2011A AMC 12 真题（P版）

1 答案 D

短信费用为 \(100 \cdot 5 = 500\) 美分 \(= \$5\)。她比 \(30\) 小时套餐多通话了 \(30\) 分钟，因此超时费用为 \(30 \cdot 10 = 300\) 美分 \(= \$3\)。

总费用为 \(\$20 + \$5 + \$3 = \$28\)。

因此，正确答案是 D。

English

The text charge is \(100 \cdot 5 = 500\) cents \(= \$5.\) She talked \(30\) minutes past the \(30\)-hour allowance, so the overage is \(30 \cdot 10 = 300\) cents \(= \$3.\)

The total is \(\$20 + \$5 + \$3 = \$28.\)

Thus, the correct answer is D.

2 答案 E

硬币 \(C\) 画成一个完整且没有中断的圆，所以没有东西盖住它，它在最上面。

继续观察其余重叠处，未被遮住的弧线显示每枚硬币在下一枚硬币上方：\(C\) 盖住 \(E\)，\(E\) 盖住 \(D\)，\(D\) 盖住 \(B\) 而 \(A\) 盖住 \(B\)，同时在其他硬币下方。因此从上到下的顺序是 \((C, E, D, A, B)\)。

因此，正确答案是 E。

English

Coin \(C\) is drawn as a complete, unbroken circle, so nothing covers it and it lies on top.

Reading the remaining overlaps, each coin’s uncovered arc shows it sits above the next: \(C\) covers \(E,\) \(E\) covers \(D,\) \(D\) covers \(B,\) and \(A\) covers \(B\) while lying under the others. This gives the top-to-bottom order \((C, E, D, A, B).\)

Thus, the correct answer is E.

3 答案 E

十四瓶可装 \(14 \cdot 35 = 490\) 毫升，这还不够。十五瓶可装 \(15 \cdot 35 = 525\) 毫升，已经足够。

所以 Jasmine 需要 \(15\) 瓶。

因此，正确答案是 E。

English

Fourteen bottles hold \(14 \cdot 35 = 490\) milliliters, which is not enough. Fifteen bottles hold \(15 \cdot 35 = 525\) milliliters, which suffices.

So Jasmine needs \(15\) bottles.

Thus, the correct answer is E.

4 答案 C

三、四、五年级的人数比为 \(4 : 2 : 1\)。加权平均数为 \[ \begin{aligned} \dfrac{4\cdot12+2\cdot15+10}{7} \\ = \dfrac{48+30+10}{7} \\ = \dfrac{88}{7} \end{aligned}\text{。}\]

因此，正确答案是 C。

English

Take the grade sizes in the ratio \(4 : 2 : 1\) for third, fourth, and fifth grades. The weighted average is \[ \begin{aligned} \dfrac{4\cdot12+2\cdot15+10}{7} \\ = \dfrac{48+30+10}{7} \\ = \dfrac{88}{7}. \end{aligned} \]

Thus, the correct answer is C.

5 答案 C

不是天鹅的鸟占总数的 \(100\% - 25\% = 75\%\)，而鹅占总数的 \(30\%\)。所求比例为 \[ \dfrac{30}{75} = \dfrac{2}{5} = 40\%\text{。}\]

因此，正确答案是 C。

English

The birds that are not swans make up \(100\% - 25\% = 75\%\) of the total, and geese are \(30\%\) of the total. The requested fraction is \[ \dfrac{30}{75} = \dfrac{2}{5} = 40\%. \]

Thus, the correct answer is C.

6 答案 A

设投中的两分球个数为 \(a\)。两分球得到 \(2a\) 分，三分球也得到同样的 \(2a\) 分。罚球个数为 \(a + 1\)，得到 \(a + 1\) 分。

总分为 \[ \begin{aligned} 2a + 2a + (a + 1) \\ = 5a + 1 = 61 \end{aligned}\text{，}\] 所以 \(a = 12\)，罚球个数为 \(a + 1 = 13\)。

因此，正确答案是 A。

English

Let \(a\) be the number of two-point shots. The two-point shots score \(2a\) points, and the three-point shots score the same \(2a\) points. The free throws number \(a + 1\) and score \(a + 1\) points.

The total is \[ \begin{aligned} 2a + 2a + (a + 1) \\ = 5a + 1 = 61, \end{aligned} \] so \(a = 12\) and the free throws number \(a + 1 = 13.\)

Thus, the correct answer is A.

