2009B AMC 10 真题（P版）

1 答案 B

如果 Jane 买了 \(b\) 个贝果，则她买了 \(5-b\) 个松饼，总价为 \[ 50(5-b)+75b = 250+25b \] 美分。当 \(250+25b\) 是 \(100\) 的倍数时，总价为整数美元，也就是 \(25b\equiv50\pmod{100}\)，所以 \(b\equiv2\pmod4\)。

在 \(0\le b\le5\) 中，唯一可能是 \(b=2\)。

所以正确答案是 B。

English

If Jane buys \(b\) bagels, she buys \(5-b\) muffins, for a total of \[ 50(5-b)+75b = 250+25b \] cents. This is a whole number of dollars when \(250+25b\) is a multiple of \(100,\) that is, when \(25b\equiv50\pmod{100},\) or \(b\equiv2\pmod4.\)

The only value with \(0\le b\le5\) is \(b=2.\)

Thus, the correct answer is B.

2 答案 C

小分数的最小公分母是 \(12\)，所以分子和分母同乘 \(12\)：\[ \dfrac{\frac13-\frac14}{\frac12-\frac13}=\dfrac{4-3}{6-4}=\dfrac12\text{。}\]

所以正确答案是 C。

English

The least common denominator of the small fractions is \(12,\) so multiply top and bottom by \(12:\) \[ \dfrac{\frac13-\frac14}{\frac12-\frac13}=\dfrac{4-3}{6-4}=\dfrac12. \]

Thus, the correct answer is C.

3 答案 C

丢失的 \(3\) 桶油漆本来可以粉刷 \(30-25=5\) 间房，所以每间房需要 \(\dfrac35\) 桶。

\(25\) 间房需要 \(\dfrac35\cdot25=15\) 桶油漆。

所以正确答案是 C。

English

The lost \(3\) cans would have painted \(30-25=5\) rooms, so each room takes \(\dfrac35\) of a can.

For \(25\) rooms she used \(\dfrac35\cdot25=15\) cans.

Thus, the correct answer is C.

4 答案 C

两条平行边相差 \(25-15=10\)，平均分到两个三角形上，所以每个等腰直角三角形的直角边为 \(5\)，面积为 \(\dfrac12\cdot5^2=\dfrac{25}{2}\)。

两块花坛总面积为 \(25\) 平方米。长方形长 \(25\)、宽 \(5\)，面积为 \(125\)。所占比例为 \(\dfrac{25}{125}=\dfrac15\)。

所以正确答案是 C。

English

The two parallel sides differ by \(25-15=10,\) split evenly between the two triangles, so each isosceles right triangle has legs of length \(5\) and area \(\dfrac12\cdot5^2=\dfrac{25}{2}.\)

Together the beds cover \(25\) square meters. The rectangle has length \(25\) and width \(5,\) so area \(125.\) The fraction is \(\dfrac{25}{125}=\dfrac15.\)

Thus, the correct answer is C.

5 答案 D

比 \(60\) 少百分之二十是 \(\dfrac45\cdot60=48\)。

若未知数为 \(n\)，则比 \(n\) 多三分之一为 \(\dfrac43 n\)，所以 \[ \dfrac43 n=48 \] \[ n=36\text{。}\]

所以正确答案是 D。

English

Twenty percent less than \(60\) is \(\dfrac45\cdot60=48.\)

If \(n\) is the unknown number, one-third more than \(n\) is \(\dfrac43 n,\) so \[ \dfrac43 n=48 \]\[ n=36. \]

Thus, the correct answer is D.

6 答案 D

因为 \(128=2^7\)，每个年龄都是 \(2\) 的幂。设双胞胎的共同年龄为 \(t\)，Kiana 的年龄为 \(k\)，则需要 \(t^2k=128\)，且 \(k\lt t\)。

写成 \(t=2^j\)。于是 \(k=2^{7-2j}\)。为了使 \(k\) 为正整数年龄，需要 \(7-2j\ge0\)，所以 \(j\le3\)。因为 Kiana 比双胞胎小，\(7-2j\lt j\)，所以 \(j\ge3\)。因此 \(j=3\)，得到 \(t=8\)、\(k=2\)。年龄和为 \(8+8+2=18\)。

所以正确答案是 D。

English

Since \(128=2^7,\) every age is a power of \(2.\) Writing the twins’ common age as \(t\) and Kiana’s as \(k,\) we need \(t^2k=128\) with \(k\lt t.\)

Write \(t=2^j.\) Then \(k=2^{7-2j}.\) For \(k\) to be a positive integer age we need \(7-2j\ge0,\) so \(j\le3.\) Because Kiana is younger than the twins, \(7-2j\lt j,\) so \(j\ge3.\) Therefore \(j=3,\) giving \(t=8\) and \(k=2.\) The sum is \(8+8+2=18.\)

Thus, the correct answer is D.

