2009A AMC 12 真题（P版）

从上午 \(10{:}34\) 到上午 \(11{:}00\) 是 \(26\) 分钟，从上午 \(11{:}00\) 到下午 \(1{:}00\) 是 \(2\) 小时，从下午 \(1{:}00\) 到下午 \(1{:}18\) 是 \(18\) 分钟。

所以飞行时间为 \(2\) 小时 \(26 + 18 = 44\) 分钟。因此 \(h = 2\)，\(m = 44\)，且 \(h + m = 46\)。

因此，正确答案是 A。

English

From \(10{:}34\) am to \(11{:}00\) am is \(26\) minutes, from \(11{:}00\) am to \(1{:}00\) pm is \(2\) hours, and from \(1{:}00\) pm to \(1{:}18\) pm is \(18\) minutes.

So the flight lasted \(2\) hours and \(26 + 18 = 44\) minutes. Thus \(h = 2,\) \(m = 44,\) and \(h + m = 46.\)

Thus, the correct answer is A.

从里面开始，\(1 + 1 = 2\)，所以 \(1 + \dfrac{1}{2} = \dfrac{3}{2}\)。然后 \(\dfrac{1}{\frac{3}{2}} = \dfrac{2}{3}\)，且 \(1 + \dfrac{2}{3} = \dfrac{5}{3}\)。

因此，正确答案是 C。

English

Starting inside, \(1 + 1 = 2,\) so \(1 + \dfrac{1}{2} = \dfrac{3}{2}.\) Then \(\dfrac{1}{\frac{3}{2}} = \dfrac{2}{3},\) and \(1 + \dfrac{2}{3} = \dfrac{5}{3}.\)

Thus, the correct answer is C.

间距是 \(\dfrac{3}{4} - \dfrac{1}{4} = \dfrac{1}{2}\)。走三分之一处就是加上 \(\dfrac{1}{3} \cdot \dfrac{1}{2} = \dfrac{1}{6}\)。

所以这个数是 \(\dfrac{1}{4} + \dfrac{1}{6} = \dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12}\)。

因此，正确答案是 B。

English

The gap is \(\dfrac{3}{4} - \dfrac{1}{4} = \dfrac{1}{2}.\) One third of the way adds \(\dfrac{1}{3} \cdot \dfrac{1}{2} = \dfrac{1}{6}.\)

So the number is \(\dfrac{1}{4} + \dfrac{1}{6} = \dfrac{3}{12} + \dfrac{2}{12} = \dfrac{5}{12}.\)

Thus, the correct answer is B.

如果四枚硬币中有一分硬币，总额不是 \(5\) 的倍数，所以不可能等于列出的五个 \(5\) 的倍数。如果没有一分硬币，每枚硬币至少值 \(5\) 分，所以总额至少是 \(20\) 分。无论哪种情况，\(15\) 都不可能。

其他数额都能达到：\(25 = 10 + 5 + 5 + 5\)，\(35 = 10 + 10 + 10 + 5\)，\(45 = 25 + 10 + 5 + 5\)，且 \(55 = 25 + 10 + 10 + 10\)。

因此，正确答案是 A。

English

If the four coins include a penny, the total is not a multiple of \(5,\) so it cannot equal any of the five listed multiples of \(5.\) If there is no penny, every coin is worth at least \(5\) cents, so the total is at least \(20\) cents. Either way, \(15\) is impossible.

The other amounts are attainable: \(25 = 10 + 5 + 5 + 5,\) \(35 = 10 + 10 + 10 + 5,\) \(45 = 25 + 10 + 5 + 5,\) and \(55 = 25 + 10 + 10 + 10.\)

Thus, the correct answer is A.

