2008B AMC 12 真题（P版）

1 答案 E

总分从 \(5 \cdot 2 = 10\)（全是两分球）到 \(5 \cdot 3 = 15\)（全是三分球）。每把一个两分球换成三分球，总分正好增加 \(1\)，所以中间的每个整数都会出现。

可能的总分是 \(10, 11, 12, 13, 14, 15\)，共有 \(6\) 个值。

因此，正确答案是 E。

English

The total ranges from \(5 \cdot 2 = 10\) (all two-pointers) to \(5 \cdot 3 = 15\) (all three-pointers). Swapping one two-pointer for a three-pointer raises the total by exactly \(1,\) so every integer in between occurs.

The possible totals are \(10, 11, 12, 13, 14, 15,\) which is \(6\) values.

Thus, the correct answer is E.

2 答案 B

将第二行和第四行反向后，数组的各行是 \(1\,2\,3\,4\)、\(11\,10\,9\,8\)、\(15\,16\,17\,18\) 和 \(25\,24\,23\,22\)。

主对角线和为 \(1 + 10 + 17 + 22 = 50\)，另一条对角线和为 \(4 + 9 + 16 + 25 = 54\)。

正差为 \(54 - 50 = 4\)。

因此，正确答案是 B。

English

Reversing the second and fourth rows gives the array with rows \(1\,2\,3\,4,\) \(11\,10\,9\,8,\) \(15\,16\,17\,18,\) and \(25\,24\,23\,22.\)

The main diagonal sums to \(1 + 10 + 17 + 22 = 50,\) and the other diagonal sums to \(4 + 9 + 16 + 25 = 54.\)

The positive difference is \(54 - 50 = 4.\)

Thus, the correct answer is B.

3 答案 C

当其他 \(20\) 名球员每人都拿最低工资 \(\$15{,}000\) 时，一名球员的工资最大。

因此最大工资为 \(\$700{,}000\) \(- 20 \cdot \$15{,}000\) \(= \$700{,}000\) \(- \$300{,}000\) \(= \$400{,}000\)。

因此，正确答案是 C。

English

One player earns the most when the other \(20\) players each receive the minimum salary of \(\$15{,}000.\)

Thus the maximum salary is \(\$700{,}000\) \(- 20 \cdot \$15{,}000\) \(= \$700{,}000\) \(- \$300{,}000\) \(= \$400{,}000.\)

Thus, the correct answer is C.

4 答案 D

因为 \(\angle AOC\)、\(\angle COD\) 和 \(\angle DOB\) 填满直径 \(\overline{AB}\) 上方的平角，所以 \[ \begin{aligned} \angle COD &= 180^\circ - 30^\circ - 45^\circ \\ &= 105^\circ \end{aligned}\text{。}\]

这个扇形占圆的比例为 \(\dfrac{105^\circ}{360^\circ} = \dfrac{7}{24}\)。

因此，正确答案是 D。

English

Since \(\angle AOC,\) \(\angle COD,\) and \(\angle DOB\) fill the straight angle over diameter \(\overline{AB},\) \[ \begin{aligned} \angle COD &= 180^\circ - 30^\circ - 45^\circ \\ &= 105^\circ. \end{aligned} \]

The sector’s share of the circle is \(\dfrac{105^\circ}{360^\circ} = \dfrac{7}{24}.\)

Thus, the correct answer is D.

5 答案 C

设 \(r\) 为玫瑰数量，\(c\) 为康乃馨数量，则 \(3r + 2c = 50\)，其中 \(r, c \ge 0\)。

因为 \(2c\) 和 \(50\) 都是偶数，\(3r\) 必须是偶数，所以 \(r\) 是偶数。最大的 \(r\) 是 \(16\)（因为 \(3 \cdot 17 \gt 50\)），所以 \(r \in \{0, 2, 4, \ldots, 16\}\)。

这给出 \(9\) 个 \(r\) 的值，每个值都确定一种花束。

因此，正确答案是 C。

English

Let \(r\) be the number of roses and \(c\) the number of carnations, so \(3r + 2c = 50\) with \(r, c \ge 0.\)

Because \(2c\) and \(50\) are even, \(3r\) must be even, forcing \(r\) to be even. The largest possible \(r\) is \(16\) (since \(3 \cdot 17 \gt 50\)), so \(r \in \{0, 2, 4, \ldots, 16\}.\)

That gives \(9\) values of \(r,\) each determining a bouquet.

Thus, the correct answer is C.

