1979 AMC 12 真题（P版）

1 答案 D

矩形 \(DEFG\) 的两条边长分别是矩形 \(ABCD\) 相应边长的一半。因此，它的面积是 \(72\) 的 \((\frac{1}{2})(\frac{1}{2})=\frac{1}{4}\)，即 \(18\)。

因此，正确答案是 D。

English

The side lengths of \(DEFG\) are one half the corresponding side lengths of \(ABCD.\) Its area is therefore \((\frac{1}{2})(\frac{1}{2})=\frac{1}{4}\) of \(72,\) which is \(18.\)

Therefore, the correct answer is D.

2 答案 D

我们有 \[ \begin{aligned} \frac1x-\frac1y&=\frac{y-x}{xy}\\ &=-\frac{x-y}{xy}\\ &=-1 \end{aligned}\text{，}\]因为 \(x-y=xy\)。

因此，正确答案是 D。

English

We have \[ \begin{aligned} \frac1x-\frac1y&=\frac{y-x}{xy}\\ &=-\frac{x-y}{xy}\\ &=-1, \end{aligned} \] because \(x-y=xy.\)

Therefore, the correct answer is D.

3 答案 C

由正方形和等边三角形可得 \(DA=AB=AE\)。此外，\[ \begin{aligned} \angle DAE &=\angle DAB+\angle BAE\\ &=90^\circ+60^\circ\\ &=150^\circ \end{aligned}\text{。}\]因此，\(\triangle DAE\) 是等腰三角形，每个底角都是 \(\frac{180^\circ-150^\circ}{2}=15^\circ\)。

因此，正确答案是 C。

English

The square and equilateral triangle give \(DA=AB=AE.\) Also \[ \begin{aligned} \angle DAE &=\angle DAB+\angle BAE\\ &=90^\circ+60^\circ\\ &=150^\circ. \end{aligned} \] Thus \(\triangle DAE\) is isosceles, and each base angle is \(\frac{180^\circ-150^\circ}{2}=15^\circ.\)

Therefore, the correct answer is C.

4 答案 E

从内向外展开，得到 \[ \begin{aligned} &x[x\{x(2-x)-4\}+10]+1\\ &\quad=x[x(2x-x^2-4)+10]+1\\ &\quad=-x^4+2x^3-4x^2\\ &\qquad+10x+1 \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Expanding from the inside gives \[ \begin{aligned} &x[x\{x(2-x)-4\}+10]+1\\ &\quad=x[x(2x-x^2-4)+10]+1\\ &\quad=-x^4+2x^3-4x^2\\ &\qquad+10x+1. \end{aligned} \]

Therefore, the correct answer is E.

5 答案 D

相同的百位数字与个位数字必须是偶数。最大的非零偶数数字是 \(8\)，十位数字可以取 \(9\)。因此最大的数是 \(898\)，其各位数字之和为 \(8+9+8=25\)。

因此，正确答案是 D。

English

The equal hundreds and units digit must be even. The largest possible nonzero such digit is \(8,\) and the tens digit can be \(9.\) Thus the largest number is \(898,\) whose digit sum is \(8+9+8=25.\)

Therefore, the correct answer is D.

6 答案 A

每个分子都比分母大一，因此原式为 \[ \begin{aligned} 6+\biggl(&\frac12+\frac14+\frac18\\ &+\frac1{16}+\frac1{32}+\frac1{64}\biggr)-7 \end{aligned}\text{。}\]括号内的等比数列之和为 \(1-\frac{1}{64}=\frac{63}{64}\)。所以原式的值为 \(6+\frac{63}{64}-7=-\frac{1}{64}\)。

因此，正确答案是 A。

English

Each numerator is one more than its denominator, so the expression is \[ \begin{aligned} 6+\biggl(&\frac12+\frac14+\frac18\\ &+\frac1{16}+\frac1{32}+\frac1{64}\biggr)-7. \end{aligned} \] The parenthesized geometric sum is \(1-\frac{1}{64}=\frac{63}{64}.\) Therefore the value is \(6+\frac{63}{64}-7=-\frac{1}{64}.\)

Therefore, the correct answer is A.

