2007B AMC 12 真题（P版）

每间卧室地面周长为 \(2(12+10)=44\) 英尺。

所以一间卧室的墙面粉刷面积为 \(44\cdot8-60=352-60=292\) 平方英尺。三间卧室共需粉刷 \(3\cdot292=876\) 平方英尺。

所以正确答案是 E。

English

The perimeter of each bedroom floor is \(2(12+10)=44\) feet.

So the wall area in one bedroom is \(44\cdot8-60=352-60=292\) square feet. Across all three bedrooms, Isabella paints \(3\cdot292=876\) square feet.

Thus, the correct answer is E.

去程用 \(\frac{120}{30}=4\) 加仑，返程用 \(\frac{120}{20}=6\) 加仑。

往返共 \(240\) 英里，用油 \(10\) 加仑，所以平均每加仑行驶的英里数为 \[ \dfrac{240}{10}=24\text{。}\]

所以正确答案是 B。

English

The trip home uses \(\frac{120}{30}=4\) gallons, and the return trip uses \(\frac{120}{20}=6\) gallons.

The round trip covers \(240\) miles on \(10\) gallons, so the average is \[ \dfrac{240}{10}=24. \]

Thus, the correct answer is B.

点 \(O\) 周围角和为 \(360^\circ\)，所以 \[ \begin{aligned} \angle AOC&=360^\circ-140^\circ-120^\circ \\ &=100^\circ \end{aligned}\text{。}\]

由圆周角定理，\(\angle ABC\) 与圆心角 \(\angle AOC\) 截同一弧 \(AC\)，所以 \[ \angle ABC=\tfrac12\angle AOC=50^\circ\text{。}\]

所以正确答案是 D。

English

The angles around \(O\) sum to \(360^\circ,\) so \[ \begin{aligned} \angle AOC&=360^\circ-140^\circ-120^\circ \\ &=100^\circ. \end{aligned} \]

By the inscribed angle theorem, \(\angle ABC\) subtends the same arc \(AC\) as the central angle \(\angle AOC,\) so \[ \angle ABC=\tfrac12\angle AOC=50^\circ. \]

Thus, the correct answer is D.

因为 \(3\) 根香蕉价格等于 \(2\) 个苹果，\(18\) 根香蕉价格等于 \(12\) 个苹果。

又因为 \(6\) 个苹果价格等于 \(4\) 个橙子，\(12\) 个苹果价格等于 \(8\) 个橙子，所以 \(18\) 根香蕉等价于 \(8\) 个橙子。

所以正确答案是 B。

English

Since \(3\) bananas cost as much as \(2\) apples, \(18\) bananas cost as much as \(12\) apples.

Since \(6\) apples cost as much as \(4\) oranges, \(12\) apples cost as much as \(8\) oranges. Therefore \(18\) bananas cost as much as \(8\) oranges.

Thus, the correct answer is B.

三道空题给 \(3\cdot1.5=4.5\) 分。

所以 Sarah 需要从答对题中得到至少 \(100-4.5=95.5\) 分。因为 \[ 15\lt\dfrac{95.5}{6}\lt16\text{，} \] 她至少要答对 \(16\) 题，此时得分为 \(16\cdot6+4.5=100.5\)。

所以正确答案是 D。

English

The three unanswered problems give \(3\cdot1.5=4.5\) points.

So Sarah needs at least \(100-4.5=95.5\) points from correct answers. Since \[ 15\lt\dfrac{95.5}{6}\lt16, \] she must solve at least \(16\) problems correctly, which would give her \(16\cdot6+4.5=100.5\) points.

Thus, the correct answer is D.

周长为 \(5+6+7=18\)，因此每只虫子在相遇前走了 \(9\) 个长度单位。

沿 \(A\to B\to C\) 方向的虫子到达 \(D\) 时走了 \(AB+BD=9\)，该点在 \(BC\) 上，且 \(AB=5\)，所以 \(BD=4\)。

所以正确答案是 D。

English

The perimeter is \(5+6+7=18,\) so each bug crawls \(9\) before they meet.

