2015B AMC 10 真题（P版）

1 答案 C

因为 \((-2)^{-2}=\frac{1}{(-2)^2}=\frac14\)，所以 \[2-\frac14=\frac74\text{。}\]

所以正确答案是 C。

English

Since \((-2)^{-2}=\frac{1}{(-2)^2}=\frac14\), the expression is \[2-\frac14=\frac74.\]

Thus, the correct answer is C.

2 答案 B

完成 \(2\) 项任务共用 \(100\) 分钟。因此在 \(2:40\) 之后还需要 \(50\) 分钟，也就是 \(3:30\)。

所以正确答案是 B。

English

The time it takes to do \(2\) tasks is \(100\) minutes. Thus, it takes \(50\) more minutes after \(2:40,\) which is \(3:30.\)

Thus, the correct answer is B .

3 答案 A

设写了两次的数为 \(x\)，写了三次的数为 \(y\)。\(2x+3y=100\)。

若 \(x=28\)，则 \(3y=100-56=44\)，所以 \(y\) 不可能是整数。因此 \(y=28\)，从而 \(2x=100-84=16\)，所以 \(x=8\)。

所以正确答案是 A。

English

Let the number written twice be \(x\), and let the number written three times be \(y\). Then \(2x+3y=100\).

If \(x=28\), then \(3y=100-56=44\), impossible for an integer \(y\). Therefore \(y=28\), and \(2x=100-84=16\), so \(x=8\).

Thus, the correct answer is A.

4 答案 C

因为 \(\frac 13 > \frac 14 > \frac 15\)，所以 Beth 吃得比 Cyril 多，Cyril 吃得比 Alex 多。因此这三人的顺序已经确定。

Dan 吃了 \[1- \dfrac 13 - \dfrac 14 - \dfrac 15 = \dfrac{13}{60}\text{。}\] 这大于 \(\frac 15\)，小于 \(\frac 14 \)，所以 Dan 排在 Cyril 和 Alex 之间，顺序为 Beth、Cyril、Dan、Alex。

所以正确答案是 C。

English

Since \(\frac 13 > \frac 14 > \frac 15,\) we know Beth ate more than Cyril and Cyril ate more than Alex. Thus, those three are in order.

The amount Dan ate is \[1- \dfrac 13 - \dfrac 14 - \dfrac 15 = \dfrac{13}{60}.\] This is greater than \(\frac 15\) and less than \(\frac 14 ,\) so Dan is in between Cyril and Alex. This makes the order Beth, Cyril, Dan, Alex.

Thus, the correct answer is C .

5 答案 B

Marta 是第 \(6\) 名，所以 Jack 是第 \(5\) 名。Jack 比 Todd 晚 \(2\) 个名次，所以 Todd 是第 \(3\) 名；Todd 比 Rand 晚 \(1\) 个名次，所以 Rand 是第 \(2\) 名。

Hikmet 比 Rand 晚 \(6\) 个名次，所以 Hikmet 是第 \(8\) 名。

所以正确答案是 B。

English

Marta finished \(6\)th, so Jack finished \(5\)th. Since Jack finished \(2\) places behind Todd, Todd finished \(3\)rd. Since Todd finished \(1\) place behind Rand, Rand finished \(2\)nd.

Hikmet finished \(6\) places behind Rand, so Hikmet finished \(8\)th.

Thus, the correct answer is B.

6 答案 E

Marley 星期一打篮球，星期三打高尔夫。若星期二不跑步，则三天跑步只能安排在星期四到星期日之间，必然有连续两天跑步，所以星期二必须跑步。

从星期四到星期日，她必须安排两天跑步、一天游泳和一天打网球。打网球不能排在跑步或游泳的后一天，所以只能在星期四。剩下两天跑步必须是星期五和星期日，因此星期六游泳。

所以正确答案是 E。

English

Marley plays basketball on Monday and golf on Wednesday. She cannot fit all three running days among Thursday, Friday, Saturday, and Sunday without having two consecutive running days, so Tuesday must be a running day.

From Thursday through Sunday, she must run twice, swim once, and play tennis once. Tennis cannot be the day after running or swimming, so tennis must be Thursday. Then the two remaining running days must be Friday and Sunday, leaving Saturday for swimming.

Thus, the correct answer is E.

