1972 AMC 12 真题（P版）

1 答案 D

对于三角形 \(\mathrm{I}\)、\(\mathrm{II}\) 和 \(\mathrm{III}\)，相应等式为 \[ \begin{aligned} 3^2+4^2&=5^2,\\ 4^2+\left(\frac{15}{2}\right)^2 &=\left(\frac{17}{2}\right)^2,\\ 7^2+24^2&=25^2 \end{aligned}\text{。}\]对于 \(\mathrm{IV}\)，\[ \begin{aligned} \left(\frac72\right)^2+\left(\frac92\right)^2 &=\frac{130}{4},\\ \left(\frac{11}{2}\right)^2&=\frac{121}{4} \end{aligned}\text{。}\]两个值并不相等。因此，恰有 \(\mathrm{I}\)、\(\mathrm{II}\) 和 \(\mathrm{III}\) 是直角三角形。

因此，正确答案是 D。

English

For triangles \(\mathrm{I},\) \(\mathrm{II},\) and \(\mathrm{III},\) the relevant equalities are \[ \begin{aligned} 3^2+4^2&=5^2,\\ 4^2+\left(\frac{15}{2}\right)^2 &=\left(\frac{17}{2}\right)^2,\\ 7^2+24^2&=25^2. \end{aligned} \] For \(\mathrm{IV},\) \[ \begin{aligned} \left(\frac72\right)^2+\left(\frac92\right)^2 &=\frac{130}{4},\\ \left(\frac{11}{2}\right)^2&=\frac{121}{4}. \end{aligned} \] These values are unequal. Thus precisely \(\mathrm{I},\) \(\mathrm{II},\) and \(\mathrm{III}\) are right triangles.

Therefore, the correct answer is D.

2 答案 B

由售价保持不变可得 \[ \begin{aligned} 1+0.01x&=0.92\\ &\quad+0.0092(x+10) \end{aligned}\text{。}\]化简得 \[ 1+0.01x=1.012+0.0092x\text{，}\]所以 \(0.0008x=0.012\)，且 \(x=15\)。

因此，正确答案是 B。

English

The fixed selling price gives \[ \begin{aligned} 1+0.01x&=0.92\\ &\quad+0.0092(x+10). \end{aligned} \] Simplifying yields \[ 1+0.01x=1.012+0.0092x, \] so \(0.0008x=0.012\) and \(x=15.\)

Therefore, the correct answer is B.

3 答案 B

直接计算，\[ \begin{aligned} x^2-x &=\left(\frac{1-i\sqrt3}{2}\right)^2\\ &\quad-\frac{1-i\sqrt3}{2}\\ &=-1 \end{aligned}\text{。}\]因此它的倒数也是 \(-1\)。

因此，正确答案是 B。

English

Directly, \[ \begin{aligned} x^2-x &=\left(\frac{1-i\sqrt3}{2}\right)^2\\ &\quad-\frac{1-i\sqrt3}{2}\\ &=-1. \end{aligned} \] Its reciprocal is therefore also \(-1.\)

Therefore, the correct answer is B.

4 答案 D

每个符合条件的 \(X\) 都包含 \(1,2\)，而 \(3,4,5\) 中的每个元素都可独立选择。因此共有 \[ 2^3=8 \] 个可能的集合。

因此，正确答案是 D。

English

Every valid \(X\) contains \(1,2,\) while each of \(3,4,5\) may be chosen independently. Hence there are \[ 2^3=8 \] possible sets.

Therefore, the correct answer is D.

5 答案 A

取便于比较的相同幂次，可得 \[ (3^{\frac{1}{3}})^6=9>8=(2^{\frac{1}{2}})^6\text{，}\]\[ (2^{\frac{1}{2}})^8=16>8=(8^{\frac{1}{8}})^8\text{，}\]以及 \[ (2^{\frac{1}{2}})^{18}=512>81=(9^{\frac{1}{9}})^{18}\text{。}\]因此，\(3^{\frac{1}{3}}\) 最大，\(2^{\frac{1}{2}}\) 第二大。

因此，正确答案是 A。

English

Raising to convenient common powers gives \[ (3^{\frac{1}{3}})^6=9>8=(2^{\frac{1}{2}})^6, \] \[ (2^{\frac{1}{2}})^8=16>8=(8^{\frac{1}{8}})^8, \] and \[ (2^{\frac{1}{2}})^{18}=512>81=(9^{\frac{1}{9}})^{18}. \] Thus \(3^{\frac{1}{3}}\) is greatest and \(2^{\frac{1}{2}}\) is next.