7 答案 E

令 \(x=n^2\)，其中 \(n\ge0\)。下一个更大的完全平方数是 \[ \begin{aligned} (n+1)^2&=n^2+2n+1\\ &=x+2\sqrt x+1 \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Let \(x=n^2\) with \(n\ge0.\) The next larger perfect square is \[ \begin{aligned} (n+1)^2&=n^2+2n+1\\ &=x+2\sqrt x+1. \end{aligned} \]

Therefore, the correct answer is E.

8 答案 C

两条射线与圆的交点所对应的方向角分别为 \(45^\circ\) 和 \(135^\circ\)，所以最小的有界区域是半径为 \(2\) 的圆的一个 \(90^\circ\) 扇形。其面积为 \[ \frac{90^\circ}{360^\circ}\pi(2)^2=\pi\text{。}\]

因此，正确答案是 C。

English

The two rays meet the circle at angles \(45^\circ\) and \(135^\circ,\) so the smallest bounded region is a \(90^\circ\) sector of the radius-\(2\) circle. Its area is \[ \frac{90^\circ}{360^\circ}\pi(2)^2=\pi. \]

Therefore, the correct answer is C.

9 答案 E

把两个根式都写成 \(2\) 的幂，得到 \[ \begin{aligned} 4^{\frac{1}{3}}8^{\frac{1}{4}} &=2^{\frac{2}{3}}2^{\frac{3}{4}}\\ &=2^{\frac{17}{12}}\\ &=2\sqrt[12]{2^5}\\ &=2\sqrt[12]{32} \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Writing both radicals as powers of \(2\) gives \[ \begin{aligned} 4^{\frac{1}{3}}8^{\frac{1}{4}} &=2^{\frac{2}{3}}2^{\frac{3}{4}}\\ &=2^{\frac{17}{12}}\\ &=2\sqrt[12]{2^5}\\ &=2\sqrt[12]{32}. \end{aligned} \]

Therefore, the correct answer is E.

10 答案 D

设 \(O\) 为中心。相邻的边心距 \(OQ_i\) 与 \(OQ_{i+1}\) 长度均为 \(2\)，夹角为 \(60^\circ\)，所以每个 \(\triangle OQ_iQ_{i+1}\) 都是边长为 \(2\) 的等边三角形。该四边形由三个这样的三角形组成，因此面积为 \[ 3\left(\frac{\sqrt3}{4}\cdot2^2\right)=3\sqrt3\text{。}\]

因此，正确答案是 D。

English

Let \(O\) be the center. Consecutive apothems \(OQ_i\) and \(OQ_{i+1}\) have length \(2\) and form a \(60^\circ\) angle, so each \(\triangle OQ_iQ_{i+1}\) is equilateral of side \(2.\) The quadrilateral is the union of three such triangles, and hence has area \[ 3\left(\frac{\sqrt3}{4}\cdot2^2\right)=3\sqrt3. \]

Therefore, the correct answer is D.

11 答案 B

分子为 \(n^2\)，分母为 \(n(n+1)\)。因而 \[ \frac{n}{n+1}=\frac{115}{116}\text{，}\]即 \(116n=115n+115\)，所以 \(n=115\)。

因此，正确答案是 B。

English

The numerator is \(n^2,\) and the denominator is \(n(n+1).\) Thus \[ \frac{n}{n+1}=\frac{115}{116}, \] which gives \(116n=115n+115,\) so \(n=115.\)

Therefore, the correct answer is B.

12 答案 B

令 \(y=\angle BAO\)。因为 \(AB=OB\)，所以 \(\angle AOB=y\)，从而 \(\angle EBO=2y\)。又因为 \(OB=OE\)，所以 \(\angle BEO=2y\)，且 \(\angle BOE=180^\circ-4y\)。另一方面，\[ \angle BOD=180^\circ-y\text{，}\]所以 \(\angle BOE=135^\circ-y\)。因此 \(180^\circ-4y=135^\circ-y\)，解得 \(y=15^\circ\)。

因此，正确答案是 B。

English

Let \(y=\angle BAO.\) Since \(AB=OB,\) \(\angle AOB=y,\) so \(\angle EBO=2y.\) Also \(OB=OE,\) hence \(\angle BEO=2y\) and \(\angle BOE=180^\circ-4y.\) On the other hand, \[ \angle BOD=180^\circ-y, \] so \(\angle BOE=135^\circ-y.\) Therefore \(180^\circ-4y=135^\circ-y,\) giving \(y=15^\circ.\)

Therefore, the correct answer is B.

