2006B AMC 12 真题（P版）

1 答案 C

因为奇数 \(k\) 时 \((-1)^k = -1\)，偶数 \(k\) 时 \((-1)^k = 1\)，各项交替为 \(-1, 1, -1, 1, \ldots\)。

共有 \(2006\) 项，可分成 \(1003\) 对，每对都是 \((-1) + 1 = 0\)。总和为 \(0\)。

因此，正确答案是 C。

English

Since \((-1)^k = -1\) for odd \(k\) and \((-1)^k = 1\) for even \(k,\) the terms alternate \(-1, 1, -1, 1, \ldots\)

There are \(2006\) terms, forming \(1003\) pairs, each equal to \((-1) + 1 = 0.\) The total is \(0.\)

Thus, the correct answer is C.

2 答案 A

因为 \(x \spadesuit y = x^2 - y^2\)，里面的值为 \(4 \spadesuit 5 = 16 - 25 = -9\)。

于是 \(3 \spadesuit (-9) = 3^2 - (-9)^2\) \(= 9 - 81 = -72\)。

因此，正确答案是 A。

English

Since \(x \spadesuit y = x^2 - y^2,\) the inner value is \(4 \spadesuit 5 = 16 - 25 = -9.\)

Then \(3 \spadesuit (-9) = 3^2 - (-9)^2\) \(= 9 - 81 = -72.\)

Thus, the correct answer is A.

3 答案 A

设 Cougars 和 Panthers 的得分分别为 \(c\) 和 \(p\)，则 \(c + p = 34\)，且 \(c - p = 14\)。

相减得 \(2p = 20\)，所以 \(p = 10\)。

因此，正确答案是 A。

English

Let \(c\) and \(p\) be the Cougars’ and Panthers’ scores. Then \(c + p = 34\) and \(c - p = 14.\)

Subtracting gives \(2p = 20,\) so \(p = 10.\)

Thus, the correct answer is A.

4 答案 A

商品价格总计 \(\$18.95\)，所以找零为 \(\$1.05\)。它占 \(\$20.00\) 的百分比为 \[ \frac{1.05}{20}\cdot100\% = 5.25\%\text{。}\]最接近的选项是 \(5\%\)。

所以正确答案是 A。

English

The prices total \(\$18.95,\) so the change is \(\$1.05.\) As a percentage of \(\$20.00,\) this is \[ \frac{1.05}{20}\cdot100\% = 5.25\%. \] The closest listed percentage is \(5\%.\)

Thus, the correct answer is A.

5 答案 A

Bob 以每小时 \(5 - 3 = 2\) 英里的相对速度缩短距离。追平 \(1\) 英里的差距需要 \[\frac{1}{2}\] 小时，也就是 \(30\) 分钟。

因此，正确答案是 A。

English

Bob closes the gap at a relative speed of \(5 - 3 = 2\) miles per hour. To cover the \(1\)-mile gap takes \[\frac{1}{2}\] hour, or \(30\) minutes.

Thus, the correct answer is A.

6 答案 B

整批柠檬水重 \(100 + 100 + 400 = 600\) 克，含有 \(25 + 386 = 411\) 卡路里。

因为 \(200\) 克是整批的三分之一，所以它含有 \[\frac{411}{3} = 137\] 卡路里。

因此，正确答案是 B。

English

The full batch weighs \(100 + 100 + 400 = 600\) grams and contains \(25 + 386 = 411\) calories.

Since \(200\) grams is one third of the batch, it has \[\frac{411}{3} = 137\] calories.

Thus, the correct answer is B.

7 答案 B

驾驶员是两位父母之一：有 \(2\) 种选择。

剩下的 \(3\) 人中任意一人可以坐副驾驶座，最后 \(2\) 人坐后排，有 \(2\) 种顺序。

总数为 \(2 \cdot 3 \cdot 2 = 12\)。

因此，正确答案是 B。

English

The driver is one of the two parents: \(2\) choices.

Any of the remaining \(3\) people can sit in the front passenger seat, and the last \(2\) people fill the back in \(2\) orders.

The total is \(2 \cdot 3 \cdot 2 = 12.\)

Thus, the correct answer is B.

