2012A AMC 12 真题（P版）

1 答案 E

小虫第一段爬了 \(|-6-(-2)| = 4\) 个单位，第二段爬了 \(|5-(-6)| = 11\) 个单位。

总距离为 \(4 + 11 = 15\)。

因此，正确答案是 E。

English

The bug crawls \(|-6-(-2)| = 4\) units on the first leg and \(|5-(-6)| = 11\) units on the second leg.

The total distance is \(4 + 11 = 15.\)

Thus, the correct answer is E.

2 答案 D

\(5\) 分钟共有 \(300\) 秒。Cagney 给 \(\dfrac{300}{20} = 15\) 个纸杯蛋糕涂糖霜，Lacey 给 \(\dfrac{300}{30} = 10\) 个纸杯蛋糕涂糖霜。

两人合起来完成 \(15 + 10 = 25\) 个纸杯蛋糕。

因此，正确答案是 D。

English

In \(5\) minutes there are \(300\) seconds. Cagney frosts \(\dfrac{300}{20} = 15\) cupcakes and Lacey frosts \(\dfrac{300}{30} = 10\) cupcakes.

Together they frost \(15 + 10 = 25\) cupcakes.

Thus, the correct answer is D.

3 答案 D

第二个盒子的体积是第一个盒子的 \(2 \times 3 \times 1 = 6\) 倍，所以它能装的黏土也是 \(6\) 倍。

因此 \(n = 6 \cdot 40 = 240\)。

因此，正确答案是 D。

English

The second box has \(2 \times 3 \times 1 = 6\) times the volume of the first, so it holds \(6\) times as much clay.

Therefore \(n = 6 \cdot 40 = 240.\)

Thus, the correct answer is D.

4 答案 C

假设共有 \(5\) 颗弹珠：\(3\) 颗蓝色，\(2\) 颗红色。红色翻倍后有 \(4\) 颗红色，而蓝色仍为 \(3\) 颗。

现在总数为 \(3 + 4 = 7\)，所以红色所占的比例是 \(\dfrac{4}{7}\)。

因此，正确答案是 C。

English

Suppose there are \(5\) marbles: \(3\) blue and \(2\) red. Doubling the red gives \(4\) red while the blue stays at \(3.\)

The total is now \(3 + 4 = 7,\) so the fraction that is red is \(\dfrac{4}{7}.\)

Thus, the correct answer is C.

5 答案 D

设蓝莓有 \(b\) 颗。则覆盆子有 \(2b\) 颗，樱桃有 \(4 \cdot 2b = 8b\) 颗，葡萄有 \(3 \cdot 8b = 24b\) 颗。

总数为 \(b + 2b + 8b + 24b = 35b = 280\)，所以 \(b = 8\)，樱桃有 \(8b = 64\) 颗。

因此，正确答案是 D。

English

Let \(b\) be the number of blueberries. Then there are \(2b\) raspberries, \(4 \cdot 2b = 8b\) cherries, and \(3 \cdot 8b = 24b\) grapes.

The total is \(b + 2b + 8b + 24b = 35b = 280,\) so \(b = 8\) and there are \(8b = 64\) cherries.

Thus, the correct answer is D.

6 答案 D

设三个数为 \(a \lt b \lt c\)。将三个两两之和相加，得到 \(2(a+b+c) = 12+17+19\) \(= 48\)，所以 \(a+b+c = 24\)。

因此 \(a = 24-19 = 5\)，\(b = 24-17 = 7\)，\(c = 24-12 = 12\)。中间的数是 \(7\)。

因此，正确答案是 D。

English

Let the numbers be \(a \lt b \lt c.\) Adding the three pairwise sums gives \(2(a+b+c) = 12+17+19\) \(= 48,\) so \(a+b+c = 24.\)

Then \(a = 24-19 = 5,\) \(b = 24-17 = 7,\) and \(c = 24-12 = 12.\) The middle number is \(7.\)

Thus, the correct answer is D.

