1992 AMC 12 真题（P版）

1 答案 B

由 \(8x+10\pi=2(4x+5\pi)\)，可得 \[ \begin{gathered} 6(8x+10\pi)\\ {}=12(4x+5\pi)\\ {}=4\bigl[3(4x+5\pi)\bigr]\\ {}=4P \end{gathered}\text{。}\]

因此正确答案是 B。

English

We have \(8x+10\pi=2(4x+5\pi),\) so \[ \begin{gathered} 6(8x+10\pi)\\ {}=12(4x+5\pi)\\ {}=4\bigl[3(4x+5\pi)\bigr]\\ {}=4P. \end{gathered} \]

Thus the correct answer is B.

2 答案 B

硬币占全部物品的 \(80\%\)，其中 \(60\%\) 的硬币是金制的。因此金制硬币所占的比例为 \[ 0.80\cdot0.60=0.48=48\%\text{，}\] 这里是相对全部物品而言的。

因此正确答案是 B。

English

Coins make up \(80\%\) of the objects, and \(60\%\) of the coins are gold. Thus gold coins make up \[ 0.80\cdot0.60=0.48=48\% \] of all the objects.

Thus the correct answer is B.

3 答案 C

由斜率条件得 \[ \frac{m-3}{1-m}=m\text{。}\]因此 \(m-3=m-m^2\)，从而 \(m^2=3\)。因为 \(m\gt0\)，所以 \(m=\sqrt3\)。

因此正确答案是 C。

English

The slope condition gives \[ \frac{m-3}{1-m}=m. \] Therefore \(m-3=m-m^2,\) so \(m^2=3.\) Since \(m\gt0,\) we obtain \(m=\sqrt3.\)

Thus the correct answer is C.

4 答案 A

\(3^a\) 是奇数。由于 \(b-1\) 是偶数，所以对于任意正整数 \(c\)，\((b-1)^2c\) 都是偶数。奇数与偶数之和总是奇数。

因此正确答案是 A。

English

The power \(3^a\) is odd. Since \(b-1\) is even, \((b-1)^2c\) is even for every positive integer \(c.\) The sum of an odd integer and an even integer is always odd.

Thus the correct answer is A.

5 答案 B

提取公因式可得 \(6\cdot6^6=6^{1+6}=6^7\)。

因此正确答案是 B。

English

Factoring gives \(6\cdot6^6=6^{1+6}=6^7.\)

Thus the correct answer is B.

6 答案 D

合并同底数幂可得 \[ \frac{x^y y^x}{y^y x^x} =x^{y-x}y^{x-y} =\left(\frac{x}{y}\right)^{y-x}\text{。}\]

因此正确答案是 D。

English

Combining like bases, \[ \frac{x^y y^x}{y^y x^x} =x^{y-x}y^{x-y} =\left(\frac{x}{y}\right)^{y-x}. \]

Thus the correct answer is D.

7 答案 B

将已知比值相乘，得到 \[ \begin{aligned} \frac wy &=\frac wx\cdot\frac xz\cdot\frac zy\\ &=\frac43\cdot6\cdot\frac23=\frac{16}{3} \end{aligned}\text{。}\]因此 \(w:y=16:3\)。

因此正确答案是 B。

English

Multiplying the given ratios, \[ \begin{aligned} \frac wy &=\frac wx\cdot\frac xz\cdot\frac zy\\ &=\frac43\cdot6\cdot\frac23=\frac{16}{3}. \end{aligned} \] Thus \(w:y=16:3.\)

Thus the correct answer is B.

8 答案 E

当 \(n\) 为奇数时，每条对角线含有 \(n\) 块瓷砖，两条对角线只共用中央的一块。因此 \(2n-1=101\)，所以 \(n=51\)。地面共有 \(51^2=2601\) 块瓷砖。

因此正确答案是 E。

English

For an odd \(n,\) each diagonal contains \(n\) tiles and the two diagonals share only the center tile. Hence \(2n-1=101,\) so \(n=51.\) The floor therefore contains \(51^2=2601\) tiles.

