2012A AMC 10 真题（P版）

1 答案 D

Cagney 每分钟完成 \(60 \div 20 = 3\) 个，Lacey 每分钟完成 \(60 \div 30 = 2\) 个。

\(5\) 分钟内，他们能完成 \[ 5(2 + 3) = 5 \cdot 5 = 25\text{。}\]

所以正确答案是 D。

English

Cagney can make \(60 \div 20 = 3\) cupcakes per minute, and Lacey can make \(60 \div 30 = 2.\)

In \(5\) minutes, they can make \[ 5(2 + 3) = 5 \cdot 5 = 25. \]

Thus, D is the correct answer.

2 答案 E

注意其中一条边仍与原正方形的边长相同。

因此一个尺寸是 \(8\)，另一个尺寸是原边长的一半，即 \(4\)。

所以正确答案是 E。

English

Note that the one of the sides remains the same as the original square.

This means that one dimension is \(8.\) The other dimension is the original side cut in half, which is \(4.\)

Thus, E is the correct answer.

3 答案 E

它爬行的距离为 \[ |-2 - (-6)| = |4| = 4 \] 即从 \(-2\) 移动到 \(-6\)。接着它爬行的距离为 \[ |-6 - 5| = |-11| = 11 \] 即从 \(-6\) 移动到 \(5\)。

总距离为 \[ 4 + 11 = 15\text{。}\]

所以正确答案是 E。

English

It crawls a distance of \[ |-2 - (-6)| = |4| = 4 \] when it moves from \(-2\) to \(-6.\) It then travels a distance of \[ |-6 - 5| = |-11| = 11 \] as it moves from \(-6\) to \(5.\)

The total distance is then \[ 4 + 11 = 15. \]

Thus, E is the correct answer.

4 答案 C

注意两个角都共用射线 \(AB\)。为了使所求角最小，让 \(D\) 位于 \(A\) 和 \(C\) 之间。

这样就有 \[ \begin{aligned} \angle CBD &= \angle ABC - \angle ABD \\ &= 4^{\circ} \end{aligned}\text{。}\]

所以正确答案是 C。

English

Note that both of the angles share the ray \(AB.\) To minimize the desired degree, we want \(D\) to be between \(A\) and \(C.\)

This would make \[ \begin{aligned} \angle CBD &= \angle ABC - \angle ABD \\ &= 4^{\circ}. \end{aligned} \]

Thus, C is the correct answer.

5 答案 B

成年母猫有 \(100 \div 2 = 50\) 只，其中带小猫的有 \(50 \div 2 = 25\) 只。

因为平均每窝有 \(4\) 只小猫，所以小猫总数为 \[ 4 \cdot 25 = 100\text{。}\]

因此成年猫和小猫的总数为 \[ 100 + 100 = 200\text{。}\]

所以正确答案是 B。

English

We have that there are \(100 \div 2 = 50\) female cats. We then have that \(50 \div 2 = 25\) cats that have kittens.

Since the average number of kittens per litter is \(4,\) the total number of kittens is \[ 4 \cdot 25 = 100. \]

The total number of cats and kittens is then \[ 100 + 100 = 200. \]

Thus, B is the correct answer.

6 答案 D

设两个数为 \(x\) 和 \(y\)，满足 \[ xy = 9 \text{ 和 } \dfrac{1}{x} = \dfrac{4}{y}\text{。}\]

由乘积条件与倒数条件得 \[ y = \dfrac{9}{x} \]\[ \dfrac{1}{x} = \dfrac{4x}{9}\text{。}\]\[ x = \dfrac{3}{2}\text{，}\] 其中 \(x\) 为正。

于是另一个数为 \[ y = 9 \div \dfrac{3}{2} = 6\text{。}\]

两数之和为 \[ 6 + \dfrac{3}{2} = \dfrac{15}{2}\text{。}\]

所以正确答案是 D。

English

Let the two numbers be \(x\) and \(y\) such that \[ xy = 9 \text{ and } \dfrac{1}{x} = \dfrac{4}{y}. \]

We get that \[ y = \dfrac{9}{x} \]\[ \dfrac{1}{x} = \dfrac{4x}{9}. \]\[ x = \dfrac{3}{2}, \] since \(x\) is positive.