6 答案 A

每次翻转表示 \(100000\) 步，所以这一年的总步数为 \[ \begin{aligned} &44 \cdot 100000 \\ &\quad {}+ 50000 = 4{,}450{,}000 \end{aligned}\text{。}\]

每英里 \(1800\) 步，所以里程为 \(\dfrac{4{,}450{,}000}{1800} \approx 2472\)，最接近 \(2500\)。

因此，正确答案是 A。

English

Each flip counts \(100000\) steps, so the year’s steps total \[ \begin{aligned} &44 \cdot 100000 \\ &\quad {}+ 50000 = 4{,}450{,}000. \end{aligned} \]

At \(1800\) steps per mile, the mileage is \(\dfrac{4{,}450{,}000}{1800} \approx 2472,\) which is closest to \(2500.\)

Thus, the correct answer is A.

7 答案 A

因为 \((y - x)^2 = (x - y)^2\)，运算的两个输入都是同一个值 \(t = (x - y)^2\)。

因此 \((x - y)^2 \$ (y - x)^2\) \(= t \$ t\) \(= (t - t)^2 = 0\)。

因此，正确答案是 A。

English

Since \((y - x)^2 = (x - y)^2,\) both inputs to the operation are the same value \(t = (x - y)^2.\)

Therefore \((x - y)^2 \$ (y - x)^2\) \(= t \$ t\) \(= (t - t)^2 = 0.\)

Thus, the correct answer is A.

8 答案 C

因为 \(AB = 4\,BD\) 且 \(AB + BD = AD\)，所以 \(5\,BD = AD\)，于是 \(BD = \tfrac{1}{5}AD\)。

同样，由 \(AC = 9\,CD\) 且 \(AC + CD = AD\) 得到 \(CD = \tfrac{1}{10}AD\)。

因为 \(B\) 和 \(C\) 的位置都从 \(A\) 起量，所以 \(\;BC = BD - CD\) \(= \tfrac{1}{5}AD - \tfrac{1}{10}AD\) \(= \tfrac{1}{10}AD\)。

因此，正确答案是 C。

English

Since \(AB = 4\,BD\) and \(AB + BD = AD,\) we have \(5\,BD = AD,\) so \(BD = \tfrac{1}{5}AD.\)

Likewise \(AC = 9\,CD\) with \(AC + CD = AD\) gives \(CD = \tfrac{1}{10}AD.\)

Because \(B\) and \(C\) both measure from \(A,\) \(\;BC = BD - CD\) \(= \tfrac{1}{5}AD - \tfrac{1}{10}AD\) \(= \tfrac{1}{10}AD.\)

Thus, the correct answer is C.

9 答案 A

设 \(O\) 为圆心，\(D\) 为 \(\overline{OC}\) 与 \(\overline{AB}\) 的交点。因为 \(C\) 是弧 \(AB\) 的中点，\(\overline{OC}\) 是弦的垂直平分线，所以 \(AD = 3\)。

在直角三角形 \(ADO\) 中，\(OD = \sqrt{5^2 - 3^2} = 4\)，所以 \(DC = OC - OD = 5 - 4 = 1\)。

然后在直角三角形 \(ADC\) 中，\(AC = \sqrt{AD^2 + DC^2}\) \(= \sqrt{3^2 + 1^2}\) \(= \sqrt{10}\)。

因此，正确答案是 A。

English

Let \(O\) be the center and \(D\) the point where \(\overline{OC}\) meets \(\overline{AB}.\) Since \(C\) is the midpoint of arc \(AB,\) \(\overline{OC}\) is the perpendicular bisector of the chord, so \(AD = 3.\)

In right triangle \(ADO,\) \(OD = \sqrt{5^2 - 3^2} = 4,\) so \(DC = OC - OD = 5 - 4 = 1.\)

Then in right triangle \(ADC,\) \(AC = \sqrt{AD^2 + DC^2}\) \(= \sqrt{3^2 + 1^2}\) \(= \sqrt{10}.\)

Thus, the correct answer is A.