7 答案 C

设最小角为 \(a\)，公差为 \(d \ge 0\)。所有角的和为 \(12a + 66d = 360\)，所以 \(2a + 11d = 60\)。

要使 \(a\) 小，就让 \(d\) 尽量大。由于 \(11d\) 必须为偶数，\(d\) 为偶数，而 \(d = 4\) 给出 \(2a = 60 - 44 = 16\)，所以 \(a = 8\)。更大的偶数 \(d\) 会使 \(a\) 非正。

因此，正确答案是 C。

English

Let \(a\) be the smallest angle and \(d \ge 0\) the common difference. The sum of the angles is \(12a + 66d = 360,\) so \(2a + 11d = 60.\)

To make \(a\) small, take \(d\) large. Since \(11d\) must be even, \(d\) is even, and \(d = 4\) gives \(2a = 60 - 44 = 16,\) so \(a = 8.\) A larger even \(d\) makes \(a\) non-positive.

Thus, the correct answer is C.

8 答案 C

对顺序 \(a, b, c, d, e\)，迭代平均数为 \[\frac{a + b + 2c + 4d + 8e}{16}\text{。}\] 后面的位置权重最大。

最大值取 \((a,b,c,d,e) = (1,2,3,4,5)\)，得到 \(\dfrac{65}{16}\)，最小值取 \((5,4,3,2,1)\)，得到 \(\dfrac{31}{16}\)。

差为 \(\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}\)。

因此，正确答案是 C。

English

For the order \(a, b, c, d, e,\) the iterative average is \[\frac{a + b + 2c + 4d + 8e}{16}.\] The later positions carry the most weight.

The largest value uses \((a,b,c,d,e) = (1,2,3,4,5),\) giving \(\dfrac{65}{16},\) and the smallest uses \((5,4,3,2,1),\) giving \(\dfrac{31}{16}.\)

The difference is \(\dfrac{65}{16} - \dfrac{31}{16} = \dfrac{34}{16} = \dfrac{17}{8}.\)

Thus, the correct answer is C.

9 答案 A

从 \(1812\) 年二月 \(7\) 日到 \(2012\) 年二月 \(7\) 日，共有 \(200 \cdot 365 = 73000\) 个普通日，再加上每个闰日。

\(200\) 年中有四分之一含闰日，但要排除 \(1900\)，因此闰日数为 \(\tfrac14 \cdot 200 - 1 = 49\)。所以总跨度为 \(73049\) 天。

因为 \(73049 = 7 \cdot 10435 + 4\)，出生日期比星期二早 \(4\) 天，也就是星期五。

因此，正确答案是 A。

English

From February \(7,\) \(1812\) to February \(7,\) \(2012\) there are \(200 \cdot 365 = 73000\) ordinary days plus one for each leap day.

One quarter of the \(200\) years contain a leap day, except \(1900,\) giving \(\tfrac14 \cdot 200 - 1 = 49\) leap days. So the span is \(73049\) days.

Since \(73049 = 7 \cdot 10435 + 4,\) the birth day was \(4\) days before a Tuesday, which is a Friday.

Thus, the correct answer is A.

10 答案 D

这条中线把原三角形分成两个面积相等的三角形，每个面积为 \(15\)。其中一个三角形有两条边长分别为 \(5\)（底边的一半）和 \(9\)（中线），它们的夹角为 \(\theta\)。

它的面积为 \(\tfrac12 \cdot 5 \cdot 9 \sin\theta = 15\)，所以 \(\sin\theta = \dfrac{2 \cdot 15}{5 \cdot 9} = \dfrac{2}{3}\)。

因此，正确答案是 D。

English

The median divides the triangle into two triangles of equal area \(15.\) One of them has the two sides of length \(5\) (half the base) and \(9\) (the median) meeting at angle \(\theta.\)

Its area is \(\tfrac12 \cdot 5 \cdot 9 \sin\theta = 15,\) so \(\sin\theta = \dfrac{2 \cdot 15}{5 \cdot 9} = \dfrac{2}{3}.\)

Thus, the correct answer is D.