Then \[ y = 9 \div \dfrac{3}{2} = 6. \]

The desired sum is then \[ 6 + \dfrac{3}{2} = \dfrac{15}{2}. \]

Thus, D is the correct answer.

7 答案 C

不妨设袋中共有 \(5\) 颗弹珠；因为只关心比例，所以可以这样假设。于是蓝色有 \(3\) 颗，红色有 \(2\) 颗。

红色数量加倍后有 \(4\) 颗红弹珠，因此红色所占比例为 \[ \dfrac{4}{4 + 3} = \dfrac{4}{7}\text{。}\]

所以正确答案是 C。

English

WLOG, let the number of marbles in the bag be \(5.\) Since we only care about ratios, we can do this. Then there are \(3\) blue marbles and \(2\) red marbles.

Doubling the red marbles gives us \(4\) of them. Then the fraction of red marbles is \[ \dfrac{4}{4 + 3} = \dfrac{4}{7}. \]

Thus, C is the correct answer.

8 答案 D

设三个数为 \(a, b, c\)，且 \(a \lt b \lt c\)。由于三个和都不同，所以这三个数也互不相等。

题意给出 \[ a + b = 12, a + c = 17\text{，} \] 以及 \[ b + c = 19\text{。} \]

把三式相加，得到 \[ 2(a + b + c) = 48\text{。} \]

因此 \[ a + b + c = 24\text{，} \] 再减去 \[ a + c = 17\text{，} \] 得到 \(b = 7\)。

所以正确答案是 D。

English

Let the three numbers be \(a, b, c\) where \(a \lt b \lt c.\) None of them are equal, since all three sums are different.

Then \[ a + b = 12, a + c = 17, \] and \[ b + c = 19. \]

Adding all three equations together gives us \[ 2(a + b + c) = 48. \]

Then \[ a + b + c = 24, \] from which we can subtract \[ a + c = 17, \] to get \(b = 7.\)

Thus, D is the correct answer.

9 答案 D

和为 \(7\) 的点数对是 \[ (2, 5), (4, 3), \text{ 和 } (6, 1)\text{。}\]

有 \[ \dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9} \] 的概率得到其中任意一个指定的点数对。

共有 \(3\) 个这样的点数对，所以和为 \(7\) 的总概率为 \[ 3 \cdot \dfrac{1}{9} = \dfrac{1}{3}\text{。}\]

所以正确答案是 D。

English

The pairs of numbers that sum to \(7\) are \[ (2, 5), (4, 3), \text{ and } (6, 1). \]

There is a \[ \dfrac{1}{3} \cdot \dfrac{1}{3} = \dfrac{1}{9} \] chance that we get any of these pairs.

There are \(3\) pairs, which means that the total probability that the rolls sum to \(7\) is \[ 3 \cdot \dfrac{1}{9} = \dfrac{1}{3}. \]

Thus, D is the correct answer.

10 答案 C

设最小扇形角为 \(a\)，等差数列的公差为 \(d\)。

于是 \[ \dfrac{12}{2}(a + a + 11d) = 12a + 66d \] 就是这个等差数列的和。

总和是 \(360\)，所以 \[ 12a + 66d = 360 \] \[ 2a + 11d = 60\text{。}\]

要使 \(a\) 尽可能小，就要让 \(d\) 尽可能大。若 \(d = 5\)，则 \(a\) 不是整数；因此 \(d = 4\)，且 \(a = 8\)。

所以正确答案是 C。

English

Let \(a\) be the smallest possible sector angle and \(d\) be the difference in the arithmetic sequence.

Then we have that \[ \dfrac{12}{2}(a + a + 11d) = 12a + 66d \] is the sum of the arithmetic sequence.

We have that this sums to \(360,\) so \[ 12a + 66d = 360 \] \[ 2a + 11d = 60. \]

We want to minimize \(a,\) so we maximize \(d.\) If \(d = 5,\) then \(a\) is not an integer, so \(d = 4\) and \(a = 8.\)

Thus, C is the correct answer.

