2009B AMC 12 真题（P版）

1 答案 B

若她买了 \(b\) 个贝果，则买了 \(5 - b\) 个松饼，总价为 \(50(5-b) + 75b = 250 + 25b\) 美分。要使总价为整数美元，它必须是 \(100\) 的倍数，于是 \(25b\equiv50\pmod{100}\)，即 \(b\equiv2\pmod4\)。因为 \(0\le b\le5\)，只有 \(b=2\) 可行，此时总价为 \(300\) 美分，即 \(= \$3\)。

所以正确答案是 B。

English

With \(b\) bagels she buys \(5 - b\) muffins, costing \(50(5-b) + 75b = 250 + 25b\) cents. For a whole number of dollars this must be a multiple of \(100.\) Thus \(25b\equiv50\pmod{100},\) or \(b\equiv2\pmod4.\) Since \(0\le b\le5,\) only \(b=2\) works, giving \(300\) cents \(= \$3.\)

Thus, the correct answer is B.

2 答案 C

丢失 \(3\) 罐使她少刷 \(5\) 间房间，所以 \(3\) 罐油漆能刷 \(5\) 间房间，每间房需要 \(\dfrac{3}{5}\) 罐。

刷 \(25\) 间房间需要 \(25 \cdot \dfrac{3}{5} = 15\) 罐。

所以正确答案是 C。

English

Losing \(3\) cans cost her \(5\) rooms, so \(3\) cans paint \(5\) rooms and each room needs \(\dfrac{3}{5}\) of a can.

For \(25\) rooms she used \(25 \cdot \dfrac{3}{5} = 15\) cans.

Thus, the correct answer is C.

3 答案 D

比 \(60\) 少百分之二十是 \(0.8 \cdot 60 = 48\)。

比 \(n\) 多三分之一是 \(\dfrac{4}{3}n\)，所以 \(\dfrac{4}{3}n = 48\)，得到 \(n = 36\)。

所以正确答案是 D。

English

Twenty percent less than \(60\) is \(0.8 \cdot 60 = 48.\)

One-third more than \(n\) is \(\dfrac{4}{3}n,\) so \(\dfrac{4}{3}n = 48\) gives \(n = 36.\)

Thus, the correct answer is D.

4 答案 C

两条平行边相差 \(25 - 15 = 10\)，所以每个三角形的直角边为 \(\dfrac{10}{2} = 5\)，面积为 \(\dfrac{1}{2} \cdot 5^2 = \dfrac{25}{2}\)。两个花坛总面积为 \(25\)。

长方形尺寸为 \(25\) 乘 \(5\)，面积为 \(125\)，花坛所占比例为 \(\dfrac{25}{125} = \dfrac{1}{5}\)。

所以正确答案是 C。

English

The parallel sides differ by \(25 - 15 = 10,\) so each triangle has legs \(\dfrac{10}{2} = 5\) and area \(\dfrac{1}{2} \cdot 5^2 = \dfrac{25}{2}.\) The two beds total \(25.\)

The rectangle measures \(25\) by \(5,\) so its area is \(125,\) and the fraction occupied is \(\dfrac{25}{125} = \dfrac{1}{5}.\)

Thus, the correct answer is C.

5 答案 D

因为 \(128 = 2^7\)，每个年龄都是 \(2\) 的幂。双胞胎年龄相同，设为 \(t\)，则 Kiana 的年龄为 \(\dfrac{128}{t^2}\)。

取 \(t = 8\) 时，Kiana 的年龄是 \(\dfrac{128}{64} = 2\)，比双胞胎小。（若双胞胎年龄更小，Kiana 就会更大，这是不允许的。）年龄和为 \(8 + 8 + 2 = 18\)。

所以正确答案是 D。

English

Since \(128 = 2^7,\) each age is a power of \(2.\) The twins share an age \(t,\) so Kiana’s age is \(\dfrac{128}{t^2}.\)

Taking \(t = 8\) gives Kiana \(\dfrac{128}{64} = 2,\) who is younger than the twins. (Smaller twins would make Kiana older, which is not allowed.) The sum is \(8 + 8 + 2 = 18.\)

Thus, the correct answer is D.

