1970 AMC 12 真题（P版）

1 答案 E

最内层根式为 \(\sqrt1=1\)，所以所给的数是 \(\sqrt{1+\sqrt2}\)。它的四次方为 \[ \begin{aligned} (1+\sqrt2)^2&=1+2\sqrt2+2\\ &=3+2\sqrt2 \end{aligned}\text{。}\]

因此，正确答案是 E。

English

The innermost radical is \(\sqrt1=1,\) so the given number is \(\sqrt{1+\sqrt2}.\) Its fourth power is \[ \begin{aligned} (1+\sqrt2)^2&=1+2\sqrt2+2\\ &=3+2\sqrt2. \end{aligned} \]

Therefore, the correct answer is E.

2 答案 A

若共同的周长为 \(p\)，则圆的半径为 \(\frac{p}{2\pi}\)，正方形的边长为 \(\frac{p}{4}\)。因此 \[ \frac{\pi\left(\frac{p}{2\pi}\right)^2}{\left(\frac p4\right)^2} =\frac{\frac{p^2}{4\pi}}{\frac{p^2}{16}}=\frac4\pi\text{。}\]

因此，正确答案是 A。

English

If the common perimeter is \(p,\) then the circle has radius \(\frac{p}{2\pi}\) and the square has side \(\frac{p}{4}.\) Therefore \[ \frac{\pi\left(\frac{p}{2\pi}\right)^2}{\left(\frac p4\right)^2} =\frac{\frac{p^2}{4\pi}}{\frac{p^2}{16}}=\frac4\pi. \]

Therefore, the correct answer is A.

3 答案 C

由 \(x=1+2^p\) 得 \(2^p=x-1\)，所以 \[ \begin{aligned} y&=1+2^{-p}\\ &=1+\frac1{x-1}=\frac{x}{x-1} \end{aligned}\text{。}\]

因此，正确答案是 C。

English

From \(x=1+2^p\) we get \(2^p=x-1,\) so \[ \begin{aligned} y&=1+2^{-p}\\ &=1+\frac1{x-1}=\frac{x}{x-1}. \end{aligned} \]

Therefore, the correct answer is C.

4 答案 B

这个和为 \[ \begin{gathered} (n-1)^2+n^2+(n+1)^2\\ =3n^2+2 \end{gathered}\text{。}\]它总是同余于 \(2\pmod3\)，所以 \(S\) 中没有元素能被 \(3\) 整除。另一方面，取 \(n=5\) 时得到 \(3(25)+2=77\)，它能被 \(11\) 整除。

因此，正确答案是 B。

English

The sum is \[ \begin{gathered} (n-1)^2+n^2+(n+1)^2\\ =3n^2+2. \end{gathered} \] It is always congruent to \(2\pmod3,\) so no member of \(S\) is divisible by \(3.\) On the other hand, taking \(n=5\) gives \(3(25)+2=77,\) which is divisible by \(11.\)

Therefore, the correct answer is B.

5 答案 D

因为 \(i^2=-1\)，且 \(i^4=1\)，所以 \[ f(i)=\frac{i^4+i^2}{i+1}=\frac{1-1}{i+1}=0\text{。}\]

因此，正确答案是 D。

English

Because \(i^2=-1\) and \(i^4=1,\) \[ f(i)=\frac{i^4+i^2}{i+1}=\frac{1-1}{i+1}=0. \]

Therefore, the correct answer is D.

6 答案 B

配方可得 \[ x^2+8x=(x+4)^2-16\text{。}\]因为 \((x+4)^2\ge0\)，所以当 \(x=-4\) 时取得最小值 \(-16\)。

因此，正确答案是 B。

English

Completing the square gives \[ x^2+8x=(x+4)^2-16. \] Since \((x+4)^2\ge0,\) the minimum occurs at \(x=-4\) and equals \(-16.\)

Therefore, the correct answer is B.