Therefore, the correct answer is A.

6 答案 C

令 \(u=3^x\)。则 \[ \begin{aligned} u^2-10u+9&=(u-1)(u-9)\\ &=0 \end{aligned}\text{。}\]因此 \(x=0\) 或 \(x=2\)，所以 \(x^2+1\) 分别为 \(1\) 或 \(5\)。

因此，正确答案是 C。

English

Let \(u=3^x.\) Then \[ \begin{aligned} u^2-10u+9&=(u-1)(u-9)\\ &=0. \end{aligned} \] Thus \(x=0\) or \(x=2,\) so \(x^2+1\) is respectively \(1\) or \(5.\)

Therefore, the correct answer is C.

7 答案 E

所求比的比值为 \[ \frac{\frac{x}{yz}}{\frac{y}{zx}}=\frac{x^2}{y^2}\text{。}\]又因为 \(yz:zx=y:x=1:2\)，所以 \(x:y=2:1\)。因此 \(x^2:y^2=4:1\)。

因此，正确答案是 E。

English

The requested ratio has quotient \[ \frac{\frac{x}{yz}}{\frac{y}{zx}}=\frac{x^2}{y^2}. \] Also \(yz:zx=y:x=1:2,\) so \(x:y=2:1.\) Therefore \(x^2:y^2=4:1.\)

Therefore, the correct answer is E.

8 答案 D

若 \(x-\log y\ge0\)，由原方程得 \(\log y=0\)，所以 \(y=1\)。若 \(x-\log y\lt0\)，则得 \(x=0\)。因此每个解都满足 \(x=0\) 或 \(y=1\)，即 \[ x(y-1)=0\text{。}\]

因此，正确答案是 D。

English

If \(x-\log y\ge0,\) the equation gives \(\log y=0,\) hence \(y=1.\) If \(x-\log y\lt0,\) it gives \(x=0.\) Therefore every solution satisfies either \(x=0\) or \(y=1,\) which is expressed by \[ x(y-1)=0. \]

Therefore, the correct answer is D.

9 答案 A

设每盒有 \(S\) 张纸和 \(E\) 个信封。两人的使用情况给出 \(S-E=50\) 和 \(E-\frac S3=50\)。两式相加得 \(\frac{2S}{3}=100\)，所以 \(S=150\)。

因此，正确答案是 A。

English

Let \(S\) and \(E\) denote sheets and envelopes per box. The two accounts give \(S-E=50\) and \(E-\frac S3=50.\) Adding yields \(\frac{2S}{3}=100,\) so \(S=150.\)

Therefore, the correct answer is A.

10 答案 D

条件 \(\lvert x-2\rvert\le7\) 给出 \(-5\le x\le9\)。条件 \(\lvert x-2\rvert\ge1\) 给出 \(x\le1\) 或 \(x\ge3\)。二者的交集为 \[ [-5,1]\cup[3,9]\text{。}\]

因此，正确答案是 D。

English

The condition \(\lvert x-2\rvert\le7\) gives \(-5\le x\le9.\) The condition \(\lvert x-2\rvert\ge1\) gives \(x\le1\) or \(x\ge3.\) Their intersection is \[ [-5,1]\cup[3,9]. \]

Therefore, the correct answer is D.

11 答案 A

由第二个方程得 \(x^2=3y-12\)。代入第一个方程，得 \[ \begin{aligned} y^2+3y-28&=(y-4)(y+7)\\ &=0 \end{aligned}\text{。}\]若 \(y=4\)，则 \(x=0\)。若 \(y=-7\)，则 \(x^2=-33\)，不存在实数 \(x\)。因此只有 \(y=4\) 可行。

因此，正确答案是 A。

English

From the second equation, \(x^2=3y-12.\) Substitution into the first gives \[ \begin{aligned} y^2+3y-28&=(y-4)(y+7)\\ &=0. \end{aligned} \] If \(y=4,\) then \(x=0.\) If \(y=-7,\) then \(x^2=-33,\) so no real \(x\) exists. Thus only \(y=4\) is possible.