7 答案 A

\[\begin{aligned} &((1\diamond2)\diamond3)-(1\diamond(2\diamond3)) \\ &= \left(\dfrac 12- \dfrac 13\right) - \left(1 \diamond \dfrac 53\right)\\ &= \left(\dfrac 12 - \dfrac 13\right) - \left(1-\dfrac 35\right)\\ &= \dfrac 16 - \dfrac 25 \\&= -\dfrac{7}{30} \end{aligned}\]

所以正确答案是 A。

English

\[\begin{aligned} &((1\diamond2)\diamond3)-(1\diamond(2\diamond3)) \\ &= \left(\dfrac 12- \dfrac 13\right) - \left(1 \diamond \dfrac 53\right)\\ &= \left(\dfrac 12 - \dfrac 13\right) - \left(1-\dfrac 35\right)\\ &= \dfrac 16 - \dfrac 25 \\&= -\dfrac{7}{30} \end{aligned}\]

Thus, the correct answer is A .

8 答案 E

第一次旋转把字母 F 移到 \(x\) 轴下方。

旋转半圈等价于依次关于两条坐标轴反射，因此最后的半圈旋转抵消了前面关于 \(y\) 轴的反射，只留下关于 \(x\) 轴的反射。

把旋转后的图形关于 \(x\) 轴反射，就得到选项 E。

所以正确答案是 E。

English

The first rotation puts the F below the \(x\)-axis.

A half turn is equivalent to reflecting in both coordinate axes. Therefore the final half turn cancels the preceding reflection in the \(y\)-axis and leaves a reflection in the \(x\)-axis.

Reflecting the rotated figure in the \(x\)-axis produces choice E.

Thus, the correct answer is E .

9 答案 B

外边界是半径为 \(3\) 的四分之一圆，所以面积为 \(\frac14\pi\cdot3^2=\frac{9\pi}{4}\)。

内边界是半径为 \(\frac32\) 的圆的右半部分，所以面积为 \(\frac12\pi\left(\frac32\right)^2=\frac{9\pi}{8}\)。

阴影面积就是二者之差，\(\frac{9\pi}{4}-\frac{9\pi}{8}=\frac{9\pi}{8}\)。

所以正确答案是 B。

English

The larger boundary is a quarter circle of radius \(3\), so its area is \(\frac14\pi\cdot3^2=\frac{9\pi}{4}\).

The inner boundary is the right half of a circle of radius \(\frac32\), so its area is \(\frac12\pi\left(\frac32\right)^2=\frac{9\pi}{8}\).

The shaded area is the difference, \(\frac{9\pi}{4}-\frac{9\pi}{8}=\frac{9\pi}{8}\).

Thus, the correct answer is B.

10 答案 C

大于 \(-2015\) 的负奇整数共有 \(1007\) 个。

负因子个数为奇数，所以乘积为负。

乘积包含因子 \(-5\)，所以是 \(5\) 的倍数，个位只能是 \(5\) 或 \(0\)。

又因为所有因子都是奇数，所以乘积不可能是偶数，个位不能为零，只能为 \(5\)。

所以正确答案是 C。

English

There are \(1007\) odd numbers greater than \(-2015.\)

Our product is of an odd number of negative numbers, so the result is negative.

Also, we multiply by \(-5\) in there, so the product is a multiple of \(5,\) making it end in \(5\) or \(0.\) None of our factors are even, so the product can’t be even.

Therefore, the product must end in \(5.\)

Thus, the correct answer is C .

11 答案 B

可用数字为 \(2,3,5,7\)。一位数有 \(4\) 个，两位数有 \(4^2=16\) 个，共 \(20\) 个选择。

所有 \(4\) 个一位选择都是质数。两位质数不能以 \(2\) 或 \(5\) 结尾，所以检查以 \(3\) 和 \(7\) 结尾的情况，得到两位质数 \(23,37,53,73\)。

因此共有 \(8\) 个质数；总选择数为 \(20\)，所以概率为 \(\frac{8}{20}=\frac25\)。

所以正确答案是 B。

English

The available digits are \(2,3,5,7\). There are \(4\) one-digit numbers and \(4^2=16\) two-digit numbers, for \(20\) total choices.

All \(4\) one-digit choices are prime. A two-digit prime cannot end in \(2\) or \(5\), so checking endings \(3\) and \(7\) gives the two-digit primes \(23,37,53,73\).

Thus \(8\) of the \(20\) choices are prime, and the probability is \(\frac{8}{20}=\frac25\).

Thus, the correct answer is B.