6 答案 C

真正不同的分组给出

这些分组给出 \((2 \times 3) + (4 \times 5) = 26\)、\(\ ((2 \times 3) + 4) \times 5 = 50\)、\(\ 2 \times (3 + (4 \times 5)) = 46\)、\(\ 2 \times (3 + 4) \times 5 = 70\)。

这四个值互不相同，所以可以得到 \(4\) 个值。

所以正确答案是 C。

English

The genuinely different groupings give

\((2 \times 3) + (4 \times 5) = 26,\) \(\ ((2 \times 3) + 4) \times 5 = 50,\) \(\ 2 \times (3 + (4 \times 5)) = 46,\) and \(\ 2 \times (3 + 4) \times 5 = 70.\)

These are all distinct, so \(4\) values can be obtained.

Thus, the correct answer is C.

7 答案 B

一月至三月后，价格变为原价的 \(1.2 \cdot 0.8 \cdot 1.25 = 1.2\) 倍。

要回到原价，四月必须乘以 \(\dfrac{1}{1.2}\)，也就是下降 \(1 - \dfrac{1}{1.2} = \dfrac{1}{6} \approx 16.7\%\)。四舍五入后，\(x = 17\)。

所以正确答案是 B。

English

After January through March the price is \(1.2 \cdot 0.8 \cdot 1.25 = 1.2\) times the original.

To return to the original, April must multiply by \(\dfrac{1}{1.2},\) a decrease of \(1 - \dfrac{1}{1.2} = \dfrac{1}{6} \approx 16.7\%.\) To the nearest integer, \(x = 17.\)

Thus, the correct answer is B.

8 答案 E

设 \(x\) 为空桶重量，\(y\) 为满桶水的水重。则 \(x + \dfrac{2}{3}y = a\)，且 \(x + \dfrac{1}{2}y = b\)。

两式相减得 \(\dfrac{1}{6}y = a - b\)，所以 \(y = 6a - 6b\)，且 \(x = b - \dfrac{1}{2}y = 4b - 3a\)。满桶总重量为 \(x + y = 3a - 2b\)。

所以正确答案是 E。

English

Let \(x\) be the bucket’s weight and \(y\) the weight of a full bucket of water. Then \(x + \dfrac{2}{3}y = a\) and \(x + \dfrac{1}{2}y = b.\)

Subtracting gives \(\dfrac{1}{6}y = a - b,\) so \(y = 6a - 6b\) and \(x = b - \dfrac{1}{2}y = 4b - 3a.\) The full weight is \(x + y = 3a - 2b.\)

Thus, the correct answer is E.

9 答案 A

直线 \(AB\) 的方程是 \(x + y = 3\)，它与 \(x + y = 7\) 平行，所以面积与 \(C\) 在该直线上的具体位置无关。

取 \(C = (7, 0)\)。此时底边 \(AC = 4\) 在 \(x\)-轴上，高为 \(3\)，面积为 \(\dfrac{1}{2} \cdot 4 \cdot 3 = 6\)。

所以正确答案是 A。

English

Line \(AB\) has equation \(x + y = 3,\) which is parallel to \(x + y = 7,\) so the area is independent of where \(C\) lies on that line.

Take \(C = (7, 0).\) Then the base \(AC = 4\) lies on the \(x\)-axis with height \(3,\) giving area \(\dfrac{1}{2} \cdot 4 \cdot 3 = 6.\)

Thus, the correct answer is A.