Therefore, the correct answer is A.

12 答案 B

设棱长为 \(f\) 英尺，即 \(12f\) 英寸。题述的数值相等关系为 \[ f^3=6(12f)^2\text{。}\]因为 \(f\gt0\)，两边除以 \(f^2\) 得 \[ f=6\cdot144=864\text{。}\]

因此，正确答案是 B。

English

Let the edge length be \(f\) feet, or \(12f\) inches. The stated numerical equality is \[ f^3=6(12f)^2. \] Since \(f\gt0,\) division by \(f^2\) gives \[ f=6\cdot144=864. \]

Therefore, the correct answer is B.

13 答案 C

因为 \(M\) 是 \(AE\) 的中点，所以它到 \(AD\) 的水平距离为 \(\frac{5}{2}\)。它到 \(BC\) 的距离为 \[ 12-\frac52=\frac{19}{2}\text{。}\]过 \(M\) 的水平线在直线 \(PQ\) 两侧截出的两个直角三角形相似，因此它们的斜边之比等于相应水平直角边之比：\[ PM:MQ=\frac52:\frac{19}{2}=5:19\text{。}\]

因此，正确答案是 C。

English

Because \(M\) is the midpoint of \(AE,\) its horizontal distance from \(AD\) is \(\frac{5}{2}.\) Its distance from \(BC\) is \[ 12-\frac52=\frac{19}{2}. \] The two right triangles cut from line \(PQ\) by the horizontal through \(M\) are similar, so their hypotenuses are in the ratio of these horizontal legs: \[ PM:MQ=\frac52:\frac{19}{2}=5:19. \]

Therefore, the correct answer is C.

14 答案 B

设 \(30^\circ\) 角所对的边长为 \(s\)，由正弦定理得 \[ \frac{s}{\sin30^\circ}=\frac8{\sin45^\circ}\text{。}\]因此 \[ s=8\frac{\frac{1}{2}}{\frac{\sqrt2}{2}}=4\sqrt2\text{。}\]

因此，正确答案是 B。

English

If \(s\) is the side opposite \(30^\circ,\) the Law of Sines gives \[ \frac{s}{\sin30^\circ}=\frac8{\sin45^\circ}. \] Therefore \[ s=8\frac{\frac{1}{2}}{\frac{\sqrt2}{2}}=4\sqrt2. \]

Therefore, the correct answer is B.

15 答案 C

若墙中有 \(N\) 块砖，两人的单独效率分别为 \(\frac{N}{9}\) 和 \(\frac{N}{10}\)。因此 \[ 5\left(\frac N9+\frac N{10}-10\right)=N\text{。}\]两边乘以 \(18\) 并化简，得 \(19N-900=18N\)，故 \(N=900\)。

因此，正确答案是 C。

English

If the wall contains \(N\) bricks, the individual rates are \(\frac{N}{9}\) and \(\frac{N}{10}.\) Thus \[ 5\left(\frac N9+\frac N{10}-10\right)=N. \] Multiplying by \(18\) and simplifying gives \(19N-900=18N,\) hence \(N=900.\)

Therefore, the correct answer is C.

16 答案 B

由数列条件得 \(x^2=3y\) 和 \(2y=x+9\)。消去 \(y\)，得 \[ \begin{aligned} 2x^2-3x-27&=(2x-9)(x+3)\\ &=0 \end{aligned}\text{。}\]由正数条件得 \(x=\frac{9}{2}\)，进而 \(y=\frac{27}{4}\)。两数之和为 \[ \frac92+\frac{27}{4}=\frac{45}{4}=11\frac14\text{。}\]

因此，正确答案是 B。

English

The progression conditions give \(x^2=3y\) and \(2y=x+9.\) Eliminating \(y\) yields \[ \begin{aligned} 2x^2-3x-27&=(2x-9)(x+3)\\ &=0. \end{aligned} \] Positivity gives \(x=\frac{9}{2},\) and then \(y=\frac{27}{4}.\) Their sum is \[ \frac92+\frac{27}{4}=\frac{45}{4}=11\frac14. \]

Therefore, the correct answer is B.