设立方体边长为 \(x\)。新长方体的三个维度是 \(x + 1\)，\(x - 1\)，和 \(x\)，因此体积为 \(x(x+1)(x-1) = x^3 - x\)。

令它等于 \(x^3 - 5\) 得 \(x^3 - x = x^3 - 5\)，所以 \(x = 5\)。

立方体体积是 \(5^3 = 125\)。

因此，正确答案是 D。

English

Let the cube have side length \(x.\) The new solid has dimensions \(x + 1,\) \(x - 1,\) and \(x,\) so its volume is \(x(x+1)(x-1) = x^3 - x.\)

Setting this equal to \(x^3 - 5\) gives \(x^3 - x = x^3 - 5,\) so \(x = 5.\)

The cube’s volume is \(5^3 = 125.\)

Thus, the correct answer is D.

因为 \(12 = 2^2 \cdot 3\)，\[ \begin{aligned} 12^{mn} &= 2^{2mn} \cdot 3^{mn} \\ &= (2^m)^{2n}(3^n)^m \\ &= P^{2n}Q^m \end{aligned}\text{。}\]

因此，正确答案是 E。

English

Since \(12 = 2^2 \cdot 3,\) \[ \begin{aligned} 12^{mn} &= 2^{2mn} \cdot 3^{mn} \\ &= (2^m)^{2n}(3^n)^m \\ &= P^{2n}Q^m. \end{aligned} \]

Thus, the correct answer is E.

相邻差相等给出 \((5x - 11) - (2x - 3)\) \(= (3x + 1) - (5x - 11)\)，即 \(3x - 8 = -2x + 12\)，所以 \(x = 4\)。

前三项是 \(5, 9, 13\)，公差为 \(4\)。

第 \(n\) 项满足 \(2009 = 5 + (n - 1)\cdot 4\)，所以 \(n - 1 = 501\)，从而 \(n = 502\)。

因此，正确答案是 B。

English

Equal consecutive differences give \((5x - 11) - (2x - 3)\) \(= (3x + 1) - (5x - 11),\) that is \(3x - 8 = -2x + 12,\) so \(x = 4.\)

The first three terms are \(5, 9, 13,\) with common difference \(4.\)

The \(n\)th term satisfies \(2009 = 5 + (n - 1)\cdot 4,\) so \(n - 1 = 501\) and \(n = 502.\)

Thus, the correct answer is B.

设这些长方形的较短边为 \(x\)，较长边为 \(y\)。外部正方形边长是 \(x + y\)，内部正方形边长是 \(y - x\)。

因为外部面积是内部面积的 \(4\) 倍，边长比是 \(\sqrt{4} = 2\)，所以 \(x + y = 2(y - x)\)。

这给出 \(y = 3x\)，因此长边与短边之比为 \(3\)。

因此，正确答案是 A。

English

Let the rectangles have shorter side \(x\) and longer side \(y.\) The outer square has side \(x + y\) and the inner square has side \(y - x.\)

Since the outer area is \(4\) times the inner area, the side ratio is \(\sqrt{4} = 2,\) so \(x + y = 2(y - x).\)

This gives \(y = 3x,\) so the ratio of longer to shorter side is \(3.\)

Thus, the correct answer is A.

注意 \(a + b + c = f(1)\)。

在 \(f(x + 3) = 3x^2 + 7x + 4\) 中取 \(x = -2\) 得 \[ \begin{aligned} f(1) &= f(-2 + 3) \\ &= 3(-2)^2 + 7(-2) + 4 \\ &= 12 - 14 + 4 = 2 \end{aligned}\text{。}\]

因此，正确答案是 D。

English

Note that \(a + b + c = f(1).\)

Using \(f(x + 3) = 3x^2 + 7x + 4\) with \(x = -2\) gives \[ \begin{aligned} f(1) &= f(-2 + 3) \\ &= 3(-2)^2 + 7(-2) + 4 \\ &= 12 - 14 + 4 = 2. \end{aligned} \]

Thus, the correct answer is D.

在 \(\triangle BCD\) 中，三角形不等式给出 \(BD + CD \gt BC\)，所以 \(BD + 5 \gt 17\)，即 \(BD \gt 12\)。

在 \(\triangle ABD\) 中，\(AB + DA \gt BD\)，所以 \(BD \lt 5 + 9 = 14\)。

唯一满足 \(12 \lt BD \lt 14\) 的整数是 \(13\)。

因此，正确答案是 C。

English

In \(\triangle BCD,\) the triangle inequality gives \(BD + CD \gt BC,\) so \(BD + 5 \gt 17\) and \(BD \gt 12.\)

In \(\triangle ABD,\) \(AB + DA \gt BD,\) so \(BD \lt 5 + 9 = 14.\)

The only integer with \(12 \lt BD \lt 14\) is \(13.\)

Thus, the correct answer is C.