10 答案 A

含有数字 \(1\) 的小时是 \(1, 10, 11, 12\)，所以 \(12\) 个小时中有 \(8\) 个小时显示正确，比例为 \(\dfrac{2}{3}\)。

若分钟的某一位是 \(1\)，则分钟显示错误：十位为一给出 \(10\text{-}19\)（\(10\) 分钟），个位为一额外给出 \(01, 21, 31, 41, 51\)（\(5\) 分钟），共 \(15\) 分钟。因此分钟正确的比例为 \(\dfrac{45}{60} = \dfrac{3}{4}\)。

因此一天中的正确显示比例为 \(\dfrac{2}{3} \cdot \dfrac{3}{4} = \dfrac{1}{2}\)。

所以正确答案是 A。

English

The hours containing a \(1\) are \(1, 10, 11, 12,\) so \(8\) of the \(12\) hours display correctly, a fraction \(\dfrac{2}{3}.\)

A minute is wrong if either digit is \(1\): the tens digit gives \(10\text{-}19\) (\(10\) minutes), and the ones digit adds \(01, 21, 31, 41, 51\) (\(5\) more), \(15\) in all. So \(\dfrac{45}{60} = \dfrac{3}{4}\) of minutes are correct.

The fraction of the day is \(\dfrac{2}{3} \cdot \dfrac{3}{4} = \dfrac{1}{2}.\)

Thus, the correct answer is A.

11 答案 D

每天鸟留下小米的 \(\dfrac{3}{4}\)，而 Millie 又加入 \(\dfrac{1}{4}\) 夸脱新小米，所以 \(n\) 天后小米量为 \[ \begin{aligned} &\dfrac{1}{4}\left(1 + \dfrac{3}{4} + \cdots + \left(\dfrac{3}{4}\right)^{n-1}\right) \\ &= 1 - \left(\dfrac{3}{4}\right)^n \end{aligned} \] 夸脱。

非小米种子总量始终是 \(\dfrac{3}{4}\) 夸脱，所以小米超过一半等价于 \(1 - \left(\dfrac{3}{4}\right)^n \gt \dfrac{3}{4}\)，即 \(\left(\dfrac{3}{4}\right)^n \lt \dfrac{1}{4}\)。

因为 \(\left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256} \gt \dfrac{1}{4}\)，而 \(\left(\dfrac{3}{4}\right)^5 = \dfrac{243}{1024} \lt \dfrac{1}{4}\)，所以首次发生在第 \(5\) 天，也就是星期五。

所以正确答案是 D。

English

Each day the birds leave \(\dfrac{3}{4}\) of the millet and Millie adds \(\dfrac{1}{4}\) quart of new millet, so after \(n\) days the millet is \[ \begin{aligned} &\dfrac{1}{4}\left(1 + \dfrac{3}{4} + \cdots + \left(\dfrac{3}{4}\right)^{n-1}\right) \\ &= 1 - \left(\dfrac{3}{4}\right)^n \end{aligned} \] quart.

The non-millet seeds always total \(\dfrac{3}{4}\) quart, so millet exceeds half when \(1 - \left(\dfrac{3}{4}\right)^n \gt \dfrac{3}{4},\) that is, \(\left(\dfrac{3}{4}\right)^n \lt \dfrac{1}{4}.\)

Since \(\left(\dfrac{3}{4}\right)^4 = \dfrac{81}{256} \gt \dfrac{1}{4}\) and \(\left(\dfrac{3}{4}\right)^5 = \dfrac{243}{1024} \lt \dfrac{1}{4},\) this first happens on day \(5,\) which is Friday.

Thus, the correct answer is D.

12 答案 E

第八项除以第五项得 \(r^3 = \dfrac{8!}{7!} = 8\)，所以 \(r = 2\)。

第五项为 \(a r^4 = 7!\)，所以 \(a = \dfrac{7!}{16} = \dfrac{5040}{16} = 315\)。

所以正确答案是 E。

English

The eighth term divided by the fifth term is \(r^3 = \dfrac{8!}{7!} = 8,\) so \(r = 2.\)

The fifth term is \(a r^4 = 7!,\) so \(a = \dfrac{7!}{16} = \dfrac{5040}{16} = 315.\)

Thus, the correct answer is E.