7 答案 B

总费用为 \(1771\) 美分，且 \(1771 = 7 \cdot 11 \cdot 23\)。写成（学生数）（每人铅笔数）（每支铅笔价格）\(= 1771\)，学生数必须是该数的一个因数，并且是 \(30\) 人中的多数，即超过 \(15\)。唯一这样的因数是 \(23\)。

因此（铅笔数）（单价）\(= 77 = 7 \cdot 11\)，且单价 \(\gt\) 铅笔数 \(\gt 1\)，所以每人买 \(7\) 支铅笔，每支 \(11\) 美分。

因此，正确答案是 B。

English

Total cents is \(1771 = 7 \cdot 11 \cdot 23.\) Writing (students)(pencils each)(cost per pencil) \(= 1771,\) the number of students is a divisor of \(1771\) that is a majority of \(30,\) hence more than \(15.\) The only such divisor is \(23.\)

Then (pencils)(cost) \(= 77 = 7 \cdot 11\) with cost \(\gt\) pencils \(\gt 1,\) forcing \(7\) pencils at \(11\) cents each.

Thus, the correct answer is B.

8 答案 C

因为 \(A + B + C = B + C + D = 30\)，所以 \(D = A\)；同理可知，这个数列以 \(3\) 为周期重复。因此第八项 \(H\) 等于 \(B\)。

由 \(A + B + C = 30\) 且 \(C = 5\)，得 \(A + H = A + B = 30 - 5 = 25\)。

因此，正确答案是 C。

English

Since \(A + B + C = B + C + D = 30,\) we get \(D = A,\) and likewise the sequence repeats with period \(3.\) Thus \(H,\) the eighth term, equals \(B.\)

From \(A + B + C = 30\) and \(C = 5,\) we have \(A + H = A + B = 30 - 5 = 25.\)

Thus, the correct answer is C.

9 答案 B

共有 \(18\) 名双胞胎成员和 \(18\) 名三胞胎成员。

双胞胎之间的握手：每个双胞胎成员与 \(18 - 2 = 16\) 名其他双胞胎成员握手，得到 \(\dfrac{18 \cdot 16}{2} = 144\)。

三胞胎之间的握手：每个三胞胎成员与 \(18 - 3 = 15\) 名其他三胞胎成员握手，得到 \(\dfrac{18 \cdot 15}{2} = 135\)。

双胞胎和三胞胎之间的握手：每个双胞胎成员与 \(18\) 名三胞胎成员中的一半握手，得到 \(18 \cdot 9 = 162\)（每次这种握手只计一次）。

总数为 \(144 + 135 + 162 = 441\)。

因此，正确答案是 B。

English

There are \(18\) twins and \(18\) triplets.

Twin-twin handshakes: each twin shakes \(18 - 2 = 16\) other twins, giving \(\dfrac{18 \cdot 16}{2} = 144.\)

Triplet-triplet handshakes: each triplet shakes \(18 - 3 = 15\) other triplets, giving \(\dfrac{18 \cdot 15}{2} = 135.\)

Twin-triplet handshakes: each twin shakes half the \(18\) triplets, giving \(18 \cdot 9 = 162\) (each such handshake counted once).

The total is \(144 + 135 + 162 = 441.\)

Thus, the correct answer is B.

10 答案 B

若直径为 \(d\)，面积 \(\lt\) 周长意味着 \(\dfrac{\pi d^2}{4} \lt \pi d\)，即 \(d \lt 4\)。因为 \(d \ge 2\)，所以点数和必须是 \(2\) 或 \(3\)。

点数和为 \(2\) 的概率是 \(\dfrac{1}{36}\)，点数和为 \(3\) 的概率是 \(\dfrac{2}{36}\)，合计 \(\dfrac{3}{36} = \dfrac{1}{12}\)。

因此，正确答案是 B。

English

For diameter \(d,\) area \(\lt\) circumference means \(\dfrac{\pi d^2}{4} \lt \pi d,\) i.e. \(d \lt 4.\) Since \(d \ge 2,\) this needs a sum of \(2\) or \(3.\)

A sum of \(2\) has probability \(\dfrac{1}{36}\) and a sum of \(3\) has probability \(\dfrac{2}{36},\) totaling \(\dfrac{3}{36} = \dfrac{1}{12}.\)

Thus, the correct answer is B.