7 答案 E

因为 \(AX=BX=AB=s\)，所以三角形 \(ABX\) 是等边三角形。因此 \(X\) 到 \(AB\) 的垂直距离为 \(\frac{s\sqrt3}{2}\)。平行边 \(AB\) 与 \(CD\) 之间的距离为 \(s\)，所以所求距离为 \[ s-\frac{s\sqrt3}{2}=\frac s2(2-\sqrt3)\text{。}\]

因此，正确答案是 E。

English

Since \(AX=BX=AB=s,\) triangle \(ABX\) is equilateral. Thus the perpendicular distance from \(X\) to \(AB\) is \(\frac{s\sqrt3}{2}.\) The distance between the parallel sides \(AB\) and \(CD\) is \(s,\) so the requested distance is \[ s-\frac{s\sqrt3}{2}=\frac s2(2-\sqrt3). \]

Therefore, the correct answer is E.

8 答案 B

利用 \(225=15^2\) 和 \(8=2^3\)，可得 \[ \begin{aligned} a&=\frac{\log_2(15^2)}{\log_2(2^3)}\\ &=\frac{2\log_2 15}{3}=\frac{2b}{3} \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Using \(225=15^2\) and \(8=2^3,\) \[ \begin{aligned} a&=\frac{\log_2(15^2)}{\log_2(2^3)}\\ &=\frac{2\log_2 15}{3}=\frac{2b}{3}. \end{aligned} \]

Therefore, the correct answer is B.

9 答案 C

由分割比可得 \[ AP=\frac25AB,\qquad AQ=\frac37AB\text{。}\]因为两点位于中点的同一侧，所以 \[ \begin{aligned} PQ&=AQ-AP\\ &=\left(\frac37-\frac25\right)AB\\ &=\frac1{35}AB \end{aligned}\text{。}\]由 \(PQ=2\) 得 \(AB=70\)。

因此，正确答案是 C。

English

The division ratios give \[ AP=\frac25AB,\qquad AQ=\frac37AB. \] Because the two points are on the same side of the midpoint, \[ \begin{aligned} PQ&=AQ-AP\\ &=\left(\frac37-\frac25\right)AB\\ &=\frac1{35}AB. \end{aligned} \] Since \(PQ=2,\) we obtain \(AB=70.\)

Therefore, the correct answer is C.

10 答案 D

我们有 \[ \begin{aligned} F&=0.4+0.081(1+0.01+\cdots)\\ &=\frac25+\frac{0.081}{0.99}\\ &=\frac25+\frac9{110}\\ &=\frac{53}{110} \end{aligned}\text{。}\]分母比分子大 \(110-53=57\)。

因此，正确答案是 D。

English

We have \[ \begin{aligned} F&=0.4+0.081(1+0.01+\cdots)\\ &=\frac25+\frac{0.081}{0.99}\\ &=\frac25+\frac9{110}\\ &=\frac{53}{110}. \end{aligned} \] The denominator exceeds the numerator by \(110-53=57.\)

Therefore, the correct answer is D.

11 答案 E

由因式定理可得 \[ \begin{aligned} -16+2h+k&=0,\\ 2-h+k&=0 \end{aligned}\text{。}\]用第一个方程减去第二个方程，得到 \(3h=18\)，所以 \(h=6\)，进而 \(k=4\)。因此 \[ |2h-3k|=|12-12|=0\text{。}\]

因此，正确答案是 E。

English

The factor theorem gives \[ \begin{aligned} -16+2h+k&=0,\\ 2-h+k&=0. \end{aligned} \] Subtracting the second equation from the first gives \(3h=18,\) so \(h=6\) and then \(k=4.\) Hence \[ |2h-3k|=|12-12|=0. \]

Therefore, the correct answer is E.

