27e}{2v} =18+\frac{18e}{v}. \] This gives \(\frac{e}{v}=2,\) and then \(n=18+18(2)=54.\)

Therefore, the correct answer is B.

29 答案 A

设三门都选的人数为 \(t\)，只选数学和英语的人数为 \(x\)，它也等于只选数学的人数。各不相交区域的人数之和为 \[x+x+6+5t+t=28\text{，}\] 所以 \(x=11-3t\)。 因为 \(t\) 为正偶数，所以只有 \(t=2\) 能使所列人数非负，此时 \(x=5\)。

因此，正确答案是 A。

English

Let \(t\) be the number taking all three and \(x\) the number taking Mathematics and English only, which also equals the Mathematics-only count. The disjoint-region total is \[x+x+6+5t+t=28,\] so \(x=11-3t.\) Since \(t\) is positive and even, \(t=2\) is the only value giving a nonnegative listed count, and \(x=5.\)

Therefore, the correct answer is A.

30 答案 B

取 \(C=(0,0),\ A=(a,0)\)， 和 \(B=(0,b)\)。 \(AB\) 与以 \(BC\) 为直径的圆的另一个交点为 \[ D=\left(\frac{ab^2}{a^2+b^2}, \frac{a^2b}{a^2+b^2}\right)\text{。}\] 圆在 \(D\) 点的切线与 \(CA\) 交于 \(F=(\frac{a}{2},0)\)。 因此 \(DF\) 平分 \(CA\)； 又由从 \(F\) 引出的切线段相等，有 \(FC=FD=FA\)，从而得到选项 C 和 E；圆及直角三角形中的角关系又能得到选项 D。

若 \(DF\) 平分 \(\angle CDA\)， 则在三角形 \(CDA\) 中，由角平分线定理必须有 \(\frac{CF}{FA}=\frac{CD}{DA}\)。 左边为 \(1\)， 而 \[ \frac{CD}{DA}=\frac{ab}{a^2}=\frac ba\text{，}\] 它不一定等于 \(1\)。 因此选项 B 无法由已知条件证明。

因此，正确答案是 B。

English

Put \(C=(0,0),\ A=(a,0),\) and \(B=(0,b).\) The second intersection of \(AB\) with the circle of diameter \(BC\) is \[ D=\left(\frac{ab^2}{a^2+b^2}, \frac{a^2b}{a^2+b^2}\right). \] The tangent at \(D\) meets \(CA\) at \(F=(\frac{a}{2},0).\) Hence \(DF\) bisects \(CA;\) also \(FC=FD=FA\) by equal tangents from \(F,\) which supplies choices C and E, and the circle/right-triangle angles supply choice D.

If \(DF\) bisected \(\angle CDA,\) the angle-bisector theorem in triangle \(CDA\) would require \(\frac{CF}{FA}=\frac{CD}{DA}.\) The left side is \(1,\) while \[ \frac{CD}{DA}=\frac{ab}{a^2}=\frac ba, \] which need not be \(1.\) Thus choice B is not provable.

Therefore, the correct answer is B.

31 答案 C

换底得 \[ \frac{\ln x}{\ln a}\frac{\ln x}{\ln b} =\frac{\ln b}{\ln a}\text{。}\] 由于各分母均不为零，\((\ln x)^2=(\ln b)^2\)。 因此 \(x=b\) 或 \(x=\frac{1}{b}\)。 因为 \(b\ne1\)， 这两个值互不相同，所以共有两个值。

因此，正确答案是 C。

English

Change of base gives \[ \frac{\ln x}{\ln a}\frac{\ln x}{\ln b} =\frac{\ln b}{\ln a}. \] Since the denominators are nonzero, \((\ln x)^2=(\ln b)^2.\) Thus \(x=b\) or \(x=\frac{1}{b}.\) These are distinct because \(b\ne1,\) so there are two values.

Therefore, the correct answer is C.

32 答案 C

以 \(\$100\) 售出时，亏损为售价的 \(x\%\)，所以 \(C=100+x\)。转售时，\(S'-100=(\frac{x}{100})S'\)，所以 \(S'=\frac{10000}{100-x}\)。因此 \[ \begin{gathered} \frac{10000}{100-x}-(100+x)\\ =\frac{x^2}{100-x} =\frac{10}{9} \end{gathered}\text{。}\]从而 \(9x^2+10x-1000=0\)，其正根为 \(x=10\)。

因此，正确答案是 C。

English

A loss of \(x\%\) of the \(\$100\) selling price means \(C=100+x.\) On resale, \(S'-100=(\frac{x}{100})S',\) so \(S'=\frac{10000}{100-x}.\) Therefore \[ \begin{gathered} \frac{10000}{100-x}-(100+x)\\ =\frac{x^2}{100-x} =\frac{10}{9}. \end{gathered} \] Thus \(9x^2+10x-1000=0,\) whose positive root is \(x=10.\)

Therefore, the correct answer is C.