11 答案 C

设 \(A = (-1, 0)\)，\(B = (1, 0)\)，则它们的切点是原点，也就是 \(\overline{AB}\) 的中点。因此 \(C = (0, 1)\)，因为圆 \(C\) 经过原点。

从 \(C\) 到 \(A\)（以及到 \(B\)）的距离为 \(\sqrt2\)。两个圆心距离为 \(\sqrt2\) 的单位圆重叠成一个透镜形，其面积为 \[ \begin{aligned} 2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) \\ {}- \tfrac{\sqrt2}{2}\sqrt{4 - 2} \\ = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1 \end{aligned}\text{。}\]

圆 \(A\) 和圆 \(B\) 只在原点相交，所以两个透镜形区域不重叠。所求面积为 \[ \pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2\text{。}\]

因此，正确答案是 C。

English

Place \(A = (-1, 0),\) \(B = (1, 0),\) so their tangency point is the origin, the midpoint of \(\overline{AB}.\) Then \(C = (0, 1),\) since \(C\) passes through the origin.

The distance from \(C\) to \(A\) (and to \(B\)) is \(\sqrt2.\) Two unit circles whose centers are \(\sqrt2\) apart overlap in a lens of area \[ \begin{aligned} 2\cos^{-1}\!\left(\tfrac{\sqrt2}{2}\right) \\ {}- \tfrac{\sqrt2}{2}\sqrt{4 - 2} \\ = 2 \cdot \tfrac{\pi}{4} - 1 = \tfrac{\pi}{2} - 1. \end{aligned} \]

Circles \(A\) and \(B\) meet only at the origin, so the two lenses do not overlap. The wanted area is \[ \pi - 2\left(\tfrac{\pi}{2} - 1\right) = 2. \]

Thus, the correct answer is C.

12 答案 D

全部相对于河水来测量。在这个参考系中，木筏静止在船出发的位置，而船相对于河水以恒定速度 \(v\) 运动，下游和上游方向都是如此。

船离开木筏，向外行驶一段时间，再以相同的相对速度返回木筏，所以去程和回程用时相等。因此到 \(B\) 的下行路程花了 \(9\) 的一半，即 \(4.5\) 小时。

因此，正确答案是 D。

English

Measure everything relative to the water. In that frame the raft is stationary at the point where the boat started, and the boat moves at its constant speed \(v\) relative to the water, both downstream and upstream.

The boat leaves the raft, travels away for some time, then returns to it at the same relative speed, so it spends equal times going and returning. Hence the outbound leg to \(B\) takes half of \(9,\) which is \(4.5\) hours.

Thus, the correct answer is D.

13 答案 B

设 \(I\) 为内心。因为 \(\overline{BI}\) 平分 \(\angle B\)，且 \(MN \parallel BC\)，由内错角可得 \(\angle MIB = \angle IBC = \angle MBI\)，所以 \(\triangle MBI\) 是等腰三角形，且 \(MB = MI\)。同理 \(NC = NI\)。

因此 \(\triangle AMN\) 的周长为 \[ \begin{gathered} AM + MN + NA \\ = AM + (MI + IN) + NA \\ = AM + MB + NC + NA \\ = AB + AC = 12 + 18 = 30 \end{gathered}\text{。}\]

因此，正确答案是 B。

English

Let \(I\) be the incenter. Because \(\overline{BI}\) bisects \(\angle B\) and \(MN \parallel BC,\) alternate angles give \(\angle MIB = \angle IBC = \angle MBI,\) so \(\triangle MBI\) is isosceles with \(MB = MI.\) Similarly \(NC = NI.\)

Therefore the perimeter of \(\triangle AMN\) is \[ \begin{gathered} AM + MN + NA \\ = AM + (MI + IN) + NA \\ = AM + MB + NC + NA \\ = AB + AC = 12 + 18 = 30. \end{gathered} \]

Thus, the correct answer is B.

14 答案 E

代入 \(x = a\)、\(y = b\)，点在抛物线上方当且仅当 \(b \gt a^3 - ab\)，即 \(b(a + 1) \gt a^3\)。

当 \(a = 1\)：\(b \gt \tfrac12\)，所有 \(9\) 个值都可行。当 \(a = 2\)：\(b \gt \tfrac83\)，所以 \(b \ge 3\) 有 \(7\) 个。当 \(a = 3\)：\(b \gt \tfrac{27}{4} = 6.75\)，所以 \(b \ge 7\) 有 \(3\) 个。当 \(a \ge 4\) 时，没有 \(b \le 9\) 可行。