12 答案 C

圆与平行边 \(AB\) 和 \(CD\) 相切。两边之间的距离就是圆的直径，所以 \(AD=BC=2r\)。设 \(R\) 为 \(AD\) 上的切点，\(Q\) 为圆心，\(M\) 为 \(AC\) 的中点。

直线 \(RQ\) 平行于 \(AB\)，并且位于 \(AB\) 与 \(CD\) 的正中间，所以 \(R\) 是 \(AD\) 的中点，且该直线经过 \(M\)。由于 \(M\) 在圆上，且 \(R,Q,M\) 共线，所以 \(RM=2r\)。因此 \(RM\) 是三角形 \(ADC\) 的中位线，从而 \(DC=2RM=4r\)。所以面积为 \[ (2r)(4r)=8r^2\text{。}\]

因此，正确答案是 C。

English

Let the circle be tangent to the parallel sides \(AB\) and \(CD.\) Their separation is its diameter, so \(AD=BC=2r.\) Let \(R\) be the tangency point on \(AD,\) let \(Q\) be the center, and let \(M\) be the midpoint of \(AC.\)

The line \(RQ\) is parallel to \(AB\) and halfway between \(AB\) and \(CD,\) so \(R\) is the midpoint of \(AD\) and the line contains \(M.\) Since \(M\) lies on the circle and \(R,Q,M\) are collinear, \(RM=2r.\) Thus \(RM\) is a midsegment of triangle \(ADC,\) so \(DC=2RM=4r.\) Therefore the area is \[ (2r)(4r)=8r^2. \]

Therefore, the correct answer is C.

13 答案 D

对于选项 D， \[ \begin{aligned} (a*b)^n&=(a^b)^n=a^{bn},\\ a*(bn)&=a^{bn} \end{aligned}\text{。}\]因此这个恒等式总是成立。其他选项会断言诸如 \(a^b=b^a\) 或 \(a^{b^c}=a^{bc}\) 之类并非恒成立的等式。

因此，正确答案是 D。

English

For choice D, \[ \begin{aligned} (a*b)^n&=(a^b)^n=a^{bn},\\ a*(bn)&=a^{bn}. \end{aligned} \] Thus that identity always holds. The other choices would assert false general identities such as \(a^b=b^a\) or \(a^{b^c}=a^{bc}.\)

Therefore, the correct answer is D.

14 答案 A

两根之差的绝对值为 \[ \sqrt{p^2-4q}\text{。}\]因此 \(p^2-4q=1\)，所以 \(p^2=4q+1\)。因为 \(p\) 为正数，故 \[ p=\sqrt{4q+1}\text{。}\]

因此，正确答案是 A。

English

The two roots differ in absolute value by \[ \sqrt{p^2-4q}. \] Hence \(p^2-4q=1,\) so \(p^2=4q+1.\) Since \(p\) is positive, \[ p=\sqrt{4q+1}. \]

Therefore, the correct answer is A.

15 答案 E

从 \((-4,5)\) 到 \((5,-1)\) 的位移为 \((9,-6)\)。因此两个三等分点为 \[ (-4,5)+\tfrac13(9,-6)=(-1,3) \] 和 \((2,1)\)。经过 \((-1,3)\) 和 \((3,4)\) 的直线斜率为 \(\frac{1}{4}\)，所以 \[ y-4=\frac14(x-3)\text{，}\]即 \(x-4y+13=0\)。

因此，正确答案是 E。

English

The displacement from \((-4,5)\) to \((5,-1)\) is \((9,-6).\) The trisection points are therefore \[ (-4,5)+\tfrac13(9,-6)=(-1,3) \] and \((2,1).\) The line through \((-1,3)\) and \((3,4)\) has slope \(\frac{1}{4},\) so \[ y-4=\frac14(x-3), \] or \(x-4y+13=0.\)

Therefore, the correct answer is E.

16 答案 C

依次应用递推式，得到 \[ \begin{aligned} F(4)&=\frac{1\cdot1+1}{1}=2,\\ F(5)&=\frac{2\cdot1+1}{1}=3 \end{aligned}\text{，}\]以及 \[ F(6)=\frac{3\cdot2+1}{1}=7\text{。}\]

因此，正确答案是 C。

English

Applying the recurrence successively, \[ \begin{aligned} F(4)&=\frac{1\cdot1+1}{1}=2,\\ F(5)&=\frac{2\cdot1+1}{1}=3, \end{aligned} \] and \[ F(6)=\frac{3\cdot2+1}{1}=7. \]

Therefore, the correct answer is C.