17 答案 E

令绳长为 \(1\)。在靠近一端的区间内，条件为 \[ 1-t\ge xt\text{，}\]所以 \(t\le\frac{1}{x+1}\)。另一端也有同样长度的区间。两区间总长度即所求概率，为 \[ \frac2{x+1}\text{。}\]

因此，正确答案是 E。

English

Take the string to have length \(1.\) Near one endpoint, the condition is \[ 1-t\ge xt, \] so \(t\le\frac{1}{x+1}.\) The same interval occurs at the other endpoint. Their total length, hence the probability, is \[ \frac2{x+1}. \]

Therefore, the correct answer is E.

18 答案 A

因为 \(AB\parallel DC\)，所以三角形 \(ABE\) 与 \(CDE\) 相似。因此 \[ AE:EC=AB:DC=2:1\text{。}\]从而 \(AC=AE+EC=3EC=11\)，所以 \[ EC=\frac{11}{3}=3\frac23\text{。}\]

因此，正确答案是 A。

English

Since \(AB\parallel DC,\) triangles \(ABE\) and \(CDE\) are similar. Hence \[ AE:EC=AB:DC=2:1. \] Thus \(AC=AE+EC=3EC=11,\) so \[ EC=\frac{11}{3}=3\frac23. \]

Therefore, the correct answer is A.

19 答案 D

第 \(k\) 项是等比数列之和 \(2^k-1\)。因此所求总和为 \[ \begin{aligned} \sum_{k=1}^n(2^k-1) &=(2^{n+1}-2)-n\\ &=2^{n+1}-n-2 \end{aligned}\text{。}\]

因此，正确答案是 D。

English

The \(k\)-th term is the geometric sum \(2^k-1.\) Therefore the desired total is \[ \begin{aligned} \sum_{k=1}^n(2^k-1) &=(2^{n+1}-2)-n\\ &=2^{n+1}-n-2. \end{aligned} \]

Therefore, the correct answer is D.

20 答案 E

取一个对边为 \(2ab\)、邻边为 \(a^2-b^2\) 的直角三角形。其斜边为 \[ \begin{aligned} &\sqrt{(2ab)^2+(a^2-b^2)^2}\\ &\qquad=\sqrt{(a^2+b^2)^2}\\ &\qquad=a^2+b^2 \end{aligned}\text{。}\]因此 \[ \sin x=\frac{2ab}{a^2+b^2}\text{。}\]

因此，正确答案是 E。

English

Use a right triangle with opposite leg \(2ab\) and adjacent leg \(a^2-b^2.\) Its hypotenuse is \[ \begin{aligned} &\sqrt{(2ab)^2+(a^2-b^2)^2}\\ &\qquad=\sqrt{(a^2+b^2)^2}\\ &\qquad=a^2+b^2. \end{aligned} \] Hence \[ \sin x=\frac{2ab}{a^2+b^2}. \]

Therefore, the correct answer is E.

21 答案 C

令 \(P=AD\cap BF\)，且 \(Q=AD\cap CE\)。在四边形 \(EFPQ\) 中，\[ E+F+\angle P+\angle Q=360^\circ\text{。}\]利用 \(P\) 和 \(Q\) 处的补角，三角形 \(BPD\) 与 \(AQC\) 分别给出 \(B+D=\angle P\) 和 \(A+C=\angle Q\)，相加得 \[ \begin{gathered} A+B+C+D+E+F\\ =360^\circ=90^\circ\cdot4 \end{gathered}\text{。}\]因此 \(n=4\)。

因此，正确答案是 C。

English

Let \(P=AD\cap BF\) and \(Q=AD\cap CE.\) In quadrilateral \(EFPQ,\) \[ E+F+\angle P+\angle Q=360^\circ. \] The triangles \(BPD\) and \(AQC\) give \(B+D=\angle P\) and \(A+C=\angle Q,\) using the supplementary angles at \(P\) and \(Q.\) Adding yields \[ \begin{gathered} A+B+C+D+E+F\\ =360^\circ=90^\circ\cdot4. \end{gathered} \] Thus \(n=4.\)

Therefore, the correct answer is C.