总数为 \(9 + 7 + 3 = 19\)，共 \(81\) 种情况，所以概率为 \(\dfrac{19}{81}\)。

因此，正确答案是 E。

English

Substituting \(x = a,\) \(y = b,\) the point is above the parabola when \(b \gt a^3 - ab,\) i.e. \(b(a + 1) \gt a^3.\)

For \(a = 1:\) \(b \gt \tfrac12,\) all \(9\) values work. For \(a = 2:\) \(b \gt \tfrac83,\) so \(b \ge 3,\) giving \(7.\) For \(a = 3:\) \(b \gt \tfrac{27}{4} = 6.75,\) so \(b \ge 7,\) giving \(3.\) For \(a \ge 4,\) no \(b \le 9\) works.

The count is \(9 + 7 + 3 = 19\) out of \(81,\) so the probability is \(\dfrac{19}{81}.\)

Thus, the correct answer is E.

15 答案 A

设底面边长为 \(s\)，底面中心在原点，顶点高度为 \(6\)。用通过顶点和两条相对底边中点的竖直平面截取。侧面在截面中表现为从 \(\left(\tfrac{s}{2}, 0\right)\) 到 \((0, 6)\) 的直线。

这条直线是 \(\tfrac{2}{s}x + \tfrac16 y = 1\)。半球与该面相切，所以原点到这条直线的距离等于半径 \(2\)：\[ \dfrac{1}{\sqrt{\tfrac{4}{s^2} + \tfrac{1}{36}}} = 2\text{。}\]

因此 \(\tfrac{4}{s^2} + \tfrac{1}{36} = \tfrac14\)，所以 \(\tfrac{4}{s^2} = \tfrac{2}{9}\)，\(s^2 = 18\)，得 \(s = 3\sqrt2\)。

因此，正确答案是 A。

English

Let the base have side \(s,\) centered at the origin, with apex at height \(6.\) Cut with the vertical plane through the apex and the midpoints of two opposite base edges. The slant face appears as the line from \(\left(\tfrac{s}{2}, 0\right)\) to \((0, 6).\)

This line is \(\tfrac{2}{s}x + \tfrac16 y = 1.\) The hemisphere is tangent to the face, so the distance from the origin to this line is the radius \(2:\) \[ \dfrac{1}{\sqrt{\tfrac{4}{s^2} + \tfrac{1}{36}}} = 2. \]

Then \(\tfrac{4}{s^2} + \tfrac{1}{36} = \tfrac14,\) so \(\tfrac{4}{s^2} = \tfrac{2}{9}\) and \(s^2 = 18,\) giving \(s = 3\sqrt2.\)

Thus, the correct answer is A.

16 答案 C

对角线按顺序连接顶点 \(A - C - E - B - D - A\)，形成一个 \(5\)-环。题目条件正是要求这个环被正常涂色。

长度为 \(n\) 的环用 \(k\) 种颜色作正常涂色的数量为 \((k-1)^n + (-1)^n (k-1)\)。取 \(n = 5\) 且 \(k = 6\)，\[ \begin{gathered} 5^5 + (-1)^5 \cdot 5 \\ = 3125 - 5 = 3120 \end{gathered}\text{。}\]

因此，正确答案是 C。

English

The diagonals connect the vertices in the order \(A - C - E - B - D - A,\) which is a \(5\)-cycle. The condition is exactly that this cycle is properly colored.

The number of proper \(k\)-colorings of a cycle of length \(n\) is \((k-1)^n + (-1)^n (k-1).\) With \(n = 5\) and \(k = 6,\) \[ \begin{gathered} 5^5 + (-1)^5 \cdot 5 \\ = 3125 - 5 = 3120. \end{gathered} \]

Thus, the correct answer is C.

17 答案 D

圆心之间的距离等于半径和：\(3\)，\(4\)，和 \(5\)，形成一个直角三角形，直角在半径为 \(1\) 的圆心处。把该圆心放在 \((0,0)\)，半径为 \(2\) 的圆心放在 \((3,0)\)，半径为 \(3\) 的圆心放在 \((0,4)\)。

切点在线段上，距离等于相应半径：\((1, 0)\)，\((0, 1)\)，以及斜边上的 \((3,0) + 2 \cdot \tfrac{(-3,4)}{5} = \left(\tfrac95, \tfrac85\right)\)。

由鞋带公式，面积为 \[ \begin{gathered} \tfrac12\left| 1\left(1 - \tfrac85\right) + 0 + \tfrac95(0 - 1) \right| \\ = \tfrac12 \cdot \tfrac{12}{5} = \tfrac65 \end{gathered}\text{。}\]