Thus the correct answer is E.

9 答案 E

每个大三角形的面积为 \[ \frac{\sqrt3}{4}(2\sqrt3)^2=3\sqrt3\text{。}\]四个重叠部分都是边长为 \(\sqrt3\) 的等边三角形，面积均为 \(\frac{3\sqrt3}{4}\)。不存在三个三角形共同重叠的区域，所以并集的面积为 \[ 5(3\sqrt3)-4\left(\frac{3\sqrt3}{4}\right)=12\sqrt3\text{。}\]

因此正确答案是 E。

English

Each large triangle has area \[ \frac{\sqrt3}{4}(2\sqrt3)^2=3\sqrt3. \] Each of the four overlaps is an equilateral triangle of side \(\sqrt3,\) hence area \(\frac{3\sqrt3}{4}.\) There are no triple overlaps, so the union has area \[ 5(3\sqrt3)-4\left(\frac{3\sqrt3}{4}\right)=12\sqrt3. \]

Thus the correct answer is E.

10 答案 D

解得 \(x=3+\frac{12}{k}\)。因此，当且仅当 \(k\) 是 \(12\) 的正因数时，\(x\) 才是整数。可能的取值为 \(1\)、\(2\)、\(3\)、\(4\)、\(6\) 和 \(12\)，共 \(6\) 个。

因此正确答案是 D。

English

Solving gives \(x=3+\frac{12}{k}.\) Thus \(x\) is an integer precisely when \(k\) is a positive divisor of \(12.\) The possibilities are \(1,\) \(2,\) \(3,\) \(4,\) \(6,\) and \(12,\) for a total of \(6.\)

Thus the correct answer is D.

11 答案 B

设 \(O\) 为共同的圆心，\(T\) 为 \(\overline{BC}\) 上的切点。由于 \(OT\perp BC\)，且 \(\angle ABC=90^\circ\)，三角形 \(OTC\) 与 \(ABC\) 相似。若大圆半径为 \(R\)，则 \(OT=\frac{R}{3}\)、\(OC=R\)，且 \(AC=2R\)。因此 \[ \frac{OT}{AB}=\frac{OC}{AC}=\frac12\text{。}\]所以 \(OT=6\)，并且 \(R=3OT=18\)。

因此正确答案是 B。

English

Let \(O\) be the common center and \(T\) the tangency point on \(\overline{BC}.\) Since \(OT\perp BC\) and \(\angle ABC=90^\circ,\) triangles \(OTC\) and \(ABC\) are similar. If the larger radius is \(R,\) then \(OT=\frac{R}{3},\) \(OC=R,\) and \(AC=2R.\) Thus \[ \frac{OT}{AB}=\frac{OC}{AC}=\frac12. \] Hence \(OT=6\) and \(R=3OT=18.\)

Thus the correct answer is B.

12 答案 C

点 \((x,y)\) 在对称后的直线上，当且仅当 \((x,-y)\) 在原直线上。因此 \(x+3y+11=0\)，即 \[ y=-\frac13x-\frac{11}{3}\text{。}\]所以 \(m+b=-\frac13-\frac{11}{3}=-4\)。

因此正确答案是 C。

English

A point \((x,y)\) is on the reflected line exactly when \((x,-y)\) is on the original line. Hence \(x+3y+11=0,\) or \[ y=-\frac13x-\frac{11}{3}. \] Therefore \(m+b=-\frac13-\frac{11}{3}=-4.\)

Thus the correct answer is C.

13 答案 C

将分子和分母同时乘以 \(ab\)，得到 \[ \frac{a^2b+a}{b+ab^2} =\frac{a(ab+1)}{b(ab+1)} =\frac ab\text{。}\]因此 \(a=13b\)。条件 \(a+b\le100\) 变为 \(14b\le100\)，所以 \(b\) 可以取从 \(1\) 到 \(7\) 的任一整数，共有 \(7\) 对。

因此正确答案是 C。

English

Multiplying the numerator and denominator by \(ab\) gives \[ \frac{a^2b+a}{b+ab^2} =\frac{a(ab+1)}{b(ab+1)} =\frac ab. \] Thus \(a=13b.\) The condition \(a+b\le100\) becomes \(14b\le100,\) so \(b\) may be any integer from \(1\) through \(7.\) There are \(7\) pairs.