11 答案 D

令 \(x\) 为 \(BC\)。注意 \(\triangle CEB\) 与 \(\triangle CDA\) 由角角判定相似（切线与过切点的半径垂直）。

于是 \[ \dfrac{x}{3} = \dfrac{8 + x}{5}\text{。} \] 交叉相乘得到 \[ 5x = 24 + 3x \]\[ x = 12\text{。} \]

所以正确答案是 D。

English

Let \(x\) be \(BC.\) Note that \(\triangle CEB\) and \(\triangle CDA\) are similar due to angle-angle (tangent lines are perpendicular to radii).

Then \[ \dfrac{x}{3} = \dfrac{8 + x}{5}. \] Cross-multiplying gives us \[ 5x = 24 + 3x \]\[ x = 12. \]

Thus, D is the correct answer.

12 答案 A

平年有 \(365\) 天，向未来移动一年，星期几会向后移动一天，因为 \[ 365 = 52 \cdot 7 + 1\text{。} \]

闰年有额外的一天，所以星期几会向后移动两天。

在 \(1812\) 年到 \(2012\) 年之间，有五十个年份是 \(4\) 的倍数，但要去掉 \(1900\) 年，所以共有 \(49\) 个闰年。

因此，往回推 \(200\) 年需要往回移动 \[ 200 + 49 = 249 = 35 \cdot 7 + 4 \] 天，也就是星期几往回移动 \(4\) 天。

星期二往前推四天是星期五，也就是 \(1812\) 年二月 \(7\) 日。

所以正确答案是 A。

English

On a typical year with \(365\) days, moving one year into the future moves the day of the week forward one since \[ 365 = 52 \cdot 7 + 1. \]

On a leap year however, the day of the week gets moved forward twice since there is an extra day.

Fifty of the years between \(2012\) and \(1812\) are multiples of \(4,\) but we have to discard \(1900,\) which leaves us with \(49\) leap years.

Therefore, moving back \(200\) years means we have to go back \[ 200 + 49 = 249 = 35 \cdot 7 + 4 \] days, which means we move back \(4\) days in the week.

This takes us back to Friday, which is the day that corresponds to February \(7,\) \(1812.\)

Thus, A is the correct answer.

13 答案 C

设排列顺序为 \[ a, b, c, d, e\text{。} \]

则迭代平均数为 \[ \dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}\]\[ = \dfrac{a + b + 2c + 4d + 8e}{16}\text{。} \]

要使结果最小，应按 \[ 5, 4, 3, 2, 1\text{，} \] 排列，得到 \[ \dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}\text{。}\]

要使结果最大，则反过来排列，得到 \[ \dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}\text{。} \]

两者之差为 \[ \dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}\text{。} \]

所以正确答案是 C。

English

Let the order of the numbers be \[ a, b, c, d, e. \]

Then the iterative average is \[ \dfrac{\dfrac{\dfrac{\dfrac{a + b}{2} + c}{2} + d}{2} + e}{2}\]\[ = \dfrac{a + b + 2c + 4d + 8e}{16}. \]

To minimize this, we make the order \[ 5, 4, 3, 2, 1, \] which gives us a sum of \[ \dfrac{5 + 4 + 6 + 8 + 8}{16} = \dfrac{31}{16}.\]

To maximize it, we have to reverse this order to get an average of \[ \dfrac{1 + 2 + 6 + 16 + 40}{16} = \dfrac{65}{16}. \]

The difference between these is \[ \dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}. \]

Thus, C is the correct answer.

14 答案 B

共有 \(15\) 行各含 \(15\) 个黑格，另有 \(16\) 行各含 \(16\) 个黑格。

第一行有 \(16\) 个黑格，其余各行按 \(15\) 个与 \(16\) 个交替。

因此共有 \(16\) 行各含 \(16\) 个黑格，其余各行含 \(15\) 个黑格。

因此黑格总数为 \[ 15^2 + 16^2 = 481\text{。}\]

所以正确答案是 B。

English

Note that there are \(15\) rows with \(15\) black tiles and \(16\) rows with \(16\) black tiles.

This can be seen by observing that the first row has \(16\) black tiles, and all the other rows alternate with \(15\) and \(16\) tiles.

Then, due to the alternating pattern, there will be a total of \(16\) rows with \(16\) tiles, and the other rows have \(15\) tiles.

The total number of black squares is then \[ 15^2 + 16^2 = 481. \]

Thus, B is the correct answer.