因此，正确答案是 D。

English

The centers are separated by the sums of radii: \(3,\) \(4,\) and \(5,\) a right triangle with the right angle at the radius-\(1\) center. Place that center at \((0,0),\) the radius-\(2\) center at \((3,0),\) and the radius-\(3\) center at \((0,4).\)

The tangency points lie on the segments at distances equal to the radii: \((1, 0),\) \((0, 1),\) and on the hypotenuse at \((3,0) + 2 \cdot \tfrac{(-3,4)}{5} = \left(\tfrac95, \tfrac85\right).\)

By the shoelace formula the area is \[ \begin{gathered} \tfrac12\left| 1\left(1 - \tfrac85\right) + 0 + \tfrac95(0 - 1) \right| \\ = \tfrac12 \cdot \tfrac{12}{5} = \tfrac65. \end{gathered} \]

Thus, the correct answer is D.

18 答案 D

恒等式 \(|x+y| + |x-y|\) \(= 2\max(|x|, |y|)\) 将条件变成 \(\max(|x|, |y|) = 1\)，即满足 \(|x| \le 1\) 且 \(|y| \le 1\) 的正方形边界。

在这个区域上，\(x^2 - 6x + y^2\) 会随着 \(x\) 变小以及 \(y^2\) 变大而增大，所以最大值在 \(x = -1\)、\(y = \pm 1\) 处取得：\[ 1 + 6 + 1 = 8\text{。}\]

因此，正确答案是 D。

English

The identity \(|x+y| + |x-y|\) \(= 2\max(|x|, |y|)\) turns the condition into \(\max(|x|, |y|) = 1,\) the boundary of the square with \(|x| \le 1\) and \(|y| \le 1.\)

On this region \(x^2 - 6x + y^2\) increases as \(x\) decreases and as \(y^2\) increases, so the maximum is at \(x = -1,\) \(y = \pm 1:\) \[ 1 + 6 + 1 = 8. \]

Thus, the correct answer is D.

19 答案 C

令 \(m = \lfloor \log_2 (N - 1) \rfloor\)，则精英人数为 \(2^{m+1} - N = 19\)，得 \(N = 2^{m+1} - 19\)。

一致性要求 \(2^m \le N - 1 = 2^{m+1} - 20\)，即 \(2^m \ge 20\)，所以 \(m \ge 5\)。

两个最小选择是 \(m = 5\)，给出 \(N = 64 - 19 = 45\)，以及 \(m = 6\)，给出 \(N = 128 - 19 = 109\)。它们的和为 \(154\)。

因此，正确答案是 C。

English

Let \(m = \lfloor \log_2 (N - 1) \rfloor,\) so the elite count is \(2^{m+1} - N = 19,\) giving \(N = 2^{m+1} - 19.\)

Consistency requires \(2^m \le N - 1 = 2^{m+1} - 20,\) i.e. \(2^m \ge 20,\) so \(m \ge 5.\)

The two smallest choices are \(m = 5\) giving \(N = 64 - 19 = 45,\) and \(m = 6\) giving \(N = 128 - 19 = 109.\) Their sum is \(154.\)

Thus, the correct answer is C.

20 答案 C

因为 \(f(1) = a + b + c = 0\)，所以 \(c = -a - b\)。因此 \[ f(7) = 48a + 6b = 6(8a + b)\text{，}\]\[ f(8) = 63a + 7b = 7(9a + b)\text{。}\]

由 \(50 \lt 6(8a + b) \lt 60\) 得 \(8a + b = 9\)，由 \(70 \lt 7(9a + b) \lt 80\) 得 \(9a + b = 11\)。相减得 \(a = 2\)，进而 \(b = -7\)，\(c = 5\)。

所以 \(f(100) = 20000 - 700 + 5\) \(= 19305\)，它满足 \(5000 \cdot 3 \lt 19305 \lt 5000 \cdot 4\)，因此 \(k = 3\)。

因此，正确答案是 C。

English

Since \(f(1) = a + b + c = 0,\) we have \(c = -a - b.\) Then \[ f(7) = 48a + 6b = 6(8a + b), \] \[ f(8) = 63a + 7b = 7(9a + b). \]

From \(50 \lt 6(8a + b) \lt 60\) we get \(8a + b = 9,\) and from \(70 \lt 7(9a + b) \lt 80\) we get \(9a + b = 11.\) Subtracting, \(a = 2,\) then \(b = -7\) and \(c = 5.\)

So \(f(100) = 20000 - 700 + 5\) \(= 19305,\) which lies in \(5000 \cdot 3 \lt 19305 \lt 5000 \cdot 4,\) giving \(k = 3.\)

Thus, the correct answer is C.