10 答案 B

设烟囱有 \(n\) 块砖。Brenda 单独工作时每小时砌 \(\tfrac{n}{9}\) 块，Brandon 每小时砌 \(\tfrac{n}{10}\) 块。一起工作时，他们的效率为 \(\tfrac{n}{9} + \tfrac{n}{10} - 10\)。

工作 \(5\) 小时完成烟囱，因此 \[ 5\left(\tfrac{n}{9} + \tfrac{n}{10} - 10\right) = n\text{。} \] 展开得 \(\tfrac{5n}{9} + \tfrac{5n}{10} - 50 = n\)，所以 \(\tfrac{95n}{90} - n = 50\)，从而 \(\tfrac{5n}{90} = 50\)。

因此 \(n = 900\)。

因此，正确答案是 B。

English

Let \(n\) be the number of bricks. Alone, Brenda lays \(\tfrac{n}{9}\) bricks per hour and Brandon lays \(\tfrac{n}{10}.\) Together, their rate is \(\tfrac{n}{9} + \tfrac{n}{10} - 10.\)

Working for \(5\) hours completes the chimney: \[ 5\left(\tfrac{n}{9} + \tfrac{n}{10} - 10\right) = n. \] Expanding, \(\tfrac{5n}{9} + \tfrac{5n}{10} - 50 = n,\) so \(\tfrac{95n}{90} - n = 50,\) giving \(\tfrac{5n}{90} = 50.\)

Hence \(n = 900.\)

Thus, the correct answer is B.

11 答案 A

水面以上的部分是一个与整座山相似的圆锥，其体积为总体积的 \(\tfrac{1}{8}\)。由于体积按长度的立方缩放，水上圆锥的高度是全高的 \(\sqrt[3]{\tfrac{1}{8}} = \tfrac{1}{2}\)。

所以水面以上的高度为 \(8000 \cdot \tfrac{1}{2} = 4000\) 英尺。

山底处的海水深度就是水下高度，等于 \(8000 - 4000 = 4000\) 英尺。

因此，正确答案是 A。

English

The part above the water is a cone similar to the whole mountain, with volume \(\tfrac{1}{8}\) of the total. Since volume scales as the cube of length, the above-water cone’s height is \(\sqrt[3]{\tfrac{1}{8}} = \tfrac{1}{2}\) of the full height.

So the above-water height is \(8000 \cdot \tfrac{1}{2} = 4000\) feet.

The ocean depth at the base is the submerged height, \(8000 - 4000 = 4000\) feet.

Thus, the correct answer is A.

12 答案 B

因为前 \(n\) 项的平均数为 \(n\)，所以它们的和为 \(n \cdot n = n^2\)。

第 \(n\) 项是相邻两个部分和的差，\(n^2 - (n-1)^2 = 2n - 1\)。

当 \(n = 2008\) 时，该项为 \(2 \cdot 2008 - 1 = 4015\)。

因此，正确答案是 B。

English

Since the mean of the first \(n\) terms is \(n,\) their sum is \(n \cdot n = n^2.\)

The \(n\)th term is the difference of consecutive sums, \(n^2 - (n-1)^2 = 2n - 1.\)

For \(n = 2008,\) the term is \(2 \cdot 2008 - 1 = 4015.\)

Thus, the correct answer is B.

13 答案 B

令 \(A = (0,0)\)、\(B = (1,0)\)、\(C = (1,1)\)、\(D = (0,1)\)，则 \(\overline{AD}\) 在 \(y\) 轴上，到 \(\overline{AD}\) 的距离就是 \(x\) 坐标。区域位于条带 \(\tfrac13 \le x \le \tfrac23\) 中，该条带在正方形内的面积为 \(\tfrac13\)。

等边三角形 \(\triangle ABE\) 有 \(E = \left(\tfrac12, \tfrac{\sqrt3}{2}\right)\)，边 \(AE\) 在 \(y = \sqrt3\,x\) 上，边 \(BE\) 在 \(y = \sqrt3(1 - x)\) 上。三角形在条带内的面积为 \[ \begin{aligned} &\int_{\frac{1}{3}}^{\frac{1}{2}} \sqrt3\,x\,dx \\ &\quad {}+ \int_{\frac{1}{2}}^{\frac{2}{3}} \sqrt3(1 - x)\,dx \\ &= 2\int_{\frac{1}{3}}^{\frac{1}{2}}\sqrt3\,x\,dx \\ &= \frac{5\sqrt3}{36} \end{aligned}\text{。}\]

因此 \[ [R] = \frac13 - \frac{5\sqrt3}{36} = \frac{12 - 5\sqrt3}{36}\text{。}\]

因此，正确答案是 B。

English

Place \(A = (0,0),\) \(B = (1,0),\) \(C = (1,1),\) \(D = (0,1),\) so \(\overline{AD}\) lies along the \(y\)-axis and distance from \(\overline{AD}\) is the \(x\)-coordinate. The region lies in the strip \(\tfrac13 \le x \le \tfrac23,\) which within the square has area \(\tfrac13.\)