7 答案 C

四个数和三个运算共有五种完整的加括号方式。逐一计算可得

\(((2\times3)+4)\times5=50\)，而 \((2\times(3+4))\times5=70\)。

\((2\times3)+(4\times5)=26\)。

\(2\times((3+4)\times5)=70\)，而 \(2\times(3+(4\times5))=46\)。

因此不同的结果是 \(26, 46, 50\) 和 \(70\)，共 \(4\) 个。

所以正确答案是 C。

English

There are five full parenthesizations of four numbers joined by three operations. Evaluating all five gives

\(((2\times3)+4)\times5=50\) and \((2\times(3+4))\times5=70.\)

\((2\times3)+(4\times5)=26.\)

\(2\times((3+4)\times5)=70\) and \(2\times(3+(4\times5))=46.\)

Thus the distinct values are \(26, 46, 50,\) and \(70,\) for a total of \(4.\)

Thus, the correct answer is C.

8 答案 B

设起始价格为 \(p\)。三月底的价格为 \[ (1.2)(0.8)(1.25)p=1.2p\text{。}\]四月降价后要回到 \(p\)，所以要减少 \(0.2p\)，占当时价格的百分比为 \[ x=100\cdot\dfrac{0.2p}{1.2p}=\dfrac{100}{6}\approx16.7\text{。}\]

四舍五入后，\(x=17\)。

所以正确答案是 B。

English

Let \(p\) be the starting price. After March the price is \[ (1.2)(0.8)(1.25)p=1.2p. \] The April drop must return it to \(p,\) so it removes \(0.2p,\) a fraction \[ x=100\cdot\dfrac{0.2p}{1.2p}=\dfrac{100}{6}\approx16.7. \]

To the nearest integer, \(x=17.\)

Thus, the correct answer is B.

9 答案 A

因为 \(\triangle ABC\) 中 \(AB=BC\)，所以 \(\angle A=\angle C\)。又 \(\angle A=\dfrac52\angle B\)，由内角和得 \[ \dfrac52\angle B+\dfrac52\angle B+\angle B=180^\circ\text{，}\]所以 \(\angle B=30^\circ\)，且 \(\angle ACB=75^\circ\)。

由对顶角，\(\angle DCE=75^\circ\)。又 \(CD=CE\)，所以三角形 \(CDE\) 为等腰三角形，满足 \[ 2\angle D+75^\circ=180^\circ\text{，}\]从而 \(\angle D=52.5^\circ\)。

所以正确答案是 A。

English

Since \(\triangle ABC\) is isosceles with \(AB=BC,\) we have \(\angle A=\angle C.\) With \(\angle A=\dfrac52\angle B,\) the angle sum gives \[ \dfrac52\angle B+\dfrac52\angle B+\angle B=180^\circ, \] so \(\angle B=30^\circ\) and \(\angle ACB=75^\circ.\)

By vertical angles \(\angle DCE=75^\circ.\) Since \(CD=CE,\) triangle \(CDE\) is isosceles, so \[ 2\angle D+75^\circ=180^\circ, \] giving \(\angle D=52.5^\circ.\)

Thus, the correct answer is A.

10 答案 E

竖直残桩高为 \(x\)，长度为 \(5-x\) 的折断部分是直角三角形的斜边，两条直角边为 \(x\) 和 \(1\)。由勾股定理，\[ \begin{aligned} x^2+1^2 &= (5-x)^2 \\ &= x^2-10x+25 \end{aligned}\text{，}\]所以 \(10x=24\)，从而 \(x=2.4\)。

所以正确答案是 E。

English

The standing stump has height \(x,\) and the snapped piece of length \(5-x\) is the hypotenuse of a right triangle with legs \(x\) and \(1.\) By the Pythagorean Theorem, \[ \begin{aligned} x^2+1^2 &= (5-x)^2 \\ &= x^2-10x+25, \end{aligned} \] so \(10x=24\) and \(x=2.4.\)

Thus, the correct answer is E.