22 答案 E

共轭根为 \(a-bi\)。因为三个根之和为 \(0\)，第三个根是 \(-2a\)。由韦达定理，\[ \begin{aligned} q&=(a+bi)(a-bi)\\ &\quad-2a(a+bi)\\ &\quad-2a(a-bi)\\ &=a^2+b^2-4a^2\\ &=b^2-3a^2 \end{aligned}\text{。}\]

因此，正确答案是 E。

English

The conjugate root is \(a-bi.\) Since the sum of the three roots is \(0,\) the third root is \(-2a.\) By Vieta’s formulas, \[ \begin{aligned} q&=(a+bi)(a-bi)\\ &\quad-2a(a+bi)\\ &\quad-2a(a-bi)\\ &=a^2+b^2-4a^2\\ &=b^2-3a^2. \end{aligned} \]

Therefore, the correct answer is E.

23 答案 D

把底边中点置于 \(O=(0,0)\)。一个下方外角点为 \(A=(1,0)\)，一个上方外角点为 \(B=(\frac{1}{2},2)\)。由对称性，最小外接圆的圆心为 \(P=(0,k)\)。在最优位置，两类外角点都在圆上，所以 \[ 1+k^2=\left(\frac12\right)^2+(2-k)^2\text{。}\]解得 \(k=\frac{13}{16}\)。因此 \[ \begin{aligned} r^2&=1+\left(\frac{13}{16}\right)^2 =\frac{425}{256},\\ r&=\frac{5\sqrt{17}}{16} \end{aligned}\text{。}\]

因此，正确答案是 D。

English

Put the midpoint of the bottom edge at \(O=(0,0).\) A lower outer corner is \(A=(1,0),\) and an upper outer corner is \(B=(\frac{1}{2},2).\) By symmetry the center of the smallest enclosing circle is \(P=(0,k).\) At the optimum both types of outer corner lie on the circle, so \[ 1+k^2=\left(\frac12\right)^2+(2-k)^2. \] This gives \(k=\frac{13}{16}.\) Hence \[ \begin{aligned} r^2&=1+\left(\frac{13}{16}\right)^2 =\frac{425}{256},\\ r&=\frac{5\sqrt{17}}{16}. \end{aligned} \]

Therefore, the correct answer is D.

24 答案 B

设实际速度和时间为 \(v\) 和 \(t\)。较快的情形给出 \[ vt=\left(v+\frac12\right)\frac{4t}{5}\text{，}\]所以 \(v=2\)。较慢的情形于是给出 \[ 2t=\frac32\left(t+\frac52\right)\text{，}\]从而 \(t=\frac{15}{2}\)。步行距离为 \[ vt=2\cdot\frac{15}{2}=15\text{。}\]

因此，正确答案是 B。

English

Let the actual speed and time be \(v\) and \(t.\) The faster case gives \[ vt=\left(v+\frac12\right)\frac{4t}{5}, \] so \(v=2.\) The slower case then gives \[ 2t=\frac32\left(t+\frac52\right), \] whence \(t=\frac{15}{2}.\) The distance was \[ vt=2\cdot\frac{15}{2}=15. \]

Therefore, the correct answer is B.

25 答案 C

可将连续四边标记为 \[ \begin{gathered} AB=25,\\ BC=39,\\ CD=52,\\ DA=60 \end{gathered}\text{。}\]因为 \[ 25^2+60^2=65^2=39^2+52^2\text{，}\]三角形 \(ABD\) 与 \(BCD\) 都是直角三角形，且有公共斜边 \(BD=65\)。二者拼成这个圆内接四边形，而直角三角形的斜边是其外接圆的直径。因此直径为 \(65\)。

因此，正确答案是 C。

English

The consecutive sides may be labeled \[ \begin{gathered} AB=25,\\ BC=39,\\ CD=52,\\ DA=60. \end{gathered} \] Since \[ 25^2+60^2=65^2=39^2+52^2, \] triangles \(ABD\) and \(BCD\) are right triangles with common hypotenuse \(BD=65.\) Their union is the cyclic quadrilateral, and a right triangle’s hypotenuse is a diameter of its circumcircle. Thus the diameter is \(65.\)

Therefore, the correct answer is C.