Equilateral \(\triangle ABE\) has \(E = \left(\tfrac12, \tfrac{\sqrt3}{2}\right),\) with side \(AE\) on \(y = \sqrt3\,x\) and side \(BE\) on \(y = \sqrt3(1 - x).\) The area of the triangle inside the strip is \[ \begin{aligned} &\int_{\frac{1}{3}}^{\frac{1}{2}} \sqrt3\,x\,dx \\ &\quad {}+ \int_{\frac{1}{2}}^{\frac{2}{3}} \sqrt3(1 - x)\,dx \\ &= 2\int_{\frac{1}{3}}^{\frac{1}{2}}\sqrt3\,x\,dx \\ &= \frac{5\sqrt3}{36}. \end{aligned} \]

Therefore \[ [R] = \frac13 - \frac{5\sqrt3}{36} = \frac{12 - 5\sqrt3}{36}. \]

Thus, the correct answer is B.

14 答案 C

周长是半径的 \(2\pi\) 倍，所以 \[ \log_{10}(b^4) = 2\pi \log_{10}(a^2)\text{。}\]

改写得 \(4\log_{10} b = 4\pi \log_{10} a\)，因此 \(\log_{10} b = \pi \log_{10} a\)。

所以 \(\log_a b = \dfrac{\log_{10} b}{\log_{10} a} = \pi\)。

因此，正确答案是 C。

English

The circumference is \(2\pi\) times the radius, so \[ \log_{10}(b^4) = 2\pi \log_{10}(a^2). \]

Rewriting, \(4\log_{10} b = 4\pi \log_{10} a,\) hence \(\log_{10} b = \pi \log_{10} a.\)

Therefore \(\log_a b = \dfrac{\log_{10} b}{\log_{10} a} = \pi.\)

Thus, the correct answer is C.

15 答案 C

凸包 \(S\) 与 \(R\) 的差只出现在正方形的四个角附近，每处形成一个小三角形空隙。每个空隙三角形有两条长度为 \(1\) 的边，也就是相邻三角形的外边。

这两条边之间的夹角为 \(360^\circ - 90^\circ - 4 \cdot 60^\circ = 30^\circ\)，所以每个空隙面积为 \[ \tfrac12 \cdot 1 \cdot 1 \cdot \sin 30^\circ = \tfrac14\text{。}\]

总面积为 \(4 \cdot \tfrac14 = 1\)。

因此，正确答案是 C。

English

The convex hull \(S\) differs from \(R\) only near the four corners of the square, where a small triangular gap forms. Each gap triangle has two sides of length \(1\) (outer edges of adjacent triangles).

The angle between those two sides is \(360^\circ - 90^\circ - 4 \cdot 60^\circ = 30^\circ,\) so each gap has area \[ \tfrac12 \cdot 1 \cdot 1 \cdot \sin 30^\circ = \tfrac14. \]

The total area is \(4 \cdot \tfrac14 = 1.\)

Thus, the correct answer is C.

16 答案 B

涂色矩形尺寸为 \((a - 2)\) 乘 \((b - 2)\)，面积是地板面积的一半，所以 \[ ab = 2(a - 2)(b - 2)\text{。}\]

展开得 \(0 = ab - 4a - 4b + 8\)，再两边加 \(8\) 得到 \((a - 4)(b - 4) = 8\)。

在 \(b \gt a \gt 0\) 下，\(8\) 的有效因数对只有 \((a - 4, b - 4) = (1, 8)\) 和 \((2, 4)\)，给出 \((a, b) = (5, 12)\) 和 \((6, 8)\)。

共有 \(2\) 种可能。

因此，正确答案是 B。

English

The painted rectangle measures \((a - 2)\) by \((b - 2)\) and has half the area of the floor, so \[ ab = 2(a - 2)(b - 2). \]

Expanding gives \(0 = ab - 4a - 4b + 8,\) and adding \(8\) yields \((a - 4)(b - 4) = 8.\)

With \(b \gt a \gt 0,\) the only valid factor pairs of \(8\) are \((a - 4, b - 4) = (1, 8)\) and \((2, 4),\) giving \((a, b) = (5, 12)\) and \((6, 8).\)

There are \(2\) possibilities.

Thus, the correct answer is B.