15 答案 B

可以用坐标几何求两条直线的交点。

令 \(A\) 为原点，\(B = (1, 0)\)。经过 \(A\) 的直线斜率为 \(-\dfrac{1}{2}\)，方程为 \[ y = -\dfrac{1}{2}x\text{。} \] 经过 \(B\) 的直线斜率为 \(2\)，\(y\) 轴截距为 \(-2\)，所以方程为 \[ y = 2x - 2\text{。} \]

联立两式，得到 \[ 2x - 2 = -\dfrac{1}{2}x \]\[ x = \dfrac{4}{5}\text{。} \]

因此点 \(C\) 的 \(y\) 坐标为 \[ -\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}\text{。} \]

三角形 \(ABC\) 的面积为 \[ \dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}\text{。} \]

所以正确答案是 B。

English

We can use coordinate geometry to figure out where the intersection of the two lines occurs.

Let \(A\) be the origin and \(B = (1, 0).\) Then the slope of the line through \(A\) is \(-\dfrac{1}{2},\) which makes the equation of the line \[ y = -\dfrac{1}{2}x. \] The slope of the line through \(B\) is \(2.\) The \(y\)-intercept is \(-2.\) This makes the equation of this line \[ y = 2x - 2. \]

Equating the equations, we get \[ 2x - 2 = -\dfrac{1}{2}x \]\[ x = \dfrac{4}{5}. \]

This makes the \(y\)-coordinate of \(C\) \[ -\dfrac{1}{2} \cdot \dfrac{4}{5} = -\dfrac{2}{5}. \]

The area of triangle \(ABC\) is then \[ \dfrac{1}{2} \cdot 1 \cdot \dfrac{2}{5} = \dfrac{1}{5}. \]

Thus, B is the correct answer.

16 答案 C

先求速度为 \(4.8\) 米/秒的跑者追上最慢跑者所需的时间。

必须有 \[ 4.8x - 4.4x = 500 \]\[ x = 1250\text{，} \] 其中 \(x\) 是较快跑者追上另一人的时间。

注意 \(4.4 \cdot 1250 = 5500\)，所以这两名跑者每次相遇都在起点。

现在要求最小的 \(t\)，使 \(t\) 是 \(1250\) 的倍数，并且最快的跑者也回到起点。

每 \(1250\) 秒，最快的跑者跑 \(1250 \cdot 5 = 6250\) 米；到 \(2500\) 秒时，他跑了 \(12500\) 米，正好是整数圈。

所以正确答案是 C。

English

Let us find the amount of time that it takes for the runner running at \(4.8\) meters per second to lap the slowest person.

We must have that \[ 4.8x - 4.4x = 500 \]\[ x = 1250, \] where \(x\) is the amount of time it takes for the faster runner to lap the other.

Note that \(4.4 \cdot 1250 = 5500,\) which means that these two runners always intersect at the starting line.

We now have to find the least time, \(t,\) such that \(t\) is a multiple of \(1250\) and the fastest runner ends up at the starting line.

Every \(1250\) seconds, the fastest runner runs \(1250 \cdot 5 = 6250\) meters. Then in \(2500\) seconds, the fastest runner runs \(12500\) meters, which is a whole number of laps.

Thus, C is the correct answer.

17 答案 C

先因式分解 \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2)\text{。} \] 消去这个因子后得到 \[ \dfrac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \dfrac{73}{3}\text{。} \]

交叉相乘并整理，得到 \[ 70a^2 - 149ab + 70b^2 = 0\text{。} \] 因为 \(b \neq 0\)，可同除以 \(b^2\)，得到 \[ 70\left(\dfrac{a}{b}\right)^2 - 149\dfrac{a}{b} + 70 = 0\text{。} \]

用二次公式，并注意 \(a \gt b\)，可得 \(\dfrac{a}{b} = \dfrac{10}{7}\)。

由于 \(a\) 和 \(b\) 互质，所以 \(a = 10\)，\(b = 7\)，差为 \(10 - 7 = 3\)。

所以正确答案是 C。

English

Recall that we can factor \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2). \] Canceling out this factor gives us that \[ \dfrac{a^2 + ab + b^2}{a^2 - 2ab + b^2} = \dfrac{73}{3}. \]

Cross-multiplying and rearranging gives us \[ 70a^2 - 149ab + 70b^2 = 0. \] Since \(b \neq 0,\) we can divide through by \(b^2\) to get \[ 70\left(\dfrac{a}{b}\right)^2 - 149\dfrac{a}{b} + 70 = 0. \]

Applying the quadratic formula and noting that \(a \gt b\) gives us that \(\dfrac{a}{b} = \dfrac{10}{7}.\)

Since \(a\) and \(b\) are relatively prime, we have that \(a = 10\) and \(b = 7.\) Their difference is \(10 - 7 = 3.\)

Thus, C is the correct answer.