26 答案 E

在弧 \(\overset{\frown}{MB}\) 上取点 \(N\)，使弧 \(\overset{\frown}{MN}\) 与 \(\overset{\frown}{CA}\) 相等，并作 \(NQ\perp AB\)。于是 \(MN=AC=x\)。其余两段等弧 \(\overset{\frown}{AM}\) 与 \(\overset{\frown}{NB}\) 在弦上的对应投影相等，所以 \(QB=AP=x+1\)。又因为 \(MNPQ\) 是矩形，故 \(PQ=MN=x\)。因此 \[ \begin{aligned} PB&=PQ+QB\\ &=x+(x+1)\\ &=2x+1 \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Choose \(N\) on arc \(\overset{\frown}{MB}\) so that arcs \(\overset{\frown}{MN}\) and \(\overset{\frown}{CA}\) are equal, and drop \(NQ\perp AB.\) Then \(MN=AC=x.\) The remaining equal arcs \(\overset{\frown}{AM}\) and \(\overset{\frown}{NB}\) give equal corresponding chord projections, so \(QB=AP=x+1.\) Also \(MNPQ\) is a rectangle, hence \(PQ=MN=x.\) Therefore \[ \begin{aligned} PB&=PQ+QB\\ &=x+(x+1)\\ &=2x+1. \end{aligned} \]

Therefore, the correct answer is E.

27 答案 D

几何平均数条件给出 \[ AB\cdot AC=12^2=144\text{。}\]因此 \[ \begin{aligned} 64&=\frac12(AB)(AC)\sin A\\ &=72\sin A \end{aligned}\text{，}\]所以 \(\sin A=\frac{8}{9}\)。

因此，正确答案是 D。

English

The geometric-mean condition says \[ AB\cdot AC=12^2=144. \] Therefore \[ \begin{aligned} 64&=\frac12(AB)(AC)\sin A\\ &=72\sin A, \end{aligned} \] so \(\sin A=\frac{8}{9}.\)

Therefore, the correct answer is D.

28 答案 E

边界上的 \(28\) 个小方格没有被完全覆盖。考虑剩余的 \(6\times6\) 内部网格，并以小方格边长为距离单位，则圆盘半径为 \(4\)。该内部网格的四个外角到中心的距离为 \[ \sqrt{3^2+3^2}=3\sqrt2\gt4 \] 所以四个角方格未被完全覆盖。其他每个内部方格都在圆盘内：其最远角点到中心的距离至多为 \[ \sqrt{3^2+2^2}=\sqrt{13}\lt4\text{。}\]因此共有 \(36-4=32\) 个方格被完全覆盖。

因此，正确答案是 E。

English

The \(28\) border squares are not fully covered. Consider the remaining \(6\times6\) interior grid and measure distances in square side lengths, so the disc has radius \(4.\) The four outer corners of this interior grid are at distance \[ \sqrt{3^2+3^2}=3\sqrt2\gt4 \] from the center, so those four corner squares are not fully covered. Every other interior square lies within the disc: its farthest possible corner is at distance at most \[ \sqrt{3^2+2^2}=\sqrt{13}\lt4. \] Hence \(36-4=32\) squares are completely covered.

Therefore, the correct answer is E.