17 答案 C

因为 \(AB\) 水平，取 \(A = (a, a^2)\)、\(B = (-a, a^2)\)，并设 \(C = (c, c^2)\)。直角不可能在 \(A\) 或 \(B\) 处（否则需要 \(c = \pm a\)），所以直角在 \(C\) 处。

令 \(d = a^2-c^2\)。计算 \(\overrightarrow{CA}\) 与 \(\overrightarrow{CB}\) 的点积，得到 \(d(d-1)=0\)。三点互异，所以 \(d \ne 0\)；因此 \(a^2-c^2=1\)。这个值就是三角形相对于 \(\overline{AB}\) 的高。

面积为 \(\tfrac12 \cdot AB \cdot \text{高}\) \(= \tfrac12 (2|a|)(1)\) \(= |a| = 2008\)，所以 \(a^2 = 2008^2 = 4{,}032{,}064\)，而 \(C\) 的 \(y\) 坐标为 \(c^2 = a^2 - 1 = 4{,}032{,}063\)。

它的数位和为 \(4 + 0 + 3 + 2 + 0 + 6 + 3 = 18\)。

所以正确答案是 C。

English

Since \(AB\) is horizontal, take \(A = (a, a^2)\) and \(B = (-a, a^2),\) and let \(C = (c, c^2).\) The right angle cannot be at \(A\) or \(B\) (that would need \(c = \pm a\)), so it is at \(C.\)

Put \(d = a^2-c^2.\) Taking the dot product of \(\overrightarrow{CA}\) and \(\overrightarrow{CB}\) gives \(d(d-1)=0.\) The points are distinct, so \(d \ne 0;\) hence \(a^2-c^2=1.\) This value is the height of the triangle above \(\overline{AB}.\)

The area is \(\tfrac12 \cdot AB \cdot \text{height}\) \(= \tfrac12 (2|a|)(1)\) \(= |a| = 2008,\) so \(a^2 = 2008^2 = 4{,}032{,}064\) and the \(y\)-coordinate of \(C\) is \(c^2 = a^2 - 1 = 4{,}032{,}063.\)

Its digit sum is \(4 + 0 + 3 + 2 + 0 + 6 + 3 = 18.\)

Thus, the correct answer is C.

18 答案 E

正方形边长为 \(\sqrt{196} = 14\)。令 \(F\) 和 \(G\) 分别为从 \(E\) 到 \(AB\) 和 \(CD\) 的垂足。于是 \(FG = 14\)，\(EF = \tfrac{2 \cdot 105}{14} = 15\)，且 \(EG = \tfrac{2 \cdot 91}{14} = 13\)。

三角形 \(EFG\) 位于垂直于底面的平面内，因此它到 \(FG\) 的高就是棱锥的高。由海伦公式，半周长 \(s = 21\)，其面积为 \(\sqrt{21 \cdot 6 \cdot 8 \cdot 7} = 84\)，所以到 \(FG\) 的高为 \(\tfrac{2 \cdot 84}{14} = 12\)。

体积为 \(\tfrac13 \cdot 196 \cdot 12 = 784\)。

因此，正确答案是 E。

English

The square has side \(\sqrt{196} = 14.\) Let \(F\) and \(G\) be the feet of the perpendiculars from \(E\) to \(AB\) and \(CD.\) Then \(FG = 14,\) \(EF = \tfrac{2 \cdot 105}{14} = 15,\) and \(EG = \tfrac{2 \cdot 91}{14} = 13.\)

Triangle \(EFG\) lies in a plane perpendicular to the base, so its altitude to \(FG\) is the pyramid’s height. By Heron’s formula with \(s = 21,\) its area is \(\sqrt{21 \cdot 6 \cdot 8 \cdot 7} = 84,\) so the altitude to \(FG\) is \(\tfrac{2 \cdot 84}{14} = 12.\)

The volume is \(\tfrac13 \cdot 196 \cdot 12 = 784.\)

Thus, the correct answer is E.

19 答案 B

令 \(\alpha = a + bi\)，\(\gamma = c + di\)。则 \(f(1) = (4 + a + c)\) \(+ (1 + b + d)i\)，且 \(f(i) = (-4 - b + c)\) \(+ (-1 + a + d)i\)。

二者都是实数，强制 \(1 + b + d = 0\) 且 \(-1 + a + d = 0\)，即 \(a = 1 - d\)、\(b = -1 - d\)。

因此 \[ \begin{aligned} &|\alpha| + |\gamma| \\ &= \sqrt{(1 - d)^2 + (1 + d)^2} \\ &\quad {}+ \sqrt{c^2 + d^2} \\ &= \sqrt{2 + 2d^2} \\ &\quad {}+ \sqrt{c^2 + d^2}\text{，} \end{aligned} \] 当 \(c = d = 0\) 时最小，值为 \(\sqrt{2}\)。

因此，正确答案是 B。

English

Let \(\alpha = a + bi\) and \(\gamma = c + di.\) Then \(f(1) = (4 + a + c)\) \(+ (1 + b + d)i\) and \(f(i) = (-4 - b + c)\) \(+ (-1 + a + d)i.\)