33 答案 D

各素因数的次数为 \[ \begin{gathered} v_2(15!)=11,\\ v_3(15!)=6,\\ v_5(15!)=3 \end{gathered}\text{。}\]因此在 \(12\) 进制中 \(k=\min(\lfloor\frac{11}{2}\rfloor,6)=5\)，在 \(10\) 进制中 \(h=\min(11,3)=3\)。所以 \(k+h=8\)。

因此，正确答案是 D。

English

The prime valuations are \[ \begin{gathered} v_2(15!)=11,\\ v_3(15!)=6,\\ v_5(15!)=3. \end{gathered} \] Hence \(k=\min(\lfloor\frac{11}{2}\rfloor,6)=5\) in base \(12,\) and \(h=\min(11,3)=3\) in base \(10.\) Thus \(k+h=8.\)

Therefore, the correct answer is D.

34 答案 B

令 \(t=x+1\geq1\)。因为 \(4x^2+8x+13=4t^2+9\)，所以原式为 \[\frac23t+\frac{3}{2t}\text{。}\]由算术平均数不小于几何平均数，该式至少为 \(2\sqrt{(\frac{2}{3})(\frac{3}{2})}=2\)，且在允许的 \(t=\frac{3}{2}\) 处取等号。

因此，正确答案是 B。

English

Let \(t=x+1\geq1.\) Since \(4x^2+8x+13=4t^2+9,\) the expression is \[\frac23t+\frac{3}{2t}.\] By AM-GM this is at least \(2\sqrt{(\frac{2}{3})(\frac{3}{2})}=2,\) with equality at \(t=\frac{3}{2},\) which is allowed.

Therefore, the correct answer is B.

35 答案 D

将长方形的四个顶点置于 \((0,0)\)、\((5,0)\)、\((5,w)\) 和 \((0,w)\)。使 \((0,0)\) 与 \((5,w)\) 重合的折痕为 \[5x+wy=\frac{25+w^2}{2}\text{。}\]因为 \(w\lt4\)，折痕与两条水平边相交。两个交点间的竖直变化量为 \(w\)，水平变化量为 \(\frac{w^2}{5}\)。因此 \[ 6=w^2+\frac{w^4}{25}\text{。}\]令 \(z=w^2\)，得 \(z^2+25z-150=0\)，所以 \(z=5\)，且 \(w=\sqrt5\)。

因此，正确答案是 D。

English

Place the rectangle at \((0,0),\) \((5,0),\) \((5,w),\) and \((0,w).\) The crease sending \((0,0)\) to \((5,w)\) is \[5x+wy=\frac{25+w^2}{2}.\] Because \(w\lt4,\) it meets the two horizontal sides. Between those intersections the vertical change is \(w\) and the horizontal change is \(\frac{w^2}{5}.\) Thus \[ 6=w^2+\frac{w^4}{25}. \] Setting \(z=w^2\) gives \(z^2+25z-150=0,\) so \(z=5\) and \(w=\sqrt5.\)

Therefore, the correct answer is D.

36 答案 E

每个新的直角三角形都有相同的锐角，所以垂线段长度构成等比数列。由于前两项为 \(a,b\)， 公比为 \(\frac{b}{a}\)。 收敛意味着 \(0\lt \frac{b}{a}\lt1\)， 因而总和为 \[\frac{a}{1-\frac{b}{a}}=\frac{a^2}{a-b}\text{。}\]

因此，正确答案是 E。

English

Each new right triangle has the same acute angle, so the perpendicular lengths form a geometric sequence. Since the first two lengths are \(a,b,\) the common ratio is \(\frac{b}{a}.\) Convergence implies \(0\lt \frac{b}{a}\lt1,\) and the sum is \[\frac{a}{1-\frac{b}{a}}=\frac{a^2}{a-b}.\]

Therefore, the correct answer is E.

37 答案 C

用质量 \(m_A=3\)、\(m_B=1\) 和 \(m_C=2\) 表示边上的比。则 \(m_E=m_A+m_B=4\)，且 \(m_D=m_B+m_C=3\)。沿 \(CE\)，有 \[ \frac{EF}{FC}=\frac{m_C}{m_E}=\frac12\text{，}\]而沿 \(AD\)，有 \[ \frac{AF}{FD}=\frac{m_D}{m_A}=1\text{。}\]两者之和为 \(\frac{3}{2}\)。

因此，正确答案是 C。

English

The side ratios are represented by masses \(m_A=3,\) \(m_B=1,\) and \(m_C=2.\) Then \(m_E=m_A+m_B=4\) and \(m_D=m_B+m_C=3.\) Along \(CE,\) \[ \frac{EF}{FC}=\frac{m_C}{m_E}=\frac12, \] while along \(AD,\) \[ \frac{AF}{FD}=\frac{m_D}{m_A}=1. \] Their sum is \(\frac{3}{2}.\)

Therefore, the correct answer is C.

