1987 AMC 12 真题（P版）

1 答案 D

展开得 \[ \begin{aligned} (1+x^2)(1-x^3) &=1+x^2-x^3\\ &\qquad-x^5 \end{aligned}\text{。}\]

所以正确答案是 D。

English

Expanding gives \[ \begin{aligned} (1+x^2)(1-x^3) &=1+x^2-x^3\\ &\qquad-x^5. \end{aligned} \]

Thus the correct answer is D.

2 答案 E

由于 \(\angle DBE=60^\circ\)，且 \(DB=EB=1\)，所以三角形 \(DBE\) 是等边三角形，从而 \(DE=1\)。原周长为 \(9\)。切去这个顶角会去掉两条单位线段，并增加一条单位线段：\[ 9-1-1+1=8\text{。}\]

所以正确答案是 E。

English

Since \(\angle DBE=60^\circ\) and \(DB=EB=1,\) triangle \(DBE\) is equilateral, so \(DE=1.\) The original perimeter is \(9.\) Cutting off the corner removes two unit segments and adds one: \[ 9-1-1+1=8. \]

Thus the correct answer is E.

3 答案 C

这些质数为 \[ 7,\ 17,\ 37,\ 47,\ 67,\ 97\text{。}\]其余候选数 \(27,57,77\) 和 \(87\) 均为合数。因此，这样的质数共有 \(6\) 个。

所以正确答案是 C。

English

The primes are \[ 7,\ 17,\ 37,\ 47,\ 67,\ 97. \] The other candidates are composite: \(27,57,77,\) and \(87.\) Hence there are \(6\) such primes.

Thus the correct answer is C.

4 答案 B

分子为 \[ 2+1+\frac12=\frac72\text{，}\]而分母为 \[ \frac14+\frac18+\frac1{16}=\frac7{16}\text{。}\]两者之商为 \(\frac{\frac{7}{2}}{\frac{7}{16}}=8\)。

所以正确答案是 B。

English

The numerator is \[ 2+1+\frac12=\frac72, \] while the denominator is \[ \frac14+\frac18+\frac1{16}=\frac7{16}. \] Their quotient is \(\frac{\frac{7}{2}}{\frac{7}{16}}=8.\)

Thus the correct answer is B.

5 答案 B

由于 \(12.5\%=\frac{1}{8}\)，总数 \(N\) 必须能被 \(8\) 整除。取 \(N=8\) 时，各频数为 \(1,0,4,2,1\)，均为整数。因此，最小可能总数为 \(8\)。

所以正确答案是 B。

English

Since \(12.5\%=\frac{1}{8},\) the total \(N\) must be divisible by \(8.\) Taking \(N=8\) gives frequencies \(1,0,4,2,1,\) all whole numbers. Thus the smallest possible total is \(8.\)

Therefore the correct answer is B.

6 答案 A

设三角形 \(ADB\) 在 \(A\) 和 \(B\) 处未标出的角分别为 \(\alpha\) 和 \(\beta\)。则 \[ \alpha+\beta+w=180^\circ\text{。}\]三角形 \(ABC\) 的内角和给出 \[ (\alpha+y)+(\beta+z)+x=180^\circ\text{。}\]两式相减，得到 \(x+y+z=w\)，所以 \(x=w-y-z\)。

所以正确答案是 A。

English

Let the unlabelled angles of triangle \(ADB\) at \(A\) and \(B\) be \(\alpha\) and \(\beta.\) Then \[ \alpha+\beta+w=180^\circ. \] The angle sum of triangle \(ABC\) gives \[ (\alpha+y)+(\beta+z)+x=180^\circ. \] Subtracting yields \(x+y+z=w,\) so \(x=w-y-z.\)

Thus the correct answer is A.

7 答案 C

若公共值为 \(k\)，则 \[ \begin{aligned} a&=k+1,& b&=k-2,\\ c&=k+3,& d&=k-4 \end{aligned}\text{。}\]因此，\(c\) 总是最大的。

所以正确答案是 C。

English

If the common value is \(k,\) then \[ \begin{aligned} a&=k+1,& b&=k-2,\\ c&=k+3,& d&=k-4. \end{aligned} \] Therefore \(c\) is always the largest.