17 答案 E

因为 \(r\gt0\)，题设条件等价于 \(p\gt q\)。于是 \(-p\lt-q\)，所以 A 错误。取 \((p,q)=(2,1)\) 可知 B 和 D 错误。取 \((p,q)=(1,-2)\) 时，\(-\frac{q}{p}=2\)，所以 C 错误。因此 A 至 D 中没有一个结论在所有情况下都成立。

因此，正确答案是 E。

English

Since \(r\gt0,\) the hypothesis is equivalent to \(p\gt q.\) Then \(-p\lt-q,\) so A is false. Taking \((p,q)=(2,1)\) makes B and D false. Taking \((p,q)=(1,-2)\) makes C false because \(-\frac{q}{p}=2.\) Thus none of A-D follows in every case.

Therefore, the correct answer is E.

18 答案 A

因为 \[ \begin{gathered} 3+2\sqrt2=(\sqrt2+1)^2,\\ 3-2\sqrt2=(\sqrt2-1)^2 \end{gathered}\text{，}\]且 \(\sqrt2+1\) 和 \(\sqrt2-1\) 都为正数，所以这个差为 \[ (\sqrt2+1)-(\sqrt2-1)=2\text{。}\]

因此，正确答案是 A。

English

Since \[ \begin{gathered} 3+2\sqrt2=(\sqrt2+1)^2,\\ 3-2\sqrt2=(\sqrt2-1)^2, \end{gathered} \] and both \(\sqrt2+1\) and \(\sqrt2-1\) are positive, the difference is \[ (\sqrt2+1)-(\sqrt2-1)=2. \]

Therefore, the correct answer is A.

19 答案 C

设首项为 \(a\)。则 \[ \frac{a}{1-r}=15,\qquad \frac{a^2}{1-r^2}=45\text{。}\]因为 \(1-r^2=(1-r)(1+r)\)，用第二个方程除以第一个方程，得到 \(\frac{a}{1+r}=3\)。因此 \[ a=15(1-r)=3(1+r)\text{。}\]解得 \(r=\frac{2}{3}\)，且 \(a=5\)。

因此，正确答案是 C。

English

Let the first term be \(a.\) Then \[ \frac{a}{1-r}=15,\qquad \frac{a^2}{1-r^2}=45. \] Since \(1-r^2=(1-r)(1+r),\) dividing the second equation by the first gives \(\frac{a}{1+r}=3.\) Thus \[ a=15(1-r)=3(1+r). \] Solving gives \(r=\frac{2}{3}\) and \(a=5.\)

Therefore, the correct answer is C.

20 答案 A

选取坐标 \[ \begin{gathered} H=(h,0),\quad B=(h,u),\\ K=(k,0),\quad C=(k,v) \end{gathered}\text{。}\]则 \[ M=\left(\frac{h+k}{2},\frac{u+v}{2}\right)\text{。}\]从 \(M\) 到 \(H\) 与 \(K\) 的水平位移互为相反数，而竖直位移相等。因此 \(MH^2=MK^2\)，所以总有 \(MH=MK\)。

因此，正确答案是 A。

English

Choose coordinates \[ \begin{gathered} H=(h,0),\quad B=(h,u),\\ K=(k,0),\quad C=(k,v). \end{gathered} \] Then \[ M=\left(\frac{h+k}{2},\frac{u+v}{2}\right). \] The horizontal displacements from \(M\) to \(H\) and \(K\) are opposites, while the vertical displacements are equal. Therefore \(MH^2=MK^2,\) so \(MH=MK\) always.

Therefore, the correct answer is A.