21 答案 A

每一步都要求 \(\sqrt{n^2 - x}\) 落在 \(f_{n-1}\) 的定义域中。追踪定义域：

\(f_1: (-\infty, 1]\)。\(f_2: \sqrt{4 - x} \in (-\infty, 1] \Rightarrow [3, 4]\)。\(f_3: \sqrt{9 - x} \in [3, 4] \Rightarrow [-7, 0]\)。\(f_4: \sqrt{16 - x} \in [-7, 0] \Rightarrow \{16\}\)（只有值 \(0\) 可能）。\(f_5: \sqrt{25 - x} = 16 \Rightarrow \{-231\}\)。

对 \(f_6\)，我们需要 \(\sqrt{36 - x} = -231\)，这是不可能的，所以定义域为空。因此 \(N = 5\)，\(c = -231\)，且 \(N + c = -226\)。

因此，正确答案是 A。

English

Each step requires \(\sqrt{n^2 - x}\) to lie in the domain of \(f_{n-1}.\) Tracking the domains:

\(f_1: (-\infty, 1].\) \(f_2: \sqrt{4 - x} \in (-\infty, 1] \Rightarrow [3, 4].\) \(f_3: \sqrt{9 - x} \in [3, 4] \Rightarrow [-7, 0].\) \(f_4: \sqrt{16 - x} \in [-7, 0] \Rightarrow \{16\}\) (only the value \(0\) is possible). \(f_5: \sqrt{25 - x} = 16 \Rightarrow \{-231\}.\)

For \(f_6\) we would need \(\sqrt{36 - x} = -231,\) impossible, so the domain is empty. Hence \(N = 5,\) \(c = -231,\) and \(N + c = -226.\)

Thus, the correct answer is A.

22 答案 C

将正方形缩放为 \([0,1]^2\)，并令 \(X=(x,y)\)。射线必须包括通向四个顶点的射线。每个小三角形的面积都是 \(\frac{1}{n}\)。以底边各段为底的三角形面积之和为 \(\frac{y}{2}\)，所以这样的三角形有 \(\frac{ny}{2}\) 个。同理，沿上、左、右三边的个数分别是 \(\frac{n(1-y)}{2}\)，\(\frac{nx}{2}\)，和 \(\frac{n(1-x)}{2}\)。

这四个数都必须是正整数。因此 \(n\) 是偶数，并且 \[ \begin{gathered} X=\left(\dfrac{2i}{n},\dfrac{2j}{n}\right), \\ 1\le i,j\le\dfrac n2-1 \end{gathered}\text{。}\]反过来，按上述数量把每条边等分，并将分点连接到 \(X\)，就会得到 \(n\) 个等面积三角形。因此这些恰好是分割点。

当 \(n=100\) 时，这些点为 \((\frac{i}{50},\frac{j}{50})\)，其中 \(1\le i,j\le49\)，共 \(49^2=2401\) 个。这样的点同时也是 \(60\) 射线分割点，当且仅当对某些整数 \(c,d\)，有 \(\frac{i}{50}=\frac{c}{30}\) 且 \(\frac{j}{50}=\frac{d}{30}\)。因此 \(i\) 和 \(j\) 都必须是 \(5\) 的倍数。每个坐标有 \(9\) 种选择，所以重合的点有 \(9^2=81\) 个。

所以所求数量为 \(2401 - 81 = 2320\)。

因此，正确答案是 C。

English

Scale the square to \([0,1]^2\) and write \(X=(x,y).\) The rays must include those through the four vertices. Every small triangle has area \(\frac{1}{n}.\) The triangles whose bases partition the bottom side together have area \(\frac{y}{2},\) so their number is \(\frac{ny}{2}.\) Similarly, the numbers along the top, left, and right sides are \(\frac{n(1-y)}{2},\) \(\frac{nx}{2},\) and \(\frac{n(1-x)}{2}.\)

These four numbers must be positive integers. Hence \(n\) is even and \[ \begin{gathered} X=\left(\dfrac{2i}{n},\dfrac{2j}{n}\right), \\ 1\le i,j\le\dfrac n2-1. \end{gathered} \] Conversely, partitioning each side into the indicated number of equal segments and joining the division points to \(X\) produces \(n\) equal-area triangles. Thus these are exactly the partitional points.