11 答案 B

Alex 获胜的概率为 \(\tfrac12\)，其余两人共享剩下的 \(\tfrac12\)。因为 Mel 获胜的可能性是 Chelsea 的两倍，\(P(\text{Mel}) = \tfrac13\)，\(P(\text{Chelsea}) = \tfrac16\)。

获胜顺序 \(AAAMMC\) 的排列数为 \(\dfrac{6!}{3!\,2!\,1!} = 60\)。概率为 \[ \begin{aligned} 60 \cdot \left(\tfrac12\right)^3 \left(\tfrac13\right)^2 \left(\tfrac16\right) &= \frac{60}{432} \\ &= \frac{5}{36} \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Since Alex wins with probability \(\tfrac12,\) the others share the remaining \(\tfrac12.\) With Mel twice as likely as Chelsea, \(P(\text{Mel}) = \tfrac13\) and \(P(\text{Chelsea}) = \tfrac16.\)

The number of orderings of the wins \(AAAMMC\) is \(\dfrac{6!}{3!\,2!\,1!} = 60.\) The probability is \[ \begin{aligned} 60 \cdot \left(\tfrac12\right)^3 \left(\tfrac13\right)^2 \left(\tfrac16\right) &= \frac{60}{432} \\ &= \frac{5}{36}. \end{aligned} \]

Thus, the correct answer is B.

12 答案 D

由对称性，设 \(A = (a, b)\)，其中 \(a \gt 0\)，且 \(B = (-a, b)\)。正方形位于切点 \((0,1)\) 上方，所以它的水平宽度为 \(2a\)，高度为 \(b - 1\)。

由于二者相等，\(2a = b - 1\)，因此 \(b = 2a + 1\)。

代入 \(a^2 + b^2 = 4\)，得到 \(5a^2 + 4a - 3 = 0\)。正根为 \(a = \dfrac{\sqrt{19} - 2}{5}\)，所以边长为 \(2a = \dfrac{2\sqrt{19} - 4}{5}\)。

因此，正确答案是 D。

English

By symmetry let \(A = (a, b)\) with \(a \gt 0\) and \(B = (-a, b).\) The square sits on the tangent point \((0,1),\) so its horizontal width is \(2a\) and its height is \(b - 1.\)

Since these are equal, \(2a = b - 1,\) giving \(b = 2a + 1.\)

Substituting into \(a^2 + b^2 = 4\) yields \(5a^2 + 4a - 3 = 0.\) The positive root is \(a = \dfrac{\sqrt{19} - 2}{5},\) so the side length is \(2a = \dfrac{2\sqrt{19} - 4}{5}.\)

Thus, the correct answer is D.

13 答案 D

设午餐时长为 \(m\) 分钟。星期一三人工作了 \(480 - m\) 分钟，星期二助手工作了 \(372 - m\) 分钟，星期三 Paula 工作了 \(672 - m\) 分钟。

若 Paula 每分钟刷 \(p\%\)，两名助手合计每分钟刷 \(h\%\)，则 \[ \begin{aligned} (p+h)(480-m) &= 50, \\ h(372-m) &= 24, \\ p(672-m) &= 26 \end{aligned}\text{。}\]

将后两个方程相加，再从第一个方程中减去，得到 \(108h - 192p = 0\)，所以 \(h = \tfrac{16}{9}p\)。解这个方程组得到 \(p = \tfrac{1}{24}\) 且 \(m = 48\)。

因此，正确答案是 D。

English

Let \(m\) be the lunch length in minutes. The three worked \(480 - m\) minutes Monday, the helpers \(372 - m\) minutes Tuesday, and Paula \(672 - m\) minutes Wednesday.

If Paula paints \(p\%\) per minute and the helpers together paint \(h\%\) per minute, then \[ \begin{aligned} (p+h)(480-m) &= 50, \\ h(372-m) &= 24, \\ p(672-m) &= 26. \end{aligned} \]

Adding the last two equations and subtracting from the first gives \(108h - 192p = 0,\) so \(h = \tfrac{16}{9}p.\) Solving the system gives \(p = \tfrac{1}{24}\) and \(m = 48.\)

Thus, the correct answer is D.