13 答案 A

当 \(x\ge0\) 时，不等式化为 \(y-x\lt x\)，即 \(y\lt2x\)。当 \(x\lt0\) 时，不等式化为 \(y-x\lt-x\)，即 \(y\lt0\)。合并两种情形，恰好得到 \(y\lt0\) 或 \(y\lt2x\)，两者可以同时成立。

因此，正确答案是 A。

English

If \(x\ge0,\) the inequality becomes \(y-x\lt x,\) or \(y\lt2x.\) If \(x\lt0,\) it becomes \(y-x\lt-x,\) or \(y\lt0.\) Combining the two cases gives exactly \(y\lt0\) or \(y\lt2x,\) with overlap allowed.

Therefore, the correct answer is A.

14 答案 C

当 \(n\ge2\) 时，第 \(n\) 项为 \[ a_n=\frac{n^2}{(n-1)^2}\text{。}\]所以 \(a_3=\frac{9}{4}\)，且 \(a_5=\frac{25}{16}\)，两者之和为 \(\frac{36}{16}+\frac{25}{16}=\frac{61}{16}\)。

因此，正确答案是 C。

English

For \(n\ge2,\) the \(n\)th term is \[ a_n=\frac{n^2}{(n-1)^2}. \] Therefore \(a_3=\frac{9}{4}\) and \(a_5=\frac{25}{16},\) whose sum is \(\frac{36}{16}+\frac{25}{16}=\frac{61}{16}.\)

Therefore, the correct answer is C.

15 答案 E

两罐总体积相等时，酒精所占比例分别为 \(\frac{p}{p+1}\) 和 \(\frac{q}{q+1}\)，水所占比例分别为 \(\frac{1}{p+1}\) 和 \(\frac{1}{q+1}\)。因而酒精与水的体积比为 \[ \frac{\frac p{p+1}+\frac q{q+1}} {\frac1{p+1}+\frac1{q+1}} =\frac{p+q+2pq}{p+q+2}\text{。}\]

因此，正确答案是 E。

English

For equal total volumes, the alcohol fractions are \(\frac{p}{p+1}\) and \(\frac{q}{q+1},\) while the water fractions are \(\frac{1}{p+1}\) and \(\frac{1}{q+1}.\) Their ratio is \[ \frac{\frac p{p+1}+\frac q{q+1}} {\frac1{p+1}+\frac1{q+1}} =\frac{p+q+2pq}{p+q+2}. \]

Therefore, the correct answer is E.

16 答案 E

等差数列条件给出 \[ 2A_2=A_1+(A_1+A_2)\text{，}\]因此 \(A_2=2A_1\)。所以大圆面积为 \(3A_1=9\pi\)，从而 \(A_1=3\pi\)。若小圆半径为 \(r\)，则 \(\pi r^2=3\pi\)，所以 \(r=\sqrt3\)。

因此，正确答案是 E。

English

The progression condition gives \[ 2A_2=A_1+(A_1+A_2), \] so \(A_2=2A_1.\) Hence the larger area is \(3A_1=9\pi,\) and \(A_1=3\pi.\) If the smaller radius is \(r,\) then \(\pi r^2=3\pi,\) so \(r=\sqrt3.\)

Therefore, the correct answer is E.