12 答案 A

点 \((x,-x)\) 到 \((5,5)\) 的距离平方为 \[ \begin{aligned} &(x-5)^2+(-x-5)^2 \\ &= 2x^2+50 \end{aligned}\text{。}\]

由 \(2x^2+50\le100\) 得 \(x^2\le25\)，所以 \(-5\le x\le5\)，共有 \(11\) 个整数。

所以正确答案是 A。

English

The squared distance from \((x,-x)\) to \((5,5)\) is \[ \begin{aligned} &(x-5)^2+(-x-5)^2 \\ &= 2x^2+50. \end{aligned} \]

Being inside or on the circle means \(2x^2+50\le100\), so \(x^2\le25\). Thus \(-5\le x\le5\), giving \(11\) integer values.

Thus, the correct answer is A.

13 答案 E

这个三角形是直角三角形，两条直角边为 \(12\) 和 \(5\)，因此斜边为 \(13\)。

两条直角边对应的高分别为 \(5\) 和 \(12\)。用面积公式 \(A = \frac{bh}2\)，其中 \(b\) 是底、\(h\) 是高。

三角形面积为 \(\frac {12\cdot 5}2 = 30\)，设斜边上的高为 \(h\)，则 \(30 = \frac{13h}2\)，所以这条高为 \(\frac{60}{13}\)。

三条高之和为 \[12+5+\dfrac{60}{13} = \dfrac{281}{13}\text{。}\]

所以正确答案是 E。

English

The triangle is a right triangle with legs of \(12\) and \(5.\) This makes the hypotenuse \(13.\)

Two of the altitudes are then \(12\) and \(5.\) Also, for any side, \(A = \frac{bh}2\) where \(b\) is the base and \(h\) is the altitude.

The area is \(\frac {12\cdot 5}2 = 30,\) so the other altitude \(h\) can be found with \(30 = \frac{13h}2.\) Thus, this altitude is \(\frac{60}{13}.\)

Therefore, the sum is \[12+5+\dfrac{60}{13} = \dfrac{281}{13} .\]

Thus, the correct answer is E .

14 答案 D

将左边因式分解，得到 \[(x-b)(2x-a-c)=0\text{。}\]因此两根是 \(b\) 和 \(\frac{a+c}{2}\)，根之和为 \[b+\frac{a+c}{2}\text{。}\]

在这个和中，\(b\) 的系数是 \(a\) 或 \(c\) 的系数的两倍，所以把最大的数字赋给 \(b\)。接下来两个最大的不同数字应赋给 \(a\) 和 \(c\)。

取 \(b=9\)、\(\{a,c\}=\{7,8\}\)，得到 \[9+\frac{7+8}{2}=16.5\text{。}\]

所以正确答案是 D。

English

Factoring the left-hand side gives \[(x-b)(2x-a-c)=0.\] Thus the roots are \(b\) and \(\frac{a+c}{2},\) whose sum is \[b+\frac{a+c}{2}.\]

The coefficient of \(b\) in this sum is twice the coefficient of either \(a\) or \(c,\) so assign the largest digit to \(b.\) The next two largest distinct digits should be \(a\) and \(c.\)

Taking \(b=9\) and \(\{a,c\}=\{7,8\}\) gives \[9+\frac{7+8}{2}=16.5.\]

Thus, the correct answer is D .

15 答案 B

若有 \(h\) 匹马和 \(c\) 头牛，则有 \(3h\) 个人、\(9h\) 只鸭和 \(4c\) 只羊，总数为 \(13h+5c\)。

除 \(47\) 外，其余选项都可表示为所需形式：\[ \begin{gathered} 41=13\cdot2+5\cdot3, \\ 59=13\cdot3+5\cdot4, \\ 61=13\cdot2+5\cdot7, \\ 66=13\cdot2+5\cdot8\text{。} \end{gathered} \] 对 \(47\)，减去 \(0,13,26,39\) 后分别得到 \(47,34,21,8\)，没有一个是 \(5\) 的倍数。

所以正确答案是 B。

English

If there are \(h\) horses and \(c\) cows, then there are \(3h\) people, \(9h\) ducks, and \(4c\) sheep. The total is therefore \(13h+5c\).

The listed values except \(47\) can be written in that form: \[ \begin{gathered} 41=13\cdot2+5\cdot3, \\ 59=13\cdot3+5\cdot4, \\ 61=13\cdot2+5\cdot7, \\ 66=13\cdot2+5\cdot8. \end{gathered} \] For \(47\), subtracting \(0,13,26,39\) leaves \(47,34,21,8\), none of which is divisible by \(5\).

Thus, the correct answer is B.