14 答案 E

每条弧在单位圆上的长度为 \(\dfrac{2\pi}{3}\)，所以对应一个 \(120^\circ\) 扇形。九个相等扇形可重新组合，使围成区域等于边长为 \(2\) 的正六边形加上一个半径为 \(1\) 的完整圆。

边长为 \(2\) 的正六边形可分成 \(6\) 个边长为 \(2\) 的等边三角形，所以面积为 \(6 \cdot \dfrac{\sqrt3}{4} \cdot 2^2 = 6\sqrt3\)。

加上单位圆面积 \(\pi\)，得到 \(\pi + 6\sqrt3\)。

因此，正确答案是 E。

English

Each arc has length \(\dfrac{2\pi}{3}\) on a unit circle, so it is a \(120^\circ\) sector. The nine equal sectors can be reassembled so that the enclosed region equals the regular hexagon of side \(2\) plus one full circle of radius \(1.\)

A regular hexagon of side \(2\) splits into \(6\) equilateral triangles of side \(2,\) so its area is \(6 \cdot \dfrac{\sqrt3}{4} \cdot 2^2 = 6\sqrt3.\)

Adding the unit circle’s area \(\pi\) gives \(\pi + 6\sqrt3.\)

Thus, the correct answer is E.

15 答案 A

旋转时，四个角格构成一个循环，四个边格构成另一个循环，而中心格保持不动。这三组相互独立。

一个位置最终仍是白色，当且仅当它和旋转到该位置的方格原来都是白色。因此四个角最终全黑，恰好等价于它们的循环颜色串中没有两个相邻的白格。允许的颜色串包括全黑的一个、恰有一个白格的 \(4\) 个，以及两个白格相对的 \(2\) 个，共有 \(7\) 种，而总数为 \(2^4=16\)。所以四个角全黑的概率是 \(\frac{7}{16}\)。四个边格同理。

中心格最终为黑色，只有它一开始就是黑色，概率为 \(\dfrac12\)。相乘可得整个方格全黑的概率为 \[\frac12 \cdot \left(\frac{7}{16}\right)^2 = \frac{49}{512}\text{。}\]

因此，正确答案是 A。

English

The four corners form one cycle under the rotation, the four edge squares form another, and the center is fixed. These three groups are independent.

A position remains white exactly when both it and the square rotated into it were originally white. Thus the corners end black exactly when their cyclic string has no adjacent pair of whites. The allowed strings are the all-black string, the \(4\) strings with one white, and the \(2\) strings with two opposite whites: \(7\) of the \(2^4=16\) possibilities. Hence the corner probability is \(\frac{7}{16}.\) The same argument applies to the four edge squares.

The center is black at the end only if it started black, with probability \(\dfrac12.\) Multiplying, the whole grid is black with probability \[\frac12 \cdot \left(\frac{7}{16}\right)^2 = \frac{49}{512}.\]

Thus, the correct answer is A.

16 答案 E

设 \(C_1\) 的半径为 \(r\)，所以 \(OX = OY = r\)。它们是 \(C_2\) 的等弦，因此在 \(Z\) 处所对角相等：\(\angle XZO = \angle OZY\)。

对三角形 \(XZO\) 和 \(YZO\) 使用余弦定理，\[ \begin{aligned} &\frac{13^2 + 11^2 - r^2}{2 \cdot 13 \cdot 11} \\ &= \frac{7^2 + 11^2 - r^2}{2 \cdot 7 \cdot 11} \end{aligned}\text{。}\]

清去分母并求解，得到 \(r^2 = 30\)，所以 \(r = \sqrt{30}\)。

因此，正确答案是 E。

English

Let \(r\) be the radius of \(C_1,\) so \(OX = OY = r.\) These are equal chords of \(C_2,\) so they subtend equal angles at \(Z:\) \(\angle XZO = \angle OZY.\)

Applying the Law of Cosines to triangles \(XZO\) and \(YZO,\) \[ \begin{aligned} &\frac{13^2 + 11^2 - r^2}{2 \cdot 13 \cdot 11} \\ &= \frac{7^2 + 11^2 - r^2}{2 \cdot 7 \cdot 11}. \end{aligned} \]

Clearing denominators and solving gives \(r^2 = 30,\) so \(r = \sqrt{30}.\)

Thus, the correct answer is E.