Thus the correct answer is C.

14 答案 E

方程 I 表示整条直线 \(y=x-2\)。方程 II 只有在 \(x\ne-2\) 时才能化简为该直线，因此少了点 \((-2,-4)\)。在方程 III 中，每个满足 \(x=-2\) 的点都使等式 \(0=0\) 成立；对其他 \(x\)，方程给出 \(y=x-2\)。所以 III 表示该直线与整条竖直直线 \(x=-2\) 的并集。三个图像各不相同。

因此正确答案是 E。

English

Equation I is the whole line \(y=x-2.\) Equation II simplifies to that line only when \(x\ne-2,\) so it omits \((-2,-4).\) In equation III, every point with \(x=-2\) satisfies \(0=0;\) for other \(x\) it gives \(y=x-2.\) Thus III is the line together with the entire vertical line \(x=-2.\) The three graphs are different.

Thus the correct answer is E.

15 答案 B

前几项为 \[ \begin{aligned} z_1&=0,\quad z_2=i,\quad z_3=-1+i,\\ z_4&=-i,\quad z_5=-1+i \end{aligned}\text{。}\]因此从 \(z_3\) 起，奇数下标项为 \(-1+i\)，偶数下标项为 \(-i\)，两者交替出现。所以 \(z_{111}=-1+i\)，它到原点的距离为 \(\sqrt{(-1)^2+1^2}=\sqrt2\)。

因此正确答案是 B。

English

The first terms are \[ \begin{aligned} z_1&=0,\quad z_2=i,\quad z_3=-1+i,\\ z_4&=-i,\quad z_5=-1+i. \end{aligned} \] Hence the sequence alternates between \(-1+i\) at odd indices and \(-i\) at even indices from \(z_3\) onward. Therefore \(z_{111}=-1+i,\) whose distance from the origin is \(\sqrt{(-1)^2+1^2}=\sqrt2.\)

Thus the correct answer is B.

16 答案 E

令 \(r=\frac{x}{y}\gt0\)。由 \(\frac{x+y}{z}=r\) 得 \(\frac{z}{y}=\frac{r+1}{r}\)。于是 \[ \frac{y}{x-z} =\frac{1}{r-\frac{r+1}{r}} =r\text{。}\]所以 \(r^2-r-1=1\)，即 \(r^2-r-2=0\)。因此 \(r=2\) 或 \(-1\)，由正值条件得 \(r=2\)。

因此正确答案是 E。

English

Let \(r=\frac{x}{y}\gt0.\) From \(\frac{x+y}{z}=r,\) we get \(\frac{z}{y}=\frac{r+1}{r}.\) Then \[ \frac{y}{x-z} =\frac{1}{r-\frac{r+1}{r}} =r. \] Thus \(r^2-r-1=1,\) or \(r^2-r-2=0.\) Hence \(r=2\) or \(-1,\) and positivity gives \(r=2.\)

Thus the correct answer is E.

17 答案 B

模 \(9\) 时，一个数与其各位数字之和同余，所以 \[ \begin{aligned} N&\equiv19+20+\cdots+92\\ &=\frac{74(19+92)}2\\ &=4107\equiv3\pmod9 \end{aligned}\text{。}\]因此 \(N\) 能被 \(3\) 整除但不能被 \(9\) 整除，所以整除它的最高次 \(3\) 的幂是 \(3^1\)。

因此正确答案是 B。

English

Modulo \(9,\) a number is congruent to its digit sum, so \[ \begin{aligned} N&\equiv19+20+\cdots+92\\ &=\frac{74(19+92)}2\\ &=4107\equiv3\pmod9. \end{aligned} \] Thus \(N\) is divisible by \(3\) but not by \(9,\) so the highest power of \(3\) dividing it is \(3^1.\)

Thus the correct answer is B.