8 答案 E

代入 \((1, 2)\)，得到 \[1 = \frac{2}{4} + a \quad\Rightarrow\quad a = \frac{1}{2}\text{，}\] 且 \[2 = \frac{1}{4} + b \quad\Rightarrow\quad b = \frac{7}{4}\text{。}\]

因此 \[a + b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4}\text{。}\]

因此，正确答案是 E。

English

Substituting \((1, 2)\) gives \[1 = \frac{2}{4} + a \quad\Rightarrow\quad a = \frac{1}{2},\] and \[2 = \frac{1}{4} + b \quad\Rightarrow\quad b = \frac{7}{4}.\]

Therefore \[a + b = \frac{1}{2} + \frac{7}{4} = \frac{9}{4}.\]

Thus, the correct answer is E.

9 答案 B

设三个数字为 \(a \lt b \lt c\)，且 \(c\) 为偶数。因为 \(a \geq 1\)，没有数字为零，并且 \(c \neq 2\)（没有两个更小的非零数字可选）。

一旦个位数字 \(c\) 固定，任意两个小于它的不同数字都能按递增顺序唯一排列。因此每个 \(c\) 的计数是 \(\binom{c-1}{2}\)。

对 \(c = 4, 6, 8\) 分别计数，得到 \[ \begin{aligned} &\binom{3}{2} + \binom{5}{2} \\ &\quad {}+ \binom{7}{2} = 3 + 10 + 21 \\ &\quad = 34 \end{aligned}\text{。}\]

因此，正确答案是 B。

English

Let the digits be \(a \lt b \lt c\) with \(c\) even. Since \(a \geq 1,\) no digit is zero, and \(c \neq 2\) (there is no room for two smaller nonzero digits).

Once the units digit \(c\) is fixed, any two distinct digits below it can be arranged in increasing order in exactly one way. So the count for each \(c\) is \(\binom{c-1}{2}.\)

For \(c = 4, 6, 8\) this gives \[ \begin{aligned} &\binom{3}{2} + \binom{5}{2} \\ &\quad {}+ \binom{7}{2} = 3 + 10 + 21 \\ &\quad = 34. \end{aligned} \]

Thus, the correct answer is B.

10 答案 A

设三边为 \(x\)、\(3x\) 和 \(15\)。三角形不等式要求 \(x + 3x \gt 15\)，所以 \(x \geq 4\)；且 \(x + 15 \gt 3x\)，所以 \(x \leq 7\)。

周长 \(4x + 15\) 在 \(x = 7\) 时最大，得到 \(7 + 21 + 15 = 43\)。

因此，正确答案是 A。

English

Let the sides be \(x,\) \(3x,\) and \(15.\) The triangle inequality requires \(x + 3x \gt 15,\) so \(x \geq 4,\) and \(x + 15 \gt 3x,\) so \(x \leq 7.\)

The perimeter \(4x + 15\) is largest when \(x = 7,\) giving \(7 + 21 + 15 = 43.\)

Thus, the correct answer is A.

11 答案 E

Joe 最后加入奶油，所以杯中保留了全部 \(2\) 盎司奶油。

JoAnn 的杯中有 \(14\) 盎司混合物，其中含 \(2\) 盎司奶油。喝掉 \(2\) 盎司会按比例移走 \(\tfrac{2}{14}\) 的所有成分，剩下的奶油为 \[2 \cdot \frac{12}{14} = \frac{12}{7}\] 盎司。

比值为 \[\frac{2}{\,\frac{12}{7}\,} = \frac{14}{12} = \frac{7}{6}\text{。}\]

因此，正确答案是 E。

English

Joe adds the cream last, so his cup holds all \(2\) ounces of cream.

JoAnn’s cup has \(14\) ounces of mixture containing \(2\) ounces of cream. Drinking \(2\) ounces removes a fraction \(\tfrac{2}{14}\) of everything, leaving \[2 \cdot \frac{12}{14} = \frac{12}{7}\] ounces of cream.

The ratio is \[\frac{2}{\,\frac{12}{7}\,} = \frac{14}{12} = \frac{7}{6}.\]

Thus, the correct answer is E.