13 答案 D

设 \(D\) 为从 \(A\) 作高所得的垂足。则 \(BD = \sqrt{13^2 - 12^2} = 5\)，\(DC = \sqrt{15^2 - 12^2} = 9\)。

若 \(D\) 在 \(B\) 与 \(C\) 之间，则 \(BC = 5 + 9 = 14\)；若三角形为钝角三角形，则 \(BC = 9 - 5 = 4\)。两值之和为 \(14 + 4 = 18\)。

所以正确答案是 D。

English

Let \(D\) be the foot of the altitude from \(A.\) Then \(BD = \sqrt{13^2 - 12^2} = 5\) and \(DC = \sqrt{15^2 - 12^2} = 9.\)

If \(D\) lies between \(B\) and \(C,\) then \(BC = 5 + 9 = 14\); if the triangle is obtuse, \(BC = 9 - 5 = 4.\) The sum is \(14 + 4 = 18.\)

Thus, the correct answer is D.

14 答案 C

五个正方形总面积为 \(5\)，所以每部分面积必须为 \(\dfrac{5}{2}\)。

从 \((a, 0)\) 到 \((3, 3)\) 的直线与坐标轴围成一个底为 \(3 - a\)、高为 \(3\) 的三角形。直线右下方的区域是这个三角形去掉一个单位正方形，所以 \[ \dfrac{3(3 - a)}{2} - 1 = \dfrac{5}{2} \] 得 \(3(3 - a) = 7\)，从而 \(a = \dfrac{2}{3}\)。

所以正确答案是 C。

English

The five squares have total area \(5,\) so each region must have area \(\dfrac{5}{2}.\)

The line from \((a, 0)\) to \((3, 3)\) together with the axes bounds a triangle of base \(3 - a\) and height \(3\); the region on the lower-right side of the line is this triangle with one unit square removed. Setting \[ \dfrac{3(3 - a)}{2} - 1 = \dfrac{5}{2} \] gives \(3(3 - a) = 7,\) so \(a = \dfrac{2}{3}.\)

Thus, the correct answer is C.

15 答案 B

每个方程都给出 \(x = \dfrac{\log(\frac{7}{3})}{\log(1 + f(r))}\)，所以正数 \(f(r)\) 越小，解越大。

当 \(0 \lt r \lt 3\) 时，有 \(\dfrac r{10}\lt r\lt2r\)，并且因为 \(r\lt100\)，还有 \(\dfrac r{10}\lt\sqrt r\)。又因为 \(r^2\lt9\lt10\)，所以 \(\dfrac r{10}\lt\dfrac1r\)。因此五个量中 \(\frac{r}{10}\) 最小，方程 (B) 的解最大。

所以正确答案是 B。

English

Each equation gives \(x = \dfrac{\log(\frac{7}{3})}{\log(1 + f(r))},\) which is largest when the positive quantity \(f(r)\) is smallest.

For \(0 \lt r \lt 3,\) we have \(\dfrac r{10}\lt r\lt2r\) and \(\dfrac r{10}\lt\sqrt r\) because \(r\lt100.\) Also \(r^2\lt9\lt10,\) so \(\dfrac r{10}\lt\dfrac1r.\) Thus \(\frac{r}{10}\) is the smallest of the five quantities, and equation (B) has the largest solution.

Thus, the correct answer is B.