For \(n=100,\) the points are \((\frac{i}{50},\frac{j}{50})\) with \(1\le i,j\le49,\) giving \(49^2=2401.\) Such a point is also \(60\)-ray partitional exactly when \(\frac{i}{50}=\frac{c}{30}\) and \(\frac{j}{50}=\frac{d}{30}\) for integers \(c,d.\) Thus \(i\) and \(j\) must both be multiples of \(5.\) There are \(9\) choices for each, so the overlap has \(9^2=81\) points.

So the count is \(2401 - 81 = 2320.\)

Thus, the correct answer is C.

23 答案 C

直接复合可得 \[ g(z)=\dfrac{Az+B}{Cz+D}\text{，}\]其中 \(A=1+a\)，\(B=a(1+b)\)，\(C=1+b\)，且 \(D=a+b^2\)。

矩阵 \(\begin{pmatrix}A&B\\C&D\end{pmatrix}\) 表示变换 \(g\)。要使 \(g\circ g\) 为恒等变换，它的平方必须是标量矩阵。比较非对角元和两个对角元，得到两种可能：或者 \(B=C=0\) 且 \(A=D\)，或者 \(A+D=0\)。第一种给出 \(b=-1\)（其中 \(a\ne-1\)）；第二种给出 \[ b^2=-(1+2a)\text{。}\]

在第二种情形中，\(|b|^2=|1+2a|\)。当 \(a\) 绕单位圆变化时，\(|1+2a|\) 的取值从 \(1\) 到 \(3\)，所以 \(1\le|b|\le\sqrt3\)。两个端点都能取得：分别取 \((a,b)=(-1,1)\) 和 \((a,b)=(1,i\sqrt3)\)。单独的情形 \(b=-1\) 也满足 \(|b|=1\)。因此所求差为 \(\sqrt3-1\)。

因此，正确答案是 C。

English

Direct composition gives \[ g(z)=\dfrac{Az+B}{Cz+D}, \] where \(A=1+a,\) \(B=a(1+b),\) \(C=1+b,\) and \(D=a+b^2.\)

The matrix \(\begin{pmatrix}A&B\\C&D\end{pmatrix}\) represents \(g.\) For \(g\circ g\) to be the identity, its square must be scalar. Comparing the off-diagonal entries and the two diagonal entries gives two possibilities: either \(B=C=0\) and \(A=D,\) or \(A+D=0.\) The first gives \(b=-1\) (with \(a\ne-1\)); the second gives \[ b^2=-(1+2a). \]

In the second case, \(|b|^2=|1+2a|.\) As \(a\) runs around the unit circle, \(|1+2a|\) ranges from \(1\) to \(3,\) so \(1\le|b|\le\sqrt3.\) Both endpoints occur: take \((a,b)=(-1,1)\) and \((a,b)=(1,i\sqrt3).\) The separate case \(b=-1\) also has \(|b|=1.\) Therefore the requested difference is \(\sqrt3-1.\)

Thus, the correct answer is C.

24 答案 C

设以 \(X\) 为圆心、半径为 \(r\) 的圆能放入其中一个四边形。如果 \(h_1,h_2,h_3,h_4\) 是 \(X\) 到四条边所在直线的距离，那么每个 \(h_i\ge r\)。将四边形分成四个三角形，得到 \[ \begin{aligned} K&=\dfrac12(14h_1+9h_2 \\ &\qquad+7h_3+12h_4) \\ &\ge21r \end{aligned}\text{。}\]

布雷特施奈德不等式表明，给定这些边长的四边形面积不超过圆内接情形：\[ \begin{aligned} K^2&\le(21-14)(21-9) \\ &\qquad\cdot(21-7)(21-12) \\ &=7\cdot12\cdot14\cdot9, \\ K&\le42\sqrt6 \end{aligned}\text{。}\]

因此 \(r\le \frac{K}{21}\le2\sqrt6\)。等号可以达到：具有这些边长的圆内接四边形也有内切圆，因为 \(14+7=9+12\)，其内切圆半径为 \(\frac{K}{21}=2\sqrt6\)。