12 答案 D

顶点式为 \(y = a(x - p)^2 + p\)。

当 \(x = 0\) 时，\(y = ap^2 + p = -p\)，所以 \(ap^2 = -2p\)，且 \(a = -\dfrac{2}{p}\)。

展开得 \(y = a x^2 - 2ap\, x + ap^2 + p\)，所以 \(b = -2ap = -2\left(-\dfrac{2}{p}\right)p = 4\)。

因此，正确答案是 D。

English

The vertex form is \(y = a(x - p)^2 + p.\)

At \(x = 0,\) \(y = ap^2 + p = -p,\) so \(ap^2 = -2p\) and \(a = -\dfrac{2}{p}.\)

Expanding, \(y = a x^2 - 2ap\, x + ap^2 + p,\) so \(b = -2ap = -2\left(-\dfrac{2}{p}\right)p = 4.\)

Thus, the correct answer is D.

13 答案 C

设菱形 \(ABCD\) 的两条对角线交于 \(O\)。它们互相垂直平分；又因为 \(\angle BAD=60^\circ\)，所以三角形 \(ABO\) 是 \(30^\circ\)–\(60^\circ\)–\(90^\circ\) 三角形。因此两条半对角线满足 \(AO=\sqrt3\,BO\)。

线段 \(BD\) 是 \(ABCD\) 的短对角线，也是相似菱形 \(BFDE\) 的长对角线。因此小菱形与大菱形的长度之比为 \(\frac{BD}{AC}=\frac{BO}{AO}=\frac{1}{\sqrt3}\)。面积按这个比的平方缩放，所以 \(BFDE\) 的面积为 \(24\cdot\dfrac13=8\)。

因此，正确答案是 C。

English

Let the diagonals of \(ABCD\) meet at \(O.\) They bisect each other at right angles, and since \(\angle BAD=60^\circ,\) triangle \(ABO\) is a \(30^\circ\)–\(60^\circ\)–\(90^\circ\) triangle. Hence the half-diagonals satisfy \(AO=\sqrt3\,BO.\)

The segment \(BD\) is the short diagonal of \(ABCD\) and the long diagonal of the similar rhombus \(BFDE.\) Thus the smaller-to-larger length ratio is \(\frac{BD}{AC}=\frac{BO}{AO}=\frac{1}{\sqrt3}.\) Areas scale by the square of this ratio, so the area of \(BFDE\) is \(24\cdot\dfrac13=8.\)

Thus, the correct answer is C.

14 答案 D

以美分计，总成本为 \(N(4B + 5J) = 253 = 11 \cdot 23\)。因为 \(N \gt 1\)，\(N\) 的可能值为 \(11\)、\(23\) 或 \(253\)。

若 \(N = 253\) 则 \(4B + 5J = 1\)；若 \(N = 23\) 则 \(4B + 5J = 11\)；二者都没有正整数解。

所以 \(N = 11\)，且 \(4B + 5J = 23\)，唯一的正整数解为 \(B = 2\)、\(J = 3\)。

果酱成本为 \(N \cdot J \cdot 5\)¢ \(= 11 \cdot 3 \cdot 5 = 165\) 美分，即 \(\$1.65\)。

因此，正确答案是 D。

English

The total cost in cents is \(N(4B + 5J) = 253 = 11 \cdot 23.\) Since \(N \gt 1,\) the value of \(N\) is \(11,\) \(23,\) or \(253.\)

If \(N = 253\) then \(4B + 5J = 1,\) and if \(N = 23\) then \(4B + 5J = 11;\) neither has a positive integer solution.

So \(N = 11\) and \(4B + 5J = 23,\) whose only positive solution is \(B = 2,\) \(J = 3.\)

The jam costs \(N \cdot J \cdot 5\)¢ \(= 11 \cdot 3 \cdot 5 = 165\) cents, or \(\$1.65.\)

Thus, the correct answer is D.