\(F_n\) 的外部正方形比 \(F_{n-1}\) 的外部正方形多 \(4\) 个菱形，而 \(F_2\) 的外部正方形有 \(4\) 个菱形，所以 \(F_n\) 的外部正方形有 \(4(n - 1)\) 个菱形。

把所有层相加，\[ \begin{gathered} 1 + 4\big(1 + 2 + \cdots + (n - 1)\big) \\ = 1 + 4\cdot\frac{(n - 1)n}{2} \\ = 1 + 2(n - 1)n \end{gathered}\text{。}\]

当 \(n = 20\) 时，它等于 \(1 + 2\cdot 19\cdot 20 = 761\)。

因此，正确答案是 E。

English

The outside square of \(F_n\) has \(4\) more diamonds than that of \(F_{n-1},\) and the outside square of \(F_2\) has \(4,\) so the outside square of \(F_n\) has \(4(n - 1)\) diamonds.

Adding all the rings, \[ \begin{gathered} 1 + 4\big(1 + 2 + \cdots + (n - 1)\big) \\ = 1 + 4\cdot\frac{(n - 1)n}{2} \\ = 1 + 2(n - 1)n. \end{gathered} \]

For \(n = 20,\) this is \(1 + 2\cdot 19\cdot 20 = 761.\)

Thus, the correct answer is E.

如果 \(N = 6\cdot(\text{数字和})\)，那么由于小于 \(1000\) 的数的数字和最大为 \(27\)，可得 \(N \le 162\)。

对于两位数，\(10t + u = 6(t + u)\)，得 \(4t = 5u\)，因而 \(t = 5\) 且 \(u = 4\)，所以 \(N = 54\)。一位数需要 \(6u = u\)，对 \(u \gt 0\) 不可能。三位数满足 \(100h + 10t + u = 6(h + t + u)\)，给出 \(94h + 4t = 5u\)，左边至少为 \(94\)，右边至多为 \(45\)，所以没有解。

因此恰好有一个数 \(54\)，满足条件。

因此，正确答案是 B。

English

If \(N = 6\cdot(\text{digit sum}),\) then since the digit sum of a number below \(1000\) is at most \(27,\) we have \(N \le 162.\)

For a two-digit number \(10t + u = 6(t + u)\) gives \(4t = 5u,\) forcing \(t = 5\) and \(u = 4,\) so \(N = 54.\) A one-digit number would need \(6u = u,\) impossible for \(u \gt 0.\) A three-digit number \(100h + 10t + u = 6(h + t + u)\) gives \(94h + 4t = 5u,\) whose left side is at least \(94\) while the right side is at most \(45,\) so there is no solution.

Hence exactly one number, \(54,\) works.

Thus, the correct answer is B.

由余弦定理，\[ \begin{aligned} AC^2 &= 10^2 + 20^2 \\ &\quad {}- 2\cdot 10\cdot 20\cos(\angle ABC) \\ &= 500 - 400\cos(\angle ABC) \end{aligned}\text{。}\]

船转过的角介于 \(45^\circ\) 和 \(60^\circ\)，所以内角 \(\angle ABC\) 介于 \(120^\circ\) 和 \(135^\circ\)。

因为 \(\cos 120^\circ = -\dfrac{1}{2}\) 且 \(\cos 135^\circ = -\dfrac{\sqrt{2}}{2}\)，\[ \begin{aligned} 700 &= 500 + 200 \\ &\le AC^2 \le 500 + 200\sqrt{2} \\ &\lt 800 \end{aligned}\text{。}\]