16 答案 C

设 Al、Bill、Cal 的数分别为 \((A,B,C)\)。需要 \(A\) 是 \(B\) 的倍数，且 \(B\) 是 \(C\) 的倍数，同时三个数互不相同。

满足条件的三元组为\[ \begin{gathered} (4,2,1),(6,2,1),(8,2,1), \\ (10,2,1),(6,3,1),(9,3,1), \\ (8,4,1),(10,5,1),(8,4,2) \end{gathered}\text{。}\] 共有 \(9\) 种有利的分配。

总分配数为 \(10\cdot9\cdot8=720\)，所以概率为 \(\frac9{720}=\frac1{80}\)。

所以正确答案是 C。

English

Let \((A,B,C)\) be the numbers assigned to Al, Bill, and Cal. We need \(A\) to be a multiple of \(B\), and \(B\) to be a multiple of \(C\), with all three numbers distinct.

The valid triples are \[ \begin{gathered} (4,2,1),(6,2,1),(8,2,1), \\ (10,2,1),(6,3,1),(9,3,1), \\ (8,4,1),(10,5,1),(8,4,2). \end{gathered} \] There are \(9\) favorable assignments.

The total number of assignments is \(10\cdot9\cdot8=720\), so the probability is \(\frac9{720}=\frac1{80}\).

Thus, the correct answer is C.

17 答案 B

这个八面体可看成两个全等棱锥，它们共用的底面是通过四个侧面中心的菱形。该菱形对角线为 \(4\) 和 \(5\)，面积为 \(\frac12\cdot4\cdot5=10\)。

两个棱锥的高都为 \(\frac32\)，即长方体高的一半，所以八面体体积为 \[2\cdot\frac13\cdot10\cdot\frac32=10\text{。}\]

所以正确答案是 B。

English

The octahedron can be viewed as two congruent pyramids whose shared base is the rhombus through the centers of the four side faces. This rhombus has diagonals \(4\) and \(5\), so its area is \(\frac12\cdot4\cdot5=10\).

Each pyramid has height \(\frac32\), half the prism’s height. Thus the total volume is \[2\cdot\frac13\cdot10\cdot\frac32=10.\]

Thus, the correct answer is B.

18 答案 D

一枚硬币最后仍为反面，当且仅当它连续 \(3\) 次都是反面，概率为 \(\frac 18\)。因此任意一枚硬币最后为正面的概率为 \(\frac 78 \)。

由于一枚硬币为正面的概率是 \(\frac78\)，且一共有 \(64\) 枚硬币，所以现在正面朝上的硬币期望数为 \[64 \cdot \dfrac 78 = 56\text{。}\]

所以正确答案是 D。

English

A coin ends as tails if and only if it has \(3\) flips that are tails, which happens with probability \(\frac 18.\) Thus, the probability of any coin being heads is \(\frac 78 .\)

Since each of the \(64\) coins ends as heads with probability \(\frac78,\) linearity of expectation gives the expected number of coins that are now heads: \[64 \cdot \dfrac 78 = 56.\]

Thus, the correct answer is D .

19 答案 C

过 \(X,Y,Z,W\) 的圆心在 \(XY\) 和 \(ZW\) 的垂直平分线上，也就是在 \(AB\) 和 \(AC\) 的垂直平分线上，所以同一点也是直角三角形 \(ABC\) 的外心。

因此圆心是斜边中点 \(O\)，其中斜边为 \(AB\)，所以 \(OA=OB=OC=6\)。令 \(a=\frac12BC\)、\(b=\frac12CA\)，则 \(a^2+b^2=6^2\)。

由 \(AB\) 上的正方形可得 \(OX^2=6^2+12^2=180\)。由 \(AC\) 上的正方形可知，对应半径也给出 \(OW^2=b^2+(a+2b)^2\)。因此 \[b^2+(a+2b)^2=180\text{。}\] 从这个方程减去 \(a^2+b^2=36\)，得到 \(b(a+b)=36\)。但同样有 \(a^2+b^2=36\)，所以 \(ab=a^2\)。由于 \(a>0\)，得到 \(a=b=3\sqrt{2}\)。

于是 \(AC=BC=6\sqrt{2}\)，所以周长为 \(12+12\sqrt{2}\)。

所以正确答案是 C。

English

The center of the circle through \(X,Y,Z,W\) lies on the perpendicular bisectors of \(XY\) and \(ZW\). These are also the perpendicular bisectors of \(AB\) and \(AC\), so the same point is the circumcenter of right triangle \(ABC\).