The bug going \(A\to B\to C\) reaches \(D\) on side \(BC,\) having traveled \(AB+BD=9.\) Since \(AB=5,\) we get \(BD=4.\)

Thus, the correct answer is D.

因为 \(AB=BC=EA\) 且 \(\angle A=\angle B=90^\circ\)，四边形 \(ABCE\) 是正方形，所以 \(\angle AEC=90^\circ\)，且 \(EC\) 等于公共边长。

又 \(CD=DE=EC\)，所以 \(\triangle CDE\) 是等边三角形，\(\angle CED=60^\circ\)。因此 \[ \begin{aligned} \angle E&=\angle AEC+\angle CED \\ &=90^\circ+60^\circ \\ &=150^\circ \end{aligned}\text{。}\]

所以正确答案是 E。

English

Because \(AB=BC=EA\) and \(\angle A=\angle B=90^\circ,\) quadrilateral \(ABCE\) is a square, so \(\angle AEC=90^\circ\) and \(EC\) equals the common side length.

Then \(CD=DE=EC,\) so \(\triangle CDE\) is equilateral and \(\angle CED=60^\circ.\) Therefore \[ \begin{aligned} \angle E&=\angle AEC+\angle CED \\ &=90^\circ+60^\circ \\ &=150^\circ. \end{aligned} \]

Thus, the correct answer is E.

\(N\) 年前 Tom 年龄为 \(T-N\)。三个孩子每人少 \(N\) 岁，所以他们当时总年龄为 \(T-3N\)。

条件给出 \[ T-N=2(T-3N)\text{，}\] 因而 \(5N=T\)，所以 \(\frac{T}{N}=5\)。

所以正确答案是 D。

English

Tom’s age \(N\) years ago was \(T-N.\) His three children were each \(N\) years younger, so their ages then totaled \(T-3N.\)

The condition gives \[ T-N=2(T-3N), \] so \(5N=T\) and \(\frac{T}{N}=5.\)

Thus, the correct answer is D.

令 \(3x-1=5\)，得 \(x=2\)。

于是 \[ f(5)=2^2+2+1=7\text{。}\]

所以正确答案是 A。

English

Setting \(3x-1=5\) gives \(x=2.\)

Then \[ f(5)=2^2+2+1=7. \]

Thus, the correct answer is A.

两名女生离开、两名男生加入，总人数不变，女生比例从 \(40\%\) 降到 \(30\%\)。

下降的 \(10\%\) 就对应这两名女生，所以全组有 \(20\) 人，起初女生人数为 \(20\) 的 \(40\%\)，即 \(8\) 人。

所以正确答案是 C。

English

Since two girls leave while two boys arrive, the total group size is unchanged. The drop from \(40\%\) to \(30\%\) girls corresponds to the two girls who left.

So those two girls are \(10\%\) of the group, meaning the group has \(20\) people. The initial number of girls was \(40\%\) of \(20,\) or \(8.\)

Thus, the correct answer is C.

令 \(x=\angle A\)，则 \(\angle B=\tfrac x2\)、\(\angle C=\tfrac x3\)、\(\angle D=\tfrac x4\)。

角和给出 \[ x+\dfrac x2+\dfrac x3+\dfrac x4=\dfrac{25x}{12}=360\text{，} \] 所以 \(x=\dfrac{12\cdot360}{25}=172.8\approx173\)。

所以正确答案是 D。

English

Let \(x=\angle A.\) Then \(\angle B=\tfrac x2,\) \(\angle C=\tfrac x3,\) and \(\angle D=\tfrac x4.\)

The angle sum gives \[ x+\dfrac x2+\dfrac x3+\dfrac x4=\dfrac{25x}{12}=360, \] so \(x=\dfrac{12\cdot360}{25}=172.8\approx173.\)

Thus, the correct answer is D.

取班上有 \(10\) 名学生，则有 \(1\) 名三年级和 \(9\) 名四年级。

全班总分为 \(10\cdot84=840\)，四年级学生贡献 \(9\cdot83=747\)。所以那名三年级学生的得分为 \[ 840-747=93\text{。}\]

所以正确答案是 C。

English

Take a class of \(10\) students, so there is \(1\) junior and \(9\) seniors.