17 答案 B

按模 \(5\) 的余数将 \(\{1, \ldots, 30\}\) 分组；每类有 \(6\) 个数。当余数组合为 \(0{+}0\)、\(1{+}4\) 或 \(2{+}3\) 时，和能被 \(5\) 整除。

因此 \(S\) 最多使用一个 \(\equiv 0\) 的数，并且在 \(\{1\}, \{4\}\) 两类中最多选一类，在 \(\{2\}, \{3\}\) 两类中最多选一类。这最多允许 \(1 + 6 + 6 = 13\) 个数。

集合 \(\{1, 2, 6, 7, 11, 12\) \(16, 17, 21, 22\) \(26, 27, 30\}\) 可以达到 \(13\)，所以最大值为 \(13\)。

因此，正确答案是 B。

English

Group \(\{1, \ldots, 30\}\) by residue modulo \(5;\) each class has \(6\) numbers. A sum is divisible by \(5\) when the residues are \(0{+}0,\) \(1{+}4,\) or \(2{+}3.\)

So \(S\) can use at most one number \(\equiv 0,\) and only one of the classes \(\{1\}, \{4\}\) and only one of \(\{2\}, \{3\}.\) That allows at most \(1 + 6 + 6 = 13\) numbers.

The set \(\{1, 2, 6, 7, 11, 12,\) \(16, 17, 21, 22,\) \(26, 27, 30\}\) achieves \(13,\) so the maximum is \(13.\)

Thus, the correct answer is B.

18 答案 A

设 \(D\) 为内心 \(I\) 到 \(BC\) 的垂足。切线段长度 \(BD = s - AC\)，其中 \(s = \tfrac12(25 + 26 + 27) = 39\)，所以 \(BD = 39 - 26 = 13\)。

由海伦公式，面积为 \(\sqrt{39 \cdot 14 \cdot 13 \cdot 12}\)，内切圆半径满足 \(r^2 = \dfrac{(s-a)(s-b)(s-c)}{s}\) \(= \dfrac{14 \cdot 13 \cdot 12}{39}\) \(= 56\)。

在直角三角形 \(BDI\) 中，\(BI^2 = r^2 + BD^2\) \(= 56 + 169\) \(= 225\)，所以 \(BI = 15\)。

因此，正确答案是 A。

English

Let \(D\) be the foot of the perpendicular from the incenter \(I\) to \(BC.\) The tangent length \(BD = s - AC,\) where \(s = \tfrac12(25 + 26 + 27) = 39,\) so \(BD = 39 - 26 = 13.\)

By Heron’s formula the area is \(\sqrt{39 \cdot 14 \cdot 13 \cdot 12},\) and the inradius satisfies \(r^2 = \dfrac{(s-a)(s-b)(s-c)}{s}\) \(= \dfrac{14 \cdot 13 \cdot 12}{39}\) \(= 56.\)

In right triangle \(BDI,\) \(BI^2 = r^2 + BD^2\) \(= 56 + 169\) \(= 225,\) so \(BI = 15.\)

Thus, the correct answer is A.

19 答案 B

将人建模为图的顶点，好友关系为边。每个人的度数相同，设为 \(n\)，其中 \(1 \le n \le 4\)。度数为 \(n\) 和 \(6 - 1 - n\) 的图互为补图，所以 \(n = 1\) 与 \(n = 4\) 配对，\(n = 2\) 与 \(n = 3\) 配对。

当 \(n = 1\) 时，图是完美匹配：有 \(5 \cdot 3 = 15\) 种。因此 \(n = 4\) 也有 \(15\) 种。

当 \(n = 2\) 时，图是若干个圈的并：要么是两个三角形 \(\left(\binom{5}{2} = 10\right)\)，要么是一个六边形 \(\left(\dfrac{6!}{12} = 60\right)\)，总计 \(70\) 种。因此 \(n = 3\) 也有 \(70\) 种。

总数为 \(15 + 15 + 70 + 70 = 170\)。

因此，正确答案是 B。

English

Model people as vertices of a graph, with edges for friendships. Everyone has the same degree \(n\) with \(1 \le n \le 4.\) The cases \(n\) and \(6 - 1 - n\) are complementary graphs, so \(n = 1\) pairs with \(n = 4\) and \(n = 2\) with \(n = 3.\)

For \(n = 1\) the graph is a perfect matching: \(5 \cdot 3 = 15\) ways. Thus \(n = 4\) also gives \(15.\)

For \(n = 2\) the graph is a union of cycles: either two triangles \(\left(\binom{5}{2} = 10\right)\) or one hexagon \(\left(\dfrac{6!}{12} = 60\right),\) totaling \(70.\) Thus \(n = 3\) also gives \(70.\)

The total is \(15 + 15 + 70 + 70 = 170.\)

Thus, the correct answer is B.