Both being real forces \(1 + b + d = 0\) and \(-1 + a + d = 0,\) i.e. \(a = 1 - d\) and \(b = -1 - d.\)

Hence \[ \begin{aligned} &|\alpha| + |\gamma| \\ &= \sqrt{(1 - d)^2 + (1 + d)^2} \\ &\quad {}+ \sqrt{c^2 + d^2} \\ &= \sqrt{2 + 2d^2} \\ &\quad {}+ \sqrt{c^2 + d^2}, \end{aligned} \] which is smallest when \(c = d = 0,\) giving \(\sqrt{2}.\)

Thus, the correct answer is B.

20 答案 B

给垃圾桶编号。在时刻 \(0\)，Michael 在 \(0\) 号桶，垃圾车在 \(1\) 号桶。Michael 在 \(40n\) 秒时到达 \(n\) 号桶。垃圾车在两个桶之间行驶 \(20\) 秒，并停 \(30\) 秒，所以它在 \(50(n - 1)\) 秒离开 \(n\) 号桶，并且（当 \(n \ge 2\) 时）在 \(50(n - 1) - 30\) 秒到达。

Michael 在 \(n\) 号桶时垃圾车也在那里，恰好当 \(50(n-1) - 30 \le 40n\) \(\le 50(n-1)\)，化简得 \(5 \le n \le 8\)。所以他们在 \(5\) 号桶（\(t = 200\)，垃圾车离开时）、\(6\) 号桶（\(t = 240\)）、\(7\) 号桶（\(t = 280\)）和 \(8\) 号桶（\(t = 320\)，垃圾车到达时）相遇。

在 \(6\) 号桶和 \(7\) 号桶之间，垃圾车（以 \(10\) 英尺/秒移动）先超过 Michael，随后又被 Michael 追上一次，额外增加一次交会。总共相遇 \(5\) 次。

因此，正确答案是 B。

English

Number the pails so Michael is at pail \(0\) and the truck at pail \(1\) at time \(0.\) Michael reaches pail \(n\) at \(40n\) seconds. The truck spends \(20\) seconds between pails and \(30\) stopped, so it leaves pail \(n\) at \(50(n - 1)\) seconds and (for \(n \ge 2\)) arrives at \(50(n - 1) - 30.\)

Michael is at pail \(n\) while the truck is there exactly when \(50(n-1) - 30 \le 40n\) \(\le 50(n-1),\) which simplifies to \(5 \le n \le 8.\) So they meet at pail \(5\) (at \(t = 200,\) as the truck departs), pail \(6\) (\(t = 240\)), pail \(7\) (\(t = 280\)), and pail \(8\) (\(t = 320,\) as the truck arrives).

Between pails \(6\) and \(7\) the truck (moving at \(10\) ft/s) pulls ahead of and is then overtaken by Michael once more, adding one crossing. In all, they meet \(5\) times.

Thus, the correct answer is B.

21 答案 E

设两个圆心为 \((a, 0)\) 和 \((b, 1)\)，其中 \(a, b \in [0, 2]\)。两个半径均为 \(1\) 的圆相交，当且仅当圆心距至多为 \(2\)：\[ \begin{aligned} &\sqrt{(a - b)^2 + 1} \le 2 \\ &\iff |a - b| \le \sqrt{3} \end{aligned}\text{。}\]

所有点对 \((a, b)\) 填满面积为 \(4\) 的正方形 \([0, 2]^2\)。失败区域 \(|a - b| \gt \sqrt3\) 是两个直角三角形，每个的两条直角边长为 \(2 - \sqrt3\)，总面积为 \((2 - \sqrt3)^2 = 7 - 4\sqrt3\)。

所以有利面积为 \(4 - (7 - 4\sqrt3) = 4\sqrt3 - 3\)，概率为 \[ \frac{4\sqrt3 - 3}{4}\text{。}\]

因此，正确答案是 E。

English

Let the centers be \((a, 0)\) and \((b, 1)\) with \(a, b \in [0, 2].\) The circles (radius \(1\) each) intersect iff the distance between centers is at most \(2:\) \[ \begin{aligned} &\sqrt{(a - b)^2 + 1} \le 2 \\ &\iff |a - b| \le \sqrt{3}. \end{aligned} \]

The pairs \((a, b)\) fill the square \([0, 2]^2\) of area \(4.\) The failing region \(|a - b| \gt \sqrt3\) is two right triangles, each with legs \(2 - \sqrt3,\) of total area \((2 - \sqrt3)^2 = 7 - 4\sqrt3.\)

So the favorable area is \(4 - (7 - 4\sqrt3) = 4\sqrt3 - 3,\) and the probability is \[ \frac{4\sqrt3 - 3}{4}. \]

Thus, the correct answer is E.