11 答案 A

\(7\) 位回文数的中间数字单独使用一次，其余外侧三对数字各出现两次。只有 \(5\) 出现奇数次，所以 \(5\) 必须在中间。

剩余数字 \(2,3,5\) 填入前三个位置，再镜像到后三个位置。共有 \(3!=6\) 种排列。

所以正确答案是 A。

English

A \(7\)-digit palindrome has the form with the middle digit used once and the outer three digits each used twice. Only \(5\) appears an odd number of times, so \(5\) must be the middle digit.

The remaining digits \(2,3,5\) fill the first three positions in some order and mirror to the last three. There are \(3!=6\) such orderings.

Thus, the correct answer is A.

12 答案 A

面积为正的三角形必须在一条直线上取两个点作底边，并在另一条直线上取一个点作顶点。

高始终是两条平行线之间的固定距离，所以面积只取决于底边长度。第一条直线上的底边长度可以是 \(1, 2\) 或 \(3\)，第二条直线上的底边长度是 \(1\)。所以不同底边长度为 \(1, 2, 3\)，共有三种可能的面积。

所以正确答案是 A。

English

A positive-area triangle uses two points on one line as its base and one point on the other line as its apex. The height is always the fixed distance between the lines, so the area depends only on the base length.

Bases on the first line can be \(1, 2,\) or \(3;\) a base on the second line is \(1.\) So the distinct base lengths are \(1, 2, 3,\) giving three possible areas.

Thus, the correct answer is A.

13 答案 C

五边形周长为 \(3+4+6+3+7=23\)。完整滚动一圈使接触点前进 \(23\)，且 \(2009=23\cdot87+8\)。

\(87\) 圈后，顶点 \(A\) 位于 \(x=23\cdot87=2001\)，顶点 \(B\) 位于 \(2004\)。继续滚动，\(C\) 在 \(2004+4=2008\) 处接触，\(D\) 在 \(2008+6=2014\) 处接触。

因为 \(2009\) 位于 \(2008\) 与 \(2014\) 之间，所以边 \(\overline{CD}\) 接触该点。

所以正确答案是 C。

English

The pentagon has perimeter \(3+4+6+3+7=23.\) One full roll advances the contact point by \(23,\) and \(2009=23\cdot87+8.\)

After \(87\) rolls, vertex \(A\) sits at \(x=23\cdot87=2001\) and \(B\) at \(2004.\) Rolling further, \(C\) touches at \(2004+4=2008\) and \(D\) at \(2008+6=2014.\)

Since \(2009\) lies between \(2008\) and \(2014,\) side \(\overline{CD}\) touches that point.

Thus, the correct answer is C.

14 答案 D

每夸脱新种子加入 \(\dfrac14\) 夸脱小米，鸟会留下已有小米的 \(\dfrac34\)。第 \(n\) 天加入后的小米量为 \[ \begin{gathered} \dfrac14\left(1+\dfrac34+\cdots+\Big(\dfrac34\Big)^{n-1}\right) \\ = 1-\Big(\dfrac34\Big)^{n} \end{gathered}\text{。}\]

其他种子总量始终为 \(\dfrac34\) 夸脱。小米超过种子总量的一半时，满足 \(1-\Big(\dfrac34\Big)^n\gt\dfrac34\)，即 \(\Big(\dfrac34\Big)^n\lt\dfrac14\)。

由于 \(\Big(\dfrac34\Big)^4=\dfrac{81}{256}\gt\dfrac14\)，而 \(\Big(\dfrac34\Big)^5=\dfrac{243}{1024}\lt\dfrac14\)，所以第一次发生在第 \(5\) 天，即星期五。

所以正确答案是 D。

English

Each quart adds \(\dfrac14\) quart of millet, and the birds leave \(\dfrac34\) of the standing millet. On day \(n\) the millet present is \[ \begin{gathered} \dfrac14\left(1+\dfrac34+\cdots+\Big(\dfrac34\Big)^{n-1}\right) \\ = 1-\Big(\dfrac34\Big)^{n}. \end{gathered} \]

The other seeds always total \(\dfrac34\) quart. Millet exceeds half when \(1-\Big(\dfrac34\Big)^n\gt\dfrac34,\) i.e. \(\Big(\dfrac34\Big)^n\lt\dfrac14.\)

Since \(\Big(\dfrac34\Big)^4=\dfrac{81}{256}\gt\dfrac14\) but \(\Big(\dfrac34\Big)^5=\dfrac{243}{1024}\lt\dfrac14,\) this first happens on day \(5,\) which is Friday.