Thus the correct answer is C.

8 答案 C

三角形 \(BCD\) 是一个 \(3\)-\(4\)-\(5\) 直角三角形，所以 \(BD=5\)。从 \(A\) 到 \(D\) 的水平位移为 \(13-3=10\)，竖直位移为 \(4\)，因此 \[ AD=\sqrt{10^2+4^2}=\sqrt{116}\text{。}\]由于 \(10\lt\sqrt{116}\lt11\)，所以 \(AD+BD\) 在 \(15\) 与 \(16\) 之间。

所以正确答案是 C。

English

Triangle \(BCD\) is a \(3\)-\(4\)-\(5\) right triangle, so \(BD=5.\) From \(A\) to \(D,\) the horizontal displacement is \(13-3=10\) and the vertical displacement is \(4,\) hence \[ AD=\sqrt{10^2+4^2}=\sqrt{116}. \] Since \(10\lt\sqrt{116}\lt11,\) the sum \(AD+BD\) is between \(15\) and \(16.\)

Thus the correct answer is C.

9 答案 B

设公差为 \(d\)。则 \(x=a+d,\ b=a+2d\)，且 \(2x=a+3d\)。因此 \[ 2(a+d)=a+3d\text{，}\]所以 \(a=d\)，且 \(b=3d\)。因此 \(\frac{a}{b}=\frac{1}{3}\)。

所以正确答案是 B。

English

Let the common difference be \(d.\) Then \(x=a+d,\ b=a+2d,\) and \(2x=a+3d.\) Thus \[ 2(a+d)=a+3d, \] so \(a=d\) and \(b=3d.\) Therefore \(\frac{a}{b}=\frac{1}{3}.\)

Thus the correct answer is B.

10 答案 D

将 \(a=bc,\ b=ca,\ c=ab\) 相乘，得到 \[ abc=(abc)^2\text{。}\]由于变量都不为零，所以 \(abc=1\)。此外，\(abc=a^2=b^2=c^2\)，所以每个变量都是 \(1\) 或 \(-1\)。乘积为 \(1\) 时，要么没有负号，要么恰有两个负号，因此共有 \(1+3=4\) 个有序三元组。

所以正确答案是 D。

English

Multiplying \(a=bc,\ b=ca,\ c=ab\) gives \[ abc=(abc)^2. \] Since none of the variables is zero, \(abc=1.\) Also \(abc=a^2=b^2=c^2,\) so each variable is \(1\) or \(-1.\) Product \(1\) allows either no negative signs or exactly two, giving \(1+3=4\) ordered triples.

Thus the correct answer is D.

11 答案 E

解方程组得 \[ x=\frac5{c+1},\qquad y=\frac{3-2c}{c+1}\text{。}\]条件 \(x\gt0\) 要求 \(c\gt-1\)。在分母为正的情况下，\(y\gt0\) 等价于 \(c\lt\frac{3}{2}\)。因此 \(-1\lt c\lt\frac{3}{2}\)。

所以正确答案是 E。

English

Solving the system gives \[ x=\frac5{c+1},\qquad y=\frac{3-2c}{c+1}. \] The condition \(x\gt0\) requires \(c\gt-1.\) With this positive denominator, \(y\gt0\) is equivalent to \(c\lt\frac{3}{2}.\) Hence \(-1\lt c\lt\frac{3}{2}.\)

Thus the correct answer is E.