Therefore the center is the midpoint \(O\) of hypotenuse \(AB\), so \(OA=OB=OC=6\). Let \(a=\frac12BC\) and \(b=\frac12CA\). Then \(a^2+b^2=6^2\).

From the square on \(AB\), \(OX^2=6^2+12^2=180\). From the square on \(AC\), the corresponding radius also gives \(OW^2=b^2+(a+2b)^2\). Hence \[b^2+(a+2b)^2=180.\] Subtracting \(a^2+b^2=36\) from this equation gives \(b(a+b)=36\). But \(a^2+b^2=36\) as well, so \(ab=a^2\). Since \(a>0\), we get \(a=b=3\sqrt{2}\).

Thus \(AC=BC=6\sqrt{2}\), and the perimeter is \(12+12\sqrt{2}\).

Thus, the correct answer is C.

20 答案 A

前两条棱可用 \(3\cdot2=6\) 种方式选择。这两条棱确定了立方体的一个起始面。

下一步必须访问该面上唯一尚未访问的顶点；否则以后到达它时，它的所有相邻顶点都已经访问过，路径将无法继续。剩余四个顶点都在对面，可以按两个环绕方向访问。

这两个方向中，恰有一个会终止在与起点不相邻的顶点。因此共有 \(6\) 条合格路径。

所以正确答案是 A。

English

The first two edges can be chosen in \(3\cdot2=6\) ways. These two edges determine an initial face of the cube. After those moves, there is one unvisited vertex on that initial face.

That remaining vertex must be visited next; otherwise Erin would later reach it after all of its neighbors had already been visited, and the path could not continue. The last four vertices are then on the opposite face and can be visited in two cyclic orders.

Of those two orders, exactly one ends at a vertex not adjacent to the starting point. Hence there are \(6\) valid paths.

Thus, the correct answer is A.

21 答案 D

设 Dash 跳 \(d+1\) 次，则台阶数 \(t\) 是 \[ \begin{gathered} 5d+1,5d+2,5d+3, \\ 5d+4,5d+5\text{。} \end{gathered} \] 中的一个。Cozy 多跳 \(19\) 次，所以 Cozy 跳 \(d+20\) 次，这意味着 \(t\) 是 \(2d+39\) 或 \(2d+40\)。

匹配这些可能性，整数解来自 \[ \begin{gathered} 5d+3=2d+39, \\ 5d+1=2d+40, \\ 5d+4=2d+40\text{。} \end{gathered} \] 对应 \(t=63,66,64\)。

因此 \(s=63+66+64=193\)，各位数字之和为 \(13\)。

所以正确答案是 D。

English

Suppose Dash takes \(d+1\) jumps. Then the number of steps \(t\) is one of \[ \begin{gathered} 5d+1,5d+2,5d+3, \\ 5d+4,5d+5. \end{gathered} \] Cozy takes \(19\) more jumps, so Cozy takes \(d+20\) jumps, which means \(t\) is either \(2d+39\) or \(2d+40\).

Matching these possibilities, the integer solutions are \[ \begin{gathered} 5d+3=2d+39, \\ 5d+1=2d+40, \\ 5d+4=2d+40. \end{gathered} \] They give \(t=63,66,64\), respectively.

Thus \(s=63+66+64=193\), and the sum of its digits is \(13\).

Thus, the correct answer is D.

22 答案 D

由对称性，\(AG=HC=HJ=1\)，且三角形 \(AFG\) 与三角形 \(BGH\) 全等，所以 \(FG=GH\)。设 \(FG=b\)、\(CD=d\)。

正五边形中的相似三角形给出 \[\frac1b=1+b\] 和 \[\frac{1+b}{1}=d\text{。}\] 第一式为 \(b^2+b-1=0\)，所以 \(b=\frac{-1+\sqrt{5}}{2}\)。于是 \(d=1+b=\frac{1+\sqrt{5}}{2}\)。

因此 \[ \begin{aligned} FG+JH+CD &= b+1+d \\ &= 1+\sqrt{5}\text{。} \end{aligned} \]

所以正确答案是 D。

English

By symmetry, \(AG=HC=HJ=1\), and triangles \(AFG\) and \(BGH\) are congruent, so \(FG=GH\). Let \(FG=b\), and let \(CD=d\).

The similar triangles in the pentagon give \[\frac1b=1+b\] and \[\frac{1+b}{1}=d.\] The first equation is \(b^2+b-1=0\), so \(b=\frac{-1+\sqrt{5}}{2}\). Then \(d=1+b=\frac{1+\sqrt{5}}{2}\).