Thus, the correct answer is D.

15 答案 E

设空桶重 \(x\)，满桶水重 \(y\)。则 \[x+\dfrac23 y=a\] 并且 \[x+\dfrac12 y=b\text{。}\]

两式相减得 \(\dfrac16 y=a-b\)，所以 \(y=6a-6b\)，且 \(x=b-\dfrac12 y=4b-3a\)。装满水时的总重量为 \[ \begin{aligned} x+y &= (4b-3a)+(6a-6b) \\ &= 3a-2b \end{aligned}\text{。}\]

所以正确答案是 E。

English

Let \(x\) be the bucket’s weight and \(y\) the weight of a full load of water. Then \[x+\dfrac23 y=a\] and \[x+\dfrac12 y=b.\]

Subtracting gives \(\dfrac16 y=a-b,\) so \(y=6a-6b,\) and \(x=b-\dfrac12 y=4b-3a.\) The full bucket weighs \[ \begin{aligned} x+y &= (4b-3a)+(6a-6b) \\ &= 3a-2b. \end{aligned} \]

Thus, the correct answer is E.

16 答案 B

设半径为 \(r\)。由对称性，\(BO\) 平分 \(60^\circ\) 角 \(ABC\)，所以 \(\angle OBC=30^\circ\)。又 \(OC\perp BC\)，三角形 \(BCO\) 是 \(30\)-\(60\)-\(90\) 三角形，斜边 \(BO=2\,OC=2r\)。

于是 \(BD=BO-OD=2r-r=r\)，所以 \(\dfrac{BD}{BO}=\dfrac{r}{2r}=\dfrac12\)。

所以正确答案是 B。

English

Let the radius be \(r.\) By symmetry \(BO\) bisects the \(60^\circ\) angle \(ABC,\) so \(\angle OBC=30^\circ.\) Since \(OC\perp BC,\) triangle \(BCO\) is a \(30\)-\(60\)-\(90\) triangle with hypotenuse \(BO=2\,OC=2r.\)

Then \(BD=BO-OD=2r-r=r,\) so \(\dfrac{BD}{BO}=\dfrac{r}{2r}=\dfrac12.\)

Thus, the correct answer is B.

17 答案 C

五个单位正方形总面积为 \(5\)，所以每个区域面积必须为 \(\dfrac52\)。

直线右下方的区域是一个直角三角形，直角边为 \(3-a\) 和 \(3\)，再减去它不包含的一个单位正方形。令其面积等于 \(\dfrac52\)，得到 \[ \dfrac{3(3-a)}{2}-1=\dfrac52\text{，}\]所以 \(3(3-a)=7\)，从而 \(a=\dfrac23\)。

所以正确答案是 C。

English

The five unit squares have total area \(5,\) so each region must have area \(\dfrac52.\)

The region to the lower right of the line is a right triangle with legs \(3-a\) and \(3,\) minus the one unit square it does not cover. Setting its area to \(\dfrac52\) gives \[ \dfrac{3(3-a)}{2}-1=\dfrac52, \] so \(3(3-a)=7\) and \(a=\dfrac23.\)

Thus, the correct answer is C.

18 答案 D

由勾股定理，\(AC=\sqrt{8^2+6^2}=10\)，所以 \(AM=5\)。直角三角形 \(AME\) 和 \(ABC\) 共用角 \(A\)，因此相似，且 \[ \dfrac{ME}{AM}=\dfrac{BC}{AB}=\dfrac68\text{，}\]得到 \(ME=\dfrac{15}{4}\)。

于是 \[ \begin{gathered} [\triangle AME]=\dfrac12\cdot AM\cdot ME \\ = \dfrac12\cdot5\cdot\dfrac{15}{4}=\dfrac{75}{8} \end{gathered}\text{。}\]

所以正确答案是 D。

English

By the Pythagorean Theorem, \(AC=\sqrt{8^2+6^2}=10,\) so \(AM=5.\) Right triangles \(AME\) and \(ABC\) share angle \(A,\) so they are similar with \[ \dfrac{ME}{AM}=\dfrac{BC}{AB}=\dfrac68, \] giving \(ME=\dfrac{15}{4}.\)

Then \[ \begin{gathered} [\triangle AME]=\dfrac12\cdot AM\cdot ME \\ = \dfrac12\cdot5\cdot\dfrac{15}{4}=\dfrac{75}{8}. \end{gathered} \]

Thus, the correct answer is D.