12 答案 D

要使选项 D 中的 \(4\) 最先打完，信 \(1,2,3,4\) 必须都已送到；信堆中紧邻下面的 \(3\) 会在取出 \(4\) 后留在最上面。此时可以送来并打完信 \(5\)，但 \(3\) 仍在 \(2\) 的上面，所以接下来不可能打信 \(2\)。因此，\(4,5,2,3,1\) 是不可能的顺序。其余每个所列顺序都可以通过交错安排送信和打字来实现。

所以正确答案是 D。

English

To type \(4\) first in choice D, letters \(1,2,3,4\) must all have been delivered, leaving \(3\) on top after \(4\) is removed. Letter \(5\) can then be delivered and typed, but \(3\) is still above \(2,\) so \(2\) cannot be typed next. Thus \(4,5,2,3,1\) is impossible. Each other listed order can be produced by interleaving deliveries and typings.

Therefore the correct answer is D.

13 答案 A

首末直径的平均值为 \[ \frac{2+10}{2}=6\text{ cm}\text{。}\]因此，总长度约为 \(600(6\pi)=3600\pi\) 厘米，即 \(36\pi\) 米。

所以正确答案是 A。

English

The average of the first and last diameters is \[ \frac{2+10}{2}=6\text{ cm}. \] Hence the total length is approximately \(600(6\pi)=3600\pi\) cm, or \(36\pi\) meters.

Thus the correct answer is A.

14 答案 B

设正方形边长为 \(2\)，并取 \(A=(0,0),\ M=(1,2)\)、\(N=(2,1)\)。则 \[ \begin{aligned} \sin\theta &=\frac{|\det((1,2),(2,1))|} {\sqrt5\sqrt5}\\ &=\frac{|1-4|}{5} =\frac35 \end{aligned}\text{。}\]

所以正确答案是 B。

English

Take the square to have side \(2,\) with \(A=(0,0),\ M=(1,2),\) and \(N=(2,1).\) Then \[ \begin{aligned} \sin\theta &=\frac{|\det((1,2),(2,1))|} {\sqrt5\sqrt5}\\ &=\frac{|1-4|}{5} =\frac35. \end{aligned} \]

Thus the correct answer is B.

15 答案 D

第二个方程可化为 \[ \begin{aligned} xy(x+y)&+(x+y)\\ &=(xy+1)(x+y)\\ &=63 \end{aligned}\text{。}\]由于 \(xy=6\)，得到 \(x+y=9\)。因此 \[ \begin{aligned} x^2+y^2 &=(x+y)^2-2xy\\ &=81-12=69 \end{aligned}\text{。}\]

所以正确答案是 D。

English

The second equation becomes \[ \begin{aligned} xy(x+y)&+(x+y)\\ &=(xy+1)(x+y)\\ &=63. \end{aligned} \] Since \(xy=6,\) we get \(x+y=9.\) Therefore \[ \begin{aligned} x^2+y^2 &=(x+y)^2-2xy\\ &=81-12=69. \end{aligned} \]

Thus the correct answer is D.

16 答案 D

由于 \(VYX\) 比 \(VYZ\) 大一，\(X\) 所对应的数字比 \(Z\) 所对应的数字大一。下一次递增使 \(VYX\) 变为 \(VVW\)，所以 \(X=4,\ W=0\)、\(Z=3\)。进位使 \(Y\) 变为 \(V\)，从而 \(Y=1,\ V=2\)。因此 \[ \begin{aligned} XYZ_5&=413_5,\\ 413_5&=4\cdot25+5+3\\ &=108 \end{aligned}\text{。}\]

所以正确答案是 D。

English

Because \(VYX\) is one more than \(VYZ,\) the digit for \(X\) is one more than the digit for \(Z.\) The next increment changes \(VYX\) to \(VVW,\) so \(X=4,\ W=0,\) and \(Z=3.\) The carry changes \(Y\) to \(V,\) leaving \(Y=1,\ V=2.\) Thus \[ \begin{aligned} XYZ_5&=413_5,\\ 413_5&=4\cdot25+5+3\\ &=108. \end{aligned} \]

Thus the correct answer is D.