18 答案 E

每段圆弧的半径都是 \(1\)，长度都是 \(\frac{2\pi}{3}\)，所以对应的各个扇形面积都相同。如图所示，这些扇形块可以重新拼接：添加到正六边形上的部分合起来恰好构成一个单位圆。

正六边形由六个边长为 \(2\) 的等边三角形拼成，所以它的面积为 \[6\left(\dfrac{\sqrt3}{4}\cdot2^2\right)=6\sqrt3\text{。}\] 再加上那个单位圆，围成的总面积为 \(\pi+6\sqrt3\)。

所以正确答案是 E。

English

Each arc has radius \(1\) and length \(\frac{2\pi}{3},\) so the corresponding circular sectors all have the same area. As the diagram shows, the sector pieces may be rearranged: everything added to the regular hexagon combines to exactly one unit circle.

The regular hexagon is made of six equilateral triangles of side \(2,\) so its area is \[6\left(\dfrac{\sqrt3}{4}\cdot2^2\right)=6\sqrt3.\] Adding the unit circle gives a total enclosed area of \(\pi+6\sqrt3.\)

Thus, E is the correct answer.

19 答案 D

设午餐休息为 \(m\) 分钟，Paula 的工作速度为每分钟 \(p\) 个百分点，两名助手的合计速度为每分钟 \(h\) 个百分点。

星期一给出 \((p+h)(480-m)=50\)。星期二给出 \(h(372-m)=24\)。星期三剩余工作量为百分之 \(26\)，所以 \(p(672-m)=26\)。

由后两式相加再减去第一式，得到 \(108h-192p=0\)，所以 \(h=\dfrac{16}{9}p\)。

代入星期一的方程得 \(\dfrac{25}{9}p(480-m)=50\)，而星期三方程为 \(p(672-m)=26\)。联立解得 \(m=48\)。

所以正确答案是 D。

English

Let the lunch break be \(m\) minutes, Paula’s rate be \(p\) percent per minute, and the helpers’ combined rate be \(h\) percent per minute.

Monday gives \((p+h)(480-m)=50\). Tuesday gives \(h(372-m)=24\). Since the remaining work on Wednesday was \(26\) percent, Wednesday gives \(p(672-m)=26\).

Adding the Tuesday and Wednesday equations and subtracting the Monday equation gives \(108h-192p=0\), so \(h=\dfrac{16}{9}p\).

Substituting into Monday gives \(\dfrac{25}{9}p(480-m)=50\), while Wednesday gives \(p(672-m)=26\). Solving these two equations gives \(m=48\).

Thus, D is the correct answer.

20 答案 A

中心格必须一开始就是黑色，贡献概率 \(\dfrac12\)。四个角在旋转下形成一个循环，四个边中点也形成另一个相同的循环。

对一个 \(4\) 循环，除非有一个白格被旋转到原本也是白格的位置，否则最后四个位置都会变黑。成功的初始颜色序列为 \(BBBB\)、\(BBBW\) 的四种旋转，以及 \(BWBW\) 的两种旋转，共 \(7\) 种，占 \(16\) 种。

边中点循环有相同的独立计数。因此概率为 \(\dfrac12\left(\dfrac7{16}\right)^2=\dfrac{49}{512}\)。

所以正确答案是 A。

English

The center square must initially be black, contributing probability \(\dfrac12\). The four corner squares form one cycle under the rotation, and the four edge-middle squares form another identical cycle.

For one \(4\)-cycle, the final four positions are all black unless a white square is rotated into a position that was also white. The successful initial colorings are \(BBBB\), the four rotations of \(BBBW\), and the two rotations of \(BWBW\), for \(7\) colorings out of \(16\).