17 答案 C

新三角形的三条边长为 \(x\)、\(y-x\) 和 \(z-y\)。由三角形不等式，\[ \begin{aligned} x&\lt(y-x)+(z-y),\\ &\Longrightarrow x\lt \frac{z}{2};\\ y-x&\lt x+(z-y),\\ &\Longrightarrow y\lt x+\frac{z}{2};\\ z-y&\lt x+(y-x),\\ &\Longrightarrow y\gt \frac{z}{2} \end{aligned}\text{。}\]因此，I 与 II 必须成立，而 III 必须不成立。

因此，正确答案是 C。

English

The new triangle has side lengths \(x,\) \(y-x,\) and \(z-y.\) Its triangle inequalities give \[ \begin{aligned} x&\lt(y-x)+(z-y),\\ &\Longrightarrow x\lt \frac{z}{2};\\ y-x&\lt x+(z-y),\\ &\Longrightarrow y\lt x+\frac{z}{2};\\ z-y&\lt x+(y-x),\\ &\Longrightarrow y\gt \frac{z}{2}. \end{aligned} \] Thus I and II must hold, while III must fail.

Therefore, the correct answer is C.

18 答案 C

我们有 \(\log_{10}5\approx1-0.301=0.699\)。因此，\[ \begin{aligned} \log_5 10&=\frac1{\log_{10}5}\\ &\approx\frac1{0.699}\\ &\approx1.431 \end{aligned}\text{。}\]在各选项中，\(\frac{10}{7}\approx1.429\) 最接近。

因此，正确答案是 C。

English

We have \(\log_{10}5\approx1-0.301=0.699.\) Therefore \[ \begin{aligned} \log_5 10&=\frac1{\log_{10}5}\\ &\approx\frac1{0.699}\\ &\approx1.431. \end{aligned} \] Of the choices, \(\frac{10}{7}\approx1.429\) is closest.

Therefore, the correct answer is C.

19 答案 A

因为 \(256=2^8\)，所以 \[ 256^{32}=2^{256}\text{。}\]因此 \((\frac{x}{2})^{256}=1\)。实数解为 \(x=2\) 和 \(x=-2\)，它们的平方和为 \(4+4=8\)。

因此，正确答案是 A。

English

Since \(256=2^8,\) \[ 256^{32}=2^{256}. \] Thus \((\frac{x}{2})^{256}=1.\) The real solutions are \(x=2\) and \(x=-2,\) and the sum of their squares is \(4+4=8.\)

Therefore, the correct answer is A.

20 答案 C

代入 \(a=\frac{1}{2}\)，得到 \(b=\frac{1}{3}\)。令 \(u=\operatorname{Arctan}a\)，\(v=\operatorname{Arctan}b\)，则 \[ \begin{aligned} \tan(u+v)&=\frac{a+b}{1-ab}\\ &=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\\ &=1 \end{aligned}\text{。}\]两个角都为正，且它们的和小于 \(\frac{\pi}{2}\)，所以 \(u+v=\frac{\pi}{4}\)。

因此，正确答案是 C。

English

Substituting \(a=\frac{1}{2}\) gives \(b=\frac{1}{3}.\) If \(u=\operatorname{Arctan}a\) and \(v=\operatorname{Arctan}b,\) then \[ \begin{aligned} \tan(u+v)&=\frac{a+b}{1-ab}\\ &=\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}\\ &=1. \end{aligned} \] Both angles are positive and their sum is less than \(\frac{\pi}{2},\) so \(u+v=\frac{\pi}{4}.\)

Therefore, the correct answer is C.

21 答案 B

若两条直角边长为 \(x\) 和 \(y\)，则直角三角形的内切圆半径恒等式为 \(x+y=h+2r\)。因此半周长为 \[ s=\frac{x+y+h}{2}=h+r\text{。}\]三角形的面积为 \(rs=r(h+r)\)，圆的面积为 \(\pi r^2\)。两者之比为 \(\frac{\pi r}{h+r}\)。

因此，正确答案是 B。

English

If the legs are \(x\) and \(y,\) then the right-triangle inradius identity is \(x+y=h+2r.\) Hence the semiperimeter is \[ s=\frac{x+y+h}{2}=h+r. \] The triangle’s area is \(rs=r(h+r),\) while the circle’s area is \(\pi r^2.\) Their ratio is \(\frac{\pi r}{h+r}.\)

Therefore, the correct answer is B.