17 答案 E

设 \(A,B,C,D\) 分别表示安、比尔、卡萝尔、迪克的分数。条件为 \[ \begin{aligned} A+C&=B+D,\\ A+B&\gt C+D,\\ D&\gt B+C \end{aligned}\text{。}\]将前两个比较关系相加，得到 \(A\gt D\)。用不等式减去等式，得到 \(B\gt C\)。最后，\(D\gt B+C\ge B\)。因此 \[ A\gt D\gt B\gt C\text{。}\]

所以正确答案是 E。

English

Let \(A,B,C,D\) be the scores of Ann, Bill, Carol, and Dick. The conditions are \[ \begin{aligned} A+C&=B+D,\\ A+B&\gt C+D,\\ D&\gt B+C. \end{aligned} \] Adding the first two comparisons gives \(A\gt D.\) Subtracting the equality from the inequality gives \(B\gt C.\) Finally, \(D\gt B+C\ge B.\) Hence \[ A\gt D\gt B\gt C. \]

Thus the correct answer is E.

18 答案 D

令书架长度为 \(1\)，代数书和几何书的厚度分别为 \(a,g\)。则 \[ \begin{aligned} Aa+Hg&=1,\\ Sa+Mg&=1,\qquad Ea=1 \end{aligned}\text{。}\]将第一个方程乘以 \(M\)，第二个方程乘以 \(H\)，再相减，得到 \[ (AM-SH)a=M-H\text{。}\]因此 \[ E=\frac1a=\frac{AM-SH}{M-H}\text{。}\]

所以正确答案是 D。

English

Let the shelf length be \(1,\) and let \(a,g\) be the algebra- and geometry-book thicknesses. Then \[ \begin{aligned} Aa+Hg&=1,\\ Sa+Mg&=1,\qquad Ea=1. \end{aligned} \] Multiplying the first equation by \(M,\) the second by \(H,\) and subtracting gives \[ (AM-SH)a=M-H. \] Hence \[ E=\frac1a=\frac{AM-SH}{M-H}. \]

Thus the correct answer is D.

19 答案 B

有理化得 \[ \sqrt{65}-\sqrt{63}=\frac2{\sqrt{65}+\sqrt{63}}\text{。}\]分母小于 \(16\)，因为它的平方为 \(128+2\sqrt{4095}\lt256\)。所以该值大于 \(\frac{2}{16}=0.125\)。此外，两个根式都大于 \(7.5\)，所以该值小于 \(\frac{2}{15}<0.134\)。因此，它与 \(0.13\) 的距离小于与其他任何选项的距离。

所以正确答案是 B。

English

Rationalizing gives \[ \sqrt{65}-\sqrt{63}=\frac2{\sqrt{65}+\sqrt{63}}. \] The denominator is less than \(16,\) because its square is \(128+2\sqrt{4095}\lt256.\) Thus the value is greater than \(\frac{2}{16}=0.125.\) Also each radical exceeds \(7.5,\) so the value is less than \(\frac{2}{15}<0.134.\) It is therefore closer to \(0.13\) than to any other choice.

Thus the correct answer is B.

20 答案 A

对每个 \(k=1,\ldots,44\)，\[ \tan k^\circ\tan(90^\circ-k^\circ)=1\text{。}\]因此，每一对对数之和都为 \(\log_{10}1=0\)。剩下的中间项为 \(\log_{10}(\tan45^\circ)=0\)。所以整个和为 \(0\)。

所以正确答案是 A。

English

For each \(k=1,\ldots,44,\) \[ \tan k^\circ\tan(90^\circ-k^\circ)=1. \] Thus each paired sum of logarithms is \(\log_{10}1=0.\) The remaining middle term is \(\log_{10}(\tan45^\circ)=0.\) Hence the whole sum is \(0.\)

Thus the correct answer is A.