22 答案 E

\(12\) 辆车停好后，有 \(4\) 个车位为空，等可能地是 \(16\) 个车位中的任意 \(4\) 个，共有 \(\binom{16}{4} = 1820\) 个等可能集合。

Auntie Em 不能停车恰好当没有两个空车位相邻。在 \(16\) 个车位中放置 \(4\) 个互不相邻的空位有 \(\binom{13}{4} = 715\) 种。

因此她能够停车的概率为 \[ 1 - \frac{715}{1820} = \frac{1105}{1820} = \frac{17}{28}\text{。}\]

因此，正确答案是 E。

English

After the \(12\) cars park, \(4\) spaces are empty, equally likely to be any \(4\) of the \(16,\) for \(\binom{16}{4} = 1820\) equally likely sets.

Auntie Em fails exactly when no two empty spaces are adjacent. The number of ways to place \(4\) non-adjacent empties among \(16\) is \(\binom{13}{4} = 715.\)

So the probability she can park is \[ 1 - \frac{715}{1820} = \frac{1105}{1820} = \frac{17}{28}. \]

Thus, the correct answer is E.

23 答案 A

所有因数的以 \(10\) 为底的对数之和，就是这些因数乘积的对数。若一个数 \(N\) 有 \(d(N)\) 个因数，则其所有因数的乘积为 \(N^{\frac{d(N)}{2}}\)。

这里 \(N = 10^n\) 有 \(d(N) = (n + 1)^2\) 个因数，所以因数乘积为 \((10^n)^{\frac{(n+1)^2}{2}}\)，其对数为 \[ \frac{n(n + 1)^2}{2} = 792\text{。}\]

因此 \(n(n + 1)^2 = 1584\) \(= 11 \cdot 144\) \(= 11 \cdot 12^2\)，得到 \(n = 11\)。

因此，正确答案是 A。

English

The sum of the base-\(10\) logs of the divisors is the log of their product. A number \(N\) with \(d(N)\) divisors has divisor product \(N^{\frac{d(N)}{2}}.\)

Here \(N = 10^n\) has \(d(N) = (n + 1)^2\) divisors, so the product is \((10^n)^{\frac{(n+1)^2}{2}}\) and its log is \[ \frac{n(n + 1)^2}{2} = 792. \]

Thus \(n(n + 1)^2 = 1584\) \(= 11 \cdot 144\) \(= 11 \cdot 12^2,\) giving \(n = 11.\)

Thus, the correct answer is A.

24 答案 C

令 \(a_n = A_0A_n\)，并令 \(c_n = a_n - a_{n-1}\) 为第 \(n\) 个等边三角形的底边。其顶点 \(B_n\) 位于底边中点上方，高度为 \(\tfrac{\sqrt3}{2}c_n\)，且在 \(y = \sqrt{x}\) 上，所以 \[ \left(\tfrac{\sqrt3}{2}c_n\right)^2 = a_{n-1} + \tfrac{c_n}{2}\text{，} \] 即 \[ \tfrac34 c_n^2 = a_{n-1} + \tfrac{c_n}{2}\text{。} \]

对前一个三角形写出相同关系并相减，得到 \(c_n = c_{n-1} + \tfrac23\)，又 \(c_1 = \tfrac23\)，所以 \(c_n = \tfrac{2n}{3}\)。求和得 \[ a_n = \sum_{k=1}^n \frac{2k}{3} = \frac{n(n + 1)}{3}\text{。} \]

需要 \(\tfrac{n(n + 1)}{3} \ge 100\)，即 \(n(n + 1) \ge 300\)。由于 \(16 \cdot 17 = 272\) 而 \(17 \cdot 18 = 306\)，最小的 \(n\) 是 \(17\)。

因此，正确答案是 C。

English

Let \(a_n = A_0A_n\) and \(c_n = a_n - a_{n-1}\) be the base of the \(n\)th equilateral triangle. Its apex \(B_n\) lies above the midpoint at height \(\tfrac{\sqrt3}{2}c_n,\) and being on \(y = \sqrt{x}\) gives \[ \left(\tfrac{\sqrt3}{2}c_n\right)^2 = a_{n-1} + \tfrac{c_n}{2}, \] i.e. \[ \tfrac34 c_n^2 = a_{n-1} + \tfrac{c_n}{2}. \]

Writing the same relation for the previous triangle and subtracting gives \(c_n = c_{n-1} + \tfrac23,\) and with \(c_1 = \tfrac23\) we get \(c_n = \tfrac{2n}{3}.\) Summing, \[ a_n = \sum_{k=1}^n \frac{2k}{3} = \frac{n(n + 1)}{3}. \]

We need \(\tfrac{n(n + 1)}{3} \ge 100,\) i.e. \(n(n + 1) \ge 300.\) Since \(16 \cdot 17 = 272\) and \(17 \cdot 18 = 306,\) the least such \(n\) is \(17.\)

Thus, the correct answer is C.