The total of all scores is \(10\cdot84=840,\) and the seniors contribute \(9\cdot83=747.\) So the junior scored \[ 840-747=93. \]

Thus, the correct answer is C.

周期长为 \(30+3+30=63\) 秒，每周期有三次变色。

Leah 看到变色当且仅当她的三秒观察区间在某次变色前的 \(3\) 秒内开始。共有 \(3\cdot3=9\) 秒的有利开始时间，在 \(63\) 秒周期中概率为 \[ \dfrac{9}{63}=\dfrac17\text{。}\]

所以正确答案是 D。

English

The cycle length is \(30+3+30=63\) seconds, with three color changes per cycle.

Leah sees a change exactly when her three-second interval starts within the \(3\) seconds before a switch. That gives \(3\cdot3=9\) favorable seconds out of \(63,\) a probability of \[ \dfrac{9}{63}=\dfrac17. \]

Thus, the correct answer is D.

设 \(s=AB\)。连接 \(P\) 与各顶点，把三角形分成 \(\triangle PAB\)、\(\triangle PBC\)、\(\triangle PCA\)，它们的面积分别为 \(\tfrac{s}{2}\)、\(s\)、\(\tfrac{3s}{2}\)。

这些面积总和为 \(3s\)，必须等于等边三角形面积 \(\tfrac{\sqrt3}{4}s^2\)。因此 \[ 3s=\dfrac{\sqrt3}{4}s^2\text{，} \] 得 \(s=\dfrac{12}{\sqrt3}=4\sqrt3\)。

所以正确答案是 D。

English

Let \(s=AB.\) Joining \(P\) to the vertices splits the triangle into \(\triangle PAB,\) \(\triangle PBC,\) and \(\triangle PCA,\) with areas \(\tfrac{s}{2},\) \(s,\) and \(\tfrac{3s}{2}.\)

Their total is \(3s,\) which must equal the area \(\tfrac{\sqrt3}{4}s^2\) of the equilateral triangle. So \[ 3s=\dfrac{\sqrt3}{4}s^2, \] giving \(s=\dfrac{12}{\sqrt3}=4\sqrt3.\)

Thus, the correct answer is D.

奇次幂项为 \(ar+ar^3+\cdots\) \(=r(a+ar^2+\cdots)\)，即偶次幂项之和的 \(r\) 倍。偶次幂项之和为 \(7-3=4\)。

因此 \(3=4r\)，得 \(r=\tfrac34\)。又 \(a=7(1-r)=\tfrac74\)，\[ a+r=\dfrac74+\dfrac34=\dfrac52\text{。}\]

所以正确答案是 E。

English

The odd-power terms are \(ar+ar^3+\cdots\) \(=r(a+ar^2+\cdots),\) that is, \(r\) times the even-power terms. The even-power terms sum to \(7-3=4.\)

So \(3=4r,\) giving \(r=\tfrac34.\) Then \(a=7(1-r)=\tfrac74,\) and \[ a+r=\dfrac74+\dfrac34=\dfrac52. \]

Thus, the correct answer is E.

四面体旋转群有 \(12\) 个元素：恒等、\(8\) 个绕顶点-对面轴的 \(3\) 阶旋转，以及 \(3\) 个绕对边中点轴的 \(2\) 阶旋转。

恒等固定全部 \(3^4=81\) 种涂色。每个顶点旋转固定一个面并轮换另外三个面，所以固定 \(3^2=9\) 种涂色；同样地，每个边旋转交换两对面，也固定 \(3^2=9\) 种。

由 Burnside 引理，可区分涂色数为 \[ \dfrac{81+8\cdot9+3\cdot9}{12}=\dfrac{180}{12}=15\text{。}\]

所以正确答案是 A。

English

The rotation group of the tetrahedron has \(12\) elements: the identity, \(8\) rotations of order \(3\) about a vertex-face axis, and \(3\) rotations of order \(2\) about an edge-midpoint axis.