21 答案 B

第一个正方形的边长为 \(21\)，所以三角形的两条直角边长均为 \(42\)。设倾斜正方形的边长为 \(s\)。位于斜边上的那条边与连接两条直角边的对边平行。这条对边的两个端点为 \((0,k)\) 和 \((k,0)\)，所以 \(s=k\sqrt2\)。平行直线 \(x+y=k\) 与 \(x+y=42\) 之间的距离也等于 \(s\)，因此 \[ s=\frac{42-k}{\sqrt2},\qquad k=\frac{s}{\sqrt2}\text{。}\]由此 \(s=14\sqrt2\)，正方形的面积为 \(s^2=392\)。

因此，正确答案是 B。

English

The first square has side \(21,\) so the triangle’s legs have length \(42.\) Let \(s\) be the side of the tilted square. Its side on the hypotenuse is parallel to the opposite side joining the legs. That opposite side has endpoints \((0,k)\) and \((k,0),\) so \(s=k\sqrt2.\) The distance between the parallel lines \(x+y=k\) and \(x+y=42\) is also \(s,\) giving \[ s=\frac{42-k}{\sqrt2},\qquad k=\frac{s}{\sqrt2}. \] Hence \(s=14\sqrt2,\) and its area is \(s^2=392.\)

Thus the correct answer is B.

22 答案 C

在竖直截面中，宽 \(24\) 厘米的洞的一半是一条长为 \(12\) 的弦段。若球的半径为 \(r\)，则球心到冰面的距离为 \(r-8\)。由此形成的直角三角形满足 \[ (r-8)^2+12^2=r^2\text{。}\]解得 \(16r=208\)，所以 \(r=13\)。

因此，正确答案是 C。

English

In a vertical cross-section, half the \(24\)-cm hole is a chord segment of length \(12.\) If the sphere radius is \(r,\) the distance from its center to the ice plane is \(r-8.\) The resulting right triangle gives \[ (r-8)^2+12^2=r^2. \] Solving yields \(16r=208,\) so \(r=13.\)

Thus the correct answer is C.

23 答案 D

整数根满足 \[ x^2=p(444-x)\text{。}\]因此 \(x\) 能被 \(p\) 整除，令 \(x=np\)。代入可得 \[ n(n+1)p=444=2^2\cdot3\cdot37\text{。}\]连续两个整数的乘积这一条件只有在 \(p=37\) 时成立，此时 \(n=3\) 或 \(-4\)。两个根是 \(111\) 和 \(-148\)，所以 \(31\lt p\le41\)。

因此，正确答案是 D。

English

An integer root satisfies \[ x^2=p(444-x). \] Thus \(x\) is divisible by \(p,\) so write \(x=np.\) Substitution gives \[ n(n+1)p=444=2^2\cdot3\cdot37. \] The consecutive product condition works only with \(p=37,\) for which \(n=3\) or \(-4.\) The roots are \(111\) and \(-148,\) so \(31\lt p\le41.\)

Thus the correct answer is D.

24 答案 B

设 \(f\) 的次数为 \(n\ge1\)，首项系数为 \(a\)。三个表达式的次数分别为 \(2n,2n,n^2\)，所以 \(n^2=2n\)，从而 \(n=2\)。比较 \(f(x^2)=[f(x)]^2\) 两边的首项系数，得到 \(a=a^2\)，因此 \(a=1\)。

令 \(f(x)=x^2+bx+c\)。那么 \[ x^4+bx^2+c=(x^2+bx+c)^2\text{。}\]三次项系数迫使 \(b=0\)，随后二次项系数迫使 \(c=0\)。因此唯一可能的函数是 \(f(x)=x^2\)，而它确实使三个表达式都相等。所以恰有一个这样的函数。

因此，正确答案是 B。

English

Let \(f\) have degree \(n\ge1\) and leading coefficient \(a.\) The three expressions have degrees \(2n,2n,n^2,\) so \(n^2=2n\) and therefore \(n=2.\) Comparing leading coefficients in \(f(x^2)=[f(x)]^2\) gives \(a=a^2,\) hence \(a=1.\)

Write \(f(x)=x^2+bx+c.\) Then \[ x^4+bx^2+c=(x^2+bx+c)^2. \] The cubic coefficient forces \(b=0,\) and then the quadratic coefficient forces \(c=0.\) Thus the only candidate is \(f(x)=x^2,\) which indeed satisfies all three expressions. There is exactly one function.