所以 \(AC^2\) 位于 \([700, 800]\) 中。

因此，正确答案是 D。

English

By the Law of Cosines, \[ \begin{aligned} AC^2 &= 10^2 + 20^2 \\ &\quad {}- 2\cdot 10\cdot 20\cos(\angle ABC) \\ &= 500 - 400\cos(\angle ABC). \end{aligned} \]

The ship turns through an angle between \(45^\circ\) and \(60^\circ,\) so the interior angle \(\angle ABC\) lies between \(120^\circ\) and \(135^\circ.\)

Since \(\cos 120^\circ = -\dfrac{1}{2}\) and \(\cos 135^\circ = -\dfrac{\sqrt{2}}{2},\) \[ \begin{aligned} 700 &= 500 + 200 \\ &\le AC^2 \le 500 + 200\sqrt{2} \\ &\lt 800. \end{aligned} \]

So \(AC^2\) lies in \([700, 800].\)

Thus, the correct answer is D.

直线 \(y = mx\) 经过顶点 \((0, 0)\)，因此它恰好在经过对边中点时平分三角形面积。对边连接 \((1, 1)\) 与 \((6m, 0)\)。其中点为 \(\left(\dfrac{6m + 1}{2}, \dfrac{1}{2}\right)\)。

要求该点满足 \(y = mx\)，得到 \[\frac{1}{2} = m\cdot\frac{6m + 1}{2}\text{，}\] 所以 \(6m^2 + m - 1 = 0\)，即 \((3m - 1)(2m + 1) = 0\)。

可能的值是 \(m = \dfrac{1}{3}\) 和 \(m = -\dfrac{1}{2}\)，它们的和为 \(-\dfrac{1}{6}\)。

因此，正确答案是 B。

English

The line \(y = mx\) passes through the vertex \((0, 0),\) so it bisects the triangle’s area exactly when it passes through the midpoint of the opposite side, joining \((1, 1)\) and \((6m, 0).\) That midpoint is \(\left(\dfrac{6m + 1}{2}, \dfrac{1}{2}\right).\)

Requiring it to satisfy \(y = mx\) gives \[\frac{1}{2} = m\cdot\frac{6m + 1}{2},\] so \(6m^2 + m - 1 = 0,\) that is \((3m - 1)(2m + 1) = 0.\)

The possible values are \(m = \dfrac{1}{3}\) and \(m = -\dfrac{1}{2},\) whose sum is \(-\dfrac{1}{6}.\)

Thus, the correct answer is B.

对于 \(4\) 的倍数 \(k\) \[ \begin{aligned} &(k + 1)i^{k+1} + (k + 2)i^{k+2} \\ &\quad {}+ (k + 3)i^{k+3} + (k + 4)i^{k+4} \\ &= (k + 1)i - (k + 2) \\ &\quad {}- (k + 3)i + (k + 4) \\ &= 2 - 2i \end{aligned}\text{。}\]

前 \(96\) 项，也就是 \(24\) 组，和为 \(24(2 - 2i) = 48 - 48i\)。

再加下一项 \(97i^{97} = 97i\) 得到 \(48 - 48i + 97i = 48 + 49i\)。所以 \(n = 97\)。

因此，正确答案是 D。

English

For \(k\) a multiple of \(4,\) \[ \begin{aligned} &(k + 1)i^{k+1} + (k + 2)i^{k+2} \\ &\quad {}+ (k + 3)i^{k+3} + (k + 4)i^{k+4} \\ &= (k + 1)i - (k + 2) \\ &\quad {}- (k + 3)i + (k + 4) \\ &= 2 - 2i. \end{aligned} \]

Summing the first \(96\) terms (that is \(24\) blocks) gives \(24(2 - 2i) = 48 - 48i.\)

Adding the next term \(97i^{97} = 97i\) yields \(48 - 48i + 97i = 48 + 49i.\) So \(n = 97.\)

Thus, the correct answer is D.