20 答案 B

展开乘积时，次数为 \(2012\) 的项来自某些因子中选取 \(x^{2^k}\)，使指数和为 \(2012\)。由于二的幂互不相同，这对应于二进制表示 \(2012 = 11111011100_2\)。

这个表示唯一，所以恰好只有一项给出 \(x^{2012}\)，它的系数是其余因子的常数项 \(2^k\) 的乘积：这些因子满足 \(k \in \{0, 1, 5\}\)。

系数为 \(2^0 \cdot 2^1 \cdot 2^5 = 2^6\)，所以 \(a = 6\)。

因此，正确答案是 B。

English

Expanding the product, a term of degree \(2012\) comes from choosing \(x^{2^k}\) from some factors so that the exponents sum to \(2012.\) Since powers of two are distinct, this corresponds to the binary representation \(2012 = 11111011100_2.\)

That representation is unique, so exactly one term gives \(x^{2012},\) and its coefficient is the product of the constants \(2^k\) from the remaining factors: those with \(k \in \{0, 1, 5\}.\)

The coefficient is \(2^0 \cdot 2^1 \cdot 2^5 = 2^6,\) so \(a = 6.\)

Thus, the correct answer is B.

21 答案 E

将两个方程相加，得到 \(2a^2 + 2b^2 + 2c^2\) \(- 2ab - 2bc - 2ca = 14\)，即 \[ \begin{aligned} &(a-b)^2 + (b-c)^2 \\ &\quad {}+ (c-a)^2 = 14 \end{aligned}\text{。}\]

将 \(14\) 写成三个平方数之和的唯一方式是 \(9 + 4 + 1\)。由于 \(a \ge b \ge c\)，有 \(a - c = 3\)，且 \((a-b, b-c) = (2,1)\) 或 \((1,2)\)。

将 \((a, b, c) = (c+3, c+1, c)\) 代入第一个方程，得到 \(3(2c+3) + 2(c+1) = 2011\)，所以 \(c = 250\)，\((a, b, c) = (253, 251, 250)\)。另一种情况没有整数解。

因此，正确答案是 E。

English

Adding the two equations gives \(2a^2 + 2b^2 + 2c^2\) \(- 2ab - 2bc - 2ca = 14,\) that is, \[ \begin{aligned} &(a-b)^2 + (b-c)^2 \\ &\quad {}+ (c-a)^2 = 14. \end{aligned} \]

The only way to write \(14\) as a sum of three squares is \(9 + 4 + 1.\) Since \(a \ge b \ge c,\) we get \(a - c = 3,\) with either \((a-b, b-c) = (2,1)\) or \((1,2).\)

Substituting \((a, b, c) = (c+3, c+1, c)\) into the first equation gives \(3(2c+3) + 2(c+1) = 2011,\) so \(c = 250\) and \((a, b, c) = (253, 251, 250).\) The other case has no integer solution.

Thus, the correct answer is E.

22 答案 C

在每个面上，所需线段都连接边的中点。一个平面截立方体时，会在各面上形成四种对称图形之一：经过中点的正方形（这样的平面有 \(3\) 个）、每条边对应一个长方形（\(12\) 个平面）、每个顶点对应一个三角形（\(8\) 个平面），或每对相对顶点对应一个正六边形（\(4\) 个平面）。