Thus the correct answer is B.

25 答案 C

对于 \(C=(u,v)\)，三角形的面积为 \[ \frac12|36v-15u|\text{。}\]因为 \(\gcd(36,15)=3\)，所以行列式的最小正值至少是 \(3\)。这个值确实可以取得，例如取 \(u=7,\ v=3\)，就有 \(36(3)-15(7)=3\)。因此最小面积为 \(\frac{3}{2}\)。

因此，正确答案是 C。

English

For \(C=(u,v),\) the area is \[ \frac12|36v-15u|. \] Since \(\gcd(36,15)=3,\) the smallest possible positive value of the determinant is at least \(3.\) It is attained, for example, by \(u=7,\ v=3,\) since \(36(3)-15(7)=3.\) Therefore the minimum area is \(\frac{3}{2}.\)

Thus the correct answer is C.

26 答案 B

令第一部分为 \(x\in[0,2.5]\)，则第二部分为 \(2.5-x\)。忽略概率为零的端点，四舍五入后的两个数之和恰为 \(3\) 的条件是 \[ \begin{gathered} \frac12\lt x\lt1\\ \text{或}\\ \frac32\lt x\lt2 \end{gathered}\text{。}\]这两个区间的总长度为 \(1\)，而样本区间的长度为 \(2.5\)。因此所求概率为 \(\frac{1}{2.5}=\frac{2}{5}\)。

因此，正确答案是 B。

English

Let the first part be \(x\in[0,2.5],\) so the second is \(2.5-x.\) Ignoring endpoints of probability zero, the rounded values sum to \(3\) exactly when \[ \begin{gathered} \frac12\lt x\lt1\\ \text{or}\\ \frac32\lt x\lt2. \end{gathered} \] These intervals have total length \(1,\) out of a sample interval of length \(2.5.\) The probability is therefore \(\frac{1}{2.5}=\frac{2}{5}.\)

Thus the correct answer is B.

27 答案 B

这三个平面恰好是两个坐标相等时的分界面。在这些平面之外，每一块都由 \(x,y,z\) 的一种严格排序决定，例如 \(x\lt y\lt z\)。\(3!=6\) 种排序中的每一种都会在立方体内部出现，所以共有 \(6\) 块。

因此，正确答案是 B。

English

The three planes are precisely the boundaries where two coordinates are equal. Away from them, each piece is determined by a strict ordering of \(x,y,z,\) such as \(x\lt y\lt z.\) Every one of the \(3!=6\) orderings occurs inside the cube, so there are \(6\) pieces.

Thus the correct answer is B.

28 答案 D

若 \(|r|=1\)，则 \(\frac{1}{r}=\overline r\)。所以所有根的倒数之和，就是所有根之和的共轭。根据韦达定理，所有根之和为 \(-a\)，这是实数，其共轭仍为 \(-a\)。

因此，正确答案是 D。

English

If \(|r|=1,\) then \(\frac{1}{r}=\overline r.\) Therefore the sum of the reciprocals is the conjugate of the sum of the roots. By Vieta’s formulas, the sum of the roots is \(-a,\) which is real. Its conjugate is still \(-a.\)

Thus the correct answer is D.

29 答案 A

设 \(N(r)\) 表示数值 \(r\) 出现时的下标。反向运用递推关系可得：\(N(r)=2N(r-1)\) 适用于 \(r\gt1\)，而 \(N(r)=N(\frac{1}{r})+1\) 适用于 \(0\lt r\lt1\)。

从 \(N(1)=1\) 开始反复运用这些关系，得到 \(N(2)=2\)、\(N(\frac{1}{2})=3\)、\(N(\frac{3}{2})=6\) 和 \(N(\frac{2}{3})=7\)。继续可得 \(N(\frac{5}{3})=14\)、\(N(\frac{8}{3})=28\)、\(N(\frac{3}{8})=29\) 和 \(N(\frac{11}{8})=58\)。