15 答案 B

两圆外切，所以 \(OP = 2 + 4 = 6\)。在四边形 \(AOPD\) 中，\(OA = 2\) 和 \(PD = 4\) 都垂直于切线 \(AD\)，因此它是直角梯形。

过 \(O\) 作平行于 \(AD\) 的直线，可形成一个直角三角形，斜边为 \(OP = 6\)，一条直角边为 \(PD - OA = 2\)，所以 \(AD = \sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt2\)。

梯形 \(AOPD\) 的面积为 \[\frac{1}{2}(2 + 4)(4\sqrt2) = 12\sqrt2\text{。}\]

由对称性，六边形 \(AOBCPD\) 由两个这样的梯形组成，所以面积为 \(2 \cdot 12\sqrt2 = 24\sqrt2\)。

因此，正确答案是 B。

English

The circles are externally tangent, so \(OP = 2 + 4 = 6.\) In quadrilateral \(AOPD,\) both \(OA = 2\) and \(PD = 4\) are perpendicular to the tangent line \(AD,\) making it a right trapezoid.

Drawing the line through \(O\) parallel to \(AD\) creates a right triangle with hypotenuse \(OP = 6\) and one leg \(PD - OA = 2,\) so \(AD = \sqrt{6^2 - 2^2} = \sqrt{32} = 4\sqrt2.\)

The trapezoid \(AOPD\) has area \[\frac{1}{2}(2 + 4)(4\sqrt2) = 12\sqrt2.\]

By symmetry the hexagon \(AOBCPD\) is made of two such trapezoids, so its area is \(2 \cdot 12\sqrt2 = 24\sqrt2.\)

Thus, the correct answer is B.

16 答案 C

距离 \(AC = \sqrt{7^2 + 1^2} = \sqrt{50}\)。正六边形边长为 \(s\) 时，相隔两个顶点的距离为 \(s\sqrt3\)，所以 \(s^2 \cdot 3 = 50\)，即 \(s^2 = \dfrac{50}{3}\)。

六边形面积为 \[\frac{3\sqrt3}{2}s^2 = \frac{3\sqrt3}{2} \cdot \frac{50}{3} = 25\sqrt3\text{。}\]

因此，正确答案是 C。

English

The distance is \(AC = \sqrt{7^2 + 1^2} = \sqrt{50}.\) In a regular hexagon with side \(s,\) the distance between vertices two apart is \(s\sqrt3,\) so \(s^2 \cdot 3 = 50,\) giving \(s^2 = \dfrac{50}{3}.\)

The hexagon’s area is \[\frac{3\sqrt3}{2}s^2 = \frac{3\sqrt3}{2} \cdot \frac{50}{3} = 25\sqrt3.\]

Thus, the correct answer is C.

17 答案 C

因为权重之和为 \(21\)，掷出 \(k\) 的概率为 \(\dfrac{k}{21}\)。

总和为 \(7\) 来自 \((1,6), (2,5), \ldots, (6,1)\)，所以概率为 \[ \begin{aligned} &\scriptsize \frac{1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1}{21^2} \\ &= \frac{56}{441} = \frac{8}{63} \end{aligned}\text{。}\]

因此，正确答案是 C。

English

Since the weights sum to \(21,\) the probability of rolling \(k\) is \(\dfrac{k}{21}.\)

A total of \(7\) comes from \((1,6), (2,5), \ldots, (6,1),\) so the probability is \[ \begin{aligned} &\scriptsize \frac{1 \cdot 6 + 2 \cdot 5 + 3 \cdot 4 + 4 \cdot 3 + 5 \cdot 2 + 6 \cdot 1}{21^2} \\ &= \frac{56}{441} = \frac{8}{63}. \end{aligned} \]

Thus, the correct answer is C.

18 答案 B

每一步都会使坐标和改变 \(1\)，所以 \(10\) 步后终点 \((a, b)\) 满足 \(a + b\) 为偶数，且 \(|a| + |b| \le 10\)。任何这样的点都可到达：先走 \(|a| + |b|\) 步到达它，再用剩下的偶数步往返。

可到达的点落在直线 \(a + b = 2k\) 上，其中 \(-5 \le k \le 5\)。每条这样的直线与该菱形恰有 \(11\) 个格点。

共有 \(11\) 条直线，每条 \(11\) 个点，因此共有 \(121\) 个点。

因此，正确答案是 B。

English

Each step changes the coordinate sum by \(1,\) so after \(10\) steps the endpoint \((a, b)\) has \(a + b\) even, and \(|a| + |b| \le 10.\) Any such point is reachable: walk \(|a| + |b|\) steps to it, then use the remaining even number of steps going out and back.