The same count applies to the edge-middle cycle, independently. Therefore the probability is \(\dfrac12\left(\dfrac7{16}\right)^2=\dfrac{49}{512}\).

Thus, A is the correct answer.

21 答案 C

注意 \(EF = \dfrac{1}{2}AD\)，因为它是 \(\triangle ABD\) 的中位线。同理，\(HG = \dfrac{1}{2}AD\)，且 \(FG = \dfrac{1}{2}BC\)。

又因为 \(\overline{EF}\) 和 \(\overline{HG}\) 垂直于 \(xy\) 平面，所以它们垂直于 \(\overline{FG}\) 和 \(\overline{EH}\)。

因此 \(EFGH\) 是长方形，且 \(EF = HG\)。有 \(EF = \dfrac{1}{2} \cdot 3 = \dfrac{3}{2}\)。

还可得 \[ FG = \dfrac{1}{2} \sqrt{1^2 + 2^2} = \dfrac{\sqrt{5}}{2}\text{。}\]

所以 \(EFGH\) 的面积为 \[ EF \cdot FG = \dfrac{3}{2} \cdot \dfrac{\sqrt{5}}{2} = \dfrac{3\sqrt{5}}{4}\text{。}\]

所以正确答案是 C。

English

Note that \(EF = \dfrac{1}{2}AD\) since it is a midsegment of \(\triangle ABD.\) Similarly, \(HG = \dfrac{1}{2}AD\) and \(FG = \dfrac{1}{2}BC.\)

We also have that \(\overline{EF}\) and \(\overline{HG}\) are perpendicular to the \(xy\)-plane, which means that they are perpendicular to \(\overline{FG}\) and \(\overline{EH}.\)

This tells us that \(EFGH\) is rectangle since \(EF = HG.\) We have \(EF = \dfrac{1}{2} \cdot 3 = \dfrac{3}{2}.\)

We also have that \[ FG = \dfrac{1}{2} \sqrt{1^2 + 2^2} = \dfrac{\sqrt{5}}{2}. \]

The area of \(EFGH\) is then \[ EF \cdot FG = \dfrac{3}{2} \cdot \dfrac{\sqrt{5}}{2} = \dfrac{3\sqrt{5}}{4}. \]

Thus, C is the correct answer.

22 答案 A

前 \(m\) 个正奇数之和为 \(m^2\)，前 \(n\) 个正偶数之和为 \(n(n+1)\)。因此 \(m^2=n(n+1)+212\)。

把它看成关于 \(n\) 的二次方程，判别式为 \(1-4(212-m^2)=4m^2-847\)，必须是一个奇平方数。设 \(p^2=4m^2-847\)，则 \((2m+p)(2m-p)=847\)。

\(847\) 的正因数配对为 \(847\cdot1\)、\(121\cdot7\) 和 \(77\cdot11\)。它们分别给出 \(p=423,57,33\)。

因为 \(n=\dfrac{-1+p}{2}\)，所以 \(n\) 的可能值为 \(211,28,16\)，它们的和为 \(255\)。

所以正确答案是 A。

English

The first \(m\) positive odd integers sum to \(m^2\), and the first \(n\) positive even integers sum to \(n(n+1)\). Thus \(m^2=n(n+1)+212\).

As a quadratic in \(n\), this has discriminant \(1-4(212-m^2)=4m^2-847\), which must be an odd square. Let \(p^2=4m^2-847\). Then \((2m+p)(2m-p)=847\).

The positive factor pairs of \(847\) are \(847\cdot1\), \(121\cdot7\), and \(77\cdot11\). They give \(p=423,57,33\), respectively.

Because \(n=\dfrac{-1+p}{2}\), the possible values of \(n\) are \(211,28,16\). Their sum is \(255\).

Thus, A is the correct answer.