19 答案 A

在 \(1\) 到 \(12\) 点中，恰好 \(1\)、\(10\)、\(11\)、\(12\) 含有数字 \(1\)，所以小时正确的时间比例为 \(\dfrac{8}{12}=\dfrac23\)。

分钟显示出错发生在十位为 \(1\)（\(10\) 到 \(19\) 分）或个位为 \(1\)（\(01,11,\dots,51\) 分）时。这样的分钟共有 \(15\) 个，而每小时共有 \(60\) 个分钟读数。所以分钟显示正确的比例为 \(\dfrac{45}{60}=\dfrac34\)。

因此整钟显示正确的比例为 \(\dfrac23\cdot\dfrac34=\dfrac12\)。

所以正确答案是 A。

English

Among the hours \(1\) through \(12,\) exactly \(1,\) \(10,\) \(11,\) \(12\) contain a \(1,\) so the hour is correct \(\dfrac{8}{12}=\dfrac23\) of the time.

A minute is displayed wrong when its tens digit is \(1\) (minutes \(10\)–\(19\)) or its units digit is \(1\) (\(01,11,\dots,51\)), which is \(15\) of the \(60\) minutes. So the minute is correct \(\dfrac{45}{60}=\dfrac34\) of the time.

The clock is correct \(\dfrac23\cdot\dfrac34=\dfrac12\) of the day.

Thus, the correct answer is A.

20 答案 B

由勾股定理，\(AC=\sqrt{1^2+2^2}=\sqrt5\)。角平分线定理给出 \[ \dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{1}{\sqrt5}\text{，}\]所以 \(DC=\sqrt5\,BD\)。

于是 \(BD+DC=2\)，\(BD(1+\sqrt5)=2\)，\[ BD=\dfrac{2}{1+\sqrt5}=\dfrac{\sqrt5-1}{2}\text{。}\]

所以正确答案是 B。

English

By the Pythagorean Theorem, \(AC=\sqrt{1^2+2^2}=\sqrt5.\) The Angle Bisector Theorem gives \[ \dfrac{BD}{DC}=\dfrac{AB}{AC}=\dfrac{1}{\sqrt5}, \] so \(DC=\sqrt5\,BD.\)

Since \(BD+DC=2,\) we have \(BD(1+\sqrt5)=2,\) so \[ BD=\dfrac{2}{1+\sqrt5}=\dfrac{\sqrt5-1}{2}. \]

Thus, the correct answer is B.

21 答案 D

任意连续四个 \(3\) 的幂之和都是 \(3^0+3^1+3^2+3^3=40\) 的倍数，而这个数能被 \(8\) 整除。

从 \(3^2\) 到 \(3^{2009}\) 的项可以分成这样的四项组，对余数贡献为 \(0\)。剩下的是 \(3^0+3^1=4\)。

所以正确答案是 D。

English

Any four consecutive powers of \(3\) sum to a multiple of \(3^0+3^1+3^2+3^3=40,\) which is divisible by \(8.\)

The terms from \(3^2\) to \(3^{2009}\) split into such blocks and contribute remainder \(0.\) What remains is \(3^0+3^1=4.\)

Thus, the correct answer is D.

22 答案 B

把顶面看作 \(2\times2\) 正方形。从 \(M\) 向远角的切线形成顶面三角形 \(A\)，其直角边为 \(1\) 和 \(2\)，面积为 \(1\)，斜边为 \(\sqrt5\)。

三角形 \(B\) 与 \(A\) 相似，但斜边为 \(2\)，所以其面积为 \(\left(\dfrac{2}{\sqrt5}\right)^2\cdot1=\dfrac45\)。蛋糕高为 \(2\)，所以体积 \(c=\dfrac45\cdot2=\dfrac85\)。

这块蛋糕上的糖霜包括顶面 \(\dfrac45\) 和它相邻的完整立方体侧面 \(2\times2=4\)，所以 \(s=\dfrac45+4=\dfrac{24}{5}\)。因此 \(c+s=\dfrac85+\dfrac{24}{5}=\dfrac{32}{5}\)。