因此，正确答案是 C。

English

Suppose a circle of radius \(r\) centered at \(X\) fits in one of the quadrilaterals. If \(h_1,h_2,h_3,h_4\) are the distances from \(X\) to the four side lines, then each \(h_i\ge r.\) Splitting the quadrilateral into four triangles gives \[ \begin{aligned} K&=\dfrac12(14h_1+9h_2 \\ &\qquad+7h_3+12h_4) \\ &\ge21r. \end{aligned} \]

Bretschneider’s inequality bounds the area of any quadrilateral with these sides by the cyclic case: \[ \begin{aligned} K^2&\le(21-14)(21-9) \\ &\qquad\cdot(21-7)(21-12) \\ &=7\cdot12\cdot14\cdot9, \\ K&\le42\sqrt6. \end{aligned} \]

Therefore \(r\le \frac{K}{21}\le2\sqrt6.\) Equality is attainable: the cyclic quadrilateral with these sides is also tangential because \(14+7=9+12,\) and its incircle has radius \(\frac{K}{21}=2\sqrt6.\)

Thus, the correct answer is C.

25 答案 D

令 \(B=\angle CBA\)，并令 \(C=\angle BCA=120^\circ-B\)。因为 \(AC\ge AB\)，所以 \(B\ge C\)，从而 \(60^\circ\le B\le90^\circ\)。标准角度公式给出 \[ \begin{aligned} \angle BOC&=2\angle A=120^\circ, \\ \angle BHC&=180^\circ-\angle A=120^\circ, \\ \angle BIC&=90^\circ+\dfrac{\angle A}{2}=120^\circ \end{aligned}\text{。}\]因此 \(B,C,O,I,H\) 位于同一个圆上。

此外，\(BC=1\) 且 \(A=60^\circ\) 确定外接圆半径 \(OB=OC=\frac{1}{\sqrt3}\)，所以 \(\triangle BCO\) 以及经过 \(B,C,O\) 的圆都是固定的。在 \(C\) 处追角可得 \[ \begin{aligned} \angle OCI&=30^\circ-\dfrac C2, \\ \angle ICH&=30^\circ-\dfrac C2 \end{aligned}\text{。}\]因此相应的弦满足 \(OI=IH\)。

五边形面积等于固定面积 \([BCO]\) 加上 \([BOIH]\)。当两个点分割固定弧 \(OB\) 时，圆内接四边形在三段连续小弧相等时面积最大（等价地，使它们的正弦和最大）。因此在最大值处有 \(OI=IH=HB\)。

在 \(\triangle BOC\) 中，\(\angle OCB=30^\circ\)。等弦使 \(\angle OCI=\angle ICH\)，且 \(\angle ICH=\angle HCB\)，所以这些角都等于 \(10^\circ\)。因此 \(30^\circ-\tfrac C2=10^\circ\)，得 \(C=40^\circ\) 和 \(B=80^\circ\)。

因此，正确答案是 D。

English

Write \(B=\angle CBA\) and \(C=\angle BCA=120^\circ-B.\) Since \(AC\ge AB,\) we have \(B\ge C,\) so \(60^\circ\le B\le90^\circ.\) The standard angle formulas give \[ \begin{aligned} \angle BOC&=2\angle A=120^\circ, \\ \angle BHC&=180^\circ-\angle A=120^\circ, \\ \angle BIC&=90^\circ+\dfrac{\angle A}{2}=120^\circ. \end{aligned} \] Hence \(B,C,O,I,H\) lie on one circle.

Also \(BC=1\) and \(A=60^\circ\) fix the circumradius \(OB=OC=\frac{1}{\sqrt3},\) so \(\triangle BCO\) and the circle through \(B,C,O\) are fixed. Angle chasing at \(C\) gives \[ \begin{aligned} \angle OCI&=30^\circ-\dfrac C2, \\ \angle ICH&=30^\circ-\dfrac C2. \end{aligned} \] Thus the corresponding chords satisfy \(OI=IH.\)

The pentagon’s area is the fixed area \([BCO]\) plus \([BOIH].\) For two points dividing a fixed arc \(OB,\) an inscribed quadrilateral has greatest area when its three consecutive subarcs are equal (equivalently, maximize the sum of their sines). Hence at the maximum \(OI=IH=HB.\)

In \(\triangle BOC,\) \(\angle OCB=30^\circ.\) Equal chords make \(\angle OCI=\angle ICH,\) \(\angle ICH=\angle HCB,\) and each of these angles is \(10^\circ.\) Therefore \(30^\circ-\tfrac C2=10^\circ,\) so \(C=40^\circ\) and \(B=80^\circ.\)

Thus, the correct answer is D.