23 答案 B

按每个人拥有的好友数分类。由于好友图既不是空图也不是完全图，这个数从 \(1\) 到 \(4\)。

注意，\(1\) 个好友和 \(2\) 个好友的情况，分别通过取补图与 \(4\) 个好友和 \(3\) 个好友的情况对应，因为确定谁是好友也就确定了谁不是好友。

情况 \(1\)：每个人有 \(1\) 个好友

这意味着 \(6\) 个人必须分成 \(3\) 对，每对中的两人互为好友。

第一个人的好友有 \(5\) 种选择，剩下 \(4\) 个人。

下一位未配对者的好友有 \(3\) 种选择，剩余的 \(2\) 个人则必须互为好友。

因此这种情况共有 \(3 \cdot 5 = 15\) 种可能。

情况 \(2\)：每个人有 \(2\) 个好友

这种情况有两种可能。第一种是分成两个三人组，每组三人彼此都是好友。

选择第一个三人组有 \(\binom{6}{3} = 20\) 种方法。由于两个组可以互换，必须除以 \(2\)，得到 \(20 \div 2 = 10\) 种配置。

第二种可能是好友关系形成一个 \(6\) 环。

六个人沿环的每一种排列都给出这样的图。选择起点会使每个图被计算 \(6\) 次，选择遍历方向又会被计算 \(2\) 次，因此不同的 \(6\) 环共有 \(\frac{6!}{6\cdot2}=60\) 个。再加上 \(10\) 个两三角形配置，本情况共有 \(10+60=70\) 种配置。

所以总配置数为 \[ 2(15 + 70) = 170\text{。}\]

所以正确答案是 B。

English

We case on the value of friends that each person has. This value ranges from \(1\) to \(4\), since the graph is neither empty nor complete.

Note that the cases for \(1\) and \(2\) friends correspond with the case for \(4\) and \(3\) friends, since choosing who are friends determines who are not friends.

Case \(1:\) everyone has \(1\) friend

This means that the \(6\) people must split up into \(3\) pairs where the people in each pair are friends.

There are \(5\) choices for the friend for the first person. This leaves \(4\) people remaining.

There are then \(3\) choices for the friend of the next unpaired person. The remaining \(2\) people are then forced to be friends.

Therefore, there are \(3 \cdot 5 = 15\) possibilities for this case.

Case \(2:\) everyone has \(2\) friends

There are two possibilities for this case. There could be two triples where everyone in a triple is friends with each other.

For this possibility, there are \(\binom{6}{3} = 20\) ways to choose the people in the first triple. We have to divide by \(2\) since we can swap the pairs. This gives us \(20 \div 2 = 10\) configurations.

The second possibility is that the friends form one \(6\)-cycle.

Every ordering of the six people around a cycle gives such a graph. Each graph is counted \(6\) times by the choice of starting person and \(2\) times by the direction of traversal, so there are \(\frac{6!}{6\cdot2}=60\) distinct \(6\)-cycles. Together with the \(10\) pairs of triangles, this case has \(10+60=70\) configurations.

The total number of arrangements is then \[ 2(15 + 70) = 170. \]

Thus, B is the correct answer.

24 答案 E

两个方程相加，得到 \[ 2(a^2 + b^2 + c^2) - 2(ab + ac + bc)\] \[ = 14\text{。}\]

把左边分组并因式分解，得到 \[ (a - b)^2 + (a - c)^2 + (b - c)^2\] \[ = 14\text{。}\]

左边每一项都是非负整数的平方。三个平方数和为 \(14\) 的唯一方式是 \(9, 4\) 和 \(1\)。

在三组差中，\(a - c\) 最大，所以 \(a - c = 3\)。

还不能确定另外两个差分别对应哪个平方。先试 \(a - b = 1\) 和 \(b - c = 2\)。

把这些值代入第一个方程，得到 \[ a^2 - (a - 1)^2 - (a - 3)^2 \]\[ + a(a - 1) = 2011\text{。}\] 化简得 \(7a = 2021\)。由于 \(2021\) 不能被 \(7\) 整除，所以应有 \(a - b = 2\)、\(b - c = 1\)。

在另一种情形下，\(b=a-2\)，\(c=a-3\)。第一个方程变为 \[\begin{aligned} a^2-(a-2)^2-(a-3)^2\\ {}+a(a-2)&=2011 \end{aligned}\text{，}\] 即 \(8a-13=2011\)。因此 \(a=253\)。

所以正确答案是 E。

English

Adding together the equations gives us \[ 2(a^2 + b^2 + c^2) - 2(ab + ac + bc)\]\[ = 14. \]