一个与两条正坐标轴都相切且半径为 \(r\) 的圆，圆心为 \((r, r)\)。与圆心 \((3, 0)\)、半径 \(1\) 的圆外切，意味着两圆心距离为 \(r + 1\)：\[(r - 3)^2 + r^2 = (r + 1)^2\text{。}\]

展开得 \(r^2 - 8r + 8 = 0\)。两个根 \(r = 4 \pm 2\sqrt{2}\) 都为正，由韦达定理，它们的和为 \(8\)。

因此，正确答案是 D。

English

A circle tangent to both positive axes with radius \(r\) has center \((r, r).\) External tangency to the circle at \((3, 0)\) of radius \(1\) means the distance between centers is \(r + 1\): \[(r - 3)^2 + r^2 = (r + 1)^2.\]

Expanding gives \(r^2 - 8r + 8 = 0.\) Both roots \(r = 4 \pm 2\sqrt{2}\) are positive, and by Vieta’s formulas their sum is \(8.\)

Thus, the correct answer is D.

对首项为 \(a\)、公比为 \(r\) 的级数，其和为 \(\dfrac{a}{1 - r} = r\)，所以 \(r^2 - r + a = 0\)。

\(r_1\) 和 \(r_2\) 都满足同一个二次方程，并且因为两个级数不同，\(r_1 \ne r_2\)，所以它们是该方程的两个不同根。由韦达定理，\(r_1 + r_2 = 1\)。

因此，正确答案是 C。

English

For a series with first term \(a\) and ratio \(r,\) the sum is \(\dfrac{a}{1 - r} = r,\) so \(r^2 - r + a = 0.\)

Both \(r_1\) and \(r_2\) satisfy this same quadratic, and since the two series are different, \(r_1 \ne r_2,\) so they are its two distinct roots. By Vieta’s formulas, \(r_1 + r_2 = 1.\)

Thus, the correct answer is C.

注意 \(I_k = 10^{k+2} + 64\) \(= 2^{k+2}5^{k+2} + 2^6\)。

当 \(k \lt 4\) 时，第一项所含的因子 \(2\) 少于 \(6\) 个，所以 \(N(k) \lt 6\)。当 \(k \gt 4\) 时，第一项可被 \(2^7\) 整除，但 \(2^6\) 这一项不能，所以 \(N(k) \lt 7\)。

当 \(k = 4\) 时，\(I_4 = 2^6(5^6 + 1)\)。因为 \(5^6 + 1\) \(= (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big)\) \(= 26\cdot 601\)，且 \(26 = 2\cdot 13\) 恰好再贡献一个因子 \(2\)，所以 \(N(4) = 7\)。

因此最大值为 \(7\)。

因此，正确答案是 B。

English

Note that \(I_k = 10^{k+2} + 64\) \(= 2^{k+2}5^{k+2} + 2^6.\)

For \(k \lt 4\) the first term has fewer than \(6\) factors of \(2,\) so \(N(k) \lt 6.\) For \(k \gt 4\) the first term is divisible by \(2^7\) but the \(2^6\) term is not, so \(N(k) \lt 7.\)

For \(k = 4,\) \(I_4 = 2^6(5^6 + 1).\) Since \(5^6 + 1\) \(= (5^2 + 1)\big((5^2)^2 - 5^2 + 1\big)\) \(= 26\cdot 601,\) and \(26 = 2\cdot 13\) contributes exactly one more factor of \(2,\) we get \(N(4) = 7.\)

So the maximum value is \(7.\)

Thus, the correct answer is B.

对边长为 \(2\) 的正多边形，令 \(O\) 为中心，\(M\) 为一条边的中点，\(N\) 为该边的一个端点。那么 \(\triangle OMN\) 在 \(M\) 处为直角，且 \(MN = 1\)，\(OM = r\)（内切圆半径），\(ON = R\)（外接圆半径）。

所以 \(R^2 - r^2 = 1\)，两圆之间的面积为 \(\pi(R^2 - r^2) = \pi\)，与边数无关。因此 \(A = B\)。

因此，正确答案是 C。

English

For a regular polygon with side length \(2,\) let \(O\) be the center, \(M\) the midpoint of a side, and \(N\) an endpoint of that side. Then \(\triangle OMN\) has a right angle at \(M,\) with \(MN = 1,\) \(OM = r\) (inradius), and \(ON = R\) (circumradius).