The reachable points lie on the lines \(a + b = 2k\) for \(-5 \le k \le 5.\) Each such line meets the diamond in exactly \(11\) lattice points.

With \(11\) lines and \(11\) points each, there are \(121\) points.

Thus, the correct answer is B.

19 答案 B

这个数的形式为 \(aabb\)，\(abab\)，或 \(baab\)。能被 \(9\) 整除说明 \(2(a + b)\) 是 \(9\) 的倍数，所以 \(a + b = 9\)。

八个不同年龄是从 \(1\) 到 \(9\) 的九个整数中的八个，所以年龄 \(4\) 和 \(8\) 中至少出现一个。因此这个数能被 \(4\) 整除。可能性变为 \(1188, 2772, 3636\)，\(5544, 6336, 7272, 9900\)。

因为末两位是 Jones 先生的年龄，\(9900\) 不可能，其余数也都不是 \(5\) 的倍数。所以孩子们的年龄不能包含 \(5\)。事实上，\(5544\) 能被 \(1, 2, 3, 4, 6, 7, 8, 9\) 整除。

所以正确答案是 B。

English

The number has the form \(aabb,\) \(abab,\) or \(baab.\) Divisibility by \(9\) means \(2(a + b)\) is a multiple of \(9,\) so \(a + b = 9.\)

The eight distinct ages are eight of the nine integers from \(1\) through \(9,\) so at least one of ages \(4\) and \(8\) occurs. Therefore the number is divisible by \(4.\) The possibilities become \(1188, 2772, 3636,\) \(5544, 6336, 7272, 9900.\)

Since the last two digits are Mr. Jones’s age, \(9900\) is impossible, and none of the others is a multiple of \(5.\) So the children’s ages cannot include \(5.\) Indeed \(5544\) is divisible by \(1, 2, 3, 4, 6, 7, 8, 9.\)

Thus, the correct answer is B.

20 答案 C

方程表示 \(\lfloor \log_{10} x \rfloor = \lfloor \log_{10} 4x \rfloor\)，也就是 \(x\) 和 \(4x\) 位于同一区间 \([10^n, 10^{n+1})\)。

这恰好在 \(10^n \le x\) 且 \(4x \lt 10^{n+1}\) 时成立，即 \(10^n \le x \lt \dfrac{10^{n+1}}{4}\)。

在 \([10^n, 10^{n+1})\) 内，有利部分所占比例为 \[ \begin{aligned} &\frac{\frac{10^{n+1}}{4} - 10^n}{10^{n+1} - 10^n} \\ &= \frac{\frac{10}{4} - 1}{10 - 1} = \frac{1}{6} \end{aligned}\text{。}\]

因为这个比例在每个这样的区间上都相同，所以总概率为 \(\dfrac{1}{6}\)。

因此，正确答案是 C。

English

The equation says \(\lfloor \log_{10} x \rfloor = \lfloor \log_{10} 4x \rfloor,\) i.e. \(x\) and \(4x\) lie in the same interval \([10^n, 10^{n+1}).\)

This holds exactly when \(10^n \le x\) and \(4x \lt 10^{n+1},\) that is \(10^n \le x \lt \dfrac{10^{n+1}}{4}.\)

Within \([10^n, 10^{n+1}),\) the favorable fraction is \[ \begin{aligned} &\frac{\frac{10^{n+1}}{4} - 10^n}{10^{n+1} - 10^n} \\ &= \frac{\frac{10}{4} - 1}{10 - 1} = \frac{1}{6}. \end{aligned} \]

Since this fraction is the same on every such interval, the overall probability is \(\dfrac{1}{6}.\)

Thus, the correct answer is C.