16 答案 B

过 \(D\) 作平行于 \(AB\) 的直线，与 \(BC\) 交于 \(E\)，于是 \(ABED\) 是平行四边形，且 \(BE = AD\)。于是 \(\angle BDE = \angle DBA = 23^\circ\)，又因为 \(\angle BDC = 46^\circ\)，所以 \(DE\) 平分 \(\angle BDC\)。

在 \(\triangle BDC\) 中由角平分线定理，\(\dfrac{EC}{BE} = \dfrac{DC}{DB}\)，所以 \[ \begin{aligned} CD &= DB \cdot \dfrac{BC - AD}{AD} \\ &= 1 \cdot \left(\dfrac{9}{5} - 1\right) \\ &= \dfrac{4}{5} \end{aligned}\text{。}\]

所以正确答案是 B。

English

Draw the line through \(D\) parallel to \(AB,\) meeting \(BC\) at \(E,\) so \(ABED\) is a parallelogram with \(BE = AD.\) Then \(\angle BDE = \angle DBA = 23^\circ,\) and since \(\angle BDC = 46^\circ,\) segment \(DE\) bisects \(\angle BDC.\)

By the angle bisector theorem in \(\triangle BDC,\) \(\dfrac{EC}{BE} = \dfrac{DC}{DB},\) so \[ \begin{aligned} CD &= DB \cdot \dfrac{BC - AD}{AD} \\ &= 1 \cdot \left(\dfrac{9}{5} - 1\right) \\ &= \dfrac{4}{5}. \end{aligned} \]

Thus, the correct answer is B.

17 答案 B

\(6\) 个面中的每个面都有 \(2\) 个等可能的条纹方向，所以共有 \(2^6 = 64\) 种配置。

环绕条纹沿着 \(3\) 对相对面中的某一对绕行。固定这样一条带，条纹经过的四个面必须对齐，概率为 \(\left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16}\)，其余两个面任意。\(3\) 条可能的带互不重叠，所以概率为 \(3 \cdot \dfrac{1}{16} = \dfrac{3}{16}\)。

所以正确答案是 B。

English

Each of the \(6\) faces has \(2\) equally likely stripe orientations, for \(2^6 = 64\) configurations.

An encircling stripe runs around one of the \(3\) pairs of opposite faces. Fixing such a band, the four faces it passes through must be aligned, with probability \(\left(\dfrac{1}{2}\right)^4 = \dfrac{1}{16},\) while the two remaining faces are free. The \(3\) possible bands are disjoint events, so the probability is \(3 \cdot \dfrac{1}{16} = \dfrac{3}{16}.\)

Thus, the correct answer is B.

18 答案 C

照片覆盖距离起点不超过 \(\dfrac{1}{8}\) 圈的弧。在 \(600\) 秒时，Rachel 已跑 \(6\tfrac{2}{3}\) 圈，距离起点还差 \(30\) 秒；四分之一圈对应 \(22.5\) 秒，所以她在 \(30 - 11.25 = 18.75\) 秒到 \(30 + 11.25 = 41.25\) 秒之间入镜，这是第 \(10\) 分钟内的时间。

在 \(600\) 秒时，Robert 距离起跑线 \(40\) 秒；四分之一圈需要 \(20\) 秒，所以他在 \(30\) 到 \(50\) 秒之间入镜。两人都入镜的时间是 \(30\) 到 \(41.25\) 秒，长度为 \(11.25\) 秒，占 \(60\) 秒的比例为 \(\dfrac{11.25}{60} = \dfrac{3}{16}\)。

所以正确答案是 C。

English

The picture covers the arc within \(\dfrac{1}{8}\) lap of the start on each side. At \(600\) s Rachel has run \(6\tfrac{2}{3}\) laps, \(30\) s short of the line; a quarter lap takes her \(22.5\) s, so she is in view between \(30 - 11.25 = 18.75\) s and \(30 + 11.25 = 41.25\) s of the \(10\)th minute.

At \(600\) s Robert is \(40\) s from the line; a quarter lap takes \(20\) s, so he is in view between \(30\) and \(50\) s. Both appear between \(30\) and \(41.25\) s, a window of \(11.25\) s out of \(60,\) giving probability \(\dfrac{11.25}{60} = \dfrac{3}{16}.\)

Thus, the correct answer is C.