29 答案 C

令 \(u=\frac{3x+x^3}{1+3x^2}\)。则 \[ \begin{aligned} \frac{1+u}{1-u} &=\frac{1+3x^2+3x+x^3} {1+3x^2-3x-x^3}\\ &=\left(\frac{1+x}{1-x}\right)^3 \end{aligned}\text{。}\]所以 \[ f(u)=3\log\left(\frac{1+x}{1-x}\right)=3f(x)\text{。}\]

因此，正确答案是 C。

English

Let \(u=\frac{3x+x^3}{1+3x^2}.\) Then \[ \begin{aligned} \frac{1+u}{1-u} &=\frac{1+3x^2+3x+x^3} {1+3x^2-3x-x^3}\\ &=\left(\frac{1+x}{1-x}\right)^3. \end{aligned} \] Consequently \[ f(u)=3\log\left(\frac{1+x}{1-x}\right)=3f(x). \]

Therefore, the correct answer is C.

30 答案 A

设纸张的高为 \(h\)。把折起的角关于折痕反射所产生的全等直角三角形给出 \[ \frac6h=\sin(2\theta)=2\sin\theta\cos\theta\text{，}\]所以 \(h=\frac{3}{\sin\theta\cos\theta}\)。它们还给出 \(\frac{L}{h}=\sec\theta\)。因此 \[ \begin{aligned} L&=h\sec\theta\\ &=\frac{3\sec\theta} {\sin\theta\cos\theta}\\ &=3\sec^2\theta\csc\theta \end{aligned}\text{。}\]

因此，正确答案是 A。

English

Let \(h\) be the sheet’s height. The congruent right triangles created by reflecting the folded corner across the crease give \[ \frac6h=\sin(2\theta)=2\sin\theta\cos\theta, \] so \(h=\frac{3}{\sin\theta\cos\theta}.\) They also give \(\frac{L}{h}=\sec\theta.\) Hence \[ \begin{aligned} L&=h\sec\theta\\ &=\frac{3\sec\theta} {\sin\theta\cos\theta}\\ &=3\sec^2\theta\csc\theta. \end{aligned} \]

Therefore, the correct answer is A.

31 答案 C

因为 \(2^6=64\equiv-1\pmod{13}\)，\[ \begin{aligned} 2^{1000}&=2^{6\cdot166+4}\\ &\equiv(-1)^{166}2^4\\ &\equiv16\\ &\equiv3\pmod{13} \end{aligned}\text{。}\]

因此，正确答案是 C。

English

Since \(2^6=64\equiv-1\pmod{13},\) \[ \begin{aligned} 2^{1000}&=2^{6\cdot166+4}\\ &\equiv(-1)^{166}2^4\\ &\equiv16\\ &\equiv3\pmod{13}. \end{aligned} \]

Therefore, the correct answer is C.

32 答案 B

由相交弦定理，\[ 2\cdot6=CE\cdot3\text{，}\]所以 \(CE=4\)。令 \(E=(0,0)\)、\(A=(-2,0)\)、\(B=(6,0)\)、\(C=(0,4)\) 且 \(D=(0,-3)\)。弦 \(AB\) 与 \(CD\) 的垂直平分线交于 \[ O=\left(2,\frac12\right)\text{。}\]因此 \[ r^2=OA^2=4^2+\left(\frac12\right)^2=\frac{65}{4}\text{，}\]直径为 \(2r=\sqrt{65}\)。

因此，正确答案是 B。

English

The intersecting-chords theorem gives \[ 2\cdot6=CE\cdot3, \] so \(CE=4.\) Put \(E=(0,0),\) \(A=(-2,0),\) \(B=(6,0),\) \(C=(0,4),\) and \(D=(0,-3).\) The perpendicular bisectors of \(AB\) and \(CD\) meet at \[ O=\left(2,\frac12\right). \] Thus \[ r^2=OA^2=4^2+\left(\frac12\right)^2=\frac{65}{4}, \] and the diameter is \(2r=\sqrt{65}.\)

Therefore, the correct answer is B.