18 答案 D

令 \(a_1=a\)、\(a_2=b\)。反复使用递推关系可得 \[ \begin{aligned} a_7&=5a+8b=120,\\ a_8&=8a+13b \end{aligned}\text{。}\]对第一个方程模 \(5\)，可知 \(b\) 是 \(5\) 的倍数。正数取值只有 \(b=5\)、\(b=10\) 或 \(b=15\)。条件 \(b\gt a\gt0\) 使得只有 \(b=10\) 和 \(a=8\) 可行。因此 \(a_8=8(8)+13(10)=194\)。

因此正确答案是 D。

English

Let \(a_1=a\) and \(a_2=b.\) Repeated use of the recurrence gives \[ \begin{aligned} a_7&=5a+8b=120,\\ a_8&=8a+13b. \end{aligned} \] Modulo \(5,\) the first equation shows that \(b\) is a multiple of \(5.\) The positive possibilities are \(b=5,\) \(b=10,\) or \(b=15.\) The conditions \(b\gt a\gt0\) leave only \(b=10\) and \(a=8.\) Hence \(a_8=8(8)+13(10)=194.\)

Thus the correct answer is D.

19 答案 D

取正方体边长为 \(2\)，则其体积为 \(8\)。每个角四面体有三条长度为 \(1\) 且两两垂直的棱，所以体积为 \(\frac16\)。八个被切去的四面体总体积为 \(\frac43\)，剩余体积为 \(\frac{20}{3}\)。所求比值为 \[ \frac{\frac{20}{3}}{8}=\frac56=83\tfrac13\%\text{，}\]最接近 \(84\%\)。

因此正确答案是 D。

English

Take the cube’s side length to be \(2.\) Its volume is \(8.\) Each corner tetrahedron has three perpendicular edges of length \(1,\) so its volume is \(\frac16.\) The eight removed tetrahedra have total volume \(\frac43,\) leaving \(\frac{20}{3}.\) The ratio is \[ \frac{\frac{20}{3}}{8}=\frac56=83\tfrac13\%, \] which is closest to \(84\%.\)

Thus the correct answer is D.

20 答案 D

设尖角与凹口处的锐角分别为 \(\alpha\) 和 \(\beta\)，则 \(\beta-\alpha=10^\circ\)。这个 \(2n\) 边形有 \(n\) 个内角 \(\alpha\) 和 \(n\) 个优角 \(360^\circ-\beta\)。因此 \[ \begin{gathered} n\alpha+n(360^\circ-\beta)\\ {}=(2n-2)180^\circ \end{gathered}\text{。}\]代入 \(\beta-\alpha=10^\circ\)，得到 \(350n=360n-360\)，所以 \(n=36\)。

因此正确答案是 D。

English

Let the acute angles at the tips and notches be \(\alpha\) and \(\beta,\) respectively, so \(\beta-\alpha=10^\circ.\) The \(2n\)-gon has \(n\) interior angles \(\alpha\) and \(n\) reflex interior angles \(360^\circ-\beta.\) Therefore \[ \begin{gathered} n\alpha+n(360^\circ-\beta)\\ {}=(2n-2)180^\circ. \end{gathered} \] Substituting \(\beta-\alpha=10^\circ\) gives \(350n=360n-360,\) so \(n=36.\)

Thus the correct answer is D.

21 答案 A

已知条件表明 \(S_1+\cdots+S_{99}=99{,}000\)。新数列的部分和依次为 \(1,1+S_1,\ldots,1+S_{99}\)。它们的和为 \[ 100+(S_1+\cdots+S_{99})=99{,}100\text{。}\]除以 \(100\)，得到新的切萨罗和 \(991\)。

因此正确答案是 A。

English

The given condition says \(S_1+\cdots+S_{99}=99{,}000.\) For the new sequence, the partial sums are \(1,1+S_1,\ldots,1+S_{99}.\) Their sum is \[ 100+(S_1+\cdots+S_{99})=99{,}100. \] Dividing by \(100\) gives the new Cesàro sum \(991.\)

Thus the correct answer is A.