25 答案 B

因为 \(AB \parallel CD\)，有 \(\angle A + \angle D = 180^\circ\)，所以 \(\angle A\) 与 \(\angle D\) 的角平分线垂直相交，即 \(\angle APD = 90^\circ\)。于是 \(\overline{AD}\) 的中点 \(M\) 是直角三角形 \(APD\) 的外心，得到 \(MP = MA = MD\)。所以 \(\angle MPA = \angle PAM = \angle PAB\)，最后一个等号使用了 \(A\) 处的角平分线。因此 \(MP \parallel AB\)。对 \(\overline{BC}\) 的中点 \(N\) 同理可得 \(QN \parallel AB\)。所以 \(M, P, Q, N\) 共线于中位线上。

中位线长为 \(\tfrac{AB + CD}{2} = 15\)，而 \(MP = \tfrac{AD}{2} = \tfrac72\)、\(QN = \tfrac{BC}{2} = \tfrac52\)。因此 \(PQ = 15 - \tfrac72 - \tfrac52 = 9\)。

作 \(AE \parallel BC\)，其中 \(E\) 位于 \(\overline{CD}\) 上，则 \(AE = 5\)，且 \(DE = CD - AB = 8\)。在 \(\triangle ADE\) 中，\(\cos(\angle AED) = \tfrac{8^2 + 5^2 - 7^2}{2 \cdot 8 \cdot 5} = \tfrac12\)，所以 \(\angle AED = 60^\circ\)，梯形高为 \(AF = 5\sin 60^\circ = \tfrac{5\sqrt3}{2}\)。

线段 \(PQ\) 位于半高处，所以六边形分成两个梯形，并且 \[ \begin{aligned} &[ABQCDP] \\ &= \frac{AF}{4}\bigl(AB + CD + 2\,PQ\bigr) \\ &= \frac{\frac{5\sqrt3}{2}}{4}(11 + 19 + 18) \\ &= 30\sqrt3 \end{aligned}\text{。}\]

所以正确答案是 B。

English

Because \(AB \parallel CD,\) \(\angle A + \angle D = 180^\circ,\) so the bisectors of \(\angle A\) and \(\angle D\) meet at right angles, \(\angle APD = 90^\circ.\) Then the midpoint \(M\) of \(\overline{AD}\) is the circumcenter of right triangle \(APD,\) giving \(MP = MA = MD.\) Thus \(\angle MPA = \angle PAM = \angle PAB,\) where the last equality uses the angle bisector at \(A.\) Therefore \(MP \parallel AB.\) The same argument gives \(QN \parallel AB\) for the midpoint \(N\) of \(\overline{BC}.\) Hence \(M, P, Q, N\) are collinear on the midline.

The midline has length \(\tfrac{AB + CD}{2} = 15,\) while \(MP = \tfrac{AD}{2} = \tfrac72\) and \(QN = \tfrac{BC}{2} = \tfrac52.\) Hence \(PQ = 15 - \tfrac72 - \tfrac52 = 9.\)

Drawing \(AE \parallel BC\) with \(E\) on \(\overline{CD}\) gives \(AE = 5\) and \(DE = CD - AB = 8.\) In \(\triangle ADE,\) \(\cos(\angle AED) = \tfrac{8^2 + 5^2 - 7^2}{2 \cdot 8 \cdot 5} = \tfrac12,\) so \(\angle AED = 60^\circ\) and the trapezoid’s height is \(AF = 5\sin 60^\circ = \tfrac{5\sqrt3}{2}.\)

The segment \(PQ\) sits at half the height, so the hexagon splits into two trapezoids and \[ \begin{aligned} &[ABQCDP] \\ &= \frac{AF}{4}\bigl(AB + CD + 2\,PQ\bigr) \\ &= \frac{\frac{5\sqrt3}{2}}{4}(11 + 19 + 18) \\ &= 30\sqrt3. \end{aligned} \]

Thus, the correct answer is B.