We can group terms and factor this to get \[ (a - b)^2 + (a - c)^2 + (b - c)^2\]\[ = 14. \]

Note that every term on the left hand side is a nonnegative square integer. The only triple of squares that add to \(14\) is \(9, 4,\) and \(1.\)

We have that \(a - c\) is the biggest difference among the three pairs. Therefore, \(a - c = 3.\)

We cannot discern which of the other terms we can match with the other squares. Let us try \(a - b = 1\) and \(b - c = 2.\)

Plugging in these values into the first equation gives us \[ a^2 - (a - 1)^2 - (a - 3)^2 \] \[ + a(a - 1) = 2011. \] Simplifying yields \(7a = 2021.\) Since \(2021\) is not divisible by \(7,\) we have that \(a - b = 2\) and \(b - c = 1.\)

In the other case, \(b=a-2\) and \(c=a-3.\) The first equation becomes \[\begin{aligned} a^2-(a-2)^2-(a-3)^2\\ {}+a(a-2)&=2011, \end{aligned}\] or \(8a-13=2011.\) Hence \(a=253.\)

Thus, E is the correct answer.

25 答案 D

这个问题可以看作几何概率问题，把区间看成坐标轴上的范围。

不妨先考虑 \[ n \geq x \geq y \geq z \geq 0\text{。}\]

满足这个顺序限制的点 \((x, y, z)\) 构成一个四面体。

这个四面体的高为 \(n\)，底面积为 \(\dfrac{n^2}{2}\)，所以体积为 \[ \dfrac{1}{3} \cdot \dfrac{n^2}{2} \cdot n = \dfrac{n^3}{6}\text{。}\]

现在加入题目中的限制。需要找出满足下面条件的区域：\[ |x - y|, |x - z|, |y - z| \geq 1\text{。}\]

根据已经规定的顺序，这些不等式可化为 \[ x - y \geq 1 \text{ 和 } y - z \geq 1\text{。}\]

这两个限制形成如下图所示的另一个四面体。

注意，在新的四面体中，所有线性尺寸都减少了 \(2\)。因此高为 \(n - 2\)，底面积为 \(\dfrac{(n - 2)^2}{2}\)。

因此体积为 \[ \dfrac{1}{3} \cdot \dfrac{(n - 2)^2}{2} \cdot (n - 2)\] \[ = \dfrac{(n - 2)^3}{6}\text{。}\]

所求概率为 \[ \dfrac{(n - 2)^3}{6} \div \dfrac{n^3}{6} = \dfrac{(n - 2)^3}{n^3}\text{。}\]

逐一检查选项可知，使这个比例大于 \(\dfrac{1}{2}\) 的最小值是 \(10\)。

所以正确答案是 D。

English

This problem lends itself to geometric probability since we can view the interval as a range on an axis.

WLOG, let \[ n \geq x \geq y \geq z \geq 0. \]

Then we have that the points \((x, y, z)\) which satisfy this restriction form a tetrahedron.

The height of this tetrahedron is \(n,\) and the base has an area of \(\dfrac{n^2}{2}.\) This makes the volume \[ \dfrac{1}{3} \cdot \dfrac{n^2}{2} \cdot n = \dfrac{n^3}{6}. \]

Now we have to apply the restrictions from the problem statement. We need to find the region where \[ |x - y|, |x - z|, |y - z| \geq 1. \]

From our ordering condition that we imposed, these inequalities reduce to \[ x - y \geq 1 \text{ and } y - z \geq 1. \]

These two restrictions form another tetrahedron as shown below.

Note that in the new tetrahedron, all the dimensions have been reduced by \(2.\) This makes the height \(n - 2\) and the base \(\dfrac{(n - 2)^2}{2}.\)

The volume is then \[ \dfrac{1}{3} \cdot \dfrac{(n - 2)^2}{2} \cdot (n - 2)\]\[ = \dfrac{(n - 2)^3}{6}. \]

The desired probability is then \[ \dfrac{(n - 2)^3}{6} \div \dfrac{n^3}{6} = \dfrac{(n - 2)^3}{n^3}. \]

Plugging in all the answer choices, we get that the smallest value such that this fraction is greater than \(\dfrac{1}{2}\) is \(10.\)

Thus, D is the correct answer.