22 答案 B

从这 \(10\) 个 \(x\) 轴上的点中选两个，再从这 \(5\) 个 \(y\) 轴上的点中选两个。在四条连接线段中，端点次序相反的两条恰好相交一次。可以选取这些点，使得没有三条线段在同一个内部点相交，因此最大值为 \[ \binom{10}{2}\binom{5}{2}=45\cdot10=450\text{。}\]

因此正确答案是 B。

English

Choose two of the \(10\) points on the \(x\)-axis and two of the \(5\) points on the \(y\)-axis. Among the four connecting segments, the two with reversed endpoint order cross exactly once. The points can be chosen so no three segments meet at one interior point, so the maximum is \[ \binom{10}{2}\binom{5}{2}=45\cdot10=450. \]

Thus the correct answer is B.

23 答案 E

余数类 \(1,2,\ldots,6,0\) 分别含有 \(8,7,7,7,7,7,7\) 个数。在每对互补余数类 \((1,6),(2,5),(3,4)\) 中，我们至多能从一个余数类取数，并且至多取一个 \(7\) 的倍数。因此 \[ \begin{aligned} |S|&\le\max(8,7)+\max(7,7)\\ &\qquad+\max(7,7)+1=23 \end{aligned}\text{。}\]取所有模 \(7\) 余 \(1\)、\(2\) 或 \(3\) 的数，再加上一个 \(7\) 的倍数，便能达到 \(23\) 个。

因此正确答案是 E。

English

The residue classes \(1,2,\ldots,6,0\) contain \(8,7,7,7,7,7,7\) numbers, respectively. We may take numbers from at most one class in each complementary pair \((1,6),(2,5),(3,4),\) and at most one multiple of \(7.\) Thus \[ \begin{aligned} |S|&\le\max(8,7)+\max(7,7)\\ &\qquad+\max(7,7)+1=23. \end{aligned} \] Taking every number whose residue modulo \(7\) is \(1,\) \(2,\) or \(3,\) together with one multiple of \(7,\) attains \(23.\)

Thus the correct answer is E.

24 答案 C

以 \(AB\) 和 \(AD\) 为基向量，则 \[ \begin{aligned} E&=\left(\frac23,0\right),\\ F&=\left(1,\frac25\right),\\ G&=\left(0,\frac25\right) \end{aligned}\text{。}\]过 \(G\) 且平行于 \(EF\) 的直线与 \(CD\) 交于 \(H=(\frac{1}{2},1)\)。用鞋带公式可得 \(EFHG\) 在该坐标系中的面积为 \(\frac{1}{2}\)。仿射坐标中的面积应乘以 \(ABCD\) 的面积 \(10\)，所以 \([EFHG]=\frac12\cdot10=5\)。

因此正确答案是 C。

English

Use \(AB\) and \(AD\) as basis vectors. Then \[ \begin{aligned} E&=\left(\frac23,0\right),\\ F&=\left(1,\frac25\right),\\ G&=\left(0,\frac25\right). \end{aligned} \] A line from \(G\) parallel to \(EF\) meets \(CD\) at \(H=(\frac{1}{2},1).\) The shoelace formula gives the coordinate area of \(EFHG\) as \(\frac{1}{2}.\) Affine coordinates scale all areas by the area \(10\) of \(ABCD,\) so \([EFHG]=\frac12\cdot10=5.\)

Thus the correct answer is C.