The identity fixes all \(3^4=81\) colorings. Each vertex rotation fixes one face and cycles the other three, so it fixes \(3^2=9\) colorings; likewise each edge rotation swaps two pairs of faces and fixes \(3^2=9.\)

By Burnside’s lemma the number of distinguishable colorings is \[ \dfrac{81+8\cdot9+3\cdot9}{12}=\dfrac{180}{12}=15. \]

Thus, the correct answer is A.

对所有 \(b\gt0\) 都有 \(b\lt10^b\)，所以 \(\log_{10}b\lt b\)。若 \(b=1\)，则 \(a=0\)，不符合条件。若 \(b\gt1\)，则 \(0\lt\dfrac{\log_{10}b}{b^2}\lt1\)，因此 \(a\) 不可能是非零整数。

所以 \(0\lt b\lt1\)，从而 \(\log_{10}b\lt0\)，且 \(a=\dfrac{\log_{10}b}{b^2}\lt0\)。排序为 \(a\lt0\lt b\lt1\lt\dfrac1b\)，所以中位数是 \(b\)。

所以正确答案是 D。

English

For \(b\gt0,\) we have \(b\lt10^b,\) hence \(\log_{10}b\lt b.\) If \(b=1,\) then \(a=0.\) If \(b\gt1,\) then \(0\lt\dfrac{\log_{10}b}{b^2}\lt1,\) so \(a\) could not be an integer.

Hence \(0\lt b\lt1,\) so \(\log_{10}b\lt0\) and \(a=\dfrac{\log_{10}b}{b^2}\lt0.\) Thus \(a\lt0\lt b\lt1\lt\dfrac1b,\) and the middle value of the sorted set is \(b.\)

Thus, the correct answer is D.

设较小平方数为 \(N^2\)，较大平方数为 \((N+1)^2\)，两者间隔为 \(2N+1\)。则 \[ \overline{abc}=N^2+\dfrac{2N+1}{3}\text{，}\]\[ \overline{acb}=N^2+\dfrac{2(2N+1)}{3}\text{。}\]

两式相减，\(\overline{acb}-\overline{abc}=9(c-b)\) \(=\dfrac{2N+1}{3}\)，所以 \(27(c-b)=2N+1\)。因为 \(\overline{acb}\) 在区间中位置更靠后，所以 \(c-b\) 为正；又因为右边是奇数，所以 \(c-b\) 是奇数。若 \(c-b\ge3\)，则 \(N\ge40\)，且 \(N^2\) 不再是三位数。

因此 \(c-b=1\)，得到 \(N=13\)。从 \(13^2=169\) 到 \(14^2=196\) 的三分之一与三分之二处分别为 \(178\) 和 \(187\)，所以 \(a+b+c=1+7+8=16\)。

所以正确答案是 C。

English

Let the smaller square be \(N^2,\) so the larger is \((N+1)^2\) and the gap is \(2N+1.\) Then \[ \overline{abc}=N^2+\dfrac{2N+1}{3}, \] \[ \overline{acb}=N^2+\dfrac{2(2N+1)}{3}. \]

Subtracting, \(\overline{acb}-\overline{abc}=9(c-b)\) \(=\dfrac{2N+1}{3},\) so \(27(c-b)=2N+1.\) Since \(\overline{acb}\) is farther along the interval, \(c-b\) is positive; and because the right side is odd, \(c-b\) is odd. If \(c-b\ge3,\) then \(N\ge40\) and \(N^2\) is not three digits.

So \(c-b=1,\) giving \(N=13.\) The points one third and two thirds of the way from \(13^2=169\) to \(14^2=196\) are \(178\) and \(187,\) so \(a+b+c=1+7+8=16.\)

Thus, the correct answer is C.