So \(R^2 - r^2 = 1,\) and the area between the circles is \(\pi(R^2 - r^2) = \pi\) for any number of sides. Hence \(A = B.\)

Thus, the correct answer is C.

在 \(\triangle AED\) 和 \(\triangle BEC\) 两边都加上 \(\triangle CED\)，可知 \(\triangle ACD\) 与 \(\triangle BCD\) 面积相等。它们共用底边 \(CD\)，所以 \(A\) 和 \(B\) 到直线 \(CD\) 的距离相等，意味着 \(AB \parallel CD\)。

因此 \(\triangle ABE \sim \triangle CDE\)，相似比为 \(\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac{3}{4}\)，所以 \(\dfrac{AE}{EC} = \dfrac{3}{4}\)。

设 \(AE = 3x\) 且 \(EC = 4x\)，则 \(7x = AC = 14\)，所以 \(x = 2\)，\(AE = 6\)。

因此，正确答案是 E。

English

Adding \(\triangle CED\) to each of \(\triangle AED\) and \(\triangle BEC\) shows \(\triangle ACD\) and \(\triangle BCD\) have equal areas. They share base \(CD,\) so \(A\) and \(B\) are equidistant from line \(CD,\) meaning \(AB \parallel CD.\)

Then \(\triangle ABE \sim \triangle CDE\) with ratio \(\dfrac{AB}{CD} = \dfrac{9}{12} = \dfrac{3}{4},\) so \(\dfrac{AE}{EC} = \dfrac{3}{4}.\)

Writing \(AE = 3x\) and \(EC = 4x,\) we get \(7x = AC = 14,\) so \(x = 2\) and \(AE = 6.\)

Thus, the correct answer is E.

因为 \(x^{12} + ax^8 + bx^4 + c = p(x^4)\)，一个值是零点当且仅当 \(x^4\) 等于 \(p\) 的某个根，即 \(2009 + 9002\pi i\)，\(2009\)，或 \(9002\)。

方程 \(x^4 = 2009 + 9002\pi i\) 有四个不同的非实根。\(x^4 = 2009\) 和 \(x^4 = 9002\) 各有两个实根和两个非实根。

所以非实零点个数为 \(4 + 2 + 2 = 8\)。

因此，正确答案是 C。

English

Since \(x^{12} + ax^8 + bx^4 + c = p(x^4),\) a value is a zero exactly when \(x^4\) equals one of the roots of \(p,\) namely \(2009 + 9002\pi i,\) \(2009,\) or \(9002.\)

The equation \(x^4 = 2009 + 9002\pi i\) has four distinct nonreal roots. Each of \(x^4 = 2009\) and \(x^4 = 9002\) has two real roots and two nonreal roots.

So the nonreal zeros number \(4 + 2 + 2 = 8.\)

Thus, the correct answer is C.

设那两张平行面为三角形。该平面经过不在这些面上的六条边的中点，形成一个边长为 \(\dfrac{1}{2}\) 的等边六边形；由对称性它也等角，因此是正六边形。

正六边形由六个等边三角形组成，所以面积为 \[6\cdot\frac{\sqrt{3}}{4}\left(\frac{1}{2}\right)^2 = \frac{3\sqrt{3}}{8}\text{。}\]

因此 \(a = 3\)，\(b = 3\)，\(c = 8\)，且 \(a + b + c = 14\)。

因此，正确答案是 E。

English

Let the two parallel faces be triangles. The plane passes through the midpoints of the six edges not on those faces, forming an equilateral hexagon of side \(\dfrac{1}{2},\) which by symmetry is also equiangular and hence regular.

A regular hexagon is six equilateral triangles, so its area is \[6\cdot\frac{\sqrt{3}}{4}\left(\frac{1}{2}\right)^2 = \frac{3\sqrt{3}}{8}.\]

Thus \(a = 3,\) \(b = 3,\) \(c = 8,\) and \(a + b + c = 14.\)

Thus, the correct answer is E.