19 答案 E

写成 \[ \begin{aligned} f(n) &= n^4 + 40n^2 + 400 - 400n^2 \\ &= (n^2 + 20)^2 - (20n)^2 \\ &= (n^2 + 20n + 20) \\ &\quad {}\cdot (n^2 - 20n + 20) \end{aligned}\text{。}\]

要使 \(f(n)\) 为质数，较小因子必须为 \(1\)。解 \(n^2 - 20n + 20 = 1\) 得 \((n - 1)(n - 19) = 0\)，所以 \(n = 1\) 或 \(n = 19\)。

此时 \(f(1) = 41\)，\(f(19) = 761\)，二者都是质数，和为 \(802\)。

所以正确答案是 E。

English

Write \[ \begin{aligned} f(n) &= n^4 + 40n^2 + 400 - 400n^2 \\ &= (n^2 + 20)^2 - (20n)^2 \\ &= (n^2 + 20n + 20) \\ &\quad {}\cdot (n^2 - 20n + 20). \end{aligned} \]

For \(f(n)\) to be prime the smaller factor must be \(1\): solving \(n^2 - 20n + 20 = 1\) gives \((n - 1)(n - 19) = 0,\) so \(n = 1\) or \(n = 19.\)

Then \(f(1) = 41\) and \(f(19) = 761\) are both prime, summing to \(802.\)

Thus, the correct answer is E.

20 答案 C

\(100\) 条边中的每一条都在靠近两个端点处各被切一次，所以 \(R\) 有 \(2 \cdot 100 = 200\) 个顶点。

在顶点 \(V_k\) 处的切割会形成一个小多边形，其边数等于 \(V_k\) 的度数。对所有顶点求和得到 \(200\)，即所有边端点的总数。每条原边的中间部分也保留下来，又增加 \(100\) 条边。因此 \(R\) 有 \(200 + 100 = 300\) 条边。

所以正确答案是 C。

English

Each of the \(100\) edges is cut once near each endpoint, so \(R\) has \(2 \cdot 100 = 200\) vertices.

The cut at vertex \(V_k\) creates a small polygon whose number of edges equals the degree of \(V_k\); summed over all vertices this is \(200,\) the total number of edge-endpoints. The middle portion of each original edge also survives, adding \(100\) edges. So \(R\) has \(200 + 100 = 300\) edges.

Thus, the correct answer is C.

21 答案 A

设 \(S_n\) 为 \(n\) 名女性的合法重新入座方式数。最右边的女性要么留在原座位，此时其余人有 \(S_{n-1}\) 种方式；要么与左边的邻座交换（这是填满最右端座位的唯一其他方式），此时剩下 \(S_{n-2}\) 种方式。

因此 \(S_n = S_{n-1} + S_{n-2}\)，且 \(S_1 = 1\)、\(S_2 = 2\)，得到斐波那契数列 \(1, 2, 3, 5, 8, 13, 21, 34, 55, 89\)。所以 \(S_{10} = 89\)。

所以正确答案是 A。

English

Let \(S_n\) be the number of valid reseatings of \(n\) women. The rightmost woman either keeps her seat, leaving \(S_{n-1}\) ways for the rest, or swaps with her left neighbor — the only other way to fill the end seat — leaving \(S_{n-2}\) ways.

Thus \(S_n = S_{n-1} + S_{n-2}\) with \(S_1 = 1\) and \(S_2 = 2,\) giving the Fibonacci values \(1, 2, 3, 5, 8, 13, 21, 34, 55, 89.\) So \(S_{10} = 89.\)

Thus, the correct answer is A.