令 \(\theta=\angle ABC\)。底面圆周长为 \(6\)，所以半径为 \(\dfrac{6}{2\pi}=\dfrac3\pi\)。圆柱高为菱形的高 \(6\sin\theta\)。

体积为 \[ \pi\left(\dfrac3\pi\right)^2(6\sin\theta)=\dfrac{54}{\pi}\sin\theta=6\text{，} \] 所以 \(\sin\theta=\dfrac{\pi}{9}\)。

所以正确答案是 A。

English

Let \(\theta=\angle ABC.\) The base circle has circumference \(6,\) so its radius is \(\dfrac{6}{2\pi}=\dfrac3\pi.\) The height of the cylinder is the rhombus altitude \(6\sin\theta.\)

The volume is \[ \pi\left(\dfrac3\pi\right)^2(6\sin\theta)=\dfrac{54}{\pi}\sin\theta=6, \] so \(\sin\theta=\dfrac{\pi}{9}.\)

Thus, the correct answer is A.

第一个平行四边形有两个顶点在 \((0,c)\) 和 \((0,d)\)，另外两个顶点的 \(x\)-坐标为 \(\pm\dfrac{c-d}{b-a}\)。因此第一个平行四边形面积为 \(\dfrac{(c-d)^2}{|b-a|}=18\)，第二个为 \(\dfrac{(c+d)^2}{|b-a|}=72\)。

于是 \((c-d)^2=18|b-a|\)，\((c+d)^2=72|b-a|\)。两式相减得 \(4cd=54|b-a|\)，即 \(2cd=27|b-a|\)。

因此 \(|b-a|\) 是偶数，取 \(\{a,b\}=\{1,3\}\) 时 \(a+b\) 最小；又 \(cd\) 是 \(27\) 的倍数，取 \(\{c,d\}=\{3,9\}\) 时 \(c+d\) 最小。这些取值满足全部条件，于是 \(a+b+c+d=1+3+3+9\) \(=16\)。

所以正确答案是 D。

English

Two vertices of the first parallelogram lie at \((0,c)\) and \((0,d),\) and the other two have \(x\)-coordinates \(\pm\dfrac{c-d}{b-a}.\) Its area works out to \(\dfrac{(c-d)^2}{|b-a|}=18.\) The same computation for the second gives \(\dfrac{(c+d)^2}{|b-a|}=72.\)

So \((c-d)^2=18|b-a|\) and \((c+d)^2=72|b-a|.\) Subtracting, \(4cd=54|b-a|,\) i.e. \(2cd=27|b-a|.\)

Thus \(|b-a|\) is even, so \(a+b\) is smallest with \(\{a,b\}=\{1,3\};\) and \(cd\) is a multiple of \(27,\) so \(c+d\) is smallest with \(\{c,d\}=\{3,9\}.\) These satisfy all conditions, giving \(a+b+c+d=1+3+3+9\) \(=16.\)

Thus, the correct answer is D.

回文数由前半部分决定。按 \(3\) 进制位数计数，长度 \(1\) 或 \(2\) 的有 \(2\) 个，长度 \(3\) 或 \(4\) 的有 \(6\) 个，长度 \(5\) 或 \(6\) 的有 \(18\) 个，长度 \(7\) 的有 \(54\) 个。

最多七位的总数为 \(2+2+6+6+18+18\) \(+54=106\)。在 \(7\) 位数中，因为 \(2007=2202100_3\)，超过它的 \(7\) 位回文数为 \(2210122\)、\(2211122\)、\(2212122\)、\(2220222\)、\(2221222\)、\(2222222\)，共有 \(6\) 个。

因此答案为 \(106-6=100\)。

所以正确答案是 A。

English

A palindrome is fixed by its first half. Counting base-\(3\) palindromes by length gives \(2\) of length \(1\) or \(2,\) \(6\) of length \(3\) or \(4,\) \(18\) of length \(5\) or \(6,\) and \(54\) of length \(7.\)

That totals \(2+2+6+6+18+18\) \(+54=106\) palindromes with at most \(7\) digits. Since \(2007=2202100_3,\) the \(7\)-digit palindromes larger than it are \(2210122,\) \(2211122,\) \(2212122,\) \(2220222,\) \(2221222,\) and \(2222222,\) which is \(6\) of them.

Therefore the count is \(106-6=100.\)

Thus, the correct answer is A.