21 答案 B

实际距离固定时，里程表读数与车轮转数成正比。因此 \[ 450r_1=440r_2, \qquad \frac{r_2}{r_1}=\frac{45}{44}\text{。}\]由 \(r_1=15\) 可知，增加的半径为 \[ \begin{aligned} r_2-r_1 &=15\left(\frac{45}{44}-1\right)\\ &=\frac{15}{44}\\ &\approx0.3409 \end{aligned}\text{。}\]精确到百分位是 \(0.34\) 英寸。

因此，正确答案是 B。

English

For a fixed true distance, the odometer reading is proportional to the number of wheel revolutions. Thus \[ 450r_1=440r_2, \qquad \frac{r_2}{r_1}=\frac{45}{44}. \] With \(r_1=15,\) the increase is \[ \begin{aligned} r_2-r_1 &=15\left(\frac{45}{44}-1\right)\\ &=\frac{15}{44}\\ &\approx0.3409. \end{aligned} \] To the nearest hundredth this is \(0.34\) inch.

Therefore, the correct answer is B.

22 答案 A

设 \(S_m=\frac{m(m+1)}{2}\)。则 \[ \begin{aligned} 2(S_{3n}-S_n) &=3n(3n+1)\\ &\quad{}-n(n+1)\\ &=8n^2+2n \end{aligned}\text{。}\]由于 \(S_{3n}-S_n=150\)，可得 \(4n^2+n=150\)。因此 \((n-6)(4n+25)=0\)，其正数解为 \(n=6\)。所以 \[ S_{4n}=S_{24}=\frac{24\cdot25}{2}=300\text{。}\]

因此，正确答案是 A。

English

Let \(S_m=\frac{m(m+1)}{2}.\) Then \[ \begin{aligned} 2(S_{3n}-S_n) &=3n(3n+1)\\ &\quad{}-n(n+1)\\ &=8n^2+2n. \end{aligned} \] Since \(S_{3n}-S_n=150,\) this gives \(4n^2+n=150.\) Hence \((n-6)(4n+25)=0,\) and the positive solution is \(n=6.\) Therefore \[ S_{4n}=S_{24}=\frac{24\cdot25}{2}=300. \]

Therefore, the correct answer is A.

23 答案 D

\(10!\) 中各质因数的指数为 \[ \begin{gathered} v_2(10!)=5+2+1=8,\\ v_3(10!)=3+1=4 \end{gathered}\text{。}\]每个因子 \(12=2^2\cdot3\) 需要两个因数 \(2\) 和一个因数 \(3\)。因此 \(12\) 进制表示的末尾零个数为 \[ \min\left(\left\lfloor\frac82\right\rfloor,4\right)=4\text{。}\]

因此，正确答案是 D。

English

The prime exponents in \(10!\) are \[ \begin{gathered} v_2(10!)=5+2+1=8,\\ v_3(10!)=3+1=4. \end{gathered} \] Each factor of \(12=2^2\cdot3\) uses two factors of \(2\) and one of \(3.\) Hence the number of trailing base-\(12\) zeros is \[ \min\left(\left\lfloor\frac82\right\rfloor,4\right)=4. \]

Therefore, the correct answer is D.

24 答案 B

设六边形的边长为 \(s\)。它的周长为 \(6s\)，所以等边三角形的边长为 \(2s\)。大三角形由 \(4\) 个边长为 \(s\) 的等边三角形组成，而六边形由 \(6\) 个这样的三角形组成。因此它们的面积之比为 \(\frac{6}{4}=\frac{3}{2}\)。由于三角形的面积为 \(2\)，所以六边形的面积为 \(3\)。

因此，正确答案是 B。

English

Let the hexagon side be \(s.\) Its perimeter is \(6s,\) so the equilateral triangle has side \(2s.\) The large triangle consists of \(4\) equilateral triangles of side \(s,\) while the hexagon consists of \(6.\) Their area ratio is therefore \(\frac{6}{4}=\frac{3}{2}.\) Since the triangle’s area is \(2,\) the hexagon’s area is \(3.\)

Therefore, the correct answer is B.