25 答案 E

取 \(B=(0,0)\)、\(A=(3,0)\)，则 \[ \begin{aligned} C&=4(\cos120^\circ,\sin120^\circ)\\ &=(-2,2\sqrt3) \end{aligned}\text{。}\]过 \(A\) 且垂直于 \(AB\) 的直线为 \(x=3\)。与 \(BC=(-2,2\sqrt3)\) 垂直的一个向量是 \((\sqrt3,1)\)，所以另一条垂线为 \(C+t(\sqrt3,1)\)。当 \(t=\frac{5}{\sqrt3}\) 时，其 \(x\) 坐标为 \(3\)。由于 \(|(\sqrt3,1)|=2\)，可得 \[ CD=2t=\frac{10}{\sqrt3}\text{。}\]

因此正确答案是 E。

English

Set \(B=(0,0)\) and \(A=(3,0).\) Then \[ \begin{aligned} C&=4(\cos120^\circ,\sin120^\circ)\\ &=(-2,2\sqrt3). \end{aligned} \] The perpendicular to \(AB\) at \(A\) is \(x=3.\) A vector perpendicular to \(BC=(-2,2\sqrt3)\) is \((\sqrt3,1),\) so the other perpendicular is \(C+t(\sqrt3,1).\) Its \(x\)-coordinate is \(3\) when \(t=\frac{5}{\sqrt3}.\) Since \(|(\sqrt3,1)|=2,\) \[ CD=2t=\frac{10}{\sqrt3}. \]

Thus the correct answer is E.

26 答案 B

三角形 \(ACD\) 和 \(BCD\) 都是等腰直角三角形，所以 \(AD=BD=\sqrt2\)、\(\angle ADB=90^\circ\)，且 \(DE=DF=2-\sqrt2\)。笑脸区域等于 \(90^\circ\) 扇形 \(EDF\) 加上全等的两个 \(45^\circ\) 扇形 \(ABE\) 和 \(BAF\)，再减去三角形 \(ABD\) 与原半圆。其面积为 \[ \begin{aligned} &\frac14\pi(2-\sqrt2)^2 +2\left(\frac18\pi\cdot2^2\right)\\ &\qquad-\frac12(2)(1)-\frac12\pi\\ &=2\pi-\pi\sqrt2-1 \end{aligned}\text{。}\]

因此正确答案是 B。

English

Triangles \(ACD\) and \(BCD\) are isosceles right triangles, so \(AD=BD=\sqrt2,\) \(\angle ADB=90^\circ,\) and \(DE=DF=2-\sqrt2.\) The smile is the \(90^\circ\) sector \(EDF,\) plus the congruent \(45^\circ\) sectors \(ABE\) and \(BAF,\) minus triangle \(ABD\) and the original semicircle. Its area is \[ \begin{aligned} &\frac14\pi(2-\sqrt2)^2 +2\left(\frac18\pi\cdot2^2\right)\\ &\qquad-\frac12(2)(1)-\frac12\pi\\ &=2\pi-\pi\sqrt2-1. \end{aligned} \]

Thus the correct answer is B.

27 答案 D

由点 \(P\) 的幂可得 \[ PA\cdot PB=PC\cdot PD\text{。}\]这里 \(PA=18\)、\(PB=8\)，且 \(PD=PC+7\)，所以 \(PC(PC+7)=144\)。因此 \(PC=9\)、\(PD=16\)。由于 \(PA=2PC\) 且 \(\angle APC=60^\circ\)，三角形 \(APC\) 在 \(C\) 处为直角，并且 \(AC=9\sqrt3\)。所以 \(AD\) 是直径，且 \[ \begin{aligned} (2r)^2&=AD^2=AC^2+CD^2\\ &=243+49=292 \end{aligned}\text{。}\]因此 \(r^2=73\)。

因此正确答案是 D。

English

Power of \(P\) gives \[ PA\cdot PB=PC\cdot PD. \] Here \(PA=18,\) \(PB=8,\) and \(PD=PC+7,\) so \(PC(PC+7)=144.\) Thus \(PC=9\) and \(PD=16.\) Since \(PA=2PC\) and \(\angle APC=60^\circ,\) triangle \(APC\) is right at \(C,\) with \(AC=9\sqrt3.\) Therefore \(AD\) is a diameter, and \[ \begin{aligned} (2r)^2&=AD^2=AC^2+CD^2\\ &=243+49=292. \end{aligned} \] Hence \(r^2=73.\)

Thus the correct answer is D.