22 答案 A

左边等于 \[ \begin{gathered} m(m+1)(m+5)\\ =m(m+1)(m+2)\\ \quad+3m(m+1) \end{gathered}\text{，}\]对每个整数 \(m\) 都能被 \(3\) 整除。右边为 \(3(9n^3+3n^2+3n)+1\)，与 \(1\pmod3\) 同余。因此不存在满足方程的整数对。

因此，正确答案是 A。

English

The left side is \[ \begin{gathered} m(m+1)(m+5)\\ =m(m+1)(m+2)\\ \quad+3m(m+1), \end{gathered} \] which is divisible by \(3\) for every integer \(m.\) The right side is \(3(9n^3+3n^2+3n)+1,\) which is congruent to \(1\pmod3.\) Therefore no integer pair satisfies the equation.

Therefore, the correct answer is A.

23 答案 C

缩放之前，取 \[ \begin{aligned} A&=(1,1,1),\\ B&=(1,-1,-1),\\ C&=(-1,1,-1),\\ D&=(-1,-1,1) \end{aligned}\text{。}\]这些棱的长度为 \(2\sqrt2\)，所以按比例因子 \(\frac{1}{2\sqrt2}\) 缩放。\(AB\) 和 \(CD\) 上的点可分别写成 \(P=(1,t,t)\) 和 \(Q=(-1,u,-u)\)。缩放前，\[ \begin{aligned} PQ^2&=4+(t-u)^2+(t+u)^2\\ &=4+2t^2+2u^2 \end{aligned}\text{，}\]在 \(t=u=0\) 时取最小值。缩放后，最小距离的平方为 \(\frac{4}{8}=\frac{1}{2}\)，所以距离为 \(\frac{\sqrt2}{2}\)。

因此，正确答案是 C。

English

Before scaling, take \[ \begin{aligned} A&=(1,1,1),\\ B&=(1,-1,-1),\\ C&=(-1,1,-1),\\ D&=(-1,-1,1). \end{aligned} \] These edges have length \(2\sqrt2,\) so scale by \(\frac{1}{2\sqrt2}.\) Points on \(AB\) and \(CD\) have forms \(P=(1,t,t)\) and \(Q=(-1,u,-u).\) Before scaling, \[ \begin{aligned} PQ^2&=4+(t-u)^2+(t+u)^2\\ &=4+2t^2+2u^2, \end{aligned} \] minimized at \(t=u=0.\) After scaling, the minimum squared distance is \(\frac{4}{8}=\frac{1}{2},\) so the distance is \(\frac{\sqrt2}{2}.\)

Therefore, the correct answer is C.

24 答案 E

取 \(B=(0,0)\) 和 \(C=(5,0)\)。因为 \(B\) 为钝角且 \(-\cos B=\frac{3}{5}\)，可取 \[ \begin{aligned} A&=4(-\frac{3}{5},\frac{4}{5})\\ &=(-\frac{12}{5},\frac{16}{5}) \end{aligned}\text{。}\]简单四边形的构型以及 \(C\) 处的钝角给出 \[ \begin{aligned} D&=C+20(\frac{4}{5},\frac{3}{5})\\ &=(21,12) \end{aligned}\text{。}\]因此，\[ \begin{aligned} AD&=\sqrt{\begin{gathered} (21+\frac{12}{5})^2\\ {}+(12-\frac{16}{5})^2 \end{gathered}}\\ &=\sqrt{(\frac{117}{5})^2+(\frac{44}{5})^2}\\ &=25 \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Put \(B=(0,0)\) and \(C=(5,0).\) Since \(B\) is obtuse and \(-\cos B=\frac{3}{5},\) we may take \[ \begin{aligned} A&=4(-\frac{3}{5},\frac{4}{5})\\ &=(-\frac{12}{5},\frac{16}{5}). \end{aligned} \] The simple configuration and the obtuse angle at \(C\) give \[ \begin{aligned} D&=C+20(\frac{4}{5},\frac{3}{5})\\ &=(21,12). \end{aligned} \] Thus \[ \begin{aligned} AD&=\sqrt{\begin{gathered} (21+\frac{12}{5})^2\\ {}+(12-\frac{16}{5})^2 \end{gathered}}\\ &=\sqrt{(\frac{117}{5})^2+(\frac{44}{5})^2}\\ &=25. \end{aligned} \]

Therefore, the correct answer is E.