22 答案 C

设 \(B = (b, b)\)、\(D = (d, kd)\)，其中 \(b, d, k\) 为正整数且 \(k \gt 1\)。面积条件给出 \((k - 1)bd = 1{,}000{,}000 = 2^6 \cdot 5^6\)。

每个平行四边形对应一个正整数有序三元组 \((k - 1, b, d)\)，其乘积为 \(2^6 \cdot 5^6\)。六个因子 \(2\) 在三个位置中的分配数为 \(\binom{6 + 2}{2} = 28\)，因子 \(5\) 也有 \(28\) 种分配，所以共有 \(28^2 = 784\) 个。

所以正确答案是 C。

English

Let \(B = (b, b)\) and \(D = (d, kd)\) with \(b, d, k\) positive integers and \(k \gt 1.\) The area is \((k - 1)bd = 1{,}000{,}000 = 2^6 \cdot 5^6.\)

Each parallelogram corresponds to an ordered triple \((k - 1, b, d)\) of positive integers with product \(2^6 \cdot 5^6.\) The six \(2\)’s distribute among the three factors in \(\binom{6 + 2}{2} = 28\) ways, and likewise the six \(5\)’s in \(28\) ways, giving \(28^2 = 784.\)

Thus, the correct answer is C.

23 答案 D

展开得 \(\left(\dfrac{3}{4} + \dfrac{3}{4}i\right)(x + iy)\) \(= \dfrac{3}{4}(x - y)\) \(+ \dfrac{3}{4}(x + y)i\)。实部和虚部都落在 \([-1, 1]\) 中，当且仅当 \(|x - y| \le \dfrac{4}{3}\) 且 \(|x + y| \le \dfrac{4}{3}\)。

在正方形 \(S\)（面积 \(4\)）内，不满足条件的区域只出现在四个角。靠近 \((1, 1)\) 的地方，直线 \(x + y = \dfrac{4}{3}\) 切去一个直角三角形，两条直角边均为 \(\dfrac{2}{3}\)，面积为 \(\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{2}{3} = \dfrac{2}{9}\)。

四个角共去掉 \(4 \cdot \dfrac{2}{9} = \dfrac{8}{9}\)，剩余面积为 \(4 - \dfrac{8}{9} = \dfrac{28}{9}\)，概率为 \(\dfrac{\frac{28}{9}}{4} = \dfrac{7}{9}\)。

所以正确答案是 D。

English

Expanding, \(\left(\dfrac{3}{4} + \dfrac{3}{4}i\right)(x + iy)\) \(= \dfrac{3}{4}(x - y)\) \(+ \dfrac{3}{4}(x + y)i.\) Both parts lie in \([-1, 1]\) iff \(|x - y| \le \dfrac{4}{3}\) and \(|x + y| \le \dfrac{4}{3}.\)

Within the square \(S\) (area \(4\)) these fail only in four corner triangles. Near \((1, 1),\) the line \(x + y = \dfrac{4}{3}\) cuts off a right triangle with legs \(\dfrac{2}{3},\) area \(\dfrac{1}{2} \cdot \dfrac{2}{3} \cdot \dfrac{2}{3} = \dfrac{2}{9}.\)

The four corners remove \(4 \cdot \dfrac{2}{9} = \dfrac{8}{9},\) leaving \(4 - \dfrac{8}{9} = \dfrac{28}{9}.\) The probability is \(\dfrac{\frac{28}{9}}{4} = \dfrac{7}{9}.\)

Thus, the correct answer is D.

24 答案 B

在 \([0, \pi],\ \cos^{-1}(\cos x) = x\)。因为 \(\sin^{-1}\) 的值域是 \([-\tfrac{\pi}{2}, \tfrac{\pi}{2}]\)，任何解都必须满足 \(x \in [0, \tfrac{\pi}{2}]\)，此时方程化为 \(\sin 6x = \sin x\)。

方程 \(\sin6x=\sin x\) 成立，当且仅当 \[ \begin{aligned} 6x&=x+2k\pi, \\ \text{或}\qquad 6x&=\pi-x+2k\pi \end{aligned} \] 对某个整数 \(k\) 成立。