所以正确答案是 B。

English

Set the top face as a \(2\times2\) square. The cut from \(M\) toward the far corner creates the top triangle \(A\) with legs \(1\) and \(2,\) so area \(1\) and hypotenuse \(\sqrt5.\)

Triangle \(B\) is similar to \(A\) but with hypotenuse \(2,\) so its area is \(\left(\dfrac{2}{\sqrt5}\right)^2\cdot1=\dfrac45.\) Since the cake has height \(2,\) the volume is \(c=\dfrac45\cdot2=\dfrac85.\)

The icing on this piece is its top (\(\dfrac45\)) plus the full cube side face it borders (\(2\times2=4\)), so \(s=\dfrac45+4=\dfrac{24}{5}.\) Therefore \(c+s=\dfrac85+\dfrac{24}{5}=\dfrac{32}{5}.\)

Thus, the correct answer is B.

23 答案 C

照片覆盖起点两侧各 \(\dfrac18\) 圈。\(600\) 秒后，Rachel 距离起点线还差 \(30\) 秒；她跑完 \(\dfrac14\) 圈需 \(22.5\) 秒，所以她在 \(30-11.25=18.75\) 秒到 \(30+11.25=41.25\) 秒之间入镜，这些时刻都在第 \(10\) 分钟内。

\(600\) 秒后，Robert 距离起点线还差 \(40\) 秒；他跑完 \(\dfrac14\) 圈需 \(20\) 秒，所以他在该分钟的 \(30\) 秒到 \(50\) 秒之间入镜。

两人同时入镜的时间为 \(30\) 秒到 \(41.25\) 秒，长度为 \(11.25\) 秒，占 \(60\) 秒的比例为 \(\dfrac{11.25}{60}=\dfrac{3}{16}\)。

所以正确答案是 C。

English

The picture spans \(\dfrac18\) lap on each side of the start. After \(600\) seconds Rachel is \(30\) seconds short of the line; running \(\dfrac14\) lap in \(22.5\) seconds, she is in view between \(30-11.25=18.75\) and \(30+11.25=41.25\) seconds of the \(10\)th minute.

After \(600\) seconds Robert is \(40\) seconds from the line; running \(\dfrac14\) lap in \(20\) seconds, he is in view between \(30\) and \(50\) seconds.

Both appear between \(30\) and \(41.25\) seconds, a window of length \(11.25\) out of \(60,\) so the probability is \(\dfrac{11.25}{60}=\dfrac{3}{16}.\)

Thus, the correct answer is C.

24 答案 A

加上镜像后，拱补成由 \(18\) 个梯形组成的对称闭环。它们的内侧边形成正 \(18\) 边形，每个内角为 \[ \dfrac{(18-2)\cdot180^\circ}{18}=160^\circ\text{。}\]

在每个内侧顶点，两个梯形的较大内角 \(x\) 与 \(160^\circ\) 之和为一周角，因此 \(x+x+160^\circ=360^\circ\)，所以 \(x=100\)。

所以正确答案是 A。

English

Adding a mirror image completes the arch into a symmetric closed loop of \(18\) trapezoids. Their inner edges form a regular \(18\)-gon, each interior angle of which is \[ \dfrac{(18-2)\cdot180^\circ}{18}=160^\circ. \]

At each inner vertex, two of the trapezoids’ larger angles \(x\) meet the \(160^\circ\) angle around a full turn: \(x+x+160^\circ=360^\circ,\) so \(x=100.\)

Thus, the correct answer is A.

25 答案 B

每个面的条纹有 \(2\) 种方向，所以共有 \(2^6=64\) 种等可能配置。

一条环绕条纹会沿 \(3\) 对相对面中的某一个方向，并经过 \(4\) 个面；给定一个方向时，这四个面的条纹方向都被确定，所以概率为 \(\left(\dfrac12\right)^4=\dfrac{1}{16}\)。

三种方向互斥，所以总概率为 \(3\cdot\dfrac{1}{16}=\dfrac{3}{16}\)。

所以正确答案是 B。

English

Each face’s stripe has \(2\) orientations, giving \(2^6=64\) equally likely configurations.

An encircling stripe runs around one of the \(3\) pairs of opposite faces. For a given band, the \(4\) faces it crosses must each be oriented to continue it, a probability of \(\left(\dfrac12\right)^4=\dfrac{1}{16}.\)

The three bands are mutually exclusive, so the probability is \(3\cdot\dfrac{1}{16}=\dfrac{3}{16}.\)

Thus, the correct answer is B.