25 答案 B

令 \(a=-\frac{1}{2}\)。因为第一次相除的余数为 \(a^8\)，所以 \[ \begin{aligned} q_1(x)&=\frac{x^8-a^8}{x-a}\\ &=x^7+ax^6+\cdots+a^7 \end{aligned}\text{。}\]因此第二次相除的余数为 \[ \begin{aligned} q_1(a)&=8a^7\\ &=8(-\frac{1}{2})^7\\ &=-\frac{1}{16} \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Let \(a=-\frac{1}{2}.\) Since the first remainder is \(a^8,\) \[ \begin{aligned} q_1(x)&=\frac{x^8-a^8}{x-a}\\ &=x^7+ax^6+\cdots+a^7. \end{aligned} \] Therefore the second remainder is \[ \begin{aligned} q_1(a)&=8a^7\\ &=8(-\frac{1}{2})^7\\ &=-\frac{1}{16}. \end{aligned} \]

Therefore, the correct answer is B.

26 答案 B

令 \(x=1\)，得到 \(f(y+1)-f(y)=y+2\)。令 \(x=y=0\)，得到 \(f(0)=-1\)。向两个方向延拓递推关系，可得对每个整数 \(n\)，\[ f(n)=\frac{n^2+3n-2}{2}\text{。}\]因而 \(f(n)=n\) 等价于 \[ \begin{aligned} n^2+n-2&=(n+2)(n-1)\\ &=0 \end{aligned}\text{。}\]整数解为 \(n=-2\) 和 \(n=1\)，排除 \(1\) 后恰好剩一个。

因此，正确答案是 B。

English

Setting \(x=1\) gives \(f(y+1)-f(y)=y+2.\) Setting \(x=y=0\) gives \(f(0)=-1.\) Extending the recurrence in both directions yields, for every integer \(n,\) \[ f(n)=\frac{n^2+3n-2}{2}. \] Thus \(f(n)=n\) is equivalent to \[ \begin{aligned} n^2+n-2&=(n+2)(n-1)\\ &=0. \end{aligned} \] The integer solutions are \(n=-2\) and \(n=1,\) and after excluding \(1\) exactly one remains.

Therefore, the correct answer is B.

27 答案 E

共有 \(11^2=121\) 个有序对。要有两个不同的正实根，必须有 \(b\lt0\)、\(c\gt0\) 且 \(b^2\gt4c\)。当 \(b=-1\) 和 \(b=-2\) 时没有可选值；当 \(b=-3\)、\(b=-4\) 和 \(b=-5\) 时，分别有 \(2\)、\(3\) 和 \(5\) 个可选的 \(c\)。因此只有 \(10\) 个有序对会产生两个不同的正实根，所求概率为 \[ 1-\frac{10}{121}=\frac{111}{121}\text{，}\]该数不在选项中。

因此，正确答案是 E。

English

There are \(11^2=121\) ordered pairs. Distinct positive roots require \(b\lt0,\) \(c\gt0,\) and \(b^2\gt4c.\) For \(b=-1\) and \(b=-2\) there are no choices; for \(b=-3,\) \(b=-4,\) and \(b=-5\) there are respectively \(2,\) \(3,\) and \(5\) choices of \(c.\) Thus only \(10\) pairs produce distinct positive roots, so the requested probability is \[ 1-\frac{10}{121}=\frac{111}{121}, \] which is not listed.

Therefore, the correct answer is E.