第一族给出 \(x=\tfrac{2k\pi}{5}\)，在区间中贡献 \(0\) 和 \(\tfrac{2\pi}{5}\)。第二族给出 \(x=\tfrac{(2k+1)\pi}{7}\)，贡献 \(\tfrac\pi7\) 和 \(\tfrac{3\pi}{7}\)。因此在 \([0,\tfrac\pi2]\) 中共有 \(4\) 个解。

所以正确答案是 B。

English

On \([0, \pi],\ \cos^{-1}(\cos x) = x.\) Since \(\sin^{-1}\) takes values in \([-\tfrac{\pi}{2}, \tfrac{\pi}{2}],\) any solution requires \(x \in [0, \tfrac{\pi}{2}],\) where the equation becomes \(\sin 6x = \sin x.\)

The equation \(\sin6x=\sin x\) holds exactly when \[ \begin{aligned} 6x&=x+2k\pi, \\ \text{or}\qquad 6x&=\pi-x+2k\pi \end{aligned} \] for some integer \(k.\)

The first family gives \(x=\tfrac{2k\pi}{5},\) contributing \(0\) and \(\tfrac{2\pi}{5}.\) The second gives \(x=\tfrac{(2k+1)\pi}{7},\) contributing \(\tfrac\pi7\) and \(\tfrac{3\pi}{7}.\) Hence there are \(4\) solutions in \([0,\tfrac\pi2].\)

Thus, the correct answer is B.

25 答案 E

\(G\) 由四个 \(5 \times 5\) 方块 \(G_1, \ldots, G_4\) 组成，每个象限一个。任何边长 \(\ge 6\) 的正方形在每个方块中恰好使用一个顶点，因为同一方块中的两点距离小于 \(6\)，而不同方块中的点之间距离至少为 \(6\)。

将每个方块按 \((\pm 5, \pm 5)\) 向内平移，可把它们叠合成同一个 \(5 \times 5\) 网格 \(G'\)（即满足 \(|x|, |y| \le 2\) 的那些点）。每个这样的正方形要么对应 \(G'\) 中的一个点，要么对应 \(G'\) 中的一个正方形。所以所求个数等于 \(G'\) 的点数，加上 \(4\) 倍的以 \(G'\) 中的点为顶点的正方形个数。

该网格中与坐标轴平行的正方形共有 \(4^2+3^2+2^2+1^2=30\) 个。对于倾斜的正方形，设它的一条边水平移动 \(a\) 个单位、竖直移动 \(b\) 个单位，其中 \(a,b\gt0\) 且 \(a+b\le4\)。对每个有序数对 \((a,b)\)，有 \((5-a-b)^2\) 种放置方式。\(a+b=2,3,4\) 对应的个数分别为 \(9,8,3\)，于是倾斜的正方形有 \(20\) 个，正方形总共有 \(50\) 个。因此所求的个数为 \(25+4\cdot50=225\)。

所以正确答案是 E。

English

\(G\) consists of four \(5 \times 5\) blocks \(G_1, \ldots, G_4,\) one in each quadrant. Any square of side \(\ge 6\) uses exactly one vertex in each block, since two points in one block are less than \(6\) apart while points in different blocks are at least \(6\) apart.

Sliding each block inward by \((\pm 5, \pm 5)\) superimposes them on one \(5 \times 5\) grid \(G'\) (points with \(|x|, |y| \le 2\)). Each such square maps to either a single point of \(G'\) or a square in \(G'.\) So the count equals the number of points of \(G'\) plus \(4\) times the number of squares with vertices in \(G'.\)

The grid has \(4^2+3^2+2^2+1^2=30\) axis-parallel squares. For a tilted square, let one side move \(a\) units horizontally and \(b\) units vertically, where \(a,b\gt0\) and \(a+b\le4.\) For each ordered pair \((a,b),\) there are \((5-a-b)^2\) placements. The totals for \(a+b=2,3,4\) are \(9,8,3,\) respectively, giving \(20\) tilted squares and \(50\) squares altogether. Therefore the required count is \(25+4\cdot50=225.\)

Thus, the correct answer is E.