Therefore \[ \begin{aligned} FG+JH+CD &= b+1+d \\ &= 1+\sqrt{5}. \end{aligned} \]

Thus, the correct answer is D.

23 答案 B

阶乘末尾零的个数等于其中因子 \(5\) 的个数。对 \(5\le n\le9\)，\(n!\) 有 \(k=1\) 个零；要使 \((2n)!\) 有 \(3\) 个零，需要 \(15\le2n\le19\)，所以 \(n=8,9\)。

对 \(10\le n\le14\)，\(n!\) 有 \(k=2\) 个零；要使 \((2n)!\) 有 \(6\) 个零，需要 \(25\le2n\le29\)，所以 \(n=13,14\)。

这就是四个最小值，所以 \(s=8+9+13+14=44\)。\(s\) 的各位数字和为 \(8\)。

所以正确答案是 B。

English

The number of trailing zeros is the number of factors of \(5\). For \(5\le n\le9\), \(n!\) has \(k=1\) zero. We need \((2n)!\) to have \(3\) zeros, which happens when \(15\le2n\le19\). Thus \(n=8,9\).

For \(10\le n\le14\), \(n!\) has \(k=2\) zeros. We need \((2n)!\) to have \(6\) zeros, which happens when \(25\le2n\le29\). Thus \(n=13,14\).

These are the four least possible values, so \(s=8+9+13+14=44\). The sum of the digits of \(s\) is \(8\).

Thus, the correct answer is B.

24 答案 D

当 Aaron 到达 \((k,-k)\) 时，他刚好完成了包含所有坐标介于 \(-k\) 与 \(k\) 之间的格点的方形螺旋。因此 \[p_{(2k+1)^2-1}=(k,-k)\text{。}\]

取 \(k=22\)，得到 \(p_{2024}=(22,-22)\)。因为 \(2024-2015=9\)，沿底边倒推时，\(x\) 坐标减去 \(9\)，所以 \[p_{2015}=(13,-22)\text{。}\]

所以正确答案是 D。

English

When Aaron reaches \((k,-k)\), he has just completed the square spiral containing all grid points with coordinates between \(-k\) and \(k\). Therefore \[p_{(2k+1)^2-1}=(k,-k).\]

With \(k=22\), this gives \(p_{2024}=(22,-22)\). Since \(2024-2015=9\), stepping backward along the bottom edge subtracts \(9\) from the \(x\)-coordinate, so \[p_{2015}=(13,-22).\]

Thus, the correct answer is D.

25 答案 B

条件为 \[abc=2(ab+ac+bc)\text{。}\] 因为 \(abc\le6bc\)，所以 \(a\le6\)。此外，\(a=1\) 和 \(a=2\) 都没有正整数解，因此只需检查 \(a=3,4,5,6\)。

当 \(a=3\) 时，\((b-6)(c-6)=36\)，得到 \((b,c)=(7,42),\) \((8,24),\) \((9,18),\) \((10,15),\) \((12,12)\)。当 \(a=4\) 时，\((b-4)(c-4)=16\)，得到 \((5,20),(6,12),(8,8)\)。

当 \(a=5\) 时，\((3b-10)(3c-10)=100\)，满足 \(a\le b\le c\) 的唯一解是 \((b,c)=(5,10)\)。当 \(a=6\) 时，\((b-3)(c-3)=9\)，满足 \(a\le b\le c\) 的唯一解是 \((6,6)\)。

所以三元组总数为 \(5+3+1+1=10\)。

所以正确答案是 B。

English

The condition is \[abc=2(ab+ac+bc).\] Since \(abc\le6bc\), we have \(a\le6\). Also \(a=1\) and \(a=2\) give no positive solutions, so test \(a=3,4,5,6\).

For \(a=3\), \((b-6)(c-6)=36\), giving \((b,c)=(7,42),\) \((8,24),\) \((9,18),\) \((10,15),\) \((12,12)\). For \(a=4\), \((b-4)(c-4)=16\), giving \((5,20),(6,12),(8,8)\).

For \(a=5\), \((3b-10)(3c-10)=100\), and the only solution with \(a\le b\le c\) is \((b,c)=(5,10)\). For \(a=6\), \((b-3)(c-3)=9\), and the only solution with \(a\le b\le c\) is \((6,6)\).

The total number of triples is \(5+3+1+1=10\).

Thus, the correct answer is B.