25 答案 E

每盎司或不足一盎司均计费，意味着计费的盎司数为 \(\lceil W\rceil\)。向下取整函数和向上取整函数满足 \[ \lceil W\rceil=-\lfloor-W\rfloor\text{。}\]因此邮资为 \[ 6\lceil W\rceil=-6\lfloor-W\rfloor\text{。}\]

因此，正确答案是 E。

English

Charging for every ounce or portion thereof means the number of charged ounces is \(\lceil W\rceil.\) The floor and ceiling functions satisfy \[ \lceil W\rceil=-\lfloor-W\rfloor. \] Thus the cost is \[ 6\lceil W\rceil=-6\lfloor-W\rfloor. \]

Therefore, the correct answer is E.

26 答案 B

第一个图形是以下两条直线的并集：\[ x+y=5,\qquad 2x-3y=-5\text{，}\]这两条直线交于 \((2,3)\)。第二个图形是以下两条直线的并集：\[ x-y=-1,\qquad 3x+2y=12\text{，}\]它们也交于 \((2,3)\)。事实上，将 \((2,3)\) 代入可知，它满足四条直线的方程。由于四条直线的斜率各不相同，不会有其他点同时属于两个图形。因此恰有一个公共点。

因此，正确答案是 B。

English

The first graph is the union of \[ x+y=5,\qquad 2x-3y=-5, \] and those two lines meet at \((2,3).\) The second graph is the union of \[ x-y=-1,\qquad 3x+2y=12, \] and these also meet at \((2,3).\) In fact, substituting \((2,3)\) satisfies all four line equations. Since the four lines have distinct slopes, no other point can belong to a line from each graph. There is exactly one common point.

Therefore, the correct answer is B.

27 答案 A

若 \(r\) 为内切圆半径，\(s\) 为半周长，则三角形的面积为 \(K=rs\)，周长为 \(2s\)。由题设等式可得 \[ rs=2s\text{。}\]因为 \(s\gt0\)，所以 \(r=2\)。

因此，正确答案是 A。

English

If \(r\) is the inradius and \(s\) the semiperimeter, the triangle’s area is \(K=rs.\) Its perimeter is \(2s.\) The given equality yields \[ rs=2s. \] Since \(s\gt0,\) we obtain \(r=2.\)

Therefore, the correct answer is A.

28 答案 A

将重心置于原点。设 \(A\) 和 \(B\) 的位置向量分别为 \(u\) 和 \(v\)。两条中线分别沿 \(u\) 和 \(v\) 的方向，所以 \(u\cdot v=0\)。由重心条件还可得 \(C=-u-v\)。

设 \(U=|u|^2\)，且 \(V=|v|^2\)。则 \[ AC^2=|2u+v|^2=4U+V=36 \] 以及 \[ \begin{aligned} BC^2&=|u+2v|^2\\ &=U+4V=49 \end{aligned}\text{。}\]将两个方程取适当倍数后相加，得到 \(U+V=17\)。最后， \[ AB^2=|u-v|^2=U+V=17\text{，}\]所以 \(AB=\sqrt{17}\)。

因此，正确答案是 A。

English

Place the centroid at the origin. Let the position vectors of \(A\) and \(B\) be \(u\) and \(v.\) The two medians lie along \(u\) and \(v,\) so \(u\cdot v=0.\) Also the centroid condition gives \(C=-u-v.\)

Let \(U=|u|^2\) and \(V=|v|^2.\) Then \[ AC^2=|2u+v|^2=4U+V=36 \] and \[ \begin{aligned} BC^2&=|u+2v|^2\\ &=U+4V=49. \end{aligned} \] Adding appropriate multiples gives \(U+V=17.\) Finally, \[ AB^2=|u-v|^2=U+V=17, \] so \(AB=\sqrt{17}.\)

Therefore, the correct answer is A.