38 答案 E

设工作效率为 \(a\)、\(b\) 和 \(c\)。由时间关系得 \[ \begin{gathered} b+c=ma,\\ c+a=nb,\\ a+b=xc \end{gathered}\text{。}\]由前两个等式，\[ \frac ba=\frac{m+1}{n+1}\] 且 \[\frac ca=\frac{mn-1}{n+1}\text{。}\]因此 \[ x=\frac{a+b}{c} =\frac{m+n+2}{mn-1}\text{。}\]

因此，正确答案是 E。

English

Let the work rates be \(a,\) \(b,\) and \(c.\) The time statements give \[ \begin{gathered} b+c=ma,\\ c+a=nb,\\ a+b=xc. \end{gathered} \] From the first two, \[ \frac ba=\frac{m+1}{n+1}\] and \[\frac ca=\frac{mn-1}{n+1}.\] Hence \[ x=\frac{a+b}{c} =\frac{m+n+2}{mn-1}. \]

Therefore, the correct answer is E.

39 答案 B

将孔的圆心置于原点。直径为 \(2\) 英寸和 \(1\) 英寸的塞规圆心分别为 \((0,-\frac{1}{2})\) 和 \((0,1)\)。 设一个新量规的半径为 \(r\)，圆心为 \((u,v)\)。 与两个塞规相切给出 \[ \begin{gathered} u^2+(v+\frac{1}{2})^2=(1+r)^2,\\ u^2+(v-1)^2=(\frac{1}{2}+r)^2 \end{gathered}\text{，}\] 而与孔内切给出 \(u^2+v^2=(\frac{3}{2}-r)^2\)。 将方程两两相减得 \(v=\frac12+\frac r3\)，且 \(v=\frac32-2r\)。 因此 \(r=\frac{3}{7}\)， 所以 \(d=2r=\frac{6}{7}\approx0.86\)。

因此，正确答案是 B。

English

Place the hole’s center at the origin. The \(2\)-inch and \(1\)-inch plug centers are \((0,-\frac{1}{2})\) and \((0,1).\) Let a new gauge have radius \(r\) and center \((u,v).\) Tangency to the two plugs gives \[ \begin{gathered} u^2+(v+\frac{1}{2})^2=(1+r)^2,\\ u^2+(v-1)^2=(\frac{1}{2}+r)^2, \end{gathered} \] while internal tangency to the hole gives \(u^2+v^2=(\frac{3}{2}-r)^2.\) Subtracting pairs of equations yields \(v=\frac12+\frac r3\) and \(v=\frac32-2r.\) Thus \(r=\frac{3}{7},\) so \(d=2r=\frac{6}{7}\approx0.86.\)

Therefore, the correct answer is B.

40 答案 D

令 \(N=x^2+3x+1\)。则 \[P=N^2-3(x-10)\text{。}\]当 \(x=10,\ P=N^2=131^2\) 时，有一个值符合条件。

若 \(x\gt10\) 且 \(P=y^2\)，则 \(N^2-y^2=3(x-10)\)。与 \(N^2\) 不同的整数平方和它的差至少为 \(2|N|-1\)，而这已经大于 \(3(x-10)\)，矛盾。若 \(x\lt10\)，则类似的界 \[ \begin{gathered} 3(10-x)=y^2-N^2\\ \geq2|y|-1\gt2|N|-1 \end{gathered} \] 只有在 \(-6\leq x\leq2\) 时才可能成立。直接代入这九个整数，所得值均不是平方数。因此 \(x=10\) 是唯一解，且 \(n=1\)。

因此，正确答案是 D。

English

Let \(N=x^2+3x+1.\) Then \[P=N^2-3(x-10).\] At \(x=10,\ P=N^2=131^2,\) so one value works.

If \(x\gt10\) and \(P=y^2,\) then \(N^2-y^2=3(x-10).\) Distinct integer squares differing from \(N^2\) differ by at least \(2|N|-1,\) which is already greater than \(3(x-10),\) a contradiction. If \(x\lt10,\) the analogous bound \[ \begin{gathered} 3(10-x)=y^2-N^2\\ \geq2|y|-1\gt2|N|-1 \end{gathered} \] can hold only for \(-6\leq x\leq2.\) Direct substitution for these nine integers gives no square. Hence \(x=10\) is the unique solution and \(n=1.\)

Therefore, the correct answer is D.