接下来，\(N(\frac{8}{11})=59\)、\(N(\frac{19}{11})=118\)、\(N(\frac{11}{19})=119\)、\(N(\frac{30}{19})=238\)、\(N(\frac{49}{19})=476\)，并且 \(N(\frac{68}{19})=952\)。最后，\(N(\frac{87}{19})=1904\)，而 \(N(\frac{19}{87})=1905\)。因此 \(n=1905\)，其各位数字之和为 \(1+9+0+5=15\)。

因此，正确答案是 A。

English

Let \(N(r)\) be the index at which the value \(r\) occurs. Reversing the recursion gives \(N(r)=2N(r-1)\) for \(r\gt1,\) and \(N(r)=N(\frac{1}{r})+1\) for \(0\lt r\lt1.\)

Starting with \(N(1)=1,\) repeated use gives \(N(2)=2,\) \(N(\frac{1}{2})=3,\) \(N(\frac{3}{2})=6,\) and \(N(\frac{2}{3})=7.\) Continuing gives \(N(\frac{5}{3})=14,\) \(N(\frac{8}{3})=28,\) \(N(\frac{3}{8})=29,\) and \(N(\frac{11}{8})=58.\)

Next, \(N(\frac{8}{11})=59,\) \(N(\frac{19}{11})=118,\) \(N(\frac{11}{19})=119,\) \(N(\frac{30}{19})=238,\) \(N(\frac{49}{19})=476,\) and \(N(\frac{68}{19})=952.\) Finally, \(N(\frac{87}{19})=1904\) and \(N(\frac{19}{87})=1905.\) Thus \(n=1905,\) whose digit sum is \(1+9+0+5=15.\)

Therefore the correct answer is A.

30 答案 E

令 \(AB=1,\ A=(0,0)\)，且 \(B=(1,0)\)。因为 \(\angle A=45^\circ\)，所以边 \(AC\) 位于直线 \(y=x\) 上。过点 \(B\)、以 \(30^\circ\) 的夹角偏离射线 \(BA\) 的直线和它相交于高度 \[ h_C=\frac1{1+\sqrt3}\text{。}\]设 \(d=AD\)。射线 \(DE\) 的斜率为 \(-\sqrt3\)，所以它与 \(y=x\) 的交点高度为 \[ h_E=\frac{\sqrt3\,d}{1+\sqrt3}\text{。}\]两部分面积相等，故 \[ \frac12d h_E=\frac12\left(\frac12h_C\right)\text{。}\]因此 \[ \begin{aligned} \frac{\sqrt3\,d^2}{2(1+\sqrt3)} &=\frac1{4(1+\sqrt3)},\\ d^2&=\frac1{2\sqrt3} =\frac1{\sqrt{12}} \end{aligned}\text{。}\]所以 \(\frac{AD}{AB}=d=\frac{1}{\sqrt[4]{12}}\)。

因此，正确答案是 E。

English

Set \(AB=1,\ A=(0,0),\) and \(B=(1,0).\) Since \(\angle A=45^\circ,\) side \(AC\) lies on \(y=x.\) The line through \(B\) making angle \(30^\circ\) with \(BA\) meets it at height \[ h_C=\frac1{1+\sqrt3}. \] Let \(d=AD.\) The ray \(DE\) has slope \(-\sqrt3,\) so its intersection with \(y=x\) has height \[ h_E=\frac{\sqrt3\,d}{1+\sqrt3}. \] Equal areas require \[ \frac12d h_E=\frac12\left(\frac12h_C\right). \] Therefore \[ \begin{aligned} \frac{\sqrt3\,d^2}{2(1+\sqrt3)} &=\frac1{4(1+\sqrt3)},\\ d^2&=\frac1{2\sqrt3} =\frac1{\sqrt{12}}. \end{aligned} \] Hence \(\frac{AD}{AB}=d=\frac{1}{\sqrt[4]{12}}.\)

Thus the correct answer is E.