29 答案 D

设现在是 \(10{:}00\) 之后 \(x\) 分钟。六分钟后，分针从 \(12\) 顺时针走过 \(x+6\) 个分钟刻度。三分钟前，时针越过 \(10\) 的位置 \(\frac{x-3}{12}\) 个分钟刻度，所以它的正对面位于 \[ 20+\frac{x-3}{12} \] 即从 \(12\) 起经过这么多个分钟刻度。因此 \[ x+6=20+\frac{x-3}{12}\text{。}\]解得 \(x=15\)，所以时间是 \(10{:}15\)。

因此，正确答案是 D。

English

Let \(x\) be the number of minutes after \(10{:}00\) now. Six minutes from now, the minute hand is \(x+6\) minute spaces clockwise from \(12.\) Three minutes ago, the hour hand was \(\frac{x-3}{12}\) minute spaces past the \(10,\) so the point opposite it was \[ 20+\frac{x-3}{12} \] minute spaces past \(12.\) Hence \[ x+6=20+\frac{x-3}{12}. \] Solving gives \(x=15,\) so the time is \(10{:}15.\)

Therefore, the correct answer is D.

30 答案 E

设 \(\angle D\) 的角平分线与 \(AB\) 交于 \(P\)。因为 \(\angle D=2\angle B\)，所以每一半都等于 \(\angle B\)。又因为 \(AB\parallel CD\)，内错角 \(\angle APD\) 也等于 \(\angle PDC\)。因此三角形 \(APD\) 为等腰三角形，并且 \[ AP=AD=a\text{。}\]此外，\(PB\parallel CD\)，且 \(BC\parallel PD\)，所以 \(PBCD\) 是平行四边形。因此 \(PB=CD=b\)。所以 \[ AB=AP+PB=a+b\text{。}\]

因此，正确答案是 E。

English

Let the bisector of \(\angle D\) meet \(AB\) at \(P.\) Because \(\angle D=2\angle B,\) each half has measure \(\angle B.\) Since \(AB\parallel CD,\) the alternate interior angle \(\angle APD\) also equals \(\angle PDC.\) Thus triangle \(APD\) is isosceles and \[ AP=AD=a. \] Also \(PB\parallel CD\) and \(BC\parallel PD,\) so \(PBCD\) is a parallelogram. Hence \(PB=CD=b.\) Therefore \[ AB=AP+PB=a+b. \]

Therefore, the correct answer is E.

31 答案 B

五位数最大的数位和为 \(45\)，所以数位和为 \(43\) 的数相对于 \(99999\) 总共减少 \(2\)。一种情况是一个数位为 \(7\)，其余数位为 \(9\)，共有 \(5\) 个数；另一种情况是两个数位为 \(8\)，其余数位为 \(9\)，共有 \(\binom52=10\) 个数。因此总共有 \(15\) 个数。

利用 \(11\) 的整除判别法，恰有 \[ 97999,\qquad 99979,\qquad 98989 \] 能被 \(11\) 整除。因此概率为 \(\frac{3}{15}=\frac{1}{5}\)。

因此，正确答案是 B。

English

The maximum five-digit digit sum is \(45,\) so a sum of \(43\) has total deficit \(2\) from \(99999.\) Either one digit is \(7\) and the others are \(9\), giving \(5\) numbers, or two digits are \(8\) and the others are \(9\), giving \(\binom52=10\) numbers. Thus there are \(15\) numbers total.

Using the divisibility-by-\(11\) test, exactly \[ 97999,\qquad 99979,\qquad 98989 \] are divisible by \(11.\) The probability is therefore \(\frac{3}{15}=\frac{1}{5}.\)

Therefore, the correct answer is B.

32 答案 C

设跑道周长为 \(2C\)。第一次相遇时，\(B\) 已行进 \(100\) 码，\(A\) 已行进 \(C-100\)，所以他们的速度之比为 \[ \frac{v_A}{v_B}=\frac{C-100}{100}\text{。}\]第二次相遇时，\(A\) 距离完成一圈还差 \(60\) 码，所以它已行进 \(2C-60\)。此时两人的路程之和为 \(3C\)，所以 \(B\) 已行进 \(C+60\)。因此 \[ \frac{C-100}{100}=\frac{2C-60}{C+60}\text{。}\]交叉相乘可得 \(C=240\)，所以跑道周长为 \(2C=480\) 码。