28 答案 D

设 \(A'\) 为以 \(B\) 和 \(C\) 为圆心的两圆在外侧的对应交点。那么 \(A'B'C'\) 与 \(ABC\) 是同心的等边三角形。设 \(K\) 为它们的共同重心，\(M\) 为 \(BC\) 的中点，则 \[ \begin{aligned} A'M&=\sqrt{r^2-1},\\ MK&=\frac{\sqrt3}{3},\\ AK&=\frac{2\sqrt3}{3} \end{aligned}\text{。}\]由相似性，\[ \begin{aligned} \frac{B'C'}2&=\frac{A'K}{AK}\\ &=\frac{\sqrt{r^2-1}+\frac{\sqrt3}{3}} {\frac{2\sqrt3}{3}} \end{aligned}\text{。}\]因此 \(B'C'=1+\sqrt{3(r^2-1)}\)。

因此，正确答案是 D。

English

Let \(A'\) be the analogous outer intersection of the circles centered at \(B\) and \(C.\) Then \(A'B'C'\) and \(ABC\) are concentric equilateral triangles. If \(K\) is their common centroid and \(M\) is the midpoint of \(BC,\) then \[ \begin{aligned} A'M&=\sqrt{r^2-1},\\ MK&=\frac{\sqrt3}{3},\\ AK&=\frac{2\sqrt3}{3}. \end{aligned} \] Similarity gives \[ \begin{aligned} \frac{B'C'}2&=\frac{A'K}{AK}\\ &=\frac{\sqrt{r^2-1}+\frac{\sqrt3}{3}} {\frac{2\sqrt3}{3}}. \end{aligned} \] Hence \(B'C'=1+\sqrt{3(r^2-1)}.\)

Therefore, the correct answer is D.

29 答案 E

令 \(t=x+\frac{1}{x}\)，\(u=x^3+x^{-3}=t^3-3t\)。因为 \(x^6+x^{-6}=u^2-2\)，所以分子为 \[ \begin{aligned} t^6-u^2&=(t^3-u)(t^3+u)\\ &=3t(t^3+u) \end{aligned}\text{。}\]分母为 \(t^3+u\)，所以 \(f(x)=3t=3(x+\frac{1}{x})\)。当 \(x\gt0\) 时，\(x+\frac{1}{x}\ge2\)，且在 \(x=1\) 时取等号。因此最小值为 \(6\)。

因此，正确答案是 E。

English

Let \(t=x+\frac{1}{x}\) and \(u=x^3+x^{-3}=t^3-3t.\) Since \(x^6+x^{-6}=u^2-2,\) the numerator is \[ \begin{aligned} t^6-u^2&=(t^3-u)(t^3+u)\\ &=3t(t^3+u). \end{aligned} \] The denominator is \(t^3+u,\) so \(f(x)=3t=3(x+\frac{1}{x}).\) For \(x\gt0,\) \(x+\frac{1}{x}\ge2,\) with equality at \(x=1.\) Thus the minimum is \(6.\)

Therefore, the correct answer is E.

30 答案 B

将 \(AB\) 经过 \(B\) 延长至 \(F\)，使 \(AF=AC=1\)。则 \(\triangle ACF\) 为等边三角形。在 \(BF\) 上取点 \(G\)，使 \(\angle BCG=20^\circ\)。由角度条件可得 \(\triangle FGC\cong\triangle ABC\)，且 \(\triangle BCG\sim\triangle DCE\)。因为 \(E\) 是 \(BC\) 的中点，所以相似比为 \(2\)，从而 \[ [BCG]=4[CDE]\text{。}\]分割这个等边三角形可得 \[ \frac{\sqrt3}{4} =2[ABC]+4[CDE]\text{。}\]两边除以 \(2\)，得到 \([ABC]+2[CDE]=\frac{\sqrt3}{8}\)。

因此，正确答案是 B。

English

Extend \(AB\) through \(B\) to \(F\) with \(AF=AC=1.\) Then \(\triangle ACF\) is equilateral. Choose \(G\) on \(BF\) so that \(\angle BCG=20^\circ.\) The angle conditions give \(\triangle FGC\cong\triangle ABC,\) while \(\triangle BCG\sim\triangle DCE.\) Since \(E\) is the midpoint of \(BC,\) the similarity scale is \(2,\) so \[ [BCG]=4[CDE]. \] Decomposing the equilateral triangle gives \[ \frac{\sqrt3}{4} =2[ABC]+4[CDE]. \] Dividing by \(2\) yields \([ABC]+2[CDE]=\frac{\sqrt3}{8}.\)

Therefore, the correct answer is B.