21 答案 C

设矩形边长为 \(x\) 和 \(y\)。点 \(A\) 在以 \(B\) 和 \(D\) 为焦点的椭圆上，所以 \(x + y = AB + AD = 2a\)。两个焦点之间的距离是矩形对角线，所以 \(\sqrt{x^2 + y^2} = 2\sqrt{a^2 - b^2}\)。

因此 \(2xy = (x + y)^2 - (x^2 + y^2)\) \(= 4a^2 - (4a^2 - 4b^2)\) \(= 4b^2\)，所以 \(xy = 2b^2\)。由矩形面积可得 \(2b^2 = 2006\)，从而 \(b^2 = 1003\)。

椭圆面积给出 \(\pi ab = 2006\pi\)，所以 \(ab = 2006\)，且 \(a = \dfrac{2006}{\sqrt{1003}} = 2\sqrt{1003}\)。

矩形周长为 \(2(x + y) = 4a = 8\sqrt{1003}\)。

因此，正确答案是 C。

English

Let the rectangle’s sides be \(x\) and \(y.\) Point \(A\) is on the ellipse with foci \(B\) and \(D,\) so \(x + y = AB + AD = 2a.\) The distance between the foci is the diagonal, so \(\sqrt{x^2 + y^2} = 2\sqrt{a^2 - b^2}.\)

Then \(2xy = (x + y)^2 - (x^2 + y^2)\) \(= 4a^2 - (4a^2 - 4b^2)\) \(= 4b^2,\) so \(xy = 2b^2.\) The area gives \(2b^2 = 2006,\) hence \(b^2 = 1003.\)

The ellipse area gives \(\pi ab = 2006\pi,\) so \(ab = 2006\) and \(a = \dfrac{2006}{\sqrt{1003}} = 2\sqrt{1003}.\)

The perimeter is \(2(x + y) = 4a = 8\sqrt{1003}.\)

Thus, the correct answer is C.

22 答案 B

因子 \(2\) 比因子 \(5\) 充足，所以 \(n\) 等于 \(a!\,b!\,c!\) 中因子 \(5\) 的个数，即 \[n = \sum_{k \ge 1}\left(\left\lfloor \tfrac{a}{5^k}\right\rfloor + \left\lfloor \tfrac{b}{5^k}\right\rfloor + \left\lfloor \tfrac{c}{5^k}\right\rfloor\right)\text{。}\]

对每个 \(k\)，\(\lfloor \frac{a}{5^k} \rfloor + \lfloor \frac{b}{5^k} \rfloor\) \(+ \lfloor \frac{c}{5^k} \rfloor\) \(\ge \lfloor \frac{2006}{5^k} \rfloor - 2\)。对 \(k = 1, 2, 3, 4\) 求和（因为 \(2006 \lt 5^5\)），得到 \[ \begin{aligned} &n \ge (401 + 80 + 16 + 3) \\ &\quad {}- 4 \cdot 2 = 492 \end{aligned}\text{。}\]

等号可以达到，例如取 \(a = b = 624\)、\(c = 758\)。所以最小值为 \(492\)。

因此，正确答案是 B。

English

Since factors of \(2\) are more plentiful than factors of \(5,\) \(n\) equals the number of factors of \(5\) in \(a!\,b!\,c!,\) namely \[n = \sum_{k \ge 1}\left(\left\lfloor \tfrac{a}{5^k}\right\rfloor + \left\lfloor \tfrac{b}{5^k}\right\rfloor + \left\lfloor \tfrac{c}{5^k}\right\rfloor\right).\]

For each \(k,\) \(\lfloor \frac{a}{5^k} \rfloor + \lfloor \frac{b}{5^k} \rfloor\) \(+ \lfloor \frac{c}{5^k} \rfloor\) \(\ge \lfloor \frac{2006}{5^k} \rfloor - 2.\) Summing over \(k = 1, 2, 3, 4\) (as \(2006 \lt 5^5\)) gives \[ \begin{aligned} &n \ge (401 + 80 + 16 + 3) \\ &\quad {}- 4 \cdot 2 = 492. \end{aligned} \]

Equality is attainable, for example with \(a = b = 624\) and \(c = 758.\) So the minimum is \(492.\)

Thus, the correct answer is B.