设 \(D,E,F\) 分别为 \(BC,CA,AB\) 的中点，设 \(X,Y,Z\) 分别为 \(AD,BE,CF\) 的中点。追踪一个始终位于两个粒子正中间的第三点：当两个粒子位于 \(A,D\) 时，它位于 \(X\)；当它们位于 \(F,C\) 时，它位于 \(Z\)；当它们位于 \(B,E\) 时，它位于 \(Y\)。

在这些时刻之间，两个粒子的位置都线性变化，所以它们的中点依次描出线段 \(XZ,ZY,YX\)。因此围成的轨迹是等边三角形 \(XYZ\)，由对称性，它与 \(\triangle ABC\) 有同一个中心 \(O\)。因为 \(Z\) 是中线 \(CF\) 的中点，\[ \begin{aligned} OZ&=OC-ZC \\ &=\dfrac23 CF-\dfrac12 CF \\ &=\dfrac16 CF \end{aligned}\text{，}\]而 \(OC=\dfrac23 CF\)。

所以外接圆半径之比为 \(\dfrac{OZ}{OC}=\dfrac14\)，面积之比为 \(\left(\dfrac14\right)^2=\dfrac{1}{16}\)。

所以正确答案是 A。

English

Let \(D,E,F\) be the midpoints of \(BC,CA,AB,\) respectively, and let \(X,Y,Z\) be the midpoints of \(AD,BE,CF,\) respectively. Track a third point halfway between the two particles. It is at \(X\) when the particles are at \(A,D;\) at \(Z\) when they are at \(F,C;\) and at \(Y\) when they are at \(B,E.\)

Between these instants both particle positions vary linearly, so their midpoint traces the segments \(XZ,ZY,YX.\) Thus the enclosed path is the equilateral triangle \(XYZ,\) which by symmetry shares the center \(O\) of \(\triangle ABC.\) Because \(Z\) is the midpoint of the median \(CF,\) \[ \begin{aligned} OZ&=OC-ZC \\ &=\dfrac23 CF-\dfrac12 CF \\ &=\dfrac16 CF, \end{aligned} \] while \(OC=\dfrac23 CF.\)

So the ratio of circumradii is \(\dfrac{OZ}{OC}=\dfrac14,\) and the area ratio is \(\left(\dfrac14\right)^2=\dfrac{1}{16}.\)

Thus, the correct answer is A.

设直角边为 \(a\le b\)。条件为 \(\tfrac12 ab=3\left(a+b+\sqrt{a^2+b^2}\right)\)，所以 \[ ab-6a-6b=6\sqrt{a^2+b^2}\text{。}\]

平方并化简得到 \(ab(ab-12a-12b+72)=0\)，所以 \((a-12)(b-12)=72\)。正整数解为 \((a,b)=(3,4)\)，\((13,84)\)，\((14,48)\)，\((15,36)\)，\((16,30)\)，\((18,24)\)，\((20,21)\)。

其中 \((3,4)\) 是伪解：它的面积为 \(6\)，而它的周长为 \(12\)，周长的三倍是 \(36\)。所以恰有 \(6\) 个三角形。

所以正确答案是 A。

English

Let the legs be \(a\le b.\) The condition is \(\tfrac12 ab=3\left(a+b+\sqrt{a^2+b^2}\right),\) so \[ ab-6a-6b=6\sqrt{a^2+b^2}. \]

Squaring and simplifying gives \(ab(ab-12a-12b+72)=0,\) hence \((a-12)(b-12)=72.\) The positive integer solutions are \((a,b)=(3,4),\) \((13,84),\) \((14,48),\) \((15,36),\) \((16,30),\) \((18,24),\) \((20,21).\)

The pair \((3,4)\) is extraneous: its area is \(6,\) while its perimeter is \(12\) and three times that is \(36.\) So exactly \(6\) triangles work.

Thus, the correct answer is A.