全部使用时得到最大值 \(k = 3 + 12 + 8 + 4 = 27\)。

完整图形由 \(24\) 条短线段和 \(12\) 条长线段组成。正方形平面所含短、长线段数为 \((0,4)\)，长方形为 \((2,2)\)，三角形为 \((3,0)\)，六边形为 \((6,0)\)。给每条短线段权重 \(1\)，每条长线段权重 \(\frac{3}{2}\)。每个平面覆盖的权重至多为 \(6\)，而全部所需线段的总权重为 \(24+(\frac{3}{2})12=42\)。因此至少需要 \(7\) 个平面。\(4\) 个六边形平面覆盖全部 \(24\) 条短线段，\(3\) 个正方形平面覆盖全部 \(12\) 条长线段，所以 \(k=7\) 可以达到。

两者之差为 \(27 - 7 = 20\)。

因此，正确答案是 C。

English

On every face, the required segments join midpoints of edges. A plane cutting the cube meets the faces in one of four symmetric shapes: a square through midpoints (\(3\) such planes), a rectangle per edge (\(12\) planes), a triangle per vertex (\(8\) planes), or a regular hexagon per pair of opposite vertices (\(4\) planes).

Using all of them gives the maximum \(k = 3 + 12 + 8 + 4 = 27.\)

The full figure consists of \(24\) short segments and \(12\) long segments. A square plane contains \((0,4),\) a rectangle \((2,2),\) a triangle \((3,0),\) and a hexagon \((6,0)\) short and long segments, respectively. Give each short segment weight \(1\) and each long segment weight \(\frac{3}{2}.\) Every plane then covers weight at most \(6,\) whereas the required segments have total weight \(24+(\frac{3}{2})12=42.\) Thus at least \(7\) planes are needed. The \(4\) hexagon planes cover all \(24\) short segments, and the \(3\) square planes cover all \(12\) long segments, so \(k=7\) is attainable.

The difference is \(27 - 7 = 20.\)

Thus, the correct answer is C.

23 答案 C

从 \((0.1, 0.7)\) 到 \((-0.1, -0.7)\) 的对角线长度为 \(\sqrt{0.2^2 + 1.4^2} = \sqrt2\)，所以 \(S\) 是面积为 \(1\) 的正方形。平移图形 \(T(v)\) 包含一个格点，当且仅当 \(v\) 位于以该格点为中心的 \(S\) 副本内部。

内部恰好包含两个格点，要求 \(v\) 位于以两个相邻格点为中心的副本的重叠区域。由周期性，答案等于一个单位格内所有这类重叠区域的总面积。

考虑以 \((0,0)\) 和 \((1,0)\) 为中心的两个副本。它们的重叠区域是一个长方形。将所给半对角线旋转四分之一圈，可得到相关顶点 \((0.7,-0.1)\) 和 \((0.3,0.1)\)。沿两组平行边找到交点，可得边长为 \(0.4\) 和 \(0.2\)，所以重叠面积是 \(0.08\)。对角位置的副本不重叠，而水平和竖直相邻的情形贡献相同。计入单位格边界后，概率为 \(2\cdot0.08=0.16\)。

因此，正确答案是 C。

English

The diagonal from \((0.1, 0.7)\) to \((-0.1, -0.7)\) has length \(\sqrt{0.2^2 + 1.4^2} = \sqrt2,\) so \(S\) is a square of area \(1.\) The translate \(T(v)\) contains a lattice point exactly when \(v\) lies inside the copy of \(S\) centered at that point.

Containing exactly two interior lattice points requires \(v\) to lie in the overlap of two copies centered at adjacent lattice points. By periodicity the answer is the total such overlap area within one unit cell.

Consider copies centered at \((0,0)\) and \((1,0).\) Their overlap is a rectangle. A quarter-turn of the listed half-diagonal gives the relevant vertices \((0.7,-0.1)\) and \((0.3,0.1).\) Following the two pairs of parallel sides through their intersections gives side lengths \(0.4\) and \(0.2,\) so the overlap area is \(0.08.\) Diagonally centered copies do not overlap, and horizontal and vertical adjacencies contribute equally. Accounting for the cell boundaries gives probability \(2\cdot0.08=0.16.\)

Thus, the correct answer is C.