因为 \(g(x) = -f(100 - x)\)，\(f\) 和 \(g\) 的图像关于点 \((50, 0)\) 互为中心对称，所以四个截距成对满足 \(x_2 + x_3 = x_1 + x_4 = 100\)。

由 \(x_3 - x_2 = 150\)，得 \(x_2 = -25\) 且 \(x_3 = 125\)。

取 \(x_1, x_3\) 为 \(f\) 的两个根，其顶点的 \(x\) 坐标为 \(h = \dfrac{x_1 + x_3}{2}\)，所以 \(x_1 = 2h - 125\)。\(f\) 的顶点落在 \(g\) 的图像上，这一条件给出 \[ \begin{aligned} 1 &= \frac{f(h)}{g(h)} \\ &= \frac{(125 - h)(h - 125)}{-(h + 25)(3h - 225)} \end{aligned}\text{，}\] 由此解得 \(h = -25 \pm 75\sqrt{2}\)。由于 \(x_1 = 2h-125\) 必须小于 \(x_2=-25\)，需要 \(h \lt 50\)，所以 \(h = -25-75\sqrt{2}\)。

又 \(x_4 = 100 - x_1\)，所以 \[ \begin{aligned} x_4 - x_1 &= 350 - 4h \\ &= 450 + 300\sqrt{2} \end{aligned}\text{。}\] 因而 \(m + n + p = 450 + 300 + 2\) \(= 752\)。

因此，正确答案是 D。

English

Because \(g(x) = -f(100 - x),\) the graphs of \(f\) and \(g\) are reflections of each other through the point \((50, 0),\) so the four intercepts pair up with \(x_2 + x_3 = x_1 + x_4 = 100.\)

With \(x_3 - x_2 = 150,\) we get \(x_2 = -25\) and \(x_3 = 125.\)

Take \(x_1, x_3\) as the roots of \(f,\) whose vertex has \(x\)-coordinate \(h = \dfrac{x_1 + x_3}{2},\) so \(x_1 = 2h - 125.\) The condition that the vertex of \(f\) lies on the graph of \(g\) gives \[ \begin{aligned} 1 &= \frac{f(h)}{g(h)} \\ &= \frac{(125 - h)(h - 125)}{-(h + 25)(3h - 225)}, \end{aligned} \] which gives \(h = -25 \pm 75\sqrt{2}.\) Since \(x_1 = 2h-125\) must be less than \(x_2=-25,\) we need \(h \lt 50,\) so \(h = -25-75\sqrt{2}.\)

Then \(x_4 = 100 - x_1,\) so \[ \begin{aligned} x_4 - x_1 &= 350 - 4h \\ &= 450 + 300\sqrt{2}. \end{aligned} \] Hence \(m + n + p = 450 + 300 + 2\) \(= 752.\)

Thus, the correct answer is D.

因为 \(\log_2 T(n + 1) = T(n)\)，每次应用 \(\log_2\) 都会从一个 \(2\) 的幂塔顶部去掉一层。

记 \(T_j=T(j)\)，并设 \(L_j\) 为对 \(B\) 恰好取 \(j\) 次 \(\log_2\) 所得的结果。前两个结果为 \[ \begin{aligned} L_1 &= A\,T_{2008}, \\ L_2 &= T_{2009}T_{2008}+T_{2007} \end{aligned}\text{。}\]

先看下界，\[ \begin{aligned} L_3 &\gt \log_2(T_{2009}T_{2008}) \\ &= T_{2008}+T_{2007} \\ &\gt T_{2008} \end{aligned}\text{。}\] 反复取对数可得，对 \(0\le k\le 2007\) 都有 \(L_{k+3} \gt T_{2008-k}\)。特别地，\(L_{2010}\gt2\)，于是 \(L_{2011}\gt1\)，进而 \(L_{2012}\gt0\)。因此 \(L_{2013}\) 有定义。

再看上界，由 \(T_{2007}\lt T_{2008}T_{2009}\) 得 \[ \begin{aligned} L_3 &\lt 1+T_{2007}+T_{2008} \\ &\lt 2T_{2008}, \\ L_4 &\lt 1+T_{2007} \lt T_{2008} \end{aligned}\text{。}\] 反复使用最后一个不等式可得，对 \(0\le k\le2007\) 都有 \(L_{k+4}\lt T_{2008-k}\)。于是 \(L_{2011}\lt2\)。结合下界即得 \(0\lt L_{2012}\lt1\)，从而 \(L_{2013}\lt0\)。因此 \(L_{2014}\) 无定义，最大的 \(k\) 是 \(2013\)。

因此，正确答案是 E。

English

Since \(\log_2 T(n + 1) = T(n),\) each application of \(\log_2\) strips one \(2\) off the top of a tower of twos.