设表达式等于整数 \(k\)。乘以 \(b\) 并减去 \(a\)，得 \(\dfrac{14b^2}{9a}=bk-a\) 为整数。由于 \(\gcd(a,b)=1\)，可得 \(14\) 是 \(a\) 的倍数。另一方面，乘以 \(9a\) 并减去 \(14b\)，得 \(\dfrac{9a^2}{b}=9ak-14b\)，所以 \(9\) 是 \(b\) 的倍数。

因此 \(a\in\{1,2,7,14\}\)，\(b\in\{1,3,9\}\)。检验互质候选，表达式为整数的只有 \((a,b)=(1,3)\)，\((2,3)\)，\((7,3)\)，\((14,3)\)。

所以共有 \(4\) 对。

所以正确答案是 A。

English

Let \(k\) be the integer value of the original expression. Multiplying by \(b\) and subtracting \(a\) gives \(\dfrac{14b^2}{9a}=bk-a,\) an integer. Since \(\gcd(a,b)=1,\) it follows that \(14\) is divisible by \(a.\) Multiplying instead by \(9a\) and subtracting \(14b\) gives \(\dfrac{9a^2}{b}=9ak-14b,\) so \(9\) is divisible by \(b.\)

Thus \(a\in\{1,2,7,14\}\) and \(b\in\{1,3,9\}.\) Checking the coprime candidates, the expression is an integer only for \((a,b)=(1,3),\) \((2,3),\) \((7,3),\) \((14,3).\)

So there are \(4\) such pairs.

Thus, the correct answer is A.

取 \(D=(-1,0,0)\)、\(E=(1,0,0)\)，令 \(\triangle ABC\) 在平面 \(z=k\gt0\) 中。由于 \(\angle CDE\) 和 \(\angle DEA\) 为直角，\(A\) 和 \(C\) 分别位于以 \(E\) 和 \(D\) 为圆心、半径为 \(2\) 的圆上；这两个圆所在的平面分别为 \(x=1\) 和 \(x=-1\)。因此 \(A=(1,y_1,k),\ C=(-1,y_2,k)\)，其中 \(y_j=\pm\sqrt{4-k^2}\)。

因为 \(\angle ABC=90^\circ\)，所以 \(AC=2\sqrt2\)，这迫使 \(y_1=-y_2\)。取 \(y_1=1\)，\(y_2=-1\)，得 \(k=\sqrt3\)，于是 \(A=(1,1,\sqrt3)\)，\(C=(-1,-1,\sqrt3)\)，而 \(B\) 为 \((1,-1,\sqrt3)\) 或 \((-1,1,\sqrt3)\)。

第一种情况下 \(BE=2\) 且 \(BE\perp DE\)；第二种情况下 \(BD=2\) 且 \(BD\perp DE\)。无论哪种，\(\triangle BDE\) 的两条垂直边长度分别为 \(2\) 和 \(2\)，面积为 \(\tfrac12(2)(2)=2\)。

所以正确答案是 C。

English

Set \(D=(-1,0,0)\) and \(E=(1,0,0),\) and let \(\triangle ABC\) lie in the plane \(z=k\gt0.\) Because \(\angle CDE\) and \(\angle DEA\) are right angles, \(A\) and \(C\) lie on radius-\(2\) circles centered at \(E\) and \(D\) in the planes \(x=1\) and \(x=-1,\) so \(A=(1,y_1,k),\ C=(-1,y_2,k)\) with \(y_j=\pm\sqrt{4-k^2}.\)

Since \(\angle ABC=90^\circ,\) \(AC=2\sqrt2,\) which forces \(y_1=-y_2.\) Taking \(y_1=1,\) \(y_2=-1,\) gives \(k=\sqrt3,\) so \(A=(1,1,\sqrt3),\) \(C=(-1,-1,\sqrt3),\) and \(B\) is \((1,-1,\sqrt3)\) or \((-1,1,\sqrt3).\)

In the first case \(BE=2\) with \(BE\perp DE;\) in the second \(BD=2\) with \(BD\perp DE.\) Either way \(\triangle BDE\) has legs \(2\) and \(2,\) so its area is \(\tfrac12(2)(2)=2.\)

Thus, the correct answer is C.