23 答案 E

将 \(\triangle ABC\) 绕 \(C\) 旋转 \(90^\circ\)，使 \(A\) 映到 \(B\)，\(P\) 映到 \(P'\)。则 \(CP' = CP = 6\)，且 \(\angle PCP' = 90^\circ\)，所以 \(\triangle PCP'\) 是等腰直角三角形，\(PP' = 6\sqrt2\)。

另外 \(BP' = AP = 11\)。因为 \((6\sqrt2)^2 + 7^2 = 72 + 49\) \(= 121 = 11^2\)，三角形 \(BPP'\) 在 \(P\) 处为直角。因此 \(\angle BPC = \angle BPP'\) \(+ \angle P'PC = 90^\circ\) \(+ 45^\circ = 135^\circ\)。

在 \(\triangle BPC\) 中使用余弦定理：\[ \begin{aligned} &BC^2 = 6^2 + 7^2 \\ &\quad {}- 2 \cdot 6 \cdot 7 \cos 135^\circ \\ &\quad = 85 + 42\sqrt2 \end{aligned}\text{。}\]

所以 \(s^2 = 85 + 42\sqrt2\)，得到 \(a = 85\)、\(b = 42\)，因此 \(a + b = 127\)。

因此，正确答案是 E。

English

Rotate \(\triangle ABC\) by \(90^\circ\) about \(C,\) sending \(A\) to \(B\) and \(P\) to \(P'.\) Then \(CP' = CP = 6\) and \(\angle PCP' = 90^\circ,\) so \(\triangle PCP'\) is an isosceles right triangle with \(PP' = 6\sqrt2.\)

Also \(BP' = AP = 11.\) Since \((6\sqrt2)^2 + 7^2 = 72 + 49\) \(= 121 = 11^2,\) triangle \(BPP'\) has a right angle at \(P.\) Hence \(\angle BPC = \angle BPP'\) \(+ \angle P'PC = 90^\circ\) \(+ 45^\circ = 135^\circ.\)

By the Law of Cosines in \(\triangle BPC,\) \[ \begin{aligned} &BC^2 = 6^2 + 7^2 \\ &\quad {}- 2 \cdot 6 \cdot 7 \cos 135^\circ \\ &\quad = 85 + 42\sqrt2. \end{aligned} \]

So \(s^2 = 85 + 42\sqrt2,\) giving \(a = 85,\) \(b = 42,\) and \(a + b = 127.\)

Thus, the correct answer is E.

24 答案 C

固定 \(y\)，把 \(\sin^2 x - \sin x \sin y + \sin^2 y = \dfrac34\) 看作关于 \(\sin x\) 的二次方程：\[ \begin{aligned} &\sin x = \frac{1}{2}\sin y \\ &\quad {}\pm \frac{\sqrt3}{2}\cos y \\ &\quad = \sin\!\left(y \pm \frac{\pi}{3}\right) \end{aligned}\text{。}\]

在 \(S\) 内，\(\sin x = \sin\!\left(y - \tfrac{\pi}{3}\right)\) 给出直线 \(x = y - \tfrac{\pi}{3}\)；而 \(\sin x = \sin\!\left(y + \tfrac{\pi}{3}\right)\) 在 \(y \le \tfrac{\pi}{6}\) 时给出 \(x = y + \tfrac{\pi}{3}\)，在 \(y \ge \tfrac{\pi}{6}\) 时给出 \(x = -y + \tfrac{2\pi}{3}\)。

这些直线把 \(S\) 分成若干区域；检验角点可知不等式只在中间带状区域成立。其面积为 \[ \begin{aligned} &\left(\frac{\pi}{2}\right)^2 - \frac{1}{2}\left(\frac{\pi}{3}\right)^2 \\ &\quad {}- 2 \cdot \frac{1}{2}\left(\frac{\pi}{6}\right)^2 \\ &\quad = \frac{\pi^2}{6} \end{aligned}\text{。}\]