24 答案 C

因为每个底数都严格介于 \(0\) 和 \(1\) 之间，函数 \(t \mapsto (\text{底数})^t\) 是递减的，而对 \(b \gt 0\)，\(t \mapsto t^b\) 是递增的。比较各项可得序列的大小顺序为 \[ \begin{aligned} &1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010} \\ &\gt a_{2011} \gt a_{2009} \\ &\gt \cdots \gt a_1 \gt 0 \end{aligned}\text{。}\]

因此在递减排列中，偶数下标项先出现，然后是奇数下标项按反向出现。某项满足 \(a_k = b_k\) 当且仅当它的位置等于它的下标；对递减排列的奇数尾部，这要求 \(2(k - 1006) = 2011 - k\)。

解得 \(3k = 4023\)，所以 \(k = 1341\)，这是唯一固定下标，因此所求和为 \(1341\)。

因此，正确答案是 C。

English

Because each base lies strictly between \(0\) and \(1,\) the function \(t \mapsto (\text{base})^t\) is decreasing, while \(t \mapsto t^b\) is increasing for \(b \gt 0.\) Comparing terms shows the sequence orders as \[ \begin{aligned} &1 \gt a_2 \gt a_4 \gt \cdots \gt a_{2010} \\ &\gt a_{2011} \gt a_{2009} \\ &\gt \cdots \gt a_1 \gt 0. \end{aligned} \]

So in the decreasing arrangement, the even-indexed terms come first, then the odd-indexed terms in reverse. A term satisfies \(a_k = b_k\) exactly when its position equals its index, which for the descending odd tail requires \(2(k - 1006) = 2011 - k.\)

Solving gives \(3k = 4023,\) so \(k = 1341,\) the unique fixed index, and the sum is \(1341.\)

Thus, the correct answer is C.

25 答案 C

因为 \(0 \le f(x) \le 1\)，每个解都位于 \([0,n]\) 内。函数 \(f\) 是周期为 \(1\) 的三角波。令 \(g(x)=xf(x)\)。对每个整数 \(a\ge1\)，函数 \(g\) 在 \([a,a+\tfrac12)\) 上从 \(a\) 递减到 \(0\)，而在 \([a+\tfrac12,a+1)\) 上从 \(0\) 递增到 \(a+1\)。第一个区间 \([0,\tfrac12)\) 是例外，但它不产生与 \(y=\frac{x}{n}\) 的交点。

计数振荡次数，在区间 \([a, a + \tfrac12)\) 和 \([a + \tfrac12, a+1)\) 上，曲线 \(y = f(g(x))\) 与直线 \(y = \tfrac{x}{n}\) 的交点总数分别为 \(2a\) 和 \(2(a+1)\)。对 \(a = 0, \ldots, n-1\) 求和，得到 \[\sum_{a=0}^{n-1}\bigl(2a + 2(a+1)\bigr) = 2n^2\] 个实数解。

满足 \(2n^2 \ge 2012\) 的最小 \(n\) 是 \(n = 32\)，因为 \(2 \cdot 31^2 = 1922\)，而 \(2 \cdot 32^2 = 2048\)。

因此，正确答案是 C。

English

Since \(0 \le f(x) \le 1,\) every solution lies in \([0,n].\) The function \(f\) is a triangular wave of period \(1.\) Put \(g(x)=xf(x).\) For each integer \(a\ge1,\) the function \(g\) decreases from \(a\) to \(0\) on \([a,a+\tfrac12),\) while it increases from \(0\) to \(a+1\) on \([a+\tfrac12,a+1).\) The first interval \([0,\tfrac12)\) is exceptional, but it contributes no intersection with \(y=\frac{x}{n}.\)

Counting the oscillations, on the intervals \([a, a + \tfrac12)\) and \([a + \tfrac12, a+1)\) the curve \(y = f(g(x))\) meets the line \(y = \tfrac{x}{n}\) a total of \(2a\) and \(2(a+1)\) times. Summing over \(a = 0, \ldots, n-1\) gives \[\sum_{a=0}^{n-1}\bigl(2a + 2(a+1)\bigr) = 2n^2\] real solutions.

The smallest \(n\) with \(2n^2 \ge 2012\) is \(n = 32,\) since \(2 \cdot 31^2 = 1922\) and \(2 \cdot 32^2 = 2048.\)

Thus, the correct answer is C.