Write \(T_j=T(j),\) and let \(L_j\) be the result of applying \(\log_2\) to \(B\) exactly \(j\) times. The first two results are \[ \begin{aligned} L_1 &= A\,T_{2008}, \\ L_2 &= T_{2009}T_{2008}+T_{2007}. \end{aligned} \]

For the lower bound, \[ \begin{aligned} L_3 &\gt \log_2(T_{2009}T_{2008}) \\ &= T_{2008}+T_{2007} \\ &\gt T_{2008}. \end{aligned} \] Repeatedly taking logarithms gives \(L_{k+3} \gt T_{2008-k}\) for \(0\le k\le 2007.\) In particular, \(L_{2010}\gt2,\) so \(L_{2011}\gt1\) and \(L_{2012}\gt0.\) Thus \(L_{2013}\) is defined.

For the upper bound, \(T_{2007}\lt T_{2008}T_{2009},\) so \[ \begin{aligned} L_3 &\lt 1+T_{2007}+T_{2008} \\ &\lt 2T_{2008}, \\ L_4 &\lt 1+T_{2007} \lt T_{2008}. \end{aligned} \] Repeating the last comparison gives \(L_{k+4}\lt T_{2008-k}\) for \(0\le k\le2007.\) Hence \(L_{2011}\lt2.\) Together with the lower bound, this yields \(0\lt L_{2012}\lt1,\) so \(L_{2013}\lt0.\) Therefore \(L_{2014}\) is undefined, and the largest possible \(k\) is \(2013.\)

Thus, the correct answer is E.

递推式正是正切加法公式，并且 \(a_1 = \tan\dfrac{\pi}{4}\)，\(a_2 = \tan\dfrac{\pi}{6}\)。

写成 \(a_n = \tan\dfrac{\pi c_n}{12}\)，其中 \(c_1 = 3\)、\(c_2 = 2\)，且 \(c_{n+2} \equiv c_n + c_{n+1} \pmod{12}\)。序列 \(c_n\) 为 \[ \begin{gathered} 3, 2, 5, 7, 0, 7, 7, 2, \\ 9, 11, 8, 7, 3, 10, 1, 11, \\ 0, 11, 11, 10, 9, 7, 4, 11, \ldots \end{gathered} \] 其周期为 \(24\)。

因为 \(2009 = 24\cdot 83 + 17\)，所以 \(c_{2009} = c_{17} = 0\)，从而 \(a_{2009} = \tan 0 = 0\)，\(|a_{2009}| = 0\)。

因此，正确答案是 A。

English

The recursion is exactly the tangent addition formula, and \(a_1 = \tan\dfrac{\pi}{4},\) \(a_2 = \tan\dfrac{\pi}{6}.\)

Writing \(a_n = \tan\dfrac{\pi c_n}{12}\) with \(c_1 = 3,\) \(c_2 = 2,\) and \(c_{n+2} \equiv c_n + c_{n+1} \pmod{12},\) the sequence \(c_n\) is \[ \begin{gathered} 3, 2, 5, 7, 0, 7, 7, 2, \\ 9, 11, 8, 7, 3, 10, 1, 11, \\ 0, 11, 11, 10, 9, 7, 4, 11, \ldots \end{gathered} \] which is periodic with period \(24.\)

Since \(2009 = 24\cdot 83 + 17,\) \(c_{2009} = c_{17} = 0,\) so \(a_{2009} = \tan 0 = 0\) and \(|a_{2009}| = 0.\)

Thus, the correct answer is A.