因此，正确答案是 C。

English

Fixing \(y,\) solve \(\sin^2 x - \sin x \sin y + \sin^2 y = \dfrac34\) as a quadratic in \(\sin x:\) \[ \begin{aligned} &\sin x = \frac{1}{2}\sin y \\ &\quad {}\pm \frac{\sqrt3}{2}\cos y \\ &\quad = \sin\!\left(y \pm \frac{\pi}{3}\right). \end{aligned} \]

Within \(S,\) \(\sin x = \sin\!\left(y - \tfrac{\pi}{3}\right)\) gives the line \(x = y - \tfrac{\pi}{3},\) while \(\sin x = \sin\!\left(y + \tfrac{\pi}{3}\right)\) gives \(x = y + \tfrac{\pi}{3}\) for \(y \le \tfrac{\pi}{6}\) and \(x = -y + \tfrac{2\pi}{3}\) for \(y \ge \tfrac{\pi}{6}.\)

These lines split \(S\) into regions; testing the corners shows the inequality holds only in the middle band. Its area is \[ \begin{aligned} &\left(\frac{\pi}{2}\right)^2 - \frac{1}{2}\left(\frac{\pi}{3}\right)^2 \\ &\quad {}- 2 \cdot \frac{1}{2}\left(\frac{\pi}{6}\right)^2 \\ &\quad = \frac{\pi^2}{6}. \end{aligned} \]

Thus, the correct answer is C.

25 答案 B

递推规则给出 \(a_n \equiv a_{n+3} \pmod 2\)，所以 \(a_2\) 与 \(a_{2006} = 1\) 的奇偶性相同，因此 \(a_2\) 是奇数。

每一项都是 \(\gcd(a_1, a_2)\) 的倍数，而 \(a_{2006} = 1\) 迫使 \(\gcd(999, a_2) = 1\)。因为 \(999 = 3^3 \cdot 37\)，所以 \(a_2\) 不能被 \(3\) 或 \(37\) 整除。

区间 \([1, 998]\) 中有 \(499\) 个奇数；去掉 \(166\) 个 \(3\) 的倍数和 \(13\) 个 \(37\) 的倍数，再加回 \(4\) 个 \(111\) 的倍数，得到 \[499 - 166 - 13 + 4 = 324\text{。}\]

每个这样的 \(a_2\) 都可行。对连续正项 \(u,v\)，更新 \((u,v)\mapsto(v,|v-u|)\) 至多每两步就会减小两者的最大值。因为初始两项都不超过 \(999\)，所以某个 \(a_N=0\) 会在 \(N\le1999\) 时出现。

每对连续项的最大公因数不变，所以零之前的两个相等项都等于 \(\gcd(999,a_2)=1\)。此后序列循环经过 \(1,1,0\)。最后，\(2006\equiv2\pmod3\)，所以 \(a_{2006}\) 与 \(a_2\) 同为奇数；在这个循环中，它必定为 \(1\)。

所以正确答案是 B。

English

The rule gives \(a_n \equiv a_{n+3} \pmod 2,\) so \(a_2\) has the same parity as \(a_{2006} = 1;\) thus \(a_2\) is odd.

Every term is a multiple of \(\gcd(a_1, a_2),\) and \(a_{2006} = 1\) forces \(\gcd(999, a_2) = 1.\) Since \(999 = 3^3 \cdot 37,\) we need \(a_2\) not divisible by \(3\) or \(37.\)

Among the odd integers in \([1, 998]\) there are \(499;\) removing the \(166\) multiples of \(3\) and \(13\) multiples of \(37,\) then adding back the \(4\) multiples of \(111,\) leaves \[499 - 166 - 13 + 4 = 324.\]

Each such \(a_2\) works. For consecutive positive terms \(u,v,\) the update \((u,v)\mapsto(v,|v-u|)\) reduces their maximum within at most two steps. Since both initial terms are at most \(999,\) some \(a_N=0\) occurs by \(N\le1999.\)

The gcd of each consecutive pair is invariant, so the equal terms immediately before that zero both equal \(\gcd(999,a_2)=1.\) The sequence then cycles through \(1,1,0.\) Finally, \(2006\equiv2\pmod3,\) so \(a_{2006}\) has the same odd parity as \(a_2;\) in this cycle it must therefore be \(1.\)

Thus, the correct answer is B.