33 答案 C

设这个商为 \(Q\)。则 \[ Q=1+\frac{99H+9T}{H+T+U}\text{。}\]若 \(U\) 小于前两位中的某个数字，交换二者会使商减小，所以个位数字必须最大。因为这个商大于 \(1\)，增大个位数字会使商减小，故 \(U=9\)。此时 \[ \frac{T+11H}{T+H+9} =1+\frac{10H-9}{T+H+9}\text{。}\]取最大的可用数字 \(T=8\)，再取最小的 \(H=1\)，即可使其最小。这个数是 \(189\)，并且 \[ \frac{189}{1+8+9}=\frac{189}{18}=10.5\text{。}\]

因此，正确答案是 C。

English

Let \(Q\) denote the quotient. Then \[ Q=1+\frac{99H+9T}{H+T+U}. \] Interchanging \(U\) with a larger digit in either earlier position decreases the quotient, so the units digit must be the largest. Increasing that units digit lowers a quotient greater than \(1,\) so \(U=9.\) Then \[ \frac{T+11H}{T+H+9} =1+\frac{10H-9}{T+H+9}. \] This is minimized by taking the largest available \(T=8\) and then the smallest \(H=1.\) The number is \(189,\) and \[ \frac{189}{1+8+9}=\frac{189}{18}=10.5. \]

Therefore, the correct answer is C.

34 答案 A

方程为 \[ \begin{aligned} 3D+T&=2H,\\ 2H^3&=3D^3+T^3 \end{aligned}\text{。}\]将它们改写为 \[ 2(H-D)=D+T \] 和 \[ \begin{gathered} 2(H-D)(H^2+HD+D^2)\\ =(D+T)\\ \qquad\cdot(D^2-DT+T^2) \end{gathered}\text{。}\]约去相等的正因子，得 \[ \begin{aligned} H^2+HD+DT-T^2&=0,\\ (H+T)(H+D-T)&=0 \end{aligned}\text{，}\]所以 \(T=H+D\)。一次方程进一步给出 \(H=4D\)，两两互质条件迫使 \[ (D,H,T)=(1,4,5)\text{。}\]所求平方和为 \(1^2+4^2+5^2=42\)。

因此，正确答案是 A。

English

The equations are \[ \begin{aligned} 3D+T&=2H,\\ 2H^3&=3D^3+T^3. \end{aligned} \] Rewrite them as \[ 2(H-D)=D+T \] and \[ \begin{gathered} 2(H-D)(H^2+HD+D^2)\\ =(D+T)\\ \qquad\cdot(D^2-DT+T^2). \end{gathered} \] Canceling the equal positive factors gives \[ \begin{aligned} H^2+HD+DT-T^2&=0,\\ (H+T)(H+D-T)&=0, \end{aligned} \] so \(T=H+D.\) The linear equation then gives \(H=4D,\) and pairwise relative primality forces \[ (D,H,T)=(1,4,5). \] The requested sum is \(1^2+4^2+5^2=42.\)

Therefore, the correct answer is A.

35 答案 D

沿正方形完成一圈需要 \(8\) 次转轴，并使三角形的方向改变一整圈的 \(\frac{2}{3}\)。因此，整个三角形恢复初始位置需要 \(3\) 圈，即 \(24\) 次转轴。其中有 \(8\) 次绕 \(P\) 旋转，所以 \(P\) 不动。其余 \(16\) 次中，八段圆弧所对圆心角为 \(120^\circ\)，另八段为 \(30^\circ\)，半径均为 \(2\)。因此路径总长为 \[ \begin{aligned} &8\left(\frac13\cdot4\pi\right) +8\left(\frac1{12}\cdot4\pi\right)\\ &\qquad=\frac{32\pi}{3}+\frac{8\pi}{3}\\ &\qquad=\frac{40\pi}{3} \end{aligned}\text{。}\]

因此，正确答案是 D。

English

One circuit of the square uses \(8\) pivots and changes the triangle’s orientation by \(\frac{2}{3}\) of a full turn. Therefore \(3\) circuits, or \(24\) pivots, are required to restore the entire triangle. In \(8\) of those pivots the rotation is about \(P,\) so \(P\) does not move. In the other \(16,\) eight arcs subtend \(120^\circ\) and eight subtend \(30^\circ,\) all with radius \(2.\) Hence the total path length is \[ \begin{aligned} &8\left(\frac13\cdot4\pi\right) +8\left(\frac1{12}\cdot4\pi\right)\\ &\qquad=\frac{32\pi}{3}+\frac{8\pi}{3}\\ &\qquad=\frac{40\pi}{3}. \end{aligned} \]

Therefore, the correct answer is D.