因此，正确答案是 C。

English

Let the circumference be \(2C.\) At the first meeting, \(B\) has traveled \(100\) yards and \(A\) has traveled \(C-100,\) so their speed ratio is \[ \frac{v_A}{v_B}=\frac{C-100}{100}. \] At the second meeting, \(A\) is \(60\) yards short of one lap, so it has traveled \(2C-60.\) By then their combined distance is \(3C,\) so \(B\) has traveled \(C+60.\) Hence \[ \frac{C-100}{100}=\frac{2C-60}{C+60}. \] Cross-multiplying gives \(C=240,\) so the circumference is \(2C=480\) yards.

Therefore, the correct answer is C.

33 答案 A

将从 \(0\) 到 \(9999\) 的整数写成带前导零的四位数。在四个数位中的每一个位置，数字 \(0\)、\(1\)、\(\ldots\)、\(9\) 都出现 \(1000\) 次。因此所有数位的总和为 \[ \begin{gathered} 4\cdot1000(0+1+\cdots+9)\\ =4000\cdot45\\ =180000 \end{gathered}\text{。}\]用 \(10000\) 取代 \(0\) 会增加 \(1\)，所以所求的和为 \(180001\)。

因此，正确答案是 A。

English

Write the integers from \(0\) through \(9999\) using four digits with leading zeros. In each of the four positions, every digit \(0,\) \(1,\) \(\ldots,\) \(9\) appears \(1000\) times. Thus their total digit sum is \[ \begin{gathered} 4\cdot1000(0+1+\cdots+9)\\ =4000\cdot45\\ =180000. \end{gathered} \] Replacing \(0\) by \(10000\) adds \(1,\) so the requested sum is \(180001.\)

Therefore, the correct answer is A.

34 答案 C

一个除数除这三个数所得余数相同，当且仅当它能整除这些数之间的差。相邻两个差为 \[ \begin{gathered} 13903-13511=392,\\ 14589-13903=686 \end{gathered}\text{。}\]因此最大的可能除数为 \[ \begin{gathered} \gcd(392,686)\\ =\gcd(392,294)\\ =\gcd(294,98)=98 \end{gathered}\text{。}\]

因此，正确答案是 C。

English

A divisor leaves the same remainder on all three numbers exactly when it divides their differences. The two successive differences are \[ \begin{gathered} 13903-13511=392,\\ 14589-13903=686. \end{gathered} \] Therefore the greatest possible divisor is \[ \begin{gathered} \gcd(392,686)\\ =\gcd(392,294)\\ =\gcd(294,98)=98. \end{gathered} \]

Therefore, the correct answer is C.

35 答案 D

设当前养老金为 \(X=k\sqrt n\)。因为 \(X+p=k\sqrt{n+a}\)，且 \(X+q=k\sqrt{n+b}\)，分别平方并利用 \(X^2=k^2n\)，得到下面前两个方程。再将它们分别乘以 \(b\) 和 \(a\)，相减后得到第三个方程：\[ \begin{gathered} 2pX+p^2=k^2a,\\ 2qX+q^2=k^2b,\\ 2X(bp-aq)=aq^2-bp^2 \end{gathered}\text{。}\]因此 \(X=\dfrac{aq^2-bp^2}{2(bp-aq)}\)。

因此，正确答案是 D。

English

Let the current pension be \(X=k\sqrt n.\) Since \(X+p=k\sqrt{n+a}\) and \(X+q=k\sqrt{n+b},\) squaring and using \(X^2=k^2n\) gives the first line below. Multiplying its equations by \(b\) and \(a,\) respectively, and subtracting gives the second: \[ \begin{gathered} 2pX+p^2=k^2a,\\ 2qX+q^2=k^2b,\\ 2X(bp-aq)=aq^2-bp^2. \end{gathered} \] Hence \(X=\dfrac{aq^2-bp^2}{2(bp-aq)}.\)

Therefore, the correct answer is D.