28 答案 B

由求根公式得 \[ z=\frac{1\pm\sqrt{21-20i}}2\text{。}\]因为 \((5-2i)^2=21-20i\)，所以两个根为 \[ \begin{aligned} \frac{1+(5-2i)}2&=3-i,\\ \frac{1-(5-2i)}2&=-2+i \end{aligned}\text{。}\]它们的实部之积为 \(3(-2)=-6\)。

因此正确答案是 B。

English

The quadratic formula gives \[ z=\frac{1\pm\sqrt{21-20i}}2. \] Since \((5-2i)^2=21-20i,\) the roots are \[ \begin{aligned} \frac{1+(5-2i)}2&=3-i,\\ \frac{1-(5-2i)}2&=-2+i. \end{aligned} \] Their real parts have product \(3(-2)=-6.\)

Thus the correct answer is B.

29 答案 D

设 \(E_n\) 和 \(O_n\) 分别为抛掷 \(n\) 次后正面次数为偶数和奇数的概率。则 \(E_n+O_n=1\)，并且 \[ \begin{aligned} E_{n+1}-O_{n+1} &=\left(\frac13-\frac23\right)\\ &\qquad\cdot(E_n-O_n) \end{aligned}\text{。}\]因为 \(E_0-O_0=1\)，所以 \(E_{50}-O_{50}=(-\frac{1}{3})^{50}=3^{-50}\)。联立和与差的方程，得到 \[ E_{50}=\frac12\left(1+\frac1{3^{50}}\right)\text{。}\]

因此正确答案是 D。

English

Let \(E_n\) and \(O_n\) be the probabilities of an even and odd number of heads after \(n\) tosses. Then \(E_n+O_n=1,\) while \[ \begin{aligned} E_{n+1}-O_{n+1} &=\left(\frac13-\frac23\right)\\ &\qquad\cdot(E_n-O_n). \end{aligned} \] Since \(E_0-O_0=1,\) we have \(E_{50}-O_{50}=(-\frac{1}{3})^{50}=3^{-50}.\) Solving the sum and difference equations gives \[ E_{50}=\frac12\left(1+\frac1{3^{50}}\right). \]

Thus the correct answer is D.

30 答案 B

取 \(A=(-46,0)\)、\(B=(46,0)\)、\(D=(-\frac{19}{2},h)\)，且 \(C=(\frac{19}{2},h)\)。由对称性，圆心必为 \(O=(0,0)\)。每条腰的水平位移为 \(\frac{73}{2}\)，所以 \[ x^2=h^2+\left(\frac{73}{2}\right)^2\text{。}\]从 \(O\) 向腰作垂线，当垂足到达上端点时，它才首次落在线段上；这个临界条件为 \(OD\perp AD\)。因此 \[ \begin{aligned} (-\frac{19}{2},h)\cdot(\frac{73}{2},h)&=0,\\ h^2&=\frac{1387}{4} \end{aligned}\text{。}\]所以 \[ m^2=\frac{1387+5329}{4}=1679\text{。}\]

因此正确答案是 B。

English

Place \(A=(-46,0),\) \(B=(46,0),\) \(D=(-\frac{19}{2},h),\) and \(C=(\frac{19}{2},h).\) Symmetry forces the circle’s center to be \(O=(0,0).\) The horizontal offset along each leg is \(\frac{73}{2},\) so \[ x^2=h^2+\left(\frac{73}{2}\right)^2. \] The perpendicular from \(O\) to a leg first has its foot on the segment when the foot reaches the upper endpoint; this boundary condition is \(OD\perp AD.\) Thus \[ \begin{aligned} (-\frac{19}{2},h)\cdot(\frac{73}{2},h)&=0,\\ h^2&=\frac{1387}{4}. \end{aligned} \] Therefore \[ m^2=\frac{1387+5329}{4}=1679. \]

Thus the correct answer is B.